11 r for 0 r 1 in order to nd the solution to the wave

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Unformatted text preview: plane as in the preceding problem. Suppose that the eigenvalues for the Laplace operator ∆ on D are λ1 , λ2 , . . . , once again. Show that the general solution to the wave equation ∂2u = ∆u, ∂t2 together with Dirichlet boundary conditions (u vanishes on ∂D) and the initial condition ∂u (x, y, 0) = 0, ∂t is ∞ u(x, t) = bn φn (x, y ) cos( −λn t), n=1 where the bn ’s are arbitrary constants. 5.8 Eigenvalues of the disk To calculate the eigenvalues of the disk, it is convenient to utilize polar coordinates r, θ in terms of which the Laplace operator is 1∂ r ∂r r ∂f ∂r + 1 ∂2f = λf, r2 ∂θ2 f |∂D = 0. (5.28) Once again, we use separation of variables and look for product solutions of the form f (r, θ) = R(r)Θ(θ), where R(a) = 0, Θ(θ + 2π ) = Θ(θ). Substitution into (5.28) yields 1d r dr r dR dr Θ+ R d2 Θ = λRΘ. r2 dθ2 144 We multiply through by r2 , d dr r r dR dr Θ+R d2 Θ = λr2 RΘ, dθ2 and divide by RΘ to obtain rd R dr r dR dr − λr2 = − 1 d2 Θ . Θ dθ2 Note that in this last equation...
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This document was uploaded on 01/12/2014.

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