, AND APPLICATION8 (Reyised end Enlarged)

Xfferential Equations and Applications

(Revised and Enlarged)

0. N. OBIKWELU, FNMS, FNSE B.Sc (Hons), M.Sc. Ph.D.

O D. 0. N. Obikwelu 2003 ISBN: 978-3664-6-X

All rights Reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior permission of the copyright owners.

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Dedicated to The Almighty God, the source of all knowledge.

ABOUT THE AUTHOR

D.O.N. Obikwelu graduated from the University of

Nigeria, Nsukka in 1975 with the Second Honours Upper

Division. He earned the Ph.D. degree from Michigan State

University, East Lansing in 1982 and was elected into the

famous Sigma XI Honour Society of America the same year.

He was a Chief EngineerIManager of Research and

Development at the Delta Steel Company, Ovwian-Aladja, and

a Senior Lecturer of Thermodynamics, Engineering

Mathematics and Mechanics of Materials at Federal University

of Technology, Owem, University of Port Harcourt and the

Shell Special Intensive Graduate Training Programme.

Dr. Obikwelu is a Registered Professional Engineer, a

Fellow of The Nigerian Metallurgical Society and a Fellow of

The Nigerian Society of Engineers, adjudged the best Fellow of

the year 2002.

He has published many research papers and authored the

following books: Differential Equations and Applications, The

Science and Engineering of Refractories and Introduction to

Computational Methods In Engineering.

He is married with children.

FOREWORD

The emphasis of this second text is on the methods of solution

of ordinary differential, difference and partial differential

equations, with variety of worked examples and applications to

real-life problems. Diferential Equations and Applications

(revised and enlarged) adds to the former text by the same

author to provide a handy text for both teachers, students and

practising engineers. A lot of knowledge will be gained by the

use of this and I commend Dr. D.O.N.Obikwelu for this

invaluable work. I recommend Diferential Equations and

Applications (revised and enlarged) to all.

Prof. U. B. C. 0. Ejike Professor of Mathematics Federal University of Technology Owerri

FOREWORD

The emphasis of this second text is on the methods of solution

of ordinary differential, difference and partial differential

equations, with variety of worked examples and applications to

real-life problems. Diflerential Equations and Applications

(revised and enlarged) adds to the former text by the same

author to provide a handy text for both teachers, students and

practising engineers. A lot of knowledge will be gained by the

use of this and I commend Dr. D.O.N.Obikwelu for this

invaluable work. I recommend D~fSerential Equations and

Applications (revised and enlarged) to all.

Prof. U. B. C . 0. Ejike Professor of Mathematics Federal University of Technology Owerri

PREFACE

Differential Equations and Applications (Revised and Enlarged) is a new and improved edition of the first Book entitled "Differential Equations and Applications" by the same author. Diflerential Equations and Applications ( Revised and Enlarged) contains more materials than the first edition. Typographical errors in the first edition which were regretted are corrected in this edition.

This course on Differential Equations is comprehensive and covers the formulation and solutions of all types of differential equations. It is designed to fit into the Engineering and Science student's tight academic programme through didactic numerous worked examples in various field to instill expected skills for the mastery of all aspects of Differential Equations. Various examples from practical situations is the author's unique way of instilling these abilities into the student who unconsciously develops the abilities and through association applies it successfully to simple and complex situations.

Differential Equations and Applications (Revised and Enlarged) covers adequately a first course on Differential Equations in the Universities and advanced aspects for the polytechnics. It is a strong foundation for Advanced Mathematics for the Engineering and Science students and is a ready handbook for the practising engineer and the mathematician.

Ordinary Differential Equations (ODE) Partial Differential Equations (PDE), Total Differential Equations (TDE), Difference Equations. Applications of Derivatives and the Jacobians, Physical and Geometrical problems and all the methods of solving differential equations from various fields are treated in a unified way. In this way the student, unconsciously, and through whole-to-part technique internalizes the concepts of each methods of solving a particular differential equation.

Singularity function in static loading of mechanical systems and physics are treated for completeness in attacking the differential equations from such fields.

Advanced undergraduate and graduate students may find the book exciting. Chapters on Laplace Transforms, Difference Equations, Applications of Derivatives and more examples on Partial Differential Equations may be read by such student at leisure to appreciate the importance of differential equations in real-life problems. Other students are advised to start reading the book from chapter one re-solving all problems under examples for good mastery of the concepts.

Important Partial Differential Equations are listed under Appendix 1 for the motivated student in his area of specialization for future applications It is my hope that Differential Equations and Applications (Revised and Enlarged) will endear Differential Equations to all Engineering and Science students.

The author immensely thanks Prof. U. B. C. 0. Ejike for reading the manuscript and making very useful contributions.

D. 0. N. OBIKWELU

vii

CONTENTS

Foreword Preface Symbols and Abbreviations

1 DEFINITIONS, PRIMITIVE AND ORIGIN OF DIFFERENTIAL EQUATIONS.

1.0 Introduction 1.1 Primitive and Origin of Differential Equations 1.2 Solutions To Differential Equation, "Existence Problem" 1.2.2 Particular Solution, General or Complete Solution Singular Solution 1.2.3 Examples .

Exercises

2 CLASSES OF ORDINARY DIFFERENTIAL EQUATIONS (ODE) FOR PURPOSES OF SOLUTIONS. 2.0 List of the classes 2.1 Separable Equations

2.2 Homogeneous Equations 2.3 Linear But Homogeneous Equations 2.4 Miscellaneous 2.5 Exact Equation 2.6 Integrating Factor 2.7 Linear Equation Of First Order 2.8 Bernoullis Equation 2.9 Equations Solvable For One Variable 2.10 Clairaut's Equation 2.1 1 Equations of Higher Orders 2.12 Numerical Methods for Solving Differential Equations 2.12.1 Euler Method 2.12.2 Taylor Series Method 2.12.3 Runge - Kutta Method 2.12.4 Picard's Method 2.13 Formation And Solution Of Differential Equations Of Physical Problems

viii

LINEAR EQUATIONS OF ORDER n 3.1 Existence and Uniqueness Problem 3.2 Operation Notation 3.3 Linear Operators 3.4 Fundamental Theorem of Linear Differential Equation 3.5 Linear Dependence and Wronskians Dependence 3.6 Linear Equations with Contant Coefficients

( I ) Non-operator techniques 3.6.1 Complementary Or Homogenous Equations 3.6.2 The particular Solution 3.6.2.2 Method of Variation of Parameters

(11) Operator Techniques: (Reduction of Order and Inverse Operators) 3.7 Short Methods for Solving Linear Equations with Constant Coefficient 3.8 Linear Equations with Variable Coefficients 3.8.1 Miscellaneous Transformation of Variable 3.8.2 Exact Equation 3.9 Operator Factorization 3.10 Variable of Parameters 3.11 Series Method (Frobenius Method) 3.12 Simultaneous Differential Equations 3.13 Singular Solution - Extraneous Loci 3.14 Application of Linear Differential Equations To Physical

Problems. Exercises

4 TOTAL DIFFERENTIAL EQUATIONS (TDE) 4.1 Definition 4.2 Intergrability 4.3 Exactness 4.4 Examples 4.5 Pairs of Total Differential Equations 4.6 Applications of Total and Simultaneous Differential Equation

Exercises

5 SERIES METHOD FOR SOLVING DIFFFERENTIAL EQUATION 5.1 Equations of Order One

Existence Theorem Examples

5.2 Linear Equations of Order Two 5.3 Singular points and intergration By Series, Recursion Formula 5.4 Indicia1 Equation 5.5 The Complete Series Solution

Exercises

4.5 The Complete Series Solution Exercises

6 LAPLACE TRANSFORM 6.0 Introduction 6.1 Existence of Laplace Transform 6.2 Inverse Laplace transform 6.3 Laplace Transform of Derivatives 6.4 Unit Step Function 6.5 Theorems of Laplace Transform 6.6 Table of Transforms (partial) 6.7 General Properties of Laplace Transforms 6.8 Partial Fractions 6.9 Examples

Exercise

7 LEGENDRE, BESSEL AND GAUSS EQUATIONS LEGENDRE AND BESSEL FUNCTIONS 7.0 Introduction 7.1 Legendre Equation 7.2 The Bessel Equation

The Bessel Functions: First and Second Kind 7.2.1 Graphs of Bassel Functions 7.2.2 Bessel Functions of Second Kind 7.2.3 Generating Function for jn(X) 7.2.4 Recurrence Formula In Bessel Functions 7.2.5 Bessel Function in Complex Plane 7.2.6 Ber, Bei, Ker Functions 7.2.7 Equations Transformed Into Bessel's Equation 7.2.8 Asmptotes for Bessel Function 7.2.9 Series Of Bessel Functions 7.2.9.1 Dirichlet Conditions 7.2.10 Frobenius Method In Bessel's Equation Solution 7.2.11 Orthogonality Of Bessel Equations 7.3 Gauss Equation and Solutions 7.4 More examples on Legendre, Bessel Equations and Other System

Like Rodrigues Formula, Hennite Polynomials, Legendre Polynomials, Sturm-houville, Eigen Value And Eigen Function Exercises

8 DIFFERENCE EQUATIONS 8.0 Introduction 8.1 A Solution Of Difference Equations 8.2 The Order Of A Difference Equation 8.3 Analogy Of Difference Equation To Differential Equat~on 8.3.1 First Order Linear Difference Equation 8.3.2 The Digamma Function 8.3.4 Linear Homogeneous Second Order Difference Equation 8.3.5 Constant Coefficients, Fibonacci Equation 8.3.6 The Non-Homogeneous Difference Equation 8.3.7 Higher Order Difference Equations 8.3.8 Non-Linear Difference Equations

Examples Exercises

9 PARTIAL DERIVATIVES AND SOME PARTIAL DERIVATIVES AND JACOBIAN 9.0 Intoduction 9.1 Notation 9.2 Definition 9.3 Higher Partial Derivatives 9.4 Differpials 9.5 Eulars Theorem on homogeneous Functions 9.6 Implicit Functions 9.7 Jacobians 9.7.1 Partial Derivatives Using Jacobians 9.7.2 Theorems of Jacobians 9.7.3 Transformations 9.8 Application of Partial Derivatives (Language Multipliers For

Maxima and Minima) 9.9 Error Calculations

Exercises

10 INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS 10.0 Partial differential equations contain one or more partial variables. 10.1 Classification 10.2 General form of Linear Partial Differential Equation 10.3 Methods of solving partial differential equation Examples 10.4 Eliurninaion of Arbitrary Constants in a Partial Differential Equation.

Exercises.

ANSWERS TO THE EXERCISES SUGGESTED FURTHER READING APPENDIX I - SOME lMPORTNG DIFFERENTIAL EQUATION INDEX

xii

Y' A,B,C, O.D.E. P.D.E. T.D.E. Cosh Sinh d/dx = D Dl FP FQ R.H.S. L.H.S.

x x-1 I- P&) Jdx)

a H(x-a) r (x-a) W

ZI D

SYMBOLS AND ABBREVIATIONS y single prime, representing dytdx Arbitrary Constants Ordinary Differential Equation Partial Differential Equation Total Differential Equation Hyperbolic Cosine Hyperbolic Sine Differential Operator dtdt Radial Components Transverse Components Right Hand Side Left Hand Side

Laplace Transform Operator

Inverse Laplace Transform Operator Gamman Function Particular Solution of Lengedre Equation Bessel Function of 1st Kind Bessel Function of 2nd Kind Euler's Constant (=0.5772156) Hankel Function of 1" Kind Hankel Function of 2nd Kind Modified Bessel Function of 1" Kind Modified Bessel Function of 2nd Kind Legendre Polynomails Legendre Functions of Legendre Polynomails Operator del = id + id + ka -- -

ax ay az Dirac delta function Heaviside function or Unit Step Function Unit Ramp Function Circular Frequency az/at Diffusion coeficient

X

=2/47d,, e-"' du Strain Velocity of Strain Rate Unit Normal Vector Laplacian of U = V (Vu) =S2dSx2 + S2dSy2+ S2u/Sz2

xiii

I)I<FINITION, PRIMITIVE AND ORIGIN OF DIFFERENTIAL EQUATIONS

1. INTRODUCTION A Differential equation is an equation relating two or more variables in terms of derivatives or differentials. These are of immense importance in Engineering of all kinds, physics and all aspects of mathematical studies . Setting up a differential equation of a physical phenomenon or a concept and solving it with well established boundary conditions is the key to inventive research. In this revised and enlarged text dy/dx will occasionally be represented by y', dZy/dx2 by yf', d3y /d~3 by y"' dny/dxn by y" where y is the dependent variable. Thus the following are differential equations.

y' = x + 5 o r d y / d x = x + 5 ............ ( 0 y" + 3 ~ ' 2 ~ = 0 or dZy/dx2 +3dy/dx + 2y = 0 . . . . . . . . . (ii) aylax = z + xaz/ay . . . . . . . . . . . . . . . . . . . . . (iii) a2z/ax2 + a2z/ay2 = x2 + xy . . . . . . . . . . . . . . . . . . . . . (iv) ( ~ ' 7 ~ + 2yf = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . ...( v) or (dZy/dx2) + 2dy/dx = 0

Examples (i) and (ii) with a single independent variable x, are called ordinary differential equations. Examples (iii) with two or more independent variables are called partial differential equations.

An nth order differential equation has a derivative of nth order which is the highest in the expression.

Example (i) has order 1 and degree 1 Example (ii) has order 2 and degree 1 Example (iv) has order 2 and degree 1 Example (v) has order 2 and degree 3

Degree of the differential equation is the highest power of the highest differential after it has been simplified by clearing radicals and fractions if any

1.1 PRIMITIVE AND ORIGIN OF DIFFERENTIAL EQUATIONS If dy/dx = 2x, then dy = 2xdx y is obtained by integration to be x2 + c Therefore y = x2 + c is the solution of the differential equation dy/dx = 2x and y = xZ + c is the primitive of the differential equation.

From mechanics, if S is the distance passed in time, t, by a body moving with uniform acceleration, a, then d2s/dt2 = a is a differential equation of order 2.

A relationship between the variables which involve n essential constants as y' = x6 + DX' + EX + ... ... ... ... (i

is a primitive of the following differential equations y = 6x5 + 2Dx + E, y" = 30 x4 + 2D, y" = 120xbbtained by differentiating (i) once, twice, thrice respectively.

The constants D and E are essential if they cannot be replaced by a smaller number of constants.

Example (i) A particle of mass m moves along a straight line (the x-axis) while subject to (i) a force proportional to its displacement x from a fixed point 0 in its path and directed towards 0 and (ii) a restoring force proportional to its velocity.

Express the total force as a differential equation.

Solution: The first force = -klx Second force = - k2dx/dt where k , and k2 are proportionally constant

Total force (mass x acceleration) m d2x/dt2 = -k,x - kzdx/dt

the differential equation is thus = m d2x/dt2 = -klx - k,dx/dt

and the solution of this differential equation is the primitive.

Example (ii) In each of the equations (a) y = x2 + A + B (b) y = ~ e " + ~ show that only one of the two arbitrary constants is essential.

Solution (a) A + B is equivalent to a single arbitrary constant thus only one arbitrary constant is involved. (b) y + A X+b = Aee eB and ~e~ is no more than a single arbitrary constant

Example (iii): One hundred grams of cane sugar is being converted to dextrose at a rate which is proportional to the amount unconverted. Find the differential equation expressing the rate of conversion after t minutes.

Solution: Let Q denote the number of grams converted in t minutes, then (100 - Q) is the number of grams unconverted, the rate of conversion is given by DQ/dt = k (100 - Q)

Where K is the proportionality constant

Example (iv): A curve is defined by the condition that at each point (x,y), its slope dy/dx is equal to twice the sum of the co-ordinates of the point, express the condition by means of a differential equation.

Fig 1

Example (v): Find the differential equation of a family of circles of fixed radius, r, with centers on the x axis

Solution: The equation of the family of circle is (X-c)~ +y2 = r ', C is arbitrary constant then 2(x-c) + 2ydyIdx = 0

x - C = -ydy/dx The required equation is y (dyld~)2 + Y2 = r2

1.0 SOLUTION TO DIFFERENTIAL EQUATIONS

1.2.0 INTRODUCTION The aim of solving a differential equation is to recover the primitive which gave rise to the equation

' 1.2.1 EXISTENCE PROBLEM gives the condition under which a

differential equation is solvable, thus a differential equation

= f(x,y) is solvable if (i) f(x,y) is continuous is single valued over a

region R of points (x,y). The function exists in the region, R

(ii) aftax, its partial derivative exists and is continuous at all points in R. aftax is read partial of F with respect to x (see chapter 9)

Region, R is defined by ( x-xo I < a , I y-yo 1 < a such that there exists in R one and only one solution

of y 1 = f(x,y) which passes through the point (Xo,yo).

6 is a very small constant

1.2.1 A particular solution of a differential equation is one obtained from the primitive by assigning definite values to the arbitrary constants. For example the primitive of the differential equation y"' = 0 is y = ~ x ~ + B x + C , i f A = B = C = O t h e n y = O , i f A = l , B = 2 , C = 5 , y = x2 + 2x + 5 and if A = 1, B + 2, C = 3, y = x2 + 2x + 3 are particular solutions, because A, B and C assume definite values.

Geometrically the primitive is the equation of a family of curve and a particular solution is the equation of one of the curves. These curves are called integral curves of the differential equation. The Primitive is commonly called a general or complete solution of the equation. A singular solution is obtained if the arbitrary constants of the ptimitive are not given specific values.

1.2.2 Examples: (i) show that y = 2x +Cex is the primitive of the differential equation dy/dx - y = 2 (1-x) and find the particular solution satisfied by x = 0, y = 3 (i.e. the equation of the integral curve passes through (0, 3)

Solution: Put y = 2x + Cex in the equation dy/dx - y = 2(1-x) thus dy/dx - 2x - Cex = 2 - 2x d y / d x = d x + C e X - 2 x + 2 = 2 + C e X therefore y = 2x + CeX is the primitive of the differential equation dy/dx - y = 2(1-x) putting the values of x = 0 and y = 3 in y = 2x + CeX, it becomes 3 = 2(0) + CeO therefore C = 3 thus the particular solution is y = 2x + 3ex

(ii) show that y = Clex + c2eZx + x is the primitive of the equation y" - 3y' + 2y = 2x - 3 and find the equation of the integral curve through the points (0,O) and (1,O).

Solntion: Y = C l e X + c2eZX + x, y1 = Clex+ 2c2e2" + 1 y" = Clex + 4c2eZX

Substitute in the differential equation y" - 3y'+ 2y = 2x - 3 to obtain C l e x + 4c2eZ" - 3 ( c l e X + 2 ~ ~ e ~ ~ + 1) + 2 ( c l e x + c2e2' + X) = 2x - 3 which si~nplifies to zero, satisfying the equation. where x = 0 , y = 0, C1 + C2 = 0

2 when x = 1, y = 0 , C l e + C2e = - 1 then C1 = -C2 = 1 and the required solution is

(e ' e ) y = x + ex - e2"

EXERCISES ON CHAPTER ONE 1. Verify and reconcile the fact that y = C1 Cos x + C2 Sin x and y =

A Cos (x + B) are primitive of y" + y = 0

2. Show that each of the following expressions is a solution of the corresponding differential equation: Classify each as a particular solution or general solution (primitive)

(i) y = 2x2 xyj = 2y (ii) x2 + y = c y y f + x = 0 (iii) y = C l e x + c2eZX y" - y = 0

2 CLASSES OF ORDINARY DIFFERENTIAL EQUATIONS

(ODE) FOR PURPOSES OF SOLUTIONS

1.0 LIST O F T H E CLASSES: As shown in 1 .O, ODE contain single independent variable and are solved by the following methods.

2.1 SEPARABLE EQUATION: Separating Variable): It is possible to arrange the terms of the differential equation in two groups each containing only one variable. The variables are therefore said to be separable and the differential equation takes the form F (x) dx + f (y) dy = 0. The

Example 1: Solve the differential equation xdy + ydx = 0

Solution: rearranging or multiplying by llxy to separate variable then dyly +dx/x = 0 Integration to solve

:.log y + log x = c1= log C

then xy = c

llxy is called an integrating factor (a factor that facilitates the solution of the differential equation)

Example 2: Solve the equation (1 + x) ydx + (1 + y) xdy = 0

Solution: Multiplying by llxy and rearranging l + x / x d x + 1 -y /dy=O orI1 - d x + I l d x + I l - d y + I 1 d y + C = O

X Y l o g x + x + l o g y + y = c l o g x y + x + y = c

19

Example 3: (a) Find the general solution of

(4x + xY2) dx + (y + x ~ ~ ) dy

(b) find the particular solution of which y(1) = 2

Solution: The equation can be rewritten thus

xdx +%= 0 (separating variables) and 1+XL 4+y

Integrating '/z In (1+ x2) '/z In (4+y2) = c,

Thus the required general solution is ( 1 + ~ ~ ) ( 4 + ~ ~ ) = C, (b) when y (1) = 2, i.e y = 2 when x = 1 then C = 16 Thus the particular solution for which y (1) = 2 is (1 + x2)(4 + y2) = 16

Example 4: Solve the boundary - value problem y' + 3y = 8

y (0) = 2 i.e. y = 2 when x = 0

We have dyldx = 8-3y, i.e. dyl8-3y = dx (separating variables)

i.e. kdy/8-3y = kdx thus - 113 In (8-3y) = X + C, Putting x = 0 and y = 2 we have -113 In 2 = CI and the required solution is '13 In (8-3y) = - 'I3 In 2 rewritten thus 1 1 l3 In (8-3y)- 13 In 2 = -x, In 2 = -3x is In (8-3yl2) = -3x

Example 5: Solve dyldx = see y tan x

Solution: dylsecy = tan x dx (separating variables) OR cos y dy = Sin xdx

Cos x

Generally variables of the equation M(x,y)dx + N(x,y)dy = o Are separable if the equation can be written in the form f l (x).gt (y)dx + f2 (x).gl (y)dy = 0 ................. (1) such that when (1) is multiplied by llf, (x).g2 (y)

equation (1) becomes fl (x)lf2 (x).dx = gl (y)/g z (y).dy = 0

Provided fz (x).gz (y) # 0

The primitive of equation (1) is obtained by integration thus

.................. Ifi (x)/f2 (x).dx +Ig1 (y)/gz (y) = C (2) is the primitive of the differential equation (1)

l/f2 (x).gz (y) is called the integrating Factor (1.F)

Multiplying by (IF) separates the variables Note: Boundary valve problem involves differential equations which should be solved to satisfy all the boundary conditions given in the statement of the problem. It is a type of particular solution for these conditions. See examples 3 (b) and (4).

2.2 HOMOGENEOUS EQUATIONS A function f (x, y) is homogeneous of degree n if f(hx,hy) = h"f(x, y) where h is a non-zero constant. e.g. (i) f (x,y) = x4 - x~~ is a homogeneous of degree 4 since f (hx,hy) = (AX) - AX)^ (hy) = h4 (x4 - x3y) = P f (x.y)

(ii) f (x,y) = eY'" + tan + tanY'" is homogeneous of degree 0 since f (hx,hy) = eY'" + tanzY/zx = eYlX + tanY/, = h0 f (x,y)

(iii) f(x,y) = xZeZ + sin x cos y is not homogeneous since f (hx,hy) = hZ - xZ + sin (Ax) cos (hy) # A" f (x,y)

The differential equation M(x,y) dx + N (x,y)dy = 0 is homogeneous if each term in the equation is of the same degree and the equation satisfies the h condition.

- Example: (i) Solve (x3 +y3) dx - 3xy2dy = 0 The equation is homogeneous of degree 3. We use transformation : vx, i.e dy = vdx + xdv to obtain x3[(1 + v3) dx - 3v2 (vdx +xdv)] = (

Using 1 as integrating factor to separate 3 x(1-2v )

the variable we have dx - 3v3dv - - 1 - 2 ~ ~

Integrating Inx + ?h In (1-2v3) = cl In x + In (1-2v3) = c OR x2 (1-zv3) =

Since v = y/x the primitive is x(l-2~~/;) = c 3 or x3 - zY = cx

(ii) Solve (2x sin h y/x +3y cos h y/x)dy = 0 The equation is homogeneous of degree 1 Using the Transformation y/x = v 2 sin h vdx - 3x cos h vdv = 0 Separating variables 2dx - 3 cos hvdv = 0 -

x sin hv

Integrating 2 In x -3 In sin hv = Inc x2 = C sin h3v and x2 = C sin h3 ylx

(iii) Solve (1 +2e"ly) dx +2edY (I-"/,) dy = 0

Solution: The equation is homogeneous of degree 0. Using transformation x = Vy, dx = Vdy + ydV

then (1 +2ev) (vdy + ydv) + 2ev (1 - v) dy = 0 (v +2ev)dy + y (1 + 2ev) dv = 0

Integrating and replacing V by "/y In y + In (V + 2ev) = Inc

and x + 2yeX" = c is the solution

2.3 LINEAR BUT NOT HOMOGENEOUS Example: (i) Solve (x+y)dx + (3x + 3y-4) dy = 0

(x + y) and 3x + 3y suggest the transformation x + y = 1 t h u s y = t - x , d y = d t - d x and we obtain tdx + (3t - 4) (dt-dx) = 0 OR (4 -2t) x + (3t - 4) dt = 0 or multiplying all through by % to give (2 - t) x + (3t - 4)dt = 0

2 Separating variables 2dx + 3t - 4 dt = 2dx - 3dt +(2/2 - t) dt = 0

2 - t

Integrating and replacing t by x + y we have 2x - 3t -2In (2-t) = cl, 2x -3(x+y)-2 In (2-x-y) =CI x + 3y + 2 In (2 -x-y) = C

2.4 MISCELLANEOUS: (i) Solve dy/dx = (y -4x12 or dy = (y - 4x1' dx

Suggested transformation y - 4x = v, dy = 4dx + dv then the equation becomes 4 dx + dv = v2dx or dx -dv/ (v2 - 4) = O((separating variables) integrating we have x + 1/4 In v + 2 = c,, In v+2 = In c - 4x

v-2 v - 2

v+2 = ~ e - ~ " and y -4x + 2 = ce-jX v - 2 y - 4 ~ - 2

(ii) Solve (2 + 2x2 Y')Y dx + ( x ~ ~ ' + 2) x dy = 0 Suggested transformation

2 x~~ = V, y = v dy = 2v dv - 4v2 dx -

x4 -- X 2

and the eqyation becomes (2 + 2v) v- dx + X(V + 2) (2v dv - 4v2 dx) = 0

-4 X

y4 7 or v (3+v)dx - x (v+2)dv = 0 Thendx- X v -1dv = 0 - -

x 3v 3 v+3

Integrating and substituting: 31nx 21nv - In (v+3) = Incl

3 ie x = clv2 (v+3) and 1 = c , xy (xZyu2 + 3) or xy (xZylJ2 + 3) = c is the solution

2.5 E X A C T E Q U A T I O N S A differential equation is exact if it can be written in the form M(xy) dx + N(x,y) dy = 0 such that a m = a N ... ... ... . (1) where a M , a N - - -- ay ax a y a x are partial derivatives

Examples: (I) (x2 -y) dx + (y2 - x) dy = 0 is exact since 2 it is in the form M(x, y) dx + N(x,y) dy and M = x2 -y, N = y - x

(ii) show that (a) (3x2 + y cos x) dx + (sin x -4y3)dy = 0 is an exact differential and

(b) Find its general solution

Solution: (a) M = 3x2 + y cos x, N = sin x - 4y3 :. a m = a N - -

ay ax

(b) Finding the solution

method 1: (Grouping term by inspection) we have the equation as 3z2 dy (y cos x dx + sin x dy)-4y3 dy = 0 i.e d(x3) + d(y sin x) -d(y4) = 0

OR d(x3 + y sin x -y4) = c Integrating, x3 + y sin x -y4 = c results which is the solution. Method 1 1 : Since Mdx + Ndy = du(x, y) is an exact differential and du = a u - dx + k d y we must have

ax a y

a~ = M and au = N - - ax ay

OR - au = 3x2 + cox x,&= sin x - 4y2 .. . . . . (2) ax ay

Integrating the first equation in (2) partially with respect to x keeping y constant, we have

~ = ~ ( 3 ~ ~ + ~ c o s x ) d x = ~ ~ + ~ s i n x + ~ ( ~ ) ........ (3) Where F(y) is the constant of the integration which may depend on y

Substituting (3) in the second equation of (2) we find y cos x + F7(y) = Sin x ay3 or F9(y) = -4y3 so that F(y) = -y4 + c, then from (3)

4 U = x 3 + y S i n x - y + c l

Thus Ndx + Ndy = du = d(x3 + y Sin x - y4 + c,) = 0

from which we must have xZ + y Sin x - y4 = c. on integration

Method 11 1 : If Mdx + Ndy = 0 is an exact equation we write Mdx + Ndy = du(x, y) =O where du is an exact differential

Thus the solution is U(xy) = c or equivalently jmdx + I(N - a/ay( j Ma x)) dy = c . . . (4) where dx indicates that the integration is to be performed with respect to x keeping y constant thus j(3xZ + y cos X) dx + Ifsin x - 4 3 ) -a / ay 1 (3xZ + y cos x) x ]dy = C i.e x3 + y Sin x + j [ ~ i n x - 4y3) -8 lay (x3 + y Sin x)] dy = C OR x3 + y sin x - y4 = C

2.6 USE OF INTEGRATING FACTORS Some differential equations of the form M(x,y)dx + N(x,y)dy = 0 which are not exact, can be made exact by an Integrating Factor for purposes of solving the equation

If M(x,y)dx + N(x,y)dy = 0 is not exact in which

by lnultiplying the equation by p, thus pM (x,y)dx + pN(x,y)dy = 0 becomes exact, hence ~ ( P M ) = a(%) F ax

p = integrating factor

Method 2.5 therefore applies in solving the differential equation. Example: (I) Show that (3xy2 + 2y)dx + (2xy2 + x)dy = 0 is not exact but that it becomes exact after multiplying by x

Solution: (i) M = 3xy2 + 2y, N = 2x2 y + x

i.e aln it aN thus the equation is not exact - - ay ax

Multiplying the equation by x, it becomes ( 3 ~ ' ~ ~ + 2y)dx + (2xy + x2)dy = 0

aM = aN = 6xZy + 2 -- ay ax Therefore the equation is now exact and method 2.5 is applied in solving the differential equation, and x is an integrating factor. By method of 4.0 the required solution is xZy + x3y" c

Integrating factors are found by inspection. The following combinations are often useful in finding integrating factors:

(9) t2 - ydx = d(y/x)

(b) xdy -,ydx = d("/,) Y-

(c) xdv - vdx = d(tan -Iy/, ) x2 + y2

(d) xdy - vdx = 'h d{ In x~y/x+y } x - Y

(e) xdx - ydy = ?h d(In x2 + y2) "2 , .,2 n 8 ,

In the next section a method of finding integrating factors for Linear Equation of first order will be presented

LINEAR EQUATIONS OF FIRST ORDER Linear Equation is of the form dyldx +P(x)y = Q(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1) Rewriting we have [P(x) y - Q(x)] dx + dy = 0 Then M = P(x)y - Q(x), N = 1 amlay = My = P(x) and ~ N I ~ X = N, = 0 Since My - Nx = P(x) - 0 = P(x) depends only on x + 1

then e-'p(x)dx is an integrating factor Multiplying (1) by e IpWx elp(x)dx (dyldx + P(x) y) = Q(x) ekx)dx which can be written thus d(el~c~)dx ,Y) = Q ( 4 e I P ( X ) ~ X

dx On integrating we have elPdx = l ~ e ~ ~ ~ ~ dx + c . . . . . .. . .. (2) y = e-lpdx J Q ~ ~ ~ ~ ~ dx + Ce -Ipdx

Example: Solve xdyldx - 2y = x2 cos4x dx . . . . . . (3)

.. Rewriting dyldx - (21x)y = x2 cos 4xdx = )/ + P(x)y = a x ) .(4) thus the integmting factor is d W d x

And ejP(x)dx = I-21x dx = -21nx ...d~(x)dx = e-?lm = elm-2 = x-2

Multiplying (4) by x'" x-'dY/dx - 2x-37 = cos 4~ Rewiting d(i4y)/dx = cos 4x (Note that d(x-'y)/dx = ~ ' ~ d ~ / d x - Integrating we have x -+=%sin4x+c o r y = Y t x 2 ~ i n 4 x + ~ x 2 The use of the term dWX should not scare you Recall the properties of exponential functions

........... dhX) = ex (3) dx

1 eXdx = ex ................ (4) ......... . 1 3ekdx = 312 8 (5)

thus to solve Linear Differential Equations first evaluate the integral j ~ d x and substitute in (2), y then results easily

Examples: (i) Solve the equation

(1 -x2) dyldx-xy= 1

Solution: Rewriting the equation in the general form of a Linear Differential Equation: y + P(x)y = a x ) we have )/-(x/lrx)y= 111-x 2

Because the equation satisfies the general form of the Linear Differential Equation, the Integrating factor to enable us solve the equation in tenndWx We notice that P(x) =-dl - x, a x ) = 111-x2 therefore j ~ d x = -jx/l-x = In (1 - x2) = In (1 - x2) ... el* . ln(l . X2)

From (2) we have (ln(1 - x'))(~) = jlll-x2 ln(1-x2)dx = dx/ln(l-x2) = sin-'x+ c :. the solution is yd(l-x2) = sin" x + c y = (sin'' x + c) I (ln(1 - x2))

(ii) Solve the Equation y '+2xy = 1 +2x2

Solution: Comparing with the general form of Linear Differential Equation, P(x) = 2x, (Q(x) = 1+2x2

:. I ~ d x = 2xdx = x2 . & P ~ X . . = ex2 = integrating factor. Substituting in (2) ex2 y = j(1 + 2x-')ex' dx = j(ex2 + 2x2ex2)dx

= xex2 + c y = x + ce-"

Example: (I) If Mdx + Ndy = 0 has an integrating factor p, which depends only on x, show that p = efP(x)dx Where f(x) = My - Nx, My = am, Nx = a N -

N 7 ax (ii) Write the condition that p depends only on y

Solution: By hypothesis: pMdx + pNdy = 0 is exact then

Since p = p ( ~ ) only. L.H.S = p amlay, RHS = p aN/ax + N dplax,

thus dp, = ('M - N) dx N - - f(x)dx

and so In p = jf(x)dx, p = &p'x'dx

(b) By interchanging M and N, x and y there will be an integrating factor p depending only on y if (My - N,)

N = g(y) SO that p =

2.8 BERNOULLIS EQUATION This equation is of the form dyldx + P(x)y = Q(x)y" ......... (1) where n + 0, 1. Use the transformation v = yl'" and the equation is similar to (2.7) and the general solution is given by ~ e ( ' - " ) ' ~ ~ " =

......... (1 -n)l Qe ('+'dx dx + c (2) If n = 0 the equation becomes linear If n = 1 it becomes separable variables

Examples Solve x d y / d ~ + y = xY3 ~ e a r r n n g i n ~ d y l d ~ + ylx = y3... ..................................... (1) This is Bernoulli equation where P(x) = l/x, Q(x) = 1, n = 3 making the transformation yl-" = V, i.e y-'= V Substituting in (2) v = y l - " - -2

- Y n = 3, P(x) = llx, Q(x) = 1 thus y-' elnx-? = -2 Jlnx-2dx + cl y - 2 ~ - 2 = -21 lnX -'dx + cl = 2x-' + c2

2 :. y- = 2x + C'X 2

y-2 = 1/(2x + c2x2) = 1/(2x + cx2)

The equation can also be solved from first principles Using the transformation yl-" = v, i.e y-2 = as n = 3 Differentiating with respect to x we have -2y3dy/dx or dyldx = -?hy3 dv/dx

The equation becomes 3

- l y d v + l = y o rdv - - 2 - = -2 2 dx x dx xy2

i.e. d_v - 2V = -2 whose d x

general solution is V = 2x + cx2, but V = y2

The solution of the equation is Y2 = 1

2x + cx2

2.9 EQUATION SOLVABLE FOR ONE VARIABLE: Example: Solve xp2 + 2px - y = 0

2 where p =y', y = xp + 2px, solving ex licitly f ~ r y dyldx = p = 2px dpldx + p2 + 2p + 2x $Idn or p(p + 1) + 2x(p + I ) d p l d ~ = 0 ................. (2)

Case 1 : p + 1 # 0 Equation (2) becomes on division by p + 1, p + 2x d p / d ~ = 0 whose solution is xP2 = C thus x = clp2 putting this in (1)

y = C + 2cIp as p =dc/x y - ~ = 2 c / d ~ / ~ y - c = 2 c / d ~ / ~ squaring (y - c)" 4cx ..........

2 i.e (y - c) = 4cx ....... (3) another form of the general solution

Case 2: p + 1 = 0 In this case p = -1 and putting in the equation 1 we find x + y = 0 as the solution. This however can be obtained From (3) by any choice of c, x + y = 0 is therefore a singular solution

2.10 CLAIRAUT'S EQUATION Solve y = p f m w h e r e p = y' Using method of (2.9) y ' = p = p + x d p l d ~ t p ~ d p l d ~ from which d p l d ~ (X fp/d p '+ 1) = 0

Case 1: d p l d ~ = 0, then p = c, general solution is y = ~ x f F + 1 = 0

Case 2: d p l d ~ # 0, then x f p -1 = 0 - -

To eliminate p note that x2 + Y2 = (p2 / P2+ 1)+(1/ p2 ) =1 The equation x2 + y2= 1 satisfies the differential equation but cannot be obtained from any choice of c and so it is a singular solution.

2.1 1 EQUATIONS OF IIIGIIER ORDER

(1) Solve y" + 2y' = 4 ........... (1) where y' = dy/dx since y is missing from the equation let y' = P, y" = dp/dx then (1) = dp/dx + 2p = 4x ....... (2) which is the linear first order equation Thus the integrating factor is esPd" Where Q(x) = 4x, P(x) = 2 Integrating factor = e6Pd" =e2"

(2) x e2" = e2" dp/dx + 2pe2X = 4xe2" This can be written as d(pe2")/dx = 4xe2" So that pe2" = 41 xe2" dx + c (on integrating) ...pe2" = 4[(x) (e2"/2) - (1) (e2"/4)] + cl :. p = dy/dx = 2x - 1 + cle2"

Integrating again we find the required solution to be 2 y = x2 - x - %cl e-2x + C2 = x - x + ~ e - ~ ' + B

(ii) Solve 1 + y" + y'2 = 0 ................ (3) Since x is missing let y = p y" = dpldx = dp/dy, dy/dx = dp/dy thus (3) = > 1 + py dpldy + p2 = 0 separating variables and integrating Ipdpl(1 + p2) + kdy/y = C1 i.e 1/21n (1 + p2) + I n 1 = C, or In [(1 + p 2 ) p ] = Cl

2 putting a = (1 + p2) y then p = dyldx =+ my separating variables ydy / = dx

Integrating -+ Ky2 = x + b Squaring gives (x + b)2 + y2 = a2 = the general solution

NUMBERICAL METHODS FOR SOLVING DIFF'ERENTIAL EQUATIONS (IN BRIEF):

Boundary value problems are amenable to approximate or numerical solution. If, for example, the boundary conditions are given thus dyldx = f(x,y) at y(xo) = yo then the following methods outlined can be used to obtain an exact solution of the problem.

STEP BY STEP OT EULER METHOD Here the differential equation is replaced by this approximation

so that y(x, + h) = y(x,) + h f(xo, yo) . . . . . .. . ... (5) By continuing in this manner we can find y(xo -t 2h), y(x, + 3h), etc to obtain an approximate solution. We choose h sufficiently small to obtain good approximation.

Example: if dyldx = 2x + y, y (0) = 1, find approximate value at y (0.5) using Euler's method with h = 0.1

Boundary condition when x = 0, y = 1.0000 so that y' = dyldx = 2x0 + yo = 2(0) + 1.000 = 1.0000 thus:

2nd line h = l(xo + 1) 3rd line h = 1 (x, - I + 1)

From equation (5) y(x, + h) = y(x) + hf (x, yo) - - Value of y in the first line + (. 1) (slope in the first line) - - 1.000 + (. 1) (1.0000) = 1 .I000 and the corresponding slop is y = 2(0.1) + 1.1000 = 1.3000

Similarly we have Value of y in third line = Value of y in second line + (0) (slope in second line) = 1.1000 + .1(1.3000) = 1.2300 and slope in third line = 2(.2) + 1.2300 = 1.6300 The remainder of the entries are obtained by continuing in this manner and we find finally y(0.5) = 1.83 15

Normal solution: The equation is linear, and the integrating factor is e-". Solving we find y 3ex - 2x - 2

Then when x = 0.5, y = 3e5 - 3 = 3(1.6487) - 3 = 1.9461 The above numerical method could be improved by using the mean of the slope corresponding to x = 0.0 and x = 0.1, using this modified slop we find the corresponding value of y to be yz. We also find the slope corresponding to this value using the given differential equation. From the improved slop, we obtain an improved average slope and from this an improved value of y, y3 is obtained. This continues until two improved values agree, then the process stops and the improved slopes are used to obtain the next line on the table. Continuing it is possible to obtain y (0.5) = 1.9483

2.122 TYLOR SERIES METHOD: By successive differentiation of the differential equation dyldx = f k y ) at y(x0) = YO

we find y' (x,), y" (x,), y"' (x,) ..... Then the solution is given by the Taylor series Y(x) = y(xo) + y' (x,)(x - x,) + y" (x,)(x -xo) + . . . . . . . . . . . . (6)

2!

assuming the series converges

Example: Solve y' = 2x + y y(0) = 1 y' = 2x + y, y" = 2 + y', y"' = y" Y" = Y"', yv = Yiv

Thusy(O)= l,y(O)= l , y " ( 0 ) = 3 , y " = 3 , y i v = 3 , y = 3 v So y(x) = y (0) + y' (0) x + y" (0) xZ + y"' (0) x3

2! 3!

When x = 0.5 1 + 0.5 + 0.375 + 0.625 + 0.0078 + ...... = 1.9461

2.123 RUNGE - KUTTA METHOD This method consists of computing: KI - - hf(x0, Yo) K2 - - hf(xo, + Mh, yo + Mkl) K3 - - hf(x0 + Mh, yo + Mkz) K4 - - hf(xo + h, yo + kd Then y(xo + h) = yo + 116 (k, + 2kz + 2k3 + k4)

Example: Solve dyldx = 2x + y, y(0) = 1 where h = 0.5, xo = 0, yo = 1, f(x, y) = 2x + y kl - - 0.5[2(0) + 11 = 0.5 k2

- - 0.5[2(.25) + 1.251 = 0.875 k3 - - 0.5[2(0.25) + 1.43751 = 0.96875 . k4

- - 0.5[2(0.5) + 1.96751 = 1 A8375

Then y(0.5) = 1 + 116 [0.5 + 1.750 + 1.9375 + 1.483751 = 1.9452 Better approximation can be made by using smaller values of h Example: if dzyldxZ - 3dyldx + 2y = x, y(0) = 1, y' (0) Using Taylor's methods.

L e t y ' = v , v ' = ~ - 2 ~ + 3 ~ By successi,ve differentiation y" = V', V" = 1 - 2y' + 3v', y"' = V", V"' = -2y"+ 3V" and we find corresponding to x = 0 y' = 0, v' = -2, y" = -2, v " = -5, V"' = - 11, yiV = -1 1, viv = -23 yv = -23, Vv = -47, yVi = -47, Vvl = -95, yV1' = -95

i.e approximately equal to y(S) = 1 - 0.25 - 0.104167 - 0.028646 - .005990 - .001020

.000147 = .6lOO3

2.124 PICARDS METHODS: This method involves integrating the differential equation dyldx = f(x,y) at y(xo)= yo and using the boundary condition, hence y(x) =

X

y(x) = Yo + lxof (u, y) dx .. . . . . . . .. . .. (7) Assuming the approximation y, (x,) = yo we obtain from (7) a nez approximation

X W Y'(X) =yo + lxof (u, Y) du Using this in (7) we obtain another approximation

the limit of this sequence, if it exists, is the required solution. Example: Solve y' = 2x + y, y(0) = 1 find approximate value of y (S) Q

Integrating and using the boundary conditions. X

= y(x) =I+ lo(2u+ y) du Let yl = 1 as first approx. we find the second approximation

X

y2 (x) = 1+ lo(2u+ y)du = 1 + x + x2 then a third approximation is y3 (x)

X

v4 (x) = 1+ JX0(2u + 1 + u + 31.1'12 + u3/3)du = I+ + 3 ~ ~ 1 2 + ~ ~ 1 2 + ~ ~ 1 1 2

X

yS (x) = 1+ Jo(2u + 1 + u + 31.1'12 + u3/3 + u4/12)du = 4 + x + 3x2/2 + x3/2 + x4/8 + x5/60)

X

y6 (x) = 1+ J0(2u + 1 + u + 3u2/2 + u3/3 + u4/8 + u5/60)du = 1 + x + 3x2/2 + x3/2 + x4/8 + xs/40 + x6/360 putting x = 0.5 y6 (0.5) = 1 + 0.5 + .375 + .0625 + ,0078 + .0008 = 1.9461

2.13 FORMULATING DIFFERENTIAL EQUATIONS AND SOLVING PHYSICAL PROBLEMS:

i. If the population of a country doubles in 50years, in how many years will it treble under the assumption that the rate of increase is proportional to the number of inhabitants?

Solution: Let y = the population at time t years and yo = the population at t = 0 And dyldt = ky or dy ly = kdt (k =Prop. Constant) Integrating In y = kt + In c or y = cekl

Att imet=O,y=y,andC=y, thus y = yoe 11

At t = 50, y = 2y0 (Doubling) 2y0= yoe" i.e e50k=2 when y = 3y0 (Trebling)

3yo = yo ek t i.e 3 = e " or (3)" = (, kt)50 = (e using the fact that e = 2 then 350 = e50kt = (e50k) = 2 i.e 350 = 2', t = 79 years

Second solution: As dyly =kdt Integrating 2y0fyodyly = 'Of0 kdt 3y0fyod~ly = k 'fO dt and In 3 = kt Then 50In3 = 50kt = t In2 and t = 50In31In2 = 79 years

ii) A 100 gallon tank is filled with brine containing 60 pounds of dissolved salt. Water runs into the tank at the rate of 2 gallons per minute and the mixture, kept uniform by stirring, runs out at the same rate. How much salt is in the tank after 1 hour? Solution: let S be the number of pounds of salt in the tank after t minutes, the concentration then being s1100 iblgallon. During an interval dt. 2 dt gallons of water flow into the tank and 2dt gallons of

h ~ n e contuning 2sl100dt = S150 dt pounds of salt flows out. Thus, the clun~c: db, ~n the amount of salt in the tank is as = S150 dt

Intcgsnting S = ~e-'" At tinic, t = 0. S = 60; hence C = 60 and S = 60e-*~' \\h.x t = GO m~riutes S = 60c-"~ = 60 (.301) - 81 pounds

(Solurio~~ is same with any unit)

iii) l'hc air in a ccrtain equipment 150cm x 50cm x 12cm tested 0.2%: Co?. E;t.cstl air containing 0.05% Co2 was then admittcd through vents at the rate o t' 9000 cubic cmlmin. Find the % Co2 after 20 minutes

Solution: 0

Volumc of equipment = 150 x 50 x 12cm = 9 0 , 0 0 0 ~ ~ . Let x = vol. of CO? in the cquiprnent at time t the concentration of Co2 then being x/90,000.dt cm3

[ h ~ i ! ~ ! : the ~nterval, dt volume of Co? entering the equipment is 9,000 (O.!M\).i)dt c cm and the amount leaving is 9000. "190,000 dt c. cm

I I(:I?u.. the change, dx in the interval is clx = 'It J(:<J (.0005 - d9000) dt = -X - 45dt

10

when t = 20, x = 45 + 135e.~ = 63 '/;I Co? 7 63190,000 = 0.0007 = 0.07%

I V . 171:ctci cerrnin conditions the constant quantity Q calorieskecvnd of heat flo\vin~ through a wall is given by 3 = I< Ad7'1dx where K = conductivity of the material

A cnt' - area of face of wall perpendiculnr to the direction ;f flow and ?' is the tenlper;~ture*xcm from that face such that T decreases as x inneases.

Find the calories of heat per hour flowing through one square meter of the wall of a refrigerator room 125cm thick for which k = 0.0025, if the temperature of the inner face is -5Oc and that of the outer face is 75Oc

Solution: Let x = the distance of a point within the wall from the outer face Q = -KA dT/dx can rewritten dT =- QIKA dx from x = 0, T = 75 to x = 125, T = -5

-5

j 7 5 d ~ = Q / K A ' " ~ ~ ~ X i.e 80 = QIKA (125) and Q = 80KN125 = 80(.0025)/125 (100)~ = 16callsec Thus the flow of heat per hour = 3600 Q = 57600 cals

v) A steam pipe 2cm in diameter is protected with a covering 6cm thick for which K = 0.0003 (a) find the heat loss per hour through a meter length of the pipe if the surface of the pipe is 2 0 0 ' ~ and that of the outer surface of the covering is 30 '~ . (b) find the temperature at a distance x >10cm from the centre of the pipe.

Solution At a distance x > lOcm from the centre of the pipe, heat is flowing across a cylindrical shell of surface area 271 x cm2 per cm of length of pipe

From problem (iv) Q = -KA dT1dx = -271xk dT/dx or 271k dT = -Qdx/x

(a) Integrating between T = 30, x = 16 and T = 200, x = 10 200 16

27tk Lo dT = - ~ j ~ ~ d x / x , 340nk = Q(ln 16 - In 10) =Q ln1.6 and Q = 340nWln 1.6 callsec. Thus heat losses per hour through a meter length of pipe is 100 ( ~ o ) ~ Q = 245,000 cals

(b) Integrating 271 kdt = -340nk .dx/x In 1.6

between T = 30, x = 16 and T = T, x = x

(Check: when x = 10, T = 30 + 170 In 1.6 = 200°C In 1.6

When x = 16, T = 30 + 0 = 30°C

vi) Find the time required for a cylindrical tank of radius 8cm and height lOcm to empty through a round hole of radius lcm in the bottom of the tank. given that water will issue from such a hole with velocity approximately V = 4.8dh. Mlsec. H being the dept of the water in the tank.

Solution: Vol. Of water which runs out per second may be thought of as that of a cylinder lcrn in radius and height V.

Hence volume which runs in time dt secs is n (11100)~ (4.8dh)dt = dloooo (4.8dh)dt

If dh = corresponding drop in the water level in the tank, the volume of water which runs out is also given by 64ndh, hence d10000(4.8dh)dt = 64ndh or dt = 64 (10000) dhldh

4.8

= -133333 dhldh, thus dt = -133333 dhldh t

Integrating between t = 0, h = 10 and t = t, h = 0, J,dt =

0

-1920Jlodh/dh and t = - 2 6 6 ~ 6 6 d h ] ~ , ~ = -266666d10secs.

vii) A spring of negligible weight hangs vertically. A mass, m, is attached to the other end. If the mass is moving with a velocity Vo ft per sec when the spring is unstretched, find the velocity V as a function of the stretch of x ft.

Solution: From Hook's Law Spring force (opposing the stretch) is proportional to the stretch. Net force on body = weight of body - Spring force, then M dvldt = mg - kx or M dvldx. dxldt = Mv dvldx = mg - kx since dddt = V

Integrating mv2 = 2mgx - kx + c When x = 0, v=v, then c = mvZ0 and mv2 = 2mgx - kx2 + mvZ0 viii) A parachutist is falling with speed l70Mlsec when his parachute opens. If the air resistance is ~ v ' / 1 0 0 l b where W is the total weight of the man and parachute, find his speed as a fq-iction of the time, t, after the parachute opened. Solution: Net force on system = Wt. of system - air resistance Then W/g dvldt = W - ~ v ' / 1 0 0 or dv/(v2 - 100) = -dt/lO

Integrating between t = 0, V = 170 and t = t, V = V t

p d v / ( v 2 - l o o ) = 1 /10rd t , 11201nv-10 1V170=- t /1~I 0 170 0 v + 10

and v = 10 .(9+8e-") / (9 - 8t-")

ix) A substance 6 is being formed by the reaction of two substances aand P in which a grams of a and b grams of P form (a + b) grams of 6. If initially there are x, grams of x and yo grams of p and none of 6 present and if the rate of formation of 6 is proportional to the product of the quantities of a and p uncombined, express the amount (z grams) of 6 formed as a function of time t.

Solution: The z grams of a formed at time t consist of az/a+b gram of a and bz/a+b gram of p

Hence at times t, there remained uncombined (x, - gram of a and (yo - bzla + b) grams of p

Then d"dt = k (x, - az/a + b)(y, - bz/a + b) =kab/(a+b)' [(a + b/a) x, - z) (a + b/b)y, - z)]

we &lain 1/(A - C) (in (A - U B - Z) -In NI3) = Kt

C;r:;c 2 : Hcl-t: tlz / (A -- z)" Kdt. Integrnti~ig S c o ~ n t = O , . Z = U t o t = t . Z = Z L t I I A - % l X u = kt 1/(A - Z) ( = kt ( and z = ~ ' k t / ( l + Akt).

0 0 x , ~ A c;ln~ilcvci. b c m has one end horizontally embedded in concrete and a

k:rcz W acting o n rhc o:iler end. Find (a) the deflection and (b) maximum

&*ticrt~on of thc 11enm assllmin; its weight is negligible.

(;I) Considcxinz the portion of the beam to the fight of x, the bending moment : ~ t s is W(L -x)

Then Ely" = W(L - x) Solvipg this subject to these conditions y' =i) at x = 0 and y = 0 at x = 0, we find y = W ( 3 L x2 - x')

6EI

(b) Max. deflection occurs at x = L and it is equal to W~ 3EI

(XI) .4 c y l ~ u l r r a l tank has 40 gallons of,a salt solution conta~niiig 2lb d i s w l \ 4 :,,ilt per gallon i e 2lblgallon. A salt solution of concentration 3lb salt per gallon flows into the tank at 4galls pr,- mlnute. How much salt is in the tank at any time if the well-stirred mixture flows out at 4 galllmin?

Solution: Let the tank contain A Lb of salt at a time t minutes , then rate of change of amount of salt = Rate of entrance - Rate of exit

dA Ib = 31b 4gaII - A Ib 4gaII -- -- - - dt min min min min min

solving dNdt = 12 . N l 0

subject to the following conditions At t = 0, A == 40gaII 2 1b.sa1t - = 80 Ib salt

gall

thus A = 120 - 4e-"O

xii) A resistor of R = 5 ohms and a condenser C = 0.02 farads are

connected in series with a battery of E = 100. If at t = 0 the chargg Q

on the condenser is 5 coulombs, find Q and the current I for

t > 0' )

Solution: R= 5ohms

Fig 3

Potential drop across R = 5 1 = 5dQ dt

Potential drop across C =Q = SOQ 0.02

Potential drop across E = -E By Kirchoff s Laws 4 d Q + 50 Q = E -

dt

d /dt(e'OL Q) = 20 elo'

Integrating and solving subject to Q = 5, at t = 0 we find Q = 2 + 3e-lo' and I= dQ1dt = -30e-lo'

1

2.14 GEOMETRICAL PROBLEMSIAPPLICATIONS 1. Rectangular coordinate:

Properties of Curve involving y' 0

9 dyldx slope of thC curve at (x, y) . ii) -dx/dy slope of the normal at (x, y)

0

iii) Y -y = dyldx (X - x) is the equation of the tangent at (x,y) where (X,Y) are the coordinates at any point on it

x - y dddy and y -x dyldx are the x and y intercepts of the tangent . *

y dddy and y dy/dx are the length of the normal end subnormal (vi ds = J(dx)' + (dy12 = dx 41 + (dy/dx)' = dy = JI + (dx/dy12 is an

element of length of arc

2b. POLAR COORDINATES Let (p.8 ) be a general point on a curve = f(8)

Fig 4 Tan cp = Pdeldcp angle between the radius vector and the part of the tangent drawn toward the initial line. ii) P tan cp = P dB = length of polar subtangent

d iii) P Cot cp = dp = length of polar subnormal

de iv) p Sin cp = J&I = length of the perpendicular from

d s the pole to the tangent

V) ds = d(dp)' + p2 (d812 = dp (dp)

vi) !'ip2 do is an element of area.

3.0 ' TRAJECTORIES Any cuwe which cuts every member of a given family of curves at a constant angle of w is called an w-trajectory of the family. .

L *

A 90' trajectory of a family is trajectofy:of the family (see fig. 5 )

called the ORTHOGONAL

Y' Fig 5

In polar coordinates, the integral curves o f the differential equation f(p, 0 - (pd0)Idp) = 0 are the orthogonal trajectories of

Problem: At each point (x, y) of curve the intercept of the tangent o n the y-axis is equal to 2xy2. Find the curve

Fig 6

Intercept of the tangent y - x dyldx = 2xy o r ydx - x d y = 2 x dx

Y Integrating xly = x2 + C o r x - x2y = C y OR from the above figure dyldx = v - 2xy2

X Problem: Find the orthogonal trajectories o f the family o f curves y = c x 2 and give a gometrical intepretation.

(a) Differential equation ui'thc t a r d y is dyldx = 2xc = 2(y/x') on ~lyldx = 2yIx

Note slope of e~rch member of the orthogonal family must be the negative reciprocal of thia slope therefore dyldx = -112yIx = -x/2y

solving we get x2 + 2y2 = K

(b) The family y -;: Cx". is a family of pp:tbolas while the orthogonal family x2 + 2 y L K is a family ellipses

,. .. . -,a Li

Fig 7

Problem: At each point (x, y) of a curve the subtangent is proportional to the square to the square of the abscissa. Find the curve if it is also passes through the (1, e)

Solution: y d d d y and y dyldx are the lengths of the subtangent and subnormal.

thus y dddy = K X ~ or d X / ~ '= Kdyly where K is the proportionality constant

Integrating K In y = - l/x+c 4 when x = l , y = e , k = - 1 + C , C = K + l The required curve is kIny = - l / x + K + 1

Problem: The areas bounded by the x-axis, a fixed ordinate x = a, variable ordinate and the part of a curve intercepted by the ordinates is revolved about the x-axis. Find the curve if the volume generated is proportional to (a) the sum of the two ordinates. (b) the difference of the two ordinates

Solution: Let A be the length of the fixed ordinate Relevant Equation of revolution = n x ~ o y % x = ~ ( y + ~ ) ............ (1) on differentiating it becomes ny2 - k dy/dx

Integrating we have y (c-nx) = k ........... ( 2 )

Using value of y from (2) to compute L.H.S of equation (1) we find np, k2 dx = k2 - k2 = k (y-A) --

(C - nx)' C - nx C - na

thus the solution is extraneous and no curve exists having property (a)

(b) Repeating the above procedure with (1) npo y2 2dx k (y - A)

we obtain on dii-fenciation ny2 = kdy/dx whose solution is y (C - nx) = k

It is seen from equation (3) that this satisfies (1) above, thus the family of curves has the required property.

EXERCISES ON CHAPTER 2 Solve the following equation 1) y dx + (1 + x)2dy = 0 2) Cot 8dp + pd8 = 0 3) (x + 2y)dx + (2x + 3y)dy = 0

In each of the following, find the particular solution indicated. 1) X dy + 2y dx = o when x = 2, y = 1

2 2 2) (x + y ) d x + x y d y = O w h e n x = l , y = - 1 3) Solve (2x2 + 3v) dx + (3x + y - 1) dy = 0 4) Solve (x4 + y4) dx - xy3 dy = 0 5) (x2 - y) dx - x dy = 0 6) Solve dyldx + 2xy = 4x 7) Solve dyldx = y + x' + 3x2 - 2x 8) Solve dyldx - y = xy5 9) Find the orthogonal trajectories of the hyperbolas xy = C

10) A body of mass m falls from rest in a medium for which the resistance (N) is proportional to the velocity (metrelsec). If the specific gravity of the medium is one fourth that of the body and if the terminal velocity is 24 metrelsec, find (a) velocity at the end of 3 secs (b) the distance travelled in 3 secs.

11) A funnel lOcm in diameter at the top and 1 cm in diameter at the bottom is 24cm deep. If initially full of water, find the time required to empty.

12) Find the time required for a square tank of side 6 metres and depth 9 metres to empty through a one cm hole in the bottom (Assume Velocity = 4.8 dh metre per sec. h being the depth of the water in the tank)

13) A steam pipe of diameter 1 meter has a jacket of insulating material (K = 0.00022) '/z metre thick. The pipe is kept at 475'F and the outside of the jacket at 75'F. Find the temperature in the jacket at a distance x metres from the centre of the pipe.

14) In a certain culture of bacteria the rate of increase is proportional to the number present.

a) if it is found that the number doubles in 4 hours, how many may be expected at the end of 12 hours

b) If there are lo4 at the end of 3 hours and 4 x lo4 at the end of 5 hours, how many were there in the beginning?

3 LINEAR DIFFERENTIAL EQUATION OF ORDER n:

A linear differential equation is generally expressed in this form: PoYr, + ply"-1 + P ~ ~ ~ - ~ -t . . . . . .. + P ,,.Iy '+ P 2 = Q . . . , ....( 1)

where Po + o, PI P2 P, and Q are functions of x or constants. A differential equation which cannot be written in this form is non- 1 inear

If Q = 0, equation (1) becomes a homogeneous linear equation. Example: xy" + 3y' - 2xy = Sin x Second Order Linear

Differential equation

'xy' - 2xy = Sin x First Order Linear Differential Equation

y (y") - x (y') + y = e*' Second Order Non-Lineax equation

xy" + 3y' - 2xy = Sin x Complete or Non-homogeneous equation

xy" + 3y'- 2xy = 0 Complementary reduced or homogeneous equation

If P,(x), PI (x) . . ... P, (x) are all constants, the equation (1) is said to have constant coefficients, otherwise equation (1) is said to have variable coefficients.

3.1 EXISTENCE AND UNIQUENESS THEOREM: If P, (x), PI (x) . ...... P, (x) and Q(x) are continuous in the interval x - xo <a and P, (x) + o then there exists one and only one solution to (1) which satisfies the conditions: y (x,) = yo y' (x,) = y', . . .yn-l (x,) = yo r 2 ) . .. . ..(2)

3.2 OPERATOR NOTATION For convenience we donate D =A = operator

dx Thus Dy, d2y D~~ ........ Dny = dy, 4 d2 d3y, dnyn

dx dx dx

D, D ~ , Dn are called differential operators and have properties like those of algebraic quantities.

We can write [ P O ( x ) ~ n + ~ ( ~ ) D"-' + .... +P,.I (x)D+P,(x)] y -R(x) ..... (3)

or briefly @ (D)y = R(x)

where @ (D) = Po (x)D " + PI (x) D + . . . .+ P,., (x) D t- Pn (x) is called operator polynominal in D.

example (ii) +3dy - 2xy = Sin x can be written

dx thus ( X D ~ + 3D - 2x)y = Sin x

LINEAR OPERATORS: An operator L is a linear opeator if for any constants A, B and function u, v, L can be applied thus:

L(Au + Bv) = AL(u) + Bl(v) . . . . . ... (40

D, D'. . . . . . and@ (D) for example are linear operators.

Problem 1 : Show that ( D ~ + 3D + 2)e4" = (D + 2) (D + 2) e4x = (D + 1) (D + 2)e4 Proof: ~ ~ 2 " + 3 ~ e ~ " + 2e4" = 16e4" + 12e4" + 2e4" = 30e4" (D + 2) (D + l)e4" = (D + 2 ) ( ~ e ~ " + e4") = (D + 2) (4e4" + e4") = (D = 2) (5e4" ) = 5 ~ e ~ " + 10e4" = 20e4" + 10e4" = 30e4" (D + 1) (D + 2)e4" = (D + 1) ( ~ e ~ " + 2e4") = (D + 1) (4e4" + 2e4" = (D +I) (be4") = 6 ~ e ~ " + 6e4" = 24e4" + 6e4" = 30e4"

This illustrates the commutative law of multiplication for operators with constant coefficients. Generally this law does not hold for operators with non-constant, that is variable) coefficients, for example xD + 1 and D - 2 are not commutative with respect to multiplication.

thus (xD + 1) (D - 20y f (D - 2) (xD + 1)y

1 3.4 FUNDAMENTAL TIIEORM ON LINEAR DIFFERENTIAL[

EQUATIONS: I ,

Find the General Solution of @ (D)y = R (x) . . ... ...( 5) where R (x) $0

This general equation is said to be complementary or homogeneous i if left hand side @ (D)y = 0 and the right hand side R (x) = 0

If Y .(x) is the general solution of this homogeneous equation @(D)y" = 0

then Yo(x) is referred to as the homogeneous or the cornplementary~ solution, then the following theorems or super position principle result:

Theorem 3 - 1 The general solution of (5) is obtained by adding the complementary/ solution Y .(x) to a particular solution, Y, (x)

Thus the general solution of a general differential equation is the sum of the complementary solution and the particular solution thus Y = Y .(x) + Y ,(x) . . . . . . . . .. (7) is the general solution of (5)

Example (iii) Solve (D - 3D + 2)y = 4x if the general solution of

(D' - 3D + 2)y = 0 is y = clex + c2e2" and a particular solution of (D' - 3D = 2) y = 4x2 is 2x2 = 6x + 7.

f Solution: The general solution of ( D ~ - 3D + 2)y = 4x2 is therefore from (7)y = clex + clex + 2x2 + 6x + 7

Note (D' - 3D + 2) y = 0 is homogeneous ( D ~ - 3D + 2)y = 4x2 is non-homo, neneous.

Thus theorem 3 - 1 requires us to find the general solution of the homogeneous equation and particuler solution of non-homogeneous ,

equation and add the two equations.

5 LINEAR DEPENDENCE AND WRONSKIANS: A set of n functions y, (x), y2 (x), ......... y, (x) is said to be linearly dependent if over an interval there exists n constants

Cl. C2, C3, ........ Cn not all zero such that

c, y, (x) + c2 y2 (x) + ........ + cn Y" (XI = 0 I

identically over the interval

Otherwise the set of functions is linearly independent.

Example (iv) If yl (x) = 2e3x, y2 (x) = 5e3' y3 (x) = e4'show that the functions are linearly dependent over an interval. If there exists constants cl = -5. c2 = 2 c3 = 0 over the interval

Solution: -5 (2e3") + 10e3' = (0) (e-4X) = - 10e3' + 10e3' + 0 = 0 therefore 2e3', 5e3" and e-4x are linearly dependent.

Example (v): ex and xex are linearly independent since clex + c2xeX = clex + c2xex = 0 separately and identically if and only if cl = 0, c2 = 0

Theorem 3 -2 The set of differential functions y 1 (x), y2(x), y3 (x) ...... yn (x) is linearly independent on an interval if and only if the determinant.

The Determinant is called WRONSKIAN of y,, y2, ....... y,. If the determinant is zero, then they are linearly dependent.

THEOREM 3- 3 (Superposition priwiple): If yl (x),y2 (x). . .y, (x) re n lined, ly independent solution d@&e n other linear equation

$, ) j Otheny=cy(x )+cy (x )+ ... : . .+cy(x)

NOTE: When a differential ecpation is Linear, : e primitive is a complete solution, otherwise ,it is not necesu the complete solution.

Examples: (i) Verify that y = e" y = xex, y = x2e%nd y = e'2x are four linearly independent solution of y" - y"' +3yV + 5y' - 2y = 0 and write the primitive

D is made up,,of constant coefficient ex only. Thus the solutions are linearly independent and the primitive is

Solution wronskian = W =

(ii) Show that the equation y" - y' 2y = 0 has two distinct solutions of the form y = eax

ex xeX x2ex e-2x

ex xe" ex x2ex + qr-" -2e-2x ex xex + 2ex x2ex + 4xex + 2ex 4ev2" , ex xex + 3ex x2ex + 6xe + 6ex -8e'2x

Solution: If y = eax. for some value of 2: is a solution, then substituting y = em in the equation satisfies the equation thus

ax y =e , y' = ae",yy" = a2eax substituling in the original equation y" - y' - 2y = 0 we have a2eax - aeaX - 2eax = eax (a2 - a - 2) = 0 which is satisfied when a = - 1, 2

:. y = e-X and y = eZX are solutions

SOLUTION OF LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS Two techniques are used, namely (i) Non - operator (ii) Operator techniques

(i) NON - OPERATOR: 1. for complementary or homogeneous equation: like

.............. .... P,(x)yn + PI ( x ) ~ "-I + Pn.1 (x)y + P, (x)y = 0 (8) Where yn = dny/dx "

To solve this we use the equation y = em", where m is constant , to obtain a solution' of (8)

If y = em" . f(D) = f(m) thus

equation (9) =P,Mn + P ~ M "-I.. + P, = 0 ............................. (10) This is called the characteristic or auxiliary equation.

Factoring (10) it becomes ............................. ... P,(m-ml)(m-m2) (m-m,)=O (1 1)

Which has roots ml, mz ..... mn

These roots are analysed thus

Case I: Real and Distinct roots:

then em", emz' ... emnx are linearly independent solution, thus from Theorem 3. 3 the required solution is

Case 2: Some roots Imaginary:

If the roots of (10) involve a + bi, also a - bi, %, a,, . . . a and a, b are real then a solution corresponding to the roots a + bi and a - bi is

y = eax (c, cos bx + c2 Sin bx) ......................................... (13)

using Euler's formulae elU = cos u + ic sin u

Case 3: Some roots are repeated: If m is a root and it occurs k times then a solution is given by

Example: i ) Solve (a) 2y" - 5y' + 2y = 0

( b ) ( 2 ~ ~ - ~ ~ - 5 ~ - 2 ) ~ = = 0

(a) The auxiliary equation is 2m2 - 5m + 2 = 0 o r ( 2 m - l ) ( m - 2 ) = O , m = % , 2

Required general solution is y = cleV% c2eZX

(b) The auxiliary equation is 2m3 - m2 - 5m - 2 = 0 or (2m+ I) (m+ I) (m-2) =Othus m=-%. -1.2

General solution y = cle'x/2 + ~ ~ e - ~ + c3eZX

ii) Solve y" + 9y = 0 or (D2 + 9)y = 0

Solution:

Auxiliary equation is m2 + 9 = 0 and m = #3i general solution is

= ~ ~ 3 ~ x + Be-3ix

= A (COS 3x + i Sin 3x) + B (cos 3x - i Sin 3x)

( using Euler's formula) or y = cl cos 3x + c2 Sin 3x

Since cos 3x, sin 3x are linearly independent, this is the general solution.

iii) Solve (D + 2)3 (D - 3) ( D ~ + 2~ + 5) = 0

I Solution: The auxiliary equation is

( m + 2 ) " m - 3 ) 4 ( m 2 + 2 m + 5 ) = ~

Roots are -2, -2, -2, 3, 3, 3, -1 +2i

The general solution is

y = (cl + c2x + c3x2) e-2x + (c4 + c5x + c6x2 + c7x3) e3' + e-'

(c8 cos 2x + cg Sin 2x)

THE PARTICULAR SOLUTION: Two important methods of

finding the particular solution (D)y = r(x) which is non-homogenous

equation are as follows:

(i) Method of Undetermined Coefficients:

Here we use a trial solution containing unknown constants

(a, b, c . . .) which are to be determined by substitution in the

given equation. The trial solution to be assumed in each

case depends on the special form of R(x) and is shown in the

following table 1. In each case f, g, p, q are given constants

and k is a positive integral.

'I'AIJLE 1: TRIAL SOLUTION K (s)

fel-

rnti ies solution

ASSUMED TRIAL SOLUTIONS Assunled Trial Solution

aepx I'cos px + g Sin px ~ , x % f ~ x " ' + ...+ fi cv ff cos px + g Sin qx) c‘l' ff-u" fflxh-l + . . . + f,.)

1:samplc: (i) Solve ( D ~ + 2D + 4)y = 8x2 + 12e-' .......................... (14)

a cos px + b Sin yx a,x%al x L ' + ...+ al eqx (a cos px + b Sin qx) e'lX (a,x% aalxL1 + . . . + at)

The complementary solution (i.e solution for homogeneous equation of the above) is e-" (c, cos d3x + c2 Sin d3x)

To get particular solution, assume the trial solution for the R.H.S. i.e. 8x2 and 12e-" to be respectively ax" bx + c and de-". Since none of this is present in the complementary solution. Substituting carefully y = ax2 + bx + c + de-" in the given equation (140 we find the L.H.S.+ 4ax2 + (4a + 4b)x + (2a + 2b + 4c) + 3dezX = 8x2 + 12e'" (that is evaluating y" + 2y' + 4 and equating to the R.H.S.).

Equating coefficients 4 2 = 8 , 4 a + b a = O 2 a + 2 b + 4 ~ = 0 , 3 d = 12

Then a = 2b = -2, c = 0, d = 4 and the particular solution is 2x2 - 2x + 4e' x

:.General solution is y = e-" (c, cos d3x + c2 Sin d3x) + 2x2 - 2x + 4e-'

Example 2: Solve (D2 + 4) y = 8 Sin 2x

The complementary solution is c, cos 2x + c2 Sin 2x. Trial solution

ought to be a cos 2x + b Sin 2x but these appear in the

complementary solution, so we multiply by x to get x(a cos 2x + b sin 2x). Substituting in the given equation.

- 4 a c o s 2 x - 4 b 2 x = 8 S i n x s o t h a t 4 a = 0 , - 4 a = 8 a n d ~ = 0 , b = -2. Required solution is: y = cl cos 2x + C2 Sin 2x - 2x sin 2x

(iii) Solve (D2 - 2D)y = ex Sin x The complementary solution is y = cl + c2ex Trial solution considering R.H.S. Y = Aex Sin x + Bex Sin x

D~~ = -2 Bex Sin x + 2Aex cos x and (D2 - 2D)y = -2 AeX Sin x -2 Bex cos x = ex Sin x

Equating coefficients -2A = 1 A = - % -2B = 0 B = O

Hence particular solution is:

y = Aex sin x + Bex cos x = ?hex Sin x

General solution is: y = C1 + C2ex - ?h sin x

(iv) Solve ( c3 + 2D2 - D - 2)y = ex + x2 Complementary function is y = Clex + C2e-' + ~ ~ e - ~ ~ Since ex occurs on the R.H.S. of the complementary function and also in the equation, we take as a particular integral:

y = ~ ~ 2 + ~ x + ~ + ~ x e X + ~ e X Then Dy = 2Ax + B + Exex + (E + F)ex

D~~ = 2A + Exex + (2E + F)ex D ' ~ = Exe" + (3E + F)ex ~ ~ ~ D ' + ~ D ~ - D - ~ ) ~ = - ~ A X ~ - ~ ( B + A ) X

+ ( 4 ~ - ~ - - 2 ~ ) + 6 ~ e " = e " + x ' Equating coefficients: -2A = 1 B + A = 0 4 A - B - 2 ~ = 0 , 6 E = 1

thus A = -% , B = %, C = -314, E = 1/16 and F is arbitrary as Czex occurs in the complement so particular solution is

thus general solution is: y = Clex + Cze-' + C3e-2" -%xv + %x - 514 + 1/6xex

(ii) METHOD OF VARIATION OF PARAMETERS: Here we replace constant coefficients with unknown functions of x, then we try to determine the unknown functions so that the equation containing the unknown functions satisfies the original equation.

Thus if for the non-homogeneous equation @ (D)y = r(x) . . . (1)we have a complementary function y = clyl (x) + czy2 (x) + . . . + c,yn (x). . . (2). Replacing the c's with unknown functions of x, the L's thus:

The method of variation of parameters consists of procedure for determining the L1 (x)yl (x) + . .. + L2 (x) so that (3) satisfies (1).

Example: (1) Solve (D' - 2D)y = ex Sin x

* Thc complementary function is C1 + c2eZX Thus y = L1 (x) + L2 (x) ezX

Obtain by differentiation Dy = 2~2e'" + (L'I + L2eZX) wesetLI +L2ezX = O ... (4) :. Dy = 2 ~ , e ~ " then DZy = 4 ~ ~ e " + 2 ~ ~ e ~ \ we set = 2Le2' R (x) = ex Sin x L2 = !hesX Sin x. L2 = %e-"Sin x + Cos X) From (4) L1= - L2eZX = !he-"in x thus L = %e-' (Sin x + Cos x) y = L1 (x) + Vie-' (Sin x - Cos x) -'/4ePx (Sin x + Cos x) = %e-%in x and general solution is: y = CI + c2eZX - !he'' Sin x

(ii) Solve (D3 + D)y = csc x Solution: y=C1+CZCosx+C3Sinx

we form y = L1+ L2 cos x + L3 Sin x Dy = (-Lz Sin x + L3 Cos x) + (Ll + Lz Cos x L3 Sin x) wesetL'l+L:!Cosx+Lg Sinx=O .......................... (1) then D, = - LZ Sin x + L3 Cos x DZy = (-L2 COS x - L3 Sin x) + (-L2 Sin + L; COS X) we set (-L Sin + L Cos x) = 0 ................................... (2) then D2y = - L2 COS x - L3 Sin x D~~ = (L2 sin x - L3 COS X) + (-IJ2 COS + L3 Sin x) we set -L2 COS x - L3 Sin x =R (x) csc x ............... (3) Adding (1) and (2) L1 = csc x and LI = - In (csc x+ Cot x) Solving (2) and (3) L13 = -1 and = Cot x so that L3 = -x and L2 = -In Sin x :.the particular solution is y =LI+L2cos x +L3sinx = - In (csc x + cot x) - cos x In sin x - x sin x and the general solution is y = cl + c2 cos x + c3 sin x - In (csc x + cot x) - cosx Insinx-xsinx.

OPERATOR TECHNIQUES: These use the methods of reductions of order and inverse operators.

Equation (8) under 3. 6-1 can be written as Q (D)y = R(x) which can be factored thus Po (D - MI) (D - M2) . . . . (D - M,)y = R(x).

Only true where MI ... M, are constants MI ... M, are the roots of equations (10) and (1 1) under 3.6 and so the complementary solution is as before.

To obtain a particular solution, let Po (D-Ml)(D-M2) ... (D- M,)y = Y, then the factorized Q(D)y becomes (D - MI) YI = R(x) which can be solved for YI. Then let Po (D - M3) ...( D - M,)y = Y2 so that (D - M2) Y2 = YI which can be solved for Y2..By continuing in this manner ,y can obtained. This method (reduction of order) yields the general solution if all the arbitrary constants are kept, while if they are omitted it yields a particular solution.

In Inverse Operator method LR(x) is defined as a particular Q(D)

f solution of y, such that Q(D)y, = R(x). We call L an inverse

Q(W

operator. Particular solutions of MD)y = R(x)~are obtained from the Table of Inverse Operators in Mathematical Handbooks. We note that>CI RI (x) + ... + Ck Rk (x)]

Q(D) =CI 1 RI(x)+ ...+ C ~ R ~ ( X ) rn 4w

Examples: (I) ~ v a l u a t e ~ e ~ ' (D - 2)

Method I: Let 1 e 4 ' = y (D - 2)

thus (D - 2)y = e4' or dyldx - 2y = e4%. Solving as first order linear equation with the integrating factor e-2X we obtain y = 1/ie4' + ce2'

Since only particular solution is needed y = %e4" Method 2: From the table of invere Operators

1 ~ ( x ) = e "je-" R(x)dx (D - M)

Example 11:

Find 1 3e-2x (D+ 1)(D+2)

Method 1 1 3e'2x = 2 (D+ 1 ) (D+2) D + 1

Method 2: Using Partial Fractions 1

SHORT METHOD FOR SOLVING LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS:

If $(D)y = R with constant coefficients has a particular solution thus

(a) If R is of the form e""

(b) If R is of the form Sin (ax t. b) or cos (as + b) y = 1 " " = Sin (ax + b) = B i n (ax + b), @(-a) f (

W 2 ) $(-a2> y = 1 eaX = cos (ax + b) = 1 c o s (ax + b). $ (-a) # 0

$(D~) $(-a2>

(c) If R is of the form x m y = 1 x = ( a + a D + ...+ a D m ) x m , a # O

obtained by expanding 1 in ascending powers of - $(Dl

D and suppressing all terms beyond Dm

Since Dn xm = 0 where n > m

(d) If R is of the form e V (x) 1 e"" V = eaX 1 V Y = - -

$(D) $(a) (e) If R is of the form xV(x)

Example: i) solve ( D ~ - 2 ~ ' - 5D + 6) y = e4X

or (D - 1) (D - 3) (D + 2)y = e4"

The complementary equation is y = clex + c2dX -t c3e-"

The particular solution is thus

(ii) Solve (D3 - 2D2 - 5D + 6)y = (e2" + 3) Complementary function is

Particular solution is

- - 1 e4" = 6 e2" (D- l)(D-3)(D+2) (D- l)(D-3)(D+2)

- - 9 = ~ e ' " 6 e2X (D- l)(D-3)(D+2) 3(1)6 I(-1)4

- - 9 = e4" 3e2" + 3 - - (- 1)(-3)(2) 18 24 2

General solution: = ~ e ~ + c e ~ " + c e ~ ~ ~ + ~ ~ ~ - & 2 x + 3

18 2 2

(iii) Solve (D' - 2D2 - 5D + 6)y = e3"

Complementary equation is y = c,ex + c2e34 + c3e-2X

but @(a) = @(3) = 0 and the short method fails

However are write bring out 1 D-3

y = 1/(D-l)(D-3)(D+2)e3" = 1/(~-3)[1/(~-1)(~+2)e~"] = 1/D-3(1/2 5 e3") = 1/10e3" dx = 1/10e3" = 1/10 e3" 0e3" e3" dx = 1/10e3" 0dx = 1/10xe3"

(from Table of Inverse operators) :. General solution is

3x y = clex + c2 e + c3 e2x + ~ e ~ ~ ' ~ ~

3.8 LINEAR EQUATIONS WITH VARIABLE COEFFICIENTS In the general differential equation thus

Po (x)d"y/dxn + PI (~)d"-'~/dx"-' + . . . + P,., (x)dy/dx + P,, (x)y = R(x)

Here P,, PI .... P,.I, P, are not constants. In this case methods of solving such equations are listed below:

3.6.1 Miscellaneous transformation of variables: a) Cauchy or Euler Equation:

This equation has the form (b, x" DrL+ b l~n - '~ I1 - l +. . .+ b1,., x D+ bll) = R(x)

where bo b, . . . b, are constants

To solve, let x =el and using the results

xD=D,, x'D~=D, (Dl-1), x3D3 =D,(D,-l)(D,-2) ..., where D, = d/dt

thus reducing the equation to one with constant coefficients. The case where R(x) = 0 can be solved by letting y = xP and determining p.

Example: (1) Let D =_d and D, => prove that if x = e' then dx dt

(a) xD = D, (b) x2D2 = D,(Dtl)

Solution: (a) Dy = dy = dy dt = dy dx = dy ex

dx dt dx d XI dt - dt

=e-' dydx = e-' D,y Then x Dy = e'Dy = D,y or XD =Dl

(ii) Solve (x'D2 + xD - 4)y = x' A

Use transformation x = et the equation becomes [D, (D,-1) + D, - 4]y = [ell3 that is using Cauchy - Euler equation or (DL, - 4)y = e3'

General solution is y = cle2' + ~ ~ e . ~ ' + 113 e3' = clx2 + c2x-' + 1/5x3

Another method Let y = X P ill the complementary equation

(x2D' + xD - 4)y = 0 we find p(p-l)xP+pxP-4xP=Oor (pP-4 )xP=0 i . e p = f 2

Thus x2 and x-%re solutions and the complementary

solution is y = klx2 + k2x-' the method of variation of parameters can be used to find the required general solution

(b) Case where one solution is known: Example: Solve ( I - X ~ ) ~ " -2y' + 2y = 0 given that y = x is a solution.

Let y = xv, then y' = xv' + v, y" = xv" + 2v' and the

given equation becomes x (1-x2)v" + (2-4x2)v' = 0 or

Integrating kdv'lv' + k(21x-211-x2)dx = cl i.e Inv' + 21nx + In (1-x2) = cl or v' = c2/x2 (1-x2) = c2 (1/x2 + 111-x2) then V = c2 (% I c + x -1 ) + c3

1-x x

and the general solution is y = vx = c2 ( d 2 In l+x - 1) + c3x

l-x

c) Reduction to Canonical Form: We can transform the general second order linear equation

Y" + P ( ~ ) Y ' + q(x)y = r(x) to the canonical form v" +f(x) v = g(x) where f(x) = q(x) - 1/4 [P(~) ]2 -Yipq x g(x) = r(x)e-mI~(x)dx

by using the substitution y = ye"RJ p'x)d"

Examples: (i) Let y = uv and choose u to obtain a corresponding

equation to y" + p(x)y' + q(x)y = r(x) with the term involving the first derivative removed. Put y = uv in the given equation uv" + (2u' + pu)v' + (u" + pu'+ qu) v = r Let 2u' + pu = 0,then u = ekp'2)dx as required,thus the above equation becomes v"+ (q-1/4p2 -1/2p9)v = rJ(p'"dx ,this is the canonical form of the given equation.

(ii) Solve 4x2 y" + 4xy' + (x2 -1)y = 0 Divide all through with 4x' and comparing with (i) above

Canonical form = v" + 114v = 0

or v = cl cos %X + c2 sin %x

then y = ve-"pn'dx = (cl cos %X + c2 sin Yix) /dx

3.8.2 EXACT EQUATIONS: - The equation [Po(x) D"+P1 (x)Dn-' + ... +Pn (x)]y = R(x) is called, exact if Po (x)D1'+P1 (x)DU-' + . . .+Pn(x) = DIP. (x)Dn-' + . . .+Pn.1 (x)] , eg [Po (x)D2+PI (x)D+. . .+P2 (x)] y=R(x) is exact This is so if and only if P", - P'l + P2 = 0 ,

Example: Solve (1-x2)y" - 3xy' - y = 1 From above P = 1 - x2, P I = -3x. P2 = -1 P", - Pl1 +P2 = 0, thus the equation is exact and can be written thus

P", - P'1 +P2 = 0, thus the equation is exact and can be written thus ~ [ ( l - x Z ) D - x]y = 1

Integrating and solving

3.9 OPERATOR FACTORIZATION (a) Show that operator

XD' + (2x + 3) D + 4 = (D+2)(xD+2) and

(b) Solve xy" +(2x+3)y1 + 4y = e2"

Solution (a) (D+2)(xD+2)y=(D+2)(xDy+2y)

= D(xDy+2y) + 2(xD+2y) = x~~~ + Dy + 2Dy + 2x Dy + 4y = [XD' + (2x+3) D +4]y

(b) The given equation can be written (D+2)(xD+2)y = e2%t (xD+2)y = Y Then (D+2)y = e2"nd Y = ?he2" cee"" We find

3.10 VARIATION OF PARAMETERS This method can be used as before when the complementary solution is known. This occurs when Po (x), P,(x) ... P,(x) are not constant but variable.

3.11 SERIES METHODS: (FROBENIUS METHOD) The equation of the type Po (x)yN + P I (x)y' + P2 (x)y = D where Po PI, P2

are polynimials can be solved thus

0

and f3 cl are constants and Z ckx k+p is the Frobenius series .ce

When (1) is put into a differential equation. An equation for p called indicicial equation and equations for the constants c, cl . . . i.e recursion formula result. By solving for P and the other constants a series solution called FRO BENIUS SERIES results.

Example: Solve 4x2 y" + 4xy' + (x2 - l)y = 0

Solution: k t y = X - c k x k + P = Zckxk+P Ck=OforK<O k? ? 0 k? 7 -

thus 4x2 ye' + 4xy' + (x2 - 1) = C4(k + P) (k + P - 1) ck x k4

z 4 ( k + p ) c k x k + ' + ~ c ~ x ~ + ~ + ~ - ~ c ~ x k + P

Writing the ser~es on the R.H.S in terms of coefficients of x ' + P, replacing summation k in the third series by k - 2 (no change on limits -w and +m )

R.H.S*z[ 4(k + P)\K + P - 1)ck + 4(k + P) ck + ck.2 - cd x k + B

= z [ 4 ( k + p12- ~ ) c ~ + c ~ - ~ ] xk+ '

Since this must be zero each coefficient must be zero thus (4 (k + p12 ck + Q - 2 = 0 . . .. . ... (1)

If k = 0 since c . ~ = 0, (4p2 - 1) C, = 0 = this is the indicia1 equation

thus co and cl are undetermined, therefore

Then the corresponding solution: Y z C k x k + ' = ~ c ~ x ~ - ~ / ~

= co cos (d2) + 2cl sin ( d 2 ) 45

NOTE: The student should compare this method with the method of Reduction to Canonical form used in solving the same differential equation earlier.

3.12 SIMULTANEOUS DIFFERENTIAL EQUATIONS: The basic procedure for solving systems of n ordinary differential equation in n dependent variables consists in obtaining, by differentiating the given equation set from which all but one of the dependent variables say x, can be eliminated. The equation resulting from the elimination is then solved for this variable x, each of the dependent variables is obtained in a similar manner.

Example - - ........ i) Solve - 2dx + - dy 4kx ye ' (1)

dt dt

Solution: Method 1 : General solution x = x(t), y = y(t)

Differentiating (2) with respect to t .............. d2x + & + dy = 0 - (3 )

dt2 dt dx

Elin~innting y in (1), (2) and (3) ( l ) x 2 + ( 2 ) x - l + ( 3 ) x 1

Solution of this equation is x = c, cos 1 + cz sin t - 1 et D + l

......... = c, cost + c2 sint -Met ( 5 ) T o eliminate x differential (I) to obtain

Eliminating x and its derivatives from (I), (2), (3), (6) From (2) we have

= - (- cl sin t + cz cos t -?h e') - 3(cl cos t + cz sin t - 1 e' D ~ + 1

= (cl - 3c2) sin t - (3cl + cz) cos t + 2e' thus x = cl cos t + c2 sin t - ?h e' y = (cl - 3c7) sin t - (3cl + c2) cos t + 2e'

METHOD 2: Using D notation ~ q u a t i o n ( 1 ) b e c o m e s 2 ( D - 2 ) x + D ( D - 1 ) ~ = e ~ ......( 7)

Equation (2) becomes (D + 3) x + y = 0 ... . . .. . . (8)

Multiply (8) by D - 1 or better we say we operate on (8) with D - 1 (or d - 1)

dt

to get (D - 1) (D + 3)x + (D - l)y = 0 Subtracting (7) to eliminate y we get [ (D - 1) (D + 3) - 2 (D -2)] x = e' or (D + l)x = e'

This is the same as (4) and the general solution follows from method 1

METHOD 3: Using determinants based on (7) and (8) thus

and 2(D - 2) D - 1 / D + 3

or ( D ~ + 1) x = et and (D + 2)y = 4e' 1 solving the two equations we have

x = c , cos t + c2 sin t - ?heL ............. (9) y = c j cos t + c4 sin - 2et ................ (10)

From method I we know that (9) and (10) contain extraneous solutions. To eliminate them (that is to reduce the number of constants) we substitute in (8) to get (c2 + 3cl + c?) cos t + (3c2 + c1 + c4) sin t = 0 for every value of t. Thus c7 = - (3cI + c2) and CJ = cl - 3c2 when put in (9) and 910), the general solution results.

(ii) Solve (D - 1) x + Dy = 2t + 1 ........... (1) (2D+ 1 ) x + 2 D y = t .............. (2)

Solution: Subtracting 2 x ( I ) from (2) we have 3x = -3t - 2 substituting x = - t - 213 in ( I ) we obtain Dy = 2t + I - (D - 1) x = t + 413 and y = ?ht2 + 4/3t + cl

The complete solution is x = - t - 2 1 3 , y = ? h t 2 + 4 / 3 t + c l

Note that

is of degree 1 in D and there is only one arbitrary constant

3.13 SINGULAR SOLUTION - EXTRANEOUS LOCI 1 .A singular solution of a differential equation satisfies the

differential equation but it is not a particular solution of the equation.

2. At each point of its locus (envelope) the number of distinct directions given by the differential equation and the number of distinct curves given by the corresponding primitive is fewer than at point off the locus.

The singular solutions are to be found by expresssing the conditions.

(i) That the differential equation (p - equation) has multiple roots and (ii) that the primitive (c - equation) has multiple roots.

Generally an equation of the flrst order does not have a singular solution. If it is of the first degree it cannot have singular solution.

If e (x,y) = 0 is a singular solution of the differential equation f (x,y,p) = 0 (p - discriminant) whose primitives is g (x,y,c) = 0 (c - discriminant), then E (x,y) is a factor of bolh discriminants. Each discriminant may have other factors which give rise to other loci associated with the primitive. Since the equation of those loci d o not satisfy the differential equation, they are called EXTRANEOUS LOCI

The simplest expression called the discriminant (involving the coefficients of an equation F(x0 = 0 whose vanishing is the condition that the equation should have multiple roots), is obtained by eliminating x in F(x) and F(x) = 0

Example: (I) Find the discriminant relations for (a) p3 + px - y = 0 (b) y = c (x - c)'

Solution: (a) Eliminating p between f(x,y,p) = p3 + px - y = 0 and a f = 3p2 + x = 0 -

a p

thus p = 3yIx substituting this in 3p2 + x = 0

that is 3(3y/x)' + x = 27 + x = 0 or 4x3 + 27y2 = 0 f

Eliminating c between 3g - cag and ag - - ac ac

in (2) we get y (4x2 - 27y) = 0

is the discriminant relation.

3 (ii) example p + px - y = 0 for singular solutions

Solution: Note that this is Clairaut Equation and the primitive is y = cx + c3

The p - and c- discriminant relation 4x3 + 27y2 = 0 is singular solution since it satisfies the differential equation.

The primitive represents a family of straight lines tangent to the semi- cubical parabola 4x3 + 27y2 = 0, the envelope.

The envelope is 4x3 + 27y2 = 0

3.14 APPLICATION OF LINEAR DIFFERENTIAL EQUATION ( 9 Motion of a pendulum: A pendulum of length ? and mass m suspended at P moves in a vertical plane through P. Disregarding all forces except that of

gravity, find its motion.

Mg Fig 8

Solution The centre of gravity, c, of the bob moves on a circle with centre P and radius. Let O (positive when measured counter clockwise) be the angle the string makes with the vertical line at time t. The only force gravity and its component along the target to the path of the bob is mg sin 9.

If s denotes the length of arc c,c then s = 1 9 and the acceleration along the arc is d2s = 1 d28

dt2 -a? Thus m. 1. d29/dt2 = -mg sin 9 or 1 d28/dt2 = -g sin 9 . . . (1) Multiplying by 2d9 / dt and integrating L (d 8 1 dt12 = 2g cos 8 + cl or de/d (2g cos 9 + c,) = f dtldl

This integral cannot be expressed in terms of elementary functions

When 8 is small sin 9 = 8. (1) becomes on rearanging d28 1 dt2 + dl. 8 = 0 whose solution is 8 = c, cosdg/l.t + c2 sind d1.t dc12 + c2' and period 2n dllg. An example of simple harmonic motion, amplitude=d(c12 +cZ2) and period = 2nd11g

@)A mass m is projected vertically upwards from 0 with initial velocity V, Find the maximum height reached assuming that the resistance of air is proportional to the velocity.

Solution Let x be distance of the mass from 0 at time t. The mass is acted upon by two forces: gravitational and the resistance.

Resistance = KV = K dx directed down - dt

Gravitational force = Mg Nat force = Mass x Acceleration, thus

o r L x + K&+g= 0 where K = mk dt2 dx

Integrating we have x = cl + c2ekt -g t ......... (2 ) - k

Differentiating (2) dx = v = -k c2e-kt - g ......... (3) - - dt k

v, = -kc2 - g/k and c = -cl = vdk + g/k2 ......... (4) putting these in (2 ) we have x =_L (g + kvo) ( I - eAkt) -q, t

k2 k

max height is reached when v = 0

(iii)

Fig 10

thus e-" = - - = , g from (4) and t i 1 In g + kvo 7-a - k cz g + kvo k g

Thus the height (max) is obtained by substituting the value pf t above in (2)

A particle p of mass 2 grams moves on the x-axis attracted toward the origin 0 with a force numerically equal to 8 x. If it is initially at rest at x = 10m, find its position at any later time assuming (a) no other forces act (b) a damping force numerically equal to 8 times the instantaneous velocity acts.

Solution: Positive direction is to the right of 0

When x > 0 the net force is To the left (i.e negative) and so is -8x When x < 0, the net force is to the right (i.e positive) and so is also - 8x

By Newton's law 2d2x = -8x or d2x + 4x = 0 dtz -z-

And x = c, cos 2t + c2 sin 2t Since x = 10cm, dx = 0 at t = 0 wc find x = 10 cos 2t -

dt

its position is shown on the graph: Fig 11

Here amplitude (max displacement from 0) is 71 secs. Frequency (number of cycles per second) is 1/71 cycles per secs. This is simple harmonic motion.

(b) The damping force is -8 - dx, regardless where the particle dt

is. Thus for example if x < 0 and - dx > 0 dt

the particle is to the left of 0 and moving to the right so the damping force must be to the left (i.e negative)

By Newton's Law

the general solution is

-2x x =.e (cl + c2t) Since x = 10, dx = 0 - dt

when t = 0, c, = 10, CL = 20

:. x = 10e.'~ (1 + 2t). The motion is non - oscillatory. The particle approaches 0 but never reaches it.

(iii) A mass m free to move along x -axis, is attracted towards the origin with a force proportional to its distance from the origin. Find the motion (a) if it starts from rest at x = x,

and (b) if it starts at x = x, with initial velocity v, moving away from the origin.

Solution: Let x denote the distance from the origin to the mass at time t

where K = mk2

Integrating x = cl sin kt + c2 cos kt ... ... ... ... ( 5 )

Differentiating ( 5 ) dx = v = -kcl sin kt + kc2 cos kt . . . (6) - dt

(a) When t = 0, x = x, and v = v, Then c l = 0 (6) and c = x from (5) and x = x cos kt

(b) when t = 0, x = x,, and v = v,. Then c2 = 0, c, = v, and x = V , sin kt + x, cos kt - k

k

In (a) the motion is simple harmonic, amplitude = x,, period - 2n: K

In (b) the motion is Simple Harmonic of amplitude equal to

and period 2n: k

(v) A chain hangs over a smooth peg, 8 metres long being on one side and 12 metres on the other. Find the time required for it to slide off (a) neglecting friction and (b) if the friction is equal to the weight of

-. 1 metre of the chain.

Solution: (a) Let mass of chain be m, length (metre) which has moved over the peg at time t be x. At time t, there are (8 - x) metres

of the chain on one side and (12 + x) metres on the other. The excess (4 +2x) metres on one side produces an unbalanced force

of (4 + 2x) %Newtons 20

Thus mZdx = (4 + 2x)mg or& = gx + 2g - dt2 20 dt2

Integrating we have

Differentiating to get velocity

When t = 0, x = 0 and v = 0 then c, = c2 = 1 and x = e('lpl")' + e-(ddlO)t -2 = 2 cosh dg/lO. t-2

Hence t =Jf cosh-I I R ( x + ~ In x + I + .iX'+vx g 2

When x = 8m has moved over the peg

t =, 10 In (5 + 2 d6) secs & (b) Here md2x = (4 + 2 x ) ~ -mg

dtl 20 20 I

Multiplying by dx/~l and integrating we have 10 (dx/dl)2 = g ~ 2 + 3gx + Cl

When t = 0, x = 0, and v = 0 and dxldt = dg/10(x2 - 3x) or dt = dlOlg.dx/ (dxZ + 3x)

83 + dx2 + 3x) + c,

(vi) An indicator of 2 henries, resistor of 16 ohms and capacity of 0.02 farads are connected in series with a battery of emf E = 100 sin 3 t. At time t = 0 the charge on the capacitor and current in the circuit are zero.

Find (a) the charge and (b) current at t > 0

SOLUTION: Let Q and I be the instantaneous charge and Ctl l .<r . l~ l at time t

By Kirchoff s Laws 2 dI + 161 + a= 100 sin 3t

dt 0.02 or since I = dQ/dt

+ 8dQ + 254 = 50 sin 3t 3 7i Solving this subject to Q = 0, dQldt = 0 at time t = we find

(a) Q =25(2 sin 3t - 3 cos 3t) +25e-4' (3 cos 3t + 2 sin 3t) 52 5 2

(b) I =* = 2 ( 2 cos 3t - 3 sin 3t) + 3 e 4 ' (17 sin 31 + 6 cos 3t) dt 62 52

The first term is the steady state current and the second which becomes negligible as time increases, is called the transient current

EXERCISES ON CHAPTER THREE

Write each of the following in operator form (a) y" +4y1+ 5y = e (b) 2y"'-2y" + y = x Invcstigate the linear dependence of ex, xex, x2ex

Solve y" - 8y' + 20y = 0

Solve D4 (D + 1)' (D2 + 4 ~ + 5 ) ~ (D4 + 4)y = 0

Solve (a) y" - 5y' + 6y = 50 sin 4x

Evaluate ( a ) L (e-'") ( b ) 1 ( 1 6 x 3 ) D + 3 D2-4

dz = 2 Sin 1 + x - dt

A mass of 20kg is suspended from a spring which is thereby stretched by 3cm. The upper end of the spring is then a motion y = 4 (Sin 2t + Cos 2t) metre. Find the equation of the motion neglecting air resistance.

Determine the shape of a uniform cable which hangs under its own weight W kg metre of length.

A horizontal beam of length L metres is fixed at both ends. Find the equation of the elastic curve and the maximum deflection if it carries a uniform load of W kg per metre of length.

4 TOTAL DIFFERENTIAL EQUAnONS

4.1 Definition:

A differential equation of general form p(x,yz ... t)dx + Q(x,y,z.. .t)dy + R(x,y.z ... t)dz ....... + S(x,y,z ,..... t)dt = 0 is called a Total Differential

Equation.

4.2 Integrability:

The total differential equation P(x,y,z.. .t)d Q(x,y,z,. . . . . . . . .t)dy + R(x,y,z ,.... t)dz .... +S(x.y,z .... t) dt =O .... (1)

is integrable if

Identically: For (3xz + 2y)dx + xdy + x2dz = 0

and by (2) we have 0 -x2 + xZ = 0, therefore the equation is integrable.

Corrollary Note: Total differential equations can be written in the form: Pdx + Qdx + Rdz = 0

where dx = aQ1az - aQlay, dv = aPIdz - ~ R I ~ X - a ~ l a y for integrability.

4.3 EXACTNESS: The condition of exactness of (1) is that:

4.4 Example: (9 Solve (x - y) dx - xdy + zdz = 0

aR=O=aR. the equation is exact

Upon regrouping we have xdx - (xdy + ydx) + zdx = 0 and integrating we have ?hx2 - xy + %z2 = k or x2 - 2xy + z2 = c which is the solution.

(ii) Solve Jdx - zdy + ydz = 0 Solution:

The equation E Pdx - Qdy + Rdz = 0 P = y2, aR I ay = 2 ~ , a m z = 0, Q= -z, aQ/ax = o, a q a z = -1 R=y,dR/ax=O,aR/dy=l . From(2)dx=-1-l ,dy=Oanddz=2y-0 Also from (Z), the expression equals zero, so the equation is integrable, but not exact, therefore find integrating factor. The integrating factor llyZ reduces the equation to: dx + ydz - zdy = 0 whose solution is: x + zly = C

ZY

(iii) Solve the homogeneous equation yzdx - z2dy - xydz = 0

Solution: The equation is integrable since:

yz(-2z + x) - z2 (-y - y) -xy (z - 0) = 0 from (2) and see corollary note under 4.2. Use the transformation x = uz, y = vz

then we have vz2 (udz +zdu) -z2(vdz - zdv) - uvzZdz = 0 Dividing by z2 and rearranging

vzdu - zdv - vdz = 0 or du - (dvlv) - (dzlz) = 0 Thenu-Inv- Inzx=Ink vz = ceU or = cedz

4.5 Pairs of total differential equation: (iv) Solve the system (y+zy)dx+(z+x)dy+(x+ y)dx=O ....... (4) (X + z)dx + ydy + xdz = 0 .. . . . .. . . (5)

Solution: Both are integrable. (4) may be written as: (ydx + xdy) + (zdy + ydz) + (xdz + zdx) = 0 and the solution is xy + yz + zx = c1 (5) may be written as: xdx + ydy + (zdx + xdz) = 0 and the solution is xz + yz + 2xz = c2 thus xy + yz + zx = cl, x2 + yZ + 2xz = c2 constitute the general solution. Geometrically, through each point in space there passes a single surface of each of the two families. Since the two surfaces on a point have a curve in common, the solution of a pair of differential equation is a family of curves. This family of curves may be given by the equation of any two families of surfaces passing through the family of curves, for example, xy + yz + zx = cl, x2 + Y2 + 2 (cI - xy - yz) = c2 also are the general solution. The solution of the simultaneous total differential equations: Pldx + Qldy + Rldz = 0 ............ (6) and P2dx + Q2dy + R2dz = 0 .......... (7) consists of a pair of equations

f(x,y,z) = cl ..... (8) g(x,y,z) = c2 .... (9)

To solve a given pair of equations, Note: (I) if (6) and (7) are both integrable, we use previous methods in 4.4, then (8) is the complete solution of (6) while (9) is the complete solution of (7) like in example (4) above. (ii) If (6) is integrable and (7) is not, then (8) is the complete solution of

(6). To obtain (9), we use (6), (7), (8) to eliminate one variable and its differential, and integrate the resulting equation.

(iv) If neither equation is integrable, we can use the method of simultaneous differential equation in chapter 3, article 8, treating two of the variables, say x and y as functions of the third variable, z

Some times it is simple to proceed thus:

and express them in symmetric form dx - - - 2 = dz - X Y z

(10) Can be written thus: Ydx = Xdy, Xdx = Zdx, Zdy = Ydz . . .. . .. (1 1) Any one can be obtained from the order two. Hence in

obtaining (10) we replace the original pair of differential equations by equivalent pair of any two of (11). If two of (11) are integrable, we proceed as in (ii). If none of (1 1) is integrable, we increase the number of possible equations.

By the principle thus: dx = dy = dz = ?,dx + mldy + nldz = ?2dx + m2dy + n2dz T PT r + T i Q - + = T + m F + n i

where ?, m, n are arbitary function of the variables such that LX + mY + nZ # 0. By a proper choice of the multipliers, it may be possible to obtain an integrable equation, say

dy = Ldx +, mdy + ndz or adx + bdy + cdz = pdx + qdy + rdz Y ~ ( + r n ~ + n ~ aX+bY+cL pX+qY+rL

In actual practice, it may be simpler at times to find by means of multipliers a second integrable equation, rather than to proceed as in (ii).

I f ;X + mY + nZ = 0, then also Xlx + mdy + ndz = 0 I f now a x + mdy + ndz = 0 is integrable, we integrate and have one of the required relations.

(v) solve the system dx + 2dy - (x +2y)dx = 0 2dx + dy + (x-y)dz = 0

Hence:

For the choice h = -113, X =-x,Y= x+~ dx = dy = dz - - -x XT 1

y, Z = 1 and the symmetric forms

From the integrable equation dx = dz we obtain - - -x 1

From the integrable equation dx = dy we obtain x2 +2xy = c2 - - -x l+v

Thus z + Inx = c, , x2 +2xy = c2 constitute the general solution

4.5.1 APPLICATION OF TOTAL AND SIMULTANEOIJS DIFFERENTIAL EQUATIONS: When a mass moves in plane subject to a force F, its acceleration continues to obey the Newton's second Law of motion F = ma. Components of the Force in the Rectangular Co-ordinates System and

Fig 12

Where Fp = radial components FO = traverse components

Fig 1.5

Example: 1. The x - component of the acceleration of a particle of unit mass, moving in a plane is equal to its ordinate and y - component is equal to twice its abscissa, find the equation of its path, given the initial conditions x = y = 0, d d d t = 2, dyldt = 4, where t = 0.

Solution: The equations of motion are:

d'x = y , d ' y = 2 x - dt' dt'

Differentiating the first twice and substituting from the second equation d4x = d'y = 2x - dt4 dt' and x = cleat + c2em + cg cosat + c4 sin a t where a4 = 2 then y = dx = Cl' (cleat + c2e." + c3 cosat + c4 sin a t )

dt' dx = a(clea' - c2e-" - cg sin a t - c4 cos a t ) - dt

and &= a3 (clem + c2e-m + c3 sin a t + c4 cos a t ) dt

using initial conditions C l +C2+ C g = 0, + C2 - C3 = O

c , - c ~ + c ~ = ~ / ~ , c , - c ~ - c ~ = ~ / ~ ~ , c ~ = ~

(ii) A projectile of mass M is fired into the air with initial velocity V, at an angle of 8 with the ground. Neglecting all forces except gravity and the resistance of the air, assumed proportional to the velocity, find the position of the projectile at time t. In the horizontal motion the projectile is affected only by the x- component of the resistance. Hence m& = -K&= -I&& K = mk

dt' dt dt

OR d'x = k& ... . . .... . . .. - (1) d2t dt

In its vertical motion, the projectile is affected by gravity and y - component of the resistance

Hence M&= -mg - rnk dy or d'y = -g -k& . . . . . ... - - (2) dtL dt dt3 dt

Integrate (1) d d d t = -kx + C1 and x = l/k C, + C2e-k' Integrate (2) d y = - g t - k y + K , - dt and y = l /k K, + kze-kt -g [(l/k)t - (~IK')]

Initial conditions: x = y = 0, & = V, cos8 dt

dyldt = V, Sin 8 when t = 0

C, = V, Cos 8 , C = 1/K V, Cos 8, K1 = V, sin 8 K2 = -1/K V, sin 8 - 1 1 ~ ' ~

Thus x = l/k (V, Cos 8) (1-eVkt) y = 1/K [(g/k + V, sin 8) (1 -e-kt) - gt]

EXERCISE ON CHAPTER FOUR

1 > Test for integrability and solve when possible. i) (y + 3z)dx + (x + 2z)dy + (3x + 2y)dz = 0 i i) dx + (x + z)dy + dz = 0

2). Solve the following pairs of equation 1) dx + dy + (x + y)dz = 0

Z(dx + dy) + )x + y)dz = 0 3) Find the family of curves orthogonal to the family of

surfaces x" y 2 + 2 z 2 = ~

4) A moving particle of mass M is attracted to a fixed point by a central force which varies inversely as the square of the distance of the particle from 0. Show that the equation of this path is conic having the fixed point as focus.

5 SERIES INTEGRATION METHOD FOR SOLVIKG DIFFERENTIAL EQUATIONS

5.J EQUATION OF ORDER ONE

The existence theorem earlier treated for the differential equation

of the form dyldx = f(x,y) . . . . . . . . . (1)

gives a s~lfficient condition Por a solution. In the proof using

power series, y is expressed in the form of a Taylor series.

Thus y = A. + A, (x-xo) + A2(x-xO) * + . . . + A,,(x ,,- xo)n . . .. (2)

Where y has been rcplaced by Ao. This series

i) Satisfies equation (1)

i i) Has the value y = yo when x = Q

iii) is convergent for all values of x sufficiently near x = xo

5.1.1 Solution of equation (1) which satisfies the condition y = yo when

x = o

(a) Assume that the solution of this series form

A. = yo + Alx + A Z X ~ + A3x3 + ... + A,lX1l in which

A. = yo and the remaining A's are constants to be determined.

(b) Substitute the assumed series in the differential equation and

proceed as in the method of undetermined coefficients discussed earlier.

Example

i) Solve y' = xZ + y in series satisfying the condition y=yowhenx=O

Solution: Since f(x,y) = x2 + y is single-valued and continuous while aftay = 1 is continuous over any range of value (x,yO enclosing (O,yo). As the conditions of the existence

Theorem are satisfied we can assume the series solution, thus

y = A. + A,X + A2x2 + A3x3 + . .. + A,x" . . .. (1)

/ With the region of convergence, this series may be differentiated term by term

/ yielding a series which converges to the derivative j.'

Hence

I y' = A, + 2A2x + 3A3xZ + ~ A ~ x ~ + . . . + ~ A ~ ~ x ~ ' ~ ~ . . .. (2)

/ and the differential equation y' - xz - y

/ = (A, + 2A2x + 3A3x2 + 4A4x5 +. . . + nAl1xu

+ . . .) - x2 - equation (1) + (Al + Ao) + (2A2 - Al)x + (3A3 - A2-1) x2 + (2& - A3)x3 + . . . + (nA, - A, )xl"' + . . . = 0

1 For this series to vanish in some region surrounding x = 0, all coefficients of each

\ power of x should vanish.

j thus i A 2 - A 1 , = 0 , A l = & = y o , 3 A 3 - A 2 - 1 = O

1 and A, = 113 + 16y0 t

These later equations are called Recursion formula and can b e used tc complete additional coefficients.

A, = USA4 = 1/60 + 1/120yo, A, = 1/6A = 11360 + 1/720yo, . . . It is also possible to obtain the coefficients as follows Since A,, = l/nA,,, A,,, = l / n - 1A11.2 An = l/n(n-1) But A,,.z = 1111-2 A,,,

Hence A,, = l/n(n-l)(n-2) .... 4. A, = [lln (n-l)(n-2) ... 4.31 (1+ %Ao) = lln! (2 + yo), n 2 3

When thc valuc of A's are substituted in the assumed series we have yo + y,x + Yzy0x2 + (113 + 1 1 6 ~ ~ ) ~ ~ + (1112 + 1/24yo)x4 + ... + lln! (2 + y0)xn+ ... y = (y + 2)ex - x2 - 2x - 2 (the required solution). If we use the integrating factor e on the differential equation (y' - x2 - y = 0), it become ye-" = jx'e-"dx = (-x2 - 2x - 2)e'X + C and = ~ e " - x - 2 ~ ~ - 2 Using initial conditions y = yo when x = x,, C = y, + 2 and y (yo + 2)cX - x2 - 2x - 2 before.

ii) Solve y' = x2 - 4x + y + 1 satisfying condition y = 3 when x = 2

Solution: Let x = v + 2 in the given differential equation and we obtain dyldv = v2 + y - 3

Adjusting the boundary conditions, x = 2, v = 0, y = 3 Assumed series solution z y = j + ~ , ~ + ~ 2 ~ 2 + ~ 3 ~ 3 + ... +A,Vn+. . . Then dyldv = Al + 2A2V + ~A,v ' + ... ~A,,v"-' :. dyldv - vZy + 3 = Al + (2A2 - Al) V + (3A3 - A ~ . ' ) V ~ + (4A4 - A,) V3 + . . . + (nA,, - A,,.1) v".' . . . = 0 equation coefficients to zero

A, = 0, 2A2 - Al = 0, AZ = 0 , 3 ~ ~ -A;' = 0 and A3 = 113

4A4-A3=OandA4= 1112 ... The recurssion formula A, = l/n A,,., yields A,, = l/n A,., = [lln(n-l)] A,,.,

i = [lln(n-l)(n-2) . . . 41 A2 = 2n!, n 1 3. Thus y = 3 + 113 v 3 + 1/12v4+ .., +2/n!Vn+ ...

I = 3 + 2 1 3 ! (x -2 )3 +2/4! (x-2)4+ ... +2/n!(x-2)n+ ... j I 5.2 LINEAR EQUATIONS OF ORDER TWO t In a differential equation thus p o (x)yW + P, (x)yt +

P 2 ( ~ ) Y = 0 . .. (1) where the P's are polynomial in x,

x = a is an ordinary point of (1) if Po (a) = 0, otherwise it is a singular point.

If x = 0 is an ordinary point, equation (1) may be solved in series about x = 0 thus

y = A (series in x) + B (series in x) in which A and B are arbitrary constants. The two series are linearly independent and both are convergent itn a region surrounding x = 0.

5.3 S I N O ~ A ~ POINTS ANb INTEGRATION BY SERIES For the equation Po (x)yW + P, (x)y'+ P, (x)y = 0 .... (1) x = 0 is singular point because Po (0) = 0

A singular point x = a of equation (1) is called regular if when equation (1) is put in the form

y" + R, (x)/x-a, y' + R2 (x)/(x-a)y = 0 R, (x) and R, (x) can be expanded in Taylor series about x = a

Examples: i) For the equation (1 + )y" + 2xy' 0 3y = 0

x = - 1 is a singular point since P(-1) = 1 + (-1) = 0

I Then y" + (R, (x)/x+ 1) y' + R2 (x)/(x+ 112

The Taylor expansion about x = -1 of R1 (x) = 2x = 2(x+1) -2

R~ (x) = -3 (x+l) F. lhus s = -1 is a singular point.

3 i i ) For the equation x y" + x ' ~ ' + y = 0, x =O is a singular point

Writing the equation in the form \I" +1 y' + l/x.\l - S X x

R1 (x) = l /x cannot be expanded in a taylor series about x=O. Thus x = 0 is not a regular singular point.

When x = 0 is a singular point of equation, there always exists a series solution of thc form.

- y = XI' C A,, x'" = A, x'" + Alxm+' + ATt2 +. . . ..

0

+ A,, s"'+". . . . (2) with A, # 0, we determine m and the A's so that (2) satisfies equation (1)

Example: Solve in series 2xy" + (x=l)y' + 3y = 0

Solulion: Here x = 0 is a regular singular point and from (2) y = Aoxl" + ~ l x ' " + ~ + A2xm+' + . . . + A, xnl+" Diffe~cntiating y1 = m&xlo-l + (rn+n) A x'" + . . . '"'" A, xl"+"

Substituting in the differential equation m (21n - l~)Aoxm-' + [(rn+l) (2m + l )Al+ (rn + 3)&]xm + [(m +2) (2rn+3)Az + (rn = 4) Al l xm-I + . . . + [(rn+n)(2m+2n-1)A, + (1n+n+2)A,.~]x "+I+' + . . . (3) Since A. # 0, the coefficient of the first term will vanish when (rnO(2m-1) = 0, this is provided m = 0 or m = ?h

Thus the series - y = &xm [ l - m+3 x + (m+3(m+4) x2 (m+ 1)(2m+l) (m+l)m+2)(2m+1(2m+3)

Satisfies the equation 2xjiW+ (x + 117 +3y = (m)(2m-l)&xm-I . . . (6)

98 The R.H.S of (60 will be zero when m = 0, m = % when m = 0 we have from (5) with & = 1, the particular solution

2 yl = 1-3 +2x -a3 + ..., 3

and when m = Y2 with & = 1, the particular solution y2=dx(1-7x+21x2-&3+ ...

6 40 80 The complete solution is then y = Ayl + By2 = A(l-3x + 2x2 - - 2x3 + . . .)

3

= B dx +...I 6 40 80

Note: The coefficient of the lowest power x in (3), also the coefficient in the right hand member of (b) has the form j(m) &. The equation j(m) = 0 is called the indical equation. The linearly independent solution of yl and y2 correspond to the distinct roots m = 0 and m = M of the indical equation.

5.4 INDICAL EQUATION: j ( ~ ) = 0 When roots m, and m2 of the indical equation are equal, the corresponding solution are indical. The complete solution is thus

When ml < m? and they differ by an integer, the greater of the roots m2 will always yield a solution while ml may or may not. In this later case we set A. = Bo (m - mi) and obtain a complete solution as

It is noteworthy that the series expanded about x = 0 which appear in these cotnplete solutions converge always in the region of the complex plane bounded by two circles centered at x = 0

Example: Solve in series 2x2y" - xy' + (x2 + l)y = 0

Solution: y = ~ o x m + ~ l x m + ' + ~ 2 ~ m + Z + ... +A,X"'+~+ ... y' = mAox "-' + (m+l) Alx " +m+2A2x '"*I + . . . + (m+n)A,x m=n-l + . . . y" = ( m - l ) ( m ) ~ ~ x " - ~ + (m+l)(rn)~~x"'-' + (m+l) (m+2)A2xm + . . . + (m+n- l)(m+n)A,x n'=nt2 + . . .

In the given differential equation, we obtain (m- 1)(2rn- 1) Aox +(m)(2m+l)Alx "+' + {[(m+2)(2m+l]A2 + &} x"'+~ + . . . + {[(m+n)(2m+2n) + 11 A, + An.2} x m-n = 0 All terms except the first two will vanish if A2, A3 . . . satisfy the recursion formula

Indical equation = I(m) - (m-1)(2m-1) = 0 m = 1, L/z

Since rn = 1, Yz will not make the second term to vanish, we therefore set AZ = 0. Thus using (1) Al = A3 = A5 = 0. then

satisfies 2 x v - x r + (x2+ l ) p = (rn-1)(2rn- 1 ) h x n and the R.H.S. = 0 when rn = 1 or rn = Yz

6 Whenm=%,A, ,= l w e h a v e y l = d x ( l - z + L - A = . . . 6 168 11088

and rn = 1 with A,, = 1 w e h a v e y 2 = x ( l - x 2 + x4 - x6 +...) ---

10 360 28080 The complete solution is thus y = Ay, = Byz

Since x = 0 is the only finite singular point, the series converge for all finite value of x.

5.5 THE COMPLETE SERIES SOLUTION The complete series solution Po (x)y" + P I (x)yl + P2(x)y = Q . . . (1) consist of the sum of the complimentary function (complete solution of (1) with Q = 0) and the particular solution of Q on he R.H.S of (1) above.

Example: 2 Solve (x2 - x)y" + 3y' - 2y = x + 3/x near x = 0

Substituting for y, y' and y", we obtain (m-4)m&xm" + [(m-3)(m+l)Al- (m-2)(m+l)&lxm +[(m-2)(m+2)A2 - ( m - l ) ( m + 2 ) ~ ~ ] x ~ + ' +. . . +[(m+n-4)(m-n)A, -(m+n+3)(m+n)A,,.l]xm+n~' + . . . = x + 3 / x 2 . . . . (2) To find the complimentary function we set LHS of (2) equal to zero and ~roceed as before The recursion formula is A, = m +n - 3 A,., . . . (3)

m + n - 4

(i.e nth term minus (n-l)lh term equal zero in the final substitution in the given differential equation).

Satisfies (x-x)y" 3y' - 2y = m (4-m)&xm-' . . .(4) The RHS of (4) will be zero when m = 0 . 4

and from m = 4 with & = 1 we have y , = x 4 ( 1 + 2 x + 3 ~ 2 + 4 ~ 3 + 5 ~ 4 + ...)

then yl = (1 +2_x +X2-y2) 3 3 3

Solution = y = C, yl + C2y2= C1 (1 + 2x + x3 +) + C2 - C_3 y2 7 7 3

To find the particular integral solution, consider each of the term on the RHS of the given differential equation.

RHS of a given differential equation = x + 3/x2

Consider the RHS of (4) and equate to the above, term by term.

Thus m(4-m)&xm-I = x, identically

i.e m(4 -m)~x~x" = x

m(4-m)Axm = x2, equating indices m = 2

With m = 2, we have 2(4-2) &= 1, so 4& = l,A = %

From the recursion Formula equation (3), putting m = 2

the recursion formula becomes

thus Al = A2 = Ag = ... 0 The corresponding particular integral solution corresponding to the term x is

x2/4

Setting m(m-4)&xm-I = 3/x2 identically and proceeding as above

We have m = -1, & = -315 and the Recursion Formula (3) for m =

Therefore the particular integral solution corresponding to 3!x"s -315~-' [ l + 3/4x = %x2 + ?hx3] l'hc required complete solution is y = A(X\ 22 = 3) +13s4 + - x c 3 -2 - -Z_x - 3L2 ( x ) 4 5 i 20 10 20

A chcck shows that the particular integral solution 2 Q y = \ - 3 - -

4 5x sntisf~rs the differential equntiotls

Since x = 1 is the only other finite singular point, the series converges in the an:iular region bounded by a circle of arbitary small radius and a circle of radius o w , both centered at x = 0.

(9 solve 2 ~ ~ ( x - l ) ~ " + x(3x+l)y1 - 2y = 0 in series convergent near x = .o

Solutio~l: Substitute u = llzy = dy/dz.dz/dx = -z2dy/dz y = 2!x3 dyldz + 1/x4 dZy/dz = ~ ' d y / d z ~ + 2x3dy/dz 'Sransforrning the given equation to 2(z - ~ ' ) d ~ ~ / d z ~ + (1.- 5z)dyIdz - 2y = 0 for which z = 0, the transformation of x = mis a regular singular point Assume series solution: y = &Zm + A~z"'+~ + A2zm+' + . . . + A,,Z'"'"+ ... and obtain condition mi2m-1) &zrn-' + [9m+1)(2m+l) A,-(2m2 +3m+2) &]Z'" + . . . + {(m+n)(2in+2n-1) A,,-[2(m+n) - (m+n)' + 11 A,.,] z"'- "-1

4 . . = O The Recursion Forrnula = A,= 2(m+n)' - (m+n) +1 A,.,

(m+n)(2m+2n- 1)

and thus the series y = & zm, (1+ 2m2+3m + 22 + 2m2+3m +2 (m+1)(2m-1) (m+1)(2m +1)

2m2 + 7m = 7x2 /(m+2)(2m+3) = . . . satifies 2 ( ~ - Z 2 ) d 9 d z 2 + (1-5z)Wdz- Zy= M(2m - ~)A~Z""' ~ & m = ~ , w i t h & = 1 w e h a v e y = 1+2z+7z /3+ 112z3/45= ... = 1 + 2 = 7 + I 1 2 + ... -

x 3x'm

and for m = 95, with A. = 1 we have y2 = Z ?4 (1+42/3 + 222'115 + 48413 15z3 + . . .)

The complete solution is y = Ay, + By2= A[ 1+2/x + 7/x2 + 112/45x5+ ...I + BX" [I + 413x + 22115~ + ...I

he series in Z converges for I Z I < 1, that is for all Z inside a circle of radius I , qentered at Z = 0. The series in x converges for 1 x I > 1, that is for a11 x outside a circle radius 1 centered at x = 0

EXERCISE ON CHAPTER FIVE

Solve in series near x = 0

(1) 2(x + x ~ ) ~ " - (x - 3x21y + y = 0 Reduction Formula = R.F = A n = -An.,

3 x2 (x+ 1 ) ~ ' ~ + x(x+ l)yt - y = 0 Singular points: x = 0. - 1, R . F = A , = - m + n = l A,,.,

m + n + l

6 LAPLACE TRANSFORM:

6.1 Introduction: Laplace transformation of a function occurs when a function, f(t), is multiplied by the e-" and the result is infcgrated with respect to t from t = O to t = D. A new function F(s) is obtained when the integral exists. The method simplifies many equations for easy solutions. Laplace transform is used to solve various linear differential equations with constants and with boundary conditions. Physical problems in engineering and allied subjects involve different linear differential equations with initial conditions. The laplace transform method is specially used in obtaining their solutions.

The new variables, the new F(s) is considered real in calculations in this chapter. However all results which hold when s is real s > so for some real value of s also hold when s is complex and real part of s >So

Hence mathematically we have X[Cf(t))] = F(s) = je-" f(t) dt . .. ( I )

X(t) is the Laplace Transform and it exists if integral exists or

converges. It does not exist if the integral does not exist or diverges.

In practice there will be real number so such that the integral (1) exist

for s > so and does not exist for s I so. The set of values of s for the

existence of the integral is "The range of convergence or existence

The symbol X is called the Laplace Transform Operator.

Then,

~ c c , f , c t ) + c2f2 (t) = C I X ~ . I ~ ~ ) + c2x{f2 (t) I ... (2)

6.1 Sufficient Condition for the existence of Laplace Transforms:

(a) Peacewise Continuity - A function , f(t) is said to be peacewise continuous in an interval if:

(i) The interval can be divided into a finite number of sub- intervals in each of which f(t) is continuous. 4

(ii) Th limits of f(t) as t approaches the end min t of each sub- interval are finite

A peacewise Continuous function is one that has only finite number of finite discontinuities shown below:

b) , Exponential Order: A funchon f(t) is said to be exponential order for t > T if we can find constants M and a such that f(t) I Me" for t >T Based on the above we have the following theorems.

Theorem 7 -1: If f(t)is peacewise continuous in every finite interval 0 < t < T and is

of exponential brder fort > T then X(f(t)) exist s a > a

Howcver if such conditions are not satisfied, X(f(t)) may still exist.

For example X(f" ) exist though t'" is not peacewise continuous in

0 1 t l T .

Theorem 7 -2 If f(t) satisfies the condition of Theorem 7 - 1

Then Lim X(f(t)) = Lim F(s) = 0 t --)m t-

thus if Lim F(s) = 0, the f(t) cannot satisfy the condition S+-

of theorem 7 - 1

6.2 Inverse Laplace Transform:

If X(f(t)) = F(s), we call f (1) the Inverse Laplace Transform of F(s)

thus: X-I (f(s)) = f(t).

Example:

Since X(t) = 1 we have x-' 1 = t - - 52 52

X-'is the Inverse Laplace Transform operator thus

Y2-' (C, F1 (s) + C2 F2 (s)) = C1 FI (t) + C2 F12(t)

6.3 Laplace transform of derivatives: Transdorms are useful for solving linear differential equations hence the need for Laplace transform of derivatives.

Theorem 7 - 3 If f (t) is continuous and have piecewise continuous derivative f(t)

in every finite interval 0 2 t 2 T and f(t) , f (t) , . . . f"-I (t) are of exponential order for t > T then

X(f'"'(t)) = S.X(f(t)) - Sfl-'f(O) - Sn-Zf(0) . .. f"-1 (0)

6.4 Unit Step Function: It {s possible to express various discontinuities in terms of the unit step function

The Unit Step Function called Heaviside's unit &ep function is Defined as

t O , t < a u (t-a) = I , t,< a as shown below:

Rg 14 Similarly,

If x-1 = [F(n)(s)] = (- 1 )"t"j" (t) Theorem 7.4 (Periodic Function): If f(t) has period P > 0 i.e f (t+p) = f (t) Then

P

Theorem 7- 5 (Integration)

P f(t) = F(s) then

~jlf(u)dul = F(s) - S

Similarly if X-1 (F(s) = f(t)

(q ; Then P- S =If(u)du

Theorem 7 - 5:

If Lim f (t) - exist and ZCf(t) = F(s) t

t+O 00

Theorem 7 -6 (Convolution Theorem) If PCf(t) = F(s), P[g(t)] = G (s) then

P kCf<u)s (t-u)dul= F (s) G(s)

Similarly if S[F(s)] = f(t). X,- 1 [C(s)] = g(t) I

then [F(s)C(s)] = I ~ f ( u ) ~ (t-u)du

We call the above integral the convolution off and g and write 1

the convolution is commutative thus f *s = s*f

6.5 OTHER THEOREMS OF LAPLACE TRANSFORM: Theorem 6.5.1 (First Transform Theorem):

If X(f(t)) = F(s) then

X(eqf(t) = F(s- a).

The inverse also holds thus if) X- 1 (F(s)) = f(t) then

XI (F(s - a)) = eq f (t)

Theorem 6.5,2:

If X(f(t) = F(s) then

X(u(t - a)f(t- a)) = e-asF(s)

also if XI (F(s)) = f (t), then XI (e-as F(s)) = U(t- a) f (t - a )

Theorem 6,5,4:

If X(f(t) = F(s) then if n = 1, 2. 3 . . . X(tn f (t)) =(- 1) ndnf = (- 1) nF(n)(s) -

dsn

Similarly if X1 (F(s)O = Cf(t) then X1 [Fen) (s)] = ( - I ) "tn f(t)

6.6 Table of Transforms: A table of functions and their transforms are provided to facilitate the determination of both direct and inverse transforms, (extracted from Mathematical Handbook, Spiegel Murry, 1968 pp 161 - 171.

91. TABLE OF SOME LAF'LACE TRANSFORMS

3. I / ~ " n = l , 2 , 3 - p-1 , o!= 1 (n - I)!

. . 7. Vs2 = a2 sin at -

a 8. S cos at

SW (n- 1)-at tW2 eat >1 - (n-I)!

eb' Sin at a

sin hat a

cosh at

13. 1 etbt Sinh at ( ~ - b ) ~ - a ~ a

s>O cos mt

s>O sin ot

s d d cosh at

s>/d sinh at

1 - (e-bt- e-") a-b

1 - (e-bt- e-") b-a

t sin at

t cos at

Sin at - at cos at

a2n-2/(~2+a2)n at/2"-' (n- l)![d.x/2(at)]n-312~n,12(at)n > ?h

General Properties of Laplace Transforms:

e - as f (s)

s2f (s) - sF(o) - F1(o) ~ " ( t )

L I

I.. .I F(u)dun P(aK)eakt

P(s) = Polynomial

Q(s) = (s-al)(s-a2) . . . (s-a,)

Degree < n

Where a,, a2, a3 . . . a, are all distinct

Table of transforms under 6.6 is explained as follows: Numbers 24 - 27 involve Singularity functions, Dirac delta function for indeterminate situations. Numbers 28 - 31 refer to J, and I, called Bessel function of order m (to be discussed).

6.8 Partial Fractions: Though the Theorems under 6.5 useful in finding especially inverse

Transforms, in partial problems, polynomials of high degree are

encountered and the use of partial fraction will be handy to solve the

problem.

6.9 Illustrative Examples: - 1. Prove that X(ent) = 1 if s > a -

s-a

M

- - e-(s-a)t - - 1 -I - S-a s-a provided s > a

0

2(a) Prove that X(tP) = T(p+ 1 ) - SF' ifs>O,p>-1

(b) Show that if p = n, a position integer then

X(tn) = n! - ~ n t l where s > 0

(a) XBtp) = [e"'tpdt = ]I 0

Let st = U. for the integral to converge s > 0 M

I = - 1 I U P S - " ~ U = T ( ~ + I ) - SP+' 0 SF', p >-1

(b) Integrating by parts 00 00 00

n p + 1) = bpe-' dx = (xP)(-ePx) \- kpx-l)(-e-x)dx 0 0

i.e T(p+ 1) = pT(p)

If p =, n, then T(n+ 1) = nT(n) = n(n- l)T(n- 1)

N(n- l)(n-2)T(n-2) = n(n-1) . . . 1 T(1)

thus r(n+l) = n! and so from (a)

Find the Laplace Transform of the following:- (a) 3e-41 (b) 4 cos st (c) - 3 4

Find the Laplace Transform for the following:- 3t4 - 2t3I2 + 6 (b) l/t2

Solution:

X(3t4 - 2t3I2 + 6) = 3X((t4)-2X(t3I2)+6v(1)

Since the integral does not converge, the Laplace Transform does not exist.

(iv) FindX(Sintcost) Solution:

Sin 2t = 2 sin t cos t, thus sin t cos t = '/z sin 2t

:.X(sin t cos t) = .X(M sin Zt) =>1 2 = 1

2 s2 +4 s2 +4

(v) (a) Prove that if Cf(t)) is peacewise continuous in every finite interval 0 5 t I T and is of exponential order t > T, then

XCf(t)) exists for s >O

(b) Also prove that if Cf(t) satisfies the condition of (a) -. above then Lim L(f(t)) = F(s) = 0

(a) Wehave 00

F(s) = XCf(t)) = fe-" (f (t)dt 0

Now since f(t) is peacewise continuous for 0 < t < T so also is e-"and thus the first integral exists. To show that the second integral exists we use the fact that

f(t) so that

For s >a and the required result is proved (b) As in pma

Since f(t) is peacewise continouous for O I t I T it is bounded, i. some constant K. From this equation (1) above we have

F(s) I je"' K.dt + m I j ~ d t + m = K+m T 7 0 -- s-a s-a s-a

Taking the limit as s+w, it follows that Lim F(s) = 0 required S--;)a,

(vi) Prove that (a) X {e2') - t+4

exist and hence that (b) l m $2tj - s+- t+4 = O

(a) In every finite interval, is continous also for all t > 0.

21 21 (t+4)

e <e - t+4 4

So t h a d is of exponential order. Thus by first part of (v) above the t+4

Laplace Transform exists (b) This follows from example (v) above and the results of (a)

(vii) Prove that if X (f(t)) = F(s)

X(g)(t> = G(s)

thenX[l f(u)g (t-u) du ] = F(s)G(s) 0

Proof: m m

F(s) = ksU f (u)du, G(s) = Je+' g(v)dv, 0 0

m m

then F(s)G(s) = [Je"" f (u)du] [Je-sv g(v)tl\ ; 0 0

= J t m t Je-" [[I =f (u)g(t-u)dudt Je-sl [J =f (u)g(~-u)tlu]r:t 0 0 0 0

Using the transformation t = u+v from the uv p l a x lo [ I I C u t plLlnc. sin t, t<n

(viii) Find X(F(t)) where f (t) =

(0; t < n

where f (t) = sin t + t - sin t; t > n

= Sin t + (t -sin t) u(t-n) = Sin t + [n + (t-n) + sin (t-n)] (t-n)

,but X(F(t)) = X6in t) + X[n + (t-n) + sin (t-n);~;t-n) p.1 18

= 1 - +e"[" [ n + 1 +l] s2+ 1 s 2 s2+1

(ix) ~ i n d X - '

' [Sj Now since X s + = cos 2t

2s + 3

~ h u s 2' L?YFFJ = 2et cos 2t + 5e1 sin 2t

= !he1 (4~0s t + 5 sin 2t) p.119

(x) Find X- ' (ln (1 +A)} S

(xi) Find X-' k ~ - 2 ) ( ~ + 1 ) ( ~ - 3 ) J

Equating coefficients 2s' - 4= A(s+l)(s-3)+ B(s-2)(s-3) + C(s-2)(s+l) by letting s=2, -1 ,3 we find A = -5 B = -L C =7-

3 6 2

thus X-I (S-2)(S+i)(S-3) I - 1

(xii) Solve y" - 3y' + 2y = 2e-', y(0) = 2, y'(0) = - 1 p.120

Solution: ' Taking Laplace Transform of the above s2y - Sy(0) - y1(0) - 3[sY - y(0) + 2Y' =L

s+ 1 Then using y'(0) = 2, y(0) = -1 and solving this algcbsni~ cqua:ion for Y, we find using partial fractions,

Y = 2s'-5s-5 =-1 - 4 7 +3 (s+2)(s-l)'(s-2) 3+

(s+ 1) (s-I) (s-2) thus taking Laplace Transform we obtain the required solution y = l i t + 4e' - 2 -

3 3

(xiii) Solve yiv + 2y" + = Sin t; y(0) =2, yt(0) = 2, y" = 3

Solution: Taking the Laplace Transform of the above equation [ S ~ Y - ~ ~ ( 1 ) - s~(-2) - ~ ( 3 ) - 01 + ~ [ S ~ Y - ~ ( 1 ) - (-2) + Y = 1

75 which can be written (s4+2s2+1)Y= 1 +s2-2s2+5s-4

s 5

Using the following

3 Sin t - 3t Cos t - 1 tZ Sin t - - 8 8

and the required solution is y = (1 + 2) Cos t - ( 2_1 - 2t + ~ t ' ) Sin t

3 8 8

Z' I"'] (s'+I>' = 2t Sin t - Sin t + t cos t

and the required solution is y = ( 1 + 3) Cost - (3 - 2t + l t 2 ) Sin t

3 8 8

(xiv) A resistor of R = IOohms, an indicator of L = 2 henries and a batteq of E volts are connected in series with a switch S, see figure below At t = 0 the switch is closed and current I = 0. Find I for t > 0 if (a) E = 40 volts (b) E = 20e - 3t (c) E = 50 Sin st

Solution: The differential equation according to Kirchoff s laws:-

R = PO ohms

Fig 15

(a) If E = 40, the LapPace Transform of (I) is

[(ST - i (ON] + 57 =a S

where i = X ( 9 , using I(O) = 0 and solving for i we find

(xvi)

(c) If E = SO Sin St

then [si - i(0)]+5i = 125 5 5

Then 1 =A eS5' - J Cos 5t +d Sin St 2 2 2

3 0 c t c 2

iff (t) = 1 2< t < 4 find Xf (t) Ot24

Solution: E 2 4 00

X f (t) = leq"'f(t)dt +le-"f (t)dt + le-"f (t)dt 2 4

Prove that (a)S(Cos wt) = s (b) X(Sin ot) = o rn -

S-o

Proof:

(xvi) Prove that ( ~ ) ~ ( C O S at ) = S (b) de(Sin ~ t ) = o ST -

S- W

Using the real and imaginary parts together we have

je-s'(~os ot dt + Sin ot) = s + io 0

C TzF

or by direct integration 00 00

J 4 a e Cos ot dt =e"'(wSin ot - s Cos wt) = . 0 s2 +02 I s2 +02

m m

Je"'Sin ot dt =e-"(wSin ot - s Cos ot) -A 0 . s2 +oZ I- s2 +02

(xvii) Solve for y(t) in the equation t

y(t) = + j y(u) Cos (t-u)du 0

Solution: Taking Laplace Transform

4

,124

The given equation is called an Integral Solution because the unknown function occurs under the integral.

(xviii) Solve - ac = D a2c at ax'

Boundary condition C(x = 0, t) go 2

c(x = 00, t) = 0 Initial Conditions c( x, 0) = 0 Taking Laplace Transform of (I), using t as an indepent variable

= S.C - c(x, 0) = Dd2c (3) dx7

m

Note: X(C(x,t)) = Ie's'.~(x,t)dt = C(x, s), ZIC(x, s) = C(x,t)

From (3) Dd2c - s.C = c(x, 0) = 0 d X 2

from initial conditions. This is an ordinary differential equation in s whose solution is:

Taking inverse transform of (4)

erf LL. % c& - 2dDt) . . ,. (5) the solution of (I)

12.5

(xix) Prove that (a) W ( x ) ] = s B[Y (x) J - Y(0) (b) g{Y"(x) ] = s2{Y(x) - 5y(0) - r ( 0 ) under suitab1e condition of Y(x)

Solution: c0

(a) L(T'(x)) = j e"" Y(x) dx 0

M + s e'" Y (x) ds )

0

M = s 1 e-'" Y (x) dx - Y (0)

0

(b) Let U(x) = Y'(x)

From above B'{U'(x)) = s B{u(x)) - U(0)

thus B'{Y"(x)J = s 9 ( Y 1 ( x ) ] - Y'(0)

= S[S .!Z{Y(x)] - Y(O)] - Y'(0)

= s2 ~ { Y ( x ) ) - sY(0) - Y'(0)

EXERCISES ON CHAPTER 6

Find the Laplace Transform of each of the following:- (i) 4ne2' (ii) (e3' - e34' (iii) Sin 2t Cos 2t

{ -1 0 S t 1 4

Find Xu@) in (a) f(t) = 1 t > 4 t + 101t>3

(b)

Does the result X(f (t) = sX(f (t) - f (t) hold for -1 t>O

(a) f (t) = dt (b) f (t) =I (c) f(t) Jt

Find (a) x' {-a s2 - (b) XI {-} Solve each of the following:- (a) y"(t) - 3yt(t) + 2y(t) = 1, y'(0) = 0 (b) ~ " ( t ) - 16y(t) + 32t; y(0) = 3, ~ ' ( 0 ) = -2 (c) yl'(t) + y'(t) = t +1

Solve the simultaneous equations:

2x (t) - y(t) = 4(1-e-') 2x(t) - y(t) = 2(1 + 3e.") Subject to x(0) = y(0) = 0

A mas is suspended from the end of a vertical spring of constant K (force required to produce a unit stretch). An external force F(t0 acts on the mass as well as a resistive force proportional to the instantaneous velocity. Assuming that x is the independent displacement of the mass at time t and that the mass starts from rest a tx=O (a) Set up a differential equation for the motion and

LEGENDRE, BESSEL AND GAUSS EQUATIO3S

These equation are differential equations whose solution gives

rise to Legendre, Bessel and Gmss functions.

Legendre and Bessel functions have important applications in

Mathematics, Physics, Elasticity, Fluid Flow, Electrical Field Theor)

cylindrical coordinates.

Generally the three equation have many interesting properties.

1. (1 -x')~" -2xy' + p(p+l)y = OjLegendre equation

11. xZy"+ xy' + (~'-k ')~ = 0 + Bessel Equation '

. . . 111. (X-X ' )~" + [y - (a + 0 + l)x]y1 [yap = 0 + Gauss Equation

. *

7.1 LEGENDRE EQUATION:

( I - X ~ ) ~ " - 2xy' + p(p+l)y = 0 ..... (1)

In chapter five, the equation of this form was solved in series

Convergent near x = 0. Under certain conditions on p, solution can

be obtained convergent near x = =

Using substitytion x = llz, (1) becomes

For which z = 0 is a regular singular point

y = A. Z"' + A~z"*' + A,z"*~ + . . . + A1,Zm+" + putting appropriate form of y in (2) we haw

(-m(m-I> + p(p+l) ) & 2"' + [-m(m+ 1) +p(p+ l)] A~z'"'

+{[-mtm+l)tm+2)+p(ptl)] A, + m(m+l) + A, ]znk'+

+ { [-m(m+n)(m+n- I)+ p(p+ l)] A,

+ (m+n-2)(m-n-1)Al,.z }Z'"+" + . . . = 0

with A = 0 , and A ,, = (m+n-2)(m+n- 1) A (m+n)(m+n- 1)-p(p+l)

satisfies the equation: (z4- z2)d21r_+ 2z3dy+ p(p+I)y = [-m(m-1) +p(p+l) AZ'"

dz- dz = (m+p)(-m+p+ 1) A ,J'"

From m = -p with A = 1 we obtain

Thus Ayl + By? is a complete solution convergent for I x I > 1, providcd that p #

95. 317, 512. or p f -312, -512, ...

II' p IS a positivc intcgcr including 0, then y which is a polynomial, up (A). I ' u t l in~ p = 0, 1, 2, 3, . ... N in (3) wc have

2 LIO (A) = I , ~ i , (x) = x, 11.4~) = x -113, 113 (x) = x3-3x15 ...

\\'here Y2k denotes the greatest 5 %k

Thc polynon~ial defined by

are called Legendre Polynomials

First few of these are:-

p4 (XI - - 1.3.5.7 XI4 (x) = 57x4 - 2 3.5x2 + l .3 - - - 4! 2.4 2.4 2.4

Thus Pp(x) is a particular solution of the Legendre Equation.

7.2 THE BESSEI, FUNCTIONS

' 2 The Bessel Equation is thus xZy + xy + (x"-k )y = 0 . . .. (1)

x = 0 is a regular singular point, k 2 0

Assumed series Solution

Substituting appropriately in (1) we have

(m2-k2)Aoxm + ((m+l) * -k2] Alxm" + { [m +2)"k2] A2+ AO] xm+' + . . 2

+{(m+n) -kl An+A,.2}x .. = 0

we take A. = 0 and An = - 1 .An-2 3- k2

1 1 -2 Then y = Aaxm{ - (m+2) -k x' + [(m+2)'-k][m+4)'-k'] .x4

2 2 2 2 2 satisfies the equation x y" +xy' + (x -k ) = (m -k )Aoxm

Noic (I) y I - y?. if k -0 (ii) y1 is I I I C ; I ~ I ~ ~ ~ ~ C S S if k is a 11cgative ii~lcges (iiij y l is ~i~c;lr~ir~glcss i f k is a positive integer. Except for these cases comple~c solution of llcssel Equation is

y = Ayl + By? convcsgcnt for all x # 0

(2) is a solution of (1)

Also J.h (x) - (-1)I'Jk (x) w h c ~ e k is a positive integer including 0.

Solution of (1 ) :is statcd in (2) also includes

and Jlcx) k=l

which are frequently used

The general solution of Bessel equation (1) is given by

y = CIJ, (x) + CzYll (x) ... . . ... (3) The solution J, (x). has a finite limit as x :~pproachcs x r o , is called

f3essel Function of first kind and order n.

The solution Yll (K) , has no finite limit (1.e unboui-tdcd) as x

approaches zero, is called the Bessel Function of the srcond kind and

order 11 or Neumann function. If the independent variable x becomes

Ax where A is constant then equation (1) becon~es

xZy' + xy' + (A2x2 - n21y = O . . ... (4)

with general solution y = CJ, (Ax) + CpY, (Ax) . . ... ( 5 )

Bessel Function of the first kind and of order of n can also be written

Where I'(n-1) is the gamma function, if n is a positive integer, I7(n=1) = n!, r (1) = 1

These functions are oscilliatory in nature (see 7.2.6) If n is half an odd integer, J,(x) can be expressed in terms of Sines and Cosines.

Replacing n by --n in (6)

7 . 2 . I?cssel Function of second kind: If n is not an integer, J, (u) and J.,, (x) are linearly independent and the generat solution becomes y =AJ,,(x)+BJ.,(r;);n+O, 1 ,2 ,3 ...... (S)

J,,(x) Cos nx - J.,,(x)

pz - J-F (s); n = Q, i, 2,3 ...... (9) Sinpx

for the case wllere n = 0, 1,2, 3 . . . we obtain the following series Expansion for Y,(x); Yn(x) being part of ~ h c soIution for the Uessel Equation.

n= 1

Y,,(x) = 2/71 {In (~12) + y } J,,(x) - 1In C (n-k-1)!(~/2)~"-' k=O

(~12) 2"" -1ln x(-1) ' [O(k) + O(n+k)]k!(n+k)! . . .. .... (10)

where y = 0.5772156 is called Euler's constant and O ( k ) = 1 + % + 1 / 3 + ...+ l/p,a>(O)=O ...... (11)

7.2.3 Gcncrating Function for J,(x): m

x/2(blh) = C J, (x)tn is called the generating function for Bessel n=-

equation of the first kind

7.2.4 Rccurrcnce Forrnulac in Bcsscl Function 1. J~+I (x) - - 2nJ,(x) - J, -(x)

X

If n Is as integer, these can be proved by using the generating function

xfS,-*&l = z 3, (xjt" n=-

Note that 3 and 4 are the same respectively as (5) and (6) The functions Yn(x) satisfy the same rules as above, with J,(x) replaced with Yn(x)

7.2.5 Bessel Function in Complex Plane: 6 ) Hankel Function of First and Second Kinds:

H,") (x) = Jn(x) + iY, (x) First Kind H,(') (x) = Jn(x) - iYn (x) Second Kind

(ii) Modified Bessel Functions of first kind, order n = In(x) = iPnJn(ix) = e-nni'2 J,(ix) If n is an integer I., = I,(x) but if n is not an integer I,(x) and I., (x) are linearly independent

(9 Modified Bessel Function of second kind order n

Sin pn ;n=O, 1,2,3, .... (13)

These functions satisfy the differential equation x ~ ~ " + xy4 - ( ~ ~ + n ' ) ~ = 0 and general solution is Y = C1Mx) + C2kn(x)

or if 11 # 0, 1, 2, 3 . . . y = AI,,(x) + BI., (x)

7.2.6 Graph of Jo(s), Jl(s), ITo Y I lo (x) and 1004 (s) .4s shown earIier the function of 3,(x) and Y,(x) are oscitlatory in nature and the amplitude of oscillation about a zero value tendmg to decrease with J ~ n x and the distance between si~ccesrive zeros of the function decreasing towards 7i. The f~mction of I,(x) and K,(ti)unlike the J,,(x) and Y,,(x) are not occillatory. I,,(x) increases exponentially with x whereas K,,(x) decreases exponentially, see Fig 16.

Fig

( 4

7.2.7 (a) Bcr, Bei Functions:

These are the real and imaginary parts of ~ , ( x e ~ " " ~ )

(b) Ker, Kei Functions:

These are the real and imaginary parts of e-n"i12~,(xem14) Thus the general solution of the equation: x"" + xy' - (ix' + n')y = 0 is: y = A{Ber,(x) + iBer,(x) 1 + B {Ker,,(x) + iKei,(x)}

Expression of these functions are found in Mathematical handbooks Fig 16b

Differential equations of the type xZy" + sy' - fix+nzj = O that have &3-, Uct, Ker and Kei functions as solution arise in Electrical Ikginccring and order fields.

7.2.S EQtrhTXOSS TRANSFORMED INTO BESSEL'S EQUATIOK: s'y" + (2k + I)xy' + (dx2 ' + P')y = 0 .................. (15) \\here k, a, I-, p are constants (15) has the 1~cneraI solution thus:

whert: k dk'-rj2 If ct = 0, the equation can be solved like Euler nr Cnuchy equation . z

7.2.9 ASYMPTOTES FOR BESSEL FUNCTIONS: For x very large we have

7.2.10. SERIES OF BISSZL FUXCTION: If f(x) satisfies the Dirichlet conditions (stated below) at every point of continuity of J(x) in the interval (0,l) i.e O< x <I, the Bessel Series Expansion of the form:

LX)

f(x) = AIJn(hlx) + A2Jn(hr,x) + .... = A,J, (h,x) ..... (IS) p= 1

A,, A?, h?, .............. are positive roots of ........ RJn(x) + SxJn(x) = 0, (R and S constants) (19)

RIS L 0, S # 0 and

At m y point of discountinuity the series on the right in equation (1 8) converges to %[f(x+O) + f(x-0)] which can replace the left side of (18)

7.2.10.1 Dirichet Condition . . . . . . . . . .. (21)

(i) f(x) is defined and simple-valued except possibly at a finite number of points in (-L, L) interval.

(i) f (x) is periodic outside (-L, L) with period 2L

(ii) f(x) and f(x) are peacewise continuous in (-L, L, then the series in (IS) with its coefficients converges to

(a) f(x) if x is a point of continuity (b) f(x+O) + f(x+O) if x is a point of continuity

2 if in (20) S = 0 so that hlA2 ..... are the roots of J, (x) = 0,

7.2.1 1 Example: Use the method of Frobenius to solve the Bessel's differential equation

x"" + xy' + (x2 - n2)y = 0

Solution:

Assumed Solution: y = &xL" where K goes from -- to and +- and c = 0 for K< O thus we have. (x' - n2)Y = Cclx I+'+' - Cck2x k+p - Cn2cLx k+D

xy' = C(k+D)c,x '+' x ~ ~ ' = C(k+P)(k+P-i)cLxL+'

'Shcn b y addition

_\:I I L+fi)(!-.+fi-l )ck + (k+P)ck + ck - n'c.J x = 0 .~nd since the coefficient of x must be zero we find.

1 ik+j3:j2 - n2cl;+ c ~ . ~ ] = O ......................... (1) Letling k = 0 in ( I ) , since c-2 = 0: the Indical equation

2 9 . ('3'-n')co = 0 or assunling c # 0, j3 = 11- l.e (3 = t 11

C ' ax 1: If j3:n (1) * k(311ih)c~ + c,.: = 0 .......... (3 ) put t in~ k = 1, 7. 3 ... successively in (2) we have

Thus ~ h c rcquired series 14 y cox" I czx'IC2 + y t .... =

s - Y 4

cox-" [ 1 - 2(2n+2) 2.4(2 -2n)(4 - n) + ] IS n = 0 both scries are indentical, if n = 1, 2, 3 .... The series ceases to cxits, but i i ' n -0, 1. 2, the two series can be shown to be linearly indzpendent so we wi-ite fo r thc gcncra; solution.

( i l ) P~.ovc (a) JS5(k) = d2tn Sin x (3) JNh(x) = 47/71 (COS x)

Proof: Jn(x) = (-1)'(~1~)"+" = ('Iz)% - + (XlZ)(J/2 - -- r =a r!T(r+ 'I?) fi) I T ( 'I?) ?!r( 'I2)

(b) J ,~ (x ) = C. (-1)'(V2P2' = ('I~)' - + (X/2)712. 1 r =O r!r(r+ 'I2) r ( 12) 1 ! r ( 312) 2!r('I2)

("/)-" 1 -x2 +x4 ...... = 2 c o s x 471 b i ! } ,L

(iii) Find the general solution of x::" + y'+ay = 0

Solution:

The equation can be written as xy" + y'+axy = 0. This 1: t i l e special case of equation (15) under 7.2.8 where k = 0, a = da, r = %, P - O thus f ~ u m equation (1 we have the solutiorl, y = cl J~ (2dax) + c2y0 (24ax).

7.2.12 Orthonality of Bessel Function 1

Prove that IXJ,(AX)J, (px)dx = ,,(A .J n ) 0 @4Jf

A - P

As noted before the general sohtion of 2 2 7 s2y" + xy' + (2. x - n-) = 0 ... (2) y = CIJn(hx) + C2Y, (hx)

~hus : y~ = J, (hx). y, = Y, (px) are solution of the equation. x'yMl + s y l + (h2x2-n') y, = O ....... (2)

and ~ ' y " ~ + x2y1?. + (h'x2 - n2) y2 = 0 ... . . .. (4) hlultiply (3) by y? and (4) by y, and substract, we have

s2[yly, - y1y"2 I + x[y2y1'1 - ylyl*l = (p2 - h ' ) ~ ' ~ ~ y ~ ....... (3 Dividing by x ( 5 ) *&[x(ydI - y~y'?)] + [y?yVI - Y ~ Y ' ~ ] = ($- h2)xylY:

dx xd X(Y?Y'I - YIP'?) I = (,u2 - AhZ)~YlY2 or - dx

By integrating and omitting the constant of integration, we have (p2 - A') f XYly, = x[y,yl, - y 1Y12\

Using YI = Jn (Ax), ~ 2 , ~2 = Jn ( p And dividing by p2 - h- # 0

then fx.I,(hx)J, (px)dx = h[Jn(px)J1, (hx) - pJ,(hx)J',(px)] $ - A L

which is the required result if h and pare any two roots of the equation thus RJ, (x) + SxJ', (x) - 0 where R and S are constants.

1 then fxJ,(hx)J, (px)dx = 0 which simply states that the functions

0 ~ X J , (px) and ~ X J , (px) are orthogonal in (0,l).

7.3 THE GUASS EQUATION

(X-X')~" + [y-(a+@+l)y1 - a+py = 0 ...... (1) assumed series, convergent near x = 0 and substitute in (1)

.... y = A oxrn' + A x m - 5 ... A .xmlr + And obtain m(m+y- l)&x"'"(m+ l)(m+ l)Al [(m+a+P) = a+p-1 + a P ] An-,lx m+n+l ........ = 0 [(m+n-l)(m+n+cx+p-1) + a @ ] & . l I ~ m+n+l ........ = 0 we take An = (mt-n-l){m+n+a+p-1) + aP]A,.I

(m+n)(rn+n+y- I ) then y = A ,,xm[l+m (rn+a+p) + ap(m+l)(m+y)x+m(m~a+p)+ ap

(m+n)(m+n+y- 1)

satisfies the equation ( x - x ~ ) ~ " + (y-(a+P+ 1)y' - a+@Y = m(m+y+ 1) A ox"-' ...... (2)

and from m = 1-y, y # 1 with & = 1 we have y2 =XI-y [ l+(a-y+l) (B y+l)x + (a-y+l)(a- y+2)(P- y+l)(P- y+2) xZ

1(2-Y) 1 .2(2-y)(-y)

The series y known as hypergeometric series converges for (x) c 1 and is represented by y, = F(a,P,y,x) y2 = x rF(a-y+l), P-y+l, 2-y,x). If y is non-integral including 0, general solution is:

(1)

(ii)

7.4

Examples: 3 Solve in series (x-x2)y" + (-I2 - 2x)y' - ?4y = 0

from eqwlion (2) under 7.3 3 a+@+1 = 2, y ='I2 up = %, therefore a = 0 = 95 and y = l2

But y, = F(a,@,y,x) = x-'. F(0,0, ?h, x) = 1 / h From equation (4) under 7.3 Complete snlutior~ is y = X( $4, 55. %, s ) + ~ f &

f'1m1 equation (2) under 7.3

~+(3+-1 = 4, y = 4 crp = 2, therefore a=1, j3 = 2 and y = 4

For cithcr y = F(1,2,3,4.r) = F(2,1,4,x)

= 1 +x/2 + 3x2/10 + x3/5 + x4/7 + 3x5/28 + ..... Asy = 4, the fourth term in y2 has zero as the denominator

However one of a - y+2 or P-y+2 in the third term is zero so that

and complete equation becomes y = AF(1. 2,4,x)+B&

X

MORE EXAMPLES ON LEGENDRE AND BESSEL' EQUATIONS AND OTHER SYSTEMS Legendre polynomials are defined by solving the Legendre equation thus (I-x2)y" - 2xy' + n(n+l)=O to get Y = Y I + , ~ 2 = C I P ~ ( 4 + C2Qn (x)

where P, (x) are Legendre polynomials Q, (x) are Legendre functions of the second kind unbounded at x = 1

of degree n P(x), the Legendre Polynomials can be expressed by the RODRIGUE'S FORMULA thus

P,(x) = 1 d n ( x - 1)" 2 "n! dx"

Example 1 Derive Rodrigues Formula Pn(x) = 1 dn(x-1)"

2"n! dx"

The Legendre Pol nomials = Pn (x) = (2n-1)(2n-3) . . .. 1 in - n(n-1)xns2 + n(n-l)(n-3)~"-~.. .. }

n! 2(n- 1) 2.4(2n- 1)(2n-3)

Integrating P, (x) n times from 0 to x we have

thus P(x) = 1 d "(x2 -1)" . . . . . . (1) 2 "n! dx"

Example (ii) Obtain the Legendre Functions of Second Kind in I

the case where n is a non-negative integer.

Solution: Legendre Function of Second Kind are the series solution of the Legendre Equation which do not terminate. Using Frobenius method to find the series solution of Legendre Equation, thus,

( l - ~ ~ ) ~ " - 2xy' + n(n+l)y = 0 Assumed series solution according to Frobenius equals

00

y = zckxktP ; ck = 0 for k < 0 a -00

n(n+ 1) = h ( n + l)ckx

Adding z[(k+P+2)(k+P+l)~~+~ -(k+P)(k+P-l)ck -2(k+B)Ck]x k+P = 0 Since the coefficient of x ';+I3 must be zero, then

(k+P+2)(k+P+l)~~+~ +[n(n+l)&+p)(k+P+ 1) CI, = 0 if k = -2 and c.2 = 0, the indical equation = P(P-l)CF 0 assuming Co #O, p = 0 or 1, we have case 1: = 0 equation (1) becomes (k+2)(k+l)Ck+z+n(n+ 1)-k(k+ l)Ck = 0 . . .. for k = - 1,O. 1,2, 3, CI is arbitrary and

so gcneral solution equals y = Co [1-n(n+ l)xZ +n(n-2)(n+l)(n+3)x4 .......I 31 , 4!

Since this leads to a solution with arbitrary constants we need not consider p = 1

For even integers n 2 0 the first of (3) terminates and gives a polynomial solution.

For odd integers n > 0, the second of (3) also terminates and gives a polynomial solution too which means that if n 2 0 the solution of Legendre Equation is polynomial equation.

So equationt3) can be expressed thus Y = Y l + Y 2 If n is even, yl terminates but y2'does not, thus y = y2 = x -(n-l)(n+2)x3 + (n-l)(n-3)(n+2)(n+4)x4 +

3! 5 ! While if n is odd

yz terminates but y, does not, thus

These series solution apart from the multiplying constants give Legendre functions of second kind.

Lagutrre PolynomiaIs are soIutions of the differentia1 equation of form xy" + (1-x)y' + ny = 0

The polynomials are given by Rodrigues formula ......... L, (x) = cxdn (xne'^) (5)

dxV

(a) Lo (x) = cx.dn (xndX) = cx.d"e-" = 1 dxn dxn

(b) L, (x) = exd (xe-") = 1-x dx

(a) Lz (x) = exd2 ~ ~ e - ~ = 2-4x+x2

Example IV: +

Use the generating function of the Hermite Polynomials to find &(x),, Hl(x), H2(x).

Solution: These Polynomials, denoted by H2(x) are solutions of the Hermites' Differential Equation: y"-2xy4+2xy= 0 ............ (6)

............... H,(x) = (- l)"ex2 (d '~-"~) (7) Ti=

The generating function

Comparing we have W x ) =l; Hl(x) = 2x, H2[x) =4x2-2

Example V:

(a) Verify the system y" + hy = 0 for y(0) = 0, y(1) = 0 .... (8) is Sturm-Liouville System

(b) Find the eigenvalues and eigenfunctions of the system (cj Show that the eigenfunctions are orthogonal

Solution: A boundary value problem of the form

a,y(a) + a2y1(a) = 0, bly(b) +bzyt(b) = 0 a,, b,, b2. a2. are constants. P(x), q(x), r(x) are possible differentiable functions, h an unspecified

parammeter independent of x, is called Strum-Liouville System of

Sturm-Liouville Value Problem. A non-trivial solution of the system

(equation) (that is the solution that is not identically zero for

particular values of h are called characteristic values or functions or

eigen values or eigen functions. Eigen functions form orthogonal set

with respect to another function, r(x) called density function which is

never negative.

(a) Equation (8) is a special case of equation (9) Where P(x) = 1, q(x) = 0, r(x) = 1, a = 0, b = 1 a, = 1, cl, = 0, = 1, Pz = 0, (where a,, a2, PI, Pz, are constants)

then the equation is a Sturm-Liouville System.

(b) General Solution: y = A Cos dhx + dhx, From the boundary conditions

y(O) = 0, :. A = O i.e y = 3 Sin & From the boundary condition y(1) = 0. B Sin dhx = B Sin d h = 0

so that B cannot be zero. Thus Sin dh = 0, then dh = mn or h =

m2n2 where m = 1, 2, 3 ..... are eigen values. The eigen functions

belonging to eigen values h = m2n2 can be represented by B, Sin

mnx. m = 0 or h = O is excluded as they correspond to zero eigen

function. The eigen function are orthogonal since

1 1

I [B, Sin mnx][B, Sin mnxldx = B,B, Isin mnx Sin nnx dx 0 0

1

B, B, [Cos(m-n)nx - Cos (m+n)nx]dx 0

I

I [B, Sin mnx12dx = 1 0

1 1

i.e B',I [B, sin2 nnx dx = B', I[ 1 - cos2 nnx dx = B', = 1 0 2 O 2

or B = 42 neglecting the negative root. Thus the set 42 Sin nnx wherem= 1,2,3. ....... is an orthonormal set.

Example VI:

(i> Evaluate lx4Jl(x) dx

Solution:

JxQJi(x) dx = ~(x2)(x2J~(x)dx] integrating by parts

= x ~ ( x ~ J ~ ( x ) ) ] - ( X ~ J ~ ( X ) ) I [ ~ X dx]

(ii) Evaluate (a) Jx"J,.,(x) dx

Solution:

(a) d(xnJ,(x) = xnJ,(x) dx dx

Cross section of the electrical coil of example VII Fig 16c

~ x a m ~ l e (VII)

Figure 16c shows a circular cross section of an electrical coil. Electrical energy is disspated within the coil at the

rate per unit volume of bo (1 + PT), where bo and p are

known positive constants. Th5dissipation rate depends

upon temperature since the resistivity of the coil material

depends upon temperature. The thermal conductivity of the

coil material is constant and the outside surface of the coil is

held at the known temperature To by a coolant. Steady-state

conditions prevail and axial temperature gradients are to be

considered negligible. Find the temperature distribution as a

function of the general radial coordinate r.

Solution: The governing partial differential equation, in the circular cylindrical coordinate system is shown below. Since steady-state conditions prevail and there is no axial eonduction (i.e, z-direction, perpendicular to the paper), and no circumferential temperature gradients (i.e., @direction). The equation for the (q"' is the generation rate per unit) circular cylindrical coordinate system thus a? + 1 c+ 1 a 2 ~ + a? + q"' = 1 s reduces to a? + 1 a~ + - qwl = o

F a 7a82 7i E a r 7% k

But q"' = bo (1 + PT); hence, defining % = b&, a 2 ~ + 1 aT - -- ar2 r ar + ~ ( i + p ~ ) = o ................ (1)

is simplified by making it homogeneous by the transformation @ = 1 + PT of the dependent variable. T h e n a = L a a n d a? =L a2@. After substitution and

ar p & 37 p a r '

rearranging. Equation (1) becomes $ + 1 84 r ar +%p@ = O

The above equation is a second-order, lineas, homogeneous ordinary differential equation with variable coefficients. Note that the substitution S = d$/dr is not appropriate because of the linear term a. in $. Multiplying ever term by rZ,

Equation (2) is Bessel's equation which bas bees soltd by finite series methods. The varies infinite series which satisfy it are called Bessel functions. The solution is

Q = AJo ( G r ) + BYo ( K p r ) . .........*.... (3

The ordinary Bessel function Jo (daOpr) and Yo ( G p r ) are zero order and of

the first and second kind, respectively. A quatitative indication of the

behaviour of these functions can be seen at various values o f e argument n in

Fig 16d. (The argument n here is identified as n = dao/3r). Values of selected

Bessel functions for a moderate range of n are given in the mathematics

I-Ianrlhnnk

Fig 16d: Graph of the ordinary Bessel functions of the first and second kind and of zero order

Replacing Q in equation (3) b a + +T as defined and rearranging, ............... TO = AJo (vbo/3r) BY^ ( G r ) - 11P (4)

Where llp in the first two terms on the right has been absorbed into the constant A and B. The boundary conditions determine the integration

constants A and B of equation (4)Bne boundary condition is at r = ro, T =

To; this leads to

T~ = ( G r o ) + B y o (Qro) - 1/P ............... (5) 154

The other boundary condition can be found from the radial s y m e t r y of the temperature distribution about the origin which leads to aT/ar = 0 at r = 0. An equivalent way of stating this, in this particular probIem, is to note that an inifinite temperature is not expected anywhere in this conduction region. Figure 16d shows that yo(0) + - -, so that to keep the temperature bounded at r = 0, B = 0. Hence, from equation ( 3 , using B = 0

A = T o + 116 ...................... (6) I Jo NaoPro) !

Combinging equations (6) and (4), the temperature in the coil as a function of r is

S DIFFERENCE EQUATION

S.0 Introduction:

IXffercnce equations which are widely used involve differences. 'They are used in solving differential equations. Summation is to Iliffel-ence as Integration is to Differentiation. Telescoping sums :1rc sums of differences thus

C AYI; = Yn - yo . . .. (1) analogus to integration of derivatives. 6=0

byl; is a difference term. k is an integer argument. (1) Can be proved by setting k at 1 ,2 , 3, ..... n - 1

S.1 A SOLUTION O F A DIFFERENCE EQUATION is a sequence of yl; values for which the equation is true, for some set of consecutive integers of k.

8.2 T H E ORDER of difference equation is the difference between the largest and smallest arguments k appearing in it. For example

.yktl = akyk + bk or Ayk = (ak - l)yn + bk.. .. . .(iii) ak, bk are functions of k. Equation (iii) has order l(that is (k+l)- k=l).

8.3 ANALOGY TO DIFFERENTIAL EQUATIONS: In differential equations. first-order equation normally has exactly one solution satisfying the initial condition yo = A, and a second-order has one solution satisfying two initial conditions yo = A, y, = B. These apply also to difference equations

Other Analogies:

( f ) First order linear difference equations are solved in terms of sums and the first order linear differential equations are solved in terms of Integrals

For example the equation yk+l = xyk + ck+,

with yo = coxn + cIxn.' + . . .. . ... ... .. + c"

Computation of this polynomial recursively from the difference equation itself is known as Homer's method for evaluating the polynomial.

(2) THE DIGAMMA FUNCTION, \V (X): y (x) is defined as

Where C is Euler's constant

It is one summation form of the solution of the first order difference equation

W(x) 4 (x+l)

The function gives the equation the character of a finite integral of 1 For integer arguments n, it follows that

(x+ 1)

Thc function reIates difference calculus to the logarithm function in differentid calculus

Other functions are relative to ~ ( x )

Gnrnmn function r and ~ ( x ) : I" (xi 1) = W(X)

1-(x+ I)

(3) LINEAR IIOMOGENEOUS SECOND ORDER DIFFERENCE EQUATION:

YI;-2 + alyk., + a2yk = 0, order E k + 2 - k = 2 has the solution family yk = cl uk + c2 vk

where uk and vk are solutions and cl and c2 are arbitrary constants. This as in differential equation is called Superposition Principle. Any solution of the equation can be expressed as such a superposition of uk and vk by proper choice of cl, c2 provided the Wronskiian Determinant.

CONSTANT COEFFICIENTS: If r,, r2 are the roots of the characteristic equation for the difference equation r2 + a, r + a2 = 0 where al and a2 are constants,

is not zero Wk = uk vk

4-I vk- 1

the solutions are

I; (i) uk = r , vl; = kr ' where a12 >4a2 (b2>4rtac) (Quadratic determinant)

k 2 (ii) uk = r k . VI; = kr where al = 4a2 rl = r2 = r

(iii) uk = in KO, vk = R~ cos K8 where aI2 < 4a2. r, = r2 = Rcos8 + isin 8 = eie

The FIBONACCI NUMBERS are solution values of difference equations like ykt2 = yktl + yk and by (I) above may be represented by real power functions. These numbers have applications in information theory.

(5) THE NON-HOMOGENEOUS DIFFERENCE EQUATION: Ykt2 + a1Ykt1 + a2Yk = bk

has family of solutions yk = ClUk + C ~ V L + Yk = yl+ y2 like in non-homogeneous differential equation

(6) HIGHER ORDER DIFFERENCE EQUATIONS ARE SOLVED JUST LIKE THOSE IN DIFFERENTIAL EQUATIONS:

e.g for the solution (fibonacci numbers) of the difference equation

k k thus yk = C, rI + c2 r2 where r, r2 = 1/2 [ l d 5 ]

initial condition yo = 0, y, = 1 then C1 = -C2 = 1

6

(7) NON-LINEAR EQUATION Direct recursive computations of solution sequences is followed as in Iinear equations

Examples:

(i) Solve the first order difference equation

yk+,= kyk + k%ecursively given initial conditions

yo = I , when k = O

When k = 1

Yl+, = y1+ l , y , = y , + 1 = y , + y o = O + 1 = 1

k = 2

y ~ + l = 2y2 + 4 = y3 = (2)(1) + 4 = 6

k = 3

y3+1 = 3y3 + 9. y4 = (3)(6) + 9 = 27

and so on

Thus the solution is

y1 = 0. y, = 1. y3 = 6. y4 = 27

(ii) Given the functions ak and bk what is the character of the solution of

the linear first order equation.

yk+~ = ak yk + bk with initial condition yo = A?

yl = a. A + bo that for k = 0

y 2 = a , y l + b r = % a , A + a , b o + b l f o r k = 1

~ 3 ~ 8 2 y z + b Z = a o a l a 7 A + a l a 2 0 b o + a 2 b , + b 2 f o r k = 2

etc

i f P n = q . a l . . a,.,

the result will be

(i) What is the character of the solution when ak = r and bk = 1

with yo = A = I? in the equation in (ii) above?

The equation simplifies to

(ii) What is the character of the solution function of

yk+l = X Y ~ + ck+1 with yo = A = c0?

Here-the solution function of equation YL+I = ak yk + bk

becomes yn = coxn + clxn" + ............ + c,

Evaluating the polynomical of argument x. involves computing y,, y2. ... ya

successively this amount to n multiplication and n additions and is equivalent

to rearanging the polynomical into

(v) Evaluate the sum n 2 for arbitrary t

x (k+t) k=l

Solution

We know x bk x Ayk = y,~ - yo (telescoping sum) k=O k= 1

This is equivalent to the solutions of difference equations

a

DIGAMM function \y(x) X 2 - c, c E Euler Constant i=l i(i+x)

A\y(x) = W + l ) - Mx)

~ ( k + t ) - ~ ( k + l - 1 ) = 1

k t - t

n n Thus 2 1- = 2 [ ~ ( k + t ) - ~ ( k + t - 1 ) = ~ ( n + t ) - ~ ( t ) ]

k=l k+t k= 1

Pt

(vi) Evaluate the series = C. Il(k+a)(k+b) in terms of diganna function i= I

Solution:

Using Partial Fractions

from previous theorem s n =L [v(n+a) - W(a> - W (n+b) + W@)l

b- 1 from the series definition in (v) and some calculation

y(n+a) - ~ ( n + b ) = (a-b) C l/(i+n+a)(i+n+b) so that n + w 1r1

this difference has limit zero /

Finally C l/(k+a)(k+b) = Lim S, = ~ ( b ) - ~ ( a ) k= I b - a

OC

(vii) Evaluate the series C 3k+1 - 1 h= 1-

k(k+ 1)2 Solution:

By partial fractions OC

I - - 6 1 % - r + r I = - - -

k k + I (k+ l )? Handling the forms together to avoid diverging of some terms we have

OC

= C [ 1 - 1 = ~ ( 1 ) - ~ ( 0 ) + ~ 9 ( 1 ) = .rr2/6 k=l - -

k(k+l) (k+112 Sums of rational terms may be treated in similar fashion. The digamma

function and its derivatives are in tables so that results such as those

obatined in the last example are readily evaluated from the tables

(iii) Solve y k + ~ - 2yk+Z + yk = 0

Solution:

First set up the characteric equation and solve Z Z Thus Yk+z = r , r - 2yk+z = 2rr Yk = r

The given equation becomes rZ - 2r + r = 0 (Characteristic

equation)

The only root is r = 1

This means that uk = 1, vk = 1 are solutions and yk = c, + c2k is a

family of solutions (Superposition Principle). Note A' yl = 0

Solve by direct computation the second order initail value problem

Yk+Z = Yk+l + Yk. YO = 0 Yk = 1

Solution:

Take k = 0, 1, 2 ... y, values 3re 1.2,3,5,8, 13,Zf . 34,55,S9, I44 ... which are known as Fibonacci numbers.

(A) 'The 'national income equation" is Y ~ + Z - 2ayk + ayk = 1

where O < a < i, Assuming a and I c o ~ s t a ~ t ~

Sult-e this eqrrztion and fir& the limiting national income far

increasing k

Solution:

The characteristic equation is

r" 2ar + n = 0 which has complex roots

r = a k d(a2 - a)= a 1 d(a - a') = daeio

where Cos 0 = da

Sin O = d(1-a)

Solution of the national income equation is non-homogeneous

To determine YL, we note that bk = 1 = constant

Substituting we find Y (1-a) = 1, Y = 1 - (1 -a)

Thus the complete solution is y, = cl (da)k Sin k8 + c2 (da)' cos k9 = Il(1-a) Since 0 <,a < 1. lim y, =L

1 -a with y, oscilating below and above this limit.

EXERCISES ON CHAPTER EIGHT

1. Solve the differece equation yk+z - 2Ayk+l+ yk = 0 In term of power functions, assuming A > I.

4. Find all solution of the homogeneous boundary value problem yk+2 + (L-2)yk = 0 Witfi YO = YN 0 Assume 0 < L c 4

5. Solve the non-linear equation Pk+l= Pd(l+Pk) by a change of variable (Hint: let yk = 1Pk)

6. Solve yk+3 + yk=O

8. Given that yk+l = ryk + k and yo = A compute y~....... y4 directly. Then discover the character of the solution function.

I'AR'l'lhL UGVIATIVES A h 4 SOME

PARTIAL DERIVATIVE AND JACOBIANS

9 0 Ixr l~OI~UCTIOS.

Partial dcr isatives are of immense use to Engirrees, Scim&zs, Ecnmmis~

and n host of n&rr pdessio~ah, $fKF z e simpk ordinary derivative of

fwctions of several variables with respect to one of the independent

variablcs. Keeping all other independent variables constant, such derivatives

nlc called the partial derivatives of the functions with respect to the variable

encountered in difl'crentiating functions of two or more variables.

9.1 NOTATION: = f (it is rcad partial off with respcct to x) -

ax

Similarly &-is partial of f with rcspcct to y ay

If z = f (x,y) we also have $ ax

Somctitnes fixing y and differetiating with respcct to a_f ax

has f, notation whilcaf has f, notation ay

9.2 DEFINITION: . * By definition &= lim f(x +A x, y) - f(x,y) . . . . .. . . . .. .. (1)

ax AX-. o nx

&' = lim f(x, Ay + Ay) -.f(x,y) - ay AY-o AY

Provided these limits exist

The derivativities evaluated at the particular point (x, y) are often indicated by

Example:

Given f(x,y) = x - 3xy + 2x - 3y + 5

find f, and fy at (2.3)

Solution:

F = 2x - 3y + (Differentiating f (x, y) with respect to x keeping y constant)

and fy {2,3) = (2)(2) - (3)(3) + 2 = 4 - 9+ 2 = -3

fy = -3x -3 (Differentiating f(x, y) with respect to y keeping x constant

and fy (2.3) = (-3)(2) -3 = -6 -3 = -9

If a function f, has continuous partial derivatives f in a region, then f must be

continuos in the region. However the existence of these partial derivatives

alone is not enough to guarantee the continuity of f. The function at any

particular point can be discontinuous.

Example:

If f(x, Y) = xyl (x + y) (x,y) + (0.0) show that

(a) f (0,O) and f (0 , 0) both exist but that

(b) f(x, y) is discontinuous at (5.0)

Solution:

(3) f (0, Of = lim f(h, 0)- f(O, 5) = lim 0 = 5 114 o -n h + o a

(b) let x = 5 and y = 0 along line y = mx in the xy plane

then linl f(x,y) = lim' mx =a Y-' 0 x+O x + m x l+m

thus the line depends on the approach and therefore does not exist.

Hence f(x, y) is discontinuous at (0,O)

Exarnplc:

find dE and _a fn t (x, y) directly from the definition (9.2) S x ay

= lim [ ~ 2 ( x + h - f x + h ) ~ + v l - 1 2 x - ~ ~ + y ] h-+O h

= lim 4 1 7 ~ ~ - 7 h - & = lirn 4 x + 2 h - y ) = 4 x - y h+ 0 h h+O

= f(x, )l) = lim f(x y i k ) - f(x - y) - ay k+O k

= Iim -kx + 2kv + k = Iim (-x + 2y + k) = x -2y k-+O k k+O

Si- &e limits exist for all points (x, y) we can write

9.3 HIGHER ORDER PARTIAL DERIVATIVES:

If f(x, y) has partial derivatives at each point (x, y) in a region, then

af and af - ax ay are themselves functions of x and y which may also have partial

derivatives

is the &I-ivative o f f trihcrt mice widx respect toy and t\vicc with respect to x. If 1;, and 5, arc continuous. t f m f,, = f ,, that is the order of differentiation is im:mtcrial.

In s!~ch n cnsc, f,,(l; 2) = 12 f),y ( I , 2) = 6 f,,(l.7-)=1;x(1, 2)= 12

If Ax = dx and Ay = dy are increments given to x and y respectively,

then A7. = f (x + Ax, y + Ay) - f(x, y) = Af

Af is called the increment in z = f(x, y)

A z - a f . A x + d f . A y + ~ ~ A x + ~ ~ A y - - dx dy

= - k . d x +&.dy + E ,dx + ~ ~ d y = Af ..... (I) dx a y

where E ,. E approach zero as Ax and Ay approach zero

is a total differentiai of the expression z = f(x, y)

It is also called differential of z or f o r the principal part of Az or Af

Note that Az # dz in general.

However if Ax = dx and Ay = dy are 'small" then dz is a close approximation of Az. The quantities dx, dy called differentials of x and y respectively need not be small.

Therefore on Differentials:

then df = a f dxl +af dx2 + ......... + af dx, ax, ax, ax,

(ii) If f(x x ... x) = a constant c, then af = O

(iii) The expression P(x, y)dx + Q(x, y)dy or briefly Pdx + Qdy is the Differential of f(x, y) if and only if &P =a

ay ax

in such a case Pdx + Qdy is called an exact differential(treated earlier in this text)

or Pdx + Qdy + Rdz is the differential of f(x, y, z)

if and only if dP = a, = a ,a = ay ax az ay ax a2

In s u c h a c a s e P d x + Q d y + R d z is a n e x a c t d i f fe ren t ia l . E x a c t d i f f e r cn t i a I c a n he e x p r e s s e d a s i n ( 2 ) a n d (3). If a s in thermodynamics, t h e va r i ab l e s o f a s y s t e m , n a m e l y (P. V a n d T ) a r c e x p r c s s c d a s f u n c r i o n s o f X w h i c h is a s t a t e f u n c t i o n o r e x a c t d i f f e r ca t i a l t h e n X = f ( P , V. T).

9.5 E U L E R S T H E O R E M O N H O M O G E N E O U S FUKCTLOKS

...... A !'unction F ( x l , x2 .x, ) is h o m o g e n e o u s o f d c g r c e P if f o r all v a l u e s o f t h e p a r a m e t e r h a n d s o m e c o n s t a n t P ( t h c d c g r e e o f t h c f u n c t i o n ) t h i s i den t i t y ex l s l s :

V i z :

F ( x , y ) = x4 + 2 x y 3 - 5 y 4 is h o m o g e n e o u s o f d e g r e e 4 , s i n c e

T h u s e u l c r ' s t h e o r e m o n h o m o g e n e o u s f u n c t i o n s s t a t e s t h a t if

...... F ( x I , xz x,,) is h o m o n o g e n e o u s o f d e g r e e P t hen

9 . 6 I M P L I C I T F U N C T I O N S :

If an c q u a t i o r ~ F ( x , y , z ) = 0 c x i s t s s u c h t h a t o n e va r i ab l e s a y z is a f y n c t i o n o f t h e t h o va r i ab l e s x a n d y then z is c a l l ed a n i m p l i c i t f u n c t i o n o f x a n d y

W h i l e in t h e c x p r c s s i o n z = f ( x , y ) , f is c a l l ed a n e x p l i c i t f u n c t i o n . t h u s

Suppose that x, y and z are variables and that z is a function of x and y which satisfies the equation x + y + z + 3xyz = 5, find & = az

& & az - ax (y constant) that is differentiating z with respect to x implicity, thus

A~SO az = 3~ + 3~ az + 3~~ az + ~ X Z = o ax x constant %X ax

and A = -y + xz ax xy+ z

9.7 JACOBIANS:

If F(u, v) and G (u, v) are differentiable in a region, the Jacobian of F and G with respect to u and v is the second order functional determinant defined by

Similarly third order determinant a(F.G.H.) - - Fu F v a ( ~ . V. W) G. G,

H, H,

is called the Jocobian of F, G and H with respect to u, v and w

9.7.1 PARTIAL DERIVATIVES USING JACOBIANS:

Jacobians are used to obtain partial derivatives of implicit functions. For example, given simultaneous equations then F(x,y,u,v)=O. Gjx,y, u,v)=O

and u and v are functions of x any y then

Note that the common denominator above must not be zero or vanish.

9.7.2 THEOREMS ON JACOBIANS:

Here it is assumed that all functions are continuously differentiable

1. For F(u,v,x,y) = 0 and G(u,v,x,y) = 0 to be solved for u and v

w, G) # m ( i . e the Jacobian) should not be identically zero in a region R

2. If x and y are functions of u and v, and u and v are functions of r and s we have

If z = f(x, y) is continuous and the first differentials f, and f, are continuous and if x and y are functions of r and s, thus x = x{r,s), )I = y (r, s) and their first differentiah all exist. Then

x is also a function of r and s

Or written like this:

Also for function of one variable : & = &J .Au dx du dx

In conclusion equation (1) is an example of Chain Rule for Jacobiana

3. If u = f(x,y) and v = f{x, y) a relationship thus

~ { u , V) = 0 exist between u and v if and only if the Jacobians

a (u,v) is identically zero - a (x Y)

Extending Jacobian concept to vector interpretations in orthogonal curvilinem- coordinates, based on transformation equ a t' ~ o n s

x = S(UI, ~ 2 , u d , y = S(UI, ~ 2 . u?), z = h(ul, u?, ~ 3 ) where f, g, h are continuous including partial derivatives. Thus dV = 13(x. y, z) I duldu2du3 = I Jacobian I duIduzdu3 = volume of a parallel

1 ~ ( U I U ~ U ~ I

piped where r is the radius vector and ul, uz, u3 are coordinates. We note that if the Jxobian vanishes there is no parallel piped.

Jacobians are useful in problerns on transformaion and mapping which establishes a correspondence between points in the uv and xy planes.

9.7.3 TRANSFORMATIONS

1. Introduction

................. (1) ' Translation w = x + P (1)

Uy this transformation figures in the z-plane are displayed or translated in

thedirection of vector P. P is a complex constant

(2) Rotation: i8 o = e ........................... (ii)

Hcre figures in this z-plane are rotated through an angle 0

O > o (anti clock wise)

0 < 0 ( clock wise)

w = az ............................ (iii)

Figures in z-plane are stretched w contracted ill the dile~tior~ z if a > 1 or a ,- I

(4) Inversion 0 = 112 ........................ .... Gv)

This is n combmation of the transformatim of translation, rotation and stretching.

( 6 ) Blinear or Fractional Transformation:

w - u z + p aa-pa#O ............ (vi)

This is a conihination of translation, ro ta~ i~m, wercnlng and inversion. In this treatment a and P are complex and angle 0 is real.

(ii) The set of equations thus

.................... .. ....... (vii)

defines a 'Transformation or mapping establishi~ip correspondence between points in uv p l a ~ ~ e and xy

plane, if for eacb point in r.!v plane there cz:x;;txds m: and only poit~t in x y plane and conversely it is

one-to-one corrt:spbndence or transformation or mapping. Fl'his i s possible if F(u,v) are co~ilinuously

differentiable w;Lh the Jacobian not identically zcro in a region.

C rich t l~c t ~ a n J w t m t ~ o n (vr) a c l o \ d legion R on the sy plane ts mapped

into a closed legion R on the uv plane then

\?'lic~c. lim dcnotcs limit as AA,, (or AA,,,,) approaches zero AA,!,, A,\, . , , arc respeclivc areas in these regions. d(u.y) is called tlic Jacobian of the --

(&!I, v)

'l'lie Jacobians d(x, y) and d(u, v) arc reciprocal . - -

acll, V ) ~NX,Y)

Il'\vc solv:: f& u and v in ternls of s, y in equation (vii) then we have U -- f ( ~ . ~ ) - g(- ;.:

9.8 APPLICAI'ION OF YARTlAL DERIVATION (LAGRANGE Mi ILTIl'IdES FOR MAXIMA AND MlNIMA

Consider I'(x,y,z subject to the conslraint condition G(x,y,z) - 0

I'rove that a necessary cod i l i on thal F(x,y,z,) luve an extreme value (minima or maxima> is that I;, G , - Fy Gz = 0

Since C;(x:y,z) = 0, and taking % - - z(xxy) FOP I;[(x,y, f(x,y)] lo liavc an extreme value (Minima or Maxima) the partial derivatives with respect to x and y inust be zero

. . . . . . . . . . . . . . . . . ~'II!!:, Fx I FxZs - 0 (i) . . . . . . . . . . . . . . . . . . a~it l Fy t FxZy 0 (ii)

Gx -tGz%x=!'.... .................. (iii) and Gy t- G zZy = 0 ............ ( I V )

Whtcti 1 olds only ~f F7, + 0 , Gz # 0

(1) Find the s lmtest distance fiom the o1ig111 oPa hypelbola

<' + 8xy + 7y2 - 225, and z=O

Solution: ,

Required lo T~nd nlininmn value o l x2 )- y2 (which is the square of

this dista~icc fi.0111 !lie origin lo any point (x, y) in the xy p l aw

subject to the constraint.

and for this no real snlulion otcurs

Case 2: L - -9 1'1-om (VI I ) and (viii) y = 2x

; ~ n d substituting in rhc hyperbola equation

kixample A ~ectangula~ box open at the top is to have a volumc of 32 cubic mccles. What milst be the dimensions so that the total sulface is ini~~imum'?

I .ct the edges (side) be x, y, 2.

Voiunle ot box = xyz = 32. . . . . . . . ............ (ix)

Sutface area of box = S = xy 4 2yz + 2xz.. ... (x)

and since z = 32 froiii (ix) " y

SY - - 2 ds = x - 64 = xy = 64 (xii) - - JY ~2

Substituting in (xi) x3 = 64 therefore, x - 4 = y

thus x = 4, y = 4, ,. = 2 Check:

A point (x y ) is a critical point (i.e niinim:mm or maximum) if and only if

then

(a) (xo,y,) is a relative max point if

A > 0 and f,, < 0 or f,, < 0 I I

( I ) ) (xo , yo) IS a relatrve nun potnl if

at x o , ~ ~ ~ at X O . ~ O

(c) (XO, YO) 1s nerthrr ~ n i n 1101 rnax point

I S A < 0 If A< 0, (xo,yo) is ca!!ec! a saddle point

Mole ~nvestrgation should be c a ~ r i e d out

A - S,, .S,, - s,: = l B ) (128) --1>0

(x') (Y')

T1ie1-efore the dirnensioris x - 4

Find the (a) error- and (b) percent el-rol- made in ~omprl t ing 1. with I = 2m

and g - 9.75ndsec instead of with the true value of 1 = 1.95111 and g =

9.81ndsec.

Error in g = Ag = dg = (9.81 - 9.75) = t 0.06

Error In 1 - A1 = dl = (2 - 1.95) = - 0.05

The ello1 In T 1s ac~ral ly AT \vIi1c11 i l l t111s c a w 1s approximately equal to dl

'fhus from the above expression for d'l' substituting for dg and dL we have

DT =n / ((2)(9.75))'12(-0.05) - n(d(2/(9.7~).' (+0.06) = -0.0444sec approx = el-rol- therefore the valu: of T for I= 2 and g = 9.75 = 2d(2/(9.75)) = 2.846sec approx

and (b) percent error (01- relative error in T) = dT/T = -0,044412.846 = - .56%

ANO'I'I-IER METHOD: Smce LnT = In 2n + %In1 - Mng for the equation T = 2ndllg

Different~ating all tlzrough term by tel-ni as usual, to get S f / T we have

d 1 - 0 I I dl - Idg ........... (2)

p c ~ cent el 1 0 1 111 'I = '/, Pel cent el r o ~ In 1 - 1/2 percent ell or ing

I Solve that u (x y z) - (x t y 2 ~ z2)-"2 sat~slies the Laplace's pal tial I)~Tfaential equation ~ ' I I t aZu I azll - o

111 t e r m of de~ivat lve with Iespect to r and s

(b) 37, = -Fy

( a ) f2x- v) x -1- y

5. Evaluate d(f, g)/ d(u,v) if Q u , v) = 3u2 - uv and G(u.v) = 2uv2 + v3

6. Fmd the extleme value of z on the surrace

7. Prove that shortest rlislance from the origin to the curve of the intersection of the surraces

xyz = a and y = bx where a > 0, b >O

'8. 'l'he diamete! of a right circu!?r is mr2sured a. 6.0 + 0.03c111, while its height is measured as 4.0 f 0.02r.111

What is the l a~ges t possible

(a) err01 and (b) pelcent ellor made in computing the volume

These differential e imtions contain one or more partial variables

du = 4aZu Linear, order of 2, (1) - - at ax2 independent variables are x and t . depe~itlent variable is u

(ii) x ' 3 k = y' 43' R Linear, o r d e ~

; I ) ?@lJ X' .+- (f2@/3 yZ 4- d2@/ d z2= O Linear, o d e r 2, dependent 1:lbles @

indepe~lr le~~t variables x, y. z

( \ ) (dz)' + (dl) - 1 Non-Lunea~ 01 d e ~ o f , (3u) ((71) ~~ldependent val ~ab lcs , u, ! dependent val able z, orcie~ 1

10 2 Cienc~al FOI m of L~neal Pa l t~a i Dilfelentlal Equalton

1 \:{: ~ ~ ( j A, D, . . . . . .. . . . . . . ... . . . . .. G may be hnctions o r x and y \ ~ u l not o.

hot? thc order of eql~ation (1) and the dimensional balancing of the tliffc~cntlals In each let m

Wlicle A is the coefficient of - ax

184

B is the coeficient of axa

C is the coefficient of b u ay'

u is dependent variable while x and y are independent variables

u = @ , A = l ,B=O,C= I, t h e n ~ ~ - 4 ~ ~ = - 4 < Oandtheequation is elliptic.

(i i) au/at = k a 2 ~ / a ~ 2 , y = t , ~ = k , B = 0 , ~ = ~ t h e n ~ 2 - 4 ~ c = ~ and equation is parabolic

(iii) a2 + 3azu/axay+ 4azu/ay2 + 5au/ax - 2au/ay + 4u = 2x - 3y

A = 1, B = 3, C = 4, B2 - 4Ac = -7 < 0 and the equation is elliptic

(iv) xa2u / ax2 + ya2u / ay2 + 3y2 au /ax = o

In the region where xy < 0, it is hyperbolic

Where xy = 0 the equation is parabolic

xy >O, equation is elliptic.

10.3 METHODS OF SOLVING PARTIAL DIFFERENTIAL EQUATIONS:

I . Definitions: (i) A solution of a partial differential equation satisfies the

equation identically and the general solution contains a number of arbitrary independent functions of the order of the equation.

(ii) A particular solution is part of general solution having particular choice of arbitrary functions

the general solution of a2$ = 2x - y is I$ = x2y - !hxy2 + F(x) + G(y) -

because the equation con ta i~~s two arbitrary independent function F(x) and G(y) if in particular

F(x) = 2 sin x and G(y) = 3y4 - 5 the general solution . therefore is

~ = x ~ ~ - ' / z x ~ ~ + 2 sinx+ 3y4-5

(iii) A singular solution is one which cannot be obtained from the general solution by the choice of particular arbitrary functions.

(iv) A boundary value problem involves a partial differential equation which should be solved to satisfy certain conditions called the boundary conditions. The existence and uniqueness of theorems relate to the existence and uniqueness of such solutions.

2. . General ~o lu t ion Method:

In this method the general solution is first obtained, then the particular solution satisfying the given boundary condition is sought. For this, the following theorems are very important

10.3 METHODS OF SOLVING PARTIAL DIFFERENTIAL EQUATIONS:

1. Definitions: (i) A solution of a partial differential equation satisfies the

equation identically and the general solution contains a number of arbitrary independent functions of the order of the equation.

(ii) A particular solution is part of general solution having particular choice of arbitrary functions

the general solution of a2$ = 2 x - y is I) = xZy - %xy? + F ( x ) + G(y) - a x a y

because the equation contair~s two arbitrary independent function F(x) and G(y) if in particular

F ( x ) = 2 sin x and G(y) = 3y4 - 5 the general solution . therefore is

1 $ = ~ ~ ~ - ' / 2 ~ ~ ~ + 2 s i n x + 3 ~ ~ - 5

(iii) A singular solution is one which cannot be obtained from the general solution by the choice of particular arbitrary functions.

(iv) A boundary value problem involves a partial differential equation which should be solved to satisfy certain conditions called the boundary conditions. The existence and uniqueness of theorems relate to the existence and uniqueness of such solutions.

2. , General Solution Method:

In this method the general solution is first obtained, then the particular solution satisfying the given boundary condition is sought. For this, the following theorems are very important

Theorem 10.1 (Super Position Principle):

If u (x), u (x), u (x) . . .. . .. . .. . ... are solution of a partial differential equation

C u (x) + c u (x) are also solution where C C C . . .. . .. . .. . .. . .. Are constants

Theorem 10.2 (Solution of Non-Homogeneous Equation):

The general solution of a 1inear.non-homogeneous pattial differential equation is obtained by adding a particular solution of the non-homogeneous equation to the general solution of the homogeneous equation. Sometimes general solution of (1) can be obtained by using the mcthods of ordinary differential equations.

Euamples:

(i) Find the general solution of 2p + 3q = 1

Solution: The auxilliary system isdx =_dy = d z (see notes next page on 2 3 1

Lagrang auxiliary system of solving partial Differential Equations).

Using

dx = & we have x - 22 = a Then & = d~ we have 3x - 2y = b - 2 1 2 3

Thus the general solution is 0 (x - 2 2 . 3 ~ - 2y) = 0

The complete solution is x - 22 = a (3x - 2y) + is a two-parameter family

of planes

The one - parameter family determined by taking $ = c? has equation

thus x - 22 = a (3x - 2y) + a2 ............... (A)

Differentiating (A) with respect to a yield

3 ~ - 2 y + 2 a = O o r a = - H ( 3 ~ - 2 y )

substituting for a in (A) we have

x - 22 = - 1/4 (3x - 2ylZ which represents the envelope, a parabolic cylinder

which is clearly a part of the general solution.

Note: (i) Standard Notation for Differentials:

Examples (iii): Find the general solution of (y - z) p + (x - y) q = z - x

" The auxiliary system is - dx = d y = d_z y - z x-y z -x

Since (y-z) + (x-y) + (z-x) = 0, dx + dy + dz = 0, x + y + z = a and

since x (y-z) + z(x-y) + y(z-x) = 0, xdx + zdy + ydz = 0, and x2 + 2yz = b (Lagrange System). The general solution= 4 (x + 2yz, x + y

+ z) = 0 and the complete solution= xZ + 2yz = a (x + y + z)+

represents a family of hyperboloids.

Note (ii): To solve partial differential equation (linear) of form:

P(x, y, 2) + Q(T, Y, 2) = Nx, Y, 2) ......... (B)

Lagrange reduced the problem of finding the general system of the

equation of the type B to that of solving an auxiliary system (called

the Lagrange system of ordinary differential equation, thus

By showing that Q (u, v) = 0 (when Q is arbitary) is the general

solution of (B) provided u = u(x, y, z) = a and v = (x, y, z) = b are

two independent solutions of (C). Here a and b are arbitary

constants and at least one of u, v must contain z (This method is

used in examples (i) and (ii) above

Example (iii)

(a) (i) Solve the equation 3 3 = xZY axay

(ii) Find the particular solution for which z (x, 0) = x2, z (1, y) = cos y

Solution: (i) Rewritting the equation as:

Integrating with respect to x

3 az = LX + q y ) ............ - (1) ay 3 Integrating with respect to y

where G(x) is arbitrary

equation (2) can be written;

cquation (3) has two arbitrary (essential) functions and is therefore a general solution

( 9 Since z (x, 0) = x" we have from (3), x2 = H (0) + G (x) or G ( s ) = x" H(0) ................ (4)

Thus z = 1 x3y' + H(y) + x% H(0) ...................... - ( 5 ) 6

Since z ( I , y) = cos y we have from (5)

Gcncrnl solution is z = 1 x3y2 + cos y - lY2 + xZ - 1 6 6

Example (iv):

(b) Solve -&I + = x2 axat ax

Solu Lion:

Rewriting, the equation becomes-h (-t& + 2u) = x2 ax ax

Integrating with respect to x, tau + 2u= 1 x2 + F(t) - a 3

or au + 2u = 1 x3 + (Ft) this is a linear equation x - & 7 - 3 t -t

Integrating t2u

= 1 t2x3 + I t~( t )dt + H(x) = 1 t2x3 + G(t) + H(x) is the required - 6 3

general solution.

(c) Find the solution of a2u + 3 a2u + a 2 u = 0 ax" ax ay2

Assume u = eax"+by

Substituting in the given equation after differentiating appropriately we get (a2 + 3ab + 2b2) eax"+by = 0 or a2 + 3ab + 2b2 = 0 r (a + b) (a + 2b) = 0

a = -b, e-bx+b~ = eb(Y-x) is also a solution for any value of b

a = - 2b, e - 2 b ~ + b ~ - M Y - 2 ~ ) . - e 1s also a solution for any value for b

Since the equation is linear and homogeneous, sums of solution are solutions also. E.g, 3e2@-"' - 2e3@-") + 5eqY-") is a solution among others. F(y-x), G(y-2x) (where F and G are arbitrary) are also solutions, and by addition the general solution is

(d) Solve a2u + #U = 10e2"+~ 37 - ay2

(e) Two parts namely homogenenous part a2u + a2u = 0 and non- ax' sv2

homogeneous part 10e2"+~

General solution for homogenous part, u = F(x + iy) + G(x - iy), for the particular solution of the,other part, we assume u = ae2X+Y, a is an unknown constant. This is a method of undetermined coefficients for the ordinary differential equation. We find a=2

:. required general solution is

u = F(x+ iy) +G(x - iy) +2eZX+'

(f) solve& - 4 &I = eZwY axZ ay2

Solution:

The solution for the homogeneous part is u = F (2x + y) + G(2x - y), for the non- homogeneous part we assume a solution thus u = ae2"y but this is already in F(2x + y) hence as in the ordinary differential equation we reassume that u = or u =ay~2x+y. Substituting we have a= % .

The general solution is U = F (2x + y) + G(2x - y) + '/4 xeZWy

3. SEPARATION OF VARIABLE METHOD:

This method makes use of Fourier series, Fourier integral, Bessel's series and

Legendre series. Here it is assumed that a solution can be expressed as a

product of unknown functions each of which depends on one of the

independent variables. In the execution of this method one side is made to

depend on one variable while the other side depends on the remaining

variable. By repeating this, the unknown functions will be determined.

Examples:

(i) Solve the bobndary value problem au + 4au ax ay

Boundary conditions: u(0,~)=8e"~ Solution:

Let u in the given equation be XY

Then the original equation becomes X'Y = 4XY' or X' = Y' 4X Y

This is obtained thus a& = aLxy) = XY ax ax

Y is obtained thus au = a (XY) = XY n -by

Then separating variables from X'Y = 4XY' we have X' = Y' ax Y

Since X depends on x only and Y depends on y only and x and y are

independent variables, each side must be constant, say C. Then X' - 4CX = 0,

Y'-eY = 0 whose solutiona are X = ~ e ~ ~ \ Y= ~ e " . A solution is thus given

by u(x,y)=XY= A B ~ ~ ( ~ ~ + ~ ) = keC(4x+Y). From boundary conditions, u ( ~ , ~ ) = k e ~ '

= 8edY which is possible if and only if k=8 and e = -3.

Then u(x,y)= 8e-3(4x+y) = 8e-12x-7y is the required solution.

(ii) Solve au = 2 2 u a t axz

given that u(0,t) = u(3,t) = 0,

u(x,O)= 5sin4nx - 3sin8nx + 2sinlOnx, I u(x,t) I <

states that u is bounded for 0 < x < 3, t > 0.

M where the last condition

Solution:

Let U = XT thus XT' = X"T an= =*separating variables). Each side X 2T

must be constant, equal to say -A2

(+h does not satisfy the bounded conditions for real h) thus X" + h Z x = 0, T' + 2 h 2 ~ = 0

The solution is X = Altos Ax + Blsin Ax, T = cle-2h2' .The solution of the

partial differential equation is thus given by u(x, t) = XT = cle-u2t (A coshx +

I3 sin Ax) = e-2A2t(~coshx + Bsinhx) since u(0, t) = 0, e-U2' (A) = 0 or A = 0

Then u(x, t) = ~e-"~'sinhx. Since u(3,t)=0, ~ e . ~ ~ ~ ' s i n 3 A = 0 . If B=O, the solution is identically zero, so we must have sin3A=0 or 3A=mn, A=m7r/3 where m=O, 21, +2... . Thus,

u(x, t)= *sin m n d 3 is a solution.

Also by the principle of superposition u(x,t)= ~ ~ e - ~ ~ ~ ~ ~ ~ * sin minx + ~ ~ e - ~ ~ , 2 ~ * sin m2nx

3 3 + B3 e - ' F 2 * sin m*x is a solution.

3 By the last boundary condition

= 5sin 4nx - 3sin 8nx + 2sin 10nx. This is possible if and only if B1

- -5,m I - - 12, BZ = -3, m2 = 24, B3 = 2. m3 = 30 .Substituting these in

the solution above, the required solution is

u(x, t)= 5e"2"2' sin 47rx - 3ee-12sn2' sin 8nx + 2e-2m2' sin 1Wx

This is a heat flow boundary valve problem in which a bar 3 unit of length with a diffussivity of 2 units, has its ends kept at zero units of temperature and its initial temperature is u(x, 0) = 5 Sin 4nx - 3 Sin 8nx + 2 Sin 10nx. Required to find the temperature at a position x, at time t, that is to find u(x, t).

10.4FOURIER SERIES METHOD FOR SOLVING PARTIAL DIFFERENTIAL EQUATIONS:

Fourier Series define a function, F(x) in the interval (+L, L) and is also defined outside the interval f(x + 2L) and f(x), that is assuming period of

f ( x ) i s 2 L = a + ( a c o s m + b s i n uu .............. (1) 2 n=l L L

a, b are Fourier coefficients evaluated thus

a = 1 f (x) cos EX dx ................. - i (2) L " L

a =_I ," f (x) cos EX dx ................. (3) L -L L

With period 2L c + 2 L

r

a = 1 J f (x) cos ~x dx ................. - (4) L -c L

a =J J f(x) cos EX dx ................. ( 3 L -c L

If c = -L (4) and (5) becomes (2) and (3)

Note that c is any real number in (4) and (5) To determine % in (1) put n = 0

The DIRICHET CONDITIONS explained in 7.2. 10. 1 of chapter 7 apply to

f(x)

(1) Examples:

Apply Fourier Series to the solution of the problem in example (ii) under

Article (3) above. If the initial temperature is 2.5'~. that is u(x, 0) = 25, replace

equation (1) in example (ii) by u(x, t)

- = C. B,,,e-, ri " sin m x . . .. . . . . . . . . .. (6)

nr=l 3 with the condition u(x,O) = 25 this is when t = 0, u(x,O) = 25. Therefore from

(6) putting t = 0 yields

expanding in Fourier Series

L 3

we have B , =2 If(x) sin m& =2 f 25sin mnxdx = 50 (1 -cos mn) L O L 3O 3 mn

Resulting to u(x. t) = f 5 0 1 -cos mn]e-m2"2" sin m- nr=l mn 3

= 100/n{e-"~" sin nx - + - 1 e-"2' sin nx + . . . } E the required solution. 3 3

The problem illustrates the importance of Fourier Series (and orthogonal series in general in solving boundary-value problems)

(ii) A circular plate of unit radius whose walls are insulated has half of its

boundari kept at constant temperature ul and the other half at constant

temperature u2. Find the steady state temperature of the plate.

Solution: In polar co-ordinates (p, $) the partial differential equation for steady flow of heat is d2u + 1 du + 1 d2u = 0 . ... . . . . ... (1)

dp 2q2 Ul,0<$<71

Boundary conditions are u(l,$) = ( u2, 71 < $ < 271

I u (p,$) I < m, that is it is bounded in the region. 4

Let the solution u (p,$) = P cP

Where P is a function of p and 0 is a function of $. Then Equation (1) becomes P" cP + 1 P' cP + I P cP' = 0

P p2 Dividing by P+ and multiplying by p2 and rearranging we have we have p2 P" + pP 'P = 0" -

P 0 Setting each side equal to h2 we have

The solutions are:

0 = Al Cos A$ + BI sin?&

P = A~P' + B2/p ' (see Cauchy or Euler equation Article 3.8

Since u (p, $) must have period 271 in cp, h must be equal to m = 0, 1,

2, 3 ..... and since u is bounded at p = 0. B2 must be zero thus u =

PO = A2pm (A1 cos m$ + B Sin rncp ) = p" (A cosmcp + B,sin mcp)

By Superposition a solution is

u(p+) = &+ C pm (A, Cos rncp + B, Sin rncp 2 IIFI

- therefore u (1,cp) = + (A, Cos rncp + B, Sin mcp)

2 ='

From Fourier series theory

X

B, =I I u(l,@) Sin m@ do n x 2n

= 1 I ul Sin m$d@ + 1 I @in m@d@ - - n O n

= (ul - u2) (1-COS m n) - mn

ca

Then u(p, @) = ul + u2 + 2 (ul - uz (1-Cos mn) pn'Sin mcp - 2 '"I mn

= u, + u2 + 2(ul + u2 { p i n cp + 1 p3 Sin 3 q + 1 pS Sin 5cp + . . ...) - - - - 2 n 3 5

1 = u, + u2 + ul - u2 tan- 2pSincp - - 2 n 1

= 2 p sinq = p sing +A p3 sin3q +I p5 sin59 + .... P 7 1 -P 1 3 5

10.5. Solution Using Fourier Integral: 1. Fourier Integral Approach: I f f (x) satisfies Dirichet condition 7.2. 10.1) in a finite interval (-L, L) and an

w

integral I ) f(x) 1 dx converges, which means that f(x) is absolutely integrable

in (--,-) then Fourier's integral theorem states that w

.............. f(x)= I (A (a) cos a x + B(a) sin a x ) d a 0

(1)

w

A (a) = 1 I f (x) cos a x dx X -- -

and B (a) = 1 I f(x) sin a x dx ................ (2) - x -

In Equation 1, R.H.S. is the Fourier Integral expansion of f(x). f(x) must be continuous at x for (1) to hold, if x is a point of discontinuity f (x) is replaced with f f(x+O) + f(x-0) as in the case of Fourier series.

2 Other forms of Fourier Integral theorem are:-

....

.............. f(x) = 1 I I f(u) cos a(x-u) du d a ..(3) - n a x u=--

if f(x) is even these results can be simplified thus;

........ f(x) = 2 1 cos a x d a I f(u) cos a u du - 0

(5 ) x O

- - I!' X'ix) 15 oc!d. f(x) = 2 J s ~ n a xda I f(u) sin a udu ........ - (6)

X O 0

iZuniplc: A sernl thin bar x 2 0 whose surf:~ce is insulated has its initial ~ c n i p ~ ~ ~ u ~ c d r s t ~ ~ b u t ~ o n equal to f(x). The bar's end is suddenly quenched to ~ c c .I: O"c (t!i:~t I S x = 0) and maintained, (a) set up the boundary value p ~ ~ b i c ~ : : i o ~ tlic tclnpcmtulc u (11.t) at any pomt x at time, t.

so11:;io::: (a) The boundary mluc problem is:

u(\,O) = f('ix). u(0,t) = 0, ( ~ ~ ( x . t ) ( < M .......... (2) Temperature must be bou~rdec! as implicd by the last boundary condition. (b) l'he mlution of (1) obtained by separation of variable is u (x, t) =

G.2 1 (A cosiLx + I3 Sin ?J)

(c) Usin: 11:c boundary condition (2) \ve find A = 0 -Li.2, . .

th~!s u(x.1) = l3c sin AX ............... (3)

Sirlcc there is no restriction on h, replacing B with B(h) will still be 3

so!u;ioi:.

( ~ i ) can be integrated from 0 to and still have a solution l ! l c x ~ ~ r c

K

u(x8.t) L- .f Ij(1.) e."." sin hxdh .............. 0

(4)

-3

Using thc condition (2), f(x) = 5 B(h) siniLxdh 0

from (6) we sec that f(x) must be an odd function "Sine" is under the intcgu!.

03 a

Therefore B(h) = 2 1 f(x) sin h xdx = 2 I f(v) sin h vdv - 71

- 71 O

Using the result in (4) we obtain u (x,t) a -

= 2 1 I f(v)e-k)'2t Sin hv Sin hx d h dv ........... - n o O

10.6. Bessel Function Method:

Bessel functions derived from the Bessel Equations have been treated in 7.2

and will be used here to solve P.D.E.

Example: A circular plate of unit radius is insulated on its plane faces. The

rim is kept at zero temperature and the initial temperature distribution in the

plate is F(p), find the temperature of the plate at any time, t.

Solution: Since temperature is independent of iP, the boundary value problem for determining u(p,t) is

= p2 + L Y at JP P P

(i)U( 1 ,t) = 0 (ii)U (p,O) = F(p) and I u(p,t) I < M

Set u = P (p) T(t) = PT in equation (1) then PT' = k (P"T +_I P'T)

P or dividing by kPT, = P" + 1 P' = - h2

kT P p P from which we have Tf+ k A 2 ~ = 0, solution = T = C, e-k)'2'and P" + 1 P' + h2p = 0 from which the

P solution = p = Al Jo (A,) + B, Yo (A,) Since U = PT is bounded at p = 0, B1 = 0

Thcn u (p. t) = Jo (hp) =u(l,t)=O from which Jo (A) = 0 2nd h = hl A?; ...... are the positive roots.Thus the solution is u (p,t) = ~ e - ~ ~ , ' Jo (h,p) ni=1 .?.3... .. 13y superposition, n solution is

ti, = I

from thc second boundary condition m

u ( ~ , o ) = F(p) = C A,,, Jo (h,,~) .. . . ... . .. . ... m = I

(2)

NOW from the series of Bessel Functions, as in Fourier series, if f(x) satisfies Dirichlet conditions then at every point of continuity of f(x) in the interval 0 < x < I , there s1iould exist the Bessel Series expansion of the form:

ca

f(x) = AIJn ( h l , ~ ) + A2Jn (&x) + AJ, (h2x) + = ..... C A J (hpx) "$1

Where hlh2 are positive roots of IiJ, (x) + S,J', (x) = 0, R S are constants with R/S 2 0; S # 0 and

I n casc S = 0, h l , h2 are roots of J, (x) = 0

Based on these treatments

rn 1

and so U(p,t) = F 2 1p F(p) JO (h,p)dp~e-*~,' Jo (Amp)

is a solution of the temperature distribution in the plate. It is also the temperature distribution in an infinitely long solid cylinder whose convex surface is kept at zero and whose initial temperature distribution is F(p).

10.7. Using Legendre Functions for solving P.D.E: Legendre functions have been treated in chapter 7 and is used in solving partial differential equations (P.D.E) Example: (i) Find the solution to Laplace equation in spherical coordinates

which do not depend on @

Solution: Laplace equation v2v = 0 for sp erical coordinates and not f (Q) thus r a r v + 1 a [sineJ=o ............. (1)

ar & sine % ae

Let v = R(r) O (8) = R(O) in (I), divide by RO and separate variables,

d r dR = 1 d Sin 8dO =A2, kh[d %%id dB)

and we find r r rdB + h 2 ~ = 0 ......... -[ dr dr 1 (2)

Equation (1) can be written r2 R" + 2rR' + h 2 ~ = 0 (4) (this is a Cauchy Equation) and the solution is R = A l r l ' + B ............. (5)

? + I

where n = -% + ='thus hZ = -n(n+l) Equation (3) with h2 = -n(n+l) can be written d (sin 8 3 + n(n+l) Sin 08 = 0 ............. - (6)

d e de if cos 8 = x then dO = dO/de = - 1 d O or Sine do= - sin2e dO -- -- - dB d d d e Sine dB d e d e

= - (1 - x2) d~ dx

So that d - (sine &3 ) = d [-(1-x2) @ = d[(l-x2) d01 dx - d e d e d e dx d x dx de

= - d [(I-x2) d 0 ] - sin 8. dx dx

Equation (7) is the Legendre Differential equation with general solution @=Az PI, (x) + B2Q, (x) .............. (8) thus O = A2 PI, (COS (3) + B2 Q, ( C O S ~ ) .......... (9) which is the solution of (6) Using (5) (8) and (9) The solution of (1) is V = RO= [A,r " + B, 1 [A2 P, (x) + B2 Q, (x)] ....... (10) 7 ' where x = cose

10.8 Laplace transform method: Laplace transforms have been treated in Chapter six, and will be used here to solve partial differential equations.

Example: Solve the boundary value problem a~ = 4 a2u ......... - (1) at 3'7

u(0,t) = 0, u (3,t) = 0, U (x,O) =losin 2nx - 6 sin 4nx

Solution: Taking Laplace Transform of the equation with respect to t c.3 .ee ee

I e-st @ddt = I e-" &32U)dt, i.e s e-" udt - U(X,O) a t ax2 o

or su - u(x,o)= 4 & ............... (2) dx2 ee

Where U = U(x,s) = h {u(x,t)}= I e" u.dt 0

Using the given condition. u(x,O) = 10 sin 2xx - 6 sin 4nx (2) Becomes 4 d2 u - su = 6 sin 4nx - losin 2nx ........ (3)

dXL h{u(O,t) = 0 h{u(3,t) = 0 o r u (O,s)=O,u(3,s)=O .......................... (4)

Solving (3) subject to condition (4) we obtain U(x,s) = 10Sin 2nx - 6Sin 4xx

S + 16n2 S + 64n2

Taking inverse Laplace transform we obtain U(x,t) = X1 {U(x,s)} = h-' {s:;6J Sin 2PX

-"'LiJ Sin 4nx = 1 0 e - l ~ ~ ~ ' Sin 2nx - 6e-64n2 ' Sin 4nx

This is the required solution. Note: In using Laplace transforms to solve partial differential equations, it is advisable to first take the transform with respect to various independent variables in the equation and then choose the variable which leads to highest simplification.

Example: A mass m is suspended from the end of a vertical spring of constant, K, (force required per unit stretch). An external force, F(t) acts on the mass as well as a resistive force proportional to the iinstantaneous velocity. Assuming that x is the displacement of the mass at time, t and that the mass starts from rest at x = 0. (a) Set up a differential equation for the motion and (b) Find x at any time t.

Solution:

(a) The resistive force is given by - p h a n d the restoring force is dt

given by -kx. By Newton's Law m d&= - p &-Kx+F(t )

dt2 dt ....... m & + p dx + Kx = F (t) (1) -

dt2 dt where x(0) = 0, x (0) = 0.. .... .(2) (b) Taking transform of (1), Using h (F(t) = F(s) and h (x) = X h (m d2x + p .dx )+ Kh (x) = hF(t)

d 7 m from the tables M [s2 f(s) - sf(o) - f' (0) ] = h ((m d f t 2

Using the initial conditions:

where R = k - - m2 4m2

R can assume values (I) R > 0 (ii) R = 0 (iii) R < 0

case (1): R > 0, Let R = o2 (3) F(s) [(s + pJ2m) + 02]

taking the universe L of this expression

Using the convolution theorem, we find from (3) As X1 (X) = x then (3) + x

Case ii: R = 0, in this case A-I - te-Pu2m 1- and the convoltion theorem in (3) yields

1

x = - 1 I ~ ( u ) ( t - ~ ) e - ~ ( " ~ " ~ du m O

Case iii: R< 0, let R = -a2

then using the convolution theorem (3) yields.

x = - 1 I F(u) e-B(t-u"m Sin a(t-u)du am O

Example: Given the clectric network below, find the clirrcnts i n the var~orls branches if h c initial cul-rrnfc nrp 7~i-o. 209

Fig 17

Kirchoff 1" law = I = 11 + 12. Kirchoff's 2"* law: Voltage drop around closed loops JKNPJ, KLMNK are respectively zero.

Thus 201 - 120 + 2 dI1 + 1011= 0 - dt

Taking Laplace of the equations

- Using h (11) = 5 , h ( G) = G, I = 5 + 12, h (I) Frl + Iz 20 ( G+ G) - 120 - + 2[s 5- (O)] + 10 5 = 0

S

-10 II - 2[s I, - I1 (o)] + 4[s G- (o)] + 20 i; = o Using I1 (0) = 0, 1% = 0 these becomes

(30+2s) + 20 = 120/s; (-10-2s)? 1 + (4s+ 20) I2 = 0

- , 1 l 2 t S 4s2! 20\, - l , ! ~ ~ ~ ~ ~ 'y11, Solving 1, = 130 + 2s 20 I 2 = 130+2s 20

- 10-2s 4 ~ + 2 -10-2s 4 ~ + 20

- - 11= 6 = 3 1 - 1 , 1 3 =-3Ll -11

~ ( ~ + 2 0 ) ts ~d ~ ( ~ + 2 0 ) 2 s ~ + 2 0 Taking the inverse Laplace

X-1 1 1 = 3(1-e-20'), X-1 12/2= 3 (1-e-'OL), (1= 11/2 + 12 = 9 (1 - e"Ot),

10.9 Elimination of Arbitrary Constants in a Partial Differential Equation:

(1) Find the differential equation of the family of spheres of radius 5cm with centres on the plane y = x.

Solution: Equation of the family = (x-a)' + (y-b) + (z-b) = 25 . . . . . . (1) a, b being arbitrary constants. Differentiating with respect to x and y partially and dividing by 2 we have (x-a) + (z-b)p = 0 and (y-a) + (z- b)q = 0 Let z -b =-m, then x -a = pm, y-a = qm. Putting in (1) we have m2 (p2 + q2 + 1 = 25, x-y = m (p - q) from the above substitution.

Therefore m =x, m2 (p2 + q2 + 1) = (x2 (P' + q2 + 1) = 25 P-q (P- 1)2

and the required differential equation is ( x -~ ) ' (P2 + q1 = 1) = 25 (p - q)2. Remember that p= az/ax, q= az/ay

( i i ) Show that the partial differential equation obtained by eliminating the arbitray conslants a and b from z = ax + by + f(a, b) is the extended Clairant equation thus z = px + py + f (p, q).

Differentiating z = ax + by + f (a, b) with respect to x and by yield p = a and q = b aid the required differential equation follows immediately. (iii) Find the differential equation arising from Q, (x+y+z, x2 +y2 - z2) = 0

Solution: I ~ e t u = x + ~ + z , v = ~ ~ + ~ ~ - z ~

thus (I (u, V) = 0

Differentiating with respect to x and y yields & (1 +p) +& (2x-2zp) = O ax av and - acp (1 + p) +& (2y - 2zq) = 0

ax av eliminating & and* we have

ax

(iv) Elimination of the arbitrary function 41 (x,y) from z = 41 (x+y)

Solution: Let x + y = u thus the given ralation is, z = 41 (u) Differentiating withrespect to x and y yields p =&cp = cp' (u) and q = cp' (u)

du Thus p = q is the resulting differential equation (v) Eliminate the abitrary function f(x) and g (y) from

z = Y f(x) + g (Y)

Solution: Differentiating partially with respect to x and y p=yff(x)=g(y) ............ (1)

.............. q = f (x) + xg' (y) (2) we cannot eliminate f , g, f', g' from these so we differentiate again r =yf'(x), s = f'(x) + g'(y), t = xg" (y) from (1) and (2) we find f'W =A [ p - g (Y) = I [q - f (x)l

Y X

is the resulting partial differential equation.

(iv) Eliminate a and b from z = (x2 + a) (y2 + b) Differentiating with respect to x and y, p = 2x (y2 + b) and q = 2y (x2 +

Then y2 + b =&x2 + a = Q and z = (x2 + a) (y2 + B) - 2x 2~

o; pq = 4xyz or we eliminate a and b thus pq = 4xy (y + b) (x + a) = 4xyz

10.10 MORE APPLICATIONS OF PARTIAL DIFFERENTIAL EQUATIONS (PDE)

(i) FICK'S FIRST AND SECOND LAWS: Adolf Fick in 1855 unraveled the diffusion or mass transfer problem

encountered in almost all fields of engineering and found out that flux of

material was proportional to the concentration gradient thus

J, = flux of mass = Ddc . . . . . ..... . . (1)

where J = the flux in kg/m2 -sec or atoms/m2 -sec

D = the proportionality constant called the diffusion or

transfer coefficient in mlsec.

c = Concentration in kg/m3 or atoms/m3

The minus sign is requircd because the flow of the materials is towards the

lower concentrations hence dcldx is negative. Hence to make J positive, a

minus sign must be used. Equation (1) is called the Fick's First Law.

To determine the parameters in equation (1) an appropriate differential

equation for the diffusion process must be developed. To do this, a mass

balance'upon a differential volume element perpendicular to the mass flow

must be developed, thus for the material transport on this element volume we

have mass in - Mass out = Accumulation .... (2) or considering some time i interval we have the rate in - rate out = rate Accumulation ..... (3)

Rate mass in = (J A), dx - differential length A = Area of plane 1. Rate mass out = rate mass in + Change in rate Mass in across the volume element

........ = J(A), + a (JA) dx - ( 5 ) ax

Rate'accumulation = a (c. A. dx) = A. d. ac ....... A) .... (6) - - at at Substituting (4), (5 ) and (6) into (3) we have a~ = ac .......... (7) x a t

is the continuity equation where A is assumed constant in the x = direction. For 3- dimentional flow we replace 6/6x with V, the del operator an it applies to all material flow processes when no reaction occurs or no material is generated within the volume element (A.dx).

Equation (8) is a partial differential equation called Fick's Second Law in which concentration, C is the dependent variable while x and t are independent variables.

If we assume that D is a constant in equation (8), that is it does not

vary with concentration, possible in small range of composition), thus

equation (8) becomes:

Daze =A .......... (9) ax" at

(9) is the equation solved with Laplace Transform Method in Chapter 6 , Article 6.9. XVIII. A direct application of solution of (9) is shown in example (1)

Example 1: (Hardening of gears and other engineering components for

application requires diffusion of interstitial solute atoms namely

carbon, into the matrix at an equilibrium temperature). An

engineering component (steel) needs to be hardened against wear by

carburizing at an equilibrium temperature. The component is packed

appropriately against graphite and heated to an equilibrium

temperature of 700°C. You are required to develop the piece, nothing

the following boundary and initial conditions:

Boundary Condition: C (x = 0, t) = Cs ........ (10)

C(X = 00, t) = 0 Initial condition: c(x,O) = 0 ................... (1 1)

And the component is long enough that no solute or hardening specie

diffuses to the end away from the graphite mass.

Solution: The linear partial differential flow equation that is needed is the Fick's Second law of diffusion (9) thus ~ a ~ c =& .......... (12)

ax? at

Using Laplace Transform technique as shown in 10.3.7 with t as the independent variable with the boundary conditions, the Laplace's transformed solution is thus,

Taking the inverse transform, the solution of (12) under condition shown in (10) becomes,

2 c (x, t)= C, [ 1 -2 1 e-" du where PI= x G O 2 7 E t

B

But error function of p = Erf(P) = 2 1 e-"* du and the solution can be 4; O

written as C (x, t) - C, [ I-erf (P)] = C, [I-erf(xI2 6 t ) . See lO(ii) next page

The engineering component already contains some solute atoms of concenption, Co < C, thkn, C (x, t) can be shown to be equal to:

In practical term, we define case depth as the effective depth, x*, hardened for wear resistance in the component. If the solute concentration along this depth is Cs12 arbirarily and the component does not contain any solute atoms initially then. C (x, t) =% = C,[ I-erf (x&2 J D ~ ) .............. (17)

2 where Xos is the case depth for arbirary concentration = O X ,

Rearranging (17) we have 0.5 = erf ( ?2& -J

........... ( I ) From the hble of error functions in the Mathematical Hand book (like table of sine and cosine) we see that the number whose error function is 0.5 is 0.477.

That is 9 , = 0.477 2 D t

equation (19) means that in any case of hardening of engineering components the case depth or thickness of case depth is proportional to J D ~ . Thus to a first order approximation an engineer knowing the diffusion coefficient, D, of a solute or carburizing specie in a given time is able to estimate the thickness of hardened layer against wear in a component like gears.

10.11 NOTATIONS AND MORE APPLICATIONS OF PDE TO ELECTRICAL AND MECHANICAL FIELDS:

(a) Error function notation: X

erfx = error function of x =2 1 e-"' du .......... (1) J n O

JJJ DI

thus = 2T0 e.' du can be expressed thus To erf x J n O 2aJ t

er f (-) = 1

(b) Gamma Function: rn

T (n) J tn-'e" dt(n > 0) .............. 0

(3) *

h ( t "} =T(P+l) if s>O, p>-1 ............... 7 1 (4)

(c) Simgularity Functions: Consider a function f(t) which has the value lit,, when 0 < t, t,, and zero

w - ................. elsewhere Jf(to dt = I fdt = 1 (8)

0

i.e., Area under the f(t) graph is unity

AS to -+ 0. f(t) over (0, to) -+ 00 and the graph of the interval (0, to) shrinks toward zero such that to

I f(t) dt retains value of unity

Dirac delta function, 6 (t) is infinite at the point t = 0 and zero eleswhere but its integral across t = 0 is unity

'h (6 (t)) = 1 as to -+ 0

The load distribution for unit load is a special case of the DJRAC delta

function or unit impulse function. An infinite force acting over zero time

interval on an area that is tending to zero, givcs rise to a unit impulse. The

limiting form of the force fuction at time t+ 0 define the unit impule function

or Dirac delta function, 6 (t), also defines mathematically a section of a beam

at x = a, as

a ( x - a ) = { O , X < a {m ,X = a .......... (9) ( 0 ,X > a and 0 x <a

I" a(x - a) dx = 0 for x<a and 1 for x >a = H(x - a) = Heaviside function.

x X X

I I a(x - a) dx dx = I H(x - a) = r(x - a) =O for x<a and x-a for x >a ... (10)

where r (x - a) is the unit ramp function.

Illustrations of Unit impulse, Unit step and Unit ramp function in Mechanical Loading.

Unit Unit Unit Impulse Step Function Function function (1: 1 Slope)

- Example 1: An elastic bar is statically loaded. Find the displacement at any cross - section if the bar is statically indeternlinate. Jf the bar is statically determinate

we use equilibrium consideration, hence ox = E E, =

or static loading do, = E a2u = -4 (f, + k,) ... .. ... - (11) dx A

d2u =-L (fx + kx) - ax2 AE where f, is surfae force per unit length along X-direction. K , is body force per unit in the x-direction. E = Young's modulus, A = cross - section area For statically indeterminate case, f, = 2Q a(x - a); k, = 0 Thus fi = -24 a (x - a), A a cross sectional area

dx2 AE E = Young's modules

integrating twice we have u, = displacement at any section of the bar = 24 [(x - a) H (x - a) + clx + c2)] AE But u, (0) = n, (2a) = 0 thus C, = 0, C? = -% therefore u, =_a [ (x - a) H (x - a)-x/2 1 .. . . . . . . . . . . (12)

AE

Example 2*: Refer to the figure in examplel*. Suppose the bar is loaded suddenly and we wish to know the displcament of any cross section at any later time.

Solution: we assume quasi-static, even though it is dependent on the viscous term, meaning that the motion is slow enough that the effects of inertial force pu, are negligible (p = mass per unit volume). Stress-Strain Law for visco- elastic material of general type (assume Kelvin and Maxwell solids) is expressed lhus

Also time - dependent strain under constant load (creep) is associated with internal viscous nature of the solid thus ox = EE, +TE, E, = Strain velocity or strain rate

E = Viscous coefficient

Putting this into (1 1)

We have E a2ux + q a3ux 5 1 (fx+ k,) ax2 P a t A

and for the example 1' loaded suddenly we have ......... & + 3 a3ux = -a &x - a} (13)

ax2 E a7at AE

(13) * U +XU' =AQ 8(x -a) E AE

This is an inhomogeneous equation with solution thus U = UH + Up = homogeneous solution +Particular solution. Trial solution: UH = c~-(~") ' this satisfies (14) when the R.H.S is zero and then qualifies as the homogeneous solution. Try Up = -20 a(x - a)

AE This satisfies (14) as it is

But U, and u', are zero at t = 0

Therefore u(x, 0)=0, thus from equation (15)

c = x a(x-a) AE

thus finallv we have

Integrating twice :jnd applying the end conditions ux (0, t) = U, (2a, we get u, =x [(x - a) I \x - a) x ] (1 - e-(w")' -

AE 2

Which indicat \ hat there is no instantaneous displacement on application of load (t = 0) and that as t increases thc dispiacements approach the elastic value. - -. -

Which indicates that there is no instantaneous displacement on application of load (t = 0) and that as t increases the displacements approach the elastic value.

When stress is removed from a loaded visco-elastic solid, the strain does not disappear immediately as it does in the elastic solid. This is Kelvin solid.

Example 3: Equation of motions for STRETCHING AND TWISTING OF BARS: (free torsional and Extensional vibration of bars). These equations simply reduce to one - dimensional wave equation when external loads (f, c, ) and body force are absent, thus we have

aZu =J aZu .......... (16) and E at2

a20 =q a%- .......... (17) ax" G at2

where p = Density E= Young's modules G = Shear Module

These equation are exactly of the same structure and will be solved exactly the same fashion. The equatuions govern elastic vibrations of a bar when fixed. Only stretching or twisting takes place.

Solution: By separation of variables ...................... Displacement, u = X (x) q (t) (18)

X function of a only q function o f t only

Put into (16) x"q = p x, -E

Dividing by x, and rearranging 2 X"E = p = -W ..................... - (19)

X

Rewriting (19) x" + d x = 0 .................... (20) E

and q + oZq = 0 ..................................... (21)

these are now two ordinary differential equations which are similar. The solution of (21) - q = c sin o t + D cos w t ............ (22) w (in radians per second) = circular frequaency = 2nf)

Putting into (18)

For a body at rest initially the velocity at every x will be zero at time, t = 0 thus u, (x, 0) = X (x) q (0) = 0 ........................ (24) or q (0) = 0 and C from (22) = 0 . . u, (x, t) = X cos o t = (Asin

Where D is absorbed into A and B

Example 4: Telegraph Equation for a long cable:-

In dealing with flow of electricity in a long insulated cable, the potential V (volts), Current i (amps), are found related by simultaneous equations thus:

av = -L ai - ax ax

ai = -c & - ax ax

Where .x = distance from one end of cable L r induction per unit length of cable

(henries per unit length) C = Capacitance to ground per unit length

(Farads per unit length) L and C are assumed constant. Eliminating i, the following wave equation is satisfied

Boundary conditions, the line is dead (no current flows) v (x,o) = a V(X, O) = o

a t v (0, t) = f(t), t > 0.

If the other end (x = L is open, i = 0 when x = L thus another condition av (1, t) = 0, (t > 0)

ax

But if the end x = L is grounded then V = 0

Where x = L, another condition V (L,t) = 0, t > 0 :. V (x,t) = { f (t - x) when t >x -

C C

((0) when t < x - C

is a solution ................................. (27)

Example 5 (Extension of example 3): If the bar described in Example 3 is of viscoelastic material of Kelvin type, the extensional vibrations are damped, hence the relevant equation is thus: aZux +I aZux = p aZux .: .............. (28) Q E a x a t E g where the second term represents the effect of strain rate in the stress equation. Using the same procedure as in Example 3, we get two coupled equations X" +&PX = 0 ...................... (29)

E

where the second term represents the effect of strain rate in the stress equation.

Using the same procedure as in Example 3, we get two coupled equations X" + & x = o ...................... (29)

E

The solution of (29) is same as that of (16) and (17) for the elastic bar. The solution of (30) is given by

......................... Q = e -02q"E c sin w* t) + D cos w*t (31) Where

(co*)~ = w2 [ l - ( q w / 2 ~ ) ~ ] = (frequency)2 with the damping ............ (32)

Thus the effect of the viscous nature of the bar is to modify the time function q but not to modify the shape function x.

Illustrations: (a) Show the time variation of the displacement at a given

section when the bar is elastic. (b) Show the displacement when the bar is viscoelastic. It is

evident that the vibration is damped and fades out with increasing time.

Equation (28) depicts the vibrations in solids which have damping character. The Kelvin viscoelastic material so well explains the vibrating solids that this model has practical applications. Examples 6: If the temperature at any point (x, y, z) of a solid at a time, t is u(x y z t ) and if the thermal condutivity, k specific heat 6 and density 2 of the solid are constant,

Show that& = kv2u Where k = k/, a t

Solution: Recall Green's theorem in space or Divergence theorem which states that the surface integral of the normal component of a vector .. ..taken over closed surface is equal

..... ( i ) to the integral of the divergence of taken over the volume enclosed by the surface:

T h u s A = A l , i , + A ~ + A 3 k a n d n = c o s u J + c o s y k Where Al A2, & A3 are continuous and have continuous partial derivative, a, p y are the angles which the normal makes with x, y, z axes respectively. S is a closed surface bounding a region of volume, V from these we have

I (a& + - aA2 + - dA3) dv =I I (Al cos u + AZ cos fi + A3 cos y)ds ax ay az S

OR J J (& + aA2 + aA3) dv = I (Al dydz + A2 dzdx +Ag dxdy) - - ax ay az S

................ (35) and in vector form 111-v * A d v = j J A. n d s .............. (36)

S

If v is the arbitrary volume lying within the solid and S denotes the surface, the quality of heat flux across or leaving the surface S per unit time is:

J (- kV,) nds .............. (37) Quantity of heat leaving the surface S

Quantity of heat entering the surface, A, per unit time is;

j (kV,) nds .............. (38) S

By the diveigence theorem '

........... J J ( k ~ , ) nds = J I J V (kV,)dv. (39) S

Rate of increase of heat is

But I j I V.(kVu)dv = j f j oz dv ................... (41) I )I a t

i.e j j j [a - a u -V.(kVu)]dv=O I a t

Since V is arbitrary and the integrand assumed continuous must be identically zero so that oz&u = V.(k V u)

at if k, o and 2 are constants &U = v . v u = k v 2 u .................... (42) at

k is the diffusivity. For steady statc rlow (that is aulat = 0 or u is independent of time, the equation reduces to the Laplace's equation V'U = 0. Compare (42) to the Fick's second Law of diffusion in one dimension, equation (9) under 10. 3. 5 where D is the diffusion coefficient.

Example 7: A string of length, L is stretched between points (0,O) and (L, 0) on the x-axis.

At time t = 0 it has a shape given by f(x). 0 < x >L as it is released from rest. Find the displacement of the string at any later time.

Solution: The equation of the vibrating string is

..................... (43)

y(x,t) is the displacement along x-axis at time, t.

Boundary Conditions: Since ends of the string are fixed at x = 0, and x = L y(0,t) = y(L,t) = 0, t > 0 initial shape of string is given by f(x) .'. y(x, 0) = f(x), 0 < x < L Since initial vdocity of the string is zero y'(x.0) =O,O<x<L To solve this boundary value problem

We set y = XT as usual. Then XT" = a2 X T o r 2 =x a? x

if -A2 is the separation constant we have " T " + h 2 a 2 ~ = 0 a n d X + h 2 x = 0

HenceT=Al s i n h a t + B 1 c o s h a t , X = A 2 s i n h x + B 2 c o s h x Thus general solution is given y(x,t) =XT=(A2 sin hx + B2 cos hx)(AI sin h a t + BI cos hat) ..(44) Considering the boundary conditions to determine Al, A2, BI, B2, with y (0,t) = 0 we have = 0 then y (x,t) from (44) = Sin hx (A sin hat + B cos hat) . . . . . . . . . (45)

with y (L,t) = 0 we have sin hL ( A sin hat + B cos hat) = 0 then sin hL = 0 and hL = Mn since the second factor must not be equal to zero. We use y, = dyldt now dyldt = y, (x,t) Sin hx (A h a cos orht - B orh sin hat ) and y, (x,O) = (sin Ax) (A ha) = 0 from which A = 0 thus from (45) y (x,t) = B s i n n x c o s s a t = B sinhx cos hat

L L With y (x,O) = f(x) and superimposing solutions we have

y(x,t) = X B, sin- cos m t , h =m mcl L L L

from the Fourier Series theory L

B, =J (f(x) sin- dx) L 0 L

from the Fourier Series theory L

B, =2 1 (f(x) sin= dx) L 0 L

D. I.

the result is y (x,t) =C 2 1 f (x) sin s d x ) sin ~x cos m t mlL 0 L L L

The terms in (46) represent natural or normal modes of vibration of the string. The frequency of the mth normal mode, fm is obtained from the term involving mnat a n d a v e n by

L - 271fm= = o r f m m a = m T

L 2L 2 p Where a = E - 4- m = 1, f l is the fundamental frequency and all frequencies are integral multiples of this fundamental frequency of the string which yields musical tone in case of musical instruments or sound in case of vibrating strings.

Example 8: A circular member of unit radius stretched over a drum and fixed at the rim of the circular drum is constantly hit by particles so that the initial displacement is F(p, @) before it relaxes to it initial position. Find the displacement of this membrane at any time.

Solution: The boundary value problem z (p,@, t) from the rest position (xy

with boundary condition z(l,@,t) = 0, z (p,@,O) = F(p,@), z,(p,@,O)=O

Let z = P(p) a(@) T(t) = P Q T . . . . . . . .... (47) Then P 0 T"

(49) x p2 ;~nd separating variable we have 7 7

p' P" + p P' +A-p- = -$" = $ - - IJ 1' T

so that cX)" c p 2 0 = 0 .............. 9 - - (50)

p ? ~ " i p' + (L- p- -.1r) P = 0 .......... (5 1) scncrtll solntion of (4S), (50) and (51) are

- T - - A , c o s h a t + B , s i n h a t ............... -

(52) 0 ............... - Ajcosp$+B2sinp@

- (53)

P - A; J, (hp) +B3 Yp(2.p) (54) ...................

A solution of the equation z (p,Q,t) is given by (52) x (53) s (54)

Sincc z must have period of 2n in variable $, CL = 111 whcrc 111 = 0, 1,2, from (53) Sincc z is bounded at p = 0, then B3 = 0

Sincc z, (p,0,0) = 0, El = 0 Thcn thc solution is u(P.@) = J,, (Ap) cos iat (A cos @ + B sin I@), since z(l,$,t)=O. Jm(A)=O so that ib=h,,,L, k= 1, 2, 3 .............. are the positive roots. By the Supcrposi~ion (summing over both m and k)

m

with t = 0, z(p, 0.0) = F (p, 0,) - I: (c, cos mg + Dm Sin mg) ....... (56) nsO

m

Where c,, = I: A,* J, (LAP) mo -

Using Fourier Series Theory in (56) 2n

C, =1 I F(p, 8) cos m$ do, for m = 1.2.3. R 0

I 2n

1 1 I ~ ( p , 0) do, for m= 0 Zi 0

2n

D, =J I F(p, $) Sin m$, m = 1.2.3. R 0

From (58) using Bessel's series Expansion 1

I 2n

= 2 f pF(p, $) J, ( A d ) cos m$ dp d$, m=l, 2,. . . NJ,l (Ld12 0 0

{ I 2n

1 2 I I PVP, 40 JO (k.d dp d$ if m 0,

N J l (bk)li 0 0

1

B& = 2 f p J, (hit& D, dp [Jnwl (uli 0

I 2rr - - 7 I J PUP, $) J,, ( L P ) sin m$ dp d$

.[JlIHl (Ld12 0 0 ifm=O. 1 .2 ...........

Putting thcsc valucs of A ,,A and B , , in (55) yield the required bolution. Unlike in Example (7) the various vibrations of the mcrnbranc arc determined by specifying particular values of m and k, f ink = frcqucncy of vibration = .a -

271

Thcsc are not integcr mutiplcs of the fundament frequency, therefore :lo musical tone will rcsult - lather noise.

!,~:ll:>plc 9: Thc vibration of a freely vibrating viscoelastic beam, simply supported. of length, L when rotatory inertial effects are neglected c:u1 bc rcprcscnted by the following differential equation thus

Find tlic gcneral solution \17hcrc m - - Mass of unit length beam

1 - - Sectional moment of inertia E - - Modules of elasticity

11 - - Viscous coefficient Solution: Separating variables as before we find

( 5 s ) can bc expressed thus XiV - h4X = 0 .............. (61)

where h4 = =' .............. (62) Ei 22

The solution of equation (59) is q = e -oq t/2E (A sinW* t + B Cos* t ................. (65)

where (w*)' = w2 [l - ( r ) ~ ) ' - 2E

Equation (59) has the following solution containing four arbitrary constants ............... x = cl sin hx + c2 cos hx + c3 sin hhx + c4 cos hhx (64)

For the simply supported beam with prescribed conditions uZ (0,t) = uZ (L,t) = Oy = bending moment

But a2uz = - M,, hence the second conditions can be dXL EI 22

expressed as d2uz (0,t) = a2uz (L,t) = 0 axZ axZ

or in terms of shape functions X(0) = X (L) = XU (L) = 0. = X"'(L) Applying each of these conditions to equation (64) we find 0 - - +C2 + C4 0 - - C, Sin hL + C2 cos hL + C3 Sinh hL + C4 cash hL 0 - - -c2h2 + c4h2 0 - - C1 h2 Sin hL + C2 h2cos hL + c3h2 Sinh hL + C4 h2cosh hL

These form a set of linear homogeneous algebraic equation in C,, C2, C3, C4 Nontrivial solution can exist if and only if the determinant of coefficients of the C, is zero thus the 4x4 matrix below: co c2 c3 c4

Sin hL cos hL Sin hL Cosh hL 10 - 1 0 1 I = 0 -sin hl hL Sin hhL cosh hL Expanding the determinant and clearing we obtain Sin hh sin h3L = 0. This can be satisfied if sinh hL = 0

Sin 1li.L = 0. The first one is valid only if h = 0 but hj = =w2 thus h = 0 means that o = 0 which indicates

EL2 That there is no motion (frequency of vibration is zero) The second condition sinhL = 0 must be applicable, thus hL = nx where n = 0, 1, 2,3 ....... Again n = 0 indicates no displacement, therefore h, IVJ where n = 1,2, 3 ........

n = 1, h = 7 i i = fundamental frequency, i.e first mode. 4 4 From (62) (hJY = n n L"

But from (63), the frequecy, o*, of the vibration is (w,,*) = EI n4 n4 ( I - q4 I Zz n4 n4

-3- 4 m r E

finally u = displacement

= C. Cn Sin nnx - d0'*2E Sin o*nt 11-1.2.3 L

The displacement for a given mode diminishes exponentially with tirne as it docs for any visco-elastic - Kelvin body.

NOTE: - Kelvin and Maxwell solids are named after two physicist who dscribcd stress-strain behavours of material in various ways.

Kelvin proposed. the time varying strain under load-creep. It is

nssocinled with the internal viscous nature of the lrslid lrimply

represented thus:

where E . is thestrain velocity or swain rate q= the viscous

coefficient

when such materials are pulled intension and stress removed the

strain does not disappear immediate, as in the elastic material but

returns to zero after a long period of time. Sunch material are

called Kelvin solids.

- Clerk Maxwell related strain rates to stress rates through the equation

If stress is held constant with time E, = 5 + oat - E rl

The strain reaches the elastic strain ox/E immediately on application of stress and continues to increase linearly with time thereafter. On removal of thestress there will be an instantaneous recovery of an amount of strain equal to o,/E but no elastic after - effect as in the Kevin type. A solid that exhibits both the Kelvin and the Max-well types is called a viscoelastic material of general type, thus

Example 10: Directional derivatives: Rove that V$ is a vector petpendicular to the surface $(x,y,z) = c Where c isr a conetant

Solution: If x = xi + yi + zk iar the parition vetor to my pint P(x,y,z) en the given eurfuee, then differentiating we have dx = dxi + dyi t dzk lie6 on the plane tangent to the ruFfae ut P, fmm the quatlen of the arurfaee $ (x, Y, 2) P e

&'I i + 30 j + 30 k . (dx is + dyj + CIA) = 0 - L.i '. cl y - a z 3

i.c Vb .d:. = 0 so that V@ is tllc pcrpcndicu1:ir ro dr and therefore to the surface.

Ko:c [!::it a u n i t nornial to the surface ? ' -1 + 4yz - 5z- = -10 nt point (3, -1,2) c:un bc c\.::!u:?:cc! thus: n vcctor norm1 to this surface = v(?_x2 + 4yz - SZ') = 4xi + 4zj + (4y - 10z)k = 12i + Sj - 24k at t lx point (3, -1,2) rhc un~t nornials to thc suiface at P - - 111 ;Si -24k - = + f3i -I- 2i - 6k) - - - - - -

-\I (12) - + (s) ' + (-24) " 7

10.12: 13iynfuuction and Eigen value: l'hcsc tcruls arc also called characteristic function and characteristic

\;iluc rc\pcct~vcly. In problems involving ordinary differentlid equations C C ~ L I I ~ c o n d ~ t ~ o r i ~ I I IL IS~ be satisfied . These conditions are specified for two or molc \.alucs of tlic independent va~iable. Such conditions as we have learnt in this text ale called boundary condltions, as distinct from initlal value problem (whcre all condit~ons are specified at one point). If the ordinary differential ecpntion of second order: y" + al(x)y' +a2(x)y =O is homogeneous and at points x =a, y=O and at x = b, y=O, this equation 1s zero, we can easily write down rhc ~cncra l solution of this equation in the form: y = clul(x) + c2u2(x) w h e ~ e ul arid uz are linearly ~ndependent solutions and cl and c2 are constants. Fionl the boundary condltions, values of the constants can be evaluated, thus: c ~ L I , ( ~ ) +c1u7(a) = 0, clul(b) + c2u2(b) = 0, solving these two equation5 s~multancously , one soluuon is cl = c2 = 0, leading to a trivial solution, y = 0, otliciwisc it wili be a non-trivial solution, y+O. Thus if the deterrmliant

u , ( 4 uda>

1 ) equals zero, i,e the determinant vanishes, then non- trivial solution exists.

In most cases one or both of the linearly independent solutions ul(x) and u2(x) or their coefficients a,(x) or a2(x) depend upon a constant parameter, A, which may take various constant values at particular situations. In such cases the value of the determinant for certain values of h will generate a general solution of the differential equation, multiplied by an arbitrary constant.. Problems of this type are known as eigen-value or characteristic value -problems and the values of h for which non-trivial solutions exist are called eigen values or characteristic values of h. The case of vibrating string, vibrating electrons and tuning forks at the tip of an open tube half-filled with liquid are good examples. In these examples there are integral values of h that produce fundamental frequencies or sound correspobding to critical speeds in rotating bodies.

EXERCISE ON CHAPTER TEN: 1. Classify The Following Equation As Elliptic, Hyperbolic

or Parabolic (i) a - & = O (ii) & + & = 4 .3 ay2 ax axay

2. Show that z(x,y) = 4eaX Cos 3y is a solution to the boundary valve problem a2z +a = 0, z (x , f l ) = 0, Z(X,O) 4e"" axZ ay2

3. If a taut, horizontal string with fixed ends vibrates in a vertical plane under the influence of gravity, show that its eauation is $ = a 2 $ - g . (g is acceleration due to gravity).

4. (a) Solve x aZz + az = 0 -- axay ay

(b) Find the partiiular solutin for which x(x,O) = x5 + x, x(2,y) = 3y4

5. solve + 2 u + a 4 ~ = 16 ax4 ax2ay2 ay4

6. Solve the boundary valve problem

If (a) f(x) = 5 Sin TCX (b) f(x) = 3 sin TCX - 2 Sin 5nx

7. A beam has it ends hinged at x = 0 and x = 1. At time t = 0, a conceritreted transverse load of constant amplitude w is suddenly applied to the midpoint. Show that the resulting transverse displacement of any point x of the beam at time t > 0 is y(x,t)

if 0 < x < L/2 while the corresponding result for L/2 is obtain by symmetry.

8. An inifinite thin bar (-rn < x< w whose surface is insulated has an initial temperature equal to f(x). Show that the temperature at any point x at any time t, is given by

\In.- 10. The tempqature of a long solid cylinder of unit radius is initially at

zero. At time t = 0, the surface is given a constant temperature u, which is than maintained. Show that the temperature of the cyliner is given by -

1 1 . Solve the boundary value problem a (x ay) = ay given that y (x,O) = f(x), y, 9x, 0) = 0, y - - ax ay 7 F

12. Find the steady - state temperature within the region bounded by two concentric spheres of radius a and 2 a if the temperature of the outer and inner sphere are u and 0 respectively

(13) solve each of the following boundary-value problem using Laplace transforms

(iii) & = & , u(0, t) = 0 u(4,t) = 0 a t ax

(nx) U(X, 0) = S i n + 3 sin nx

2

y ( l + r ) = C

iJ = C COS 0

x2+4xy+3+c

1

x-y = 4

.\': ; ?&2 = j

!ks4+!hY2-)'+ 3xy=c1

:,f = 4x41ns + csJ

ly = \-'n + c

y = 2. + cc-X2

3y = s3 + Gx - 4lnx + Cx

11);' = -x + lh + ce4x 2 ' O,111o~ona1 trajcctojics are family of curves (hyperbolas) y - t = C

42 nxtrics

13.7 scc

137min . T = 75 - 400 (lns)An 2

(a) S tinm tlic original number (b) 10418 bacteria.

6. (a> e- 2x (b) -4x3-6x

7. x = 2e2' - 3et + e-'. y = 3e' - 3e-', z = e2' - e" - 3e + 5 - 2 cos t

z 8. x = -0.13 cos dl28t - 0.73 sin d128t i-4.13 (sin 2t + cos 2t)

9. y =H (cos h (H/w)x - I), a catenary W

where H = T cos 8, T sin 8 = w, tan8 = w/H, T 2 tension

Max deflection y max = ~ 1 ~ 1 3 8 4 ~ 1 occurring at an end of the internal 0 I x I 1

CHAPTER FOUR 1. (I) xy + 2yz + 3xz = C ( i i ) y + ln(x + z ) = C

4. P = 1 is an equation of a conic l+ e cos (8 + a )

having 0 as focus 1 - - cI2/k2

I:HAPTER FOUR I . (I) xy + 2yz + 3xz = C (ii) y + ln(x + z ) = C

4. P = 1 is an equation of a conic 1+_ e cos (0 + a )

having 0 as focus 1 - - c,'A?

CHAPTER FIVE

1. 2 3 y = ( ~ d x + Bx) (1 - x +x -x .......................... 1

converges for x < 1

Converges for all finite values of x

3. Y -. A X ( ~ - ~ / ~ + X ~ / ~ - X ~ / ~ ~ + ...)+ B X - ' ( ~ +x) Converges in the annular region bounded by a circle of arbitrarily small values and a circle of radius one both centred at x = 0

x = 1 Jo F(u) e-B('-uwm Sin w (t - u) du - wm

(iii) R<O.R=-a

1

x = 1 Jo F(U)~-~'( ' - 'Izm Sinh a(t - u).& a m

<I

CHAPTER SEVEN /

3. It is a special case of

d [P'"' 91 + [q(x) + hr(x)J = 0 where p(x) = 1 , q(x) = 0 - dx dx

011 - - 1, a2 = 0, $, = 1 , Pz = 0 and thus is a sturmliouville

system.

I

x = I ~ , F ( U ) ~ - ~ ' ( ' - ~ ) ~ ~ Sinh a(t - u) du am

CHAPTER SEVEN

3. It is a special case of

'

- d [P'") dyl + [q(x) + hr(x)] = 0 where p(x) = 1, q(x) = 0 - dx dx

(41 = 1, a2 = 0, o1 = 1, o2 = 0 and thus is a sturmliouville system.

CHAPTER EIGHT

2. Roots of the characteristic equation (for a12 < 4a2) become

Where A = cos 8, R = 1, uk =sin ke. vk = cos ke And the family of solutions

yk = c , sin ke + cz cos k0 is available

3. Since a I 2 > 4a. There are two real roots

Using the parameter. L in the difference equation

L,, = 4 sin2 (nW2N) leads to the family of solutions eigen functions.

~ k ( ~ ) = C, Sin (nnk/N).

For other L values only solutions yk 0 exist.

6. yk = CI (-ilk + c2 sin (71W3) + c3 cos (nW3)

7. yk = ~ ~ [ c l si&+ c2 cos zck + c3 sin 3zck + c4 cos W] 4 4 4 4

CHAPTER NINE

2. 1 (2aZu + 3 aZu - 2 aZu) 23 3r2 XS ds"

assuming 1 I S continuous partial derivatives

6. Maximu ,. 5, minimum = -5

8. (a) ! ?cm3 (b) 1.5%

6. yk=c, (ellk + c2 sin (nk/3) + c3 cos (nk/3)

7. y k = A'[c, sin&+ c2 cos zk + c3 sin 371.k + c4'cos 3&] 4 4 4 4

CHAPTER NINE

assuming u has continuous partial derivatives

6. Maximum = 5, minimum = -5

(a) 1 .70cm3 (b) 1.5%

CHAPTER TEN

1. (a) Hyperbolic (b) Hyperbolic (c) Elliptic (d) Parabolic

4. (a) xz = F(x) + G(y) (b) xz = x6 + x2 + by4 - 68

5. u = F (x + iy) + G(x -iy) + xH(x + iy) + xJ(x -iy) + 4(x2 + y 2)z

6. (a) y = 5 sin nx Sin 2nt -2F

(b) y = 3 sin 2nx Sin 4nt - 1 sin 5nx Sin 10nt - 2 5 5n

(c) u = 6e -"'I4 Sin (912) + 3 e - ~ ' Sin x

SUGGESTED FURTHER READING

Hilderbrand Francis B. Advanced Calculus Second Edition Prentice - Hall Inc., New Jersey (1976)

Rogers G. L Mechanics of Solids John - Wiley & Sons Sidney (1964)

Esmanbal Fundamental Concepts of Inorganic Chemistry McGraw Hill Book Company Ltd (1958)

Sherwood T. el a1 Chemical Engineering Series McGraw Hill Inc. (1975)

Spieiel M. R. Advanced Shaum's Series in Mathematics MaGraw Hill Book Company

Murray H. P. and Morraey C. B Modern Mathematical Analysis Addision -Wiley Publishing Company, Mass, U.S.A. (1964)

Spiegel M. R Advanced Calculus McGraw Hill Book Company, Sidney (1963)

Shield F. Numerical Analysis

Shanms Outline Series in Mathematics McGraw Hill Company Sidney (1968)

Ayers F. Theory and Practice of Differential Equations Shaum's Series McGraw Hill book Company Sydney (1952)

Diffusion Papers Presented at the Seminar American Society for Metals Oct. 14 - 15 1972

D. 0 . N. Obikwelu Lectures on Carburizing Mechanism Anambra State University of Technology (1983)

D. 0 . N. Obikwelu Lectures on Diffusion Phenomenon

. Physical Metallurgy I1 Federal University of Technology Owerri, 1997

APPENDIX

SOME IMPORTANT PARTIAL DIFFERENTIAL EOUATIONS

........................ 1. Poisson's Equation: v2q = h (i)

v2 - - if + if+ if ax2 dy2 dz2 = Laplacian operator.

In rectangular to-ordinates of x - y - z space

cp - .................................. - q(x,y,z) (ii)

Equation (I) is satisfied by

) The velocity potential of an incompressible irrotational, ideal fluid with continuously distributed sources or sinks.

(b) Steady state temperature distribution due to distributed heat sources.

(c) A stress function involved in the elastic torsion of prismatic bars with a suitable prescribed function, h.

(1) Wave Equation:

............................................ (iii)

This equation is encountered in the study of the propagation of waves with velocity, C, independent of the wave length. It is satisfies by

(a) the components of the electric or magnetic vector in electromagnetic theory

(b) componebts of the displacement in the theory of elastic vibrations suitably chosen.

(c) The velocity potential in the theory of sound (acoustics) for a perfect gas.

(d) Acoustic waves in water

(e) Compressional waves in steel

(f) Light waves in a Vacuum

3. Heat Conduction:

This is satisfied by

(1) Diffusion in solids

(ii) Temperature at a point of homogeneous body a is a constant

4. Telegraph Equation:

The equationis one - dimensional and is satisfied by the potential in a telegraph cable where h = LC and p = Rc. If leakage is neglected, L is inductance, C capacitance and R resistance per unit length

5. Vibrating String Equation:

s2y = a2aZy st' axL ........................................................... (vi>

It applies to the small transverse vibrations of a taut, flexible string such as violin string initially along the x-axis and set in motion at a point, X, and time t, producing a displacement y(x,t) is a displacement along the y - axis at point x and time t.

a2 = Tlp where T = the constant tension in the string p is mass per unit length of the string. It is assumed that the string vibrates because of its elasticity only, no extension forces.

6. Lapace's Equation

v2v = 0 .......................................... (vii)

This equation occurs in many fields, namely heat conduction, where v is the steady state temperature (temperature after a long time has elapsed and is equivalent to

au - at = 0, in the theory of gravitation or electricity.

v represents the gravitational or electric potential respectively, in the case of equation (vii) is called the potential equation

7. Longitudinal Vibrations of a Beam

a 2 ~ - - C~EU - at2 ax2 .......................... (viii)

C - - gW7. g is the acceleration due to gravity.

8. Transverse Vibration of a Beam:

E is modulus of elasticity I is moment of mertia g is acceleration due to gravity.

If an external force (transverse) F (x.t) is applied

b4 = &is replaced with b2. F(x, t)El ax4

E is modulus of elasticity I is moment of mertia g is acceleration due to gravity.

If an external force (transverse) F (x,t) is applied <

b4 = fi is replaced with b2. F(x, t)El ax4

9. Bi - Laplacian Equation

v4q = v2v2q =&+ 22lp- + &p = 0 ....( ix) ax4 ax2aY2 ay4

involved in many two dimensional problems in thetheory of elasticity. 10. Steady State Temperature Distribution in a Rectangular Plate:

11. Steady State Temperature Distribution in a Circular Annulus:

Laplace's Equation in plane using polar co-ordinates becomes

12. Vibration of a Circular Membrane:

a2w + i ao 1 a2w = 1 a20 ......... (xii) - 37 r ar r2 ae2 c v

In polar co-ordinates.

14. Pulsating Cylinder: (Aerody namics of non-uniform motion):

Here the Lateral boundary of an infinite circular cylinder surrounded by ideal imcompressible fluid is caused to pulsate radially with circular frequency w, such that the radial velocity of the boundary varies periodically about a zero value with a maximum velocity V,

From hydrodynamic the velocity, V is here the gradient of velocity potential q which satisfies the wave equation.

vZq = Lihp .......................... ...( xi v) c2 at2

c is the velocity of sound in the fluid

OR in cylindrical co-ordinates

(xiv) d %'- + = * r ar at

V radial velocity = ac~ - ar

15. Wave Equation for particles in optics and acoustics applied to

the motion of electrons atoms by Schrodinger.

Where cp = amplitude function for the particle in three directions.

m - - mass of the particle

h - - planks constant

E - - Total energy

v - - Potential energy

The solution of (xy) obtained in spherical polar coordinates yield the

electron distribution in space, and are of the form cp = F(r) S1(8,@). E - V = Kinetic energy.

In chemical Engineering, a case of falling film is encounter in wetted gas

absorbers, condensers and the process equipment's where a liquid is fed at the

top of a vertical tube or plate at say q cubic centimeter per second along each

Centimeter of the perimeter or width of the top edge. At low liquid rates the

flow down the wall is lamillar and a parabolic velocity profile is established.

If liquid fed has solute concentration C,, and the surface of the falling film has

constant concentration of C1, concentration at any point (x, y) in the film is got

by solving this differential equation:

UX 1 velocity at a distance y from the wall - - Pg (YY - % Y)

P

P - - Liquid velocity

g - - Local gravity

P - - Liquid density

Kinematic viscosity = Cl/p

Yo a= Film thickness

17. Mechanical systems:

(i) Compatibility Equation: + a2ay

ay2 axay ..... ..... .. ... (xvii)

E EY strain along x, y axes

a - - shear strain on xy plane

(ii) For the axially loaded bar made of Kelvin material: the displacement, u(x, t) at any section of the bar at time, t is represented by partial

differential equation thus

A = Cross section area

E = Modulus of elasticity

"I = Viscous coefficient

F, = Surface force per unit length x-direction

K, = Body force per unit length x-direction

(iii) Twisting of circular visco elastic bar

*xe = Yxe - a2ex - - x- atax ......... (xviii)

e x - - Angle of twist of x-face

YXO - - Shear strain

INDEX 'but.. - Accumula~on 212 Arbitrary constant 15, 209 'Auxiliary equation 55 Auxiliary system 187

Bacteria 49 Beams 42 Bernoulli's equation 30 Bessel equation 13 1 Bessel 13 Bessel function 134

Canonical 65 Carburizing 2 13 Cauchy equation 197 c-discriminant 75 Characteristic equation 162 Clairaut's equation 3 1 Complementary 58 ComplemAtary solution of B .

Ordinary differential equations 19 Constant coefficient 156 Continuity equation 2 13 Convolution 108 Convolution Theorem 107

Degree 14 Dirichet 139, 195 I;riscriminant, Gauss 75,143 Divergence 223 Diagamma 160 Diagram function 155 Differential 168 Difference equation 154 Dirac function 21 6

Eigen values 234 Eigen functions 158,234

Eigen 234 Elliptic 185 Euler 134, 170 Euler method 33 Error calculation 180 Error function 214 Existence theorem 16 Exact equation 24

Fibonacci numbers 157 Fick's laws 212 Fourier integral 199 Fourier series 195 Frequency 227 Frobenius 139

Gauss 143

Hankel 135 Hyperbolic 185 Hypergeometric series 143 Homogeneous 21

Integrating factor 19 Indicia] equation 97

Jacobian 171.. 174 Jacobians 176

Kei. Ker 137

Lagrange multipliers 177 Lagrange system 188 Laguerre polynomials 147 Laplace transformation 104,205

Laplace 104 Legendre equation 128 Legendre 203 ,

Legendre polynomial 145 Linear ordinary differential equations of order 1 52

of order n 50 Linear partial differential equation 183 Loci. Extranous 74

Numerical solutions 33 Non-operator 55 Non-homogeneous 187,191

Operator 55 Orthogonal 46 Orthogonality 141, 148

Pendulum 77 Primitive 14 Particular solution 17

Rodrigues formla 145 Runge-Kutta 35

Saddle point 180 Separation of variables 19, 192 Series 95, 195 Sturm-Liouville system 48 Superposition principle 187

Taylor 34 Taylor series 34 Telegraph equation 248 Telescoping sums 154

Undetermined coefficients 57

Variables separable 19, 192 Variation of parameter 60,69 Variable coefficients 50

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Wronskian 53

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