Help with a PDE

Question

Recently I've come across the following partial differential equation for $p(x,y)$: $$ p(x,y)-A\left(\frac{\partial^2p(x,y)}{\partial x^2}+\frac{\partial^2p(x,y)}{\partial y^2}\right)+B=0 $$

where $A$ and $B$ are constant. As can be seen, the term multiplying $A$ is the Laplacian of the function. The boundary conditions for the equation are given by:

$$ p(1,y)=p(y,2)=0 $$

$$ \frac{\partial p(0,y)}{\partial x} = \frac{\partial p(x,0)}{\partial y} = 0 $$

My question is: Is it possible to find a general solution to it, so I don't have to adopt any numerical methods to solve it? I really appreciate any help you can provide. Thank you in advance!

Answer 1

Set $q(x, y) = p(x, y)+B$. Then,

$$ Δ q = \frac{\partial^2 q(x, y)}{\partial x^2}+\frac{\partial^2 q(x, y)}{\partial y^2} = \frac{\partial^2 p(x, y)}{\partial x^2}+\frac{\partial^2 p(x, y)}{\partial y^2} = Δ p, $$

and we have the following PDE for $q(x, y)$.

$$ \Delta q(x, y) = \frac{1}{A}q(x, y), $$

which can be solved, for example, by the separation of variables.

Answer 2

The first thing I would do is let $u(x,y)= p(x,y)- B$ so that $A\left(\frac{\partial^2 p}{\partial x^2}+ \frac{\partial^2 p}{\partial y^2}\right)= P- B$ becomes $A\left(\frac{\partial^2 u}{\partial x^2}+ \frac{\partial^2 u}{\partial y^2}\right)= u$.

Now I would try $u= X(x)Y(y)$ - that is, $u$ is the product of a function of $x$ only and a function of $y$ only. Then the equation becomes $Y\frac{d^2X}{dx^2}+ X\frac{d^2Y}{d^2y}= \frac{XY}{A}$.

Finally, divides both sides by $XY$ giving $\frac{1}{X}\frac{d^2X}{dx^2}+ \frac{1}{Y}\frac{d^2Y}{dy^2}= \frac{1}{A}$. This must be true for all $x$ and $y$. If we were to hold $y$ fixed while changing $x$, both $\frac{1}{Y}\frac{d^2Y}{dy^2}$ and and $\frac{1}{A}$ are constant so that $\frac{1}{X}\frac{d^2X}{dx^2}$ must be constant for all $x$. Similarly, if we hold $x$ constant while changing $y$, $\frac{1}{Y}\frac{d^2Y}{dy^2}$ must be constant for all $y$.

That is we must have $\frac{1}{X}\frac{d^2X}{dx^2}= P$ and $\frac{1}{Y}\frac{d^2Y}{dy^2}= Q$ where $P$ and $Q$ are constants such that $P+ Q= A$.

Those two ordinary differential equations, $\frac{d^2X}{dx^2}- PX= 0$ and $\frac{d^2Y}{dy^2}- QY= 0$ are easy to solve. They are linear homogeneous equations with constant coefficients with characteristic equations $r^2- P= 0$ and $r^2- Q= 0$ so characteristic roots $\pm\sqrt{P}$ and $\pm\sqrt{Q}$ so solutions$P(x)= Me^{\sqrt{Px}}+ Ne^{-\sqrt{Px}}$ and $Q(y)= Re^{\sqrt{Qy}}+ Se^{-\sqrt{-\sqrt{Qy}}}$ so that $P(x)Q(y)= (Me^{\sqrt{Px}}+ Ne^{-\sqrt{Px}})(Re^{\sqrt{(A- P)y}}+ Se^{-\sqrt{-\sqrt{(A- P)y}}})$.

How we would continue from that depends on the boundary conditions. If, for example, we are given a function of $x$ on the lines $(x, a)$ and $(x, b)$, and a function of $y$ on $(c, y)$ and $(d, y)$ then we make $P$ depend on an integer value and add all such functions. This leads to a solution in the form of a generalized Fourier series.