If we don't care about uniqueness, can we relax the coercivity condition in Lax-Milgram theorem?

Question

Let $(H, \langle \cdot, \cdot \rangle)$ be a real Hilbert space and $\|\cdot \|$ its induced norm. Let $a: H \times H \to \mathbb R$ be a bilinear form. We say that

• $a$ is coercive IFF there is $C>0$ such that $a(u, u) \ge C \|u \|^2$.
• $a$ is continuous IFF there is $C>0$ such that $|a(u, v)| \le C \|u \| \|v \|$.

Lax–Milgram theorem says that if $a$ is a continuous coercive bilinear form on $H$, then for each $\varphi \in H^*$ there is a unique $u \in H$ such that $$ a(u, v) = \varphi (v), \quad \forall v \in H. $$

Assume that we only want the existence and not necessarily the uniqueness. Can we relax any of above assumtions?

My question arises when I look for the solution of Sturm–Liouville problem $-(pU')' + qU = f$ with a boundary condition of only one endpoint, i.e.,

Let $I$ be the open interval $(0, 1)$. Assume that $p \in C^1(\bar I)$ and $q \in$ $C(\bar I)$ such that $p(x) \geq \alpha \in \mathbb R_{>0}$ and $q(x) \geq 0$ for all $x \in \bar I$. Let $f \in L^2 (I)$. We want to find $U \in H^1(I)$ such that $U(0)=1$ and that for each $V \in C_c^\infty (I)$, $$ \int_I pU' V' + \int_I qUV = \int_I fV. $$

Answer 1

This is more suited for math.SE but I'll still post the answer here, although the post will likely be closed soon. You can simply apply Lax-Milgram in the Hilbert space $$ H=\Big\{f\in H^1(0,1): \qquad f(0)=0\Big\} $$ equipped with the usual norm $$ \|f\|:=\int_0^1 |f'|^2. $$ With the left boundary condition $f(0)=0$ this is clearly equivalent to the full $H^1$ norm due to Poincaré's inequality. Looking for a solution in the form $U(x)=1+u(x)$ your problem is clearly equivalent to solving $$ \begin{cases} -(pu')'+q u =g \\ u(0)=0 \end{cases} \qquad \text{with }g=f-q\in L^2. $$ Then the bilinear form $a(u,v)=\int p u'v'$ is continuous, $\alpha$-coercive, and the linear form $\phi(v)=\int gv$ is clearly continuous. This gives uniqueness altogether.