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[SOLVED] Checking E-L eq problem

dwsmith

Well-known member
Feb 1, 2012
1,673
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
Jan 26, 2012
4,197
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
Feb 1, 2012
1,673
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
Jan 26, 2012
4,197
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 $[-1,1]$.