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?

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

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

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


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