Calculus of variations with inequality and non-integral constraints

I have a question on solving an optimization problem with calculus of variations.

I am attempting to maximize the functional

$$J[y] = \displaystyle\int_a^b F(x,y,y') \, \mathrm{d}x, \tag{1}$$

but my constraint is not in integral form; it is an inequality

$$0 \leq g(x,y,y') \leq 1, \tag{2}$$

Is it possible to solve this with calculus of variations?

I have rewritten the constraint to the form of $g(x,y,y')-g(x,y,y')^2 \geq 0$ and I think I should use Lagrange multiplier in form of $\lambda(x)$ and $\lambda(x) \geq 0$ as follows:

$$L(y)=\int_a^b F(x,y,y')+\lambda(x)(g(x,y,y')-g(x,y,y')^2) \, \mathrm{d}x$$

In this way my problem got very complicated and non-solvable, can anybody guide me to the correct or better solution. Thanks in advance and any suggestion will be appreciated.

• I'm confused. You want to maximize the set of functions that satisfy that inequality, right? May 25, 2016 at 16:06
• Hi, I want to find the function $y(x)$ that maximize the $J[y]$ and also satisfy the constraint. May 25, 2016 at 16:09
• If I recall correctly you should be able to use a substitution to force g(x,y,y') to be bounded. May 25, 2016 at 16:22

We can introduce slack functions $s_1, s_2$ such that the inequality constraints $g (x, y, y') \geq 0$ and $g (x, y, y') \leq 1$ can be replaced by the following equality constraints

$$g (x, y, y') - s_1^2 (x, y, y') = 0, \qquad \qquad g (x, y, y') + s_2^2 (x, y, y') = 1$$

The Lagrangian then becomes

$$F (x, y, y') + \lambda_1 (x) \left( g (x, y, y') - s_1^2 (x, y, y')\right) + \lambda_2 (x) \left( g (x, y, y') + s_2^2 (x, y, y') - 1\right)$$

I do not know what the corresponding Euler-Lagrange equations would be, though.

• Thanks for your answer, but after using this Lagrangian and solving the Euler-Lagrange equation, two constraint are omitted and only the $s_1^2(x) + s_2^2(x)=1$ remains that cannot be used to solve the equation. May 26, 2016 at 11:15
• @Mostafa Did you include the slack functions in the Lagrangian? I suppose the Lagrangian should be something like $$\mathcal{L} (x,y,\lambda,s_1,s_2,y')$$ May 26, 2016 at 11:17
• It's correct, but adding s1 and s2 to lagrangian only give me the result of $\lambda_1 (x)*2 s_1(x)=0$ and $\lambda_2 (x)*2 s_2(x)=0$ I Don't know how to use them to solve my equations May 26, 2016 at 14:01