# Is the minimum of this functional $C^{\infty}$?

The problem:

Let us define $$\mathscr{F}(u)=\int_0^1(u'(x))^4-e^x\sin (u(x))\, \mathrm{d}x$$ for $$u \in W^{1,1}([0,1])$$ such that $$u(0)=A$$ and $$u(1)=B$$. It don't matters what $$A$$ or $$B$$ are, because I am interested in the reasoning.

I am asked to study if the minimum $$u$$ is such that $$u \in C^{\infty}([0,1])$$ or eventually $$u \in C^{\infty}([0,1]-E)$$ where $$E$$ is a closed and negligible subset of $$[0,1]$$, i.e. it is closed and $$\mu(E)=0$$.

An attempt:

First of all, thanks to Ioffe's theorem, we know that $$\mathscr{F}$$ is sequentially weakly lower semi continuous in $$W^{1,1}([0,1])$$. Further more, because $$(u'(x))^4-e^x\sin (u(x)) \geq (u'(x))^4-e$$, a minimum $$u \in W^{1,1}$$ exists.

But what can be said about the regularity of $$u$$? I know the Tonelli’s partial regularity theorem but I cannot directly apply it here because $$F_{pp}$$ is not defined positive but we only have $$F_{pp} \geq 0$$. I know that if $$u'(x) \neq 0$$ then $$u$$ is $$C^{\infty}$$ in a neighborhood of $$x$$.

My question is: $$u \in C^{\infty}([0,1])$$?

Further more I would like to understand when can we apply Tonelli’s theorem if $$F_{pp} \geq 0$$ but not $$F_{pp} > 0$$ i.e. if $$F_{pp}$$ is positive semidefinite but not positive definite.

Remark: I tried first solving the related Question.

• If you partition the Euler-Lagrange equations as $p(x)=u'(x)^3$, $p'(x)=-\frac14e^x\cos(u(x))$, then this looks like a continuous first order system if one takes the signed real cube root. However, at $p=0$ it is not Lipschitz. But it is sub-linear, so global existence for IVP is not a problem, uniqueness is in doubt and what influence this has on the solvability of BVP I do not know. Jul 27, 2020 at 16:22
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• If $u$ is continuously differentiable and its supremum is larger than $\pi$, then we can define $\overline{u} := u\wedge \frac{\pi}{2} := \min(u,\frac{\pi}{2})$. In that case note that $(\overline{u}'(x))^4 \leq (u'(x))^4$ for almost all $x \in [0,1]$ and $\sin(\overline{u}(x)) \geq \sin(u(x))$ for all $x \in [0,1]$. So $\mathscr{F}(u') \leq \mathscr{F}(u)$.  Could you show that $\int (\overline{u}'(x))^4\,dx \leq \int (u'(x))^4\,dx$ when $u'$ is a weak derivative? If so, that would imply that $\mathscr{F}(\overline{u}) \leq \mathscr{F}(u)$ for all $u \in W^{1,1}$. Nov 13, 2020 at 1:20
• It's been a while since I worked with Sobolev spaces so I've forgotten all the interpolation and density results. Is $C^1$ dense in $W^{1,1}$ in some topology? If so, that might be enough. Nov 13, 2020 at 1:23

$$F(u)=\int_{0}^{1}(u^{\prime}(x))^{4}-e^{x}\sin(u(x))\,dx.$$ If $$u$$ is a minimum, then for every $$v\in C_{0}^{1}([0,1])$$ and every $$t\in\mathbb{R}$$, $$g(t):=F(u+tv)\geq F(u)=g(0),$$ and so $$0=g^{\prime}(0)=\int_{0}^{1}4(u^{\prime}(x))^{3}v^{\prime}(x)-e^{x} \cos(u(x))v(x)\,dx.$$ It follows that the weak derivative of $$4(u^{\prime}(x))^{3}$$ is the function $$-e^{x}\cos(u(x))$$. Thus, $$4(u^{\prime})^{3}\in W^{1,1}((0,1))$$. By a characterization of Sobolev functions of one variables, we have that $$4(u^{\prime})^{3}(x)=-\int_{0}^{x}e^{t}\cos(u(t))\,dt$$ for a.e. $$x\in\lbrack0,1]$$. It follows that $$u^{\prime}(x)=\left( -\frac{1}{4}\int_{0}^{x}e^{t}\cos(u(t))\,dt\right) ^{1/3}=:f(x)$$ Since the right-hand side is continuous, the function $$u^{\prime}$$ has a continuous representative. Since $$u\in W^{1,1}((0,1))$$, we have that $$u(x)=\int_{0}^{x}u^{\prime}(t)\,dt=\int_{0}^{x}f(t)\,dt$$ for a.e. $$x\in\lbrack0,1]$$. Define $$\bar{u}(x):=\int_{0}^{x}f(t)\,dt.$$ Then $$\bar{u}$$ is a representative of $$u$$ of class $$C^{1}$$ and $$\bar {u}^{\prime}(x)=f(x)$$ for all $$x\in\lbrack0,1]$$. So $$\bar{u}^{\prime}(x)=\left( -\frac{1}{4}\int_{0}^{x}e^{t}\cos (u(t))\,dt\right) ^{1/3}=\left( -\frac{1}{4}\int_{0}^{x}e^{t}\cos(\bar {u}(t))\,dt\right) ^{1/3}.$$ Consider the set $$V=\{x\in(0,1):\,\bar{u}^{\prime}(x)\neq0\}$$. Since $$\bar {u}^{\prime}\$$is continuous, this set is open. If $$x\in V$$, we can differentiate to obtain $$\bar{u}^{\prime\prime}(x)=-\frac{1}{3}\left( -\frac{1}{4}\int_{0}^{x} e^{t}\cos(\bar{u}(t))\,dt\right) ^{-2/3}\frac{1}{4}e^{t}\cos(\bar{u}(x)).$$ Since the right-hand side is $$C^{2}$$ in $$V$$, we can keep differentiating to show that $$\bar{u}$$ is $$C^{\infty}$$ in $$V$$. It remains to study what happens in $$[0,1]\setminus V$$.
Define $$h(x):=(\bar{u}^{\prime}(x))^{3}=-\frac{1}{4}\int_{0}^{x}e^{t}\cos(\bar {u}(t))\,dt.$$ Since the right hand-side is of class $$C^{2}$$, we have that $$h^{\prime}(x)=-\frac{1}{4}e^{x}\cos(\bar{u}(x))$$ for all $$x\in\lbrack0,1]$$.
If $$h(x)=0$$, and $$\bar{u}(x)\neq\frac{\pi}{2} +2k\pi$$, then $$h^{\prime}(x)\neq0$$. In particular, $$x$$ is isolated.
If $$\bar{u}(x_{0})=\frac{\pi}{2}$$ and $$\bar{u}^{\prime}(x_{0})=0$$, then I am not sure about what to do.
If we had uniqueness of the ODE, then I would be tempted to say that the constant solution $$\bar{u}=\frac{\pi}{2}$$ is the only solution (this would require $$A=B=\frac{\pi}{2}$$). But it is not obvious to me why there is uniqueness, since we have $$\bar{u}(t)=A+\int_{0}^{x}\bar{u}^{\prime}(t)\,dt=A+\int_{0}^{x}h^{1/3}(t)\,dt,$$ and so $$h^{\prime}(x)=-\frac{1}{4}e^{x}\cos\left( A+\int_{0}^{x}h^{1/3}% (t)\,dt\right) .$$ The ODE $$p^{\prime}=p^{1/3}%$$ does not have a unique solution exactly at points $$p=0$$.