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

#### dwsmith

##### Well-known member
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
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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