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Questionable ODE Solutions

  1. Feb 18, 2014 #1
    Find the eigenvalues λ, and eigenfunctions u(x), associated with the following homogeneous ODE problem:

    [tex] {u}''\left ( x \right )+2{u}'\left ( x \right )+\lambda u\left ( x \right )=0\; ,\; \; u\left ( 0 \right )=u\left ( 1 \right )=0 [/tex]

    Solution:

    Try [itex] u\left ( x \right )=Ae^{rx} [/itex], which gives roots [itex] r=-1\pm \sqrt{1-\lambda } [/itex]. Solution is altered with [itex] \lambda <1\; ,\; \; \lambda =1\; ,\; \; \lambda >1 [/itex]

    For the first case [itex] \lambda <1 [/itex]:

    General solution:
    [tex] u\left ( x \right )=Ae^{\left ( -1+\sqrt{1-\lambda } \right )x}+Be^{\left ( -1-\sqrt{1-\lambda } \right )x} [/tex]
    [tex] u\left ( x \right )=C\cosh \left ( -1+\sqrt{1-\lambda } \right )x+D\sinh \left ( -1-\sqrt{1-\lambda } \right )x [/tex]

    Applying boundaries: (this is where my question lies - how to correctly apply BCs)

    [itex] u\left ( 0 \right )=0 \; \; \Rightarrow \; \; C+D=0 [/itex] (some cases i've seen the conclusion that only [itex] C=0 [/itex]). Do i assume that as [itex] \cosh [/itex] is never zero that [itex]C=0[/itex] and therefore it must be that [itex]D=0[/itex]. Or do i only take [itex]C=0[/itex] and then apply the second BC to see what happens to [itex]D[/itex]?
    The latter (assuming [itex]C=0[/itex]) gives [itex]D\sinh \left ( -1-\sqrt{1-\lambda } \right )=0[/itex]. So either [itex]D=0[/itex] or [itex]\sinh \left ( -1-\sqrt{1-\lambda } \right )=i\pi n[/itex].

    I'm confused by the rules for the BCs. Can anyone point out how to proceed? Thanks
     
    Last edited: Feb 18, 2014
  2. jcsd
  3. Feb 18, 2014 #2

    ehild

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    What does (x) mean? Is it the independent variable, and λ is also function of x, or the whole equation is multiplied by x, and λ is a constant?


    ehild
     
  4. Feb 18, 2014 #3

    Simon Bridge

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    To apply a BC u(x=a)=b, you take the general expression for u(x), and everywhere you see an "x" you write in an "a", and at the end of the line you write "=b".

    In your case, for u(0)=0 you write down:
    $$C\cosh (-1+\sqrt{1-\lambda})(0)+D\sinh (-1-\sqrt{1-\lambda})(0) = 0$$ ... and simplify.

    You don't need to make assumptions - you know the values of the cosh and sinh parts at x=0 and if you don't you can look them up.

    Then repeat for u(1)=0 ...

    Note: I had the same puzzle as ehild over ##\lambda(x)## ... I assumed it was a typo considering the solution and focussed on the answer to your question.
    Have you checked your general solution against the DE?
     
    Last edited: Feb 18, 2014
  5. Feb 18, 2014 #4
    Sorry, that was meant to be λu(x). I've corrected the original post.
     
  6. Feb 18, 2014 #5
    Thanks Simon.
    With the first BC i'm seeing that the result doesn't actually tell me anything about the coefficients as they're multiplied by zero. Is that correct? So i still know nothing about C and D. Assuming i got some kind of positive result from the first BC how does that impact on use of the second?. For instance if i got C=0 then do i ditch the first term in the general solution when i use the second BC?
     
  7. Feb 18, 2014 #6

    ehild

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    No need to complicate the solution with cosh and sinh.

    With the notation a=√(1-λ)>0, the general solution is u(x)=e-x(Aeax+Be-ax)

    At x=0, u(0)=A+B=0 --->B=-A

    At x=1, u(1)=e-1(Aea+Be-a)=0---> A(ea-e-a)=0

    Is it possible if a>0??

    ehild
     
  8. Feb 18, 2014 #7

    HallsofIvy

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    A lot of people consider cosh and sinh to be less complicated than [itex]e^x[/itex] and [itex]e^{-x}[/itex]! Especially when one of the boundary conditions is at x=0. cosh(0)= 1 and sinh(0)= 0.
     
  9. Feb 18, 2014 #8
    Ok. I'm happy with that. So [itex] A\left ( e^{a}-e^{-a} \right )=0 [/itex] gives [itex] a=i\pi n [/itex] and we can get the eigenvalues from there [itex] \lambda = 1-\pi ^2 n^2 [/itex]

    So, how do i write the general solution now, given that we haven't found anything out about the coefficients A and B. Would it be, [tex] u\left ( x \right )=Ae^{i\pi n}-Be^{-i\pi n} [/tex]

    You wrote
    Did you mean a<0?
     
  10. Feb 18, 2014 #9

    ehild

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    I mean a>0 which means λ<1. Is it possible to fulfil the BC-s with λ<1? As you started to discuss that case.

    And yes, you can get a solution zero at both boundaries when λ>1, that is, √(1-λ) is imaginary.

    By the way, your formula with the cosh and sinh is not correct. You should have written
    [tex]u(x)=e^{-x}(C\cosh(\sqrt{1-\lambda}x)+Dsinh(\sqrt{1-\lambda}x)[/tex]

    ehild
     
    Last edited: Feb 18, 2014
  11. Feb 18, 2014 #10
    Sorry, weren't we just doing the case for λ<1?
     
  12. Feb 18, 2014 #11

    pasmith

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    That should be
    [tex]
    u(x) = Ce^{-x} \cosh((1-\lambda)^{1/2}x) + De^{-x} \sinh((1 - \lambda)^{1/2}x)
    [/tex]

    The solution to the eigenvalue problem [itex]v'' + \mu v = 0[/itex] with [itex]v(0) = v(1) = 0[/itex] is [itex]\mu_n = n^2\pi^2[/itex] with [itex]v_n(x) = \sin(n\pi x)[/itex]. This is relevant to your problem, because the substitution [itex]u(x) = e^{-x} v(x)[/itex] reduces your ODE to
    [tex]
    v'' + (\lambda - 1)v = 0
    [/tex]
    with [itex]v(0) = v(1) = 0[/itex].
     
  13. Feb 19, 2014 #12
    Thanks for that.
    When we apply the first boundary u(0)=0 doesn't that yield that C=D=0?
     
  14. Feb 19, 2014 #13

    ehild

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    When λ<1
    at the first boundary, x=0, [itex]u(0)=C\cosh(0)=C\rightarrow C=0[/itex]
    at the second boundary,x=1, [itex]u(1)=D\sinh(\sqrt{1-\lambda})=0 \rightarrow D=0[/itex]

    ehild
     
  15. Feb 20, 2014 #14
    Why is the first boundary not [itex]u(0)=C\cosh(0)+D\sinh(0)\rightarrow C=D=0[/itex]?
    This is what i don't understand. Someone else was also telling me that the boundaries are applied to find a result for one coefficient at a time. Why is this?
     
  16. Feb 20, 2014 #15

    pasmith

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    [itex]\sinh(0) = 0[/itex], so [itex]D \sinh(0) = 0[/itex] for any [itex]D[/itex].
     
  17. Feb 20, 2014 #16

    Simon Bridge

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    Because that would be incorrect algebra. The relationship you get is not C=D.

    Since sinh(0)=0, the expression collapses to u(0)=C. That's just normal algebra.
    You also know that u(0)=0, therefore C=0. Notice how, by itself, this tells you nothing about the value of D. Which is why you need another relation.

    If, however, the expression was more like u(x)=Ccosh(x)+D(sinh(x)-1)
    then u(0)=0 would get you 0=C-D so that C=D.
    But that is not what you have. Not even close.

    Exactly what he said:
    Also think about how simultaneous equations work.
    This is the same thing.
     
  18. Feb 21, 2014 #17
    Ok, just to be really clear... As a result of the boundary conditions we have
    [tex] u(0)=0 \Rightarrow C\cosh(0)+D\sinh(0)=0 [/tex]and
    [tex] u(1)=0 \Rightarrow C\cosh \left ( -1+\sqrt{1-\lambda } \right )+D\sinh \left ( -1-\sqrt{1-\lambda } \right )=0 [/tex]
    I understand your point about solving simultaneously but how to we infer, for the first boundary, that only C=0? Given that [itex] \sinh (0)=0 [/itex], surely then D=0 also. So [itex] \cosh (0) [/itex] is undefined and [itex] \sinh (0)=0 [/itex] so the resultant equation for [itex] u(0)=0 [/itex] is [itex] 0=0 [/itex]. Therefore both C and D must be zero???
     
  19. Feb 21, 2014 #18

    ehild

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    the second formula is wrong. It should be

    [tex]u(x)=e^{-x}(C\cosh(\sqrt{1-\lambda}x)+Dsinh(\sqrt{1-\lambda}x))[/tex]

    ehild
     
    Last edited: Feb 21, 2014
  20. Feb 21, 2014 #19
    Can you show me how this is wrong? I don't get it.
     
  21. Feb 21, 2014 #20

    ehild

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    See Post #11

    ehild
     
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