Question in finding Green's function

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  • #1
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Main Question or Discussion Point

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
[tex]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)[/tex]
[tex]\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) [/tex]
[tex]\hbox{Where}\;\lambda_{mn}=(\frac{m\pi}{a})^2+(\frac {n\pi}{b})^2[/tex]
Skipping a few steps:

[tex]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[/tex]

[tex]\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)[/tex]

For Poisson eq with zero boundary
[tex]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[/tex]
[tex]\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] [/tex]

[tex]=\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)}[/tex]


The book gave the next step:

[tex]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)}[/tex]

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
[tex]\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) [/tex]
 
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Answers and Replies

  • #2
DrClaude
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##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
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##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.
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
DrClaude
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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
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
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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
$$
\nabla^2 u=-\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}E_{ij}\lambda_{ij}\sin\left(\frac{i\pi}{a}x\right)\sin\left(\frac{j\pi}{b}y\right)
$$
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
DrClaude
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3,423
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.
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.)
 
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