Simple question on calculus of variations: critical point of functional subject to constraint

Question

Let $V$ be the set of smooth functions $f:[0,1]\to \mathbb{R}$ such that $\int_0^1 f(t) dt =k$. If $F:V\to\mathbb{R}$ is given by $F(f) = \int_0^1 f(t)^2 dt$, show that the only critical point of $F$ is the constant function $f(t)=k$.

Answer 1

Let $W = \{ \phi \in C^\infty [0,1] | \phi(0) = 0,\ \ \phi(1) = k \}$. Let $j: W \to V$ be given by $j(\phi) = \phi'$. It should be clear that $j$ is a linear bijection.

Let $L(x,v) = v^2$, and define $\tilde{F} (\phi) = F(j(\phi))$, and note that $\tilde{F} (\phi) = \int_0^1 L(\phi(t), \phi'(t)) dt$.

Let $h = \phi_1-\phi_2$, where $\phi_k \in W$. Then we have, for $t>0$, $\frac{\tilde{F}(\phi+t h) - \tilde{F}(\phi)}{t} = \frac{F(j(\phi)+t j(h)) - F(j(\phi))}{t} $, and so $d \tilde{F}(\phi , h) = d F(j(\phi), j(h))$. It follows that $\phi$ is critical for $\tilde{F}$ iff $j(\phi)$ is critical for $F$. Hence we can focus on $\tilde{F}$ since it fits into the Euler Lagrange framework.

The Euler Lagrange equation gives $\frac{\partial L(\phi(t),\phi'(t))}{\partial x} = \frac{d}{dt} \frac{\partial L(\phi(t),\phi'(t))}{\partial v}$, for $t \in [0,1]$. Evaluating gives $2 \phi''(t) = 0$, which has a solution $\phi(t) = \phi(0) + t \phi'(0)$. Since $\phi \in W$, we have $\phi(t) = t k$.

It follows that $j(\phi) = \phi'$ is critical for $F$, and since $j(\phi)(t) = k$, we have the desired result.

An alternative approach would be to use Lagrange multipliers. Let $X = C[0,1]$, and define $G(f) = \int_0^1 f(t) dt$. Then both $F$ and $G$ are Fréchet differentiable and $DF(f)(h) = 2 \int_0^1 f(t)h(t) dt$, $DG(f)(h) = \int_0^1 h(t) dt$. Then we can look at the relaxed problem of looking for critical points of $F$ subject to $G(f) = k$. (The problem is relaxed in that I have expanded to domain to $X$ which is a Banach space.) Then the critical points satisfy $DF(f)(h) + \lambda DG(f)(h) = 0$ for all $h$. This gives $\int_0^1 (2f(t)+\lambda)h(t) dt = 0$ for all $h \in X$. The usual argument is applied to conclude that $2f(t)+\lambda = 0$ for all $t \in [0,1]$. It follows that $f$ is constant, hence $f(t) = k$ for all $t \in [0,1]$. Since $f \in V$, it must be a critical point for the original problem.