I am trying to find the extremals of the functional
$$\mathcal{F}(u) \, := \, \int_0^1 (u(t)^2 – u(t)u'(t)) \, dt $$
where $u \in C^1([0, 1], \mathbb{R}). $
Solving the Euler-Lagrange equations, it seems to me that the only extremals is $u \equiv 0$.
My attempt: $\, $ The Lagrangian is
$F(t, y, z) = y^2 – yz$, which gives for the partial derivatives
$\frac{\partial{F}}{{\partial y}} = 2y-z$ and $\frac{\partial{F}}{{\partial z}} = – y$. Then the Euler-Lagrange equation $\frac{\partial{F}}{{\partial y}}(t, u, u') – \frac{d}{dt} \frac{\partial{F}}{{\partial z}}(t, u, u') = 0$ is equivalent to
$$2u(t) – u'(t) – \frac{d}{dt}[- u(t)] = 2u(t) = 0.$$
Am I wrong? Any help would be appreciated, thanks!
Best Answer
If $u(0) = u(1) = 0$, then $$ \int^1_0 u(t)u'(t) ~\mathrm{d}t = u^2(t) \bigg \vert^{t=1}_{t=0} - \int^1_0 u'(t)u(t)~\mathrm{d}t = - \int^1_0 u'(t)u(t)~\mathrm{d}t \iff \int^1_0 u(t)u'(t)~\mathrm{d}t = 0 $$ using integration by parts. So in this case we have $$ \mathcal{F}(u) = \int^1_0~u^2(t)~\mathrm{d}t. $$
Now it is obvious that $u \equiv 0$ is the unique minimizer.
If we only assume that $u(0) = 0$, then $u_n(x) := nx$ is admissible for all $n \in \mathbb{N}$. Now $$ \mathcal{F}(u_n) = -\frac{n^2}{6} \overset{n \rightarrow \infty}{\longrightarrow} - \infty, $$ which proves that $\mathcal{F}$ does not attain its minimum in this case.
The Euler-Lagrange-equation helps you with neither of the two problems because it is a necessary and not a sufficient condition.