Boundary Value Problem

MathematicalPhysics

I need some help starting off on this question.

Electrostatic potential [tex]V (x,y)[/tex] in the channel [tex]- \infty < x < \infty, 0 \leq y \leq a[/tex] satisfies the Laplace Equation

[tex]\frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2}= 0[/tex]

the wall [tex]y = 0[/tex] is earthed so that

[tex]V (x,0) = 0[/tex]

while the potential on the wall [tex]y = a[/tex]

[tex]V (x,a) = V_0 \cos{kx}[/tex] where [tex]V_0 , k[/tex] are positive constants.

By seeking a soln of an appropriate form, find [tex]V (x,y)[/tex] in the channel.

arildno

When in doubt, try separation of variables!
V(x,y)=F(x)G(y).
Inserting this into Laplace, yields, by rearrangement:
[tex]\frac{G''(y)}{G(y)}=-\frac{F''(x)}{F(x)}[/tex]

MathematicalPhysics

Thanks, I was trying to get it in the form [tex]V (x,y) = F (a) V_0 \cos{kx}[/tex] then sub the values into the Laplace eqn. Is that going about it the wrong way?

arildno

You DO mean (in my notation!)
[tex]V(x,y)=G(y)V_{0}\cos(kx)[/tex]???
If so, then it will work.
Note that the separation of variables method in this case implies:
[tex]G''(y)=k^{2}G(y)[/tex]

MathematicalPhysics

Hmm im not really keeping up with you here sorry. How can I find things out about F(x) and G(y) (in your notation) when we have just introduced them?

--------edit----------

oh yes sorry im with you, then differentiate V (x,y) = G(y) V_0 cos (kx) to get the different bits to go in the Laplace eqn? ahh!

arildno

Your lightbulb switched on in your edit, I see..
Hint:
You should be able to see that G(y)=ASinh(ky), where A is some constant, and Sinh() the hyperbolic sine function.

MathematicalPhysics

Now I've got G''(y) - k^2 G (y) = 0, still on the right track yeh?

arildno

Yes, you are.

MathematicalPhysics

From this how do I get to G(y)=ASinh(ky)?

arildno

Note that:
[tex]\frac{d}{dy}ASinh(ky)=kACosh(ky),\frac{d^{2}}{dy^{2}}ASinh(ky)=k^{2}ASi nh(ky)[/tex]
This shows that ASinh(ky) is a solution for G(y).
Similarly, you may show that BCosh(ky) is another solution for G(y).
Your general solution is therefore:
G(y)=ASinh(ky)+BCosh(ky)
Apply the boundary condition at y=0 to prove that B=0

MathematicalPhysics

Thanks for being so patient. How can I show that ASinh(ky) is a soln for G(y)? Im used to looking at the differential eqn and substituting

G(y) = e^{ry} so G'(y) = re^{ry} and G''(y) = r^2. e^{ry}

Is there no such substitution to show that ASinh(ky) & BCosh(ky) are solns?

arildno

Your approach is ABSOLUTELY correct!
Do you agree with the following result, using your method:
[tex]r=\pm{k}[/tex]??

MathematicalPhysics

Yeh which gives gen soln G (y) = Ae^{ky} + Be^{-ky}

but the boundary condition at y=0 doesnt allow for B so we are left with

G(y) = Ae^{ky} ?

arildno

A bit too fast there..
Let's write the general solution for G(y) as follows:
[tex]G(y)=K_{+}e^{ky}+K_{-}e^{-ky}[/tex]
where the K's are constants to be determined by boundary conditions.
Prior to that step, however, let's rewrite the general solution as:
[tex]G(y)=(K_{+}+K_{-})\frac{e^{ky}+e^{-ky}}{2}+(K_{+}-K_{-})\frac{e^{ky}-e^{-ky}}{2}[/tex]
1. We now set [tex]B=K_{+}+K_{-},A=K_{+}-K_{-}[/tex]
(Clearly, A and B are as arbitrary as the K's!)
2) We now recognize:
[tex]Cosh(ky)=\frac{e^{ky}+e^{-ky}}{2}[/tex]
[tex]Sinh(ky)=\frac{e^{ky}-e^{-ky}}{2}[/tex]
Or, we may rewrite G(y) as:
[tex]G(y)=ASinh(ky)+BCosh(ky)[/tex]

MathematicalPhysics

right now I need to find

B: iv done this and basically its because cosh can never be zero which implies B must be zero.

A: at y=a V(x,a) = V_0 cos{kx}

therefore for y=a, G(y)=1

ASinh(ka) = 1 is that right?

hmm im not sure about this.

arildno

Yep, so your solution is:
[tex]V(x,y)=V_{0}\frac{Sinh(ky)\cos(kx)}{Sinh(ka)}[/tex]

MathematicalPhysics

Thank you so much, I understand much more now than I did even yesterday!