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

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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

Homework Helper
1,735
408
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 [itex]y = \pm b[/itex]. (Unless the boundary is actually at [itex]y = \pm b/2[/itex], in which case it is correct.)

Setting [itex]\phi = \sum_{n=1}^\infty Y_n(y) \sin(n\pi x/a)[/itex] I find that [itex]Y_n = 0[/itex] for even [itex]n[/itex] and [tex]
Y_n'' - \frac{n^2 \pi^2 G_{zx}}{a^2 G_{zy}}Y_n = -\frac{8 \theta G_{zx}}{n\pi}[/tex] for odd [itex]n[/itex]. So far we agree.

The solution of this which satisfies the boundary condition on [itex]y = \pm b[/itex] is [tex]
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)
[/tex] which is basically [itex]\phi_{mine}[/itex] 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}}}##.
[itex]\phi_{sol}[/itex] has the same issue: It doesn't satisfy the boundary condition on [itex]y = \pm b[/itex] because of the 2 in the denominator of the [itex]\cosh \left(\frac{b \pi k \mu}{2 a}\right)[/itex] expression. But it does vanish at [itex]y = \pm b/2[/itex].

##\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 [itex]y[/itex] suggests that the boundary is indeed at [itex]y = \pm \frac12 b[/itex] rather than [itex]y = \pm b[/itex].

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

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

If you apply the boundary condition at [itex]y = b[/itex] and integrate from [itex]-b/2[/itex] to [itex]b/2[/itex], then you get something like [tex]
\frac{1}{n^4}\left(b - \frac{1}{Cn}\sinh(\tfrac12 Cnb)\mathrm{sech}(Cnb)\right)[/tex] which should converge because the exponential terms are asymptotic to [itex]e^{-Cnb/2}[/itex]. If you do it the other way around the exponential terms are asymptotic to [itex]e^{Cnb/2}[/itex], 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 [itex]\tan(k \pi x/a) = (-1)^{(k-1)/2}[/itex] for every [itex]x[/itex], 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 [itex]\partial \phi/\partial x[/itex] should vanish at [itex]x = 0, a[/itex] rather than [itex]\phi[/itex]. That of course results in a different series expansion for [itex]-2\theta[/itex].
 
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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?
 
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pasmith

Homework Helper
1,735
408
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.
 
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