Mathematical Methods for Physics and Engineering - Matematica.NET

17.5 SUPERPOSITION OF EIGENFUNCTIONS: GREEN’S FUNCTIONS(√ )1Now, the boundary conditions require that B =0andsin − λ π =0,andso4√1− λ = n, where n =0, ±1, ±2,....4Therefore, the independent eigenfunctions that satisfy the boundary conditions arey n (x) =A n sin nx,where n is any non-negative integer, and the corresponding eigenvalues are λ n = 1 − 4 n2 .The normalisation condition further requires∫ π( ) 1/2 2A 2 n sin 2 nx dx =1 ⇒ A n = .0πComparison with (17.51) shows that the appropriate Green’s function is therefore givenbyG(x, z) = 2 ∞∑ sin nx sin nz1π − .n=0 4 n2Case (i). Using (17.50), the solution with f(x) =sin2x is given byy(x) = 2 ∫ (π ∞)∑ sin nx sin nz1π 0− sin 2zdz= 2 ∞∑n=0 4 n2 πn=0Now the integral is zero unless n = 2, in which case it is∫ π0sin 2 2zdz= π 2 .∫sin nx π1− sin nz sin 2zdz.4 n2 0Thusy(x) =− 2 sin 2x ππ 15/4 2 = − 4 sin 2x15is the full solution for f(x) =sin2x. This is, of course, exactly the solution found by usingthe methods of chapter 15.Case (ii). The solution with f(x) =x/2 isgivenby∫ ()π2∞∑ sin nx sin nz zy(x) =10 π − n=0 4 n2 2 dz = 1 ∞∑πn=0The integral may be evaluated by integrating by parts. For n ≠0,∫ [π∫z cos nzπcos nzz sin nz dz = − + dznn0=−π cos nπn] π00[ ] π sin nz+n 2 0∫sin nx π1− z sin nz dz.4 n2 0= − π(−1)n .nFor n = 0 the integral is zero, and thus∞∑y(x) = (−1) n+1 sin nxn ( 1− n2) ,n=14is the full solution for f(x) =x/2. Using the methods of subsection 15.1.2, the solutionis found to be y(x) =2x − 2π sin(x/2), which may be shown to be equal to the abovesolution by expanding 2x − 2π sin(x/2) as a Fourier sine series. ◭571