MHB Can any twice-differentiable function satisfy these conditions?

[Dustinsfl]

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]

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

 

[Dustinsfl]

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]

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 $[-1,1]$.