# Solving 2D Poisson problem with a single series

yungman
Solving 2D Poisson problem with a single series!!

Conventional solution of $$\nambla^2u(x,y)=f(x,y)$$ 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:

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

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 (1) 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:

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

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

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

Thanks

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## Answers and Replies

yungman

Anyone please? You don't have to have the answer, I just want to hear your opinions.

I can't even find anything on the web, it would be very helpful if you can just guide me where to look for the information.

Is it important to do this in single series? I have 4 other books and none mentioned this, in fact, they don't even have much on Poisson equation. I am studying advance electromagnetic, Poisson's equation is quite important for me.

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yungman

hosun

The dummy variable does not make any difference.I think your calculation is the same as that on your book.But the only difference is: that one on your book has already put the boundary conditions "u=0 at y=0 and b" into the two intergration constants arisen from the two integrations.
By I think it is better to work on your own way: use your formula, and put the boundary conditions into the constant,verify that they are in fact the same.

yungman

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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hosun

I get the same problem before when the first time I saw this formula.
Why $$f(y)=\int_y^b b_m(s) sinh(s)ds$$ is not trival..

It is because the term $$f(y)=\int_0^y b_m(s) sinh(b-s)ds$$ has been combined with the integration constant(or from the constant of homogenous solution) which equals to $$f(y)=\int_0^b -b_m(s) sinh(b-s)ds$$. It is better to derive it step by step in order to see how it comes out.

yungman

I think I might found the problem. I decided to move on because I was stuck for a while on this. I went ahead and start working on a problem and found out the formula used in the problemd is

$$\int_y^b b_m(s) sinh(b-s)ds$$

So far, I have put in use b_m(s)=s and worked out both ways, the solutions are -ve of each other and the constant is different.

I am still verifying this, I'll post the final result. I hate that the book have typos!!!! THis is the first typos so far!!!!