Solution. By Cauchy–Schwarz and Poincaré with \(C = d/\pi \), \(\int fw \leq \|f\|\,\|w\| \leq C\|f\|\,\|w\|_E\), so \(E[w] \geq \frac 12\|w\|_E^2 - C\|f\|\,\|w\|_E \geq -\frac 12C^2\|f\|^2\), the minimum of the quadratic \(\frac 12s^2 - C\|f\|s\).

Solution. \(u = \frac 12x(1 - x)\), with \(\int u'^2 = \int (\frac 12 - x)^2 = \frac 1{12}\) and \(\int u = \frac 1{12}\), so \(E[u] = \frac 1{24} - \frac 1{12} = -\frac 1{24} = -0.04167\). For \(w\): \(\int w'^2 = \frac {\pi ^2}4\cdot \frac 12 = \frac {\pi ^2}8\) and \(\int w = \frac 1\pi \), so \(E[w] = \frac {\pi ^2}{16} - \frac 1\pi = 0.6169 - 0.3183 = 0.2986\), far above, because the amplitude is wrong. The best multiple \(c\sin \pi x\) minimises \(\frac {\pi ^2}4c^2 - \frac {2c}\pi \), at \(c = 4/\pi ^3\), with energy \(-4/\pi ^4 = -0.04106\), just above \(-\frac 1{24}\).

Solution. Multiply by \(v\) vanishing at the ends and integrate by parts: \(\int _0^1au'v'\,dx = \int _0^1fv\,dx\) for all \(v \in H^1_0(0, 1)\). The form \(a(u, v) = \int au'v'\) is symmetric with \(a(v, v) \geq a_0\int v'^2 = a_0\|v\|_E^2\), so it is coercive with \(\alpha = a_0\), and bounded by \(a_1\). Lax–Milgram (or Riesz with this inner product) gives a unique weak solution, even when \(a\) is discontinuous, as for a rod made of two materials.

Solution. \(\int |\nabla w|^2 = \int _0^14r^2\cdot 2\pi r\,dr = 2\pi \) and \(\int w^2 = 2\pi \int _0^1(1 - r^2)^2r\,dr = \frac \pi 3\), so \(R = 6\), against \(j_{01}^2 = 5.783\): an upper bound \(3.7\%\) high, and for the frequency \(\sqrt R\), only \(1.9\%\).

Solution. The weak form is \(\int \nabla u\cdot \nabla v = \int fv\) for all \(v \in H^1(\Omega )\), with no boundary condition on \(v\) (the Neumann condition is natural: it comes out of the integration by parts). Taking \(v = 1\) gives \(0 = \int f\). If \(u\) is a solution so is \(u + c\), since \(\nabla c = 0\); on the subspace of functions with mean zero the form is coercive by the Poincaré–Wirtinger inequality, and the solution is unique there.

Solution. \(\frac {d}{dt}\int u_x^2 = 2\int u_xu_{xt} = -2\int u_{xx}u_t = -2\kappa \int u_{xx}^2 \leq 0\), integrating by parts with \(u_t = 0\) at the ends. Hence \(\int u_x(x, t)^2 \leq \int g'^2\). Integrating the identity \(\frac {d}{dt}\int u^2 = -2\kappa \int u_x^2\) from \(0\) to \(t\) and using the monotonicity gives the smoothing estimate \(2\kappa t\int u_x(x, t)^2 \leq \int g^2\): the gradient becomes finite at once, even for rough data.