I have a controlled stochastic process described by

$\dot{x} = X(x) + u(t) + \eta_x(t),$

where $u$ is the control, $\eta_x$ is a white noise with zero mean. The equation is called Langevin equation. The initial value

$x(0) = x_0$

is given. It is also given that

$u(t) = A \sin(\omega t).$

The problem is to find the values of $A$ and $\omega$ which provide the constraints

$x(t_k) = x_k, ~~~ k = 1, ..., K,$

at the given instants $t_k$, $x_k$ are also given.

Let me fix $K = 2, t_1 = 1, x_1 = 0.5$ and $t_2 = 2, x_2 = 1$. What should the algorithm for finding $A$ and $\omega$ be look like?

Currently, I have this piece which works with fixed $A$ and $\omega$:

\[Rho]t = 1;
\[Delta]t = 9.062;
\[Gamma]t = 6.63 10^-3;
A = 1;
\[Omega] = 0.1;

sol = RandomFunction[
        t] == \[DifferentialD]t*(\[Rho]t (1 - \[Gamma]t \
Sinh[\[Delta]t x[t]]) + A Sin[\[Omega] t]) + \[DifferentialD]\[Eta]x[
         t]}, x[t], {x, 0}, {t, 0}, \[Eta]x \[Distributed] 
     WienerProcess[]], {0, 1, 0.01}];

Now I have stucked on

  1. how to involve a white noise with zero mean instead of the Wiener
    proc (this somehow does not work)?
  2. how to impose the intermediate constraints above?

Your help on either part is highly appreciated.

  • $\begingroup$ In which sense $x(t_k) = x_k$ if $x(t_k)$ is a random variable ? $\endgroup$ – b.gates.you.know.what Mar 14 '18 at 11:11
  • $\begingroup$ I am not sure about the correct mathematics behind, but it simply means that the trajectory of the particle must pass through some given points ($x_k$). $\endgroup$ – Asatur Khurshudyan Mar 14 '18 at 11:16
  • 1
    $\begingroup$ I think the WhiteNoiseProcess issue of the question may be a bug. It seams that WhiteNoiseProcess only works in RandomFunction for integer time steps. RandomFunction[ WhiteNoiseProcess[UniformDistribution[{-1, 1}]], {0, 1, .1}] fails but RandomFunction[ WhiteNoiseProcess[UniformDistribution[{-1, 1}]], {0, 10}] succeeds. Please report to WRI. $\endgroup$ – Edmund Mar 17 '18 at 22:22
  • $\begingroup$ @Edmund You are right, that works! Thanks. I was thinking about the intermediate constraints. What if I define two different processes with initial values at the constraints? Then the control $u$ will be a piecewise function and the problem will be to provide the terminal condition at each piece. Does it simplify the problem? $\endgroup$ – Asatur Khurshudyan Mar 18 '18 at 1:33

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