= Solution
Fix an admissible $u$. The preceding result gives $\int_0^bu'^2\,dx\geq\lambda^2\int_0^bu^2\,dx$ for every $\lambda^2<\pi^2/b^2$. Letting $\lambda^2$ increase to $\pi^2/b^2$ proves
$$
\boxed{\int_0^bu'^2dx\geq\frac{\pi^2}{b^2}\int_0^bu^2dx.}
$$
Equality is attained by
$$
\boxed{u(x)=A\sin(\pi x/b),\qquad A\in\mathbb R.}
$$
Indeed the <derivative> and function squared integrals are $A^2\pi^2/(2b)$ and $A^2b/2$. These are all equality cases: an equality function minimizes the nonnegative threshold <quadratic form>, so its <first variation> vanishes. The resulting weak equation $u''+(\pi/b)^2u=0$ makes $u$ smooth and, with the endpoint conditions, yields precisely these multiples. This is the <Wirtinger inequality>.
Back to article page
