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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 $f(t)=k$.

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The solution to the other question finds the minimum, but the question asked is to show that the only critical point is the given $f$. This involves showing that the Euler Lagrange equation is satisfied (and involves formulating the problem appropriately). The other solution does not address this. Perhaps the other solution is sufficient for the OP? –  copper.hat Apr 8 '13 at 17:38

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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.

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Thank you very much, it's nice to see how the Euler-Lagrange equations come in to this problem. –  user55225 Apr 9 '13 at 9:25

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