On Balder's Existence Theorem for Infinite-Horizon Optimal Control Problems

TL;DR

This paper extends Balder's 1983 existence theorem for infinite-horizon optimal control problems to improper integrals, relaxing uniform integrability conditions and providing a simpler proof approach with an illustrative example.

Contribution

It generalizes Balder's theorem by weakening integrability conditions and offers a new, simpler proof method for the existence result.

Findings

Extended existence theorem to improper integrals

Relaxed uniform integrability conditions

Provided an illustrative example

Abstract

Balder's well-known existence theorem (1983) for infinite-horizon optimal control problems is extended to the case when the integral functional is understood as an improper integral. Simultaneously, the condition of strong uniform integrability (over all admissible controls and trajectories) of the positive part $\max\{f_0,0\}$ of the utility function (integrand) $f_0$ is relaxed to the requirement that the integrals of $f_0$ over intervals $[T,T']$ be uniformly bounded from above by a function $\omega(T,T')$ such that $\omega(T,T')\to 0$ as $T,T'\to\infty$. This requirement was proposed by A.V. Dmitruk and N.V. Kuz'kina (2005); however, the proof in the present paper does not follow their scheme but is instead derived in a rather simple way from the auxiliary results of Balder himself. An illustrative example is also given.

Equations

$$I(x,u):={\int }_0^{\infty }{f}_0(t,x(t),u(t))dt\to \max ,$$

$$\dot{x}(t)=f(t,x(t),u(t))\text{for a.e. }t\in {\mathbb{R}}_{+}:=[0,+\infty ),$$

$$x(t)\in A(t),u(t)\in U(t,x(t))\text{for a.e. }t\in {\mathbb{R}}_{+},$$

$${\int }_0^{\infty }g(t)dt:={\int }_0^{\infty }{g}^{+}(t)dt-{\int }_0^{\infty }{g}^{-}(t)dt,\text{where}{g}^{\pm }:=\max \{\pm g,0\},$$

$$J(x,u):=\underset{T\to +\infty }{\lim }{\int }_0^T{f}_0(t,x(t),u(t))dt$$

$$J(x,u)\to \max$$

$$\begin{gathered} \underset{T\to +\infty }{\overline{\lim }}\underset{T^{\prime }>T}{\sup }\underset{(x,u)\in {\Omega }_{\alpha }}{\sup }{\int }_T^{T^{\prime }}{f}_0(t,x(t),u(t))dt\le \underset{T\to +\infty }{\overline{\lim }}\underset{T^{\prime }>T}{\sup }\underset{g\in F_{0,\alpha }^{+}}{\sup }{\int }_T^{T^{\prime }}g(t)dt \\ \le \underset{h\in L_1({\mathbb{R}}_{+})}{\inf }\underset{T\to +\infty }{\overline{\lim }}\underset{T^{\prime }>T}{\sup }\underset{g\in F_{0,\alpha }^{+}}{\sup }\left({\int }_{[T,T^{\prime }]\cap C_{g,h}}g(t)dt+{\int }_{[T,T^{\prime }]\setminus C_{g,h}}h(t)dt\right) \\ \le \underset{h\in L_1({\mathbb{R}}_{+})}{\inf }\underset{T\to +\infty }{\overline{\lim }}\underset{T^{\prime }>T}{\sup }\underset{g\in F_{0,\alpha }^{+}}{\sup }{\int }_{C_{g,h}}g(t)dt+0=\underset{h\in L_1({\mathbb{R}}_{+})}{\inf }\underset{g\in F_{0,\alpha }^{+}}{\sup }{\int }_{C_{g,h}}g(t)dt. \end{gathered}$$

$$\underset{(x,u)\in \Omega }{\sup }{\int }_T^{T^{\prime }}{f}_0(t,x(t),u(t))dt\le \omega (T,T^{\prime })\forall T,T^{\prime }:T^{\prime }>T\ge 0;$$

$$\underset{T^{\prime }>T}{\sup }\underset{(x,u)\in \Omega }{\sup }{\int }_T^{T^{\prime }}{f}_0(t,x(t),u(t))dt\le \tilde{\omega }(T)\forall T\ge 0.$$

$$\hat{\omega }(T):={\left(\underset{T^{\prime }>T}{\sup }\underset{(x,u)\in \Omega }{\sup }{\int }_T^{T^{\prime }}{f}_0(t,x(t),u(t))dt\right)}^{+},T\ge 0.$$

$$\hat{\omega }(T)\le \underset{(x,u)\in \Omega }{\sup }{\int }_0^{T_1}{f}_0^{+}(t,x(t),u(t))dt+\hat{\omega }(T_1)\le \underset{c>0}{\inf }\underset{g\in F_0^{T_1,+}}{\sup }\left({\int }_{C_{g,c}^{T_1}}g(t)dt+c{T}_1\right)+1.$$

$$\underset{g\in F_0^{T_1,+}}{\sup }{\int }_{C_{g,c_1}^{T_1}}g(t)dt\le 1.$$

$$\begin{gathered} \underset{T\to +\infty }{\overline{\lim }}{\int }_0^T{f}_0(t,x(t),u(t))dt=\underset{T_1\to +\infty }{\underline{\lim }}\underset{T\to +\infty }{\overline{\lim }}\left({\int }_0^{T_1}+{\int }_{T_1}^T\right)f_0(t,x(t),u(t))dt \\ \le \underset{T_1\to +\infty }{\underline{\lim }}{\int }_0^{T_1}{f}_0(t,x(t),u(t))dt+\underset{T_1\to +\infty }{\overline{\lim }}\underset{T>T_1}{\sup }{\int }_{T_1}^T{f}_0(t,x(t),u(t))dt \\ \le \underset{T\to +\infty }{\underline{\lim }}{\int }_0^T{f}_0(t,x(t),u(t))dt, \end{gathered}$$

$$I(x_0,u_{\ast })\ge \underset{k\to \infty }{\overline{\lim }}I(x_k,u_k).$$

$$f_0^N(t,\chi ,\upsilon ):=\{\begin{array}{ll}{f}_0(t,\chi ,\upsilon ),& t\in [N-1,N),(t,\chi ,\upsilon )\in \mathcal{U},\\ 0,& t\in {\mathbb{R}}_{+}\setminus [N-1,N),(t,\chi ,\upsilon )\in \mathcal{U},\end{array}$$

$$I_N(x_0,u_{N\ast })\ge \underset{k\to \infty }{\overline{\lim }}{I}_N(x_k,u_k).$$

$$J(x_{0},u_{*})=\lim_{K\to\infty}\sum_{N=1}^{K}I_{N}(x_{0},u_{*})=\lim_{K\to\infty}\sum_{N=1}^{K}I_{N}(x_{0},u_{N*})\geq\lim_{K\to\infty}\sum_{N=1}^{K}\varlimsup_{k\to\infty}I_{N}(x_{k},u_{k})\geq\\ \geq\lim_{K\to\infty}\,\varlimsup_{k\to\infty}\sum_{N=1}^{K}I_{N}(x_{k},u_{k})\geq\lim_{K\to\infty}\,\varlimsup_{k\to\infty}\bigl{(}J(x_{k},u_{k})-\widetilde{\omega}(K)\bigr{)}=\varlimsup_{k\to\infty}J(x_{k},u_{k}),$$

$${\int }_0^{\infty }\frac{u(t)}{t+1}dt\to \max ,$$

$$\dot{x}(t)=u(t)\text{for a.e. }t\in {\mathbb{R}}_{,}$$

$$x(t)\in [-t,t]\cap [-1,1],u(t)\in [-1,1]\text{for a.e. }t\in {\mathbb{R}}_{+}.$$

$$\begin{gathered} {\int }_T^{T^{\prime }}\frac{u(t)}{t+1}dt={\int }_T^{T^{\prime }}\frac{\dot{x}(t)}{t+1}dt=\frac{x(T^{\prime })}{T^{\prime }+1}-\frac{x(T)}{T+1}+{\int }_T^{T^{\prime }}\frac{x(t)}{(t+1{)}^2}dt \\ \le \frac{1}{T^{\prime }+1}+\frac{1}{T+1}+{\int }_T^{T^{\prime }}\frac{dt}{(t+1{)}^2}=\frac{2}{T+1}\forall T>0. \end{gathered}$$

$$\underset{T\to \infty }{\lim }{\int }_0^T\frac{u(t)}{t+1}dt=\underset{T\to \infty }{\lim }\left(\frac{x(T)}{T+1}+{\int }_0^T\frac{x(t)}{(t+1{)}^2}dt\right)=\underset{T\to \infty }{\lim }{\int }_0^T\frac{x(t)}{(t+1{)}^2}dt,$$

$$u_0(t)=\cos t,x_0(t)=\sin t,t\ge 0.$$

$${\int }_0^{\infty }{\left(\frac{u_0(t)}{t+1}\right)}^{+}dt={\int }_0^{\infty }\frac{\max \{\cos t,0\}}{t+1}dt=+\infty ;$$

Full text

As was already mentioned, the only condition in Theorem A that regulates the behavior of system (1)–(3) at infinity is condition (xi). At the same time, in many optimal economic growth problems it seems more natural to define the value of the objective functional not in the sense of (4) but rather in the limit sense

[TABLE]

provided that the limit exists (see, e.g., [4, 9]). We will also follow this definition, in which case problem (1)–(3) is replaced by the problem

[TABLE]

subject to conditions (2) and (3).

As noticed in [10], instead of condition (xi) one can consider the condition

1. (xii)

$\varlimsup_{T\to+\infty}\sup_{T^{\prime}>T}\sup_{(x,u)\in\Omega}\int_{T}^{T^{\prime}}\!f_{0}(t,x(t),u(t))\,dt\leq 0$.

It is easy to see that for admissible pairs in $\Omega_{\alpha}$ condition (xii) is weaker555On the whole it would be incorrect to say that condition (xii) is weaker than condition (xi), because condition (xii) is considered for the set $\Omega$ while condition (xi) is considered only for the subset $\Omega_{\alpha}\subset\Omega$. Therefore, formally none of the conditions implies the other.

than condition (xi). Indeed,

[TABLE]

However, below we will still need a local version of condition (xi), namely,

1. (xi*′*)

for every $T>0$, the set of functions $F_{0}^{T,+}:=\{f_{0}^{+}(\kern 1.0pt\cdot\kern 1.0pt,x(\cdot),u(\cdot))|_{[0,T]}\mid(x,u)\in\Omega\}$ is uniformly integrable on $[0,T]$.

(In [10], due to the continuity and compact-valuedness of the functions and set-valued maps considered there, condition (xi*′*) holds automatically.)

Let us make the following important observation.

1. (xii*′*)

there is a continuous function $\omega\colon\mathbb{R}_{+}^{2}\to\mathbb{R}_{+}$ such that $\omega(T,T^{\prime})\to 0$ as $T,T^{\prime}\to\infty$ and

[TABLE] 2. (xii*′′*)

there is a continuous function $\widetilde{\omega}\colon\mathbb{R}_{+}\to\mathbb{R}_{+}$ such that $\widetilde{\omega}(T)\to 0$ as $T\to\infty$ and

[TABLE]

[TABLE]

Due to condition (xii) we have $\lim_{T\to\infty}\widehat{\omega}(T)=0$. Therefore, there is a $T_{1}$ such that $\widehat{\omega}(T)\leq 1$ for $T\geq T_{1}$. Let us show that this function is bounded for all $T\geq 0$. For $T<T_{1}$ we have

[TABLE]

Due to condition (xi*′*) there is a constant $c_{1}>0$ such that

[TABLE]

Then $\widehat{\omega}(T)\leq c_{1}T_{1}+2$ for all $T\geq 0$.

Put $\widehat{\omega}_{1}(T):=\sup_{T^{\prime}\geq T}\widehat{\omega}(T^{\prime})$ for $T\geq 0$. Then $\widehat{\omega}_{1}$ is a bounded monotonically nonincreasing function on $\mathbb{R}_{+}$ that tends to zero as $T\to\infty$.

Finally, put $\widetilde{\omega}(T):=\int_{T-1}^{T}\widehat{\omega}_{1}(t^{+})\,dt$ (recall that $t^{+}=\max\{t,0\}$). It is clear that $\widetilde{\omega}$ is a continuous function on $\mathbb{R}_{+}$ that satisfies all requirements formulated in condition (xii*′′*). ∎

One of the important corollaries to condition (xii) is that the value of the functional $J(\cdot,\cdot)$ is defined on any admissible pair. For completeness, we will give a slightly shorter proof of this fact than in [10].

[TABLE]

where we have used condition (xii) at the last step. At the same time, the limit does not exceed $\widetilde{\omega}(0)$ for a continuous function $\widetilde{\omega}\colon\mathbb{R}_{+}\to\mathbb{R}_{+}$. ∎

Now we formulate our main result. By analogy with the set $\Omega_{\alpha}$, we introduce the set $\widetilde{\Omega}_{\alpha}:=\{(x,u)\in\Omega\mid J(x,u)\geq\alpha\}$ for $\alpha\in\mathbb{R}$.

The main role in the proof is played by another result of Balder.

to a function $x_{0}\in\mathrm{AC}_{\textup{loc}}^{n}(\mathbb{R}_{+})$ and the set of functions $\{f_{0}^{+}(\kern 1.0pt\cdot\kern 1.0pt,x_{k}(\cdot),u_{k}(\cdot))\}_{k=1}^{\infty}$ is strongly uniformly integrable on $\mathbb{R}_{+}$. Then there exists a Lebesgue measurable function $u_{*}\colon\mathbb{R}_{+}\to\mathbb{R}^{m}$ such that $(x_{0},u_{*})\in\Omega$ and*

[TABLE]

For $N\in\mathbb{N}$ introduce the function

[TABLE]

and consider problem (1)–(3) with $f_{0}^{N}$ instead of $f_{0}$. Denote the corresponding functional (in which the integral is actually taken over the interval $[N-1,N)$) by $I_{N}$. Assume first that for every $N\in\mathbb{N}$ all hypotheses of Theorem B hold for this truncated problem (with the objective functional $I_{N}$) and for our sequence $\{(x_{k},u_{k})\}_{k=1}^{\infty}$. Then there exists a Lebesgue measurable function $u_{N*}\colon\mathbb{R}_{+}\to\mathbb{R}^{m}$ such that $(x_{0},u_{N*})\in\Omega$ and

[TABLE]

Put $u_{*}(t):=u_{N*}(t)$ for $t\in[N-1,N)$, $N\in\mathbb{N}$. Clearly, $(x_{0},u_{*})\in\Omega$ and

[TABLE]

where $\widetilde{\omega}$ is a function from condition (xii*′′*).

Thus, $(x_{0},u_{*})$ is the required admissible pair. It remains to explain why the conclusion of Theorem B holds for the truncated problem with functional $I_{N}$. Among all hypotheses of Theorem B, only conditions (ix) and (x) need to be checked for $t\notin[N-1,N)$. The validity of condition (ix) follows from the fact that a projection of a convex set is convex. However, the validity of condition (x) is fairly difficult to prove (moreover, we are even not sure that it takes place).

To overcome this difficulty, we proceed as follows. Note that in the above reasoning the values of $u_{N*}(t)$ are used only for $t\in[N-1,N)$. Therefore, we can arbitrarily vary the sequence $\{(x_{k},u_{k})\}_{k=1}^{\infty}$ and the parameters of problem (1)–(3) outside the interval $[N-1,N)$. In particular, we can set $f(t,\cdot\kern 1.0pt,\cdot\kern 1.0pt)=0$, $A(t)=\mathbb{R}^{n}$, $U(t,\cdot\kern 1.0pt)=\{0\}$ and $u_{k}(t)=0$ for $t\notin[N-1,N)$, as well as $x_{k}(t)=x_{k}(N-1)$ for $0\leq t<N-1$ and $x_{k}(t)=x_{k}(N)$ for $t\geq N$. For the problem thus modified (still with the functional $I_{N}$) the validity of all hypotheses of Theorem B is undoubtable, and we get the desired function $u_{N*}$ on $[N-1,N)$. ∎

[TABLE]

It is clear that $x(0)=0$ and the absolute value of the integrand in (7) is bounded by $1/(t+1)$ for every admissible pair $(x,u)$. All local conditions (i)–(x) and (xi*′) are satisfied. Let us show that condition (xii′′*) also holds:

[TABLE]

Thus, if we consider the functional (7) as an improper integral, i.e., in the sense of (5), then we can apply Theorem 1, which guarantees the existence of an optimal solution.

This optimal solution can easily be found explicitly. Indeed, since

[TABLE]

it suffices to maximize $x(t)$ at every $t$ (which is possible here), i.e., set $u_{*}(t)=1$ for $t<1$ and $u_{*}(t)=0$ for $t\geq 1$. The corresponding optimal trajectory is $x_{*}(t)=\min\{t,1\}$.

Since the integrand is positive, by Remark 1 this solution is also optimal in the case when the objective functional is understood in the sense of (4). Let us show that nevertheless Theorem A is inapplicable in this case for any $\alpha$ (except for $\alpha$ equal to the exact value $\alpha_{*}$ ($=\ln 2$) of the functional on the optimal solution, but in this case the theorem is almost worthless, because the set $\Omega_{\alpha_{*}}$ consists of a single admissible pair). The reason is that condition (xi) of strong uniform integrability does not hold for $\alpha<\alpha_{*}$. Let us demonstrate this.

Consider first an admissible pair with $u(t)=\cos t$, i.e., the pair

[TABLE]

Then

[TABLE]

i.e., no family of functions containing the integrand in (11) can be strongly uniformly integrable.

To show that condition (xi) is violated even for admissible pairs for which the value of the functional (in any sense) is close to the optimal value, it suffices to construct such an admissible pair from pieces:

• •

first, on a sufficiently large interval $[0,T_{1}]$ with $T_{1}=\pi/2+2\pi k$, $k\in\mathbb{N}$, use the optimal control $u_{*}$ and follow the optimal trajectory $x_{*}$;

• •

second, on a sufficiently large interval $[T_{1},T_{2}]$, use the control $u_{0}$ and follow the trajectory $x_{0}$ (since $x_{0}(T_{1})=1=x_{*}(T_{1})$, we can switch from one trajectory to the other);

• •

for $t>T_{2}$, use the control $u=0$.

Due to the vanishing control on the last interval, the value of the functional (in any sense) on such an admissible pair is finite. By virtue of estimate (10) (note that replacing $u$ with $-u$ changes the trajectory $x$ to $-x$, so estimate (10) also holds for the absolute value of the integral on the left-hand side), the value of the functional (in any sense) on such a pair differs from the optimal value by at most $4/(T_{1}+1)$. Choosing a sufficiently large $T_{2}$ (depending on $T_{1}$), we can make the integral analogous to (11) as large as desired. This means that condition (xi) of strong uniform integrability does not hold for $\Omega_{\alpha}$ for any $\alpha<\alpha_{*}$.