How do I decide if the solution of the E-L equations is a minimizer?

Question

I am just starting with variational calculus and I am trying to determine if a minimizer of the following functional $$I[u]=\int_1^2x^2u^{'2}(x)dx$$ exists in the following class of functions: $$\mathscr{C}=\{u \in C^1([1,2];\mathbb{R}), u(1)=1,u(2)=1/2\}$$ using the Euler-Lagrange equations:

The Euler Lagrange equation yields:

$$0= (x^2u'(x))'.$$ Integrating two times and imposing boundary conditions I get $u(x)=1/x$ which lives in $\mathscr{C}$.

Now my question is is this enough to say that this is a minimizer?

I read online that if The Euler Lagrange equations hold, thats a necessary condition for a weak extremal, not sufficient. So I am unsure how to continue from here, how do I determine if this is really an extremal and in particular a minimizer?

Answer 1

Hints:

1. Yes, a stationary configuration is not necessarily a minimum.

2. To prove that OP's stationary configuration $u_{\ast}(x)= 1/x$ is a minimum,

   • one way is to prove that the Hessian functional is positive semi-definite $$\begin{align} H[v]~:=~& \frac{1}{2}\int_{[1,2]}\!dx \int_{[1,2]}\!dy ~v(x)\left.\frac{\delta^2I[u]}{\delta u(x)\delta u(y)}\right|_{u=u_{\ast}}v(y)\cr ~=~&\underbrace{I[v]}_{\geq 0},\qquad v~\in~C^1_c(]1,2[;\mathbb{R}). \end{align}$$ where the 1st and 2nd functional derivatives are $$ \frac{\delta I[u]}{\delta u(x)} ~=~-\frac{d}{dx}(2x^2 \frac{d}{dx}u(x)) ,$$ and $$\begin{align} \frac{\delta^2I[u]}{\delta u(y)\delta u(x)} ~=~&-\frac{d}{dx}(2x^2 \frac{d}{dx}\delta(x\!-\!y)) \cr ~=~&\frac{d}{dx}\frac{d}{dy}(2xy\delta(x\!-\!y)).\end{align}$$

   • Alternatively, in OP's case, it is conceptionally simpler to just decompose an arbitrary configuration as $$u~=~u_{\ast}+v,$$ where the fluctuation part $v \in C^1([1,2];\mathbb{R})$ satisfies $v(1)=0=v(2)$. Now show that $$I[u]~=~I[u_{\ast}]~+~\underbrace{I[v]}_{\geq 0}.$$

Answer 2

For the more classical/standard problem of a function optimization, you would distinguish the stationary points with the help of the sign of the second-order derivative. In the same manner, you can study the second-order functional derivative of this linear functional to determine whether the solution $u_*(x) = 1/x$ is a maximum or a mimimum (or neither of them).

Let's recall that the functional derivative of $$ I[u(x)] = \int_1^2 L(x,u(x),u'(x)) \,\mathrm{d}x \quad\mathrm{with}\quad L = x^2u'^2 $$ is given by the integral of $$ \frac{\delta L}{\delta u} = \left(\frac{\partial}{\partial u} - \frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial}{\partial u'}\right)L = -2(x^2u')' = -4xu' - 2x^2u''. $$ Since this quantity depends on $u''$, the "next" functional derivative is calculated a bit differently, namely as follows : $$ \frac{\delta^2 L}{\delta u^2} = \left(\frac{\partial}{\partial u} - \frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial}{\partial u'} + \frac{\mathrm{d}^2}{\mathrm{d}x^2}\frac{\partial}{\partial u''}\right) (-4xu'-2x^2u'') = 4-4 = 0 $$ In consequence, one has $\frac{\delta^2 I}{\delta u^2}[u_*(x)] = 0$, that is why $u_*(x) = 1/x$ is neither a minimum nor a maximum, but the functional analog of a point of inflection $-$ as $x_* = 0$ for the function $f(x) = x^3$ for instance.