Dynamic Marketing Model: Optimization of Retailer’s Role

Abstract

We study a vertical control distribution channel in which a manufacturer sells a single kind of good to a retailer. The state variables are the cumulative sales and the retailer’s motivation. The manufacturer chooses wholesale price discount while retailer chooses pass-through. We assume that the wholesale price discount increases the retailer’s sale motivation thus improving sales. In contrast to previous settings, we focus on the maximization of retailer’s profit with respect to pass-through. The arising problem is linear with respect to both cumulative sales and the retailer’s motivation, while it is quadratic with respect to wholesale price discount and pass-through. We obtain a complete description of optimal strategies and optimal trajectories. In particular, we demonstrate that the number of switches for change in the type of optimal policy is no more than one.

A Appendix

1.1 A.1 Assumptions About the Input Data of the Model

In this Section we show (4) and (5).

The Choice of Pass-Through Upper Boundary. The choice of $B_2$ is carried out from the following considerations. It is clear that the retailer can not receive a negative profit. We will assume that in the “most unfavorable” case for the retailer, i.e., when $\beta (t)=B_2 \ \forall t \in [t_1, t_2],$ the profit is zero, i.e.,

$$\begin{aligned} \alpha p\int \limits _{t_1}^{t_2}(1-B_2)\dot{x}(t)dt= \alpha p(1-B_2)(x(t_2)-x(t_1))=0. \end{aligned}$$

Hence, since $x(t_1)=0,$ we obtain $\alpha p(1-B_2)=0.$ Therefore, $B_2=1.$

Thus, (4) holds.

The Concavity of Under the Constant Pass-Through. Let the control $\beta (t)$ be constant, i.e., $\beta (t)=\overline{\beta }\ \forall t\in [t_1,t_2].$ This means that the wholesale price is constant throughout the sales period. Then it makes sense to assume that buyers “forget” about the discount provided by the firm (and retailer). We assume that in this case the function x(t) is concave, i.e., $\ddot{x}(t)\le 0 \ \forall t\in [t_1,t_2].$ This happens when the market is mature and saturation occurs. Moreover, since x(t) is accumulated sales, we assume that the coefficients of the model are such that $\dot{x}(t)\ge 0.$ The latter should be performed, in particular, for any constant control.

Consider the system

$$\begin{aligned} \begin{array}{lcr} \dot{x}(t)=-\theta x(t)+\delta M(t)+\eta _\alpha \overline{\beta }, \\ \dot{M}(t)=\gamma \dot{x}(t)+\varepsilon \varOmega , \\ x(t_1)=0, \ M(t_1)=\overline{M}. \end{array} \end{aligned}$$

(11)

We get

$$\begin{aligned} \ddot{x}(t)=-\theta \dot{x}(t)+\delta \dot{M}(t)= -a\dot{x}(t) + \delta \varepsilon \varOmega , \end{aligned}$$

(12)

If $a=0$ then it is not possible to obtain concavity of x(t) due to (12).

Let $a\ne 0.$ Then we get from (11)

$$\begin{aligned} x(t)=\displaystyle \frac{C(\overline{\beta })}{a^2}\cdot (1-e^{a(t_1-t)})+\frac{\delta \varepsilon \varOmega }{a}(t-t_1), \end{aligned}$$

(13)

where $C(\overline{\beta })=a\delta \overline{M}+b\overline{\beta }-c.$

Due to (13), $\ddot{x}(t)=-C(\overline{\beta })e^{a(t_1-t)}.$ So the cumulative sales function is concave for every $\overline{\beta } \in [B_1,1]$ if and only if $C(\overline{\beta })\ge 0 \ \forall \overline{\beta } \in [B_1,1],$ i.e.,

$$\begin{aligned} a\delta \overline{M}+b\overline{\beta }\ge c \ \forall \overline{\beta } \in [B_1,1]. \end{aligned}$$

(14)

Note that (14) can be only if the condition

$$\begin{aligned} a>0 \end{aligned}$$

(15)

holds. Moreover if (15) holds then (14) is true if and only if

$$\begin{aligned} a\delta \overline{M}+b B_1\ge c \end{aligned}$$

(16)

holds. Now we get (5) by substituting (6) to (16). Remark that assumption (16) implies (15).

1.2 A.2 Solution of Retailer’s Problem

Equivalent Problem. Let us introduce the new control

$$\begin{aligned} u(t)=\beta (t)-1\le 0 \end{aligned}$$

and the new state variable

$$\begin{aligned} y(t)=-\theta x(t)+\delta M(t)+\eta _\alpha . \end{aligned}$$

Then the Retailer’s problem can be rewritten as

$$\begin{aligned} \begin{array}{l} -\alpha p \int \limits _{t_1}^{t_2}u(t)(y(t)+\eta _\alpha u(t))dt \longrightarrow max, \\ \dot{x}(t)=y(t)+\eta _\alpha u(t), \\ \dot{y}(t)= -a y(t)-b u(t)+c, \\ x(t_1)=0, \ \ \ y(t_1)=y_1, \\ u(t)\in [u_1,0]\subseteq [-1,0], \\ \end{array} \end{aligned}$$

where $u_1=B_1-1<0, \ \ y_1=\delta \overline{M}+\eta _\alpha >0$ while definition of a, b and c see in (6), (7) and (8). Note that

$$\begin{aligned} a y_1+b u_1-c = a\left( \delta \overline{M}+\eta _\alpha B_1\right) -c. \end{aligned}$$

So due to (16) we get

$$\begin{aligned} a y_1+b u_1-c \ge 0. \end{aligned}$$

(17)

Solution of Equivalent Problem. Consider the Hamiltonian function

$$\begin{aligned} H=-p\alpha (y(t)+\eta _\alpha u(t))u(t)+z_0(t)(y(t)+\eta _\alpha u(t))+z(t)( -a y(t)-b u(t)+c) \end{aligned}$$

and the Lagrangian function

$$\begin{aligned} L=H+\mu _1(t)(u(t)-u_1)-\mu _2(t)u(t). \end{aligned}$$

Due to the Pontryagin Maximum Principle one has that if $u^*(t)$ is the optimal control and $x^*(t), y^*(t)$ are optimal state variables, then continuous and piece-wise continuously differentiable functions $z_0(t)$ and z(t) and piece-wise continuous functions $\mu _1(t)$ and $\mu _2(t)$ must exist such that (cf. p. 546 of [1])

$$\begin{aligned} \dot{z_0}(t)= - \frac{\partial L^*}{\partial x}, \ \ z_0(t_2)=0, \ \ \dot{z}(t)= - \frac{\partial L^*}{\partial y}, \ \ z(t_2)=0, \ \ \frac{\partial L^*}{\partial u}=0, \end{aligned}$$

(18)

where, as usual, $L^*=L(y^*,x^*,u^*,z,z_0,\mu _1,\mu _2)$; moreover slackness complementary conditions

$$\begin{aligned} \mu _1(t)\ge 0, \ \ \mu _1(t)(u^*(t)-u_1)=0, \ \ \mu _2(t)\ge 0, \ \ \mu _2(t)u^*(t)=0 \end{aligned}$$

(19)

hold. Note that in our case $\displaystyle \frac{\partial L^*}{\partial x}\equiv 0.$ Therefore from (18) we get $z_0(t)\equiv 0.$ So equations $\dot{z}(t)= - \displaystyle \frac{\partial L^*}{\partial y}$ and $\displaystyle \frac{\partial L^*}{\partial u}=0$ in (18) are

$$\begin{aligned} \dot{z}(t)=a z(t)+ p\alpha u^*(t), \end{aligned}$$

(20)

$$\begin{aligned} \mu _1(t)-\mu _2(t)=p\alpha y^*(t)+b z(t)+2p\alpha \eta _\alpha u^*(t). \end{aligned}$$

(21)

Lemma 1

Functions $\mu _1(t)$ and $\mu _2(t)$ are continuous.

Proof

(Cf. p. 546 of [1].). Function $u^*$ is continuous since the Hamiltonian function H is strictly concave w.r.t. u (see e.g. p. 86 of [18]). So also $\mu _1(t)-\mu _2(t)$ are continuous due to (21). From (19) we get $\mu _1(t)(u^*(t)-u_1)=\mu _2(t)u^*(t)$ therefore $(\mu _1(t)-\mu _2(t))u^*(t)=\mu _1u_1.$ Thus, functions $\mu _1(t)$ and $\mu _2(t)$ are continuous.

Lemma 2

$\mu _2(t)=0 \ \forall t \in [t_1,t_2].$

Proof

(Cf. pp. 546–547 of [1].). Suppose, conversely, that $\exists \ \tilde{t} \in (t_1,t_2): \mu _2(\tilde{t})>0.$ Due to Lemma 1, an interval exists where $\mu _2$ is positive. Let $(\rho _1,\rho _2) \subset (t_1,t_2)$ be the interval of maximum length such that

$$\begin{aligned} \mu _2(t)>0, \ t \in (\rho _1,\rho _2). \end{aligned}$$

(22)

Due to (19), $u^*(t)=\mu _1(t)=0 \ \forall t \in (\rho _1,\rho _2).$ Hence in $(\rho _1,\rho _2)$ (21) becomes

$$\begin{aligned} \mu _2(t)=-p\alpha y^*(t)-b z(t)=-p\alpha \left( C_y e^{-a t}+\frac{c}{a}\right) -b C_z e^{a t}, \end{aligned}$$

(23)

where

$$\begin{aligned} C_y=\left( y(\rho _1) -\displaystyle \frac{c}{a}\right) e^{a\rho _1}, \ \ C_z=z(\rho _2)e^{-a\rho _2}. \end{aligned}$$

(24)

Let us consider the all possible cases.

• Let $\rho _1=t_1,\rho _2=t_2.$ Since $y(t_1)=y_1>0, z(t_2)=0,$ we get from (23), (24) and (15)

  $$\begin{aligned} \mu _2(t)=-p\alpha \left( y_1 e^{a (t_1-t)}+\frac{c}{a}\cdot \left( 1-e^{a (t_1-t)}\right) \right) <0, \ t \in [t_1,t_2], \end{aligned}$$

  (25)

  But (25) contradicts (19).

• Let $\rho _1=t_1,\rho _2<t_2.$ Then

  $$\begin{aligned} \mu _2(t_1)>0, \ \mu _2(\rho _2)=0 \end{aligned}$$

  (26)

  and, moreover, for $t \in [t_1,\rho _2]$ we get

  $$\begin{aligned} \mu _2(t)=-p\alpha \left( y_1 e^{a (t_1-t)}+\frac{c}{a}\cdot \left( 1-e^{a (t_1-t)}\right) \right) -a\eta _\alpha z(\rho _2)e^{a(t-\rho _2)}. \end{aligned}$$

  (27)

  Hence to keep (22) we need

  $$\begin{aligned} z(\rho _2)<0. \end{aligned}$$

  (28)

  We get from (27)

  $$\begin{aligned} \dot{\mu }_2(t)=a^2 \cdot \left( p\alpha \left( a y_1 -c\right) e^{a (t_1-t)} -\eta _\alpha z(\rho _2)e^{a(t-\rho _2)}\right) , \ t \in [t_1,\rho _2]. \end{aligned}$$

  (29)

  From (17), (28) and (29) we get $\dot{\mu }_2(t)>0, \ t \in [t_1,\rho _2],$ which contradicts (26).

• Let $\rho _1<t_1,\rho _2=t_2.$ Then

  $$\begin{aligned} \mu _2(\rho _1)=0, \ \mu _2(t_2)>0 \end{aligned}$$

  (30)

  and

  $$\begin{aligned} \mu _2(t)=-p\alpha \left( y(\rho _1)e^{a(\rho _1-t)} +\frac{c}{a}\cdot \left( 1-e^{a(\rho _1-t)} \right) \right) , \ t \in [\rho _1,t_2]. \end{aligned}$$

  (31)

  Moreover, (31) becomes due to (30)

  $$\begin{aligned} \mu _2(t)=-\frac{p\alpha c}{a}\cdot \left( 1-e^{a(\rho _1-t)} \right) , \ t \in [\rho _1,t_2]. \end{aligned}$$

  (32)

  From (32) we get $\dot{\mu }_2(t)<0, \ t \in [\rho _1,t_2],$ which contradicts (30).

• Let $\rho _1<t_1,\rho _2<t_2.$ Then

  $$\begin{aligned} \mu _2(\rho _1)=\mu _2(\rho _2)=0. \end{aligned}$$

  (33)

  Moreover

  $$\begin{aligned} \ddot{\mu }_2(t)=a^2\left( -p\alpha C_y e^{-a t}-a\eta _\alpha C_z e^{a t}\right) =a^2\left( \mu _2(t)+\frac{\delta \varepsilon \varOmega }{a}\right) , \ t \in (\rho _1,\rho _2). \end{aligned}$$

  Hence, due to (15) and (22), we get $\ddot{\mu }_2(t)>0, \ t \in (\rho _1,\rho _2),$ i.e., function $\mu _2(t)$ is strictly convex in $[\rho _1,\rho _2]$ which contradicts (22) and (33).

Thus, (19) and (21) are

$$\begin{aligned} \mu _1(t)\ge 0, \ \ \mu _1(t)(u^*(t)-u_1)=0, \end{aligned}$$

(34)

$$\begin{aligned} \mu _1(t)=p\alpha y^*(t)+b z(t)+2p\alpha \eta _\alpha u^*(t). \end{aligned}$$

(35)

Lemma 3

Let for some $\rho _0<\rho _1<\rho _2<\rho _3$ be

$$\begin{aligned} \mu _1(t) \ \left\{ \begin{array}{ll} =0 \ &{} t\in [\rho _0,\rho _1], \\ >0 \ &{} t\in (\rho _1,\rho _2), \\ =0 \ &{} t\in [\rho _2,\rho _3]. \\ \end{array} \right. \end{aligned}$$

(36)

Then

$$\begin{aligned} u^*(t)=\left\{ \begin{array}{ll} E_2t^2+D_1t+D_0 \ &{} t\in [\rho _0,\rho _1], \\ u_1 \ &{} t\in [\rho _1,\rho _2], \\ E_2t^2+E_1t+E_0 \ &{} t\in [\rho _2,\rho _3], \\ \end{array} \right. \end{aligned}$$

(37)

where $E_2=\displaystyle \frac{a c}{4\eta _\alpha }$ while $D_1,D_0,E_1,E_0$ are some constants.

Proof

Due to (34) and (35), we get from (36)

$$\begin{aligned} u^*(t)=\left\{ \begin{array}{ll} -\displaystyle \frac{1}{2p\alpha \eta _\alpha }\cdot \left( p\alpha y^{*}(t)+b z(t)\right) \ &{} t\in [\rho _0,\rho _1], \\ u_1 \ &{} t\in [\rho _1,\rho _2], \\ -\displaystyle \frac{1}{2p\alpha \eta _\alpha }\cdot \left( p\alpha y^{*}(t)+b z(t)\right) \ &{} t\in [\rho _2,\rho _3].\\ \end{array} \right. \end{aligned}$$

Hence $y^{*}(t),z(t)$ and $u^*(t)$ in $[\rho _0,\rho _1]\cup [\rho _2,\rho _3]$ satisfy

$$\begin{aligned} \left\{ \begin{array}{l} \dot{y}^*(t)= -a y^*(t)-b u^*(t)+c, \\ \dot{z}(t)=a z(t)+ p\alpha u^*(t), \\ u^*(t)= -\displaystyle \frac{1}{2p\alpha \eta _\alpha }\cdot \left( p\alpha y^{*}(t)+bz(t)\right) . \\ \end{array} \right. \end{aligned}$$

(38)

In $[\rho _0,\rho _1]$ and in $[\rho _2,\rho _3],$ the solution of (38) has a form

$$\begin{aligned} \left\{ \begin{array}{l} y^{*}(t)=-\displaystyle \frac{ac}{4}\cdot t^{2}+C_{1}\cdot t+C_{2}, \\ z(t)=\displaystyle \frac{p\alpha }{b}\cdot \left( -\displaystyle \frac{ac}{4}\cdot t^{2}+\left( C_{1}-c\right) \cdot t+\displaystyle \frac{2C_{1}-2c a+aC_{2}}{a}\right) , \\ u^{*}(t)=\displaystyle \frac{1}{b}\cdot {\displaystyle \left( \frac{a^{2}c}{4}\cdot t^{2}+{\displaystyle \frac{a\left( c-2C_{1}\right) }{2}\cdot t-\left( C_{1}-c+aC_{2}\right) }\right) }. \\ \end{array} \right. \end{aligned}$$

(39)

Hence (37) holds.

Note that $y^{*}(t)$ and z(t) in $[\rho _1,\rho _2]$ satisfy

$$\begin{aligned} \left\{ \begin{array}{l} \dot{y}^*(t)= -a y^*(t)-b u_1+c, \\ \dot{z}(t)=a z(t)+ p\alpha u_1, \\ \end{array} \right. \end{aligned}$$

(40)

i.e.,

$$\begin{aligned} \left\{ \begin{array}{l} y^{*}(t)=C_y\cdot e^{-at}+\displaystyle \frac{c-bu_{1}}{a}, \\ z(t)= C_z e^{at}-\displaystyle \frac{p\alpha u_{1}}{a}, \\ \end{array} \right. \end{aligned}$$

(41)

where

$$\begin{aligned} C_y=\left( y(\rho _1) -\displaystyle \frac{c-b u_1}{a}\right) e^{a\rho _1}, \ \ C_z=\left( z(\rho _2)+\frac{p\alpha u_1}{a}\right) e^{-a\rho _2}. \end{aligned}$$

Lemma 4

Either $\mu _1(t)=0 \ \forall t \in [t_1,t_2]$ or only one interval $(\rho _1,\rho _2) \subset [t_1,t_2]$ exists such that $\mu _1(t)>0, \ t \in (\rho _1,\rho _2).$

Proof

Due to Lemma 3, if for some $t_1 \le \rho _0<\rho _1<\rho _2<\rho _3\le t_2$ (37) holds, then (see (39)) $u^{*}(t)=u_1 \ \forall t \in [\rho _1,\rho _2]$ while function $u^{*}(t)$ is strictly larger than $u_1$ and strictly convex in $[\rho _0,\rho _1] \cup [\rho _2,\rho _3].$ Hence function $u^{*}(t)$ strictly decreases in $[\rho _0,\rho _1]$ and strictly increases in $[\rho _2,\rho _3].$ Hence function $u^{*}(t)$ strictly decreases in $[t_1,\rho _1]$ and strictly increases in $[\rho _2,t_2].$ Hence $\mu _1(t)=0 \ \forall t \in [t_1,\rho _1] \cup [\rho _2,t_2].$

From Lemmas 3 and 4 we get

Corollary 1

The optimal control $u^*$ has the form

$$\begin{aligned} u^*(t)=\left\{ \begin{array}{ll} E_2t^2+D_1t+D_0 \ &{} t\in [t_1,\rho _1], \\ u_1 \ &{} t\in [\rho _1,\rho _2], \\ E_2t^2+E_1t+E_0 \ &{} t\in [\rho _2,t_2], \\ \end{array} \right. \end{aligned}$$

(42)

where $E_2=\displaystyle \frac{a^2 c}{4b}$ while $D_1,D_0,E_1,E_0$ are some constants.

Lemma 5

The optimal control can has no more than one switch.

Proof

Let, by contradiction,

$$\begin{aligned} t_1<\rho _1<\rho _2<t_2 \end{aligned}$$

(43)

and

$$\begin{aligned} y^{*}\left( t\right) =\left\{ \begin{array}{ll} -\displaystyle \frac{ac}{4}\cdot t^{2}+C_{1}\cdot t+C_{2} \ &{} t\in [t_1,\rho _1], \\ y^{*}\left( \rho _{1}\right) \cdot e^{a\left( \rho _{1}-t\right) }+\displaystyle \frac{c-bu_{1}}{a}\cdot \left( 1-e^{a\left( \rho _{1}-t\right) }\right) \ &{} t\in [\rho _1,\rho _2], \\ -\displaystyle \frac{ac}{4}\cdot t^{2}+D_{1}\cdot t+D_{2} \ &{} t\in [\rho _2,t_2],\\ \end{array} \right. \end{aligned}$$

$$\begin{aligned} z\left( t\right) =\left\{ \begin{array}{ll} \displaystyle \frac{p\alpha }{b}\cdot \left( -\frac{ca}{4}\cdot t^{2}+\left( C_{1}-c\right) \cdot t+\left( C_{1}-c\right) \cdot \frac{2}{a}+C_{2}\right) \ &{} t\in [t_1,\rho _1], \\ z(\rho _{2})\cdot e^{a\left( t-\rho _{2}\right) }-\displaystyle \frac{p\alpha u_{1}}{a}\cdot \left( 1-e^{a\left( t-\rho _{2}\right) }\right) \ &{} t\in [\rho _1,\rho _2], \\ \displaystyle \frac{p\alpha }{b}\cdot \left( -\frac{ca}{4}\cdot t^{2}+\left( D_{1}-c\right) \cdot t+\left( D_{1}-c\right) \cdot \frac{2}{a}+D_{2}\right) \ &{} t\in [\rho _2,t_2],\\ \end{array} \right. \end{aligned}$$

$$\begin{aligned} u^{*}\left( t\right) =\left\{ \begin{array}{ll} \displaystyle \frac{1}{b}\cdot \left( \frac{a^{2}c}{4}\cdot t^{2}-\left( C_{1}-\frac{c}{2}\right) a\cdot t-\left( C_{1}-c+aC_{2}\right) \right) \ &{} t\in [t_1,\rho _1], \\ u_{1} \ &{} t\in [\rho _1,\rho _2], \\ \displaystyle \frac{1}{b}\cdot \left( \frac{a^{2}c}{4}\cdot t^{2}-\left( D_{1}-\frac{c}{2}\right) a\cdot t-\left( D_{1}-c+aD_{2}\right) \right) \ &{} t\in [\rho _2,t_2].\\ \end{array} \right. \end{aligned}$$

Since $y^{*}\left( t_1\right) =y_1$ and $z\left( t_2\right) =0$ we get

$$\begin{aligned} \left\{ \begin{array}{ll} C_{2}=y_{1}+\displaystyle \frac{ac}{4}\cdot \left( t_{1}\right) ^{2}-C_{1}\cdot t_{1}, &{} \\ D_{2}=\displaystyle \frac{ca}{4}\cdot \left( t_{2}\right) ^{2}-\left( D_{1}-c\right) \cdot t_{2}-\left( D_{1}-c\right) \cdot \frac{2}{a}. &{}\\ \end{array} \right. \end{aligned}$$

(44)

Due to (44) and continuity of $y^{*}$ and z we get

$$\begin{aligned} \left\{ \begin{array}{ll} y^{*}\left( \rho _{1}\right) =y_{1}-\displaystyle \frac{ac}{4}\cdot \left( \left( \rho _{1}\right) ^{2}-\left( t_{1}\right) ^{2}\right) +C_{1}\cdot \left( \rho _{1}-t_{1}\right) , &{} \\ z\left( \rho _{2}\right) = \displaystyle \frac{p\alpha }{b}\cdot \left( -\frac{ca}{4}\cdot \left( \rho _{2}+t_{2}\right) +D_{1}-c\right) \cdot \left( \rho _{2}-t_{2}\right) . &{}\\ \end{array} \right. \end{aligned}$$

(45)

Due to (44), (45) and continuity of $u^{*}$ we get

$$\begin{aligned} \left\{ \begin{array}{ll} C_{1}=\displaystyle \frac{1}{z_{1}}\cdot \left( \frac{a^{2}c}{4}\cdot \left( \left( \rho _{1}\right) ^{2}-\left( t_{1}\right) ^{2}\right) +\frac{ac}{2}\cdot \rho _{1}-ay_{1}-bu_{1}+c\right) , &{} \\ D_{1}=\displaystyle \frac{1}{z_{2}}\cdot \left( \frac{a^{2}c}{4}\cdot \left( \left( t_{2}\right) ^{2}-\left( \rho _{2}\right) ^{2}\right) +\frac{ac}{2}\cdot \left( 2t_{2}-\rho _{2}\right) +bu_{1}+c\right) . &{}\\ \end{array} \right. \end{aligned}$$

(46)

where

$$\begin{aligned} \left\{ \begin{array}{ll} z_{1}=a\left( \rho _{1}-t_{1}\right) +1>1, &{} \\ z_{2}=a\left( t_{2}-\rho _{2}\right) +1>1. &{}\\ \end{array} \right. \end{aligned}$$

(47)

Due to (44)–(47) and continuity of $y^{*}$ and z we get

$$\begin{aligned} \left\{ \begin{array}{ll} e^{a\left( \rho _{1}-\rho _{2}\right) }=-\displaystyle \frac{z_{2}+2+{\displaystyle \frac{L_{2}}{z_{2}}}}{z_{1}-2+{\displaystyle \frac{L_{1}}{z_{1}}}}, &{} \\ e^{a\left( \rho _{1}-\rho _{2}\right) }=-\displaystyle \frac{z_{1}+2+{\displaystyle \frac{L_{1}}{z_{1}}}}{z_{2}-2+{\displaystyle \frac{L_{2}}{z_{2}}}}, &{}\\ \end{array} \right. \end{aligned}$$

(48)

where $L_{1}:=1+4\cdot {\displaystyle \frac{ay_{1}+bu_{1}-c}{c}}$ and $L_{2}=1+{\displaystyle \frac{4bu_{1}}{c}}$. Note that

$$\begin{aligned} L_{1}\ge 1 \end{aligned}$$

(49)

due to (17). Let us rewrite (48) as

$$\begin{aligned} \left\{ \begin{array}{ll} e^{a\left( \rho _{1}-\rho _{2}\right) }=-\displaystyle \frac{z_{2}+2+{\displaystyle \frac{L_{2}}{z_{2}}}}{z_{1}-2+{\displaystyle \frac{L_{1}}{z_{1}}}}, &{} \\ \left( z_{1}+{\displaystyle \frac{L_{1}}{z_{1}}}\right) =\pm \left( z_{2}+{\displaystyle \frac{L_{2}}{z_{2}}}\right) . &{}\\ \end{array} \right. \end{aligned}$$

(50)

Let $z_{1}+{\displaystyle \frac{L_{1}}{z_{1}}}=-\left( z_{2}+{\displaystyle \frac{L_{2}}{z_{2}}}\right) .$ Then due to (50) we get $e^{a\left( \rho _{1}-\rho _{2}\right) }=1$ in contradiction with (43). Let $z_{1}+{\displaystyle \frac{L_{1}}{z_{1}}}=z_{2}+{\displaystyle \frac{L_{2}}{z_{2}}}.$ Then due to (50) we get

$$\begin{aligned} e^{a\left( \rho _{1}-\rho _{2}\right) }=-\frac{\left( z_{1}+1\right) ^{2}+L_{1}-1}{\left( z_{1}-1\right) ^{2}+L_{1}-1}. \end{aligned}$$

Since $z_{1}>1$ and $a\left( \rho _{1}-\rho _{2}\right) <0$, we get $\left( z_{1}+1\right) ^{2}+L_{1}-1>0>\left( z_{1}-1\right) ^{2}+L_{1}-1$ and moreover $\left( z_{1}+1\right) ^{2}+L_{1}-1<-\left( z_{1}-1\right) ^{2}-L_{1}+1,$ i.e., $L_{1}<-\left( z_{1}\right) ^{2}<-1$, in contradiction with (49).

Due to Lemma 5, only two cases are possible, namely $\rho _2=t_2$ and $\rho _1=t_1$.

Lemma 6

(i) If $\rho _2=t_2$ then the optimal control is

$$\begin{aligned} u^{*}\left( t\right) = {\left\{ \begin{array}{ll} \left( {\displaystyle \frac{a^{2}c}{4b}}\cdot t+T_{1}\left( \rho _{1}\right) \right) \cdot \left( t-\rho _{1}\right) +u_{1}, &{} t\in \left[ t_{1},\rho _{1}\right] , \\ u_{1}, &{} t\in \left[ \rho _{1},t_{2}\right] , \end{array}\right. } \end{aligned}$$

where

$$\begin{aligned} T_{1}\left( \rho _{1}\right) =\displaystyle \frac{a}{4b}\cdot \left( \frac{c+4\left( ay_{1}+bu_{1}-c\right) }{q_{1}+1}-c\cdot \left( at_{1}-1\right) \right) \end{aligned}$$

while $q_{1}=a\left( \rho _{1}-t_{1}\right) \in \left[ 0;a\left( t_{2}-t_{1}\right) \right]$ is the root of the equation

$$\begin{aligned} \displaystyle e^{q_{1}-a(t_{2}-t_{1})}=\displaystyle \frac{\displaystyle \frac{c}{4}\cdot \left( q_{1}\right) ^{2}+cq_{1}+ay_{1}+bu_{1}}{-bu_{1}\left( q_{1}+1\right) }. \end{aligned}$$

(ii) If $\rho _1=t_1$ then the optimal control is

$$\begin{aligned} u^{*}\left( t\right) = {\left\{ \begin{array}{ll} u_{1}, &{} t\in \left[ t_{1},\rho _{1}\right] , \\ \left( \displaystyle \frac{a^{2}c}{4b}\cdot t+T_{2}\left( \rho _{2}\right) \right) \left( t-\rho _{2}\right) +B_{1}, &{} t\in \left[ \rho _{1},t_{2}\right] , \end{array}\right. } \end{aligned}$$

where

$$\begin{aligned} T_{2}\left( \rho _{2}\right) =-\displaystyle \frac{a}{4b}\cdot \left( \displaystyle \frac{c+4b\left( B_{1}-1\right) }{q_{2}}+c\cdot \left( at_{2}+1\right) \right) \end{aligned}$$

while $q_2=a\left( t_{2}-\rho _{2}\right) \in \left[ 0;a\left( t_{2}-t_{1}\right) \right]$ is the root of the equation

$$\begin{aligned} e^{q_{2}-a\left( t_{2}-t_{1}\right) }=-\displaystyle \frac{{\displaystyle \frac{c}{4}}\cdot \left( q_{2}\right) ^{2}+cq_{2}+bu_{1}+c}{\left( q_{2}+1\right) \left( ay_{1}+bu_{1}-c\right) }. \end{aligned}$$

Proof

As in the proof of Lemma 5, we can write the form of $y^*(t), z(t)$ and $u^*(t)$. But now we have only one switch. To finish the proof we need only use the conditions $y^*\left( t_1\right) =y_1, z\left( t_2\right) =0,$ continuity of optimal trajectories and optimal control, and straightforward calculations.