Mathematical Methods for Physics and Engineering - Matematica.NET

EIGENFUNCTION METHODS FOR DIFFERENTIAL EQUATIONSand we assume that we may interchange the order of summation and integration,then (17.49) can be written as∫ {b ∑ ∞ [ ] } 1y(x) =ŷ n (x)ŷ ∗λn(z) f(z) dz.nan=0The quantity in braces, which is a function of x and z only, is usually writtenG(x, z), and is the Green’s function for the problem. With this notation,wherey(x) =G(x, z) =∫ baG(x, z)f(z) dz, (17.50)∞∑n=01λ nŷ n (x)ŷ ∗ n(z). (17.51)We note that G(x, z) is determined entirely by the boundary conditions and theeigenfunctions ŷ n , and hence by L itself, and that f(z) depends purely on theRHS of the inhomogeneous equation (17.44). Thus, for a given L and boundaryconditions we can establish, once and for all, a function G(x, z) that will enableus to solve the inhomogeneous equation for any RHS. From (17.51) we also notethatG(x, z) =G ∗ (z,x). (17.52)We have already met the Green’s function in the solution of second-order differentialequations in chapter 15, as the function that satisfies the equationL[G(x, z)] = δ(x − z) (and the boundary conditions). The formulation givenabove is an alternative, though equivalent, one.◮Find an appropriate Green’s function for the equationy ′′ + 1 y = f(x),4with boundary conditions y(0) = y(π) =0.Hence,solvefor(i) f(x) =sin2x and (ii)f(x) =x/2.One approach to solving this problem is to use the methods of chapter 15 and finda complementary function and particular integral. However, in order to illustrate thetechniques developed in the present chapter we will use the superposition of eigenfunctions,which, as may easily be checked, produces the same solution.The operator on the LHS of this equation is already Hermitian under the given boundaryconditions, and so we seek its eigenfunctions. These satisfy the equationy ′′ + 1 y = λy.4This equation has the familiar solution(√ ) (√ )1y(x) =A sin − λ 1x + B cos − λ x.4 4570