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Oscillating OED: Difference between revisions

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For a single parameter <math>p</math> the original initial value problem is given by
For a single parameter <math>p</math> the original initial value problem is given by
<math>
<math>
   \dot{x}(t) =: f(t, p) = 0.2 + 0.8 \cdot t + 0.3 \cdot (\sin(p \cdot t) + \cos(p \cdot t) \cdot p \cdot t) - 2.5 \cdot \sin(50 \cdot t), \quad x(0) = x_0.
   \dot{x}(t) = f(t, p) = 0.2 + 0.8 \cdot t + 0.3 \cdot (\sin(p \cdot t) + \cos(p \cdot t) \cdot p \cdot t) - 2.5 \cdot \sin(50 \cdot t), \quad x(0) = x_0.
</math>
</math>



Revision as of 08:19, 27 January 2026

Oscillating OED
State dimension: 1
Differential states: 4
Discrete control functions: 1


The Oscillating OED problem looks for an optimal measurement strategy to determine a single parameter in a one-dimensional ODE model, where we can directly measure the single state.

The optimal integer control functions shows bang bang behavior.

Mathematical formulation

For a single parameter p the original initial value problem is given by x˙(t)=f(t,p)=0.2+0.8t+0.3(sin(pt)+cos(pt)pt)2.5sin(50t),x(0)=x0.

We assume both x0 and tf to be fixed and are only interested in when to measure, with an upper bound M on the measuring time. We can measure the state directly, i.e. h(x(t))=x(t).

Now we formulate the OED problem:

miny,G,F,z,wtrace(F1(tf))subject toy˙(t)=f(t,p)G˙(t)=fp(y(t),p)F˙(t)=w(t)(hy(y(t))G(t))T(hy(y(t))G(t))z˙(t)=w(t),y(0)=y0G(0)=0F(0)=0,z(0)=0w(t)𝒲z(tf)M

Parameters

These fixed values are used within the model:

x0=0.1;tf=2;𝒲=[0,1];M=0.2;p=15

Reference Solutions

Here is one local solution to the above control problem.

Miscellaneous and Further Reading

This problem was introduced by Sebastian Sager.