MATHEMATICAL METHOD IN SCIENCE AND ENGINEERING Episode 16 pptx

We now consider the Volterra equation and differentiate it with respect to z as 18.5 SOLUTION OF INTEGRAL EQUATIONS Because the unknown function appears under a n integral sign, integr

SOLUTION OF INTEGRAL EQUATIONS 553

Using f(z) in Equation (18.35) we can also write

Y(z) = z2 - 2f(z), (18.38) which when substituted back into Equation (18.37) gives a differential equa- tion to be solved for f(z):

(18.39) the solution of which is

Finally substituting this into Equation (18.38) gives us the solution for the integral equation as

y(z) = 1 - c e - x z (18.40) Because an integral equation also contains the boundary conditions, constant

of integration is found by substituting this solution [Eq (18.40)] into the integral Equation (18.35) as C = 1

We now consider the Volterra equation

and differentiate it with respect to z as

18.5 SOLUTION OF INTEGRAL EQUATIONS

Because the unknown function appears under a n integral sign, integral equa- tions are in general more difficult t o solve than differential equations How- ever, there are also quite a few techniques that one can use in finding their solutions In this section we introduce some of the most commonly used tech- niques

554 INTEGRAL EQUATIONS

18.5.1 Method of Successive Iterations: Neumann Series

Consider a Fredholm equation given as

SOLUTION OF INTEGRAL EQUATIONS 555

and if the inequality

(18.53)

1

B

is true, where 1x1 < - , and C is a constant the same for all x in the interval

[a, b], then the following sequence is uniformly convergent in the interval [a, b]:

556 INTEGRAL EQUATIONS

18.5.2

By using the nth term of the Neumann sequence as our solution we will have committed ourselves to the error given by

Error Calculation in Neumann Series

Example 18.4 E m r calculation an Neumann series: For the integral equation

When the kernel is given in the form

Solution for the Case of Separable Kernels

n

K(x,t) = Mj(x)Nj(t), n is a finite number, (18.65)

3 = 1

SOLUTION OF INTEGRAL EQUATIONS 557

it is called separable or degenerate In such cases we can reduce the solution

of an integral equation to the soliition of a linear system of equations Let us

write a Fredholm equation with a separable kernel as

cj a3

(1 - X U l l ) C l - X a 1 2 ~ 2 - X a 1 3 ~ - - XU~,C, = bl

- X C L ~ ~ C ~ + (1 - X C L ~ ~ ) C ~ - X ~ 2 3 C 3 - - XU~,C, = b2

C1 - X a n 2 c 2 - X U ~ ~ C : ~ - + (1 - Xann)cn = b, (18.74)

Example 18.5 The case of sepamble kernels: Consider the homogeneous

Fredholm equation given as

(18.76) where

Using Equation (18.75) we write

to find the eigenvalues as

As in the eigenvalue problems in linear algebra, we have only obtained

the ratio, cI/cz, of these constants Because Equation (18.76) is homo- geneous, normalization is arbitrary Choosing c1 as one, we can write the solutions of Equation (18.76) as

y2(z) = J 1 3(1 - gz) for = 'J" (18.83)

SOLUTION OF INTEGRAL EQUATIONS 559

When Equation (18.74) is inhomogeneous, the solution can still be found

by using the techniques of linear algebra We will come back to the

subject of integral equations and eigenvalue problems shortly

18.5.4

Sometimes it may be possible to free the unknown function under the integral sign, thus making the solution pcssible

Solution of Integral Equations by Integral Transforms

(Z -t) and the range of the integral is from co to +m we can use the Fourier transform method

Fourier Transform Method:

Example 18.6 Fourier transform method: Consider the integral equa-

tion

We take the Fourier transform of this equation to write

where tilde means the Fourier transform, which is defined as

(18.87)

which after taking the inverse transform will give us the solution in terms of a definite integral:

(1 8.88)

560 INTEGRAL EQUATIONS

18.5.4.2 Laplace Transform Method: The Laplace transform method

is useful when the kernel is a function of (z - t ) and the range of the integral

is from 0 to X For example, consider the integral equation

y(z) = 1 + L‘ y(u)sin(a - u)du (18.89)

We take the Laplace transform of this equation to write

L [y(z)] = E [I] + E [I’ y(u)sin(a: - u)du 1 (18.90)

After using the convolution theorem:

(18.91)

where F ( s ) and G(s) indicate the Laplace transforms of $(z) and g(z), r e

spectively, we obtain the Laplace transform of the solution as

INTEGRAL EQUATIONS AND ElGEN VAL UE PROBLEMS (HIL BERT-SCHMID T Tff EOR Y) 561

For the eigenvalue X i we write

Multiplying Equation (18.96) by Xjy;(z) and Equation (18.97) by Xiyi(z),

and integrating over x in the interval [a, b] we obtain two equations

and

If the kernel satisfies the relation

K * ( z , t ) = K(t,z), Equation (18.99) becomes

Subtracting Equations (18.98) and (18.101) we obtain

In the case of degenerate eigenvalues, using the Gram-Schmidt orthogonaliza- tion method we can always choose the eigenvectors as orthogonal Thus we can write

Summary: For a linear integral operator

(18.109)

18.6.3 Completeness of the Eigenfunction Set

Proof of the completeness of the eigenfunction set is rather technical for our

purposes and can be found in Courant and llilbert (chapter 3, vol 1, p 136)

We simply quote the following theorem:

Expansion theorem: Every continuous function F ( z ) , which can be repre- sented as the integral transform of a piecewise continuous function G(z) and with respect to the real and symmetric kernel K ( z , 2’) as

F ( z ) = 1 K ( z , z’)G(z’)dx’, (18.110)

INTEGRAL EQUATIONS AND EIGENVALUE PROBLEMS (HILBERT-SCHMIDT THEORY) 563

can be expanded in a series in the eigenfunctions of K ( z , 2’); this series converges uniformly and absolutely

This conclusion is also true for Hermitian kernels We can now write

(18.111) where the coefficients a,, are found by using the orthogonality relation as

s,” F (z) Y (x) cia: = c J: anYn (x) Y (x> dx,

Substituting these coefficients back into Equation (18.111) we get

(18.113)

(18.114) This gives us a formal expression for the completeness of {ym (z)} as

(18.115) Keep in mind that in general {yi(z)} do not form a complete set Not just any function, but only the functions that can be generated by the integral transform [Eq (18.1 lo)] can be expanded in terms of them

Let us now assume that a given Hermitian kernel can be expanded in terms

of the eigenfunction set {yi(x)} as

564 INTEGRAL EQUATIONS

We now take the Hermitian conjugate of the eigenvalue equation

yi (z) = xi J K ( z , z’)yi (z’)dz’, (18.119) t,o write

yt*(z) = xi yf(z’)K*(.’,z)dz’ (18.120)

(18.121) (18.122)

We now suhstitute Equation (18.122) into Equation (18.118) and solve for

18.7 EIGENVALUE PROBLEM FOR T H E NON-HERMITIAN

18.1 Find the solution of the integral equation

y(t) = 1 + l y(u)sin(t - u)du

Check your answer by substituting into the above integral equation

18.2 Show that the following differential equation and boundary conditions:

y”(z) - y(z) = 0, y(0) = 0 and y’(0) = I,

are equivalent to the integral equation

y(1) = 0 and y(-I) = 1

18.4 Using the Neumann series method solve the integral equation

18.5

functions:

For the following integral equation find the eigenvalues and the eigen-

2T

y(5) = x 1 cos(5 - z’)y(z’)d5’

18.6 To show that the solution of the integral equation

F Z

y(5) = 1 + x2 ( 5 - z’)y(z’)dz’

.I0

566 INTEGRAL EQUATIONS

is given as

y(z) = cosh AX

a) First convert the integral equation into a differential equation and then solve

b) Solve by using Neumann series

c) Solve by using the integral transform method

18.7

integral equation By using different methods of your choice find the solution of the

y(z) = z + X .I’ zz’y(d)dz’

Answer: y(z) = 3 2 / ( 3 - A)

18.8

of motion Consider the damped harmonic oscillator problem, where the equation is given as

a) Using the boundary conditions z(0) = z o and Z(0) = 0 show that z(t)

satisfies the integral equation

equation of motion and the boundary conditios are given as

Obtain an integral equation for the anharmonic oscillator, where the

First show that a Neumann series solution exists and then find it

18.11 Using the Neumann series method find the solution of

1

y(z) = x2 + 6 1 (z + t)y(t)dt

I9

GREEN’S FUNCTIONS

Green’s functions are among the most versatile mathematical tools They provide a powerful tool in solving differential equations They are also very useful in transforming differential equations into integral equations, which are preferred in certain cases like the scattering problems Propagator interpre- tation of Green’s functions is also very useful in quantum field theory, and with their path integral representation they are the starting point of modern perturbation theory In this chapter, we introduce the basic features of both the time-dependent and the timeindependent Green’s functions, which have found a wide range of applications in science and engineering

19.1 TIME-INDEPENDENT GREEN’S FUNCTIONS

19.1.1 Green’s Functions in One Dimension

We start with the differential equation

where L is the Sturm-Liouville operator

(19 I)

(19.2)

567

We now define a function G(z, 0, which for a given [ E [a, b] reduces to

G l ( z ) when z < [ and to G2(z) when z > [, and also has the following properties:

i) Both Gl(z) and G ~ ( x ) satisfy

(19.8)

TIME-INDEPENDENT GREEN ‘5 FUNCTIONS 569

where G(z,I) is called the Green’s function If c)(z,y([)) does not depend

on y(x) explicitly, then finding the Green’s function is tantamount to solv- ing the problem For the cases where 4(z,y(I)) depends explicitly on y(x), then Equation (19.8) becomes the integral equation version of the problem de- fined by the differential equation plus the homogeneous boundary conditions [Eqs (19.1-19.3)] Before we prove the equivalence of Equations (19.8) and (19.1-19.3), we show how a Green’s function can be constructed However,

we first drive a useful result called Abel’s formula

570 GREEN'S FUNCTIONS

This Green's function satisfies conditions (i) and (ii) For conditions (iii) and (iv) we require c1 and c2 to satisfy the equations

c 2 4 E ) - C l U ( F ) = 0 (19.11) and

1

cpu'(J) - c * u ' ( [ ) = -

For a unique solution of these equations we have to satisfy the condition

where W [ u , ~ ] is called the Wronskian of the solutions u(z) and ~ ( z ) When these solutions are linearly independent, W [ u , ~ ] is different from zero and according to Abel's formula W [u, u] is equal to -, where A is a constant independent of [ Equations (19.11) and (19.12) can now be solved for c1 and

c2 to yield

A P(E)

(19.14) Now the Green's function becomes

TIME-INDEPENDENT GREEN'S FUNCTIONS 571

where we have used the formula

(19.26)

572 GREEN’S FUNCTIONS

19.1.4

To find the differential equation that the Green’s function satisfies, we operate

on y(x) in Equation (19.16) with € to write

The Differential Equation That the Green’s Function Satisfies

which is the defining equation for the Dirac-delta function 6 (x - 0 Hence

we obtain the differential equation for the Green’s function as

.LG(z, t) = 6 (Z - [) (19.28) Along with the homogeneous boundary conditions

Equation (19.28) is the defining equation for G(z, I )

19.1.5 Single- Point Boundary Conditions

We have so far used the boundary conditions in Equation (19.3), which are also called the twepoint boundary conditions In mechanics we usually encounter single-point boundary conditions, where the position and the velocity are given a t some initial time We first write the Green’s function satisfying the homogeneous singlepoint boundary conditions G(z0, x’) = 0 and G’(z0, x’) =

TlME-INDEPENDENT GREEN 5 FUNCTIONS 573

Following the steps of the method used for two-point boundary conditions (see Problem 19.4), we can find the constants c1 and c2, and construct the Green’s function as

where W[yl (x’), yz(z’)] is the Wronskian

Now the differential equation

-CY(Z) = +(.>,

Y(Z0) = Yo and Y’(Z0) = YA

with the given singlepoint boundary conditions

is equivalent to the integral equation

19.1.6 Green’s Function for the Operator d 2 / d x 2

The Helmholtz equation in one dimension is written as

(19.33)

where l e ~ is a constant Using the homogeneous boundary conditions

y(0) = O and y(L) =0, (19.34)

we integrate Equation (19.33) between (0, x) to write

574 GREEN'S FUNCTIONS

This leads us to the following integral equation for y(x):

(19.38)

d2 dx2

To identify the Green's function for the operator L = -, we rewrite this as

TIME-INDEPENDENT GREEN’S FUNCTIONS 575

For a different set of boundary conditions one must construct a new Green’s function

Example 19.1 Green’s function f o r the f = & operator: Wehaveob tained the Green’s function [Eq (19.42)] for the operator L = d2/dx2

with the boundary conditions y(0) = y(L) = 0 Ransverse waves on

a uniform string of fixed length L with both ends clamped rigidly are described by

where f ( z , y) represents external forces acting on the string Using the Green’s function for the d2/dx2 operator we can convert this into an integral equation as

19.1.7

In the presence of inhomogeneous boundary conditions we can still use the Green’s function obtained for homogeneous boundary conditions and modify the solution [Eq (19.8)] as

Green’s Functions for lnhomogeneous Boundary Conditions

b

Y k ) = P(.> + 1 G(.’ 0 4 (0 4’ (19.49) where y(z) now satisfies

with the inhomogeneous boundary conditions Operating on Equation (19.49) with f and using the relation between the Green’s functions and the Dirac- delta function [Eq (19.28)], we obtain a differential equation to be solved for P(x) as

(19.51)

Because the second term in Equation (19.49) satisfies the homogeneous bound- ary conditions, P( ) must satisfy the inhomogeneous boundary conditions

576 GREEN’S FUNCTIONS

Existence of P(z) is guaranteed by the existence of G(z, E) The equivalence

of this approach with our previous method can easily be seen by defining a

new unknown function

which satisfies the homogeneous boundary conditions

Example 19.2 Inhomogeneow boundary conditions: Equation of m e tion of a simple plane pendulum of length 1 is given as

We have already obtained the Green’s function for the 8 / d x 2 operator

for the homogeneous boundary conditions [Eq (19.42)] We now solve

Because Cp( t ) is

4(t) = -wi sin e(t), (19.60)

we can write the differential equation [Eq (19.55)] plus the inhomoge

neous boundary conditions [Eq (19.56)] as an integral equation:

Example 19.3 Green’s function: We now consider the differential equa- tion

2 8 Y dY

x - + x- + (k2x2 - 1) y = 0

TIME-INDEPENDENT GREENS FUNCTIONS 577

with the boundary conditions given as

y(0) = 0 and y(L) = 0 (19.63)

We write this differential equation in the form

and define the L operator as

(19.64)

(19.65) where

p(z) = z, q ( 2 ) = , 1 r ( z ) = 2 (19.66)

X The general solution of