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?
[tex] au_{xx} + bu_{yy} + c u_{xy} = u + k[/tex]
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 ([itex] C^2 - 4AB [/itex] 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 [tex]
\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
[/tex] This can be diagonalized to [tex]
\lambda_1 \frac{\partial^2 u}{\partial s^2} + \lambda_2 \frac{\partial^2 u}{\partial t^2}[/tex] where [itex]s[/itex] and [itex]t[/itex] are linear combinations of [itex]x[/itex] and [itex]y[/itex]. Setting [itex]\phi = u + k[/itex] now yields [tex]
\lambda_1 \frac{\partial^2 \phi}{\partial s^2} + \lambda_2 \frac{\partial^2 \phi}{\partial t^2} = \phi[/tex] which is separable.