Integral control for population management

Abstract

We present a novel management methodology for restocking a declining population. The strategy uses integral control, a concept ubiquitous in control theory which has not been applied to population dynamics. Integral control is based on dynamic feedback—using measurements of the population to inform management strategies and is robust to model uncertainty, an important consideration for ecological models. We demonstrate from first principles why such an approach to population management is suitable via theory and examples.

Appendix

1.1 The constants $\upgamma ,\kappa ,\mu _g$ and $\nu _g$

$\bullet$ We first prove the inequality (3.24). In the proceeding arguments, for a sequence $v$ we use the notation $\hat{v}$ to denote the $Z$-transform of $v$ given by

$$\begin{aligned} \hat{v}(z) = \sum _{j=0}^\infty v(j) z^{-j}, \end{aligned}$$

defined for all complex $z$ where the summation converges absolutely. The step response of the linear system (3.5) is the output of (3.5) subject to zero initial state ($x^0 = 0$) and constant input $\tilde{u} =1$ and is given by

$$\begin{aligned} s(0) = 0, \quad s(t) = \sum _{j=0}^{t-1} c^TA^j b, \quad t= 1,2,\dots . \end{aligned}$$

Assumption (A1) ensures that $s(t) \rightarrow G(1)$ as $t \rightarrow \infty$. Furthermore, a calculation shows that $s$ has $Z$-transform

$$\begin{aligned} \hat{s}(z) = \frac{zG(z)}{z-1}, \quad z \in \mathbb {C}, \quad \vert z \vert >1\,. \end{aligned}$$

Under the assumptions that $A,b,c^T \ge 0$ and (A2) it follows that $s(t) \ge 0$ and is non-decreasing. We define the step response error

$$\begin{aligned} e(t) = s(t) - G(1), \quad t =0,1,2,\dots , \end{aligned}$$

which is consequently non-positive, non-decreasing and converges to 0. Furthermore, the $Z$-transform of $e$ satisfies

$$\begin{aligned} \frac{\hat{e}(z)}{z} = \frac{G(z) - G(1)}{z-1}, \end{aligned}$$

(5.1)

for every complex $z$ with modulus greater than one. Since $G$ is differentiable at $z=1$ we note that

$$\begin{aligned} \lim _{z \rightarrow 1} \hat{e}(z) = \lim _{z \rightarrow 1} \frac{\hat{e}(z)}{z} = \lim _{z \rightarrow 1} \frac{G(z) - G(1)}{z-1} = G^{\prime }(1)\,. \end{aligned}$$

(5.2)

As $z \mapsto \tfrac{\hat{e}(z)}{z}$ is continuous outside of the unit disc the above shows that we can extend $z \mapsto \tfrac{\hat{e}(z)}{z}$ continuously to $z=1$ with

$$\begin{aligned} \hat{e}(1) =\frac{\hat{e}(1)}{1} = G^{\prime }(1)\,. \end{aligned}$$

(5.3)

We now use (5.3) and the property that $e(t) \le 0$ for every $t$ to show that for any complex $z$ with modulus one,

$$\begin{aligned} -G^{\prime }(1)&= -\hat{e}(1) = -\sum _{k=0}^\infty e(k) = \sum _{k=0}^\infty \vert e(k) \vert \cdot \vert z^{-(k+1)} \vert \ge \left| \frac{\hat{e} (z)}{z} \right| \ge \mathrm{Re } \,\left( \frac{-\hat{e} (z)}{z} \right) \nonumber \\&\quad = \mathrm{Re} \, \left[ \frac{G(1) - G(z)}{z-1}\right] \nonumber \\&\quad = -\frac{G(1)}{2} - \mathrm{Re} \, \left[ \frac{ G(z)}{z-1}\right] . \end{aligned}$$

(5.4)

Rearranging (5.4) gives

$$\begin{aligned} \mathrm{Re}\, \left[ \frac{G(z)}{z-1} \right] \ge G^{\prime }(1) - \frac{G(1)}{2}, \quad \text {for all complex } z \text { with modulus one,} \end{aligned}$$

which implies that

$$\begin{aligned} \inf _{\vert z \vert =1 } \mathrm{Re}\, \left[ \frac{G(z)}{z-1}\right] =\inf _{\theta \in [0,2\pi )} \mathrm{Re}\, \left[ \frac{G(\mathrm {e}^{\mathrm {i}\theta })}{\mathrm {e}^{\mathrm {i}\theta }-1}\right] \ge G^{\prime }(1) - \frac{G(1)}{2}. \end{aligned}$$

(5.5)

From Coughlan and Logemann (2009) we have that $\upgamma$ satisfies $-\infty <\upgamma \le \tfrac{-G(1)}{2} <0$ as $G(1) >0$, and so

$$\begin{aligned} 0 > \upgamma \ge \inf _{\theta \in [0,2\pi )} \mathrm{Re} \left[ \frac{ G(\mathrm {e}^{\mathrm {i}\theta })}{\mathrm {e}^{\mathrm {i}\theta } -1} \right] \ge G^{\prime }(1) - \frac{G(1)}{2}\,, \end{aligned}$$

(5.6)

where we have used the estimate (5.5). It is clear from $\upgamma \le \tfrac{-G(1)}{2}$ and (5.6) that $G^{\prime }(1) <0$ and consequently inequality (5.6) is equivalent to

$$\begin{aligned} 0 < -\upgamma = \vert \upgamma \vert \le -G^{\prime }(1) + \frac{G(1)}{2} = \vert G^{\prime }(1) \vert + \frac{G(1)}{2} {=:} \frac{1}{\kappa }, \end{aligned}$$

which implies (3.24).

$\bullet$ The constants $\mu _g$ and $\nu _g$ in (3.16) and (3.20) are given by

$$\begin{aligned} \mu _g {:=} \sum _{j=0}^\infty \left| \begin{bmatrix}c^T&0 \end{bmatrix}\begin{bmatrix}A&\quad b \\ -gc^T&\quad 1 \end{bmatrix}^j\begin{bmatrix}0 \\ g \end{bmatrix}\right| \quad \text {and} \quad \nu _g {:=} \sum _{j=0}^\infty \left| \begin{bmatrix}c^T&0 \end{bmatrix}\begin{bmatrix}A&\quad b \\ -gc^T&\quad 1 \end{bmatrix}^j\begin{bmatrix}b \\ 0 \end{bmatrix}\right| \,, \end{aligned}$$

respectively, which are both finite since by assumption $g >0$ is such that has $\rho (A_g) <1$.

$\bullet$ We now derive the inequality (3.29). For complex $z$ with modulus greater than or equal to one the transfer function $G$ given by (3.11) of the linear system (3.5) can be written as

$$\begin{aligned} G(z) = \sum _{j =0}^\infty g_j z^{-j}, \quad \text {where} \quad g_j = \left\{ \begin{aligned}&0,&j=0, \\&c^T A^{j-1}b,&j \ge 1\,. \end{aligned} \right. \end{aligned}$$

We define $z \mapsto \tilde{G}(z) {:=} zG(z)$ and introduce the constant

$$\begin{aligned} \tilde{\upgamma }{:=}\sup _{q\ge 0}\left\{ \inf _{\theta \in [0,2\pi )}\mathrm{Re} \left[ \left( \frac{q}{\mathrm {e}^{i\theta }} +\frac{1}{\mathrm {e}^{i\theta }-1}\right) \tilde{G}(\mathrm {e}^{i\theta })\right] \right\} \,. \end{aligned}$$

We know that $-\infty < \tilde{\upgamma } \le -\tfrac{\tilde{G}(1)}{2} = -\tfrac{G(1)}{2}$. By inspection of the definition of $\tilde{G}$, the constant $\tilde{\upgamma }$, and ${\upgamma }_0$ in (3.29) we see that

$$\begin{aligned} \tilde{\upgamma }&= {\upgamma }_0, \end{aligned}$$

(5.7)

$$\begin{aligned} \tilde{G}^{\prime }(z)&= G(z) + zG^{\prime }(z) \quad \text {and so} \quad \tilde{G}^{\prime }(1) = G(1) + G^{\prime }(1)\,. \end{aligned}$$

(5.8)

We note from (5.8) that

$$\begin{aligned} \tilde{G}^{\prime }(1) = G(1) + G^{\prime }(1) = \sum _{j =1}^\infty g_j z^{-j} - \sum _{j =1}^\infty j g_j z^{-j} = \sum _{j =1}^\infty (1 -j)g_j z^{-j} \le 0\,, \end{aligned}$$

(5.9)

and consequently we can apply the estimate (3.24) to $\tilde{G}$ to yield that

$$\begin{aligned} \frac{2}{\tilde{G}(1) + 2 \vert \tilde{G}^{\prime }(1)\vert } \le \frac{1}{\vert \tilde{\upgamma } \vert }\,. \end{aligned}$$

(5.10)

In light of (5.7), (5.8) and the following definition of $\kappa _0$, (5.10) implies that

$$\begin{aligned} \kappa _0 : = \frac{2}{G(1) + 2 \vert G(1) + G^{\prime }(1) \vert } = \frac{2}{\tilde{G}(1) + 2 \vert \tilde{G}^{\prime }(1)\vert } \le \frac{1}{\vert \tilde{\upgamma } \vert } = \frac{1}{\vert {\upgamma }_0 \vert }\,, \end{aligned}$$

as required. Finally, as $G(1) >0$, it is clear from (5.9) that $G^{\prime }(1) \le \tilde{G}^{\prime }(1) \le 0$ and thus

$$\begin{aligned} \vert G(1) + G^{\prime }(1) \vert = \vert \tilde{G}^{\prime }(1) \vert < \vert G^{\prime }(1)\vert \,. \end{aligned}$$

(5.11)

From inequality (5.11) we deduce that

$$\begin{aligned} \kappa = \frac{2}{G(1) + 2 \vert G^{\prime }(1) \vert } < \frac{2}{G(1) + 2 \vert G(1) + G^{\prime }(1) \vert } = \kappa _0\,. \end{aligned}$$

1.2 Proof of Theorem 2

Let $(x,u)$ denote the solution of

$$\begin{aligned} \left. \begin{aligned} x(t+1)&= Ax(t)+bu(t),\,\,&x(0)=x^0, \\ u(t+1)&= u(t)+g(r-(1+\varepsilon (t))c^Tx(t)),&u(0)=u^0, \end{aligned}\right\} \quad t = 0,1,2,\dots , \end{aligned}$$

(5.12)

the integral control system (3.10) with proportional observation errors $\varepsilon (t)$. When $\varepsilon$ is a sequence of random variable then so are $x$ and $u$. We let

$$\begin{aligned} x_*= (I-A)^{-1}b\frac{r}{G(1)}, \quad u_*= \frac{r}{G(1)}, \end{aligned}$$

which are equilibria of (3.10) as in (3.12). For notational convenience we define the random variable

$$\begin{aligned} z(t) {:=} \begin{bmatrix}x(t) - x_*\\ u(t) - u_* \end{bmatrix}, \quad t =0,1,2,\dots , \end{aligned}$$

(5.13)

a vector with $n+1$ components. A short calculation using (3.12) and (5.12) demonstrates that $z(t)$ has dynamics given by

$$\begin{aligned} z(t+1) = \left[ \begin{bmatrix}A&\quad b \\ -gc^T&\quad 1 \end{bmatrix} - \begin{bmatrix}0 \\ 1 \end{bmatrix}g\varepsilon (t) \begin{bmatrix}c^T&0 \end{bmatrix}\right] z(t) - \begin{bmatrix}0 \\ 1 \end{bmatrix}gr \varepsilon (t), \quad t = 0,1,2,\dots . \end{aligned}$$

(5.14)

We introduce the notation

$$\begin{aligned} A_g {:=} \begin{bmatrix}A&\quad b \\ -gc^T&\quad 1 \end{bmatrix}, \quad D {:=} \begin{bmatrix}0_{n \times 1} \\ 1 \end{bmatrix}, \quad E {:=} \begin{bmatrix}c^T&0 \end{bmatrix}, \end{aligned}$$

where recall that $0_{n \times 1}$ is a column vector of $n$ zeros. With this notation (5.14) can be more concisely expressed as

$$\begin{aligned} z(t+1) = \left[ A_g - g\varepsilon (t) DE\right] z(t) - Dgr \varepsilon (t), \quad t = 0,1,2,\dots . \end{aligned}$$

(5.15)

Letting $\overline{z(t)} = \mathbb {E}(z(t))$ denote the expectation of $z(t)$, we take expectations in (5.15) to yield that

$$\begin{aligned} \overline{z(t+1)} = A_g \overline{z(t)}, \quad t = 0,1,2,\dots , \end{aligned}$$

(5.16)

where we have used the facts that expectation is linear, $\overline{\varepsilon (t)} =0$ and that $\varepsilon (t)$ and $z(t)$ are independent. We are assuming that the gain parameter $g>0$ is such that $\rho (A_g) <1$, and hence from (5.16) we conclude that

$$\begin{aligned} \lim _{t \rightarrow \infty } \overline{z(t)} = 0, \quad \text {and thus} \quad \lim _{t \rightarrow \infty } \overline{y(t)} = \lim _{t \rightarrow \infty } \begin{bmatrix}c^T&0 \end{bmatrix}\overline{z(t)} +r =r, \end{aligned}$$

establishing claim (1). We now consider the covariance

$$\begin{aligned} \mathrm{cov} \, (z(t), z(t))&= \mathbb {E}\left( (z(t) - \overline{z(t)})(z(t) - \overline{z(t)})^T \right) = \mathbb {E}(z(t)z^T(t)) - \overline{z(t)} \cdot \overline{z^T(t)} \nonumber \\&{=:} C(t) - \overline{z(t)} \cdot \overline{z^T(t)}, \quad t = 0,1,2,\dots , \end{aligned}$$

(5.17)

where $z^T(t) = (z(t))^T$. We focus on the quantity $C(t)$, which (appealing to (5.15)) has dynamics

$$\begin{aligned} C(t+1)&= \mathbb {E}(z(t+1)z^T(t+1)) \nonumber \\&= \mathbb {E}\left( \left[ [A_g - g\varepsilon (t) DE] z(t) - Dg r\varepsilon (t)\right] \left[ [A_g - g\varepsilon (t) DE] z(t) - Dgr \varepsilon (t)\right] ^T \right) \nonumber \\&= \mathbb {E}(A_g z(t) (A_g z(t))^T) \nonumber \\&\quad +\, \underbrace{\mathbb {E}(A_g z(t) z^T(t)(-g\varepsilon (t))(DE)^T) + \mathbb {E}((A_g z(t) z^T(t)(-g\varepsilon (t))(DE)^T)^T)}_{=0} \nonumber \\&\quad +\, \underbrace{\mathbb {E}(A_g z(t)\varepsilon (t) D^Tgr) + \mathbb {E}((A_g z(t)\varepsilon (t) D^Tgr)^T)}_{=0} \nonumber \\&\quad +\, g^2 {\sigma }^2 \mathbb {E}(DE z(t) z^T(t) (DE)^T ) \nonumber \\&\quad +\, g^2r^2{\sigma }^2DD^T + \mathbb {E}(D(-g\varepsilon (t))Ez(t)\varepsilon (t)(-D^Tgr)) \nonumber \\&\quad +\, \mathbb {E}((D(-g\varepsilon (t))Ez(t)\varepsilon (t)(-D^Tgr))^T), \end{aligned}$$

(5.18)

for $t = 0,1,2,\dots$. Equation (5.18) simplifies to

$$\begin{aligned} C(t+1)&= A_g C(t) A_g^T + g^2 {\sigma }^2 (DE)C(t) (DE)^T + g^2 {\sigma }^2 r^2 DD^T\nonumber \\&\quad + rg^2{\sigma }^2 DE \overline{z(t)} D^T + rg^2{\sigma }^2 (DE)^T D\overline{z^T(t)}, \quad t =0,1,2,\dots . \end{aligned}$$

(5.19)

Define $A_1 {:=} A_g$, $A_2 {:=} g \sigma DE$ and write

$$\begin{aligned}&c(t)\, {:=}\, \mathrm{vec}\,C(t),\quad p(t) {:=} \mathrm{vec}\,\left( g^2 {\sigma }^2 r^2 DD^T + rg^2{\sigma }^2 DE \overline{z(t)} D^T \nonumber \right. \\&\quad \left. +\, rg^2{\sigma }^2 (DE)^T D\overline{z^T(t)}\right) , \end{aligned}$$

(5.20)

where if $x_i$ are the columns of the $n \times n$ matrix $X = [x_1, x_2, \dots , x_n]$ then

$$\begin{aligned} \mathrm{vec}\,X {:=} \begin{bmatrix}x_1^T&x_2^T&\dots&x_n^T \end{bmatrix}^T \in {\mathbb {R}}^{n^2}. \end{aligned}$$

Arguing now as in Ran and Reurings (2002), the matrix difference equation (5.19) can be written as the $(n+1)^2 \times (n+1)^2$ linear system

$$\begin{aligned} c(t+1) = \left( \sum _{i=1}^2 A_i \otimes A_i \right) c(t) + p(t), \quad t =0,1,2,\dots , \end{aligned}$$

(5.21)

where $\otimes$ denotes the Kronecker product. Using the fact that $\overline{z(t)} \rightarrow 0$ as $t \rightarrow \infty$ it follows from (5.20) that

$$\begin{aligned} \lim _{t\rightarrow \infty } p(t) = \mathrm{vec}\,(g^2 {\sigma }^2 r^2 DD^T) {=:} p_{\infty }. \end{aligned}$$

Consequently, if

$$\begin{aligned} \rho \left( \sum _{i=1}^2 A_i \otimes A_i \right) = \rho ( A_1 \otimes A_1 + A_2 \otimes A_2)<1, \end{aligned}$$

(5.22)

then for any initial condition $c(0)$ the solution $c$ of (5.21) converges to a finite limit $c_{\infty }$ satisfying

$$\begin{aligned} c_{\infty } = \left( \sum _{i=1}^2 A_i \otimes A_i \right) c_{\infty } + p_{\infty }. \end{aligned}$$

(5.23)

Assuming that (5.22) holds, defining $C_{\infty }$ as the matrix such that

$$\begin{aligned} c_{\infty } = \mathrm{vec}\, C_{\infty }, \end{aligned}$$

we have from (5.23) that $C_{\infty }$ must satisfy

$$\begin{aligned} C_{\infty } = A_g C_{\infty } A_g^T + g^2 {\sigma }^2 (DE)C_{\infty } (DE)^T + g^2 {\sigma }^2 r^2 DD^T. \end{aligned}$$

Furthermore, as $C(t)$ converges to $C_{\infty }$ the iterative scheme (5.19) provides a method for approximating $C_{\infty }$.

It remains to find a characterisation of the stability condition (5.22). Recalling that for square matrices $X,Y$

$$\begin{aligned} \sigma (X\otimes Y) = \left\{ \lambda \mu \, : \, \lambda \in \sigma (X), \mu \in \sigma (Y) \right\} , \end{aligned}$$

we have that

$$\begin{aligned} \rho (A_1 \otimes A_1) = \rho (A_g \otimes A_g) = \rho (A_g)^2 <1, \end{aligned}$$

and thus we can view $A_1 \otimes A_1 + A_2 \otimes A_2$ as a structured perturbation of $A_1 \otimes A_1$. Therefore we can characterise the condition (5.22) by appealing to stability radius arguments (Hinrichsen and Pritchard 1986a, b). A calculation shows that $A_2 \otimes A_2$ is a rank one perturbation, namely

$$\begin{aligned}&A_2 \otimes A_2 = g^2 {\sigma }^2 \begin{bmatrix}0&\quad 0 \\ c^T&\quad 0 \end{bmatrix}\otimes \begin{bmatrix}0&\quad 0 \\ c^T&\quad 0 \end{bmatrix} = g^2 {\sigma }^2 \begin{bmatrix}0_{(n^2 + 2n) \times 1} \\ 1 \end{bmatrix}\begin{bmatrix} (c^T&\quad 0) \otimes (c^T&\quad 0) \end{bmatrix}\\&\quad {=:} g^2 {\sigma }^2 \tilde{D} \tilde{E}. \end{aligned}$$

Hence, condition (5.22) is satisfied if, and only if,

$$\begin{aligned} {\sigma }^2 g^2 < \frac{1}{ {\displaystyle \mathop {\max }_{ \vert z \vert =1 }}\, \vert \tilde{E}(zI- A_g \otimes A_g)^{-1}\tilde{D} \vert }, \end{aligned}$$

which is equivalent to the condition (3.18). We can now take the limit as $t \rightarrow \infty$ in (5.17) and use that $\overline{z(t)}$ converges to zero to deduce that

$$\begin{aligned} \lim _{t \rightarrow \infty } \mathrm{cov} \, (z(t), z(t)) = \lim _{t \rightarrow \infty } C(t) = C_{\infty }. \end{aligned}$$

(5.24)

The variance of the output satisfies

$$\begin{aligned} \mathrm{var}\,y(t)&= \mathrm{var}\,(y(t) -r) = \mathrm{var}\left( \begin{bmatrix}c^T&\quad 0 \end{bmatrix}z(t)\right) = \mathrm{cov} \left( \begin{bmatrix}c^T&\quad 0 \end{bmatrix}z(t), \begin{bmatrix}c^T&\quad 0 \end{bmatrix}z(t) \right) \nonumber \\&= \begin{bmatrix}c^T&0 \end{bmatrix} \mathrm{cov} \left( z(t), z(t) \right) \begin{bmatrix}c \\ 0 \end{bmatrix}, \quad t =0,1,2, \dots . \end{aligned}$$

(5.25)

Therefore taking limits in (5.25) and invoking (5.24) we have that

$$\begin{aligned} \lim _{t \rightarrow \infty } \mathrm{var}\,y(t) = \begin{bmatrix}c^T&\quad 0 \end{bmatrix} C_{\infty } \begin{bmatrix}c \\ 0 \end{bmatrix} <\infty , \end{aligned}$$

proving claim (2).

1.3 More general input nonlinearities

We comment further on Remarks 2 and 3. Theorem 3 applies when $\phi$ in (3.21) is replaced by any function $\phi : \mathbb {R}\rightarrow \mathbb {R}$ that satisfies a so-called Lipschitz condition, namely

(A3) :

there exists $l>0$ such that $0\le \phi (v)-\phi (w)\le l(v-w)$ for all $v\ge w$.

The constant $l$ in assumption (A3) is called the Lipschitz constant of $\phi$ and, for example, the function $\phi$ in (3.21) satisfies (A3) with $l =1$.

For a function $\phi : \mathbb {R}\rightarrow \mathbb {R}$ and a set $X \subseteq \mathbb {R}$ we let $\mathrm{im }\, \phi$ and $\phi ^{-1}(X)$ denote the image of $\phi$ and preimage of $X$ of under the function $\phi$ respectively.

In this more general setting, Theorem 3 can be restated as: Assume that (3.22) satisfies (A1)–(A3). Then, for every $r \in \mathbb {R}$ such that $r/G(1)\in \mathrm{im}\,\phi$, every $g\in (0,1/|\upgamma l|)$ and all initial conditions $(x^0,u^0)\in {\mathbb {R}}^n\times \mathbb {R}$, statements (1), (2) and (3) hold. Moreover, if additionally $\phi ^{-1}(r/G(1))$ is a singleton then $(x^r,u^r)$ is a globally asymptotically stable equilibrium of (3.22).

The adaptive integral control result, Theorem 4, can be restated as: Assume that (3.26) satisfies assumptions (A1)-(A3). Then, for every $r\in \mathbb {R}$ such that $r/G(1)\in \mathrm{im}\,\phi$, and all initial conditions $(x^0,u^0,h^0)\in {\mathbb {R}}^n\times \mathbb {R}\times (0,\infty )$,

• (1) $\displaystyle {\lim _{t\rightarrow \infty }}u(t)=\displaystyle {\frac{r}{G(1)}}$,

• (2) $\displaystyle {\lim _{t\rightarrow \infty }}x(t)=x^r{:=}(I-A)^{-1}b\displaystyle {\frac{r}{G(1)}}$,

• (3) $\displaystyle {\lim _{t\rightarrow \infty }}y(t)=\lim _{t\rightarrow \infty } c^Tx(t)=r$.

Moreover, if $\phi ^{-1}(r/G(1))$ is a singleton, then

• (4) the non-increasing gain $k(t)=1/h(t)$ converges to a positive limit as $t\rightarrow \infty$,

• (5) $\displaystyle {\lim _{t\rightarrow \infty }}w(t)=w^r$, where $\phi (w^r)=\displaystyle {\frac{r}{G(1)}}$.

1.4 Proof of Theorem 5

By assumption $k>0$ is chosen so that $A - kbc^T$ is componentwise nonnegative. Since $A, b$ and $c^T$ are also componentwise nonnegative we clearly have that $A \ge A- kbc^T$ (the inequality is understood componentwise) and so Berman and Plemmons (1994, p. 27) implies that

$$\begin{aligned} 0 \le \rho (A-kbc^T) \le \rho (A) <1. \end{aligned}$$

We deduce that assumption (A1) holds for $A - kbc^T$. Moreover, one can show that the transfer function of $(A-kbc^T, b,c^T)$ is

$$\begin{aligned} z \mapsto G_k(z) = \frac{G(z)}{1+kG(z)}, \quad \text {so that} \quad G_k(1) = \frac{G(1)}{1+kG(1)} >0, \end{aligned}$$

implying that assumption (A2) applies to $(A-kbc^T, b,c^T)$. Therefore, Theorem 1 now applies to the feedback system (3.30), that is the original integral control system (3.10) with $A$ replaced by $A-kbc^T$. It is straightforward to demonstrate that the equilibria $(x^{*}, u^{*})$ of (3.30) are the same as those of (3.10).

1.5 IPM example

Following Briggs et al. (2010) we take $\varOmega = [\mathrm {e}^{-0.5}, \mathrm {e}^{3.5}]$, so that $\alpha =\mathrm {e}^{-0.5} \sim 0.6$ and $\beta =\mathrm {e}^{3.5} \sim 33$. The kernel $k$ is divided into

$$\begin{aligned} k(y,x) = p(y,x) + f(y,x), \end{aligned}$$

where $p$ denotes the survival component and $f$ denotes the reproductive component. These have respective decompositions

$$\begin{aligned} p(y,x) = s(x)(1-f_p(x))g(y,x), \quad \text {and} \quad \text{ f }(y,x) = P_eJ(y)s(x)f_p(x)S(x). \end{aligned}$$

The functions $s,f_p,g,J, S$ and constant $P_e$ are as in (Briggs et al. (2010), Table 1), where a biological interpretation is also provided. For our simulations we have altered $s$, $S$ and $P_e$ to

$$\begin{aligned} s(x) = 0.7\frac{\mathrm {e}^{0.85x -0.62}}{1+ \mathrm {e}^{0.85x -0.62}}, \quad S(x) = \mathrm {e}^{1.85x +0.37}, \quad P_e = 0.05. \end{aligned}$$

We have made these alterations so that the population is declining, and we can apply our results.

Finite element approximations are one method of reducing the infinite-dimensional IPM to a finite-dimensional difference equation by discretising the spatial domain. That is, the function space $L^1(\varOmega )$ is approximated by an indexed sequence of finite-dimensional subspaces which get ‘closer’ to $L^1(\varOmega )$ as the index $N$ increases. In what follows we give a very brief description of how finite elements is used to derive an approximation of the IPM and refer the reader to the texts by Johnson (1987) or Brenner and Scott (1994) for a thorough treatment.

For an integer $N$, the interval $[\alpha , \beta ]$ is partitioned into $N$ subintervals with $N+1$ equally spaced endpoints $s_i$ defined by

$$\begin{aligned} s_i = \alpha + \frac{(i-1)(\beta -\alpha )}{N}, \quad 1 \le i \le N+1. \end{aligned}$$

In particular $s_1 = \alpha$ and $s_{N+1} =\beta$. The $N+1$ ‘hat’ or ‘tent’ functions $\delta _i$ are defined by

$$\begin{aligned} \delta _i(s) = \left\{ \begin{aligned}&\frac{s-s_{i-1}}{s_i - s_{i-1}}&\quad s \in [s_{i-1}, s_i], \\&\frac{s_{i+1}-s}{s_{i+1} - s_i}&\quad s \in [s_i, s_{i+1}], \\&0&\quad \text {otherwise,} \end{aligned} \right. \quad 1\le i \le N+1, \end{aligned}$$

(5.26)

where $s_0 = s_1 = \alpha$ and $s_{N+2} = s_{N+1} = \beta$. The hat functions are more readily understood visually, and some examples are plotted in Fig. 15.

Fig. 15

Three sample hat functions defined by (5.26) with $\alpha = 0$, $\beta = 1$ and $N=10$. The functions $\delta _1$, $\delta _5$ and $\delta _{11}$ are plotted in solid, dashed and dashed–dotted lines respectively

Loosely speaking, the finite element method assumes that functions in $L^1(\varOmega )$ are well approximated by a linear combination of finitely many of the $\delta _i$ functions. And so, supposing that $n$ is a solution of the IPM (3.5), using (3.32), (3.34) and (3.35), with input $u$ and output $y$ then for any continuous function $v$ the following equation is satisfied

$$\begin{aligned} \int _{\xi \in \varOmega } v(\xi ) \left[ n(\xi ,t+1) - (An)(\xi ,t) - b(\xi )u(t)\right] \,d\xi =0, \quad t =0,1,2,\dots . \end{aligned}$$

(5.27)

We assume that $v$ and $n$ can be written as a linear combination of the $\delta _i$, that is, as

$$\begin{aligned} v(t,\xi ) =\sum _{i=1}^{N+1} v_i(t) \delta _i(\xi ), \quad n(t,\xi ) =\sum _{j=1}^{N+1} n_j(t) \delta _j(\xi ), \end{aligned}$$

(5.28)

for some coefficients $v_i$ and $n_j$. Substituting (5.28) into (5.27) and simplifying gives the following matrix equation

$$\begin{aligned} M\mathtt{n }(t+1) = D \mathtt{n }(t) + J u(t), \quad t = 0,1,2,\dots , \end{aligned}$$

(5.29)

where $\mathtt{n }(t) = \begin{bmatrix}n_1(t)&\dots&n_{N+1}(t) \end{bmatrix}^T$ and the matrices $M,D$ and vector $J$ have components given by

$$\begin{aligned} \left. \begin{aligned} M_{ij}&= \int _{\xi \in \varOmega } \delta _i(\xi ) \delta _j(\xi ) \, d\xi ,&D_{ij}&= \int _{\xi \in \varOmega } \delta _i(\xi ) \int _{s \in \varOmega } k(\xi ,s) \delta _j(s) \,ds d\xi , \\ J_i&= \begin{bmatrix}J_1&\dots&J_{N+1} \end{bmatrix}^T,&J_i&=\int _{\xi \in \varOmega } \delta _i(\xi ) b(\xi ) \, d\xi , \\ \end{aligned} \right\} \quad 1 \le i,j \le N+1. \end{aligned}$$

It is straightforward to see that the matrix $M$ is invertible; if $q \in \mathbb {C}^{N+1}$ has $i^\mathrm{th}$ component $q_i$ then we see that

$$\begin{aligned} \overline{q}^T Mq = \sum _{i,j=1}^{N+1} \overline{q_i} M_{ij} q_j = \int _{\xi \in \varOmega } \left\| \sum _{i=1}^{N+1} q_i \delta _i(\xi ) \right\| ^2 \, d\xi \ge 0\,. \end{aligned}$$

(5.30)

Furthermore, if $\overline{q}^T Mq =0$ then as $\xi \mapsto \sum _{i=1}^{N+1} q_i \delta _i(\xi )$ is continuous it follows from (5.30) that

$$\begin{aligned} \sum _{i=1}^{N+1} q_i \delta _i(\xi ) = 0, \quad \forall \, \xi \in \varOmega \quad \Rightarrow \quad q_i =0, \quad \forall \, i \in \{1,2,\dots , N+1\}\,, \end{aligned}$$

and thus $q=0$, proving that $M$ is invertible.

When the output is of the form

$$\begin{aligned} y(t) = \int _{\xi _1}^{\xi _2} n(s,t) \,ds, \quad t = 0,1,2,\dots \,, \end{aligned}$$

(5.31)

where $\xi _1 < \xi _2$ denote the range of stage-classes observed, then substituting (5.28) into Eq.  (5.31) gives $y(t) = F\mathtt{n }(t)$, where the row vector $F = \begin{bmatrix}F_1&\dots&F_{N+1} \end{bmatrix}$ has components

$$\begin{aligned} F_i = \int _{\xi _1}^{\xi _2} \delta _i(s) \,ds, \quad 1 \le i \le N+1. \end{aligned}$$

Therefore, we have the following system with $N+1$ states

$$\begin{aligned} \left. \begin{aligned} \mathtt{n }(t+1)&= M^{-1}D \mathtt{n }(t) + M^{-1}J u(t), \\ y(t)&= F\mathtt{n }(t), \end{aligned} \right\} \quad t = 0,1,2,\dots , \end{aligned}$$

(5.32)

which is an approximation of the IPM (3.5) and can be readily implemented. The matrix $M$ and vector $F$ can be found analytically, whilst $D$ and $J$ generally need to be computed numerically. This can be achieved using quadrature, or for example the Matlab functions integral and integral2. In principle, larger $N$ gives rise to a closer approximation, but clearly adds complexity to simulations. We denote by $G_N$ the transfer function of (5.32) so that the steady-state gain of (5.32) is

$$\begin{aligned} G_N(1) = F(I - M^{-1}D)^{-1}M^{-1}J, \end{aligned}$$

(whenever $\rho (M^{-1}D) <1$). For our example we worked on the log of the interval $[\alpha , \beta ]$, as this gave better results. As such the above goes through with $s_1 = -0.5$, $s_{N+1} = 3.5$. Figure 16 plots both the spectral radius of $M^{-1}D$ and the steady state gain $G_N(1)$ for increasing $N$. The figure suggests that both converge for $N \ge 10$ and thus we choose $N=12$ for the simulations in Fig. 14. Furthermore, this suggests that the model in Example 9 satisfies both assumptions (A1 $^{\prime }$) and (A2 $^{\prime }$).

Fig. 16

Spectral radius in solid-crossed and steady-state gain in solid-circled of the finite element approximations (5.32) of the IPM model of platte thistle of Example 9