If $f\in L^2$, has the equation $u_{xx}=f$ an unique solution $u\in H^2\cap H_0^1$?

Question

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.

Answer 1

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 $0$.