Using Lax-Milgram for linear ODEs

Question

Consider an Sturm-Liouville deferential equation as: $$Lu=(pu')'+qu$$ and differential equation as: $$Lu+f=0$$ where $u(a)=u(b)=0$. We can convert the problem into a Lax-Milgram form for $f\in C[a,b]$ as follows: $$B(u,\phi)=<f,\phi>, \forall \phi \in C_c^\infty(a,b)$$ Assuming all the necessary conditions of Lax-Milgram theorem hold, I am wondering in which space the unique solution $u^*$ evolves?

In addition, for $f\in L_2[a,b]$, we can again construct a Lax-Milgram form as: $$B(u,v)=F(v), \forall v\in H^1[a,b]$$ Again assume all necessary conditions hold, what is the space of solution?

Answer 1

Let $L$ be the differential operator given by $$Lu=(pu')'+qu$$ A weak solution of the Sturm–Liouville problem $$\left\{\begin{aligned} Lu+f=0&\quad\text{on}\quad (a,b)\\ u(a)=u(b)=0& \end{aligned}\right.\tag{1}$$ is a function $u^*\in H_0^1(a,b)$ such that $$-\int_a^bp(u^*)'v'+\int_a^bqu^*v=-\int_a^bfv,\qquad\forall\ v\in H_0^1(a,b)$$

Theorem: Assume $q\in C([a,b])$ and $p\in C^1([a,b])$ with $p(x)\leq \alpha<0$ for all $x\in(a,b)$.

• If $f\in L_2(a,b)$, then the problem $(1)$ has an unique weak solution $u^*\in H_0^1(a,b)$. Furthermore, $u^*\in H^2(a,b)$.
• If $f\in C([a,b])$, then the weak solution $u^*$ of $(1)$ belongs to $C^2([a,b])$ and in fact is a classical solution of $(1)$.

Proof: Brezis book, pages 223-224.

Remark: In the proof we consider the bilinear form $B:H_0^1(a,b)\times H_0^1(a,b)\to\mathbb{R}$ and the functional linear $F:H_0^1(a,b)\to\mathbb{R}$ given by $$B(u,v)=-\int_a^bpu'v'+\int_a^bquv,\qquad F(v)=-\int_a^bfv$$