# Finding Complex Eigenvalues

1. Nov 18, 2013

### trap101

Find the complex eignevalues of the first derivative operator d/dx subject to the single boundary condition X(0) = X(1).

So this has to do with PDEs and separation of variables:

I get to the point of using the BC and I am left with an expression:

1 = eλ, this is where my issue falls. How do I convert it into a complex eignvalue and then also get that eigenfunction? I know how to do this when dealing with the Fourier series in terms of sines and cosine, but complex has me baffled and I know our professor is going to ask us about the complex form on our test in 2 days, he's been talking about it all semester.

2. Nov 18, 2013

### Dick

Well, $e^0=1$ and $e^{2 \pi i}=1$. Does that give you any ideas?

3. Nov 18, 2013

### trap101

erck...kinda sorta: I suppose I could set λ = 2$\pi$i and the only way that equals 0 is when it is 2n$\pi$i, but that is me reaching, the issue I think falls with me not even knowing how to manipulate the complex representation too well.

How is $e^{2 \pi i}=1$? Should I take it in terms of euler's formula to see the relation?

4. Nov 18, 2013

### Dick

Yes, use Euler's formula. If $\lambda=a+bi$ with a and b real then $e^{a+bi}=e^a e^{bi}=e^a (\cos(b)+i\sin(b))$. Set that equal to 1 and figure out what a and b can be.

5. Nov 18, 2013

### trap101

Ok thanks. I actually had another quaetion with regards to this problem. In finding the solution I followed a procedure they did in the book:

X'(x) = λX and then solved for the solution, but my question is how or why did they put the function in this form first?

6. Nov 18, 2013

### Dick

If X(x) is an eigenfunction of an operator O with eigenvalue λ, then O(X(x))=λX(x). That's the definition of eigenfunction and eigenvalue. Now put O=d/dx.

7. Nov 18, 2013

### trap101

Ohhhh, now if I consolidate this with what I previoiusly studied in linear algebra:

that would mean AX = λX, but in this case we're dealing with functions so that A matrix is now an operator. Awesome. Aha moments.... thanks.

8. Nov 18, 2013

### Dick

Yes, it's same concept as eigenvalues in linear algebra.