该文研究超音速流越过弯曲坡面的反问题, 即我们要确定一个弯曲的坡面, 使得当超音速流越过它时产生的激波在事先给定的位置.关于超音速流越过弯曲坡面的直接问题, 读者可以参考文献[1-7].李大潜和王利彬研究了反活塞问题解的整体存在性及唯一性^[8-9].在参考文献[10]中, 王利彬研究了定常等熵无旋流越过弯曲坡面的反问题, 得到了该问题解的整体存在性及唯一性.该文将研究由TSD方程刻画的超音速流越过弯曲坡面的反问题.解决此类激波控制问题的关键是我们需要确定一个弯曲的坡面, 使得当TSD超音速流越过它时产生的激波在事先给定的位置.这类反问题在航空工业领域应用广泛.

众所周知, TSD方程为^[11-12]

$ \begin{equation}\label{tsd} \left\{ \begin{array}{c} u u_x+v_y=0, \\ v_x-u_y=0, \\ \end{array} \right. \end{equation} $

(1.1)

其中$u$和$v$是流体速度的两个分量.当$u<0$时, 方程组(1.1)有两个互异的实特征根

$ \begin{equation}\label{tsdtz} \lambda_1=-\sqrt{-\frac{1}{u}}, ~~~\lambda_2=\sqrt{-\frac{1}{u}}. \end{equation} $

(1.2)

我们首先考虑速度为$(u_0, v_0)$的超音速流越过直坡面的情形.若给定直激波$S:y=(\tan\beta_0)x$ (这里$\beta_0$满足$\beta_0<\beta_{ext}$), 我们可以确定一个通过原点的直边界$y=(\tan\theta_0)x $ $(x\geq0)$^[1]和流体越过激波后的超音速解$(u, v)=(u_1, v_1)$.这里$(u, v)=(u_1, v_1)$满足如下的边界条件^[12].

在直激波边界上, 由R-H条件知

$ \begin{equation}\label{tsd2} v =v_{0}-\sqrt{-\frac{1}{2}(u-u_{0})^2(u+u_{0})}, ~~~~ 0>u>u_{0} \end{equation} $

(1.3)

和

$ \begin{equation}\label{tsd3} \frac{{\rm d}y}{{\rm d}x}=\frac{u-u_0}{v_0-v}, \end{equation} $

(1.4)

同时, 在直坡面边界$y=(\tan\theta_0)x$上有

$ \begin{equation}\label{tsd1} v=(\tan\theta_0)u, \end{equation} $

(1.5)

其中$\theta_0$不超过坡面最大角$\theta_{ext}$, $\beta_{ext}$表示坡面角为$\theta_{ext}$时的激波角^[1].

接下来我们考虑超音速流越过弯曲坡面的情形.令$y=g(x)\in C^2 (x\geq0)$($g(0)=0$且$g'(0)=\tan\beta_0$)为我们事先给定的弯曲激波的位置, 当$\beta_0<\beta_{ext}$时, 我们可以确定一个弯曲坡面$y=f(x)\in C^2 (x\geq0)$($f(0)=0$且$f'(0)=\tan\theta_0$), 使得速度为$(u_0, v_0)$的超音速流越过它时产生的激波在事先给定的位置.越过激波后流体的状态可以由(1.3)-(1.4)式确定且在弯曲坡面边界上满足

$ \begin{equation}\label{wqbj} f'(x)=\frac{v}{u}. \end{equation} $

(1.6)

在本文中, 我们将证明如下定理.

定理1.1  假设$\beta_0<\beta_{ext}$, 存在合适的小量$\epsilon>0$, 使得若$g(x)\in C^2(g(0)=0, g'(0)=\tan\beta_0)$满足

$ \begin{equation}\label{thm1} |g'(x)-g'(0)|\leq \epsilon, ~~~\forall x\geq0 \end{equation} $

(1.7)

和

$ \begin{equation}\label{thm2} |g''(x)|\leq \frac{\epsilon}{1+x}, ~~~\forall x\geq0, \end{equation} $

(1.8)

则在角状区域

$ \begin{equation}\label{thm3} \Omega=\{(x, y)|f(x)\leq y\leq g(x), x\geq0\} \end{equation} $

(1.9)

中我们可以唯一地确定一个弯曲的坡面$y=f(x)\in C^2$, 使得流体越过它时产生的激波正好是$y=g(x)$.这里$f(0)=0, f'(0)=\tan\theta_0$满足

$ \begin{equation}\label{thm6} |f'(x)-f'(0)|\leq K_1 \epsilon, ~~~\forall x\geq0 \end{equation} $

(1.10)

和

$ \begin{equation}\label{thm7} |f''(x)|\leq \frac{K_2 \epsilon}{1+x}, ~~~\forall x\geq 0, \end{equation} $

(1.11)

其中$K_1$和$K_2$是不依赖于$\epsilon$的正常数.

注1.1  定理1.1相应的直接问题是:给定弯曲坡面$y=f(x)\in C^2 (x\geq0)(f(0)=0, $ $f'(0)=\tan\theta_0)$, 如果$\theta_0<\theta_{ext}$, 当速度为$(u_0, v_0)$的超音速流越过它时会产生激波$y=g(x)(g(0)=0)$.我们可以通过解区域$\Omega$上方程组$(1.1)$的自由边值问题得到上述直接问题的解^{[1, 13]}, 其中在$y=f(x)$上满足边界条件$(1.6)$, 在$y=g(x)$上满足R-H条件$(1.3)$和$(1.4)$.

在$(x, y)$平面上原点邻域内, 上述自由边值问题存在唯一的局部超音速解$(u, v)=(u(x, y), $ $v(x, y))\in C^1$, 且在原点处有$(u, v)=(u_1, v_1)$且激波$y=g(x)\in C^2$的方向为^{[4, 13]}

$ \begin{equation}\label{fbp1} g'(0)=\tan\beta_0=\frac{u_1-u_0}{v_0-v_1}. \end{equation} $

(1.12)

引入黎曼不变量^[6]

$ \begin{equation}\label{Riemann} \left\{ \begin{array}{l} r=v+\frac{2}{3}(-u)^{\frac{3}{2}}, \\[3mm] s=v-\frac{2}{3}(-u)^{\frac{3}{2}}, \end{array} \right. \end{equation} $

(2.1)

方程组(1.1)可以等价为区域$\Omega$上的方程组

$ \begin{equation}\label{Riemann1} \left\{ \begin{array}{l} \frac{\partial r}{\partial x}+\bar{\lambda_1}(r, s)\frac{\partial r}{\partial y}=0, \\[3mm] \frac{\partial s}{\partial x}+\bar{\lambda_2}(r, s)\frac{\partial s}{\partial y}=0, \\ \end{array} \right. \end{equation} $

(2.2)

其中

$ \begin{equation}\label{Riemann5} \bar{\lambda_1}(r, s)=- \Big[\frac{3}{4}(r-s)\Big]^{-\frac{1}{3}}, ~~\bar{\lambda_2}(r, s)= \Big[\frac{3}{4}(r-s)\Big]^{-\frac{1}{3}}. \end{equation} $

(2.3)

在定理1.1的假设下, 由(1.3)-(1.4)式可知, 在$y=g(x)$上有

$ \begin{equation}\label{gtj} \left\{ \begin{array}{l} u=u(x)=\frac{-2}{g'(x)^2}-u_0, \\[3mm] v=v(x)=\frac{2}{g'(x)^3}+\frac{2u_0}{g'(x)}+v_0. \\ \end{array} \right. \end{equation} $

(2.4)

由(2.1)式知, 在$y=g(x)$上, $r=r(x)$和$s=s(x)$($r(0)=r_0$, $s(0)=s_0$)满足

$ \begin{equation}\label{rs1} |r(x)-r_0|, ~|s(x)-s_0|\leq K_3 \epsilon \end{equation} $

(2.5)

和

$ \begin{equation}\label{rs2} |r'(x)|, ~|s'(x)|\leq \frac{K_4 \epsilon}{1+x}, \end{equation} $

(2.6)

其中$(r_0, s_0)$是方程组$(2.1)$相应于$(u_1, v_1)$的解, $K_3$和$K_4$是不依赖于$\epsilon$的正常数.为了证明定理1.1, 我们需要在区域$\Omega$上解方程组(2.2)的广义柯西问题

$ \begin{equation}\label{gkc1} \left\{ \begin{array}{l} \frac{\partial r}{\partial x}+\bar{\lambda_1}(r, s)\frac{\partial r}{\partial y}=0, \\[3mm] \frac{\partial s}{\partial x}+\bar{\lambda_2}(r, s)\frac{\partial s}{\partial y}=0, \\[2mm] y=g(x):~r=r(x), ~s=s(x), ~~\forall x\geq 0. \end{array} \right. \end{equation} $

(2.7)

由(1.6), (1.12), (2.1)和(2.3)式可知

$ \begin{equation}\label{wtz1} \bar{\lambda_1}(r_0, s_0)=\lambda_1(u_1, v_1)=-\sqrt{-\frac{1}{u_1}}, \end{equation} $

(2.8)

$ \begin{equation}\label{wtz2} f'(0)=\frac{v_1}{u_1}, \end{equation} $

(2.9)

$ \begin{equation}\label{wtz3} g'(0)=-\frac{u_1-u_0}{v_1-v_0}=-\frac{u_1-u_0}{-\sqrt{-\frac{1}{2}(u_1-u_0)^2(u_1+u_0)}}=\sqrt{\frac{2}{-(u_1+u_0)}}, \end{equation} $

(2.10)

$ \begin{equation}\label{wtz4} \bar{\lambda_2}(r_0, s_0)=\lambda_2(u_1, v_1)=\sqrt{-\frac{1}{u_1}}. \end{equation} $

(2.11)

利用(2.8)-(2.11)式和$u_1>u_0$, 我们得到

$ \begin{equation}\label{wtz} \bar{\lambda_1}(r_0, s_0)<f'(0)<g'(0)<\bar{\lambda_2}(r_0, s_0), \end{equation} $

(2.12)

上式表明无特征线从原点处进入区域$\Omega$.由(2.12)式可知, 广义柯西问题(2.7)有唯一的局部$C^1$解$(r, s)=(r(x, y), s(x, y))$^[13].在本文中, 我们暂时假定在解$(r, s)=(r(x, y), s(x, y))$的任一给定的存在区间上有

$ \begin{equation}\label{duoyu} |r(x, y)-r_0|, ~|s(x, y)-s_0|\leq \delta, \end{equation} $

(2.13)

其中$\delta>0$是一个合适的小量.在证明的结尾处, 我们将说明上述假设的合理性.

令

$ \begin{equation}\label{bjbh} \left\{ \begin{array}{l} \tilde{y}=g(x)-y, \\ \tilde{x}=x. \\ \end{array} \right. \end{equation} $

(2.14)

由(2.12), (1.9)和(2.13)式知, 区域$\Omega$上的广义柯西问题(2.7)可变为区域$\tilde{\Omega}=\{(\tilde{x}, \tilde{y})|\tilde{y}\geq 0, $ $ \tilde{x}\geq h(\tilde{y})\}$上的柯西问题

$ \begin{equation}\label{gkc2} \left\{ \begin{array}{l} \frac{\partial \tilde{r}}{\partial \tilde{y}}+\tilde{\lambda_1}(r, s)\frac{\partial \tilde{r}}{\partial \tilde{x}}=0, \\[3mm] \frac{\partial \tilde{s}}{\partial \tilde{y}}+\tilde{\lambda_2}(r, s)\frac{\partial \tilde{s}}{\partial \tilde{x}}=0, \\[2mm] \tilde{y}=0:~(\tilde{r}, \tilde{s})=(r(\tilde{x}), s(\tilde{x})), ~~\forall \tilde{x}\geq 0, \end{array} \right. \end{equation} $

(2.15)

其中

$ \begin{equation}\label{b1} (\tilde{r}(\tilde{x}, \tilde{y}), \tilde{s}(\tilde{x}, \tilde{y}))=(r(\tilde{x}, g(\tilde{x})-\tilde{y}), s(\tilde{x}, g(\tilde{x})-\tilde{y})), \end{equation} $

(2.16)

$ \begin{equation}\label{b2} \tilde{\lambda_1}(\tilde{x}, \tilde{r}, \tilde{s})=\frac{1}{g'(\tilde{x})-\bar{\lambda_1}(\tilde{r}, \tilde{s})}, \end{equation} $

(2.17)

$ \begin{equation}\label{b3} \tilde{\lambda_2}(\tilde{x}, \tilde{r}, \tilde{s})=\frac{1}{g'(\tilde{x})-\bar{\lambda_2}(\tilde{r}, \tilde{s})}, \end{equation} $

(2.18)

且$\tilde{x}\geq h(\tilde{y})\in C^2$ ($h(0)=0$)由

$ \begin{equation}\label{b4} \tilde{y}=g(\tilde{x})-f(\tilde{x}) \end{equation} $

(2.19)

决定, 其中$y=f(x)$是弯曲坡面的方程.由$(u, v)$平面的激波极线我们可以得到$f'(0)=\tan\theta_0$.在文章的最后我们将整体地描述$y=f(x)$.

利用(1.7), (2.12)和(2.13)式, 我们得到

$ \begin{equation}\label{b5} \tilde{\lambda_2}(\tilde{x}, \tilde{r}, \tilde{s})<\tilde{\lambda_1}(\tilde{x}, \tilde{r}, \tilde{s})<\frac{1}{g'(\tilde{x})-f'(\tilde{x})}, \end{equation} $

(2.20)

由(2.5)-(2.6)式可得

$ \begin{equation}\label{rs11} |r(\tilde{x})-r_0|, ~|s(\tilde{x})-s_0|\leq K_3 \epsilon, ~~~\forall \tilde{x}\geq 0 \end{equation} $

(2.21)

和

$ \begin{equation}\label{rs21} |r'(\tilde{x})|, ~|s'(\tilde{x})|\leq \frac{K_4 \epsilon}{1+\tilde{x}}, ~~~\forall \tilde{x}\geq 0. \end{equation} $

(2.22)

显然, 在区域$\tilde{\Omega}(\delta_0)=\{(\tilde{x}, \tilde{y})|0\leq \tilde{y}\leq\delta_0, \tilde{x}\geq h(\tilde{y}) \}$上, (2.15)式有唯一的$C^1$解$(\tilde{r}, \tilde{s})=(\tilde{r}(\tilde{x}, \tilde{y}), \tilde{s}(\tilde{x}, \tilde{y})) =(r(\tilde{x}, g(\tilde{x})-\tilde{y}), s(\tilde{x}, g(\tilde{x})-\tilde{y}))$, 其中$\delta_0>0$是一个合适的小量^[13].为了得到$\tilde{\Omega}$区域上$C^1$解的整体存在性, 我们需要在任意给定的存在区域$\tilde{\Omega}(Y)=\{(\tilde{x}, \tilde{y})|0\leq \tilde{y}\leq Y, $ $ \tilde{x}\geq h(\tilde{y}) \}$上得到解$(\tilde{r}(\tilde{x}, \tilde{y}), \tilde{s}(\tilde{x}, \tilde{y}))$的$C^1$范数的一致先验估计, 其中$Y>0$.

由(2.15)和(2.21)式, 我们得到

$ \begin{equation}\label{rs111} |\tilde{r}(\tilde{x}, \tilde{y})-r_0|, ~|\tilde{s}(\tilde{x}, \tilde{y})-s_0|\leq K_3 \epsilon, ~~~\forall (\tilde{x}, \tilde{y})\in\tilde{\Omega}(Y). \end{equation} $

(2.23)

下面我们希望得到区域$\tilde{\Omega}(Y)$上$\frac{\partial\tilde{r}}{\partial\tilde{x}}$, $\frac{\partial\tilde{r}}{\partial\tilde{y}}$, $\frac{\partial\tilde{s}}{\partial\tilde{x}}$和$\frac{\partial\tilde{s}}{\partial\tilde{y}}$的$C^0$范数的一致先验估计.因为$(2.15)$式中的方程组只依赖于$\tilde{x}$, 为此我们引入

$ \begin{equation}\label{xin1} R={\rm e}^{q(\tilde{r}, \tilde{s})}\frac{\partial \tilde{r}}{\partial \tilde{y}}, \end{equation} $

(2.24)

其中$q(\tilde{r}, \tilde{s})\in C^{1}$满足

$ \begin{equation}\label{xin2} \frac{\partial q}{\partial \tilde{s}}=\frac{1}{\bar{\lambda_{1}}(\tilde{r}, \tilde{s})-\bar{\lambda_{2}}(\tilde{r}, \tilde{s})}\frac{\partial\bar{\lambda_{1}}}{\partial\tilde{s}}. \end{equation} $

(2.25)

由(2.15)-(2.18)式, 我们很容易得到

$ \begin{equation}\label{xin3} \left\{ \begin{array}{l} \frac{\partial R}{\partial \tilde{y}}+\tilde{\lambda_{1}}(\tilde{x}, \tilde{r}, \tilde{s})\frac{\partial R}{\partial \tilde{x}}={\rm e}^{-q(\tilde{r}, \tilde{s})}\frac{\partial\bar{\lambda_{1}}(\tilde{r}, \tilde{s})}{\partial\tilde{r}}\tilde{\lambda_{1}}(\tilde{x}, \tilde{r}, \tilde{s})R^{2}, \\[2mm] \tilde{y}=0:R=-{\rm e}^{q(r(\tilde{x}), s(\tilde{x}))}\tilde{\lambda_{1}}(\tilde{x}, r(\tilde{x}), s(\tilde{x}))r'(\tilde{x}), ~~\forall \tilde{x}\geq0 . \\ \end{array} \right. \end{equation} $

(2.26)

由(2.20)式可知, 经过任意给定点$(\tilde{x}, \tilde{y})=(\beta, 0)(\beta\geq0)$的每条特征线在有限的时间内会与区域$\tilde{\Omega}$的边界$\tilde{x}\geq h(\tilde{y})(\tilde{y}\geq0)$相交.结合(2.12)式可知, $\tilde{\Omega}\in\{(\tilde{x}, \tilde{y})\mid\tilde{x}\geq \frac{1}{2(g'(0)-f'(0))}\tilde{y}, \tilde{y}\geq0\}$.令$\tilde{x}=\tilde{x_{1}}(\tilde{y};\beta)$为过点$(\beta, 0)$的特征线, 其斜率为$\tilde{\lambda_{1}}(\tilde{x}, \tilde{r}, \tilde{s})$且与区域$\tilde{\Omega}$的边界相交于$(\tilde{x_{1}}(\tilde{Y};\beta), \tilde{Y})$, 其中$\tilde{x}= \frac{1}{2(g'(0)-f'(0))}\tilde{y}$.由(1.7), (2.18)和(2.23)式知, 对合适的小量$\delta>0$, 我们有

$ \begin{eqnarray}\label{xin4} \frac{\tilde{Y}}{2(g'(0)-f'(0))}&=&\tilde{x_{1}}(\tilde{Y};\beta)=\beta+\int_{0}^{\tilde{Y}}\tilde{\lambda_{1}}(\tilde{x}, \tilde{r}, \tilde{s}) (\tilde{x_{1}}(\tau;\beta), \tau){\rm d}\tau\\ &\leq & \beta+\frac{\tilde{Y}}{2(g'(0)-\bar{\lambda_1}(r_0, s_0))}. \end{eqnarray} $

(2.27)

于是我们得到

$ \begin{equation}\label{xin5} \tilde{Y}\leq M_0\beta, \end{equation} $

(2.28)

其中

$ \begin{equation}\label{xin6} M_0=\frac{2(g'(0)-f'(0))(g'(0)-\bar{\lambda_1}(r_0, s_0))}{f'(0)-\bar{\lambda_1}(r_0, s_0)}>0. \end{equation} $

(2.29)

另一方面, 我们有

$ \begin{eqnarray}\label{xin7} \frac{\tilde{Y}}{2(g'(0)-f'(0))}&=&\tilde{x_{1}}(\tilde{Y};\beta)=\beta+\int_{0}^{\tilde{Y}}\tilde{\lambda_{1}}(\tilde{x}, \tilde{r}, \tilde{s}) (\tilde{x_{1}}(\tau;\beta), \tau){\rm d}\tau\\ &\geq & \beta+\frac{2\tilde{Y}}{g'(0)-\bar{\lambda_1}(r_0, s_0)}. \end{eqnarray} $

(2.30)

于是我们得到

$ \begin{equation}\label{xin8} \tilde{Y}\geq M_1\beta, \end{equation} $

(2.31)

其中

$ \begin{equation}\label{xin9} M_1=\frac{2(g'(0)-f'(0))(g'(0)-\bar{\lambda_1}(r_0, s_0))}{4f'(0)-\bar{\lambda_1}(r_0, s_0)-3g'(0)}>0. \end{equation} $

(2.32)

利用(2.15)式, 在$\tilde{x}=\tilde{x_{1}}(\tilde{y};\beta)$上有

$ \begin{equation}\label{xin10} \tilde{r}(\tilde{x}, \tilde{y})=\tilde{r}(\tilde{x_{1}}(\tilde{y};\beta), \tilde{y})=r(\beta) \end{equation} $

(2.33)

和

$ \begin{equation}\label{xin11} \tilde{s}(\tilde{x}, \tilde{y})=\tilde{s}(\tilde{x_{1}}(\tilde{y};\beta), \tilde{y}). \end{equation} $

(2.34)

由(2.26)式可知, 在$\tilde{x}=\tilde{x_{1}}(\tilde{y};\beta)$上有

$ \begin{equation}\label{xin12} R(\tilde{x}, \tilde{y})=R(\tilde{x_{1}}(\tilde{y};\beta), \tilde{y})=\frac{-{\rm e}^{q(r(\beta), s(\beta))}\tilde{\lambda_1}(\beta, r(\beta), s(\beta))r'(\beta)}{1+B}, \end{equation} $

(2.35)

其中

$ \begin{eqnarray}\label{xin13} B&=&\int_{0}^{\tilde{y}}\frac{\partial\bar{\lambda_1}}{\partial \tilde{r}}(r(\beta), \tilde{s}(\tilde{x_{1}}(\tau;\beta), \tau)) {\rm e}^{q(r(\beta), s(\beta))-q(r(\beta), \tilde{s}(\tilde{x_{1}}(\tau;\beta), \tau))}\tilde{\lambda_1}(\beta, r(\beta), s(\beta))\\ &&\cdot\tilde{\lambda_1}(\tilde{x_{1}}(\tau;\beta), r(\beta), \tilde{s}(\tilde{x_{1}}(\tau;\beta), \tau))r'(\beta){\rm d}\tau. \end{eqnarray} $

(2.36)

于是, 由(2.24)式我们得到

$ \begin{equation}\label{xin14} \frac{\partial \tilde{r}}{\partial \tilde{y}}(\tilde{x_{1}}(\tilde{y};\beta), \tilde{y})=\frac{-{\rm e}^{q(r(\beta), s(\beta))-q(r(\beta), \tilde{s}(\tilde{x_{1}}(\tilde{y};\beta), \tilde{y}))} \tilde{\lambda_1}(\beta, r(\beta), s(\beta))r'(\beta)}{1+B}. \end{equation} $

(2.37)

注意到(2.21), (2.23)和(2.33)-(2.34)式, 我们有

$ \begin{equation}\label{xin15} {\rm e}^{2\mid q\mid}\mid \tilde{\lambda_1}\mid\leq M_2, ~~\Big| \frac{\partial\bar{\lambda_1}}{\partial \tilde{r}}\Big| {\rm e}^{2\mid q\mid}\tilde{\lambda_1}^2\leq M_3, \end{equation} $

(2.38)

其中$M_2$和$M_3$是两个不依赖于$\epsilon$的正数.我们可以选择适当小的$\epsilon>0$使得

$ \begin{equation}\label{xin16} K_4M_0M_3\epsilon<\frac{1}{2}. \end{equation} $

(2.39)

于是, 注意到(2.22)式, 由(2.37)式我们得到

$ \begin{equation}\label{xin17}\Big| \frac{\partial \tilde{r}}{\partial \tilde{y}}(\tilde{x_{1}}(\tilde{y};\beta), \tilde{y})\Big|\leq M_2\frac{K_4\epsilon}{1+\beta}\Big(1-M_3\frac{K_4\epsilon}{1+\beta}\tilde{Y}\Big)^{-1}. \end{equation} $

(2.40)

于是, 利用(2.28)和(2.39)式, 我们得到

$ \begin{equation}\label{xin18}\Big| \frac{\partial \tilde{r}}{\partial \tilde{y}}(\tilde{x_{1}}(\tilde{y};\beta), \tilde{y})\Big|\leq M_2\frac{K_4\epsilon}{1+\beta}(1-K_4M_0M_3\epsilon)^{-1} \leq2K_4M_2\frac{\epsilon}{1+\beta}\leq\frac{K_5\epsilon}{1+\tilde{y}}, ~~0\leq \tilde{y}\leq Y. \end{equation} $

(2.41)

由(1.7), (2.12), (2.21), (2.33)和(2.34)式可知, 在$\tilde{x}=\tilde{x_{1}}(\tilde{y};\beta)$上有

$ \begin{equation}\label{xin19} 2(g'(0)-\bar{\lambda_1}(r_0, s_0))<g'(\tilde{x})-\bar{\lambda_1}(\tilde{r}, \tilde{s})<\frac{1}{2}(g'(0)-\bar{\lambda_1}(r_0, s_0)). \end{equation} $

(2.42)

于是, 注意到(2.18)和(2.42)式, 我们得到

$ \begin{equation}\label{xin20} \Big|\frac{\partial \tilde{r}}{\partial \tilde{x}(\tilde{x_{1}}(\tilde{y};\beta), \tilde{y})}\Big|\leq\frac{K_6\epsilon}{1+\tilde{y}}, ~~0\leq \tilde{y}\leq Y. \end{equation} $

(2.43)

最后, 我们得到

$ \begin{equation}\label{xin21} \Big|\frac{\partial \tilde{r}}{\partial \tilde{x}}(\tilde{x}, \tilde{y})\Big|, \Big| \frac{\partial \tilde{r}}{\partial \tilde{y}}(\tilde{x}, \tilde{y})\Big| \leq\frac{K_7\epsilon}{1+\tilde{y}}, ~~\forall(\tilde{x}, \tilde{y})\in \tilde{\Omega}(Y). \end{equation} $

(2.44)

类似地, 我们有

$ \begin{equation}\label{xin22} \Big|\frac{\partial \tilde{s}}{\partial \tilde{x}}(\tilde{x}, \tilde{y})\Big|, \Big|\frac{\partial \tilde{s}}{\partial \tilde{y}}(\tilde{x}, \tilde{y})\Big| \leq\frac{K_8\epsilon}{1+\tilde{y}}, ~~\forall(\tilde{x}, \tilde{y})\in \tilde{\Omega}(Y). \end{equation} $

(2.45)

于是, 在区域$\tilde{\Omega}$上, 我们得到了(2.15)式的唯一整体$C^{1}$解$(\tilde{r}, \tilde{s})=(\tilde{r}(\tilde{x}, \tilde{y}), \tilde{s}(\tilde{x}, \tilde{y}))$.注意到(2.41)式, 在区域$\Omega$上, 我们得到了广义柯西问题($2.7$)的唯一整体$C^{1}$解

$ \begin{equation}\label{xin23} (r, s)=(r(x, y), s(x, y))=(\tilde{r}(x, g(x)-y), \tilde{s}(x, g(x)-y)). \end{equation} $

(2.46)

由(2.23)式我们很容易得到

$ \begin{equation}\label{xin24} |r(x, y)-r_0|, ~|s(x, y)-s_0|\leq K_9 \epsilon, ~~\forall (x, y)\in \Omega, \end{equation} $

(2.47)

这表明我们的假设(2.13)式是合理的.而且, 注意到(1.7)式, 由(2.44)和(2.45)式可得

$ \begin{equation}\label{xin25} \Big|\frac{\partial r}{\partial x}(x, y)\Big|, \Big| \frac{\partial r}{\partial y}(x, y)\Big|, \Big| \frac{\partial s}{\partial x}(x, y)\Big|, \Big| \frac{\partial s}{\partial y}(x, y)\Big|\leq\frac{K_{10}\epsilon}{1+g(x)-y}, ~~\forall (x, y)\in \Omega. \end{equation} $

(2.48)

在上式中, $r$和$s$是$x$和$f(x)$的函数, 于是区域$\Omega=\{(x, y)\mid x\geq 0, f(x)\leq y\leq g(x)\}$的边界$y=f(x)$可以由

$ \begin{equation}\label{xin26} \left\{ \begin{array}{l} f'(x)=-\frac{\sqrt[3]{2}(r+s)}{(3(r-s))^{\frac{2}{3}}}(x, f(x)), \\[2mm] f(0)=0 \end{array} \right. \end{equation} $

(2.49)

得到.注意到(1.7)和(2.47)-(2.48)式, 我们可以唯一地确定$y=f(x)\in C^2$满足(1.10)和(1.11)式.进而由$(u, v)$平面的激波极线可知, $f'(0)=\tan\theta_0$.

于是, 定理1.1得证.