Partial differential equation by separation. How to multiply solution sets?

Question

I have the following PDE: $${ \partial^2 u \over \partial x^2} + { \partial^2 u \over \partial y^2} = u$$

I was instructed to use separation of variables, so: $ u = X(x) Y(y) $ and the separation constant is $ \lambda $

Separating leads me to: $$ {\ddot{X} \over -X } = {\ddot{Y} - Y \over Y} = \lambda $$

Doing $ {\ddot{X} \over -X } = \lambda$, the solution set of $X$ is: $$ A_1\cos\bigl(\sqrt{-\lambda}x \bigr) + A_2\sin\bigl(\sqrt{-\lambda}x\bigr) ; \lambda >0 $$ $$ A_1 + A_2x \thinspace ; \lambda=0 $$ $$A_1e^{\sqrt{\lambda}x} + A_2e^{-\sqrt{\lambda}x} ; \lambda<0$$

Doing $ {\ddot{Y} - Y \over Y } = \lambda$, the solution set of $Y$ is:

$$ B_1e^{\sqrt{\lambda+1}y} + B_2e^{-\sqrt{\lambda+1}y} ;\lambda>-1 $$ $$ B_1 + B_2y \thinspace ; \lambda=-1 $$ $$ B_1\cos\bigl(\sqrt{\lambda+1}y\bigr) + B_2\sin\bigl(\sqrt{\lambda+1}y\bigr) ;\lambda<-1 $$

Now, first of all, are those right? And, if they are, how do I multiply the solution sets to find $ u(x,y) $?

Answer 1

You correctly found some particular $X(x,\lambda)=e^{\pm(-\lambda)^{1/2}x}$ and $Y(y,\lambda)=e^{\pm(\lambda+1)^{1/2}y}$. This includes the complex cases involving sinusoidal terms. The product of two particular solutions is a particular solution of the PDE :

$$u(x,y,\lambda)=e^{\pm(-\lambda)^{1/2}x\:\pm(\lambda+1)^{1/2}y}$$ Any linear combination of these particular solutions with any coefficient is a solution of the PDE. As a consequence a more general form of solution can be expressed as : $$u(x,y)=\sum_{\forall\lambda}C_\lambda \:e^{\pm(-\lambda)^{1/2}x\:\pm(\lambda+1)^{1/2}y}$$ $C_\lambda$ are arbitrary coefficients. If one want eliminate some terms from the infinite sum, one set to zero the respective coefficients. So the above solution includes the cases of finite sums.

Generalisation on integral form : $$u(x,y)=\int_{\forall\lambda}f(\lambda) \:e^{\pm(-\lambda)^{1/2}x\:\pm(\lambda+1)^{1/2}y}d\lambda$$ $f(\lambda)$ is an arbitrary function. This function can be a piecewise function and can be continue or not insofar the integration is on the defined parts. So the above solution includes the cases of the sum of distinct functions.

One cannot go further in determining the coefficients (discret or functions) without knowing the boundary conditions. Of course this would be the most difficult part of the task. Fortunately other methods than separation of variables enable to solve more easily some PDE problems.