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Homework Help: Reduction of PDE to an ODE by means of linear change of variables

  1. Jan 12, 2009 #1
    1. The problem statement, all variables and given/known data

    So it's been a really long time since I've done any ode/linear algebra and would like some help with this problem.

    Derive the general solution of the given equation by using an appropriate change of variables

    2[tex]\delta[/tex]u/[tex]\delta[/tex]t + 3[tex]\delta[/tex]u/[tex]\delta[/tex]x = 0

    The thing that I'm really confused about is how do we decide what an appropriate change of variable is? Is there a general rule that I should go by?

    3. The attempt at a solution

    none yet because I'm not sure how to find the appropriate change of variable

    Thank you
  2. jcsd
  3. Jan 12, 2009 #2


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    (2*d/dt+3*d/dx)u=0. Suppose u is a function only of 3t-2x?
  4. Jan 12, 2009 #3
    I understand what that means but the book seems to want a different way to approach this. In their example, they used [tex]\delta[/tex]u/[tex]\delta[/tex]t + [tex]\delta[/tex]u/[tex]\delta[/tex]x = 0.

    Next, they made two linear change of variables equations i think...

    [tex]\alpha [/tex] = ax+bt and [tex]\beta[/tex] = cx + dt
    where a,b,c,d will be chosen appropriately. Then they used chain rule in 2 dimension giving:

    [tex]\delta[/tex]u/[tex]\delta[/tex]x=[tex]\delta[/tex]u/[tex]\delta[/tex][tex]\alpha[/tex]*[tex]\delta[/tex][tex]\alpha[/tex]/[tex]\delta[/tex]x + [tex]\delta[/tex]u/[tex]\delta[/tex][tex]\beta[/tex]*[tex]\delta[/tex][tex]\beta[/tex]/[tex]\delta[/tex]t

    and then they did the same for [tex]\delta[/tex]u/[tex]\delta[/tex]t

    which gives them (a+b)[tex]\delta[/tex]u/[tex]\delta[/tex][tex]\alpha[/tex]+(c+d)[tex]\delta[/tex]u/[tex]\delta[/tex][tex]\beta[/tex] = 0

    Then they assumed a =1, b=0, c=1, d=-1 which gives them:

    and they were able to find the general solution from there.

    So I guess what I meant is how were they able to determine what [tex]\alpha[/tex] and [tex]\beta[/tex] are?
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