Separation of variables possible in this problem?

[DuckAmuck]

TL;DR Summary

Second order linear partial differential equation

Is it possible to use separation of variables on this equation?
$$au_{xx} + bu_{yy} + c u_{xy} = u + k$$
Where u is a function of x and y, abck are constant.
I tried the u(x,y) = X(x)Y(y) type of separation but I think something more clever is needed.

Thank you.

 

[jasonRF]

Yes, something more clever will be required because presence of the $u_{xy}$ term as well as the nonhomogeneous term $k$ mean that straightforward separation of variables will not work. Getting rid of $k$ is easy: let $u(x,y) = f(x,y) - k$ and the equation for $f$ is homogeneous. Dealing with the $u_{xy}$ term isn't so easy, but sometimes a trick applies. First try separation of variables so you have an equation with terms such as $\frac{X^{\prime\prime}(x)}{X(x)}$, etc. Then, differentiate that equation with respect to $y$; the resulting equation might be separable.

Of course, even if the equation is separable you still need separable boundary conditions along curves of constant $x$ and $y$ in order for separation of variables to apply to your problem.

jason

 

[Dr Transport]

By looking at the discriminant ($C^2 - 4AB$ in this case) you can make a variable substitution and get a separable equation.

 

[pasmith]

Yes, after making a change of variables.

The left hand side is $$\begin{pmatrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} \end{pmatrix} \begin{pmatrix} a & \frac12 b \\ \frac12 b & c \end{pmatrix} \begin{pmatrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \end{pmatrix} u$$ This can be diagonalized to $$\lambda_1 \frac{\partial^2 u}{\partial s^2} + \lambda_2 \frac{\partial^2 u}{\partial t^2}$$ where $s$ and $t$ are linear combinations of $x$ and $y$. Setting $\phi = u + k$ now yields $$\lambda_1 \frac{\partial^2 \phi}{\partial s^2} + \lambda_2 \frac{\partial^2 \phi}{\partial t^2} = \phi$$ which is separable.