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Sturm-Liouville Orthogonality Proof

  1. Feb 20, 2009 #1
    1. The problem statement, all variables and given/known data

    A set of eigenfunctions yn(x) satisfies the Sturm-Liouville equation #1 with boundary conditions #2. The function g(x) = 0. Show that the derivatives un(x) = yn'(x) are also orthogonal functions. Determine the weighting function w(x) for these functions. What boundary conditions are required for orthogonality?

    2. Relevant equations

    #1: [tex]\frac{d}{dx}[/tex]f(x)[tex]\frac{dy}{dx}[/tex] - g(x)[tex]\frac{dy}{dx}[/tex] + [tex]\lambda[/tex]w(x)y = 0 where w(x) [tex]\geq[/tex] 0 and a [tex]\leq[/tex] x [tex]\leq[/tex] b

    #2: [tex]\alpha[/tex]1y + [tex]\beta[/tex]1[tex]\frac{dy}{dx}[/tex] = 0 at x = a and [tex]\alpha[/tex]2y + [tex]\beta[/tex]2[tex]\frac{dy}{dx}[/tex] = 0 at x = b

    [tex]\alpha[/tex] and [tex]\beta[/tex] are both constants and cannot both equal 0.

    There is also the orthogonality relation [tex]\int y_{n}w(x)y_{m}dx = 0[/tex]


    3. The attempt at a solution

    I guess what I'm most confused about here is what the problem is asking me to show. By saying the set "un(x) = yn'(x) are orthogonal functions", what exactly are they orthogonal to? I'm assuming it means that [tex]\int u_{n}w(x)u_{m}dx = 0[/tex], provided that m [tex]\neq[/tex] n. Is this correct or am I way off?

    Since g(x) = 0, the DE becomes [tex]\frac{d}{dx}[/tex]f(x)[tex]\frac{dy}{dx}[/tex] + [tex]\lambda[/tex]w(x)y = 0. From the boundary conditions I can tell that y(a) = - [tex]\frac{\alpha_{1}}{\beta_{1}}[/tex]u(a) and y(b) = - [tex]\frac{\alpha_{2}}{\beta_{2}}[/tex]u(b), but I don't know how this would help me at all. Can somebody please point me in the right direction?

    Thanks; Conor.
     
  2. jcsd
  3. Feb 21, 2009 #2

    lanedance

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    Homework Helper

    might be on the wrong track, but here's a guess...

    based on the wording of the question i think you need to show the u's are orthogonal, as you've written, but to some weighting function yet to be determined

    how about trying to show the u's constitute their own sturm-liouville problem? existence and orthogonality properties of the solutions would show the un are orthogonal?

    did this and got a weighting function ~(d/dx(f'/w)+lambda)... though not sure how to show this is >0 for all x...
     
  4. Feb 21, 2009 #3
    Yeah, that's what I thought to, but how did you show that they form their own Sturm-Liouville problem? Did you replace y(x) with u(x) in #1? When you sub the u(x)'s into eq'n #1, the last term contains [tex]\int u(x)dx[/tex], because u = y'. I'm curious to see how you got that weighting function.

    I did it a little differently, but I'm not sure if I'm cheating. If you take eqn #1 in both n and m (two different indicies), multiply both by ym and ym (respectively), add the two equations and then integrate over all x you get...

    [tex]\int\left[y_{m}\frac{d}{dx}f\frac{dy_{n}}{dx} + y_{n}\frac{d}{dx}f\frac{dy_{m}}{dx}\right]dx = - \left( \lambda_{n} - \lambda_{m} \right) \int y_{n}w(x)y_{m}dx = 0[/tex] (because of orthogonality of y)​

    It turns out that the first and second terms in the integration are equal due to symmetry in m and n, making the integral equal to...

    [tex]2\left(\frac{\beta_{1}}{\alpha_{1}}u_{n}(a)f(a)u_{m}(a) - \frac{\beta_{2}}{\alpha_{2}}u_{n}(b)f(b)u_{m}(b)\right) - 2\int u_{n}f(x)u_{m}dx = 0[/tex]​

    ...where I've made use of the boundary conditions and integration by parts. The additional boundary condition the problem is asking for in this case could be...

    [tex]\frac{\beta_{1}}{\alpha_{1}}u_{n}(a)f(a)u_{m}(a) = \frac{\beta_{2}}{\alpha_{2}}u_{n}(b)f(b)u_{m}(b)[/tex]​

    ...in which case [tex]\int u_{n}f(x)u_{m}dx = 0[/tex] and the [tex]u_{n}[/tex] are orthogonal, and f(x) is the weighting function. I am concerned, however, that I did not obtain an actual expression for f(x), as the problem may have been asking. If you have time, please show me how you got your weighting function. Thanks for your help!
     
  5. Feb 21, 2009 #4

    lanedance

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    Homework Helper

    nice idea,

    don't even need to add the 2 equations, Also i would try and write the boundary conditions independent of m,n maybe this could be done by

    y'(a) = [tex]\sqrt{f(a)\beta 1 \alpha 2 / (f(b)\beta 2 \alpha 1)[/tex] y'(b)


    not convinced with what i did now as i don't think its in correct SL form

    but method was
    - sub in u =y'
    - divide by w & then differentiate
    - try to re-arrange into SL form
     
  6. Feb 21, 2009 #5
    Yeah good call, I guess you don't need to add them. I'll try to rewrite those boundary conditions.
     
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