Chem Differential Eq HW Solutions Fall 2011 41

Chem Differential Eq HW Solutions Fall 2011 41 - e-n 2 t...

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Section 3.5 The One Dimensional Heat Equation 41 Solutions to Exercises 3.5 1. Multiply the solution in Example 1 by 78 100 to obtain u ( x, t )= 312 π ± k =0 e - (2 k +1) 2 t 2 k +1 sin(2 k +1) x. 5. We have u ( x, t )= ± n =1 b n e - ( ) 2 t sin( nπx ) , where b n =2 ² 1 0 x sin( nπx ) dx =2 ³ - x cos( nπx ) + sin( nπx ) n 2 π 2 ´ µ µ µ 1 0 = - 2 cos =2 ( - 1) n +1 . So u ( x, t )= 2 π ± n =1 ( - 1) n +1 e - ( ) 2 t sin( nπx ) n . 9. (a) The steady-state solution is a linear function that passes through the points (0 , 0) and (1 , 100). Thus, u ( x ) = 100 x . (b) The steady-state solution is a linear function that passes through the points (0 , 100) and (1 , 100). Thus, u ( x ) = 100. This is also obvious: If you keep both ends of the insulated bar at 100 degrees, the steady-state temperature will be 100 degrees. 13. We have u 1 ( x )= - 50 π x + 100. We use (13) and the formula from Exercise 10, and get (recall the Fourier coefficients of f from Exercise 3) u ( x, t )= - 50 π x + 100 + ± n =1 ³ 132 π sin( n π 2 ) n 2 - 100 2 - ( - 1) n ·´
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Unformatted text preview: e-n 2 t sin nx. 17. Fix t > 0 and consider the solution at time t = t : u ( x, t ) = ∞ ± n =1 b n sin nπ L xe-λ 2 n t . We will show that this series converges uniformly for all x (not just 0 ≤ x ≤ L ) by appealing to the Weierstrass M-test. For this purpose, it suffices to establish the following two inequalities: (a) µ µ b n sin nπ L xe-λ 2 n t µ µ ≤ M n for all x ; and (b) ∑ ∞ n =1 M n < ∞ . To establish (a), note that | b n | = µ µ µ µ 2 L ² L f ( x ) sin nπ L x dx µ µ µ µ ≤ 2 L ² L µ µ µ f ( x ) sin nπ L x µ µ µ dx (The absolute value of the integral is ≤ the integral of the absolute value.) ≤ 2 L ² L | f ( x ) | dx = A (because | sin u | ≤ 1 for all u ) ....
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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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