Find the first positive eigenvalue $\lambda$ of the boundary value problem over $x\in [0,\frac{1}{2}]$. $$y''-\lambda y'+\frac{2\lambda-1}{x}y=0, \quad y(0)=y(\tfrac{1}{2})=0.$$

My approach: I have tried to use Frobenius Theorem because $x=0$ is a regular-singular point and also the indicial equation implies that the eigenfunction (non-trivial solution) will not a similar form of a Bessel function.

I have managed to use self-adjoint properties but the differential operator of the left hand side turns out to be non self-adjoint.

  • $\begingroup$ This appears to be from a textbook, what textbook was it? I'm curious as I haven't been exposed to this area but it seems awfully close to some of the things I enjoy learning about $\endgroup$ – frogeyedpeas Dec 31 '15 at 1:47
  • 1
    $\begingroup$ I made this question on my own while I am reading Appell's manuscripts on orthogonal polynomials. $\endgroup$ – user31899 Dec 31 '15 at 4:08

When writing $y(x) = x w(z)$ with $z= \lambda x$, the differential equation is transformed into $$ z w''(z)+ (2-z) w'(z) -(\lambda^{-1} -1) w(z) =0$$ which is Kummer's equation. The regular solution to this equation (fulfilling $y(0)=0$) is $$ w(z) = {}_1 F_1(\lambda^{-1} -1; 2; z).$$ The first positive eigenvalue, corresponds to the first zero of the function $$ f(\lambda) = w(\lambda/2).$$

Numerics shows that this is situated at $$\lambda \approx 4.60571.$$

  • $\begingroup$ Can you derive the general solution of the ODE using other methods? For example, power series solutions or integrating factors or considering the ODE as a Sturm-Liouville equation? $\endgroup$ – johnny09 Dec 31 '15 at 2:49
  • $\begingroup$ Could you clarify why we choose $f(\lambda) = w(\lambda/2)$? $\endgroup$ – user31899 Dec 31 '15 at 3:37
  • $\begingroup$ @user31899: we want to fulfil the boundary condition $y(x=1/2)=0$. Now $x=1/2$ translates to $z=\lambda/2$ and thus we need to have $w(\lambda/2) =0 $. $\endgroup$ – Fabian Dec 31 '15 at 3:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.