Robust

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Biochemical process

We consider a biochemical process model \(Y\) of continuous culture fermentation. Let \(Y = (X, S, P) \in \mathbb{R}^3\) satisfies the differential system: \begin{equation*} \begin{cases} \overset{.}{X}(t) = -D X(t) + \mu(t) X(t)\\ \overset{.}{S}(t) = D \big( S_f(t) - S(t) \big) - \frac{\mu(t)X(t)}{Y_{x/s}}\\ \overset{.}{P}(t) = -D P + \big( \alpha\mu(t) + \beta \big) X(t)\\ \end{cases} \end{equation*}

where \(X\) denotes the biomass concentration, \(S\) the substrate concentration, and \(P\) the product concentration of a continuous fermentation process. The model is controlled by \(S_f \in [S_f^\mathit{min}, S_f^\mathit{max}]\). While the dilution rate \(D\), the biomass yield \(Y_{x/s}\) , and the product yield parameters \(\alpha\) and \(\beta\) are assumed to be constant and thus independent of the actual operating condition, the specific growth rate \(\mu : \mathbb{R} \rightarrow \mathbb{R}\) of the biomass is a function of the states:

\begin{equation*} \mu(t) = \mu_m\frac{\left( 1-\frac{P(t)}{P_m}\right) S(t)}{K_m + S(t)+\frac{S(t)^2}{K_i}} \end{equation*}

The goal is to maximize the average productivity presented by the cost function:

\begin{equation*} J_k = \frac{1}{T}\int^T_0DP(t)dt \end{equation*}

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Spin

We consider a 1/2 spin particle system \(X\). Let \(X = (x, y, z) \in \mathbb{R}^3\) satisfies the differential system:

\begin{equation*} \overset{.}{X(t)} = A(\vec{\omega}) X(t) + D \end{equation*}

where \(D = (0, 0, \gamma)^\top)\) and \(A(\vec{\omega})\) is a \(3 \times 3\) matrix:

\begin{equation*} A(\vec{\omega}) = \begin{pmatrix} -\Gamma & -\omega & \omega_y(t)\\ \omega & -\Gamma & -\omega_x(t)\\ -\omega_y(t) & \omega_x(t) & -\gamma \\ \end{pmatrix} \end{equation*}

with \(\vec{\omega}(t) = (\omega_x(t), \omega_y(t)) \). The model is controlled by \(\vec{\omega}(t)\) where \(\omega_y(t) = 0\). The goal is to maximize the normalized figure of merit:

\begin{equation*} F_{opt} = \frac{1}{\sqrt{1+T_c}}\sqrt{x(T_c)^2 + y(T_c)^2} \end{equation*}

where \(T_d\) and \(T_c\) denote the detection time (fixed by the experimental setup) and the control time, respectively.

Results Source code Readme
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Van Der Pol

We consider Van Der Pol system with: \begin{equation*} \begin{cases} \overset{.}{x}(t) = 10 (y + x - 1/3 x^3)\\ \overset{.}{y}(t) = b_0 - x - 0.75y \end{cases} \end{equation*} In the right, we present some simulations of \(y(x(t))\) first with \(b_0=0.3\) and then with \(b_0=\sqrt(2)/3\).
In the two cases, the blue curve designates the simulations of \(y(x(t))\), the red zone indicates the expansive area (\(\lambda>0\)) and the black point shows the fixed point at the saddle-node bifurcation.
In the first figure (when \(b_0=0.3\)), x and y form a limit cycle. In the second figure (when \(b_0=\sqrt(2)/3\approx0.471\) at the saddle­‐node (SN) bifurcation), the limit cycle collides with the fixed black point \(\approx (0.97,-0.66)\) and disappears.

Results Source code Readme
Design: TemplateMo