## Boundary Value Problem

I need some help starting off on this question.

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

$$\frac{\partial^2 V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2}= 0$$

the wall $$y = 0$$ is earthed so that

$$V (x,0) = 0$$

while the potential on the wall $$y = a$$

$$V (x,a) = V_0 \cos{kx}$$ where $$V_0 , k$$ are positive constants.

By seeking a soln of an appropriate form, find $$V (x,y)$$ in the channel.
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 Recognitions: Gold Member Homework Help Science Advisor When in doubt, try separation of variables! V(x,y)=F(x)G(y). Inserting this into Laplace, yields, by rearrangement: $$\frac{G''(y)}{G(y)}=-\frac{F''(x)}{F(x)}$$
 Thanks, I was trying to get it in the form $$V (x,y) = F (a) V_0 \cos{kx}$$ then sub the values into the Laplace eqn. Is that going about it the wrong way?

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## Boundary Value Problem

You DO mean (in my notation!)
$$V(x,y)=G(y)V_{0}\cos(kx)$$???
If so, then it will work.
Note that the separation of variables method in this case implies:
$$G''(y)=k^{2}G(y)$$
 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!
 Recognitions: Gold Member Homework Help Science Advisor 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.
 Now I've got G''(y) - k^2 G (y) = 0, still on the right track yeh?
 Recognitions: Gold Member Homework Help Science Advisor Yes, you are.

 Quote by MathematicalPhysics Now I've got G''(y) - k^2 G (y) = 0, still on the right track yeh?
From this how do I get to G(y)=ASinh(ky)?
 Recognitions: Gold Member Homework Help Science Advisor Note that: $$\frac{d}{dy}ASinh(ky)=kACosh(ky),\frac{d^{2}}{dy^{2}}ASinh(ky)=k^{2}ASi nh(ky)$$ 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
 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?

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 Quote by 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?
Do you agree with the following result, using your method:
$$r=\pm{k}$$??

 Quote by arildno Your approach is ABSOLUTELY correct! Do you agree with the following result, using your method: $$r=\pm{k}$$??

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} ?
 Recognitions: Gold Member Homework Help Science Advisor A bit too fast there.. Let's write the general solution for G(y) as follows: $$G(y)=K_{+}e^{ky}+K_{-}e^{-ky}$$ where the K's are constants to be determined by boundary conditions. Prior to that step, however, let's rewrite the general solution as: $$G(y)=(K_{+}+K_{-})\frac{e^{ky}+e^{-ky}}{2}+(K_{+}-K_{-})\frac{e^{ky}-e^{-ky}}{2}$$ 1. We now set $$B=K_{+}+K_{-},A=K_{+}-K_{-}$$ (Clearly, A and B are as arbitrary as the K's!) 2) We now recognize: $$Cosh(ky)=\frac{e^{ky}+e^{-ky}}{2}$$ $$Sinh(ky)=\frac{e^{ky}-e^{-ky}}{2}$$ Or, we may rewrite G(y) as: $$G(y)=ASinh(ky)+BCosh(ky)$$
 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.
 Recognitions: Gold Member Homework Help Science Advisor Yep, so your solution is: $$V(x,y)=V_{0}\frac{Sinh(ky)\cos(kx)}{Sinh(ka)}$$
 Thank you so much, I understand much more now than I did even yesterday!

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