Finally we can rewrite this as du au dt where a n2 p

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Unformatted text preview: + 2 sin 2xe−4t + 7 sin 3xe−9t . Example 2. Suppose that we want to find the function u(x, t), defined for 0 ≤ x ≤ π and t ≥ 0, which satisfies the initial-value problem: ∂u ∂2u , = ∂t ∂x2 u(0, t) = u(π, t) = 0, where h(x) = u(x, 0) = h(x), x, for 0 ≤ x ≤ π/2, π − x, for π/2 ≤ x ≤ π . We saw in Section 3.3 that the Fourier sine series of h is h(x) = 4 4 4 4 sin x − sin 3x + sin 5x − sin 7x + . . . , π 9π 25π 49π and hence u(x, t) = 4 4 4 sin xe−t − sin 3xe−9t + sin 5xe−25t − . . . . π 9π 25π 89 Exercises: 4.2.1. Find the function u(x, t), defined for 0 ≤ x ≤ π and t ≥ 0, which satisfies the following conditions: ∂u ∂2u , u(0, t) = u(π, t) = 0, u(x, 0) = sin 2x. = ∂t ∂x2 You may assume that the nontrivial solutions to the eigenvalue problem f (x) = λf (x), f (0) = 0 = f (π ) are λ = −n2 , f (x) = bn sin nx, for n = 1, 2, 3, . . . , where the bn ’s are constants. 4.2.2. Find the function u(x, t), defined for 0 ≤...
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