During my self study to the calculus of variations I come across this problem. Because of my search, I know what I wanted to do but I need some help to do them.
The function $f:[-1,1] \times \mathbb R \to \mathbb R$ is defined by
$$
f(x,\xi) = \left[ w(x) \xi – 2 x w(x) \sin\left(\frac{\pi}{x}\right) + \pi w(x) \cos\left(\frac{\pi}{x}\right) \right]^2,
$$
where
$$
w(x) =
\begin{cases}
e^{- \frac{1}{x^2}},& x\not = 0\\
0, & x=0
\end{cases}.
$$
In this problem we consider the regularity of solutions of the Euler-Lagrange equation corresponding to the one-dimensional minimization problem
$$
\inf_{u \in X} I[u],
$$
where $X= \big\{ u \in \operatorname{Lip}[-1,1]: u(-1) = u(1)=0 \big\}$ and
$$
I[u] = \int_{-1}^1 f(x, u'(x)) dx.
$$
I want to prove the so-called regularity theorem; Let $\overline{u} \in \operatorname{Lip}[a,b]$ satisfy the integral Euler equation for the functional $\displaystyle \int_a^b f(x,u(x),u'(x)) dx$, where for every $x \in [a,b]$ the function $\xi \to f(x,\overline{u}(x),\xi)$ is strictly convex. Then $\overline{u}$ lies in $C^1 [a,b]$.
As far as reached, to prove the theorem we can solve the following sub-problems, to explain why the regularity theorem does not apply to this minimization problem.
The steps are:
- Showing that $f$ is infinitely differentiable, $\xi \mapsto f(x,\xi)$ is convex and $f_{\xi \xi} (x,\xi) > 0$ holds for all $x$ except for $x=0$.
- Showing that the function
$$\overline{u}(x) = \begin{cases} x^2 \sin \frac \pi x,& x \not = 0\\ 0, & x=0\end{cases}$$
yields the minimum of the minimization problem. Also confirm that this function is Lipschitz continuous on $[-1,1]$. - And we show that there is no other minimizer except for the function $\overline{u}$ above.
- And we need to show that $\overline{u}$ does not belong to $C^1 ([-1,1])$.
The main step is to explain why the regularity theorem does not apply
My Proof:
First we prove that the function $w(x)$ is infinitely differentiable. The function to be infinitely differentiable, one would need to show that in the limit the function $w(x)$, and all its derivatives, go to zero as $x$ goes to $0$. Since $x \mapsto \frac{1}{x}$ is smooth for $x \neq 0$ and $x \mapsto e^x$ is smooth, it is clear that $w$ is smooth for $x \neq 0$.
Suppose $x \ne 0$, then ${w}^{(k)}$ has the form ${w}^{(k)}(x) = e^{-{1 \over x^2}} p_k({1 \over x})$ for some polynomial $p_k$. This is clearly true for $k=0$, so suppose it is true for $k=0,…,n$. Then ${w}^{(n)}(x) = e^{-{1 \over x^2}} p_n({1 \over x})$ and the chain rule gives
$$
\begin{split}
{w}^{(n+1))}(x) & = {w}^{(1)}(x) p_n\left({1 \over x}\right) – {w}^{(0)}(x) p_n'\left({1 \over x}\right) \left({1 \over x^2}\right) \\
& = e^{-{1 \over x^2}} \left[{2 \over x^3}p_n\left({1 \over x}\right)-p_n'\left({1 \over x}\right) \left({1 \over x^2}\right) \right].
\end{split}
$$
If $p_{n+1}(y) = 2 y^3p_n(y)-p_n'(y) y^2 $, then
$$
{w}^{(n+1)}(x) = e^{-{1 \over x^2}}
p_{n+1}\left( {1 \over x} \right),
$$ and so the result is true for all $n$. If $x \neq 0$, we have
$$
e^{-{1 \over x^2}} = {1 \over {e^{1 \over x^2}}}\quad \text{ and }\quad e^{1 \over x^2} \ge \sum_{k=0}^n {1 \over k!} {1 \over x^{2k}}
$$
and thus
$$
e^{-{1 \over x^2}} \le {x^{2n} \over \sum_{k=0}^n {1 \over k!} {x^{2(n-k)}}} \le {x^{2n} \over n!}.
$$
Suppose $p$ is a polynomial of degree $d$. Then for any $n$ we see that there is some constant $K$ such that $|e^{-{1 \over x^2}} p({1\over x})| \le K |x|^{2n-d}$ whenever $0 <|x| \le 1$. In particular, there is some $K$ such that
$|e^{-{1 \over x^2}} p({1\over x})| \le K x^2$ for all $0 < |x| \le 1$. We have ${w}^{(0)}(x) \le x^2$ for all $x$, and so ${w}$ is continuous at $x=0$. Since
$|{w}^{(0)}(x) – {w}^{(0)}(0) -0| \le x^2$, we see that ${w}^{(0)}$ is differentiable at $x=0$, and ${w}^{(1)}(0) = 0$.
Now suppose ${w}^{(k)}$ is differentiable at $x=0$ and ${w}^{(k)}(0) = 0$ for $k=0,…,n$. Then
$|{w}^{(n)}(x) – {w}^{(n)}(0) -0| \le K x^2$ for some $K$ and $|x| \le 1$. Hence
${w}^{(n)}$ is differentiable at $x=0$, and ${w}^{(n+1)}(0) = 0$. With a similar argument we find that the function
$$
\cos {\pi \over x} + i \sin \frac \pi x= \begin{cases}
e^{- \frac{\pi}{x}},& x\not = 0\\
0, & x=0
\end{cases}$$
is infinitely differentiable. Therefore, $\cos{\frac \pi x}$ and $\sin \frac{\pi}{x}$ are smooth functions. Since the summation and product of smooth functions is smooth function then $f$, defined above, is smooth.
Next step is showing $f(x,\xi)$ is convex w.r.t. $\xi$. For notational simplicity we put
$$g=2 x \sin \frac{\pi}{x} + \pi \cos\frac{\pi}{x}.,$$
then for fix $x \in [-1,1]$, we have $f(\xi)= w^2(\xi-g)^2$ is a convex function, for let $\lambda\in [0,1]$, then
$$
\begin{split}
f&(x,\lambda \xi_1 + (1-\lambda) \xi_2) – \lambda \xi_1 f( \xi_1) – (1-\lambda) f(\xi_2) \\
& = w^2(\lambda \xi_1 + (1-\lambda) \xi_2 -g)^2 – \lambda w^2( \xi_1-g)^2 – (1-\lambda) w^2(\xi_2-g)^2 \\
& = w^2(\lambda (\xi_1-g) + (1-\lambda) (\xi_2 -g))^2 – \lambda w^2( \xi_1-g)^2 – (1-\lambda) w^2(\xi_2-g)^2 \\
&=w^2(\lambda^2 (\xi_1-g)^2 +2\lambda(1-\lambda)(\xi_1-g)(\xi_2 -g) + (1-\lambda)^2 (\xi_2 -g)^2 -\lambda( \xi_1-g)^2 -(1-\lambda)(\xi_2-g)^2 )\\
& = w^2( (\lambda^2-\lambda)(\xi_1-g)^2 + 2\lambda(1-\lambda)(\xi_1-g)(\xi_2 -g) + ((1-\lambda)^2 – (1-\lambda) ) (\xi_2 -g)^2)\\
& = w^2(\lambda(1-\lambda)(\xi_1-g)^2 +2\lambda(1-\lambda)(\xi_1-g)(\xi_2 -g) + \lambda(1-\lambda) (\xi_2 -g)^2)\\
& = \lambda(1-\lambda)w^2( (\xi_1-g)^2 + 2)(\xi_1-g)(\xi_2 -g) + (\xi_2 -g)^2) \\
&= \lambda(1-\lambda)w^2((\xi_1-g) + (\xi_2 -g))^2 \geq 0
\end{split}
$$
(The last inequality is as a product of two non-negative numbers and two square numbers is non-negative)
Could you please help with the other steps!
Best Answer
$\newcommand\ol\overline$The steps were:
Show that $f$ is infinitely differentiable, $\xi \mapsto f(x,\xi)$ is convex and $f_{\xi \xi} (x,\xi) > 0$ holds for all $x$ except for $x=0$.
Show that the function \begin{equation} \overline{u}(x) = \begin{cases} x^2 \sin \frac \pi x,& x \not = 0\\ 0, & x=0\end{cases} \tag{0} \end{equation} yields the minimum of the minimization problem. Also confirm that this function is Lipschitz continuous on $[-1,1]$.
Show that there is no other minimizer except for the function $\overline{u}$ above.
Show that $\overline{u}$ does not belong to $C^1 ([-1,1])$.
Step 1: You have already checked that $f$ is infinitely differentiable. Next, \begin{equation*} f(x,\xi)=w(x)^2(\xi - g(x))^2, \tag{1} \end{equation*} where \begin{equation*} g(x):=2 x \sin\frac{\pi}{x} - \pi \cos\frac{\pi}{x}. \end{equation*} Note that $f(x,\xi)$ is so far undefined at $x=0$ (since $g(x)$ is so far undefined at $x=0$). So, let $g(0):=0$ and $f(0,\xi):=0$, so that (1) holds even for $x=0$. Then clearly $\xi \mapsto f(x,\xi)$ is convex and $f_{\xi \xi} (x,\xi)=2w(x)^2 > 0$ for all $x\ne0$.
Steps 2 and 3: We have $I[\ol u]=0$, since \begin{equation*} I[u] = \int_{-1}^1 w(x)^2(u'(x) - g(x))^2\, dx \end{equation*} and $\ol u'=g$. Also, if $u\in X\setminus\{\ol u\}$, then the Lebesgue measure of the set $\{x\colon u'(x)\ne g(x)\}=\{x\colon u'(x)\ne\ol u'(x)\}$ is $>0$ and hence $I[u]>0$. So, $\ol u$ is the only minimizer of $I[u]$.
Also, $|\ol u'(x)|=|g(x)|\le2|x|+\pi\le2+\pi$ for $x\in[-1,1]\setminus\{0\}$ and $\ol u$ is continuous. So, $\ol u$ is Lipschitz continuous on $[-1,1]$.
Step 4: By the definition of the derivative and (0), $\ol u'(0)=0$. However, $\ol u'(x)$ does not converge as $x\to0$. Indeed, otherwise, $-\pi \cos\frac{\pi}{x}=g(x)-2 x \sin\frac{\pi}{x}=\ol u'(x)-2 x \sin\frac{\pi}{x}$ would converge as $x\to0$, which is clearly not so. So, $\ol u\notin C^1 ([-1,1])$.
This completes the steps.