The problem:
Let $(T\phi)(x)=-\phi ''(x) + x\phi(x)$ be a linear operator acting on the space of functions $$V:=\left\{ {\phi:[0,1] \to \mathbb{R}}; \phi(0)=\phi(1)=0\right\}.$$ Estimate the smallest eigenvalue of T.
Okay so this question was totally left field. Reading in my book, I found no true way of making this as general as I assume it needs to be and it does not do a good job of explaining Sturm-Liouville Theory (this equation is in regular Sturm-Liouville form). Here's what I've attempted:
Obeserve that our equation can re-written as $$-(\phi ')' + x\phi$$ where it is now clear that we're dealing with a regular Sturm-Liouville equation ($[s(x)\phi']' +q(x)\phi +\lambda^2 p(x)\phi)$. Let's find a lower bound for overall equation by multiplying through $\phi$ into our original equation and then integrating on our interval: $$\int_{0}^1 [-\phi''\phi+x\phi^2]dx$$ Performing integration by parts we end up obtaining $$\int_{0}^1 [(\phi')^2+x\phi^2]dx$$ Notice that on our interval, our integral will never be negative $$\implies \int_{0}^1 [(\phi')^2+x\phi^2]dx \geq 0$$ so $0$ is a lower bound.
Now recall that Rayleigh's method is estimating the first eigenvalue, where we end up with the Rayleigh quotient $$\lambda _1^2 \leq \frac{N(y)}{D(y)}$$ where in the most general case we have $$N(y)=\int_{lower}^{upper} [(s(x)\phi')^2+q(x)\phi^2]dx$$ $$D(y)=\int_{lower}^{upper} [p(x) \phi^2]dx$$.
So where do I proceed from here? I feel like if I just consider a function $y(x)$ that satisfies our boundary value conditions, there's always a chance we can find another function that has smaller eigenvalue. Any help would be greatly appreciated
The eigenfunctions to the operator $T$ are those satisfying
$$-\phi'' + x\phi = \lambda\phi$$
where $\lambda$ is the eigenvalue.
Switching variables to $y = x-\lambda$ the equation becomes
$$\frac{d^2\phi}{dy^2} = \phi y$$
This is the equation for the Airy functions (http://en.wikipedia.org/wiki/Airy_function). The general solution to the eigenvalue problem can be written
$$\phi(x) = a Ai(x-\lambda) + b Bi(x-\lambda) $$
Now imposing the boundary condition at $x=0$
$$0 = a Ai(-\lambda) + b Bi(-\lambda) $$
Certainly $a=b=0$ is a solution (for any $\lambda$), I will neglect this. This leads to
$$\phi(x) = C\left[Ai(x-\lambda)Bi(-\lambda) - Ai(-\lambda) Bi(x-\lambda)\right]$$
for some constant $C\not = 0$. Now for the second boundary condition to be satisfied we need
$$Ai(1-\lambda)Bi(-\lambda) = Ai(-\lambda)Bi(1-\lambda)$$
It can be shown, using the properties of the Airy functions, that this equation has a smallest solution which numerically turns out to be $\lambda \approx 10.3$.