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Reducing a PDE to an ODE Using a Change of Coordinates

  1. Aug 31, 2013 #1
    I've been studying Walter A. Strauss' Partial Differential Equations, 2nd edition in an attempt to prepare for my upcoming class on Partial Differential Equations but this problem has me stumped. I feel like it should be fairly simple, but I just can't get it.

    10. Solve ##u_{x} + u_{y} + u = e^{x+2y}## with ##u(x,0) = 0##. (Partial Differential Equations, 2nd ed. by Walter A. Strauss, pg. 10)

    This section features only a few exercises and I assume they are to be solved using the methods presented in this section (primarily because they appear so early in the text). This section introduces two types of equations, the constant coefficient equation (which the above question is an example of) and the variable coefficient equation.

    To solve the constant coefficient equation Strauss introduces the Coordinate Method. In general, the equation ##au_{x} + bu_{y} = 0## can be simplified by using the substitutions ##x' = ax + by## and ##y' = bx - ay## (note: these are not the derivatives of x and y, but are simply new variables - I'm sticking with the notation used in the text). By using this change of variables the equation ##au_{x} + bu_{y}## is reduced to ##(a^{2} + b^{2})u_{x'}##.

    If I apply this same method to the above exercise, I find:

    ##u_{x} + u_{y} + u = e^{x+2y}##
    ##2u_{x'} + u = ???##

    My difficulty lies in writing ##e^{x+2y}## in terms of only ##x'## and ##y'##. I feel like I'm missing something fairly obvious but can't just give up on it.

    Otherwise, I have reduced this to an ODE solvable by using the method of integrating factors.

    Any help would be greatly appreciated! Thanks!
     
  2. jcsd
  3. Aug 31, 2013 #2
    Just going off of what you have it seems that you have taken a=b=1.

    So we have the equations:

    x'= x + y
    and
    y'= x - y

    from here we can solve for x and 2y

    x= (x' + y')/2
    and
    2y= x' - y'

    so the RHS simplifies to

    exp[(x' + y')/2 + x' - y']= exp[(3x'-y')/2]

    which may be beneficial to you to write as

    exp[(3x'-y')/2] = exp[3x'/2]*exp[-y'/2]

    At this point I will leave the problem to you, mainly because I am not sure how the change of variables will help you in this case.

    But best of luck.

    Oh, and you should have one more initial condition, otherwise your solution will just be the general answer.
     
  4. Aug 31, 2013 #3
    Thank you very much! I'm not sure why I was unable to solve for x' and y' - I'll blame it on being tired ;)

    I am only interested in finding the general solution so all the information provided is enough.
     
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