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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.

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

1 Answer 1

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Mostafa
    May 26, 2016 at 11:15
  • $\begingroup$ @Mostafa Did you include the slack functions in the Lagrangian? I suppose the Lagrangian should be something like $$\mathcal{L} (x,y,\lambda,s_1,s_2,y')$$ $\endgroup$ May 26, 2016 at 11:17
  • $\begingroup$ 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 $\endgroup$
    – Mostafa
    May 26, 2016 at 14:01

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