# How to show existence and uniqueness of a SL problem with von Neumann BCs

Let $f\in C[0,1]$ be a continuous function and consider for $x\in(0,1)$ the Sturm-Liouvile problem $$-u''(x)+x\cdot u(x)=f(x) \tag1$$ where $u'(0)=u'(1)=0.$

I need to show that for any $f\in C[0,1]$ there is a unique $u\in C^2[0,1]$ that satisfies (1).

Is there someone who knows a good book where I can find this result?

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Hint: this question is probably related to the Airy function: en.wikipedia.org/wiki/Airy_function – Siminore Jun 11 '12 at 12:22

Uniqueness is easy, as your equation is linear.

Suppose $u,v$ are both solutions. Then it suffices to show that the function $w = u-v$, being their difference, vanishes everywhere. By subtracting the equations satisfied by $u$ and $v$, we get

$$- w'' + x w = 0$$

with boundary condition $w'(0) = w'(1) = 0$. Multiply your equation by $w$ and integrate, you get

$$0 = \int_0^1 w(-w'' + xw) \mathrm{d}x = \int_0^1 (w')^2 + xw^2 \mathrm{d}x$$

where we integrated by parts and used the boundary condition that $w'(0) = w'(1) = 0$. But as $x\in[0,1]$ and hence the integrand is manifestly non-negative, for the total integral to be 0 it is necessary that

$$(w')^2 + xw^2 = 0 \implies (w')^2 = w^2 = 0 \implies w \equiv 0$$

Existence can be obtained as an application of the variation principle. One can consult Gelfand and Fomin, Calculus of Variations for an exposition. The usual way would be to consider the Hilbert space $H$ of functions as the completion in the $H^1$ norm (both $u$ and $u'$ measured in $L^2$) of $C^1$ functions with vanishing derivatives at the boundary. One first show that on $H$ the functional $\int_0^1 (u')^2 + x u^2 - 2fu \mathrm{d}x$ is bounded from below, further more any local minimum must have bounded $H^1$ norm. This implies that we can find a bounded minimising sequence, which necessarily has a weakly converging subsequence which also converges strongly in $L^2$. With the observation that if $u_n \rightharpoonup u$ then $\|u\| \leq \liminf \|u_n\|$ with equality only when $u_n\to u$, we see that the fact that the sequence is minimising implies that $u_n \to u$ strongly. Hence we have an $H^1$ weak solution. Using the Sobolev embedding theorems, we have that $H^1[0,1]\hookrightarrow C^0[0,1]$ and hence the solution must be continuous. Using the equation this implies the solution must be $C^2$.

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