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Sturm-Liouville Orthogonality of Eigenfunctions

  1. Nov 5, 2015 #1
    1. The problem statement, all variables and given/known data
    Consider the following Sturm-Liouville Problem:
    [tex]\dfrac{d^2y(x)}{dx^2} + {\lambda}y(x)=0, \ (a{\geq}x{\leq}b)[/tex]
    with boundary conditions
    [tex]a_1y(a)+a_2y{\prime}(a)=0, \ b_1y(b)+b_2y{\prime}(b)=0[/tex]
    and distinguish three cases:
    [tex]a_1=b_1, a_2{\neq}0, b_2{\neq}0[/tex][tex]a_2=b_2=0, a_1{\neq}0, b_1{\neq}0[/tex][tex]a_21{\neq}0, a_2{\neq}0, b_1{\neq}0, b_2{\neq}0[/tex]

    a) Without determining the actual eigenvalues and eigenfunctions, prove, in detail, the orthogonality of non-degenerate (i.e. having different eigenvalues) eigenfunctions in all three cases.
    2. Relevant equations
    Suppose we have a solution with eigen functions [tex]y_j,y_m[/tex] such that their eigen values are [tex]\lambda_j,\lambda_m, \lambda_m\neq{\lambda}_j[/tex]
    Then they are orthogonal if:
    [tex]\int_{a}^{b}dxy_j(x)y_m(x)=0, [/tex]

    3. The attempt at a solution
    So I am okay with doing this integral and proving the orthogonality of the two eigen functions. But what i do not understand is what a1, a2, b1 and b2 are and how I implement the special cases.

    Basically I solve the integral and I am left with

    [tex][y_{j}p(x)y_m{\prime}(x)-y_m(x)p(x)y_j{\prime}(x)]^{b}_{a}-\int^{b}_{a}[y_{j}{\prime}(x)p(x)y_{m}{\prime}(x)-y_{m}{\prime}(x)p(x)y_{j}{\prime}(x)][/tex]
    where clearly the second part is zero and the first part should be zero with boundary conditions. But like I said I have no idea what the a1, a2, b1, b2 mean.

    Any help would be appreciated
     
  2. jcsd
  3. Nov 6, 2015 #2

    andrewkirk

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    They just serve to identify three alternative sets of boundary conditions, which will give different solutions to the DE. Taking it just one step further, the alternative sets of boundary conds are:

    (1) ##
    a_1y(a)+a_2y{\prime}(a)=0, \ a_1y(b)+b_2y{\prime}(b)=0;
    ##
    (2) ##
    a_1y{\prime}(a)=0, \ b_1y{\prime}(b)=0
    ##
    (3) ##
    a_1y(a)+a_2y{\prime}(a)=0, \ b_1y(b)+b_2y{\prime}(b)=0
    ##
    and none of the ##a_k,b_k## coefficients may be zero except ##a_1##.

    You said the first term in your result 'should be zero with boundary conditions'. Have you done the calcs to test that?
     
  4. Nov 6, 2015 #3
    A couple things I want to add
    General Sturm Liouville equation:
    [tex]\frac{d}{dx}p(x)\frac{dy}{dx}-s(x)y(x)+{\lambda}w(x)y(x)=0[/tex]
    from my understanding, in our problem p(x) = 1 and s(x)=0 and w(x)=1.
    and so the last bit of that integral for us is:
    [tex][y_j(x)y{\prime}_m(x)-y_m(x)y{\prime}_j(x)]_a^b[/tex]

    And the reason I know this should be zero is because that entire integral should be 0 due to orthogonality. The problem I am having is, I guess, testing this using those boundary conditions. I don't know if I'm just missing something stupid but the constants (a1, b1, a2, b2) are completely throwing me off with this question.

    Also I made a mistake writing the first case and third case, this is how it should be:
    [tex](1)a_1=b_1=0,a_2\neq0,b_2\neq0[/tex]
    [tex](3)a_1\neq0,a_2\neq0,b_1\neq0,b_2\neq0[/tex]
     
  5. Nov 7, 2015 #4

    vela

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    You're probably overthinking it. For example, if ##a_1 = 0## and ##a_2\ne 0##, you have ##a_2 y'(a) = 0##. Since ##a_2 \ne 0##, you can divide by it, so ##y'(a) = 0##.
     
  6. Nov 7, 2015 #5
    Yeah that's what I ended up doing. And for case 3 I rearranged the boundary conditions so that [tex]y=\frac{a_2}{a_1}y\prime[/tex] and similarly with the b2, b1 and was able to get it to zero.

    Thanks for your help guys.
     
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