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Math Help - Eigenvalues of y''+2y'+\lambda y=0

  1. #1
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    Eigenvalues of y''+2y'+\lambda y=0

    \varphi''+2\varphi'+\lambda\varphi=0
    \varphi(0)=0
    \varphi'(L)=0

    \displaystyle m^2+2m+\lambda=0\Rightarrow m= -1\pm i\sqrt{\lambda-1}

    \varphi=e^{-x}\left[ C_1\cos(x\sqrt{\lambda-1})+C_2\sin\left(x\sqrt{\lambda-1}) \right]

    \varphi'=-e^{-x}\left[ C_1\cos(x\sqrt{\lambda-1})+C_2\sin\left(x\sqrt{\lambda-1}\right)+e^{-x}\left[C_2\sqrt{\lambda-1}\cos(x\sqrt{\lambda-1})-C_1\sqrt{\lambda-1}\sin\left(x\sqrt{\lambda-1}\right)\right]

    \varphi_1(0): \ C_1=1, \ \ \varphi_1'(0): \ C_2=0
    \varphi_1=e^{-x}\cos\left(x\sqrt{\lambda-1}\right)

    \displaystyle\varphi_2(0): \ C_1=0, \ \ \varphi_2'(0): \ C_2=\frac{1}{\sqrt{\lambda-1}}

    \displaystyle\varphi_2=e^{-x}\frac{\sin\left(x\sqrt{\lambda-1}\right)}{\sqrt{\lambda-1}}

    \displaystyle\varphi(x)=e^{-x}\left[A\cos\left(x\sqrt{\lambda-1}\right)+B\frac{\sin\left(x\sqrt{\lambda-1}\right)}{\sqrt{\lambda-1}}\right]

    \displaystyle\varphi(0): \ A=0, \ \ \varphi'(L): \ Be^{-L}\left[\cos\left(L\sqrt{\lambda-1}\right)-\frac{\sin\left(L\sqrt{\lambda-1}\right)}{\sqrt{\lambda-1}}\right]=0

    \displaystyle\tan\left(L\sqrt{\lambda-1}\right)=\sqrt{\lambda-1}

    Is this correct and is this solvable?

    Also, the Latex \right] isn't showing up in a few of the lines but it is there.
    Last edited by CaptainBlack; February 26th 2011 at 09:32 PM. Reason: fogot '
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  2. #2
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    Yes, that is correct. (Strictly speaking you should show that [itex]\lambda[/itex] must be greater than 1 to be an eigenvalue but that is fairly obvious.)

    As for "solvable", yes, but not by any algebraic means.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Yes, that is correct. (Strictly speaking you should show that [itex]\lambda[/itex] must be greater than 1 to be an eigenvalue but that is fairly obvious.)

    As for "solvable", yes, but not by any algebraic means.
    How can I find the \lambda_n then?
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  4. #4
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    Here's a plot of the tangent function and just plain ol' x. You can see that the solutions get closer and closer to odd-integer multiples of pi over two. The only way to find the solutions is numerically. Note, however, that you do have one exact known solution at lambda = 1.
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  5. #5
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    Quote Originally Posted by Ackbeet View Post
    Here's a plot of the tangent function and just plain ol' x. You can see that the solutions get closer and closer to odd-integer multiples of pi over two. The only way to find the solutions is numerically. Note, however, that you do have one exact known solution at lambda = 1.
    I don't know how I am supposed to find a representation for the solution that fits:

    \displaystyle u(x,t)=\sum_{n=0}^{\infty}a_n\exp(t\lambda_n)f(x;\  lambda_n)
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  6. #6
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    Just write out the solution just as you've done, and explain that the lambda's satisfy the equation you've given. I don't think you can do any more.
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  7. #7
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    Quote Originally Posted by Ackbeet View Post
    Just write out the solution just as you've done, and explain that the lambda's satisfy the equation you've given. I don't think you can do any more.
    I don't know what lambda is.
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  8. #8
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    I'm saying you don't have to know exactly, necessarily. Just say that \lambda_{n} is the nth solution of the equation \tan(L\sqrt{\lambda-1})=\sqrt{\lambda-1}, and leave it at that when you present your answer.
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  9. #9
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    Quote Originally Posted by Ackbeet View Post
    I'm saying you don't have to know exactly, necessarily. Just say that \lambda_{n} is the nth solution of the equation \tan(L\sqrt{\lambda-1})=\sqrt{\lambda-1}, and leave it at that when you present your answer.
    What about solving for a_n\text{?}

    Or is that just left as a_n
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  10. #10
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    Hmm. Typically, you'd find those via Fourier analysis, right? Trouble is, I'm not sure that your eigenfunctions here are orthogonal. If you look at the original DE, multiplying through by exp(2x) gets you a Stürm-Liouville form, to be sure. Your q(x)=0, and your weight function is exp(2x). So, that tells me that your eigenfunctions are orthogonal with respect to the weighted inner product

    \displaystyle\langle \phi_{n}|\phi_{m}\rangle=\int_{0}^{L}\phi_{n}\phi_  {m}e^{2t}\,dt. So you could do Fourier analysis that way.

    Do you follow?
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  11. #11
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    Quote Originally Posted by Ackbeet View Post
    Hmm. Typically, you'd find those via Fourier analysis, right? Trouble is, I'm not sure that your eigenfunctions here are orthogonal. If you look at the original DE, multiplying through by exp(2x) gets you a Stürm-Liouville form, to be sure. Your q(x)=0, and your weight function is exp(2x). So, that tells me that your eigenfunctions are orthogonal with respect to the weighted inner product

    \displaystyle\langle \phi_{n}|\phi_{m}\rangle=\int_{0}^{L}\phi_{n}\phi_  {m}e^{2t}\,dt. So you could do Fourier analysis that way.

    Do you follow?
    But

    \displaystyle\phi_n=\tan(L\sqrt{\lambda-1})=\sqrt{\lambda-1}

    How can that be plugged in when it has equality?

    Unless it should be re-written as

    \displaystyle\frac{\tan(L\sqrt{\lambda-1})}{\sqrt{\lambda-1}}-1
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  12. #12
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    No, no. Your \phi_{n}'s are the eigenfunctions that you found in the OP:

    \varphi_{n}(x)=e^{-x}\left[A\cos(x\sqrt{\lambda_{n}-1})+B\,\dfrac{\sin(x\sqrt{\lambda_{n}-1})}{\sqrt{\lambda_{n}-1}}\right].

    Stürm-Liouville theory gives you that those eigenfunctions are orthogonal w.r.t. to the weight function exp(2x). Hence, if you have an initial condition like

    f(x)=u(x,0), and you have the representation

    u(x,t)=\displaystyle\sum_{n=0}^{\infty}a_{n}e^{t\l  ambda_{n}}\varphi_{n}(x), then you're trying to solve the equation

    f(x)=\displaystyle\sum_{n=0}^{\infty}a_{n}\varphi_  {n}(x)

    for the a_{n}'s. So do your Fourier analysis: multiply both sides by both your arbitrary eigenfunction \varphi_{m} and the weight function e^{2x}, and integrate:

    \displaystyle \int_{0}^{L}f(x)\varphi_{m}(x)\,e^{2x}\,dx=\sum_{n  =0}^{\infty}a_{n}\int_{0}^{L}\varphi_{n}(x)\varphi  _{m}(x)\,e^{2x}\,dx.

    Because you have orthogonality of the eigenfunctions w.r.t. this weighted inner product, the RHS collapses down nicely to the following:

    \displaystyle \int_{0}^{L}f(x)\varphi_{m}(x)\,e^{2x}\,dx=a_{m}\i  nt_{0}^{L}\varphi_{m}^{2}(x)\,e^{2x}\,dx.

    Hence, it follows that

    a_{m}=\displaystyle\frac{\int_{0}^{L}f(x)\varphi_{  m}(x)\,e^{2x}\,dx}{\int_{0}^{L}\varphi_{m}^{2}(x)\  ,e^{2x}\,dx}.

    That is what I call Fourier analysis.

    Does that make sense?

    [EDIT]: Double-check the signs of your DE with the form of the Stürm-Liouville type DE. I may have gotten a sign wrong, which might affect the weight function. The basic idea, however, is sound.
    Last edited by Ackbeet; February 26th 2011 at 05:51 PM. Reason: Sign error possibility.
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