I am solving an exercise concerning the Airy eigenvalue problem $$ -y''+xy =\lambda x, \quad y(0)=y(1)=0, \quad (*) $$ which (among other things) asks me to prove that all eigenvalues are positive. I would guess that there are simpler ways to reach this result, but I would like to make the argument below work.
In a previous part of the exercise, I was asked to find two linearly independent solutions of the homogeneous Airy equation $$ y''=xy, $$ which I found to be $$ y_1(x)=\sum_{k=0}^\infty \frac{x^{3k+1}}{\prod_{n=1}^k(3n+1)(3n)}, \quad y_2(x)=\sum_{k=0}^\infty \frac{x^{3k}}{\prod_{n=1}^k(3n)(3n-1)}. $$ Here I use the convention $\prod_{n=1}^0 a_n = 1$.
Now after applying a translation in $(*)$, one can show that for an eigenvalue $\lambda$, the corresponding eigenvector is given by $$ u(x) = Ay_1(x-\lambda)+By_2(x-\lambda), $$ where the constants are chosen according to the boundary conditions.
To show that $\lambda$ must be positive, I was expecting to be able to show that if $\lambda \leq 0$, we must have $A=B=0$ so that $u$ cannot be an eigenvector. I was able to make this work for $\lambda = 0$, but the case $\lambda < 0$ seems tougher.
If $\lambda < 0$, I let $\lambda = -\omega$ where $\omega>0$ for convenience, and then use the boundary conditions to conclude that $$\begin{cases} 0 = u(0) = Ay_1(\omega) + By_2(\omega)\\ 0 = u(1) = Ay_1(1+\omega) + By_2(1+\omega) \end{cases}.$$ To complete the argument it would therefore suffice to prove that $$ \det \begin{pmatrix} y_1(\omega) & y_2(\omega)\\ y_1(1+\omega) & y_2(1+\omega) \end{pmatrix} \not=0. $$ Is there an elegant way to do this?
The BVP is
$$ -y'' + xy = \lambda y $$
Multiply through by $y$ and integrate over $(0,1)$
$$ -\int_0^1 yy'' dx + \int_0^1 x y^2\ dx = \lambda\int_0^1y^2 \ dx $$
If $y$ satisfies the boundary conditions, then from integration by parts
$$ \int_0^1 yy'' dx = -\int_0^1 (y')^2\ dx $$
Therefore
$$ \int_0^1 (y')^2 dx + \int_0^1 xy^2 \ dx = \lambda \int_0^1 y^2\ dx $$
Since all the integrals are positive, this can only be satisfied if $\lambda > 0$