0

Considering the Sturm-Liouville problem $$-y'' = \lambda y, $$ with $y(0)=0$ and $y'(1)-y(1)=0$, and where we let $\lambda_0 < \lambda_1 < \lambda_2 <...$ be the eigenvalues in increasing order, I want to show that $\lambda_0 = 0$ and that $(\pi n)^2 < \lambda_n < ((n+\frac{1}{2})\pi)^2$ for $n \geq 1$.

I've found that the general solution is $y(x) = A\cos(\sqrt{\lambda} x) + B\sin(\sqrt{\lambda} x)$. So $y(0) = 0$ gives $A=0$ and we're left with $y(x) = B\sin(\sqrt{\lambda}x)$. Then $y'(1)-y(1)=0$ gives $\tan (\sqrt{\lambda}) = \sqrt{\lambda}$.

Is this enough to say that $\lambda$ must be nonnegative, because otherwise $\sqrt{\lambda}$ would be undefined (or at least not real, and the eigenvalues of a Sturm-Liouville problem must be real)?

Clearly $\tan (\sqrt{\lambda}) = \sqrt{\lambda}$ is satisfied by $\lambda = 0$, so if the above was enough to show that $\lambda$ is nonnegative, then we have found that $\lambda_0 = 0$.

But then I am pretty lost on how to show that $(\pi n)^2 < \lambda_n < ((n+\frac{1}{2})\pi)^2$ for $n \geq 1$.

I know that $\tan (k)$is strictly increasing on each interval $\left((n-\frac{1}{2}\pi), (n+\frac{1}{2})\pi \right)$, and positive when $k \in \left(n\pi, (n+\frac{1}{2})\pi \right)$. But I can't seem to figure out why this implies that $\lambda_n \in \left((n\pi)^2, ((n+\frac{1}{2})\pi)^2\right)$. Why couldn't we have for example $\lambda_1 \in \left((2\pi)^2,(\frac{5\pi}{2})^2\right)$?

rickss
  • 137
  • Instead of considering $\tan x=x$, use the equivalent $x=n\pi+\arctan x$ for $x\in ((n-0.5)\pi,(n+0.5)\pi)$. Or consider directly the function $f(x)=\sin x-x\cos x$ and what values it has at the points $x=k\frac\pi2$. – Lutz Lehmann Mar 21 '19 at 10:53

1 Answers1

2

First, if $\lambda$ were negative, your general solution would not be given by $$ y(x) = A\cos(\sqrt\lambda x) + B\sin(\sqrt\lambda x), $$ but rather $$ y(x) = Ae^{\sqrt\lambda x} + Be^{-\sqrt\lambda x}, $$ so you should eliminate this case separately (which should not be too hard by using the boundary conditions to conclude that $y\equiv 0$, and so it cannot be an eigenvector). You should also consider $\lambda = 0$ separately, in which case you have the solution $$ y(x) = Ax+B, $$ but in this case you will be able to conclude that $y$ is in fact an eigenvector.

As for your question regarding why $\lambda_n \in \left((n\pi)^2, ((n+1/2)\pi)^2\right)$, observe that the image of $(n\pi, (n+1/2)\pi)$ under $t \mapsto \tan t$ is $\mathbb{R}_+$, so that the graphs of the continuous functions $$t \mapsto t \quad\text{and}\quad t\mapsto \tan t$$ must intersect at least once in each of the intervals $(n\pi, (n+1/2)\pi)$ (essentially as a consequence of the intermediate value theorem applied to the function $t\mapsto \tan t-t$)

Can you conclude on your own that they intersect exactly once in each of these intervals?

MSDG
  • 7,143