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    Show that the operator:

    L=-(1-x^2)d2/dx2+2xd/dx is self adjoint if both y(1) and y(-1) are required to be finite.

    I've somehow done some integration by parts over an arbitrary interval and shown that the integral of f*(x)[Lg(x)] is equal to the integral of
    [Lf(x)]*g(x) + some boundary terms. But I don't understand the requirement that y(1) and y(-1) must be finite for L to be self adjoint, looking at the operator I can see that x=+(-1) are going to be important points and applying the operator to y will give an ODE of the form: y''+p(x)y'+q(x)y=0 which means p(x) and q(x) will be singular and non-differentiable at x=1 and -1, but I don't see how to tie this in with the requirement above.

    Cheers.
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    For the operator to be self adjoint you require <Ly,w>=<y,Lw> so you need the boundary terms to be zero using the fact that y is finite at +/-1
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    (Original post by thebadgeroverlord)
    For the operator to be self adjoint you require <Ly,w>=<y,Lw> so you need the boundary terms to be zero using the fact that y is finite at +/-1
    I'm pretty sure this is incorrect. If the boundary terms sum to zero this indicates that the operator is Hermitian as well as being self adjoint.
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    Hermitian and self adjoint are the same thing. The operator incudes the boundary conditions aswell
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    Ok, the last part of this question asks me to find the solution of the equation Ly=x^3 in terms of p1 and p3, where:

    p1=x and p3=1/2(5x^3-3x) are the first 2 odd Legendre polynomials.

    I've expressed x^3 in terms of the eigenfunction expansion of the Legendre polynomials above if thats necessary. But how do I solve this inhomogenous ODE and get a solution in terms of p1 and p3? Cheers.
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    After thinking more about this, I think I must find the solution to the ODE and then just express it in terms of the p1 and p2 using an eigenfunction expansion?
 
 
 
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Updated: January 24, 2010
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