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By Asmar N.H.

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Example text

17. Part (a) is straightforward as in Example 2. We omit the details that lead to the separated equations: T − kT = 0, X − kX = 0, X (0) = −X(0), X (1) = −X(1), where k is a separation constant. (b) If k = 0 then X = 0 ⇒ X = ax + b, X (0) = −X(0) ⇒ a = −b X (1) = −X(1) ⇒ a = −(a + b) ⇒ 2a = −b; ⇒ a = b = 0. 6 Heat Conduction in Bars: Varying the Boundary Conditions 45 So k = 0 leads to trivial solutions. (c) If k = α2 > 0, then X − µ2 X = 0 ⇒ X = c1 cosh µx + c2 sinh µx; X (0) = −X(0) ⇒ µc2 = −c1 X (1) = −X(1) ⇒ µc1 sinh µ + µc2 cosh µ = −c1 cosh µ − c2 sinh µ ⇒ µc1 sinh µ − c1 cosh µ = −c1 cosh µ − c2 sinh µ ⇒ µc1 sinh µ = −c2 sinh µ c1 ⇒ µc1 sinh µ = sinh µ.

Taking the bounded solutions only, we get Rn(r) = r4n. Thus the product solutions are r4n sin 4θ and the series solution of the problem is of the form ∞ bn r4n sin 4nθ. u(r, θ) = n=1 To determine bn, we use the boundary condition: ∞ ur (r, θ) r=1 = sin θ bn 4nr4n−1 sin 4nθ ⇒ r=1 = sin θ n=1 ∞ ⇒ bn 4n sin 4nθ = sin θ n=1 ⇒ 4nbn = 2 π/4 π/4 sin θ sin 4nθ dθ.

This is a big advantage of the d’Alembert’s solution over the Fourier series solution. 3, 1 5 Ticks Show GraphicsArray Partition tt, 4 37 ; 9. 3, which tells us that the time period of motion is T = 2L c . So, in the case of Exercise 1, T = 2π, and in the case of Exercise 5, T = 2. You can also obtain these results directly by considering the formula for u(x, t). In the case of Exercise 1, u(x, t) = 12 sin(πx + t) + sin(πx − t) so u(x, t + 2π) = 12 sin(πx + t2π) + sin(πx − t2π) = u(x, t). In the case of Exercise 5, use the fact that f ∗ and G are both 2-periodic.

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Partial differential equations with Fourier series and BVP. Student solutions manual by Asmar N.H.


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