# Singular extremal of a constrained variational problem

Consider the following constrained variational problem: $$\min_{u \in H^{1}(I)} \{\mathcal{F}(u) : u(\pm 1) = 1, \mathcal{G}(u) = 1/3 \},$$ where $$I = [-1, 1] \subseteq \mathbb{R}, H^1 (I) := H^{1, 2}(I)$$ is a one-dimensional Sobolev space and $$\mathcal{F}(u) := \int_I |u'(x)|^2 dx, \;\;\;\;\mathcal{G}(u) := \int_I \left(x^2 u(x) -\frac{u(x)^3}{3} \right)dx$$ are respectively the objective and constraint functionals.

If $$u$$ is a $$\mathcal{G}$$-regular extremal, then the Euler-Lagrange equation leads to the following boundary value problem: $$u''(x) = \lambda(u^2(x)-x^2), \;\;\; u(\pm 1) = 1.$$ However, it turns out that there is also a $$\mathcal{G}$$-singular extremal (for which $$\frac{d}{d\epsilon} \mathcal{G}(u + \epsilon \psi)|_{\epsilon = 0} = 0 \; \forall \psi \in C_c^{\infty}(\mathring{I}))$$, namely $$u(x) = |x|$$. The question is the following: is $$u(x) = |x|$$ a minimum of the described problem?

The approaches I have tried involve trying to study the previous differential equation or searching for ways to modify the singular solution to get lower values for the energy, but neither way has yielded results.

Here's a partial answer. Extended functional \begin{align} S[u] ~=~& F[u] +\lambda(G[u]-\frac{1}{3}), \cr F[u]~=~&\frac{1}{2}\int_I\!\mathrm{d}x~u^{\prime 2}, \cr G[u]~=~&\int_I\!\mathrm{d}x~(x^2u-u^3/3). \end{align}\tag{1} \begin{align}u(x)~=~&|x|+v(x), \cr u^{\prime}(x)~=~&{\rm sgn}(x)+v^{\prime}(x).\end{align}\tag{2} \begin{align} F[|\cdot|+v]~=~&\frac{1}{2}\int_I\!\mathrm{d}x(1+2{\rm sgn}(x)v^{\prime}+ v^{\prime 2})\cr ~\stackrel{\rm IBP}{=}~&1+\frac{1}{2}\int_I\!\mathrm{d}x(-4\delta(x)v+ v^{\prime 2})\cr ~=~&1-2v(0)+\frac{1}{2}\int_I\!\mathrm{d}x~v^{\prime 2}. \end{align}\tag{3} \begin{align} 0~\stackrel{?}{=}~G[|\cdot|+v]-\frac{1}{3}~=~&\int_I\!\mathrm{d}x((|x|+v)x^2-(|x|x^2+3x^2v+3|x|v^2+v^3)/3) \cr ~=~&-\int_I\!\mathrm{d}x (\underbrace{|x|v^2}_{\geq 0}+v^3/3). \cr \end{align}\tag{4} We see that an infinitesimal deformation $$v$$ maintains the constraint, it vanishes $$v\equiv 0$$. So $$u=|\cdot|$$ is at least a local minimum.
• Sorry I don't understand why an infinitesimal deformation that maintains the constraint must vanish: can't the term $v^3/3$ inside the integral compensate? May 15 at 13:35
• Oh yeah, that makes sense. So this proves that $|\cdot|$ is isolated right? May 15 at 14:04
• $\uparrow$ Right. May 15 at 14:11
• @LorenzoCatani One should be way more careful with Qmechanic's "isolated" claim. In $C([0,1])$ it is just false and in $H^1$, which is the natural space for the problem, it is not at all obvious, even if true (I haven't tried to figure that out). The problem is that the "infinitesimal deformation" may also have "infinitesimal support" near the origin and then the "higher order" argument fails dramatically. May 22 at 15:21