On the Infinite-Horizon Optimal Control of Age-Structured Systems

Abstract

The paper presents necessary optimality conditions of Pontryagin’s type for infinite-horizon optimal control problems for age-structured systems with state- and control-dependent boundary conditions. Despite the numerous applications of such problems in population dynamics and economics, a “complete” set of optimality conditions is missing in the existing literature, because it is problematic to define in a sound way appropriate transversality conditions for the corresponding adjoint system. The main novelty is that (building on recent results by Aseev and the second author) the adjoint function in the Pontryagin principle is explicitly defined, which avoids the necessity of transversality conditions. The result is applied to several models considered in the literature.

Appendix

Below we give the proofs of Lemmas 4.1, 5.1–5.4.

Proof of Lemma 4.1

For a fixed $t > 0$, the integrand in (27) can be estimated using (28), (A2) and (A3) as follows:

$$\begin{aligned} |\psi (\theta )\, R(\theta ,t)|&= \left| \int _0^{\omega } [g_yX(\theta +s,s) \varPhi (\theta ) + g_z(\theta ,s)] \mathrm{d}s\, R(\theta ,t) \right| \nonumber \\&\le \int _0^{\omega } \left[ \rho (\theta +s) \,\vert X(\theta +s,s) \vert \, \vert \varPhi (\theta ) \vert + \rho (\theta ) \right] \mathrm{d}s \, \vert R(\theta ,t) \vert \le \lambda (\theta , t). \end{aligned}$$

(45)

The first claim of Lemma 4.1 follows from the integrability of $\lambda (\cdot ,t)$. The local boundedness of $\hat{\zeta }$ follows from the local boundedness of $\psi$ and $\int _t^\infty \lambda (\theta ,t) \mathrm{d}\theta$.

Now consider $\hat{\xi }$. Denote $\xi _1(t,a) := \int _a^{\omega } g_y X(t-a+s,s) \mathrm{d}s X^{-1}(t,a)$, which is the first term in (26), including the multiplication with $X^{-1}$. First, we show that $\xi _1$ is Lipschitz continuous along the characteristic lines $t-a = \mathrm{const}$:

$$\begin{aligned}&\hat{\xi }_1(t+\varepsilon ,a+\varepsilon ) - \hat{\xi }_1(t,a) = \dots \\&\quad \int _{a+\varepsilon }^{\omega } g_yX(t-a+s,s) \mathrm{d}s \, X^{-1}(t+\varepsilon ,a+\varepsilon )\\&\qquad -\int _a^{\omega } g_y X(t-a+s,s)\mathrm{d}s \, X^{-1}(t,a) = \dots \\&\quad \int _{a+\varepsilon }^{\omega } \! g_yX(t-a+s,s) \mathrm d s \left[ X^{-1}(t+\varepsilon ,a+\varepsilon ) \! - \! X^{-1}(t,a) \right] \\&\qquad + \! \int _a^{a+\varepsilon } \! \! g_yX(t-a+s,s) \mathrm d s \, X^{-1}(t,a). \end{aligned}$$

The Lipschitz continuity follows from the Lipschitz continuity of $X^{-1}$ along the characteristic lines and the local boundedness of $g_y X$ and $X^{-1}$. Applying the differentiation $\fancyscript{D}$ to $\xi _1$ and using the expression (14) for $\fancyscript{D}X^{-1}$, we obtain

$$\begin{aligned} - \fancyscript{D}\xi _1(t,a)&= \int _a^{\omega } g_y X(t-a+s,s) \mathrm{d}s\, (X^{-1}(t,a) F(t,a)) + g_y(t,a) \\&= \xi _1(t,a) F(t,a) + g_y(t,a). \end{aligned}$$

The proof that the second term, $\xi _2$, in the definition of $\hat{\xi }$ in (22) is Lipschitz continuous along the characteristic lines $t-a = \mathrm{const}$ and satisfies the equation $-\fancyscript{D}\xi _2(t,a) = \xi _2(t,a) F(t,a) + \hat{\zeta }(t) H(t,a)$ is similar and therefore omitted. Then, $\hat{\xi }= \xi _1 + \xi _2$ belongs to the space $\fancyscript{A}(D)$ and satisfies (22).

The definition (27) of $\hat{\zeta }$ has the form (9), with $T = \infty$. We know that $K(t,s) = 0$ for $t \not \in [s,s+\omega ]$. Moreover, from (45) and (A3) we obtain that the integral in (9) with $T = \infty$ is locally bounded in $t$. Then, we may apply the implication in the end of Sect. 3.1, which claims that

$$\begin{aligned} \hat{\zeta }(t) = \psi (t) + \int _t^\infty \hat{\zeta }(\theta ) \, K(\theta ,t) \mathrm{d}\theta = \psi (t) + \int _t^{t+\omega } \hat{\zeta }(\theta ) \, K(\theta ,t) \mathrm{d}\theta . \end{aligned}$$

Inserting the expressions (20) and (28) for $\psi$ and $K$, respectively, we obtain that

$$\begin{aligned} \hat{\zeta }(t) \!&= \! \left[ \int _0^\omega g_y X(t+x,x) \mathrm{d}x \!+\! \int _t^{t+\omega } \hat{\zeta }(\theta ) \, HX(\theta ,\theta -t) \mathrm{d}\theta \right] \varPhi (t) \!+\! \int _0^{\omega } g_z(t,a) \mathrm{d}a \\&= \hat{\xi }(t,0) \, \varPhi (t) + \int _0^{\omega } g_z(t,a), \end{aligned}$$

that is, (23) is fulfilled by $(\hat{\xi },\hat{\zeta })$. The proof is complete. $\square$

Proof of Lemma 5.1

We represent

$$\begin{aligned} \Delta _\tau (u_\alpha )&= \int _{\tau -\alpha }^\tau \int _0^\omega [(g(t,a,y_{\alpha },z_{\alpha },u_{\alpha }) - g(t,a,z_{\alpha },u_{\alpha }))\nonumber \\&+ (g(t,a,z_{\alpha },u_{\alpha }) - g(t,a,u_{\alpha })) \\&+ \, (g(t,a,u_{\alpha })- g(t,a))] \mathrm{d}a \mathrm{d}t \, =: \, I_1 + I_2 + I_3. \end{aligned}$$

We remind that $\Vert \Delta y\Vert _{L_\infty (D_\tau )} + \Vert \Delta z\Vert _{L_\infty ([0,\tau ])} \le c_0 \,\alpha$ (see (38)). Moreover, $u_\alpha (t,a) \not = \hat{u}(t,a)$ only on the set $B(\tau ,b;\alpha )$ (where $u_\alpha (t,a) = u$), and $\tilde{y}_\alpha (t,a) = \hat{y}(t,a)$ except on a set of measure $2 \alpha ^2$. Using, in addition, that $g$ is Lipschitz continuous with respect to $(y,z)$ in the domain where $(y_\alpha ,z_\alpha )$ and $(\hat{y}, \hat{z})$ take values for $(t,a) \in D_\tau$, we apparently have $|I_1| = o(\alpha ^2)$. For $I_3$ we have

$$\begin{aligned} I_3 = \iint _{B(\tau ,b;\alpha )} (g(t,a,u)- g(t,a)) \mathrm{d}a \mathrm{d}t = \alpha ^2 (g(\tau ,b,u)- g(\tau ,b)) + o(\alpha ^2) \end{aligned}$$

due to the Lebesgue property of $(\tau ,b)$; see the beginning of “Part 2 for $u$” in Sect. 5. Finally,

$$\begin{aligned} I_2 =&\int _{\tau -\alpha }^\tau \left( \int _0^\omega g_z(t,a,u_\alpha (t,a))\, \Delta z(t) \mathrm{d}a + o(\alpha ;t) \right) \mathrm{d}t \\ =&\int _{\tau -\alpha }^\tau \int _0^\omega g_z(t,a)\, \Delta z(t) \mathrm{d}a \mathrm{d}t + o(\alpha ^2), \end{aligned}$$

where here and below $o(\alpha ;t)/\alpha$ converges to zero uniformly in $t$ in the interval of interest (in this case, it is $[0,\tau ]$). For $\Delta z$ we have

$$\begin{aligned} \Delta z(t)&= \int _0^\omega \left( H^\sharp (t,a,y_\alpha (t,a),u_\alpha (t,a)) - H^\sharp (t,a)\right) \mathrm{d}a \nonumber \\&= \int _0^\omega \left( H^\sharp (t,a,u_\alpha (t,a)) - H^\sharp (t,a)\right) \mathrm{d}a + o(\alpha )\end{aligned}$$

(46)

$$\begin{aligned}&= \int _{b-\alpha }^b \left( H^\sharp (t,a,u) - H^\sharp (t,a)\right) \mathrm{d}a + o(\alpha ). \end{aligned}$$

(47)

Inserting this in the expression for $I_2$ and changing the order of integration, we obtain

$$\begin{aligned} I_2&= \iint _{B(\tau ,b;\alpha )} \int _0^\omega g_z(t,s) \mathrm{d}s \,(H^\sharp (t,a,u)- H^\sharp (t,a)) \mathrm{d}a \mathrm{d}t + o(\alpha ^2) \\&= \alpha ^2 \int _0^\omega g_z(\tau ,s) \mathrm{d}s \,(H^\sharp (\tau ,b,u)- H^\sharp (\tau ,b)) + o(\alpha ^2), \end{aligned}$$

where for the last equality we use the Lebesgue property of $(\tau ,b)$; see the beginning of “Part 2 for $u$” in Sect. 5.

Summing the obtained expressions for $I_1$, $I_2$, and $I_3$ we obtain the claim of the lemma. $\square$

Proof of Lemma 5.2

We remind the inequalities $2 \alpha < \tau$, $2 \alpha < b$, and $2 \alpha < \omega - b$ posed for $\alpha$ in “Part 2 for $u$” in Sect. 5. Observe that $\Delta y(\tau ,a) = 0$ for all $a$ except for $a \in [0,\alpha ]\cup [b-\alpha ,b+\alpha ]$. Therefore, we consider the integral in the formulation of the lemma separately on these two intervals.

Beginning with $[0,\alpha ]$, we represent

$$\begin{aligned}&\int _0^\alpha \hat{\xi }(\tau ,a) \, \Delta y(\tau ,a) \mathrm{d}a \\&\quad = \! \int _0^{\alpha } \left[ \hat{\xi }(\tau -a,0) + \int _0^a \fancyscript{D}\hat{\xi }(\tau -a+s,s) \mathrm{d}s\right] \\&\qquad \times \Bigg [\Delta y(\tau -a,0) \\&\qquad + \int _0^a \fancyscript{D}\Delta y(\tau -a+s,s)\mathrm{d}s\Bigg ] \mathrm{d}a. \end{aligned}$$

We remind that $\Vert \Delta y\Vert _{L_\infty (D_\tau )} \le c_0 \,\alpha$. For any $t \ge 0$ and $s \in [0,\alpha ]$, we have $u_\alpha (t,s) = \hat{u}(t,s)$, hence $|\fancyscript{D}\Delta y(t,s)| = |F(t,s)\, \Delta y(t,s)| \le c \,\alpha$. Moreover, due to Lemma 4.1 and the local boundedness of $F$, $\hat{\xi }$, $\hat{\zeta }$, $H$, and $g_y$, we have $\Vert \fancyscript{D}\hat{\xi }\Vert _{L_\infty (D_\tau )} \le c$. Hence,

$$\begin{aligned} \int _0^\alpha \hat{\xi }(\tau ,a) \, \Delta y(\tau ,a) \mathrm{d}a = \int _0^{\alpha } \hat{\xi }(\tau -a,0)\, \Delta y(\tau -a,0) \mathrm{d}a + o(\alpha ^2). \end{aligned}$$

Since $\Delta y(\tau -a,0) = \varPhi (\tau -a) \, \Delta z(\tau -a)$, using representation (47) and changing the integration variable, we obtain that

$$\begin{aligned} \int _0^\alpha \hat{\xi }(\tau ,a) \, \Delta y(\tau ,a) \mathrm{d}a&= \iint _{B(\tau ,b;\alpha )} \hat{\xi }(t,0) \,\varPhi (t) \, (H^\sharp (t,a,u)\nonumber \\&-H^\sharp (t,a)) \mathrm{d}a \mathrm{d}t + o(\alpha ^2). \end{aligned}$$

Due to the Lebesgue point property of $(\tau ,b)$, the expression in the right-hand side equals the second term in the right-hand side in the assertion of the lemma.

Now, we consider $E := \int _{b-\alpha }^{b+\alpha } \hat{\xi }(\tau ,a)\, \Delta y(\tau ,a) \mathrm{d}a$. For $a$ in the interval of integration

$$\begin{aligned} \Delta y(\tau ,a) = \Delta y(\tau -\alpha ,a-\alpha ) + \int _0^\alpha \fancyscript{D}\Delta y(\tau -x,a-x) \mathrm{d}x, \end{aligned}$$

and the first term in the right-hand side is zero (due to $a - \alpha \ge 0$). Then, $E$ is equal to

$$\begin{aligned} \!&\! \int _{b-\alpha }^{b+\alpha } \! \int _0^\alpha \! \! \hat{\xi }(\tau ,a) \! \left[ F^\sharp (\tau -x,a-x,y_\alpha (\tau -x,a-x),u_\alpha (\tau -x,a-x))\right. \\&\quad \left. - F^\sharp (\tau -x,a-x)\right] \! \mathrm d x \mathrm{d}a \\&= \int _{\tau -\alpha }^{\tau } \int _{b-\tau +t-\alpha }^{b-\tau +t+\alpha } \hat{\xi }(\tau ,\tau -t+s) \left[ F^\sharp (t,s,u_\alpha (t,s)) - F^\sharp (t,s)\right] \mathrm{d}s \mathrm{d}t + o(\alpha ^2), \end{aligned}$$

where we passed to the new variables $t = \tau - x$ and $s = a-x$ and used that $\Vert \Delta y\Vert _{L_\infty (D_\tau )} \le c_0 \,\alpha$. Note that, if $s < b - \alpha$ or $s > b$, then the last integrand is zero, since $u_\alpha (t,s)= \hat{u}(t,s)$. Otherwise $u_\alpha (t,s) = u$. Then

$$\begin{aligned} E = \iint _{B(\tau ,b;\alpha )} \hat{\xi }(\tau ,\tau -t+s) \, \left[ F^\sharp (t,s,u) - F^\sharp (t,s)\right] \mathrm{d}s \mathrm{d}t + o(\alpha ^2). \end{aligned}$$

Using the Lebesgue property of $(\tau ,b)$ (see the beginning of “Part 2 for $u$” in Sect. 5), we obtain the first term in the right-hand side in the assertion of the lemma. $\square$

Proof of Lemma 5.3

By the definition of $v_\alpha$ we have $|v_\alpha (t) - \hat{v}(t)| \le c \,\alpha$. According to Proposition 3.2, $\Vert \Delta y(t,\cdot ) \Vert _{L_1([0,\omega ])} \le c \alpha ^2$. Moreover,

$$\begin{aligned} \Delta z(t) = \int _0^{\omega } \alpha H^\sharp _v(t,a)(v-\hat{v}(t)) \mathrm{d}a + o(\alpha ;t). \end{aligned}$$

Then, we obtain

$$\begin{aligned} \Delta _{\tau }(v_{\alpha })&= \int _{\tau -\alpha }^{\tau } \int _0^{\omega } [g(t,a,z_{\alpha }(t),v_\alpha (t)) - g(t,a,v_\alpha (t)) + g(t,a,v_\alpha (t)) \\&- g(t,a) ] \mathrm{d}a \mathrm{d}t + o(\alpha ^2)\\&= \int _{\tau -\alpha }^{\tau } \int _0^{\omega } [g_z(t,a) \Delta z(t) + g(t,a,v_\alpha (t)) - g(t,a)] \mathrm{d}a \mathrm{d}t + o(\alpha ^2) \\&= \int _{\tau -\alpha }^{\tau } \left\{ \int _0^{\omega } g_z(t,s) \mathrm{d}s \int _0^{\omega } H^\sharp _v(t,a) \mathrm{d}a + \int _0^{\omega } g_v(t,a) \mathrm{d}a \right\} \nonumber \\&\times \,\alpha (v-\hat{v}(t)) \mathrm{d}t+ o(\alpha ^2), \end{aligned}$$

which proves the claim of the lemma due to the Lebesgue property of $\tau$ (see the beginning of “Part 2 for $v$” in Sect. 5). $\square$

Proof of Lemma 5.4

The proof uses similar arguments as that of Lemma 5.2, and therefore is somewhat shortened. From (3), we have

$$\begin{aligned} \Delta y(t,0) = \varPhi (t,\hat{v}) \Delta z(t) + \alpha \varPhi ^\sharp _v(t)(v-\hat{v}(t)) + o(\alpha ;t). \end{aligned}$$

For $t=\tau$ and $a < \alpha$,

$$\begin{aligned} \Delta y(\tau ,a) = \Delta y(\tau -a,0) + \int _{-a}^0 \fancyscript{D}\Delta y(\tau +s,a+s) \mathrm{d}s = \Delta y(\tau -a,0) + o(\alpha ;a), \end{aligned}$$

while for $a > \alpha$,

$$\begin{aligned} \Delta y(\tau ,a) =&\int _{-\alpha }^0 [F^\sharp (\tau +s,a+s,y_{\alpha },v_{\alpha }) - F^\sharp (\tau +s,a+s)] \mathrm{d}s\\ =&\int _{-\alpha }^0 \alpha F^\sharp _v(\tau +s,a+s)(v-\hat{v}(\tau +s)) \mathrm{d}s + o(\alpha ;a). \end{aligned}$$

This is obtained by splitting the difference in two parts: one for the difference between $y_{\alpha }$ and $\hat{y}$, and one for $v_{\alpha }$ and $\hat{v}$. Due to Proposition 3.2, the difference with respect to $y$ can be estimated by $o(\alpha ;a)$.

Consider the integral $\int _0^{\omega } \hat{\xi }(\tau ,a) \Delta y(\tau ,a) \mathrm{d}a$ on $[0,\alpha [$. Because $\vert \Delta y(\tau ,a)\vert \le c \alpha$, and the above representation,

$$\begin{aligned} \int _0^{\alpha } \hat{\xi }(\tau ,a) \Delta y(\tau ,a) \mathrm{d}a = \int _0^{\alpha } \hat{\xi }(\tau -a,0) \Delta y(\tau -a,0) \mathrm{d}a + o(\alpha ^2). \end{aligned}$$

Since $\tau$ is a Lebesgue point, using the representation for $\Delta y(\tau ,0)$ and for $\Delta z(\tau )$ from the proof of the Lemma 5.3 we obtain the expression

$$\begin{aligned} \alpha ^2 \hat{\xi }(\tau ,0) \varPhi (\tau ) \int _0^{\omega } H^\sharp _v(\tau ,a) (v-\hat{v}(\tau )) \mathrm{d}a + \alpha ^2 \hat{\xi }(\tau ,0) \varPhi ^\sharp _v(\tau ) (v-\hat{v}(\tau )) + o(\alpha ^2). \end{aligned}$$

With the representation for $\Delta y(\tau ,a)$ for $a > \alpha$, and the absolute continuity of $\hat{\xi }$ along the characteristic lines, we obtain

$$\begin{aligned} \int _{\alpha }^{\omega } \hat{\xi }(\tau ,a) \Delta y(\tau ,a) \mathrm{d}a =&\int _{\alpha }^{\omega } \int _{\tau -\alpha }^{\tau } \hat{\xi }(t,a-\tau +t) \alpha F^\sharp _v(t,a-\tau +t)(v-\hat{v}(t)) \mathrm{d}t \mathrm{d}a \\&+ o(\alpha ^2) \\ =&\; \alpha \int _{\alpha }^{\omega -\alpha } \int _{\tau -\alpha }^{\tau } \hat{\xi }(t,a) \,F^\sharp _v(t,a) (v-\hat{v}(t)) \mathrm{d}t \mathrm{d}a + o(\alpha ^2)\\ =&\;\alpha \int _{\tau -\alpha }^{\tau } \int _0^{\omega } \hat{\xi }(t,a)\, F^\sharp _v(t,a) (v-\hat{v}(t)) \mathrm{d}a \mathrm{d}t + o(\alpha ^2). \end{aligned}$$

Since $\tau$ is a Lebesgue point, this implies the claim of the lemma. $\square$