If $f\in L^2$, has the equation $u_{xx}=f$ an unique solution $u\in H^2\cap H_0^1$? Let $-\infty<a<b<+\infty$ and $f\in L^2(a,b)$.
Is it possible to prove that the equation $u_{xx}=f$ has an unique solution $u\in H^2(a,b)\cap H_0^1(a,b)$? If so, how can we prove it?
Thanks.
 A: You can write down the formula for $g$ directly. First,
$$F(x)=\int_a^x f(t)\,dt$$
is in $H^1(a,b)$ by construction. Second, 
$$G(x)=\int_a^x F(t)\,dt$$
is in $H^2(a,b)$ by construction. Finally, 
$$u(x) = G(x) - \frac{x-a}{b-a}G(b)$$ has the same second derivative as $G$ (namely $f$) and satisfies the boundary conditions. 
Uniqueness follows from integration by parts. If $u,v$ are two solutions, then $w=u-v$ has zero second derivative, which implies
$$
\int_a^b (w')^2 = -\int_a^b ww'' = 0 
$$
Hence $w$ is constant, and by virtue of the boundary conditions the constant is $0$.
A: The bilinear form
\begin{align}
B:H^1_0(a,b)\times H_0^1(a,b)&\longrightarrow \mathbb{R}\\
(w,v)&\longmapsto\int_a^b w_x(x)v_x(x)\ dx
\end{align}
is continuous and coercive. Furthermore, the linear functional
\begin{align}
\Lambda: H_0^1(a,b)&\longrightarrow \mathbb{R}\\
v &\longmapsto-\int_a^b f(x)v(x)\ dx
\end{align}
is continuous. Thus, by the Lax-Milgram Theorem (Brezis, page 140), there exists an unique $u\in H_0^1(a,b)$ such that
$$\int_a^b u_x(x)v_x(x)\ dx=B(u,v)=\Lambda(v)=-\int_a^bf(x) v(x)\ dx,\quad\forall\ v\in H_0^1(a,b).$$
The last equality shows that $u_x\in H^2(a,b)$ with $u_{xx}=f$. Hence $u$ is the desired solution.
Remark: a similar argument can be applied to more general equations like $-u_{xx}+cu=f$ with $c\geq 0$ constant.
