= Solution
{c}
Taking $w=z$ in the dual problem, and using $e^x\geq1$, gives
$$
\|z\|_{H^1}^2\leq a(z,z)=\int_0^1z\psi
\leq\|z\|_2\|\psi\|_2\leq\|z\|_{H^1}\|\psi\|_2.
$$
Hence $\|z\|_{H^1}\leq\|\psi\|_2$. Testing with smooth compactly supported functions identifies $-z''+e^xz=\psi$ in <distributions>. Therefore $z''\in L^2$ and $z\in H^2(0,1)$, with
$$
\boxed{\|z''\|_2\leq e\|z\|_2+\|\psi\|_2\leq(e+1)\|\psi\|_2.}
$$
<integration by parts> with arbitrary endpoint test values also gives the natural <Neumann boundary conditions> $z'(0)=z'(1)=0$.
Back to article page
