Guaranteed Optimal Reachability Control of Reaction-Diffusion Equations Using One-Sided Lipschitz Constants and Model Reduction

Abstract

We show that, for any spatially discretized system of reaction-diffusion, the approximate solution given by the explicit Euler time-discretization scheme converges to the exact time-continuous solution, provided that diffusion coefficient be sufficiently large. By “sufficiently large”, we mean that the diffusion coefficient value makes the one-sided Lipschitz constant of the reaction-diffusion system negative. We apply this result to solve a finite horizon control problem for a 1D reaction-diffusion example. We also explain how to perform model reduction in order to improve the efficiency of the method.

Appendices

Appendix 1: Proof of Lemma 1

Proof

It is easy to check that $0< \alpha _u< 1$ when $\frac{|\lambda _u|G_u}{4}<1$.

Let $t^* := G_u(1-\alpha _u)$. Let us first prove $\delta _{e_0}(t)\leqslant e_0$ for $t=t^*$. We have:

$$-\frac{1}{2}|\lambda _{u}| G_{u} +(2+\frac{1}{2}|\lambda _{u}|G_{u})\alpha _u-\alpha _u^2 = 0.$$

Hence:

$$\frac{1}{2G_{u}(1-\alpha _u)}\lambda _{u} G_{u}^2(1-\alpha _u)^2 +2\alpha _u-\alpha _u^2 = 0,$$

i.e.

$$\frac{1}{2t^*}\lambda _{u} (t^*)^2 +2\alpha _u-\alpha _u^2 = 0.$$

We have: $-\frac{1}{4G_u^2t^*}\lambda _{u} (t^*)^4e^{\lambda _ut^*}\geqslant 0$. It follows:

$$\frac{1}{2t^*}\lambda _{u} (t^*)^2 +2\alpha _u-\alpha _u^2-\frac{1}{4G_u^2t^*}\lambda _{u} (t^*)^4 e^{\lambda _ut^*}\geqslant 0.$$

Hence:

$$1+\frac{1}{2t^*}\lambda _{u} (t^*)^2 -\frac{1}{G_{u}^2}((t^*)^2+\frac{1}{4t^*}\lambda _{u} (t^*)^4e^{\lambda _ut^*}) \geqslant 0.$$

By multiplying by $t^*$:

$$(t^*+\frac{1}{2}\lambda _{u} (t^*)^2) -\frac{1}{G_{u}^2}((t^*)^3+\frac{1}{4}\lambda _{u} (t^*)^4e^{\lambda _ut^*}) \geqslant 0.$$

Since $G=\sqrt{3}|\lambda _u| e_0/C_u$:

$$e_0^2(t^*+\frac{1}{2}\lambda _{u} (t^*)^2) +\frac{C_u^2}{\lambda _u^2}(-\frac{1}{3}(t^*)^3-\frac{1}{12}\lambda _{u} (t^*)^4e^{\lambda _ut^*}) \geqslant 0.$$

By multiplying by $\lambda _u$:

$$e_0^2(\lambda _ut^*+\frac{1}{2}\lambda _{u}^2 (t^*)^2) +\frac{C_u^2}{\lambda _u^2}(-\frac{1}{3}\lambda _u(t^*)^3-\frac{1}{12}\lambda _{u}^2 (t^*)^4e^{\lambda _ut^*}) \leqslant 0.$$

Note that, in the above formula, the subexpression $\lambda _ut^*+\frac{1}{2}\lambda _{u}^2 (t^*)^2$ is such that:

$$\lambda _ut^*+\frac{1}{2}\lambda _{u}^2 (t^*)^2\geqslant e^{\lambda _u t^*}-1$$

since $e^{\lambda _u t^*}-1=\lambda _ut^*+\frac{1}{2}\lambda _{u}^2 (t^*)^2e^{\lambda \theta }\leqslant \lambda _ut^*+\frac{1}{2}\lambda _{u}^2 (t^*)^2$.

On the other hand, the subexpression $-\frac{1}{3}\lambda _u(t^*)^3-\frac{1}{12}\lambda _{u}^2 (t^*)^4e^{\lambda _ut^*}$ is such that:

$$-\frac{1}{3}\lambda _u(t^*)^3-\frac{1}{12}\lambda _{u}^2 (t^*)^4e^{\lambda _ut^*}\geqslant \frac{2 t^*}{\lambda _{u}}+(t^*)^2+\frac{2}{\lambda _{u}^2}(1-e^{\lambda _{u} t^*})$$

since

$\frac{2 t^*}{\lambda _{u}}+(t^*)^2+\frac{2}{\lambda _{u}^2}(1-e^{\lambda _{u} t^*})$

$=\frac{2 t^*}{\lambda _{u}}+(t^*)^2+\frac{2}{\lambda _{u}^2}(-\lambda _u t^*-\frac{1}{2}\lambda _u^2 (t^*)^2-\frac{1}{6}\lambda _u^3(t^*)^3-\frac{1}{24}\lambda _u^4(t^*)^4 e^{\lambda _u\theta }$

$=\frac{2}{\lambda _u^2}(-\frac{1}{6}\lambda _u^3(t^*)^3-\frac{1}{24}\lambda _u^4(t^*)^4 e^{\lambda _u\theta })$ for some $0\leqslant \theta \leqslant t^*$

$=-\frac{1}{3}\lambda _u(t^*)^3-\frac{1}{12}\lambda _u^2(t^*)^4 e^{\lambda _u\theta }$

$\leqslant -\frac{1}{3}\lambda _u(t^*)^3-\frac{1}{12}\lambda _u^2(t^*)^4e^{\lambda _ut^*}.$

It follows:

$$e_0^2(e^{\lambda _{u} t^*}-1)+\frac{C_{u}^2}{\lambda _{u}^2}(\frac{2 t^*}{\lambda _{u}}+(t^*)^2+\frac{2}{\lambda _{u}^2}(1-e^{\lambda _{u} t^*}))\leqslant 0.$$

$$e_0^2e^{\lambda _{u} t^*}+\frac{C_{u}^2}{\lambda _{u}^2}(\frac{2 t^*}{\lambda _{u}}+(t^*)^2+\frac{2}{\lambda _{u}^2}(1-e^{\lambda _{u} t^*}))\leqslant e_0^2.$$

i.e.

$$(\delta _{e_0}^u(t^*))^2\leqslant e_0^2.$$

Hence: $\delta _{e_0}^u(t^*)\leqslant e_0.$ It remains to show: $\delta _{e_0}^u(t)\leqslant e_0$ for $t\in [0,t^*]$.

Consider the 1rst and 2nd derivative $\delta '(\cdot )$ and $\delta ''(\cdot )$ of $\delta (\cdot )$. We have:

$\delta '(t)=\lambda _{u}e_0^2e^{\lambda _{u}t}+\frac{C_{u}^2}{\lambda _{u}^2}(2t+\frac{2}{\lambda _{u}}-\frac{2}{\lambda _{u}}e^{\lambda _{u}t})$

$\delta ''(t)=\lambda _{u}^2e_0^2e^{\lambda _{u}t}+\frac{C_{u}^2}{\lambda _{u}^2}(2-2e^{\lambda _{u}t}).$

Hence $\delta ''(t)>0$ for all $t\geqslant 0$. On the other hand, for $t=0$, $\delta '(t)=\lambda _u e_0^2<0$, and for t sufficiently large, $\delta '(t)>0$. Hence, $\delta '(\cdot )$ is strictly increasing and has a unique root. It follows that the equation $\delta (t)=e_0$ has a unique solution $t^{**}$ for $t>0$. Besides, $\delta (t)\leqslant e_0$ for $t\in [0,t^{**}]$, and $\delta (t)\geqslant e_0$ for $t\in [t^{**},+\infty )$. Since we have shown: $\delta (t^*)\leqslant e_0$, it follows $t^*\leqslant t^{**}$ and $\delta (t)\leqslant e_0$ for $t\in [0,t^{*}]$.    $\Box$

Appendix 2: Numerical Results

See Fig. 5.

Fig. 5.

Simulation of the controllers for $\sigma =1$.

Fig. 6.

Simulation of the controllers for $\sigma =0.5$.

Table 1. Value $\Vert y_i^{\pi ^\varepsilon } (T) - y_i^f \Vert$ for $\sigma =1$ and $\sigma =0.5$ ($T=2$, $i=1,2$).

Table 2. Projection value $\Vert P {y}^{\pi ^\varepsilon }_2(T)-y_1^f\Vert$ for $\sigma =1$, $\sigma =0.5$ ($T=2$).