Initial Value Problems

Ali jan

M. Sc Previous (Semester-II)

Outline

Taylor Series

RungeKutta

Method

Stability

Introduction

• The general form of the 1st order differential equation is

• Integrating the above equation we get a constant ‘c’, to find that value, we must know an initial condition, that is the value of ‘y’ at ‘x’. Let the initial value is ‘x=a’

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• An ordinary differential equation of order ‘n’:

• Can always transform into ‘n’ first order equation:

• The equivalent first-order equations are:

• The solution now requires the knowledge ‘n’ auxiliary conditions. If there conditions are specified at the same value of ‘x=a’, the problem is said to be an initial value problem. The auxiliary conditions, called initial conditions:

• For example,

Taylor Series

• Taylor series method is use to gain high accuracy . Its basis is the truncated Taylor series for y about x, it is also a formula for numerical integration:

• The last term determines the order of integration, the “truncation error”, due to the terms omitted from the series:

• Using the finite difference approximation:

• We obtain the more useful form:

• The Taylor function implements the Taylor series method of integration of order four. It can handle any number of first-order differential equation:

• The user is required to supply the function “derivative” that returns the 4 x n matrix:

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Graphical Representation

Working of Taylor Series

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• Example 7.1

h=0.1

• The Taylor Series

• Differentiation of the differential equation, yields

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at x=0 and y=1

With h=0.1 and substituting the values of derivatives, the above equation becomes

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The approximate truncation error:

Therefore

The analytic solution of the differential equation:

• Example 7.2

• From x = 0 to 2 and h=0.25.

• With the notation the equivalent first order equations and the initial conditions are

• Repeated differentiation of the differential equation yields

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Thus the derivative array that has to be computed is

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1. from numpy import *

2. def taylor(x, y, n, h):

3. X = [ ]

4. Y = [ ]

5. X.append(x)

6. Y.append(y)

7. for i in range (n) :

8. D = f(x,y)

9. H = 1.0

10. for j in range (4) :

11. H = H * h/(j+1)

12. y = y + D[j] *H

13. x = x + h

14. X.append(x)

15. Y.append(y)

16. print array(X), array(Y)

17. return array (X), array(Y)

Code of Taylor Series

18. def f (x,y):

19. D = zeros ((4,2))

20. D [0] = [ y[1], -0.1*y[1]-x ]

21. D [1]=[ D[0,1], 0.01*y[1]+0.1*x-1.0 ]

22. D [2]=[ D[1,1], -0.00*y[1]-0.01*x+0.1 ]

23. D [3]=[ D[2,1], 0.0001*y[1]+0.001*x-0.01 ]

24. return D

25. taylor( 0.0, array ([0.0,1.0]) ,8 , 0.25 )

The results for Taylor series are:

x y[0] y[1]

0 0 1

0.25 0.24431315 0.94432131

0.5 0.46713137 0.82829196

0.75 0.65355402 0.65340186

1 0.78904832 0.42110415

1.25 0.85944028 0.13281605

1.5 0.85090579 -0.21008015

1.75 0.74996208 -0.60623633

2 0.54345919 -1.05433763

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• The need for Runge-Kutta Method increased, because of a drawback of Taylor series.

• Taylor series requires repeated differentiation of the dependent variables.

• There is extra work of coding each of the derivatives.

• Runge-Kutta method is used to eliminate the need for repeated differentiation of the differential equations.

Runge-Kutta Method

Runge-Kutta Graph

Runge-Kutta method of Fourth orderRunge-Kutta Method of Fourth Order

Example no. 7.4

From x = 0 to 2 and h=0.25.

With the notation the equivalent first order equations and the initial conditions are

Code of Runge-Kutta1. from numpy import *

2. from math import *

3. def run_kutta(x, y, h,n):

4. def run_kut4(x,y,h):

5. k0=h*f(x,y)

6. k1=h*f(x+h/2.0, y+k0/2.0)

7. k2=h*f(x+h/2.0, y+k1/2.0)

8. k3=h*f(x+h, y+k2)

9. return (k0 + 2.0*k1 + 2.0*k2 + k3)/6.0

10. X=[]

11. Y=[]

12. X.append(x)

13. Y.append(y)

14. while x<n:

15. y=y+run_kut4(x,y,h)

16. x=x+h

17. X.append(x)

18. Y.append(y)

19. print array(X), array(Y)

20. return array(X), array(y)

Code of Runge-Kutta

21. def f(x,y):

22. f=zeros((2))

23. f[0]=y[1]

24. f[1]=-0.1*y[1]-x

25. return f

26. run_kutta (0.0,array([0.0,1.0]),0.25,2.0)

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The results for Runge-Kutta:

x y[0] y[1]

0 0 1

0.25 0.24431315 0.94432131

0.5 0.46713137 0.82829196

0.75 0.65355402 0.65340186

1 0.78904832 0.42110415

1.25 0.85944028 0.13281605

1.5 0.85090579 -0.21008015

1.75 0.74996208 -0.60623633

2 0.54345919 -1.05433763

• A numerical integration is said to be stable if the effects of local errors do not accumulate catastrophically.

• If method is unstable, the global error will increase exponentially, eventually causing numerical overflow.

• Stability has nothing to do with accuracy, in fact, an inaccurate method can be very stable.

Stability

• Consider the linear problem

• Where ‘λ’ is a positive constant. The numerical solution:

• Substituting we get

• If “І1- λ І>1”, the method is clearly unstable since ‘y’ increases in every integration step. Thus it is stable only if “І1- λ І<1”.

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Miscellaneous

Example 7.6

A Spacecraft is launched at an altitude ‘H=772’ km above sea level with the speed ‘v0=6700 m/s’ in the direction shown. The differential equations describing the motion of the spacecraft are

‘r’ and ‘θ’ are polar coordinates of spacecraft.G= 6.672 × 10−11 m3kg−1s−2 = universal gravitational constantMe= 5.9742 × 1024 kg = mass of the earthRe = 6378.14 km = radius of the earth at sea level

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Let

then

So, we have

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Code1. from numpy import *2. from math import *3. def integrate(x, y, h,n):4. def run_kut4(x,y,h):5. k0=h*f(x,y)6. k1=h*f(x+h*0.5, y+k0*0.5)7. k2=h*f(x+h*0.5, y+k1*0.5)8. k3=h*f(x+h, y+k2)9. return (k0 + 2.0*k1 + 2.0*k2 + k3)/6.010. X=[]11. Y=[]12. X.append(x)13. Y.append(y)14. while x<n:15. y=y+run_kut4(x,y,h)16. x=x+2*h17. X.append(x)18. Y.append(y)19. print array(X), array(Y)20. return array(X), array(y)

21.f=zeros((4))22. f[0]=y[1]23. f[1]=(y[0]*(y[3]**2) - 3.9860e14)/(y[0]**2)24. f[2]=y[3]25. f[3]=(-2.0*y[1]*y[3])/y[0]26. return f27.Integrate (0.0,array([7.15014e6,0,0,0.937045e-3]),50.,1200.)

Result

X=t(s) r=y[0](m) r’=y[1](m) θ =y[2](rad) θ’ =y[3](rad)

0 7.15E+06 0.00E+00 0.00E+00 9.37E-04

100 7.14E+06 -3.90E+02 4.69E-02 9.40E-04

200 7.11E+06 -7.83E+02 9.40E-02 9.47E-04

300 7.06E+06 -1.18E+03 1.42E-01 9.61E-04

400 6.99E+06 -1.58E+03 1.90E-01 9.80E-04

500 6.90E+06 -2.00E+03 2.40E-01 1.01E-03

600 6.79E+06 -2.42E+03 2.91E-01 1.04E-03

700 6.66E+06 -2.86E+03 3.44E-01 1.08E-03

800 6.51E+06 -3.32E+03 3.99E-01 1.13E-03

900 6.33E+06 -3.80E+03 4.57E-01 1.20E-03

1000 6.13E+06 -4.32E+03 5.19E-01 1.28E-03

1100 5.90E+06 -4.87E+03 5.85E-01 1.38E-03

1200 5.64E+06 -5.47E+03 6.57E-01 1.51E-03

Result