Solving this PDE?

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  • #1
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Homework Statement



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

Homework Equations



How do I go about solving this PDE ??

The Attempt at a Solution



Please help !!!!!!!!!!!
 

Answers and Replies

  • #2
lanedance
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what is X a function of X = X(a,b,x)?
 
  • #3
HallsofIvy
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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.
 
  • #4
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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
 
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  • #5
lanedance
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so now X = X(a) and it becomes a nonlinear ordinary differential equation, not a partial?
 

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