The eigenvalues must satisfy:

The eigenfunction is:

Is this much correct?

Thanks.

Results 1 to 7 of 7

- Mar 16th 2011, 05:47 PM #1

- Joined
- Mar 2010
- From
- Florida
- Posts
- 3,093
- Thanks
- 9

- Mar 19th 2011, 02:54 PM #2

- Joined
- Mar 2010
- From
- Florida
- Posts
- 3,093
- Thanks
- 9

- Mar 19th 2011, 03:31 PM #3

- Joined
- Apr 2005
- Posts
- 18,961
- Thanks
- 2740

Where did and suddenly come from?

Are they and ?

If so, you cannot, generally, do that. But here it does not hurt.

Then . Again, the exponential is not 0 so we must have either B= 0, which would mean the solution was identically equal to 0, and so not an eigenfunction, or cos(L\sqrt{\lambda})= 0.

The eigenvalues must satisfy:

The eigenfunction is:

Is this much correct?

Thanks.

- Mar 19th 2011, 03:33 PM #4

- Joined
- Mar 2010
- From
- Florida
- Posts
- 3,093
- Thanks
- 9

My book solves all PDEs in the same manner. I don't know why it is done that way but it works in all cases.

Here is what the book says about 1 and 2.

The theorem in the theory of ODE which explicitly states the proposition above is: if the coefficients are continuous on the interval [a,b] and is different from zero throughout the interval, then for any fixed in the interval [a,b] and for any constants . The DE has one and only one solution satisfying at the initial conditions .

Moreover, and are continuous and continuously differentiable functions of both x for and all lambda.

It is convenient to single out the two special solutions of which satisfy the DE

.

We call 1 and 2 the basic solutions at x nought.

- Mar 19th 2011, 07:32 PM #5

- Joined
- Mar 2010
- From
- Florida
- Posts
- 3,093
- Thanks
- 9

- Mar 20th 2011, 11:42 AM #6

- Joined
- Mar 2010
- From
- Florida
- Posts
- 3,093
- Thanks
- 9

- Mar 21st 2011, 03:11 PM #7

- Joined
- Mar 2010
- From
- Florida
- Posts
- 3,093
- Thanks
- 9