# Evaluating Fourier coefficients to complete a Laplace equation solution

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.

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

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

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

• en.wikipedia.org/wiki/… – Paul Sinclair Feb 17 at 0:09
• @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 ? – Indrasis Mitra Feb 17 at 1:36
• Ypu are integrating them from $0$ to $L$. What happens when you integrate those two terms over a this range? – Paul Sinclair Feb 17 at 1:49
• @PaulSinclair $\cos(\frac{m\pi x}{L})\sin(\frac{n\pi x}{L})$ .On integrating ove this range they just result in $0$ ? – Indrasis Mitra Feb 17 at 1:59
• Exactly. $\cos m\theta$ and $\sin n\theta$ are orthogonal to each other even when $m = n$. – Paul Sinclair Feb 17 at 2:05

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 $$0$$.

• 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. – Indrasis Mitra Feb 17 at 17:35
• I have added this information to the original question. Thankyou anyway. Your comments and answer really made a lot of muddy points clear. – Indrasis Mitra Feb 17 at 17:42
• 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 $0$.) – Paul Sinclair Feb 17 at 17:49
• Sorry, my bad. It is $\frac{\partial T}{\partial x}$ at $(x,y,-w)$. In words : the derivative of $T$ evaluated at $(x,y,-w)$ – Indrasis Mitra Feb 17 at 18:00
• Added another attempt to math.stackexchange.com/questions/3117489/… . the exponential is still what causes problems. have a look if you get time. – Indrasis Mitra Feb 23 at 16:27