Potential between two parallel planes

[noamriemer]

Hello ! I have an exam this Wednesday... your help will be appreciated...

There are two types of questions I can't figure out how to answer...
the first is this one:

Find the potential between two parallel planes. The first is in x=0, the second in x=L.
$\Phi(x=L)=\Phi_0 |sin(ky)|$
and
$\Phi (x=0)=0$

What I thought I should do is look for a solution of the sort:

$X(x)=Ae^{\sqrt{k^2+l^2}}+Be^{-\sqrt{k^2+l^2}}$
$Y(y)=Csin(ky)+Dcos(ky)$
$Z(z)=Esin(lz)+Fcos(lz)$

Then, I should start checking which of the terms should vanish.

But when I looked at the published solution, it was completely different:

$|sin ky|=\sum_{n=0}^{\infty}{A_ncos(2kny)}$

$\rightarrow |sin(ky)|=\frac{2}{\pi}+\sum{\frac{4}{\pi(1-4n^2)}}cos(2kny)$

So now we seek a solution for Laplas eq. this way:

$\Phi(x,y)=C_0x+\sum_{n=1}^\infty{C_n(x)cos(2kny)}$
and:
$\frac{d^2}{dx^2}C_n =(2kn)^2C_n$

I don't understand this solution at all.
Why should one expand this function, |sin(ky)| ?
and what is the last equation:
$\frac{d^2}{dx^2}C_n =(2kn)^2C_n$

Thank you so much!

 

[jambaugh]

Hello ! I have an exam this Wednesday... your help will be appreciated...

There are two types of questions I can't figure out how to answer...
the first is this one:

Find the potential between two parallel planes. The first is in x=0, the second in x=L.
$\Phi(x=L)=\Phi_0 |sin(ky)|$
and
$\Phi (x=0)=0$

What I thought I should do is look for a solution of the sort:

$X(x)=Ae^{\sqrt{k^2+l^2}}+Be^{-\sqrt{k^2+l^2}}$
$Y(y)=Csin(ky)+Dcos(ky)$
$Z(z)=Esin(lz)+Fcos(lz)$

Then, I should start checking which of the terms should vanish.

That isn't the most general solution to Laplace's equation (which with your factorization reverts to: $X^{''}/X + Y^{''}/Y + Z^{''}/Z = 0$:
$X'' = aX$, $Y''=bY$, $Z''=cZ$, $a+b+c=0$.
Your set of solutions doesn't cover the case where say $a=0, b+c=0$. You can try to enumerate every combination of cases but it is better to start with your boundary conditions for some guidance. You see that [itex]Y(y)[/tex] is going to be non-trivial and start there.<br /> <br /> It looks like the published solution simply expresses the expansion of Y solutions using coefficients which are functions of the other independent variable(s) and re-applies the Laplace equation.<br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> But when I looked at the published solution, it was completely different: <br /> <br /> $|sin ky|=\sum_{n=0}^{\infty}{A_ncos(2kny)}$<br /> <br /> $\rightarrow |sin(ky)|=\frac{2}{\pi}+\sum{\frac{4}{\pi(1-4n^2)}}cos(2kny)$<br /> <br /> So now we seek a solution for Laplas eq. this way:<br /> <br /> $\Phi(x,y)=C_0x+\sum_{n=1}^\infty{C_n(x)cos(2kny)}$<br /> and:<br /> $\frac{d^2}{dx^2}C_n =(2kn)^2C_n$<br /> <br /> I don't understand this solution at all. <br /> Why should one expand this function, |sin(ky)| ?<br /> and what is the last equation: <br /> $\frac{d^2}{dx^2}C_n =(2kn)^2C_n$<br /> <br /> Thank you so much! </div> </div> </blockquote>Firstly since the boundary conditions and original equation are symmetric under z translation the solution is independent of z.<br /> <br /> The text is expanding solutions of $Y^{''}=bY$ in terms of the cosine expansion of the boundary condition (using the fact that it is an even function of y and independent of z). (One might also have tried a hyperbolic cosine expansion if the boundary condition where even and asymptotically exponential. Cosh is the even component of exp.)<br /> <br /> By the linearity of Lap.Eqn. Each term in the expansion $\Phi(x,y) = C_0(x) + C_1(x)\cos(2ky) + \cdots$, must independently satisfy Laplace's eqn.<br /> <br /> You get that last equation for all terms even the first term where n=0 which is the exceptional case I mentioned above.<br /> <br /> Matching your attempt:<br /> [itex]\Phi(x,y) = \sum_{some\,a,b} X_a(x)Y_b(y),\quad a+b=0[/tex]<br /> $=\sum_n C_n(x)\cos(2kny)$<br /> with $b = -(2kn)^2$ and so $a = (2kn)^2$.[/itex][/itex]

 

[noamriemer]

Thank you for the detailed answer!