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Math Help - Eigenvalues in PDE

  1. #1
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    Eigenvalues in PDE

    After some calculations in a PDE I have the following equations:

    X'' + \lambda X = 0

    T' + \lambda k T = 0

    For the first equation, there are 3 possible cases:
    (i) \lambda is positive

    (ii) \lambda is negative

    (iii) \lambda equals 0.

    I also have the boundary conditions X'(0) = 0, X'(L) = 0

    I was hoping someone could answer some general questions I have about these cases.

    First of all, by starting with case (iii) we have:

    X'' = 0 \ \Rightarrow \ X' = A \ \Rightarrow X = Ax + B

    Then with BC's I get

    X'(0) = 0 = A \ \Rightarrow \ \A = 0

    so X = B is a solution.

    Also, T' = 0 \ \Rightarrow \ T = C

    so u(x,t) = BC......my question here is if \lambda = 0 leads to a solution as just shown, does this automatically mean that a_0 need to be included in my fourier series solution to the PDE?

    Now for case (i):

    X'' + \lambda X = 0 leads to solutions

    X = A \cos (\sqrt{\lambda} x) + B \sin (\sqrt{\lambda} x)

    X' = - \sqrt{\lambda} A \sin (\sqrt{\lambda} x) + \sqrt{\lambda} B \cos (\sqrt{\lambda} x)

    so with BC's

    X'(0) = \sqrt{\lambda} B = 0 \ \Rightarrow B = 0

     X'(L) = - \sqrt{\lambda} A \sin (\sqrt{\lambda} L) = 0

    Now A = 0 leads to a trivial solution, so

    \sqrt{\lambda} L = n \pi for n = 0,1,...

    and \lambda = \frac{n^2 \pi^2}{L^2}

    so X_n = A \cos (\frac{n \pi x}{L})

    For the third case,

    X'' - \lambda X = 0

    X = A \cosh (\sqrt{\lambda} x) + B \sinh (\sqrt{\lambda} x)

    X' = \sqrt{\lambda} A \sinh (\sqrt{\lambda} x) + \sqrt{\lambda} B \cosh (\sqrt{\lambda} x)

    Now I get

    X'(0) = \sqrt{\lambda} B = 0 \ \Rightarrow \ B = 0

    X'(L) = \sqrt{\lambda} A \sinh (\sqrt{\lambda} L) = 0

    Apparently this should be a trivial solution, but I don't understand how it is different from case (ii). I would get

    \lambda = \frac{n^2 \pi^2}{L^2}

    so X_n = A \cosh (\frac{n \pi x}{L})

    What am I missing here? How can I have two values for X_n?

    Thanks to anyone who can help me understand this.
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  2. #2
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    From this equation
    <br />
X'(L) = \sqrt{\lambda} A \sinh (\sqrt{\lambda} L) = 0<br />

    and as
    <br />
\sinh (\sqrt{\lambda} L) \ne 0<br />

    if follows that
    <br />
A= 0.<br />

    Please read here (example 4):
    Pauls Online Notes : Differential Equations - Solving the Heat Equation
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  3. #3
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    I have really confused myself now....

    Quote Originally Posted by zzzoak View Post
    From this equation
    <br />
X'(L) = \sqrt{\lambda} A \sinh (\sqrt{\lambda} L) = 0<br />

    First of all, I made a mistake. I should have had:

     X'(L) = \sqrt{- \lambda} A \sinh (\sqrt{- \lambda} L) = 0



    and as
    <br />
\sinh (\sqrt{\lambda} L) \ne 0<br />

    if follows that  A= 0.
    So why can't  \sinh (\sqrt{- \lambda} L)  = 0 ?

    The way I'm thinking is that for case 1, we have

    X'(L) = - \sqrt{\lambda} A \sin (\sqrt{\lambda} L) = 0

    and then we can have \sin (\sqrt{\lambda} L) = 0 for \lambda = \frac{n^2 \pi^2}{L^2} because we are looking for non-trivial solutions.

    And then in case 2, we have

     X'(L) = \sqrt{- \lambda} A \sinh (\sqrt{- \lambda} L) = 0

    Now \sinh 0 = \sin 0 = 0, and they only difference I can see is that \lambda is positive in case 1 and negative in case 2...

    so why can't you still end up with the possibility that  \sinh (\sqrt{- \lambda} L)  = 0 ? Is it because that \lambda is negative?

    Is it because perhaps \sqrt{- \lambda } L \ne 0?
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by OliviaB View Post
    I have really confused myself now....




    First of all, I made a mistake. I should have had:

     X'(L) = \sqrt{- \lambda} A \sinh (\sqrt{- \lambda} L) = 0





    So why can't  \sinh (\sqrt{- \lambda} L)  = 0 ?

    The way I'm thinking is that for case 1, we have

    X'(L) = - \sqrt{\lambda} A \sin (\sqrt{\lambda} L) = 0

    and then we can have \sin (\sqrt{\lambda} L) = 0 for \lambda = \frac{n^2 \pi^2}{L^2} because we are looking for non-trivial solutions.

    And then in case 2, we have

     X'(L) = \sqrt{- \lambda} A \sinh (\sqrt{- \lambda} L) = 0

    Now \sinh 0 = \sin 0 = 0, and they only difference I can see is that \lambda is positive in case 1 and negative in case 2...

    so why can't you still end up with the possibility that  \sinh (\sqrt{- \lambda} L)  = 0 ? Is it because that \lambda is negative?

    Is it because perhaps \sqrt{- \lambda } L \ne 0?
    I think you are forgetting what the \sinh(x) function is. First it only has ONE zero and it is at zero. It is not periodic like the sine function.

    Remember \displaystle \sinh(x)=\frac{e^{x}-e^{-x}}{2}.

    Does this answer your question?
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  5. #5
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    Quote Originally Posted by TheEmptySet View Post

    Remember \displaystle \sinh(x)=\frac{e^{x}-e^{-x}}{2}.

    Does this answer your question?
    Bingo. Thank-you... I wasn't thinking.
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