amath 350 notes

8/11/2019 AMATH 350 Notes

1/47

AMATH 350 notes: Differential Equations in Business andEconomics

Johnew Zhang

December 12, 2012

Contents

1 Basic Concept & Terminology 3

2 First-Order ODEs 4

3 Mathematical Models of Population Growth 53.1 The Malthusian Model (T. Robert Malthus) . . . . . . . . . . . . . . . . . . 53.2 The Logistic Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

4 The Principle of Dimensional Homogeneity 6

5 First-Order Linear ODEs 75.1 Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

5.1.1 Homogeneous Equations . . . . . . . . . . . . . . . . . . . . . . . . . 95.1.2 The Form y = G(ax + by) . . . . . . . . . . . . . . . . . . . . . . . . 95.1.3 Bernoulli Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

5.2 Graphing Families of Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 10

6 Higher-order Linear ODEs 116.1 Existence & Uniqueness Theorem . . . . . . . . . . . . . . . . . . . . . . . . 12

6.1.1 Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126.2 The Principle of Superposition (1st-order version) . . . . . . . . . . . . . . . 12

6.2.1 Special Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136.3 Linear Independence of Functions . . . . . . . . . . . . . . . . . . . . . . . . 136.4 Homogeneous Linear DEs Finding Solutions . . . . . . . . . . . . . . . . . . 156.5 Solving Homogeneous Constant-Coefficient Linear ODEs . . . . . . . . . . . 15

6.5.1 Generalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176.5.2 Solving Inhomogenuous Linear Equation . . . . . . . . . . . . . . . . 17

1


2/47

6.6 Variation of Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196.6.1 First Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196.6.2 Second Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

6.7 The Reduction of Order technique . . . . . . . . . . . . . . . . . . . . . . 20

7 Boundary Value Problems (BVP) 227.1 System of ODEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247.2 Homogeneous Systems with Constant Coefficients Method of Solution . . . 267.3 The 2D Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267.4 Generalizations for Higher-Order Systems . . . . . . . . . . . . . . . . . . . 287.5 Inhomogeneous Linear Vector DEs . . . . . . . . . . . . . . . . . . . . . . . 287.6 Variation of Parameters for vector Des . . . . . . . . . . . . . . . . . . . . . 29

8 The 3 Most Famous PDEs 308.1 The Heat Equation (or Conduction Equation) . . . . . . . . . . . . . . . . . 30

8.2 The Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318.3 Laplaces Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318.4 First-order Linear PDEs and Partial Integration . . . . . . . . . . . . . . . 318.5 First-Order Linear PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338.6 Second-Order Linear PDEs . . . . . . . . . . . . . . . . . . . . . . . . . . . 348.7 Method of Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

8.7.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358.7.2 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

8.8 Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378.9 The Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

8.9.1 Demonstration of Concept . . . . . . . . . . . . . . . . . . . . . . . . 418.9.2 The Convolution Theorem . . . . . . . . . . . . . . . . . . . . . . . . 42

8.10 Solution of the Heat Equation for an Infinitely-Long Bar . . . . . . . . . . . 428.11 The Black-Scholes PDE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

8.11.1 Eliminating the Stochastic Term . . . . . . . . . . . . . . . . . . . . 45

2


3/47

1 Basic Concept & Terminology

Differential Equation A differential equation is an equation relating a function to itsown derivatives.

Solving a DE means finding the functions which satisfy it.

For example,dy

dx= 2y+ 1

A solution is

y= 5e2x 12

We can verify this easily: dydx = 10e2x while 2y+ 1 = 2(5e2x 12 ) + 1 = 10e2x

DEs usually have multiple solutions.

General Solution The general solution is an expression which represents all (or nothingall) of the solution.

Singular Solution Occasionally there are singular solutions, which dont fit general pat-tern.

For example, consider the DE y = y. The general solution is y= C ex.

Consider the equation,dy

dx= 4xy2

has general solution y= 12x2+c . However,y= 0 is also a solution.

Hence, a singular solution is a solution that does not match the general patterns.

Partial differential equations In multivariable calculus we encounter partial differentialequations (PDEs), which involve partial derivatives. However, some other are veryhard to solve. Consider if dydx =x, then y =

12 x

2 +C. For example, Ut =Uxx whereu= u(x, t). But ifux= x, then u(x, t) =

12 x

2 + g(t).

Order The order of a DE is the order of the highest-order derivative. For example,y+3yy= 0 is a third-order ODE. (y)3xy= x2 is first-order ODE.ut= ux+sinxis a first-order PDE.

boundary conditions In application we will usually be seeking one particular solution.To determine this we need (for first-order ODE) the value of the function at one point

(at initial condition). For an nth-order ODE, we need n inputs to solve the equation.If they are all given at the same value of x we call them initial conditions (e.g.y(0), y(0), y(0), etc.). If they are given at different values we call them boundaryconditions. For example,y = y, y(0) = 0, y(1) = 0. Solution: y= Asin(nx).

3


4/47

Linear An important distinction: a linear ODE has the structure + f2(x)y+ f1(x)y+f0(x)y= f(x).

2 First-Order ODEsMost general expression: F(x,y,y) = 0. We will consider only equations which can besolved for y: dydx =f(x, y)

Theorem. Existence& uniqueness of solutions: The initial value problem dydx =f(x, y), y(x0) =y0 has a unique solution, defined on some interval aroundx0, if f(x, y) and fy(x, y) arecontinuous within some rectangle containing(x0, y0).

For example, y =x2y+exy has a unique solution for any initial condition. However,y = (xy)2/3 may not have unique solutions for initial condition with y= 0.

Separable equations Suppose

dy

dx =f(x, y). Iff(x, y) can be factored as g(x)h(y), thenthe equation is separable, and can be solved thus:

dy

dx=g(x)h(y)

Divide by h(y) : 1h(y)dydx =g(x)

Integrate on x:

1h(y)

dydxdx=

g(x)dx

Now dydxdxis just dy , so 1

h(y)dy=

g(x)dx

(as a shortcut : dydx =g(x)h(y) = dyh(y) =g(x)dx = dyh(y) =

g(x)dx)

Example Solve the initial value problem dydx =ex+y, y(0) = 0.

dy

dx=exey = dy

ey =exdx =

eydy=

exdx = ey =ex + c1

Solve for y:

ey =

ex

c1 =

y= ln(

ex

c1) =

y=

ln(

ex

c1) =

y= ln[

1

c ex

](c=

c1)

Apply the initial condition: we will get c = 2 and hence y= ln[ 12ex ]

4


5/47

The only dander here is overlooking singular solutions. For example, find the generalsolution to

dy

dx= 4xy2

it is easy to get the solution

y= 12x2+c y= 0y= 0 y= 0

(watch out the singular solutions)

Well here has an example that does not satisfy the theorem. Consider the differentialequation

dy

dx= (xy)2/3

Notice that f(x, y) = (xy)2/3 is continuous but fy(x, y) = 2

3 x2/3y1/3 is not when

y= 0.

Solve:

dyy2/3

= x2/3dx = y= (1

5x5/3 + c)3( ify

= 0)

3 Mathematical Models of Population Growth

Strictly speaking, differential equations should apply only to continuous processes, butquantities which change sequentially can be approximated by continuous functions (popu-lations, prices, etc.)

Suppose we wish to predict future values of a population, P(t). We will assume thatP(t) is differentiable. To find P(t), we will make assumptions about dPdt.

3.1 The Malthusian Model (T. Robert Malthus)

The most basic assumptions under ideal circumstances (unlimited food, space, etc.), therate of change of population should be proportionate to the population itself. That is,dPdt =rP for some r R. To complete the model we need an initial condition: P(0) =P0.

Solve: dPP =

rdt = ln P = rt + C1 = P = Cert.From the initial condition,

we can solve C = P0. To determine the rate r, we need some point in the future for thepopulation. For example IfP(10) = 2.4P0, then r = ln 2.4/10.

3.2 The Logistic Model

Of course, Malthus realized that resources were not unlimited. to improve on the model,he suggested the concept of a maximum sustainable population (a carrying capacity)

which well call K. With this, we should have

dP

dt = 0 when P =K;

dP

dt K;

dP

dt rP when P K

5


6/47

Malthus suggested the Logistic Equation:

dP

dt =rP(1 P

K)

Solution: this is separable:

dP

P(1 PK)=rdt =

KdP

P(K P) =

rdt =

1

P +

1

K PdP =

rdt

Hence we get ln P ln |K P| = rt+C1 = ln | PKP| = rt+C1 = PKP = C2ert.Eventually, we get

P = KC2e

rt

1 + C2ert =

K

C3ert + 1(C3=

1

C2)

To apply the initial condition that P(0) = P0 = C2 = PKP0 , so C3 = KP0P0 , andP = K

(KP0P0

)ert+1. The inflection points occur when P = K/2. How do we know this?

Consider: dP

dt =rP(1 P

K)

Differentiate with respect to t:

d2P

dt2 =r

dP

dt(1 P

K) rP(1

K

dP

dt) =r

dP

dt[1 2P

K ]

sod2P

dt2 =r2P[1 P

K][1 2P

K ]

and so d2Pdt2 = 0 when P = 0, K, or K/2

4 The Principle of Dimensional Homogeneity

A useful check on our calculations is the realization that in any equation, the units of eachterm must be the same. e.g., In the equation dPdt =rP.

dPdt has dimensions of

populationtime

ThereforerPmust also have dimensions of populationtime so r must have dimensions of 1time (r

is a frequency!) (we say [r] =T1)Now consider the solution: P(t) =P0e

rt. We see thatert must be dimensionless. Why?Consider that

ex = 1 + x +x2

2! +

x3

3! +

This cannot make sense if x has dimension!Now consider the logistic equation:

dP

dt =rP(1 P

K)

6


7/47

where populationtime = ( 1

time )(population)(dimensionless) and its solution:

P(t) = KP0

(K

P0)ert + P0

Comments: Angle is dimensionless because = Sr

5 First-Order Linear ODEs

Consider

a1(x)dy

dx+ a0(x)y= f(x)

As motivation, consider the example

xdy

dx+ y= ex

This is just ddx(xy) =ex so xy =

exdx= ex + C soy= 1xe

x + CxTo solve a1(x)

dydx+ a0(x)y= f(x), we

1. write it in standard form:dy

dx+p(x)y= q(x)

2. Multiply through by an unknown function(x) (an integrating factor)

(x)dy

dx+ (x)p(x)y= (x)q(x)

3. Now assume that the LHS is ddx((x)y(x))

This requires thaty+ py= y + y

so (x)p(x)y(x) = (x)y(x). That is ddx = p, so d

=

p(x)dx and so ln =p(x)dx + Cand so (x) =e

p(x)dx+C

4. The DE will now bed

dx(y) =q

and so (x)y(x) = (x)q(x)dxandy(x) = 1(x) (x)q(x)dx

7


8/47

Example 1

Solve dydx+ xy= x.Solution: This is linear in standard form so we identify p(x) =x and calculate (x) =

exdx =e

1

2x

2

Factor this in

e1

2x2 dy

dx+ xe

1

2x2y= xe

1

2x2

This is ddx(e1

2x2y) =xe

1

2x2

soe1

2x2y=

xe

1

2x2dx= e

1

2x2 + C

= y= 1 + Ce 12x2

Example 2

xy 2y= x3 cos x(assume x >0)Solution: In standard form we have

y 2x

y= x2 cos x

so(x) =e 2

xdx =e2 lnx

2=eln(x

2) = 1x2This gives us

1

x2y 2

x3y= cos x

That is ddx( 1x2 y) = cos x = yx2 = sin x + C

Example 3

Solve the IVP dxdt + x=

1 + cos2 t, x(1) = 4.

Solution: Here (t) =e1dt

=et

so we have

etdx

dt + etx= et

1 + cos2 t

That is ddt(etx) =et

1 + cos2 t

= etx= et1 + cos2 tdtWe may write this as

etx=

t1

e

1 + cos2 d+ C

= x(t) =et t

1e

1 + cos2 d+ Cet

Use the initial condition x(1) = 4.4 =e1 0 + Ce1 = c= 4e

sox(t) =ett

1e

1 + cos2 d+ 4e1t

8


9/47


10/47

Here is another example: dydx =

x + y 1Letu= x + y. Then du

dx

= 1 + dy

dx

. The DE becomes

du

dx 1 = u 1 = du

dx=

u =

u1/2du=

dx

= 2u= x + c1 = u= ( x2

+ c2)2

= x + y= ( x2

+ c2)2 = y= 1

4(x + c)2 x

5.1.3 Bernoulli Equation

A Bernoulli equation has the form

dy

dx+ p(x)y= q(x)yn

We can solve these by letting v = y1n =y(n1)

Example: Solve dydx 5y= 52 xy3Lety= y2, then dydx = 2y3dydxThis is easiest to use if we first divide the original equation by y3.

1

y3dy

dx 5

y2 = 5

2x = 1

2

dv

dx 5v= 5

2x

i.e. dv

dx+ 10v= 5x.

We need u(x) =e

10dx =e10x

e10xdv

dx+ 10e10xv= 5xe10x = d

dx(e10xv) = 5xe10x

= e10xv = 12

xe10x 120

e10x + c = 1y2

=1

2x 1

20+ ce10x or y = 0

5.2 Graphing Families of Solutions

We can gain information about solutions directly from the DE. Example: suppose dydx =y2

4

We can see immediately that y = 2 are equilibrium solutions , andy = 0 anywhereelse.

10


11/47

Also,y > 0 when|y| >2; u < 0 when|y|


12/47

6.1 Existence & Uniqueness Theorem

The IVP consisting of equation above and the n initial condition y(x0) = p0, y(x0) =p1, , y(n1)(x0) =pn1 has a unique solution or an interval I containing x0. If

1. an, an1, , a0 and F are continuous on I2. an(x) = 0 anywhere on I

6.1.1 Operator

Writing equations like (1) is tiresome so we have a more concise notation.

Definition. An operator is a mapping from functions to functions.

For example, we can speak of the differential operator - D which produces the derivativeof the input.

Df(x) =f(x)

D(x2 + 7) = 2x

The identity operator is I: I f(x) =f(x). We can combine these to produce new operators.For example Letting D2 =D D, we could write

(D2 + 5I)f(x) =D2f+ 5If=D(Df) + 5f=Df(x) + 5f(x) =f(x) + 5f(x)

If we give the operator D2 + 5Ia new name say (D), then we can write

y+ 5y= cos x

as(D)[y] = cos xor just [y] = cos x

Similarly, by letting

= an(x)Dn + an1(x)Dn1 + + a1(x)D+ a0(x)I

we can write equation (1) as[y] =F(x)

6.2 The Principle of Superposition (1st-order version)

Let y1(x) be a solution to the equation y+ p(x)y =F1(x) and let y2(x) be a solution to

the equationy+p(x)y= F2(x).Theny1+ y2 is a solution to the equation

y+ p(x)y= F1(x) + F2(x)

.

12


13/47

Proof. Ify1 and y2 are as described then

(y1+ y2)+p(x)(y1+ y2) =y 1+p(x)y1+ y

2+p(x)y2= F1(x) + F2(x)

6.2.1 Special Cases

1. Ifyn is a solution to y+ p(x)y = 0 and yp is a solution to y+ p(x)y =f(x), then

yn+ yp is also a solution to y+p(x)y= f(x)

2. Ify1 and y2 are both solutions to y+ p(x)y = 0, then so is y1+ y2. Furthermore,any linear combination ofy1 and y2 will also be a solution.

Proof. Ify = c1y1+ c2y2, then

y+ p(x)y= c1y1+ c2y2+p(x)y1c1+ c2p(x)y2

=c1(y1+p(x)y1) + c2(y

2+ p(x)y2) = 0

These results generalize naturally to higher orders.

6.3 Linear Independence of Functions

Definition. The functionf1(x), f2(x), , fn(x)are linearly dependent on an interval I ifthere exist constantsc1, c2, , cn (not all zero) such that

c1f1(x) + c2f2(x) +

+ cnfn(x) = 0

for allx I.For example, ex, ex, sinh x (cosh x = 12 (e

x +ex), sinh x= 12 (ex ex)) is a linearly

dependent set.Otherwise, linearly independent. For example, 1, x , x2 is a linearly independent set,

since c1+ c2x+c3x2 = 0, x = c1 = c2 = c3 = 0.Another example, cos1 x, sin1 x, 1

are linearly dependent, since cos1 x + sin1 x= 2

Definition. The Wronskian of the n functionsf1, f2, , fn is the determinant of thennmatrix

f1(x) f2(x) fn(x)f1(x) f2(x) fn(x)... ... . . . ...

f(n1)1 (x) f

(n1)2 (x) f(n1)n (x)

13


14/47

We will denote this by W(x) or W(f1, f2, , fn)Example: Iff1 = sin x and f2 = e

2x then W(x) = det(

sin x e2x

cos x 2e2x

) = 2e2x sin x

e2x cos x

Theorem. If f1 and f2 are linearly dependent on an interval I, then W(f1, f2) = 0 forsomex0 IProof. Suppose W is nonzero for all x I, and suppose thatc1f1+c2f2= 0. Differentiatinggives c1f

1+ c2f

2= 0, so we may write

f1 f2f1 f

2

c1c2

=

00

. SinceW= 0 this matrix is

invertible, so c1 = c2 = 0, meaning that f1 and f2 are linearly independent. The converseis not generally true. However, it does hold in the context of linear ODEs!.

Theorem. Suppose y1 & y2 are solutions to the equations y+ p(x)y + q(x)y = 0 and

p(x) andq(x) are continuous on an interval I. IfW(y1, y2) = 0 for somex0 I, theny1andy2 are linearly dependent.

Proof. Assume thatc1y1 + c2y2= 0 for allx I(we will construct a non-zero c1, c2) Thenc1y1+ c2y

2 = 0 so

y1 y2y1 y

2

c1c2

=

00

. Since W = 0 for some x0 I, there exists a

non-zero solution [c1, c2]T whenx= x0.

Now let u(x) =c1y1+ c2y2. Then u(x) =c1y1+ c2y

2 and sou(x0) = 0 and u

(x) = 0.This means that u(x) is a solution to the IVP u+ p(x)y + q(x)y = 0 with y(x0) = 0and y(x0) = 0. By inspection, the function y = 0 is a solution, and by the Existenceand Uniqueness Theorem the solution is unique so u(x) is the zero function. Thus wevefound nonzero c1, c2 such that c1y1+c2y2 = 0 for all x I, so y1 and y2 are linearlyindependent.

(soy1&y2 are linearly dependent if and only ifW(x) = 0 for a some x0 I)Theorem. Ify1 andy2 are solutions to a linear DE, thenW(t) is either zero everywhereor nowhere.

Proof. Abels Formula: W(x) =W(x0)e x

x0p()d

.

Proof. Ify +p(x)y1 + q(x)y1= 0 and y0+p(x)y

2 + q(x)y2= 0, multiply the first equation

byy2 and multiply the second by y1 and subtract them.

(y1y2 y1 y2) +p(x)(y1y2 y1y2) = 0

HenceW(x) +p(x)W(x) = 0

so we have W(x) +p(x)W(x) = 0. Hence W(x) =c1ep(x)dx =c1e

xx0

p()d

Settingx= x0, gives c1= W(x0).

14


15/47

6.4 Homogeneous Linear DEs Finding Solutions

1. The Existence and Uniqueness theorem tells us that a second-order linear ODE needsto be accompanied by two ICs. The general solution need s two arbitrary constant.

2. The principle of superposition tells us that if y1 and y2 are solutions to a linearhomogeneous DE, then so is c1y1+ c2y2. This suggests the following:

Theorem. Suppose p(x) and q(x) are continuous on I. If y1 and y2 are linearlyindependent solutions to the DEy+p(x)y+ q(x)y= 0, then the general solution isy= c1y1+ c2y2.

Proof. Let (x) be any solution (we will show that c1, c2 such that = c1y1 +c2y2)Consider the IVP consisting of the DE and the ICs y(x0) = (x0) and y(x0) = (x0) Ifwe match the proposed general soliton y = c1y1+ c2y2 to these ICs, we

c1y1(x0) + c2y2(x0) = (x0)

c1y(x0) + c2y2(x0) =

(x0)

That is y1(x0) y2(x0)y1(x0) y

2(x0)

c1c2

=

(x0)(x0)

SinceW(y1, y2) = 0, we can apply the inverse matrix to find c1&c2. Hence (x) is a linearcombination ofy1&y2.

6.5 Solving Homogeneous Constant-Coefficient Linear ODEs

We have difficulty solving y + p(x)y+ q(x)y= 0, unless p(x)&q(x) are constants.Example: Consider y =y.We have y1 =ex and y2 = e1, and so the general solutionis y= c1e

x + c2ex

Example: Consider y = y. We have y1 = sin x and y2 = cos x and so y =c1cos x+c2sin x.

Clever Observation

cos x, sin x are related to ex. Eulers formal: eix = cos x+i sin x. In fact,eix( and eix)are also solutions to y = y, so the general solution can be written as

y= c1eix + c2e

ix

Perhaps every equation y + py + qy = 0 has exponential solutions? Lets assume thaty= emx. then y = memx and y = m2emx

15


16/47

Substituting these into the DE gives,

m2emx +pmemx + qemx = 0

= m2

+pm + q= 0

We call this the characteristic equation of the DE. Ifm = p

p24q

2 then emx is a

solutionCase I: Distinct Real Roots. If we have solutionm1, m2, thene

m1x andem2x are linearlyindependent solutions, so y = c1e

m1x + c2em2x is the general solution.

Eg. Suppose y y 2y = 0. The characteristic equation is m2 m 2 = 0 s om= 1, 2.

= y= c1ex + c2e2x

Case II: Complex Conjugate Roots.Here we can still say that the general solution isy= c1e

m1x + c2em2x butm1, m2= i

However, we can assume that y is real (we know that real solutions exist) so lets rewritethis :

y= c1e(+i)x + c2e

(i)x

=ex[c1eix + c2e

ix ]=ex[c1(cos x + i sin x) + c2(cos x i sin x)]=ex[(c1+ c2)cos x + (c1 c2)i sin x]

since this is real, (c1+ c2) Rand (c1 c2)i R. Lets rename them!

y= ex[A cos x + B sin x]

Eg/ Suppose y + 2y+ 5y= 0. The characteristics equation is

m2 + 2m + 5 = 0 = (m + 1)2 + 4 = 0 = m= 1 2i

= y= ex[c1cos 2x + c2sin 2x]Case III: Repeated Real Roots. In this case our guess (ofemx) has yielded only one

solution. We need a second one.DAlemberts Method of discovery: Having 2 identical roots should not be much dif-

ferent from having 2 nearly identical roots. So, suppose we have two roots, m and m+.Then the general solution is y = c1e

mx + c2e(m+)x

As 0 it appears that the solutions all merge. However, if we have c1 = 1 andc2=

1 then y=

1 e

(m+)x 1 emx =emx[ex1 ]

Now lim0 ex1 = lim0

xex

1 =x

16


17/47

Therefore a second (linearly independent) solution isxemx, and the general solution is

y= c1emx + c2xe

mx

For example, for y+ 6y+ 9y = 0, we have m2 + 6m+ 9 = 0 i.e. (m+ 3)2 = 0, som= 3 and so y = C1e3x + c2xe3x

6.5.1 Generalization

For nth-order equations

y(n) + bn1y(n1) + + b1y+ b0y= 0

, the characteristic equation is

mn + bn1mn1 + + b1m + b0= 0

. For every distinct real root m, emx is a solution. For every complex pair i,ex cos xand ex sin x are solutions. If we have repeated roots, we multiply by x, repeatedly ifnecessary.

Examples

1. y4y+7y6y= 0. Characteristic equationm34m2+7m6 = (m2)(m22m+3) = 0. Hence m= 1 2i. Hence we have y= c1e2x + c2ex cos

2x + c3e

x sin

2x

2. y+ 3y+ 3y+ y= 0. Characteristics equation ism3 + 3m2 + 3m+1 = (m+ 1)3 = 0,som= 1, 1, 1. The general solution is y= c1ex + c2xex + c3x2ex

3. y(4) + 2y+ y= 0 Characteristics equation is m4 + 2m2 + 1 = 0. i.e. (m2 + 1)2 = 0,som= i,i, i, iThe general solution is y = c1cos x + c2sin x + c3x cos x + c4x sin x.

6.5.2 Solving Inhomogenuous Linear Equation

From our discussion of the theory of linear DEs. If we can solve the corresponding homo-geneous solution (call the solution yh) and we can find one solution to the full DE (call ityp), then the general solution will be y = yh+ yp.

How do we find yp? One option: The Method of Undetermined Coefficient

17


18/47

Example 1

Consider the DE y+ y 6y = e4x. What kind of function could satisfy this? Maybe amultiple ofe4x? We guess that yp = Ae

4x. Then yp = 4Ae4x, and yp = 16Ae

4x so the DE

becomes16Ae4x + 4Ae4x 6Ae4x =e4x = 14A= 1

soA= 114 = 114 e4x is a solution.Now for the homogeneous problem we have

y+ y 6y= 0= m2 + m 6 = 0

(m + 3)(m 2) = 0 = m= 3, 2 so yh= c1e3x + c2e2x

Example 2

Considery+ y 6y= 6x2

? We tryyp = Ax

2 + Bx + C

yp = 2Ax + B

yp = 2A

Substitution into the DE gives

2A + (2Ax + B) 6(Ax2 + Bx + C) = 6x2

= 6Ax2 + (2A 6B)x + (2A + B 6C) = 6x2

=

A=

1, B=

1

3

, C=

7

18soyp= x2 13 x 718 is a solution. Example: y+ y 4y= cos 2x

We tryyp = A cos2x + B sin2x

yp= 2A sin2x + 2B cos2xyp = 4A cos2x 4B sin2x

We find4A cos2x 4B sin2x2A sin2x + 2B cos2x

4A cos2x

4B sin2x= cos 2x

= A= 217

, B= 1

34

= yp= 217cos2x + 134sin2x is a solution.

18


19/47

Summary

Inhomogeneous Term Trial function

ekx Aekx

xn

Anxn+ + A0sin kx or cos kx A cos kx + B sin kx

1. IfF(x) is a sum of functions, we try the sum of the corresponding trial functions.

For example, for y + y+ y= e3x + x, we will try yp= Ae3x + Bx + C.

2. If F(x) is a product of functions, we guess the product of the corresponding trialf)unctions but (for example y + y + y = x2ex wed try yp = (Ax2 +Bx+ C)ex)one constant will be redundant, and must be omitted. Another example will bey+ y+ y= x cos x, wed try yp= (Ax + B)(Ccos x + D sin x)

3. There is one other special case. Consider y y 2y = e2x. Try yp = Ae2x,yp = 2Ae

2x

,y p = 4Ae2x

. The DE becomes 4Ae2x

2Ae2x

2Ae2x

=e2x

, (i.e. 0 = 1)What happened? e2x is one of the solutions of the homogeneous problem, so it cantsolve the inhomogeneous one!. What else might work? We try Axe2x. Ifyp= Axe

2x,then y p= Ae

2x + 2Axe2x and yp = 4Ae2x + 4Axe2x.

Plug into DE.

4Ae2x + 4Axe2x Ae2x 2Axe2x 2Axe2x =e2x

soA= 13 and yp= 13 xe

2x is a solution.

This works in general:

(a) If any term in the usual trial function is a solution to the homogeneous prob-

lem, then multiply it by x. repeat if necessary. If this duplicates another termin your trial function, then multiply that term by x as well.

E.g. Fory + 2y+ y= ex, we haveyh = c1ex+ c2xex so we tryyp= Ax2ex

Comment: It is advisable to always find yn first.

Comment 2: For other forcing terms we will probably not be able to guess thecorrect form. For example y+ y+ y= 1x

6.6 Variation of Parameters

6.6.1 First Order

Consider y+ xy = 1

x . Find solution to homogeneous: y+ xy = 0 = yh = C e1

2x2

tofindyp we try: yp=u(x)e 1

2x2

yp= ue

1

2x2 xue 12x2

19


20/47

u= 1

xe1

2x2 = u(x) =

1

xe1

2x2 = u(x) =

xx0

1

te1

2t2dt + C

yp = e 1

2x2

xx0

1t e

1

2t2dt = y= yp+ yh

Note: we could keep the +C in u to get the general solution.

6.6.2 Second Order

Now suppose y+ py + qy = F. Suppose we can find yn = c1y1+ c2y2, so we try yp =u1y1 +u2y2and y

p= u

1y1 +u1y

1 +u

2y2 +u2y

2. If we differentiate again, will get very messy

so we impose a second condition: u1y1+ u2y2 = 0. Hencey

p =u

1y1+ u1y

1+ u

2y2+ u2y

2 .

Therefore u1y1 + u1y

1 + u

2y2 +u2y

2 + pu1y

1 + pu2y

2 +qu1y1 +qu2y2 = F. That is

u1y1+ u

2y2= F. Therefore we can solve u

1 andu

2.

Example: y+ qy = 9sec2(3t) (for t (0, 6 )). Firstly solvey+ 9y = 0 = yh =c1cos 3t+c2sin 3t. Then we try y = u1cos 3t+u2sin 3t = u1cos 3t+u2sin 3t= 0 and

u1sin 3t + u

2sin 3t= 3 sec

2 3t. Therefore,u2= 3 sec 3t =

u2= ln

|sec3t + tan 3t

|+ c2

and u1 = sec3t + c1

6.7 The Reduction of Order technique

The same trick (variation of parameters) can be used to find full solutions to homogeneousproblems, when we know one solution. When used in this context, it is called the reductionof order technique.

Example: Considerx2y+ 2xy 2y= 0 (forx >0)

we might be able to spot one solution: y1= x. To find the second solution, we write

y2= u(x)y1= xuTheny2=xu

+ uy2 =xu

+ 2u

Plug this into the DE:

x2(xu+ 2u) + 2x(xu+ u) 2xu= 0= x3u+ 4x2u = 0 = xu+ 4u = 0 (A first-order DE for u)

For convenience, letv= u, so x dvdx+ 4v = 0

=

dv

v = 4

dx

x =

ln v =

4 ln x =

v= x4 =

u=

13

x3

Therefore a second solution is y2= x(13 x3) or just y2= 1x2 The general solution to thisDE is y= c1x +

c2x2

20


21/47

Example: Considery(4) + 4y+ 6+ 4y+ y= 0

The characteristic equation ism4 +4m3 +6m2 +4m+1 = 0 (i.e. (m+1)4 = 0) soy1= ex is

one solution so supposey = uex

Theny = uex

uex

andy = uex

2uex

+ uex

.Next,y = uex 3uex + 3uex uex. Lastly y = uex 4uex + 6uex 4uex + uex. Then DE becomes

uex = 0

This implies u = 0, u = c1, u = c1x + c2, u = c1x2

2 + c2x + c3, .So the general solution is y = (c1x

3 + c2x2 + c3x + c4)ex This even works for inhomo-

geneous equations!Example:

x2y+ xy y= 72x5Notice that a solution to the homogeneous problem isy1 = x. To find the general solution,we have two options:

1. Use reduction of order to findyn and then variance of parameters.

2. Use reduction of order directly!

We will try both:

1. Considerx2y+ xy y= 0Supposey2= ux Theny

2= u

x + uand y 2 =ux + 2u. This impliesx2(ux + 2u) +

x(ux + u) ux= 0. Hence xu+ 3u = 0. Let v= u so xv + 3v= 0. Hencedv

v =

3dxx

= ln v = 3 ln x = y2= 1x

Thereforeyh = c1x + c2x Now, forx

2y + xy

y= 75x5, we try y = u

1x + u2x. We

need to solveu1y1+ u

2y2= 0

u1y1+ u

2y2= F(x)( This requires the DE to be in standard form!)

That is

u1x +u2

x= 0 (1)

u1 u2x2

= 72x3 (2)

Then(1)/x + (2) = u1= 36x3 = u1= 9x4 + c1

1/x 2 = u2= 36x5

= u2= 6x6

+ c2Hence

y= c1x +c2

x + 9x5 6x5 =c1x + c2

x + 3x5

21


22/47

2. Use reduction of order directly:

Assume that y = u(x)xsoy = ux + u and y = ux + 2u. Substitute into DE,

x2(ux + 2u) + x(ux + u)

ux= 72x5

= x3u+ 3x2u = 72x5 = xu+ 3u = 72x3

Letv = u : x dvdx + 3v= 72x3. That is

dv

dx+

3

xv = 72x2

Integrating factor u(x) =e 3

xdx =e3 ln x =x3. Hence x3dvdx + 3x

2v= 72x5.

x3v = 12x6 + C = v = 12x3 + Cx3

= u= 3x4 Cx2

+ C1

= y= C1x + C2x

+ 3x5

7 Boundary Value Problems (BVP)

A boundary value problem involves a DE with conditions specified at different points. E.G.y+ y= 0, y(0) = 0, y(1) = 1. There is no E/U Theorem for such problems.

Usually there are no solutions, and when they do exist there are often infinitely manyof them.

Examples:

1. y+ 2y = 0, y(0) = 0, y(1) = 1. The general solution is y = C1cos x+C2sin xIfy(0) = 0, then C1 = 0. If y(1) = 1, then C1 =1. Hence this problem has nosolutions.

2. y+ 2y= 0, y(0) = 0, y(1) = 0.

y= C1cos x + C2sin x

again. C1= 0, C1= 0. Therefore C2 is free. Every multiple of sin x is a solution!

The characteristic equation is m2 + k = 0.

Case 1: IfK


23/47

Hencec2= 0, and y = 0 is the only solution for k 0, let K=2.y+ 2y= 0

General solution is y = c1cos x+ c2sin x. Therefore, y(0) = 0 = c1 = 0,y(1) = 0 = c2sin = 0. Therefore either c2= 0 (so y = 0 again). or = n, n=1, 2, (i.e. k= n22) in which case the solutions are

y= Csin nx

Terminology: The numbers

n22 =2, 42, 92.

are called the eigenvalues of the BVP. The functions sin nx= sin x sin2x, are calledthe eigenfunctions. There BVPs are referred to as eigenvalue problems.

Example: Find the eigenvalues & eigenfunctions of the BVP

y+ 2y+ ky = 0, y(0) =y(1) = 0

Solution: the characteristics equations is m2 + 2m + k = 0 som=

1

1

k.

IfK 1, then

y= ex[C1cos

k 1x + C2sin

k 1x]

y(0) = 0 = C1 = 0, y(1) = 0 = C2e1 sin

k 1 = 0. Hence C2 = 0 ork= 1 =n (i.e. k = n22 + 1, n Z. (These are the eigenvalues)

The eigenfunctions are y= ex

sin nx

23


24/47

7.1 System of ODEs

In some applications, we encounter coupled equations; the rates of change of two quan-tities may each depend on both quantities. Example: Suppose we have two populationsL

one predator and one prey. Letx9t) be the population of prey. Let y(t) be the populationof the predators. We may argue that

dx

dt =r1x 1xy

wherer1 is natural growth rate of prey, without predation and 1 is the death rate of preycaused by interaction with predators. and

dy

dt = r2y+ 2xy

where r2 is the natural death rate of predators in absence of prey and 2 is the growthrate for predators due to interaction with prey.

Unfortunately, we cant solve this system, because its non-linear. Well stick to linearsystems, such as

x(t) =a1x + b1+ c1z+ f1(t)

y(t) =a2x + b2y+ c2z+ f2(t)

z(t) =a3x + b3y+ c3z+ f3(t)

Linear systems can be expressed as vector DEs. Let

x(t) =

x(t)y(t)z(t)

Then

x(t) =

a1 b1 c1a2 b2 c2

a3 b3 c3

x(t) +

f1(t)f2(t)

f3(t)

or justx = Ax= Ax + fOur analysis of these problems will be shaped by 2 theorems.

Theorem. Any higher-order linear ODE, or system of higher-order Des, can be convertedinto a system of first-order linear ODEs

Proof. (for a pair of second-order equations) Suppose

x+p1x+ q1x= f1(t)

and y+p2y+ q2y= f2(t)

24


25/47

If we let x(t) =

xx

yy

, then

x=

x

x

y

y

=

x

p1x q1x + f1(t)y

p2y q2y+ f2(t)

=

0 1 0 0q1 p1 0 0

0 0 0 10 0 q2 p2

x +

0f1(t)

0f2(t)

Implication: we really only need to study first-order vector equations.

Theorem. (partial converse - not in notes) Every system of first-order linear constant-coefficient ODEs can be converted into a higher-order linear constant-coefficient ODE.

Proof. (2D-Case) Suppose x = ax + by+ f(t) (1) and y = cx + dy+ g(t) (2) . Solve (1)

for y : y = xb abx f(t)b = y = xb abx

f(t)b . Hence (2) becomes

x (a + d)x+ (ad bc)x= f df+ bg

Implication: The theory of linear constant-coefficient ODEs will carry over to vector DEs.

In particular, we haveExistence & Uniqueness: The IVP x =Ax+ f(t), x(0) =xhas a unique solution

on an interval I provided that f(t) is continuous on I. (This holds even ifA = A(t), aslong as A(t) is continuous). The Principle of Superposition applies. Ifx1 and x2 aresolutions tox = Axthen so is c1x1+ c2x2 (etc.).

An n-dimensional homogeneous system will require n linearly independent solutions,and hence n arbitrary constants.

The solution to an inhomogeneous system can be found as x= xh+ xp. The Wronskian must be modified.

W( f1, ,fn) =det[ f1, , fn]

Example: Suppose the functions, x1 = e2t

10

5

, x2 =e

2t

11

1

, x3 = e

3t

01

0

are

solutions to a linear vector DE. Are they independent? Check W(t) = e3t(15e4t) =0 = yes.

25


26/47

7.2 Homogeneous Systems with Constant Coefficients Method of Solu-tion

Consider the DEx = Ax(i.e. the system x(t) =a11x + a21y and y(t) =a12x + a22y)

Recalled that the ODE y = ay has solution y = Ceax

we guess that x = Ax hassolution x =vemt, for some m R and v R2. This givesx = mvemt so plugging thisinto the DE gives

mvemt =Avemt

so Av = mv. That is, x= vemt is a solution to x =Ax, if m is an eigenvalue of A and vis a corresponding eigenvector.

Recall: Av= v can be rewritten as (A I)v = 0. This will have a nonzero solution if and only (A I)1 does not exist (otherwise

v= (A I)10 = 0).

A matrix is invertible if and only if its determinant is non-zero.

Therefore we need to solve det(A I) = 0 for . We call this the characteristicequation of the DE.

Once we have found the eigenvalues , we can find corresponding eigenvectors bysolving (A I)v = 0 forv (there will be one free variable here).

Ann n matrix will have n eigenvectors if multiplicity is counted. The eigenvectors corresponding to distinct eigenvalues will be linearly independent. Complex eigenvalues always occur in conjugate pairs.

If an eigenvalue has multiplicity m, it will possess anywhere from 1 to m linearlyindependent eigenvectors.

7.3 The 2D Case

Case 1: Distinct Real eigenvalues

Example: Solve x = Ax, A=

2 32 1

(that is x(t) = 2x + 3y and y(t) = 2x + y)

Find the eigenvectors

For = 4: (AI)v = 0 becomes2 3

2 3

v1v2

=

00

so 2v1 3v2 = 0,

i.e. v2 = 2

3 v1. Setting v1 = 3, we get v =3

2

. Therefore one solution to the DE is

x1=

32

e4t.

26


27/47

For =1: (A I)v = 0 =

3 32 2

v1v2

=

00

so v1 +v2 = 0, and we

may use v = 1

1. Therefore a second solution is x2 =

1

1 e

t. Hence the general

solution isx= c1

32

e4t+c2

11

et. Of course we may have initial conditions. Suppose

x(0) =

11

. Then

11

=c1

32

+ c2

11

= c1 = 25 , c2= 15 . After all, we can plugc1, c2 in and get the general solution.

Case 2: Complex Eigenvalues:

Example: Solve x =2 1

3 4

x. Find : A I =2 1

3 4

Hence

det(A

I) = ( + 3)2 +2. Hence =

3

2i. Findv: x1=

1

1 +

i e(3+

2i)t. We

break this into its real and imaginary parts: x1= e3t[cos

2t+i sin

2t][

11

+i

0

2

].

Thereforex1 = e3t[

cos

2t

cos 2t 2sin 2t

+ i

sin

2t

2cos

2t sin 2t

]

From this we can conclude immediately that the general solution is

x= e3t[c1

cos

2t

cos 2t 2sin 2t

+ c2

sin

2t

2cos

2t sin 2t

]

Case 3: Repeated Eigenvalues:

Considerx = 3 41 1x.A I=

3 41 1

so det(A I) = (+ 1)2. Hence=1 (repeated). Eigenvectors? (A I)v =0 =2 41 2

v1v2

=

00

. Therefore our eigenvector is v =

21

. Hence one solution is

x1 =

21

et. For the second solution, try multiplying by t? Trying x2 = tetv, gives

x2= etv+ tetv. sox = Ax = etv+ tetv= Atetv. Hencev= 0 (butv= 0!)

We need more constants! It turns out that the right guess is x2= tet

v + et

w. Thisgives x2 = etv + tetv + et wsox = Ax = etv + tetv + et w= Atetv + Aet w.

Therefore, A w = w+v. This implies (A I) w = v. Note: observe that if we apply

27


28/47

(A I) to this we get (A I)2 w= (A I)v= 0. w is called a generalized eigenvectorof A. Back to our example: we need to solve

2 41 2w1

w2 = 2

1

Therefore w=

10

. The general solution is x= c1x1 +c2x2= c1e

t

21

+ c2[te

t

21

+

et1

0

]

7.4 Generalizations for Higher-Order Systems

If A is n nit has n eigenvalues Each distant real eigenvalue will give us one solution

Each pair of complex eigenvalues will give us two solutions. If an eigenvalue is repeated k times, it may still have k linearly independent eigen-

vectors, in which case we obtain k independent solutions

If an eigenvalue has multiplicity k but has < kindependent eigenvectors, we will needgeneralized eigenvectors.

For example, if has multiplicity 2 and 1 eigenvectorv, then one solution is uet andanother is utet + vet, where (A I)v = u.

Another example is that if has multiplicity 3 and 1 eigenvector u, then the solutionsareuet,utet + vet where (A I)v= u, 12 ut2et + vtet + wet where (A I) w= u.

Here is another example, if has multiplicity 3 and 2 eigenvectors, u1 and u2, then 3solutions areu1et, u2et,vtet + wet, where v is some linear combination ofu1 and u2,and (A I) w= v. (v will become obvious when we write this down!).

Example: Solve x =

1 2 30 1 0

2 1 2

x. We find = 1, 1, 4. and v =

16

4

, 30

2

,10

1

sox(t) =c1etv1+ c2e

tv2+ c3e4tv3.

7.5 Inhomogeneous Linear Vector DEs

The method of undetermined coefficients can be applied to vector DEs.

Example: Consider: x = 1 10 2

x+ te2t

3e2t. We find xh = c1et1

0 +c2e2t1

1.

We guessxp = [a + bt]e2t. This gives xp = be

2t 2[a + bt]e2t. The DE becomese2t[b 2a 2bt] =A[a + bt]e2t + f=e2t[Aa + Abt] + f

28


29/47

That is

e2t

b1 2a1 2b1tb2 2a2 2b2t

=e2t

1 10 2

a1a2

+ te2t

1 10 2

b1b2

+

t3

e2t

=e2t

a1+ a2+ (b1+ b2+ 1)t2a2+ 3 + 2b2t

Therefore,a2 =34 , a1 = 536 , b1 =13 , b2 = 0. Therefore, x= c1et

10

+c2e

2t

11

+

e2t

53634

+ te2t

130

. Note:variation of parameters (next lecture) will usually be

quicker, unless f(t) is of a simple form. For example, if I seex = Ax +

a1a2

. I will always

use undetermined coefficient.If for linear homogeneous ODEs, Wronskian is zero if only if the solution is dependent.

7.6 Variation of Parameters for vector Des

Supposex = Ax +F(t).Ifh = c1x1 + c2x2, then we tryx= u1x1 + u2x2 and plug this intoDE:

u1x1+ u1x1+ u

2x2+ u2x

2= Au1x1+ Au2x2+ F

Thereforeu1x1+ u2x2=

F. In component form:

u1x11+ u2x21= F1

u1x12+ u2x22= F2

We will always be able to solve this for u1 and u2, because the columns of

x11 x21

x12 x22

are x1 and x2, which are linearly independent. So, we will get u1 = G1(t), u

2 = G2(t),

x= [

G1(t)dt]x1+ [

G2(t)dt]x2Note: If we omit the constants of integration we will have a particular solution. If we

include them, we get the general solution.Example (MOUC example revisited)

x =

1 10 2

x + e2t

t3

xh = c1et

10

+ c2e

2t

11

We tryx= u1et

10

+ u2e2t

11

which means

u1et + u2e

2t =te2t

29


30/47

u2e2t = 3e2t

From the second one, we get u2= 3e4t, sou2= 34 e

4t + c2. The first one becomes

u1=

(t 3)e3tdt= 1

3 (t 3)e3t + 1

3 e3tdt=

8

9 e3t

1

3 te3t + c1

so x = ( 89 e3t 13 te3t +c1)et

10

+ (c2 34 e4t)e2t

11

(same as last time, but in a

different form!)

Example: Solve x = Ax + F, where A=

1 41 1

and F=

et

0

Solution: First, find xh = c1e3t

21

+ c2e

t2

1

Tryx= u1e3t

21+ u2e

t

21 and solve

u1(2e3t) + u2(2et) =et

andu1(e

3t) + u2(et) = 0

Hence solve u1= 14 e4t = u1= 116 e4t + c1. Also u2= t4 + c2.

8 The 3 Most Famous PDEs

8.1 The Heat Equation (or Conduction Equation)

Consider a metal bar of length L, with uniform cross-sectional area and on insulated surface.Let u(x, t) be the temperature at a distance x from one end at time t. It can be shownthough a careful analysis of the physics of heat transfer that u should obey the equation.

u

t =

2u

x2

Rough Explanation: Suppose the temperature profile looks a wave curve at time t.Where this graph is concave up the temperature will initially increase, and where it isconcave down it will decrease! To solve the Heat equation, we need on initial conditionu(x, 0) =f(x). We will also need boundary conditions, and there are several possibilitiesfor these.

1. If we imagine fixing the ends of the bar in ice, then weed haveu(0, t) = 0, u(L, t) = 0.

2. If we insulate the ends, we need instead ux(0, t) = 0, ux(L, t) = 0.

30


31/47

3. If we leave the ends expressed, the conditions turn out to involve both u andux.

4. We could have different conditions at the two ends.

5. We could even imagine an infinitely long bar, in which case we require only that ube bounded as x .Note: For a metal plate, we have the 2D Heat equation. ut= (uxx+ uyy)

8.2 The Wave Equation

Now consider a string of length L, under tension. Letu(x, t) be the vertical displacementof each point on the string at time t. Then utt=

2uxx.Rough Explanation: the vertical acceleration of each point on the string is determined

by the concavity at that point. With this, we need ICs

u(x, 0) =(x)

ut(x, 0) =(x)

(We could set the string in motion either by plucking it ((x) = 0) or striking it ((x) = 0)).For BCs, the simplest would be u(0, t) = 0, u(L, t) = 0.

8.3 Laplaces Equation

This is (in 2D) uxx+ uyy = 0. This can be thought of as a kind of smoothness condition.For example, In 1D it is just f(x) = 0 ( = f(x) =Ax + B).The BCs must be closed.For example, we might have u(x, 0) = 0, u(x, 1) = 0, u(0, y) = 0, u(1, y) = sin y.

8.4 First-order Linear PDEs and Partial Integration

Just as simple ODEs can be solved by integration (e.g. y = 2x = y =x2 +c). Somesimple PDEs can be solved by Partial integration. For example, suppose uy =ey(whereu= u(x, y). We integrate with respect to y to obtain

u(x, y) =ey + g(x)

The constant of integration is an arbitrary function of x! To evaluate it, we need aninitial condition for each value of x. There could all be given at y = 0, but they could bespecified along a curve in the xy plane.

For example, Ifu(x, 0) = 11+x2 , then 11+x2 = 1 + g(x) sog(x) = 11+x2 1, andu(x, y) =1

1+x2 1 + ey. We essentially have an ODE for each value of x here. The solution canbe viewed as propagating along the lines x = c. When this happens, we call the lines

31


32/47

characteristics. Eg, if, instead, weve given the IC, u = 0 along y = x2 (i.e. u(x, x2) = 0)then 0 =ex

2

+ g(x) sog (x) = ex2 and u(x, y) =ey ex2 .Partial integration is only possible if both ux and uy appear in the PDE. However, we

can always eliminate one of them through a change of variable if the equation is linear:

a(x, y)ux+ b(x, y)uy+ c(x, y)u= f(x, y)

Example: Suppose uy = ux, u(x, 0) =ex2 If we let= x 2y, = y

then we may writeZ=u(x, y) = u(, )

z

x=

z

x

z

y

=z

y

+z

d

dyThat is,u

x=

u

x

z

y =

u

y+

u

d

dy

Since= x2y, = y,ux= uand uy = 2u+u. The PDEuy = 2ux, therefore becomes(2u+ u = 2u or just u = 0 = u(, ) =g(). Returning to the original variables,u(x, y) = g(x, 2y). This can be matched to the IC: u(x, 0) = ex

2

= ex2 = g(x).Henceu(x, y) =e(x2y)

2

. Check?

ux=

2(x

2y)e(x2y)

2

uy = 4(x 2y)e(x2y)2

souy = 2ux.Question: How do we choose the new variables? Consider the ODE dydx = f(x, y), If

its general solution is written as (x, y) =k then xy =dydx (that is

xy

= f(x, y))Proof. Just differentiate (x, y) =k implicitly! (w.r to x):

x+ ydy

dx= 0 = dy

dx=

xy


We will let = (x, y), = (x, y).

32


33/47

8.5 First-Order Linear PDEs


We will let = (x, y) and = (x, y) to replaceu(x, y) with u(, ). By the chain rule,

ux= ux+ ux

uy = uy+ y

so the PDE becomes

a(x, y)[ux+ ux] + b(x, y)[uy+ uy] + c(x, y)u= f(x, y)

Rearranging, we have

[a(x, y)x+ b(x, y)y]u+ [a(x, y)x+ b(x, y)y]u+ c(x, y)u= f(x, y)

The goal is to make either u or u disappear. Lets eliminate u, by setting

a(x, y)x+ b(x, y)y = 0

That is assuming that y= 0, xy =b(x,y)a(x,y)

Our Lemma now suggests that we should choose such that(x, y) =k is the general

solution to the ODE dydx = b(x,y)a(x,y) . What about ?Our only other constraint is that the

transformation (x, y) (, ) is invertible. Therefore its Jacobian must be non-zero.(, )

(x, y)=det[

x yx y

] =xy yx= 0

If we just let = x, then this is satisfied!Example Solve the IVP xux+ yuy = 3u and u(x, 1) = 1 x2, where x >0, y 1.

Solution: We start by solving

dy

dx =

y

x = dyy = dxx = ln y = ln x+ C1 soln y ln x= C1.This tells us that we may use = x, = ln y ln x. Thenu2= ux+ ux = u 1x u

and uy = uy+ uy = 1y u.

The PDE becomes (xu u) + u = 3u = u = 3uReturning to the original variables, we have

u(x, y) =f(ln y ln x)x3 =f(lnyx

)x3 =g(y

x)x3 =x3 ln

x

y

Finally, since u(x, 1) = 1 x2, we have 1 x2 =x3 ln xso h(x) = 1x2x3 . Hence

u(x, y) =x3[1 (xy )2

(xy )3 ] =y3(1

x2

y2) =y3

x2y

Comment: This is known as the method of characteristics. In these problems, informationfrom the K can be viewed as being carried along the curves (xy) =K.

33


34/47

8.6 Second-Order Linear PDEs

Consider

a(x, y)Uxx+ b(x, y)Uxy+ c(x, y)Uyy + d(x, y)Ux+ e(x, y)Uy+ f(x, y)U=g(x, y)

We will introduce (x, y) and (x, y) to convert this toA(, )U + B(, )U+ C(, )U + D(, )U+ E(, )U+ F(, )U=G(, )

The goal is to eliminate one or more of the second order terms (make A, B, or C = 0)What are those functions?

This is tedious. I will get you started. We need Ux, Uy, Uxx, etc. in terms ofU, etc.

Ux= Ux+ Ux

Therefore

Uxx= x

(Ux) = x

[Ux+ Ux]

=

x[Ux+ U

x[x] +

x[U]x+ U

x[x]

= [Ux+ Ux]x+ Uxx+ [Ux+ Ux]x+ Uxx

= (x)2U + 2xxU+ (x)

2U + Uxx+ Uxx

Repeating this for Uxy andUyy , plugging these into the original PDE and rearranging,we get

A(, ) =a(x, y)(x)2 + b(x, y)xy =c(x, y)(y)

2

B(, ) = 2axx+ b(xy+ yx) + cyy

C(, ) =a(x, y)(x)2 + b(x, y)xy =c(x, y)(y)

2

To eliminate the U term, we set A = 0,

A(, ) =a(x, y)(x)2 + b(x, y)xy =c(x, y)(y)

2 = 0

Assuming that y= 0, we can rewrite this as

a(xy

)2 + b(xy

) + c= 0

Thereforex

y=

b b2 4ac2a

Similarly we can eliminate the U) term by requiring that y= 0 and xy =bb24ac

2aThree cases arise, dividing second-order linear PDEs into 3 classes

34


35/47

1. Hyperbolic Equations: Ifb2 4ac >0, then we can choose = and =2 where1= k1, is the solution to

xy =

bb24ac

2a ;

2= k2 is the solution to xy =

b+b24ac

2a

This will eliminate both U and U leaving after divisor by B.

U+ (,,U , U, U) = 0

This is the canonical form of hyperbolic PDE

E.G. Uxy =Uis hyperbolic

2. Parabolic Equations

Ifb2 4ac= 0, then we can choose = , where = k is the solution to dydx = b2a . Wecannot also eliminate U . However since

xy

= b2a , the U term disappear. HenceB= 0. Lastly, the canonical form of a parabolic PDE is

U + (,,U , U, U) = 0

The classification of a PDE has implications on the kind of boundary conditions requiredand the method of solutions used.

8.7 Method of Characteristics

8.7.1 Example

uxy+ uy =xy, u(x, 0) = 0, u(0, y) = sin yCan just integrate with respect to y. Hence ux+ u =

1

2

xy2 +f(x), then this is justODE, integrating factor is ex.

Hence

exu=

1

2xexy2 + exf(x) =

1

2y2(xex ex + h(y)) +

exf(x)dx

u(x, y) =1

2y2(x 1 + h1(y)

ex ) + ex[F(x) + h2(y)]

whereF = exf.Thereforeu(x, y) = 12 y

2(x 1) + g(x) + exh(y) is the general solution.u(x, 0) = 0 = g(x) + exh(0),u(0, y) = sin y= 12 y2 + g(0)+ h(y). sog(x) = exh(0)

and h(y) = sin y+ 12 y

2

g(0) = g(0) = h(0).sou(x, y) = 12 y

2(x 1) + ex sin y+ 12 y2ex

35


36/47

8.7.2 Example

Consider the wave equation utt= 2uxx subject to u(x, 0) =(x), ut(x, 0) = 0.

Starting by solving

dtdx

= b b2

4ac2a

= 1

sodx= dtHence x= t + c so= x + tandt= x t

utt= 2u 22ut+ 2utt

uxx= u + 2ut+ utt

thenutt= 2uxx = t = 0

Continuing with the 1D Wave equation problem

Utt= 2Uxx

U(x, 0) =(x)

ut(x, 0) = 0

A clarification: unless we restrict the domain of(x) and add BCs, this problem is for aninfinitely long string!.

We let = x + t, t= x tand this converts the PDE to

ut = 0

We can integrate this:U(, t) =F() + G(t)

That is,u(x, t) =F(x + t) + G(x t)

Enforce the ICs:u(x, 0) =(x) = (x) =F(x) + G(x)ut(x, 0) = 0 = 0 =F(x) G(x)

Now(x) =F(x) + G(x)

Therefore

F(x) =G(x) =1

2(x)

Therefore,

F(x) =1

2(x) + c1

36


37/47

G(x) =1

2(x) + c2

F+ G=

so c1+ c2= 0

Finally, then u(x, t) = 12 [(x + t) + (x t)]We have u(x, 2) =

12 [(x + 2) + (x 2)]

This method (method of characteristics) worlds on many hyperbolic equations (but notall, consideruxy+ ux+ uy = 1); on some parabolic equations (e.g. Uxx+ Ux= 1) but noton the heat equation

ut= uxx

; on no elliptic equations.

8.8 Separation of Variables

possibly the most powerful method in general we assume that u(x, y) can be factorized asu= F(x)G(y)

Example: Solving the heat equation.Consider the initial/boundary value problem

Ut= Uxx

u(x, 0) = 20 sin(3x)

u(0, t) = 0, u(L, t) = 0

Consider a bar of length L, with its ends fixed in ice, and initial temperature profile.

We assume that u(x, t) =F(x)G(t). Then

ut = F(x)G(t)

uxx= F(x)G(t)

sout= uxx = F(x)G(t) =F(x)G(t)

Next separate the variables:G(t)G(t)

=F(x)

F(x)

The key realization: For this equation to hold true, for all x and all y, both sides mustequal a constant.

G(t)G(t)

=F(x)

F(x) =c, c R

37


38/47

This gives two ODEs:G(t) =cG(t)

F(x) =cF(x)

We will start with the second one. We need side conditions: u(0, t) = 0 = F(0)G(t) = 0Also,u(1, t) = 0 = F(1)G(t) = 0 = F(1) = 0. We have a BVP for F(x):

F(x) cF(x) = 0, F(0) =F(1) = 0You can verify that nontrivial solutions only exist ifc


39/47

We might instead need the Full Fourier Series.

f(x) =

n=1[Ancos nx + Bnsin nx]

If we have an infinitely long bar, we end up with a Fourier Integral instead:

f(x) =

0

[a(w)cos wx + b(w)sin wx]dx

which can be expressed in complex form as

f(x) = 1

2

c(w)eiwxdx

There are formulas for all of the coefficients, An, Bn, a(w), b(w), c(w). Specifically,

c(w) =

f(x)eiwxdx

8.9 The Fourier Transform

Definition. The Fourier Transform of a functionf : R R is defined as

F(f(x)) =

f(x)eiwxdx

provided that this integer converges. We will also use the notation f(w).

Example:

FindF(f(x)), iff(x) =

0 ifx 0.

Solution: F(f(x)) = f(x)e

iwxdx=

0 eaxeiwxdx=

0 e

(a+iw)xdx= 1a+iwComment: The FT is just one of many integral transforms. The expression

f(w) =

f(x)K(w, x)dx

The functionL(w, x) is called the kernel of the transformation.Note: This often fails to exist!For example: Iff(x) = 1, then

F{f} =F{1} =

eiwxdx=

0eiwxdx +

0

eiwxdx

Above does not exist.

39


40/47

Theorem. Letf : R R. If1. f is piecewise continuous onR, and

2. f is absolutely integerable onR

. (That means |f(x)|dx converges)

Thenf(w) exists.

Proof. First note that|f(x)eiwt|dx =

|f(x)|dx since|eiwt| = 1. If this con-

verges, then we can apply the triangle inequality:

|

f(x)eiwxdx|

|f(x)eiwx|dx

then it converges.

Note that for f to be absolutely integrable we must have limx f(x) = 0, we must

have limxx f(x) = 0. In fact,xn, sin x, cos x, ex, ln x, etc, do not have Fourier Transforms.Theorem. (Linearity) Letf : R R andg: R R. Let, R. Iff andg exist, thenF{f+ g} =f+ g.Proof. By definition, it can be written as an integral form so by the linearity of integration,it is true.

Theorem. If f is continuous and absolutely integrable, then the Fourier transform is in-vertible and in fact

f(x) =F1{f(w)} = 12

f(w)eiwxdw

(this is the formula for the Fourier Integral representation off(x)!)

Theorem. Let f be differentiable onR, with transformf(w), ThenF{f(x)} =iwf(w).Proof. F{f(x)} = f(x)eiwxdx= f(x)eiwt|+ iwf(x)eiwxdx

Now since f(w) exists, f(x)e

iwxdx converges, so limx f(x)eiwx = 0 s oF{f(x)} =iw f(x)eiwxdx= iwf(w)

In general, F{f(x)} = (iw)nf(w) and we can even say that F{xa f(t)dt} = 1iw f(w).These results suggest that the F.T. can convert ODEs to algebraic equations.

40


41/47

8.9.1 Demonstration of Concept

Suppose y + ay+ by= f(x). Apply F.T.,

F{y+ ay+ by} = f(w)

Hence F{y} +aF{y} +bF{y} = f(w). Therefore,w2y+ aiwy+ by = f(w). Lastly,y= f(w)w2iawb . That implies y= F

1{ f(w)w2iawb}.

Theorem. The Time-Shifting Property: Letf(t) be continuous and absolutely integrablewithF{f(t)} = f(w). ThenF{f(t a)} = f(w)eiwa and so F1{f(w)eiwa} =f(t a)Proof. F{f(t a)} = f(t a)eiwtdt= f()eiw(+a)dt= eiwaf(w)

Example: Consider the unidirectional wave equation,

ut+ ux= 0

The strategy is to eliminate the x derivative by using Fourier Transform in the x variable.That is, we define F{u(x, t)} = u(w, t) = u(x, t)eiwxdx

With this, F{ux} =iwu. Meanwhile,

F{ut} =

u

teiwxdx=

t

ueiwxdx= ut

Apply f to the PDE:ut+ x= 0

= F{ut+ ux} =F{0}= ut+ (iw)u= 0

This is essentially an ODE for u(t): the solution is

u(w, t) = G(w)eiwt

Now we just need the inverse transform:

u(x, t) =F1{G(w)eiwt} =G(x t)

by the time shifting properties.Note:

1. Weve assumed that u exists, which probably isnt true for this PDE! (We needu = 0 as x ). Nevertheless, weve got the correct result.

41


42/47

2. Evaluating F1 is usually harder. We will need this.

Definition: The convolution of two functions f and g is f g(x) = f(x)g()d.

Note this wont always exist, but a sufficient condition is that f and g be causal,

meaning thatf(x) = 0 when x


43/47

So, let u(w, t) = F{u(xt)}= u(x, t)eiwxdx. Thereforeut ut, uxx w2u. Also,the ICu(x, 0) =f(x) implies that u(w, 0) = f(w). (weve already used the BCs in assumingthat uexists). This implies ut= w2u,u(w, 0) = f(w). Therefore, u(w, t) =H(w)ew2t .Enforce the IC: u(x, 0) = f(w) =

H(w) = f(w), so u(w, t) = f(w)e

w2t

Now, u(x, t) =F1{f(w)ew2t} =F1{f(w)} F1{ew2t} =f(x) F1{ew2t}Use the formula for F1:

F1{ew2t} = 12

ew2teiwxdw=

1

2

e(tw2ixw)dw

= 1

2

et[w2 ix

tw]

dw= 1

2

et[(w ix

2t)2+ x

2

42t2]

dw

= 1

2e x2

4t

et(w ix

2t)2

dw Letv =

t(w ix2t

)

= 1

2e x2

4t

ev2 dv

t

= 1

2

te

x2

4t

ev2

dv

Now e

v2dv is know to be

so this is 12t

ex2/4t . You can find the proof on

my website.Finally, u(x, t) =f(x) 1

2t

ex2/4t = 1

2t

f(x t)et

2/4tdt

We can simplify this a bit: Let y = 2t

so dy = dt2t

and y as .Therefore

u(x, t) = 1

2

t

f(x + 2

ty)ey2

(2tdy) = 1

f(x + 2

ty)ey2

dy

Example: Suppose f(x) =

1 |x|


44/47

1. We will assume that trading is continuous and that assets are infinitely divisible.

2. We will ignore transaction costs.

3. We permit short selling4. We will assume that our assets pay no dividend

5. Two more we will need soon.

(a) We will assume that it is always possible to invest money at a constant interestrate r.

(b) We will assume that no arbitrage exists. That is, the values of financial instru-ments will automatically (and instantaneously) adjust to put them in balancewith investments at rate r.

Stock price:

S itself depends on time: S= S(t). Furthermore, this dependence is partly random.Change in stock price in a given time interval made up with two parts, deterministic

growth/decay and a random part. That is S= fdeterministic+ frandom. We will assumefdet to be Malthusian:

fdet= St

, so that without the random component wed have S=St. Therefore S=C et.We will assume frandom to be stochastic:

frandom= SW(t)

where W(t) is a random variable with mean 0 and variance t (Wiener Process) so

S= St + Sw(t). We call u the growth rate and is the volatility.Customarily this is written (letting t 0). Hence

dS=Sdt + SdW(t)

(this is a stochastically differential equation).Assume that the value of the option is F(S, t). Assume that the value of the stock is

S(t), satisfyingdS=uSdt + SdW(t)

(i.e. S=St+SW(t)) where W(t) is a random variable with a time-dependentprobability density function such that the mean is 0 and the variance is t. It an be shown

that [W(t)]2 has mean tand variance 2(t)2. Now recall that a functionf(x, y) cam beexpanded in a Taylor series as f(x, y) =f(x0, y0)+ fx(x0, y0)(xx0)+ fy(x0, y0)(yy0 +

44


45/47

Assuming thatF(S, t) is smooth, then we can take an arbitrary point (S0, t0) and write

F =F

SS+

F

t +

1

2

2F

S2S2 +

2F

StSt +

1

2

2F

t2t2 +

Now since S= St+SW(t). F =FSSt+FSSW(t)+Ftt+12 FSS

2(t)2+FSS

2tW(t) + 12 FSS2S2[W(t)]2 + FStS(t)

2 + FStStW(t) +12 Ftt(t)

2 + Next, if t is small, then we neglect terms of order (t)2. Furthermore, since W(t)

has mean 0 and variance t, we can also neglect, the term of order tW(t). Also since[W(t)]2 has mean tand variance 2(t)2, we can say that [W(t)]2 t, and so

F FSSt + FSSW(t) + Ftt +12

FSS2S2t

8.11.1 Eliminating the Stochastic Term

The key step: if we assume that arbitrage is always instantly eliminated then it must bepossible to cancel out the random contribution to F. Here is the Black-Scholes argument:

Consider a portfolio constructed by buying one option and selling unit of stocks.(assuming short-selling and divisible stocks!)

The portfolio has value(t) =F S

Over the interval t, the change in is = F S= [SFS S]W(t) + [Ft+SFS+

12

2FSS S]tNow, if we set = FS, then the stochastic component disappears. We are left with

= [Ft+1

22S2FSS]t

so ddt =Ft+ 12

2S2FSSFinally, if arbitrage cannot exist, then the value of must equal the value of a simple

investment at interest rate r, so ddt =rHencer = Ft+

12

2S2FSS so

1

22S2FSS+ rSFS+ Ft rF = 0

Hence we get our Blakc-Scholes equation for European call option.Note that u has disappeared, which implies that the value of the option is independent

of the growth rate of the underlying stock - only the volatility matter.

45


46/47

8.12 Boundary Conditions

What domain should we consider? We must haveS 0, want to consider t [0, T] andwant to know f when t = 0.

Consider the value of the option at the strike time T. IfS(T) > K, then owner canmake profitS(T) K; ifS(T)< K, then the option is worthless

F(ST) =

S T S(T)> K0 S(T)< K

= max{S K, 0} = [S T]+

If S is over 0, ds= uSdt + Sdv(t), so S remains 0. S(T) = 0, F= 0, and F(0, t) = 0.On the other hand, if S ever becomes very large then we know we will be in the money attime T soF =S K since S Kand limS F(S, t) =S.

Summary:

Ft = 12

2S2FSS rSfS+ rf

F(S, T) = [S K]+

F(0, t) = 0, F(, t) =SNote PDE is parabolic.

Could normalize the coefficient ofFSSby letting t = 12 2t. Convert our final condition into a time initial condition by letting t = T t (use

t= 12 (T t))

The equal dimensional equation ax2

y+bxy + cy = 0 can be transformed into aconstant-coefficient equation by letting x= et.

Can convert PDE into a constant coefficient PDE by letting S=ex (x= ln S) IN the final condition,F(S, T) = [S K]T, eliminate K: LetF =kv, thenF(S, T) =

[ 1k 1]+. Also scale S by K, let S=kex.LetS= K ex (x= ln S/K)Lett= T 22 (= 12 2(T t))LetF =K v.

Ft=F

t = 1

2k2

v

t

46


47/47

FS=K

S

v

x

FSS= K

S2(vxx vx)

Then we simply plug in all those into the BS equation and get vt = vxx + (2r2

1)vx 2r2

vLetting= 2r

2, vt= vxx+ ( 1)vx v

47

amath 350 notes

Documents