- 2

- 1

$$\frac{1}{G_{zx}}\frac{\partial ^2\phi (x,y)}{\partial^2 y}+\frac{1}{G_{zy}}\frac{\partial ^2\phi (x,y)}{\partial^2 x}=-2 \theta$$

where ##G_{zy}##, ##G_{zx}##, ##\theta##, ##a##, and ##b## are constants and with BCs ##\phi (0,y)=\phi (a,y)=0## and ##\phi (x,-b)=\phi (x,b)=0##. The solution that I have sets ##\phi (x,y)=\sum _{k=1,3,5\text{...}}^{\infty } Y(y) \sin \left(\frac{\pi k}{a}x\right)## and expands ##-2 \theta## in a Fourier sine series in the interval between 0 and a so we end up with a second order differential equation,

$$\frac{Y''(y)}{G_{zx}}-\frac{\pi ^2 k^2 Y(y)}{a^2 G_{zy}}=-\frac{8 \theta }{\pi k}$$

I tried several times to solve it by hand but end up making mistakes. Therefore I resorted to Mathematica and it gives me the following solution,

$$\phi_{mine}(x,y)=-\sum_{k=1,3,5,...}^{\infty}\frac{8 a^2 G_{zy} \theta \left(\text{sech}\left(\frac{\pi b k}{2 a} \frac{\sqrt{G_{zx}}}{\sqrt{G_{zy}}} \right) \cosh \left(\frac{\pi k y}{a} \frac{\sqrt{G_{zx}}}{\sqrt{G_{zy}}}\right)-1\right)}{\pi ^3 k^3}\sin \left(\frac{\pi k x}{a}\right)$$

This is almost exactly what is written in the solution that I have, which is

$$\phi_{sol}(x,y)=\frac{8}{\pi^3} G_{zy} a^2 \sum_{k=1,3,5,...}^{\infty}\frac{(-1)^{(k-1)/2}}{k^3}\left( 1-\frac{\cosh \left(\frac{\pi k \mu }{a}y\right)}{\cosh \left(\frac{b \pi k \mu}{2 a}\right)} \right)\cos \left(\frac{\pi k}{a}x\right)$$

where ##\mu=\sqrt{\frac{G_{zx}}{G_{zy}}}##.

##\phi_{sol}(x,y)## is missing ##\theta## but I suspect that it might be a typo. I know that ##\phi_{mine}(x,y)## is wrong because the next step in the process involves working out a constant ##\beta## where

$$\beta=\frac{2 \int_{-b/2}^{b/2} \left(\int_0^a \phi (x,y) \, dx\right) \, dy}{G_{zx} a b^3}$$

I get that ##\phi_{sol}(x,y)## converges to a value while ##\phi_{mine}(x,y)## goes to infinity. Therefore I'm doing something wrong but I'm not sure what. I've been trying to figure out the difference between my answer and the solution and all I can find is that somehow

$$\sin \left(\frac{\pi k x}{a}\right)=\cos\left(\frac{\pi k x}{a}\right) (-1)^{(k-1)/2}$$

for ##k=1,3,5,...## Is it possible to change ##\sin \left(\frac{\pi k x}{a}\right)## into ##\cos\left(\frac{\pi k x}{a}\right) (-1)^{(k-1)/2}## when only odd values of ##k## are used? I've never seen this and when I plot them they give different curves so they don't seem to be equivalent.