2.1 Preliminary characteristic decompositions

The matrix form of system (1.3) is

(2.1)\begin{equation} \left(\begin{array}{ccccc} c^2-U^2 & -UV \\ 0 & -1 \end{array}\right) \left(\begin{array}{c} u \\ v \end{array}\right)_\xi+ \left(\begin{array}{ccccc} -UV & c^2-V^2 \\ 1 & 0 \end{array}\right) \left(\begin{array}{c} u \\ v \end{array}\right)_\eta=0. \end{equation}

The two eigenvalues of (2.1) are given in (1.5) and the corresponding left eigenvectors are $\ell _\pm =(1,\,\Lambda _\mp )$. Performing a standard procedure gives the characteristic forms of (2.1)

(2.2)\begin{equation} \left\{ \begin{array}{@{}l} \partial^+ u+\Lambda_-\partial^+ v=0,\\ \partial^- u+\Lambda_+\partial^- v=0, \end{array} \right. \quad \partial^\pm =\partial_\xi+\Lambda_\pm\partial_\eta. \end{equation}

Following the work [Reference Li and Zheng31], we introduce the pseudo-flow angle $\theta$ and pseudo-Mach angle $\omega$ as follows

(2.3)\begin{equation} \tan\theta=\frac{V}{U}, \quad \sin\omega=\frac{c}{\sqrt{U^2+V^2}}. \end{equation}

Moreover, we denote

(2.4)\begin{equation} \alpha: =\theta+\omega, \quad \beta: =\theta-\omega, \end{equation}

and use the expression of $\Lambda _\pm$ to get

(2.5)\begin{equation} \tan\alpha=\Lambda_+, \quad \tan\beta=\Lambda_-. \end{equation}

In other words, the angles $\alpha$ and $\beta$ are, respectively, the inclination angles of positive and negative characteristic curves. In view of (2.3) and the pseudo-Bernoulli law (1.4), the functions $(c,\,u,\,v)$ can be expressed in terms of $\phi,\, \theta,\, \omega$

(2.6)\begin{equation} c=\sqrt{\frac{-2\phi\kappa\varpi^2}{\kappa+\varpi^2}}, \quad u=\xi-c\frac{\cos\theta}{\varpi},\quad v=\eta-c\frac{\sin\theta}{\varpi},\quad \kappa=\frac{\gamma-1}{2}. \end{equation}

Here and below, the mixed variables $\omega$ and $\varpi :=\sin \omega$ are used for convenience. Obviously, the sonic curve $\{(\xi,\,\eta ): M(\xi,\,\eta )=1\}$ now is $\{(\xi,\,\eta ): \varpi (\xi,\,\eta )=1\}$.

Further introduce the normalized directional derivatives along the characteristics

(2.7)\begin{equation} \bar{\partial}^+=\cos\alpha\partial_\xi+\sin\alpha\partial_\eta, \quad \bar{\partial}^-=\cos\beta\partial_\xi+\sin\beta\partial_\eta, \quad \bar{\partial}^{0}=\cos\theta\partial_\xi+\sin\theta\partial_\eta. \end{equation}

Consequently

(2.8)\begin{equation} \displaystyle \partial_\xi=\cos\theta\bar\partial^0 -\frac{\sin\theta}{2\varpi}(\bar{\partial}^+-\bar{\partial}^-), \quad \displaystyle \partial_\eta=\sin\theta\bar\partial^0 +\frac{\cos\theta}{2\varpi}(\bar{\partial}^+-\bar{\partial}^-), \quad \displaystyle \bar{\partial}^0=\frac{\bar{\partial}^+{+}\bar{\partial}^-}{2\cos\omega}. \end{equation}

In terms of the variables $(\theta,\, \varpi )$, we thus obtain a new system by (2.2) and (2.7)

(2.9)\begin{equation} \left\{ \begin{array}{@{}ll} \displaystyle \bar\partial^+\theta+\dfrac{\cos\omega}{\kappa+\varpi^2}\bar\partial^+\varpi =\dfrac{\varpi^2}{c}\cdot\dfrac{\kappa-1+2\varpi^2}{\kappa+\varpi^2}, \\ \displaystyle \bar\partial^-\theta-\dfrac{\cos\omega}{\kappa+\varpi^2}\bar\partial^-\varpi =-\dfrac{\varpi^2}{c}\cdot\dfrac{\kappa-1+2\varpi^2}{\kappa+\varpi^2}. \end{array} \right. \end{equation}

Furthermore, applying (1.4), (2.4), (2.6), and (2.7) leads to the equations of the pseudo-velocity potential $\phi$

(2.10)\begin{equation} \bar\partial^0\phi=-\frac{c}{\varpi},\quad \bar\partial^\pm\phi=-\frac{c\cos\omega}{\varpi}. \end{equation}

Denote

(2.11)\begin{equation} R=\frac{\bar\partial^+c}{c},\quad S=\frac{\bar\partial^-c}{c}, \end{equation}

one employs the pseudo-Bernoulli's law (1.4) again to get the relations between $\varpi$ and $(R,\,S)$

(2.12)\begin{equation} \bar\partial^+\varpi=\frac{\varpi(\kappa+\varpi^2)}{\kappa}R-\frac{\cos\omega\varpi^2}{c},\quad \bar\partial^-\varpi=\frac{\varpi(\kappa+\varpi^2)}{\kappa}S-\frac{\cos\omega\varpi^2}{c}. \end{equation}

The variables $c$ and $\varpi$ also satisfy

(2.13)\begin{equation} \bar\partial^0c=\frac{\kappa(c\bar\partial^0\varpi+\varpi^2)}{\varpi(\kappa+\varpi^2)}. \end{equation}

With the aid of the commutator relation between $\bar \partial ^+$ and $\bar \partial ^-$ [Reference Li and Zheng31]

(2.14)\begin{equation} \bar{\partial}^-\bar{\partial}^+-\bar{\partial}^+\bar{\partial}^-=\frac{\cos(2\omega)\bar{\partial}^-\alpha-\bar{\partial}^+\beta}{\sin(2\omega)}\bar{\partial}^+{+}\frac{\cos(2\omega)\bar{\partial}^+\beta-\tilde{\partial}^-\alpha}{\sin(2\omega)}\bar{\partial}^-, \end{equation}

we have the characteristic decompositions for the variable $c$

(2.15)\begin{equation} \begin{cases} \displaystyle \bar{\partial}^-R= R\Bigg\{-\frac{2\cos\omega\varpi}{c}+\frac{(\kappa+1)(R+S)}{2\kappa\cos^2\omega}-\frac{\kappa+2\varpi^2}{\kappa}S \Bigg\},\\ \displaystyle \bar{\partial}^+S= S\Bigg\{-\frac{2\cos\omega\varpi}{c}+\frac{(\kappa+1)(R+S)}{2\kappa\cos^2\omega}-\frac{\kappa+2\varpi^2}{\kappa}R \Bigg\}, \end{cases} \end{equation}

Set

(2.16)\begin{equation} \overline{R}=\varpi\sqrt{\kappa+\varpi^2}R,\quad \overline{S}=-\varpi\sqrt{\kappa+\varpi^2}S, \end{equation}

which together with (2.15) yields

(2.17)\begin{equation} \begin{cases} \displaystyle \bar{\partial}^-\overline{R}=\overline{R}\Bigg\{\frac{\kappa+1}{2\kappa\varpi\sqrt{\kappa+\varpi^2}}\cdot\frac{\overline{R} -\overline{S}}{\cos^2\omega} -\frac{\varpi(3\kappa+4\varpi^2)\cos\omega}{c(\kappa+\varpi^2)}\Bigg\},\\ \displaystyle \bar{\partial}^+\overline{S}=\overline{S}\Bigg\{\frac{\kappa+1}{2\kappa\varpi\sqrt{\kappa+\varpi^2}}\cdot\frac{\overline{R} -\overline{S}}{\cos^2\omega} -\frac{\varpi(3\kappa+4\varpi^2)\cos\omega}{c(\kappa+\varpi^2)}\Bigg\}. \end{cases} \end{equation}

For later use, we list some relations by (2.8), (2.9), (2.12), and (2.16) here

(2.18)\begin{align} & \begin{cases} \displaystyle \bar\partial^+\theta=-\frac{\cos\omega \overline{R}}{\kappa\sqrt{\kappa+\varpi^2}}+\frac{\varpi^2}{c},\\ \displaystyle \bar\partial^-\theta=-\frac{\cos\omega \overline{S}}{\kappa\sqrt{\kappa+\varpi^2}}-\frac{\varpi^2}{c}, \end{cases} \end{align}

(2.19)\begin{align} & \begin{cases} \displaystyle \varpi_\xi=\cos\theta\frac{\sqrt{\kappa+\varpi^2}}{\kappa}\overline{W} -\cos\theta\frac{\varpi^2}{c} -\sin\theta \frac{\sqrt{\kappa+\varpi^2}}{2\kappa\varpi}(\overline{R}+\overline{S}), \\ \displaystyle \varpi_\eta=\sin\theta\frac{\sqrt{\kappa+\varpi^2}}{\kappa}\overline{W} -\sin\theta\frac{\varpi^2}{c} +\cos\theta \frac{\sqrt{\kappa+\varpi^2}}{2\kappa\varpi}(\overline{R}+\overline{S}), \end{cases} \end{align}

and

(2.20)\begin{equation} \phi_\xi\varpi_\eta-\phi_\eta\varpi_\xi=-\frac{c\sqrt{\kappa+\varpi^2}}{2\kappa\varpi^2}(\overline{R}+\overline{S}), \end{equation}

where $\overline {W}=(\overline {R}-\overline {S})/2\cos \omega$.

2.2 The problem and the main result in terms of angle variables

In this subsection, we formulate the problem in terms of the angle variables $(\theta,\, \varpi )$. Given a smooth curve $\widehat {AB}: \eta =\varphi (\xi ) (\xi \in [\xi _1,\,\xi _2])$ in the $(\xi,\,\eta )$ plane, we assign the boundary data for $(\rho,\, u,\, v)$ on $\widehat {AB}$, $(\rho,\, u,\, v)(\xi,\,\varphi (\xi ))=(\hat {\rho },\, \hat {u},\, \hat {v})(\xi )$ such that

(2.21)\begin{equation} \begin{array}{ll} \hat{\rho}(\xi)>0,\quad \hat{v}(\xi)-\varphi(\xi)=\varphi'(\xi)(\hat{u}(\xi)-\xi), & \forall\ \xi\in[\xi_1,\xi_2], \\ (\hat{u}(\xi)-\xi)^2+(\hat{v}(\xi)-\varphi(\xi))^2>A\gamma\hat{\rho}^{\gamma-1}(\xi), & \forall\ \xi\in[\xi_1,\xi_2), \\ (\hat{u}(\xi_2)-\xi_2)^2+(\hat{v}(\xi_2)-\varphi(\xi_2))^2=A\gamma\hat{\rho}^{\gamma-1}(\xi_2) & . \end{array} \end{equation}

It follows by (2.21) that the curve $\widehat {AB}$ is a pseudo-streamline and the flow is supersonic on $\widehat {AB}\setminus \{B\}$ and sonic at point $B$. From (2.3), one obtains the data of $(c,\,\theta,\,\varpi )$ on $\widehat {AB}$

(2.22)\begin{equation} \begin{aligned} \displaystyle c(\xi,\varphi(\xi)) & =\sqrt{A\gamma\hat{\rho}^{\gamma-1}(\xi)}=:\hat{c}(\xi),\\ \displaystyle \theta(\xi,\varphi(\xi)) & =\arctan\Bigg(\frac{\hat{v}(\xi)-\varphi(\xi)}{\hat{u}(\xi)-\xi}\Bigg)=:\hat{\theta}(\xi), \\ \displaystyle \varpi(\xi,\varphi(\xi)) & =\frac{\hat{c}(\xi)}{\sqrt{(\hat{u}(\xi)-\xi)^2+(\hat{v}(\xi)-\varphi(\xi))^2}} =:\hat{\varpi}(\xi), \end{aligned}\quad\ \forall\ \xi\in[\xi_1,\xi_2]. \end{equation}

Combining with (2.21) and (2.22) gets

(2.23)\begin{equation} \hat{\theta}(\xi)=\arctan\varphi'(\xi)\ (\forall\ \xi\in[\xi_1,\xi_2]),\quad \hat{\varpi}(\xi)<1\ (\forall\ \xi\in[\xi_1,\xi_2)),\quad \hat{\varpi}(\xi_2)=1. \end{equation}

We suppose that the functions $(\hat {c},\, \hat {\theta },\, \hat {\varpi })$ satisfy the following compatibility condition by (2.13)

(2.24)\begin{equation} \cos\hat{\theta}\hat{c}' =\frac{\kappa(\hat{c}\cos\hat{\theta}\hat{\varpi}'+\hat{\varpi}^2)}{\hat{\varpi}(\kappa+\hat{\varpi}^2)},\quad \forall\ \xi\in[\xi_1,\xi_2]. \end{equation}

Obviously, Problem 1.1 in terms of angle variables can be restated as

By extracting the features of the problem described in figure 1, we further assume that the functions $\varphi (\xi )$ and $\hat {\varpi }(\xi )$ satisfy

(2.25)\begin{equation} \begin{array}{c} \varphi'(\xi), \hat{\varpi}(\xi)\in C^2([\xi_1,\xi_2]), \\ \varphi'(\xi_2)<0, \quad \varphi''(\xi_2)<0, \quad \hat{\varpi}'(\xi_2)>0. \end{array} \end{equation}

Since we are looking for a solution near point $B$, without loss of generality, we can replace (2.25) with the following

(2.26)\begin{equation} \begin{array}{ll} \varphi'(\xi), \hat{\varpi}(\xi)\in C^2([\xi_1,\xi_2]), & \\ \varphi_0\leq-\varphi'(\xi), -\varphi''(\xi), \hat{\varpi}'(\xi)\leq\varphi_1, & \forall\ \xi\in[\xi_1,\xi_2],\\ \end{array} \end{equation}

for some positive constants $\varphi _0$ and $\varphi _1$. Otherwise, a suitable point $A_1$ on $\widehat {AB}$ can be selected to replace $A$ such that (2.26) is valid on $\widehat {A_1B}$.

Our main conclusion theorem 1.2 can be restated in the following theorem.

2.3 The boundary data for $(\overline {R},\, \overline {S})$

In this subsection, we derive the data of $(\overline {R},\, \overline {S})$ on the pseudo-streamline $\widehat {AB}$, which need to be used later.

Since $\widehat {AB}$ is a pseudo-streamline, we have by (2.7), (2.23), and (2.26)

(2.27)\begin{equation} \bar\partial^0\theta|_{\widehat{AB}}=\cos\hat{\theta}\hat{\theta}'=\frac{\cos\hat{\theta}\varphi''}{1+(\varphi')^2}<0,\quad \bar\partial^0\varpi|_{\widehat{AB}}=\cos\hat{\theta}\hat{\varpi}'>0. \end{equation}

Making use of (2.8) and (2.9) yields

(2.28)\begin{equation} \begin{aligned} \bar\partial^+\varpi & =-(\kappa+\varpi^2)\bar\partial^0\theta+\cos\omega\bar\partial^0\varpi, \\ \bar\partial^-\varpi & =(\kappa+\varpi^2)\bar\partial^0\theta+\cos\omega\bar\partial^0\varpi, \end{aligned} \end{equation}

which together with (2.12) and (2.16) arrives at

(2.29)\begin{equation} \begin{aligned} \displaystyle \overline{R} & =\frac{\kappa}{\sqrt{\kappa+\varpi^2}} \Bigg(-(\kappa+\varpi^2)\bar\partial^0\theta+\cos\omega\bar\partial^0\varpi +\frac{\cos\omega\varpi^2}{c}\Bigg), \\ \displaystyle \overline{S} & =\frac{\kappa}{\sqrt{\kappa+\varpi^2}} \Bigg(-(\kappa+\varpi^2)\bar\partial^0\theta-\cos\omega\bar\partial^0\varpi -\frac{\cos\omega\varpi^2}{c}\Bigg). \end{aligned} \end{equation}

Combining with (2.27) and (2.29), one gets the boundary data of $(\overline {R},\, \overline {S})$

(2.30)\begin{equation} \begin{aligned} \displaystyle \overline{R}|_{\widehat{AB}} & =\frac{\kappa}{\sqrt{\kappa\!+\!\hat{\varpi}^2}} \Bigg(-\frac{(\kappa\!+\!\hat{\varpi}^2)\cos\hat{\theta}\varphi''}{1+(\varphi')^2} \!+\!\sqrt{1-\hat{\varpi}^2} \frac{\hat{\varpi}^2+\hat{c}\cos\hat{\theta}\hat{\varpi}'}{\hat{c}}\Bigg)(\xi)=:\hat{a}(\xi), \\ \displaystyle \overline{S}|_{\widehat{AB}} & =\frac{\kappa}{\sqrt{\kappa\!+\!\hat{\varpi}^2}} \Bigg(-\frac{(\kappa\!+\!\hat{\varpi}^2)\cos\hat{\theta}\varphi''}{1\!+\!(\varphi')^2} -\sqrt{1-\hat{\varpi}^2} \frac{\hat{\varpi}^2+\hat{c}\cos\hat{\theta}\hat{\varpi}'}{\hat{c}}\Bigg)(\xi)=:\hat{b}(\xi), \end{aligned} \end{equation}

for $\xi \in [\xi _1,\,\xi _2]$. It easily seen by (2.26) and the fact $\varpi (\xi _2)=1$ that there exists a number $\xi _0\in [\xi _1,\,\xi _2)$ such that $\hat {a}(\xi )>0$ and $\hat {b}(\xi )>0$ for all $\xi \in [\xi _0,\,\xi _2]$. Denote the point $(\xi _0,\,\varphi (\xi _0))$ by $Q$.

In addition, for $\overline {W}=(\overline {R}-\overline {S})/2\cos \omega$, we deduce by (2.11), (2.13), and (2.16)

(2.31)\begin{equation} \overline{W}=\frac{\varpi\sqrt{\kappa+\varpi^2}}{c}\bar\partial^0c =\frac{\kappa}{\sqrt{\kappa+\varpi^2}}\Bigg(\bar\partial^0\varpi+\frac{\varpi^2}{c}\Bigg). \end{equation}

Hence by (2.27) and (2.31) one acquires the boundary data of $\overline {W}$ on $\widehat {AB}$

(2.32)\begin{equation} \overline{W}|_{AB}=\frac{\kappa}{\sqrt{\kappa+\hat{\varpi}^2}} \Bigg(\cos\hat{\theta}\hat{\varpi}'+\frac{\hat{\varpi}^2}{\hat{c}}\Bigg)(\xi)=:\hat{d}(\xi). \end{equation}

Obviously, one has $\hat {d}(\xi )>0$ for all $\xi \in [\xi _1,\,\xi _2]$.

Summing up (2.26), (2.30), and (2.32), we obtain the boundary conditions $(\overline {R},\,\overline {S},\,\overline {W})$ on $\widehat {QB}$

(2.33)\begin{equation} \begin{array}{ll} \hat{a}(\xi), \hat{b}(\xi)\in C^0([\xi_{0},\xi_2])\cap C^1([\xi_{0},\xi_2)), \hat{d}(\xi)\in C^1([\xi_{0},\xi_2]), \\ 0<\hat{m}_0\leq \hat{a}(\xi), \hat{b}(\xi), \hat{d}(\xi)\leq\hat{M}_0, & \forall\ \xi\in[\xi_{0},\xi_2], \\ \hat{a}(\xi)-\hat{b}(\xi)=2\sqrt{1-\hat{\varpi}^2}\hat{d}(\xi), & \forall\ \xi\in[\xi_{0},\xi_2], \end{array} \end{equation}

for some constants $\hat {m}_0$ and $\hat {M}_0$. Moreover, there hold by (2.30) and (2.32)

(2.34)\begin{equation} \begin{aligned} & \displaystyle \sqrt{1-\hat{\varpi}^2}\hat{a}'(\xi)=-\hat{\varpi}\hat{\varpi}'\hat{d}+(1-\hat{\varpi}^2)\hat{d}' -\sqrt{1-\hat{\varpi}^2}\Bigg(\frac{\kappa\sqrt{\kappa+\hat{\varpi}^2}\cos\hat\theta \varphi''}{1+(\varphi')^2}\Bigg)', \\ & \displaystyle \sqrt{1-\hat{\varpi}^2}\hat{b}'(\xi)=\hat{\varpi}\hat{\varpi}'\hat{d}+(1-\hat{\varpi}^2)\hat{d}' -\sqrt{1-\hat{\varpi}^2}\Bigg(\frac{\kappa\sqrt{\kappa+\hat{\varpi}^2}\cos\hat\theta \varphi''}{1+(\varphi')^2}\Bigg)', \end{aligned} \end{equation}

for $\xi \in [\xi _0,\,\xi _2)$.

3.1 The problem in a partial hodograph plane

We first derive the boundary value of the pseudo-velocity potential $\phi$ defined in (1.4) on $\widehat {AB}$. From (2.6) one obtains

(3.1)\begin{equation} \phi=-\frac{c^2(\kappa+\varpi^2)}{2\kappa\varpi^2}, \end{equation}

then

(3.2)\begin{equation} \phi|_{\widehat{AB}}=-\frac{\hat{c}^2(\kappa+\hat{\varpi}^2)}{2\kappa\hat{\varpi}^2}(\xi)=:\hat{\phi}(\xi),\ \ \xi\in[\xi_1,\xi_2]. \end{equation}

Moreover, it follows by (2.10) that

(3.3)\begin{equation} \hat{\phi}'(\xi)=\frac{\bar\partial^0\phi}{\cos\theta}\bigg|_{\widehat{AB}} =-\frac{\hat{c}(\xi)}{\hat{\varpi}(\xi)\cos\hat{\theta}(\xi)} =-\frac{\hat{c}(\xi)\sqrt{1+(\varphi'(\xi))^2}}{\hat{\varpi}(\xi)}<0, \end{equation}

for all $\xi \in [\xi _1,\,\xi _2]$, which means that $\hat {\phi }(\xi )$ a strictly decreasing function.

Now introduce the following transformation

(3.4)\begin{equation} t=\cos\omega(\xi,\eta),\quad z=\phi(\xi,\eta)-\hat{\phi}(\xi_2). \end{equation}

Recalling the facts $\hat {\varpi }'(\xi )>0$ by (2.26) and $\hat {\phi }'(\xi )<0$ by (3.3), we find that the image of the arc $\widehat {BQ}$ in the $(z,\,t)$-plane is a smooth increasing curve $\widehat {B'Q'}: z=\tilde {z}(t) (t\in [0,\,t_0])$ defined through a parametric $\xi$

(3.5)\begin{equation} t=\sqrt{1-\hat\varpi^2(\xi)},\quad z=\hat{\phi}(\xi)-\hat{\phi}(\xi_2),\ (\xi\in[\xi_0,\xi_2]). \end{equation}

Here the number $t_0=\sqrt {1-\hat {\varpi }^2(\xi _0)}\in (0,\,1)$. Moreover, we denote the inverse function of the function $z=\hat {\phi }(\xi )-\hat {\phi }(\xi _2)$ by $\xi =\hat {\xi }(z)$ ($z\in [0,\,z_0]$), where $z_0=\hat {\phi }(\xi _0)-\hat {\phi }(\xi _2)>0$. Hence, combining with (2.30) and (2.32), one can achieve the boundary data of $(\overline {R},\, \overline {S},\, \overline {W})$ on $\widehat {B'Q'}$

(3.6)\begin{equation} \begin{aligned} & \displaystyle\overline{R}|_{\widehat{B'Q'}}=\hat{a}(\hat{\xi}(z))=:\hat{a}(z),\quad \overline{S}|_{\widehat{B'Q'}}=\hat{b}(\hat{\xi}(z))=:\hat{b}(z),\\ & \overline{W}|_{\widehat{B'Q'}}=\hat{d}(\hat{\xi}(z))=:\hat{d}(z), \end{aligned} \quad \forall\ z\in[0,z_0]. \end{equation}

Furthermore we also have by (2.33)

(3.7)\begin{equation} \begin{array}{c} \hat{a}(z), \hat{b}(z), \hat{d}(z)\in C^1([0,z_0]), \\ \hat{m}_0\leq \hat{a}(z), \hat{b}(z), \hat{d}(z)\leq\hat{M}_0,\quad \forall\ z\in[0,z_0]. \end{array} \end{equation}

By (2.6), the sound speed $c$ in terms of the coordinate variables $(z,\,t)$ is

(3.8)\begin{equation} c=c(z,t)=\sqrt{\frac{-2\kappa(1-t^2)(z+\hat{\phi}(\xi_2))}{\kappa+1-t^2}}, \end{equation}

which is a known smooth bounded positive function. Subsequently, the operators $\bar {\partial }^\pm$ are

(3.9)\begin{equation} \begin{aligned} \displaystyle \bar{\partial}^+ & =\Bigg\{-\frac{\sqrt{(\kappa+1-t^2)(1-t^2)}\cdot\overline{R}}{\kappa t}+\frac{\sqrt{(1-t^2)^3}}{c}\Bigg\}\partial_t-\frac{ct}{\sqrt{1-t^2}}\partial_z,\\ \displaystyle \bar{\partial}^- & =\Bigg\{\frac{\sqrt{(\kappa+1-t^2)(1-t^2)}\cdot\overline{S}}{\kappa t}+\frac{\sqrt{(1-t^2)^3}}{c}\Bigg\}\partial_t-\frac{ct}{\sqrt{1-t^2}}\partial_z. \end{aligned} \end{equation}

Thus we can deduce the system of variables $(\overline {R},\,\overline {S})(z,\,t)$ by (2.17)

(3.10)\begin{equation} \begin{cases} \displaystyle \partial_-\overline{R}=\frac{(\kappa+1)\overline{R}}{(1-t^2)\sqrt{\kappa+1-t^2}T_1} \cdot\frac{\overline{R}-\overline{S}}{2 t}- \frac{\kappa(3\kappa+4-4t^2)}{c(\kappa+1-t^2)T_1}\cdot\overline{R}t^2,\\ \displaystyle \partial_+\overline{S}=\frac{(\kappa+1)\overline{S}}{(1-t^2)\sqrt{\kappa+1-t^2}T_2} \cdot\frac{\overline{S}-\overline{R}}{2 t}+ \frac{\kappa(3\kappa+4-4t^2)}{c(\kappa+1-t^2)T_2}\cdot\overline{S}t^2, \end{cases} \end{equation}

where

(3.11)\begin{equation} \partial_\pm=\partial_t+\lambda_\pm\partial_z,\quad \lambda_-=-\frac{\kappa c}{(1-t^2)T_1}t^2,\quad \lambda_+=\frac{\kappa c}{(1-t^2)T_2}t^2, \end{equation}

and

(3.12)\begin{equation} T_1=\sqrt{\kappa+1-t^2}\overline{S}+\frac{\kappa(1-t^2)}{c}t,\quad T_2=\sqrt{\kappa+1-t^2}\overline{R}-\frac{\kappa(1-t^2)}{c}t. \end{equation}

Obviously, system (3.10) is a closed system for $(\overline {R},\,\overline {S})(z,\,t)$.

For the degenerate boundary value problem (3.10), (3.6), we have the following theorem.

3.2 A strong determinate domain and the $C^0$-estimates

In order to prove theorem 3.1, we need to construct a strong determinate domain for the nonlinear system (3.10) and then establish a prior estimates of solutions in this domain.

Set $\delta _0=\min \{t_0,\,1/\sqrt {2}\}$ and denote

\[ \hat{c}_0=\sqrt{\frac{-\kappa(z_0+\hat\phi(\xi_2))}{\kappa+1}},\quad \hat{c}_1=\sqrt{\frac{-2\kappa\hat\phi(\xi_2)}{\kappa+\frac{1}{2}}}. \]

Then for any $(z,\,t)\in [0,\,z_0]\times [0,\,\delta _0]$ one has by (3.8)

(3.13)\begin{equation} \hat{c}_0\leq c(z,t)\leq \hat{c}_1. \end{equation}

Now denote

(3.14)\begin{equation} \widehat{M}=1+\frac{4\sqrt{\kappa}(3\kappa+4)}{\hat{c}_0(\kappa+\frac{1}{2})\hat{m}_0},\quad \delta_1=\min\Bigg\{\delta_0, \frac{\hat{c}_0\hat{m}_0}{4\sqrt{\kappa}}, \frac{\ln 2}{\widehat{M}}\Bigg\}, \end{equation}

such that

(3.15)\begin{equation} \delta_1\leq\frac{\hat{c}_0\hat{m}_0}{4\sqrt{\kappa}},\quad e^{\widehat{M}\delta_1}\leq 2. \end{equation}

The reasons for the selections of $\widehat {M}$ and $\delta _1$ are as follows. If $(\overline {R},\,\overline {S})$ satisfy the inequality

(3.16)\begin{equation} \frac{1}{2}\hat{m}_0\leq \overline{R},\overline{S}\leq 2\hat{M}_0, \end{equation}

then we obtain by (3.12) and (3.15)

(3.17)\begin{align} |T_1|,|T_2|& \geq \sqrt{\kappa+1-t^2}\cdot\frac{1}{2}\hat{m}_0-\frac{\kappa(1-t^2)t}{\hat{c}_0} \nonumber\\ & \geq \frac{\sqrt{\kappa}}{2}\hat{m}_0-\frac{\kappa\delta_1}{\hat{c}_0}\geq \frac{\sqrt{k}\hat{m}_0}{4}, \end{align}

for $t\in [0,\,\delta _1]$. Thus by (3.14) and (3.17), the coefficients of the last two terms in (3.10) satisfy

(3.18)\begin{equation} \bigg|-\frac{\kappa(3\kappa+4-4t^2)}{c(\kappa+1-t^2)T_1}\bigg|, \bigg|\frac{\kappa(3\kappa+4-4t^2)}{c(\kappa+1-t^2)T_2}\bigg|\leq \frac{\kappa(3\kappa+4)}{\hat{c}_0(\kappa+\frac{1}{2})\frac{\sqrt{\kappa}\hat{m}_0}{4}}<\widehat{M}. \end{equation}

Moreover, thanks to the expressions of $\lambda _\pm$ in (3.11), one finds by (3.17) again that

(3.19)\begin{equation} \displaystyle \frac{-\lambda_-}{t^2}, \frac{\lambda_+}{t^2}\leq \frac{\kappa\hat{c}_1}{\frac{1}{2}\cdot\frac{\sqrt{\kappa}\hat{m}_0}{4}}=\frac{8\sqrt{\kappa}\hat{c}_1}{\hat{m}_0} =:\tilde{M}. \end{equation}

We now derive the slope of curve $\widehat {B'Q'}\cap \{t\leq \delta _1\}$ defined by (3.5). Performing a direct calculation and using (3.3) leads to

(3.20)\begin{align} \tilde{z}'(t)& =\hat{\phi}'(\xi) \cdot\Bigg(-\frac{\sqrt{1-\hat{\varpi}^2(\xi)}}{\hat\varpi(\xi)\cdot\hat\varpi'(\xi)}\Bigg) \nonumber\\ & =-\frac{\hat{c}(\xi)\sqrt{1+(\varphi'(\xi))^2}}{\hat{\varpi}(\xi)}\cdot \Bigg(-\frac{\sqrt{1-\hat{\varpi}^2(\xi)}}{\hat\varpi(\xi)\cdot\hat\varpi'(\xi)}\Bigg) =\frac{\hat{c}\sqrt{1+(\varphi')^2}}{\hat{\varpi}^2\hat\varpi'}t. \end{align}

Recalling (2.26), we denote

\[ \tilde{m}=\frac{\hat{c}_0\sqrt{1+\varphi_{0}^2}}{\varphi_1}, \]

then

(3.21)\begin{equation} \frac{\hat{c}\sqrt{1+(\varphi')^2}}{\hat{\varpi}^2\hat\varpi'}(\xi) \geq \tilde{m},\quad\forall\ \xi\in[\xi_1, \xi_2]. \end{equation}

Let $\delta =\min \{\delta _1,\,\tilde {m}/\tilde {M}\}$, that is

(3.22)\begin{equation} \delta=\min\Bigg\{t_0, \frac{1}{\sqrt{2}}, \frac{\hat{c}_0\hat{m}_0}{4\sqrt{\kappa}}, \frac{\ln 2}{\widehat{M}}, \frac{\tilde{m}}{\tilde{M}}\Bigg\}. \end{equation}

Now consider the curve $z=\bar {z}(t)$ defined by

(3.23)\begin{equation} \bar{z}(t)=\tilde{z}(\delta)-\frac{\tilde{M}}{3}\delta^3+\frac{\tilde{M}}{3}t^3,\quad \forall\ t\in[0,\delta]. \end{equation}

By (3.20)–(3.22) one deduces

\begin{align*} \bar{z}(0)& =\tilde{z}(\delta)-\frac{\tilde{M}}{3}\delta^3 =\int_{0}^\delta \frac{\hat{c}\sqrt{1+(\varphi')^2}}{\hat{\varpi}^2\hat\varpi'}t\ {\rm d}t -\frac{\tilde{M}}{3}\delta^3 \\ & \geq \int_{0}^\delta \tilde{m}t\ {\rm d}t -\frac{\tilde{M}}{3}\delta^3 =\frac{\tilde{m}}{2}\delta^2-\frac{\tilde{M}}{3}\delta^3>\frac{\tilde{m}-\tilde{M}\delta}{3}\delta^2\geq0, \end{align*}

which indicates that the intersection point of the curve $z=\bar {z}(t)$ and the line $t=0$ is to the right of point $B'$. Denote the point $(\bar {z}(0),\,0)$ by $C'$, the point $(\tilde {z}(\delta ),\,\delta )$ by $E'$ and the domain $B'C'E'$ by $\Omega$. According to the construction process, the domain $\Omega$ is a strong determinate domain for system (3.10) if (3.16) holds. We shall construct a solution for the degenerate boundary value problem (3.10), (3.6) in the whole domain $\Omega$. It is worthwhile to comment by (3.22) that the solution domain $\Omega$ is dependent only on the given initial constants but independent of the construction process of solutions.

Since system (3.10) is strictly hyperbolic and its coefficients are smooth near point $E'$, then by the classical local existence theory [Reference Li and Yu34] we have the existence of $C^1$ solutions near $E'$ in $\Omega$. In addition, this local solution satisfies the inequality in (3.16) by the initial conditions in (3.7). We next establish the a prior estimate (3.16) on $\Omega$. Let $\varepsilon \in (0,\,\delta ]$ be an arbitrary constant. Denote the domain $\Omega \cap \{(z,\,t)|\ t\geq \varepsilon \}$ by $\Omega _\varepsilon$. We have the following lemma

Lemma 3.2 Let (3.7) hold and $(\overline {R},\,\overline {S})(z,\,t)$ be a $C^1$ solution of problem (3.10), (3.6) in the domain $\Omega _\varepsilon$. Then the solution $(\overline {R},\,\overline {S})(z,\,t)$ satisfies

(3.24)\begin{equation} \frac{1}{2}\hat{m}_0\leq \overline{R}(z,t),\overline{S}(z,t)\leq 2\hat{M}_0, \quad \forall\ (z,t)\in\Omega_\varepsilon. \end{equation}

Proof. Introduce

(3.25)\begin{equation} \hat{R}=\overline{R}e^{-\widehat{M}t},\quad \hat{S}=\overline{S}e^{-\widehat{M}t}, \end{equation}

then by (3.7) and (3.15)

(3.26)\begin{equation} \hat{R}|_{\widehat{E'B'}},\hat{S}|_{\widehat{E'B'}}\geq \hat{m}_0e^{-\widehat{M}\delta}> \frac{\hat{m}_0}{2}. \end{equation}

From (3.10), the equations for $(\hat {R},\,\hat {S})$ are

(3.27)\begin{equation} \begin{cases} & \displaystyle \partial_-\hat{R}=\frac{(\kappa+1)e^{\widehat{M}t}\hat{R}}{(1-t^2)\sqrt{\kappa+1-t^2}T_1} \cdot\frac{\hat{R}-\hat{S}}{2 t}- I_1\hat{R}, \\ & \displaystyle \partial_+\hat{S}=\frac{(\kappa+1)e^{\widehat{M}t}\hat{S}}{(1-t^2)\sqrt{\kappa+1-t^2}T_1} \cdot\frac{\hat{S}-\hat{R}}{2 t}- I_2\hat{S}, \end{cases} \end{equation}

where

\[ I_1=\widehat{M}+ \frac{\kappa(3\kappa+4-4t^2)}{c(\kappa+1-t^2)T_1}t^2,\quad I_2=\widehat{M}- \frac{\kappa(3\kappa+4-4t^2)}{c(\kappa+1-t^2)T_2}t^2. \]

We now consider the level set of $t=\varepsilon ' (\varepsilon '\in [\varepsilon,\,\delta ])$ and move it down from $t=\delta$ to $t=\varepsilon$. Suppose that a point $P$ is the first time so that either $\hat {R}=\hat {m}_0/2$ or $\hat {S}=\hat {m}_0/2$ in the closed region bounded by $\widehat {E'B'}$, $\widehat {E'C'}$ and $t=t_P$. From the point $P$, we draw the negative and positive characteristic curves up to the boundary $\widehat {E'B'}$ at points $P_-$ and $P_+$, respectively. Without the loss of generality, we assume that $\hat {S}(P)=\hat {m}_0/2$ and then there has $\hat {R}>\hat {m}_0/2$ and $\hat {S}>\hat {m}_0/2$ on $\widehat {PP_+}\setminus \{P\}$. Due to the above assumptions, one has

(3.28)\begin{equation} \partial_+\hat{S}|_{P}\geq0. \end{equation}

On the other hand, we note by (3.18) that $I_2\geq 0$ on $\widehat {PP_+}$, which together with (3.27) yields

\[ \partial_+\hat{S}|_{P}=\Bigg(\frac{(\kappa+1)e^{\widehat{M}t}\hat{S}}{(1-t^2)\sqrt{\kappa+1-t^2}T_1}\Bigg) \bigg|_{P} \cdot\frac{\frac{\hat{m}_0}{2}-\hat{R}|_{P}}{2 t_P}- (I_2\hat{S})|_{P}< 0, \]

which leads to a contradiction with (3.28). Hence we have

\[ \hat{R}, \hat{S}\geq\frac{\hat{m}_0}{2},\quad \forall\ (z,t)\in\Omega_\varepsilon, \nonumber \]

which along with (3.25) arrives at

(3.29)\begin{equation} \overline{R}, \overline{S}\geq\frac{\hat{m}_0}{2}e^{\widehat{M}t}\geq \frac{\hat{m}_0}{2},\quad \forall\ (z,t)\in\Omega_\varepsilon. \end{equation}

To derive the upper bounded of $(\overline {R},\, \overline {S})$, we introduce

(3.30)\begin{equation} \widehat{R}=\overline{R}e^{\widehat{M}t},\quad \widehat{S}=\overline{S}e^{\widehat{M}t}, \end{equation}

and then

(3.31)\begin{equation} \begin{cases} \displaystyle \partial_-\widehat{R}=\frac{(\kappa+1)e^{-\widehat{M}t}\widehat{R}}{(1-t^2)\sqrt{\kappa+1-t^2}T_1} \cdot\frac{\widehat{R}-\widehat{S}}{2 t}+ I_2\widehat{R}, \\ \displaystyle \partial_+\widehat{S}=\frac{(\kappa+1)e^{-\widehat{M}t}\widehat{S}}{(1-t^2)\sqrt{\kappa+1-t^2}T_1} \cdot\frac{\widehat{S}-\widehat{R}}{2 t}+ I_1\widehat{S}. \end{cases} \end{equation}

On the boundary $\widehat {E'B'}$, one sees that

(3.32)\begin{equation} \widehat{R}|_{\widehat{E'B'}}, \widehat{S}|_{\widehat{E'B'}}\leq \hat{M}_0e^{\widehat{M}\delta}< 2\hat{M}_0. \end{equation}

Following the previous symbols, if $\widehat {S}(P)=2\hat {M}_0$ and $\widehat {R}<2\hat {M}_0$ and $\widehat {S}<2\hat {M}_0$ on $\widehat {PP_+}\setminus \{P\}$, then

(3.33)\begin{equation} \partial_+\widehat{S}|_P\leq0. \end{equation}

On the other hand, we apply the equation for $\widehat {S}$ in (3.31) to achieve

\[ \displaystyle \partial_+\widehat{S}|_P=\Bigg(\frac{(\kappa+1)e^{-\widehat{M}t}\widehat{S}}{(1-t^2)\sqrt{\kappa+1-t^2}T_1}\Bigg) \bigg|_P \cdot\frac{2\hat{M}_0-\widehat{R}|_P}{2 t_P}+ (I_1\widehat{S})|_P>0, \]

which contradicts to the inequality in (3.33). Thus we obtain

\[ \widehat{R}, \widehat{S}\leq 2\hat{M}_0,\quad \forall\ (z,t)\in\Omega_\varepsilon, \]

and then

(3.34)\begin{equation} \overline{R}, \overline{S}\leq 2\hat{M}_0e^{-\widehat{M}t}\leq 2\hat{M}_0,\quad \forall\ (z,t)\in\Omega_\varepsilon. \end{equation}

Combining with (3.29) and (3.34) completes the proof of the lemma.

3.3 Existence of global solutions in $\Omega$

In order to extend the local $C^1$ solution near point $E'$ to the whole domain $\Omega$, we need derive the a priori $C^1$ estimates for the degenerate problem (3.10), (3.6).

Owning to lemma 3.2, if the solution of problem (3.10), (3.6) exists, it is feasible and convenient to introduce

(3.35)\begin{equation} \widetilde{R}=\frac{1}{\overline{R}},\quad \widetilde{S}=\frac{1}{\overline{S}}. \end{equation}

In terms of variables $(\widetilde {R},\,\widetilde {S})$, system (3.10) can be transformed into

(3.36)\begin{equation} \begin{cases} \displaystyle \widetilde{R}_{t}-\frac{\kappa c\widetilde{S}t^{2}}{(1-t^{2})T_{3}}\widetilde{R}_{z}= \frac{\kappa+1}{(1-t^{2})\sqrt{\kappa+1-t^{2}}T_{3}}\cdot\frac{\widetilde{R}-\widetilde{S}}{2t}+ \frac{\kappa(3\kappa+4-4t^{2})}{c(\kappa+1-t^{2})T_{3}}\widetilde{R}\widetilde{S}t^{2},\\ \displaystyle \widetilde{S}_{t}+\frac{\kappa c\widetilde{R}t^{2}}{(1-t^{2})T_{4}}\widetilde{S}_{z}= \frac{\kappa+1}{(1-t^{2})\sqrt{\kappa+1-t^{2}}T_{4}}\cdot\frac{\widetilde{S}-\widetilde{R}}{2t}- \frac{\kappa(3\kappa+4-4t^{2})}{c(\kappa+1-t^{2})T_{4}}\widetilde{R}\widetilde{S}t^{2}, \end{cases} \end{equation}

or arranged to

(3.37)\begin{equation} \begin{cases} \displaystyle \tilde\partial_-\widetilde{R}=\frac{\widetilde{R}-\widetilde{S}}{2t}+H_{11}(\widetilde{R}-\widetilde{S})+H_{12}t,\\ \displaystyle \tilde\partial_+\widetilde{S}=\frac{\widetilde{S}-\widetilde{R}}{2t}+H_{21}(\widetilde{R}-\widetilde{S})+H_{22}t, \end{cases} \end{equation}

where

(3.38)\begin{equation} \begin{aligned} \displaystyle T_{3} & =\sqrt{\kappa+1-t^{2}}+\frac{\kappa (1-t^{2})}{c(z,t)}\widetilde{S}t, \quad T_{4}=\sqrt{\kappa+1-t^{2}}-\frac{\kappa (1-t^{2})}{c(z,t)}\widetilde{R}t, \\ \displaystyle H_{11} & =-\frac{\kappa(1-t^2)\widetilde{S}}{2cT_3},\qquad H_{21}=-\frac{\kappa(1-t^2)\widetilde{R}}{2cT_4}, \\ \displaystyle H_{12} & =\frac{(\kappa+2-t^2)(\widetilde{R}-\widetilde{S})}{2(1-t^2)\sqrt{\kappa+1-t^2}T_3} +\frac{\kappa(3\kappa+4-4t^2)\widetilde{R}\widetilde{S}t}{c(\kappa+1-t^2)T_3}, \\ \displaystyle H_{22} & =\frac{(\kappa+2-t^2)(\widetilde{S}-\widetilde{R})}{2(1-t^2)\sqrt{\kappa+1-t^2}T_4} -\frac{\kappa(3\kappa+4-4t^2)\widetilde{R}\widetilde{S}t}{c(\kappa+1-t^2)T_4}, \end{aligned} \end{equation}

and

(3.39)\begin{equation} \tilde\partial_\pm=\partial_t+\tilde\lambda_\pm\partial_z,\quad \tilde\lambda_-=-\frac{\kappa c\widetilde{S}}{(1-t^2)T_3}t^2,\quad \tilde\lambda_+=\frac{\kappa c\widetilde{R}}{(1-t^2)T_4}t^2. \end{equation}

Now applying the commutator relation as follows

(3.40)\begin{equation} \tilde\partial_-\tilde\partial_+-\tilde\partial_+\tilde\partial_-=\frac{\tilde\partial_-\tilde\lambda_+-\tilde\partial_+\tilde\lambda_-}{\tilde\lambda_+-\tilde\lambda_-} (\tilde\partial_+-\tilde\partial_-), \end{equation}

we obtain the equations of $\tilde \partial _+\widetilde {R}$ and $\tilde \partial _-\widetilde {S}$

(3.41)\begin{equation} \tilde\partial_-\tilde\partial_+\widetilde{R} =\tilde\partial_+\tilde\partial_-\widetilde{R} +\frac{\tilde\partial_-\tilde\lambda_+-\tilde\partial_+\tilde\lambda_-}{\tilde\lambda_+-\tilde\lambda_-} (\tilde\partial_+\widetilde{R}-\tilde\partial_-\widetilde{R}), \end{equation}

and

(3.42)\begin{equation} \tilde\partial_+\tilde\partial_-\widetilde{S} =\tilde\partial_-\tilde\partial_+\widetilde{S} +\frac{\tilde\partial_-\tilde\lambda_+-\tilde\partial_+\tilde\lambda_-}{\tilde\lambda_+-\tilde\lambda_-} (\tilde\partial_-\widetilde{S}-\tilde\partial_+\widetilde{S}). \end{equation}

Performing a tedious but straightforward calculation yields

(3.43)\begin{equation} \frac{\tilde\partial_-\tilde\lambda_+-\tilde\partial_+\tilde\lambda_-}{\tilde\lambda_+-\tilde\lambda_-}=\frac{2}{t}+h, \end{equation}

where

\begin{align*} h& =\frac{(1-t^2)T_3T_4}{\kappa c(T_3\widetilde{R}+T_4\widetilde{S})}\Bigg\{ \frac{\kappa^2(\widetilde{R}^2-\widetilde{S}^2)}{2T_3T_4} +\frac{\kappa^2(\widetilde{R}-\widetilde{S})(\widetilde{R}T_{3}^2+\widetilde{S}T_{4}^2)}{2T_{3}^2T_{4}^2} \\ & \quad+[H_{11}(\widetilde{R}-\widetilde{S})+H_{12}t]\cdot\Bigg(\frac{\kappa c}{(1-t^2)T_4}+\frac{\kappa^2t\widetilde{R}}{T_{4}^2}\Bigg) \\ & \quad+[H_{21}(\widetilde{R}-\widetilde{S})+H_{22}t]\cdot\Bigg(\frac{\kappa c}{(1-t^2)T_3}-\frac{\kappa^2t\widetilde{S}}{T_{3}^2}\Bigg) +\frac{\kappa c_t\widetilde{R}}{(1-t^2)T_4} \\ & \quad+\frac{\kappa c_t\widetilde{S}}{(1-t^2)T_3} +\frac{\kappa c\widetilde{R}[2tT_4+(1-t^2)T_{4t}]}{(1-t^2)^2T_{4}^2} +\frac{\kappa c\widetilde{S}[2tT_3-(1-t^2)T_{3t}]}{(1-t^2)^2T_{3}^2} \Bigg\}. \end{align*}

Here

\begin{align*} T_{3t}& =-\frac{t}{\sqrt{\kappa+1-t^2}} +\frac{\kappa(1-3t^2)\widetilde{S}}{c}-\frac{\kappa(1-t^2)\widetilde{S}tc_t}{c^2}, \\ T_{4t}& =-\frac{t}{\sqrt{\kappa+1-t^2}} -\frac{\kappa(1-3t^2)\widetilde{R}}{c}+\frac{\kappa(1-t^2)\widetilde{R}tc_t}{c^2}. \end{align*}

Furthermore, differentiating (3.37) give

(3.44)\begin{equation} \tilde\partial_+\tilde\partial_-\widetilde{R}=G_{11}\tilde\partial_+\widetilde{R} +G_{12}, \end{equation}

where

\begin{align*} G_{11}& =\frac{1}{2t}+H_{11}+tH_{12\widetilde{R}}, \\ G_{12}& =\Bigg(-\frac{1}{2t}-H_{11}+(\widetilde{R}-\widetilde{S})H_{11\widetilde{S}}+tH_{12\widetilde{S}}\Bigg)\cdot \Bigg(\frac{\widetilde{S}-\widetilde{R}}{2t}+H_{21}(\widetilde{R}-\widetilde{S})+H_{22}t\Bigg) \\ & \quad-\frac{\widetilde{R}-\widetilde{S}}{2t^2} +(\widetilde{R}-\widetilde{S})(H_{11t}+\tilde\lambda_+H_{11z}) +H_{12} +tH_{12t} +tH_{12z}\tilde\lambda_+, \end{align*}

and

(3.45)\begin{equation} \tilde\partial_-\tilde\partial_+\widetilde{S}=G_{21}\tilde\partial_-\widetilde{S} +G_{22}, \end{equation}

where

\begin{align*} G_{21}& =\frac{1}{2t}-H_{21}+tH_{22\widetilde{S}}, \\ G_{12}& =\Bigg(-\frac{1}{2t}+H_{21}+(\widetilde{R}-\widetilde{S})H_{21\widetilde{R}}+tH_{22\widetilde{R}}\Bigg)\cdot \Bigg(\frac{\widetilde{R}-\widetilde{S}}{2t}+H_{11}(\widetilde{R}-\widetilde{S})+H_{12}t\Bigg) \\ & \quad-\frac{\widetilde{S}-\widetilde{R}}{2t^2} +(\widetilde{R}-\widetilde{S})(H_{21t}+\tilde\lambda_-H_{21z}) +H_{22} +tH_{22t} +tH_{22z}\tilde\lambda_-. \end{align*}

We now put (3.43), (3.44) and (3.43), (3.45) into (3.41) and (3.42), respectively, to achieve

(3.46)\begin{equation} \begin{cases} \displaystyle \tilde\partial_-\tilde\partial_+\widetilde{R}=\widetilde{G}_{11}\tilde\partial_+\widetilde{R} +\widetilde{G}_{12},\\ \displaystyle \tilde\partial_+\tilde\partial_-\widetilde{S}=\widetilde{G}_{21}\tilde\partial_-\widetilde{S} +\widetilde{G}_{22}, \end{cases} \end{equation}

where

\begin{align*} \widetilde{G}_{11}& =G_{11}+\frac{2}{t}+h,\quad \widetilde{G}_{21}=G_{21}+\frac{2}{t}+h, \\ \widetilde{G}_{12}& =G_{12}-\Bigg(\frac{2}{t}+h\Bigg) \cdot \Bigg(\frac{\widetilde{R}-\widetilde{S}}{2t}+H_{11}(\widetilde{R}-\widetilde{S})+H_{12}t\Bigg), \\ \widetilde{G}_{22}& =G_{22}-\Bigg(\frac{2}{t}+h\Bigg) \cdot \Bigg(\frac{\widetilde{S}-\widetilde{R}}{2t}+H_{21}(\widetilde{R}-\widetilde{S})+H_{22}t\Bigg). \end{align*}

It is noted by lemma 3.2 and the exact expressions of $T_3,\, T_4,\, H_{ij}$ in (3.38) that the terms $h$ and $H_{ijI}\ (I=\widetilde {R},\,\widetilde {S},\,z,\,t)$ are uniformly bounded. Thus there exists a uniform positive constant $K$ such that

(3.47)\begin{equation} |\widetilde{G}_{11}|,|\widetilde{G}_{21}|\leq \frac{K}{t},\quad |\widetilde{G}_{12}|,|\widetilde{G}_{22}|\leq \frac{K}{t^2}. \end{equation}

Therefore, for any point $(z,\,t)\in \Omega _\varepsilon$, we draw the positive and negative characteristic curves up to the boundary $\widehat {E'B'}$ and integrate system (3.46) along the characteristic curves to obtain by (3.47)

(3.48)\begin{equation} \big|\tilde\partial_+\widetilde{R}\big|,\big|\tilde\partial_-\widetilde{S}\big|\leq \frac{K}{\varepsilon}, \end{equation}

for some positive constant $K$, independent of $\varepsilon$. Combining with (3.37) and (3.48) and employing lemma 3.2 concludes

(3.49)\begin{equation} \big|\tilde\partial_\pm\widetilde{R}\big|,\big|\tilde\partial_\pm\widetilde{S}\big|\leq \frac{K}{\varepsilon}, \end{equation}

for some positive constant $K$, independent of $\varepsilon$. Moreover, one recalls (3.39) to acquire

\[ \partial_t=\frac{\widetilde{R}T_3\tilde\partial_-{+}\widetilde{S}T_4\tilde\partial_+}{\widetilde{R}T_3+\widetilde{S}T_4},\quad \partial_z=\frac{(1-t^2)T_3T_4}{\kappa c(\widetilde{R}T_3+\widetilde{S}T_4)}\cdot\frac{\tilde\partial_+-\tilde\partial_-}{t^2}, \]

which together with (3.49) and (3.24) leads to

(3.50)\begin{equation} \|(\widetilde{R}, \widetilde{S})\|_{C^1(\Omega_\varepsilon)}\leq\frac{K}{\varepsilon^3}. \end{equation}

Hence we have the $C^1$ estimates of solutions by (3.35) and

Lemma 3.3 Let (3.7) hold and $(\overline {R},\,\overline {S})(z,\,t)$ be a $C^1$ solution of problem (3.10), (3.6) in the domain $\Omega _\varepsilon$. Then the solution $(\overline {R},\,\overline {S})(z,\,t)$ satisfies

(3.51)\begin{equation} \|(\overline{R},\overline{S})\|_{C^1(\Omega_\varepsilon)}\leq\frac{K}{\varepsilon^3}, \end{equation}

where $K$ is a suitable positive constant, independent of $\varepsilon$.

In view of lemmas 3.2 and 3.3, we can use the classical technique to extend the local $C^1$ solution near point $E'$ to the domain $\Omega$. In fact, for any $\varepsilon >0$, the level set of $t$ can be taken as the ‘Cauchy supports’ in the domain $\Omega _\varepsilon$ and the extension step size is dependent only on the boundary data and the $C^0$, $C^1$ norms of $(\overline {R},\,\overline {S})$ which are uniformly bounded in $\Omega _\varepsilon$. Due to the compactness of $\Omega _\varepsilon$, one can complete the extension process in a finite number of steps. By the arbitrariness of $\varepsilon$, we thus obtain the $C^1$ solution in $\Omega \setminus \{t=0\}$.

3.4 The uniformity of solutions

In this subsection, we extend the solution constructed in theorem 3.4 to the degenerate line $t=0$. The key is to establish the uniform regularity of the terms $(\overline {R}_z,\, \overline {S}_z)$ or equivalently $(\widetilde {R}_z,\,\widetilde {S}_z)$. For this purpose, we introduce new variables $(X,\,Y)$

(3.52)\begin{equation} X=\tilde\partial_+\widetilde{R}-\tilde\partial_-\widetilde{R},\quad Y=\tilde\partial_+\widetilde{S}-\tilde\partial_-\widetilde{S}, \end{equation}

so that by (3.39)

(3.53)\begin{equation} \widetilde{R}_z=\frac{X}{\tilde\lambda_+-\tilde\lambda_-}=\frac{(1-t^2)T_3T_4}{\kappa c(\widetilde{R}T_3+\widetilde{S}T_4)}\cdot\frac{X}{t^2},\quad \widetilde{S}_z=\frac{Y}{\tilde\lambda_+-\tilde\lambda_-}=\frac{(1-t^2)T_3T_4}{\kappa c(\widetilde{R}T_3+\widetilde{S}T_4)}\cdot\frac{Y}{t^2}. \end{equation}

Next we derive the equations for $(X,\,Y)$. Using the commutator relation (3.40), one arrives at

(3.54)\begin{align} \tilde\partial_-X& =\tilde\partial_-\tilde\partial_+\widetilde{R}-\tilde\partial_-\tilde\partial_-\widetilde{R} \nonumber\\ & = \big(\tilde\partial_+\tilde\partial_-\widetilde{R}-\tilde\partial_-\tilde\partial_-\widetilde{R}\big) +\frac{\tilde\partial_-\tilde\lambda_+-\tilde\partial_+\tilde\lambda_-}{\tilde\lambda_+-\tilde\lambda_-} (\tilde\partial_+\widetilde{R} -\tilde\partial_-\widetilde{R}) \nonumber\\ & = \frac{\tilde\partial_-\tilde\lambda_+-\tilde\partial_+\tilde\lambda_-}{\tilde\lambda_+-\tilde\lambda_-}X +(\tilde\lambda_+-\tilde\lambda_-)(\tilde\partial_-\widetilde{R})_z. \end{align}

A similar argument gives

(3.55)\begin{equation} \tilde\partial_+Y=\frac{\tilde\partial_-\tilde\lambda_+-\tilde\partial_+\tilde\lambda_-}{\tilde\lambda_+-\tilde\lambda_-}Y +(\tilde\lambda_+-\tilde\lambda_-)(\tilde\partial_+\widetilde{S})_z. \end{equation}

By direct calculations, one applies (3.37) and (3.53) to achieve

(3.56)\begin{align} (\tilde\lambda_+-\tilde\lambda_-)(\tilde\partial_-\widetilde{R})_z& =\frac{X-Y}{2t}+(H_{11}+tH_{12\widetilde{R}})X +[-H_{11}+H_{11\widetilde{S}}(\widetilde{R}-\widetilde{S})+tH_{12\widetilde{S}}]Y \nonumber\\ & \quad+(\tilde\lambda_+-\tilde\lambda_-)[H_{11z}(\widetilde{R}-\widetilde{S})+tH_{12z}], \end{align}

and

(3.57)\begin{align} (\tilde\lambda_+-\tilde\lambda_-)(\tilde\partial_+\widetilde{S})_z& =\frac{Y-X}{2t}+(-H_{21}+tH_{22\widetilde{R}})Y +[H_{21}+H_{21\widetilde{R}}(\widetilde{R}-\widetilde{S})+tH_{22\widetilde{R}}]X \nonumber\\ & \quad+(\tilde\lambda_+-\tilde\lambda_-)[H_{21z}(\widetilde{R}-\widetilde{S})+tH_{22z}]. \end{align}

Combining with (3.43) and (3.54)–(3.57), we obtain the equations for $(X,\,Y)$

(3.58)\begin{equation} \begin{cases} \displaystyle \tilde\partial_-X=\Bigg(\frac{5}{2t}+f_1\Bigg)X +\Bigg(-\frac{1}{2t}+f_2\Bigg)Y+f_3t^2,\\ \displaystyle \tilde\partial_+Y=\Bigg(\frac{5}{2t}+g_1\Bigg)Y +\Bigg(-\frac{1}{2t}+g_2\Bigg)X+g_3t^2, \end{cases} \end{equation}

where

\begin{align*} f_1& =h+H_{11}+tH_{12\widetilde{R}},\quad g_1=h-H_{21}+tH_{22\widetilde{S}}, \\ f_2& =-H_{11}+H_{11\widetilde{S}}(\widetilde{R}-\widetilde{S})+tH_{12\widetilde{S}}, \quad g_2=H_{21}+H_{21\widetilde{R}}(\widetilde{R}-\widetilde{S})+tH_{22\widetilde{R}}, \\ f_3& =\frac{\kappa c(\widetilde{R}T_3+\widetilde{S}T_4)}{(1-t^2)T_3T_4}\cdot [H_{11z}(\widetilde{R}-\widetilde{S})+tH_{12z}], \\ g_3& =\frac{\kappa c(\widetilde{R}T_3+\widetilde{S}T_4)}{(1-t^2)T_3T_4}\cdot [H_{21z}(\widetilde{R}-\widetilde{S})+tH_{22z}]. \end{align*}

It is noted by lemma 3.2 that the functions $f_i,\, g_i (i=1,\,2,\,3)$ are uniform bounded in $\Omega$, that is

(3.59)\begin{equation} |h|, |f_i|, |g_i|\leq \widehat{K},\quad \forall\ (z,t)\in\Omega, \end{equation}

for some uniform positive constant $\widehat {K}$.

Next we discuss the properties of $\widetilde {R}_z$ and $\widetilde {S}_z$ on the curve $\widehat {B'E'}$ near point $B'$. We differentiate $\widetilde {R}$ along $\widehat {B'E'}$ and apply (3.20) to get

(3.60)\begin{equation} \widetilde{R}_t|_{\widehat{B'E'}}+\frac{\hat{c}\sqrt{1+(\varphi')^2}}{\hat{\varpi}^2\hat\varpi'}t\cdot \widetilde{R}_z|_{\widehat{B'E'}}=-\frac{\hat{a}'(\xi)\hat{\xi}'(t)}{\hat{a}^2}, \end{equation}

which together with the fact $\hat {\xi }'(t)=-\sqrt {1-\hat {\varpi }^2}/\hat {\varpi }\hat {\varpi }'(\xi )$ by (3.5), one has

(3.61)\begin{equation} \widetilde{R}_t|_{\widehat{B'E'}}+\frac{\hat{c}\sqrt{1+(\varphi')^2}}{\hat{\varpi}^2\hat\varpi'}t\cdot \widetilde{R}_z|_{\widehat{B'E'}}= \frac{\sqrt{1-\hat{\varpi}^2}\hat{a}'}{\hat{\varpi}\hat{\varpi}'\hat{a}^2}. \end{equation}

Inserting the first relationship in (2.34) into (3.61) leads to

(3.62)\begin{align} \widetilde{R}_t|_{\widehat{B'E'}}\!+\!\frac{\hat{c}\sqrt{1\!+\!(\varphi')^2}}{\hat{\varpi}^2\hat\varpi'}t\cdot \widetilde{R}_z|_{\widehat{B'E'}}=-\frac{\hat{d}}{\hat{a}^2} +\frac{t^2\hat{d}'}{\hat{\varpi}\hat{\varpi}'\hat{a}^2} -\frac{t}{\hat{\varpi}\hat{\varpi}'\hat{a}^2} \Bigg(\frac{\kappa\sqrt{\kappa\!+\!\hat{\varpi}^2}\cos\hat\theta \varphi''}{1+(\varphi')^2}\Bigg)'. \end{align}

On the other hand, we use the equation of $\widetilde {R}$ in (3.37), the relations in (3.35) and the boundary data in (3.6) to acquire

(3.63)\begin{align} \widetilde{R}_t|_{\widehat{B'E'}} -\frac{\kappa \hat{c}t^2}{(1-t^2)\hat{T_3}\hat{b}}\widetilde{R}_z|_{\widehat{B'E'}}& =\frac{\widetilde{R}-\widetilde{S}}{2t}\bigg|_{\widehat{B'E'}} +\hat{H}_{11}(\widetilde{R}-\widetilde{S})|_{\widehat{B'E'}}+\hat{H}_{12}t \nonumber\\ & =-\frac{\hat{d}}{\hat{a}\hat{b}} +\hat{H}_{11}\cdot\frac{-2t}{\hat{a}\hat{b}}\hat{d}+\hat{H}_{12}t, \end{align}

where $\hat {T}_3,\, \hat {H}_{11}$ and $\hat {H}_{12}$ are the boundary values of $T_3$, $H_{11}$ and $H_{12}$ on $\widehat {B'E'}$, respectively. Combing (3.62) and (3.63) yields

(3.64)\begin{align} & \Bigg(\frac{\hat{c}\sqrt{1+(\varphi')^2}}{\hat{\varpi}^2\hat\varpi'} +\frac{\kappa \hat{c}t}{(1-t^2)\hat{T_3}\hat{b}}\Bigg)\widetilde{R}_z|_{\widehat{B'E'}} \nonumber\\ & \quad=\frac{1}{t}\Bigg\{\frac{\hat{d}(\hat{a}-\hat{b})}{\hat{a}^2\hat{b}} +\frac{\hat{d}'}{\hat{\varpi}\hat{\varpi}'\hat{a}^2}t^2 -\frac{t}{\hat{\varpi}\hat{\varpi}'\hat{a}^2} \Bigg(\frac{\kappa\sqrt{\kappa+\hat{\varpi}^2}\cos\hat\theta \varphi''}{1+(\varphi')^2}\Bigg)' +\frac{2\hat{H}_{11}\hat{d}}{\hat{a}\hat{b}}t-\hat{H}_{12}t\Bigg\} \nonumber\\ & \quad= \frac{2\hat{d}^2}{\hat{a}^2\hat{b}} +\frac{\hat{d}'t}{\hat{\varpi}\hat{\varpi}'\hat{a}^2} -\frac{1}{\hat{\varpi}\hat{\varpi}'\hat{a}^2} \Bigg(\frac{\kappa\sqrt{\kappa+\hat{\varpi}^2}\cos\hat\theta \varphi''}{1+(\varphi')^2}\Bigg)' +\frac{2\hat{H}_{11}\hat{d}t}{\hat{a}\hat{b}}-\hat{H}_{12}, \end{align}

which implies that the boundary value $\widetilde {R}_z$ is uniformly bounded on the curve $\widehat {B'E'}$. Here we employed the facts that $\hat {T}_3$ is uniformly positive, $\hat {H}_{11}$ and $\hat {H}_{12}$ are uniformly bounded, and the relationship $\hat {a}-\hat {b}=2t\hat {d}$. The uniform boundedness of the value $\widetilde {S}_z$ on $\widehat {B'E'}$ can be similarly verified.

Based on the uniform boundedness of $\widetilde {R}_z$ and $\widetilde {S}_z$ on $\widehat {B'E'}$, we have

Proof. Set

(3.65)\begin{equation} \widetilde{X}=\frac{X}{t^{2-\nu}},\quad \widetilde{Y}=\frac{Y}{t^{2-\nu}}. \end{equation}

According to lemma 3.2 and (3.53), we only need to show the uniform boundedness of $\widetilde {X}$ and $\widetilde {Y}$ in the whole domain $\Omega$. Thanks to (3.58), the governing equations for $\widetilde {X}$ and $\widetilde {Y}$ are

\[ \left\{ \begin{array}{@{}l} \displaystyle \tilde\partial_-\widetilde{X}=\Bigg(\dfrac{1+2\nu}{2t}+f_1\Bigg)\widetilde{X} +\Bigg(-\dfrac{1}{2t}+f_2\Bigg)\widetilde{Y}+f_3t^\nu,\\ \displaystyle \tilde\partial_+\widetilde{Y}=\Bigg(\dfrac{1+2\nu}{2t}+g_1\Bigg)\widetilde{Y} +\Bigg(-\dfrac{1}{2t}+g_2\Bigg)\widetilde{X}+g_3t^\nu, \end{array} \right. \]

or equivalently

(3.66)\begin{equation} \left\{ \begin{array}{@{}l} \displaystyle \tilde\partial_-\Bigg(t^{-\tfrac{1}{2}-\nu}\widetilde{X}\Bigg)=t^{-\tfrac{3}{2}-\nu}\Bigg(tf_1\widetilde{X} +\dfrac{f_2t-1}{2}\widetilde{Y} +f_3t^{1+\nu}\Bigg),\\ \displaystyle \tilde\partial_+\Bigg(t^{-\tfrac{1}{2}-\nu}\widetilde{Y}\Bigg)=t^{-\tfrac{3}{2}-\nu}\Bigg(tg_1\widetilde{Y} +\dfrac{g_2t-1}{2}\widetilde{X} +g_3t^{1+\nu}\Bigg). \end{array} \right. \end{equation}

Due to the uniform boundedness of $\widetilde {R}_z$ and $\widetilde {S}_z$ on $\widehat {B'E'}$, it is easy to know by (3.53) that $\widetilde {X}$ and $\widetilde {Y}$ are uniformly bounded on $\widehat {B'E'}$. Set

(3.67)\begin{equation} \widehat{C}=\max_{\widehat{B'E'}}\{|\widetilde{X}|,\ |\widetilde{Y}|\},\quad \bar{\varepsilon}=\min\Bigg\{\bar\delta,\ \frac{\nu}{4\widehat{K}}\Bigg\}>0. \end{equation}

Then the region $\Omega$ is divided into two parts $\Omega _1:=\Omega \cap \{t\leq \bar \varepsilon \}$ and $\Omega _2:=\Omega \cap \{t\geq \bar \varepsilon \}$. It is obvious that $(\widetilde {X},\,\widetilde {Y})$ are uniform bounded in the whole domain $\overline {\Omega _2}$, where $\overline {\Omega _2}$ is the closure of $\Omega _2$. Denote

(3.68)\begin{equation} M=1+2\max\{\widehat{C},\ \max_{\overline{\Omega_2}}\{|\widetilde{X}|, |\widetilde{Y}|\}\}. \end{equation}

We shall show that

(3.69)\begin{equation} \max_{\Omega_1}\{|\widetilde{X}(z,t)|,\ |\widetilde{Y}(z,t)|\}< M, \end{equation}

which leads to the uniform boundedness of $\widetilde {X}$ and $\widetilde {Y}$ in the whole domain $\Omega$. Clearly, by the chosen of $M$ in (3.68), one has

(3.70)\begin{equation} |\widetilde{X}|, |\widetilde{Y}|<\frac{1}{2}M, \quad \forall\ (z,t)\in(\widehat{B'E'}\cap\{t\leq \bar\varepsilon\})\cup (\Omega\cap\{t=\bar{\varepsilon}\}). \end{equation}

Now we use the contradiction argument to show (3.69). We move the level set of $t$ down from $t=\bar {\varepsilon }$ to $t=0$ and suppose a point $P(z,\,t)\in \Omega _1$ is the first time so that, without the loss of generality, $|\widetilde {X}(P)|=M$. From the point $P$, we draw the negative characteristic curve up to the boundary curve $\widehat {B'E'}\cap \{t\leq \bar \varepsilon \}$ or the line $t=\bar {\varepsilon }$ at point $P_-(z_-,\,t_-)$. See figure 3 for the illustration. Then there has $|\widetilde {X}|\leq M$ and $|\widetilde {Y}|\leq M$ on the negative characteristic from $P$ to $P_-$. Integrating the equation for $\widetilde {X}$ in (3.66) from $P$ to $P_-$ gives

\[ t_-^{-\frac{1}{2}-\nu}\widetilde{X}(P_-) -t^{-\frac{1}{2}-\nu}\widetilde{X}(P) =\int_{t}^{t_-} s^{-\frac{3}{2}-\nu}\Bigg(sf_1\widetilde{X}+\frac{sf_2-1}{2}\widetilde{Y}+f_3s^{1+\nu}\Bigg)\ {\rm d}s, \]

from which and the chosen of $\bar \varepsilon$ in (3.67), we get

(3.71)\begin{align} M& =|\widetilde{X}(P)|=t^{\frac{1}{2}+\nu}\bigg|t_-^{-\frac{1}{2}-\nu}\widetilde{X}(P_-)\nonumber\\ & \quad -\int_{t}^{t_-} s^{-\frac{3}{2}-\nu}\Bigg(sf_1\widetilde{X}+\frac{sf_2-1}{2}\widetilde{Y}+f_3s^{1+\nu}\Bigg)\ {\rm d}s\bigg| \nonumber\\ & \quad\leq \Bigg(\frac{t}{t_-}\Bigg)^{\frac{1}{2}+\nu}|\widetilde{X}(P_-)| + t^{\frac{1}{2}+\nu}\nonumber\\ & \quad\int_{t}^{t_-} s^{-\frac{3}{2}-\nu}\Bigg(s|f_1|\cdot|\widetilde{X}|+\frac{s|f_2|+1}{2}|\widetilde{Y}|+|f_3|s^{1+\nu}\Bigg)\ {\rm d}s \nonumber\\ & \leq \frac{1}{2}M \Bigg(\frac{t}{t_-}\Bigg)^{\frac{1}{2}+\nu}+ t^{\frac{1}{2}+\nu}\int_{t}^{t_-} s^{-\frac{3}{2}-\nu}\Bigg(\bar{\varepsilon}\widehat{K}\cdot M +\frac{\bar{\varepsilon}\widehat{K}+1}{2}M+\widehat{K}\bar{\varepsilon}^{1+\nu}\Bigg)\ {\rm d}s \nonumber\\ & \leq \frac{1}{2}M \Bigg(\frac{t}{t_-}\Bigg)^{\frac{1}{2}+\nu} + t^{\frac{1}{2}+\nu}\int_{t}^{t_-} s^{-\frac{3}{2}-\nu}\Bigg(\frac{\nu}{4} M +(\frac{\nu}{8}+\frac{1}{2})M+\frac{\nu}{4}M\Bigg)\ {\rm d}s \nonumber\\ & = \frac{1}{2}M \Bigg(\frac{t}{t_-}\Bigg)^{\frac{1}{2}+\nu} + t^{\frac{1}{2}+\nu}\Bigg(\frac{1}{2}+\frac{5\nu}{8}\Bigg)M\cdot\frac{1}{\frac{1}{2}+\nu}\Bigg(t^{-\frac{1}{2}-\nu} -t_-^{-\frac{1}{2}-\nu}\Bigg) \nonumber\\ & = M\Bigg(\frac{1}{2}-\frac{\frac{1}{2}+\frac{5\nu}{8}}{\frac{1}{2}+\nu}\Bigg) \Bigg(\frac{t}{t_-}\Bigg)^{\frac{1}{2}+\nu} +\frac{\frac{1}{2}+\frac{5\nu}{8}}{\frac{1}{2}+\nu}M\leq \frac{\frac{1}{2}+\frac{5\nu}{8}}{\frac{1}{2}+\nu}M< M, \end{align}

a contradiction. Here we have used the facts $t_-\leq \bar \varepsilon$ and $|\widetilde {X}(P_-)|\leq \frac {1}{2}M$ by (3.70). If the point $P(z,\,t)\in \Omega _1$ is the first point so that $|\widetilde {Y}(P)|=M$, we can similarly derive a contradiction. Therefore, one has $|\widetilde {X}(z,\,t)|< M$ and $|\widetilde {Y}(z,\,t)|< M$ for any point $(z,\,t)\in \Omega _1$, that is (3.69) holds. The proof of the lemma is completed.

Figure 3. The region $B'C'D'$.

Based on lemma 3.5, we have the uniform boundedness of $\overline {W}=(\overline {R}-\overline {S})/2t$.

Proof. Set

\[ \widetilde{W}=\frac{\widetilde{R}-\widetilde{S}}{2t}=-\frac{\overline{W}}{\overline{R}\cdot\overline{S}}, \]

from which and (3.24), we only need to establish the uniform boundedness of $\widetilde {W}$. With the aid of (3.37) and (3.65), the equation for $\widetilde {W}$ can be easily deduced

(3.72)\begin{equation} \begin{aligned} & \displaystyle \tilde\partial_-\widetilde{W}=(H_{11}-H_{21})\widetilde{W}+\frac{1}{2}(H_{12}-H_{22}) +\frac{1}{2}t^{1-\nu}\widetilde{Y}, \\ & \displaystyle \tilde\partial_+\widetilde{W}=(H_{11}-H_{21})\widetilde{W}+\frac{1}{2}(H_{12}-H_{22}) +\frac{1}{2}t^{1-\nu}\widetilde{X}. \end{aligned} \end{equation}

Note by (3.6) and (3.7) that the boundary data of $\widetilde {W}$ on $\widehat {B'E'}$ is

\[ \widetilde{W}|_{\widehat{B'E'}}=-\frac{\hat{d}(z)}{\hat{a}(z)\hat{b}(z)}, \]

which is uniformly bounded. Then the uniform boundedness of $\widetilde {W}$ can be obtained by integrating the equation for $\widetilde {W}$ along the negative characteristic curve from a point $(z,\,t)$ in $\Omega$ to the boundary $\widehat {B'E'}$ and employing the uniform boundedness of $H_{ij}$ and $(\widetilde {X},\,\widetilde {Y})$. The proof of the lemma is finished.

Based on the properties of solutions in remark 3.7, we can establish the uniform regularity of $(\overline {R},\, \overline {S},\, \overline {W})$.

Proof. Let $P_1(z_1,\,0)$ and $P_2(z_2,\,0)$ $(z_1< z_2)$ be any two points on the degenerate curve $\widehat {B'C'}$. We draw the positive characteristic from $(z_1,\,0)$ and positive characteristic from $(z_2,\,0)$ and denote the intersection point of these two characteristics by $P_m(z_m,\,t_m)$. Recalling the expressions of $\tilde \lambda _\pm$ in (3.39), one obtains the relations for $t_m$ and $z_m$

\[ z_m=z_1+\int_{0}^{t_m}\frac{\kappa c\widetilde{R}t^2}{(1-t^2)T_4}\ {\rm d}t =z_2-\int_{0}^{t_m}\frac{\kappa c\widetilde{S}t^2}{(1-t^2)T_3}\ {\rm d}t, \]

which together with (3.24) yields

(3.73)\begin{equation} \underline{K}t_m\leq|z_2-z_1|^{\frac{1}{3}}\leq \overline{K}t_m, \end{equation}

for some uniform positive constants $\underline {K}$ and $\overline {K}$. Since $\widetilde {W}$ is uniformly bounded, then we use (3.37) and lemma 3.5 to acquire

(3.74)\begin{equation} |\tilde\partial_-\widetilde{R}|,|\tilde\partial_+\widetilde{S}|, |\widetilde{R}_t|, |\widetilde{S}_t|, t^{\frac{1}{2}}|\widetilde{R}_z|, t^{\frac{1}{2}}|\widetilde{S}_z|\leq \overline{M}_1, \end{equation}

for some constant $\overline {M}_1>0$. Therefore one has by Remark 2 and (3.73)–(3.74)

(3.75)\begin{align} |\widetilde{R}(z_1,0)-\widetilde{R}(z_2,0)|& =|\widetilde{S}(z_1,0)-\widetilde{R}(z_2,0)| \nonumber\\ & \quad\leq |\widetilde{S}(z_1,0)-\widetilde{S}(z_m,t_m)|+|\widetilde{S}(z_m,t_m)\nonumber\\& \quad -\widetilde{R}(z_m,t_m)| +|\widetilde{R}(z_m,t_m)-\widetilde{R}(z_2,0)| \nonumber\\ & \quad\leq \overline{M}_1t_m +2\sup_{\Omega}|\widetilde{W}| \cdot t_m +\overline{M}_1 t_m\nonumber\\& \quad\leq \frac{2\overline{M}_1+2\sup_{\Omega}|\widetilde{W}|}{\overline{K}}|z_2-z_1|^{\frac{1}{3}}. \end{align}

Now for any two points $P_1(z_1,\,t_1)$ and $P_2(z_2,\,t_2)$ $(z_1\leq z_2,\, 0\leq t_1\leq t_2)$ in the region $\Omega$, if $t_1\geq (z_2-z_1)$, then by (3.74)

(3.76)\begin{align} |\widetilde{R}(z_1,t_1)-\widetilde{R}(z_2,t_2)|& \leq |\widetilde{R}(z_1,t_1)-\widetilde{R}(z_2,t_1)| +|\widetilde{R}(z_2,t_1)-\widetilde{R}(z_2,t_2)| \nonumber\\ & \leq \sup_{\Omega}|\widetilde{R}_z|\cdot|z_1-z_2|+ \sup_{\Omega}|\widetilde{R}_t|\cdot|t_1-t_2| \nonumber\\ & \leq \overline{M}_1t_{1}^{-\frac{1}{2}}|z_1-z_2|+ \overline{M}_1|t_1-t_2| \!\leq\! \overline{M}_1|z_1\!-\!z_2|^{\frac{1}{2}}\!+\! \overline{M}_1|t_1-t_2| \nonumber\\ & \leq 2\overline{M}_1|(z_1,t_1)-(z_2,t_2)|^{\frac{1}{3}}, \end{align}

and if $t_1<(z_2-z_1)$, then by (3.75) and (3.74) again

(3.77)\begin{align} & |\widetilde{R}(z_1,t_1)-\widetilde{R}(z_2,t_2)| \nonumber\\ & \quad\leq |\widetilde{R}(z_1,t_1)-\widetilde{R}(z_1,0)| +|\widetilde{R}(z_1,0)-\widetilde{R}(z_2,0)| \nonumber\\ & \qquad+ |\widetilde{R}(z_2,0)-\widetilde{R}(z_2,t_1)| +|\widetilde{R}(z_2,t_1)-\widetilde{R}(z_2,t_2)| \nonumber\\ & \quad\leq \sup_{\Omega}|\widetilde{R}_t|\cdot t_1 +\frac{2\overline{M}_1+2\sup_{\Omega}|\widetilde{W}|}{\overline{K}}|z_2-z_1|^{\frac{1}{3}} + \sup_{\Omega}|\widetilde{R}_t|\cdot t_1 +\sup_{\Omega}|\widetilde{R}_t|\cdot |t_1-t_2| \nonumber\\ & \quad\leq \Bigg(3\overline{M}_1+\frac{2\overline{M}_1+2\sup_{\Omega}|\widetilde{W}|}{\overline{K}}\Bigg) |(z_1,t_1)-(z_2,t_2)|^{\frac{1}{3}}. \end{align}

We combine (3.76) and (3.77) to achieve the uniform $C^{\frac {1}{3}}$ continuity of $\widetilde {R}$ in the whole domain $\Omega$. A similar argument yields the uniform regularity for $\widetilde {S}$. The function $\widetilde {W}$ can be handled in a similar way by (3.72). The proof of the lemma is ended by the relationship between $(\overline {R},\,\overline {S},\,\overline {W})$ and $(\widetilde {R},\,\widetilde {S},\,\widetilde {W})$ and Lemma 3.2.

Thanks to lemma 3.8, we can improve the uniform regularity of $(\overline {R},\, \overline {S},\, \overline {W})$.

Proof. It suffices to show that $(\widetilde {R},\, \widetilde {S})$ are uniformly $C^{1-\nu }$ continuous and $\widetilde {W}$ is uniformly $C^{\frac {2-\nu }{3}}$ continuous for any $\nu \in (0,\,1)$. Let $M^*$ be a uniform positive constant such that for $i,\,j=1,\,2$

(3.78)\begin{equation} \begin{array}{r} \displaystyle \bigg|\dfrac{\kappa c\widetilde{S}}{(1-t^2)T_3}\bigg|, \bigg|\dfrac{\kappa c\widetilde{R}}{(1-t^2)T_4}\bigg|, |2tH_{i1}+1|\leq M^*, \\ |\widetilde{W}|, |H_{ij\widetilde{R}}|, |H_{ij\widetilde{S}}|, |H_{ijc}c_z|\leq M^*. \end{array} \end{equation}

The above inequalities hold by lemma 3.2 and the exact expressions of $T_3,\, T_4$ and $H_{ij}$ in (3.38). Moreover, in view of lemma 3.5, we see that there exists a uniform positive constant $\overline {M}_2$ depending on the parameter $\nu$ such that

(3.79)\begin{equation} |\widetilde{X}|, |\widetilde{Y}|, t^{\nu}|\widetilde{R}_z|, t^{\nu}|\widetilde{S}_z|\leq \overline{M}_2. \end{equation}

We first improve the uniform regularity of $\widetilde {W}$. Recalling the first equation of $\widetilde {W}$ in (3.72) leads to

(3.80)\begin{equation} \widetilde{W}_t=\frac{\kappa c\widetilde{S}\cdot t}{2(1-t^2)T_3}(\widetilde{R}_z-\widetilde{S}_z) +(H_{11}-H_{21})\widetilde{W} +\frac{1}{2}(H_{12}-H_{22}) +\frac{1}{2}t^{1-\nu}\widetilde{Y}. \end{equation}

Integrating (3.80) from $(z_i,\, 0)$ to $(z_i,\, t_m)\ (i=1,\,2)$ yield

(3.81)\begin{align} & \widetilde{W}(z_i,0)=\widetilde{W}(z_i,t_m) \nonumber\\ & \quad- \int_{0}^{t_m}\Bigg\{\frac{\kappa c\widetilde{S}(t^\nu\widetilde{R}_z-t^\nu\widetilde{S}_z)}{2(1-t^2)T_3}t^{1-\nu} +(H_{11}-H_{21})\widetilde{W}\nonumber\\& \quad +\frac{1}{2}(H_{12}-H_{22}) +\frac{1}{2}t^{1-\nu}\widetilde{Y}\Bigg\}\ {\rm d}t, \end{align}

from which one gets

(3.82)\begin{align} & |\widetilde{W}(z_1,0)-\widetilde{W}(z_2,0)|\leq|\widetilde{W}(z_1,t_m)-\widetilde{W}(z_2,t_m)| \nonumber\\ & \quad+ \int_{0}^{t_m}\Bigg\{\bigg|\frac{\kappa c\widetilde{S}(t^\nu\widetilde{R}_z-t^\nu\widetilde{S}_z)}{2(1-t^2)T_3}(z_1,t)\bigg| +\bigg|\frac{\kappa c\widetilde{S}(t^\nu\widetilde{R}_z-t^\nu\widetilde{S}_z)}{2(1-t^2)T_3}(z_2,t)\bigg|\Bigg\}t^{1-\nu}\ {\rm d}t \nonumber\\ & \quad+ \int_{0}^{t_m} |(H_{11}-H_{21})(z_1,t)|\cdot|\widetilde{W}(z_1,t)-\widetilde{W}(z_2,t)| \ {\rm d}t \nonumber\\ & \quad+ \int_{0}^{t_m} |\widetilde{W}(z_2,t)|\cdot|(H_{11}-H_{21})(z_1,t)-(H_{11}-H_{21})(z_2,t)| \ {\rm d}t \nonumber\\ & \quad+ \int_{0}^{t_m} \frac{1}{2}|(H_{12}-H_{22})(z_1,t)-(H_{12}-H_{22})(z_2,t)| \ {\rm d}t \nonumber\\ & \quad+ \int_{0}^{t_m} \frac{1}{2}\big(|\widetilde{Y}(z_1,t)|+|\widetilde{Y}(z_2,t)|\big) t^{1-\nu}\ {\rm d}t. \end{align}

For the first term of the right-hand side of (3.82), we find by (3.79) and (3.73) that

(3.83)\begin{align} |\widetilde{W}(z_1,t_m)-\widetilde{W}(z_2,t_m)|& \leq \frac{|\widetilde{R}(z_1,t_m)-\widetilde{R}(z_2,t_m)| +|\widetilde{S}(z_1,t_m)-\widetilde{S}(z_2,t_m)|}{t_m} \nonumber\\ & \leq\frac{|t_{m}^\nu\widetilde{R}_z(z',t_m)|\cdot|z_1-z_2| +|t_{m}^\nu\widetilde{S}_z(z'',t_m)|\cdot|z_1-z_2|}{t_{m}^{1+\nu}} \nonumber\\ & \leq 2\overline{M}_2\frac{|z_1-z_2|}{t_{m}^{1+\nu}}\leq 2\overline{M}_2\overline{K}^{1+\nu}|z_1-z_2|^{\frac{2-\nu}{3}}. \end{align}

Moreover, since the functions $(\widetilde {R},\, \widetilde {S},\, \widetilde {W})(z,\,t)$ are uniformly $C^{\frac {1}{3}}$ continuous by lemma 3.8, then one obtains

(3.84)\begin{equation} |\widetilde{R}(z_1,t)-\widetilde{R}(z_2,t)|, |\widetilde{S}(z_1,t)-\widetilde{S}(z_2,t)|, |\widetilde{W}(z_1,t)-\widetilde{W}(z_2,t)|\leq \overline{M}_3|z_1-z_2|^{\frac{1}{3}}, \end{equation}

for some uniform positive constant $\overline {M}_3$. It follows by (3.78) and (3.84) that for $i=1,\,2$

(3.85)\begin{align} & |(H_{1i}-H_{2i})(z_1,t)-(H_{1i}-H_{2i})(z_2,t)| \nonumber\\ & \quad\leq|H_{1i\widetilde{R}}-H_{2i\widetilde{R}}|\cdot|\widetilde{R}(z_1,t)-\widetilde{R}(z_2,t)| +|H_{1i\widetilde{S}}-H_{2i\widetilde{S}}|\cdot|\widetilde{S}(z_1,t)-\widetilde{S}(z_2,t)| \nonumber\\ & \qquad + |H_{1ic}-H_{2ic}|\cdot|c_z|\cdot|z_1-z_2 | \nonumber\\ & \quad\leq 4M^*\overline{M}_3|z_1-z_2|^{\frac{1}{3}} +2M^*|z_1-z_2|. \end{align}

We now put (3.83)–(3.84) into (3.82) and make use of (3.78)–(3.79) and (3.73) to acquire

(3.86)\begin{align} & |\widetilde{W}(z_1,0)-\widetilde{W}(z_2,0)|\leq 2\overline{M}_2\overline{K}^{1+\nu}|z_1-z_2|^{\frac{2-\nu}{3}} + \int_{0}^{t_m} (2M^*+1)\overline{M}_2 t^{1-\nu}\ {\rm d}t \nonumber\\ & \quad+ \int_{0}^{t_m} 2M^*\overline{M}_3 |z_1-z_2|^{\frac{1}{3}}\ {\rm d}t\nonumber\\& \quad + \int_{0}^{t_m} \Bigg(M^*+\frac{1}{2}\Bigg) \Bigg(4M^*\overline{M}_3|z_1-z_2|^{\frac{1}{3}} +2M^*|z_1-z_2|\Bigg)\ {\rm d}t \nonumber\\ & \quad\leq 2\overline{M}_2\overline{K}^{1+\nu}|z_1-z_2|^{\frac{2-\nu}{3}} + \frac{(2M^*+1)\overline{M}_2}{2-\nu} t_{m}^{2-\nu} + 2M^*\overline{M}_3 |z_1-z_2|^{\frac{1}{3}}t_{m} \nonumber\\ & \quad+ (M^*+1) \Bigg(4M^*\overline{M}_3|z_1-z_2|^{\frac{1}{3}} +2M^*|z_1-z_2|\Bigg)t_m\leq \overline{M}_4|z_1-z_2|^{\frac{2-\nu}{3}}, \end{align}

for some uniform positive constant $\overline {M}_4$ depending on $\nu$. Based on (3.86), one can repeat the same process as in lemma 3.8 to get that the function $\widetilde {W}$ is uniformly $C^{\frac {2-\nu }{3}}$ continuous in the whole domain $\Omega$.

For the functions $\widetilde {R}$ and $\widetilde {S}$, we recall (3.37) to achieve

(3.87)\begin{equation} \begin{aligned} \displaystyle \widetilde{R}_{t} & =\frac{\kappa c\widetilde{S}t^{2}}{(1-t^{2})T_{3}}\widetilde{R}_{z} + (2tH_{11}+1)\widetilde{W} +H_{12}t, \\ \displaystyle \widetilde{S}_{t} & =-\frac{\kappa c\widetilde{R}t^{2}}{(1-t^{2})T_{4}}\widetilde{S}_{z} +(2tH_{21}-1) \widetilde{W} +H_{22}t. \end{aligned} \end{equation}

One integrates the equation of $\widetilde {R}$ in (3.87) from $(z_i,\, 0)$ to $(z_i,\, t_m)\ (i=1,\,2)$ to find that

(3.88)\begin{equation} \widetilde{R}(z_i, 0)=\widetilde{R}(z_i, t_m) - \int_{0}^{t_m}\Bigg\{\frac{\kappa c\widetilde{S}t^{2}\widetilde{R}_{z}}{(1-t^{2})T_{3}}+ (2tH_{11}+1)\widetilde{W} +H_{12}t \Bigg\}\quad (z_i, t)\ {\rm d}t, \end{equation}

from which we have

(3.89)\begin{align} & |\widetilde{R}(z_1, 0)-\widetilde{R}(z_2, 0)| \leq|\widetilde{R}(z_1, t_m)-\widetilde{R}(z_2, t_m)| \nonumber\\ & \quad+ \int_{0}^{t_m}\Bigg\{\bigg|\frac{\kappa c\widetilde{S}}{(1-t^2)T_3}(z_1,t)\bigg|\cdot|t^{\nu}\widetilde{R}_z(z_1,t)| \nonumber\\& \quad + \bigg|\frac{\kappa c\widetilde{S}}{(1-t^2)T_3}(z_2,t)\bigg|\cdot|t^{\nu}\widetilde{R}_z(z_2,t)|\Bigg\} t^{2-\nu}\ {\rm d}t \nonumber\\ & \quad+ \int_{0}^{t_m}\Bigg\{ |2tH_{11}(z_1,t)+1|\cdot |\widetilde{W}(z_1,t)-\widetilde{W}(z_2,t)| \nonumber\\& \quad +2t|\widetilde{W}|\cdot|H_{11}(z_1,t)-H_{11}(z_2,t)|\Bigg\} \ {\rm d}t \nonumber\\ & \quad+ \int_{0}^{t_m}|H_{12}(z_1,t)-H_{12}(z_2,t)|t \ {\rm d}t. \end{align}

Thanks to (3.78) and (3.79), we obtain

(3.90)\begin{align} & |\widetilde{R}(z_1, t_m)-\widetilde{R}(z_2, t_m)|=|\widetilde{R}_z(z', t_m)|\cdot|z_1-z_2| =|t_{m}^\nu\widetilde{R}_z(z', t_m)|\cdot\frac{|z_1-z_2|}{t_{m}^\nu} \nonumber\\ & \quad\leq \overline{M}_2 \frac{|z_1-z_2|}{t_{m}^\nu} \leq \overline{M}_2 \overline{K}^\nu\frac{|z_1-z_2|}{|z_1-z_2|^\nu} = \overline{M}_2 \overline{K}^\nu|z_1-z_2|^{1-\nu}, \end{align}

and

(3.91)\begin{equation} \bigg|\frac{\kappa c\widetilde{S}}{(1-t^2)T_3}(z_1,t)\bigg|\cdot|t^{\nu}\widetilde{R}_z(z_1,t)| + \bigg|\frac{\kappa c\widetilde{S}}{(1-t^2)T_3}(z_2,t)\bigg|\cdot|t^{\nu}\widetilde{R}_z(z_2,t)|\leq 2M^*\overline{M}_2. \end{equation}

In addition, it suggests by (3.78) that for $i=1,\,2$

(3.92)\begin{align} |H_{1i}(z_1,t)-H_{1i}(z_2,t)|& \leq |H_{1i\widetilde{R}}|\cdot|\widetilde{R}(z_1,t)-\widetilde{R}(z_2,t)| \nonumber\\ & \quad+|H_{1i\widetilde{S}}|\cdot|\widetilde{S}(z_1,t)-\widetilde{S}(z_2,t)| +|H_{1ic}|\cdot|c(z_1,t)-c(z_2,t)| \nonumber\\ & \leq 2M^*\overline{M}_2\frac{|z_1-z_2|}{t^\nu} +M^*|z_1-z_2|. \end{align}

Due to the uniform $C^{\frac {2-\nu }{3}}$-continuity of $\widetilde {W}$, one gets

(3.93)\begin{equation} |\widetilde{W}(z_1,t)-\widetilde{W}(z_2,t)|\leq \overline{M}_5|z_1-z_2|^{\frac{2-\nu}{3}}, \end{equation}

for some uniform positive constant $\overline {M}_5$ depending on $\nu$. Inserting (3.90)–(3.93) into (3.89) and employing (3.73) and (3.78) concludes

(3.94)\begin{align} & |\widetilde{R}(z_1, 0)-\widetilde{R}(z_2, 0)| \leq \overline{M}_2\overline{K}^\nu|z_1-z_2|^{1-\nu} +\int_{0}^{t_m}2M^*\overline{M}_2t^{2-\nu}\ {\rm d}t \nonumber\\ & \qquad+ \int_{0}^{t_m}\Bigg\{M^*\cdot \overline{M}_5|z_1-z_2|^{\frac{2-\nu}{3}}\nonumber\\& \quad +t\cdot(2M^*+1)\Bigg(2M^*\overline{M}_2\frac{|z_1-z_2|}{t^\nu} +M^*|z_1-z_2|\Bigg)\Bigg\}\ {\rm d}t \nonumber\\ & \quad= \overline{M}_2\overline{K}^\nu|z_1-z_2|^{1-\nu} +\frac{2M^*\overline{M}_2}{3-\nu}t_{m}^{3-\nu}+ M^* \overline{M}_5|z_1-z_2|^{\frac{2-\nu}{3}}t_m \nonumber\\ & \qquad+ (2M^*+1)M^*(2\overline{M}_2+1)|z_1-z_2| \nonumber\\ & \quad\leq \overline{M}_2\overline{K}^\nu|z_1-z_2|^{1-\nu} +\frac{M^*\overline{M}_2}{\underline{K}^{3-\nu}}|z_1-z_2|^{1-\frac{\nu}{3}}+ \frac{M^*\overline{M}_5}{\underline{K}}|z_1-z_2|^{1-\frac{\nu}{3}} \nonumber\\ & \qquad+ (2M^*+1)M^*(2\overline{M}_2+1)|z_1-z_2| \leq \overline{M}_6|z_1-z_2|^{1-\nu}, \end{align}

for some uniform positive constant $\overline {M}_6$. From (3.94), we can use similar arguments as in Lemma 3.8 to get that the function $\widetilde {R}$ is uniformly $C^{1-\nu }$ continuous in the whole domain $\Omega$. Analogously, the function $\widetilde {S}$ is also uniformly $C^{1-\nu }$ continuous. The proof of the lemma is complete.

Finally, we draw the positive characteristic from the point $C'(\bar {z}(0),\,0)$ up to the boundary $\widehat {B'E'}$ and denote the intersection point by $D'(\tilde {z}(t_d),\,t_d)$. Taking $\bar \delta =t_d$ finishes the proof of theorem 3.1.

4.1 Inversion

Thanks to theorem 3.1, we obtain the functions $(\overline {R},\,\overline {S})(z,\,t)$ defined on the whole region $B'C'D'$. To acquire a solution in the self-similar $(\xi,\,\eta )$ plane, it is necessary to establish the coordinate functions $\xi =\xi (z,\,t)$ and $\eta =\eta (z,\,t)$ and discuss their inversion. We first construct the function $\theta (z,\,t)$. Applying (2.18) and (3.9) gives

(4.1)\begin{equation} \partial_-\theta=-\frac{\kappa t}{\sqrt{1-t^2}T_1}\Bigg(\frac{t \overline{S}}{\kappa\sqrt{\kappa+1-t^2}}+\frac{\varpi^2}{c}\Bigg). \end{equation}

For any point $(z^*,\,t^*)$ in the region $B'C'D'$, we draw the negative characteristic $z=z_-(t;t^*,\,z^*)\ (t\geq t^*)$ up to the boundary $\widehat {B'D'}$ at a unique point $(\tilde {z}(\hat {t}^*),\, \hat {t}^*)$ satisfying

(4.2)\begin{equation} \begin{cases} \displaystyle \frac{{\rm d}z_-(t;t^*,z^*)}{{\rm d}t}=-\frac{\kappa ct^2}{(1-t^2)T_1}(z_-(t;t^*,z^*), t), \\ z_-(t^*;t^*,z^*)=z^*, \end{cases} \quad z_-(\hat{t}^*;t^*,z^*)=\tilde{z}(\hat{t}^*). \end{equation}

We integrate (4.1) along the negative characteristic $z=z_-(t;t^*,\,z^*)$ from the point $(z^*,\,t^*)$ to the point $(\tilde {z}(\hat {t}^*),\, \hat {t}^*)$ and use the boundary data of $\theta$ on $\widehat {B'D'}$ to get

(4.3)\begin{equation} \theta(z^*,t^*)=\hat{\theta}(\hat{\xi}(\tilde{z}(\hat{t}^*))) +\int_{t^*}^{\hat{t}^*}\frac{\kappa t}{\sqrt{1-t^2}T_1}\Bigg(\frac{t \overline{S}}{\kappa\sqrt{\kappa+1-t^2}}+\frac{\varpi^2}{c}\Bigg)(z_-(t;t^*,z^*), t)\ {\rm d}t \end{equation}

Hence we have the function $\theta (z,\,t)$ defined on the whole region $B'C'D'$.

Recalling the coordinate transformation (3.4) and employing (2.19)–(2.20), we see that

(4.4)\begin{equation} \xi_t=\frac{-c\sin\theta t}{(1-t^2)J},\quad \eta_t=\frac{c\cos\theta t}{(1-t^2)J},\quad \xi_z=\frac{\varpi_\eta}{J},\quad \eta_z=\frac{-\varpi_\xi}{J}, \end{equation}

where

(4.5)\begin{equation} \begin{aligned} & \displaystyle J=\phi_\xi\varpi_\eta-\phi_\eta\varpi_\xi =-\frac{c\sqrt{\kappa+1-t^2}}{2\kappa(1-t^2)}(\overline{R}+\overline{S}), \\ & \displaystyle \varpi_\xi=\cos\theta\frac{\sqrt{\kappa+1-t^2}}{\kappa}\overline{W} -\cos\theta\frac{1-t^2}{c} -\sin\theta \frac{\sqrt{\kappa+1-t^2}}{2\kappa\sqrt{1-t^2}}(\overline{R}+\overline{S}), \\ & \displaystyle \varpi_\eta=\sin\theta\frac{\sqrt{\kappa+1-t^2}}{\kappa}\overline{W} -\sin\theta\frac{1-t^2}{c} +\cos\theta \frac{\sqrt{\kappa+1-t^2}}{2\kappa\sqrt{1-t^2}}(\overline{R}+\overline{S}). \end{aligned} \end{equation}

Therefore it suggests that

(4.6)\begin{equation} \partial_-\xi=\frac{\kappa t(\cos\theta t+\sin\theta\sqrt{1-t^2})}{\sqrt{1-t^2}T_1},\quad \partial_-\eta=\frac{\kappa t(\sin\theta t-\cos\theta\sqrt{1-t^2})}{\sqrt{1-t^2}T_1}. \end{equation}

For any point $(z^*,\,t^*)$ in the region $B'C'D'$, one thus finds by integrating (4.6) along the negative characteristic that

(4.7)\begin{equation} \begin{cases} \displaystyle \xi(z^*,t^*)=\hat{\xi}(\tilde{z}(\hat{t}^*))-\int_{t^*}^{\hat{t}^*} \frac{\kappa t(\cos\theta t+\sin\theta\sqrt{1-t^2})}{\sqrt{1-t^2}T_1}(z_-(t;t^*,z^*), t)\ {\rm d}t, \\ \displaystyle \eta(z^*,t^*)=\varphi(\hat{\xi}(\tilde{z}(\hat{t}^*)))-\int_{t^*}^{\hat{t}^*} \frac{\kappa t(\sin\theta t-\cos\theta\sqrt{1-t^2})}{\sqrt{1-t^2}T_1}(z_-(t;t^*,z^*), t)\ {\rm d}t, \end{cases} \end{equation}

The arbitrariness of $(z^*,\,t^*)$ indicates that the expressions in (4.7) define two functions $\xi =\xi (z,\,t)$ and $\eta =\eta (z,\,t)$ on the whole region $B'C'D'$. Furthermore, the mapping $(z,\,t)\mapsto (\xi,\,\eta )$ is a global one-to-one mapping, which comes from the following fact by (4.5)

\[ (\phi_\xi,\phi_\eta)\cdot(\varpi_\eta, -\varpi_\xi)=J=-\frac{c\sqrt{\kappa+1-t^2}}{2\kappa(1-t^2)}(\overline{R}+\overline{S})<0. \]

The above inequality implies that $\phi$ is a strictly decreasing function along each level curve of $(1-\varpi )\geq 0$.

4.2 Proof of theorem 2.2

Owning to the global one-to-one property of the mapping $(z,\,t)\mapsto (\xi,\,\eta )$, we then obtain the functions $t=\check {t}(\xi,\,\eta )$ and $z=\check {z}(\xi,\,\eta )$. Moreover, one also has

(4.8)\begin{equation} \check{t}_\xi=\frac{\eta_z}{j},\quad \check{t}_\eta=\frac{-\xi_z}{j}, \quad \check{z}_\xi=\frac{-\eta_t}{j}, \quad \check{z}_\eta=\frac{\xi_t}{j}, \end{equation}

where $j=\xi _t\eta _z-\eta _t\xi _z$ and $(\xi _t,\, \xi _z)$, $(\eta _t,\,\eta _z)$ are given in (4.4). Now we define the functions $(c,\,\theta,\,\varpi )$ in terms of variables $(\xi,\,\eta )$ as follows

(4.9)\begin{equation} c=c(\check{z}(\xi,\eta), \check{t}(\xi,\eta)),\quad \theta=\theta(\check{z}(\xi,\eta), \check{t}(\xi,\eta)), \ \varpi=\sqrt{1-\check{t}^2(\xi,\eta)},\ \forall\ (\xi,\eta)\in \overline{BCD}, \end{equation}

where the region $BCD$ is bounded by the curves $\widehat {BC}$, $\widehat {CD}$, and $\widehat {BD}$. Here the curve $\widehat {BC}$ is defined by

(4.10)\begin{equation} \widehat{BC}=\{(\xi,\eta)|\ \varpi(\xi,\eta)=1,\ \xi\in[\xi_*,\xi_2]\}, \end{equation}

where $\xi _*=\xi (\bar {z}(0),\,0)$, and $\widehat {CD}$ is defined by

(4.11)\begin{equation} \widehat{CD}=\{(\xi,\eta)|\ \check{z}(\xi,\eta)=z_+(\check{t}(\xi,\eta); \bar\delta, \tilde{z}(\bar\delta)),\ \xi\in[\xi_{**},\xi_*]\}, \end{equation}

where the function $z_+(t;\bar \delta,\, \tilde {z}(\bar \delta ))$ is the solution of the following ODE problem

(4.12)\begin{equation} \begin{cases} \displaystyle \frac{{\rm d}z_+(t; \bar\delta, \tilde{z}(\bar\delta))}{{\rm d}t}=\frac{\kappa ct^2}{(1-t^2)T_2}(z_+(t; \bar\delta, \tilde{z}(\bar\delta)),t), \\ z_+(\bar\delta; \bar\delta, \tilde{z}(\bar\delta))=\tilde{z}(\bar\delta), \end{cases} \quad t\in[0,\bar\delta], \end{equation}

and the number $\xi _{**}$ satisfies

\[ \check{z}(\xi_{**},\varphi(\xi_{**}))=z_+(\check{t}(\xi_{**},\varphi(\xi_{**})); \bar\delta, \tilde{z}(\bar\delta)). \]

The coordinates of points $C$ and $D$ in $(\xi,\,\eta )$ plane are $(\xi _*,\, \eta (\bar {z}(0),\,0))$ and $(\xi _{**},\,\varphi (\xi _{**}))$, respectively. According to the construction of $(\xi (z,\,t),\, \eta (z,\,t))$, we can see that the functions $(c(\xi,\,\eta ),\, \theta (\xi,\,\eta ),\,\varpi (\xi,\,\eta ))$ defined in (4.9) satisfy the boundary conditions in (2.22). From (4.9), we then define the functions $(\omega,\, \alpha,\, \beta )(\xi,\,\eta )$ as follows

(4.13)\begin{equation} \begin{cases} \omega(\xi,\eta)=\arcsin\varpi(\xi,\eta),\\ \alpha(\xi,\eta)=\theta(\xi,\eta)+\omega(\xi,\eta),\quad \beta(\xi,\eta)=\theta(\xi,\eta)-\omega(\xi,\eta), \end{cases} \quad\forall\ (\xi,\eta)\in \overline{BCD}. \end{equation}

For the regularity of $(c(\xi,\,\eta ),\,\theta (\xi,\,\eta ),\,\varpi (\xi,\,\eta ))$ defined in (4.9), we have

Proof. For the function $\theta (\xi,\,\eta )$, we note by (2.8) and (2.18) that

(4.14)\begin{equation} \begin{cases} \displaystyle \theta_\xi=-\frac{\cos\theta(\overline{R}+\overline{S})}{2\kappa\sqrt{\kappa+1-t^2}} +\frac{\sin\theta}{\sqrt{1-t^2}}\Bigg(\frac{t(\overline{R}-\overline{S})} {2\kappa\sqrt{\kappa+1-t^2}}-\frac{1-t^2}{c}\Bigg), \\ \displaystyle \theta_\eta=-\frac{\sin\theta(\overline{R}+\overline{S})}{2\kappa\sqrt{\kappa+1-t^2}} -\frac{\cos\theta}{\sqrt{1-t^2}}\Bigg(\frac{t(\overline{R}-\overline{S})} {2\kappa\sqrt{\kappa+1-t^2}}-\frac{1-t^2}{c}\Bigg). \end{cases} \end{equation}

Due to (4.14) and (4.5), it suffices to show that $(\overline {R},\,\overline {S},\, \overline {W})(\xi,\,\eta )$ are uniformly $C^{\mu }$-continuous for $\mu \in (0,\,1/3)$ on the whole region $BCD$. These regularity results come from lemma 3.9 and the following assertion that, if $I(z,\,t)$ is a $C^{2\tilde {\mu }}$ function on the whole region $B'C'D'$, then the function $\tilde {I}(\xi,\,\eta ):=I(\check {z}(\xi,\,\eta ),\, \check {t}(\xi,\,\eta ))$ is uniformly $C^{\tilde {\mu }}$-continuous on the whole region $BCD$.

To prove the above assertion, we assume that $(\xi ',\,\eta ')$ and $(\xi '',\,\eta '')$ are two points in $BCD$ and $(z',\,t')$ and $(z'',\,t'')$ are the corresponding two points in $B'C'D'$, and consider the differences

(4.15)\begin{equation} |z'-z''|=|\phi(\xi',\eta')-\phi(\xi'',\eta'')|\leq M_1(|\xi''-\xi'|+|\eta''-\eta'|), \end{equation}

and

(4.16)\begin{align} |t'-t''|^2& \leq|t'-t''|\cdot|t'+t''|=|t'^2-t''^2|=|\varpi^2(\xi',\eta') -\varpi^2(\xi'',\eta'')| \nonumber\\ & \leq 2|\varpi(\xi',\eta') -\varpi(\xi'',\eta'')|\leq M_2(|\xi''-\xi'|+|\eta''-\eta'|), \end{align}

where

\[ M_1=\displaystyle\max_{(\xi,\eta)\in BCD}\frac{c}{\varpi}(\xi,\eta),\quad M_2=2\max\Bigg\{\max_{(\xi,\eta)\in BCD}|\varpi_\xi|, \max_{(\xi,\eta)\in BCD}|\varpi_\eta|\Bigg\}. \]

Here $M_1$ and $M_2$ are two uniform positive constants by (4.5) and lemmas 3.2, 3.6. Combining with (4.15) and (4.16) and using the $C^{2\tilde {\mu }}$-continuity gives

\begin{align*} & |\tilde{I}(\xi',\eta')-\tilde{I}(\xi'',\eta'')|=|I(z',t')-I(z'',t'')| \\ & \quad\leq M_3 |(z',t')-(z'',t'')|^{2\tilde{\mu}} \leq M_3\Bigg(|z'-z''|^2 +|t'-t''|^2\Bigg)^{\tilde{\mu}} \\ & \quad\leq M_3|(\xi',\eta')-(\xi'',\eta'')|^{\tilde{\mu}}, \end{align*}

for some uniformly constant $M_3>0$, which means that the function $\tilde {I}(\xi,\,\eta )$ is uniformly $C^{\tilde {\mu }}$-continuous on the whole region $BCD$. According to lemma 3.9, we find that $(\overline {R},\,\overline {S})(\xi,\,\eta )$ are uniformly $C^{(1-\nu )/2}$-continuous and $\overline {W}(\xi,\,\eta )$ is uniformly $C^{(2-\nu )/6}$-continuous on the whole region $BCD$ for any $\nu \in (0,\,1)$. Thus $(\overline {R},\,\overline {S},\, \overline {W})(\xi,\,\eta )$ are uniformly $C^{\mu }$-continuous for $\mu \in (0,\,1/3)$. Hence, by (4.5) and (4.14), the functions $\theta (\xi,\,\eta )$ and $\varpi (\xi,\,\eta )$ are uniformly $C^{1,\mu }$-continuous for $\mu \in (0,\,1/3)$ on the whole region $BCD$. The uniform regularity of $c(\xi,\,\eta )$ follows from the expression of $c$ in (2.6).

Finally, for the sonic curve $\widehat {BC}$, we find by employing (4.5) again that

\[ \varpi_{\xi}^2+\varpi_{\eta}^2=\Bigg(\frac{\sqrt{\kappa+1-t^2}\cdot\overline{W}}{\kappa}-\frac{1-t^2}{c}\Bigg)^2 + \frac{(\kappa+1-t^2)(\overline{R}+\overline{S})^2}{4\kappa^2(1-t^2)}, \]

which together with lemmas 3.2 and 3.6 yields

\[ 0< C_1\leq \varpi_{\xi}^2+\varpi_{\eta}^2\leq C_2, \]

for some uniformly positive constants $C_1$ and $C_2$. Hence the curve $\widehat {BC}$ is $C^{1,\mu }$-continuous for $\mu \in (0,\,1/3)$ and the proof of the lemma is completed.

For the curve $\widehat {CD}$, there has

Proof. We differentiate the equality $\check {z}(\xi,\,\eta )=z_+(\check {t}(\xi,\,\eta ); \bar \delta,\, \tilde {z}(\bar \delta ))$ with respect to $\xi$ and use (4.12) to see that

\[ \check{z}_\xi+\check{z}_\eta\frac{{\rm d}\eta}{{\rm d}\xi}=\frac{\kappa c\check{t}^2}{(1-\check{t}^2)T_2}\Bigg(\check{t}_\xi +\check{t}_\eta\frac{{\rm d}\eta}{{\rm d}\xi}\Bigg). \]

Putting (4.8) into above arrives at

\[ \Bigg(\xi_t +\frac{\kappa c\check{t}^2}{(1-\check{t}^2)T_2}\xi_z\Bigg)\frac{{\rm d}\eta}{{\rm d}\xi}=\eta_t+ \frac{\kappa c\check{t}^2}{(1-\check{t}^2)T_2}\eta_z, \]

which combined with (4.4) leads to

(4.17)\begin{equation} \displaystyle \frac{{\rm d}\eta}{{\rm d}\xi}=\frac{\displaystyle \eta_t+\frac{\kappa c\check{t}^2}{(1-\check{t}^2)T_2}\eta_z}{\displaystyle \xi_t +\frac{\kappa c\check{t}^2}{(1-\check{t}^2)T_2}\xi_z} =-\frac{T_2\cos\theta-\kappa \check{t}\varpi_\xi}{T_2\sin\theta-\kappa \check{t}\varpi_\eta}. \end{equation}

Furthermore, it suggests by (4.5), (4.13) and the expression of $T_2$ in (3.12) that

\begin{align*} & T_2\cos\theta-\kappa \check{t}\varpi_\xi= \Bigg\{\sqrt{\kappa+1-\check{t}^2}R-\frac{\kappa(1-\check{t}^2)\check{t}}{c}\Bigg\}\cos\theta \\ & \qquad- \kappa \check{t}\Bigg\{\cos\theta\frac{\sqrt{\kappa+1-\check{t}^2}}{\kappa}\cdot\frac{\overline{R}-\overline{S}}{2\check{t}} -\cos\theta\frac{1-\check{t}^2}{c} -\sin\theta\frac{\sqrt{\kappa+1-\check{t}^2}}{2\kappa\sqrt{1-\check{t}^2}}(\overline{R}+\overline{S})\Bigg\} \\ & \quad= \frac{\sqrt{\kappa+1-\check{t}^2}}{2\kappa\sqrt{1-\check{t}^2}}\Bigg(\sqrt{1-\check{t}^2}\cos\theta +\check{t}\sin\theta\Bigg)=\frac{\sqrt{\kappa+1-\check{t}^2}}{2\kappa\sqrt{1-\check{t}^2}}\sin(\theta+\omega) \\ & \quad= \frac{\sqrt{\kappa+1-\check{t}^2}}{2\kappa\sqrt{1-\check{t}^2}}\sin\alpha, \end{align*}

and

\begin{align*} & T_2\sin\theta-\kappa \check{t}\varpi_\eta = \Bigg\{\sqrt{\kappa+1-\check{t}^2}R-\frac{\kappa(1-\check{t}^2)\check{t}}{c}\Bigg\}\sin\theta \\ & \qquad- \kappa \check{t}\Bigg\{\sin\theta\frac{\sqrt{\kappa+1-\check{t}^2}}{\kappa}\cdot\frac{\overline{R}-\overline{S}}{2\check{t}} -\sin\theta\frac{1-\check{t}^2}{c} +\cos\theta\frac{\sqrt{\kappa+1-\check{t}^2}}{2\kappa\sqrt{1-\check{t}^2}}(\overline{R}+\overline{S})\Bigg\} \\ & \quad= \frac{\sqrt{\kappa+1-\check{t}^2}}{2\kappa\sqrt{1-\check{t}^2}}\Bigg(\sqrt{1-\check{t}^2}\sin\theta -\check{t}\cos\theta\Bigg)=-\frac{\sqrt{\kappa+1-\check{t}^2}}{2\kappa\sqrt{1-\check{t}^2}}\cos(\theta+\omega) \\ & \quad=-\frac{\sqrt{\kappa+1-\check{t}^2}}{2\kappa\sqrt{1-\check{t}^2}}\cos\alpha. \end{align*}

We insert the above into (4.17) to achieve

\[ \displaystyle \frac{{\rm d}\eta}{{\rm d}\xi}=\frac{\sin\alpha}{\cos\alpha}=\Lambda_+, \]

which implies that $\widehat {CD}$ is a positive characteristic curve of system (2.9). The proof of the lemma is finished.

Finally, we have

Proof. We here just check the first equation of (2.9), the second one can be handled similarly. It follows directly by (2.7), (4.5), (4.13), and (4.14) that

(4.18)\begin{align} & \bar\partial^+\theta=\cos\alpha\theta_\xi+\sin\alpha\theta_\eta \nonumber\\ & \quad=\cos(\theta+\omega)\Bigg\{-\frac{\cos\theta(\overline{R}+\overline{S})}{2\kappa\sqrt{\kappa+1-t^2}} +\frac{\sin\theta}{\sqrt{1-t^2}}\Bigg(\frac{t(\overline{R}-\overline{S})} {2\kappa\sqrt{\kappa+1-t^2}}-\frac{1-t^2}{c}\Bigg)\Bigg\} \nonumber\\ & \qquad+ \sin(\theta+\omega)\Bigg\{-\frac{\sin\theta(\overline{R}+\overline{S})}{2\kappa\sqrt{\kappa+1-t^2}} -\frac{\cos\theta}{\sqrt{1-t^2}}\Bigg(\frac{t(\overline{R}-\overline{S})} {2\kappa\sqrt{\kappa+1-t^2}}-\frac{1-t^2}{c}\Bigg)\Bigg\} \nonumber\\ & \quad= \frac{t(\overline{R}-\overline{S})[\cos(\theta+\omega)\sin\theta-\sin(\theta+\omega)\cos\theta]} {2\kappa\sqrt{1-t^2}\sqrt{\kappa+1-t^2}}\nonumber\\& \quad -\frac{(\overline{R}+\overline{S})[\cos(\theta+\omega)\cos\theta+\sin(\theta+\omega)\sin\theta]} {2\kappa\sqrt{\kappa+1-t^2}} \nonumber\\ & \qquad+ \frac{\sqrt{1-t^2}[\sin(\theta+\omega)\cos\theta-\cos(\theta+\omega)\sin\theta]}{c} \nonumber\\ & \quad=-\frac{t\overline{R}}{\kappa\sqrt{\kappa+1-t^2}}+\frac{1-t^2}{c} =-\frac{\cos\omega\overline{R}}{\kappa\sqrt{\kappa+\varpi^2}}+\frac{\varpi^2}{c}, \end{align}

and

(4.19)\begin{align} & \bar\partial^+\varpi=\cos\alpha\varpi_\xi+\sin\alpha\varpi_\eta \nonumber\\ & \quad=\cos(\theta+\omega)\Bigg\{\cos\theta\frac{\sqrt{\kappa+1-t^2}}{\kappa}\overline{W} -\cos\theta\frac{1-t^2}{c} -\sin\theta \frac{\sqrt{\kappa+1-t^2}}{2\kappa\sqrt{1-t^2}}(\overline{R}+\overline{S})\Bigg\} \nonumber\\ & \qquad+ \sin(\theta+\omega)\Bigg\{\sin\theta\frac{\sqrt{\kappa+1-t^2}}{\kappa}\overline{W} -\sin\theta\frac{1-t^2}{c} +\cos\theta \frac{\sqrt{\kappa+1-t^2}}{2\kappa\sqrt{1-t^2}}(\overline{R}+\overline{S})\Bigg\} \nonumber\\ & \quad=\Bigg(\frac{\sqrt{\kappa+1-t^2}}{\kappa}\cdot\frac{\overline{R}-\overline{S}}{2t} -\frac{1-t^2}{c}\Bigg)[\cos(\theta+\omega)\cos\theta+\sin(\theta+\omega)\sin\theta] \nonumber\\ & \qquad+ \frac{\sqrt{\kappa+1-t^2}(\overline{R}+\overline{S})}{2\kappa\sqrt{1-t^2}} [\sin(\theta+\omega)\cos\theta-\cos(\theta+\omega)\sin\theta] \nonumber\\ & \quad= \frac{\sqrt{\kappa+1-t^2}\overline{R}}{\kappa}-\frac{t(1-t^2)}{c} =\frac{\sqrt{\kappa+\varpi^2}\cdot\overline{R}}{\kappa}-\frac{\varpi^2\cos\omega}{c}. \end{align}

We apply (4.18) and (4.19) to calculate

\begin{align*} \bar\partial^+\theta+\frac{\cos\omega}{\kappa+\varpi^2}\bar\partial^+\varpi & =\Bigg(-\frac{\cos\omega\overline{R}}{\kappa\sqrt{\kappa+\varpi^2}}+\frac{\varpi^2}{c}\Bigg)\nonumber\\& \quad + \frac{\cos\omega}{\kappa+\varpi^2}\cdot \Bigg(\frac{\sqrt{\kappa+\varpi^2}\cdot\overline{R}}{\kappa}-\frac{\varpi^2\cos\omega}{c}\Bigg) \\ & = \frac{\varpi^2}{c} -\frac{\varpi^2\cos^2\omega}{c(\kappa+\varpi^2)} =\frac{\varpi^2}{c}\cdot\frac{\kappa-1+2\varpi^2}{\kappa+\varpi^2}. \end{align*}

This is the desired first equation of (2.9), which ends the proof of the lemma.

We sum up lemmas 4.1–4.4 to complete the proof of theorem 2.2. Based on (4.9) and (2.6), we define the functions $(\rho,\, u,\, v)(\xi,\,\eta )$ as follows

\[ \rho=\Bigg(\frac{c^2(\xi,\eta)}{A\gamma}\Bigg)^{{1}/{\gamma-1}}, \quad u=\xi-c(\xi,\eta)\frac{\cos\theta(\xi,\eta)}{\varpi(\xi,\eta)},\quad v=\eta-c(\xi,\eta)\frac{\sin\theta(\xi,\eta)}{\varpi(\xi,\eta)}. \]

It is obvious by lemma 4.1 that $(\rho,\, u,\, v)(\xi,\,\eta )$ are uniformly $C^{1,\mu }$-continuous for $\mu \in (0,\,1/3)$ on the whole region $BCD$. Furthermore, one can check that the functions $(\rho,\, u,\, v)(\xi,\,\eta )$ defined above satisfy the 2-D isentropic pseudo-steady Euler equations (1.2). The proof of theorem 1.2 is completed.