How to get a converging solution for a second order PDE?

[enea19]

I have been struggling with a problem for a long time. I need to solve the second order partial differential equation
$$\frac{1}{G_{zx}}\frac{\partial ^2\phi (x,y)}{\partial^2 y}+\frac{1}{G_{zy}}\frac{\partial ^2\phi (x,y)}{\partial^2 x}=-2 \theta$$
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.

 

[pasmith]

I have been struggling with a problem for a long time. I need to solve the second order partial differential equation
$$\frac{1}{G_{zx}}\frac{\partial ^2\phi (x,y)}{\partial^2 y}+\frac{1}{G_{zy}}\frac{\partial ^2\phi (x,y)}{\partial^2 x}=-2 \theta$$
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,
$$\phi_{mine}(x,y)=-\sum_{k=1,3,5,...}^{\infty}\frac{8 a^2 G_{zy} \theta \left(\text{sech}\left(\frac{\pi b k}{2 a} \frac{\sqrt{G_{zx}}}{\sqrt{G_{zy}}} \right) \cosh \left(\frac{\pi k y}{a} \frac{\sqrt{G_{zx}}}{\sqrt{G_{zy}}}\right)-1\right)}{\pi ^3 k^3}\sin \left(\frac{\pi k x}{a}\right)$$

I don't know where the 2 in the denominator of the sech term comes from; it results in this solution not satisfying the boundary condition on y = \pm b. (Unless the boundary is actually at y = \pm b/2, in which case it is correct.)

Setting \phi = \sum_{n=1}^\infty Y_n(y) \sin(n\pi x/a) I find that Y_n = 0 for even n and <br /> Y_n&#039;&#039; - \frac{n^2 \pi^2 G_{zx}}{a^2 G_{zy}}Y_n = -\frac{8 \theta G_{zx}}{n\pi} for odd n. So far we agree.

The solution of this which satisfies the boundary condition on y = \pm b is <br /> Y_n = \frac{8\theta a^2 G_{zy}}{n^3 \pi^3} \left(1 - \cosh\left( \frac{n\pi G_{zx}^{1/2}}{aG_{zy}^{1/2}} y \right)\mathrm{sech}\left(\frac{n\pi G_{zx}^{1/2}}{aG_{zy}^{1/2}} b\right)\right)<br /> which is basically \phi_{mine} but without the erroneous 2 in the denominator of the sech factor.

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} has the same issue: It doesn't satisfy the boundary condition on y = \pm b because of the 2 in the denominator of the \cosh \left(\frac{b \pi k \mu}{2 a}\right) expression. But it does vanish at y = \pm b/2.

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

The domain in y suggests that the boundary is indeed at y = \pm \frac12 b rather than y = \pm b.

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.

If you apply the boundary condition at y = \pm b and integrate from -b to b then you end up summing terms of the form (b - \tanh(Cnb)/(Cn))/n^4 for some constant C, which will converge because you can bound it above by (b + 1/(Cn))/n^4 and below by (b - 1/(Cn))/n^4, both of which converge.

If you apply the boundary condition at y = \pm b/2 and integrate from -b/2 to b/2, then you'll get the same thing for a different value of C.

If you apply the boundary condition at y = b and integrate from -b/2 to b/2, then you get something like <br /> \frac{1}{n^4}\left(b - \frac{1}{Cn}\sinh(\tfrac12 Cnb)\mathrm{sech}(Cnb)\right) which should converge because the exponential terms are asymptotic to e^{-Cnb/2}. If you do it the other way around the exponential terms are asymptotic to e^{Cnb/2}, and the sum won't converge.

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?

That would mean that \tan(k \pi x/a) = (-1)^{(k-1)/2} for every x, which is not the case.

I've never seen this and when I plot them they give different curves so they don't seem to be equivalent.

Expanding in cosines rather than sines suggests that the boundary condition is that \partial \phi/\partial x should vanish at x = 0, a rather than \phi. That of course results in a different series expansion for -2\theta.

 

[enea19]

Thanks for your answer! Oh my, I realized that I wrote the wrong limits! I'm really sorry, you are right the limits for $y$ are indeed $\pm b/2$. Apologies for the confusion and thanks again for the answer.

If it's not too much bother, where do you think the $\cos\left(\frac{\pi k x}{a}\right) (-1)^{(k-1)/2}$ comes from?

 

[pasmith]

Thanks for your answer! Oh my, I realized that I wrote the wrong limits! I'm really sorry, you are right the limits for $y$ are indeed $\pm b/2$. Apologies for the confusion and thanks again for the answer.

If it's not too much bother, where do you think the $\cos\left(\frac{\pi k x}{a}\right) (-1)^{(k-1)/2}$ comes from?

I have no idea. Expanding as a series in cosines makes no sense, since they don't satisfy the given boundary conditions.