Differential Equations Lecture Work Solutions 250

Differential Equations Lecture Work Solutions 250 - x = c t...

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3. a. The set of ODEs to solve is dx dt = c du dt =0 The characteristics are: x = x 0 + ct The solution of the second ODE is u ( x, t )=con s tan t= u ( x 0 , 0) = sin x 0 Substitute for x 0 ,weget u ( x, t )=s in ( x ct ) b. For x>c t the solution is u ( x, t )= f ( x ct ) But f ( x ) is de±ned only for positive values of the independent variable x , therefore f ( x ct ) cannot be used for x<c t . In this case ( x<c t ) we must use the condition u (0 ,t )= g ( t ) The characteristics for which x 0 < 0i sg ivenby x
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Unformatted text preview: x = c ( t − t ) and u (0 , t ) = g ( t ) = g t − x c -4-2 2 4 6 8-2-1 1 2 3 4 x t x - ct > 0 x - ct < 0 x-ct=0 u(x,0)=f(x) u(0,t)=g(t) Figure 48: Domain and characteristics for problem 3b The solution is therefore given by u ( x, t ) = f ( x − ct ) f o r x − ct > g t − x c for x − ct < 250...
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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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