4
$\begingroup$

I want to solve the following partial differential equation:

$$2\frac{\partial^2u}{\partial x^2} - \frac{\partial ^2u}{\partial x\,\partial y} - \frac{\partial ^2u}{\partial y^2} = 0$$

It's hyperbolic, so I converted it to canonical form, using $E=x+y$ and $N=x-2y$

This lead me to the equation:

$$9\frac{\partial ^2u}{\partial E\,\partial N}=0$$

...which should be far easier to solve, but I just cannot get my head around it for some reason. Can someone please explain the method for solving this?

$\endgroup$
  • 1
    $\begingroup$ Try integrating $u$ twice, once in $E$ and in $N$ each. $\endgroup$ – Christopher A. Wong Apr 9 '13 at 10:21
0
$\begingroup$

The solution simply is

$$u(E,N) = f(E) + g(N)$$

Think about it: differentiating with respect to $E$ leaves $f'(E)$. Then differentiating with respect to $N$ gives zero. Works vice-versa too.

That means that

$$u(x,y) = f(x+y) + g(x-2 y)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.