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It's a problem abstracted from a real engineering project.
I want to find the best curve $y=y(x)$, $x \in [0,1]$: $y$ doesn't have to be a continuous function.
The constraint is $$ L=\int_{0}^{1} \frac{1}{y^3} dx =\text{const.}, $$ which represents the energy consumption of the system.
The objective function is $$ J[y]=\int_{0}^{1} \left[ L \int_{0}^{z} \frac{1}{y^4} dx - \int_{0}^{z} \frac{1}{y^3} dx \int_{0}^{1} \frac{1}{y^4} dx \right] dz $$ which instead represents the thrust force of the system.
The meaning of this question is find out what's the maximum force exerted by a specific system when its energy consumption is limited.

Questions

  • Does the objective function has a finite maximum?
  • If the objective function has a finite maximum, what's the expression of $y$?
  • If the objective function does not have a finite maximum, what's the expression of $y$?
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  • $\begingroup$ A possible condition can be added $y(1)=1$. $\endgroup$
    – Yangong Wu
    Commented Mar 11, 2020 at 17:31
  • $\begingroup$ Is $y$ positive, or not necessarily? $\endgroup$
    – fedja
    Commented Mar 11, 2020 at 23:10
  • $\begingroup$ From the view of engineering, y must be greater than 0. $\endgroup$
    – Yangong Wu
    Commented Mar 12, 2020 at 8:30

1 Answer 1

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By introducing a few more variables and constraints you can also write it into a form which can partially be solved with Pontryagin's maximum principle:

\begin{align} \max_y & \int_0^1 L_1\,x_1(t) - L_2\,x_2(t)\,dt \\ \text{s.t.}\, & \frac{d\,x_1(t)}{dt} = \frac{1}{y^4(t)},\ \frac{d\,x_2(t)}{dt} = \frac{1}{y^3(t)}, \\ & x_1(0) = 0,\ x_2(0) = 0, \\ & x_1(1) = L_2,\ x_2(1) = L_1. \end{align}

After solving this problem one would still be free to choose $L_2$, but this would just be a scalar optimization problem, which should be easier to solve. The resulting Hamiltonian associated with the problem can be written as

$$ H(x,y,\lambda) = \frac{\lambda_1}{y^4} + \frac{\lambda_2}{y^3} + L_2\,x_2 - L_1\,x_1 $$

The dynamics of the co-states are given by

$$ \frac{d\,\lambda_1(t)}{dt} = L_1, \\ \frac{d\,\lambda_2(t)}{dt} = -L_2. $$

The expression for $y$ can be found by maximizing the Hamiltonian. However, depending on the values of $\lambda_1$ and $\lambda_2$ this does not necessarily have to satisfy the solution $H_y = 0$, with $H_y$ the partial derivative of $H$ with respect to $y$.

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  • $\begingroup$ Thank you, this is the right answer. I checked it with numerical methods. $\endgroup$
    – Yangong Wu
    Commented Jun 21, 2020 at 14:29

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