0
$\begingroup$

I need to solve the following BVP:

$\Delta u - 1/\delta * u = R(x,y)$,

where $\Delta$ is the laplacian operator. and boundary conditions:

$u(0,y)=u(L,y)=u(x,0)=u(x,L)=0$

where $L=1$, $\delta=0.01$, and $R(x,y)=1$.

I understand the basic concept of separation of variables, i.e. rewriting $u(x,y)$ as the product of some function of $x, X(x)$, and some function of $y, Y(y)$. Then by plugging $u(x,y)=X(x)Y(y)$ into the PDE, obtaining two ODEs that are much easier to solve. My solution started as...

$X''Y+XY''-1/\delta*XY=0$

$X''/X=(1/\delta*Y-Y'')/Y=-\lambda^2$

$X(x) = Asin\lambda x + Bcos\lambda y$

$Y(y) = C_1 e^{\sqrt{1/\delta+\lambda^2}y} + C_2 e^{-\sqrt{1/\delta+\lambda^2}y}$

eventually...

$u(x,y) = \sum{A_n(e^{\sqrt{1/\delta+\lambda^2}y} + e^{-\sqrt{1/\delta+\lambda^2}y})sin{(n \pi x)}}$

However, this was merely for the homogeneous case, not the BVP as given. And in addition, it seemed counter intuitive to me that the functions $X(x)$ and $Y(y)$ should be of different forms. Why would $X(x)$ be composed of sinusoidal functions and $Y(y)$ be compased of exponential terms if $u(x,y)$ is taken to the second partial derivative in both the x and y directions?

$\endgroup$
0
+100
$\begingroup$

I am a little suspicious of your solution to the homogeneous problem, since $u(x, y) \equiv 0$ is a solution to this problem too (with the boundary conditions) --- is there a reason in particular that this is not the unique solution to your problem? Have you actually computed the $A_n$?

As for your harder question: how to solve the inhomogeneous problem: this requires the use of Green's functions as well as the use of a Sommerfeld radiation condition As seen here. This is almost too much to explain here, since Green's functions usually occupy at least a quarter, if not three of applied math graduate school.

A good book for practical use of Green's function is Stakgold - this is what I learned from.

$\endgroup$
0
$\begingroup$

My application of separation of variables was incorrect:

$\Delta u - 1/\delta * u = R(x,y)$,

Consider the homogeneous case:

$X''Y+XY''-1/\delta*XY=0$

$\frac{X''Y}{XY}+\frac{XY''}{XY}-1/\delta*\frac{XY}{XY}=0$

$\frac{X''}{X}+\frac{Y''}{Y}+1/\delta=0$

Because X is only a function of x and Y is only a function of y, the only way for the above to be true is if $\frac{X''}{X}$ and $\frac{Y''}{Y}$ are constants.

$\frac{X''}{X}=-\lambda^2$

$\frac{Y''}{Y}=-\gamma^2$

$X(x) = Asin\lambda x + Bcos\lambda y$

$X(x) = Csin\gamma x + Dcos\gamma y$

eventually, given the boundary conditions, the general solution is...

$u(x,y) = \sum{A_n sin(n\pi x) sin(n\pi y)}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.