Solving 2D Poisson problem with single series solution.

**yungman** · March 18th 2010, 11:02 AM

Conventional solution of Poission’s equation involve solution $u(x,y)= \sum_{n=1}^{\infty} \sum_{m=1}^{\infty}E_{mn} sin(\frac{m\pi}{a}x) sin(\frac{n\pi}{a}y)$

This is a two series solution which is tedious to solve.

The book PDE by Asmar suggested a method of solving Poisson problem with one series by lumping function of y into the coefficient $E_{mn}$ by using:

$u(x,y)= \sum_{m=1}^{\infty} E_m(y)sin(\frac{m\pi}{a}x)$ .AND

The book gave the final equation of:

$E_m(y) = (\frac{-1}{\frac{m\pi}{a} sinh(\frac{m\pi}{a}b)})[ h_1(y)\int_0^y h_2(s)b_m(s)ds \;\;+\;\; h_2(y)\int_y^b h_1(s)b_m(s)ds]$

where the associate homogeneous solution is:

$h_1(y) = sinh(\frac{m\pi}{a}(b-y)) \;\;and\;\; h_2(y) = sinh(\frac{m\pi}{a}y)$

This book claimed this is by using variation of parameters using h1 and h2 as y1 and y2 obtain from solving the associate homogeneous equation.

I use the standard variation of parameter and cannot get the same answer. Can someone point me to a web site to verify the book? I hate to say the book is wrong but I did triple verified and fail. I have not manage to find anything on this from 4 other text book nor on the web to even talk about single series solution.

Is it really important to use single series rather than two series because I have not problem doing in the convensional way using two series, it is very easy to understand. It is the book trying to be simple and jump steps that I don't agree with their formula all all.

I think $b_m(s)$ in should not be integrated like this in the definite integral. Because it really a function of y, not a function of (b-y)

I think the solution using variation of parameters should be:

$E_m(y) = (\frac{-1}{\frac{m\pi}{a} sinh(\frac{m\pi}{a}b)})[ h_1(y)\int h_2(y)b_m(y)dy \;\;+\;\; h_2(y)\int h_1(y)b_m(y)dy]$

$\Rightarrow E_m(y) = (\frac{-1}{\frac{m\pi}{a} sinh(\frac{m\pi}{a}b)})[ sinh(\frac{m\pi}{a}(b-y)) \int sinh(\frac{m\pi}{a}y) b_m(y)dy$

. $\;\;+\;\;sinh(\frac{m\pi}{a}y) \int sinh(\frac{m\pi}{a}(b-y)) b_m(y)dy]$

Please tell me whether I am correct, no more of the dummy variable "s".

Thanks

**shawsend** · March 19th 2010, 07:05 AM

It's annoying for me too Yung. There is a way out of this quagmire of course: Solve it for a particular case and then back-substitute the solution into the PDE and see if satisfies it. You can do this in Mathematica very nicely.

This is what I did which is inadequate: Let $b_m(y)=\frac{2}{a}\int_0^a f(x,y)\sin(m \pi x/a)dx$

Then using variation of parameters, I obtain for the y-part:

$E_m(y)=\frac{1}{2m\pi/a}\left(e^{m\pi y/a}\int e^{-m\pi s/a}b_m(s)ds-e^{-m\pi y/a}\int e^{m\pi s/a} b_m(s)ds\right)$

which is similar to what you posted but I do not understand how to set up the limits of integration and again, in order to confirm the answer, I would figure out how to set up the limits, then solve if for a particular IBVP, back-substitute the solution into the PDE and see if my solution or the book's solution satisfies it.

**yungman** · April 6th 2010, 12:06 PM

I know the original question is confusing. I simplify to the essence of my question. It is actually a question of integration with change of variable. Below is the simplified version where f(y) is part of the original question.

Given:
$f(y)=\int b_m(y)sinh(b-y)dy \;\;for\;\;0<x<b$

Solve for f(y)

Convension way is:

$f(y)=\int_0^y b_m(s) sinh(b-s)ds$

But the book gave:

$f(y)=\int_y^b b_m(s) sinh(s)ds$

The rest of the original equation is very straight forward, only this part. I don't see the connection.

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