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

Conventional solution of [tex]\nambla^2u(x,y)=f(x,y)[/tex] involve solution [tex]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)[/tex]

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 [tex]E_{mn}[/tex] by using:

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

The book gave the final equation of:

[tex]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][/tex] (1)

where the associate homogeneous solution is:

[tex]h_1(y) = sinh(\frac{m\pi}{a}(b-y)) \;\;and\;\; h_2(y) = sinh(\frac{m\pi}{a}y) [/tex]

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 [tex]b_m(s)[/tex] 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:

[tex]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][/tex]

[tex]\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][/tex]

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

Thanks

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