Validity of Introducing a Slack Function when Extremizing a Functional using Calculus of Variations

Question

I want to solve the following problem: $$\begin{aligned} \min_{f(x)} F[f]&=\int_0^R\frac{x}{1+[f'(x)]^2}dx \\ G[f] &= \int_0^R xf(x)dx=V\\ f(x)&\geq 0 \ \ \ (\forall x \in [0,R]) \end{aligned}$$ where $F$ is the objective functional, $G$ is an integral constraint, $f$ has the constraint that it is nonnegative on $[0,R]$, and $R$ and $V$ are fixed real constants. I am familiar with the method of Lagrange multipliers being used to solve the problem if the inequality constraint is ignored, i.e. forming the Lagrangian function $$\mathcal{L}(x,f,f')=\frac{x}{1+[f'(x)]^2}+\lambda xf(x)$$ then extremizing $$\int_0^R \mathcal{L}(x,f,f')dx$$ by solving the corresponding Euler-Lagrange Equation (ELE) $$\frac{\partial\mathcal{L}}{\partial f}-\frac{d}{dx}\frac{\partial \mathcal{L}}{\partial f'}=0.$$ My issue arises when trying to incorporate the inequality constraint into this problem. Browsing the site for answers, I stumbled upon this post which incorporates a similar inequality constraint. The OP mentions that single inequality constraints can be reduced to an equality constraint by introducing the slack function $s$ and transforming the inequality into $$f(x)=s^2(x)$$ to ensure nonnegativity. I assume then we merely substitute $s^2$ for $f$ and $2ss'$ for $f'$ wherever they occur in $\mathcal{L}(x,f,f')$ (call the new lagrangian function $\mathcal{L}^\star$) and its ELE. In other words, the problem should be solved by now finding the minimum of $$\int_0^R \mathcal{L}^\star(x,s,s')dx=\int_0^R \frac{x}{1+[2s(x)s'(x)]^2} +\lambda x s^2(x) dx$$ by solving the ELE of $\mathcal{L^\star(x,s,s')}$ to find the optimal function $$s^\star=s^\star(x, \lambda).$$ Then, we can theoretically find the unknown constants by analyzing the constraints and square the final function to find the optimal nonnegative function $f^\star$. However, in the aforementioned post, user @Red shoes states

you are not allowed to do those kinda changes of variables! Do you know anything about constraint qualifications?

which I interpreted to mean that such formations of slack variables is not allowed (there was no further elaboration).

My question is whether my method of tackling the problem is valid or not, and if the latter, how I would solve such problems with inequality constraints. Furthermore, it would be great if anyone could shed some light on what the user meant (I do not know what constraint qualifications are).

Answer 1

I have a very similar problem — a combination of an integral constraint with an inequality constraint on a function — so I've been searching online for a solution. Even though I do not have a full solution to your problem, I thought I'd share what I was able to find.

For an optimization problem $$\min_{u}\int_a^b F(x, u, u') \, dx$$ subject to constraint $$G(x, u) \le 0$$ the following equations should hold: $$\left(\frac{d}{dx}\frac{\partial}{\partial u'} + \frac{\partial}{\partial u}\right)\left(F+\mu(x)G\right) = 0$$ $$G(x, u) \le 0$$ $$\mu(x) \geq 0$$ $$\mu(x)G(x, u)= 0 \Rightarrow \text{ either } \mu = 0, G<0, \text{ or } \mu >0, G = 0.$$

For a minimization problem with an integral constraint $$\int_a^b g(x, u, u') \, dx = l$$ we should consider a new functional $$L(x, u, u', \lambda) = \int_a^b \left(F(x, u, u') + \lambda g(x, u, u')\right) \, dx$$ then write Euler-Lagrange equations $$\left(\frac{d}{dx}\frac{\partial}{\partial u'} + \frac{\partial}{\partial u}\right) L(x, u, u', \lambda) =0$$ From this it should be possible to find minimizing $u$, which will depend on the parameter $\lambda$. $\lambda$ is then found from the integral constraint.

What I think we should do in our problem where both constraints are present is to consider the following functional: $$L(x, u, u', \lambda) = \int_a^b \left(F(x, u, u') + \mu(x)G(x, u) + \lambda g(x, u, u')\right) \, dx $$ and the following system of equations: $$\left(\frac{d}{dx}\frac{\partial}{\partial u'} + \frac{\partial}{\partial u}\right)\left(F+\mu(x)G + \lambda g(x, u, u')\right) = 0$$ $$G(x, u) \le 0$$ $$\mu(x) \geq 0$$ $$\mu(x)G(x, u)= 0 \Rightarrow \text{ either } \mu = 0, G<0, \text{ or } \mu >0, G = 0.$$

In your problem, however, the term $\frac{d}{dx}\frac{\partial}{\partial f'}\left(\frac{x}{1+(f'(x))^2}\right)$ gets pretty complicated, but maybe with some additional considerations about higher derivatives of $f$ you can solve it!

This is the source of information about separate problems.