State-Constrained Control-Affine Parabolic Problems I: First and Second Order Necessary Optimality Conditions

Abstract

In this paper we consider an optimal control problem governed by a semilinear heat equation with bilinear control-state terms and subject to control and state constraints. The state constraints are of integral type, the integral being with respect to the space variable. The control is multidimensional. The cost functional is of a tracking type and contains a linear term in the control variables. We derive second order necessary conditions relying on the concept of alternative costates and quasi-radial critical directions. The appendix provides an example illustrating the applicability of our results.

Appendices

Appendix A: Strong Solutions of the Heat Equation

We consider the heat equation with Dirichlet boundary condition:

$$ \dot y - {\Delta} y = f \text{in \textit{Q}}, y(x,0)= y_0(x); y = h \text{on \ensuremath{{\Sigma}}}. $$

(A.1)

We have the following result, see Lieberman [29, Thm 7.32, p. 182]:

Theorem A.1

Let r ≥ 2, w ∈ W^2,1,r(Q) and f ∈ L^r(Q). Setting y₀ := w(⋅, 0) and h := τ_Σw (trace of w over Σ), equation Eq. A.1has a unique solution y ∈ W^2,1,r(Q). In addition there exists C > 0 such that

$$ \| y \|_{W^{2,1,r}(Q)} \leq C \left( \| f\|_{L^r(Q)} + \| w \|_{W^{2,1,r}(Q)} \right). $$

(A.2)

Corollary A.2

Given r ≥ 2, $y_{0}\in {W^{1,r}_{0}({\Omega })\cap W^{2,r}({\Omega })}$ and f ∈ L^r(Q), equation Eq. A.1has, for h = 0, a unique solution y ∈ W^2,1,r(Q) that satisfies

$$ \| y \|_{W^{2,1,r}(Q)} \leq C \left( \| f\|_{L^r(Q)} + \| y_0 \|_{W^{2,r}({\Omega})} \right). $$

(A.3)

Proof

Apply Theorem A.1 with w(x,t) := y₀(x). It is clear that w ∈ W^2,1,r(Q) and that w has trace y₀ at time 0 and zero trace over Σ. The conclusion follows. □

By the standard Sobolev embeddings, we have the continuous inclusion

$$ W^{2,1,r}(Q) \subset W^{1,r}(Q) \subset L^{\infty}(Q), \quad \text{if \ensuremath{r>n+1}.} $$

(A.4)

This allows to prove the following.

Theorem A.3

Assume that $u\in L^{\infty }(0,T)$, $y_{0}\in {W^{1,r}_{0}({\Omega })\cap W^{2,r}({\Omega }) }$ and f ∈ L^r(Q), with r > n + 1. Then the state Eq. 2.1 has a unique solution y[u, y₀, f] in W^2,1,r(Q), and the mapping y[u, y₀, f] is of class $C^{\infty }$ from $L^{\infty }(0,T) \times {W^{1,r}_{0}({\Omega })\cap W^{2,r}({\Omega }) }\times L^{r}({\Omega })$ into W^2,1,r(Q).

Proof

We have that g := −Δy₀ belongs to L^r(Ω). Let $y^{\pm }_{0}$ be the unique solution of $-{\Delta } y^{\pm }_{0} =g^{\pm }$ in Ω, where $g^{+}:=\max \limits (g,0)$ and $g^{-}:=-\min \limits (g,0)$, with homogeneous Dirichlet condition on the boundary. Set $f^{+} := \max \limits (f,0)$ and $f^{-}:=-\min \limits (f,0)$. Denote by y⁺ (resp., y⁻) the solution of the state Eq. 2.1 when (y₀,f) is $(y_{0}^{+},f^{+})$ (resp. $(y_{0}^{-},f^{-})$). By the monotonicity results in Lemma 2.3, we have that − y⁻≤ y ≤ y⁺. Now let y⁺⁺, y⁻⁻ denote the solutions of the state Eq. 2.1 when (y₀,f) is $(y_{0}^{+},f^{+})$, $(y_{0}^{-},f^{-}),$ respectively and, in addition, γ = 0. We claim that − y⁻⁻≤−y⁻≤ y ≤ y⁺ ≤ y⁺⁺. Indeed, for z ∈ Y, set $H_{u} z := \dot z - {\Delta } z - z {\sum }_{i} u_{i} b_{i}$. Then

$$ H_u y^{+} = f^+-\gamma (y^+)^3 \leq f^+ = H_u y^{++}. $$

(A.5)

Since y⁺ and y⁺⁺ have the same initial conditions, it follows that y⁺ ≤ y⁺⁺. In an analogous way, it can be proved that − y⁻⁻≤−y⁻.

Since $y_{0}^{\pm } \in {W^{1,r}_{0}({\Omega })\cap W^{2,r}({\Omega }) }$ and f^±∈ L^r(Q), by Corollary A.2, y⁺⁺ and y⁻⁻ belong to W^2,1,r(Q) and, therefore, since r > n + 1, they are also elements of $L^{\infty }(Q)$. So, $y \in L^{\infty }(Q)$. Consequently, H_uy = f − γy³ ∈ L^r(Ω) and, by Theorem A.1 again, y ∈ W^2,1,r(Q).

We recall that, for r > n + 1, Y_r denotes the set of elements of W^2,1,r(Q) with zero trace on Σ, and ${Y^{0}_{r}}$ denotes the trace of Y_r at time zero. Endowed with the “trace norm”, ${Y^{0}_{r}}$ is a Banach space that contains $W^{1,r}_{0}({\Omega }) \cap W^{2,r}({\Omega })$ in view of the proof of the above Corollary A.2 (by Lions [24, p. 20], ${Y^{0}_{r}}$ is a subset of W^{2 − 2/r,r}(Ω)). That (u,y₀,f)↦y[u,y₀,f] is of class $C^{\infty }$ is a consequence of the Implicit Function Theorem applied to the mapping F from $Y_{r}\times L^{\infty }(0,T) \times {Y_{r}^{0}} \times L^{r}(Q)$ into $L^{r}(Q)\times {Y^{0}_{r}}$, defined by $$ F(y,u,y_0,f) := (H_u y + \gamma y^3, y(0)- y_0). $$

$$ F(y,u,y_0,f) := (H_u y + \gamma y^3, y(0)- y_0). $$

(A.6)

The key step is to prove that the partial derivative D_yF is bijective; this can be done easily, taking advantage of the fact that $W^{2,1,r}(Q) \subset L^{\infty }(Q)$ when r > n + 1. □

Appendix B: An Example

Since we made a number of hypotheses about the optimal trajectory, especially at junction points, it is useful to give an example where these hypotheses are satisfied. For that purpose we discuss a particular case in which the original optimal control problem can be reduced to the optimal control of a scalar ODE.

Let Ω = (0, 1), and denote by $c_{1}(x):=\sqrt 2 {\sin \limits } \pi x$ the first (normalized) eigenvector of the Laplace operator.

We assume that γ = 0, the control is scalar (m = 1), b₀ ≡ 0 and b₁ ≡ 1 in Ω, and that f ≡ 0 in Q. Then the state equation with initial condition c₁ reads

$$ \dot y(x,t) - {\Delta} y(x,t) = u(t) y(x,t); \quad (x,t) \in (0,1)\times (0,T), \quad y(x,0) = c_1(x),\quad x\in {\Omega}. $$

(B.1)

It is easily seen that the state satisfies y(x,t) = y₁(t)c₁(x), where y₁ is solution of

$$ \dot y_1(t) + \pi^2 y_1(t) = u(t) y_1(t); \quad t \in (0,T), \quad y_1(0) = y_{10}=1. $$

(B.2)

We set T = 3 and consider the state constraint Eq. 3.17 with q = 1 and d₁ := − 2, and the cost function Eq. 2.5 with α₁ = 0.

The state constraint reduces to

$$ y_1(t) \leq 2,\quad t\in [0,3]. $$

(B.3)

As target functions take y_dT := c₁ and $y_{d}(x,t) := \hat {y}_{d}(t) c_{1}(x)$ with

$$ \hat{y}_d(t) := \left\{ \begin{array}{lll} 1.5e^{t} &\quad \text{for }t\in (0, \log 2),\\ 3 &\quad \text{for }t\in (\log 2, 1),\\ 4- t &\quad \text{for }t\in (1, 3). \end{array}\right. $$

(B.4)

We assume that the lower and upper bounds for the control are $\check {u} := -1$ and $\hat {u} :=\pi ^{2}+1$. We will check that the optimal control is

$$ \bar{u}(t) := \left\{ \begin{array}{ll} \hat{u} & \quad \text{for } t\in (0, \log 2), \\ \pi^2 & \quad \text{for } t\in (\log 2, 2),\\ \pi^2 -1/\hat{y}_d & \quad \text{for } t\in (2, 3). \end{array}\right. $$

(B.5)

Thus, for the optimal state we have

$$ \bar{y}_1(t) := \left\{ \begin{array}{lll} e^t & \quad\text{for } t\in (0, \log 2), &\\ 2 & \quad\text{for } t\in (\log 2,2), &\\ 4-t & \quad \text{for } t\in (2, 3). & \end{array}\right. $$

(B.6)

The above control is feasible. The trajectory $(\bar {u},\bar {y})$ is optimal since for any t ∈ (0,T), the state $\bar {y}_{1}(t)$ has the best possible value (in order to approach $\hat y_{d}$ and minimize the cost function) that respects the state constraint.

Let us check Hypothesis 4.1 for this example. Conditions 1 and 2 are obviously satisfied. For the constraint qualification in Condition 3 consider the linearized state equation with unique z₁[v]:

$$ \dot z_1=(\bar{u} - \pi^2) z_1 + v\bar{y}_1; \quad z_1(0)=0, $$

(B.7)

with $v(t):= \check {u}-\bar {u}(t) < 0$. One easily checks that z₁[v](t) < 0 for all t > 0. Hence, we can find ε > 0 such that

$$ g_1(\bar{y}(\cdot,t)) + g_1^{\prime}(\bar{y}(\cdot,t))z_1[v](\cdot,t) = \bar{y}_1(t) -2 + z_1(t) < -\varepsilon,\quad \text{for all } t\in (0,T). $$

(B.8)

Conditions 4 holds, since

$$ M(t)=\bar M_1(t)={\int\limits}_{\Omega} c_1(x) \bar{y}(x,t)\text{d} x=\bar{y}_1(t)>0\quad \text{for } t \in (0,T). $$

(B.9)

For Condition 5 we have

$$ \operatorname{dist}(t,I^C_1)= \begin{cases} \log 2 -t & \text{for } t \in (0, \log 2), \\ 0 & \text{for } t\in (\log 2, 2), \\ t- 2 & \text{for } t \in (2,3), \end{cases} $$

(B.10)

and hence,

$$ g_1(\bar{y}(\cdot,t))=\bar{y}_1(t)-2 \le - \operatorname{dist}(t,I^{C}_1). $$

(B.11)

Conditions 6 and 8 hold by the choice of the control in Eq. B.5. Condition 7 holds by definition.

We solve this problem numerically using BOCOP [30] and get the optimal control and state given in Fig. 1.

Fig. 1

Optimal control and state for the example

We now discuss the second order optimality condition for this example. The costate equation is

$$ -\dot p +Ap = c_1 (\bar{y}_1-\hat y_d) + c_1\dot{\mu}_1,\quad p(\cdot,T)= \bar{y}(T)-y_{dT} = 0 $$

(B.12)

with A as defined in Eq. 2.20. Since $\bar {y}$ and y_d are colinear to c₁, it follows that p(x,t) = p₁(t)c₁(x), and

$$ -\dot p_1 + \pi^2 p_1 = \bar{u} p_1 + \bar{y}_1 - \hat{y}_{d} + \dot{\mu}_1; \quad p_1(3)=0. $$

(B.13)

Over (2,3), $\dot {\mu }_{1}=0$ (sate constraint not active) and $\bar {y}_{1} = \hat {y}_{d}$, therefore p₁ and p identically vanish. Over $(\log 2,2)$, $\bar {u}$ is out of bounds and therefore

$$ 0 = {\int\limits}_{\Omega} p(x,t) \bar{y}(x,t) = p_1(t) \bar{y}_1(t) {\int\limits}_{\Omega} c_1(x)^2 = 2p_1(t). $$

(B.14)

It follows that p₁ and p also vanish on $(\log 2,2)$ and that

$$ \dot \mu_1 = -(\bar{y}_1 - \hat{y}_{d} ) >0, \quad \text{a.a. }t\in (\log 2,2). $$

(B.15)

Over $(0,\log 2),$ the control attains its upper bound, then

$$ -\dot p_{1} = p_{1} - \frac{1}{2} e^{t} $$

(B.16)

with final condition $p_{1}(\log 2)=0$, so that

$$ p_1(t) = \frac{e^{t}}{4} - e^{-t}. $$

(B.17)

As expected, p₁ is negative.

Next, the linearized state equation at $(\bar {u},\bar {y})$ reads

$$ \dot z - {\Delta} z = \bar{u} z + v \bar{y}; \quad z(\cdot,0) = 0. $$

(B.18)

Since $\bar {y}=\bar {y}_{1}(t) c_{1}(x)$, we deduce that z = z₁(t)c₁(x), with z₁ solution of

$$ \dot z_1 + \pi^{2} z = \bar{u} z_{1} + v \bar{y}_{1}; \quad z_{1}(0) = 0. $$

(B.19)

Therefore if (v,z) satisfy the linearized state equation

$$ \begin{aligned} {\mathcal Q}[p](z,v) &= {\int\limits}_{Q} (z^2 + p v z)\text{d} x \text{d} t + {\int\limits}_{\Omega} z(x,T)^2 \text{d} x = {\int\limits}_0^3 (z_1(t)^2 + p_1(t) v(t) z_1(t) ) \text{d} t + z_1(3)^2. \end{aligned} $$

(B.20)

If in addition v is a critical direction, since v = 0 and z₁ = 0 a.e. on (0, 2), and p₁(t) = 0 on (2, 3), we get

$$ \begin{aligned} {\mathcal Q}[p](z,v) &= {\int\limits}_2^3 z_1(t)^2 \text{d} t + z_1(3)^2. \end{aligned} $$

(B.21)

Thus, ${\mathcal Q}$ is non-negative for any critical directions (z[v],v), in accordance with the second-order necessary condition of Theorem 4.7.