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Homework Help: Help me with difficult differential equation

  1. Jun 1, 2012 #1
    Believe me or not, ive just received a mail from my differential equations teacher asking me for help to solve this DE in wolphram alpha or maple but i think im doing something wrong because they wont solve it.

    It seems to be a

    Partial (completely) non linear diferential equation:

    u(x,y) = 2a2(ux)2 - 2bux-c2uy

    Thank you for your help guys, you are awesome ;)
  2. jcsd
  3. Jun 2, 2012 #2
    Well, there are many solutions. In order to get a unique solution you need to first specify what the values of u will be over a "non-characteristic" curve.

    Example. Find the solution if u(x,0) = g(x).

    We use the method of characteristics and end up with a bunch of equations:

    1. [itex] x = x_0 + 2bt + 4a^2g'(x_0)(e^{-t}-1) [/itex], and
    2. [itex] y = c^2t [/itex]

    You have to solve those equations for t and x_0 in terms of x and y. Then you substitute into
    $$ u = 2a^2g'(x_0)^2e^{-2t} - (2bg'(x_0)+c^2h(x_0))e^{-t}.$$
    Here, h(x) satisfies the equation [itex] g(x)+c^2h(x)+2bg'(x)-2a^2g'(x)^2=0.[/itex]

    Crystal clear, right?

    Ok, I'll do a special case so you can at least take home one bona fide solution. Lets suppose the initial condition is u(x,0)= x. So [itex] g(x)=x,\, g'(x)=1,\, h(x)=(2a^2-2b-x)/c^2.[/itex]

    From equation 2, we find that t = y/c^2. Substituting that into equation 1, we find [itex] x-x_0 = (2b/c^2)y +4a^2(e^{-y/c^2} - 1) [/itex]. Therefore,
    $$x_0 = x - (2b/c^2)y -4a^2(e^{-y/c^2} - 1).$$
    So we can plug in our expressions for t and x_0 into the formula for u and obtain the solution.

    $$ u(x,y) = (x-(2b/c^2)y-4a^2(e^{-y/c^2}-1)-2a^2)e^{-y/c^2} +2a^2e^{-2y/c^2}.$$

    [EDIT] I found an error and corrected it, but there might be (probably are) other errors. You'd have to plug that back into the original equation to see.
    Last edited: Jun 2, 2012
  4. Jun 2, 2012 #3
    The reason Wolfram alpha did not work is that it solves Ordinary Differential equations. This is a partial differential equation. As such is it more complex. If you want to refer your teacher to a reference on this, here is a link that explains the theory (but does not offer any concrete examples that I noticed):

  5. Jun 6, 2012 #4
    Well it worked! You are awesome! Thank you a lot.
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