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PDE: Method of characteristics question

  1. Sep 28, 2009 #1
    The problem statement, all variables and given/known data
    x ut +ux =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 Lc={(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)?

    The attempt at a solution
    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=x2 +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
     
  2. jcsd
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