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|nd        = 1
|nd        = 1
|nx        = 3
|nx        = 3
|nw        = 5
|nw        = 1
|nre      = 3
}}
}}


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<math>
<math>
\begin{array}{llclr}
\begin{array}{llclr}
  \displaystyle \min_{u} & \int_0^{t_f} (x_0(t) - 1.5)^2 + (x_1(t) - 1)^2 + (x_2(t) - 1)^2 dt \\[1.5ex]
  \displaystyle \min_{u} & \int_0^{t_f} && (x_0(t) - 1.5)^2 + (x_1(t) - 1)^2 + (x_2(t) - 1)^2 \ dt \\[1.5ex]
  \mbox{s.t.}  
  \mbox{s.t.}  
  & \dot{x}_0 & = &  x_0(t) - x_0(t) x_1(t) - x_0(t) x_2(t), \\
  & \dot{x}_0(t) & = &  x_0(t) - x_0(t) x_1(t) - x_0(t) x_2(t), \\
  & \dot{x}_1 & = & - x_1(t) + x_0(t) x_1(t) - c_1 x_1(t) u(t),  \\
  & \dot{x}_1(t) & = & - x_1(t) + x_0(t) x_1(t) - c_1 x_1(t) u(t),  \\
  & \dot{x}_2 & = & -x_2(t) + \alpha x_0(t) x_2(t) - c_2 x_2(t) u(t),  \\[1.5ex]
  & \dot{x}_2(t) & = & -x_2(t) + \alpha x_0(t) x_2(t) - c_2 x_2(t) u(t),  \\[1.5ex]
  & x(0) &=& x_0, \\
  & x(0) &=& x_0, \\
  & u(t) &\in& [0,1],
  & u(t) &\in& [0,1], \\
  & \alpha &>&  1.
  & \alpha &>&  1.
\end{array}  
\end{array}  
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<math>
<math>
\begin{array}{rcl}
\begin{array}{rcl}
[t_0, t_f] &=& [0, 12],\\
[t_0, t_f] &=& [0, 40],\\
(c_{0,1}, c_{1,1}) &=& (0.2, 0.1),\\
(c_{1}, c_{2}) &=& (0.1, 0.4),\\
(c_{0,2}, c_{1,2}) &=& (0.4, 0.2),\\
x_0 &=& (1.5, 0.5, 1) \text{ or } (1.5, 1, 0.5),\\
(c_{0,3}, c_{1,3}) &=& (0.01, 0.1),\\
\alpha &=& 1.2.
(c_{0,4}, c_{1,4}) &=& (0, 0),\\
(c_{0,5}, c_{1,5}) &=& (-0.1, -0.2).
\end{array}
\end{array}
</math>
</math>
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== Reference Solutions ==
== Reference Solutions ==


If the problem is relaxed, i.e., we demand that <math>w(t)</math> is in the continuous interval <math>[0, 1]</math> rather than being binary, the optimal solution can be determined by means of direct optimal control.
<gallery caption="Reference solution plots" widths="400px" heights="240px" perrow="2">
 
  Image:LV_Shared_init_1.png| Local optimum a direct approach for start values <math>x_0 = (1.5, 0.5, 1)</math>.
The optimal objective value of the relaxed problem with  <math> n_t=12000, \, n_u=150  </math> is <math>x_2(t_f) =0.345768563</math>. The objective value of the solution with binary controls obtained by Combinatorial Integral Approximation (CIA) is <math>x_2(t_f) =0.348617982</math>. 
  Image:LV_Shared_init_2.png| Local optimum a direct approach for start values <math>x_0 = (1.5, 1, 0.5)</math>.
 
<gallery caption="Reference solution plots" widths="180px" heights="140px" perrow="2">
  Image:Lotka_abs_fish_relaxed_12000_80.pdf| Optimal relaxed controls and states determined by an direct approach with ampl_mintoc (Radau collocation)  and <math>n_t=12000, \, n_u=150</math>.
  Image:Lotka_abs_fish_CIA_states_12000_80.pdf| Differential states determined by an direct approach (Radau collocation) with ampl_mintoc and <math>n_t=12000, \, n_u=150</math>. The relaxed controls were approximated by Combinatorial Integral Approximation.
Image:Lotka_abs_fish_CIA_controls_12000_80.pdf| Binary control determined by an direct approach (Radau collocation) with ampl_mintoc and <math>n_t=12000, \, n_u=150</math>. The relaxed controls were approximated by Combinatorial Integral Approximation. The fishing control shows a lot of chattering.
</gallery>
</gallery>




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[[Category:ODE model]]
[[Category:ODE model]]
[[Category:Tracking objective]]
[[Category:Tracking objective]]
[[Category:Chattering]]
[[Category:Sensitivity-seeking arcs]]
[[Category:Sensitivity-seeking arcs]]
[[Category:Population dynamics]]
[[Category:Population dynamics]]

Latest revision as of 13:43, 28 November 2025

LV Shared Resource
State dimension: 1
Differential states: 3
Discrete control functions: 1


This Lotka Volterra problem with explicit inclusion of a shared resource is a variant of the Lotka Volterra fishing problem. Its dynamics are given via a three-dimensional ODE model.

Mathematical formulation

The optimal control problem is given by

minu0tf(x0(t)1.5)2+(x1(t)1)2+(x2(t)1)2 dts.t.x˙0(t)=x0(t)x0(t)x1(t)x0(t)x2(t),x˙1(t)=x1(t)+x0(t)x1(t)c1x1(t)u(t),x˙2(t)=x2(t)+αx0(t)x2(t)c2x2(t)u(t),x(0)=x0,u(t)[0,1],α>1.

Parameters

These fixed values are used within the model.

[t0,tf]=[0,40],(c1,c2)=(0.1,0.4),x0=(1.5,0.5,1) or (1.5,1,0.5),α=1.2.

Reference Solutions