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Solving this PDE?

  1. Jul 8, 2011 #1
    1. The problem statement, all variables and given/known data

    [tex] 2\frac{\partial^2X}{\partial a \partial b} + \frac{\partial X}{\partial a}(x^4-1) = 0 [/tex]

    2. Relevant equations

    How do I go about solving this PDE ??

    3. The attempt at a solution

    Please help !!!!!!!!!!!
     
  2. jcsd
  3. Jul 8, 2011 #2

    lanedance

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    Re: Pde

    what is X a function of X = X(a,b,x)?
     
  4. Jul 9, 2011 #3

    HallsofIvy

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    Re: Pde

    If you have written that correctly and X really is a function of the three variables x, a, and b, then let [itex]U= \partial X/\partial b[/itex].

    The problem becomes
    [tex]2\frac{\partial U}{\partial a}+ (x^4- 1)U= 0[/tex]

    which, since we have differentiation with respect to a only, is the same as
    [tex]2\frac{dU}{da}= (1- x^4)U[/tex]
    where we are treating x and b as constants. This is a separable equation:
    [tex]2\frac{dU}{U}= (1- x^4)da[/tex]

    [tex]2ln(U)= (1- x^4)a+f(b, x)[/tex]
    [tex]\frac{dX}{db}= U= F(b, x)e^{((1-x^4)a)/2}[/tex]
    (F(b,x) is [itex]e^{f(b,x)}[/itex] and is simply an arbitrary differentiable function of b and x.


    Now, just integrate again. Your "constant of integration" will be an arbitrary differentiable function of a and x.
     
  5. Jul 9, 2011 #4
    Re: Pde

    There was a mistake in what I originally posted. The PDE to be solved is simpler:

    [tex] \frac{\partial^2X}{\partial a^2} + (X^4-1)\frac{\partial X}{\partial a} = 0 [/tex]


    Would really appreciate your input. Thanks
     
    Last edited by a moderator: Jul 10, 2011
  6. Jul 9, 2011 #5

    lanedance

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    Re: Pde

    so now X = X(a) and it becomes a nonlinear ordinary differential equation, not a partial?
     
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