Differential Equations Lecture Work Solutions 238

Differential Equations Lecture Work Solutions 238 - + L = x...

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b. For x 0 < 0then f ( x 0 ) = 1 and the solution is u ( x, t )=1 on x = x 0 +2 t or u ( x, t )=1 on x< 2 t For x 0 >L then f ( x 0 ) = 2 and the solution is u ( x, t )=2 on x> 4 t + L For 0 <x 0 <L then f ( x 0 )=1+ x 0 /L and the solution is u ( x, t )=1+ x 0 L on x =2 t 1+ x 0 L ± + x 0 That is x 0 = x 2 t 2 t + L L Thus the solution on this interval is u ( x, t )=1+ x 2 t 2 t + L = 2 t + L + x 2 t 2 t
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Unformatted text preview: + L = x + L 2 t + L Notice that u is continuous.-5 5 10 15 20 25 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 x=2t x=4t+L u=(x+L)/(2t+L) u=1 u=2 Figure 47: Solution for problem 4 238...
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This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.

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