I have a question: With what positive function must the operator $L(y)=x^2y''+y' $ be multiplied, to be self-adjoint? With what weight are then the "eigenfunctions" of the multiplied operator orthogonal, with boundary conditions y(1)=0=y(2) ?

So from the weight equation : $w(x)=\frac{1}{a(x)}e^{\int \frac{b(x)}{a(x)}dx}$

Where in general $ L y =a(x)y''+b(x)y'+c(y)$

My solution: the function should be multiplied by $\frac{1}{x^2}e^{\int 1/x^2} $ and the weight is then $\frac{1}{x^2}e^{\int 1/x^2} $ , is this correct, or am I wrong ?

  • $\begingroup$ You are correct. $\endgroup$ – DisintegratingByParts Sep 14 '16 at 22:23
  • $\begingroup$ What about the bessel diff. equation $x^2y''+xy'+(k^2x-\nu^2)y=0$ , depending on whether I take $k^2$ as a eigenvalue and or $\nu^2$ as a eigenvalue, I get two different weights : $x$ and $1/x$, are they both correct? $\endgroup$ – user3633438 Sep 14 '16 at 22:52
  • $\begingroup$ $x^2y''+xy'+(k^2x^2-\nu^2)y=0$ , I meant.. $\endgroup$ – user3633438 Sep 14 '16 at 23:02
  • $\begingroup$ You are correct that either $k^2$ or $\nu^2$ can be treated as eigenvalues, and the weight functions are different. Notice, however, that $Lf = x^2f''+xf'-\nu^2f$ has the property that $Lf(\alpha x)=(Lf)(\alpha x)$, which allows you to set $k=1$ if $k\ne 0$. So you can consider solutions $f_{\nu}(kx)$ if $k \ne 0$. $\endgroup$ – DisintegratingByParts Sep 15 '16 at 2:14
  • $\begingroup$ Thank you! This really helped me out! $\endgroup$ – user3633438 Sep 15 '16 at 4:08

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