# Boundary Value Problem

1. Oct 16, 2004

### MathematicalPhysics

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.

2. Oct 16, 2004

### arildno

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

3. Oct 16, 2004

### MathematicalPhysics

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?

4. Oct 16, 2004

### arildno

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:
$$G''(y)=k^{2}G(y)$$

5. Oct 16, 2004

### 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!

Last edited: Oct 16, 2004
6. Oct 16, 2004

### arildno

Hint:
You should be able to see that G(y)=ASinh(ky), where A is some constant, and Sinh() the hyperbolic sine function.

7. Oct 16, 2004

### MathematicalPhysics

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

8. Oct 16, 2004

### arildno

Yes, you are.

9. Oct 17, 2004

### MathematicalPhysics

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

10. Oct 17, 2004

### arildno

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

11. Oct 17, 2004

### 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?

12. Oct 17, 2004

### arildno

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

13. Oct 17, 2004

### 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} ?

14. Oct 17, 2004

### arildno

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:
$$G(y)=ASinh(ky)+BCosh(ky)$$

15. Oct 17, 2004

### 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?

16. Oct 17, 2004

### arildno

$$V(x,y)=V_{0}\frac{Sinh(ky)\cos(kx)}{Sinh(ka)}$$