Liquidity management with decreasing returns to scale and secured credit line

Abstract

This paper examines the dividend and investment policies of a cash constrained firm, assuming a decreasing-returns-to-scale technology and adjustment costs. We extend the literature by allowing the firm to draw on a secured credit line both to hedge against cash-flow shortfalls and to invest/disinvest in a productive asset. We formulate this problem as a two-dimensional singular control problem and use both a viscosity solution approach and a verification technique to get qualitative properties of the value function. We further solve quasi-explicitly the control problem in two special cases.

Appendix

1.1 6.1 Proof of Theorem 5.4

1. Supersolution property. Let $(\bar{x},\bar{k}) \in S$ and $\varphi\in C^{2}(\mathbb{R}_{+}^{2})$ be such that $(\bar{x},\bar{k})$ is a minimum of $V^{*}-\varphi$ in a neighbourhood $B_{\varepsilon}(\bar {x},\bar{k})$ of $(\bar{x},\bar{k})$ with $\varepsilon$ small enough to ensure $B_{\varepsilon}\subset S$ and $V^{*}(\bar{x},\bar{k}) = \varphi(\bar{x},\bar{k})$.

First, consider the admissible control $\hat{\pi} = (\hat{Z}, \hat{I})$, where the shareholders decide to never invest or disinvest, while the dividend policy is defined by $\hat{Z}_{t} = \eta$ for $t \geq0$, with $0 \leq\eta \leq \varepsilon$. Define the exit time $\tau_{\varepsilon}=\inf\{t \geq0: (X_{t}^{\bar{x}},K_{t}^{\bar{k}}) \notin\overline{B}_{\varepsilon}(\bar {x},\bar{k}) \}$. We notice that $\tau_{\varepsilon}< \tau_{0}$ for $\varepsilon$ small enough. From the dynamic programming principle, we have

$$\begin{aligned} \varphi(\bar{x},\bar{k}) = V^{*}(\bar{x},\bar{k}) \geq& \mathbb{E}\bigg[\int_{0}^{\tau_{\varepsilon}\wedge h} e^{-r t}d\hat {Z}_{t} + e^{-r(\tau_{\varepsilon}\wedge h)} V^{*}\big(X_{\tau_{\varepsilon}\wedge h}^{\bar{x}},K_{\tau_{\varepsilon}\wedge h}^{\bar{k}}\big) \bigg] \\ \geq& \mathbb{E}\bigg[\int_{0}^{\tau_{\varepsilon}\wedge h} e^{-r t}d\hat{Z}_{t} + e^{-r(\tau_{\varepsilon}\wedge h)} \varphi\big(X_{\tau_{\varepsilon}\wedge h}^{\bar{x}},K_{\tau_{\varepsilon}\wedge h}^{\bar{k}}\big) \bigg]. \end{aligned}$$

(6.1)

Applying Itô’s formula to the process $(e^{-r t}\varphi(X_{t}^{\bar{x}},K_{t}^{\bar{k}}))$ between 0 and $\tau_{\varepsilon}\wedge h$ and taking expectations, we obtain

$$\begin{aligned} \mathbb{E}\big[e^{-r (\tau_{\varepsilon}\wedge h)} \varphi(X_{\tau _{\varepsilon}\wedge h}^{\bar{x}},K_{\tau_{\varepsilon}\wedge h}^{\bar{k}}) \big] =& \varphi(\bar {x},\bar{k}) + \mathbb{E}\bigg[\int_{0}^{\tau_{\varepsilon}\wedge h} e^{-r t} \mathcal{L} \varphi (X_{t}^{\bar{x}},K_{t}^{\bar{k}}) dt \bigg] \\ &{} + \mathbb{E}\bigg[ \sum_{0< t \leq\tau_{\varepsilon}\wedge h} e^{-r t} \big(\varphi(X_{t}^{\bar{x}},K_{t}^{\bar{k}}) - \varphi(X_{t-}^{\bar{x}},K_{t}^{\bar {k}})\big) \bigg]. \end{aligned}$$

(6.2)

Combining (6.1) and (6.2), we have

$$\begin{aligned} &\mathbb{E}\bigg[\int_{0}^{\tau_{\varepsilon}\wedge h} e^{-r t}(- \mathcal{L}) \varphi (X_{t}^{\bar{x}},K_{t}^{\bar{k}}) dt \bigg] - \mathbb{E}\bigg[\int_{0}^{\tau_{\varepsilon}\wedge h} e^{-r t}d\hat {Z}_{t} \bigg] \\ &{}- \mathbb{E}\bigg[ \sum_{0< t \leq\tau_{\varepsilon}\wedge h} e^{-r t} \big(\varphi (X_{t}^{\bar{x}},K_{t}^{\bar{k}}) - \varphi(X_{t-}^{\bar{x}},K_{t}^{\bar {k}})\big) \bigg] \geq0. \end{aligned}$$

(6.3)

⋆:

First take $\eta= 0$. We then observe that $X$ is continuous on $[\![0, \tau_{\varepsilon}\wedge h]\!]$ and only the first term of (6.3) is nonzero. By dividing the above inequality by $h$ and letting $h \rightarrow0$, we conclude that $- \mathcal{L} \varphi(\bar{x},\bar{k}) \geq0$.

⋆:

Now take $\eta> 0$ in (6.3). We see that $\hat{Z}$ jumps only at $t = 0$ with size $\eta$, so that

$$\mathbb{E}\bigg[\int_{0}^{\tau_{\varepsilon}\wedge h} e^{-r t}( - \mathcal{L} \varphi )(X_{t}^{\bar{x}},K_{t}^{\bar{k}}) dt \bigg] - \eta-\big(\varphi(\bar{x} - \eta,\bar{k}) - \varphi(\bar {x},\bar {k})\big) \geq0. $$

By sending $h \rightarrow0$ and then dividing by $\eta$ and letting $\eta\rightarrow0$, we obtain

$$\frac{\partial\varphi}{\partial x}(\bar{x},\bar{k}) - 1 \geq0. $$

Second, consider the admissible control $\bar{\pi} = (\bar{Z}, \bar{I})$, where the shareholders decide to never pay out dividends, while the investment/disinvestment policy is defined by $\bar{I}_{t}=\eta\in\mathbb{R}$ for $t$ ≥ 0, with $0<|\eta| \leq \varepsilon$. Define again the exit time $\tau_{\varepsilon}=\inf\{t \geq0: (X_{t}^{\bar{x}},K_{t}^{\bar{k}}) \notin\overline{B}_{\varepsilon}(\bar {x},\bar {k}) \}$. Proceeding analogously as in the first part and observing that $\bar{I}$ jumps only at $t=0$, we get

$$\mathbb{E}\bigg[\int_{0}^{\tau_{\varepsilon}\wedge h} e^{-r t}( - \mathcal{L} \varphi )(X_{t}^{\bar{x}},K_{t}^{\bar{k}}) dt \bigg] - \big(\varphi(\bar{x} - \gamma|\eta|,\bar{k}+\eta) - \varphi (\bar {x},\bar{k})\big) \geq0. $$

Assuming first $\eta>0$, by sending $h \rightarrow0$, and then dividing by $\eta$ and letting $\eta\rightarrow0$, we obtain

$$\gamma\frac{\partial\varphi}{\partial x}(\bar{x},\bar{k})-\frac {\partial\varphi}{\partial k}(\bar{x},\bar{k}) \ge0. $$

When $\eta<0$, we get in the same manner

$$\gamma\frac{\partial\varphi}{\partial x}(\bar{x},\bar{k})+\frac {\partial\varphi}{\partial k}(\bar{x},\bar{k}) \ge0. $$

This proves the required supersolution property.

2. Subsolution property. We prove the subsolution property by contradiction. Suppose that the claim is not true. Then there exist $(\bar{x},\bar{k}) \in S$ and a neighbourhood $B_{\varepsilon}(\bar {x},\bar{k})$ of $\bar{x},\bar{k}$, included in $S$ for $\varepsilon$ small enough, a $C^{2}$ function $\varphi$ with $(\varphi- V^{*})(\bar{x},\bar{k})= 0$ and $\varphi\geq V^{*}$ on $B_{\varepsilon}(\bar{x},\bar{k})$, and $\eta> 0$ such that we have, for all $(x,k) \in B_{\varepsilon}(\bar{x},\bar{k})$,

$$\begin{aligned} - \mathcal{L}\varphi(x,k) > \eta, \end{aligned}$$

(6.4)

$$\begin{aligned} \frac{\partial\varphi}{\partial x}(x,k) - 1 > \eta, \end{aligned}$$

(6.5)

$$\begin{aligned} \bigg(\gamma\frac{\partial\varphi}{\partial x}-\frac{\partial \varphi }{\partial k}\bigg)(x,k) > \eta, \end{aligned}$$

(6.6)

$$\begin{aligned} \bigg(\gamma\frac{\partial\varphi}{\partial x}+\frac{\partial \varphi }{\partial k}\bigg)(x,k) > \eta. \end{aligned}$$

(6.7)

For any admissible control $\pi$, the exit time $\tau_{\varepsilon}= \inf\{ t \geq0: (X_{t}^{\bar {x}},K_{t}^{\bar{k}}) \notin B_{\varepsilon}(\bar{x},\bar{k}) \}$ satisfies $\tau_{\varepsilon}< \tau_{0}$. Applying Itô’s formula to the process $(e^{-r t}\varphi(X_{t}^{\bar{x}},K_{t}^{\bar{k}}))$ between 0 and $\tau_{\varepsilon}{-}$, we have

$$\begin{aligned} \mathbb{E}\big[e^{-r \tau_{\varepsilon}-}\varphi(X_{\tau _{\varepsilon}-},K_{\tau _{\varepsilon}-})\big] =& \varphi(\bar{x},\bar{k}) - \mathbb{E}\bigg[\int _{0}^{\tau _{\varepsilon}-}e^{-ru}\mathcal{L}\varphi du\bigg] \\ &{}+ \mathbb{E}\bigg[\int_{0}^{\tau_{\varepsilon}-}e^{-r u}\bigg(-\gamma \frac {\partial\varphi}{\partial x} + \frac{\partial\varphi}{\partial k}\bigg)dI_{u}^{c,+}\bigg] \\ &{}+ \mathbb{E}\bigg[\int_{0}^{\tau_{\varepsilon}-}e^{-r u}\bigg(-\gamma \frac {\partial\varphi}{\partial x} - \frac{\partial\varphi}{\partial k}\bigg)dI_{u}^{c,-}\bigg] \\ &{}- \mathbb{E}\bigg[\int_{0}^{\tau_{\varepsilon}-}e^{-r u}\frac {\partial \varphi }{\partial x}dZ^{c}_{u}\bigg] \\ &{}+ \mathbb{E}\bigg[\sum_{0 < s < \tau_{\varepsilon}}e^{-r s}\big(\varphi (X_{s},K_{s}) - \varphi(X_{s-},K_{s-})\big)\bigg]. \end{aligned}$$

Using (6.4)–(6.7), we obtain

$$\begin{aligned} V^{*}(\bar{x},\bar{k}) = \varphi(\bar{x},\bar{k}) \geq& \eta\mathbb{E} \bigg[ \int_{0}^{\tau_{\varepsilon}-} e^{-r u} du + e^{-r \tau_{\varepsilon}-}\varphi\big(X_{\tau_{\varepsilon}-},K_{\tau _{\varepsilon}-}\big)\bigg]\\ &{}+ \eta\mathbb{E}\bigg[\int_{0}^{\tau_{\varepsilon}-}e^{-r u}dI_{u}^{c,+}\bigg] + \eta\mathbb{E}\bigg[\int_{0}^{\tau_{\varepsilon}-}e^{-r u}dI_{u}^{c,-}\bigg]\\ &{}+ (1+\eta)\mathbb{E}\bigg[\int_{0}^{\tau_{\varepsilon}-}e^{-r u}dZ^{c}_{u}\bigg]\\ &{}- \mathbb{E}\bigg[\sum_{0 < s < \tau_{\varepsilon}}e^{-r s}\big(\varphi (X_{s},K_{s}) - \varphi(X_{s-},K_{s-})\big)\bigg]. \end{aligned}$$

Note that $\Delta X_{s} = -\Delta Z_{s} - \gamma( \Delta I_{s}^{+} + \Delta I_{s}^{-} )$, $\Delta K_{s} = \Delta I_{s}^{+} - \Delta I_{s}^{-}$ and by the mean value theorem, there is some $\theta\in(0,1)$ such that

$$\begin{aligned} &\varphi(X_{s},K_{s}) - \varphi(X_{s-},K_{s-}) \\ &\quad= \frac{\partial\varphi}{\partial x}(X_{s-} + \theta\Delta X_{s}, K_{s-} + \theta\Delta K_{s}) \Delta X_{s} \\ & \qquad{} +\frac{\partial\varphi}{\partial k}(X_{s-} + \theta \Delta X_{s}, K_{s-} + \theta\Delta K_{s})\Delta K_{s}\\ &\quad= \frac{\partial\varphi}{\partial x}(X_{s-} + \theta\Delta X_{s}, K_{s-} + \theta\Delta K_{s})\big(-\Delta Z_{s} - \gamma( \Delta I_{s}^{+} + \Delta I_{s}^{-} ) \big) \\ &\qquad{}+ \frac{\partial\varphi}{\partial k}(X_{s-} + \theta \Delta X_{s}, K_{s-} + \theta\Delta K_{s})(\Delta I_{s}^{+} - \Delta I_{s}^{-})\\ &\quad= - \frac{\partial\varphi}{\partial x}(X_{s-} + \theta\Delta X_{s}, K_{s-} + \theta\Delta K_{s})\Delta Z_{s}\\ &\qquad{}+ \bigg(- \gamma\frac{\partial\varphi}{\partial x}(X_{s-} + \theta\Delta X_{s}, K_{s-} + \theta\Delta K_{s}) \\ & \qquad\qquad+ \frac{\partial\varphi}{\partial k}(X_{s-} + \theta\Delta X_{s}, K_{s-} + \theta\Delta K_{s})\bigg) \Delta I_{s}^{+}\\ &\qquad{}+ \bigg(- \gamma\frac{\partial\varphi}{\partial x}(X_{s-} - \theta\Delta X_{s}, K_{s-} + \theta\Delta K_{s}) \\ &\qquad\qquad+ \frac{\partial\varphi}{\partial k}(X_{s-} + \theta\Delta X_{s}, K_{s-} + \theta\Delta K_{s})\bigg) \Delta I_{s}^{-}. \end{aligned}$$

Because $(X_{s}+\theta\Delta X_{s}, K_{s}+\theta\Delta K_{s}) \in B_{\varepsilon}(\bar {x},\bar{k})$, we use (6.5)–(6.7) again to get

$$- \big(\varphi(X_{s},K_{s}) - \varphi(X_{s-},K_{s-})\big) \geq(1+\eta) \Delta Z_{s} + \eta\Delta I_{s}^{+} + \eta\Delta I_{s}^{-}. $$

Therefore,

$$\begin{aligned} V^{*}(\bar{x},\bar{k}) \geq& \mathbb{E}\big[e^{-r \tau_{\varepsilon}-}\varphi (X_{\tau_{\varepsilon}-},K_{\tau_{\varepsilon}-})\big] + \mathbb {E}\bigg[\int _{0}^{\tau_{\varepsilon}-}e^{-r u} dZ_{u}\bigg]\\ &{}+ \eta\bigg( \mathbb{E} \bigg[ \int_{0}^{\tau_{\varepsilon}-} e^{-r u} du \bigg] + \mathbb{E}\bigg[\int_{0}^{\tau_{\varepsilon}-}e^{-ru}dI_{u}^{+}\bigg] + \mathbb {E}\bigg[\int_{0}^{\tau_{\varepsilon}-}e^{-r u}dI_{u}^{-}\bigg] \\ &{} + \mathbb{E}\bigg[\int_{0}^{\tau_{\varepsilon}-}e^{-r u}dZ_{u}\bigg] \bigg). \end{aligned}$$

Notice that while $(X_{\tau_{\varepsilon}-}, K_{\tau_{\varepsilon}-}) \in B_{\varepsilon}(\bar {x},\bar{k})$, $(X_{\tau_{\varepsilon}},K_{\tau_{\varepsilon}})$ is either on the boundary $\partial B_{\varepsilon}(\bar{x},\bar{k})$ or outside of $\bar{B}_{\varepsilon}(\bar{x},\bar{k})$. However, there is some random variable $\alpha$ valued in $[0,1]$ such that

$$\begin{aligned} (X^{(\alpha)},K^{(\alpha)}) =& (X_{\tau_{\varepsilon}-},K_{\tau _{\varepsilon}-}) + \alpha(\Delta X_{\tau_{\varepsilon}}, \Delta K_{\tau_{\varepsilon}})\\ =& (X_{\tau_{\varepsilon}-},K_{\tau_{\varepsilon}-}) + \alpha(- \Delta Z_{\tau _{\varepsilon}} - \gamma\Delta I^{+}_{\tau_{\varepsilon}} - \gamma\Delta I^{-}_{\tau _{\varepsilon}}, \Delta I^{+}_{\tau_{\varepsilon}}- \Delta I^{-}_{\tau_{\varepsilon}}) \end{aligned}$$

is in $\partial B_{\varepsilon}(\bar{x},\bar{k})$. Proceeding analogously as above, we show that

$$\varphi(X^{(\alpha)},K^{(\alpha)}) - \varphi(X_{\tau_{\varepsilon}-},K_{\tau_{\varepsilon}-}) \leq- \alpha\big((1+\eta) \Delta Z_{\tau_{\varepsilon}} + \eta \Delta I_{\tau_{\varepsilon}}^{+} + \eta\Delta I_{\tau_{\varepsilon}}^{-}\big). $$

Observe that

$$(X^{(\alpha)},K^{(\alpha)}) = (X_{\tau_{\varepsilon}},K_{\tau _{\varepsilon}}) + (1-\alpha )(\Delta Z_{\tau_{\varepsilon}} + \gamma\Delta I^{+}_{\tau _{\varepsilon}} + \gamma \Delta I^{-}_{\tau_{\varepsilon}}, -\Delta I^{+}_{\tau_{\varepsilon}} + \Delta I^{-}_{\tau_{\varepsilon}}). $$

Starting from $(X^{(\alpha)},K^{(\alpha)})$, the strategy that consists of investing $(1-\alpha)\Delta I_{\tau_{\varepsilon}}^{+}$ or disinvesting $(1-\alpha)\Delta I_{\tau_{\varepsilon}}^{-}$, depending on the sign of $K^{(\alpha )}-K_{\tau_{\varepsilon}}$, and paying out $(1-\alpha)\Delta Z_{\tau _{\varepsilon}}$ as dividends leads to $(X_{\tau_{\varepsilon}},K_{\tau_{\varepsilon}})$, and therefore

$$V^{*}(X^{(\alpha)},K^{(\alpha)}) - V^{*}(X_{\tau_{\varepsilon}},K_{\tau _{\varepsilon}}) \geq (1- \alpha) \Delta Z_{\tau_{\varepsilon}}. $$

Using $\varphi(X^{(\alpha)},K^{(\alpha)}) \geq V^{*}(X^{(\alpha )},K^{(\alpha)})$, we deduce

$$\varphi(X_{\tau_{\varepsilon}-},K_{\tau_{\varepsilon}-}) - V^{*}(X_{\tau_{\varepsilon}}, K_{\tau_{\varepsilon}}) \geq(1 + \alpha\eta) \Delta Z_{\tau _{\varepsilon}} + \alpha \eta(\Delta I^{+}_{\tau_{\varepsilon}} + \Delta I^{-}_{\tau _{\varepsilon}}). $$

Hence,

$$\begin{aligned} V^{*}(\bar{x},\bar{k}) \geq& \eta\bigg( \mathbb{E}\bigg[ \int _{0}^{\tau _{\varepsilon}-} e^{-r u} du \bigg] + \mathbb{E}\bigg[\int _{0}^{\tau_{\varepsilon }-}e^{-r u}dI_{u}^{+}\bigg] +\mathbb{E}\bigg[\int_{0}^{\tau _{\varepsilon}-}e^{-r u}dI_{u}^{-}\bigg] \\ & \phantom{\eta\Big(}+ \mathbb{E}\bigg[\int_{0}^{\tau _{\varepsilon}-}e^{-r u}dZ_{u}\bigg] + \mathbb{E}\big[e^{-r \tau_{\varepsilon}}\alpha (\Delta Z_{\tau _{\varepsilon}} + \gamma\Delta I_{\tau_{\varepsilon}}^{+} + \gamma \Delta I_{\tau _{\varepsilon }}^{-})\big]\bigg) \\ &{}+ \mathbb{E}\big[e^{-r \tau_{\varepsilon}}V^{*}(X_{\tau _{\varepsilon}},K_{\tau _{\varepsilon }})\big] + \mathbb{E}\bigg[\int_{0}^{\tau_{\varepsilon}}e^{-r u}dZ_{u}\bigg]. \end{aligned}$$

(6.8)

We now claim there is $c_{0} > 0$ such that for any admissible strategy,

$$\begin{aligned} c_{0} \leq&\mathbb{E} \bigg[ \int_{0}^{\tau_{\varepsilon}-} e^{-r u} du + \int _{0}^{\tau_{\varepsilon}-}e^{-r u}dI_{u}^{+} + \int_{0}^{\tau _{\varepsilon}-}e^{-r u}dI_{u}^{-} + \int_{0}^{\tau_{\varepsilon}-}e^{-r u}dZ_{u}\bigg] \\ &{}+ \mathbb{E}\big[e^{-r \tau_{\varepsilon}}\alpha(\Delta Z_{\tau _{\varepsilon }} + \gamma\Delta I_{\tau_{\varepsilon}}^{+} + \gamma\Delta I_{\tau _{\varepsilon }}^{-})\big]. \end{aligned}$$

(6.9)

Let us consider the $C^{2}$ function $\phi(x,k) = c_{0}(1-\frac{(x-\bar {x})^{2}}{\varepsilon^{2}})$ with

$$0 < c_{0} \leq\min\bigg\{ \frac{\varepsilon}{2}, \frac{\varepsilon }{2 \gamma}, \frac {1}{r}, \frac{\varepsilon^{2}}{\sigma_{n}^{2}\bar{\beta}^{2}}, \frac {\varepsilon }{2d_{\max}}\bigg\} , $$

where

$$d_{\max} = \sup\bigg\{ \frac{|\beta(k)\mu- \alpha ((k-x)^{+})|}{\varepsilon}: (x,k) \in B_{\varepsilon}(\bar{x},\bar{k})\bigg\} >0. $$

This function satisfies

$$\left\{ \begin{aligned} &\phi(\bar{x},\bar{k}) = c_{0},\\ &\phi= 0 \quad \text{for } (x,k) \in\partial B_{\varepsilon},\\ &\min\bigg\{ 1 - \mathcal{L}\phi, 1 -\gamma\frac{\partial\phi }{\partial x} + \frac{\partial\phi}{\partial k}, 1 -\gamma\frac {\partial\phi}{\partial x} - \frac{\partial\phi}{\partial k}, 1 - \frac{\partial\phi}{\partial x}\bigg\} \geq0\quad \text{for } (x,k) \in B_{\varepsilon}. \end{aligned} \right. $$

Applying Itô’s formula, we have

$$\begin{aligned} 0 < c_{0} =& \phi(\bar{x},\bar{k}) \\ \leq&\mathbb{E}\big[e^{-r \tau_{\varepsilon}-}\phi(X_{\tau _{\varepsilon }-},K_{\tau _{\varepsilon}-})\big] + \mathbb{E}\bigg[\int_{0}^{\tau _{\varepsilon}-}e^{-r u}du\bigg] \\ &{}+ \mathbb{E}\bigg[\int_{0}^{\tau_{\varepsilon}-}e^{-r u}dI^{+}_{u}\bigg]+ \mathbb {E}\bigg[\int_{0}^{\tau_{\varepsilon}-}e^{-r u}dI^{-}_{u}\bigg]+ \mathbb {E}\bigg[\int _{0}^{\tau_{\varepsilon}-}e^{-r u}dZ_{u}\bigg]. \end{aligned}$$

(6.10)

Noting that $\frac{\partial\phi}{\partial x} \leq1$ and $\frac {\partial\phi}{\partial k}= 0$, we have

$$\phi(X_{\tau_{\varepsilon}-},K_{\tau_{\varepsilon}-}) - \phi (X^{(\alpha )},K^{(\alpha )}) \leq X_{\tau_{\varepsilon}-} - X^{(\alpha)} = \alpha(\Delta Z_{\tau _{\varepsilon }} + \gamma\Delta I^{+}_{\tau_{\varepsilon}} + \gamma\Delta I^{-}_{\tau_{\varepsilon}}). $$

Plugging this into (6.10) with $\phi(X^{(\alpha )},K^{(\alpha )})=0$, we obtain

$$\begin{aligned} c_{0} \leq&\mathbb{E} \bigg[ \int_{0}^{\tau_{\varepsilon}-} e^{-r u} du + \int _{0}^{\tau_{\varepsilon}-}e^{-r u}dI_{u}^{+} + \int_{0}^{\tau _{\varepsilon}-}e^{-r u}dI_{u}^{-} + \int_{0}^{\tau_{\varepsilon}-}e^{-r u}dZ_{u}\bigg]\\ &{}+ \mathbb{E}\big[e^{-r \tau_{\varepsilon}}\alpha(\Delta Z_{\tau _{\varepsilon }} + \gamma\Delta I_{\tau_{\varepsilon}}^{+} + \gamma\Delta I_{\tau _{\varepsilon}}^{-}) \big]. \end{aligned}$$

This proves the claim (6.9). Finally, by taking the supremum over $\pi$ and using the dynamic programming principle, (6.8) implies $V^{*}(\bar{x},\bar{k}) \geq V^{*}(\bar{x},\bar{k}) + \eta c_{0}$, which is a contradiction.

3. Uniqueness. Suppose $u$ is a continuous subsolution and $w$ a continuous supersolution of (5.1) on $S$ satisfying the boundary conditions

$$u(x,0)\le w(x,0), \qquad u(\gamma k,k)\le w(\gamma k,k)\quad\mbox{for } (x,k) \in S, $$

and the linear growth condition

$$|u(x,k)|+|w(x,k)|\le C_{1} +C_{2}(x+k) \quad\text{for } (x,k) \in S, $$

for some positive constants $C_{1}$ and $C_{2}$. We show by adapting some standard arguments that $u \le w$.

Step 1. We first construct a strict supersolution of (5.1) with a perturbation of $w$. Set

$$h(x,k) = A + B x + Ck + Dxk + Ex^{2} + k^{2} $$

with

$$ A = \frac{1+\mu\bar{\beta} B + \sigma^{2} \bar{\beta}^{2}E }{r} + C_{1} $$

(6.11)

and

$$B = 2 + \frac{1+C}{\gamma} + \frac{2\mu\bar{\beta}E}{r},\qquad C = \frac{\mu\bar{\beta}D}{r},\qquad D = 2\gamma E,\qquad E = \frac{1}{\gamma^{2}}, $$

and define for $\lambda\in[0,1]$ on $S$ the continuous function

$$w^{\lambda}=(1-\lambda)w+\lambda h. $$

Because

$$\begin{aligned} \frac{\partial h}{\partial x} - 1 &= B+Dk+2Ex - 1 \geq1, \\ \gamma\frac{\partial h}{\partial x} - \frac{\partial h}{\partial k} &= \gamma(B+Dk+2Ex)-(C+Dx+2k) \geq1, \\ \gamma\frac{\partial h}{\partial x} + \frac{\partial h}{\partial k} &= \gamma(B+Dk+2Ex)+ (C+Dx+2k)\geq1 \end{aligned}$$

and

$$\begin{aligned} -\mathcal{L}h = &-\Big(\beta(k)\mu- \alpha\big((k-x)^{+}\big)\Big)(B+Dk+2Ex) \\ &{}- \frac{\sigma^{2}\beta(k)^{2}}{2}2E + r (A + Bx + Ck + Dxk + Ex^{2} + k^{2})\\ \geq& \big(r A - \beta(k)\mu B - \sigma^{2} \beta(k)^{2}E\big) + \big(r B - 2 \mu\beta(k)E\big)x + \big(rC- \mu\beta(k)D\big)k\\ \geq&1, \end{aligned}$$

we have that

$$\min\bigg\{ -\mathcal{L} h, \frac{\partial h}{\partial x} - 1, \gamma \frac{\partial h}{\partial x} - \frac{\partial h}{\partial k}, \gamma \frac{\partial h}{\partial x} + \frac{\partial h}{\partial k}\bigg\} \geq1 $$

which implies that $w^{\lambda}$ is a strict supersolution of (5.1). To prove this point, one only needs to take $\bar{x}$ and $\varphi\in C^{2}$ such that $\bar{x}$ is a minimum of $w^{\lambda}- \varphi$ and notice that $\bar{x}$ is also a minimum of $w^{\lambda}- \varphi_{2}$ with $\varphi_{2} = \frac{\varphi- \lambda h}{1-\lambda}$, which allows us to use that $w$ is a viscosity supersolution of (5.1).

Step 2. In order to prove the strong comparison result, it suffices to show that for every $\lambda\in[0,1]$,

$$ \sup_{S} (u-w^{\lambda}) \le0. $$

(6.12)

Assume by way of contradiction that there exists $\lambda$ such that

$$ \sup_{S} (u-w^{\lambda}) > 0. $$

(6.13)

Because $u$ and $w$ have linear growth, we have

$$\displaystyle{\lim_{|(x,k)|\to\infty}}(u-w^{\lambda})(x,k)= -\infty. $$

Using the boundary conditions

$$\begin{aligned} u(x,0)-w^{\lambda}(x,0) = & (1- \lambda) \big(u(x,0) -w(x,0)\big) \\ &{}+ \lambda\big( u(x,0) - (A + Bx + E x^{2})\big), \\ \le& \lambda\big(u(x,0) - (A + Bx + E x^{2})\big),\\ u(\gamma k,k)-w^{\lambda}(\gamma k,k) \le& \lambda\Big(u(\gamma k,k) \\ &\phantom{\lambda\big(}{} - \big( A + (B \gamma+ C) k + ( D \gamma + E \gamma^{2} +1 )k^{2}\big)\Big) \end{aligned}$$

and the linear growth condition, it is always possible to find $C_{1}$ in (6.11) such that both expressions above are negative and the maximum in (6.13) is reached inside the domain $S$. By continuity of the functions $u$ and $w^{\lambda}$, there exists a pair $(x_{0},k_{0})$ with $x_{0} \ge\gamma k_{0}$ such that

$$M=\sup_{S} (u-w^{\lambda})=(u-w^{\lambda})(x_{0},k_{0}). $$

For $\epsilon> 0$, consider the functions

$$\begin{aligned} \varPhi_{\epsilon}(x,y,k,\ell) &= u(x,k) - w^{\lambda}(y,\ell) - \phi _{\epsilon}(x,y,k,\ell),\\ \phi_{\epsilon}(x,y,k,\ell) &= \frac{1}{2\epsilon}\big(|x-y|^{2}+|k-\ell |^{2}\big) + \frac{1}{4}\big(|x-x_{0}|^{4}+|k-k_{0}|^{4}\big). \end{aligned}$$

By standard arguments in the comparison principle of the viscosity solution theory (see Pham [26, Sect. 4.4.2]), the function $\varPhi_{\varepsilon}$ attains a maximum in $(x_{\epsilon},y_{\epsilon},k_{\epsilon},\ell_{\epsilon})$, which converges (up to a subsequence) to $(x_{0}, k_{0},x_{0},k_{0})$ when $\varepsilon$ goes to zero. Moreover,

$$\begin{aligned} \lim_{\epsilon\rightarrow\infty}\frac{|x_{\epsilon}-y_{\epsilon}|^{2}+|k_{\epsilon}-\ell_{\epsilon}|^{2}}{2\epsilon} = 0. \end{aligned}$$

Applying Theorem 3.2 in Crandall et al. [7], we get the existence of symmetric square matrices $M_{\varepsilon}$, $N_{\varepsilon}$ of size 2 such that

$$\begin{aligned} (p_{\varepsilon}, M_{\varepsilon}) \;\in\; J^{2,+} u(x_{\varepsilon},k_{\varepsilon}), \\ (q_{\varepsilon}, N_{\varepsilon}) \;\in\; J^{2,-} w^{\lambda}(y_{\varepsilon},\ell_{\varepsilon}) \end{aligned}$$

and

$$\begin{aligned} \left( \textstyle\begin{array}{cc} M_{\varepsilon}& 0 \\ 0 & -N_{\varepsilon}\end{array}\displaystyle \right) \; \leq\; D^{2} \phi_{\varepsilon}(x_{\varepsilon}, k_{\varepsilon},y_{\epsilon},\ell_{\epsilon}) + \varepsilon\big(D^{2} \phi_{\epsilon}(x_{\varepsilon}, k_{\varepsilon},y_{\epsilon},\ell _{\epsilon})\big)^{2}, \end{aligned}$$

(6.14)

where

$$\begin{aligned} p_{\varepsilon}&=D_{x,k}\phi_{\epsilon}(x_{\epsilon},k_{\epsilon},y_{\epsilon},\ell _{\epsilon}) = \bigg(\frac{x_{\epsilon}- y_{\epsilon}}{\epsilon} + (x_{\epsilon}- x_{0})^{3},\frac{k_{\epsilon}- \ell_{\epsilon}}{\epsilon} + (k_{\epsilon}- k_{0})^{3}\bigg),\\ q_{\varepsilon}&=-D_{y,\ell}\phi_{\epsilon}(x_{\epsilon},k_{\epsilon},y_{\epsilon},\ell _{\epsilon}) = \bigg(\frac{x_{\epsilon}- y_{\epsilon}}{\epsilon},\frac {k_{\epsilon}- \ell_{\epsilon}}{\epsilon}\bigg) \end{aligned}$$

and

$$D^{2} \phi_{\varepsilon}(x_{\varepsilon}, k_{\varepsilon},y_{\epsilon},\ell_{\epsilon}) = \frac {1}{\epsilon} \left( \textstyle\begin{array}{c@{\quad}c} I_{2} & -I_{2}\\ -I_{2} & I_{2} \end{array}\displaystyle \right) + \left( \textstyle\begin{array}{c@{\quad}c@{\quad}c@{\quad}c} 3(x_{\epsilon}- x_{0})^{2} & 0& 0& 0\\ 0 & 3(k_{\epsilon}- k_{0})^{2} & 0& 0\\ 0&0&0&0\\ 0&0&0&0 \end{array}\displaystyle \right), $$

so that

$$\begin{aligned} &D^{2} \phi_{\varepsilon}(x_{\varepsilon}, k_{\varepsilon},y_{\epsilon},\ell_{\epsilon}) + \varepsilon \big(D^{2} \phi_{\epsilon}(x_{\varepsilon}, y_{\varepsilon},k_{\epsilon},\ell _{\epsilon})\big)^{2} \\ &= \frac{3}{\epsilon} \left( \textstyle\begin{array}{c@{\quad}c} I_{2} & -I_{2}\\ -I_{2} & I_{2} \end{array}\displaystyle \right)M_{\epsilon}\\ &\phantom{=:}+ \left( \textstyle\begin{array}{c@{\quad}c@{\quad}c@{\quad}c} 9(x_{\epsilon}- x_{0})^{2}(1 + \epsilon(x_{\epsilon}- x_{0})^{2}) & 0& 0& 0\\ 0 & 9(k_{\epsilon}- k_{0})^{2}(1 + \epsilon(k_{\epsilon}- k_{0})^{2}) & 0& 0\\ 0&0&0&0\\ 0&0&0&0 \end{array}\displaystyle \right). \end{aligned}$$

Equation (6.14) implies that

$$\begin{aligned} &\text{tr} \bigg(\frac{\sigma^{2}\beta(k_{\epsilon})^{2}}{2}M_{\epsilon}- \frac {\sigma^{2}\beta(\ell_{\epsilon})^{2}}{2}N_{\epsilon}\bigg) \\ & \leq\frac{3\sigma^{2}}{2\epsilon}\big(\beta(k_{\epsilon})^{2} - \beta (\ell _{\epsilon})^{2}\big) + \frac{9\sigma^{2} \beta(k_{\epsilon})^{2}}{2} (x_{\epsilon}- x_{0})^{2}\big(1+\epsilon(x_{\epsilon}- x_{0})^{2}\big). \end{aligned}$$

(6.15)

Because $u$ and $w^{\lambda}$ are respectively a subsolution and a strict supersolution, we have

$$\begin{aligned} \min\bigg\{ &-\Big(\beta(k_{\epsilon})\mu- \alpha\big((k_{\epsilon}-x_{\epsilon})^{+}\big)\Big) \bigg(\frac{x_{\epsilon}-y_{\epsilon}}{\epsilon }+(x_{\epsilon}-x_{0})^{3}\bigg) \\ & -\text{tr}\bigg(\frac{\sigma^{2} \beta(k_{\epsilon})^{2}}{2}M_{\epsilon}\bigg) + r u(x_{\epsilon},k_{\epsilon}), \frac{x_{\epsilon}-y_{\epsilon}}{\epsilon }+(x_{\epsilon}-x_{0})^{3} - 1,\\ &\gamma\bigg(\frac{x_{\epsilon}-y_{\epsilon}}{\epsilon}+(x_{\epsilon}-x_{0})^{3}\bigg) - \bigg(\frac{k_{\epsilon}-\ell_{\epsilon}}{\epsilon }+(k_{\epsilon}-k_{0})^{3}\bigg),\\ &\gamma\bigg(\frac{x_{\epsilon}-y_{\epsilon}}{\epsilon}+(x_{\epsilon}-x_{0})^{3}\bigg) + \bigg(\frac{k_{\epsilon}-\ell_{\epsilon}}{\epsilon }+(k_{\epsilon}-k_{0})^{3}\bigg) \bigg\} \leq0 \end{aligned} $$

and

$$\begin{aligned} \min\bigg\{ &-\Big(\beta(\ell_{\epsilon})\mu- \alpha\big((\ell _{\epsilon}-y_{\epsilon})^{+}\big)\Big) \frac{x_{\epsilon}-y_{\epsilon}}{\epsilon} - \text{tr}\bigg(\frac{\sigma^{2} \beta(\ell_{\epsilon})^{2}}{2}N_{\epsilon}\bigg) + r w^{\lambda}(y_{\epsilon},\ell_{\epsilon}), \\ &\frac{x_{\epsilon}-y_{\epsilon}}{\epsilon} - 1,\gamma\frac {x_{\epsilon}-y_{\epsilon}}{\epsilon}- \frac{k_{\epsilon}-\ell_{\epsilon}}{\epsilon },\gamma\frac{x_{\epsilon}-y_{\epsilon}}{\epsilon}+ \frac{k_{\epsilon}-\ell _{\epsilon}}{\epsilon} \bigg\} \geq\lambda. \end{aligned}$$

(6.16)

We then distinguish the following four cases:

Case 1.:

If $\frac{x_{\epsilon}-y_{\epsilon}}{\epsilon }+(x_{\epsilon}-x_{0})^{3} - 1 \leq0$, we get from (6.16) that $\lambda + (x_{\epsilon}- x_{0})^{3} \leq0$, yielding a contradiction when $\epsilon$ goes to 0.

Case 2.:

If $\gamma(\frac{x_{\epsilon}-y_{\epsilon}}{\epsilon }+(x_{\epsilon}-x_{0})^{3}) - (\frac{k_{\epsilon}-\ell_{\epsilon}}{\epsilon }+(k_{\epsilon}-k_{0})^{3}) \leq0$, we get from (6.16) that $\lambda+ \gamma((x_{\epsilon}-x_{0})^{3} -(k_{\epsilon}-k_{0})^{3}) \leq0$, yielding a contradiction when $\epsilon$ goes to 0.

Case 3.:

If $\gamma(\frac{x_{\epsilon}-y_{\epsilon}}{\epsilon }+(x_{\epsilon}-x_{0})^{3}) + (\frac{k_{\epsilon}-l_{\epsilon}}{\epsilon }+(k_{\epsilon}-k_{0})^{3}) \leq0$, we get from (6.16) that $\lambda+ \gamma((x_{\epsilon}-x_{0})^{3} +(k_{\epsilon}-k_{0})^{3} )\leq 0$, yielding a contradiction when $\epsilon$ goes to 0.

Case 4.:

If

$$\begin{aligned} &-\Big(\beta(k_{\epsilon})\mu- \alpha\big((k_{\epsilon}-x_{\epsilon})^{+}\big)\Big) \bigg(\frac{x_{\epsilon}-y_{\epsilon}}{\epsilon}+(x_{\epsilon}-x_{0})^{3}\bigg) \\ &- \text{tr}\bigg(\frac{\sigma^{2} \beta(k_{\epsilon})^{2}}{2}M_{\epsilon}\bigg) + r u(x_{\epsilon},k_{\epsilon}) \leq0, \end{aligned}$$

we deduce from

$$-\Big(\beta(\ell_{\epsilon})\mu- \alpha\big((\ell_{\epsilon}-y_{\epsilon})^{+}\big)\Big) \frac{x_{\epsilon}-y_{\epsilon}}{\epsilon} - \text{tr}\bigg(\frac{\sigma^{2} \beta(\ell_{\epsilon})^{2}}{2}N_{\epsilon}\bigg) + r w^{\lambda}(y_{\epsilon},\ell_{\epsilon}) \geq\lambda $$

that

$$\begin{aligned} &\frac{x_{\epsilon}-y_{\epsilon}}{\epsilon}\Big(\mu\big(\beta(\ell _{\epsilon})-\beta(k_{\epsilon})\big)+\alpha\big((k_{\epsilon}-x_{\epsilon})^{+}\big)-\alpha\big((\ell_{\epsilon}-y_{\epsilon})^{+}\big)\Big)\\ &-\text{tr}\bigg(\frac{\sigma^{2}\beta(k_{\epsilon})^{2}}{2}N_{\epsilon}\bigg) + \text{tr}\bigg(\frac{\sigma^{2}\beta(k_{\epsilon})^{2}}{2}N_{\epsilon}\bigg) \\ &-\Big(\beta(k_{\epsilon})\mu- \alpha\big((k_{\epsilon}-x_{\epsilon})^{+}\big)\Big) (x_{\epsilon}-x_{0})^{3} \\ &+ r\big(u(x_{\epsilon},k_{\epsilon}) - w^{\lambda}(y_{\epsilon},\ell _{\epsilon})\big)\leq-\lambda. \end{aligned}$$

Using (6.15), we get

$$\begin{aligned} &\frac{x_{\epsilon}-y_{\epsilon}}{\epsilon}\Big(\mu\big(\beta(\ell _{\epsilon})-\beta(k_{\epsilon})\big)+\alpha\big((k_{\epsilon}-x_{\epsilon})^{+}\big)-\alpha\big((\ell_{\epsilon}-y_{\epsilon})^{+}\big)\Big)\\ &-\Big(\beta(k_{\epsilon})\mu- \alpha\big((k_{\epsilon}-x_{\epsilon})^{+}\big)\Big) (x_{\epsilon}-x_{0})^{3} + r\big(u(x_{\epsilon},k_{\epsilon}) - w^{\lambda}(y_{\epsilon},\ell_{\epsilon})\big) \\ &\leq-\lambda+ \frac{3\sigma^{2}}{2\epsilon}\big(\beta(k_{\epsilon})^{2} - \beta(\ell_{\epsilon})^{2}\big) + \frac{9\sigma^{2} \beta(k_{\epsilon})^{2}}{2} (x_{\epsilon}- x_{0})^{2}\big(1+\epsilon(x_{\epsilon}- x_{0})^{2}\big). \end{aligned} $$

By sending $\varepsilon$ to zero and using the continuity of $u$, $w^{\gamma}_{i}$, $\alpha$ and $\beta$ we obtain the required contradiction, namely $r M \leq-\lambda$.

We have thus shown for every $\lambda\in[0,1]$ that (6.12) holds, i.e., $\sup_{S} (u - w^{\lambda}) \leq0$. This implies by letting $\lambda\to0$ the strong comparison result $u \leq w$ for any subsolution $u$ and supersolution $w$. Clearly, this strong comparison result implies uniqueness. This ends the proof of Theorem 5.4.  □

1.2 6.2 Proof of Proposition 5.5

Because $\beta$ is concave and $\beta'$ goes to 0, the existence of $a$ is equivalent to assuming

$$ \sigma^{2}\beta'(0) \ge\frac{\mu}{1-\delta}. $$

(6.17)

Let us define the function $w_{A}$ for $A>0$ as the unique solution on $(a, \infty)$ of the Cauchy problem

$$\mu\beta(x)w_{A}'(x) + \frac{\sigma^{2}\beta(x)^{2}}{2}w_{A}''(x) - r w_{A}(x) = 0, $$

with $w_{A}(x)=Ax^{\delta}$ for $0 \le x \le a$ and $w_{A}$ differentiable at $a$.

Remark 6.1

The above Cauchy problem is well defined with the condition that $w_{A}$ is differentiable at $a$. Moreover, it is easy to check, using the definition of $a$, that the function $w_{A}$ is also $C^{2}$. Because the spread $\alpha$ is high, the shareholders optimally choose not to tap the credit line, but rather adjust costlessly their level of investment.

Lemma 6.2

For every $A>0$, the function $w_{A}$ is increasing.

Proof

Clearly, $w_{A}$ is increasing and therefore positive on $[0,a]$. If we define $c$ by $c=\min\{x> a:w'_{A}(c)=0\}$, then $w_{A}(c)>0$ because $w_{A}$ is increasing and positive in a left neighbourhood of $c$. Thus, according to the differential equation, we have $w_{A}''(c)\ge0$ which implies that $w_{A}$ is also increasing in a right neighbourhood of $c$. Therefore, $w'_{A}$ cannot become negative. □

Lemma 6.3

For every $A>0$, there is some $b_{A}$ such that $w''_{A}(b_{A})=0$ and $w_{A}$ is a concave function on $(a, b_{A})$.

Proof

Assume by way of contradiction that $w_{A}''$ does not vanish. Using (5.6) and (5.5), we have

$$\frac{\sigma^{2}\beta^{2}(a)}{2} w_{A}''(a)=-rAa^{\delta}. $$

Therefore, we equivalently assume that $w_{A}''<0$. This implies that $w_{A}'$ is strictly decreasing and bounded below by 0 by Lemma 6.2; therefore $w_{A}$ is an increasing concave function. Therefore, ${\lim_{x\to\infty}}w_{A}'(x)$ exists and is denoted by $\ell$. Letting $x \to\infty$ in the differential equation, we obtain, because $\beta$ has a finite limit,

$$\frac{\sigma^{2}\bar{\beta}^{2}}{2}\lim_{x\to\infty}w_{A}''(x)=r\lim _{x\to \infty}w_{A}(x)-\mu\bar{\beta} \ell. $$

Therefore, either $\lim_{x\to\infty}w_{A}(x)$ is $+\infty$, from which we get a contradiction, or finite, from which we get ${\lim_{x\to \infty }}w_{A}''(x)=0$ by the mean value theorem. In the second case, differentiating the differential equation, we have

$$\begin{aligned} &\mu\beta'(x)w_{A}'(x)+\mu\beta(x)w_{A}''(x) + \sigma^{2}\beta'(x)\beta (x)w_{A}''(x) \\ &+ \frac{\sigma^{2}\beta(x)^{2}}{2}w_{A}'''(x)-r w_{A}'(x) = 0. \end{aligned}$$

(6.18)

Proceeding analogously, we obtain that ${\lim_{x\to\infty }}w_{A}'''(x)=0$ and thus $\ell=0$. Coming back to the differential equation, we get

$$0=r\lim_{x\to\infty}w_{A}(x), $$

contradicting the fact that $w_{A}$ is increasing. Setting $b_{A}:=\inf\{x \ge a:w_{A}''(x)=0\}$ allows us to conclude. □

Lemma 6.4

There exists $A^{*}$ such that $w'_{A^{*}}(b_{A^{*}})=1$.

Proof

For every $A>0$, we have

$$ \mu\beta(b_{A})w'_{A}(b_{A})=rw_{A}(b_{A}). $$

(6.19)

Let $A_{1}=\frac{\mu\bar{\beta}}{ra^{\delta}}$. Lemma 6.2 yields

$$ w_{A_{1}}(b_{A_{1}})\ge w_{A_{1}}(a) =\frac{\mu\bar{\beta}}{r} \ge\frac{\mu\beta(b_{A_{1}})}{r}. $$

Therefore, (6.19) yields $w'_{A_{1}}(b_{A_{1}})\ge1$. On the other hand, let $A_{2}=\frac{a^{1-\delta}}{\delta}$. By construction, $w'_{A_{2}}(a)=1$ and thus $w'_{A_{2}}(b_{A_{2}}) \le1$ by the concavity of $w_{A}$ on $(0,b_{A})$. Thus, there is some $A^{*} \in[\min (A_{1},A_{2}),\max(A_{1},A_{2})]$ such that $w'_{A^{*}}=1$. □

Hereafter, we set $b=b_{A^{*}}$.

Lemma 6.5

We have $\mu\beta'(b)\le r$.

Proof

Differentiating the differential equation for $w_{A}$ and plugging in $x=b$, we get

$$\frac{\sigma^{2}\beta(b)^{2}}{2}w_{A}'''(b)+\mu\beta'(b)-r=0. $$

Because $w''_{A^{*}}$ is increasing in a left neighbourhood of $b$, we have $w_{A}'''(b) \ge0$, implying the result. □

Let us define

$$v(x)= \left\{ \textstyle\begin{array}{l@{\quad}l} w_{A^{*}}(x),& x\le b,\\ x-b+\frac{\mu\beta(b)}{r},& x \ge b. \end{array}\displaystyle \right. $$

We are in a position to prove the following result.

Proposition 6.6

The shareholders’ value is $v$.

Proof

We have to check that $(v,b)$ satisfies the HJB free boundary problem

$$ \max\big\{ \max_{k} \mathcal{L}_{k} v(x),1-v'(x)\big\} =0. $$

(6.20)

By construction, $v$ is a $C^{2}$ concave function on $(0, \infty)$ satisfying $v'\ge1$. It remains to check that $\max_{k}\mathcal {L}_{k}v(x)\le0$. For $x >b$, we have

$$\mathcal{L}_{k}v(x)=\mu\beta(k) -\alpha\big((k-x)^{+}\big)-\mu\beta (b) -r(x-b). $$

If $k\le x$, concavity of $\beta$ and Lemma 6.5 imply

$$ \mathcal{L}_{k}v(x)= \mu\big(\beta(x)-\beta(b)\big)-r(x-b) \le\big(\mu\beta'(b)-r\big)(x-b) \le0. $$

If $k \ge x$, we differentiate $\mathcal{L}_{k}v(x)$ with respect to $k$ and obtain, using again concavity of $\beta$ and convexity of $\alpha$, that

$$\frac{\partial\mathcal{L}_{k}v(x)}{\partial k}=\mu\beta'(k)-\alpha '(k-x)\le\mu\beta'(0)-\alpha'(0) \le0. $$

Therefore, $\mathcal{L}_{k}v(x) \le\mathcal{L}_{x}v(x)\le0$.

Let $x < b$. Because $v$ is concave, the same argument as in the previous lines shows that

$$\frac{\partial\mathcal{L}_{k} v(x)}{\partial k}\le0 \quad\mbox{ for } k \ge x $$

and therefore

$$\max_{k\ge0}\mathcal{L}_{k}v(x)=\max_{k \le x}\mathcal{L}_{k}v(x). $$

The first order condition gives for $0 \leq k < x$ that

$$ \frac{\partial}{\partial k}(\mathcal{L}_{k}v)(x) = \mu\beta'(k) v'(x) + \sigma^{2}\beta'(k)\beta(k)v''(x)= \beta'(k)\big(\mu v'(x) + \sigma ^{2}\beta (k)v''(x)\big). $$

Thus for $0< x< a$, we have

$$\frac{\partial}{\partial k}(\mathcal{L}_{k}v)(x)= \beta '(k)A^{*}x^{\delta -2}\delta\big(\mu x+ \sigma^{2}\beta(k)(\delta-1)\big), $$

which gives

$$\left\{ \begin{aligned} \frac{\partial}{\partial k}(\mathcal{L}_{k}v)(x) \bigg|_{k=0}> 0,\\ \frac{\partial}{\partial k}(\mathcal{L}_{k}v)(x)\bigg|_{k=x} < 0. \end{aligned} \right. $$

Therefore the maximum $k^{*}(x)$ of $\mathcal{L}_{k}v(x)$ lies in the interior of the interval $[0,x]$ and satisfies

$$\beta\big(k^{*}(x)\big) = \frac{\mu x}{\sigma^{2}(1-\delta)} \quad \text{for all } 0 < x < a. $$

Hence, for $x \le a$, we have by construction

$$ \max_{0\leq k\leq x}\mathcal{L}_{k}v(x) = \frac{\mu^{2}x}{\sigma ^{2}(1-\delta )}A^{*}\delta x^{\delta-1} + \frac{\sigma^{2}\mu^{2}x^{2}}{2\sigma ^{4}(1-\delta )^{2}}A^{*}\delta(\delta-1)x^{\delta-2}-r A^{*}x^{\delta}=0 . $$

Now fix $x \in(a,b)$. Note that ${\frac{\partial}{\partial k}}(\mathcal{L}_{k}v)(x)$ has the same sign as $\mu v'(x) + \sigma^{2}\beta(k)v''(x)$ because $\beta$ is strictly increasing. Moreover, since $v$ is concave and $\beta$ increasing, we have

$$\min_{0 \leq k \leq x} \mu v'(x) + \sigma^{2}\beta(k)v''(x) = \mu v'(x) + \sigma^{2}\beta(x)v''(x). $$

Thus, it suffices to prove $\mu v'(x) + \sigma^{2}\beta(x)v''(x) \geq0$ for $x \in(a,b)$, or equivalently, because $\beta$ is a positive function, that the function $\phi$ defined as

$$\phi(x) = \mu\beta(x) v'(x) + \sigma^{2}\beta(x)^{2}v''(x) $$

is positive. We make a proof by contradiction, assuming there is some $x$ such that $\phi(x) < 0$. As $\phi(a) = 0$ by (5.5) and $\phi(b) > 0$, there is some $x_{1} \in[a,b]$ such that

$$\left\{ \begin{aligned} \phi(x_{1}) &< 0,\\ \phi'(x_{1}) &= 0. \end{aligned} \right. $$

Using the differential equation (6.18) satisfied by $v'$, we obtain

$$\phi'(x_{1}) = \big(2 r - \mu\beta'(x_{1})\big)v'(x_{1}) - \mu\beta(x_{1}) v''(x_{1}) = 0, $$

from which we deduce that

$$\begin{aligned} \phi(x_{1}) &= \mu\beta(x_{1}) v'(x_{1}) + \sigma^{2}\beta(x_{1})^{2}v''(x_{1}) \\ &= \mu\beta(x_{1}) v'(x_{1}) + \frac{\sigma^{2}\beta(x_{1})}{\mu} \big(2 r - \mu\beta'(x_{1})\big)v'(x_{1}) \\ &= \beta(x_{1}) v'(x_{1}) \bigg( \mu+ \frac{2r\sigma^{2}}{\mu} - \sigma^{2} \beta'(x_{1}) \bigg). \end{aligned} $$

But $x_{1} \geq a$ and thus $\beta'(x_{1}) \leq\beta'(a)$. Moreover, by the definition of $a$, we have $\sigma^{2}\beta'(a) \le\frac{ \mu }{1-\delta}$. Therefore, (5.6) yields

$$\begin{aligned} \phi(x_{1}) &\geq\beta(x_{1}) v'(x_{1}) \bigg(\mu+ \frac{2r\sigma ^{2}}{\mu} - \frac{\mu}{1-\delta}\bigg)\\ &\geq\beta(x_{1}) v'(x_{1}) \bigg( \frac{2r\sigma^{2}}{\mu} - \mu\frac {\delta}{1-\delta}\bigg)\\ &\geq\beta(x_{1}) v'(x_{1}) \bigg( \frac{2r\sigma^{2}}{\mu} - \mu\frac {2r\sigma^{2}}{\mu^{2} + 2r\sigma^{2}}\frac{\mu^{2}+2r\sigma^{2}}{\mu^{2}} \bigg)\\ &= 0, \end{aligned} $$

which is a contradiction. □

To complete the characterization of the shareholders’ value when the spread is high, we have to study the optimal policy when (6.17) is not fulfilled. We expect that $a=0$ in that case, which means that for all $x$, the manager should invest all the cash in the productive asset. Thus we are interested in the solutions to

$$ \mu\beta(x)w'(x) + \frac{\sigma^{2}\beta(x)^{2}}{2}w''(x) - r w(x) = 0 $$

(6.21)

such that $w(0)=0$.

Proposition 6.7

Suppose that the functions $x \mapsto\frac{x}{\beta(x)}$ and $x \mapsto\frac{x^{2}}{\beta(x)^{2}}$ are analytic in 0 with a radius of convergence $R$. Then the solutions $w$ to (6.21) such that $w(0)=0$ are given by

$$w(x) = \sum_{k=0}^{\infty}A_{k} x^{k+y_{1}} $$

with

$$A_{k} = \frac{1}{-I(k+y_{1})}\sum_{j=0}^{k-1}\frac{(j+y_{1})p^{(k-j)}(0) + q^{(k-j)}(0)}{(k-j)!}A_{j}, \quad\textit{for all}\ k \geq1, $$

where the functions $p$ and $q$ are

$$\left\{ \begin{aligned} p(x) &= \frac{2\mu x}{\sigma^{2}\beta(x)},\\ q(x) &= - \frac{2r x^{2}}{\sigma^{2} \beta(x)^{2}}, \end{aligned} \right. $$

the function $I$ is given by

$$I(y) = \mu\beta'(0)y+\frac{\sigma^{2}}{2}\beta'(0)^{2}y(y-1)-r $$

and $y_{1}$ is the positive root of $I$ given by

$$y_{1} = \frac{-\mu+ \frac{\sigma^{2}}{2}\beta'(0) + \sqrt{(\mu- \frac {\sigma^{2}}{2}\beta'(0))^{2} + 2r \sigma^{2}}}{\sigma^{2}\beta'(0)}. $$

The radius of convergence of $w$ is at least equal to $R$.

Proof

This result is given by the Fuchs’ theorem [1, Sect. 9.5]. □

Note that choosing $A_{0} > 0$ is enough to characterize a unique solution $w$ of (6.21) because the $(A_{k})_{k\geq1}$ are given by a recurrence relation. Moreover, because $\mu\beta'(0) \geq r$, we have $y_{1} < 1$. As a consequence, if we choose $A_{0}>0$, we have

$$\lim_{x \rightarrow0} w'(x) = +\infty\quad\mbox{ and }\quad\lim_{x \rightarrow0} w^{\prime\prime}(x) = -\infty. $$

Thus, proceeding analogously as in Lemma 6.3, we can prove the existence of $b$ such that $w''(b) = 0$ for all $A_{0} > 0$. Now, observe that $b$ does not depend on $A_{0}$ so that we can choose $A_{0} = A^{*}$ in order to have $w'(b) = 1$. Hence, we have built a concave solution $w^{*}$ to (6.21) with $w^{*}(0) = 0$, $(w^{*})'(b) = 1$ and $(w^{*})''(b) = 0$. We extend $w^{*}$ linearly on $[b,\infty)$ as usual to obtain a $C^{2}$ function on $[0,\infty)$.

Proposition 6.8

The shareholders’ value is $w^{*}$.

Proof

It suffices to check that $w^{*}$ satisfies the free boundary problem (6.20). By construction, $w^{*}$ is a $C^{2}$ concave function on $(0,\infty)$. Because $(w^{*})'(b) = 1$, we have

$$(w^{*})'(x) \geq1,\quad\forall x \in(0,b] $$

and

$$(w^{*})'(x) = 1, \quad\forall x \geq b. $$

On $[b, \infty)$, we have

$$\begin{aligned} \max_{k\geq0}\mathcal{L}_{k} w^{*}(x) = \max_{k \geq0}\Big(&\mu\beta (k) - \alpha\big((k-x)^{+}\big) - \mu\beta(b) + r (b - x)\Big)\\ =\max\Big\{ &\max_{k\leq x}\mu\beta(k) - \mu\beta(b) + r (b - x),\\ &\max_{k\geq x}\big(\mu\beta(k) - \alpha(k-x) \big)- \mu\beta (b) + r (b - x)\Big\} . \end{aligned} $$

Using that $\beta$ is concave increasing, $\alpha$ is convex and $\alpha '(0+)> \mu\beta'(0+)$, we have

$$\max_{k\geq0}\mathcal{L}_{k} w^{*}(x) =\mu\beta(x) - \mu\beta(b) + r (b - x). $$

Then using the concavity of $\beta$,

$$\max_{k\geq0}\mathcal{L}_{k}w^{*}(x) \leq0 \quad\text{for all } x \geq b. $$

It remains to show that for every $x< b$

$$\max_{k\geq0}\mathcal{L}_{k}w^{*}(x) = 0. $$

Using that $\beta$ is concave, $\alpha$ is convex, $\alpha'(0)> \mu \beta '(0)$ and $w^{*}$ is concave increasing, we have for all $k>x$ that

$$\frac{\partial}{\partial k}(\mathcal{L}_{k} w^{*})(x) = \big(\mu\beta'(k) - \alpha'(k-x)\big)(w^{*})'(x) + \sigma^{2}\beta'(k)\beta(k)(w^{*})''(x) \leq0. $$

Thus,

$$\max_{k\geq0}\mathcal{L}_{k} w^{*} (x) = \max_{0\leq k\leq x}\mathcal{L}_{k} w^{*} (x). $$

Moreover, for $0 < k < x$,

$$\begin{aligned} \frac{\partial}{\partial k}(\mathcal{L}_{k} w^{*})(x) &= \mu\beta'(k) (w^{*})'(x) + \sigma^{2}\beta'(k)\beta(k)(w^{*})''(x)\\ &= \beta'(k)\big(\mu(w^{*})'(x) + \sigma^{2}\beta(k)(w^{*})''(x)\big). \end{aligned} $$

We expect for all $x \in(0,b]$ and all $k \leq x$ that

$$\frac{\partial}{\partial k}(\mathcal{L}_{k} w^{*} )(x) \geq0. $$

Notice that $\beta'(k) \geq0$ and

$$\min_{0 \leq k \leq x} \mu(w^{*})'(x) + \sigma^{2}\beta(k)(w^{*})''(x) = \mu (w^{*})'(x) + \sigma^{2}\beta(x)(w^{*})''(x), $$

because $(w^{*})''(x) \leq0$ and $\beta$ is increasing. Thus it is enough to prove that for every $x< b$,

$$\mu(w^{*})'(x) + \sigma^{2}\beta(x)(w^{*})''(x) \geq0, $$

or equivalently, using $\beta\geq0$,

$$\phi(x) = \mu\beta(x) (w^{*})'(x) + \sigma^{2}\beta(x)^{2} (w^{*})''(x) \geq0 \quad\text{for } x< b. $$

We make a proof by contradiction, assuming the existence of $x$ such that $\phi(x) < 0$. In a neighbourhood of 0, we have

$$(w^{*})'(x) \sim A^{*} y_{1}x^{y_{1} -1} $$

and

$$(w^{*})''(x) \sim A^{*} y_{1}(y_{1}-1)x^{y_{1}-2}. $$

From this we deduce, because $\beta(x)x^{y_{1}-1}\le\beta'(0)x^{y_{1}}$, that

$$\begin{aligned} \lim_{x\rightarrow0} \beta(x)(w^{*})'(x) &= 0,\\ \lim_{x\rightarrow0} \beta(x)^{2}(w^{*})''(x) &= 0, \end{aligned}$$

yielding

$$\lim_{x\rightarrow0} \phi(x) = 0. $$

But $\phi(b) > 0$; thus there is $x_{1} \in(0,b)$ such that

$$\left\{ \begin{aligned} \phi(x_{1}) &< 0,\\ \phi'(x_{1}) &= 0. \end{aligned} \right. $$

To get the expression of $\phi'$, we differentiate (6.21) to obtain

$$\phi'(x) = \big(2 r - \mu\beta'(x)\big) (w^{*})'(x) - \mu\beta(x) (w^{*})''(x). $$

Then using that $\phi''(x_{1}) = 0$, we get that

$$\big(2 r - \mu\beta'(x)\big) (w^{*})'(x_{1}) = \mu\beta(x_{1}) (w^{*})''(x_{1}), $$

from which we deduce

$$\begin{aligned} \phi(x_{1}) &= \mu\beta(x_{1}) (w^{*})'(x_{1}) + \sigma^{2}\beta (x_{1})^{2}(w^{*})''(x_{1}) \\ &= \mu\beta(x_{1}) (w^{*})'(x_{1}) + \frac{\sigma^{2}\beta(x_{1})}{\mu} \big(2 r - \mu\beta'(x_{1})\big)(w^{*})'(x_{1}) \\ &= \beta(x_{1}) (w^{*})'(x_{1}) \bigg( \mu+ \frac{2r\sigma^{2}}{\mu} - \sigma ^{2} \beta'(x_{1}) \bigg). \end{aligned} $$

Now, remember that $x_{1} > 0$ and thus, by the concavity of $\beta$, we have $\beta'(x_{1}) \leq\beta'(0)$. Furthermore, $\beta'(0) \leq\frac{\mu^{2}+2r \sigma^{2}}{\sigma^{2}\mu}$ when (5.5) is not fulfilled. Hence,

$$\phi(x_{1}) \geq\beta(x_{1}) (w^{*})'(x_{1}) \bigg(\mu+ \frac{2r\sigma ^{2}}{\mu } - \frac{\mu^{2} + 2r \sigma^{2}}{\mu}\bigg) \geq0 $$

which yields a contradiction and ends the proof. □