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By Asmar N.H.

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Example text

So u(x y) = T if and only if 2xy = T if and only T if y = 2x , which shows that the isotherms lie on hyperbolas centered at the origin. 13. We follow the steps in Example 4 (with α = in Θ and R. The solution in Θ is Θn (θ) = sin(4nθ), π 4) and arrive at the same equation n = 1, 2, . . , and the equation in R is r2R + rR − (4n)2 R = 0. The indicial equation for this Euler equation is ρ2 − (4n)2 = 0 ⇒ ρ = ±4n. 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.

We apply (2), with a = b = 1: ∞ ∞ u(x, y) = Emn sin mπx sin nπy, n=1 m=1 where −4 π 2 (m2 + n2 ) Emn = 1 1 x sin mπx sin nπy dx dy 0 0 = 1−(−1)n nπ 1 1 −4 x sin mπx dx sin nπy dy π 2 (m2 + n2 ) 0 0 −4 1 − (−1)n x cos(m π x) sin(m x) − + 4 2 2 π (m + n ) n m m2 π 4 1 − (−1)n (−1)m . π 4 (m2 + n2 ) n m = = = 1 0 Thus u(x, y) = 8 π4 ∞ ∞ k=0 m=1 (−1)m sin mπx sin((2k + 1)πy). (m2 + (2k + 1)2)m(2k + 1) 5. We will use an eigenfunction expansion based on the eigenfunctions φ(x, y) = sin mπx sin nπy, where ∆π(x, y) = −π 2 m2 +n2 sin mπx sin nπy.

3nJ4 (α3n) Hence ∞ u(r, θ, t) = r3 sin 3θ − 2 sin(3θ) n=1 (where α3nr = s) J3 (α3nr) −α23n t . 4 1. Since f is already given by its Fourier series, we have from (4) u(r, θ) = r cos θ = x. 5. Let us compute the Fourier coefficients of f. We have a0 = an = bn = 100 π 50 π π/4 dθ = 0 π/4 100 π cos nθ dθ = 0 π/4 sin nθ dθ = − 0 25 ; 2 100 sin nθ nπ 100 cos nθ nπ π = 0 π = 0 100 nπ sin ; nπ 4 100 nπ (1 − cos ). nπ 4 Hence ∞ 25 100 1 nπ nπ f(θ) = + sin cos nθ + (1 − cos ) sin nθ ; 2 π n=1 n 4 4 and ∞ u(r, θ) = 25 100 1 nπ nπ + sin cos nθ + (1 − cos ) sin nθ rn .

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


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