Chem Differential Eq HW Solutions Fall 2011 87

Chem Differential Eq HW Solutions Fall 2011 87 - Indeed,...

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Section 5.4 The Helmholtz, Poisson, Heat, and Wave Equations 87 and so the series expansion becomes: for 0 <r< 1, 1= ± j =1 4( - 1) j +1 π sin jπr jπr Y 0 , 0 ( θ, φ ) = ± j =1 4( - 1) j +1 π sin jπr jπr 1 2 π = ± j =1 2( - 1) j +1 sin jπr jπr . It is interesting to note that this series is in fact a half range sine series expansion.
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Unformatted text preview: Indeed, multiplying both sides by r , we get r = 2 j =1 (-1) j +1 sin jr j (0 &lt; r &lt; 1) , which is a familiar sines series expansion (compare with Example 1, Section 2.4)....
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