I showed that $u(x) = \frac{x^2}{2}$ is a potential minimizer for the functional $\int_0^2 \frac{n}{2}u'(x)^2-nu(x) \, dx$ in $C^2[0,2]$ with $u(0) = 0$ and $u(2)=2$ where $n$ is a positive constant (using Euler-Lagrange equation). I am now trying to prove that this is actually a weak local minimizer and am unsure how to proceed. By weak minimizer I mean using the $C^1$ norm.

I've been thinking about somehow using the first variation of a functional.

Thanks in advance for your help.

  • $\begingroup$ have you tried using the direct method of calculus of variations? But I'm not sure if this can work for smooth functions (weak convergence is maybe not as good as in sobolev spaces) $\endgroup$ Mar 22, 2013 at 6:45
  • $\begingroup$ Unless I have missed something, if $L(u,v) = n(\frac{1}{2}v^2-u)$, the Euler Lagrange equation results in $u(x) = \frac{1}{2}x(4-x)$. Is the minus sign in $-n u$ correct? $\endgroup$
    – copper.hat
    Mar 22, 2013 at 7:18
  • $\begingroup$ Also, what do you mean by a weak local minimizer? Weak often means that the space of perturbations is restricted (eg, smooth, or some more restrictive differentiability requirements). However here the space is $C^2$, which is sufficient to produce a strong minimizer? $\endgroup$
    – copper.hat
    Mar 22, 2013 at 7:21
  • 4
    $\begingroup$ Please do not delete questions with good answers. People have devoted time and effort to commenting and answering. Deleting these questions is not fair to them. $\endgroup$
    – robjohn
    Mar 22, 2013 at 18:48
  • 6
    $\begingroup$ Please do not delete or vandalize questions with good answers. Changing the question orphans the answers and keeps others from benefiting from the question. $\endgroup$
    – robjohn
    Mar 22, 2013 at 19:33

1 Answer 1


Let $L(u,v) = \frac{n}{2} v^2-n u$. Let $C = \{u \in C^2[0,2] | u(0) =0, \, u(2) = 2 \}$. Let $\hat{u}(x) = \frac{1}{2}x (4-x)$. We note that $\hat{u}$ satisfies both the boundary conditions and the Euler–Lagrange equation, ie, $\frac{\partial L(\hat{u}(x), \hat{u}'(x))}{\partial u} = -n = n \hat{u}''(x) = \frac{d}{dx}\frac{\partial L(\hat{u}(x), \hat{u}'(x))}{\partial v}$.

Note that $C$ is convex, and $L$ is convex. Since $u \mapsto u'$ is linear, it follows that the function $u \mapsto L(u,u')$ is convex, and hence the functional $J$, with $J(u) = \int_0^2 L(u(x),u'(x)) dx$, is convex.

The problem is $P: \ \min_{u \in C} J(u)$. Since the objective and constraints are convex, $\hat{u}$ solves $P$ iff $dJ(\hat{u}, u-\hat{u}) \ge 0$ for all $u \in C$.

The directional derivative is given by $dJ(u,\delta) = \int_0^2 ( \frac{\partial L(u(x), u'(x))}{\partial u} \delta(x) + \frac{\partial L(u(x), u'(x))}{\partial v} \delta'(x) ) dx$.

It is straightforward to show that $dJ(\hat{u},\delta) = 0$ for any $\delta \in C^1[0,2]$ such that $\delta(0) = \delta(2) = 0$, hence $\hat{u}$ is a minimizer of $P$.

Since $\hat{u} \in C^2[0,2]$ (so that the the Euler Lagrange equation is satisfied), the optimality condition $dJ(\hat{u},\delta) = 0$ shows that the constraint set $C$ may be expanded to $C_{\text{relaxed}} = \{u \in C^1[0,2] | u(0) =0, \, u(2) = 2 \}$.


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