# Question in finding Green's function

1. Sep 20, 2013

### yungman

Consider $\nabla^2 u(x,y)=f(x,y)$ in rectangular region bounded by (0,0),(0,b),(a,b)(a,0). And $u(x,y)=0$ on the boundary. Find Green's function $G(x,y,x_0,y_0)$.

For Poisson's eq, let
$$u(x_0,y_0)=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}E_{mn}\sin\left(\frac{m\pi}{a}x_0\right)\sin\left(\frac{n\pi}{b}y_0\right)$$
$$\Rightarrow\;\nabla^2 u=-\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}E_{mn}\lambda_{mn}\sin\left(\frac{m\pi}{a}x\right)\sin\left(\frac{n\pi}{b}y\right)$$
$$\hbox{Where}\;\lambda_{mn}=(\frac{m\pi}{a})^2+(\frac {n\pi}{b})^2$$
Skipping a few steps:

$$E_{mn}=-\frac{4}{ab\lambda_{mn}}\int_0^a\int_0^b \nabla^2 u\;\sin\left(\frac{m\pi}{a}x\right)\sin\left(\frac{n\pi}{b}y\right)\;dydx$$

$$\Rightarrow\;u(x_0,y_0)=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}E_{mn} \sin\left(\frac{m\pi}{a}x_0\right) \sin\left(\frac{n\pi}{b}y_0\right)= \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \left[- \frac{4}{ab\lambda_{mn}}\int_0^b\int_0^b \nabla^2 u\;\sin\left(\frac{m\pi}{a}x\right)\sin\left(\frac{m\pi}{a}x\right) \;dydx \right] \sin\left(\frac{m\pi}{a}x_0\right)\;\sin\left(\frac{n\pi}{b}y_0\right)$$

For Poisson eq with zero boundary
$$u(x_0,y_0)=\frac{1}{2\pi}\int_0^a\int_0^b\; \nabla^2 u\;G(x,y,x_0,y_0)\;dydx$$
$$\Rightarrow\;u(x_0,y_0)=\frac{1}{2\pi}\int_0^a\int_0^b\; \nabla^2 u\;G(x,y,x_0,y_0)\;dydx= \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \left[\left(-\frac{4}{ab\lambda_{mn}}\int_0^b\int_0^a \nabla^2 u\;\sin\left(\frac{m\pi}{a}x\right)\sin\left(\frac{n\pi}{b}y\right) \;dydx\right) \sin\left(\frac{m\pi}{a}x_0\right)\;\sin\left(\frac{n\pi}{b}y_0\right) \right]$$

$$=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \int_0^b \int_0^a \nabla^2 u\;\frac{-4}{ab\lambda_{mn}}\sin\left(\frac{m\pi}{a}x\right)\sin\left(\frac{n\pi}{b}y\right) \sin\left(\frac{m\pi}{a}x_0\right)\;\sin\left(\frac{n\pi}{b}y_0\right)dydx\; \hbox{ (1)}$$

The book gave the next step:

$$u(x_0,y_0)=\int_0^a\int_0^b\; \nabla^2u \left[ \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}-\frac{4}{ab\lambda_{mn}}\sin\left(\frac{m\pi}{a}x\right)\sin\left(\frac{n\pi}{b}y\right)\sin\left(\frac{m\pi}{a}x_0\right)\;\sin\left(\frac{n\pi}{b}y_0\right)\right] \;dydx \;\hbox{ (2)}$$

Compare (1) and (2) above, How can you move the $\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}$ inside the integral and pass $\nabla^2u$ where
$$\nabla^2 u=-\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}E_{mn}\lambda_{mn}\sin\left(\frac{m\pi}{a}x\right)\sin\left(\frac{n\pi}{b}y\right)$$

Last edited by a moderator: Sep 21, 2013
2. Sep 21, 2013

### Staff: Mentor

$m$ and $n$ are dummy indices. The two sums, i.e., the one in the equation of $\nabla^2 u$ and the one in the equation for $u(x_0,y_0)$ are distinct. Changing $m$ and $n$ to $m'$ and $n'$ or $k$ and $l$ in the equation for $\nabla^2 u$ will not change the result.

Therefore, in the equation for $u(x_0,y_0)$, $\nabla^2 u$ is independent of $m,n$ and can be taken outside the summation.

3. Sep 21, 2013

### yungman

But $E_{mn}$ and $\lambda_{mn}$ is dependent on $m,n$.

as$\nabla^2 u=-\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}E_{mn}\lambda_{mn}\sin\left(\frac{m\pi}{a}x\right)\sin\left(\frac{n\pi}{b}y\right)$
Thanks

4. Sep 21, 2013

### Staff: Mentor

And that is why there is a sum over $m$ and $n$ in there. But $\nabla^2 u$ is a function of $x$ and $y$ only. If you are still confused, just write
$$\nabla^2 u=-\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}E_{ij}\lambda_{ij}\sin\left(\frac{i\pi}{a}x\right)\sin\left(\frac{j\pi}{b}y\right)$$

5. Sep 21, 2013

### yungman

Thanks
So what you are saying

$\nabla^2 u=-\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}E_{ij}\lambda_{ij}\sin\left(\frac{i\pi}{a}x\right)\sin\left(\frac{j\pi}{b}y\right)$

is just one big lump totally independent to the summation of (m,n). So far as the whole function, $\nabla^2u$ is a constant.

6. Sep 21, 2013

### Staff: Mentor

Independent of the indices of the summation, yes, but still dependent on $x$ and $y$. (It is the Laplacian of a function $u(x,y)$, after all.)