Solving PDE: How to go about it?

[coverband]

Homework Statement

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

Homework Equations

How do I go about solving this PDE ??

The Attempt at a Solution

Please help !

 

[lanedance]

what is X a function of X = X(a,b,x)?

 

[HallsofIvy]

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

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

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

2ln(U)= (1- x^4)a+f(b, x)
\frac{dX}{db}= U= F(b, x)e^{((1-x^4)a)/2}
(F(b,x) is e^{f(b,x)} 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.

 

[coverband]

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

\frac{\partial^2X}{\partial a^2} + (X^4-1)\frac{\partial X}{\partial a} = 0Would really appreciate your input. Thanks

 

[lanedance]

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