# Least positive eigenvalue of the BVP $y''-\lambda y'+\frac{2\lambda-1}{x}y=0$, $y(0) = y(1/2) = 0$

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.

• 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 – frogeyedpeas Dec 31 '15 at 1:47
• I made this question on my own while I am reading Appell's manuscripts on orthogonal polynomials. – 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 $$\lambda \approx 4.60571.$$
• Could you clarify why we choose $f(\lambda) = w(\lambda/2)$? – user31899 Dec 31 '15 at 3:37
• @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 $w(\lambda/2) =0$. – Fabian Dec 31 '15 at 3:40