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While solving a PDE problem involving the Laplace equation in 3D, I arrive at the following summation relation when i substitute the only non-homogeneous boundary condition available

$$ \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}A_{n,m}\cos(\frac{n\pi x}{L})\cos(\frac{m\pi y}{l})\sqrt{\frac{n^2\pi^2}{L^2}+\frac{m^2\pi^2}{l^2}}\sinh\bigg(w\sqrt{\frac{n^2\pi^2}{L^2}+\frac{m^2\pi^2}{l^2}}\bigg) = p_h\bigg(\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}A_{n,m}\cos(\frac{m\pi y}{l})\cos(\frac{n\pi x}{L})\cosh\bigg(w\sqrt{\frac{n^2\pi^2}{L^2}+\frac{m^2\pi^2}{l^2}}\bigg) - \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{A_{n,m}b_h}{b_h^2 + n^2 \pi^2 }\cos(\frac{m\pi y}{l})\cosh\bigg(w\sqrt{\frac{n^2\pi^2}{L^2}+\frac{m^2\pi^2}{l^2}}\bigg)\bigg[b_h\cos(\frac{n\pi x}{L}) + n\pi\sin(\frac{n\pi x}{L}) \bigg]\bigg) $$

I need to find the Fourier coefficients here, denoted by $A_{n,m}$.

Known

As a starting point i know that if $\cos(\frac{m\pi x}{L})$ and $\cos(\frac{n\pi y}{l})$ is multiplied on the RHS and LHS of the whole equation i could use the known orthogonality of

$\int_{0}^{L} \cos(\frac{n\pi x}{L})\cos(\frac{m\pi y}{L})$ and $\int_{0}^{L} \cos(\frac{n\pi x}{l})\cos(\frac{m\pi y}{l})$

But the problem is, I will get a term like $\int_{0}^{L} \cos(\frac{m\pi x}{L})\sin(\frac{n\pi x}{L})$.

I do not know what to do with this term. Additionally, I have another doubt as to how the $A_{n,m}$ occurring on both sides of the equation are to be handled. Also, I have done 2D problems but how to handle this 3D case ? Is a double integration required ?

Note

The question asked has its origin as follows:-

From a solution of the form

$$ T(x,y,z)=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}A_{n,m}\cos(\frac{n\pi x}{L})\cos(\frac{m\pi y}{l})\cosh\left(\sqrt{\frac{n^2\pi^2}{L^2}+\frac{m^2\pi^2}{l^2}}z\right). $$

with bc(s) as:

$\frac{\partial T(0,y,z)}{\partial x}=\frac{\partial T(L,y,z)}{\partial x}=0 $

$\frac{\partial T(x,0,z)}{\partial y}=\frac{\partial T(x,l,z)}{\partial y}=0$

$$\frac{\partial T(x,y,-w)}{\partial z}=p_h\bigg( T(x,y,-w) - \frac{e^\frac{-b_h x}{L}b_h}{L}\int e^\frac{b_h x}{L}T\mathrm{d}x \bigg) $$

$ \frac{\partial T(x,y,0)}{\partial z} = 0. $

Extra information From the physical problem available at hand the following is too known:-

$$\frac{\partial T(0,y,-w)}{\partial z}=p_h\bigg( T(0,y,-w) - T_i\bigg) $$


Attempt

Multiplying both sides by $\cos(\frac{k\pi x}{L})$ and $\cos(\frac{j\pi y}{l})$ assuming that the series converges, converting the summation to an integration, first for $x$ and then for $y$, followed by using orthogonality.

$$ \sum\sum A_{n,m}\sqrt{()}\sinh(w\sqrt{()})\bigg(\frac{L}{2} \vee n=k\bigg)\bigg(\frac{l}{2} \vee m=j\bigg) = p_h\bigg[\sum\sum A_{n,m}\cosh(w\sqrt{()})\bigg(\frac{L}{2} \vee n=k\bigg)\bigg(\frac{l}{2} \vee m=j\bigg)- \sum\sum\frac{A_{n,m}b_h^2}{b_h^2 + n^2 \pi^2}\cosh(w\sqrt{()})\bigg(\frac{L}{2} \vee n=k\bigg)\bigg(\frac{l}{2} \vee m=j\bigg)\bigg] $$

leads to

$$ A_{k,j}\sqrt{()}\sinh(w\sqrt{()})\bigg(\frac{L}{2}\bigg)\bigg(\frac{l}{2} \bigg) = p_h\bigg[A_{k,j}\cosh(w\sqrt{()})\bigg(\frac{L}{2}\bigg)\bigg(\frac{l}{2}\bigg)-\frac{A_{k,j}b_h^2}{b_h^2 + n^2 \pi^2}\cosh(w\sqrt{()})\bigg(\frac{L}{2}\bigg)\bigg(\frac{l}{2}\bigg)\bigg] $$

Hence, $A_{k,j}$ needs to be evaluated. But still I have the problem of $A_{k,j}$ cancelling out from the resulting equation.

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  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – Paul Sinclair Feb 17 at 0:09
  • $\begingroup$ @PaulSinclair I followed the link you suggested. I have thought about writing $\cos(\frac{m\pi x}{L})\sin(\frac{n\pi x}{L})$ as $\frac{1}{2}\bigg(\sin(\frac{m\pi x}{L} + \frac{n\pi x}{L}) +\sin(\frac{m\pi x}{L} - \frac{n\pi x}{L})\bigg)$. But now the arguments of these $\sin$ functions are different from the rest of the equation. Can you elaborate a bit more ? $\endgroup$ – Indrasis Mitra Feb 17 at 1:36
  • $\begingroup$ Ypu are integrating them from $0$ to $L$. What happens when you integrate those two terms over a this range? $\endgroup$ – Paul Sinclair Feb 17 at 1:49
  • $\begingroup$ @PaulSinclair $\cos(\frac{m\pi x}{L})\sin(\frac{n\pi x}{L})$ .On integrating ove this range they just result in $0$ ? $\endgroup$ – Indrasis Mitra Feb 17 at 1:59
  • $\begingroup$ Exactly. $\cos m\theta$ and $\sin n\theta$ are orthogonal to each other even when $m = n$. $\endgroup$ – Paul Sinclair Feb 17 at 2:05
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I admit that I had failed to check this at the start: you have $A_{m,n}$ on both sides. One solution to your original equation is just $A_{m,n} = 0$ for all $m,n$. And in fact, that is the only solution.

To see this, multiply by $\sin\left(\frac{k\pi x}L\right)$ and take the $\int_0^{2L}\ \ dx$ integral of both sides (note the $2L$ - you have to integrate over a full period, and that is $2L$, not $L$). Every term with $\cos\left(\frac{n\pi x}L\right)$ in it becomes $0$. And the only $\sin\left(\frac{n\pi x}L\right)$ term that survives is $n = k$. Now, $\int_0^{2L}\sin^2\left(\frac{k\pi x}L\right)\,dx = L$, so the result is

$$0 = -p_h\left[\sum_{m=1}^{\infty}\frac{A_{k,m}b_h}{b_h^2 + k^2 \pi^2 }\cos(\frac{m\pi y}{l})\cosh\left(w\sqrt{\frac{k^2\pi^2}{L^2}+\frac{m^2\pi^2}{l^2}}\right)k\pi L \right]$$

Then we multiply through by $\cos\left(\frac{j\pi y}l\right)$ and take the $\int_0^{2l}\ \ dy$ integral, which similarly leaves us with

$$0 =-p_h\left[\frac{A_{k,j}b_h}{b_h^2 + k^2 \pi^2 }l\cosh\left(w\sqrt{\frac{k^2\pi^2}{L^2}+\frac{j^2\pi^2}{l^2}}\right)k\pi L \right]$$ $\cosh$ is never $0$. If we assume that $p_h, b_h, L$, and $l$ are all non-zero, this solves to $A_{k,j} = 0$.

Either you have an error in deriving the equation posted, or else your solution is $0$.

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  • $\begingroup$ Thankyou Paul. I will recheck my derivations. Actually the only non-homogeneous BC in the problem (if you see the note in the question) has $A_{m,n}$ in all the terms which is what causes the problem. Although i do have an extra information which tells that at $(0,y,-w)$ (instead of $(x,y,-w)$ in the original problem) the bc is $\frac{\partial T(0,y,-w)}{\partial x} = p_h(T(0,y,-w)-T_i)$ where $T_i$ is a constant.This does remove the $A_{m,n}$ cancelling out problem. But now this reduces the bc to an edge type condition in a 3D problem. $\endgroup$ – Indrasis Mitra Feb 17 at 17:35
  • $\begingroup$ I have added this information to the original question. Thankyou anyway. Your comments and answer really made a lot of muddy points clear. $\endgroup$ – Indrasis Mitra Feb 17 at 17:42
  • $\begingroup$ By $\frac{\partial T(x,y,-w)}{\partial z}$, do you mean $\frac{\partial T}{\partial z}(x,y,-w)$, or are any or all of $x, y, w$ supposed to be a function of $z$? (This is an order of operations question - the way you wrote it, T is evaluated at (x,y,-w) first, then the result is differentiated by $z$. But unless the other variables are functions of $z$, the differentiation will be $0$.) $\endgroup$ – Paul Sinclair Feb 17 at 17:49
  • $\begingroup$ Sorry, my bad. It is $\frac{\partial T}{\partial x}$ at $(x,y,-w)$. In words : the derivative of $T$ evaluated at $(x,y,-w)$ $\endgroup$ – Indrasis Mitra Feb 17 at 18:00
  • $\begingroup$ Added another attempt to math.stackexchange.com/questions/3117489/… . the exponential is still what causes problems. have a look if you get time. $\endgroup$ – Indrasis Mitra Feb 23 at 16:27

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