Download Solutions to Math. 412 (Fulling) Test C: Laplace's Equation and Wave Equation and more Exams Differential Equations in PDF only on Docsity! Math. 412 (Fulling) 19 November 2004 Test C – Solutions Calculators may be used for simple arithmetic operations only! 1. (55 pts.) (a) Solve Laplace’s equation, ∂ 2u ∂x2 + ∂ 2u ∂y2 = 0 , in a square, 0 < x < π and 0 < y < π , with the boundary conditions u(0, y) = 0 , u(π, y) = f(y) , u(x, 0) = 0 , ∂u ∂y (x, π) + 2u(x, π) = 0 . In the process you will discover a sequence of eigenfunctions and eigenvalues, which you should name {φn(y)} and {ωn2} . Describe the ωn qualitatively but don’t expect to find their exact numerical values. (Also, don’t bother to evaluate the normalization integral.) Look for separated solutions X(x)Y (y) (postponing the nonhomogeneous BC). X ′′ X = −Y ′′ Y = ω2. Assuming for the moment that ω2 > 0 , the two problems are X′′ = ω2X, X(0) = 0 — hence X(x) = sinh(ωx) , Y ′′ = −ω2Y, Y (0) = 0, Y ′(π) + 2Y (π) = 0 . From this we get Y (y) = sin(ωy), −1 2 ω = tan(πω) . Graphing the two sides of the eigenvalue equation (for positive ω ), we see that there are intersections on each branch of the tangent function except the first. That is, there are eigenvalues ωn 2 , n = 1, 2, . . . , with ωn slightly greater than (n + 1 2 )π . Let’s write Yn(y) = sin(ωny) and φn(y) for the normalized eigenfunctions, Yn(y)/‖Yn‖ . (Here ‖Yn‖2 = ∫ π 0 Yn(y) 2 dy , and you were not expected to evaluate this integral.) The general theorem about Sturm–Liouville problems (with +2 in the Robin boundary condition at the top end of the interval) guarantees that negative and zero eigenvalues don’t occur. This can also be shown directly. We can now write the solution of the original problem as u(x, y) = ∞∑ n=1 cnφn(y) sinh(ωnx) 412C-F04 Page 2 and impose the boundary data f(y) = ∞∑ n=1 cnφn(y) sinh(ωnπ) . Since the basis functions are normalized, we can calculate immediately cn = ∫ π 0 φn(y)f(y) dy sinh(ωnπ) . Alternatively, in terms of the Yn you have u(x, y) = ∞∑ n=1 CnYn(y) sinh(ωnx) and Cn = ∫ π 0 Yn(y)f(y) dy ‖Yn‖2 sinh(ωnπ) . (b) Express the solution in terms of a Green function, so that u(x, y) = ∫ π 0 G(x, y, z)f(z) dz, with a formula for G in terms of the eigenfunctions φn . Write the cn integral with the variable of integration called z instead of y , then insert into the formula for u . You arrive at G(x, y, z) = ∞∑ n=1 sinh(ωnx) sinh(ωnπ) φn(y)φn(z) . If you use Y instead of φ , you need to divide by ‖Yn‖2 . (c) What are the orthogonality and completeness relations satisfied by the eigenfunctions? (You can answer (c) on abstract grounds even if you have trouble with (a).)∫ π 0 φn(y)φm(y) dy = δnm , ∞∑ n=1 φn(y)φn(z) = δ(y − z) . (Since the eigenfunctions are real, no complex conjugations are necessary, but it doesn’t hurt to put them in.) In terms of Y , ∫ π 0 Yn(y)Ym(y) dy = δnm‖Yn‖2 , ∞∑ n=1 Yn(y)Yn(z) ‖Yn‖2 = δ(y − z) . A check: Set x = π in the Green function formula: δ(y − z) = G(π, y, z) = ∞∑ n=1 φn(y)φn(z) .
