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Finding Complex Eigenvalues

  1. Nov 18, 2013 #1
    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.
     
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  3. Nov 18, 2013 #2

    Dick

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    Well, ##e^0=1## and ##e^{2 \pi i}=1##. Does that give you any ideas?
     
  4. Nov 18, 2013 #3
    erck...kinda sorta: I suppose I could set λ = 2[itex]\pi[/itex]i and the only way that equals 0 is when it is 2n[itex]\pi[/itex]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?
     
  5. Nov 18, 2013 #4

    Dick

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    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.
     
  6. Nov 18, 2013 #5
    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?
     
  7. Nov 18, 2013 #6

    Dick

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    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.
     
  8. Nov 18, 2013 #7

    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.
     
  9. Nov 18, 2013 #8

    Dick

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