Solutions to Math 412-501 Exam 2: Heat Equation and Laplace's Equation, Exams of Differential Equations

Download Solutions to Math 412-501 Exam 2: Heat Equation and Laplace's Equation and more Exams Differential Equations in PDF only on Docsity! Math 412-501 October 20, 2006 Exam 2: Solutions Problem 1 (50 pts.) Solve the heat equation in a rectangle 0 < x < π, 0 < y < π, ∂u ∂t = ∂2u ∂x2 + ∂2u ∂y2 subject to the initial condition u(x, y, 0) = (sin 2x + sin 3x) sin y and the boundary conditions u(0, y, t) = u(π, y, t) = 0, u(x, 0, t) = u(x, π, t) = 0. Solution: u(x, y, t) = e−5t sin 2x sin y + e−10t sin 3x sin y. We search for the solution of the initial-boundary value problem as a superposition of solutions u(x, y, t) = φ(x)h(y)G(t) with separated variables of the heat equation that satisfy the boundary conditions. Substituting u(x, y, t) = φ(x)h(y)G(t) into the heat equation, we obtain φ(x)h(y)G ′(t) = φ′′(x)h(y)G(t) + φ(x)h′′(y)G(t), G ′(t) G(t) = φ′′(x) φ(x) + h′′(y) h(y) . Since any of the expressions G ′(t) G(t) , φ′′(x) φ(x) , and h′′(y) h(y) depend on one of the variables x, y, t and does not depend on the other two, it follows that each of these expressions is constant. Hence φ′′(x) φ(x) = −λ, h ′′(y) h(y) = −µ, G ′(t) G(t) = −(λ + µ), where λ and µ are constants. Then φ′′ = −λφ, h′′ = −µh, G ′ = −(λ + µ)G. Conversely, if functions φ, h, and G are solutions of the above ODEs for the same values of λ and µ, then u(x, y, t) = φ(x)h(y)G(t) is a solution of the heat equation. Substituting u(x, y, t) = φ(x)h(y)G(t) into the boundary conditions, we get φ(0)h(y)G(t) = φ(π)h(y)G(t) = 0, φ(x)h(0)G(t) = φ(x)h(π)G(t) = 0. It is no loss to assume that neither φ nor h nor G is identically zero. Then the boundary conditions are satisfied if and only if φ(0) = φ(π) = 0, h(0) = h(π) = 0. To determine φ, we have an eigenvalue problem φ′′ = −λφ, φ(0) = φ(π) = 0. This problem has eigenvalues λn = n 2, n = 1, 2, . . . . The corresponding eigenfunctions are φn(x) = sinnx. 1 To determine h, we have the same eigenvalue problem h′′ = −µh, h(0) = h(π) = 0. Hence the eigenvalues are µm = m 2, m = 1, 2, . . . . The corresponding eigenfunctions are hm(y) = sinmy. The function G is to be determined from the equation G ′ = −(λ + µ)G. The general solution of this equation is G(t) = c0e −(λ+µ)t, where c0 is a constant. Thus we obtain the following solutions of the heat equation satisfying the boundary conditions: unm(x, y, t) = e −(λn+µm)tφn(x)hm(y) = e −(n2+m2)t sinnx sinmy, n, m = 1, 2, 3, . . . A superposition of these solutions is a double series u(x, y, t) = ∞ ∑ n=1 ∞ ∑ m=1 cnme −(n2+m2)t sinnx sinmy, where cnm are constants. To determine the coefficients cnm, we substitute the series into the initial condition u(x, y, 0) = (sin 2x + sin 3x) sin y: (sin 2x + sin 3x) sin y = ∞ ∑ n=1 ∞ ∑ m=1 cnm sinnx sinmy. It is easy to observe that c2,1 = c3,1 = 1 while the other coefficients are equal to 0. Therefore u(x, y, t) = e−5t sin 2x sin y + e−10t sin 3x sin y. Problem 2 (50 pts.) Solve Laplace’s equation inside a quarter-circle 0 < r < 1, 0 < θ < π/2 (in polar coordinates r, θ) subject to the boundary conditions u(r, 0) = 0, u(r, π/2) = 0, |u(0, θ)| < ∞, u(1, θ) = f(θ). Solution: u(r, θ) = ∑∞ n=1 cnr 2n sin 2nθ, where cn = 4 π ∫ π/2 0 f(θ) sin 2nθ dθ, n = 1, 2, . . . Laplace’s equation in polar coordinates (r, θ): ∂2u ∂r2 + 1 r ∂u ∂r + 1 r2 ∂2u ∂θ2 = 0. We search for the solution of the boundary value problem as a superposition of solutions u(r, θ) = h(r)φ(θ) with separated variables of Laplace’s equation that satisfy the three homogeneous boundary conditions. Substituting u(r, θ) = h(r)φ(θ) into Laplace’s equation, we obtain h′′(r)φ(θ) + 1 r h′(r)φ(θ) + 1 r2 h(r)φ′′(θ) = 0, r2h′′(r) + rh′(r) h(r) = −φ ′′(θ) φ(θ) . 2