Appendix
Proof of Theorem 2
Note that
$$\begin{aligned} A=\begin{bmatrix} -\lambda _1&\lambda _2\\ \lambda _1&-\lambda _2 \\ \end{bmatrix}. \end{aligned}$$
Then the coupled homogeneous ODEs reduce to the following system
$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{1}{2}x^2\sigma _1^2P_1''(x)+\mu _1xP_1'(x)-r_1P_1(x)+c+\lambda _1\big (P_2(x)-P_1(x)\big )=0,\\ P_1(v_1)=\min (v_1,c/r),\\ \lim \limits _{x\rightarrow \infty }P_1(x)=c/r. \end{array}\right. } \end{aligned}$$
$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{1}{2}x^2\sigma _2^2P_2''(x)+\mu _2xP_2'(x)-r_2P_2(x)+c+\lambda _2\big (P_1(x)-P_2(x)\big )=0,\\ P_2(v_2)=\min (v_2,c/r),\\ \lim \limits _{x\rightarrow \infty }P_2(x)=c/r. \end{array}\right. } \end{aligned}$$
For simplicity, we further assume that \(v_1\le v_2\le c/r\), as if \(v_i\ge c/r\), then \(P_i(v_i)=c/r\), and moreover \(P_i(x)=c/r\) for all \(x\ge v_i\) (as the equity holders will declare bankruptcy immediately at time 0). Note that the value of the default-free perpetual bond is \(c/r\). Intuitively, if the barrier \(v_i\ge c/r\), the equity holders choose to default immediately at time 0. To obtain the optimal values \(v_i\), we also impose the smooth pasting conditions at \(v_i\). Therefore
$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{1}{2}x^2\sigma _1^2P_1''(x)+\mu _1xP_1'(x)-r_1P_1(x)+c+\lambda _1\big (P_2(x)-P_1(x)\big )=0\\ P_1(v_1)=v_1\\ {\lim }_{x\rightarrow \infty }P_1(x)=c/r,\\ P_1'(x)\big |_{x=v_1}=1. \end{array}\right. } \end{aligned}$$
(4)
$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{1}{2}x^2\sigma _2^2P_2''(x)+\mu _2xP_2'(x)-r_2P_2(x)+c+\lambda _2\big (P_1(x)-P_2(x)\big )=0\\ P_2(v_2)=v_2\\ {\lim }_{x\rightarrow \infty }P_2(x)=c/r,\\ P_2'(x)\big |_{x=v_2}=1. \end{array}\right. } \end{aligned}$$
(5)
Therefore, when \(v_1\le v_2\le c/r\), we can write the coupled ODEs as follows. For \(x\in [0,v_1]\), we have
$$\begin{aligned} P_1(x)=P_2(x)=x. \end{aligned}$$
(6)
For \(x\in [v_1, v_2]\), we have
$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{1}{2}x^2\sigma _1^2P_1''(x)+\mu _1xP_1'(x)-r_1P_1(x)+c+\lambda _1\big (P_2(x)-P_1(x)\big )=0\\ P_2(x)=x. \end{array}\right. } \end{aligned}$$
(7)
For \(x\in [v_2, \infty )\), we have
$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{1}{2}x^2\sigma _1^2P_1''(x)+\mu _1xP_1'(x)-r_1P_1(x)+c+\lambda _1\big (P_2(x)-P_1(x)\big )=0,\\ \frac{1}{2}x^2\sigma _2^2P_1''(x)+\mu _2xP_2'(x)-r_2P_2(x)+c+\lambda _2\big (P_1(x)-P_2(x)\big )=0. \end{array}\right. } \end{aligned}$$
(8)
Now (8) has an characteristic function
$$\begin{aligned} g_1(\beta )g_2(\beta )=\lambda _1\lambda _2, \end{aligned}$$
(9)
where
$$\begin{aligned} g_1(\beta )&=\lambda _1+r_1-\mu _1\beta -\frac{1}{2}\sigma _1^2\beta (\beta -1),\\ g_2(\beta )&=\lambda _2+r_2-\mu _2\beta -\frac{1}{2}\sigma _2^2\beta (\beta -1). \end{aligned}$$
This characteristic function has four distinct roots \(\beta _1<\beta _2<0<\beta _3<\beta _4\). To obtain a particular solution of (8), we consider
$$\begin{aligned} {\left\{ \begin{array}{ll} -r_1P^*_1+c+\lambda _1(P^*_2-P^*_1)=0\\ -r_2P^*_2+c+\lambda _2(P^*_1-P^*_2)=0. \end{array}\right. } \end{aligned}$$
This system reduces to
$$\begin{aligned} \begin{bmatrix} -(r_1+\lambda _1)&\lambda _1 \\ -(r_2+\lambda _2)&\lambda _2 \\ \end{bmatrix} \begin{bmatrix} P^*_1 \\ P^*_2 \\ \end{bmatrix} = \begin{bmatrix} -c \\ -c \\ \end{bmatrix}, \end{aligned}$$
and hence
$$\begin{aligned} \begin{bmatrix} P^*_1 \\ P^*_2 \\ \end{bmatrix}&= \begin{bmatrix} -(r_1+\lambda _1)&\lambda _1 \\ -(r_2+\lambda _2)&\lambda _2 \\ \end{bmatrix}^{-1} \begin{bmatrix} -c \\ -c \\ \end{bmatrix}\\&=\frac{1}{r_2\lambda _1-r_1\lambda _2} \begin{bmatrix} \lambda _2&-\lambda _1 \\ r_2+\lambda _2&-(r_1+\lambda _1) \\ \end{bmatrix} \begin{bmatrix} -c \\ -c \\ \end{bmatrix}\\&=\frac{1}{r_2\lambda _1-r_1\lambda _2} \begin{bmatrix} c(\lambda _1-\lambda _2) \\ c\big ((r_1+\lambda _1)-(r_2+\lambda _2)\big ) \\ \end{bmatrix}, \end{aligned}$$
Thus a particular solution of (8) is
$$\begin{aligned} \begin{aligned} P^*_1&=\frac{c(\lambda _1-\lambda _2)}{r_2\lambda _1-r_1\lambda _2}\\ P^*_2&=\frac{c\big ((r_1+\lambda _1)-(r_2+\lambda _2)\big )}{r_2\lambda _1-r_1\lambda _2}. \end{aligned} \end{aligned}$$
(10)
Therefore, the general form of the solution to (8) is
$$\begin{aligned} P_1(x)&=P^*_1+A_1x^{\beta _1}+A_2x^{\beta _2}+A_3x^{\beta _3}+A_4x^{\beta _4},\\ P_2(x)&=P^*_2+B_1x^{\beta _1}+B_2x^{\beta _2}+B_3x^{\beta _3}+B_4x^{\beta _4}, \end{aligned}$$
with \(B_i=l_iA_i\) and \(l_i=l(\beta _i)=g_1(\beta _i)/\lambda _1=\lambda _2/g_2(\beta _i)\).
When \(x\rightarrow \infty \), \(P_1(x)\) and \(P_2(x)\) are both bounded. Thus \(A_3=A_4=B_3=B_4=0\), and the solution is
$$\begin{aligned} \begin{aligned} P_1(x)&=P^*_1+A_1x^{\beta _1}+A_2x^{\beta _2},\\ P_2(x)&=P^*_2+B_1x^{\beta _1}+B_2x^{\beta _2}. \end{aligned} \end{aligned}$$
(11)
Next we solve (7). The first equation is
$$\begin{aligned} \frac{1}{2}x^2\sigma _1^2P_1''(x)+\mu _1xP_1'(x)-r_1P_1(x)+c+\lambda _1\big (x-P_1(x)\big )=0. \end{aligned}$$
(12)
This is an inhomogeneous equation, and thus the solution can be written as
$$\begin{aligned} P_1(x)=C_1x^{\gamma _1}+C_2x^{\gamma _2}+\phi (x), \end{aligned}$$
(13)
where \(\phi (x)\) is a particular solution and \(\gamma _1\) and \(\gamma _2\) are the two roots of
$$\begin{aligned} \frac{1}{2}\sigma _1^2\gamma (\gamma -1)+\mu _1\gamma -r_1-\lambda _1=0. \end{aligned}$$
(14)
To obtain \(\phi (x)\), we assume that \(\phi (x)=ax+b\) and substitute it to Eq. (12). This yields
$$\begin{aligned} a\mu _1x-r_1(ax+b)+c-\lambda _1(ax+b)+\lambda _1 x=0, \end{aligned}$$
or
$$\begin{aligned} x\big ((\mu _1-r_1-\lambda _1)a+\lambda _1\big )+c-r_1b-\lambda _1b=0. \end{aligned}$$
Thus \(a=\lambda _1/(r_1+\lambda _1-\mu _1)\), \(b=c/(r_1+\lambda _1)\), and thus
$$\begin{aligned} \phi (x)=\frac{\lambda _1}{r_1+\lambda _1-\mu _1}x+\frac{c}{r_1+\lambda _1}. \end{aligned}$$
(15)
Now we solve for the coefficients \(A_i\), \(B_i\), \(C_i\) and the optimal values \(v_1\) and \(v_2\). Using the boundary and smooth pasting conditions for \(P_2(x)\) (see(5), (8)) at \(v_2\) with \(x\in [v_2,\infty )\),
$$\begin{aligned} {\left\{ \begin{array}{ll} P^*_2+B_1v_2^{\beta _1}+B_2v_2^{\beta _2}=v_2,\\ \beta _1B_1v_2^{\beta _1-1}+\beta _2B_2v_2^{\beta _2-1}=1. \end{array}\right. } \end{aligned}$$
or
$$\begin{aligned} {\left\{ \begin{array}{ll} l_1A_1v_2^{\beta _1}+l_2A_2v_2^{\beta _2}=v_2-P^*_2,\\ \beta _1l_1A_1v_2^{\beta _1}+\beta _2l_2A_2v_2^{\beta _2}=v_2. \end{array}\right. } \end{aligned}$$
(16)
Similarly, using the boundary and smooth pasting conditions for \(P_1(x)\) (see (4), (7)) at \(v_2\) with \(x\in [v_1,v_2]\),
$$\begin{aligned} {\left\{ \begin{array}{ll} P^*_1+A_1v_2^{\beta _1}+A_2v_2^{\beta _2}=C_1v_2^{\gamma _1}+C_2v_2^{\gamma _2}+\phi (v_2),\\ \beta _1A_1v_2^{\beta _1}+\beta _2A_2v_2^{\beta _2}=\gamma _1C_1v_2^{\gamma _1}+\gamma _2C_2v_2^{\gamma _2}+v_2\phi '(v_2). \end{array}\right. } \end{aligned}$$
(17)
Using the boundary and smooth pasting conditions for \(P_1(x)\) (see (4), (7)) at \(v_1\) with \(x\in [v_1,v_2]\)
$$\begin{aligned} {\left\{ \begin{array}{ll} C_1v_1^{\gamma _1}+C_2v_1^{\gamma _2}+\phi (v_1)=v_1,\\ \gamma _1C_1v_1^{\gamma _1}+\gamma _2C_2v_1^{\gamma _2}+v_1\phi '(v_1)=v_1. \end{array}\right. } \end{aligned}$$
(18)
Combining these equations, we can obtain the solutions \(v_1\) and \(v_2\). From (18), we have
$$\begin{aligned} F_1(v_1):= \begin{bmatrix} 1&1 \\ \gamma _1&\gamma _2 \end{bmatrix}^{-1} \begin{bmatrix} v_1-\phi (v_1)\\ v_1-v_1\phi '(v_1) \end{bmatrix} =\begin{bmatrix} C_1v_1^{\gamma _1}\\ C_2v_1^{\gamma _2} \end{bmatrix}. \end{aligned}$$
(19)
From (16), we have
$$\begin{aligned} \begin{bmatrix} l_1&l_2 \\ \beta _1l_1&\beta _2l_2 \end{bmatrix}^{-1} \begin{bmatrix} v_2-P^*_2 \\ v_2 \end{bmatrix} =\begin{bmatrix} A_1v_2^{\beta _1}\\ A_2v_2^{\beta _2} \end{bmatrix}. \end{aligned}$$
(20)
Then, using (17) and (20), we have
$$\begin{aligned} \begin{bmatrix} 1&1 \\ \beta _1&\beta _2 \end{bmatrix} \begin{bmatrix} l_1&l_2 \\ \beta _1l_1&\beta _2l_2 \end{bmatrix}^{-1} \begin{bmatrix} v_2-P^*_2 \\ v_2 \end{bmatrix}&=\begin{bmatrix} 1&1 \\ \beta _1&\beta _2 \end{bmatrix} \begin{bmatrix} A_1v_2^{\beta _1}\\ A_2v_2^{\beta _2} \end{bmatrix}\nonumber \\&= \begin{bmatrix} C_1v_2^{\gamma _1}+C_2v_2^{\gamma _2}+\phi (v_2)-P^*_2\\ \gamma _1C_1v_2^{\gamma _1}+\gamma _2C_2v_2^{\gamma _2}+v_2\phi '(v_2) \end{bmatrix}. \end{aligned}$$
(21)
This equation can be rewritten as
$$\begin{aligned} \begin{bmatrix} 1&1 \\ \beta _1&\beta _2 \end{bmatrix} \begin{bmatrix} l_1&l_2 \\ \beta _1l_1&\beta _2l_2 \end{bmatrix}^{-1} \begin{bmatrix} v_2-P^*_2 \\ v_2 \end{bmatrix} -\begin{bmatrix} \phi (v_2)-P^*_1 \\ v_2\phi '(v_2) \end{bmatrix} = \begin{bmatrix} 1&1 \\ \gamma _1&\gamma _2 \end{bmatrix} \begin{bmatrix} C_1v_2^{\gamma _1}\\ C_2v_2^{\gamma _2} \end{bmatrix}. \end{aligned}$$
Therefore
$$\begin{aligned} F_2(v_2)&:=\begin{bmatrix} 1&1 \\ \gamma _1&\gamma _2 \end{bmatrix}^{-1} \left( \begin{bmatrix} 1&1 \\ \beta _1&\beta _2 \end{bmatrix} \begin{bmatrix} l_1&l_2 \\ \beta _1l_1&\beta _2l_2 \end{bmatrix}^{-1} \begin{bmatrix} v_2-P^*_2 \\ v_2 \end{bmatrix} -\begin{bmatrix} \phi (v_2)-P^*_1 \\ v_2\phi '(v_2) \end{bmatrix}\right) \nonumber \\&= \begin{bmatrix} C_1v_2^{\gamma _1}\\ C_2v_2^{\gamma _2} \end{bmatrix}. \end{aligned}$$
(22)
Hence, from (19) and (22) we have the equation
$$\begin{aligned} \begin{bmatrix} v_1^{-\gamma _1}&0\\ 0&v_1^{-\gamma _2} \end{bmatrix}F_1(v_1) =\begin{bmatrix} C_1\\ C_2 \end{bmatrix} =\begin{bmatrix} v_2^{-\gamma _1}&0\\ 0&v_2^{-\gamma _2} \end{bmatrix}F_2(x_2). \end{aligned}$$
(23)
In particular, for \(\phi (x)\) of the form in (15), we have from (19) and (22)
$$\begin{aligned} F_1(v_1)=\begin{bmatrix} 1&1 \\ \gamma _1&\gamma _2 \end{bmatrix}^{-1} \begin{bmatrix} v_1-\frac{\lambda _1}{r_1+\lambda _1-\mu _1}v_1-\frac{c}{r_1+\lambda _1}\\ v_1-v_1\frac{\lambda _1}{r_1+\lambda _1-\mu _1} \end{bmatrix}, \end{aligned}$$
(24)
and
$$\begin{aligned} F_2(v_2)&=\begin{bmatrix} 1&1 \\ \gamma _1&\gamma _2 \end{bmatrix}^{-1} \left( \begin{bmatrix} 1&1 \\ \beta _1&\beta _2 \end{bmatrix} \begin{bmatrix} l_1&l_2 \\ \beta _1l_1&\beta _2l_2 \end{bmatrix}^{-1} \begin{bmatrix} v_2-P^*_2 \\ v_2 \end{bmatrix}\right. \nonumber \\&\quad \left. -\begin{bmatrix} \frac{\lambda _1}{r_1+\lambda _1-\mu _1}v_2+\frac{c}{r_1+\lambda _1}-P^*_1 \\ v_2\frac{\lambda _1}{r_1+\lambda _1-\mu _1}. \end{bmatrix}\right) . \end{aligned}$$
(25)
Substituting \(F_1(v_1)\) and \(F_2(v_2)\) into (23), we can obtain the values for \(v_1\) and \(v_2\). Now we derive coefficients \(A_i\), \(B_i\), and \(C_i\). From (16), the coefficients \(A_1\) and \(A_2\) are give by
$$\begin{aligned} \begin{bmatrix} A_1 \\ A_2 \end{bmatrix} =\begin{bmatrix} l_1v_2^{\beta _1}&l_2v_2^{\beta _2} \\ \beta _1l_1v_2^{\beta _1}&\beta _2l_2v_2^{\beta _2} \end{bmatrix}^{-1} \begin{bmatrix} v_2-P^*_2 \\ v_2 \end{bmatrix}, \end{aligned}$$
(26)
and
$$\begin{aligned} \begin{bmatrix} B_1 \\ B_2 \end{bmatrix} =\begin{bmatrix} l_1A_1 \\ l_2A_2 \end{bmatrix}. \end{aligned}$$
(27)
From (18), we have
$$\begin{aligned} \begin{bmatrix} C_1 \\ C_2 \end{bmatrix}&=\begin{bmatrix} v_1^{\gamma _1}&v_1^{\gamma _2} \\ \gamma _1v_1^{\gamma _1}&\gamma _2v_1^{\gamma _2} \end{bmatrix}^{-1} \begin{bmatrix} v_1-\phi (v_1) \\ v_1-v_1\phi '(v_1) \end{bmatrix} \nonumber \\&=\begin{bmatrix} v_1^{\gamma _1}&v_1^{\gamma _2} \\ \gamma _1v_1^{\gamma _1}&\gamma _2v_1^{\gamma _2} \end{bmatrix}^{-1} \begin{bmatrix} v_1-\frac{\lambda _1}{r_1+\lambda _1-\mu _1}v_1-\frac{c}{r_1+\lambda _1} \\ v_1-v_1\frac{\lambda _1}{r_1+\lambda _1-\mu _1} \end{bmatrix}. \end{aligned}$$
(28)
With these coefficients, the value functions \(P_1(x)\) and \(P_2(x)\) become
$$\begin{aligned} P_1(x)={\left\{ \begin{array}{ll} A_1x^{\beta _1}+A_2x^{\beta _2}+\frac{c(\lambda _1-\lambda _2)}{r_2\lambda _1-r_1\lambda _2} \quad &{}\text {if }x>v_2,\\ C_1x^{\gamma _1}+C_2x^{\gamma _2}+\phi (x) &{}\text {if }v_1\le x\le v_2\\ x &{}\text {if }x\le v_1, \end{array}\right. } \end{aligned}$$
and
$$\begin{aligned} P_2(x)={\left\{ \begin{array}{ll} B_1x^{\beta _1}+B_2x^{\beta _2}+\frac{c\big ((r_1+\lambda _1)-(r_2+\lambda _2)\big )}{r_2\lambda _1-r_1\lambda _2} \quad &{}\text {if }x>v_2,\\ x &{}\text {if }x\le v_2, \end{array}\right. } \end{aligned}$$
where
$$\begin{aligned} \phi (x)=\frac{\lambda _1}{r_1+\lambda _1-\mu _1}x+\frac{c}{r_1+\lambda _1}. \end{aligned}$$
\(\square \)
Proof of Theorem 3
The coupled homogeneous ODEs reduce to the following system
$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{1}{2}x^2\sigma _1^2P_1''(x)+\mu _1xP_1'(x)-r_1P_1(x)+c-c'+\lambda _1\big (P_2(x)-P_1(x)\big )=0,\\ P_1(v_1)=\min ((1-\alpha )v_i,(c-c')/r),\\ \lim _{x\rightarrow \infty }P_1(x)=(c-c')/r. \end{array}\right. } \end{aligned}$$
$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{1}{2}x^2\sigma _2^2P_2''(x)+\mu _2xP_2'(x)-r_2P_2(x)+c-c'+\lambda _2\big (P_1(x)-P_2(x)\big )=0,\\ P_2(v_2)=\min ((1-\alpha )v_i,(c-c')/r),\\ \lim _{x\rightarrow \infty }P_2(x)=(c-c')/r. \end{array}\right. } \end{aligned}$$
For simplicity, we assume that \((1-\alpha )v_1\le (1-\alpha )v_2\le (c-c')/r\), as if \((1-\alpha )v_i\ge (c-c')/r\), then \(P_i(v_i)=\min ((1-\alpha )v_i,(c-c')/r)=(c-c')/r\), and moreover \(P_i(x)=(c-c')/r\) for all \(x\ge v_i\). Note that the value of the default-free perpetual bond is \((c-c')/r\). Intuitively, if the barrier \(v_i\ge (c-c')/r\), the equity holders choose to default immediately. To obtain the optimal values \(v_i\), we also impose the smooth pasting conditions at \(v_i\). Therefore
$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{1}{2}x^2\sigma _1^2P_1''(x)+\mu _1xP_1'(x)-r_1P_1(x)+c-c'+\lambda _1\big (P_2(x)-P_1(x)\big )=0\\ P_1(v_1)=(1-\alpha )v_1\\ {\lim }_{x\rightarrow \infty }P_1(x)=(c-c')/r,\\ P_1'(x)\big |_{x=v_1}=1-\alpha . \end{array}\right. } \end{aligned}$$
(29)
$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{1}{2}x^2\sigma _2^2P_2''(x)+\mu _2xP_2'(x)-r_2P_2(x)+c-c'+\lambda _2\big (P_1(x)-P_2(x)\big )=0\\ P_2(v_2)=(1-\alpha )v_2\\ {\lim }_{x\rightarrow \infty }P_2(x)=(c-c')/r,\\ P_2'(x)\big |_{x=v_2}=1-\alpha . \end{array}\right. } \end{aligned}$$
(30)
Therefore, when \(v_1\le v_2\le (c-c')/r\), we can write the coupled ODEs as follows. For \(x\in [0,v_1]\), we have
$$\begin{aligned} P_1(x)=P_2(x)=(1-\alpha )x. \end{aligned}$$
(31)
For \(x\in [v_1, v_2]\), we have
$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{1}{2}x^2\sigma _1^2P_1''(x)+\mu _1xP_1'(x)-r_1P_1(x)+c-c'+\lambda _1\big (P_2(x)-P_1(x)\big )=0\\ P_2(x)=(1-\alpha )x. \end{array}\right. } \end{aligned}$$
(32)
For \(x\in [v_2, \infty )\), we have
$$\begin{aligned} {\left\{ \begin{array}{ll} \frac{1}{2}x^2\sigma _1^2P_1''(x)+\mu _1xP_1'(x)-r_1P_1(x)+c-c'+\lambda _1\big (P_2(x)-P_1(x)\big )=0,\\ \frac{1}{2}x^2\sigma _2^2P_1''(x)+\mu _2xP_2'(x)-r_2P_2(x)+c-c'+\lambda _2\big (P_1(x)-P_2(x)\big )=0. \end{array}\right. } \end{aligned}$$
(33)
Now (33) has an characteristic function
$$\begin{aligned} g_1(\beta )g_2(\beta )=\lambda _1\lambda _2, \end{aligned}$$
(34)
where
$$\begin{aligned} g_1(\beta )&=\lambda _1+r_1-\mu _1\beta -\frac{1}{2}\sigma _1^2\beta (\beta -1),\\ g_2(\beta )&=\lambda _2+r_2-\mu _2\beta -\frac{1}{2}\sigma _2^2\beta (\beta -1). \end{aligned}$$
This characteristic function has four distinct roots \(\beta _1<\beta _2<0<\beta _3<\beta _4\). To obtain a particular solution of (8), we consider
$$\begin{aligned} {\left\{ \begin{array}{ll} -r_1Q^*_1+c-c'+\lambda _1(Q^*_2-Q^*_1)=0\\ -r_2Q^*_2+c-c'+\lambda _2(Q^*_1-Q^*_2)=0. \end{array}\right. } \end{aligned}$$
This system reduces to
$$\begin{aligned} \begin{bmatrix} -(r_1+\lambda _1)&\lambda _1 \\ -(r_2+\lambda _2)&\lambda _2 \\ \end{bmatrix} \begin{bmatrix} Q^*_1 \\ Q^*_2 \\ \end{bmatrix} = \begin{bmatrix} c'-c \\ c'-c \\ \end{bmatrix}, \end{aligned}$$
and hence
$$\begin{aligned} \begin{bmatrix} Q^*_1 \\ Q^*_2 \\ \end{bmatrix}&= \begin{bmatrix} -(r_1+\lambda _1)&\lambda _1 \\ -(r_2+\lambda _2)&\lambda _2 \\ \end{bmatrix}^{-1} \begin{bmatrix} c'-c \\ c'-c \\ \end{bmatrix}\\&=\frac{1}{r_2\lambda _1-r_1\lambda _2} \begin{bmatrix} \lambda _2&-\lambda _1 \\ r_2+\lambda _2&-(r_1+\lambda _1) \\ \end{bmatrix} \begin{bmatrix} c'-c \\ c'-c \\ \end{bmatrix}\\&=\frac{1}{r_2\lambda _1-r_1\lambda _2} \begin{bmatrix} (c-c')(\lambda _1-\lambda _2) \\ (c-c')\big ((r_1+\lambda _1)-(r_2+\lambda _2)\big ) \\ \end{bmatrix}, \end{aligned}$$
Thus a particular solution of (8) is
$$\begin{aligned} \begin{aligned} Q^*_1&=\frac{(c-c')(\lambda _1-\lambda _2)}{r_2\lambda _1-r_1\lambda _2}\\ Q^*_2&=\frac{(c-c')\big ((r_1+\lambda _1)-(r_2+\lambda _2)\big )}{r_2\lambda _1-r_1\lambda _2}. \end{aligned} \end{aligned}$$
(35)
Therefore the general form of the solution to (33) is
$$\begin{aligned} P_1(x)&=Q^*_1+A_1x^{\beta _1}+A_2x^{\beta _2}+A_3x^{\beta _3}+A_4x^{\beta _4},\\ P_2(x)&=Q^*_2+B_1x^{\beta _1}+B_2x^{\beta _2}+B_3x^{\beta _3}+B_4x^{\beta _4}, \end{aligned}$$
with \(B_i=l_iA_i\) and \(l_i=l(\beta _i)=g_1(\beta _i)/\lambda _1=\lambda _2/g_2(\beta _i)\).
When \(x\rightarrow \infty \), \(P_1(x)\) and \(P_2(x)\) are both bounded. Thus \(A_3=A_4=B_3=B_4=0\), and the solution is
$$\begin{aligned} \begin{aligned} P_1(x)&=Q^*_1+A_1x^{\beta _1}+A_2x^{\beta _2},\\ P_2(x)&=Q^*_2+B_1x^{\beta _1}+B_2x^{\beta _2}. \end{aligned} \end{aligned}$$
(36)
Next we solve (32). The first equation is
$$\begin{aligned} \frac{1}{2}x^2\sigma _1^2P_1''(x)+\mu _1xP_1'(x)-r_1P_1(x)+c-c'+\lambda _1\big ((1-\alpha )x-P_1(x)\big )=0. \end{aligned}$$
(37)
This is an inhomogeneous equation, and thus the solution can be written as
$$\begin{aligned} P_1(x)=C_1x^{\gamma _1}+C_2x^{\gamma _2}+\phi (x), \end{aligned}$$
(38)
where \(\phi (x)\) is a particular solution and \(\gamma _1\) and \(\gamma _2\) are the two roots of
$$\begin{aligned} \frac{1}{2}\sigma _1^2\gamma (\gamma -1)+\mu _1\gamma -r_1-\lambda _1=0. \end{aligned}$$
(39)
To obtain \(\phi (x)\), we assume that \(\phi (x)=ax+b\) and substitute it to Eq. (37). This yields
$$\begin{aligned} a\mu _1x-r_1(ax+b)+c-c'+\lambda _1(1-\alpha )x-\lambda _1(ax+b)=0, \end{aligned}$$
or
$$\begin{aligned} x\big ((\mu _1-r_1-\lambda _1)a+\lambda _1(1-\alpha )\big )+c-c'-r_1b-\lambda _1b=0. \end{aligned}$$
Thus \(a=\lambda _1(1-\alpha )/(r_1+\lambda _1-\mu _1)\), \(b=(c-c')/(r_1+\lambda _1)\), and thus
$$\begin{aligned} \phi (x)=\frac{\lambda _1(1-\alpha )}{r_1+\lambda _1-\mu _1}x+\frac{c-c'}{r_1+\lambda _1}. \end{aligned}$$
(40)
Now we solve for the coefficients \(A_i\), \(B_i\), \(C_i\) and the optimal values \(v_1\) and \(v_2\). Using the boundary and smooth pasting conditions for \(P_2(x)\) (see(30), (33)) at \(v_2\) with \(x\in [v_2,\infty )\),
$$\begin{aligned} {\left\{ \begin{array}{ll} Q^*_2+B_1v_2^{\beta _1}+B_2v_2^{\beta _2}=(1-\alpha )v_2,\\ \beta _1B_1v_2^{\beta _1-1}+\beta _2B_2v_2^{\beta _2-1}=(1-\alpha ). \end{array}\right. } \end{aligned}$$
or
$$\begin{aligned} {\left\{ \begin{array}{ll} l_1A_1v_2^{\beta _1}+l_2A_2v_2^{\beta _2}=(1-\alpha )v_2-Q^*_2,\\ \beta _1l_1A_1v_2^{\beta _1}+\beta _2l_2A_2v_2^{\beta _2}=(1-\alpha )v_2. \end{array}\right. } \end{aligned}$$
(41)
Similarly, using the boundary and smooth pasting conditions for \(P_1(x)\) (see (29), (32)) at \(v_2\) with \(x\in [v_1,v_2]\),
$$\begin{aligned} {\left\{ \begin{array}{ll} Q^*_1+A_1v_2^{\beta _1}+A_2v_2^{\beta _2}=C_1v_2^{\gamma _1}+C_2v_2^{\gamma _2}+\phi (v_2),\\ \beta _1A_1v_2^{\beta _1}+\beta _2A_2v_2^{\beta _2}=\gamma _1C_1v_2^{\gamma _1}+\gamma _2C_2v_2^{\gamma _2}+v_2\phi '(v_2). \end{array}\right. } \end{aligned}$$
(42)
Using the boundary and smooth pasting conditions for \(P_1(x)\) (see (29), (32)) at \(v_1\) with \(x\in [v_1,v_2]\)
$$\begin{aligned} {\left\{ \begin{array}{ll} C_1v_1^{\gamma _1}+C_2v_1^{\gamma _2}+\phi (v_1)=(1-\alpha )v_1,\\ \gamma _1C_1v_1^{\gamma _1}+\gamma _2C_2v_1^{\gamma _2}+v_1\phi '(v_1)=(1-\alpha )v_1. \end{array}\right. } \end{aligned}$$
(43)
Combining these equations, we can obtain the solutions \(v_1\) and \(v_2\). From (43), we have
$$\begin{aligned} F_1(v_1):= \begin{bmatrix} 1&1 \\ \gamma _1&\gamma _2 \end{bmatrix}^{-1} \begin{bmatrix} (1-\alpha )v_1-\phi (v_1)\\ (1-\alpha )v_1-v_1\phi '(v_1) \end{bmatrix} =\begin{bmatrix} C_1v_1^{\gamma _1}\\ C_2v_1^{\gamma _2} \end{bmatrix}. \end{aligned}$$
(44)
From (41), we have
$$\begin{aligned} \begin{bmatrix} l_1&l_2 \\ \beta _1l_1&\beta _2l_2 \end{bmatrix}^{-1} \begin{bmatrix} (1-\alpha )v_2-Q^*_2 \\ (1-\alpha )v_2 \end{bmatrix} =\begin{bmatrix} A_1v_2^{\beta _1}\\ A_2v_2^{\beta _2} \end{bmatrix}. \end{aligned}$$
(45)
Then, using (42) and (45), we have
$$\begin{aligned} \begin{bmatrix} 1&1 \\ \beta _1&\beta _2 \end{bmatrix} \begin{bmatrix} l_1&l_2 \\ \beta _1l_1&\beta _2l_2 \end{bmatrix}^{-1} \begin{bmatrix} (1-\alpha )v_2-Q^*_2 \\ (1-\alpha )v_2 \end{bmatrix}&=\begin{bmatrix} 1&1 \\ \beta _1&\beta _2 \end{bmatrix} \begin{bmatrix} A_1v_2^{\beta _1}\\ A_2v_2^{\beta _2} \end{bmatrix}\nonumber \\&= \begin{bmatrix} C_1v_2^{\gamma _1}+C_2v_2^{\gamma _2}+\phi (v_2)-Q^*_1\\ \gamma _1C_1v_2^{\gamma _1}+\gamma _2C_2v_2^{\gamma _2}+v_2\phi '(v_2) \end{bmatrix}. \end{aligned}$$
(46)
This equation can be rewritten as
$$\begin{aligned} \begin{bmatrix} 1&1 \\ \beta _1&\beta _2 \end{bmatrix} \begin{bmatrix} l_1&l_2 \\ \beta _1l_1&\beta _2l_2 \end{bmatrix}^{-1} \begin{bmatrix} (1-\alpha )v_2-Q^*_1 \\ (1-\alpha )v_2 \end{bmatrix} -\begin{bmatrix} \phi (v_2)-Q^*_1 \\ v_2\phi '(v_2) \end{bmatrix} \!=\! \begin{bmatrix} 1&1 \\ \gamma _1&\gamma _2 \end{bmatrix} \begin{bmatrix} C_1v_2^{\gamma _1}\\ C_2v_2^{\gamma _2} \end{bmatrix}. \end{aligned}$$
Therefore
$$\begin{aligned} F_2(v_2)&:=\begin{bmatrix} 1&1 \\ \gamma _1&\gamma _2 \end{bmatrix}^{-1} \left( \begin{bmatrix} 1&1 \\ \beta _1&\beta _2 \end{bmatrix} \begin{bmatrix} l_1&l_2 \\ \beta _1l_1&\beta _2l_2 \end{bmatrix}^{-1} \begin{bmatrix} (1-\alpha )v_2-Q^*_2 \\ (1-\alpha )v_2 \end{bmatrix}\right. \nonumber \\&\quad \left. -\begin{bmatrix} \phi (v_2)-Q^*_1 \\ v_2\phi '(v_2) \end{bmatrix}\right) \nonumber \\&= \begin{bmatrix} C_1v_2^{\gamma _1}\\ C_2v_2^{\gamma _2} \end{bmatrix}. \end{aligned}$$
(47)
Hence, from (44) and (47) we have the equation
$$\begin{aligned} \begin{bmatrix} v_1^{-\gamma _1}&0\\ 0&v_1^{-\gamma _2} \end{bmatrix}F_1(v_1) =\begin{bmatrix} C_1\\ C_2 \end{bmatrix} =\begin{bmatrix} v_2^{-\gamma _1}&0\\ 0&v_2^{-\gamma _2} \end{bmatrix}F_2(x_2). \end{aligned}$$
(48)
In particular, for \(\phi (x)\) of the form in (40), we have from (44) and (47)
$$\begin{aligned} F_1(v_1)=\begin{bmatrix} 1&1 \\ \gamma _1&\gamma _2 \end{bmatrix}^{-1} \begin{bmatrix} (1-\alpha )v_1-\frac{\lambda _1(1-\alpha )}{r_1+\lambda _1-\mu _1}v_1-\frac{c-c'}{r_1+\lambda _1}\\ (1-\alpha )v_1-v_1\frac{\lambda _1(1-\alpha )}{r_1+\lambda _1-\mu _1} \end{bmatrix}, \end{aligned}$$
(49)
and
$$\begin{aligned} F_2(v_2)&=\begin{bmatrix} 1&1 \\ \gamma _1&\gamma _2 \end{bmatrix}^{-1} \left( \begin{bmatrix} 1&1 \\ \beta _1&\beta _2 \end{bmatrix} \begin{bmatrix} l_1&l_2 \\ \beta _1l_1&\beta _2l_2 \end{bmatrix}^{-1} \begin{bmatrix} (1-\alpha )v_2-Q^*_2 \\ (1-\alpha )v_2 \end{bmatrix}\right. \nonumber \\&\quad \left. -\begin{bmatrix} \frac{\lambda _1(1-\alpha )}{r_1+\lambda _1-\mu _1}v_2+\frac{c-c'}{r_1+\lambda _1}-Q^*_1\\ (1-\alpha )v_2\frac{\lambda _1}{r_1+\lambda _1-\mu _1}. \end{bmatrix}\right) . \end{aligned}$$
(50)
Substituting \(F_1(v_1)\) and \(F_2(v_2)\) into (48), we can obtain the values for \(v_1\) and \(v_2\). Now we derive coefficients \(A_i\), \(B_i\), and \(C_i\). From (41), the coefficients \(A_1\) and \(A_2\) are give by
$$\begin{aligned} \begin{bmatrix} A_1 \\ A_2 \end{bmatrix} =\begin{bmatrix} l_1v_2^{\beta _1}&l_2v_2^{\beta _2} \\ \beta _1l_1v_2^{\beta _1}&\beta _2l_2v_2^{\beta _2} \end{bmatrix}^{-1} \begin{bmatrix} (1-\alpha )v_2-Q^*_2 \\ (1-\alpha )v_2 \end{bmatrix}, \end{aligned}$$
(51)
and
$$\begin{aligned} \begin{bmatrix} B_1 \\ B_2 \end{bmatrix} =\begin{bmatrix} l_1A_1 \\ l_2A_2 \end{bmatrix}. \end{aligned}$$
(52)
From (43), we have
$$\begin{aligned} \begin{bmatrix} C_1 \\ C_2 \end{bmatrix}&=\begin{bmatrix} v_1^{\gamma _1}&v_1^{\gamma _2} \\ \gamma _1v_1^{\gamma _1}&\gamma _2v_2^{\gamma _2} \end{bmatrix}^{-1} \begin{bmatrix} (1-\alpha )v_1-\phi (v_1) \\ (1-\alpha )v_1-v_1\phi '(v_1) \end{bmatrix}\end{aligned}$$
(53)
$$\begin{aligned}&=\begin{bmatrix} v_1^{\gamma _1}&v_1^{\gamma _2} \\ \gamma _1v_1^{\gamma _1}&\gamma _2v_1^{\gamma _2} \end{bmatrix}^{-1} \begin{bmatrix} (1-\alpha )v_1-\frac{\lambda _1}{r_1+\lambda _1-\mu _1}v_1-\frac{c-c'}{r_1+\lambda _1} \\ (1-\alpha )v_1-v_1\frac{\lambda _1}{r_1+\lambda _1-\mu _1} \end{bmatrix}. \end{aligned}$$
(54)
With these coefficients, the value functions \(P_1(x)\) and \(P_2(x)\) become
$$\begin{aligned} P_1(x)={\left\{ \begin{array}{ll} A_1x^{\beta _1}+A_2x^{\beta _2}+\frac{(c-c')(\lambda _1-\lambda _2)}{r_2\lambda _1-r_1\lambda _2} \quad &{}\text {if }x>v_2,\\ C_1x^{\gamma _1}+C_2x^{\gamma _2}+\phi (x) &{}\text {if }v_1\le x\le v_2\\ (1-\alpha )x &{}\text {if }x\le v_1, \end{array}\right. } \end{aligned}$$
and
$$\begin{aligned} P_2(x)={\left\{ \begin{array}{ll} B_1x^{\beta _1}+B_2x^{\beta _2}+\frac{(c-c')\big ((r_1+\lambda _1)-(r_2+\lambda _2)\big )}{r_2\lambda _1-r_1\lambda _2} \quad &{}\text {if }x>v_2,\\ (1-\alpha )x &{}\text {if }x\le v_2, \end{array}\right. } \end{aligned}$$
where
$$\begin{aligned} \phi (x)=\frac{\lambda _1(1-\alpha )}{r_1+\lambda _1-\mu _1}x+\frac{c-c'}{r_1+\lambda _1}. \end{aligned}$$
\(\square \)
