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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?

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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 $\lambda > 0$

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  • $\begingroup$ Very nice argument! However I asked specifically if it is possible to make the argument presented in my post work. $\endgroup$ – MisterRiemann Mar 9 at 15:35
  • $\begingroup$ @MisterRiemann Are you allowed to use the well-known properties of the Airy functions $A(x)$ and $B(x)$? such as they're both postive for $\forall x>0$, etc $\endgroup$ – Dylan Mar 9 at 15:40
  • $\begingroup$ This is the first time I am encountering them, so I do not feel comfortable using any non-trivial properties without seeing a proof first. (I would guess that $A(x)$ and $B(x)$ are the functions $y_1$ and $y_2$ from my post?) $\endgroup$ – MisterRiemann Mar 9 at 15:44
  • $\begingroup$ They're the functions described here. $A(x)$ decays to $0$ while $B(x)$ blows up at $x\to \infty$ $\endgroup$ – Dylan Mar 9 at 15:45
  • $\begingroup$ I see. I would prefer not to invoke any of their properties in that case, if possible. Is there really no simple way to directly show that the determinant from my post is non-zero? $\endgroup$ – MisterRiemann Mar 9 at 15:47

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