Chem Differential Eq HW Solutions Fall 2011 49

Chem Differential Eq HW Solutions Fall 2011 49 - n = 7 A 7...

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Section 3.8 Laplace’s Equation in Rectangular Coordinates 49 Solutions to Exercises 3.8 1. The solution is given by u ( x, y ) = n =1 B n sin( nπx ) sinh( nπy ) , where B n = 2 sinh(2 ) 1 0 x sin( nπx ) dx = 2 sinh(2 ) - x cos( n π x ) n π + sin( n π x ) n 2 π 2 1 0 = 2 sinh(2 ) - ( - 1) n n π = 2 sinh(2 ) ( - 1) n +1 n π . Thus, u ( x, y ) = 2 π n =1 ( - 1) n +1 n sinh(2 ) sin( nπx ) sinh( nπy ) . 5. Start by decomposing the problem into four subproblems as described by Figure 3. Let u j ( x, y ) denote the solution to problem j ( j = 1 , 2 , 3 , 4). Each u j consists of only one term of the series solutions, because of the orthogonality of the sine functions. For example, to compute u 1 , we have u 1 ( x, y ) = n =1 A n sin nπx sinh[ (1 - y )] , where A n = 2 sinh 1 0 sin 7 πx sin nπx dx. Since the integral is 0 unless n = 7 and, when
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Unformatted text preview: n = 7, A 7 = 2 sinh 7 π ² 1 sin 2 7 πx dx = 1 sinh7 π . Thus u 1 ( x, y ) = 1 sinh7 π sin 7 πx sinh[7 π (1-y )] . In a similar way, appealing to the formulas in the text, we ±nd u 2 ( x, y ) = 1 sinh π sin πx sinh( πy ) u 3 ( x, y ) = 1 sinh 3 π sinh[3 π (1-x )] sin(3 πy ) u 4 ( x, y ) = 1 sinh 6 π sinh6 πx sin(6 πy ); u ( x, y ) = u 1 ( x, y ) + u 2 ( x, y ) + u 3 ( x, y ) + u 4 ( x, y ) = 1 sinh 7 π sin7 πx sinh[7 π (1-y )] + 1 sinh π sin( πx ) sinh( πy ) + 1 sinh 3 π sinh[3 π (1-x )] sin(3 πy ) + 1 sinh6 π sinh(6 πx ) sin(6 πy )...
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