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On the last step of solving a three-dimensional Laplace equation,($\nabla^2T=0$) with BC(s) as $T(0,y,z) = T(L,y,z) = T_a$,

$T(x,0,z) = T(x,l,z) = T_a$,

$\frac{\partial T(x,y,0)}{\partial z} = p_c\bigg(e^{-b_cy/l}\left[T_{ci} + \frac{b_c}{l}\int_0^y e^{b_cs/l}T(x,s,z)ds\right] - T(x,y,0)\bigg)$

$\frac{\partial T(x,y,w)}{\partial z} = p_h\bigg(e^{-b_hx/L}\left[T_{hi} + \frac{b_h}{l}\int_0^x e^{b_hs/l}T(s,y,z)ds\right] - T(x,y,w)\bigg)$

On using a solution form

$T(x,y,z)=\sum_{m,n=1}^{\infty}(A_{nm}e^{\gamma z} + B_{nm}e^{-\gamma z})\cos(\frac{n\pi x}{L})\cos(\frac{m\pi y}{l}).$

I need to evaluate the coefficients $A_{nm}$ and $B_{nm}$ .

On applying the $z=0$ BC, I obtain the following equation:

$$ \frac{1}{p_c}\sum_{n,m=1}^\infty\sin(\frac{n\pi x}{L})\sin(\frac{m\pi y}{l})\gamma(A_{nm}-B_{nm}) = \delta e^{\frac{-b_c y}{l}} + P-Q+R-S \rightarrow (\mathrm{1}) $$

$P =\sum_{n,m=1}^\infty\frac{b_c^2}{b_c^2+(m\pi)^2}(A_{nm}+B_{nm})\sin(\frac{n\pi x}{L})\sin(\frac{m\pi y}{l})$

$Q =\sum_{n,m=1}^\infty \frac{b_c m\pi}{b_c^2+(m\pi)^2}(A_{nm}+B_{nm})\sin(\frac{n\pi x}{L})\cos(\frac{m\pi y}{l})$

$R = \sum_{n,m=1}^\infty \frac{b_c m\pi}{b_c^2+(m\pi)^2}(A_{nm}+B_{nm})e^{\frac{-b_c y}{l}}$

$S = \sum_{n,m=1}^\infty(A_{nm}+B_{nm})\sin(\frac{n\pi x}{L})\sin(\frac{m\pi y}{l})$

Here ,$\gamma^2 = (\frac{n\pi}{L})^2 + (\frac{m\pi}{l})^2$ and $p_c,\delta=(T_{ci} - T_a)$ are constants.

Obviously, there would be another equation in $A_{nm}$ and $B_{nm}$ from the $z=w$ BC.My question pertains to the fact as to how should I handle the $e^{\frac{-b_c y}{l}}$ term while calculating the value of the Fourier coefficients.

My Attempt added herein might make the situation moe clearer


Attempt

I can multiply both sides of the equation $\mathrm{(1)}$ with $\int_0^L\sin(\frac{k\pi x}{L})$ and $\int_0^l\sin(\frac{j\pi y}{l})$ and then use the principle of orthogonality to finally arrive at

For some $n=k$ and $m=j$, $\mathrm{(1)}$ becomes

$$ \frac{1}{p_c}\frac{L}{2}\frac{l}{2}\gamma(A_{kj}-B_{kj})=\delta e^{\frac{-b_c y}{l}}\int_0^L\sin(\frac{k\pi x}{L})\mathrm{d}x\int_0^l\sin(\frac{j\pi y}{l})\mathrm{d}y+\underbrace{\frac{b_c^2}{b_c^2+(j\pi)^2}(A_{kj}+B_{kj})\frac{L}{2}\frac{l}{2}}_{P} - \overbrace{0}^{Q} + \underbrace{\sum_{n,m=1}^\infty \frac{b_c m\pi}{b_c^2+(m\pi)^2}(A_{nm}+B_{nm})e^{\frac{-b_c y}{l}}\int_0^L\sin(\frac{k\pi x}{L})\mathrm{d}x\int_0^l\sin(\frac{j\pi y}{l})\mathrm{d}y}_{R} - \overbrace{(A_{kj}+B_{kj})\frac{L}{2}\frac{l}{2}}^{S} $$


Fourier expansion of $e^{\frac{-b_c y}{l}}$ in the interval $y \in [0,l]$

I expanded $e^{\frac{-b_c y}{l}}$

$$e^{\frac{-b_c y}{l}} = \frac{(1-e^{-b_c})}{b_c} + \sum_{n=1}^{\infty}\bigg[\frac{2b_c(1-e^{-b_c})}{(b_c)^2 + (2n\pi)^2}\cos\bigg(\frac{2n\pi y}{l}\bigg) + \frac{4n\pi(1-e^{-b_c})}{(b_c)^2 + (2n\pi)^2}\sin\bigg(\frac{2n\pi y}{l}\bigg)\bigg]$$

I guess, I need to substitute this in $(1)$ for $e^{\frac{-b_c y}{l}}$ ?


Any suggestion on how to handle the exponential term while evaluating the Fourier coefficients would be extremely helpful

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  • $\begingroup$ isn't this is a duplicate of mathoverflow.net/q/323684/11260 ? $\endgroup$ – Carlo Beenakker Feb 26 at 7:46
  • $\begingroup$ @CarloBeenakker Yes sir it might seem so. But I reached these equations with Dirichlet conditions and for the previous question I already posted two attempts. it seemed not useful to keep on adding more to the same problem making it more cumbersome. $\endgroup$ – Indrasis Mitra Feb 26 at 7:50

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