# Finding V for a 2D box using separation of variables.

• maherelharake
In summary, the homework statement is that there is a two dimensional box with one side at potential V0. The other three sides are grounded. The box is a square with top and bottom at y=a/2 and –a/2 and sides at x=±a/2. Find V(x,y) (it should contain cos and sinh).
maherelharake

## Homework Statement

We have a 2-dimensional box with only one side at a potential V0. The other 3 sides are grounded. The box is a square with top and bottom at y=a/2 and –a/2 and sides at x=±a/2. Find V(x,y) (it should contain cos and sinh).

## The Attempt at a Solution

I need help with this whole problem. Maybe if I get a start, I can work through it. Thanks.

What do you know that might be relevant? Specifically, what differential equation must be satisfied by the electric potential within some space?

Hmmm, well I know Laplace's equation. In the book they also convert Laplace's into two ordinary differential equations.
V(x,y)=(Aekx+Be-kx)(Csin(ky) + Dcos(ky))

Right, you would use Laplace's equation. Are you familiar with the methods used to solve that equation? (I see you posted a component of the general solution)

If so, find the specific values of A, B, C, and D that correspond to the boundary conditions given.

I'm not sure what you mean by the methods. I think the only one our book/prof discussed was separation of variables. This method leads to what I posted. I hope that is the equation I can use for this problem. Is it not?

Yes, separation of variables was basically what I was talking about. You can use separation of variables to separate Laplace's 2D equation into two one-variable differential equations. Then you will need to use some other technique(s) to solve each of those one-variable equations, and combine the solutions of the one-variable equations to get the solution of the full 2D Laplace equation. That's what you posted. So yes, that is what you use for this problem.

More accurately, if I remember correctly, the general solution of Laplace's equation will be a sum of what you posted for different values of k.
$$V(x,y) = \sum_i[A_i e^{k_i x} + B_i e^{-k_i x}][C_i \sin(k_iy) + D_i \cos(k_iy)]$$
You will need to find which choices of the coefficients will make that solution satisfy the given boundary conditions.

Ok I will try this and let you know how I do in the morning. Thanks.

Ah ok I didn't make it very far. I set up all the boundary conditions, and tried to solve for them. I was reading through my book and found that when the situation is symmetric with respect to x, A=B. I currently have the equation...
V(x,y)=cosh(kx)(Csin(ky)+Dcos(ky))

I'm not sure how to get this to satisfy my boundary conditions. By the way, the conditions I got are:
V=V0 when y=a/2
V=0 when y=-a/2
V=0 when x=a/2
V=0 when x=-a/2

It is admittedly tricky. I haven't done this sort of thing in a while so I don't remember if there is a systematic method.

Anyway, for starters, you have to remember that the solution for V is a sum of products with different values of k, as I wrote in my previous post. Finding out that $A_k = B_k$ is a good start. The next thing I'd think about is what values of k are allowed.

Consider a fixed value of x. At that fixed x, the potential will form some function in y. Your expression for that function is basically a Fourier series,
$$V(x_0, y) = \sum_i C_i' \cos(k_iy) + D_i' \sin(k_iy)$$
where
$$C_i' = 2A_i C_i$$
and similarly for Di'. Now, when you're writing a Fourier series for a function that has a domain of width a, do you know what the possible values of k are? (Hint: this is equivalent to writing the series for a function that is periodic with period a.)

I'm not sure how you realized C'=2AC

As far as the question you asked, would it be k=nPi/a? They had this in the book too.

maherelharake said:
I'm not sure how you realized C'=2AC
I didn't realize it, I defined it. It's just to make the equation look more like a Fourier series. If it bothers you, we can certainly go back to writing 2AiCi.
maherelharake said:
As far as the question you asked, would it be k=nPi/a? They had this in the book too.
Almost. I think you're off by a factor of 2.

Oh ok I see it now. I think it's supposed to be k=2nPi/a. So is the next step using Fourier's trick?

I can't seem to get all of the conditions to be satisfied. I am getting D has to be equal to 0, but I am unsure if that is correct.

I have one more update. My professor changed the original question to this:

Show that after applying x boundary conditions, the general result is a sum over terms Vn(x,y) = cos(kn x) * (An cosh (kn y) + Bn sinh (kn y)).

Finally, find the result for An and Bn in terms of a and V0He also hinted that properties of sinh and cosh are important.

OK, well in light of the updated question, have you been able to show that the general result is a sum over terms of that form?

Hmm not really. I thought we were supposed to worry about the "sum over terms" part towards the end. Maybe I have confused the order of how to find the solution.

I have gotten nowhere with this problem. It's due Thursday morning for me. Any help would be appreciated, if not I understand.

Sorry about that, I got rather busy myself. I'll see if someone else can take a look at it.

But as far as showing that the general solution is a sum over terms, you will have to apply the technique of separation of variables. After that, use the domain of the solution (i.e. the fact that it is of width a) to determine the allowed values of the separation constant. (Actually I guess you already did that earlier on this thread)

Sorry I couldn't help you out - I asked if any of the other homework helpers would be able to help you but I guess nobody else had time either.

Anyway, now that the assignment is over, maybe you could ask your instructor. This sort of thing would be easier to explain in person. Of course, if you'd like to continue working through it here (so that you can be prepared if you see a problem like this in the future), I'd be happy to keep going, as I have time.

I actually have another problem that is due tomorrow, that should be easier than this one. Maybe I will start a new thread?

## 1. What is the concept of separation of variables?

The concept of separation of variables is a mathematical technique used to solve differential equations with multiple variables by breaking down the equation into simpler, single variable equations.

## 2. How does separation of variables apply to finding V for a 2D box?

In the context of finding V for a 2D box, separation of variables is used to break down the partial differential equation into two separate ordinary differential equations: one for the x-axis and one for the y-axis. These equations can then be solved individually and combined to find the solution for the entire 2D box.

## 3. Why is finding V for a 2D box important in science?

Finding V for a 2D box is important in science because it allows us to model and understand physical systems with two dimensions, such as the electric potential in a 2D space or the temperature distribution in a 2D object. This can provide valuable insights and predictions for various scientific phenomena.

## 4. What are the steps involved in using separation of variables to find V for a 2D box?

The steps involved in using separation of variables to find V for a 2D box include identifying the partial differential equation, applying separation of variables to break it down into two ordinary differential equations, solving the individual equations, and combining the solutions to find the general solution for the 2D box.

## 5. Are there any limitations to using separation of variables to find V for a 2D box?

Yes, there are limitations to using separation of variables to find V for a 2D box. This method is only applicable to linear partial differential equations and may not work for more complex equations. Additionally, in some cases, the solutions obtained through separation of variables may not satisfy all boundary conditions, requiring further modifications to the method.

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