Exercise (1.4) of Renardy and Rogers: "An introduction to partial differential equations" asks to
Show that there is an infinite family of minimizers of $$ J(u) = \int_0^1 (1-u'(t)^2)^2\,dt $$ over the set of all piecewise $C^1$ functions satisfying $u(0)=u(1)=0$.
Now the corresponding Euler-Lagrange equation is
$$ \frac{d}{dt} \left[4 (1-u'(t)^2) u'(t) \right] = 0. $$
I would argue that the equation above has no solutions satisfying the given boundary conditions. Integrating once leaves us with
$$ (1-u'(t)^2) u'(t) = C $$
and, depending on $C$, there might be zero to three solutions. However, any solution, if it exists, requires $u'(t)=\mathrm{const}$ and, consequently, $u(t)$ is linear and hence cannot satisfy the boundary conditions (except for the trivial solution).
I suspect that the key to the solution lies in the fact that we allow piecewise $C^1$ functions. However, I cannot see how we could employ that. Doesn't any discontinuity in the first derivative show up as a ($\delta$) source term in the Euler-Lagrange equation?
I'd be grateful for any pointers. Thanks heaps!
Without solving the Euler-Lagrange equation, it's clear that $J(u)$ is minimized when $1-(u'(t))^2=0$, or $u'(t)=\pm 1$. Here are some examples of piecewise $C^1$ functions satisfying $u(0)=u(1)=0$ and such that $u'(t)=\pm 1$: \begin{align} u_{1}(t)&= \begin{cases} t,&\text{if}\,\,0\leq t\leq\frac{1}{2}, \\ -t+1,&\text{if}\,\,\frac{1}{2}<t\leq 1, \end{cases} \\ u_{2}(t)&= \begin{cases} t,&\text{if}\,\,0\leq t\leq \frac{1}{4}, \\ -t+\frac{1}{2},&\text{if}\,\,\frac{1}{4}<t\leq \frac{3}{4}, \\ t-1,&\text{if}\,\,\frac{3}{4}<t\leq 1, \end{cases} \\ u_{3}(t)&= \begin{cases} t,&\text{if}\,\,0\leq t\leq \frac{1}{4}, \\ -t+\frac{1}{2},&\text{if}\,\,\frac{1}{4}<t\leq \frac{1}{2}, \\ t-\frac{1}{2},&\text{if}\,\,\frac{1}{2}<t\leq\frac{3}{4}, \\ -t+1,&\text{if}\,\,\frac{3}{4}<t\leq 1. \end{cases} \end{align}