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PDE: Method of characteristics question
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[QUOTE="mike1111, post: 2368765, member: 194874"] [b]Homework Statement [/b] x u[SUB]t[/SUB] +u[SUB]x[/SUB] =0 intial condition u(x,0)=f(x) 1. Find the characteristics curves 2. What area of the xt-plane do u expect a solution 3. Find solution when f(x)=cos x 4.Now u(x,0)=f(x) (again), Find the level curves of u i.e for each c find the set L[SUB]c[/SUB]={(x,t):u(x,t)=f(c)} 5. Show there is not solution for u(x,0)=sin x 6. For what function is there a solution for u(x,0)=f(x), Then what the soution for u(x,y)? [b]The attempt at a solution[/b] Can anyone check my solution to tell me if this is right and advise me on how to do this correctly. I think it wrong leading to incorrect answers everywhere else. 1. dt/dx = x y=x[SUP]2[/SUP] +C 2. only on the x=0 since it can't apss throught the charactistics curve more than once 3. From initial x(0,s) =s x(k,s) =s+k t(0,s)=0 t(k,s) =xk u(0,s)= cos s u(k,s) =cos s u= cos (x/(x-y)) 4. Not sure what 4 is asking can anyone head point me in the right direction? 5. I dont see how this is any different from the cos equation in part 3, I end up getting u= sin(x/(x-y)) by the same method, but there not meant to be a solution 6. Any help is appreciated [/QUOTE]
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PDE: Method of characteristics question
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