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[SOLVED] Min of an Integral lagrange multipliers for E-L

dwsmith

Well-known member
Feb 1, 2012
1,673
Find the minimum value of \(\int_0^1y^{'2}dx\) subject to the conditions \(y(0) = y(1) = 0\) and \(\int_0^1y^2dx = 1\).

Let \(f = y^{'2}\) and \(h = y^2\).
Then
\begin{align*}
G[y(x)] &= \int_0^1[f - \lambda h]dx\\
&= \int_0^1\left[y^{'2} - \lambda y^2\right]dx
\end{align*}
The Euler-Lagrange equation is then
\begin{align*}
-2y\lambda - \frac{d}{dx}\left(2y'\right) &= 0\\
y\lambda + y'' &= 0
\end{align*}
We have the equation for a simple harmonic oscillators,
\(y(x) = A\cos\left(x\sqrt{\lambda}\right) +
B\sin\left(x\sqrt{\lambda}\right)\).
\begin{alignat*}{3}
y(0) &= A &{}= 0\\
y(1) &= B\sin\big(\sqrt{\lambda}\big) &{}= 0\\
\sin\big(\sqrt{\lambda_n}\big) &= 0\\
\lambda_n &= \pi^2n^2 &&\quad (\text{where \(n\in\mathbb{Z}\).})
\end{alignat*}
So our equation is \(y_n(x) = B\sin\big(\pi nx\big)\).
We can now use our constraint integral.
\begin{align*}
\int_0^1B\sin^2\big(\pi nx\big)dx &= 1\\
\frac{B}{2}\int_0^1(1 - \cos\big(2\pi nx\big))dx &=
\frac{B}{2}\left[x - \frac{1}{\pi n}\sin\big(2\pi n x\big)\right|_0^1\\
B &= 2
\end{align*}
Therefore, our equation is \(y_n(x) = 2\sin\big(\pi n x\big)\).

So then the minimum is
\[
\int_0^1y^{'2}dx = 2\pi^2n^2.
\]

Is this correct?
 

M R

Active member
Jun 22, 2013
51
Perfect question for my revision. Thanks. :)

I did the question before I looked at your solution and I only have a couple of small differences.

I explicitly considered \(\displaystyle \lambda\le 0\) to show that those solutions to the ELE don't satisfy the boundary conditions.

Also at the end you can say that n=1 will give the minimum as n=0 doesn't satisfy the constraint.

So \(\displaystyle y=\sqrt{2} \sin(\pi x)\) and the minimum value is \(\displaystyle 2\pi^2\).

As I say, I'm just revising this for an exam myself so I'm not an expert.

Edit: I think you forgot to square B when you evaluated the constraint integral.
 
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