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Thread: Eigenvalues in PDE

  1. #1
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    Eigenvalues in PDE

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

    $\displaystyle X'' + \lambda X = 0$

    $\displaystyle T' + \lambda k T = 0$

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

    (ii) $\displaystyle \lambda$ is negative

    (iii) $\displaystyle \lambda$ equals 0.

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

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

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

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

    Then with BC's I get

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

    so $\displaystyle X = B$ is a solution.

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

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

    Now for case (i):

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

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

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

    so with BC's

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

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

    Now $\displaystyle A = 0$ leads to a trivial solution, so

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

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

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

    For the third case,

    $\displaystyle X'' - \lambda X = 0$

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

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

    Now I get

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

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

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

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

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

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

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

    and as
    $\displaystyle
    \sinh (\sqrt{\lambda} L) \ne 0
    $

    if follows that
    $\displaystyle
    A= 0.
    $

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

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

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



    and as
    $\displaystyle
    \sinh (\sqrt{\lambda} L) \ne 0
    $

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

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

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

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

    And then in case 2, we have

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

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

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

    Is it because perhaps $\displaystyle \sqrt{- \lambda } L \ne 0$?
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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:

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





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

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

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

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

    And then in case 2, we have

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

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

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

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

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

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

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

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