# Solving this PDE?

## 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

lanedance
Homework Helper

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

HallsofIvy
Homework Helper

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.

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

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

Would really appreciate your input. Thanks

Last edited by a moderator:
lanedance
Homework Helper

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