# 1 a2 one simply types in sol ndsolve xt xt

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Unformatted text preview: 2g +2 ∂x2 ∂y Hence if the restriction of f to ∂D vanishes, f ∆gdxdy = − D ∇f · ∇gdxdy. D Similarly, if the restriction of g to ∂D vanishes, g ∆f dxdy = − D ∇f · ∇gdxdy. D Thus if f and g lie in V , f , ∆g = − ∇f · ∇gdxdy = g , ∆f . D In particular, if (5.32) holds, then µ f , g = f , ∆g = ∆f, g = λ f , g , and hence (λ − µ) f , g = 0. 151 dxdy. It follows immediately that either λ − µ = 0 or f , g = 0, and the lemma is proven. Now let us focus on the special case in which D is a circular disk. Recall the problem that we want to solve, in terms of polar coordinates: Find u(r, θ, t), 0 &lt; r ≤ 1, so that ∂2u 1∂ = ∂t2 r ∂r r ∂u ∂r u(r, θ + 2π, t) = u(r, θ, t), + 1 ∂2u , r2 ∂θ2 u well-behaved near r = 0, u(1, θ, t) = 0, ∂u (r, θ, 0) = 0. ∂t It follows from the argument in the preceding section that the general solution to the homogeneous linear part of the problem can be expressed in terms of the eigenfunctions u(r, θ, 0) = h(r, θ), fn,k (r, θ) = Jn (αn,k r/a) cos(nθ), gn,k = Jn (αn,k r/a) sin(nθ). Indeed, the general solution must be a superposition of the products of these eigenfunctions with periodic functions of t of frequencies −λn,k /2π . Thus this general solution is of the form ∞ u(r, θ, t) = a0,k g0,k (r, θ) cos(α0,k t) k=1 ∞ ∞ + [an,k fn,k (r, θ) + bn,k gn,k (r, θ)] cos(αn,k t). (5.33) n=1 k=1 We need to determine the an,k ’s and the bn,k ’s so that the inhomoge...
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