I have to solve this:

I've tried separation of variables, and got

What next?

Is that even the right way to go?

Do I have to homogenize the conditions first?

Thank you.

Printable View

- September 6th 2008, 01:41 PMgeorgellaplace equation
I have to solve this:

I've tried separation of variables, and got

What next?

Is that even the right way to go?

Do I have to homogenize the conditions first?

Thank you. - September 6th 2008, 05:19 PMmr fantastic
- September 6th 2008, 05:30 PMThePerfectHacker
- September 6th 2008, 05:56 PMmr fantastic
- September 7th 2008, 12:41 AMgeorgel
First of all, thank you for your responses.

This is what I did:

,

According to my textbook, we use _squared_ because it's easier to work with it later, and the minus is according to the theorem of non-negativity of eigenvalues (I'm not sure that's the correct english term.) so basically, I'm checking and .

gives me:

, that is, .

gives me .

So, I have , which is not a trivial solution. I get the same when I do it like you said, .

What did I do wrong?

:

Characteristic equation:

gives me . So, .

I'm really confused, please help. - September 7th 2008, 01:10 AMmr fantastic
OK, now I'm confused. I might have let TPH change my mind incorrectly.

By do you mean:

1. , or

2. .

I thought you meant 2. since there are only two boundary conditions given. But I changed my mind after reading TPH's post (after all, is a commonly used symbol for the Laplacian operator).

So which is it? If it's 1. then there are no trivial solutions because there aren't enough boundary conditions to get rid of all the arbitrary constants.

If it's 2. I'll reply later since i have things to do at the moment. - September 7th 2008, 01:29 AMgeorgel
I meant: .

We've had a several exercises like this one in class, but there's always been more boundary conditions, and that's what's confusing me.

I've found a similar exercise in another textbook, but there it says to solve it on . But even with that information, I still don't know what to do next.. - September 7th 2008, 07:04 AMThePerfectHacker
Here your equations lead to: .

Thus, .

This means, .

Note which means .

To have non-trivial solutions we require .

And which means .

But this is impossible because LHS is non constant.

It means #1, there is no need to solve #2 by seperation of variables.

Because #2 has a general solution where is a differenciable function. - September 7th 2008, 07:39 AMgeorgel