I have been struggling with a problem for a long time. I need to solve the second order partial differential equation
[math] \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[/math] [math][/math]
where [math]G_{zx}[/math], [math]G_{zy}[/math], [math]\theta[/math], [math]a[/math], and [math]b[/math] are constants and with BCs [math]\phi (0,y)=\phi (a,y)=0[/math] and [math]\phi (x,-b)=\phi (x,b)=0[/math]. The solution that I have sets [math]\phi (x,y)=\sum _{k=1,3,5\text{...}}^{\infty } Y(y) \sin \left(\frac{\pi k}{a}x\right)[/math] and expands [math]-2 \theta[/math] in a Fourier sine series in the interval between 0 and a so we end up with a second order differential equation,
[math]\frac{Y''(y)}{G_{zx}}-\frac{\pi ^2 k^2 Y(y)}{a^2 G_{zy}}=-\frac{8 \theta }{\pi k}[/math]
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,
[math]\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)[/math]
This is almost exactly what is written in the solution that I have, which is
[math]\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)[/math]
where [math]\mu=\sqrt{\frac{G_{zx}}{G_{zy}}}[/math]. [math]\phi_{sol}(x,y)[/math] is missing \theta but I suspect that it might be a typo.
I know that unfortunately [math]\phi_{mine}(x,y)[/math] is wrong because the next step in the process involves working out a constant [math]\beta[/math] where
[math]\beta=\frac{2 \int_{-b/2}^{b/2} \left(\int_0^a \phi (x,y) \, dx\right) \, dy}{G_{zx} a b^3}[/math]
When I work out [math]\beta[/math] I get that [math]\phi_{sol}(x,y)[/math] converges to a value while [math]\phi_{mine}(x,y)[/math] 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
[math]\sin \left(\frac{\pi k x}{a}\right)=\cos\left(\frac{\pi k x}{a}\right) (-1)^{(k-1)/2}[/math]
for [math]k=1,3,5,...[/math] Is it possible to change [math]\sin \left(\frac{\pi k x}{a}\right)[/math] into [math]\cos\left(\frac{\pi k x}{a}\right) (-1)^{(k-1)/2}[/math] when only odd values of [math]k[/math] are used? I've never seen this and when I plot them they give different curves so they don't seem to be equivalent.
[math] \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[/math] [math][/math]
where [math]G_{zx}[/math], [math]G_{zy}[/math], [math]\theta[/math], [math]a[/math], and [math]b[/math] are constants and with BCs [math]\phi (0,y)=\phi (a,y)=0[/math] and [math]\phi (x,-b)=\phi (x,b)=0[/math]. The solution that I have sets [math]\phi (x,y)=\sum _{k=1,3,5\text{...}}^{\infty } Y(y) \sin \left(\frac{\pi k}{a}x\right)[/math] and expands [math]-2 \theta[/math] in a Fourier sine series in the interval between 0 and a so we end up with a second order differential equation,
[math]\frac{Y''(y)}{G_{zx}}-\frac{\pi ^2 k^2 Y(y)}{a^2 G_{zy}}=-\frac{8 \theta }{\pi k}[/math]
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,
[math]\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)[/math]
This is almost exactly what is written in the solution that I have, which is
[math]\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)[/math]
where [math]\mu=\sqrt{\frac{G_{zx}}{G_{zy}}}[/math]. [math]\phi_{sol}(x,y)[/math] is missing \theta but I suspect that it might be a typo.
I know that unfortunately [math]\phi_{mine}(x,y)[/math] is wrong because the next step in the process involves working out a constant [math]\beta[/math] where
[math]\beta=\frac{2 \int_{-b/2}^{b/2} \left(\int_0^a \phi (x,y) \, dx\right) \, dy}{G_{zx} a b^3}[/math]
When I work out [math]\beta[/math] I get that [math]\phi_{sol}(x,y)[/math] converges to a value while [math]\phi_{mine}(x,y)[/math] 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
[math]\sin \left(\frac{\pi k x}{a}\right)=\cos\left(\frac{\pi k x}{a}\right) (-1)^{(k-1)/2}[/math]
for [math]k=1,3,5,...[/math] Is it possible to change [math]\sin \left(\frac{\pi k x}{a}\right)[/math] into [math]\cos\left(\frac{\pi k x}{a}\right) (-1)^{(k-1)/2}[/math] when only odd values of [math]k[/math] are used? I've never seen this and when I plot them they give different curves so they don't seem to be equivalent.