System of second order linear coupled pde with constant coefficient

[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?

 

[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.

 

[HallsofIvy]

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

 

[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}