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?

Thanks in advance!

  • $\begingroup$ Yes, it is because of the discontinuity as the question specifies that the extremal must be in $C^2[-1,1]$. $\endgroup$ – Cardioid_Ass_22 Apr 21 '19 at 17:22
  • $\begingroup$ Ok, is there any proof/lemma that a extremal has to be continuous on the interval? $\endgroup$ – wroomwrooom Apr 21 '19 at 18:31
  • $\begingroup$ I was pointing out the differentiability class requirement of the question. $\endgroup$ – Cardioid_Ass_22 Apr 21 '19 at 20:08
  1. In this answer we will not use the Euler-Lagrange (EL) equation at all!

  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.

  3. For functions
    $$y~\in~C^1([-1,1]) \tag{B}$$ 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.

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