System of second order linear coupled pde with constant coefficient

1. Oct 11, 2012

galuoises

Someone know how to uncouple this system of pde?

Δu$_{1}$(x) + a u$_{1}$(x) + b u$_{2}$(x) =f(x)
Δu$_{2}$(x) + c u$_{1}$(x) + d u$_{2}$(x) =g(x)

a,b,c,d are constant.

I would like to find a solution in one, two, three dimension, possibily in terms of Green function...someone could help me, please?

2. Oct 13, 2012

Ben Niehoff

Let

$$y_1 = u_1 + \lambda_1 u_2, \qquad y_2 = u_1 + \lambda_2 u_2$$
where $\lambda_1, \lambda_2$ are constants. For the right choice of constants, the equations will separate when written in terms of $y_1, y_2$.

$\lambda_{1,2}$ will have to solve a quadratic equation that involves a, b, c, d, hence generically you will get two roots.

3. Oct 14, 2012

HallsofIvy

Staff Emeritus
Why do you refer to this as a "PDE" when you have only the single independent variable, x?

4. Oct 14, 2012

galuoises

Thank you so much Ben Niehoff!

Sorry for the notation for the variable x, HallsofIvy, I intended it is a vector
x$\equiv$(x,y,z)
and
Δ$\equiv$$\partial_{xx}+\partial_{yy}+\partial_{zz}$