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