Carleman Estimates and Controllability for a Degenerate Structured Population Model

Abstract

In this paper we study the null controllability property for a single population model in which the population y depends on time t, space x, age a and size $\tau$. Moreover, the diffusion coefficient k is degenerate at a point of the domain or both extremal points. Our technique is essentially based on Carleman estimates. The $\tau$ dependence requires us to modify the weight for the Carleman estimates, and accordingly the proof of the observability inequality. Thanks to this observability inequality we obtain a null controllability result for an intermediate problem and finally for the initial system through suitable cut off functions.

Appendix

1.1 Proof of Proposition 4.2

Let us consider a smooth function $\xi : [0,1] \rightarrow \mathbb R$ such that

$$\begin{aligned} {\left\{ \begin{array}{ll} 0 \le \xi (x) \le 1, &{} \text { for all } x \in [0,1], \\ \xi (x) = 1 , &{} x \in \omega ', \\ \xi (x)=0, &{} x \in (0,1)\setminus \omega . \end{array}\right. } \end{aligned}$$

Then, integrating by parts one has

$$\begin{aligned} \begin{aligned} 0 =&\int _0^T\!\!\!\frac{d}{dt}\left( \int _{Q_{A, \tau _2,1}}(\xi e^{s\psi })^2v^2dxd \tau da\right) dt \\&= \int _Q 2s\psi _t(\xi e^{s\psi })^2v^2 + 2(\xi e^{s\psi })^2v(-v_a-gv_\tau -\mathcal A_0 v+ \mu v+f) \,dxd \tau dadt \\&=\; 2s\int _Q \psi _t(\xi e^{s\psi })^2v^2dxd\tau da dt\\&\quad + 2s\int _Q \psi _a(\xi e^{s\psi })^2v^2dxd\tau da dt+ 2\int _Q \left( \xi ^2e^{2s\psi }\sigma \right) _x\eta vv_xdxd\tau da dt \\&\quad + 2\int _Q (\xi ^2e^{2s\psi }k)v_x^2dxd\tau da dt+2\int _Q \xi ^2e^{2s\psi }\mu v^2dxd\tau da dt \\&\quad +2\int _Q \xi ^2e^{2s\psi }fvdxd\tau da dt\\&\quad + 2s \int _Q \psi _\tau (\xi e^{s\psi })^2gv^2dxd\tau da dt + \int _Q \xi ^2 e^{2s\psi }g_\tau v^2dxd\tau da dt. \end{aligned} \end{aligned}$$

Hence, using Young’s inequality

$$\begin{aligned}&2\int _Q (\xi ^2e^{2s\psi }k)v_x^2dxd\tau da dt \\&\quad = -2s\int _Q \psi _t\left( \xi e^{s\psi }\right) ^2v^2dxd\tau da dt- 2s\int _Q \psi _a(\xi e^{s\psi })^2v^2dxd\tau da dt \\&\qquad -2\int _Q\frac{ \left( \xi ^2e^{2s\psi }\sigma \right) _x}{\xi e^{s\psi }\sqrt{\sigma }}\xi e^{s\psi }\sqrt{\sigma }\eta vv_x\,dxd\tau da dt-2\int _Q \xi ^2e^{2s\psi }\mu v^2dxd\tau da dt\\&\qquad -2\int _Q \xi ^2e^{2s\psi }fvdxd\tau da dt\\&\qquad -2s \int _Q \psi _\tau (\xi e^{s\psi })^2gv^2dxd\tau da dt -\int _Q \xi ^2 e^{2s\psi }g_\tau v^2dxd\tau da dt \\&\quad \le -2s\int _Q \psi _t(\xi e^{s\psi })^2v^2dxd\tau da dt - 2s\int _Q \psi _a(\xi e^{s\psi })^2v^2dxd\tau da dt\\&\qquad -2s \int _Q \psi _\tau (\xi e^{s\psi })^2gv^2dxd\tau da dt -\int _Q \xi ^2 e^{2s\psi }g_\tau v^2dxd\tau da dt\\&\qquad + 4\int _Q \left( \xi e^{s\psi }\sqrt{\sigma }\,\right) _x^2\eta v^2dxd\tau da dt+\int _Q (\xi ^2e^{2s\psi }k)v_x^2dxd\tau da dt \\&\qquad +(2\Vert \mu \Vert _{L^\infty (Q)}+1)\int _Q \xi ^2v^2dxd\tau da dt +\int _Q \xi ^2e^{2s\psi }f^2dxd\tau da dt. \end{aligned}$$

It follows that,

$$\begin{aligned} \begin{aligned}&\int _Q (\xi ^2e^{2s\psi }k)v_x^2dxd\tau da dt\\&\quad \le -2s\int _Q \psi _t(\xi e^{s\psi })^2v^2dxd\tau da dt - 2s\int _Q \psi _a(\xi e^{s\psi })^2v^2dxd\tau da dt\\&\qquad -2s \int _Q \psi _\tau (\xi e^{s\psi })^2gv^2dxd\tau da dt -\int _Q \xi ^2 e^{2s\psi }g_\tau v^2dxd\tau da dt\\&\qquad + 4\int _Q \left( \xi e^{s\psi }\sqrt{\sigma }\,\right) _x^2\eta v^2dxd\tau da dt \\&\qquad +(2\Vert \mu \Vert _{L^\infty (Q)}+1)\int _Q \xi ^2v^2dxd\tau da dt +\int _Q \xi ^2e^{2s\psi }f^2dxd\tau da dt. \end{aligned} \end{aligned}$$

Thus,

$$\begin{aligned}&\inf _{\omega '}\{k\}\int _{Q_{T,A, \tau _2}}\int _{\omega '}e^{2s\psi }v_x^2dxd\tau da dt \nonumber \\&\quad \le \left( \sup _{\omega \times (0,T)}\Big \{\left| 4\left( \xi e^{s\psi }\sqrt{\sigma }\,\right) _x^2-2s(\psi _t+\psi _a+ \psi _\tau g+g_\tau ) (\xi e^{s\psi })^2\right| \Big \} +\, 2\Vert \mu \Vert _{L^\infty (Q)}+1\right) \\&\qquad \int _{Q_{T,A, \tau _2}}\int _{\omega }v^2dxd\tau da dt +\int _Q f^2e^{2s\psi } dxd\tau da dt. \end{aligned}$$