我们首先来看$\alpha$调和映照流, 其方程可以写作
$$\left\{\begin{aligned}
\partial_t u^\beta-\Delta u^\beta-(\alpha-1) \frac{u_{kl}^{\beta\gamma}u_k^\gamma u_l^\beta}{1+\lvert \nabla u \rvert^2}&=\Gamma^\beta(u)(\nabla u,\nabla u),\quad x\in M;\\
u(\cdot,0)&=u_0\in C^\infty(M), \quad x\in M;\\
u^n(x,t)&=0, \quad (x,t)\in\partial M\times[0,T];\\
\frac{\partial u^\beta}{\partial\nu}&=0,\quad (x,t)\in \partial M\times [0,T],\quad \beta=1,2,\ldots, n-1.
\end{aligned}\right.$$
我们将考虑其线性形式, 即
$$
\partial_t u^\beta-\Delta u^\beta-(\alpha-1) \frac{u_{kl}^{\beta\gamma}w_k^\gamma w_l^\beta}{1+\lvert \nabla w \rvert^2}=\Gamma^\beta(w)(\nabla w,\nabla w),\quad x\in M.
$$
改写成标准的形式
$$
\mathcal{L}u=f,
$$
即
$$
l_{\beta\gamma}u^\gamma_{kl}=f_\beta(x,t),
$$
其中
$$
l_{\beta\gamma}(x,t,\partial_x,\partial_t)=\delta_{\beta\gamma}\partial_t-\left( \delta_{kl}\delta_{\beta\gamma}+(\alpha-1) \frac{w_k^\gamma w_l^\beta}{1+\lvert \nabla w \rvert^2} \right)\partial_k\partial_l,
$$
以及
$$
f_\beta(x,t)=\Gamma^\beta(w)(\nabla w,\nabla w).
$$
这些系数满足[1,p. 8]的条件, 即
1.
$$\begin{aligned}
l_{\beta\gamma}(x,t,i\xi\lambda,p\lambda^2)&=\delta_{\beta\gamma}p\lambda^2-\left( \delta_{kl}\delta_{\beta\gamma}+(\alpha-1)\frac{w_k^\gamma w_l^\beta}{1+\lvert \nabla w \rvert^2} \right)(i\xi^k\lambda)(i\xi^l\lambda)\\
&=\left( \delta_{\beta\gamma}p+\delta_{\beta\gamma}\lvert \xi \rvert^2+(\alpha-1)\frac{w_k^\gamma w_l^\beta \xi^k\xi^l}{1+\lvert \nabla w \rvert^2} \right)\lambda^2
:=l_{\beta\gamma}^0(x,t,i\xi,p)\lambda^2.
\end{aligned}$$
其中, 以$l_{\beta\gamma}^0$作为系数的算子$\mathcal{L}_0$称为原算子$\mathcal{L}$的主部. 在我们的情形由于没有低阶导数项, $l_{\beta\gamma}^0=l_{\beta\gamma}$.
2. 抛物性条件. 即以$l_{\beta\gamma}^0(x,t,i\xi, p)$为$(\beta,\gamma)$元素的矩阵$L_0$的行列式作为$p$的多形式之根$p_s$满足
$$
\mathrm{Re}p_s\leq-\delta \lvert \xi \rvert^2,\quad \delta>0.
$$
事实上, 若令
$$
X=\sqrt{\alpha-1}\frac{\xi\cdot\nabla w}{\sqrt{1+\lvert \nabla w \rvert^2}},
$$
则
$$
L(x,t,i\xi,p)=\det L_0(x,t,i\xi,p)=\left\lvert (p+\lvert \xi \rvert^2)E +XX^T \right\rvert,
$$
其特征根为
$$
p_1=-\lvert \xi \rvert^2-(\alpha-1)\frac{\xi^kw_k^\gamma \xi^lw_l^\gamma}{1+\lvert \nabla w \rvert^2},\quad
p_2=p_3=\cdots=p_n=-\lvert \xi \rvert^2.
$$
注意到
$$
\frac{\xi^kw_k^\gamma \xi^lw_l^\gamma}{1+\lvert \nabla w \rvert^2}
\leq \lvert \xi \rvert^2.
$$
可见, 我们可取$\delta=1$.
3. 边界条件. 在$\Gamma=\partial\Omega\times(0,T)$上, 我们的一般边界条件是
$$
\mathcal{B}u=\Phi,
$$
即
$$
\sum_{\beta=1}^n B_{q\beta}(x,t,\partial_x,\partial_t)u^\beta(x,t)=\Phi_q(x,t),\quad q=1,2,\ldots, br,
$$
注意到对我们的情形, $b=1$, $r=n$, 其中$2b$就是抛物方程的阶, 而$r$就是$\det L_0$作为$p$的多形式的最高次幂. 根据我们的边值条件, 我们知道
$$
(B_{q\beta}(x,t,\partial_x,\partial_t))_{n\times n}=\left(
\begin{array}{cccc}
\sum_{k=1}^m\nu^k\partial_k&0&\cdots &0\\
0&\sum_{k=1}^m\nu^k\partial_k&\cdots &0\\
\vdots&\vdots&\cdots&\vdots\\
0 &0 &\sum_{k=1}^m\nu^k\partial_k&0\\
0 &0 &0 &1 \\
\end{array}
\right),
$$
其中,
$$
\nu=(\nu^1,\ldots, \nu^m),\quad m=\dim M,
$$
是$\partial\Omega\times\left\{ t \right\}$在$(x,t)$处的内法向. 特别地,
$$
B_{q\beta}(x,t,i\xi\lambda,p\lambda^{2b})=\left\{
\begin{array}{cl}
i\sum_{k=1}^m\nu^k \xi^k\lambda,&q=\beta=1,2,\ldots, n-1,\\
1,&q=\beta=n,\\
0,&\text{otherwise}.
\end{array}
\right.
$$
注意到$B_{q\beta}$都是齐次多形式, 故其主部恰好等于自身, 即$B_{q\beta}^0=B_{q\beta}$.
4. 互补条件(complementary condition).
- 对边界条件, 我们的互补条件如下: 假设$\mathrm{Re}p\geq -\delta_1\lvert \zeta \rvert^2$, 其中$0<\delta_1<\delta$且$\lvert p \rvert^2+\lvert \zeta \rvert^4>0$, 注意这里$p$为复数. 则可以证明
$$
L(x,t,i(\zeta+\tau\nu),p)
$$
作为$\tau$的函数有$n$个虚部为正的根$\tau_s^+$, 以及虚部为负的根$\tau_s^-$, 其中$\zeta=\zeta(x)$是$x\in\partial\Omega$处的切向量, 而$\nu=\nu(x)$为$x\in\partial\Omega$处的单位内法向量.
事实上, 在我们的情形, 直接验证知
$$
L(x,t,i(\zeta+\tau\nu),p)=0
$$
的根分别为
$$\begin{aligned}
\tau^\pm_2&=\cdots=\tau^\pm_n=\pm i\sqrt{p+\lvert \zeta \rvert^2},\\
\tau^\pm_1&=\frac{\frac{2(\alpha-1)\nabla_\nu w\cdot\nabla_\zeta w}{1+\lvert \nabla w \rvert^2}\pm\sqrt{\left( \frac{2(\alpha-1)(\nabla_\nu w\cdot\nabla_\zeta w)}{1+\lvert \nabla w \rvert^2} \right)^2-4\left( 1+\frac{(\alpha-1)\lvert \nabla_\nu w \rvert^2}{1+\lvert \nabla w \rvert^2} \right)\left( p+\lvert \zeta \rvert^2+\frac{(\alpha-1)\lvert \nabla_\zeta w \rvert^2}{1+\lvert \nabla w \rvert^2} \right)}}{2\left( 1+\frac{(\alpha-1)\lvert \nabla_\nu w \rvert^2}{1+\lvert \nabla w \rvert^2} \right)}\\
&=\pm i\sqrt{p+\lvert \zeta \rvert^2}+o(\alpha-1).
\end{aligned}$$
对2-值函数$\sqrt{p+\lvert \zeta \rvert^2}$我们可以选取其中一个分支, 使得其实部是大于零的. 为方便起见, 我们令$\tau_0=i\sqrt{p+\lvert \zeta \rvert^2}$.
现在, 我们令
$$
L_0^+(\zeta,\tau,p)=(\tau-\tau_1^+)\cdots(\tau-\tau_n^+)=(\tau-\tau_1^+)(\tau-\tau_0)^{n-1},
$$
以及
$$
R(\zeta,\tau,p)=B_0(i\zeta,i\tau,p)\hat L_0(i\zeta,i\tau,p),
$$
其中
$$
B_0(i\zeta,i\tau,p)=B_0(x_0,t_0,i(\zeta+\tau\nu),p)=\mathrm{diag}(\tau,\cdots,\tau,1)_{n\times n},\quad (x_0,t_0)\in\partial\Omega\times(0,T),
$$
而
$$
L_0(i\zeta,i\tau,p)=L_0(x_0,t_0,i(\zeta+\tau\nu),p)=(p+\lvert \zeta \rvert^2+\tau^2)E+X_\tau^TX_\tau,
$$
其中
$$
X_\tau=\sqrt{\frac{\alpha-1}{1+\lvert \nabla w \rvert^2}} (\nabla_\zeta w+\tau\nabla_\nu w).
$$
注意到$E+X_\tau^TX_\tau$的伴随矩阵
$$
(E+X_\tau^TX_\tau)^*=(1+X_\tau X_\tau^T)E-X_\tau^TX_\tau.
$$
或者
$$
(\lambda E+X_\tau^TX_\tau)^*=(\lambda^{n-1}+\lambda^{n-2} X_\tau X_\tau^T)E-\lambda^{n-2} X_\tau^TX_\tau.
$$
因此
$$
\hat L_0(i\zeta,i\tau,p)=\left( (p+\lvert \zeta \rvert^2+\tau^2)^{n-1}+(p+\lvert \zeta \rvert^2+\tau^2)^{n-2}X_\tau X_\tau^T \right)E
-(p+\lvert \zeta \rvert^2+\tau^2)^{n-2}X_\tau^T X_\tau.
$$
注意到
$$
\begin{aligned}
X_\tau X_\tau^T&=\lvert X_\tau \rvert^2=(\alpha-1)\frac{\lvert \nabla_\zeta w+\tau \nabla_\nu w \rvert^2}{1+\lvert \nabla w \rvert^2}\\
&=(\alpha-1)\frac{1}{1+\lvert \nabla w \rvert^2}\left( \tau^2\lvert \nabla_\nu w \rvert^2+2\tau\nabla_\zeta w\nabla_\nu w+\lvert \nabla_\nu w \rvert^2 \right).
\end{aligned}
$$
我们首先对$n=2$来进行验证. 此时
$$
L_0^+(i\zeta,i\tau,p)=(\tau-\tau_0)(\tau-\tau_1^+),
$$
而
$$
B_0(i\zeta,i\tau,p)=\mathrm{diag}(\tau,1),
$$
以及
$$
\begin{aligned}
\hat L_0(i\zeta,i\tau,p)&=\left( p+\lvert \zeta \rvert^2+\tau^2+(\alpha-1)\frac{\tau^2\lvert \nabla_\nu w \rvert^2+2\tau\nabla_\zeta w\nabla_\nu w^T+\lvert \nabla_\nu w \rvert^2}{1+\lvert \nabla w \rvert^2} \right)E\\
&\qquad-(\alpha-1)\frac{\tau^2\nabla_\nu w^T\nabla_\nu w+2\tau\nabla_\nu w^T\nabla_\zeta w+\nabla_\zeta w^T\nabla_\zeta w}{1+\lvert \nabla w \rvert^2}.
\end{aligned}$$
因此若记
$$
R(i\zeta,i\tau,p) :=B_0(i\zeta,i\tau,p)\hat L_0(i\zeta,i\tau,p),
$$
则
$$
\begin{aligned}
R_{11}&=\tau\left( p+\lvert \zeta \rvert^2+\tau^2+(\alpha-1)\frac{\tau^2\lvert \nabla_\nu w \rvert^2+2\tau\nabla_\zeta w\nabla_\nu w^T+\lvert \nabla_\nu w \rvert^2}{1+\lvert \nabla w \rvert^2} \right)\\
&\qquad-(\alpha-1)\tau\frac{\tau^2\nabla_\nu w^1\nabla_\nu w^1+2\tau\nabla_\nu w^1\nabla_\zeta w^1+\nabla_\zeta w^1\nabla_\zeta w^1}{1+\lvert \nabla w \rvert^2}\\
R_{12}&=-(\alpha-1)\tau \frac{\tau^2\nabla_\nu w^1\nabla_\nu w^2+2\tau\nabla_\nu w^1\nabla_\zeta w^2+\nabla_\zeta w^1\nabla_\zeta w^2}{1+\lvert \nabla w \rvert^2}\\
R_{21}&=-(\alpha-1)\frac{\tau^2\nabla_\nu w^2\nabla_\nu w^1+2\tau\nabla_\nu w^2\nabla_\zeta w^1+\nabla_\zeta w^2\nabla_\zeta w^1}{1+\lvert \nabla w \rvert^2}\\
R_{22}&=\left( p+\lvert \zeta \rvert^2+\tau^2+(\alpha-1)\frac{\tau^2\lvert \nabla_\nu w \rvert^2+2\tau\nabla_\zeta w\nabla_\nu w^T+\lvert \nabla_\nu w \rvert^2}{1+\lvert \nabla w \rvert^2} \right)\\
&\qquad-(\alpha-1)\frac{\tau^2\nabla_\nu w^2\nabla_\nu w^2+2\tau\nabla_\nu w^2\nabla_\zeta w^2+\nabla_\zeta w^2\nabla_\zeta w^2}{1+\lvert \nabla w \rvert^2}.
\end{aligned}
$$
注意到
$$
p+\lvert \zeta \rvert^2+\tau_1^{+2}+(\alpha-1)\frac{\lvert \nabla_{\zeta+{\tau_1^+}\nu}w \rvert^2}{1+\lvert \nabla w \rvert^2}=0,
$$
因此
$$
\begin{aligned}
&p+\lvert \zeta \rvert^2+\tau^2+(\alpha-1)\frac{\tau^2\lvert \nabla_\nu w \rvert^2+2\tau\nabla_\zeta w\nabla_\nu w^T+\lvert \nabla_\nu w \rvert^2}{1+\lvert \nabla w \rvert^2}\\
&\qquad=\tau^2-\tau_1^{+2}+(\alpha-1)\frac{(\tau^2-\tau_1^{+2})\lvert \nabla_\nu w \rvert^2+2(\tau-\tau_1^+)\nabla_\zeta w\nabla_\nu w^T}{1+\lvert \nabla w \rvert^2}.
\end{aligned}
$$
令
$$\begin{aligned}
a&=(\alpha-1)\frac{|\nabla_\nu w|^2}{1+\lvert \nabla w \rvert^2},\\
a_{11}&=(\alpha-1)\frac{|\nabla_\nu w^1|^2}{1+\lvert \nabla w \rvert^2},\quad
a_{12}=a_{21}=(\alpha-1)\frac{\nabla_\nu w^1\nabla_\nu w^2}{1+\lvert \nabla w \rvert^2},\quad
a_{22}=(\alpha-1)\frac{|\nabla_\nu w^2|^2}{1+\lvert \nabla w \rvert^2},\\
b&=(\alpha-1)\frac{\nabla_\nu w\nabla_\zeta w^T}{1+\lvert \nabla w \rvert^2},\\
b_{11}&=(\alpha-1)\frac{\nabla_\nu w^1\nabla_\zeta w^1}{1+\lvert \nabla w \rvert^2},\quad
b_{12}=(\alpha-1)\frac{\nabla_\nu w^1\nabla_\zeta w^2}{1+\lvert \nabla w \rvert^2},\quad
b_{21}=(\alpha-1)\frac{\nabla_\nu w^2\nabla_\zeta w^1}{1+\lvert \nabla w \rvert^2},\quad
b_{22}=(\alpha-1)\frac{\nabla_\nu w^2\nabla_\zeta w^2}{1+\lvert \nabla w \rvert^2},\\
c&=(\alpha-1)\frac{\lvert \nabla_\zeta w \rvert^2}{1+\lvert \nabla w \rvert^2},\\
c_{11}&=(\alpha-1)\frac{\lvert \nabla_\zeta w^1 \rvert^2}{1+\lvert \nabla w \rvert^2},\quad
c_{12}=c_{21}=(\alpha-1)\frac{ \nabla_\zeta w^1\nabla_\zeta w^2}{1+\lvert \nabla w \rvert^2},\quad
c_{22}=(\alpha-1)\frac{\lvert \nabla_\zeta w^2 \rvert^2}{1+\lvert \nabla w \rvert^2},
\end{aligned}$$
则
$$\begin{aligned}
R_{11}&=\tau \left( (\tau^2-\tau_1^{+2})(1+a)+2(\tau-\tau_1^+)b \right)-a_{11}\tau^3-2b_{11}\tau^2-c_{11}\tau,\\
R_{12}&=-a_{12}\tau^3-2b_{12}\tau^2-c_{12}\tau,\\
R_{21}&=-a_{21}\tau^2-2b_{21}\tau-c_{21},\\
R_{22}&=\left( (\tau^2-\tau_1^{+2})(1+a)+2(\tau-\tau_1^+)b \right)-a_{22}\tau^2-2b_{22}\tau-c_{22}.
\end{aligned}$$
因此, 模掉$L_0^+=(\tau-\tau_1^+)(\tau-\tau_0)$, 我们得到
$$
\begin{aligned}
R_{11}’&=\tau\left(\tau_0^2(1+a-a_{11})+\tau_0\tau_1^+(1+a-a_{11})-a_{11}\tau_1^{+2}-2b_{11}(\tau_0+\tau_1^+)+2b\tau_0-c_{11}\right)\\
&\qquad-\tau_0\tau_1^+\left( (1+a-a_{11})(\tau_0+\tau_1^+)-2b_{11}+2b \right),\\
R_{12}’&=-\tau\left( a_{12}(\tau_0^2+\tau_0\tau_1^++\tau_1^+)+2b_{12}(\tau_0+\tau_1^+)+c_{12} \right)\\
&\qquad-\tau_0\tau_1^+\left( a_{12}(\tau_0+\tau_1^+)+2b_{12} \right),\\
R_{21}’&=-\tau\left( a_{12}(\tau_0+\tau_1^+)+2b_{21} \right)\\
&\qquad+a_{12}\tau_0\tau_1^+-c_{12},\\
R_{22}’&=\tau\left( (1+a-a_{22})(\tau_0+\tau_1^+)-2b_{22}+2b \right)\\
&\qquad-\tau_1^+\left( (1+a)\tau_1^++(1+a-a_{22})\tau_0^++2b \right)-c_{22}.
\end{aligned}
$$
我们的边界互补条件要求
$$
\mathrm{rank}R’=2.
$$
这在$\alpha=1$的情形, 只需要$\tau_0\neq0$即可.
- 对初值条件, 我们的互补条件如下: 注意到, 我们的初值条件改写成一般形式为
$$
\mathcal{C}u|_{t=0}=\Psi,
$$
即
$$
\sum_{\beta=1}^n C_{q\beta}(x,\partial_x,\partial_t)u^\beta|_{t=0}=\psi_q(x), \quad q=1,2,\ldots,n.
$$
在我们的情形, $C_{q\beta}=\delta_{q\beta}$, 因此初值算子$\mathcal{C}$的主部$\mathcal{C}^0=\mathcal{C}$. 如果我们记以$C^0_{q\beta}=\delta_{q\beta}$为元素的矩阵为$C^0$, 则
$$
C^0(x,i\xi,p)=E_n.
$$
回忆, $L_0$为$l_{\beta\gamma}^0(x,t,i\xi,p)$作为$(\beta,\gamma)$元素的矩阵. 在我们的情形,
$$
L_0(x,0,0,p)=(\delta_{\beta\gamma}p)_{n\times n}=p E_n.
$$
若记$\hat L_0(x,t,i\xi,p)$为$L_0(x,t,i\xi,p)$的伴随矩阵为$C^0$,
则在我们的情形
$$
\hat L_0(x,0,0,p)=p^{n-1}E_n.
$$
而所谓的初值条件的互补性条件即是说矩阵
$$
C_0(x,0,p)\hat L_0(x,0,0,p)=p^{n-1}E_n
$$
的行向量模掉首项系数为1的多形式(即$\det L_0(x,0,0,0)$的最高次项)$p^n$是线性无关的.