Turnpike property and convergence rate for an investment and consumption model

Abstract

We discuss the turnpike property for optimal investment and consumption problems. We find there exists a threshold value that determines the turnpike property for investment policy. The threshold value only depends on the Sharpe ratio, the riskless interest rate and the discount rate. We show that if utilities behave asymptotically like power utilities and satisfy some simple relations with the threshold value, then the turnpike property for investment holds. There is in general no turnpike property for consumption policy. We also provide the rate of convergence and illustrate the main results with examples of power and non-HARA utilities and numerical tests.

Appendix

Derivation of optimal trading strategy (2.1) and budget equation (2.2) We may use a standard stochastic control method to solve problem (1.1). Define the value function u by

$$\begin{aligned} u(x,t)=\sup _{\pi , c} E\left[ \int _t^T e^{-\delta (s-t)} U_2(c_s)ds +e^{-\delta (T-t)}U_1(X_T)\bigg | X_t=x\right] \end{aligned}$$

for $(x,t)\in R_+\times [0,T]$. Then u satisfies the HJB equation

$$\begin{aligned} {\partial u\over \partial t}+\max _{\pi } \left\{ \frac{1}{2}\sigma ^2x^2\pi ^2 u_{xx}+x\pi \sigma \theta u_x\right\} +rxu_x-\delta u+\max _{c\ge 0}\{U_2(c)-cu_x\}=0, \end{aligned}$$

(5.1)

for $(x,t)\in R_+\times [0,T)$ with the terminal condition $u(x,T)=U_1(x)$ for $x\in R_+$, where $u_x(x,t)$ is the partial derivative of u with respect to x and evaluated at (x, t) (we have omitted (x, t) in equation (5.1) to simplify notations), ${\partial \over \partial t}u$ and $u_{xx}$ are defined similarly.

Let $V_i$ be the dual functions of $U_i$, $i=1,2$, defined in (1.5). Then $V_i$ are nonnegative (since $U_i(0)=0$), continuously differentiable, strictly decreasing, strictly convex functions.

The optimal investment and consumption strategies in the HJB equation (5.1) are given by

$$\begin{aligned} \pi ^*(x,t)=- \frac{\theta }{\sigma }\frac{u_x}{xu_{xx}},\ \ \ c^*(x,t)=-V'_2 (u_x). \end{aligned}$$

(5.2)

The HJB equation (5.1) can be written as

$$\begin{aligned} {\partial u\over \partial t}-\frac{1}{2}\theta ^2{u_x^2 \over u_{xx}}+rxu_x-\delta u+V_2(u_x)=0. \end{aligned}$$

(5.3)

Equation (5.3) is a fully nonlinear PDE and is in general difficult to solve. In the literature equation (5.3) is solved with some trial-and-error method, which works when $U_1$ and $U_2$ are the same power utility functions. Bian et al. [2] and Bian and Zheng [3] apply the dual stochastic control method to solve problem (1.1) and show that the HJB equation (5.1) has a classical solution that can be represented in terms of the dual function of the corresponding dual control problem. The value function v(y, t) of the dual control problem satisfies a linear PDE

$$\begin{aligned} {\partial v\over \partial t} + {1\over 2}\theta ^2 y^2 v_{yy} -(r-\delta )yv_y - \delta v +V_2(y) =0 \end{aligned}$$

(5.4)

for $(y,t)\in R_+\times [0,T)$, with the terminal condition $v(y,T)=V_1(y)$. Its solution v is continuous on $R_+\times [0,T]$ and $C^{2,1}$ on $R_+\times [0,T)$, and v is strictly decreasing and strictly convex in y for fixed $t<T$. A classical solution to the HJB equation (5.1) is given by

$$\begin{aligned} u(x,t) = v(y, t) + x y, \end{aligned}$$

where y is the unique solution to the equation

$$\begin{aligned} x=-v_y(y,t). \end{aligned}$$

This leads to $u_x(x,t)=y$ and $u_{xx}=-\,1/v_{yy}$. From (5.2) we conclude that the optimal amount of investment A(x, t) and the optimal consumption rate C(x, t) are given by

$$\begin{aligned} A(x,t)=x\pi ^*(x,t)=\frac{\theta }{\sigma }y v_{yy}(y,t) ,\quad C(x,t)= c^*(x,t)=-V'_2 (y). \end{aligned}$$

To simplify the notation, we define $\tau =T-t$, the time horizon. Then $T\rightarrow \infty$ is equivalent to $\tau \rightarrow \infty$. We still use t to represent a time horizon variable, instead of $\tau$.

It is easy to verify that $v_y(y,t)$ is the solution of the initial value problem

$$\begin{aligned} w_t-\frac{1}{2}\theta ^2 y^2 w_{yy}+(r-\delta -\theta ^2)yw_y+ \delta w=V_2^{'}(y) \end{aligned}$$

(5.5)

for $(y,t)\in R_+\times R_+$ with initial condition $v_y(y,0)=V'_1(y),\ y\in R_+$. By Poisson’s formula, the solution of equation (5.5) is given by

$$\begin{aligned} v_y(y,t)=-I_1(y,t)-\int _{0}^tI_2(y,\tau )d\tau ,\ yv_{yy}(y,t)=J_1(y,t)+\int _{0}^tJ_2(y,\tau )d\tau , \end{aligned}$$

(5.6)

which proves (2.1) and (2.2).

Preliminaries of mathematical proofs We need some technical results in proofs. Simple calculus gives that for any constant A,

$$\begin{aligned} \frac{1}{2 \sqrt{\pi }}\int _{-\infty }^{\infty } e^{-\frac{\eta ^2}{4}-A\eta }d\eta =e^{A^2 } \end{aligned}$$

(5.7)

Hence for any $q<1$,

$$\begin{aligned} \frac{e^{\beta t}}{2\sqrt{\pi }}\int _{-\infty }^\infty e^{-\frac{\eta ^2}{4}-(\alpha -1) a\sqrt{t}\eta }\left( \frac{\eta }{2a\sqrt{t}}+\alpha -1\right) e^{(q-1)a\sqrt{t}\eta }d\eta =(q-1)e^{\lambda (q) t} \end{aligned}$$

(5.8)

and

$$\begin{aligned} \frac{e^{\beta t}}{2\sqrt{\pi }}\int _{-\infty }^\infty e^{-\frac{\eta ^2}{4}-(\alpha -1) a\sqrt{t}\eta }\left| \frac{\eta }{2a\sqrt{t}}+\alpha -1\right| e^{(q-1)a\sqrt{t}\eta }d\eta \le \left[ 1-q+\frac{1}{a\sqrt{\pi t}}\right] e^{\lambda (q) t}. \end{aligned}$$

(5.9)

For any $y>0$, the convexity of $V_i$ implies that

$$\begin{aligned} V_i\left( \frac{y}{2}\right) \ge V_i(y)-\frac{y}{2} V_i'(y), \end{aligned}$$

which, together with the decreasing property and the nonnegativity of V, gives

$$\begin{aligned} 0\le -yV_i'(y) \le 2 V_i\left( \frac{y}{2}\right) \le 2V_i\left( \frac{1}{2}\right) , \end{aligned}$$

(5.10)

for all $y\ge 1$.

We need the following algebraic inequality in the proof:

$$\begin{aligned} |x^c-y^c|\le (1+c)\max \{x, y\}^{c-1}|x-y| \end{aligned}$$

(5.11)

for all $x,y\ge 0$ and $0<c<\infty$. (5.11) can be proved as follows: assume $x>y>0$ and let $z=x/y$. Then $z>1$ and (5.11) is equivalent to $z^c -1 \le (1+c)(z^c-z^{c-1})$. Let $g(z)=z^c -1-(1+c)(z^c-z^{c-1})$. Then $g'(z)=cz^{c-1}-(1+c)(cz^{c-1}-(c-1)z^{c-2})=-c^2(1-z)z^{c-2}-z^{c-2}\le 0$ for $z\ge 1$. So g is decreasing and $g(z)\le g(1)=0$ for $z\ge 1$.

We have, for any $q<\bar{q}<1$, $\lambda =\lambda (q), \bar{\lambda }=\lambda (\bar{q})$

$$\begin{aligned} e^{\bar{\lambda } t}y^{\bar{q}-1}=(e^{\lambda t}y^{q-1})^{\frac{\bar{q}-1}{q-1}}e^{\left( \bar{\lambda }-\lambda \frac{\bar{q}-1}{q-1}\right) t}. \end{aligned}$$

(5.12)

and

$$\begin{aligned} \bar{\lambda }-\lambda \frac{\bar{q}-1}{q-1}=(\bar{q}-q)\left[ \frac{1}{2}\theta ^2(\bar{q}-1)+\frac{r}{q-1}\right] <0. \end{aligned}$$

(5.13)

Proof of Theorem 2.3

Since $U'_i(x)= x^{p_i-1}$ for $i=1,2$, the dual functions are given by $V'_i(y)=-y^{q_i-1}$ where $q_i=p_i/(p_i-1)$. Substituting $V'_i(y)$ into (5.6), also noting (5.7), we have

$$\begin{aligned} v_y(y,t)=-e^{\lambda _1 t} y^{q_1-1} -\frac{e^{\lambda _2 t }-1}{\lambda _2} y^{q_2-1}. \end{aligned}$$

Hence we derive (2.4) from (2.1), where y is the solution to the equation (2.5).

We next complete the proof by directly discussing three cases.

Case 1 When $q_1<q^*$ or $q_2\le q^*$, we have $\lambda _1>0$ or $\lambda _2\ge 0$. We know that the solution $y=u_x$ to equation (2.5) must tend to $\infty$ as t tends to $\infty$ (otherwise the right side of (2.5) tends to $\infty$). We can find the dominating term in (2.5) as t tends $\infty$ for different $q_1$ and $q_2$.

Assume $q=q_1=q_2 \le q^*$. It is easy to see from (2.1) and (2.5) that $A(x,t)=\frac{\theta }{\sigma } (1-q)x$ and $C(x,t)=-V'_2 (u_x(x,t))=(u_x(x,t))^{q_2-1}=E(t)x$, where $E(t)=(e^{\lambda _1 t} +\frac{e^{\lambda _2 t }-1}{\lambda _2})^{-1}$. This implies $R(t)C(x,t)=x$.

Assume $q_1<\min \{q_2, q^*\}$, then $\lambda _1>\max \{\lambda _2, 0\}$. Noting that

$$\begin{aligned} \frac{e^{\lambda _2 t }-1}{\lambda _2}=\frac{1-e^{-|\lambda _2| t}}{|\lambda _2|} e^{\lambda ^+_2 t}\le te^{\lambda ^+_2 t}, \end{aligned}$$

we obtain

$$\begin{aligned} \frac{e^{\lambda _2 t }-1}{\lambda _2}y^{q_2-1}\le \left( e^{\lambda _1 t}y^{q_1-1}\right) ^{\frac{q_2-1}{q_1-1}}te^{(\lambda ^+_2-\lambda _1\frac{q_2-1}{q_1-1}) t} \le x^{\frac{q_2-1}{q_1-1}}te^{(\lambda ^+_2-\lambda _1\frac{q_2-1}{q_1-1}) t}. \end{aligned}$$

(5.14)

From (2.5) and (5.13), we deduce that $\lim _{t\rightarrow \infty }\frac{e^{\lambda _2 t }-1}{\lambda _2}y^{q_2-1}=0$ and $\lim _{t\rightarrow \infty }e^{\lambda _1 t}y^{q_1-1}=x$. Hence from (2.4)

$$\begin{aligned} \lim _{t\rightarrow \infty }A(x,t)=\frac{\theta }{\sigma }(1-q_1)x. \end{aligned}$$

We have by (2.4)

$$\begin{aligned} e^{\lambda _1\frac{q_2-1}{q_1-1} t}C(x,t)=\left( e^{\lambda _1 t}y^{q_1-1}\right) ^{\frac{q_2-1}{q_1-1}}, \end{aligned}$$

which implies $\lim _{t\rightarrow \infty } R(t) C(x,t)=x^{\frac{q_2-1}{q_1-1}}$.

Assume $q_2< q_1$ and $q_2\le q^*$, then $\lambda _2> \lambda _1$ and $\lambda _2\ge 0$. We have

$$\begin{aligned} \frac{\lambda _2 }{1-e^{-\lambda _2 t}}\le E:=\frac{\lambda _2 }{1-e^{-\lambda _2 }} \end{aligned}$$

(5.15)

for $t\ge 1$. Then by (2.5), for $t\ge 1$

$$\begin{aligned} e^{\lambda _1 t}y^{q_1-1} = \left( \frac{e^{\lambda _2 t }-1}{1-e^{-\lambda _2 t}} y^{q_2-1}\right) ^{\frac{q_1-1}{q_2-1}} e^{(\lambda _1-\lambda _2\frac{q_1-1}{q_2-1}) t}\le (Ex)^{\frac{q_1-1}{q_2-1}} e^{(\lambda _1-\lambda _2\frac{q_1-1}{q_2-1}) t}. \end{aligned}$$

(5.16)

We deduce from (2.5) and (5.13) $\lim _{t\rightarrow \infty }e^{\lambda _1 t}y^{q_1-1}=0$ and $\lim _{t\rightarrow \infty }\frac{e^{\lambda _2 t }-1}{\lambda _2}y^{q_2-1}=x$. Furthermore, we get

$$\begin{aligned} \lim _{t\rightarrow \infty }A(x,t)=\frac{\theta }{\sigma }(1-q_2)x, \end{aligned}$$

and $\frac{e^{\lambda _2 t }-1}{\lambda _2}C(x,t)= \frac{e^{\lambda _2 t }-1}{\lambda _2}y^{q_2-1}$, which gives $\lim _{t\rightarrow \infty } R(t)c(x,t)=x$.

Case 2 Since $q_1> q^*$ and $q_2>q^*$, we have $\lambda _1, \lambda _2< 0$. From (2.5), we deduce that $y=u_x(x,t)$ must have upper bound in t (otherwise, we would have $x=0$, a contradiction). Therefore $x=\lim _{t\rightarrow \infty }(-1/\lambda _2)u_x(x,t)^{q_2-1}$. Taking the limit in (2.4) leads to the desired results.

Case 3 Since $q_1= q^*$ and $q_2>q^*$, we have $\lambda _1=0$ and $\lambda _2<0$. Again from (2.5), we deduce that $y=u_x(x,t)$ must have upper bound. Let Y be the unique solution to the nonlinear equation

$$\begin{aligned} x=Y^{q^*-1}-\frac{1}{\lambda _2}Y^{q_2-1}, \end{aligned}$$

then we have $\lim _{t\rightarrow \infty }u_x=Y$. Taking the limit in (2.4) leads to the desired results. $\square$

Proof of Theorem 2.4

From (1.4) and (5.10), we conclude that there is constant $\bar{q}\in (\max \{q_1,q_2,0\}, 1)$, for any fixed $\epsilon >0$, there is $K_\epsilon >0$, such that

$$\begin{aligned} |V'_i(y)+ y^{q_i-1}|\le \epsilon |V'_i(y)| + K_\epsilon y^{\bar{q}-1}. \end{aligned}$$

(5.17)

for $i=1, 2$. By (5.17)(i=1) and (5.7), we get

$$\begin{aligned} |I_1(y,t)-y^{q_1-1}e^{\lambda _1 t}|\le \epsilon I_1(y,t)+K_\epsilon y^{\bar{q}-1}e^{\bar{\lambda } t}, \end{aligned}$$

for all $(y,t)\in R_+\times (0,\infty )$. Taking $\epsilon =1$ and denoting $\bar{K}=K_1$, we get, by (5.12) and (5.13)

$$\begin{aligned} y^{q_1-1}e^{\lambda _1 t}\le 2 I_1(y,t)+\bar{K} \left( y^{q_1-1}e^{\lambda _1 t}\right) ^{\frac{\bar{q}-1}{q_1-1}}. \end{aligned}$$

Since $0<\frac{\bar{q}-1}{q_1-1}<1$, we see that $(y^{q_1-1}e^{\lambda _1 t})^\frac{q_1-\bar{q}}{q_1-1} \le 2 I_1(y,t)+ \bar{K}$ if $y^{q_1-1}e^{\lambda _1 t}\ge 1$. Hence

$$\begin{aligned} y^{q_1-1}e^{\lambda _1 t}\le \left[ 2 I_1(y,t)+1+ \bar{K} \right] ^{\frac{q_1-1}{q_1-\bar{q}}}. \end{aligned}$$

Similarly, we obtain for $t\ge 1$

$$\begin{aligned} y^{q_2-1}\frac{e^{\lambda _2 t}-1}{\lambda _2}\le \left[ 2 \int _{0}^tI_2(y,\tau )d\tau +1+ \frac{\bar{K}E^{\bar{q}-1\over q_2-1}}{|\bar{\lambda }|} \right] ^{\frac{q_2-1}{q_2-\bar{q}}}. \end{aligned}$$

Let $H_1(t)=(u_x(x,t))^{q_1-1}e^{\lambda _1 t}$, $H_2(t)=(u_x(x,t))^{q_2-1}\frac{e^{\lambda _2 t}-1}{\lambda _2}$, $M_1(t)=(u_x(x,t))^{\bar{q}-1}e^{\bar{\lambda } t}$, $M_2(t)=(u_x(x,t))^{\bar{q}-1}\frac{e^{\bar{\lambda } t}-1}{\bar{\lambda }}$, $H(t)=H_1(t)+H_2(t)$ and $M(t)=M_1(t)+M_2(t)$. Then by (2.2)

$$\begin{aligned} H(t)\le H:=\left[ 2x+1+ \bar{K}\right] ^{\frac{q_1-1}{q_1-\bar{q}}}+\left[ 2x +1+ \frac{\bar{K}E^{\bar{q}-1 \over q_2-1}}{|\bar{\lambda }|} \right] ^{\frac{q_2-1}{q_2-\bar{q}}}. \end{aligned}$$

(5.18)

Recalling $E(t)=(e^{\lambda _1 t}+\frac{e^{\lambda _2 t}-1}{\lambda _2})^{-1}$. For $q_1<q^*$ or $q_2 \le q^*$, we see $\lambda _1>0$ or $\lambda _2 \ge 0$. Hence $E(t)\le 1$ for $t\ge 1$ and $\lim _{t\rightarrow \infty }E(t)=0$. We have for all $y>0$,

$$\begin{aligned} y^{q_1-1}e^{\lambda _1 t}+y^{q_2-1}\frac{e^{\lambda _2 t}-1}{\lambda _2}\ge E(t)^{-1}\min \{ y^{q_1-1},y^{q_2-1}\} . \end{aligned}$$

Hence

$$\begin{aligned} u_x(x,t) \ge \min \left\{ (E(t)H(t))^{\frac{1}{q_1-1}}, (E(t)H(t))^{\frac{1}{q_2-1}}\right\} . \end{aligned}$$

(5.19)

This implies, for $t\ge 1$

$$\begin{aligned} u_x(x,t)\ge H^{\frac{1}{\max \{q_1,q_2\}-1}} (E(t))^{\frac{1}{\min \{q_1,q_2\}-1}}, \end{aligned}$$

(5.20)

and $\lim _{t\rightarrow \infty }u_x(x,t)=\infty$. Using (5.12) and (5.13), we derive the following estimates:

$$\begin{aligned} M(t) \le \tilde{M}(x,t):=H^{\frac{\bar{q}-1}{q_1-1}}e^{(\bar{\lambda }-\lambda _1\frac{\bar{q}-1}{q_1-1}) t} + \frac{1}{|\bar{\lambda }|} H^{\frac{\bar{q}-1}{\max \{q_1,q_2\}-1}}(E(t))^{\frac{\bar{q}-1}{\min \{q_1,q_2\}-1}}. \end{aligned}$$

(5.21)

By (5.17), (2.2) and (5.7), we get

$$\begin{aligned} |H(t)-x|\le \epsilon H(t) + K_\epsilon M(t). \end{aligned}$$

(5.22)

Noting that (5.13), (5.21), and letting $t\rightarrow \infty$ and then $\epsilon \rightarrow 0$ in (5.22), we deduce the limiting form of (2.5)

$$\begin{aligned} \lim _{t\rightarrow \infty }(u_x(x,t))^{q_1-1}e^{\lambda _1 t}+(u_x(x,t))^{q_2-1}\frac{e^{\lambda _2 t}-1}{\lambda _2}=x. \end{aligned}$$

(5.23)

By (5.17),(2.1), (5.8) and (5.9), we conclude that

$$\begin{aligned}&\left| yv_{yy}(y,t)-(1-q_1)y^{q_1-1}e^{\lambda _1 t}-(1-q_2)y^{q_2 -1}\frac{e^{\lambda _2 t}-1}{\lambda _2}\right| \\&\quad \le \epsilon \left[ \left( 1-q_1+\frac{1}{a\sqrt{\pi }}\right) y^{q_1-1}e^{\lambda _1 t}+\left( 1-q_2+\frac{2Ee^{\lambda ^+_2}+1}{a\sqrt{\pi }}\right) y^{q_2-1}\frac{e^{\lambda _2 t}-1}{\lambda _2}\right] \\&\qquad + K_\epsilon \left[ \left( 1-\bar{q}+\frac{1}{a\sqrt{\pi }}\right) y^{\bar{q}-1}e^{\bar{\lambda } t}+\left( 1-\bar{q}+\frac{2Ee^{\lambda ^+_2}+1}{a\sqrt{\pi }}\right) y^{\bar{q}-1}\frac{e^{\bar{\lambda } t}-1}{\bar{\lambda }}\right] \end{aligned}$$

for all $(y,t)\in R_+\times [1,\infty )$. Let

$$\begin{aligned} \bar{D}=\max \left\{ 1-q_1+\frac{1}{a\sqrt{\pi }}, 1-q_2+\frac{2Ee^{\lambda ^+_2}+1}{a\sqrt{\pi }}\right\} . \end{aligned}$$

Noting $q_1,q_2<\bar{q}$, for fixed $x>0$, we obtain

$$\begin{aligned} \left| A(x,t)-\frac{\theta }{\sigma }[(1-q_1)H_1(t)+(1-q_2)H_2(t)]\right| \le \frac{\theta \bar{D}}{\sigma }(\epsilon H(t) +K_\epsilon M(t)). \end{aligned}$$

(5.24)

From estimates (5.18) and (5.21), letting $t\rightarrow \infty$ and then $\epsilon \rightarrow 0$ in (5.24), we derive

$$\begin{aligned} \lim _{t\rightarrow \infty }\left[ A(x,t)-\frac{\theta }{\sigma }[(1-q_1)H_1(t)+(1-q_2)H_2(t)]\right] =0. \end{aligned}$$

(5.25)

Next, dividing the cases: $q_1=q_2 \le q^*$, $q_1<\min \{q_2, q^*\}$ and $q_2< q_1, q_2\le q^*$ as in Theorem 2.3, we derive the turnpike property (1.7) from (5.23) and (5.25). To derive (2.7), noting $C(x,t)=-V'_2(u_x(x,t))$, (2.6), (5.23) and $\lim _{t\rightarrow \infty } u_x(x,t)=\infty$, we have

$$\begin{aligned} \lim _{t\rightarrow \infty } R(t)^{\frac{{\hat{q}}-1}{q_2-1}} C(x,t) = \lim _{t\rightarrow \infty } \left( R(t)^{1\over q_2-1}u_x(x,t)\right) ^{{\hat{q}}-1}= x^{\frac{ {\hat{q}}-1}{\min \{q_1,q_2\}-1}}. \end{aligned}$$

The last equality is derived by discussing three cases for $q_1$ and $q_2$, the same as that in deriving (1.7). $\square$

Proof of Theorem 2.5

The proof is similar to that of Theorem 2.4. With strengthened assumptions, we can give better estimations to some inequalities in the proof of Theorem 2.4. From (5.10) we conclude that assumption (2.9) holds for all $y>0$ with a changed constant L. This implies that (5.22) becomes

$$\begin{aligned} |H(t)-x|\le LM(t), \end{aligned}$$

(5.26)

and (5.18) becomes

$$\begin{aligned} H(t)\le H:=(x+1+ L)^{\frac{q_1-1}{q_1-\bar{q}}}+\left[ x +1+ \frac{LE^{\bar{q}-1 \over q_2-1}}{|\bar{\lambda }|} \right] ^{\frac{q_2-1}{q_2-\bar{q}}}. \end{aligned}$$

(5.27)

We have also the estimate (5.21). (5.24) becomes

$$\begin{aligned} \left| A(x,t)-\frac{\theta }{\sigma }[(1-q_1)H_1(t)+(1-q_2)H_2(t)]\right| \le \frac{D\theta }{\sigma }L M(t), \end{aligned}$$

(5.28)

where $D= 1-\bar{q}+\frac{2Ee^{\lambda ^+_2}+1}{a\sqrt{\pi }}$. Note that $\tilde{q}< \min \{\hat{q}, q_2\}$, (2.10) holds for all $y>0$. Setting $c:=\frac{{\hat{q}} -1}{\min \{q_1,q_2\}-1}$, we have, from (5.11)

$$\begin{aligned} \left| R(t)^{\frac{{\hat{q}}-1}{q_2-1}} C(x,t)-x^c\right| \le R(t)^{\frac{{\hat{q}}-1}{q_2-1}}\left| V'_2(u_x)+u_x^{{\hat{q}}-1}\right| +\left| R(t)^{\frac{{\hat{q}}-1}{q_2-1}} u_x^{{\hat{q}}-1}-x^c\right| . \end{aligned}$$

(5.29)

Noting that $R(t)^{\frac{{\hat{q}}-1}{q_2-1}}\le (E(t))^{\frac{{\hat{q}}-1}{q_2-1}}$, we have, from (2.10) and (5.20)

$$\begin{aligned} R(t)^{\frac{{\hat{q}}-1}{q_2-1}}|V'_2(u_x)+u_x^{{\hat{q}}-1}|\le LR(t)^{\frac{{\hat{q}}-1}{q_2-1}} u_x^{\tilde{q}-1}\le L H^{\frac{\tilde{q}-1}{\max \{q_1,q_2\}-1}} (E(t))^{\frac{\tilde{q}-{\hat{q}}}{\min \{q_1,q_2\}-1}}. \end{aligned}$$

This gives the estimate for the first term in (5.29).

For $q=q_1=q_2\le q^*$, we deduce from (5.28) and (5.26),

$$\begin{aligned} \left| A(x,t)-\frac{\theta }{\sigma }(1-q)x\right| \le \frac{\theta }{\sigma }L(1-q+D)M(t). \end{aligned}$$

Let $G=G(x,c)=(1+c)H^{c-1}$ if $c>1$ and $G=G(x,c)=(1+c)x^{c-1}$ if $c\le 1$. It is clear that $x<H$ since $\frac{q_1-1}{q_1-\bar{q}}>1$ from (5.27). Noting $R(t)u_x^{q-1}=H(t)$ and $x \le \max \{H(t),x\} \le H$, we deduce the estimate for the second term in (5.29) by (2.10), (5.26) and (5.11),

$$\begin{aligned} |R(t)^c u_x^{{\hat{q}}-1}-x^c| =|(R(t)u_x^{q-1})^c - x^c|\le G(x,c)|H(t)-x| \le LG(x,c)M(t). \end{aligned}$$

We deduce convergence rate by (5.21).

If $q_1<\min \{q_2, q^*\}$, we have, by (5.14),

$$\begin{aligned} H_2(t)\le H^{\frac{q_2-1}{q_1-1}}te^{(\lambda ^+_2-\lambda _1\frac{q_2-1}{q_1-1}) t}, \end{aligned}$$

and by (5.26) and (5.28),

$$\begin{aligned} \left| A(x,t)-\frac{\theta }{\sigma }(1-q_1)x\right| \le \frac{\theta }{\sigma }L(1-q_1+D)M(t)+(q_2-q_1)\frac{\theta }{\sigma }H_2(t). \end{aligned}$$

Note that $R(t)=e^{\frac{q_2-1}{q_1-1}\lambda _1 t}$ in this case, we have

$$\begin{aligned} |R(t)^c u_x^{{\hat{q}}-1}-x^c| = |H_1(t)^c - x^c| \le G(x,c)|H_1(t) -x| \le G(x,c)(LM(t)+H_2(t)). \end{aligned}$$

If $q_2< q_1, q_2\le q^*$, we have, by (5.16), for $t\ge 1$

$$\begin{aligned} H_1(t)\le (EH)^{\frac{q_1-1}{q_2-1}}e^{(\lambda _1-\lambda _2\frac{q_1-1}{q_2-1}) t}, \end{aligned}$$

where $E:=\frac{\lambda _2}{1- e^{-\lambda _2}}$. We deduce from (5.26) and (5.28)

$$\begin{aligned} \left| A(x,t)-\frac{\theta }{\sigma }(1-q_2)x\right| \le \frac{\theta }{\sigma }L(1-q_2+D)M(t)+(q_1-q_2)\frac{\theta }{\sigma } H_1(t). \end{aligned}$$

Noting that $R(t)=\frac{e^{\lambda _2 t}-1}{\lambda _2}$, from (5.26),

$$\begin{aligned} |R(t)^c u_x^{{\hat{q}}-1}-x^c| = |(H_2(t))^c -x^c| \le G(x,c)|H_2(t) -x| \le G(x,c)(LM(t)+H_1(t)). \end{aligned}$$

Note that E(t) converges to 0 exponentially as $t\rightarrow \infty$ if $q_1<q^*$ or $q_2 < q^*$, and polynomially if $q_1\ge q^*$ and $q_2=q^*$. All other terms converge to 0 exponentially. Combining all discussions above, we have proved the results. $\square$

Proof of Theorem 2.6

In this case, $\lambda _1\le 0$, $\lambda _2< 0$. From (1.4) and (5.10), we conclude that there is constant $\bar{q}\in (\max \{q_1,q_2,0\}, 1)$, and for any fixed $\epsilon >0$, there is $K_\epsilon >0$, such that

$$\begin{aligned} \left| V'_i(y)+ y^{q_i-1}\right| \le \epsilon y^{q_i-1} + K_\epsilon y^{\bar{q}-1}. \end{aligned}$$

(5.30)

Let $h(y,t)=\int _{0}^t I_2(y,\tau )d\tau$; then from (5.30) ($i=2$)

$$\begin{aligned} \left| h(y,t)-y^{q_2-1}\frac{e^{\lambda _2 t}-1}{\lambda _2}\right| \le \epsilon y^{q_2-1}\frac{e^{\lambda _2 t}-1}{\lambda _2}+ K_\epsilon y^{\bar{q}-1}\frac{e^{\bar{\lambda } t}-1}{\bar{\lambda }}. \end{aligned}$$

(5.31)

Hence h(y, t) is increasing and bounded above in t. We deduce that $\lim _{t\rightarrow \infty }h(y,t)=h(y)$, which implies that h(y) is well defined and

$$\begin{aligned} \left| h(y)-\frac{1}{|\lambda _2|}y^{q_2-1}\right| \le \epsilon \frac{1}{|\lambda _2|}y^{{q_2}-1}+ K_{\epsilon } \frac{1}{|\bar{\lambda }|}y^{\bar{q}-1}. \end{aligned}$$

(5.32)

Since $|V'_2(y)|$ is decreasing, we conclude that h(y) is strictly decreasing, $h(\infty )=0$ and $h(0)=\infty$ by (5.32). Hence equation (2.13) admits a unique solution for any fixed $x>0$. We denote this unique solution by $Y=Y(x)$.

Since $h(y)-h(y,t)=\int _{t}^\infty I_2(y,\tau )d\tau$ and $yh'(y)-yh_y(y,t)=\int _{t}^\infty J_2(y,\tau )d\tau$, we have from (5.30) ($i=2$)

$$\begin{aligned} \left| h(y)-h(y,t)-\frac{1}{|\lambda _2|}y^{q_2-1}e^{\lambda _2 t}\right| \le \epsilon y^{q_2-1}\frac{e^{\lambda _2 t}}{|\lambda _2|}+ K_\epsilon y^{\bar{q}-1}\frac{e^{\bar{\lambda } t}}{|\lambda |}, \end{aligned}$$

(5.33)

and

$$\begin{aligned} \left| yh'(y)-yh_y(y,t)-(1-q_2)y^{q_2-1}\frac{e^{\lambda _2 t}}{|\lambda _2|}\right| \le \left( 1-q_2+\frac{1}{a\sqrt{\pi }}\right) \left[ \epsilon y^{q_2-1}\frac{e^{\lambda _2 t}}{|\lambda _2|}+ K_\epsilon y^{\bar{q}-1}\frac{e^{\bar{\lambda } t}}{|\lambda |}\right] .\nonumber \\ \end{aligned}$$

(5.34)

By (5.30)(i=1) and (5.7), we get

$$\begin{aligned} \left| v_y(y,t)+y^{q_1-1}e^{\lambda _1 t}+h(y,t)\right| \le \epsilon y^{q_1-1}e^{\lambda _1 t}+ K_\epsilon y^{\bar{q}-1}e^{\bar{\lambda } t}. \end{aligned}$$

Hence for fixed $x>0$

$$\begin{aligned} \left| u_x(x,t)^{q_1-1}e^{\lambda _1 t}+h(u_x(x,t),t)-x\right| \le \epsilon u_x(x,t)^{q_1-1}e^{\lambda _1 t} + K_\epsilon u_x(x,t)^{\bar{q}-1}e^{\bar{\lambda } t}. \end{aligned}$$

(5.35)

Note that $\lambda _1=0$ if $q_1=q^*$, $\lambda _1<0$ if $q_1>q^*$ and $\bar{\lambda }<0$. Taking $\epsilon =\frac{1}{2}$, we see from (5.35) and (5.31) that there exists a constant $\bar{Y}=\bar{Y}(x)$ such that $u_x(x,t)\le \bar{Y}$. Using (5.31) and taking $\epsilon =\frac{1}{2}$, we deduce $\lim _{y\rightarrow 0} h(y,t)=0$ uniformly in t when $t\ge 1$. Hence there exists a lower bound $\underline{Y}=\underline{Y}(x)$, $u_x(x,t)\ge \underline{Y}$. Now assume, for any sequence ${t_k}$, $\lim _{k\rightarrow \infty }t_k=\infty$ and $\lim _{k\rightarrow \infty }u_x(x,t_k)=\hat{Y}$. Then (5.33) yields $\lim _{k\rightarrow \infty }h(u_x(x,t_k),t_k)=h(\hat{Y})$. Letting $k\rightarrow \infty$ and then $\epsilon \rightarrow 0$ in (5.35) with $t=t_k$, we conclude $\hat{Y}^{q_1-1}1_{\{q_1=q^*\}}+h(\hat{Y})=x$ and $\hat{Y}=Y$, by the uniqueness of solution for equation (2.13). Therefore, $\lim _{t\rightarrow \infty }u_x(x,t)=Y$. This yields (2.12) directly.

By (5.30), (5.6) and (5.7), we get

$$\begin{aligned} \left| yv_{yy}(y,t)-(1-q_1)y^{q_1-1}e^{\lambda _1 t}-yh_y(y,t)\right| \le \left( 1-q_1+\frac{1}{a\sqrt{\pi }}\right) \left[ \epsilon y^{q_1-1}e^{\lambda _1 t} + K_\epsilon y^{\bar{q}-1}e^{\bar{\lambda } t}\right] , \end{aligned}$$

hence

$$\begin{aligned}&\left| A(x,t)-\frac{\theta }{\sigma }[(1-q_1)(u_x(x,t))^{q_1-1}e^{\lambda _1 t}-u_x(x,t)h_y(u_x(x,t),t)]\right| \\&\quad \le \left( 1-q_1+\frac{1}{a\sqrt{\pi }}\right) \left[ \epsilon u_x(x,t)^{q_1-1}e^{\lambda _1 t} + K_\epsilon u_x(x,t)^{\bar{q}-1}e^{\bar{\lambda } t}\right] . \end{aligned}$$

Noting $\lambda _2, \bar{\lambda } <0$, it yields (2.11) by (5.34).

If $V_2$ satisfies $V_2'(y)=-y^{q_2-1}$ with $q_2<0$, then a simple calculus shows that when $q_2>q^*$, we have $h(y)=-(1/\lambda _2)y^{q_2-1}$ and $yh'(y)=(q_2-1)h(y)$. If $q_1>q^*$, then relations (2.11)-(2.13) give (2.14). $\square$

Proof of Theorem 2.7

From (1.9), for any fixed $0<\gamma <1-\max \{q_1,q_2\}$, there is $0<Y=Y_\gamma <1$ such that

$$\begin{aligned} |R_i(y)-(1-q_i)|\le \gamma ,\ y\le Y. \end{aligned}$$

Since $R_i(y)=-yV_i''(y)/V_i'(y)$, we have

$$\begin{aligned} V_i'(y)=V_i'(Y)e^{\int _y^Y\frac{R_i(\eta )}{\eta }d\eta } \end{aligned}$$

for $y>0$. Hence, there exist positive constants $l=l_\gamma$ and $L=L_\gamma$, such that

$$\begin{aligned} l y^{q_i+\gamma -1}\le |V'_i(y)|\le L y^{q_i-\gamma -1}, \,y\le Y. \end{aligned}$$

(5.36)

We now give some estimates for $I_1(u_x,t)$ and $\int _{0}^t I_2(y,\tau )d\tau |_{y=u_x}$. Suppose $q_1<\min \{q_2, q^*\}$, we choose $\gamma <\min \{\frac{q_2-q_1}{2}, q^*-q_1\}$. Then

$$\begin{aligned} I_1\ge & {} \frac{e^{\beta t}}{2\sqrt{\pi }}\int _{-\infty }^{\frac{1}{a\sqrt{t}} \ln \frac{Y}{y}} e^{-\frac{\eta ^2}{4}-(\alpha -1) a\sqrt{t}\eta }|V_1'\left( ye^{a\sqrt{t}\eta }\right) |d\eta \\\ge & {} \frac{l y^{q_1+\gamma -1}e^{\lambda (q_1+\gamma ) t}}{2\sqrt{\pi }}\int _{-\infty }^{\frac{1}{a\sqrt{t}} \ln \frac{Y}{y}+2(\alpha -q_1-\gamma )a\sqrt{t}} e^{-\frac{\xi ^2}{4}}d\xi . \end{aligned}$$

Suppose that $\ln \frac{Y}{y}+2(\alpha -q_1-\gamma )a^2 t\ge 0$; then we get $x\ge I_1 \ge \frac{l}{2}u_x^{q_1+\gamma -1}e^{\lambda (q_1+\gamma ) t}$. Otherwise, $\ln \frac{Y}{y}+2(\alpha -q_1-\gamma )a^2 t\le 0$, we have $u_x \ge Y e^{2(\alpha -q_1-\gamma )a^2 t}$. Noting that $\alpha -q\ge \alpha -q^*>0$ for $q\le q^*$ and $\lambda (q)+2(q-1)(\alpha -q)a^2=-[a^2(q-1)^2+r]<0$, we conclude that

$$\begin{aligned} u_x(x,t)^{q_1+\gamma -1}e^{\lambda (q_1+\gamma ) t}\le \max \left\{ \frac{2x}{l}, Y^{q_1+\gamma -1} \right\} . \end{aligned}$$

(5.37)

Next assume $q_2<\min \{q_1, q^*\}$, we chose $\gamma <\min \{\frac{q_1-q_2}{2}, q^*-q_2\}$. Similarly, we have

$$\begin{aligned} \int _{0}^t I_2(y,\tau )d\tau \ge \int _{0}^t \frac{l y^{q_2+\gamma -1}e^{\lambda (q_2+\gamma )\tau }}{2\sqrt{\pi }}d\tau \int _{-\infty }^{\frac{1}{a\sqrt{\tau }} \ln \frac{Y}{y}+2(\alpha -q_2-\gamma )a\sqrt{\tau }} e^{-\frac{\xi ^2}{4}}d\xi . \end{aligned}$$

Let $k_0=\frac{\lambda (q_2+\gamma )}{2a^2(1 -q_2-\gamma )(\alpha -q_2-\gamma )}\in (0,1)$ and $E_0=\frac{\lambda (q_2+\gamma )}{1-e^{(k_0-1)\lambda (q_2+\gamma )}}>0$. Assume $\ln \frac{Y}{y}+2(\alpha -q_2-\gamma )a^2 k_0 t\ge 0$, then $x\ge \frac{l}{2E_0}u_x^{q_2+\gamma -1}e^{\lambda (q_2+\gamma ) t}$ for $t\ge 1$. Otherwise, we conclude that $u_x \ge Y e^{2(\alpha -q_2-\gamma )a^2 k_0 t}$. We get, for $t\ge 1$,

$$\begin{aligned} u_x(x,t)^{q_2+\gamma -1}e^{\lambda (q_2+\gamma ) t}\le \max \left\{ \frac{2\bar{E}x}{l}, Y^{q_2+\gamma -1}\right\} . \end{aligned}$$

(5.38)

Since $q_2+\gamma <q^*$, we have $\lambda (q_2+\gamma )>0$ and $e^{\lambda (q_2+\gamma )t}\rightarrow \infty$ as $t\rightarrow \infty$. Noting $q_2+\gamma -1<0$ and (5.38), we conclude that $\lim _{t\rightarrow \infty }u_x(x,t)=\infty$. On the other hand, from (5.10), we obtain with a changed L

$$\begin{aligned} |V'_i(y)|\le L(y^{q_i-\gamma -1}+y^{\bar{q}-1}), \ y>0. \end{aligned}$$

Hence we deduce upper bounds

$$\begin{aligned} I_1\le K[y^{q_1-\gamma -1}e^{\lambda (q_1-\gamma ) t}+y^{\bar{q}-1}e^{\bar{\lambda } t}] \end{aligned}$$

(5.39)

and

$$\begin{aligned} \int _{0}^t I_2(y,\tau )d\tau \le K\left[ y^{q_2-\gamma -1}\frac{e^{\lambda (q_2-\gamma ) t}-1}{\lambda (q_2-\gamma )}+y^{\bar{q}-1}\frac{e^{\bar{\lambda } t}-1}{\bar{\lambda }}\right] . \end{aligned}$$

(5.40)

Note that

$$\begin{aligned} yV''(y)=-R(y)V'(y)= (q-1)V'(y)+(1-q-R(y))V'(y). \end{aligned}$$

(5.41)

From (2.15) and (5.10), we conclude that for any fixed $\epsilon >0$, there is $K=K_\epsilon >0$, such that

$$\begin{aligned} |(1-q-R(y))V'(y)|\le \epsilon |V'(y)| + K y^{\bar{q}-1}. \end{aligned}$$

(5.42)

Substituting (5.41) into (5.6), we get by (5.42)

$$\begin{aligned}&\left| y v_{yy}(y,t)-(1-q_1)I_1-(1-q_2)\int _{0}^t I_2(y,\tau )d\tau \right| \nonumber \\&\quad \le \epsilon \left( I_1+\int _{0}^t I_2(y,\tau )d\tau \right) + K_\epsilon y^{\bar{q}-1}\left( e^{\bar{\lambda } t}+\frac{e^{\bar{\lambda } t}-1}{\bar{\lambda }}\right) . \end{aligned}$$

(5.43)

Letting $y=u_x$, $t\rightarrow \infty$ and then $\epsilon \rightarrow 0$ in (5.43), also noting $\bar{q}>q^*$, $\bar{\lambda }<0$ and $\lim _{t\rightarrow \infty }u_x(x,t)=\infty$, we obtain

$$\begin{aligned} \lim _{t\rightarrow \infty } \left[ A(x,t)-\frac{\sigma }{\theta }\left( (1-q_1)I_1+(1-q_2)\int _{0}^t I_2(y,\tau )d\tau \right) \right] =0. \end{aligned}$$

If $q_1=q_2 < q^*$, we deduce (1.7) directly. Assume $q_1<\min \{q_2, q^*\}$. By (5.37) and (5.40), we get $\lim _{t\rightarrow \infty }\int _{0}^t I_2(u_x,\tau )d\tau =0$, then (1.7). Assume $q_2<\min \{q_1, q^*\}$. By (5.38) and (5.39), we get $\lim _{t\rightarrow \infty }I_1=0$ and (1.7). $\square$

Proof of Theorem 2.8

From representation theorem for functions of regular variation ([4], Theorem 1.3.1 and equation (1.5.1)), we conclude that (5.36). Assume $q_1\ne q_2$, we chose $\gamma <\min \{\frac{|q_1-q_2|}{2}, 1-q_1, 1-q_2\}$ in (5.36). Hence, we get $\lim _{y\rightarrow 0} {V_2'(y)\over V'_1(y)} = 0$ if $q_1 < q_2$ and $\lim _{y\rightarrow 0} {V_1'(y)\over V'_2(y)} = 0$ if $q_2 < q_1$.

We claim that, for any fixed $t_0>0$, $v_y(y,t_0)$ satisfies (1.9) with index $1-\min \{q_1,q_2\}$ and (2.15). Suppose $q_1 <q_2$. From Potter’s bound ([4], Theorem 1.5.6.iii) (see also Back et al. [1], equation (73)), (5.6) and (5.10), for any fixed $t_0>0$, we get, by the dominated convergence theorem, $\lim _{y\rightarrow 0} {v_y(y,t_0)\over V'_1(y)} = e^{\lambda _1 t_0}$ and $\lim _{y\rightarrow 0} {yv_{yy}(y,t_0)\over V'_1(y)} = (q_1-1)e^{\lambda _1 t_0}$. We have

$$\begin{aligned} \lim _{y\rightarrow 0}R(y,t_0):= {-yv_{yy}(y,t_0)\over v_y(y,t_0)} = 1-q_1. \end{aligned}$$

Hence $v_y(y,t_0)$ satisfies (1.9) with index $1-q_1$. By (5.6), (5.10) and Potter’s bound, we deduce (2.15). Other cases can be prove similarly. We have shown that $u_x(x,t_0)$ satisfies (1.10) with index $1-\max \{p_1, p_2\}$. Consider the following problem: find a solution $u=\bar{u}(x,t)$ satisfying the equation

$$\begin{aligned} -{\partial u\over \partial t}-\frac{1}{2}\theta ^2{u_x^2 \over u_{xx}}+rxu_x-\delta u+V_2(u_x)=0 \end{aligned}$$

(5.44)

with the initial condition $\bar{u}(x,0)=u(x,t_0)$ for $x\in R_+$, where t represent a time horizon variable. Note that $u(x,t_0)$ is a utility function. Then we get $\bar{u}(x,t)=u(x,t+t_0)$ and $\bar{A}(x,t)=A(x,t+t_0)$ for all $x\in R_+$ by the uniqueness of solution to the Cauchy problem of the Bellman equation. From Theorem 2.7 and the dual representation, we deduce $\lim _{t\rightarrow \infty }\bar{A}(x,t)=\frac{\theta }{\sigma }(1-\min \{q_1,q_2\})x$, and the turnpike property (1.7) holds. $\square$