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Solution Set 7 1. Consider waves in a resistant medium that satisfy the problem utt = c2 uxx − rut for 0 < x < L u(0, t) = u(L, t) = 0 u(x, 0) = φ(x) ut (x, 0) = ψ(x) where r is a constant, 0 < r < 2πc/L. Write down the series expansion of the solution. Answer Separation gives T 00 + rT 0 X 00 = = −λ. c2 T X X is a solution of the standard Dirichlet BVP. Therefore, Xn(x) = C sin βn x with βn = nπ/l, n = 1, 2, ... is a solution. The T equation becomes T 00 + rT + c2 βn3 T = 0. The characteristic equation ofpthis ODE is s2 + rs + c2 βn2 = 0. It has two distinct roots s = (−r ± r2 − 4c2 βn2 )/2 = −r/2 ± iωn where ωn = p (2nπc/l)2 − r2 /2 is a real positive number since 0 < r < 2πc/l and n ≥ 1. Therefore, Tn = e−rt/2 (A cos ωn t + B sin ωn t). The series solution is ∞ X −rt/2 u(x, t) = e (An cos ωn t + Bn sin ωn t) sin βn x. n=1 with φ(x) = ∞ X An sin βn x, ψ(x) = n=1 ∞ X r (− An + Bn ωn ) sin βn x. 2 n=1 2. Use the method of subtraction to solve utt = 9uxx for 0 < x < 1 with u(0, t) = h, u(1, t) = k, where h and k are given constants, and u(x, 0) = 0, ut (x, 0) = 0. 1 Answer Let U(x, t) = (1 − x)h + xk = h + (k − h)x. Let v = u − U. Then, v satisfies the equation vtt = 9vxx with the Dirichlet BC v(0, t) = v(1, t) = 0 and initial conditions v(x, 0) = u(x, 0) − U(x, 0) = −h − (k − h)x and vt (x, 0) = ut (x, 0) − Ut (x, 0) = 0. The series solution for this Dirichlet BC is ∞ X v(x, t) = (An cos 3nπt + Bn sin 3nπt) sin nπx. n=1 First, note that vt (x, 0) = 0 ⇒ Bn = 0. 1 1 sin2 nπxdx = , the remaining coefficients can be computed as 2 0 Z 1 h 1 An = −2 [h+(k−h)x] sin nπxdx = −2 [1 − (−1)n ] − (k − h) (−1)n nπ nπ 0 2 = − [h + k(−1)n+1 ]. nπ As Z Therefore, ∞ 2X u(x, t) = h + (k − h)x − [h + k(−1)n+1 ] cos 3nπt sin nπx. π n=1 3. Use the method of descent to derive d’Alembert’s formula for the one-dimensional Cauchy problem from Poisson’s formula for two space dimensions. Answer Poisson’s formula for two space dimensions is "Z Z # 1 ∂ φ(ξ, η) p u(x, y, t) = dξdη 2πc ∂t (ct)2 − (ξ − x)2 − (η − y)2 Dt 1 + 2πc ZZ Dt ψ(ξ, η) p (ct)2 − (ξ − x)2 − (η − y)2 dξdη where Dt is a disk centered at (x, y) with radius ct. To make a onedimensional problem, let’s consider points on (x, 0). Furthermore, u, φ and ψ are made independent of y (or η inside the double integral). The formula can be written as 2 "Z Z # 1 ∂ φ(ξ) p dξdη 2πc ∂t (ct)2 − (ξ − x)2 − η 2 Dt ZZ 1 ψ(ξ) p + dξdη. 2πc Dt (ct)2 − (ξ − x)2 − η 2 As an example, consider the how the double integral for φ can be handled. The integral inside the rectangular bracket can be written as ! Z x+ct Z +√(ct)2−(ξ−x)2 1 p φ(ξ) dη dξ. √ 2 (ct)2 − (ξ − x)2 − η 2 x−ct − (ct) −(ξ−x)2 p Let a = (ct)2 − (ξ − x)2 and η = a sin θ. The integal insider the large bracket can be evaluated as Z +π/2 1 p a cos θdθ = π. 2 2 2 −π/2 a − a sin Therefore, ZZ Z x+ct φ(ξ) p dξdη = π φ(ξ)dξ. (ct)2 − (ξ − x)2 − η 2 Dt x−ct u(x, t) = This gives the solution of CP as Z x+ct Z 1 ∂ 1 x+t u(x, t) = φ(ξ) dξ + ψ(ξ) dξ 2c ∂t x−ct 2c x−t Z 1 1 x+ct = (φ(x + t) + φ(x − t)) + ψ(ξ) dξ, 2 2c x−ct d’Alembert’s solution. 4. Solve the problem: utt − c2 (uxx + uyy + uzz ) = 0 ut=0 = x2 + yz, ut t=0 = 0. Answer Kirchhoff-Poisson’s solution gives: Z 2πZ π 1 ∂ u= t (x + ct sin θ cos φ)2 + (y + ct sin θ sin φ)(z + ct cos θ) sin θdθdφ 4π ∂t 0 0 ∂ 1 = x2 t + c2 t3 + yzt = x2 + c2 t2 + yz ∂t 3 3