Stabilization of Structured Populations via Vector Target-Oriented Control

Abstract

In contrast with unstructured models, structured discrete population models have been able to fit and predict chaotic experimental data. However, most of the chaos control techniques in the literature have been designed and analyzed in a one-dimensional setting. Here, by introducing target-oriented control for discrete dynamical systems, we prove the possibility to stabilize a chosen state for a wide range of structured population models. The results are illustrated with introducing a control in the celebrated LPA model describing a flour beetle dynamics. Moreover, we show that the new control allows to stabilize periodic solutions for higher-order difference equations, such as the delayed Ricker model, for which previous target-oriented methods were not designed.

Appendix

The next result shows that VTOC and MVTOC are topologically conjugate; thus, they have the same dynamics. We recall that two maps $\phi$ and $\psi$ are topologically conjugate if there is a homeomorphism h such that $\phi \circ h= h \circ \psi$.

Lemma 5

Assume $\mathbf f :D \rightarrow D$, with D convex and $\mathbf {T}\in D$. Then the difference Eqs. (5) and (6) are topologically conjugate.

Proof

We are going to show that the maps defining the difference Eqs. (5) and (6) are topologically conjugate. We begin by defining such maps.

Consider the map $\varphi (\mathbf {x})=c\mathbf {T}+(1-c)\mathbf {x}$ from D to $\varphi (D)$. Moreover, since D is convex $\varphi (D) \subset D$ and the map $\mathbf {f}\circ \varphi :D \rightarrow D$ is well defined. Clearly, map $\mathbf {f}\circ \varphi$ defines the recurrent relation in Eq. (5). On the other hand, note that after the first iterate the solutions of (6) belong to $\varphi (D)$. Therefore, we have that after the first iterate the map that defines the recurrence given by (6) is $\varphi \circ \mathbf {f} :\varphi (D) \rightarrow \varphi (D)$.

It is easy to check that $\varphi$ is a homeomorphism from D to $\varphi (D)$ and obviously $(\varphi \circ \mathbf {f})\circ \varphi =\varphi \circ ( \mathbf {f}\circ \varphi )$. Hence, $\mathbf {f}\circ \varphi$ and $\varphi \circ \mathbf {f}$ are topologically conjugate. $\square$

Proof of Lemma 1

An equilibrium $\mathbf {x}^{*}$ of VMTOC is a fixed point of the map

$$\begin{aligned} \mathbf {g(x)}= c \mathbf {T} + (1-c) \mathbf {f(x)}. \end{aligned}$$

(17)

Since $\mathbf {T}\in \mathbb {R}^d_{+}\setminus \{\mathbf {0} \}$ and $\mathbf {f}:\mathbb { R}^d_+\rightarrow \mathbb { R}^d_+$, we have for every $c \in (0, 1)$ that $c \mathbf {T} + (1-c) \mathbf {f(0)}\in \mathbb {R}^d_{+}\setminus \{\mathbf {0} \}$ and

$$\begin{aligned} \Vert \mathbf {g(0)}\Vert = \Vert c \mathbf {T} + (1-c) \mathbf {f(0)}\Vert > \Vert \mathbf {0}\Vert . \end{aligned}$$

Hence, by the continuity of $\mathbf {g}$ and the norm, for each fixed $c \in (0, 1)$ it is possible to find $0<m<M$ such that

$$\begin{aligned} \Vert \mathbf {g(x)} \Vert \ge \Vert \mathbf {x}\Vert , \quad \mathbf {x}\in {\mathbb {R}}^d_{+} , \Vert \mathbf {x}\Vert =m. \end{aligned}$$

On the other hand, we have for any $\mathbf {x}\in {\mathbb {R}}^d_{+}$ with $\Vert \mathbf {x}\Vert \ge \max \{M,\Vert \mathbf {T}\Vert \}$,

$$\begin{aligned} \Vert \mathbf {g}(\mathbf {x})\Vert&= \Vert c \mathbf {T} + (1-c) \mathbf {f}(\mathbf {x})\Vert \le \Vert c \mathbf {T} \Vert + \Vert (1-c) \mathbf {f}(\mathbf {x})\Vert \\&= c\Vert \mathbf {T} \Vert + (1-c)\Vert \mathbf {f}(\mathbf {x})\Vert \le c\Vert \mathbf {x} \Vert + (1-c)\Vert \mathbf {x}\Vert = \Vert \mathbf {x}\Vert . \end{aligned}$$

Thus, by Krassnosel’skiĭ fixed-point theorem for cone-compressing operators (see e.g., Granas and Dugundji 2013, Theorem 7.12), the map $\mathbf {g}$ has at least one fixed point ${\mathbf x}^{*}$ in the set

$$\begin{aligned} \left\{ \mathbf {x}\in {\mathbb {R}}^d_+ : 0<\Vert \mathbf x\Vert <\max \{M,\Vert \mathbf {T}\Vert \} \right\} \end{aligned}$$

for every $c \in (0, 1)$.

Finally, the last statement of the lemma follows from noticing that an equilibrium $\mathbf {x}^*$ of VMTOC satisfies

$$\begin{aligned} \mathbf {x}^*=c \mathbf {T} + (1-c) \mathbf {f}(\mathbf {x}^*). \end{aligned}$$

$\square$

Proof of Lemma 2

If $\mathbf {K}=\mathbf {f}(\mathbf {K})$, we take $\mathbf {T}_\mathbf {K}=\mathbf {K}$ and any $c_\mathbf {K}\in (0,1)$. Let $\mathbf {K} \ne \mathbf {f}(\mathbf {K})$. Consider the set of points

$$\begin{aligned} R = \{\mathbf {f}(\mathbf {K})+ \alpha (\mathbf {K}-\mathbf {f}(\mathbf {K})): \alpha >1\}, \end{aligned}$$

which is the ray in the direction $\mathbf {K}-\mathbf {f}(\mathbf {K})$ starting at $\mathbf {K}$. Using that $\mathbf {K} \not \in \partial D$, we have that $R\cap D\ne \emptyset$. Therefore, there exists $\alpha >1$ such that $\mathbf {T_K}:=\alpha \mathbf {K}+(1-\alpha )\mathbf {f(K)} \in D$. Fixing $c_{\mathbf {K}}=\alpha ^{-1}\in (0,1)$ and noticing that

$$\begin{aligned} \mathbf {g}(\mathbf {K})=c_{\mathbf K} \mathbf {T_K}+ (1-c_\mathbf {K})\mathbf {f(K)}= & {} \alpha ^{-1} \left( \alpha \mathbf {K}+(1-\alpha )\mathbf {f(K)} \right) + (1-\alpha ^{-1})\mathbf {f(K)} \\= & {} \mathbf {K}+ \frac{1-\alpha }{\alpha } \mathbf {f(K)} + \frac{\alpha -1}{\alpha } \mathbf {f(K)} = \mathbf {K} \end{aligned}$$

concludes the proof. $\square$

Proof of Lemma 3

We note that it is sufficient to consider only controls $c \in (0,1)$, since for $c=0$ we have the original map. Let $\phi _1(\mathbf {x})=c_1 \mathbf {T}_1+(1-c_1)\mathbf {x}$ be applied first, and next another argument transformation $\phi _2(\mathbf {x})=c_2 \mathbf {T}_2+(1-c_2)\mathbf {x}$. Then

$$\begin{aligned} \phi _2\left( \phi _1(\mathbf {f}(\mathbf {x})) \right)= & {} c_2\mathbf {T}_2+(1-c_2) \left[ c_1\mathbf {T}_1+(1-c_1) \mathbf {f}(\mathbf {x})\right] \\= & {} c_2 \mathbf {T}_2+(1-c_2)c_1 \mathbf {T}_1+(1-c_1)(1-c_2)\mathbf {f}(\mathbf {x})= c\mathbf {T}+(1-c) \mathbf {f}(\mathbf {x}), \end{aligned}$$

where

$$\begin{aligned} c=c_1(1-c_2)+c_2, \quad \text{ and } \quad \mathbf {T}=\frac{c_1(1-c_2)}{c_1(1-c_2)+c_2} \mathbf {T}_1+\frac{c_2}{c_1(1-c_2)+c_2} \mathbf {T}_2. \end{aligned}$$

Since $c_1,c_2 \in (0,1)$, also $c=c_1(1-c_2)+c_2 \in (0,1)$. On the other hand, the vector $\mathbf {T}=\alpha \mathbf {T}_1+(1-\alpha ) \mathbf {T}_2$, where $\alpha = c_1(1-c_2)/c\in (0,1)$, therefore by the convexity of D the target $\mathbf {T} \in D$. $\square$

Proof of Theorem 1

Recall that, by linearization, any equilibrium $\mathbf {p}_c$ of VMTOC is asymptotically stable if the spectral radius of the Jacobian matrix $J\mathbf g$ of $\mathbf {g(x)}=c\mathbf {T}+(1-c)\mathbf {f(x)}$ at $\mathbf {p}_c$ is smaller than 1.

By a well-known bound for the eigenvalues of a matrix (see, e.g., Horn and Johnson 2012, Corollary 6.1.5), we have that any eigenvalue $\lambda$ of $J\mathbf g(\mathbf {p}_c)$ satisfies

$$\begin{aligned} |\lambda | \le (1-c) \min \left\{ \max _{j=1,\ldots , d} \sum _{i=1}^{d} \left| \frac{\partial f_{j}}{\partial x_i}(\mathbf {p}_c) \right| , \max _{j=1,\ldots ,d}\sum _{i=1}^{d} \left| \frac{\partial f_{i}}{\partial x_j}(\mathbf {p}_c) \right| \right\} . \end{aligned}$$

Therefore, for $c\in (c^*,1)$ the eigenvalues of $J\mathbf g(\mathbf {p}_c)$ have modulus smaller than 1. $\square$

Proof of Theorem 2

If L in (9) satisfies $L \in (0,1)$ and $\mathbf {x}_n$ is a solution of (6) with an initial condition $\mathbf {x}_0 \in D$ and $c \in [0,1)$, then for $n\in \mathbb {N}$

$$\begin{aligned} \Vert \mathbf {x}_{n+1} - \mathbf {K} \Vert = \Vert c\mathbf {K} +(1-c) \mathbf {f}(\mathbf {x}_{n}) - \mathbf {K} \Vert =(1-c)\Vert \mathbf {f}(\mathbf {x}_{n})-\mathbf {K} \Vert \le L \Vert \mathbf {x}_{n}-\mathbf {K} \Vert . \end{aligned}$$

For any $\varepsilon \in (0, \Vert \mathbf {x}_0 -\mathbf {K} \Vert )$, we have

$$\begin{aligned} \Vert \mathbf {x}_n - \mathbf {K} \Vert \le L^n \Vert \mathbf {x}_0 -\mathbf {K} \Vert < \varepsilon \hbox { whenever } n> \ln \left( \frac{\varepsilon }{\Vert \mathbf {x}_0 -\mathbf {K} \Vert } \right) \bigg / \ln L, \end{aligned}$$

so $\mathbf {x}_n \rightarrow \mathbf {K}$ as $n \rightarrow \infty$.

Next, let $L\ge 1$. Denote

$$\begin{aligned} c^*= \left\{ \begin{array}{ll} 0, &{} L=1, \\ {\displaystyle 1 -\frac{1}{L}}, &{} L>1, \end{array} \right. \end{aligned}$$

(18)

and assume that $c \in (c^*,1) \subseteq (0,1)$. Then $(1-c)L<1$, denote $\theta =(1-c)L \in (0,1)$. We have $1-c=\theta /L$, $c\mathbf {T}=c\mathbf {K}$ and

$$\begin{aligned} \Vert \mathbf {x}_{n+1}-\mathbf {K} \Vert= & {} \Vert (1-c)\mathbf {f}(\mathbf {x}_n)+c\mathbf {K-K}\Vert = \Vert (1-c)[\mathbf {f}(\mathbf {x}_n)-\mathbf {K}]\Vert \\= & {} (1-c)\Vert \mathbf {f}(\mathbf {x}_n) -\mathbf {K} \Vert \nonumber \\= & {} \frac{\theta }{L} \Vert \mathbf {f}(\mathbf {x}_n)-\mathbf {K} \Vert \le \frac{\theta }{L} L \Vert \mathbf {x}_n-\mathbf {K} \Vert = \theta \Vert \mathbf {x}_n-\mathbf {K} \Vert . \nonumber \end{aligned}$$

By induction,

$$\begin{aligned} \Vert \mathbf {x}_{n+1}-\mathbf {K} \Vert \le \theta ^n \Vert \mathbf {x}_0-\mathbf {K} \Vert . \end{aligned}$$

Therefore, $\Vert \mathbf {x}_{n+1}-\mathbf {K} \Vert< \varepsilon < \Vert \mathbf {x}_0 -\mathbf {K} \Vert$ for any $\displaystyle n > \ln \left( \frac{\varepsilon }{\Vert \mathbf {x}_0 -\mathbf {K} \Vert } \right) \bigg / \ln \theta$. Thus, $\lim \limits _{n \rightarrow \infty } \mathbf {x}_n=\mathbf {K}$. Moreover, $\Vert \mathbf {x}_{n}-\mathbf {K} \Vert$ decays at least geometrically, which concludes the proof. $\square$

Proof of Lemma 4

Let $M>0$ be an upper bound of $\mathbf {f}$:

$$\begin{aligned} \Vert \mathbf {f(x)} \Vert \le M, ~~\mathbf {x} \in D. \end{aligned}$$

(19)

Since $\mathbf {f}$ is locally Lipschitz continuous, for $\mathbf x$ in the intersection of the ball $\Vert \mathbf { x-K} \Vert \le \Vert \mathbf {K}\Vert$ with D, there is a constant $\tilde{L}>0$ such that $\Vert \mathbf {f(x)-K} \Vert \le \tilde{L} \Vert \mathbf {x-K}\Vert$. Next, let $\mathbf {x}\in D$, $\Vert \mathbf {x-K} \Vert > \Vert \mathbf {K}\Vert$. Then, by (19),

$$\begin{aligned} \Vert \mathbf {f(x)-K} \Vert&\le \Vert \mathbf {f(x)}\Vert + \Vert \mathbf {K} \Vert \le M + \Vert \mathbf {K} \Vert \\&= \left( \frac{M}{\Vert \mathbf {K} \Vert }+1 \right) \Vert \mathbf {K} \Vert < \left( \frac{M}{\Vert \mathbf {K} \Vert }+1 \right) \Vert \mathbf {x-K} \Vert . \end{aligned}$$

Finally, choosing

$$\begin{aligned} L= \max \left\{ \tilde{L}, \frac{M}{\Vert \mathbf {K} \Vert }+1 \right\} , \end{aligned}$$

(20)

we obtain that inequality (9) is satisfied for any $\mathbf {x} \in D$. $\square$

Proof of Theorem 3

By Lemma 2 there exist $c_{\mathbf K}\in (0,1)$ and $\mathbf {T_K}\in D$ such that $\mathbf {K}$ is an equilibrium of $\mathbf {g}(\mathbf {x})=c_\mathbf {K} \mathbf {T_K}+(1-c_\mathbf {K})\mathbf {g}(\mathbf {x})$. By Lemma 4, $\mathbf {f}$ satisfies (9) with some constant $L_1$ instead of L. Then

$$\begin{aligned} \Vert \mathbf {g}(\mathbf {x}) - \mathbf {K} \Vert= & {} \Vert \mathbf {g}(\mathbf {x}) - \mathbf {g}(\mathbf {K}) \Vert = \Vert c_\mathbf {K} \mathbf {T_K}+(1-c_\mathbf {K})\mathbf {f(x)} - c_\mathbf {K} \mathbf {T_K}+(1-c_\mathbf {K})\mathbf {f(K)} \Vert \\= & {} (1-c_\mathbf {K}) \Vert \mathbf {f(x) - f(K)} \Vert \le (1-c_{\mathbf {K}}) L_1 \Vert \mathbf {x - K} \Vert , \end{aligned}$$

thus condition (9) holds for $\mathbf {g}$ with $L=(1-c_{\mathbf K})L_1$.

Now, applying Theorem 2, we obtain that there exists $c^* \in [0,1)$ such that for $\hat{c} \in (c^*,1)$ and $\mathbf {T=K}$, all solutions of (6) with $\mathbf {g}$ instead of $\mathbf {f}$, $\hat{c}$ instead of c and $\mathbf {x}_0 \in D$ converge to $\mathbf {K}$. By Lemma 3, a combination of two VMTOCs is a VMTOC. Thus, if we choose $c=c_\mathbf {K}(1-\hat{c})+\hat{c}$, where $\hat{c} \in (c^*,1)$, $c_\mathbf {K}$ is defined above, and

$$\begin{aligned} \mathbf {T}=\frac{c_\mathbf {K}(1-\hat{c})}{c_\mathbf {K}(1-\hat{c})+\hat{c}} \mathbf {T_K}+\frac{\hat{c}}{c_\mathbf {K}(1-\hat{c})+\hat{c}} \mathbf {K}, \end{aligned}$$

we get that $\mathbf {K}$ is a global attractor of the combination of VMTOCs.$\square$