I have been struggling with a problem for a long time. Solving a second order partial differential equation using Fourier half series in sine with the help of Mathematica gives me $$ \phi_{mine}(x,y)=-\sum_{k=1,3,5,...}^{\infty}\frac{8 a^2 G_{zy} \theta \left(\text{sech}\left(\frac{\pi b k}{2 a} \frac{\sqrt{G_{zx}}}{\sqrt{G_{zy}}} \right) \cosh \left(\frac{\pi k y}{a} \frac{\sqrt{G_{zx}}}{\sqrt{G_{zy}}}\right)-1\right)}{\pi ^3 k^3}\sin \left(\frac{\pi k x}{a}\right) $$ where $G_{zy}$, $G_{zx}$, $\theta$, $a$, and $b$ are constants.

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's a typo. 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 $k=1,3,5,...$ I've never seen this and when I plot them they give different curves so they don't seem to be equivalent.

Plot when k=1 Plot when k=3

As can be seen, $a$ determines the amplitude of the periodic functions so that $ \sin \left(\frac{\pi k x}{a}\right)$ is always $0$ at -1 and 1. $\cos\left(\frac{\pi k x}{a}\right) (-1)^{(k-1)/2}$ on the other hand flips from $1$ to $-1$ at $\pm a$.

The next step in the process involves working out the constant $\beta$, $$ \beta=\frac{2 \int_0^b \left(\int_0^a \phi (x,y) \, dx\right) \, dy}{G_{zx} a b^3} $$ 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. Is it possible to change $\sin \left(\frac{\pi k x}{a}\right)$ into $\cos\left(\frac{\pi k x}{a}\right) (-1)^{(k-1)/2}$?

  • $\begingroup$ No, that's wrong. I corrected it, thanks for pointing it out. $\endgroup$ – enea19 Apr 8 at 1:07

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