Chem Differential Eq HW Solutions Fall 2011 47

Chem Differential Eq HW Solutions Fall 2011 47 - In a...

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Section 3.6 Heat Conduction in Bars: Varying the Boundary Conditions 47 Thus the functions are orthogonal. We still have to verify the orthogonality when one of the X n ’s is equal to e - x . This can be done by modifying the argument that we just gave. (f) Superposing the product solutions, we Fnd that u ( x, t )= c 0 e - x e t + ± n =1 c n T n ( t ) X n ( x ) . Using the initial condition, it follows that u ( x, 0) = f ( x )= c 0 e - x + ± n =1 c n X n ( x ) . The coefficients in this series expansion are determined by using the orthog- onality of the X n ’s in the usual way. Let us determine c 0 . Multiplying both sides by e - x and integrating term by term, it follows from the orthogonality of the X n that ² 1 0 f ( x ) e - x dx = c 0 ² 1 0 e - 2 x dx + ± n =1 c n =0 ³ ´µ ² 1 0 X n ( x ) e - x dx. Hence ² 1 0 f ( x ) e - x dx = c 0 ² 1 0 e - 2 x dx = c 0 1 - e - 2 2 . Thus c 0 = 2 e 2 e 2 - 1 ² 1 0 f ( x ) e - x dx.
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Unformatted text preview: In a similar way, we prove that c n = 1 κ n ² 1 f ( x ) X n ( x ) dx where κ n = ² 1 X 2 n ( x ) dx . This integral can be evaluated as we did in Exercise 15 or by straightforward computations, using the explicit formula for the X n ’s, as follows: ² 1 X 2 n ( x ) dx = ² 1 ( nπ cos nπx-sin nπx ) 2 dx = ² 1 ( n 2 π 2 cos 2 nπx + sin 2 nπx-2 nπ cos( nπx ) sin( nπx ) ) dx = =( n 2 π 2 ) / 2 ³ ´µ ¶ ² 1 n 2 π 2 cos 2 nπx dx + =1 / 2 ³ ´µ ¶ ² 1 sin 2 nπx dx-2 nπ =0 ³ ´µ ¶ ² 1 cos( nπx ) sin( nπx ) dx = n 2 π 2 + 1 2 ....
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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