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Given the following functional with fixed parameters $a>0$ and $b>1$, $$F[x;u,u']: = \int_{1}^b\left(x^3-\frac{ax}{u^2}\right)(u')^2~\mathrm{d}x$$ Suppose the space of functions satisfy $u(1) = u(b) = 1$ and $u\geq \frac{1}{x^2}$. Does $F$ have a minimizer? If so, is the minimizer unique/global?

Note: I derived the Euler-Lagrange equation of $F$ under the proposed boundary conditions to be $$\left(x^2-\frac{a}{u^2}\right)u''+2xu'+\frac{a}{u^3}(u')^2=0.$$

It is clear that $u\equiv 1$ is a solution to the above equation subject to the boundary conditions $u(1) = u(b) = 1$ and the constraint $u\geq \frac{1}{x^2}$, but I am not sure about whether it is a minimizer of $F$ (I tried to manipulate the second variation of $F$ but it takes an indefinite sign).

I am also not sure whether the above differential equation have any non-constant solution.

Rodrigo de Azevedo
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user31899
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    A minimizer over what space? I'm guessing once differentiable functions, but it could also be Sobolev spaces? Also I'm guessing $a \leq 1$? In that case $u=1$ is a minimizer, since the integral is always $\geq 0$. – Niko Gruben Aug 01 '25 at 11:36
  • Yes, $u=1$ is the global minimizer if $a\leq 1$, but it is not clear for cases with $a> 1$. Only smooth differentiable functions are considered. I defined the functional as I am curious of how the $x^3u'^2$ term competes with $\frac{ax}{u^2}u'^2$ for different values of $a$. – user31899 Aug 02 '25 at 00:36

1 Answers1

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Here is a partial answer with some ideas that may help OP.

  1. If we substitute $t=x^{-2}$ then OP's functional simplifies to $$\begin{align} \frac{F[u]}{4}~=~&\frac{1}{2}\int_{b^{-2}}^1 \mathrm{d}t~\left(1-\frac{at}{u^2}\right)\dot{u}^2\cr ~=~&\int_{b^{-2}}^1 \mathrm{d}t~L(t,u,\dot{u}),\cr L(t,u,\dot{u}) ~:=~& \frac{1}{2}m(t,u)\dot{u}^2,\cr m(t,u)~:=~& 1-\frac{at}{u^2},\cr b~>~&1,\end{align}\tag{A} $$ with conditions $$ u(b^{-2})~=~1~=~u(1) \qquad\text{and}\qquad \forall t\in[b^{-2},1]:~u(t)~\geq~ t.\tag{B} $$

  2. Clearly, $F$ is manifestly non-negative if $a\leq b^{-2}$, so then $u\equiv 1$ is a minimizer $F[u\!\equiv\!1]=0$, cf. above comment by Niko Gruben.

  3. The momentum is $$ p(t,u,\dot{u})~:=~\frac{\partial L(t,u,\dot{u})}{\partial \dot{u}}~=~m(t,u)\dot{u}.\tag{C} $$ The energy is $$h(t,u,\dot{u})~:=~ \dot{u}p(t,u,\dot{u})- L(t,u,\dot{u}) ~=~L(t,u,\dot{u}).\tag{D} $$

  4. OP's functional $F$ has a scaling symmetry $$ t~\longrightarrow~ \lambda^2 t, \qquad u~\longrightarrow~ \lambda u.\tag{E} $$

  5. It is possible to use Noether's theorem to obtain a 1st integral to the 2nd-order EL equation in a manner similar to e.g. this Phys.SE post. The conserved Noether charge is $$ \begin{align} Q(t,u,\dot{u})~=~&u p(t,u,\dot{u})-2t h(t,u,\dot{u})\cr ~=~&(u-2t\dot{u})p(t,u,\dot{u})+2t L(t,u,\dot{u})\cr ~=~&m(t,u)\dot{u}(u-t\dot{u}) \cr ~=~&\left( 1-\frac{at}{u^2}\right)\dot{u}(u-t\dot{u}) .\end{align}\tag{F} $$

Qmechanic
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