In order to prove the uniqueness of the solution to (1)-(2), we need the following two lemmas.

### Lemma 1

$$\begin{aligned} \sup_{m_{\rho}\in{\mathcal{M}}_{0}} E \biggl[ \int_{0}^{+\infty }e^{-r\rho}(S_{\rho}-K)^{+}\,dm_{\rho}\Bigl| m_{0^{-}}=M-a, S_{0}=s \biggr]=a V_{\infty}(s), \end{aligned}$$

(5)

*where*
\({\mathcal{M}}_{0}\)
*is the same as above*, *and*
\(V_{\infty}(s)\)
*is the price of a standard perpetual American call option*, *that is*, [6, 7]

$$\begin{aligned} &\textit{if } r=\alpha,\quad V_{\infty}(s)=s; \\&\textit{if } r>\alpha,\quad V_{\infty}(s)=\left \{ \textstyle\begin{array}{l@{\quad}l} (\frac{s}{S^{*}} )^{\lambda_{+}}(S^{*}-K), &0< s\leq S^{*},\\ s-K,&s> S^{*}, \end{array}\displaystyle \right . \end{aligned}$$

*where*

$$\begin{aligned} s=e^{z},\qquad \lambda_{+}=\frac{-\mu+\sqrt{\mu^{2}+2r\sigma^{2}}}{\sigma ^{2}},\qquad S^{*}=\frac{\lambda_{+} K}{\lambda_{+}-1}>K, \end{aligned}$$

\(S^{*}\)
*is the optimal exercise boundary of the standard perpetual American call options*.

### Proof

By Remark 3 in [8], when the utility function is taken as \(U (y) =y\) (that is, without utility function), the value function in the case of an unrestricted exercise is equal to the one in the case of a block exercise. That is, when there is no utility function, unrestricted exercise is equivalent to block exercise. Thus

$$\begin{aligned} &\sup_{m_{\rho}\in{\mathcal{M}}_{0}} E \biggl[ \int_{0}^{+\infty }e^{-r\rho}(S_{\rho}-K)^{+}\,dm_{\rho}\Bigl| m_{0^{-}}=M-a, S_{0}=s \biggr] \\ &\quad=\sup_{\rho\in{\mathcal{F}}_{0}} E \bigl[a e^{-r\rho }(S_{\rho}-K)^{+} | S_{0}=s \bigr] \\ &\quad= a V_{\infty}(s), \end{aligned}$$

where \({\mathcal{F}}_{0}\) is defined by the following expression:

$$\begin{aligned} {\mathcal{F}}_{0}\equiv\bigl\{ \rho \vert \rho\in[0,+\infty) \textit{ is a stopping time} \bigr\} . \end{aligned}$$

This completes the proof. □

### Lemma 2

*Suppose*
\(v(s,x,a)\)
*is provided by* (3), *and*
\(u (Z, \tau) \)
*is given by* (4), *then*

$$ -1\leq u(z,\tau)\leq-e^{-\tau V_{\infty}(e^{z})}, \quad(z,\tau)\in Q_{\infty}, $$

(6)

*where*
\(Q_{\infty}=(-\infty,+\infty)\times(0,A]\), \(s=e^{z}\), \(V_{\infty}(s)=V_{\infty}(e^{z}) \)
*represents the price of a standard perpetual American call option*, *the specific expression of which is given by Lemma*
1.

### Proof

By (3) and \(U(y)=-e^{-\gamma y}\) (\(y\geq0\)), it is clear that \(v(s,x,z)\geq-e^{-\gamma x}\).

By (4), we have \(u(z,\tau)\geq-1\).

By (3), we obtain

$$\begin{aligned} v(s,x,a) \leq& \sup_{m_{\rho}\in{\mathcal{M}}_{0}} U \biggl(E \biggl[x+ \int_{0}^{+\infty}e^{-r\rho}(S_{\rho}-K)^{+}\,dm_{\rho}\Bigl| m_{0^{-}}=M-a, S_{0}=s \biggr] \biggr) \\ \leq& U \biggl(\sup_{m_{\rho}\in{\mathcal{M}}_{0}} E \biggl[x+ \int_{0}^{+\infty}e^{-r\rho}(S_{\rho}-K)^{+}\,dm_{\rho}\Bigl| m_{0^{-}}=M-a, S_{0}=s \biggr] \biggr) \\ =& U \bigl(x+a V_{\infty}(s) \bigr)= U \bigl(x+a V_{\infty } \bigl(e^{z}\bigr) \bigr), \end{aligned}$$

where the Jensen inequality is used in the first inequality, and equation (5) in Lemma 1 is applied in the second inequality. \(V_{\infty}(e^{z})\) is the price of a standard perpetual American call option.

By (4) and \(U(y)=-e^{-\gamma y}\) (\(y\geq0\)), we get

$$\begin{aligned} u(z,\tau) \leq -e^{-\gamma a V_{\infty}(e^{z})}= -e^{-\tau V_{\infty}(e^{z})}. \end{aligned}$$

This completes the proof. □

Now to prove the uniqueness of the solution to the definite solution problem (1)-(2) which satisfies (6).

The definite solution problem (1)-(2) which satisfies (6) is equivalent to the following variational problem:

$$\begin{aligned} &\mathcal{L}u \equiv r\tau\frac{\partial u}{\partial\tau}-\frac {\sigma^{2}}{2} \frac{\partial^{2} u}{ \partial z^{2}}-\mu\frac{\partial u}{\partial z}\geq0, \quad(z,\tau )\in Q_{\infty}, \end{aligned}$$

(7)

$$\begin{aligned} &\mathcal{ B}u \equiv\frac{\partial u}{\partial\tau}+\bigl(e^{z}- K \bigr)^{+}u\geq0, \quad(z,\tau)\in Q_{\infty}, \end{aligned}$$

(8)

$$\begin{aligned} &\mathcal{ L}u\cdot\mathcal{B}u=0,\quad (z,\tau)\in Q_{\infty}, \end{aligned}$$

(9)

$$\begin{aligned} &u(z,0)=-1,\quad -\infty< z< +\infty, \end{aligned}$$

(10)

$$\begin{aligned} &-1\leq u(z,\tau)\leq-e^{-\tau V_{\infty}(e^{z})}, \quad(z,\tau)\in Q_{\infty}. \end{aligned}$$

(11)

### Remark 2

For the continuous-time optimization problem, the corresponding variational inequality is usually obtained by formal derivation (not strictly derived). Is the solution of the variational inequality really the value function of the original problem? It needs to be further verified, that is, one has to prove a so-called ‘verification theorem’. By a verification theorem, we can prove that when it is sufficiently smooth, the solution of the variational inequality (7)-(11) is equal to the value function of the corresponding singular stochastic control problem (obtained by an identical deformation of (3)).

### Remark 3

Using the standard method in [4], we can also prove that the value function of the corresponding singular stochastic control problem (obtained by identical deformation of (3)) is a viscosity solution of the variational inequality (7)-(11).

### Theorem 1

*Assume that*
\(r>\alpha\)
*and the parameters*
*A*, *r*, *δ*, *σ*, *γ*, *K*
*are all positive constants*. *For any*
\(\rho>0\)
*and any*
\(\varepsilon \in (0,A)\), *let*
\(Q_{\infty}=(-\infty,+\infty)\times(0,A]\), \(Q_{\rho}=(-\rho,\rho )\times(0,A]\), *and*
\(Q_{\varepsilon }\rho=(-\rho,\rho)\times[\varepsilon ,A]\). *Then the problem* (7)-(11) *has a unique solution*
\(u(z,\tau)\)
*satisfying*

$$\begin{aligned}& u(z,\tau)\in W^{2,1}_{\infty}(Q_{\rho})\cap C(\overline{Q}_{\infty}),\\& u_{\tau}\in C^{\frac{1}{2},\frac{1}{4}}(Q_{\varepsilon \rho}),\qquad u_{z\tau}\in L^{2}(Q_{\rho}),\qquad u_{\tau\tau}\in L^{2}(Q_{\varepsilon }\rho). \end{aligned}$$

### Proof

Suppose there are two solutions \(u_{1}\) and \(u_{2}\) of (7)-(11) in \(W^{2,1}_{\infty,\mathrm{loc}}(Q_{\infty})\cap C(\overline{Q}_{\infty})\). Denote \(w=u_{1}-u_{2}\). In order to prove \(w=0\), we need to first prove \(w\leq0\) in \(\overline{Q}_{\infty}\).

If \(z\rightarrow+\infty\), *i.e.*
\(s\rightarrow+\infty\), by (3) and \(U(y)=-e^{-\gamma y}\), we have \(v(s,x,a)\rightarrow0\). By (4), we get \(u(z,\tau)\rightarrow0\), so \(w(z,\tau)\rightarrow0\) holds.

If \(z\rightarrow-\infty\), by (11) and the expression of \(V_{\infty }(e^{z})\), we have \(u(z,\tau)\rightarrow-1\), then \(w(z,\tau)\rightarrow0\) also holds.

Therefore

$$ \lim_{z\rightarrow\infty}w(z,\tau)=0,\quad \mbox{uniformly convergent for $\tau \in[0, A]$}. $$

(12)

By (12), \(\forall\varepsilon>0\), ∃ a sufficiently large \(R_{\varepsilon}>0\), *s.t.*

$$ \bigl|w(z,\tau)\bigr|\leq\varepsilon, \quad|z|\geq R_{\varepsilon}, 0 \leq\tau\leq A. $$

(13)

By (10), we get

$$ w(z,0)=0, \quad-\infty< z< \infty. $$

(14)

Set

$$ \mathcal{O}_{\varepsilon}=(-R_{\varepsilon},R_{\varepsilon})\times(0,A]. $$

(15)

Suppose \((z_{\ast},\tau_{\ast})\) is a maximum point of \(w(z,\tau)\) in \(\mathcal{O}_{\varepsilon}\), then

$$ \frac{\partial w}{\partial\tau}(z_{\ast},\tau_{\ast})\geq0. $$

(16)

Now we prove

$$ w(z_{\ast},\tau_{\ast})\leq\varepsilon. $$

(17)

(1) If \((z_{\ast},\tau_{\ast})\in\partial_{p} {\mathcal{O}_{\varepsilon}}\) (\(\partial_{p} {\mathcal{O}_{\varepsilon}}\) is the parabolic boundary of \(\mathcal{O}_{\varepsilon}\)), then by (13)-(14), (17) holds.

(2) If \((z_{\ast},\tau_{\ast})\in{\mathcal{O}_{\varepsilon}} \) and \({\mathcal{B}}u_{1}(z_{\ast},\tau_{\ast})=0\), then by (8), we have

$$ {\mathcal{B} }w(z_{\ast},\tau_{\ast})={\mathcal{B}}(u_{1}-u_{2}) (z_{\ast},\tau_{\ast})\leq0. $$

(18)

If \({\mathcal{B}}u_{1}(z_{\ast},\tau_{\ast})=0\), then \((z_{\ast},\tau_{\ast})\) belongs to stopping region (or exercise region), so \(z_{\ast}>\ln K\), *i.e.*, \((e^{z*} -K)^{+}\neq0\). By (8), (16), and (18), we obtain

$$\begin{aligned} w(z_{\ast},\tau_{\ast})= (u_{1}-u_{2}) (z_{\ast},\tau_{\ast})\leq-\frac {1}{(e^{z_{\ast}}-K)^{+}} \frac{\partial w}{\partial\tau}(z_{\ast},\tau _{\ast})\leq0. \end{aligned}$$

Thus \(w(z_{\ast},\tau_{\ast})\leq\varepsilon\), *i.e.*, (17) holds.

(3) If \((z_{\ast},\tau_{\ast})\in{\mathcal{O}_{\varepsilon}}\) and \({\mathcal{B}}u_{1}(z_{\ast},\tau_{\ast})>0\), let

$$\begin{aligned} \widehat {\mathcal{O}}_{\varepsilon}= \bigl\{ (z,\tau)\in{\mathcal {O}_{\varepsilon}} | {\mathcal{B}}u_{1}(z,\tau)>0 \bigr\} . \end{aligned}$$

Then \(\widehat {\mathcal{O}}_{\varepsilon}\) is an open set of \(\mathcal {O}_{\varepsilon}\), and \((z_{\ast},\tau_{\ast})\in{ \widehat {\mathcal {O}}_{\varepsilon}}\). By (7)-(9), we get

$$\begin{aligned} \mathcal{L}u_{1}(z,\tau)=0\quad\mbox{in } \widehat {\mathcal{O}}_{\varepsilon}\subseteq Q_{\infty},\qquad \mathcal{L}u_{2}(z,\tau)\geq0 \quad\mbox{in } \widehat {\mathcal {O}}_{\varepsilon}\subseteq Q_{\infty}. \end{aligned}$$

Therefore

$$\begin{aligned} \mathcal{L}w=\mathcal{L}u_{1}-\mathcal{L}u_{2}=r\tau \frac{\partial w}{\partial\tau}-\frac{\sigma^{2}}{2} \frac{\partial^{2} w}{ \partial z^{2}}-\mu\frac{\partial w}{\partial z} \leq0, \quad(z,\tau )\in{ \widehat {\mathcal{O}}_{\varepsilon}}. \end{aligned}$$

On the parabolic boundary of \(\widehat {\mathcal{O}}_{\varepsilon}\),

$$\partial_{p} {\widehat {\mathcal{O}}_{\varepsilon}}\equiv\partial \widehat { \mathcal{O}}_{\varepsilon}\cap\{\tau< A\}, $$

we have

In fact, let \((z_{0},\tau_{0})\) be a maximum point of \(w(z,\tau)\) on the parabolic boundary \(\partial_{p} {\widehat {\mathcal{O}}_{\varepsilon}}\). Because

$$\begin{aligned} \widehat {\mathcal{O}}_{\varepsilon}= \bigl\{ (z,\tau)\in{\mathcal {O}_{\varepsilon}} | {\mathcal{B}}u_{1}(z,\tau)>0 \bigr\} =(-R_{\varepsilon},a)\times(0,A], \end{aligned}$$

for some \(a\in(-R_{\varepsilon},R_{\varepsilon})\), we have \(\tau\in (0,A) \) on \(\partial_{p} {\widehat {\mathcal{O}}_{\varepsilon}}\equiv \partial \widehat {\mathcal{O}}_{\varepsilon}\cap\{\tau< A\}\). Then

$$\frac{\partial w}{\partial\tau}(z_{0},\tau_{0})=0. $$

\({\mathcal{B}}u_{1}(z,\tau)=0\) on \(\partial_{p} {\widehat {\mathcal {O}}_{\varepsilon}}\), so \({\mathcal{B}}u_{1}(z_{0},\tau_{0})=0\). Similar to the proof of (2), \(w(z_{0},\tau_{0})\leq\varepsilon\). Thus \(w\leq\varepsilon\) on \(\partial_{p} {\widehat {\mathcal {O}}_{\varepsilon}}\).

Then by the maximum principle, \(w\leq\varepsilon\) on \(\widehat {\mathcal {O}}_{\varepsilon}\). By \((z_{\ast},\tau_{\ast})\in \widehat {\mathcal{O}}_{\varepsilon}\), (17) holds.

In summary, (17) holds, so \(w\leq\varepsilon\) on \(\mathcal {O}_{\varepsilon}\). Combining with (13), we have \(w\leq\varepsilon\) on \(\overline {Q}_{\infty}\). Let \(\varepsilon\rightarrow0\), we get \(w\leq0\) on \(\overline {Q}_{\infty}\). Similarly, we can prove \(w\geq0\) on \(\overline {Q}_{\infty}\) by exchanging \(u_{1}\) and \(u_{2}\). Thus \(w=0\) on \(\overline {Q}_{\infty}\), *i.e.*, \(u_{1}=u_{2}\) on \(\overline {Q}_{\infty}\).

This completes the proof. □