Separation of Variables Problem

[Slusho]

Homework Statement

A square is made up of four plates with a potential of zero on the top and bottom plates at (x,L/2) and (x,-L/2), and a potential of cos(πy/L)+cos(3πy/L). Find the potential and electric fields inside the square.

The Attempt at a Solution

I start with V=(Ae^kx+Be^-kx)(Csin(ky)+Dcos(ky))
Then I am supposed to use the boundary conditions stated above to determine values for A,B,C,D, and k. My book has an example of this. However, I can't figure out how to use these conditions to find the values. Could someone point me in the right direction?

 

[vela]

Homework Statement

A square is made up of four plates with a potential of zero on the top and bottom plates at (x,L/2) and (x,-L/2), and a potential of cos(πy/L)+cos(3πy/L).

The problem statement doesn't make sense. You seem to have omitted some words.

The Attempt at a Solution

I start with V=(Ae^kx+Be^-kx)(Csin(ky)+Dcos(ky))
Then I am supposed to use the boundary conditions stated above to determine values for A,B,C,D, and k. My book has an example of this. However, I can't figure out how to use these conditions to find the values. Could someone point me in the right direction?

What's stopping you from finding the values? Can you at least write down the equations corresponding to the boundary conditions at the top and bottom plate?

 

[Slusho]

Sorry, it was late when I wrote this. I've bolded the change:

1. Homework Statement
A square is made up of four plates with a potential of zero on the top and bottom plates at (x,L/2) and (x,-L/2), and a potential of cos(πy/L)+cos(3πy/L) on the left and right plates at (-L/2,y) and (L/2,y). Find the potential and electric fields inside the square.

The boundary conditions are those potentials on the plates (ex: V(x,L/2)=0).

 

[SammyS]

Sorry, it was late when I wrote this. I've bolded the change:

Homework Statement

A square is made up of four plates with a potential of zero on the top and bottom plates at (x,L/2) and (x,-L/2), and a potential of cos(πy/L)+cos(3πy/L) on the left and right plates at (-L/2,y) and (L/2,y). Find the potential and electric fields inside the square.

The boundary conditions are those potentials on the plates (ex: V(x,L/2)=0).

The Attempt at a Solution

I start with V=(Aekx+Be-kx)(Csin(ky)+Dcos(ky))
Then I am supposed to use the boundary conditions stated above to determine values for A,B,C,D, and k. My book has an example of this. However, I can't figure out how to use these conditions to find the values. Could someone point me in the right direction?

What was the general method demonstrated in the textbook?

 

[Slusho]

The textbook does not have a general solution. It gives an example and works through it. Therefore, because it has different boundary conditions, it is not applicable to this exact problem. They plug in the values from a single boundary condition at a time into the equation I gave to determine the coefficients and k. When I do this in this problem, I am unable to determine them.

 

[SammyS]

The textbook does not have a general solution. It gives an example and works through it. Therefore, because it has different boundary conditions, it is not applicable to this exact problem.

In general, what was the process that was used in their example ?

 

[Slusho]

I just edited my above reply, which should answer your question.

 

[SammyS]

The textbook does not have a general solution. It gives an example and works through it. Therefore, because it has different boundary conditions, it is not applicable to this exact problem. They plug in the values from a single boundary condition at a time into the equation I gave to determine the coefficients and k. When I do this in this problem, I am unable to determine them.

OK.

So what equations you get when plugging in the left & right boundary conditions?

 

[Slusho]

For the right boundary condition, I get cos(πy/L)+cos(3πy/L)=(Ae^kL/2+Be^-kL/2)(Csin(ky)+Dcos(ky))
And the left is the same except the signs of the exponents are switched.

 

[SammyS]

For the right boundary condition, I get cos(πy/L)+cos(3πy/L)=(Ae^kL/2+Be^-kL/2)(Csin(ky)+Dcos(ky))
And the left is the same except the signs of the exponents are switched.

Good.

It looks like you need to use super position. It's clear the you need two distinct values for k.

 

[Slusho]

Okay, so the example problem in the book has k=nπ/a, with n=1,2,3,...
For the equation to work here, (Ae^kL/2+Be^-kL/2) would have to equal 1 (not sure how that would work), C=0, D=1, and k=nπ/L, where n=1,3
But wouldn't the right side of the equation need a summation ∑ over n? And doesn't n usually just keep going forever like in the example?

 

[vela]

Okay, so the example problem in the book has k=nπ/a, with n=1,2,3,...
For the equation to work here, (Ae^kL/2+Be^-kL/2) would have to equal 1 (not sure how that would work), C=0, D=1, and k=nπ/L, where n=1,3

That's not quite correct. You can't assert (Ae^kL/2+Be^-kL/2) =1 and D=1. The best you can say is their product equals 1.

But wouldn't the right side of the equation need a summation ∑ over n? And doesn't n usually just keep going forever like in the example?

The coefficients generally depend on the value of $k$. You can have different values of C and D for the different values of $n$, so for the terms that don't appear, you have $C_n = D_n=0$.

If I were you, I'd start with the other two boundary conditions because they involve setting the potential to 0, and as usual, zeroes make things simpler. They will let you deduce the allowed values of $k$.

 

[SammyS]

Okay, so the example problem in the book has k=nπ/a, with n=1,2,3,...
For the equation to work here, (Ae^kL/2+Be^-kL/2) would have to equal 1 (not sure how that would work), C=0, D=1, and k=nπ/L, where n=1,3
But wouldn't the right side of the equation need a summation ∑ over n? And doesn't n usually just keep going forever like in the example?

It looks like for this problem, n = 1, 3 are all that are needed, as far as the solution for k is concerned.

To find A and B use both right and left boundary conditions. You should find that A = B .

The result can be written with exponential functions or more compactly with hyperbolic functions such as sinh, cosh, and sech .

 

[SammyS]

It looks like for this problem, n = 1, 3 are all that are needed, as far as the solution for k is concerned.

To find A and B use both right and left boundary conditions. You should find that A = B .

The result can be written with exponential functions or more compactly with hyperbolic functions such as sinh, cosh, and sech .

Added in edit:

We should have done what vela says above, first.

 

[Slusho]

Okay, I used both right and left boundaries, dividing one by the other, to see that A=B. This allowed me to simplify (Ae^kL/2+Be^-kL/2) to be 2cosh(kx).
Now I tried going to the other boundary conditions: 0=2cosh(kx)(Csin(kL/2)+Dcos(kL/2)) which lead to tan(kL/2)=-D/C which is where I'm stuck now.

 

[SammyS]

Okay, I used both right and left boundaries, dividing one by the other, to see that A=B. This allowed me to simplify (Ae^kL/2+Be^-kL/2) to be 2cosh(kx).
Now I tried going to the other boundary conditions: 0=2cosh(kx)(Csin(kL/2)+Dcos(kL/2)) which lead to tan(kL/2)=-D/C which is where I'm stuck now.

If A = B, that seems fine.

But A or B will remain in your solution. (Ae^kL/2+Be^-kL/2) becomes 2Acosh(kL/2) .

The boundary conditions at x = ±L/2 dictate that C = 0 . So your previous result for k seem to me to be fine. It makes the solution match the boundary condition at y = ±L/2 .

You have chosen D = 0 , so all that has to be done is to get A which you are close to getting.

Added in Edit:

That should have been D = 1 .
.

 

[Slusho]

Thanks for catching that. I accidentally dropped the A.

It makes sense to me that C=0, as we only have cos terms on the left of cos(πy/L)+cos(3πy/L)=2Acosh(kL/2)(Csin(ky)+Dcos(ky)) but what about the 2Acosh(kL/2) term? If k=nπ/L then that would become 2Acosh(π/2) or 2Acosh(3π/2).

Also, I don't understand what you mean by "You have chosen D=0". D must be nonzero for the left and right boundary conditions since the left side of that equation is nothing but cos terms, right?

 

[SammyS]

Thanks for catching that. I accidentally dropped the A.

It makes sense to me that C=0, as we only have cos terms on the left of cos(πy/L)+cos(3πy/L)=2Acosh(kL/2)(Csin(ky)+Dcos(ky)) but what about the 2Acosh(kL/2) term? If k=nπ/L then that would become 2Acosh(π/2) or 2Acosh(3π/2).

Also, I don't understand what you mean by "You have chosen D=0". D must be nonzero for the left and right boundary conditions since the left side of that equation is nothing but cos terms, right?

That D = 0 was a typo. You set it to 1 which is fine.

What's left is 2Acosh(kL/2) = 1

 

[Slusho]

Okay, so that gives me A=sech(nπ/2)/2 which means V(x,y)=∑sech(nπ/2)cosh(nπx/L)cos(nπy/L) (summed over n=1,3).
And E=∑(nπ/L)sech(nπ/2)[sinh(nπx/L)cos(nπy/L[PLAIN]http://mathworld.wolfram.com/images/equations/Hat/Inline1.gif-cosh(nπx/L)sin(nπy/L)ŷ]