AMS Classification: 60G40; 62P20; 91B99; 91B28; 35B37; 49L25.
Introduction
In recent decades, stochastic differential games have attracted considerable attention due to their numerous applications in economics, finance, engineering, and operational research. Among these problems, zero-sum two-player switching games constitute an important class of stochastic control problems in which competing players influence the evolution of a system by switching between different operational regimes. Such models combine the strategic interactions of differential games with the flexibility of regime-switching control, providing a natural framework for studying decision-making under uncertainty when switching actions incur costs.
A zero-sum switching game describes a competitive situation in which the gain of one player is exactly the loss of the other. In the stochastic setting, the players’ decisions affect both the current payoff and the future evolution of the system, leading to highly coupled optimization problems. The strategic complexity increases significantly when several operating regimes are available, since each player must determine not only the optimal timing of a switch but also the most advantageous regime to select while anticipating the opponent’s actions.
The theory of regime-switching models originated with the pioneering work of [11 ] and was further developed by [10 ] for the analysis of economic and financial systems exhibiting structural changes. Since then, regime-switching models have become a standard tool for describing systems whose dynamics evolve according to a finite collection of states. When combined with stochastic control and game-theoretic considerations, they give rise to switching games in which players modify the regime of the system through costly interventions.
Switching control and switching games have been extensively studied in the literature. Important contributions include the works of Bensoussan and Lions [3 ] , Øksendal and Sulem [15 ] , Pham [18 ] , and Bayraktar and Yao [1 ] , among others. In general, the value functions of these games are characterized as viscosity solutions of systems of variational or quasi-variational inequalities, whose analysis relies on sophisticated techniques from stochastic control theory. Existence and uniqueness results have been established in a broad setting, and several qualitative properties of the associated Nash equilibria have been investigated.
Despite these advances, obtaining explicit solutions remains a difficult task. Most existing works focus on the characterization of the value function through variational inequalities, while closed-form analytical solutions are available only in a limited number of special cases. The difficulty increases rapidly as the number of regimes grows because the number of possible switching decisions and the interactions between the players become considerably more intricate.
In [7 ] , we obtained an explicit solution for a zero-sum two-player switching game in the case of two regimes, providing an analytical characterization of the value functions and the corresponding optimal switching strategies. The present paper extends that analysis to the considerably more challenging case of three regimes. The transition from two to three regimes substantially enriches the strategic structure of the game by introducing additional switching possibilities and more complex interactions between the players.
The principal novelty of this paper is the derivation of an explicit solution for the three-regime zero-sum two-player switching game. To the best of our knowledge, explicit analytical solutions for this class of problems are scarce in the literature due to the significant mathematical complexity generated by the increased number of regimes. Our results show that moving beyond the two-regime setting gives rise to a richer strategic structure, in which the interaction between multiple switching opportunities fundamentally shapes the equilibrium strategies. The three-regime framework therefore provides new structural insights into the geometry of the switching regions and the resulting equilibrium policies, while substantially broadening the scope of the explicit analysis.
From an application point of view, the three-regime model is relevant in numerous contexts. In finance, it can describe markets evolving through bearish, neutral, and bullish phases. In economics, it models strategic decisions under different macroeconomic environments, while in engineering it applies to systems operating under several modes with different performance characteristics. The additional regime allows for a richer representation of realistic systems and more flexible modeling of strategic interactions.
The remainder of the paper is organized as follows. Section 2 introduces the mathematical framework and formulates the zero-sum switching game together with the standing assumptions. Section 3 derives the associated quasi-variational inequalities and characterizes the switching regions. Section 4 presents the explicit solution of the three-regime switching game and establishes the corresponding equilibrium strategies, in section 5 we give a numerical procedure to compute the value function in case we know the qualitative structure of the switching regions and in section 6 we illustrate our results by numerical simulations.
The problem consists of a two-player game. The first player (Player I) and the second player (Player II) both make decisions based on the value of a state process X : = ( X t ) t ≥ 0 X := (X_t)_{t \ge0} X := ( X t ) t ≥ 0
Each player operates under several modes or regimes. Let D : = { 1 , 2 , … , d } \mathcal{D} := \{1,2,\ldots,d\} D := { 1 , 2 , … , d } { 1 , … , d } \{1,\ldots,d\} { 1 , … , d }
The switching costs are known from both players, and at any given time, each player knows the current regime of his counterpart.
The regime configuration of both players is modeled by a two-dimensional càdlàg process I t I_t I t D × D \mathcal{D} \times\mathcal{D} D × D I t I_t I t f I t f_{I_t} f I t
The state process X X X R + ∗ \mathbb{R}_{+}^{*} R + ∗
We denote by J J J J J J ξ : = ( ξ n ) n ≥ 1 \xi:= (\xi_n)_{n \ge1} ξ := ( ξ n ) n ≥ 1 τ : = ( τ n ) n ≥ 1 \tau:= (\tau_n)_{n \ge1} τ := ( τ n ) n ≥ 1 J J J η : = ( η m ) m ≥ 1 \eta:= (\eta_m)_{m \ge1} η := ( η m ) m ≥ 1 ρ : = ( ρ m ) m ≥ 1 \rho:= (\rho_m)_{m \ge1} ρ := ( ρ m ) m ≥ 1
Switching from regime i i i j j j c i j c_{ij} c ij χ i j \chi_{ij} χ ij
The payoff of the game is given by
J ( x , ξ , η ) = E [ ∫ 0 ∞ e − r s f ( X s μ s , ν s , μ s , ν s ) d s − ∑ n ≥ 1 e − r τ n C ( ξ n − 1 , ξ n ) + ∑ m ≥ 1 e − r ρ m χ ( η m − 1 , η m ) ] , (1) J(x,\xi,\eta) = \mathbb{E}\left[\int_{0}^{\infty} e^{-rs} f\left(X_s^{\mu_s,\nu_s},\mu_s,\nu_s\right)\,\mathrm{d}s - \sum_{n \ge1} e^{-r\tau_n} C(\xi_{n-1},\xi_n) + \sum_{m \ge1} e^{-r\rho_m}\chi(\eta_{m-1},\eta_m)\right], \tag*{(1)} J ( x , ξ , η ) = E [ ∫ 0 ∞ e − r s f ( X s μ s , ν s , μ s , ν s ) d s − n ≥ 1 ∑ e − r τ n C ( ξ n − 1 , ξ n ) + m ≥ 1 ∑ e − r ρ m χ ( η m − 1 , η m ) ] , (1) where r > 0 r > 0 r > 0
The regime processes of the players are defined by
μ t = ∑ n ≥ 1 ξ n 1 { τ n ≤ t < τ n + 1 } , ν t = ∑ m ≥ 1 η m 1 { ρ m ≤ t < ρ m + 1 } . \mu_t = \sum_{n \ge1} \xi_n \mathbf{1}_{\{\tau_n \le t < \tau_{n+1}\}}, \qquad\nu_t = \sum_{m \ge1} \eta_m \mathbf{1}_{\{\rho_m \le t < \rho_{m+1}\}}. μ t = n ≥ 1 ∑ ξ n 1 { τ n ≤ t < τ n + 1 } , ν t = m ≥ 1 ∑ η m 1 { ρ m ≤ t < ρ m + 1 } . Hence, by definition, we have I t = ( μ t , ν t ) I_t = (\mu_t,\nu_t) I t = ( μ t , ν t )
State dynamics
In this work, the process X t X_t X t
d X t = b ( X t , μ t , ν t ) d t + σ ( X t , μ t , ν t ) d W t , X 0 = x , I 0 − = ( i , j ) . (2) \mathrm{d}X_t = b(X_t,\mu_t,\nu_t)\,\mathrm{d}t + \sigma(X_t,\mu_t,\nu_t)\,\mathrm{d}W_t,\qquad X_0 = x,\qquad I_{0-} = (i,j). \tag*{(2)} d X t = b ( X t , μ t , ν t ) d t + σ ( X t , μ t , ν t ) d W t , X 0 = x , I 0 − = ( i , j ) . (2) where
b ( X t , μ t , ν t ) = b I t X t , σ ( X t , μ t , ν t ) = σ I t X t . b(X_t,\mu_t,\nu_t) = b_{I_t}X_t,\qquad\sigma(X_t,\mu_t,\nu_t) = \sigma_{I_t}X_t. b ( X t , μ t , ν t ) = b I t X t , σ ( X t , μ t , ν t ) = σ I t X t . Throughout the paper, we refer to b I t b_{I_t} b I t σ I t \sigma_{I_t} σ I t X I 0 x X_{I_0}^{x} X I 0 x X x X^x X x
Assumptions
Throughout the paper, d d d D = { 1 , … , d } \mathcal{D} = \{1,\ldots,d\} D = { 1 , … , d }
[H1] The functions b : R + ∗ × D × D → R + b : \mathbb{R}_{+}^{*} \times\mathcal{D} \times\mathcal{D} \to\mathbb{R}_{+} b : R + ∗ × D × D → R + σ : R + ∗ × D × D → R + \sigma: \mathbb{R}_{+}^{*} \times\mathcal{D} \times\mathcal{D} \to\mathbb{R}_{+} σ : R + ∗ × D × D → R + C > 0 C > 0 C > 0 x , x ′ ∈ R + ∗ x,x' \in\mathbb{R}_{+}^{*} x , x ′ ∈ R + ∗ i , j ∈ D i,j \in\mathcal{D} i , j ∈ D
∣ σ ( x , i , j ) − σ ( x ′ , i , j ) ∣ + ∣ b ( x , i , j ) − b ( x ′ , i , j ) ∣ ≤ C ∣ x − x ′ ∣ . (3) \left|\sigma(x,i,j)-\sigma(x',i,j)\right|+\left|b(x,i,j)-b(x',i,j)\right| \le C|x-x'|. \tag*{(3)} ∣ σ ( x , i , j ) − σ ( x ′ , i , j ) ∣ + ∣ b ( x , i , j ) − b ( x ′ , i , j ) ∣ ≤ C ∣ x − x ′ ∣. (3) Moreover, they satisfy the linear growth condition:
∣ σ ( x , i , j ) ∣ + ∣ b ( x , i , j ) ∣ ≤ C ( 1 + ∣ x ∣ ) . |\sigma(x,i,j)|+|b(x,i,j)| \le C(1+|x|). ∣ σ ( x , i , j ) ∣ + ∣ b ( x , i , j ) ∣ ≤ C ( 1 + ∣ x ∣ ) . [H2] The function f : R + ∗ × D × D → R + f : \mathbb{R}_{+}^{*} \times\mathcal{D} \times\mathcal{D} \to\mathbb{R}_{+} f : R + ∗ × D × D → R + C > 0 C > 0 C > 0 x , x ′ ∈ R + ∗ x,x' \in\mathbb{R}_{+}^{*} x , x ′ ∈ R + ∗ i , j ∈ D i,j \in\mathcal{D} i , j ∈ D
∣ f ( x , i , j ) ∣ ≤ C ( 1 + ∣ x ∣ ) , ∣ f ( x , i , j ) − f ( x ′ , i , j ) ∣ ≤ C ∣ x − x ′ ∣ . (4) |f(x,i,j)| \le C(1+|x|),\qquad|f(x,i,j)-f(x',i,j)| \le C|x-x'|. \tag*{(4)} ∣ f ( x , i , j ) ∣ ≤ C ( 1 + ∣ x ∣ ) , ∣ f ( x , i , j ) − f ( x ′ , i , j ) ∣ ≤ C ∣ x − x ′ ∣. (4) [H3] For all i , j ∈ D i,j \in\mathcal{D} i , j ∈ D c i j c_{ij} c ij χ i j \chi_{ij} χ ij
c i k < c i j + c j k , χ i k < χ i j + χ j k , j ≠ i , k . (5) c_{ik} < c_{ij} + c_{jk}, \qquad\chi_{ik} < \chi_{ij} + \chi_{jk}, \qquad j \ne i,k. \tag*{(5)} c ik < c ij + c j k , χ ik < χ ij + χ j k , j = i , k . (5) This ensures that direct switching is cheaper than indirect switching through an intermediate regime.
Note that switching costs may be negative, but condition (2.5) prevents arbitrage by repeated switching. In particular,
0 < c i j + c j i , 0 < χ i j + χ j i . (6) 0 < c_{ij} + c_{ji}, \qquad0 < \chi_{ij} + \chi_{ji}. \tag*{(6)} 0 < c ij + c j i , 0 < χ ij + χ j i . (6) We may alternatively write c ( i , j ) c(i,j) c ( i , j ) c i j c_{ij} c ij f ( x , i , j ) f(x,i,j) f ( x , i , j ) f i j ( x ) f_{ij}(x) f ij ( x )
[H4] The discount rate r r r
r > max i , j ∈ D b i j . r > \max_{i,j \in\mathcal{D}} b_{ij}. r > i , j ∈ D max b ij . This assumption is mainly technical and ensures integrability properties needed for the explicit solution.
[H5] The sequences of cumulative switching costs
c n = ∑ m = 1 n e − r τ m c ( ξ m − 1 , ξ m ) , χ n = ∑ l = 1 n e − r ρ l χ ( η l − 1 , η l ) , c_n = \sum_{m=1}^{n} e^{-r\tau_m}c(\xi_{m-1},\xi_m), \qquad\chi_n = \sum_{l=1}^{n} e^{-r\rho_l}\chi(\eta_{l-1},\eta_l), c n = m = 1 ∑ n e − r τ m c ( ξ m − 1 , ξ m ) , χ n = l = 1 ∑ n e − r ρ l χ ( η l − 1 , η l ) , converge almost surely, and their limits satisfy
lim n → ∞ c n ∈ L 1 , lim n → ∞ χ n ∈ L 1 . (7) \lim_{n \to\infty} c_n \in L^1, \qquad\lim_{n \to\infty} \chi_n \in L^1. \tag*{(7)} n → ∞ lim c n ∈ L 1 , n → ∞ lim χ n ∈ L 1 . (7) Preliminary Results
Existence of solutions for the SDE
Under assumption [H1], the stochastic differential equation (2.2) admits a unique strong solution.
Viscosity solutions
Let F F F
F ( x , v , D x v , D x x 2 v ) = 0 , (8) F(x,v,D_xv,D^2_{xx}v) = 0, \tag*{(8)} F ( x , v , D x v , D xx 2 v ) = 0 , (8) where D x v D_xv D x v D x x 2 v D^2_{xx}v D xx 2 v x x x
We assume that F F F x x x O ⊂ R + \mathcal{O} \subset\mathbb{R}^{+} O ⊂ R +
Definition 2.1 (Viscosity solution) (i) Supersolution. A function v v v x ˉ ∈ O \bar{x} \in\mathcal{O} x ˉ ∈ O ϕ ∈ C 2 \phi\in C^2 ϕ ∈ C 2 v − ϕ v-\phi v − ϕ x ˉ \bar{x} x ˉ
F ( x ˉ , v ( x ˉ ) , D x ϕ ( x ˉ ) , D x x 2 ϕ ( x ˉ ) ) ≥ 0. F(\bar{x},v(\bar{x}),D_x\phi(\bar{x}),D^2_{xx}\phi(\bar{x})) \ge0. F ( x ˉ , v ( x ˉ ) , D x ϕ ( x ˉ ) , D xx 2 ϕ ( x ˉ )) ≥ 0. (ii) Subsolution. A function v v v x ˉ ∈ O \bar{x} \in\mathcal{O} x ˉ ∈ O ϕ ∈ C 2 \phi\in C^2 ϕ ∈ C 2 v − ϕ v-\phi v − ϕ x ˉ \bar{x} x ˉ
F ( x ˉ , v ( x ˉ ) , D x ϕ ( x ˉ ) , D x x 2 ϕ ( x ˉ ) ) ≤ 0. F(\bar{x},v(\bar{x}),D_x\phi(\bar{x}),D^2_{xx}\phi(\bar{x})) \le0. F ( x ˉ , v ( x ˉ ) , D x ϕ ( x ˉ ) , D xx 2 ϕ ( x ˉ )) ≤ 0. (iii) Solution. A function is a viscosity solution if it is both a subsolution and a supersolution.
Equivalently, let J 2 , + v ( x ) J^{2,+}v(x) J 2 , + v ( x ) J 2 , − v ( x ) J^{2,-}v(x) J 2 , − v ( x ) v v v x x x ( q , X ) (q,X) ( q , X )
v ( y ) ≤ v ( x ) + ⟨ q , y − x ⟩ + 1 2 ⟨ X ( y − x ) , y − x ⟩ + o ( ∣ y − x ∣ 2 ) , v(y) \le v(x) + \langle q,y-x\rangle+ \frac{1}{2}\langle X(y-x),y-x\rangle+ o(|y-x|^2), v ( y ) ≤ v ( x ) + ⟨ q , y − x ⟩ + 2 1 ⟨ X ( y − x ) , y − x ⟩ + o ( ∣ y − x ∣ 2 ) , (resp.
v ( y ) ≥ v ( x ) + ⟨ q , y − x ⟩ + 1 2 ⟨ X ( y − x ) , y − x ⟩ + o ( ∣ y − x ∣ 2 ) . v(y) \ge v(x) + \langle q, y-x\rangle+ \frac{1}{2}\langle X(y-x),y-x\rangle+ o(|y-x|^2). v ( y ) ≥ v ( x ) + ⟨ q , y − x ⟩ + 2 1 ⟨ X ( y − x ) , y − x ⟩ + o ( ∣ y − x ∣ 2 ) . Then:
v v v ⟺ F ( x , v , q , X ) ≥ 0 ∀ ( q , X ) ∈ J 2 , − v ( x ) \Longleftrightarrow F(x,v,q,X) \ge0 \quad\forall(q,X) \in J^{2,-}v(x) ⟺ F ( x , v , q , X ) ≥ 0 ∀ ( q , X ) ∈ J 2 , − v ( x )
v v v ⟺ F ( x , v , q , X ) ≤ 0 ∀ ( q , X ) ∈ J 2 , + v ( x ) \Longleftrightarrow F(x,v,q,X) \le0 \quad\forall(q,X) \in J^{2,+}v(x) ⟺ F ( x , v , q , X ) ≤ 0 ∀ ( q , X ) ∈ J 2 , + v ( x )
Existence of the value function
Let us define an admissible strategy in the two-player game.
Definition 2.2 (Admissible strategy) Let ( Ω , F , P ) (\Omega,\mathcal{F},\mathbb{P}) ( Ω , F , P ) d d d W = ( W t ) t ≥ 0 W=(W_t)_{t\geq0} W = ( W t ) t ≥ 0
F t 0 : = σ ( W s , s ≤ t ) , F = ( F t ) t ≥ 0 \mathcal{F}^0_t := \sigma(W_s,s\leq t), \qquad\mathcal{F}=(\mathcal{F}_t)_{t\geq0} F t 0 := σ ( W s , s ≤ t ) , F = ( F t ) t ≥ 0 being the completed filtration.
We consider two players, I and II, acting on the system through switching strategies.
An admissible strategy for Player I is a sequence
δ = ( τ m , ξ m ) m ≥ 0 , \delta=(\tau_m,\xi_m)_{m\geq0}, δ = ( τ m , ξ m ) m ≥ 0 , and for Player II:
ν = ( ρ n , η n ) n ≥ 0 , \nu=(\rho_n,\eta_n)_{n\geq0}, ν = ( ρ n , η n ) n ≥ 0 , where:
(i) ( τ m ) (\tau_m) ( τ m ) ( ρ n ) (\rho_n) ( ρ n ) τ m → ∞ \tau_m\to\infty τ m → ∞ ρ n → ∞ \rho_n\to\infty ρ n → ∞
(ii) ( ξ m ) (\xi_m) ( ξ m ) ( η n ) (\eta_n) ( η n ) D \mathcal{D} D ξ m \xi_m ξ m F τ m \mathcal{F}_{\tau_m} F τ m η n \eta_n η n F ρ n \mathcal{F}_{\rho_n} F ρ n
We now define the upper and lower value functions.
Definition 2.3 (Upper and lower value functions) Let A i \mathcal{A}^i A i B j \mathcal{B}^j B j
A nonanticipative strategy for Player I is a mapping
ϑ : ⋃ j ∈ D B j → A i , \vartheta:\bigcup_{j\in\mathcal{D}}\mathcal{B}^j\to\mathcal{A}^i, ϑ : j ∈ D ⋃ B j → A i , such that if two strategies b , b ′ b,b' b , b ′ τ \tau τ ϑ ( b ) \vartheta(b) ϑ ( b ) ϑ ( b ′ ) \vartheta(b') ϑ ( b ′ ) τ \tau τ
Similarly, a nonanticipative strategy for Player II is a mapping
ϱ : ⋃ i ∈ D A i → B j \varrho:\bigcup_{i\in\mathcal{D}}\mathcal{A}^i\to\mathcal{B}^j ϱ : i ∈ D ⋃ A i → B j satisfying the analogous property.
Let Γ i \Gamma^i Γ i Δ j \Delta^j Δ j
The upper value is
V ‾ i j = inf ϱ ∈ Δ j sup φ ∈ A i J ( x , φ , ϱ ( φ ) ) , (9) \overline{V}_{ij}=\inf_{\varrho\in\Delta^j}\sup_{\varphi\in\mathcal{A}^i}J(x,\varphi,\varrho(\varphi)), \tag*{(9)} V ij = ϱ ∈ Δ j inf φ ∈ A i sup J ( x , φ , ϱ ( φ )) , (9) @page 5/29 starts_mid=1 ends_mid=1 (resp.
v ( y ) ≥ v ( x ) + ⟨ q , y − x ⟩ + 1 2 ⟨ X ( y − x ) , y − x ⟩ + o ( ∣ y − x ∣ 2 ) . v(y) \ge v(x) + \langle q, y-x\rangle+ \frac{1}{2}\langle X(y-x),y-x\rangle+ o(|y-x|^2). v ( y ) ≥ v ( x ) + ⟨ q , y − x ⟩ + 2 1 ⟨ X ( y − x ) , y − x ⟩ + o ( ∣ y − x ∣ 2 ) . Then:
v v v ⟺ F ( x , v , q , X ) ≥ 0 ∀ ( q , X ) ∈ J 2 , − v ( x ) \Longleftrightarrow F(x,v,q,X) \ge0 \quad\forall(q,X) \in J^{2,-}v(x) ⟺ F ( x , v , q , X ) ≥ 0 ∀ ( q , X ) ∈ J 2 , − v ( x )
v v v ⟺ F ( x , v , q , X ) ≤ 0 ∀ ( q , X ) ∈ J 2 , + v ( x ) \Longleftrightarrow F(x,v,q,X) \le0 \quad\forall(q,X) \in J^{2,+}v(x) ⟺ F ( x , v , q , X ) ≤ 0 ∀ ( q , X ) ∈ J 2 , + v ( x )
Existence of the value function
Let us define an admissible strategy in the two-player game.
Definition 2.2 (Admissible strategy) Let ( Ω , F , P ) (\Omega,\mathcal{F},\mathbb{P}) ( Ω , F , P ) d d d W = ( W t ) t ≥ 0 W=(W_t)_{t\geq0} W = ( W t ) t ≥ 0
F t 0 : = σ ( W s , s ≤ t ) , F = ( F t ) t ≥ 0 \mathcal{F}^0_t := \sigma(W_s,s\leq t), \qquad\mathcal{F}=(\mathcal{F}_t)_{t\geq0} F t 0 := σ ( W s , s ≤ t ) , F = ( F t ) t ≥ 0 being the completed filtration.
We consider two players, I and II, acting on the system through switching strategies.
An admissible strategy for Player I is a sequence
δ = ( τ m , ξ m ) m ≥ 0 , \delta=(\tau_m,\xi_m)_{m\geq0}, δ = ( τ m , ξ m ) m ≥ 0 , and for Player II:
ν = ( ρ n , η n ) n ≥ 0 , \nu=(\rho_n,\eta_n)_{n\geq0}, ν = ( ρ n , η n ) n ≥ 0 , where:
(i) ( τ m ) (\tau_m) ( τ m ) ( ρ n ) (\rho_n) ( ρ n ) τ m → ∞ \tau_m\to\infty τ m → ∞ ρ n → ∞ \rho_n\to\infty ρ n → ∞
(ii) ( ξ m ) (\xi_m) ( ξ m ) ( η n ) (\eta_n) ( η n ) D \mathcal{D} D ξ m \xi_m ξ m F τ m \mathcal{F}_{\tau_m} F τ m η n \eta_n η n F ρ n \mathcal{F}_{\rho_n} F ρ n
We now define the upper and lower value functions.
Definition 2.3 (Upper and lower value functions) Let A i \mathcal{A}^i A i B j \mathcal{B}^j B j
A nonanticipative strategy for Player I is a mapping
ϑ : ⋃ j ∈ D B j → A i , \vartheta:\bigcup_{j\in\mathcal{D}}\mathcal{B}^j\to\mathcal{A}^i, ϑ : j ∈ D ⋃ B j → A i , such that if two strategies b , b ′ b,b' b , b ′ τ \tau τ ϑ ( b ) \vartheta(b) ϑ ( b ) ϑ ( b ′ ) \vartheta(b') ϑ ( b ′ ) τ \tau τ
Similarly, a nonanticipative strategy for Player II is a mapping
ϱ : ⋃ i ∈ D A i → B j \varrho:\bigcup_{i\in\mathcal{D}}\mathcal{A}^i\to\mathcal{B}^j ϱ : i ∈ D ⋃ A i → B j satisfying the analogous property.
Let Γ i \Gamma^i Γ i Δ j \Delta^j Δ j
The upper value is
V ‾ i j = inf ϱ ∈ Δ j sup φ ∈ A i J ( x , φ , ϱ ( φ ) ) , (10) \overline{V}_{ij}=\inf_{\varrho\in\Delta^j}\sup_{\varphi\in\mathcal{A}^i}J(x,\varphi,\varrho(\varphi)), \tag*{(10)} V ij = ϱ ∈ Δ j inf φ ∈ A i sup J ( x , φ , ϱ ( φ )) , (10) and the lower value is
V ‾ i j = sup φ ∈ Γ i inf β ∈ B j J ( x , φ ( β ) , β ) . \underline{V}_{ij}=\sup_{\varphi\in\Gamma_i}\inf_{\beta\in B_j}J(x,\varphi(\beta),\beta). V ij = φ ∈ Γ i sup β ∈ B j inf J ( x , φ ( β ) , β ) . Proof 2.1 We refer to [7 ] for further properties of the value functions.
System of quasi-variational inequalities and switching regions
We now state the Hamilton–Jacobi–Bellman–Isaacs (HJBI) system of quasi-variational inequalities.
Using the dynamic programming principle, the value function satisfies:
max { min [ r v i j ( x ) − L i j v i j ( x ) − f i j ( x ) , v i j ( x ) − M i j [ v ] ( x ) , v i j ( x ) − N i j [ v ] ( x ) ] } = 0. (11) \max\left\{\min\left[rv_{ij}(x)-\mathcal{L}_{ij}v_{ij}(x)-f_{ij}(x),v_{ij}(x)-M_{ij}[v](x),v_{ij}(x)-N_{ij}[v](x)\right]\right\}=0. \tag*{(11)} max { min [ r v ij ( x ) − L ij v ij ( x ) − f ij ( x ) , v ij ( x ) − M ij [ v ] ( x ) , v ij ( x ) − N ij [ v ] ( x ) ] } = 0. (11) min { max [ r v i j ( x ) − L i j v i j ( x ) − f i j ( x ) , v i j ( x ) − N i j [ v ] ( x ) , v i j ( x ) − M i j [ v ] ( x ) ] } = 0. (12) \min\left\{\max\left[rv_{ij}(x)-\mathcal{L}_{ij}v_{ij}(x)-f_{ij}(x),v_{ij}(x)-N_{ij}[v](x),v_{ij}(x)-M_{ij}[v](x)\right]\right\}=0. \tag*{(12)} min { max [ r v ij ( x ) − L ij v ij ( x ) − f ij ( x ) , v ij ( x ) − N ij [ v ] ( x ) , v ij ( x ) − M ij [ v ] ( x ) ] } = 0. (12) where
M i j [ v ] ( x ) = max k ≠ i { v k j ( x ) − c i k } , N i j [ v ] ( x ) = min l ≠ j { v i l ( x ) + χ j l } , M_{ij}[v](x)=\max_{k\ne i}\{v_{kj}(x)-c_{ik}\},\qquad N_{ij}[v](x)=\min_{l\ne j}\{v_{il}(x)+\chi_{jl}\}, M ij [ v ] ( x ) = k = i max { v k j ( x ) − c ik } , N ij [ v ] ( x ) = l = j min { v i l ( x ) + χ j l } , and
L i j v i j ( x ) = 1 2 Tr [ σ i j σ i j ∗ D 2 v i j ( x ) ] + ⟨ b i j , D v i j ( x ) ⟩ . \mathcal{L}_{ij}v_{ij}(x)=\frac{1}{2}\operatorname{Tr}\left[\sigma_{ij}\sigma_{ij}^{*}D^2v_{ij}(x)\right]+\langle b_{ij},Dv_{ij}(x)\rangle. L ij v ij ( x ) = 2 1 Tr [ σ ij σ ij ∗ D 2 v ij ( x ) ] + ⟨ b ij , D v ij ( x )⟩ . The viscosity sub/supersolution definitions follow from standard arguments and are omitted here for brevity.
Proposition 3.1 The upper and lower value functions coincide, and the value function of the stochastic differential game is given by
V i j ( x ) = V ‾ i j ( x ) = V ‾ i j ( x ) , ∀ i , j ∈ D , x ∈ R + . V_{ij}(x)=\underline{V}_{ij}(x)=\overline{V}_{ij}(x),\qquad\forall i,j\in\mathcal{D},\ x\in\mathbb{R}_{+}. V ij ( x ) = V ij ( x ) = V ij ( x ) , ∀ i , j ∈ D , x ∈ R + . Consequently, (2.9) and (2.10) admit a unique continuous solution of polynomial growth.
Proof 3.1 We refer the reader to [6 ] . □
We now introduce the switching and continuation regions.
For Player I:
S i ξ = { x > 0 : V i ⋅ ( x ) = max k ∈ D ∖ { i } ( V k ⋅ ( x ) − c i k ) } , S_i^\xi=\left\{x>0:V_{i\cdot}(x)=\max_{k\in\mathcal{D}\setminus\{i\}}\left(V_{k\cdot}(x)-c_{ik}\right)\right\}, S i ξ = { x > 0 : V i ⋅ ( x ) = k ∈ D ∖ { i } max ( V k ⋅ ( x ) − c ik ) } , C i ξ = { x > 0 : V i ⋅ ( x ) > max k ∈ D ∖ { i } ( V k ⋅ ( x ) − c i k ) } . C_i^\xi=\left\{x>0:V_{i\cdot}(x)>\max_{k\in\mathcal{D}\setminus\{i\}}\left(V_{k\cdot}(x)-c_{ik}\right)\right\}. C i ξ = { x > 0 : V i ⋅ ( x ) > k ∈ D ∖ { i } max ( V k ⋅ ( x ) − c ik ) } . For Player II:
S j η = { x > 0 : V ⋅ j ( x ) = min l ∈ D ∖ { j } ( V ⋅ l ( x ) + χ j l ) } , S_j^\eta=\left\{x>0:V_{\cdot j}(x)=\min_{l\in\mathcal{D}\setminus\{j\}}\left(V_{\cdot l}(x)+\chi_{jl}\right)\right\}, S j η = { x > 0 : V ⋅ j ( x ) = l ∈ D ∖ { j } min ( V ⋅ l ( x ) + χ j l ) } , C j η = { x > 0 : V ⋅ j ( x ) < min l ∈ D ∖ { j } ( V ⋅ l ( x ) + χ j l ) } . C_j^\eta=\left\{x>0:V_{\cdot j}(x)<\min_{l\in\mathcal{D}\setminus\{j\}}\left(V_{\cdot l}(x)+\chi_{jl}\right)\right\}. C j η = { x > 0 : V ⋅ j ( x ) < l ∈ D ∖ { j } min ( V ⋅ l ( x ) + χ j l ) } . Remark 3.1 The smooth-fit property (see [17 ] ) still holds in this framework: the value function is C 1 C^1 C 1 C 2 C^2 C 2
We finally define
H i j [ v ] ( x ) = r v ( x ) − L i j v ( x ) − f i j ( x ) . H_{ij}[v](x)=rv(x)-\mathcal{L}_{ij}v(x)-f_{ij}(x). H ij [ v ] ( x ) = r v ( x ) − L ij v ( x ) − f ij ( x ) . Lemma 3.1 Let C C C S ξ S^\xi S ξ S η ⊂ R + ∗ S^\eta\subset\mathbb{R}_{+}^{*} S η ⊂ R + ∗
C = R + ∗ ∖ ( S ξ ∪ S η ) , C=\mathbb{R}_{+}^{*}\setminus(S^\xi\cup S^\eta), C = R + ∗ ∖ ( S ξ ∪ S η ) , where S ξ ∪ S η S^\xi\cup S^\eta S ξ ∪ S η R + ∗ \mathbb{R}_+^* R + ∗ i , j ∈ D i,j \in\mathcal{D} i , j ∈ D
Let Z \mathcal{Z} Z h ξ h^\xi h ξ h η h^\eta h η R + ∗ \mathbb{R}_+^* R + ∗
H i j [ Z ] ( x ) = 0. (13) \mathcal{H}^{ij}[\mathcal{Z}](x)=0. \tag*{(13)} H ij [ Z ] ( x ) = 0. (13) Z = h ξ \mathcal{Z}=h^\xi Z = h ξ S ξ S^\xi S ξ Z ≤ h η \mathcal{Z}\leq h^\eta Z ≤ h η S η S^\eta S η Z ∈ C 1 \mathcal{Z}\in C^1 Z ∈ C 1 ∂ S ξ \partial S^\xi ∂ S ξ Z \mathcal{Z} Z
H i j [ Z ] ( x ) ≥ 0 in int ( S ξ ) . \mathcal{H}^{ij}[\mathcal{Z}](x)\geq0 \quad\text{in } \operatorname{int}(S^\xi). H ij [ Z ] ( x ) ≥ 0 in int ( S ξ ) . min { H i j [ Z ] ( x ) , Z ( x ) − h ξ ( x ) } ≤ 0 in int ( S η ) . \min\left\{\mathcal{H}^{ij}[\mathcal{Z}](x),\mathcal{Z}(x)-h^\xi(x)\right\}\leq0 \quad\text{in } \operatorname{int}(S^\eta). min { H ij [ Z ] ( x ) , Z ( x ) − h ξ ( x ) } ≤ 0 in int ( S η ) . Then Z \mathcal{Z} Z
max { min [ H i j [ Z ] ( x ) , Z ( x ) − h ξ ( x ) ] , Z ( x ) − h η ( x ) } = 0 on R + ∗ . (14) \max\left\{\min\left[\mathcal{H}^{ij}[\mathcal{Z}](x),\mathcal{Z}(x)-h^\xi(x)\right],\mathcal{Z}(x)-h^\eta(x)\right\}=0 \quad\text{on } \mathbb{R}_+^*. \tag*{(14)} max { min [ H ij [ Z ] ( x ) , Z ( x ) − h ξ ( x ) ] , Z ( x ) − h η ( x ) } = 0 on R + ∗ . (14) Proof 3.2 Let x ˉ ∈ R + ∗ \bar{x}\in\mathbb{R}_+^* x ˉ ∈ R + ∗
(a) Case x ˉ ∈ C \bar{x}\in C x ˉ ∈ C
Since Z ∈ C 2 ( C ) \mathcal{Z}\in C^2(C) Z ∈ C 2 ( C )
h ξ ≤ Z ≤ h η on C , h^\xi\leq\mathcal{Z}\leq h^\eta\quad\text{on } C, h ξ ≤ Z ≤ h η on C , it follows that Z \mathcal{Z} Z C C C
(b) Case x ˉ ∈ S ξ \bar{x}\in S^\xi x ˉ ∈ S ξ
We focus on the boundary case x ˉ ∈ ∂ S ξ \bar{x}\in\partial S^\xi x ˉ ∈ ∂ S ξ
(i) Disjoint boundaries ∂ S ξ ∩ ∂ S η = ∅ \partial S^\xi\cap\partial S^\eta=\varnothing ∂ S ξ ∩ ∂ S η = ∅
Without loss of generality, assume that x ˉ \bar{x} x ˉ S ξ S^\xi S ξ ϵ > 0 \epsilon>0 ϵ > 0
( x ˉ − ϵ , x ˉ ) ⊂ C , (\bar{x}-\epsilon,\bar{x})\subset C, ( x ˉ − ϵ , x ˉ ) ⊂ C , where Z ∈ C 2 \mathcal{Z}\in C^2 Z ∈ C 2
Let ϕ ∈ C 2 \phi\in C^2 ϕ ∈ C 2 x ˉ \bar{x} x ˉ Z − ϕ \mathcal{Z}-\phi Z − ϕ Z ∈ C 1 \mathcal{Z}\in C^1 Z ∈ C 1 S ξ S^\xi S ξ
Z ′ ( x ˉ ) = ϕ ′ ( x ˉ ) . \mathcal{Z}'(\bar{x})=\phi'(\bar{x}). Z ′ ( x ˉ ) = ϕ ′ ( x ˉ ) . By Taylor's formula,
Z ( x ˉ − λ ) = Z ( x ˉ ) − λ ∫ 0 1 Z ′ ( x ˉ − t λ ) d t , \mathcal{Z}(\bar{x}-\lambda)=\mathcal{Z}(\bar{x})-\lambda\int_0^1\mathcal{Z}'(\bar{x}-t\lambda)\,dt, Z ( x ˉ − λ ) = Z ( x ˉ ) − λ ∫ 0 1 Z ′ ( x ˉ − t λ ) d t , ϕ ( x ˉ − λ ) = ϕ ( x ˉ ) − λ ∫ 0 1 ϕ ′ ( x ˉ − t λ ) d t . \phi(\bar{x}-\lambda)=\phi(\bar{x})-\lambda\int_0^1\phi'(\bar{x}-t\lambda)\,dt. ϕ ( x ˉ − λ ) = ϕ ( x ˉ ) − λ ∫ 0 1 ϕ ′ ( x ˉ − t λ ) d t . Since Z − ϕ \mathcal{Z}-\phi Z − ϕ x ˉ \bar{x} x ˉ
∫ 0 1 ( ϕ ′ ( x ˉ − t λ ) − Z ′ ( x ˉ − t λ ) ) d t ≥ 0 , ∀ λ ∈ ( 0 , ϵ ) . (15) \int_0^1\left(\phi'(\bar{x}-t\lambda)-\mathcal{Z}'(\bar{x}-t\lambda)\right)\,dt\geq0,\quad\forall\lambda\in(0,\epsilon). \tag*{(15)} ∫ 0 1 ( ϕ ′ ( x ˉ − t λ ) − Z ′ ( x ˉ − t λ ) ) d t ≥ 0 , ∀ λ ∈ ( 0 , ϵ ) . (15) Using (13) , we have H i j [ Z ] ( x ) = 0 \mathcal{H}^{ij}[\mathcal{Z}](x)=0 H ij [ Z ] ( x ) = 0 ( x ˉ − ϵ , x ˉ ) (\bar{x}-\epsilon,\bar{x}) ( x ˉ − ϵ , x ˉ ) Z ′ ′ ( x ˉ − ) \mathcal{Z}^{\prime\prime}(\bar{x}^{-}) Z ′′ ( x ˉ − )
r Z ( x ˉ ) − b i j x ˉ Z ′ ( x ˉ ) − 1 2 σ i j 2 x ˉ 2 Z ′ ′ ( x ˉ − ) − f i j ( x ˉ ) = 0. (16) r\mathcal{Z}(\bar{x})-b_{ij}\bar{x}\mathcal{Z}^{\prime}(\bar{x})-\frac{1}{2}\sigma_{ij}^{2}\bar{x}^{2}\mathcal{Z}^{\prime\prime}(\bar{x}^{-})-f_{ij}(\bar{x})=0. \tag*{(16)} r Z ( x ˉ ) − b ij x ˉ Z ′ ( x ˉ ) − 2 1 σ ij 2 x ˉ 2 Z ′′ ( x ˉ − ) − f ij ( x ˉ ) = 0. (16) Letting λ → 0 \lambda\to0 λ → 0 (15) , we obtain
ϕ ′ ′ ( x ˉ ) ≤ Z ′ ′ ( x ˉ − ) . \phi^{\prime\prime}(\bar{x})\leq\mathcal{Z}^{\prime\prime}(\bar{x}^{-}). ϕ ′′ ( x ˉ ) ≤ Z ′′ ( x ˉ − ) . Substituting into (16) yields
H i j [ Z ] ( x ˉ ) ≥ 0 , \mathcal{H}^{ij}[\mathcal{Z}](\bar{x})\geq0, H ij [ Z ] ( x ˉ ) ≥ 0 , which proves the supersolution property.
(ii) Non-disjoint boundaries ∂ S ξ ∩ ∂ S η ≠ ∅ \partial S^{\xi}\cap\partial S^{\eta}\ne\varnothing ∂ S ξ ∩ ∂ S η = ∅
Consider x ˉ ∈ ∂ S ξ ∩ ∂ S η \bar{x}\in\partial S^{\xi}\cap\partial S^{\eta} x ˉ ∈ ∂ S ξ ∩ ∂ S η
Z = h ξ = h η . \mathcal{Z}=h^{\xi}=h^{\eta}. Z = h ξ = h η . If H i j [ Z ] ( x ˉ ) ≥ 0 \mathcal{H}^{ij}[\mathcal{Z}](\bar{x})\geq0 H ij [ Z ] ( x ˉ ) ≥ 0
max { min [ H i j [ Z ] , Z − h ξ ] , Z − h η } = 0. \max\left\{\min\left[\mathcal{H}^{ij}[\mathcal{Z}],\mathcal{Z}-h^{\xi}\right],\mathcal{Z}-h^{\eta}\right\}=0. max { min [ H ij [ Z ] , Z − h ξ ] , Z − h η } = 0. If H i j [ Z ] ( x ˉ ) ≤ 0 \mathcal{H}^{ij}[\mathcal{Z}](\bar{x})\leq0 H ij [ Z ] ( x ˉ ) ≤ 0
max { min [ H i j [ Z ] , Z − h ξ ] , Z − h η } = 0. \max\left\{\min\left[\mathcal{H}^{ij}[\mathcal{Z}],\mathcal{Z}-h^{\xi}\right],\mathcal{Z}-h^{\eta}\right\}=0. max { min [ H ij [ Z ] , Z − h ξ ] , Z − h η } = 0. Thus, Z \mathcal{Z} Z (14) on the intersection.
(c) Case x ˉ ∈ S η \bar{x}\in S^{\eta} x ˉ ∈ S η
The proof follows by symmetric arguments as in case (b).
Therefore, Z \mathcal{Z} Z (14) . □
Explicit Solution of the Switching Problem
In this part, we solve the switching game problem in the three-regime case (d = 3 d=3 d = 3 v i j ( x ) v_{ij}(x) v ij ( x )
\max\left\{\min\left[\mathcal{H}[v_{00](x),v_{00}(x)-M_{00}[v](x)\right],v_{00}(x)-N_{00}[v](x)\right\}=0,} \tag*{(17)}
max { min [ H [ v 01 ] ( x ) , v 01 ( x ) − M 01 [ v ] ( x ) ] , v 01 ( x ) − N 01 [ v ] ( x ) } = 0 , (18) \max\left\{\min\left[\mathcal{H}[v_{01}](x),v_{01}(x)-M_{01}[v](x)\right],v_{01}(x)-N_{01}[v](x)\right\}=0, \tag*{(18)} max { min [ H [ v 01 ] ( x ) , v 01 ( x ) − M 01 [ v ] ( x ) ] , v 01 ( x ) − N 01 [ v ] ( x ) } = 0 , (18) max { min [ H [ v 02 ] ( x ) , v 02 ( x ) − M 02 [ v ] ( x ) ] , v 02 ( x ) − N 02 [ v ] ( x ) } = 0 , (19) \max\left\{\min\left[\mathcal{H}[v_{02}](x),v_{02}(x)-M_{02}[v](x)\right],v_{02}(x)-N_{02}[v](x)\right\}=0, \tag*{(19)} max { min [ H [ v 02 ] ( x ) , v 02 ( x ) − M 02 [ v ] ( x ) ] , v 02 ( x ) − N 02 [ v ] ( x ) } = 0 , (19) max { min [ H [ v 10 ] ( x ) , v 10 ( x ) − M 10 [ v ] ( x ) ] , v 10 ( x ) − N 10 [ v ] ( x ) } = 0 , (20) \max\left\{\min\left[\mathcal{H}[v_{10}](x),v_{10}(x)-M_{10}[v](x)\right],v_{10}(x)-N_{10}[v](x)\right\}=0, \tag*{(20)} max { min [ H [ v 10 ] ( x ) , v 10 ( x ) − M 10 [ v ] ( x ) ] , v 10 ( x ) − N 10 [ v ] ( x ) } = 0 , (20) max { min [ H [ v 11 ] ( x ) , v 11 ( x ) − M 11 [ v ] ( x ) ] , v 11 ( x ) − N 11 [ v ] ( x ) } = 0 , (21) \max\left\{\min\left[\mathcal{H}[v_{11}](x),v_{11}(x)-M_{11}[v](x)\right],v_{11}(x)-N_{11}[v](x)\right\}=0, \tag*{(21)} max { min [ H [ v 11 ] ( x ) , v 11 ( x ) − M 11 [ v ] ( x ) ] , v 11 ( x ) − N 11 [ v ] ( x ) } = 0 , (21) max { min [ H [ v 12 ] ( x ) , v 12 ( x ) − M 12 [ v ] ( x ) ] , v 12 ( x ) − N 12 [ v ] ( x ) } = 0 , (22) \max\left\{\min\left[\mathcal{H}[v_{12}](x),v_{12}(x)-M_{12}[v](x)\right],v_{12}(x)-N_{12}[v](x)\right\}=0, \tag*{(22)} max { min [ H [ v 12 ] ( x ) , v 12 ( x ) − M 12 [ v ] ( x ) ] , v 12 ( x ) − N 12 [ v ] ( x ) } = 0 , (22) max { min [ H [ v 20 ] ( x ) , v 20 ( x ) − M 20 [ v ] ( x ) ] , v 20 ( x ) − N 20 [ v ] ( x ) } = 0 , (23) \max\left\{\min\left[\mathcal{H}[v_{20}](x),v_{20}(x)-M_{20}[v](x)\right],v_{20}(x)-N_{20}[v](x)\right\}=0, \tag*{(23)} max { min [ H [ v 20 ] ( x ) , v 20 ( x ) − M 20 [ v ] ( x ) ] , v 20 ( x ) − N 20 [ v ] ( x ) } = 0 , (23) max { min [ H [ v 21 ] ( x ) , v 21 ( x ) − M 21 [ v ] ( x ) ] , v 21 ( x ) − N 21 [ v ] ( x ) } = 0 , (24) \max\left\{\min\left[\mathcal{H}[v_{21}](x),v_{21}(x)-M_{21}[v](x)\right],v_{21}(x)-N_{21}[v](x)\right\}=0, \tag*{(24)} max { min [ H [ v 21 ] ( x ) , v 21 ( x ) − M 21 [ v ] ( x ) ] , v 21 ( x ) − N 21 [ v ] ( x ) } = 0 , (24) max { min [ H [ v 22 ] ( x ) , v 22 ( x ) − M 22 [ v ] ( x ) ] , v 22 ( x ) − N 22 [ v ] ( x ) } = 0. (25) \max\left\{\min\left[\mathcal{H}[v_{22}](x),v_{22}(x)-M_{22}[v](x)\right],v_{22}(x)-N_{22}[v](x)\right\}=0. \tag*{(25)} max { min [ H [ v 22 ] ( x ) , v 22 ( x ) − M 22 [ v ] ( x ) ] , v 22 ( x ) − N 22 [ v ] ( x ) } = 0. (25) We choose the following structure for the running payoff:
f i j ( x ) = x γ , 0 < γ < 1. (26) f_{ij}(x)=x^{\gamma},\qquad0<\gamma<1. \tag*{(26)} f ij ( x ) = x γ , 0 < γ < 1. (26) As shown in [7 ] , the function
V ^ i j ( x ) = E [ ∫ 0 ∞ e − r t f i j ( X t x , i j ) d t ] , \widehat{V}_{ij}(x) = \mathbb{E}\left[\int_{0}^{\infty} e^{-rt} f_{ij}\left(X_{t}^{x,ij}\right)\,dt\right], V ij ( x ) = E [ ∫ 0 ∞ e − r t f ij ( X t x , ij ) d t ] , is a particular solution of
H [ v i j ] ( x ) = 0. (27) \mathcal{H}[v_{ij}](x) = 0. \tag*{(27)} H [ v ij ] ( x ) = 0. (27) In particular, when f i j ( x ) = x γ f_{ij}(x) = x^{\gamma} f ij ( x ) = x γ
V ^ i j ( x ) = K i j x γ , K i j = 1 r − b i j γ + 1 2 σ i j 2 γ ( 1 − γ ) > 0 , i , j ∈ { 0 , 1 , 2 } . \widehat{V}_{ij}(x) = K_{ij}x^{\gamma}, \qquad K_{ij} = \frac{1}{r-b_{ij}\gamma+\frac{1}{2}\sigma_{ij}^{2}\gamma(1-\gamma)} > 0,\qquad i,j\in\{0,1,2\}. V ij ( x ) = K ij x γ , K ij = r − b ij γ + 2 1 σ ij 2 γ ( 1 − γ ) 1 > 0 , i , j ∈ { 0 , 1 , 2 } . Under the above assumptions, the general solution of (27) is given by
v ( x ) = A i j x m i j + + B i j x m i j − , v(x) = A_{ij}x^{m_{ij}^{+}} + B_{ij}x^{m_{ij}^{-}}, v ( x ) = A ij x m ij + + B ij x m ij − , where A i j , B i j A_{ij}, B_{ij} A ij , B ij
m i j + = − b i j σ i j 2 + 1 2 + ( − b i j σ i j 2 + 1 2 ) 2 + 2 r σ i j 2 > 1 , (28) m_{ij}^{+} = -\frac{b_{ij}}{\sigma_{ij}^{2}}+\frac{1}{2}+\sqrt{\left(-\frac{b_{ij}}{\sigma_{ij}^{2}}+\frac{1}{2}\right)^{2}+\frac{2r}{\sigma_{ij}^{2}}}>1, \tag*{(28)} m ij + = − σ ij 2 b ij + 2 1 + ( − σ ij 2 b ij + 2 1 ) 2 + σ ij 2 2 r > 1 , (28) m i j − = − b i j σ i j 2 + 1 2 − ( − b i j σ i j 2 + 1 2 ) 2 + 2 r σ i j 2 < 0. (29) m_{ij}^{-} = -\frac{b_{ij}}{\sigma_{ij}^{2}}+\frac{1}{2}-\sqrt{\left(-\frac{b_{ij}}{\sigma_{ij}^{2}}+\frac{1}{2}\right)^{2}+\frac{2r}{\sigma_{ij}^{2}}}<0. \tag*{(29)} m ij − = − σ ij 2 b ij + 2 1 − ( − σ ij 2 b ij + 2 1 ) 2 + σ ij 2 2 r < 0. (29) We also assume that
K a b = K c d ⟺ { m a b − = m c d − } ∩ { m a b + = m c d + } . K_{ab}=K_{cd}\Longleftrightarrow\{m_{ab}^{-}=m_{cd}^{-}\}\cap\{m_{ab}^{+}=m_{cd}^{+}\}. K ab = K c d ⟺ { m ab − = m c d − } ∩ { m ab + = m c d + } . With regard to (28) , (29) , and linear growth property of the value functions (see [7 ] ), we guess that if both players are in their continuation region in the neighborhood of 0 0 0 B i j = 0 B_{ij}=0 B ij = 0 + ∞ +\infty + ∞ A i j = 0 A_{ij}=0 A ij = 0
Theorem 4.1 Case where K i j K_{ij} K ij
Let v i j ( x ) = V ^ i j ( x ) v_{ij}(x)=\widehat{V}_{ij}(x) v ij ( x ) = V ij ( x ) i , j ∈ { 0 , 1 , 2 } i,j\in\{0,1,2\} i , j ∈ { 0 , 1 , 2 } v i j v_{ij} v ij
S i j ξ = ∅ , C i j ξ = ( 0 ; + ∞ ) , S i j η = ∅ , C i j η = ( 0 ; + ∞ ) . S_{ij}^{\xi}=\varnothing,\qquad C_{ij}^{\xi}=(0;+\infty),\qquad S_{ij}^{\eta}=\varnothing,\qquad C_{ij}^{\eta}=(0;+\infty). S ij ξ = ∅ , C ij ξ = ( 0 ; + ∞ ) , S ij η = ∅ , C ij η = ( 0 ; + ∞ ) . Proof 4.1 Since V ^ i j ( x ) \widehat{V}_{ij}(x) V ij ( x ) (27) , it follows that
H [ v i j ] ( x ) = 0. \mathcal{H}[v_{ij}](x)=0. H [ v ij ] ( x ) = 0. Moreover,
v i j ( x ) − M [ v i j ] ( x ) = V ^ i j ( x ) − ( V ^ i j ( x ) − c i j ) = c i j , v_{ij}(x)-M[v_{ij}](x)=\widehat{V}_{ij}(x)-\left(\widehat{V}_{ij}(x)-c_{ij}\right)=c_{ij}, v ij ( x ) − M [ v ij ] ( x ) = V ij ( x ) − ( V ij ( x ) − c ij ) = c ij , and since c i j > 0 c_{ij}>0 c ij > 0 we have v i j ( x ) − M [ v i j ] ( x ) ≥ 0 v_{ij}(x)-M[v_{ij}](x)\geq0 v ij ( x ) − M [ v ij ] ( x ) ≥ 0
Similarly,
v i j ( x ) − N [ v i j ] ( x ) = V ^ i j ( x ) − ( V ^ i i ( x ) + χ j i ) = − χ j i , v_{ij}(x)-N[v_{ij}](x)=\widehat{V}_{ij}(x)-\left(\widehat{V}_{ii}(x)+\chi_{ji}\right)=-\chi_{ji}, v ij ( x ) − N [ v ij ] ( x ) = V ij ( x ) − ( V ii ( x ) + χ j i ) = − χ j i , and since χ j i > 0 \chi_{ji}>0 χ j i > 0 we obtain v i j ( x ) − N [ v i j ] ( x ) ≤ 0 v_{ij}(x)-N[v_{ij}](x)\leq0 v ij ( x ) − N [ v ij ] ( x ) ≤ 0
Therefore, all the conditions of Lemma 3.1 are satisfied, which implies that v i j ( x ) v_{ij}(x) v ij ( x )
Theorem 4.2 Case where K 0 j = C s t 1 < K i j = C s t 2 K_{0j}=Cst_1<K_{ij}=Cst_2 K 0 j = C s t 1 < K ij = C s t 2 i = 1 , 2 i=1,2 i = 1 , 2 j = 0 , 1 , 2 j=0,1,2 j = 0 , 1 , 2
Let v i j ( x ) v_{ij}(x) v ij ( x )
v 0 j ( x ) = ( V ^ 0 j ( x ) + A 0 j x m 0 j + ) 1 x < x 0 j + ( V ^ 1 j ( x ) − min { c 01 , c 02 } ) 1 x ≥ x 0 j , j = 0 , 1 , 2 (30) v_{0j}(x)=\left(\widehat{V}_{0j}(x)+A_{0j}x^{m_{0j}^{+}}\right)\mathbf{1}_{x<x_{0j}}+\left(\widehat{V}_{1j}(x)-\min\{c_{01},c_{02}\}\right)\mathbf{1}_{x\geq x_{0j}},\quad j=0,1,2 \tag*{(30)} v 0 j ( x ) = ( V 0 j ( x ) + A 0 j x m 0 j + ) 1 x < x 0 j + ( V 1 j ( x ) − min { c 01 , c 02 } ) 1 x ≥ x 0 j , j = 0 , 1 , 2 (30) v i j ( x ) = V ^ i j ( x ) , i = 1 , 2 , j = 0 , 1 , 2. (31) v_{ij}(x)=\widehat{V}_{ij}(x),\quad i=1,2,\ j=0,1,2. \tag*{(31)} v ij ( x ) = V ij ( x ) , i = 1 , 2 , j = 0 , 1 , 2. (31) where x i j x_{ij} x ij A i j A_{ij} A ij C 1 C^1 C 1
( K 1 j − K 0 j ) ( x 0 j ) γ = m 0 j + m 0 j + − γ min { c 01 , c 02 } , A 0 j = ( K 1 j − K 0 j ) γ m 0 j + ( x 0 j ) γ − m 0 j + . (K_{1j}-K_{0j})(x_{0j})^\gamma=\frac{m_{0j}^{+}}{m_{0j}^{+}-\gamma}\min\{c_{01},c_{02}\},\qquad A_{0j}=(K_{1j}-K_{0j})\frac{\gamma}{m_{0j}^{+}}(x_{0j})^{\gamma-m_{0j}^{+}}. ( K 1 j − K 0 j ) ( x 0 j ) γ = m 0 j + − γ m 0 j + min { c 01 , c 02 } , A 0 j = ( K 1 j − K 0 j ) m 0 j + γ ( x 0 j ) γ − m 0 j + . Such v i j ( x ) v_{ij}(x) v ij ( x )
S 0 j ξ = [ x 0 j , + ∞ ) , C 0 j ξ = ( 0 , x 0 j ) , S 0 j ξ = ∅ , C 0 j ξ = ( 0 , + ∞ ) , j = 0 , 1 , 2. S_{0j}^{\xi}=[x_{0j},+\infty),\quad C_{0j}^{\xi}=(0,x_{0j}),\quad S_{0j}^{\xi}=\varnothing,\quad C_{0j}^{\xi}=(0,+\infty),\quad j=0,1,2. S 0 j ξ = [ x 0 j , + ∞ ) , C 0 j ξ = ( 0 , x 0 j ) , S 0 j ξ = ∅ , C 0 j ξ = ( 0 , + ∞ ) , j = 0 , 1 , 2. S i j ξ = ∅ , C i j ξ = ( 0 , + ∞ ) , S i j η = ∅ , C i j η = ( 0 , + ∞ ) . i = 1 , 2 , j = 0 , 1 , 2. S_{ij}^{\xi}=\varnothing,\quad C_{ij}^{\xi}=(0,+\infty),\quad S_{ij}^{\eta}=\varnothing,\quad C_{ij}^{\eta}=(0,+\infty).\quad i=1,2,\quad j=0,1,2. S ij ξ = ∅ , C ij ξ = ( 0 , + ∞ ) , S ij η = ∅ , C ij η = ( 0 , + ∞ ) . i = 1 , 2 , j = 0 , 1 , 2. We can see that by design
v 00 ( x ) = v 01 ( x ) = v 02 ( x ) . v_{00}(x)=v_{01}(x)=v_{02}(x). v 00 ( x ) = v 01 ( x ) = v 02 ( x ) . Proof 4.2 We only verify the QVI for v 00 v_{00} v 00 v 10 v_{10} v 10
Step 1. Verification for v 00 v_{00} v 00
v 00 ( x ) − M 00 [ v ] ( x ) = v 00 ( x ) − max ( v 10 ( x ) − c 01 , v 20 ( x ) − c 02 ) . v_{00}(x)-M_{00}[v](x)=v_{00}(x)-\max(v_{10}(x)-c_{01},v_{20}(x)-c_{02}). v 00 ( x ) − M 00 [ v ] ( x ) = v 00 ( x ) − max ( v 10 ( x ) − c 01 , v 20 ( x ) − c 02 ) . max ( v 10 ( x ) − c 01 , v 20 ( x ) − c 02 ) = v 10 ( x ) − min { c 01 , c 02 } , \max(v_{10}(x)-c_{01},v_{20}(x)-c_{02})=v_{10}(x)-\min\{c_{01},c_{02}\}, max ( v 10 ( x ) − c 01 , v 20 ( x ) − c 02 ) = v 10 ( x ) − min { c 01 , c 02 } , then
v 00 ( x ) − M 00 [ v ] ( x ) = V ^ 00 ( x ) + A 00 x m 00 + − V ^ 10 ( x ) + min { c 01 , c 02 } . v_{00}(x)-M_{00}[v](x)=\widehat{V}_{00}(x)+A_{00}x^{m_{00}^{+}}-\widehat{V}_{10}(x)+\min\{c_{01},c_{02}\}. v 00 ( x ) − M 00 [ v ] ( x ) = V 00 ( x ) + A 00 x m 00 + − V 10 ( x ) + min { c 01 , c 02 } . Define
h 1 ( x ) = V ^ 00 ( x ) + A 00 x m 00 + − V ^ 10 ( x ) + min { c 01 , c 02 } . h_1(x)=\widehat{V}_{00}(x)+A_{00}x^{m_{00}^{+}}-\widehat{V}_{10}(x)+\min\{c_{01},c_{02}\}. h 1 ( x ) = V 00 ( x ) + A 00 x m 00 + − V 10 ( x ) + min { c 01 , c 02 } . A straightforward computation shows that h 1 ′ ′ ( x ) > 0 h_1''(x)>0 h 1 ′′ ( x ) > 0 h 1 ′ ( x ) h_1'(x) h 1 ′ ( x ) h 1 ′ ( x 00 ) = 0 h_1'(x_{00})=0 h 1 ′ ( x 00 ) = 0 h 1 ′ ( x ) < 0 h_1'(x)<0 h 1 ′ ( x ) < 0 x < x 00 x<x_{00} x < x 00 h 1 ( x 00 ) = 0 h_1(x_{00})=0 h 1 ( x 00 ) = 0 h 1 ( x ) > 0 h_1(x)>0 h 1 ( x ) > 0 x < x 00 x<x_{00} x < x 00
v 00 ( x ) − M 00 [ v ] ( x ) > 0. v_{00}(x)-M_{00}[v](x)>0. v 00 ( x ) − M 00 [ v ] ( x ) > 0. Next,
v 00 ( x ) − N 00 [ v ] ( x ) = v 00 ( x ) − min ( v 01 ( x ) + χ 01 , v 02 ( x ) + χ 02 ) = − min ( χ 01 , χ 02 ) < 0. v_{00}(x)-N_{00}[v](x)=v_{00}(x)-\min(v_{01}(x)+\chi_{01},v_{02}(x)+\chi_{02})=-\min(\chi_{01},\chi_{02})<0. v 00 ( x ) − N 00 [ v ] ( x ) = v 00 ( x ) − min ( v 01 ( x ) + χ 01 , v 02 ( x ) + χ 02 ) = − min ( χ 01 , χ 02 ) < 0. Thus, both obstacle inequalities are satisfied, and by Lemma 3.1, v 00 v_{00} v 00
For x ≥ x 00 x \ge x_{00} x ≥ x 00 v 00 ( x ) = V ^ 10 ( x ) − min ( c 01 , c 02 ) v_{00}(x) = \widehat{V}_{10}(x) - \min(c_{01},c_{02}) v 00 ( x ) = V 10 ( x ) − min ( c 01 , c 02 )
v 00 ( x ) − M 00 [ v ] ( x ) = 0. v_{00}(x) - M_{00}[v](x) = 0. v 00 ( x ) − M 00 [ v ] ( x ) = 0. Moreover, a straightforward calculus shows that
H [ v 00 ] ( x ) = K 10 − K 00 K 00 x γ − c 02 > 0. \mathcal{H}[v_{00}](x) = \frac{K_{10} - K_{00}}{K_{00}}x^\gamma- c_{02} > 0. H [ v 00 ] ( x ) = K 00 K 10 − K 00 x γ − c 02 > 0. Next,
v 00 ( x ) − N 00 [ v ] ( x ) = − min ( χ 01 , χ 02 ) < 0. v_{00}(x) - N_{00}[v](x) = -\min(\chi_{01},\chi_{02}) < 0. v 00 ( x ) − N 00 [ v ] ( x ) = − min ( χ 01 , χ 02 ) < 0. Hence, v 00 v_{00} v 00
Step 2. Verification for v 10 v_{10} v 10 We have H [ v 10 ] ( x ) = 0 \mathcal{H}[v_{10}](x) = 0 H [ v 10 ] ( x ) = 0 x > 0 x > 0 x > 0
Moreover,
v 10 ( x ) − M 10 [ v ] ( x ) = v 10 ( x ) − max ( v 00 ( x ) − c 10 , v 20 ( x ) − c 12 ) . v_{10}(x) - M_{10}[v](x) = v_{10}(x) - \max(v_{00}(x) - c_{10}, v_{20}(x) - c_{12}). v 10 ( x ) − M 10 [ v ] ( x ) = v 10 ( x ) − max ( v 00 ( x ) − c 10 , v 20 ( x ) − c 12 ) . h 2 ( x ) = v 10 ( x ) − v 00 ( x ) + c 10 . h_2(x) = v_{10}(x) - v_{00}(x) + c_{10}. h 2 ( x ) = v 10 ( x ) − v 00 ( x ) + c 10 . Then
h 2 ( x ) = ( K 10 − K 00 ) x γ − A 00 x m 00 + + c 10 . h_2(x) = (K_{10} - K_{00})x^\gamma- A_{00}x^{m_{00}^{+}} + c_{10}. h 2 ( x ) = ( K 10 − K 00 ) x γ − A 00 x m 00 + + c 10 . Since h 2 ′ ′ ( x ) < 0 h_2^{\prime\prime}(x) < 0 h 2 ′′ ( x ) < 0 h 2 ′ h_2^{\prime} h 2 ′
Using
h 2 ′ ( x 00 ) = 0 , h_2^{\prime}(x_{00}) = 0, h 2 ′ ( x 00 ) = 0 , we obtain
h 2 ′ ( x ) > 0 for x < x 00 , h_2^{\prime}(x) > 0 \text{ for } x < x_{00}, h 2 ′ ( x ) > 0 for x < x 00 , and hence h 2 h_2 h 2
Since h 2 ( 0 ) = c 10 > 0 h_2(0) = c_{10} > 0 h 2 ( 0 ) = c 10 > 0
v 10 ( x ) − M 10 [ v ] ( x ) > 0. v_{10}(x) - M_{10}[v](x) > 0. v 10 ( x ) − M 10 [ v ] ( x ) > 0. v 10 ( x ) − M 10 [ v ] ( x ) = min ( min { c 01 , c 02 } + c 10 , c 12 ) > 0. v_{10}(x) - M_{10}[v](x) = \min(\min\{c_{01},c_{02}\} + c_{10},c_{12}) > 0. v 10 ( x ) − M 10 [ v ] ( x ) = min ( min { c 01 , c 02 } + c 10 , c 12 ) > 0. Finally,
v 10 ( x ) − N 10 [ v ] ( x ) = − min ( χ 01 , χ 02 ) < 0. v_{10}(x) - N_{10}[v](x) = -\min(\chi_{01},\chi_{02}) < 0. v 10 ( x ) − N 10 [ v ] ( x ) = − min ( χ 01 , χ 02 ) < 0. Thus, v 10 v_{10} v 10
Theorem 4.3 Case where K i 0 = C s t 1 < K i j = C s t 2 K_{i0} = Cst_1 < K_{ij} = Cst_2 K i 0 = C s t 1 < K ij = C s t 2 i = 0 , 1 , 2 i = 0,1,2 i = 0 , 1 , 2 j = 1 , 2 j = 1,2 j = 1 , 2
Let v i j ( x ) v_{ij}(x) v ij ( x )
v i 0 ( x ) = V ^ i 0 ( x ) , i = 0 , 1 , 2 (32) v_{i0}(x) = \widehat{V}_{i0}(x), \quad i = 0,1,2 \tag*{(32)} v i 0 ( x ) = V i 0 ( x ) , i = 0 , 1 , 2 (32) v i j ( x ) = ( V ^ i j ( x ) + A i j x m i j + ) 1 x < x i j + ( V ^ i 0 ( x ) + χ j 0 ) 1 x ≥ x i j , i = 0 , 1 , 2 , j = 1 , 2. (33) v_{ij}(x) = \left(\widehat{V}_{ij}(x) + A_{ij}x^{m_{ij}^{+}}\right)1_{x<x_{ij}} + \left(\widehat{V}_{i0}(x) + \chi_{j0}\right)1_{x\geq x_{ij}}, \quad i = 0,1,2,\ j = 1,2. \tag*{(33)} v ij ( x ) = ( V ij ( x ) + A ij x m ij + ) 1 x < x ij + ( V i 0 ( x ) + χ j 0 ) 1 x ≥ x ij , i = 0 , 1 , 2 , j = 1 , 2. (33) Where x i j x_{ij} x ij A i j A_{ij} A ij C 1 C^1 C 1
( K i j − K i 0 ) ( x i j ) γ = m i j + m i j + − γ χ j 0 , A i j = ( K i 0 − K i j ) γ m i j + ( x i j ) γ − m i j + . (K_{ij} - K_{i0})(x_{ij})^\gamma= \frac{m_{ij}^{+}}{m_{ij}^{+} - \gamma}\chi_{j0}, \qquad A_{ij} = (K_{i0} - K_{ij})\frac{\gamma}{m_{ij}^{+}}(x_{ij})^{\gamma-m_{ij}^{+}}. ( K ij − K i 0 ) ( x ij ) γ = m ij + − γ m ij + χ j 0 , A ij = ( K i 0 − K ij ) m ij + γ ( x ij ) γ − m ij + . Such v i j ( x ) v_{ij}(x) v ij ( x )
With:
S i 0 ξ = ∅ , C i 0 ξ = ( 0 ; + ∞ ) , S i 0 η = ∅ , C i 0 η = ( 0 ; + ∞ ) , i = 0 , 1 , 2 S_{i0}^{\xi} = \varnothing,\quad C_{i0}^{\xi} = (0;+\infty),\quad S_{i0}^{\eta} = \varnothing,\quad C_{i0}^{\eta} = (0;+\infty),\quad i = 0,1,2 S i 0 ξ = ∅ , C i 0 ξ = ( 0 ; + ∞ ) , S i 0 η = ∅ , C i 0 η = ( 0 ; + ∞ ) , i = 0 , 1 , 2 S i j ξ = ∅ , C i j ξ = ( 0 ; + ∞ ) , S i j η = [ x i j , ∞ ) , C i j η = ( 0 ; x i j ) , i = 0 , 1 , 2 , j = 1 , 2. S_{ij}^{\xi} = \varnothing,\quad C_{ij}^{\xi} = (0;+\infty),\quad S_{ij}^{\eta} = [x_{ij},\infty),\quad C_{ij}^{\eta} = (0;x_{ij}),\quad i = 0,1,2,\quad j = 1,2. S ij ξ = ∅ , C ij ξ = ( 0 ; + ∞ ) , S ij η = [ x ij , ∞ ) , C ij η = ( 0 ; x ij ) , i = 0 , 1 , 2 , j = 1 , 2. We can see that, by design,
v 01 ( x ) = v 11 ( x ) = v 21 ( x ) v_{01}(x) = v_{11}(x) = v_{21}(x) v 01 ( x ) = v 11 ( x ) = v 21 ( x ) and
v 02 ( x ) = v 12 ( x ) = v 22 ( x ) . v_{02}(x) = v_{12}(x) = v_{22}(x). v 02 ( x ) = v 12 ( x ) = v 22 ( x ) . Proof 4.3 We only verify the QVI for v 00 v_{00} v 00 v 20 v_{20} v 20
Step 1. Verification for v 00 v_{00} v 00
Since
v 00 ( x ) = V ^ 00 ( x ) , v_{00}(x) = \widehat{V}_{00}(x), v 00 ( x ) = V 00 ( x ) , we immediately have
H [ v 00 ] ( x ) = 0. \mathcal{H}[v_{00}](x) = 0. H [ v 00 ] ( x ) = 0. Furthermore,
v 00 ( x ) − M 00 [ v ] = v 00 ( x ) − max { v 10 ( x ) − c 01 , v 20 ( x ) − c 02 } . v_{00}(x) - M_{00}[v] = v_{00}(x) - \max\{v_{10}(x) - c_{01}, v_{20}(x) - c_{02}\}. v 00 ( x ) − M 00 [ v ] = v 00 ( x ) − max { v 10 ( x ) − c 01 , v 20 ( x ) − c 02 } . Given that
v 10 ( x ) = v 20 ( x ) , v_{10}(x) = v_{20}(x), v 10 ( x ) = v 20 ( x ) , then
v 00 ( x ) − M 00 [ v ] = v 00 ( x ) − v 10 ( x ) + min { c 01 , c 02 } = min { c 01 , c 02 } > 0. v_{00}(x) - M_{00}[v] = v_{00}(x) - v_{10}(x) + \min\{c_{01},c_{02}\} = \min\{c_{01},c_{02}\} > 0. v 00 ( x ) − M 00 [ v ] = v 00 ( x ) − v 10 ( x ) + min { c 01 , c 02 } = min { c 01 , c 02 } > 0. This proves
v 00 ( x ) − M 00 [ v ] > 0. v_{00}(x) - M_{00}[v] > 0. v 00 ( x ) − M 00 [ v ] > 0. Next,
v 00 ( x ) − N 00 [ v ] = v 00 ( x ) − min { v 01 ( x ) + χ 01 , v 02 ( x ) + χ 02 } = − min { χ 01 , χ 02 } < 0. v_{00}(x) - N_{00}[v] = v_{00}(x) - \min\{v_{01}(x) + \chi_{01}, v_{02}(x) + \chi_{02}\} = -\min\{\chi_{01},\chi_{02}\} < 0. v 00 ( x ) − N 00 [ v ] = v 00 ( x ) − min { v 01 ( x ) + χ 01 , v 02 ( x ) + χ 02 } = − min { χ 01 , χ 02 } < 0. Hence, v 00 v_{00} v 00
Step 2. Verification for v 11 v_{11} v 11
H [ v 11 ] ( x ) = 0. \mathcal{H}[v_{11}](x) = 0. H [ v 11 ] ( x ) = 0. Moreover,
v 11 ( x ) − M 11 [ v ] = v 11 ( x ) − max { v 01 ( x ) − c 10 , v 21 ( x ) − c 12 } = min { c 10 , c 12 } > 0. v_{11}(x) - M_{11}[v] = v_{11}(x) - \max\{v_{01}(x) - c_{10}, v_{21}(x) - c_{12}\} = \min\{c_{10}, c_{12}\} > 0. v 11 ( x ) − M 11 [ v ] = v 11 ( x ) − max { v 01 ( x ) − c 10 , v 21 ( x ) − c 12 } = min { c 10 , c 12 } > 0. Next,
v 11 ( x ) − N 11 [ v ] = v 11 ( x ) − min { v 10 ( x ) + χ 10 ; v 12 ( x ) + χ 12 } . v_{11}(x) - N_{11}[v] = v_{11}(x) - \min\{v_{10}(x) + \chi_{10}; v_{12}(x) + \chi_{12}\}. v 11 ( x ) − N 11 [ v ] = v 11 ( x ) − min { v 10 ( x ) + χ 10 ; v 12 ( x ) + χ 12 } . Let us determine
m ( x ) = min { v 10 ( x ) + χ 10 ; v 12 ( x ) + χ 12 } . m(x) = \min\{v_{10}(x) + \chi_{10}; v_{12}(x) + \chi_{12}\}. m ( x ) = min { v 10 ( x ) + χ 10 ; v 12 ( x ) + χ 12 } . x 11 x_{11} x 11 x 12 x_{12} x 12 χ 10 \chi_{10} χ 10 χ 20 \chi_{20} χ 20
Assume that x 11 < x 12 x_{11} < x_{12} x 11 < x 12
Let
h 3 ( x ) = V ^ 12 ( x ) + A 12 x m 12 + − V ^ 10 ( x ) − χ 20 , h_3(x) = \widehat{V}_{12}(x) + A_{12}x^{m_{12}^{+}} - \widehat{V}_{10}(x) - \chi_{20}, h 3 ( x ) = V 12 ( x ) + A 12 x m 12 + − V 10 ( x ) − χ 20 , and
h 4 ( x ) = v 12 ( x ) + χ 12 − ( v 10 ( x ) + χ 10 ) = V ^ 12 ( x ) + A 12 x m 12 + + χ 12 − V ^ 10 ( x ) − χ 10 . \begin{aligned}
h_4(x) &= v_{12}(x) + \chi_{12} - (v_{10}(x) + \chi_{10}) \\
&= \widehat{V}_{12}(x) + A_{12}x^{m_{12}^{+}} + \chi_{12} - \widehat{V}_{10}(x) - \chi_{10}.
\end{aligned} h 4 ( x ) = v 12 ( x ) + χ 12 − ( v 10 ( x ) + χ 10 ) = V 12 ( x ) + A 12 x m 12 + + χ 12 − V 10 ( x ) − χ 10 . We can see that
h 4 ( x ) = h 3 ( x ) + χ 20 + χ 12 − χ 10 . (34) h_4(x) = h_3(x) + \chi_{20} + \chi_{12} - \chi_{10}. \tag*{(34)} h 4 ( x ) = h 3 ( x ) + χ 20 + χ 12 − χ 10 . (34) A straightforward calculus shows that h 4 ′ ′ ( x ) < 0 h_4^{\prime\prime}(x) < 0 h 4 ′′ ( x ) < 0
Hence, h 4 ′ ( x ) h_4^{\prime}(x) h 4 ′ ( x )
h 4 ′ ( x 12 ) = 0 , h_4^{\prime}(x_{12}) = 0, h 4 ′ ( x 12 ) = 0 , it follows that h 4 ( x ) h_4(x) h 4 ( x )
We have, from 2.32,
h 4 ( 0 ) = χ 12 − χ 10 and h 4 ( x 12 ) = χ 20 + χ 12 − χ 10 . h_4(0) = \chi_{12} - \chi_{10} \quad\text{and} \quad h_4(x_{12}) = \chi_{20} + \chi_{12} - \chi_{10}. h 4 ( 0 ) = χ 12 − χ 10 and h 4 ( x 12 ) = χ 20 + χ 12 − χ 10 . m ( x ) = v 10 ( x ) + χ 10 . m(x) = v_{10}(x) + \chi_{10}. m ( x ) = v 10 ( x ) + χ 10 . In this case,
v 11 ( x ) − N 11 [ v ] = v 11 ( x ) − v 10 ( x ) − χ 10 = V ^ 11 ( x ) + A 11 x m 11 + − V ^ 10 ( x ) − χ 10 . \begin{aligned}
v_{11}(x) - N_{11}[v] &= v_{11}(x) - v_{10}(x) - \chi_{10} \\
&= \widehat{V}_{11}(x) + A_{11}x^{m_{11}^{+}} - \widehat{V}_{10}(x) - \chi_{10}.
\end{aligned} v 11 ( x ) − N 11 [ v ] = v 11 ( x ) − v 10 ( x ) − χ 10 = V 11 ( x ) + A 11 x m 11 + − V 10 ( x ) − χ 10 . Let
h 5 ( x ) = V ^ 11 ( x ) + A 11 x m 11 + − V ^ 10 ( x ) − χ 10 . h_5(x)=\widehat{V}_{11}(x)+A_{11}x^{m_{11}^{+}}-\widehat{V}_{10}(x)-\chi_{10}. h 5 ( x ) = V 11 ( x ) + A 11 x m 11 + − V 10 ( x ) − χ 10 . A straightforward calculus shows that h 5 ′ ′ ( x ) < 0 h_5''(x)<0 h 5 ′′ ( x ) < 0 hence h 5 ′ ( x ) h_5'(x) h 5 ′ ( x ) is decreasing.
Using
h 5 ′ ( x 11 ) = 0 , h_5'(x_{11})=0, h 5 ′ ( x 11 ) = 0 , it follows that
h 5 ′ ( x ) > 0 , x < x 11 . h_5'(x)>0,\quad x<x_{11}. h 5 ′ ( x ) > 0 , x < x 11 . Hence, h 5 ( x ) h_5(x) h 5 ( x ) is increasing.
By using
h 5 ( x 11 ) = 0 , h_5(x_{11})=0, h 5 ( x 11 ) = 0 , we prove that
v 11 ( x ) − N 11 [ v ] < 0. v_{11}(x)-N_{11}[v]<0. v 11 ( x ) − N 11 [ v ] < 0. * If χ 12 < χ 10 \chi_{12}<\chi_{10} χ 12 < χ 10 there is a x ∗ x^* x ∗ such that:
m ( x ) = { V ^ 12 ( x ) + A 12 x m 12 + + χ 12 , x < x ∗ , v 10 ( x ) + χ 10 , x ≥ x ∗ . m(x)=
\begin{cases}
\widehat{V}_{12}(x)+A_{12}x^{m_{12}^{+}}+\chi_{12}, & x<x^*,\\
v_{10}(x)+\chi_{10}, & x\geq x^*.
\end{cases} m ( x ) = { V 12 ( x ) + A 12 x m 12 + + χ 12 , v 10 ( x ) + χ 10 , x < x ∗ , x ≥ x ∗ . Then, for x < x ∗ x<x^* x < x ∗
h 6 ( x ) = V 11 ( x ) + A 11 x m 11 + − V 12 ( x ) − A 12 x m 12 + − χ 12 = ( A 11 − A 12 ) x m 11 + − χ 12 \begin{aligned}
h_6(x)&=V_{11}(x)+A_{11}x^{m_{11}^{+}}-V_{12}(x)-A_{12}x^{m_{12}^{+}}-\chi_{12}\\
&=(A_{11}-A_{12})x^{m_{11}^{+}}-\chi_{12}
\end{aligned} h 6 ( x ) = V 11 ( x ) + A 11 x m 11 + − V 12 ( x ) − A 12 x m 12 + − χ 12 = ( A 11 − A 12 ) x m 11 + − χ 12 h 6 ( x ) = − h 4 ( x ) + V ^ 11 ( x ) + A 11 x m 11 + − V ^ 10 ( x ) − χ 10 . h_6(x)=-h_4(x)+\widehat{V}_{11}(x)+A_{11}x^{m_{11}^{+}}-\widehat{V}_{10}(x)-\chi_{10}. h 6 ( x ) = − h 4 ( x ) + V 11 ( x ) + A 11 x m 11 + − V 10 ( x ) − χ 10 . It is easy to see that
h 7 ( x ) = V ^ 11 ( x ) + A 11 x m 11 + − V ^ 10 ( x ) − χ 10 h_7(x)=\widehat{V}_{11}(x)+A_{11}x^{m_{11}^{+}}-\widehat{V}_{10}(x)-\chi_{10} h 7 ( x ) = V 11 ( x ) + A 11 x m 11 + − V 10 ( x ) − χ 10 is increasing for x < x 11 x<x_{11} x < x 11
Using
h 7 ( 0 ) = − χ 10 and h 7 ( x 11 ) = 0 , h_7(0)=-\chi_{10}\quad\text{and}\quad h_7(x_{11})=0, h 7 ( 0 ) = − χ 10 and h 7 ( x 11 ) = 0 , we obtain that
for x < x 11 x<x_{11} x < x 11 This implies that h 7 ( x ∗ ) < 0 h_7(x^*)<0 h 7 ( x ∗ ) < 0
Hence, h 6 ( x ) < 0 h_6(x)<0 h 6 ( x ) < 0 for x < x ∗ x<x^* x < x ∗ which proves
v 11 ( x ) − N 11 [ v ] < 0. v_{11}(x)-N_{11}[v]<0. v 11 ( x ) − N 11 [ v ] < 0. For x ≥ x ∗ x\geq x^* x ≥ x ∗ the proof for v 11 ( x ) − N 11 [ v ] < 0 v_{11}(x)-N_{11}[v]<0 v 11 ( x ) − N 11 [ v ] < 0 is identical to that for
{ χ 12 > χ 10 } ∩ { x 11 < x 12 } . \{\chi_{12}>\chi_{10}\}\cap\{x_{11}<x_{12}\}. { χ 12 > χ 10 } ∩ { x 11 < x 12 } . Assume now that x 12 ≤ x 11 x_{12} \le x_{11} x 12 ≤ x 11
Then we have
m ( x ) = V ^ 10 ( x ) + min { χ 10 , χ 12 + χ 20 } for x 12 ≤ x ≤ x 11 = V ^ 10 ( x ) + χ 10 . \begin{aligned}
m(x) = \widehat{V}_{10}(x) + \min\{\chi_{10},\chi_{12}+\chi_{20}\}\quad\text{for }x_{12} \le x \le x_{11} \\
&= \widehat{V}_{10}(x) + \chi_{10}.
\end{aligned} m ( x ) = V 10 ( x ) + min { χ 10 , χ 12 + χ 20 } for x 12 ≤ x ≤ x 11 = V 10 ( x ) + χ 10 . Hence, the proof for v 11 ( x ) − N 11 [ v ] < 0 v_{11}(x)-N_{11}[v] < 0 v 11 ( x ) − N 11 [ v ] < 0 is identical to that for
{ χ 12 > χ 10 } ∩ { x 11 < x 12 } . \{\chi_{12} > \chi_{10}\} \cap\{x_{11} < x_{12}\}. { χ 12 > χ 10 } ∩ { x 11 < x 12 } . For x < x 12 x < x_{12} x < x 12
m ( x ) = v 10 ( x ) + χ 10 . m(x) = v_{10}(x) + \chi_{10}. m ( x ) = v 10 ( x ) + χ 10 . And the proof for v 11 ( x ) − N 11 [ v ] < 0 v_{11}(x)-N_{11}[v] < 0 v 11 ( x ) − N 11 [ v ] < 0 is identical to that for
{ χ 12 > χ 10 } ∩ { x 11 < x 12 } . \{\chi_{12} > \chi_{10}\} \cap\{x_{11} < x_{12}\}. { χ 12 > χ 10 } ∩ { x 11 < x 12 } . m ( x ) = { V ^ 12 ( x ) + A 12 x m 12 + + χ 12 , x < x ∗ , v 10 ( x ) + χ 10 , x ≥ x ∗ . m(x) =
\begin{cases}
\widehat{V}_{12}(x) + A_{12}x^{m_{12}^{+}} + \chi_{12}, & x < x^{*},\\
v_{10}(x) + \chi_{10}, & x \ge x^{*}.
\end{cases} m ( x ) = { V 12 ( x ) + A 12 x m 12 + + χ 12 , v 10 ( x ) + χ 10 , x < x ∗ , x ≥ x ∗ . The proof follows the same arguments as for
{ χ 12 < χ 10 } ∩ { x 11 < x 12 } . \{\chi_{12} < \chi_{10}\} \cap\{x_{11} < x_{12}\}. { χ 12 < χ 10 } ∩ { x 11 < x 12 } . Therefore, the QVI is satisfied on ( 0 , x 11 ) (0,x_{11}) ( 0 , x 11 )
v 11 ( x ) − M 11 [ v ] = v 11 ( x ) − max { v 01 ( x ) − c 10 , v 21 ( x ) − c 12 } = V ^ 10 ( x ) + χ 10 − max { v 01 ( x ) − c 10 , v 21 ( x ) − c 12 } = V ^ 10 ( x ) + χ 10 − max { V ^ 00 ( x ) + χ 10 − c 10 , V ^ 10 ( x ) + χ 10 − c 12 } = min { c 10 , c 12 } > 0. \begin{aligned}
v_{11}(x)-M_{11}[v] = v_{11}(x)-\max\{v_{01}(x)-c_{10},v_{21}(x)-c_{12}\} \\
&= \widehat{V}_{10}(x)+\chi_{10}-\max\{v_{01}(x)-c_{10},v_{21}(x)-c_{12}\} \\
&= \widehat{V}_{10}(x)+\chi_{10}-\max\{\widehat{V}_{00}(x)+\chi_{10}-c_{10},\widehat{V}_{10}(x)+\chi_{10}-c_{12}\} \\
&= \min\{c_{10},c_{12}\} > 0.
\end{aligned} v 11 ( x ) − M 11 [ v ] = v 11 ( x ) − max { v 01 ( x ) − c 10 , v 21 ( x ) − c 12 } = V 10 ( x ) + χ 10 − max { v 01 ( x ) − c 10 , v 21 ( x ) − c 12 } = V 10 ( x ) + χ 10 − max { V 00 ( x ) + χ 10 − c 10 , V 10 ( x ) + χ 10 − c 12 } = min { c 10 , c 12 } > 0. Moreover,
H [ v 11 ] ( x ) = K 10 − K 11 K 11 x γ + χ 10 , \mathcal{H}[v_{11}](x)=\frac{K_{10}-K_{11}}{K_{11}}x^{\gamma}+\chi_{10}, H [ v 11 ] ( x ) = K 11 K 10 − K 11 x γ + χ 10 , is negative by a straightforward calculus.
Hence,
min [ H [ v 11 ] ( x ) , v 11 ( x ) − M 11 [ v ] ( x ) ] < 0. \min\left[\mathcal{H}[v_{11}](x),v_{11}(x)-M_{11}[v](x)\right] < 0. min [ H [ v 11 ] ( x ) , v 11 ( x ) − M 11 [ v ] ( x ) ] < 0. Next,
v 11 ( x ) − N 11 [ v ] = v 11 ( x ) − min { v 10 ( x ) + χ 10 ; v 12 ( x ) + χ 12 } . v_{11}(x)-N_{11}[v]=v_{11}(x)-\min\{v_{10}(x)+\chi_{10};v_{12}(x)+\chi_{12}\}. v 11 ( x ) − N 11 [ v ] = v 11 ( x ) − min { v 10 ( x ) + χ 10 ; v 12 ( x ) + χ 12 } . If x 12 < x 11 x_{12}<x_{11} x 12 < x 11
v 11 ( x ) − N 11 [ v ] = V ^ 10 ( x ) + χ 10 − min { V ^ 10 ( x ) + χ 10 ; V ^ 10 ( x ) + χ 20 + χ 12 } = 0. v_{11}(x)-N_{11}[v]=\widehat{V}_{10}(x)+\chi_{10}-\min\{\widehat{V}_{10}(x)+\chi_{10};\widehat{V}_{10}(x)+\chi_{20}+\chi_{12}\}=0. v 11 ( x ) − N 11 [ v ] = V 10 ( x ) + χ 10 − min { V 10 ( x ) + χ 10 ; V 10 ( x ) + χ 20 + χ 12 } = 0. Else, for x 11 < x < x 12 x_{11}<x<x_{12} x 11 < x < x 12
v 11 ( x ) − N 11 [ v ] = V ^ 10 ( x ) + χ 10 − min { V ^ 10 ( x ) + χ 10 ; V ^ 12 ( x ) + A 12 x m 12 + + χ 12 } . v_{11}(x)-N_{11}[v]=\widehat{V}_{10}(x)+\chi_{10}-\min\{\widehat{V}_{10}(x)+\chi_{10};\widehat{V}_{12}(x)+A_{12}x^{m_{12}^{+}}+\chi_{12}\}. v 11 ( x ) − N 11 [ v ] = V 10 ( x ) + χ 10 − min { V 10 ( x ) + χ 10 ; V 12 ( x ) + A 12 x m 12 + + χ 12 } . Define
h 9 ( x ) = V ^ 12 ( x ) + A 12 x m 12 + + χ 12 − V ^ 10 ( x ) − χ 10 . h_{9}(x)=\widehat{V}_{12}(x)+A_{12}x^{m_{12}^{+}}+\chi_{12}-\widehat{V}_{10}(x)-\chi_{10}. h 9 ( x ) = V 12 ( x ) + A 12 x m 12 + + χ 12 − V 10 ( x ) − χ 10 . A straightforward calculus shows that h 9 h_{9} h 9
h 9 ( 0 ) = χ 12 + χ 20 − χ 10 > 0. h_{9}(0)=\chi_{12}+\chi_{20}-\chi_{10}>0. h 9 ( 0 ) = χ 12 + χ 20 − χ 10 > 0. Hence,
min { V ^ 10 ( x ) + χ 10 ; V ^ 12 ( x ) + A 12 x m 12 + + χ 12 } = V ^ 10 ( x ) + χ 10 . \min\{\widehat{V}_{10}(x)+\chi_{10};\widehat{V}_{12}(x)+A_{12}x^{m_{12}^{+}}+\chi_{12}\}=\widehat{V}_{10}(x)+\chi_{10}. min { V 10 ( x ) + χ 10 ; V 12 ( x ) + A 12 x m 12 + + χ 12 } = V 10 ( x ) + χ 10 . For x > x 12 x>x_{12} x > x 12
min { v 10 ( x ) + χ 10 ; v 12 ( x ) + χ 12 } = min { V ^ 10 ( x ) + χ 10 ; V ^ 10 ( x ) + χ 20 + χ 12 } = V ^ 10 ( x ) + χ 10 . \begin{aligned}
\min\{v_{10}(x)+\chi_{10};v_{12}(x)+\chi_{12}\}
&=\min\{\widehat{V}_{10}(x)+\chi_{10};\widehat{V}_{10}(x)+\chi_{20}+\chi_{12}\}\\
&=\widehat{V}_{10}(x)+\chi_{10}.
\end{aligned} min { v 10 ( x ) + χ 10 ; v 12 ( x ) + χ 12 } = min { V 10 ( x ) + χ 10 ; V 10 ( x ) + χ 20 + χ 12 } = V 10 ( x ) + χ 10 . In all cases,
v 11 ( x ) − N 11 [ v ] = 0. v_{11}(x)-N_{11}[v]=0. v 11 ( x ) − N 11 [ v ] = 0. Hence, v 11 v_{11} v 11 ( x 11 , ∞ ) (x_{11},\infty) ( x 11 , ∞ ) ( 0 , ∞ ) (0,\infty) ( 0 , ∞ ) v 01 v_{01} v 01 v 02 v_{02} v 02 v 10 v_{10} v 10 v 12 v_{12} v 12 v 21 v_{21} v 21 v 22 v_{22} v 22
Theorem 4.4 Case where
K i j = C s t 1 < K 2 j = C s t 2 , i = 0 , 1 , j = 0 , 1 , 2. K_{ij}=Cst_{1}<K_{2j}=Cst_{2},\qquad i=0,1,\qquad j=0,1,2. K ij = C s t 1 < K 2 j = C s t 2 , i = 0 , 1 , j = 0 , 1 , 2. Let the candidate value functions be defined by
v i j ( x ) = ( V ^ i j ( x ) + A i j x m i j + ) 1 x < x i j + ( V ^ 2 j ( x ) − c i 2 ) 1 x ≥ x i j , i = 0 , 1 , j = 0 , 1 , 2. (35) v_{ij}(x)=\left(\widehat{V}_{ij}(x)+A_{ij}x^{m_{ij}^{+}}\right)1_{x<x_{ij}}+\left(\widehat{V}_{2j}(x)-c_{i2}\right)1_{x\geq x_{ij}},\qquad i=0,1,\qquad j=0,1,2. \tag*{(35)} v ij ( x ) = ( V ij ( x ) + A ij x m ij + ) 1 x < x ij + ( V 2 j ( x ) − c i 2 ) 1 x ≥ x ij , i = 0 , 1 , j = 0 , 1 , 2. (35) v 2 j ( x ) = V ^ 2 j ( x ) , j = 0 , 1 , 2. (36) v_{2j}(x)=\widehat{V}_{2j}(x),\qquad j=0,1,2. \tag*{(36)} v 2 j ( x ) = V 2 j ( x ) , j = 0 , 1 , 2. (36) Where x i j x_{ij} x ij A i j A_{ij} A ij
( K 2 j − K i j ) ( x i j ) γ = m i j + m i j + − γ c i 2 , A i j = ( K 2 j − K i j ) γ m i j + ( x i j ) γ − m i j + . (K_{2j}-K_{ij})(x_{ij})^{\gamma}=\frac{m_{ij}^{+}}{m_{ij}^{+}-\gamma}c_{i2},\qquad A_{ij}=(K_{2j}-K_{ij})\frac{\gamma}{m_{ij}^{+}}(x_{ij})^{\gamma-m_{ij}^{+}}. ( K 2 j − K ij ) ( x ij ) γ = m ij + − γ m ij + c i 2 , A ij = ( K 2 j − K ij ) m ij + γ ( x ij ) γ − m ij + . Such v i j ( x ) v_{ij}(x) v ij ( x )
S 2 j ξ = ∅ , C 2 j ξ = ( 0 ; + ∞ ) , S 2 j η = ∅ , C 2 j η = ( 0 ; + ∞ ) , j = 0 , 1 , 2. S_{2j}^{\xi} = \varnothing,\quad C_{2j}^{\xi} = (0;+\infty),\quad S_{2j}^{\eta} = \varnothing,\quad C_{2j}^{\eta} = (0;+\infty),\quad j=0,1,2. S 2 j ξ = ∅ , C 2 j ξ = ( 0 ; + ∞ ) , S 2 j η = ∅ , C 2 j η = ( 0 ; + ∞ ) , j = 0 , 1 , 2. S i j ξ = [ x i j , ∞ ) , C i j ξ = ( 0 ; x i j ) , S i j η = ∅ , C i j η = ( 0 ; + ∞ ) , i = 0 , 1 , j = 0 , 1 , 2. S_{ij}^{\xi} = [x_{ij},\infty),\quad C_{ij}^{\xi} = (0;x_{ij}),\quad S_{ij}^{\eta} = \varnothing,\quad C_{ij}^{\eta} = (0;+\infty),\quad i=0,1,\quad j=0,1,2. S ij ξ = [ x ij , ∞ ) , C ij ξ = ( 0 ; x ij ) , S ij η = ∅ , C ij η = ( 0 ; + ∞ ) , i = 0 , 1 , j = 0 , 1 , 2. Proof 4.4 We only verify the QVI for v 00 v_{00} v 00 v 20 v_{20} v 20
Step 1. Verification for v 00 v_{00} v 00
H [ v 00 ] ( x ) = 0. \mathcal{H}[v_{00}](x)=0. H [ v 00 ] ( x ) = 0. Moreover,
v 00 ( x ) − M 00 [ v ] = v 00 ( x ) − max { v 10 ( x ) − c 01 , v 20 ( x ) − c 02 } . v_{00}(x)-M_{00}[v]=v_{00}(x)-\max\{v_{10}(x)-c_{01},v_{20}(x)-c_{02}\}. v 00 ( x ) − M 00 [ v ] = v 00 ( x ) − max { v 10 ( x ) − c 01 , v 20 ( x ) − c 02 } . Let us determine
n ( x ) = max { v 10 ( x ) − c 01 , v 20 ( x ) − c 02 } . n(x)=\max\{v_{10}(x)-c_{01},v_{20}(x)-c_{02}\}. n ( x ) = max { v 10 ( x ) − c 01 , v 20 ( x ) − c 02 } . x 00 x_{00} x 00 x 10 x_{10} x 10 c 02 c_{02} c 02 c 12 c_{12} c 12
Assume that x 00 < x 10 x_{00}<x_{10} x 00 < x 10
Let
h 10 ( x ) = V ^ 10 ( x ) + A 10 x m 10 + − V ^ 20 ( x ) + c 12 . h_{10}(x)=\widehat{V}_{10}(x)+A_{10}x^{m_{10}^{+}}-\widehat{V}_{20}(x)+c_{12}. h 10 ( x ) = V 10 ( x ) + A 10 x m 10 + − V 20 ( x ) + c 12 . and
h 11 ( x ) = v 10 ( x ) − c 01 − ( v 20 ( x ) − c 02 ) = V ^ 10 ( x ) + A 10 x m 10 + − c 01 − V ^ 20 ( x ) + c 02 . \begin{aligned}
h_{11}(x)&=v_{10}(x)-c_{01}-(v_{20}(x)-c_{02})\\
&=\widehat{V}_{10}(x)+A_{10}x^{m_{10}^{+}}-c_{01}-\widehat{V}_{20}(x)+c_{02}.
\end{aligned} h 11 ( x ) = v 10 ( x ) − c 01 − ( v 20 ( x ) − c 02 ) = V 10 ( x ) + A 10 x m 10 + − c 01 − V 20 ( x ) + c 02 . We can see that
h 11 ( x ) = h 10 ( x ) − c 01 + c 02 − c 12 . (37) h_{11}(x)=h_{10}(x)-c_{01}+c_{02}-c_{12}. \tag*{(37)} h 11 ( x ) = h 10 ( x ) − c 01 + c 02 − c 12 . (37) A straightforward calculus shows that h 11 ′ ′ ( x ) > 0 h_{11}^{\prime\prime}(x)>0 h 11 ′′ ( x ) > 0
Hence, h 11 ′ ( x ) h_{11}^{\prime}(x) h 11 ′ ( x )
h 11 ′ ( x 10 ) = 0 , h_{11}^{\prime}(x_{10})=0, h 11 ′ ( x 10 ) = 0 , h 11 ( x ) h_{11}(x) h 11 ( x )
We have, from 2.35,
h 11 ( 0 ) = c 02 − c 01 and h 11 ( x 10 ) = − c 01 + c 02 − c 12 < 0. h_{11}(0)=c_{02}-c_{01}\quad\text{and}\quad h_{11}(x_{10})=-c_{01}+c_{02}-c_{12}<0. h 11 ( 0 ) = c 02 − c 01 and h 11 ( x 10 ) = − c 01 + c 02 − c 12 < 0. ⋆ If c 02 < c 01 c_{02}<c_{01} c 02 < c 01 h 11 ( x ) < 0 h_{11}(x)<0 h 11 ( x ) < 0 x < x 00 x<x_{00} x < x 00
n ( x ) = V ^ 20 ( x ) − c 02 . n(x)=\widehat{V}_{20}(x)-c_{02}. n ( x ) = V 20 ( x ) − c 02 . In this case,
Let
h 12 ( x ) = V ^ 00 ( x ) + A 00 x m 00 + − V ^ 20 ( x ) + c 02 . h_{12}(x)=\widehat{V}_{00}(x)+A_{00}x^{m_{00}^{+}}-\widehat{V}_{20}(x)+c_{02}. h 12 ( x ) = V 00 ( x ) + A 00 x m 00 + − V 20 ( x ) + c 02 . A straightforward calculus shows that h 12 ′ ′ ( x ) > 0 h_{12}^{\prime\prime}(x) > 0 h 12 ′′ ( x ) > 0 h 12 ′ ( x ) h_{12}^{\prime}(x) h 12 ′ ( x )
Using
h 12 ′ ( x 00 ) = 0 , h_{12}^{\prime}(x_{00}) = 0, h 12 ′ ( x 00 ) = 0 , it follows that h 12 ′ ( x ) < 0 h_{12}^{\prime}(x) < 0 h 12 ′ ( x ) < 0 h 12 ( x ) h_{12}(x) h 12 ( x )
Given that
h 12 ( x 00 ) = 0 , h_{12}(x_{00}) = 0, h 12 ( x 00 ) = 0 , we get the desired result.
⋆ If c 02 > c 01 c_{02} > c_{01} c 02 > c 01 x ∗ x^{*} x ∗
n ( x ) = { v 20 ( x ) − c 02 , x ≥ x ∗ , V ^ 10 ( x ) + A 10 x m 10 + − c 01 , x < x ∗ . n(x) =
\begin{cases}
v_{20}(x) - c_{02}, & x \ge x^{*}, \\
\hat{V}_{10}(x) + A_{10}x^{m_{10}^{+}} - c_{01}, & x < x^{*}.
\end{cases} n ( x ) = { v 20 ( x ) − c 02 , V ^ 10 ( x ) + A 10 x m 10 + − c 01 , x ≥ x ∗ , x < x ∗ . Then, for x ≥ x ∗ x \ge x^{*} x ≥ x ∗ v 00 ( x ) − M 00 [ v ] > 0 v_{00}(x) - M_{00}[v] > 0 v 00 ( x ) − M 00 [ v ] > 0
{ c 02 < c 01 } ∩ { x 00 < x 10 } . \{c_{02} < c_{01}\} \cap\{x_{00} < x_{10}\}. { c 02 < c 01 } ∩ { x 00 < x 10 } . For x < x ∗ x < x^{*} x < x ∗
h 13 ( x ) = V ^ 00 ( x ) + A 00 x m 00 + − V ^ 10 ( x ) − A 10 x m 10 + + c 10 = ( A 00 − A 10 ) x m 00 + + c 10 . \begin{aligned}
h_{13}(x) &= \hat{V}_{00}(x) + A_{00}x^{m_{00}^{+}} - \hat{V}_{10}(x) - A_{10}x^{m_{10}^{+}} + c_{10} \\
&= (A_{00} - A_{10})x^{m_{00}^{+}} + c_{10}.
\end{aligned} h 13 ( x ) = V ^ 00 ( x ) + A 00 x m 00 + − V ^ 10 ( x ) − A 10 x m 10 + + c 10 = ( A 00 − A 10 ) x m 00 + + c 10 . If ( A 00 − A 10 ) > 0 (A_{00} - A_{10}) > 0 ( A 00 − A 10 ) > 0
Else, h 13 ′ ( x ) < 0 h_{13}^{\prime}(x) < 0 h 13 ′ ( x ) < 0 h 13 ( x ) h_{13}(x) h 13 ( x )
Let
h 14 ( x ) = V ^ 10 ( x ) + A 10 x m 10 + − V ^ 20 ( x ) − c 10 + c 02 . h_{14}(x) = \hat{V}_{10}(x) + A_{10}x^{m_{10}^{+}} - \hat{V}_{20}(x) - c_{10} + c_{02}. h 14 ( x ) = V ^ 10 ( x ) + A 10 x m 10 + − V ^ 20 ( x ) − c 10 + c 02 . We can see that
h 13 ( x ) = − h 14 ( x ) + V ^ 00 ( x ) + A 00 x m 00 + − V ^ 20 ( x ) + c 02 = − h 14 ( x ) + h 12 ( x ) . \begin{aligned}
h_{13}(x) &= -h_{14}(x) + \hat{V}_{00}(x) + A_{00}x^{m_{00}^{+}} - \hat{V}_{20}(x) + c_{02} \\
&= -h_{14}(x) + h_{12}(x).
\end{aligned} h 13 ( x ) = − h 14 ( x ) + V ^ 00 ( x ) + A 00 x m 00 + − V ^ 20 ( x ) + c 02 = − h 14 ( x ) + h 12 ( x ) . Since
h 13 ( x ∗ ) = h 12 ( x ∗ ) > 0 , h_{13}(x^{*}) = h_{12}(x^{*}) > 0, h 13 ( x ∗ ) = h 12 ( x ∗ ) > 0 , this proves that
v 00 ( x ) − M 00 [ v ] > 0. v_{00}(x) - M_{00}[v] > 0. v 00 ( x ) − M 00 [ v ] > 0. Assume now that x 10 ≤ x 00 x_{10} \le x_{00} x 10 ≤ x 00
Then we have
n ( x ) = V ^ 20 ( x ) − min { c 01 + c 12 , c 02 } for x 10 ≤ x ≤ x 00 = V ^ 20 ( x ) − c 02 . \begin{aligned}
n(x) &= \hat{V}_{20}(x) - \min\{c_{01} + c_{12}, c_{02}\} \quad\text{for } x_{10} \le x \le x_{00} \\
&= \hat{V}_{20}(x) - c_{02}.
\end{aligned} n ( x ) = V ^ 20 ( x ) − min { c 01 + c 12 , c 02 } for x 10 ≤ x ≤ x 00 = V ^ 20 ( x ) − c 02 . Hence, the proof for v 00 ( x ) − M 00 [ v ] > 0 v_{00}(x)-M_{00}[v]>0 v 00 ( x ) − M 00 [ v ] > 0
{ c 02 < c 01 } ∩ { x 00 < x 10 } ∩ { x < x ∗ } . \{c_{02}<c_{01}\}\cap\{x_{00}<x_{10}\}\cap\{x<x^*\}. { c 02 < c 01 } ∩ { x 00 < x 10 } ∩ { x < x ∗ } . For x < x 10 x<x_{10} x < x 10
n ( x ) = v 10 ( x ) − c 01 . n(x)=v_{10}(x)-c_{01}. n ( x ) = v 10 ( x ) − c 01 . And the proof for v 00 ( x ) − M 00 [ v ] > 0 v_{00}(x)-M_{00}[v]>0 v 00 ( x ) − M 00 [ v ] > 0
{ c 02 < c 01 } ∩ { x 00 < x 10 } ∩ { x < x ∗ } . \{c_{02}<c_{01}\}\cap\{x_{00}<x_{10}\}\cap\{x<x^*\}. { c 02 < c 01 } ∩ { x 00 < x 10 } ∩ { x < x ∗ } . n ( x ) = { v 20 ( x ) − c 02 , x < x ∗ , V ^ 10 ( x ) + A 10 x m 10 + − c 01 , x ≥ x ∗ . n(x)=
\begin{cases}
v_{20}(x)-c_{02}, & x<x^*,\\
\hat{V}_{10}(x)+A_{10}x^{m_{10}^{+}}-c_{01}, & x\ge x^*.
\end{cases} n ( x ) = { v 20 ( x ) − c 02 , V ^ 10 ( x ) + A 10 x m 10 + − c 01 , x < x ∗ , x ≥ x ∗ . The proof for v 00 ( x ) − M 00 [ v ] > 0 v_{00}(x)-M_{00}[v]>0 v 00 ( x ) − M 00 [ v ] > 0
{ c 02 < c 01 } ∩ { x 00 < x 10 } . \{c_{02}<c_{01}\}\cap\{x_{00}<x_{10}\}. { c 02 < c 01 } ∩ { x 00 < x 10 } . Next,
v 00 ( x ) − N 00 [ v ] = v 00 ( x ) − min { v 01 ( x ) + χ 01 , v 02 ( x ) + χ 02 } . v_{00}(x)-N_{00}[v]=v_{00}(x)-\min\{v_{01}(x)+\chi_{01},v_{02}(x)+\chi_{02}\}. v 00 ( x ) − N 00 [ v ] = v 00 ( x ) − min { v 01 ( x ) + χ 01 , v 02 ( x ) + χ 02 } . Using the same reasoning as for v 00 ( x ) − M 00 [ v ] > 0 v_{00}(x)-M_{00}[v]>0 v 00 ( x ) − M 00 [ v ] > 0
v 00 ( x ) − N 00 [ v ] < 0. v_{00}(x)-N_{00}[v]<0. v 00 ( x ) − N 00 [ v ] < 0. Therefore, the QVI is satisfied on ( 0 , x 00 ) (0,x_{00}) ( 0 , x 00 )
H [ v 00 ] ( x ) = K 20 − K 00 K 00 x r − c 02 . \mathcal{H}[v_{00}](x)=\frac{K_{20}-K_{00}}{K_{00}}x^r-c_{02}. H [ v 00 ] ( x ) = K 00 K 20 − K 00 x r − c 02 . A straightforward calculus shows that H [ v 00 ] ( x ) \mathcal{H}[v_{00}](x) H [ v 00 ] ( x )
Moreover:
v 00 ( x ) − M 00 [ v ] = v 00 ( x ) − max { v 10 ( x ) − c 01 , v 20 ( x ) − c 02 } . v_{00}(x)-M_{00}[v]=v_{00}(x)-\max\{v_{10}(x)-c_{01},v_{20}(x)-c_{02}\}. v 00 ( x ) − M 00 [ v ] = v 00 ( x ) − max { v 10 ( x ) − c 01 , v 20 ( x ) − c 02 } . If x 00 > x 10 x_{00}>x_{10} x 00 > x 10
v 00 ( x ) − M 00 [ v ] = V ^ 20 ( x ) − c 02 − max { V ^ 20 ( x ) − c 01 − c 12 , V ^ 20 ( x ) − c 02 } = c 02 − c 02 = 0. \begin{aligned}
v_{00}(x)-M_{00}[v]&=\hat{V}_{20}(x)-c_{02}-\max\{\hat{V}_{20}(x)-c_{01}-c_{12},\hat{V}_{20}(x)-c_{02}\}\\
&=c_{02}-c_{02}=0.
\end{aligned} v 00 ( x ) − M 00 [ v ] = V ^ 20 ( x ) − c 02 − max { V ^ 20 ( x ) − c 01 − c 12 , V ^ 20 ( x ) − c 02 } = c 02 − c 02 = 0. And
v 00 ( x ) − N 00 [ v ] = v 00 ( x ) − min { v 01 ( x ) + χ 01 , v 02 ( x ) + χ 02 } = − max { χ 01 , χ 02 } < 0. \begin{aligned}
v_{00}(x)-N_{00}[v]&=v_{00}(x)-\min\{v_{01}(x)+\chi_{01},v_{02}(x)+\chi_{02}\}\\
&=-\max\{\chi_{01},\chi_{02}\}<0.
\end{aligned} v 00 ( x ) − N 00 [ v ] = v 00 ( x ) − min { v 01 ( x ) + χ 01 , v 02 ( x ) + χ 02 } = − max { χ 01 , χ 02 } < 0. If x 00 < x 10 x_{00}<x_{10} x 00 < x 10 x 00 < x < x 10 x_{00}<x<x_{10} x 00 < x < x 10
v 00 ( x ) − M 00 [ v ] = V ^ 20 ( x ) − c 02 − max { v 10 ( x ) − c 01 , v 20 ( x ) − c 02 } . v_{00}(x)-M_{00}[v]=\widehat{V}_{20}(x)-c_{02}-\max\{v_{10}(x)-c_{01},v_{20}(x)-c_{02}\}. v 00 ( x ) − M 00 [ v ] = V 20 ( x ) − c 02 − max { v 10 ( x ) − c 01 , v 20 ( x ) − c 02 } . It suffices to show that, on ( x 00 , x 10 ) (x_{00},x_{10}) ( x 00 , x 10 )
n ( x ) = max { v 10 ( x ) − c 01 , v 20 ( x ) − c 02 } = v 20 ( x ) − c 02 . n(x)=\max\{v_{10}(x)-c_{01},v_{20}(x)-c_{02}\}=v_{20}(x)-c_{02}. n ( x ) = max { v 10 ( x ) − c 01 , v 20 ( x ) − c 02 } = v 20 ( x ) − c 02 . From above, we know that if c 02 < c 01 c_{02}<c_{01} c 02 < c 01
Else,
n ( x ) = { v 20 ( x ) − c 02 , x ≥ x ∗ , V ^ 10 ( x ) + A 10 x m 10 + − c 01 , x < x ∗ . n(x)=
\begin{cases}
v_{20}(x)-c_{02}, & x\geq x^{*},\\
\widehat{V}_{10}(x)+A_{10}x^{m_{10}^{+}}-c_{01}, & x<x^{*}.
\end{cases} n ( x ) = { v 20 ( x ) − c 02 , V 10 ( x ) + A 10 x m 10 + − c 01 , x ≥ x ∗ , x < x ∗ . Let’s show that x ∗ < x 00 x^{*}<x_{00} x ∗ < x 00
h 11 ( x ) = v 10 ( x ) − c 01 − ( v 20 ( x ) − c 02 ) = V ^ 10 ( x ) + A 10 x m 10 + − c 01 − V ^ 20 ( x ) + c 02 = V ^ 10 ( x ) + A 10 x m 10 + + V ^ 00 ( x ) + A 00 x m 00 + − ( V ^ 00 ( x ) + A 00 x m 00 + ) − c 01 − V ^ 20 ( x ) + c 02 = ( A 10 − A 00 ) x m 00 + + V ^ 00 ( x ) + A 00 x m 00 + − c 01 − V ^ 20 ( x ) + c 02 . \begin{aligned}
h_{11}(x)&=v_{10}(x)-c_{01}-(v_{20}(x)-c_{02})\\
&=\widehat{V}_{10}(x)+A_{10}x^{m_{10}^{+}}-c_{01}-\widehat{V}_{20}(x)+c_{02}\\
&=\widehat{V}_{10}(x)+A_{10}x^{m_{10}^{+}}+\widehat{V}_{00}(x)+A_{00}x^{m_{00}^{+}}\\
&\quad-(\widehat{V}_{00}(x)+A_{00}x^{m_{00}^{+}})-c_{01}-\widehat{V}_{20}(x)+c_{02}\\
&=(A_{10}-A_{00})x^{m_{00}^{+}}+\widehat{V}_{00}(x)+A_{00}x^{m_{00}^{+}}-c_{01}-\widehat{V}_{20}(x)+c_{02}.
\end{aligned} h 11 ( x ) = v 10 ( x ) − c 01 − ( v 20 ( x ) − c 02 ) = V 10 ( x ) + A 10 x m 10 + − c 01 − V 20 ( x ) + c 02 = V 10 ( x ) + A 10 x m 10 + + V 00 ( x ) + A 00 x m 00 + − ( V 00 ( x ) + A 00 x m 00 + ) − c 01 − V 20 ( x ) + c 02 = ( A 10 − A 00 ) x m 00 + + V 00 ( x ) + A 00 x m 00 + − c 01 − V 20 ( x ) + c 02 . x 00 < x 10 ⟹ A 10 < A 00 x_{00}<x_{10}\Longrightarrow A_{10}<A_{00} x 00 < x 10 ⟹ A 10 < A 00
( A 10 − A 00 ) x 00 m 00 + < 0. (A_{10}-A_{00})x_{00}^{m_{00}^{+}}<0. ( A 10 − A 00 ) x 00 m 00 + < 0. On the other hand, f ( x 00 ) = − c 01 f(x_{00})=-c_{01} f ( x 00 ) = − c 01
f ( x ) = V 00 ( x ) + A 00 x m 00 + − c 01 − V ^ 20 ( x ) + c 02 . f(x)=V_{00}(x)+A_{00}x^{m_{00}^{+}}-c_{01}-\widehat{V}_{20}(x)+c_{02}. f ( x ) = V 00 ( x ) + A 00 x m 00 + − c 01 − V 20 ( x ) + c 02 . We deduce that
h 11 ( x 00 ) < 0. h_{11}(x_{00})<0. h 11 ( x 00 ) < 0. Using the fact that h 11 h_{11} h 11 ( 0 , x 10 ) (0,x_{10}) ( 0 , x 10 ) h 11 ′ ( x ∗ ) = 0 h'_{11}(x^{*})=0 h 11 ′ ( x ∗ ) = 0
Finally, for v 00 ( x ) − N 00 [ v ] v_{00}(x)-N_{00}[v] v 00 ( x ) − N 00 [ v ]
v 00 ( x ) − N 00 [ v ] = v 00 ( x ) − min { v 01 ( x ) + χ 01 ; v 02 ( x ) + χ 02 } = − max { χ 01 , χ 02 } < 0. \begin{aligned}
v_{00}(x)-N_{00}[v]&=v_{00}(x)-\min\{v_{01}(x)+\chi_{01};v_{02}(x)+\chi_{02}\}\\
&=-\max\{\chi_{01},\chi_{02}\}<0.
\end{aligned} v 00 ( x ) − N 00 [ v ] = v 00 ( x ) − min { v 01 ( x ) + χ 01 ; v 02 ( x ) + χ 02 } = − max { χ 01 , χ 02 } < 0. Hence, we deduce that v 00 v_{00} v 00 ( x 00 , ∞ ) (x_{00},\infty) ( x 00 , ∞ ) ( 0 , ∞ ) (0,\infty) ( 0 , ∞ )
Step 2. Verification for v 20 v_{20} v 20
Since
v 20 ( x ) = V ^ 20 ( x ) , v_{20}(x)=\widehat{V}_{20}(x), v 20 ( x ) = V 20 ( x ) , we immediately have
H [ v 20 ] ( x ) = 0. \mathcal{H}[v_{20}](x)=0. H [ v 20 ] ( x ) = 0. Furthermore,
v 20 ( x ) − M 20 [ v ] = v 20 ( x ) − max { v 00 ( x ) − c 20 , v 10 ( x ) − c 21 } . v_{20}(x) - M_{20}[v] = v_{20}(x) - \max\{v_{00}(x) - c_{20}, v_{10}(x) - c_{21}\}. v 20 ( x ) − M 20 [ v ] = v 20 ( x ) − max { v 00 ( x ) − c 20 , v 10 ( x ) − c 21 } . Assume that
max { v 00 ( x ) − c 20 , v 10 ( x ) − c 21 } = v 00 ( x ) − c 20 . \max\{v_{00}(x) - c_{20}, v_{10}(x) - c_{21}\} = v_{00}(x) - c_{20}. max { v 00 ( x ) − c 20 , v 10 ( x ) − c 21 } = v 00 ( x ) − c 20 . v 20 ( x ) − M 20 [ v ] = V ^ 20 ( x ) − V ^ 00 ( x ) − A 00 x m 00 + + c 20 . v_{20}(x) - M_{20}[v] = \hat{V}_{20}(x) - \hat{V}_{00}(x) - A_{00}x^{m_{00}^{+}} + c_{20}. v 20 ( x ) − M 20 [ v ] = V ^ 20 ( x ) − V ^ 00 ( x ) − A 00 x m 00 + + c 20 . Let
h 15 ( x ) = V ^ 20 ( x ) − V ^ 00 ( x ) − A 00 x m 00 + + c 20 . h_{15}(x) = \hat{V}_{20}(x) - \hat{V}_{00}(x) - A_{00}x^{m_{00}^{+}} + c_{20}. h 15 ( x ) = V ^ 20 ( x ) − V ^ 00 ( x ) − A 00 x m 00 + + c 20 . A straightforward calculus shows that h 15 ( x ) h_{15}(x) h 15 ( x ) is increasing on ( 0 , x 00 ) (0, x_{00}) ( 0 , x 00 ) Given that h 15 ( 0 ) = c 20 > 0 h_{15}(0) = c_{20} > 0 h 15 ( 0 ) = c 20 > 0 we deduce that
v 20 ( x ) − M 20 [ v ] > 0. v_{20}(x) - M_{20}[v] > 0. v 20 ( x ) − M 20 [ v ] > 0. v 20 ( x ) − M 20 [ v ] = v 20 ( x ) − { ( v 20 ( x ) − c 02 ) − c 20 } = c 02 + c 20 > 0. \begin{aligned}
v_{20}(x) - M_{20}[v] &= v_{20}(x) - \{(v_{20}(x) - c_{02}) - c_{20}\} \\
&= c_{02} + c_{20} > 0.
\end{aligned} v 20 ( x ) − M 20 [ v ] = v 20 ( x ) − {( v 20 ( x ) − c 02 ) − c 20 } = c 02 + c 20 > 0. We deduce that
v 20 ( x ) − M 20 [ v ] > 0. v_{20}(x) - M_{20}[v] > 0. v 20 ( x ) − M 20 [ v ] > 0. Else, assume that
max { v 00 ( x ) − c 20 , v 10 ( x ) − c 21 } = v 10 ( x ) − c 21 . \max\{v_{00}(x) - c_{20}, v_{10}(x) - c_{21}\} = v_{10}(x) - c_{21}. max { v 00 ( x ) − c 20 , v 10 ( x ) − c 21 } = v 10 ( x ) − c 21 . For x < x 10 x < x_{10} x < x 10
v 20 ( x ) − M 20 [ v ] = V ^ 20 ( x ) − V ^ 10 ( x ) − A 10 x m 10 + + c 21 . v_{20}(x) - M_{20}[v] = \hat{V}_{20}(x) - \hat{V}_{10}(x) - A_{10}x^{m_{10}^{+}} + c_{21}. v 20 ( x ) − M 20 [ v ] = V ^ 20 ( x ) − V ^ 10 ( x ) − A 10 x m 10 + + c 21 . Let
h 16 ( x ) = V ^ 20 ( x ) − V ^ 10 ( x ) − A 10 x m 10 + + c 21 . h_{16}(x) = \hat{V}_{20}(x) - \hat{V}_{10}(x) - A_{10}x^{m_{10}^{+}} + c_{21}. h 16 ( x ) = V ^ 20 ( x ) − V ^ 10 ( x ) − A 10 x m 10 + + c 21 . A straightforward calculus shows that h 16 ( x ) h_{16}(x) h 16 ( x ) is increasing on ( 0 , x 00 ) (0, x_{00}) ( 0 , x 00 ) Given that h 16 ( 0 ) = c 21 > 0 h_{16}(0) = c_{21} > 0 h 16 ( 0 ) = c 21 > 0 we deduce that
v 20 ( x ) − M 20 [ v ] > 0. v_{20}(x) - M_{20}[v] > 0. v 20 ( x ) − M 20 [ v ] > 0. For x ≥ x 10 x \geq x_{10} x ≥ x 10
v 20 ( x ) − M 20 [ v ] = v 20 ( x ) − { ( v 20 ( x ) − c 12 ) − c 21 } = c 12 + c 21 > 0. \begin{aligned}
v_{20}(x) - M_{20}[v] &= v_{20}(x) - \{(v_{20}(x) - c_{12}) - c_{21}\} \\
&= c_{12} + c_{21} > 0.
\end{aligned} v 20 ( x ) − M 20 [ v ] = v 20 ( x ) − {( v 20 ( x ) − c 12 ) − c 21 } = c 12 + c 21 > 0. We deduce that
v 20 ( x ) − M 20 [ v ] > 0. v_{20}(x) - M_{20}[v] > 0. v 20 ( x ) − M 20 [ v ] > 0. Finally,
v 20 ( x ) − N 20 [ v ] = v 20 ( x ) − min { v 21 ( x ) + χ 01 , v 22 ( x ) + χ 22 } = v 20 ( x ) − v 21 ( x ) − max { χ 01 , χ 02 } = − max { χ 01 , χ 02 } < 0. \begin{aligned}
v_{20}(x) - N_{20}[v] &= v_{20}(x) - \min\{v_{21}(x) + \chi_{01}, v_{22}(x) + \chi_{22}\} \\
&= v_{20}(x) - v_{21}(x) - \max\{\chi_{01}, \chi_{02}\} \\
&= -\max\{\chi_{01}, \chi_{02}\} < 0.
\end{aligned} v 20 ( x ) − N 20 [ v ] = v 20 ( x ) − min { v 21 ( x ) + χ 01 , v 22 ( x ) + χ 22 } = v 20 ( x ) − v 21 ( x ) − max { χ 01 , χ 02 } = − max { χ 01 , χ 02 } < 0. Hence, v 20 v_{20} v 20 ( 0 , + ∞ ) (0,+\infty) ( 0 , + ∞ )
The proofs for v 01 v_{01} v 01 v 02 v_{02} v 02 v 10 v_{10} v 10 v 11 v_{11} v 11 v 12 v_{12} v 12
The conclusion follows from Lemma 3.1. □
Theorem 4.5 Case where
K i j = C s t 1 < K i 2 = C s t 2 , i = 0 , 1 , 2 , j = 0 , 1 , K_{ij}=Cst_1<K_{i2}=Cst_2,\qquad i=0,1,2,\qquad j=0,1, K ij = C s t 1 < K i 2 = C s t 2 , i = 0 , 1 , 2 , j = 0 , 1 , v i j ( x ) = V ^ i j ( x ) , i = 0 , 1 , 2 , j = 0 , 1 , (38) v_{ij}(x)=\widehat{V}_{ij}(x),\qquad i=0,1,2,\qquad j=0,1, \tag*{(38)} v ij ( x ) = V ij ( x ) , i = 0 , 1 , 2 , j = 0 , 1 , (38) v i 2 ( x ) = ( V ^ i 2 ( x ) + A i 2 x m i 2 + ) 1 { x < x i 2 } + ( v i 0 ( x ) + min { χ 20 , χ 21 } ) 1 { x ≥ x i 2 } , i = 0 , 1 , 2. (39) v_{i2}(x)=\left(\widehat{V}_{i2}(x)+A_{i2}x^{m_{i2}^{+}}\right)\mathbf{1}_{\{x<x_{i2}\}}+\left(v_{i0}(x)+\min\{\chi_{20},\chi_{21}\}\right)\mathbf{1}_{\{x\geq x_{i2}\}},\qquad i=0,1,2. \tag*{(39)} v i 2 ( x ) = ( V i 2 ( x ) + A i 2 x m i 2 + ) 1 { x < x i 2 } + ( v i 0 ( x ) + min { χ 20 , χ 21 } ) 1 { x ≥ x i 2 } , i = 0 , 1 , 2. (39) where x i 2 x_{i2} x i 2 A i 2 A_{i2} A i 2
( K i 0 − K 2 j ) ( x i 2 ) γ = m i 2 + m i 2 + − γ c i 2 , A i 2 = ( K i 0 − K 2 j ) γ m i 2 + ( x i 2 ) γ − m i j + . (K_{i0}-K_{2j})(x_{i2})^{\gamma}=\frac{m_{i2}^{+}}{m_{i2}^{+}-\gamma}c_{i2},\qquad A_{i2}=(K_{i0}-K_{2j})\frac{\gamma}{m_{i2}^{+}}(x_{i2})^{\gamma-m_{ij}^{+}}. ( K i 0 − K 2 j ) ( x i 2 ) γ = m i 2 + − γ m i 2 + c i 2 , A i 2 = ( K i 0 − K 2 j ) m i 2 + γ ( x i 2 ) γ − m ij + . Such v i j ( x ) v_{ij}(x) v ij ( x )
With:
S i j ξ = ∅ , C i j ξ = ( 0 , + ∞ ) , S i j η = ∅ , C i j η = ( 0 , + ∞ ) , j = 0 , 1 , 2 , S_{ij}^{\xi}=\varnothing,\qquad C_{ij}^{\xi}=(0,+\infty),\qquad S_{ij}^{\eta}=\varnothing,\qquad C_{ij}^{\eta}=(0,+\infty),\qquad j=0,1,2, S ij ξ = ∅ , C ij ξ = ( 0 , + ∞ ) , S ij η = ∅ , C ij η = ( 0 , + ∞ ) , j = 0 , 1 , 2 , S i 2 ξ = [ x i 2 , ∞ ) , C i 2 ξ = ( 0 , x i 2 ) , S i 2 η = ∅ , C i 2 η = ( 0 , + ∞ ) , i = 0 , 1 , j = 0 , 1 , 2. S_{i2}^{\xi}=[x_{i2},\infty),\qquad C_{i2}^{\xi}=(0,x_{i2}),\qquad S_{i2}^{\eta}=\varnothing,\qquad C_{i2}^{\eta}=(0,+\infty),\qquad i=0,1,\qquad j=0,1,2. S i 2 ξ = [ x i 2 , ∞ ) , C i 2 ξ = ( 0 , x i 2 ) , S i 2 η = ∅ , C i 2 η = ( 0 , + ∞ ) , i = 0 , 1 , j = 0 , 1 , 2. Proof 4.5 We only verify the QVI for v 00 v_{00} v 00 v 02 v_{02} v 02
Step 1. Verification for v 00 v_{00} v 00
Since
v 00 ( x ) = V ^ 00 ( x ) , v_{00}(x)=\widehat{V}_{00}(x), v 00 ( x ) = V 00 ( x ) , we immediately have
H [ v 00 ] ( x ) = 0. \mathcal{H}[v_{00}](x)=0. H [ v 00 ] ( x ) = 0. Furthermore,
v 00 ( x ) − M 00 [ v ] = v 00 ( x ) − max { v 10 ( x ) − c 01 , v 20 ( x ) − c 02 } . = v 00 ( x ) − v 10 ( x ) + min { c 01 , c 02 } = min { c 01 , c 02 } > 0. \begin{aligned}
v_{00}(x)-M_{00}[v]=v_{00}(x)-\max\{v_{10}(x)-c_{01},v_{20}(x)-c_{02}\}. \\
&=v_{00}(x)-v_{10}(x)+\min\{c_{01},c_{02}\} \\
&=\min\{c_{01},c_{02}\}>0.
\end{aligned} v 00 ( x ) − M 00 [ v ] = v 00 ( x ) − max { v 10 ( x ) − c 01 , v 20 ( x ) − c 02 } . = v 00 ( x ) − v 10 ( x ) + min { c 01 , c 02 } = min { c 01 , c 02 } > 0. Finally,
v 00 ( x ) − N 00 [ v ] = v 00 ( x ) − min { v 01 ( x ) + χ 01 , v 02 ( x ) + χ 02 } . v_{00}(x)-N_{00}[v]=v_{00}(x)-\min\{v_{01}(x)+\chi_{01},v_{02}(x)+\chi_{02}\}. v 00 ( x ) − N 00 [ v ] = v 00 ( x ) − min { v 01 ( x ) + χ 01 , v 02 ( x ) + χ 02 } . If the minimum is attained at
v 01 ( x ) + χ 01 , v_{01}(x)+\chi_{01}, v 01 ( x ) + χ 01 , then
v 00 ( x ) − N 00 [ v ] = v 00 ( x ) − v 01 ( x ) − χ 01 = − χ 01 < 0. v_{00}(x)-N_{00}[v]=v_{00}(x)-v_{01}(x)-\chi_{01}=-\chi_{01}<0. v 00 ( x ) − N 00 [ v ] = v 00 ( x ) − v 01 ( x ) − χ 01 = − χ 01 < 0. Otherwise, for x < x 02 x<x_{02} x < x 02
v 00 ( x ) − N 00 [ v ] = v 00 ( x ) − V ^ 02 ( x ) − A 02 x m 02 + − χ 02 . v_{00}(x)-N_{00}[v]=v_{00}(x)-\widehat{V}_{02}(x)-A_{02}x^{m_{02}^{+}}-\chi_{02}. v 00 ( x ) − N 00 [ v ] = v 00 ( x ) − V 02 ( x ) − A 02 x m 02 + − χ 02 . Define
G ( x ) = V ^ 00 ( x ) − V ^ 02 ( x ) − A 02 x m 02 + − χ 02 . G(x)=\widehat{V}_{00}(x)-\widehat{V}_{02}(x)-A_{02}x^{m_{02}^{+}}-\chi_{02}. G ( x ) = V 00 ( x ) − V 02 ( x ) − A 02 x m 02 + − χ 02 . A direct computation shows that
G ′ ′ ( x ) > 0. G''(x)>0. G ′′ ( x ) > 0. Hence, G ′ G' G ′
G ′ ( x 02 ) = 0 , G'(x_{02})=0, G ′ ( x 02 ) = 0 , it follows that
G ′ ( x ) < 0 , x < x 02 . G'(x)<0,\qquad x<x_{02}. G ′ ( x ) < 0 , x < x 02 . Using
G ( x 02 ) = − min { χ 20 , χ 21 } − χ 02 < 0 , G ( 0 ) = − χ 02 < 0 , G(x_{02})=-\min\{\chi_{20},\chi_{21}\}-\chi_{02}<0,\qquad G(0)=-\chi_{02}<0, G ( x 02 ) = − min { χ 20 , χ 21 } − χ 02 < 0 , G ( 0 ) = − χ 02 < 0 , we conclude that
G ( x ) < 0 , x < x 02 , G(x)<0,\qquad x<x_{02}, G ( x ) < 0 , x < x 02 , which proves
v 00 ( x ) − N 00 [ v ] < 0. v_{00}(x)-N_{00}[v]<0. v 00 ( x ) − N 00 [ v ] < 0. For x > x 02 x>x_{02} x > x 02
v 00 ( x ) − N 00 [ v ] = v 00 ( x ) − V 00 ( x ) − χ 02 − min { χ 20 , χ 21 } < 0. v_{00}(x)-N_{00}[v]=v_{00}(x)-V_{00}(x)-\chi_{02}-\min\{\chi_{20},\chi_{21}\}<0. v 00 ( x ) − N 00 [ v ] = v 00 ( x ) − V 00 ( x ) − χ 02 − min { χ 20 , χ 21 } < 0. Hence, v 00 v_{00} v 00
Step 2. Verification for v 02 v_{02} v 02
H [ v 02 ] ( x ) = 0. \mathcal{H}[v_{02}](x)=0. H [ v 02 ] ( x ) = 0. Moreover,
v 02 ( x ) − M 02 [ v ] = v 02 ( x ) − max { v 12 ( x ) − c 01 , v 22 ( x ) − c 02 } . = min { c 01 , c 02 } > 0. \begin{aligned}
v_{02}(x)-M_{02}[v]&=v_{02}(x)-\max\{v_{12}(x)-c_{01},v_{22}(x)-c_{02}\}.\\
&=\min\{c_{01},c_{02}\}>0.
\end{aligned} v 02 ( x ) − M 02 [ v ] = v 02 ( x ) − max { v 12 ( x ) − c 01 , v 22 ( x ) − c 02 } . = min { c 01 , c 02 } > 0. Next,
v 02 ( x ) − N 02 [ v ] = v 02 ( x ) − min { v 01 ( x ) + χ 21 , v 00 ( x ) + χ 20 } , = v 02 ( x ) − v 01 ( x ) − min { χ 21 , χ 20 } , = V ^ 02 ( x ) + A 02 x m 02 + − V ^ 01 ( x ) − min { χ 21 , χ 20 } . \begin{aligned}
v_{02}(x)-N_{02}[v]&=v_{02}(x)-\min\{v_{01}(x)+\chi_{21},v_{00}(x)+\chi_{20}\},\\
&=v_{02}(x)-v_{01}(x)-\min\{\chi_{21},\chi_{20}\},\\
&=\widehat{V}_{02}(x)+A_{02}x^{m_{02}^{+}}-\widehat{V}_{01}(x)-\min\{\chi_{21},\chi_{20}\}.
\end{aligned} v 02 ( x ) − N 02 [ v ] = v 02 ( x ) − min { v 01 ( x ) + χ 21 , v 00 ( x ) + χ 20 } , = v 02 ( x ) − v 01 ( x ) − min { χ 21 , χ 20 } , = V 02 ( x ) + A 02 x m 02 + − V 01 ( x ) − min { χ 21 , χ 20 } . Let
h 17 ( x ) = V ^ 02 ( x ) + A 02 x m 02 + − V ^ 01 ( x ) − min { χ 21 , χ 20 } . h_{17}(x)=\widehat{V}_{02}(x)+A_{02}x^{m_{02}^{+}}-\widehat{V}_{01}(x)-\min\{\chi_{21},\chi_{20}\}. h 17 ( x ) = V 02 ( x ) + A 02 x m 02 + − V 01 ( x ) − min { χ 21 , χ 20 } . h 17 ′ ( x ) > 0 h'_{17}(x)>0 h 17 ′ ( x ) > 0 h 17 h_{17} h 17
Using
h 17 ( x 02 ) = 0 , h_{17}(x_{02})=0, h 17 ( x 02 ) = 0 , we deduce that
v 02 ( x ) − N 02 [ v ] < 0. v_{02}(x)-N_{02}[v]<0. v 02 ( x ) − N 02 [ v ] < 0. Therefore, the QVI is satisfied on ( 0 , x 02 ) (0,x_{02}) ( 0 , x 02 )
H [ v 02 ] ( x ) = K 00 − K 02 K 02 x γ + min { χ 21 , χ 20 } , \mathcal{H}[v_{02}](x)=\frac{K_{00}-K_{02}}{K_{02}}x^\gamma+\min\{\chi_{21},\chi_{20}\}, H [ v 02 ] ( x ) = K 02 K 00 − K 02 x γ + min { χ 21 , χ 20 } , which is negative by definition of the switching threshold.
Hence,
min { H [ v 02 ] ( x ) , v 02 ( x ) − M 02 [ v ] ( x ) } < 0. \min\left\{\mathcal{H}[v_{02}](x),v_{02}(x)-M_{02}[v](x)\right\}<0. min { H [ v 02 ] ( x ) , v 02 ( x ) − M 02 [ v ] ( x ) } < 0. Finally,
v 02 ( x ) − N 02 [ v ] = v 02 ( x ) − min { v 01 ( x ) + χ 21 , v 00 ( x ) + χ 20 } . = v 02 ( x ) − v 01 ( x ) − min { χ 21 , χ 20 } . = V ^ 01 ( x ) + min { χ 21 , χ 20 } − V ^ 01 ( x ) − min { χ 21 , χ 20 } = 0. \begin{aligned}
v_{02}(x)-N_{02}[v]&=v_{02}(x)-\min\{v_{01}(x)+\chi_{21},v_{00}(x)+\chi_{20}\}.\\
&=v_{02}(x)-v_{01}(x)-\min\{\chi_{21},\chi_{20}\}.\\
&=\widehat{V}_{01}(x)+\min\{\chi_{21},\chi_{20}\}-\widehat{V}_{01}(x)-\min\{\chi_{21},\chi_{20}\}=0.
\end{aligned} v 02 ( x ) − N 02 [ v ] = v 02 ( x ) − min { v 01 ( x ) + χ 21 , v 00 ( x ) + χ 20 } . = v 02 ( x ) − v 01 ( x ) − min { χ 21 , χ 20 } . = V 01 ( x ) + min { χ 21 , χ 20 } − V 01 ( x ) − min { χ 21 , χ 20 } = 0. Hence, v 02 v_{02} v 02 ( x 02 , ∞ ) (x_{02},\infty) ( x 02 , ∞ ) ( 0 , ∞ ) (0,\infty) ( 0 , ∞ )
A Numerical Procedure for Computing the Value Functions
In this section, we propose a numerical procedure for computing the value functions when the qualitative structure of the switching regions is known, while the numerical values of the thresholds are not explicitly available.
Notation and Structure of the Switching Regions
Let us first introduce the following notation. For i , j , k , l i,j,k,l i , j , k , l
S i k ξ , j : = S_{ik}^{\xi,j}:= S ik ξ , j := i i i k k k j j j
S j l η , i : = S_{jl}^{\eta,i}:= S j l η , i := j j j l l l i i i
To illustrate our approach, we restrict attention to the particular case in which ( μ 0 , ν 0 ) = ( 0 , 0 ) (\mu_0,\nu_0)=(0,0) ( μ 0 , ν 0 ) = ( 0 , 0 )
S 01 ξ , 0 = ( 0 , x 01 0 ] , S 10 ξ , 0 = [ x 10 0 , + ∞ ) , (40) S_{01}^{\xi,0}=(0,x_{01}^{0}],\qquad S_{10}^{\xi,0}=[x_{10}^{0},+\infty), \tag*{(40)} S 01 ξ , 0 = ( 0 , x 01 0 ] , S 10 ξ , 0 = [ x 10 0 , + ∞ ) , (40) S 01 η , 0 = ( 0 , y 01 0 ] , S 12 η , 0 = [ y 12 0 , + ∞ ) . (41) S_{01}^{\eta,0}=(0,y_{01}^{0}],\qquad S_{12}^{\eta,0}=[y_{12}^{0},+\infty). \tag*{(41)} S 01 η , 0 = ( 0 , y 01 0 ] , S 12 η , 0 = [ y 12 0 , + ∞ ) . (41) All other switching regions are assumed to be empty.
Without loss of generality, we assume that the thresholds satisfy
x 01 0 < y 01 0 < x 10 0 < y 12 0 . (42) x_{01}^{0}<y_{01}^{0}<x_{10}^{0}<y_{12}^{0}. \tag*{(42)} x 01 0 < y 01 0 < x 10 0 < y 12 0 . (42) Auxiliary Functionals
For a , b , x > 0 a,b,x>0 a , b , x > 0
R 1 ( x , a ) : = E [ e − r τ a ] , (43) R_1(x,a):=\mathbb{E}\left[e^{-r\tau_a}\right], \tag*{(43)} R 1 ( x , a ) := E [ e − r τ a ] , (43) R 2 ( x , a , b ) : = E [ e − r τ a b ] , (44) R_2(x,a,b):=\mathbb{E}\left[e^{-r\tau_{ab}}\right], \tag*{(44)} R 2 ( x , a , b ) := E [ e − r τ ab ] , (44) R 3 ( x , a , b ) : = E [ e − r τ a 1 { τ a < τ b } ] . (45) R_3(x,a,b):=\mathbb{E}\left[e^{-r\tau_a}\mathbf{1}_{\{\tau_a<\tau_b\}}\right]. \tag*{(45)} R 3 ( x , a , b ) := E [ e − r τ a 1 { τ a < τ b } ] . (45) Where τ a \tau_a τ a τ b \tau_b τ b
We also introduce the following expected discounted running-payoff functionals:
F 1 ( a ) : = E [ ∫ 0 τ a e − r s f ( X s x ) d s ] . (46) F_1(a) := \mathbb{E}\left[\int_0^{\tau_a} e^{-rs}f(X_s^x)\,\mathrm{d}s\right]. \tag*{(46)} F 1 ( a ) := E [ ∫ 0 τ a e − r s f ( X s x ) d s ] . (46) F 2 ( a , b ) : = E [ ∫ τ a τ b e − r s f ( X s x ) d s ] . (47) F_2(a,b) := \mathbb{E}\left[\int_{\tau_a}^{\tau_b} e^{-rs}f(X_s^x)\,\mathrm{d}s\right]. \tag*{(47)} F 2 ( a , b ) := E [ ∫ τ a τ b e − r s f ( X s x ) d s ] . (47) F 3 ( a , b ) : = E [ ∫ 0 τ a ∧ τ b e − r s f ( X s x ) d s ] . (48) F_3(a,b) := \mathbb{E}\left[\int_0^{\tau_a\wedge\tau_b} e^{-rs}f(X_s^x)\,\mathrm{d}s\right]. \tag*{(48)} F 3 ( a , b ) := E [ ∫ 0 τ a ∧ τ b e − r s f ( X s x ) d s ] . (48) The computation of these expectation functionals can be found in the Appendix of [18 ] .
Case x < x 01 0 x < x_{01}^0 x < x 01 0
Assume that
x < x 01 0 . x < x_{01}^0. x < x 01 0 . In this case, the optimal strategy can be described as follows:
Initial regime ( μ 0 , ν 0 ) = ( 0 , 0 ) (\mu_0,\nu_0)=(0,0) ( μ 0 , ν 0 ) = ( 0 , 0 ) c 01 c_{01} c 01 χ 01 \chi_{01} χ 01 J 11 J_{11} J 11 ( μ 0 , ν 0 ) = ( 1 , 1 ) (\mu_0,\nu_0)=(1,1) ( μ 0 , ν 0 ) = ( 1 , 1 )
J 00 ( x ) = − c 01 + χ 01 + J 11 ( x ) , and then v 00 ( x ) = − c 01 + χ 01 + V ^ 11 ( x ) . J_{00}(x)=-c_{01}+\chi_{01}+J_{11}(x), \qquad\text{and then} \qquad v_{00}(x)=-c_{01}+\chi_{01}+\widehat{V}_{11}(x). J 00 ( x ) = − c 01 + χ 01 + J 11 ( x ) , and then v 00 ( x ) = − c 01 + χ 01 + V 11 ( x ) . Case x 01 0 < x < y 01 0 x_{01}^0<x<y_{01}^0 x 01 0 < x < y 01 0
Assume that
x 01 0 < x < y 01 0 . x_{01}^0<x<y_{01}^0. x 01 0 < x < y 01 0 . In this case, the optimal strategy can be described as follows.
Initial regime ( μ 0 , ν 0 ) = ( 0 , 0 ) (\mu_0,\nu_0)=(0,0) ( μ 0 , ν 0 ) = ( 0 , 0 ) χ 01 \chi_{01} χ 01 y 12 0 y_{12}^0 y 12 0 τ y 12 0 \tau_{y_{12}^0} τ y 12 0 χ 12 \chi_{12} χ 12 ( μ 0 , ν 0 ) = ( 0 , 2 ) (\mu_0,\nu_0)=(0,2) ( μ 0 , ν 0 ) = ( 0 , 2 )
J 00 ( x ) = E [ ∫ 0 τ y 12 0 e − r t f ( X t x ) d t + e − r τ y 12 0 ( χ 12 + J 02 ( x ) ) ] . J_{00}(x)=\mathbb{E}\left[\int_0^{\tau_{y_{12}^0}}e^{-rt}f(X_t^x)\,\mathrm{d}t+e^{-r\tau_{y_{12}^0}}\left(\chi_{12}+J_{02}(x)\right)\right]. J 00 ( x ) = E [ ∫ 0 τ y 12 0 e − r t f ( X t x ) d t + e − r τ y 12 0 ( χ 12 + J 02 ( x ) ) ] . There is no switching in the state ( μ 0 , ν 0 ) = ( 0 , 2 ) (\mu_0,\nu_0)=(0,2) ( μ 0 , ν 0 ) = ( 0 , 2 ) J 02 ( x ) = V ^ 02 ( x ) J_{02}(x)=\widehat{V}_{02}(x) J 02 ( x ) = V 02 ( x )
We get
v 00 ( x , y 12 0 ) = χ 01 + F 1 ( y 12 0 ) + ( χ 12 + J 02 ( y 12 0 ) ) R 1 ( x , y 12 0 ) . v_{00}(x,y_{12}^0)=\chi_{01}+F_1(y_{12}^0)+\left(\chi_{12}+J_{02}(y_{12}^0)\right)R_1(x,y_{12}^0). v 00 ( x , y 12 0 ) = χ 01 + F 1 ( y 12 0 ) + ( χ 12 + J 02 ( y 12 0 ) ) R 1 ( x , y 12 0 ) . Player 2 aims at minimizing the payoff, so
v 00 ( x ) = min y 12 0 [ χ 01 + F 1 ( y 12 0 ) + ( χ 12 + J 02 ( y 12 0 ) ) R 1 ( x , y 12 0 ) ] . v_{00}(x)=\min_{y_{12}^0}\left[\chi_{01}+F_1(y_{12}^0)+\left(\chi_{12}+J_{02}(y_{12}^0)\right)R_1(x,y_{12}^0)\right]. v 00 ( x ) = y 12 0 min [ χ 01 + F 1 ( y 12 0 ) + ( χ 12 + J 02 ( y 12 0 ) ) R 1 ( x , y 12 0 ) ] . Case y 01 0 < x < x 10 0 y_{01}^0<x<x_{10}^0 y 01 0 < x < x 10 0
Assume that
y 01 0 < x < x 10 0 . y_{01}^0<x<x_{10}^0. y 01 0 < x < x 10 0 . In this case, the optimal strategy can be described as follows. Initial regime ( μ 0 , ν 0 ) = ( 0 , 0 ) (\mu_0,\nu_0)=(0,0) ( μ 0 , ν 0 ) = ( 0 , 0 ) y 01 0 y_{01}^{0} y 01 0 τ y 01 0 \tau_{y_{01}}^{0} τ y 01 0 χ 01 \chi_{01} χ 01 y 12 0 y_{12}^{0} y 12 0 τ y 12 0 \tau_{y_{12}}^{0} τ y 12 0 χ 12 \chi_{12} χ 12 ( μ 0 , ν 0 ) = ( 0 , 2 ) (\mu_0,\nu_0)=(0,2) ( μ 0 , ν 0 ) = ( 0 , 2 )
J 00 ( x ) = E [ ∫ 0 τ y 01 0 e − r t f ( X t x ) d t + e − r τ y 01 0 ( χ 01 + J 01 ( y 01 0 ) ) ] . J_{00}(x)=\mathbb{E}\left[\int_{0}^{\tau_{y_{01}}^{0}}e^{-rt}f(X_t^x)\,\mathrm{d}t+e^{-r\tau_{y_{01}}^{0}}\left(\chi_{01}+J_{01}(y_{01}^{0})\right)\right]. J 00 ( x ) = E [ ∫ 0 τ y 01 0 e − r t f ( X t x ) d t + e − r τ y 01 0 ( χ 01 + J 01 ( y 01 0 ) ) ] . J 01 ( y 01 0 ) = E [ ∫ 0 τ y 12 0 e − r t f ( X t x ) d t + e − r τ y 12 0 ( χ 12 + J 02 ( y 12 0 ) ) ] . J_{01}(y_{01}^{0})=\mathbb{E}\left[\int_{0}^{\tau_{y_{12}}^{0}}e^{-rt}f(X_t^x)\,\mathrm{d}t+e^{-r\tau_{y_{12}}^{0}}\left(\chi_{12}+J_{02}(y_{12}^{0})\right)\right]. J 01 ( y 01 0 ) = E [ ∫ 0 τ y 12 0 e − r t f ( X t x ) d t + e − r τ y 12 0 ( χ 12 + J 02 ( y 12 0 ) ) ] . Thus, the computation of J 01 J_{01} J 01 J 01 J_{01} J 01 J 00 J_{00} J 00
Case x 10 0 < x < y 12 0 x_{10}^{0}<x<y_{12}^{0} x 10 0 < x < y 12 0
Assume that
x 10 0 < x < y 12 0 . x_{10}^{0}<x<y_{12}^{0}. x 10 0 < x < y 12 0 . In this case, the optimal strategy can be described as follows.
Initial regime ( μ 0 , ν 0 ) = ( 0 , 0 ) (\mu_0,\nu_0)=(0,0) ( μ 0 , ν 0 ) = ( 0 , 0 ) y 01 0 y_{01}^{0} y 01 0 τ y 01 0 \tau_{y_{01}}^{0} τ y 01 0 χ 01 \chi_{01} χ 01 y 12 0 y_{12}^{0} y 12 0 τ y 12 0 \tau_{y_{12}}^{0} τ y 12 0 χ 12 \chi_{12} χ 12 ( μ 0 , ν 0 ) = ( 0 , 2 ) (\mu_0,\nu_0)=(0,2) ( μ 0 , ν 0 ) = ( 0 , 2 )
J 00 ( x ) = E [ ∫ 0 τ y 01 0 e − r t f ( X t x ) d t + e − r τ y 01 0 ( χ 01 + J 01 ( y 01 0 ) ) ] . J_{00}(x)=\mathbb{E}\left[\int_{0}^{\tau_{y_{01}}^{0}}e^{-rt}f(X_t^x)\,\mathrm{d}t+e^{-r\tau_{y_{01}}^{0}}\left(\chi_{01}+J_{01}(y_{01}^{0})\right)\right]. J 00 ( x ) = E [ ∫ 0 τ y 01 0 e − r t f ( X t x ) d t + e − r τ y 01 0 ( χ 01 + J 01 ( y 01 0 ) ) ] . J 01 ( y 01 0 ) = E [ ∫ 0 τ y 12 0 e − r t f ( X t x ) d t + e − r τ y 12 0 ( χ 12 + J 02 ( y 12 0 ) ) ] . J_{01}(y_{01}^{0})=\mathbb{E}\left[\int_{0}^{\tau_{y_{12}}^{0}}e^{-rt}f(X_t^x)\,\mathrm{d}t+e^{-r\tau_{y_{12}}^{0}}\left(\chi_{12}+J_{02}(y_{12}^{0})\right)\right]. J 01 ( y 01 0 ) = E [ ∫ 0 τ y 12 0 e − r t f ( X t x ) d t + e − r τ y 12 0 ( χ 12 + J 02 ( y 12 0 ) ) ] . Case x > y 12 0 x>y_{12}^{0} x > y 12 0
Assume that
x > y 12 0 . x>y_{12}^{0}. x > y 12 0 . In this case, the optimal strategy can be described as follows.
Initial regime ( μ 0 , ν 0 ) = ( 0 , 0 ) (\mu_0,\nu_0)=(0,0) ( μ 0 , ν 0 ) = ( 0 , 0 ) y 01 0 y_{01}^{0} y 01 0 τ y 01 0 \tau_{y_{01}}^{0} τ y 01 0 χ 01 \chi_{01} χ 01 y 12 0 y_{12}^{0} y 12 0 τ y 12 0 \tau_{y_{12}}^{0} τ y 12 0 χ 12 \chi_{12} χ 12 ( μ 0 , ν 0 ) = ( 0 , 2 ) (\mu_0,\nu_0)=(0,2) ( μ 0 , ν 0 ) = ( 0 , 2 )
J 00 ( x ) = E [ ∫ 0 τ y 01 0 e − r t f ( X t x ) d t + e − r τ y 01 0 ( χ 01 + J 01 ( y 01 0 ) ) ] . J_{00}(x)=\mathbb{E}\left[\int_{0}^{\tau_{y_{01}}^{0}}e^{-rt}f(X_t^x)\,\mathrm{d}t+e^{-r\tau_{y_{01}}^{0}}\left(\chi_{01}+J_{01}(y_{01}^{0})\right)\right]. J 00 ( x ) = E [ ∫ 0 τ y 01 0 e − r t f ( X t x ) d t + e − r τ y 01 0 ( χ 01 + J 01 ( y 01 0 ) ) ] . J 01 ( y 01 0 ) = E [ ∫ 0 τ y 12 0 e − r t f ( X t x ) d t + e − r τ y 12 0 ( χ 12 + J 02 ( y 12 0 ) ) ] . J_{01}(y_{01}^{0})=\mathbb{E}\left[\int_{0}^{\tau_{y_{12}}^{0}}e^{-rt}f(X_t^x)\,\mathrm{d}t+e^{-r\tau_{y_{12}}^{0}}\left(\chi_{12}+J_{02}(y_{12}^{0})\right)\right]. J 01 ( y 01 0 ) = E [ ∫ 0 τ y 12 0 e − r t f ( X t x ) d t + e − r τ y 12 0 ( χ 12 + J 02 ( y 12 0 ) ) ] . Graphic Illustration
In this section, we shall provide illustrations of the players’ strategies. We rely on the case where K i j = C s t 1 < K i 2 = C s t 2 K_{ij}=Cst_1<K_{i2}=Cst_2 K ij = C s t 1 < K i 2 = C s t 2 i = 0 , 1 , 2 i=0,1,2 i = 0 , 1 , 2 j = 0 , 1 j=0,1 j = 0 , 1 v 02 v_{02} v 02 v 12 v_{12} v 12 R 2 R_2 R 2 ν 0 = 2 \nu_0=2 ν 0 = 2 R 0 R_0 R 0 μ 0 = 0 \mu_0=0 μ 0 = 0 R 1 R_1 R 1 μ 0 = 1 \mu_0=1 μ 0 = 1 χ 20 − χ 21 \chi_{20}-\chi_{21} χ 20 − χ 21
Figure 1. μ 0 = 0 \mu_0=0 μ 0 = 0 ν 0 = 2 \nu_0=2 ν 0 = 2 χ 20 − χ 21 < 0 \chi_{20}-\chi_{21}<0 χ 20 − χ 21 < 0
Figure 2. μ 0 = 0 \mu_0=0 μ 0 = 0 ν 0 = 2 \nu_0=2 ν 0 = 2 χ 20 − χ 21 > 0 \chi_{20}-\chi_{21}>0 χ 20 − χ 21 > 0
Figure 3. μ 0 = 1 \mu_0=1 μ 0 = 1 ν 0 = 2 \nu_0=2 ν 0 = 2 χ 20 − χ 21 < 0 \chi_{20}-\chi_{21}<0 χ 20 − χ 21 < 0
Figure 4. μ 0 = 1 \mu_0=1 μ 0 = 1 ν 0 = 2 \nu_0=2 ν 0 = 2 χ 20 − χ 21 > 0 \chi_{20}-\chi_{21}>0 χ 20 − χ 21 > 0
Conclusion
In this paper, we have investigated a two-player zero-sum stochastic differential game with three regimes, focusing on the impact of regime switching on the value function and the optimal strategies of the players. By formulating the problem in terms of the associated Hamilton–Jacobi–Bellman–Isaacs system, we have provided a framework for analyzing the interactions between the different regimes and the resulting switching decisions.
Under suitable assumptions on the model coefficients and switching structure, we have characterized the value function using the viscosity solutions approach and derived explicit representations for a number of relevant configurations. The analysis also provides a qualitative description of the optimal value function in the case where we know the structure of the switching regions without knowing the switching thresholds explicitly. In particular, the results illustrate how the strategic interaction between the two players and the possibility of moving between regimes jointly determine the structure of the optimal policies.
The three-regime setting considered in this work provides a useful framework that captures a richer class of switching mechanisms than the standard two-regime case, while still allowing for an explicit analysis of several configurations. The results obtained here can therefore serve as a basis for studying more general regime-switching stochastic games.
Several directions for future research are possible. One natural extension would be to consider a larger number of regimes and more general switching structures, for which the characterization of the switching regions may become substantially more involved. Another interesting direction would be to develop numerical procedures for computing the switching thresholds when explicit expressions are no longer available. Finally, it would be of interest to investigate extensions involving asymmetric information, non-zero-sum interactions, or regime-dependent diffusion coefficients and switching costs.
Declarations
Conflict of interest. The authors have not disclosed any competing interests.
