Partial differential equations (ACM30220) Assignment 3 Issue Date: 4 November 2014 Due Date: 14 November 2014 1. Using d’Alembert’s formula, or any other means, solve the wave equation utt = uxx , −∞ < x < ∞, subject to the initial conditions u(x, t = 0) = f (x) = 0, ( 1 |x| ≤ L, ut (x, t = 0) = g(x) = . 0 otherwise [5 marks] 2. Solve the wave equation 1 ∂ 2u ∂ 2u ∂ 2u = + c2 ∂t2 ∂x2 ∂y 2 in two space dimensions (x, y) ∈ (0, L) × (0, L), subject to the following boundary conditions: u(x, y = 0) = u(x, y = L) = 0, u(x = 0, y) = u(x = L, y) = 0, and the following initial conditions: u(x, y, t = 0) = f (x, y), ut (x, y, t = 0) = g(x, y). Hints: Attempt a separable solution u(x, y, t) = X(x)Y (y)T (t) and show that u(x, y, t) = = ∞ X ∞ X n=1 m=1 ∞ X ∞ X n=1 m=1 where Xn (x)Ym (y)Tnm (t), sin nπx L sin mπy L [Anm cos (knm ct) + Bnm sin (knm t)] , π√ 2 n + m2 . L and Bnm by double integration. knm = Determine the constants Anm 1 [10 marks] Partial differential equations The wave equation 3. Consider Bessel’s equation of integer order, r2 d2 y dy + r + (k 2 r2 − m2 )y = 0, 2 dr dr m ∈ Z. (1) The general solution to this equation is ym (r) = αJm (kr) + βKm (kr), (2) where Jm (·) and Km (·) are known power series. The function Jm (·) is well behaved for all values of its argument, while Km (·) has a singularity at zero. (a) Consider solutions ym (r) to Bessel’s equation in an interval R1 < r < R2 , such that ym (R1 ) = ym (R2 ) = 0. (3) Argue that the condition (3) implies that αm Jm (kR1 ) + βm K(kR1 ) = 0, αm Jm (kR2 ) + βm K(kR2 ) = 0 and hence Jm (kR1 )K(kR2 ) − K(kR1 )Jm (kR2 ) = 0. Argue further that this gives rise to a (possibly infinite) set of eigenvalues kmn : Jm (kmn R1 )K(kmn R2 ) − Km (kmn R1 )Jm (kmn R2 ) = 0, p 2 + β 2 = 1. with associated eigenvectors (αmn , βmn ), with αmn mn (b) Let Imn (r) = αmn Jm (kmn r) + βmn Km (kmn r). Argue that the general solution to Bessel’s equation (1) with boundary conditions (3) is X ym (r) = Cn Imn (r), n where the Cn ’s are arbitrary constants. (c) Demonstrate that the Imn functions are orthogonal, in the sense that Z R2 rImn (r)Imn′ (r) dr = 0, R1 n 6= n. 2 Partial differential equations The wave equation Hints for (c): • Consider dImn d2 Imn 2 +r + (kmn r2 − m2 )Imn = 0, 2 dr dr 2 dImn′ 2 d Imn′ 2 2 2 r + r + (kmn ′ r − m )Imn′ = 0. 2 dr dr r2 (4a) (4b) Take [Eq. (4a)]Imn′ -[Eq. (4b)]Imn and show that it can be re-written as dImn dImn′ d Imn′ − Imn r dr dr dr dImn dImn′ 2 2 + r Imn′ − Imn + r2 (kmn − kmn ′ )Imn Imn′ = 0. dr dr 2 • Show further that this can be re-written as dImn′ dImn d 2 2 − Imn r Imn′ + r(kmn − kmn ′ )Imn Imn′ = 0. dr dr dr • Integrate the result from r = R1 to r = R2 and use the fact that Imn (R1 ) = Imn (R2 ) = 0. [10 marks] 4. Solve the wave equation 1 ∂ 2u = ∇2 u c2 ∂t2 in two dimensions, in polar coordinates, with ∂ 1 ∂2 1 ∂ 2 r + 2 2, ∇ = r ∂r ∂r r ∂ϕ in the annulus R1 < r < R2 , and subject to the following boundary and initial conditions: u(R1 , ϕ, t) = u(R2 , ϕ, t) = 0, u(r, ϕ, 0) = f (r, ϕ), ut (r, ϕ, 0) = g(r, ϕ). [5 marks] 3
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