Path-Dependent Hamilton–Jacobi Equations: The Minimax Solutions Revised

Abstract

Motivated by optimal control problems and differential games for functional differential equations of retarded type, the paper deals with a Cauchy problem for a path-dependent Hamilton–Jacobi equation with a right-end boundary condition. Minimax solutions of this problem are studied. The existence and uniqueness result is obtained under assumptions that are weaker than those considered earlier. In contrast to previous works, on the one hand, we do not require any properties concerning positive homogeneity of the Hamiltonian in the impulse variable, and on the other hand, we suppose that the Hamiltonian satisfies a Lipshitz continuity condition with respect to the path variable in the uniform (supremum) norm. The progress is related to the fact that a suitable Lyapunov–Krasovskii functional is built that allows to prove a comparison principle. This functional is in some sense equivalent to the square of the uniform norm of the path variable and, at the same time, it possesses appropriate smoothness properties. In addition, the paper provides non-local and infinitesimal criteria of minimax solutions, their stability with respect to perturbations of the Hamiltonian and the boundary functional, as well as consistency of the approach with the non-path-dependent case. Connection of the problem statement under consideration with some other possible statements (regarding the choice of path spaces and derivatives used) known in the theory of path-dependent Hamilton–Jacobi equations is discussed. Some remarks concerning viscosity solutions of the studied Cauchy problem are given.

Appendices

Proof of Proposition 3

Let a functional $\varphi : [0, T] \times C([- h, T], \mathbb {R}^n) \rightarrow \mathbb {R}$ be non-anticipative and continuous and satisfy boundary condition (9). By Proposition 2, the functional $\varphi$ is a minimax solution of problem (8), (9) if and only if it possesses properties (M*) and (M_*). Observe that together properties (M*) and (M_*) are nothing more than property (MC) for the particular characteristic complex $(\mathbb {R}^n, E) \in \mathcal {E}^*(H) \cap \mathcal {E}_*(H)$, defined in accordance with (20). Thus, it remains to show that if $\varphi$ satisfies (MC) for an upper $(Q, E^*) \in \mathcal {E}^*(H)$ and a lower $(P, E_*) \in \mathcal {E}_*(H)$ characteristic complexes, then $\varphi$ satisfies both (M*) and (M_*). Below, we prove property (M*) only, while property (M_*) is verified in a similar way. Moreover, note that, proving (M*), we use only the facts that $\varphi$ is non-anticipative and lower semicontinuous and the part of (MC) that concerns the upper characteristic complex $(Q, E^*)$.

Arguing by contradiction, assume that there are $(t, x(\cdot )) \in [0, T) \times C([- h, T], \mathbb {R}^n)$, $s \in \mathbb {R}^n$, and $\tau \in (t, T]$ such that, for any characteristic $(y(\cdot ), z(\cdot )) \in CH(t, x(\cdot ), s)$, the inequality $\varphi (\tau , y(\cdot )) - \varphi (t, x(\cdot )) > z(\tau )$ holds. Then, there exists $\varepsilon > 0$ such that

$$\begin{aligned} \min _{(y(\cdot ), z(\cdot )) \in CH(t, x(\cdot ), s)} \big ( \varphi (\tau , y(\cdot )) - z(\tau ) \big ) > \varphi (t, x(\cdot )) + \varepsilon . \end{aligned}$$

(37)

Here, the minimum is attained owing to lower semicontinuity of the functional $\varphi$ and compactness of the set of characteristics $CH(t, x(\cdot ), s)$ in $C([- h, T], \mathbb {R}^n) \times C([- h, T], \mathbb {R})$ (see, e.g., [24, Proposition 4.1]). Denote

$$\begin{aligned} \beta (\xi ) = \varphi (t, x(\cdot )) + \varepsilon (\xi - t) / (\tau - t), \quad \xi \in [t, \tau ], \end{aligned}$$

(38)

and put

$$\begin{aligned} \xi _0 = \sup \big \{ \xi \in [t, \tau ]: \, \min _{(y(\cdot ), z(\cdot )) \in CH(t, x(\cdot ), s)} \big ( \varphi (\xi , y(\cdot )) - z(\xi ) \big ) \le \beta (\xi ) \big \}. \end{aligned}$$

(39)

Observe that $\varphi (t, y(\cdot )) - z(t) = \varphi (t, x(\cdot )) = \beta (t)$ for any $(y(\cdot ), z(\cdot )) \in CH(t, x(\cdot ), s)$, since $\varphi$ is non-anticipative and initial condition (21) is fulfilled. Therefore, the set under supremum in relation (39) is not empty. Further, lower semicontinuity of $\varphi$ and compactness of $CH(t, x(\cdot ), s)$ imply that this supremum is actually attained. In particular, it follows from assumption (37) and the equality $\beta (\tau ) = \varphi (t, x(\cdot )) + \varepsilon$ that $\xi _0 < \tau$. So, let us choose a characteristic $(y_0(\cdot ), z_0(\cdot )) \in CH(t, x(\cdot ), s)$ from the condition

$$\begin{aligned} \varphi (\xi _0, y_0(\cdot )) - z_0(\xi _0) \le \beta (\xi _0). \end{aligned}$$

(40)

Based on property (C.4) of the upper characteristic complex $(Q, E^*)$, take $q \in Q$ such that

$$\begin{aligned} \min _{(f, g) \in E^*(\xi _0, y_0(\cdot ), q)} (\langle s, f \rangle - g) \ge H(\xi _0, y_0(\cdot ), s) - \varepsilon / (2 (\tau - t)). \end{aligned}$$

(41)

Consider the set $CH^*(\xi _0, y_0(\cdot ), q)$ (see Sect. 3.2 for the definition). Due to properties (C.1) to (C.3) of $(Q, E^*)$, the set $CH^*(\xi _0, y_0(\cdot ), q)$ is compact in $C([- h, T], \mathbb {R}^n) \times C([- h, T], \mathbb {R})$ (see, e.g., [24, Proposition 4.1]). Hence, in view of choice (41) of q and also since the Hamiltonian H is continuous by (B.1), functional (7) is non-anticipative, and $(Q, E^*)$ possesses property (C.2), there exists $\delta \in (0, \tau - \xi _0]$ such that, for every $(y(\cdot ), z(\cdot )) \in CH^*(\xi _0, y_0(\cdot ), q)$, the inequality below is valid:

$$\begin{aligned} \langle s, \dot{y}(\xi ) \rangle - \dot{z}(\xi ) \ge H(\xi , y(\cdot ), s) - \varepsilon / (\tau - t) \text { for a.e. } \xi \in [\xi _0, \xi _0 + \delta ]. \end{aligned}$$

(42)

Recalling that $\varphi$ satisfies condition (MC), let us choose $(y^*(\cdot ), z^*(\cdot )) \in CH^*(\xi _0, y_0(\cdot ), q)$ such that

$$\begin{aligned} \varphi (\xi _0 + \delta , y^*(\cdot )) - \varphi (\xi _0, y_0(\cdot )) \le z^*(\xi _0 + \delta ). \end{aligned}$$

(43)

Consider the function

$$\begin{aligned} z^{**}(\xi ) = {\left\{ \begin{array}{ll} z_0(\xi ) &{} \text{ if } \xi \in [0, \xi _0], \\ \displaystyle z_0(\xi _0) + \int _{\xi _0}^{\xi } \big ( \langle s, \dot{y}^*(\eta ) \rangle - H(\eta , y^*(\cdot ), s) \big ) \, \mathrm {d}\eta &{} \text{ if } \xi \in (\xi _0, T]. \end{array}\right. } \end{aligned}$$

Relying on the inclusions $(y_0(\cdot ), z_0(\cdot )) \in CH(t, x(\cdot ), s)$ and $(y^*(\cdot ), z^*(\cdot )) \in CH^*(\xi _0, y_0(\cdot ), q)$, property (C.3) of $(Q, E^*)$ and taking into account that functional (7) is non-anticipative, we obtain $(y^*(\cdot ), z^{**}(\cdot )) \in CH(t, x(\cdot ), s)$. Besides, owing to relation (42) and the equality $z^*(\xi _0) = 0$, we have

$$\begin{aligned} z^{**}(\xi _0 + \delta )= & {} z_0(\xi _0) + \int _{\xi _0}^{\xi _0 + \delta } \big ( \langle s, \dot{y}^*(\eta ) \rangle - H(\eta , y^*(\cdot ), s) \big ) \, \mathrm {d}\eta \\&\ge z_0(\xi _0) + z^*(\xi _0 + \delta ) - \varepsilon \delta / (\tau - t), \end{aligned}$$

and, therefore, by virtue of (38), (40), and (43), we derive

$$\begin{aligned} \begin{array}{c} \varphi (\xi _0 + \delta , y^*(\cdot )) - z^{**}(\xi _0 + \delta ) \\ \le \varphi (\xi _0 + \delta , y^*(\cdot )) - z^*(\xi _0 + \delta ) - z_0(\xi _0) + \varepsilon \delta / (\tau - t) \\ \le \varphi (\xi _0, y_0(\cdot )) - z_0(\xi _0) + \varepsilon \delta / (\tau - t) \le \beta (\xi _0) + \varepsilon \delta / (\tau - t) = \beta (\xi _0 + \delta ). \end{array} \end{aligned}$$

Thus, for the number $\xi _0 + \delta$, we get

$$\begin{aligned} \min _{(y(\cdot ), z(\cdot )) \in CH(t, x(\cdot ), s)} \big ( \varphi (\xi _0 + \delta , y(\cdot )) - z(\xi _0 + \delta ) \big ) \le \beta (\xi _0 + \delta ). \end{aligned}$$

Since $\xi _0 < \xi _0 + \delta \le \tau$ by construction, we come to a contradiction to definition (39) of $\xi _0$. This completes the proof of Proposition 3.

Proof of Co-invariant Smoothness of the Functional V from (27)

In this appendix, we provide a direct proof of the facts that the functional $V: [0, T] \times C([- h, T], \mathbb {R}^n) \rightarrow \mathbb {R}$ from (27) is ci-smooth and that its ci-derivatives $\partial _t V$ and $\nabla V$ are given by (28). Consider the functionals

$$\begin{aligned} V_1(t, x(\cdot )) = {\left\{ \begin{array}{ll} \displaystyle \frac{\big (\Vert x(\cdot )\Vert _{[- h, t]}^2 - \Vert x(t)\Vert ^2\big )^2}{\Vert x(\cdot )\Vert _{[- h, t]}^2} &{} \text{ if } \Vert x(\cdot )\Vert _{[- h, t]} > 0, \\ 0 &{} \text{ if } \Vert x(\cdot )\Vert _{[- h, t]} = 0, \end{array}\right. } \quad V_2(t, x(\cdot )) = \Vert x(t)\Vert ^2, \end{aligned}$$

(44)

where $(t, x(\cdot )) \in [0, T] \times C([- h, T], \mathbb {R}^n)$. Observe that the representation $V = V_1 + V_2$ holds, and, moreover, the functional $V_2$ is ci-smooth and, for every $(t, x(\cdot )) \in [0, T) \times C([- h, T], \mathbb {R}^n)$, the equalities $\partial _t V_2 (t, x(\cdot )) = 0$ and $\nabla V_2(t, x(\cdot )) = 2 x(t)$ are valid (see also, e.g., [19, Sect. 1]). Thus, it remains to show that the functional $V_1$ is ci-smooth and that its ci-derivatives are calculated as follows:

$$\begin{aligned} \partial _t V_1(t, x(\cdot )) = 0, \quad \nabla V_1(t, x(\cdot )) = {\left\{ \begin{array}{ll} \displaystyle - \frac{4 \big (\Vert x(\cdot )\Vert _{[- h, t]}^2 - \Vert x(t)\Vert ^2\big )}{\Vert x(\cdot )\Vert _{[- h, t]}^2} x(t) &{} \text{ if } \Vert x(\cdot )\Vert _{[- h, t]} > 0, \\ 0 &{} \text{ if } \Vert x(\cdot )\Vert _{[- h, t]} = 0 \end{array}\right. } \end{aligned}$$

(45)

for any $(t, x(\cdot )) \in [0, T) \times C([- h, T], \mathbb {R}^n)$.

Let us fix a point $(t, x(\cdot )) \in [0, T) \times C([- h, T], \mathbb {R}^n)$ and prove that the functional $V_1$ is ci-differentiable at this point (see Sect. 2.2 for the definition) and that formulas (45) hold. Fix $y(\cdot ) \in {\text {Lip}}(t, x(\cdot ))$ and take $L > 0$ such that $\Vert y(\tau ) - y(\xi )\Vert \le L |\tau - \xi |$ for any $\tau$, $\xi \in [t, T]$.

Firstly, let us suppose that $\Vert x(t)\Vert < \Vert x(\cdot )\Vert _{[- h, t]}$. Then, there exists $\delta \in (0, T - t]$ satisfying the condition $L \delta \le \Vert x(\cdot )\Vert _{[- h, t]} - \Vert x(t)\Vert$. In view of (3), for every $\tau \in [t, t + \delta ]$, we derive

$$\begin{aligned} \Vert y(\tau )\Vert \le \Vert y(\tau ) - y(t)\Vert + \Vert x(t)\Vert \le L \delta + \Vert x(t)\Vert \le \Vert x(\cdot )\Vert _{[- h, t]}, \end{aligned}$$

and, therefore, $\Vert y(\cdot )\Vert _{[- h, \tau ]} \le \Vert x(\cdot )\Vert _{[- h, t]}$. Consequently, we get $\Vert y(\cdot )\Vert _{[- h, \tau ]} = \Vert x(\cdot )\Vert _{[- h, t]} > 0$, $\tau \in [t, t + \delta ]$. Thus, according to (44), we obtain

$$\begin{aligned}&V_1(\tau , y(\cdot )) - V_1(t, x(\cdot ))\\&\quad = \frac{\big (\Vert x(\cdot )\Vert _{[- h, t]}^2 - \Vert y(\tau )\Vert ^2\big )^2 - \big (\Vert x(\cdot )\Vert _{[- h, t]}^2 - \Vert x(t)\Vert ^2\big )^2}{\Vert x(\cdot )\Vert _{[- h, t]}^2}, \quad \tau \in [t, t + \delta ]. \end{aligned}$$

Let us fix $\tau \in [t, t + \delta ]$ and estimate the numerator of the obtained expression. Based on the difference of squares formula and the equality $\Vert y(\tau )\Vert ^2 = \Vert x(t)\Vert ^2 + 2 \langle x(t), y(\tau ) - x(t) \rangle + \Vert y(\tau ) - x(t)\Vert ^2$, we derive

$$\begin{aligned} \begin{array}{c} \big (\Vert x(\cdot )\Vert _{[- h, t]}^2 - \Vert y(\tau )\Vert ^2\big )^2 - \big (\Vert x(\cdot )\Vert _{[- h, t]}^2 - \Vert x(t)\Vert ^2\big )^2 \\ = - \Vert y(\tau ) - x(t)\Vert ^2 \big (2 \Vert x(\cdot )\Vert _{[- h, t]}^2 -\Vert y(\tau )\Vert ^2 - \Vert x(t)\Vert ^2\big )\\ + 2 \langle x(t), y(\tau ) - x(t) \rangle \big (\Vert y(\tau )\Vert ^2 - \Vert x(t)\Vert ^2 \big ) \\ - 4 \langle x(t), y(\tau ) - x(t) \rangle \big (\Vert x(\cdot )\Vert _{[- h, t]}^2 - \Vert x(t)\Vert ^2 \big ) = A_1 + A_2 + A_3. \end{array} \end{aligned}$$

Since $\Vert y(\tau )\Vert \le \Vert x(\cdot )\Vert _{[- h, t]}$, for the first two terms $A_1$ and $A_2$, we have

$$\begin{aligned} |A_1| = \Vert y(\tau ) - x(t)\Vert ^2 \big (2 \Vert x(\cdot )\Vert _{[- h, t]}^2 - \Vert y(\tau )\Vert ^2 - \Vert x(t)\Vert ^2\big ) \le 2 L^2 (\tau - t)^2 \Vert x(\cdot )\Vert _{[- h, t]}^2 \end{aligned}$$

and

$$\begin{aligned}&|A_2| \le 2 \Vert x(t)\Vert \Vert y(\tau ) - x(t)\Vert ^2 \big (\Vert y(\tau )\Vert + \Vert x(t)\Vert \big ) \\&\quad \le 4 \Vert x(\cdot )\Vert _{[- h, t]}^2 \Vert y(\tau ) - x(t)\Vert ^2 \le 4 L^2 (\tau - t)^2 \Vert x(\cdot )\Vert _{[- h, t]}^2. \end{aligned}$$

Thus, we get

$$\begin{aligned} \begin{array}{c} \displaystyle \bigg | V_1(\tau , y(\cdot )) - V_1(t, x(\cdot )) + \frac{4 \big (\Vert x(\cdot )\Vert _{[- h, t]}^2 - \Vert x(t)\Vert ^2 \big )}{\Vert x(\cdot )\Vert _{[- h, t]}^2} \langle x(t), y(\tau ) - x(t) \rangle \bigg | \\ \displaystyle = \bigg | V_1(\tau , y(\cdot )) - V_1(t, x(\cdot )) - \frac{A_3}{\Vert x(\cdot )\Vert _{[- h, t]}^2} \bigg | \\ = \dfrac{|A_1 + A_2|}{\Vert x(\cdot )\Vert _{[- h, t]}^2} \le 6 L^2 (\tau - t)^2, \quad \tau \in [t, t + \delta ]. \end{array} \end{aligned}$$

Hence, in the considered case, the functional $V_1$ is ci-differentiable at $(t, x(\cdot ))$ and formulas (45) hold.

Now, suppose that $\Vert x(t)\Vert = \Vert x(\cdot )\Vert _{[- h, t]}$. Then, $V_1(t, x(\cdot )) = 0$ according to (44). Let us show that, in this case, for every $\tau \in [t, T]$, the inequality below is valid:

$$\begin{aligned} V_1(\tau , y(\cdot )) \le 4 L^2 (\tau - t)^2. \end{aligned}$$

(46)

If $\Vert y(\cdot )\Vert _{[- h, \tau ]} = 0$, then $V_1(\tau , y(\cdot )) = 0$, and inequality (46) is fulfilled. Therefore, we can assume further that $\Vert y(\cdot )\Vert _{[-h, \tau ]} > 0$ and, consequently,

$$\begin{aligned} V_1(\tau , y(\cdot )) = \frac{\big (\Vert y(\cdot )\Vert _{[- h, \tau ]}^2 - \Vert y(\tau )\Vert ^2\big )^2}{\Vert y(\cdot )\Vert _{[- h, \tau ]}^2}. \end{aligned}$$

(47)

In view of the relations (see also (3))

$$\begin{aligned}&\Vert y(\cdot )\Vert _{[- h, \tau ]} = \max \big \{ \Vert x(\cdot )\Vert _{[- h, t]}, \max _{\xi \in [t, \tau ]} \Vert y(\xi )\Vert \big \}\\&\quad = \max \big \{ \Vert x(t)\Vert , \max _{\xi \in [t, \tau ]} \Vert y(\xi )\Vert \big \} = \max _{\xi \in [t, \tau ]} \Vert y(\xi )\Vert , \end{aligned}$$

there exists $\xi ^{(\tau )} \in [t, \tau ]$ such that $\Vert y(\cdot )\Vert _{[- h, \tau ]} = \Vert y(\xi ^{(\tau )})\Vert$. So, we get

$$\begin{aligned} 0 \le \Vert y(\cdot )\Vert _{[- h, \tau ]} - \Vert y(\tau )\Vert \le \Vert y(\xi ^{(\tau )}) - y(\tau )\Vert \le L |\xi ^{(\tau )} - \tau | \le L (\tau - t), \end{aligned}$$

which yields

$$\begin{aligned}&\big (\Vert y(\cdot )\Vert _{[- h, \tau ]}^2 - \Vert y(\tau )\Vert ^2\big )^2 \\&\quad = \big (\Vert y(\cdot )\Vert _{[- h, \tau ]} - \Vert y(\tau )\Vert \big )^2 \big (\Vert y(\cdot )\Vert _{[- h, \tau ]} + \Vert y(\tau )\Vert \big )^2 \le 4 L^2 (\tau - t)^2 \Vert y(\cdot )\Vert _{[- h, \tau ]}^2. \end{aligned}$$

In accordance with (47), the obtained estimate proves the desired inequality (46). Based on this inequality, we derive $|V_1(\tau , y(\cdot )) - V_1(t, x(\cdot ))| = V_1(\tau , y(\cdot )) \le 4 L^2 (\tau - t)^2$ for every $\tau \in [t, T]$, and, thus, the functional $V_1$ is ci-differentiable at $(t, x(\cdot ))$ and formulas (45) hold in the second case as well.

Finally, let us turn to the question about continuity of $V_1$ and $\nabla V_1$. Observe that, if follows directly from definition (2) of the metric ${\text {dist}}$ that the mappings $[0, T] \times C([- h, T], \mathbb {R}^n) \ni (t, x(\cdot )) \mapsto \Vert x(\cdot )\Vert _{[- h, t]} \in \mathbb {R}$ and $[0, T] \times C([- h, T], \mathbb {R}^n) \ni (t, x(\cdot )) \mapsto x(t) \in \mathbb {R}^n$ are continuous. This implies continuity of $V_1$ and $\nabla V_1$ at each point $(t, x(\cdot ))$ such that $\Vert x(\cdot )\Vert _{[- h, t]} > 0$. Further, for any $(t, x(\cdot )) \in [0, T] \times C([- h, T], \mathbb {R}^n)$, we have $0 \le \Vert x(\cdot )\Vert _{[- h, t]}^2 - \Vert x(t)\Vert ^2 \le \Vert x(\cdot )\Vert _{[- h, t]}^2$, which gives $V_1(t, x(\cdot )) \le \Vert x(\cdot )\Vert _{[- h, t]}^2$ and $\Vert \nabla V_1(t, x(\cdot ))\Vert \le 4 \Vert x(t)\Vert$. These estimates provide continuity of $V_1$ and $\nabla V_1$ at the remaining points $(t, x(\cdot ))$, when $\Vert x(\cdot )\Vert _{[- h, t]} = 0$. The proof is complete.