MATHEMATICAL METHOD IN SCIENCE AND ENGINEERING Episode 16 pptx

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 - ce-xz. (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 r* (18.41) (18.42) where we have used Equation (18.5). Eliminating the integral between these two formulas we obtain Y’@) - (A + l)y(z) = 9’b) - 9(z>- (18.43) The boundary condition to be imposed on this differential equation follows from integral equation (18.41) as y(0) = g(0). 18.5 SOLUTION OF INTEGRAL EQUATIONS Because the unknown function appears under an integral sign, integral equa- tions are in general more difficult to 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 b f(.) = g(.) + J’ K(z, t)f(W. (18.44) a We start the Neumann sequence by taking the first term as fo(.) = d.). (18.45) Using this as the approximate solution of Equation (18.44) we write b fl(Z) = dz) + / K(., t)fo(t)dt. (18.46) a We keep iterating like this to construct the Neumann sequence as fob) = dz) (18.47) (18.50) This gives us the Neumann series solution as f(.) = g(z) + x K(z, z’)g(.’)d.’ + x2 lb dz’ lb &’)K(z,z’)K(.’, .”)g(d’) + . 1” (18.51) If we take 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]: {fz} = fo,f1,f2, ,fn, -+ f(x). (18.54) The limit of this sequence, that is, f(z), is the solution of Equation (18.44) and it is unique. Example 18.3. Neumann sequence: For the integral equation (18.55) we start the Neumann sequence by taking So(.) = x2 and continue to write: f](x)=x2+-S_l(t-x)t2dr 1' 2 -x2 - 3' X 21' t f2(x) = x + - (t - X)(t2 - $dt 2 1' x1 39 - - 22 - - - - 1' t1 f3(x) = 22 + 2 s,(t - x)(t2 - - - -)& 39 (18.56) Obviously, in this case the solution is of the form f(x) = x2 + C'X + c2. (18.57) Substituting this [Eq. (18.57)] into Equation (18.55) and comparing the coefficients of equal powers of x we obtain C1 = -4 and C2 = -&; thus the exact solution in this case is given as 1 4 f(X) = x2 - z - 1 12' - (18.58) 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. Emr calculation an Neumann series: For the integral equation (18.60) x O<xSt t tlxll, { I((., t) = since Equations (18.51-18.54) tell us that the Neumann sequence is conver- gent. Taking fo(z) = 1, we find the first three terms as fo(x) = 1, fi (x) = 1 + (1/10)2 - (1/20)2, f2(~) = 1 + (31/300)~ - (1/20)x2 - If we take the solution as f(x) = f2b) (18.62) 1/600)z3 + ( 1/2400)x4. (18.63) the error in the entire interval will be less than = 0.0001. (18.64) 18.5.3 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 n r "h 1 If we define the quantity inside the square brackets as lbN3(t)Y(t)dt = Cj, Equation (18.66) becomes n d.)=f(4fXCc&w. j=1 (18.66) (18.67) (18.68) After the coefficients cj are evaluated, this will give us the solution y(z). To find these constants we multiply Equation (18.68) with N;(z) and integrate to get n where and b azj = 1 Ni(Z)Mj(Z)dZ. We now write Equation (18.69) as a matrix equation: b = C-XAC, b = (1-XA)c. (A = ~ij) (18.69) (18.70) (18.71) (18.72) (18.73) This gives us a system of n linear equations to be solved for the n coefficients cj a3 (1 - XUll)Cl - Xa12~2 - Xa13~ - . . . - XU~,C, = bl -XCL~~C~ + (1 - XCL~~)C~ - X~23C3 - . . . - XU~,C, = b2 C1 - Xan2c2 - XU~~C:~ - . . . + (1 - Xann)cn = b, . (18.74) 558 INTEGRAL EQUATIONS When the Fredholm equation is homogeneous (f(z) = 0) all bi are zero; thus for the solution to exist we must have det[I-XA] = 0. (18.75) Solutions of this equation give the eigenvalues Xi. Substituting these eigen- values into Equation (18.74) we can solve for the values of ci. Example 18.5. The case of sepamble kernels: Consider the homogeneous Fredholm equation given as (18.76) where MI(%) = 1, M2(z) = %, Nl(t) = 2t, N2(t) = 1 (18.77) and with A written as 0 4/3 A=[2 0 1- Using Equation (18.75) we write to find the eigenvalues as (18.78) = 0, (18.79) x1,2 = f- (18.80) Substituting these into Equation (18.74) we find two relations between the c1 and the c2 values as c1 fc2g =o. (18.81) 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 13 (1 - gz) for = 'J" (18.83) 25 2 2' 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 18.5.4.1 When the kernel is a function of (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 (18.84) (18.85) (18.86) In writing Equation (18.85) we have also used the convolution theorem: J-00 J-00 which indicates that the Fourier transform of the convolution, f * g, of two functions is the product of their Fourier transforms. We now solve (18.85) for F(k) to find (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 + y(u)sin(a - u)du. (18.89) L‘ We take the Laplace transform of this equation to write L [y(z)] = E [I] + E y(u)sin(a: - u)du . (18.90) [I’ 1 After using the convolution theorem: r PX 1 (18.91) where F(s) and G(s) indicate the Laplace transforms of $(z) and g(z), re spectively, we obtain the Laplace transform of the solution as 1 +s2 Y(s) = - 53 . Taking the inverse Laplace transform, we obtain the solution: X2 y(z) = 1 + T. (18.92) (18.93) (18.94) 18.6 INTEGRAL EQUATIONS AND EIGENVALUE PROBLEMS (HILBERT-SCHMIDT THEORY) In the Sturm-Liouville theory we have defined eigenvalue problems using linear differential operators. We are now going to introduce the Hilbert-Schmidt theory, where an eigenvalue problem is defined in terms of linear integral operators. 18.6.1 Using the Fredholm equation of the second kind, we can define an eigenvalue problem as Eigenvalues Are Real for Hermitian Operators (18.95) INTEGRAL EQUATIONS AND ElGEN VAL UE PROBLEMS (HIL BERT-SCHMID T Tff EOR Y) 561 For the eigenvalue Xi we write (18.96) where yi(t) denotes the corresponding eigenfunction. Similarly, we write Equation (18.95) for another eigenvalue Xj and take its complex conjugate as (18.97) 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 (18.98) (18.99) (18.100) (18.101) (18.102) Kernels satisfying relation (18.100) are called Hermitian. For i = j Equation (18.102) becomes (18.103) 6 Since J, lyi(~)1~ dz # 0, Hermitian operators have real eigenvalues. 562 INTEGRAL EQUATIONS 18.6.2 Orthogonality of Eigenfunctions Using the fact that eigenvalues are real, for z # j Equation (18.102) becomes b (Xj - Xi)/ y,*(z)y;(s)& = 0. a (18.104) For distinct (nondegenerate) eigenvalues this gives JdbY;(z)Yi(z)dz = 0, (Xj # Xi). (18.105) This means that the eigenfunctions for the distinct eigenvalues are orthogonal. 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 b E = dtK(x,t), (18.106) (18.107) we can define an eigenvalue problem as b Y4Z) = xi K(z, t)Yi(t)dt. (18.108) For Hermitian kernels satisfying K*(z, t) = K(t, z), eigenvalues are real and the eigenfunctions are orthogonal; hence after a suitable normaliza- tion we can write: (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) [...]... scattering problems Propagator interpretation 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. .. solution exists and then find it 18.11 1 Using the Neumann series method find the solution of y(z) = x2 +6 1 (z +t ) y ( t ) d t 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... 19.108) Using this Green’s function, solution of the inhomogeneous Helmholtz equation (19.107) is written as where f(x) represents the driving force in wave motion Note that in this case the operator is defined as d2 L=-+k2 dx2 o- Green’s function for this operator can also he obtained by direct construction, that is, by determining the u and the solutions in Equation (19.15) as sin b x and sin ko(x... (19. 116) Because the eigenvalues are continuous we use the Fourier transforms of y(z) and f ( z ) as 1 1 f(z) = V G " dk'g(k')eikfZ -" ( 19.117) and (19.118) Their inverse Fourier transforms are (19.119) and (19.120) Using these in Equation (19. 116) we get (19.121) which gives us (19.122) Substituting this in Equation (19.118) we obtain (19.123) Writing g ( k ' ) explicitly this becomes (19.124) TIME-INDEPENDENT... = sin ko (x- z’) 2 b 1 = -ssinb(z’-x) 2kO (19.130) Note that for the (x- x’)< 0 case the Cauchy principal value is - z ~ f ( u ) In the following cases we add small imaginary pieces, &ZE, to the two roots, f k o and -ko, of the denominator in Equation (19.125), thus moving them away from the real axis We can now use the Cauchy integral theorems t o evaluate the integral (19.125) and then obtain the... 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 19.1.2 Abel’s Formula Let u(z) and ~ ( xbe two linearly independent solutions of Q ( x )= 0, so that ) we can write and respectively Multiplying the first equation by v and the second by u and then subtracting gives us After expanding and rearranging, we can write... equation and boundary conditions: y”(z) - y(z) = 0, y(0) = 0 and y’(0) = I, are equivalent to the integral equation y(5) = z + lz(5 - z’)y(z’)dd 18.3 Write the following differential equation and boundary conditions as an integral equation: Y ” ( 2 ) - 51(2)= 0, y(1) = 0 and y(-I) = 1 18.4 Using the Neumann series method solve the integral equation 18.5 For the following integral equation find the eigenvalues... eigenvalues and the eigenfunctions: y(5) = x 18.6 1 2T cos(5 - z’)y(z’)d5’ To show that the solution of the integral equation FZ y(5) = 1 + x2.I0 ( 5 - z’)y(z’)dz’ 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 By using different methods... closed in the upper or lower z-planes, respectively There are also other ways to treat these singular points in the complex plane, thus giving us a collection of Green's functions each satisfying a different boundary condition, which we study in the following example Example 19.5 Helmholtz equation in the continuum limit: We now evaluate the Green's function given in Equation (19.125) by using the... obtain a closed expression for G(z, d)as G(x, = x‘) sin kox sin ICg(x’ - L ) , x 2’ b sin koL (19.109) TIME-INDEPENDENT GREEN’S FUNCTIONS 19.1.10 583 Green’s Functions and the Dirac-Delta Function Let us operate on the Green’s function [Eq (19.95)] with the d: operator: (19.110) Because E is a linear operator acting on the variable x, we can write (19.111) Using . Xj and take its complex conjugate as (18.97) 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. 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 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.