# Show that a functional has no extremal

I'm studying variations of calculus and it is quite new for me and I had problem with one exercise in my book. I was given a functional $$J(y)=\int_{-1}^{1}x^4\big(y'\big)^2 dx$$ and I'm supposed to show that is has no extremals in $$C^2[-1,1]$$ which satisfy the boundary condition $$y(-1)=-1$$ and $$y(1)=1$$.

I know that if y is an extremal it has to satisfy the Euler-Lagrange equation $$\dfrac{d}{dx}\Big(\dfrac{\partial f}{\partial y'}\Big)-\dfrac{\partial f}{\partial y} =0$$

for all $$x \in [-1,1]$$

So I have that $$f(x,y,y')=x^4y'^2$$ and when using E.L equation I get: $$xy''+4y'=0$$ Ansatz $$g=y' \rightarrow g'=y''$$ which then gives the separable differential equation: $$\dfrac{dg}{g}=-\dfrac{4}{x}dx$$ Solving this I end up with $$y=\dfrac{c_1}{x^3}+c_2$$ and by using the boundary condition I get that $$\begin{cases}c_1=1\\ c_2=0 \end{cases}$$ Therefore the extremal should be given by: $$y=\dfrac{1}{x^3}$$

However, I'm not sure how this shows that it has no extremals, is it because of discontinuity at $$0$$? Because the Euler-Lagrange equation is satisfied, so shouldn't this be the extremal of $$J$$? Or have I done something wrong / misinterpreted the theory, if so what is it?

• Yes, it is because of the discontinuity as the question specifies that the extremal must be in $C^2[-1,1]$. – Cardioid_Ass_22 Apr 21 '19 at 17:22
• Ok, is there any proof/lemma that a extremal has to be continuous on the interval? – wroomwrooom Apr 21 '19 at 18:31
• I was pointing out the differentiability class requirement of the question. – Cardioid_Ass_22 Apr 21 '19 at 20:08

2. OP's functional $$J[y]~\equiv~\int_{[-1,1]} \!dx ~(x^2y^{\prime})^2~\geq~ 0, \tag{A}$$ is non-negative. But it is not bounded from above, so there is no maximum.
$$y~\in~C^1([-1,1]) \tag{B}$$ it follows that $$J[y]~=~0\quad\Leftrightarrow \quad y\text{ is constant}. \tag{C}$$ However, constant functions cannot satisfy the boundary conditions (BCs) $$y(\pm 1)~=\pm 1 .\tag{D}$$
4. On the other hand, it is easy$$^1$$ to construct a sequence $$(y_n)_{n\in\mathbb{N}}$$ of functions in $$C^1([-1,1])$$ that satisfies the BCs, and converge $$x$$-pointwise (and hence uniformly) to a constant function, so that $$J[y_n]\to 0$$ for $$n\to \infty$$. This shows that there is no minimum in $$C^1([-1,1])$$ satisfying the BCs. $$\Box$$
$$^1$$ Think of a $$C^1$$-function that is mostly constant but modified close to the boundary to accommodate the BCs.