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.

  • 1
    $\begingroup$ 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. $\endgroup$ Jul 27, 2020 at 16:22
  • 1
    $\begingroup$ Some elementary comments: 1) The stationary path satisfies $12u'^2u''+e^x\cos u=0$. 2) The Jacobi equation is $\frac{2v'}{u'^2}+\frac{v''}{u''}+v\tan u=0$ with initial conditions $v(0)=0$ and $v'(0)=1$. $\endgroup$
    – TheSimpliFire
    Jul 29, 2020 at 12:35
  • 1
    $\begingroup$ 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 $u \in W^{1,1}$. $\endgroup$ Nov 13, 2020 at 1:20
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 13, 2020 at 1:23

1 Answer 1


$$ F(u)=\int_{0}^{1}(u^{\prime}(x))^{4}-e^{x}\sin(u(x))\,dx. $$ 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 $p=0$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .