Claymathorgmillennium exercise 541 show that if the

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Unformatted text preview: ison with (5.12) shows that b11 = 1, b21 = 3, b32 = 7, and all the other bmn ’s must be zero. Thus the solution to the initial value problem in this case is u(x, y, t) = sin x sin ye−2t + 3 sin 2x sin ye−5t + 7 sin 3x sin 2ye−13t . Here is another example. Suppose that a = b = π and h(x, y ) = p(x)q (y ), where x, for 0 ≤ x ≤ π/2, π − x, for π/2 ≤ x ≤ π, p(x) = for 0 ≤ y ≤ π/2, for π/2 ≤ y ≤ π. y, π − y, q (y ) = In this case, 2 π π π p(x)q (y ) sin mxdx sin nydy 0 0 2 2 π = = 2 2 π bmn = π π p(x) sin mxdx q (y ) sin nydy 0 2 0 π /2 π (π − x) sin mxdx x sin mxdx + 0 π/2 π /2 π (π − y ) sin nydy . y sin nydy + 0 π/2 The integration can be carried out just like we did in Section 3.3 to yield 2 π bmn = 2 2 sin(mπ/2) m2 = 2 sin(nπ/2) n2 16 1 1 sin(mπ/2) sin(nπ/2) . π 2 m2 n 2 Thus in this case, we see that h(x, y ) = 16 π2 ∞ m,n=1 11 sin(mπ/2) sin(nπ/2) sin mx sin my, m2 n 2 and hence u(x, y, t) = 16 π2 ∞ m,n=1 2 2 11 sin(mπ/2) sin(nπ/2) sin mx sin mye−(m +n )t . m2 n 2 127 Exercises: 5.3.1. Solve the following initial value problem for the heat equation in a square region: Fi...
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