Chem Differential Eq HW Solutions Fall 2011 76

# Chem Differential Eq HW Solutions Fall 2011 76 - 76 Chapter...

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Unformatted text preview: 76 Chapter 4 Partial Diﬀerential Equations in Polar and Cylindrical Coordinates So the coeﬃcients are Aj 2 = = α j J 3/2 ( α j ) 2 jπ J3/2(jπ) 2 = sin jπ jπ 2 π jπ jπ − cos jπ 2 j = (−1)j −1 and the Bessel series expansion becomes, or 0 < x < 1, √ ∞ 2 J 1/2 ( α j x ) . j (−1)j −1 x= j =1 (c) Writing J1/2(x) in terms of sin x and simplifying, this expansion becomes √ ∞ (−1)j −1 2 J 1/2 ( α j x ) j (−1)j −1 x= 2 j j =1 ∞ = j =1 Upon multiplying both sides by x= 2 π ∞ j =1 ∞ 2 π = j =1 2 sin αj π αj (−1)j −1 sin(jπx) √ . j x √ x, we obtain (−1)j −1 sin(jπx) j for 0 < x < 1, which is the familiar Fourier sine series (half-range expansion) of the function f ( x) = x. 25. By Theorem 2 with p = 1, we have Aj = 2 J2 (α1,j )2 1 J1 (α1,j x) dx 1 2 α1,j = 2 α1,j J2 (α1,j )2 = 2 [−J0 (s) α1,j J2 (α1,j )2 = −2 J0 (α1,j ) − J0 ( α1,j J2 (α1,j )2 = −2 J0 (α1,j ) − J0 ( α1,j J0 (α1,j )2 α1,j 2 J1(s) ds (let α1,j x = s) α1,j (by (8) with p = 1) α1,j 2 α1,j ) 2 α1,j ) 2 , where in the last equality we used (6) with p = 1 at x = α1,j (so J0(α1,j )+J2 (α1,j ) = 0 or J0 (α1,j ) = −J2 (α1,j )). Thus, for 0 < x < 1, ∞ f (x) = −2 α1,j 2) −2 J0(α1,j ) − J0( α1,j J0(α1,j )2 j =1 J1(α1,j x). ...
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