# [SOLVED]Checking E-L eq problem

#### dwsmith

##### Well-known member
I want to show for all twice differentialable functions there are no functions that take a min or max with the following conditions $$y(-1) = -1$$, $$y(1) = 1$$, and
$\int_{-1}^1x^2y^{'2}dx.$

From the E-L eq, we have $$2x^2y' = c$$.
So
$y(x) = -\frac{c}{6} + d$
Using the conditions, we get
$\begin{pmatrix} \frac{1}{6} & 1 & -1\\ -\frac{1}{6} & 1 & 1 \end{pmatrix}\Rightarrow \begin{pmatrix} \frac{1}{6} & 1 & -1\\ 0 & 2 & 0 \end{pmatrix}$
Therefore, we have an inconsistent system since no two tuples satisfy the bottom equation.
Thus, no twice differentiable function with the given conditions takes on a minimum or maximum.

#### Ackbach

##### Indicium Physicus
Staff member
I think from $2x^{2}y'=c$, the next step would be
$$y'= \frac{c}{2x^{2}}=(c/2)x^{-2} \quad \implies \quad y=- \frac{c}{2x}+d.$$

#### dwsmith

##### Well-known member
I think from $2x^{2}y'=c$, the next step would be
$$y'= \frac{c}{2x^{2}}=(c/2)x^{-2} \quad \implies \quad y=- \frac{c}{2x}+d.$$
Not everyone can integrate but after that error, the result analysis would be the same correct?

#### Ackbach

##### Indicium Physicus
Staff member
If you just apply the BC's, you might have
\begin{align*}
1&=-c+d\\
-1&=c+d.
\end{align*}
From here, you could say that $d=0$ and $c=-1$, and hence $y=1/x$. The problem is that this function is not in the collection of functions in which you were interested - at least not on the interval $[-1,1]$. That is, $1/x$ is not twice-differentiable on $[-1,1]$.