严格椭圆算子的Schauder内估计
1.齐次方程的内估计
本节我们研究一般线性算子的内估计:
\[\begin{equation*}
Lu=a^{ij}(x)D_{ij}u+b^i(x)D_iu+c(x)u=f(x),a^{ij}=a^{ji}
\end{equation*}
\]
本节中我们始终要求算子\(L\)是严格椭圆的,即存在\(\lambda>0\)使得:
\[a^{ij}(x)\xi_i\xi_j\ge \lambda|\xi|^2,\forall x\in \Omega,\xi\in \mathbb{R}^n
\]
在研究一般的方程之前,我们先解决齐次方程的内估计:
\[\begin{equation*}
L_0u=A^{ij}D_{ij}u=f(x),A^{ij}=A^{ji}
\end{equation*}
\]
其中\([A^{ij}]\)是常数矩阵且满足严格椭圆条件(其实满足一致椭圆条件.)我们先建立如下的引理:
引理1:齐次方程的内估计:设\(u\in C^2(\Omega),f\in C^{\alpha}(\Omega)\)在\(\mathbb{R}^n\)中的开集满足\(L_0u=f\),则:
\[\begin{equation*}
|u|_{2, \alpha ; \Omega}^* \leqslant C\left(|u|_{0 ; \Omega}+|f|_{0, \alpha ; \Omega}^{(2)}\right)
\end{equation*}
\]
(2):设\(\Omega\)是\(\mathbb{R}^n_+\)中的开子集,并且\(x_n=0\)上有边界部分,设\(u\in C^2(\Omega)\cap C^0(\Omega\cup T)\),\(f\in C^{\alpha}(\Omega\cup T)\),且在\(\Omega\)上满足\(L_0u=f\),在\(T\)上有\(u=0\),则:
\[\begin{equation*}
|u|_{2, \alpha ; \Omega \cup T}^* \leqslant C\left(|u|_{0 ; \Omega}+|f|_{0, \alpha ; \Omega \cup T}^{(2)}\right),
\end{equation*}
\]
其中\(C=C(n,\lambda,\Lambda,\alpha)\).
证明: 由于\(A^{ij}\)是正定矩阵,因此存在存在正交矩阵\(P\)使得:
\[P'AP=\mathrm{diag}\{\lambda_1,\cdots,\lambda_n\}
\]
令\(D=\{1/\sqrt{\lambda_1},\cdots,1/\sqrt{\lambda_n}\}\),作变换:\(z=xPD\),则我们有:
\[L_0u=f\iff \Delta_zu=\tilde{f}(z)
\]
同时注意到由于\(P\)是正交矩阵,因此:\(|Px|=|x|\)所以:
\[\Lambda^{-\frac{1}{2}}|x| \leqslant|x \mathbf{Q}| \leqslant \lambda^{-\frac{1}{2}}|x| .
\]
因此我们有:\(u(x)\to \tilde{u}(z),\Omega\to \tilde{\Omega}\).
\[\begin{aligned}
& c^{-1}|v|_{k, \alpha ; \Omega}^* \leqslant|\tilde{v}|_{k, \alpha ; \tilde{\Omega}}^* \leqslant c|v|_{k, \alpha ; \Omega}^*, \quad k=0,1,2, \ldots, \quad 0 \leqslant \alpha \leqslant 1, \\
& c^{-1}|v|_{0, \alpha ; \Omega}^{(k)} \leqslant|\widetilde{v}|_{0, \alpha ; \Omega}^{(k)} \leqslant c|v|_{0, \alpha ; \Omega}^{(k)}, \quad
\end{aligned}
\]
其中\(c=c(k,n,\lambda,\Lambda)\).因此利用Poisson方程的内估计我们得到:
\[|\tilde{u}|_{2, \alpha ; \tilde{\Omega}}^* \leqslant C\left(|u|_{0 ; \tilde{\Omega}}+|f|_{0, \alpha ; \tilde{\Omega}}^{(2)}\right)
\]
故:
\[\begin{aligned}
|u|_{2, a ; \Omega}^* & \leqslant C \mid \tilde{u}_{2, \alpha ; \tilde{\Omega}}^* \leqslant C\left(|\tilde{u}|_{0 ; \tilde{\Omega}}+|\tilde{f}|_{0, \alpha ; \tilde{\Omega}}^{(2)}\right) \\
& \leqslant C\left(|u|_{0 ; \Omega}+|f|_{0, \alpha ; \Omega}^{(2)}\right) .
\end{aligned}
\]
2.Schauder内估计
2.1:预备知识
为了后边的叙述方便,我们这里引入Holder空间的内插定理(不加证明).
内插定理
设\(u\in C^{2,\alpha}(\Omega)\),其中\(\Omega\)是\(\mathbb{R}^n\)的开子集,则对任一给定的\(\varepsilon>0\),存在\(C=C(\varepsilon)\)使得:
\[\begin{aligned}
& {[u]_{j, \beta ; \Omega}^* \leqslant C|u|_{0 ; \Omega}+\varepsilon[u]_{2, \alpha ; \Omega}^*,} \\
& \\
& |u|_{j, \beta ; \Omega}^* \leqslant C|u|_{0 ; \Omega}+\varepsilon[u]_{2, \alpha ; \Omega}^*,
\end{aligned}
\]
其中$ j=0,1,2 ; 0 \leqslant \alpha, \beta \leqslant 1, j+\beta<2+\alpha$
当\(\Omega\)是有界集时,我们也可以将拟范数\([\cdot]^*\)换为\([\cdot]\).为了得到更强的结论,我们定义:
\[ \begin{aligned}
& {[f]_{k, 0 ; \Omega}^{(\sigma)}=[f]_{k ; \Omega}^{(\sigma)}=\sup _{\substack{x \in \Omega \\
|\beta|=k}} d_x^{k+\sigma}\left|D^\beta f(x)\right| ;} \\
& {[f]_{k, \alpha ; \Omega}^{(\sigma)}=\sup _{\substack{x, y \in \Omega \\
\text { |. } \\
|\beta|=k}} d_{x, y}^{k+\alpha+\sigma} \frac{\left|D^\beta f(x)-D^\beta f(y)\right|}{|x-y|^\alpha}, \quad 0<\alpha \leqslant 1 \text {; }} \\
& |f|_{k ; \Omega}^{(\sigma)}=\sum_{j=0}^k[f]_{j ; \Omega}^{(\sigma)} \\
& |f|_{k, \alpha ; \Omega}^{(\sigma)}=|f|_{k ; \Omega}^{(\sigma)}+[f]_{k, \alpha ; \Omega}^{(\sigma)} . \\
&
\end{aligned}
\]
注意到:
\[\boxed{|f g|_{0, \alpha ; \Omega}^{(\sigma+\tau)} \leqslant|f|_{0, \alpha ; \Omega}^{(\sigma)}|g|_{0, \alpha ; \Omega}^{(\tau)} .}
\]
2.2:Schauder内估计
Schauder内估计设\(\Omega\)是\(\mathbb{R}^n\)中的开子集,\(u\in C^{2,\alpha}\)是方程\(Lu=f\)的有界解,并且\([a^{ij}]\)是严格椭圆的,假设存在\(\Lambda\)使得:
\[ \left|a^{i j}\right|_{0, \alpha ; \Omega}^{(0)},\left|b^i\right|_{0, \alpha ; \Omega}^{(1)},|c|_{0, \alpha ; \Omega}^{(2)} \leqslant \Lambda .
\]
则我们有:
\[ \begin{equation*}
|u|_{2, \alpha ; \Omega}^* \leqslant C\left(|u|_{0 ; \Omega}+|f|_{0, \alpha ; \Omega}^{(2)}\right),
\end{equation*}
\]
其中\(C=C(n,\lambda,\Lambda,\alpha)\).
证明: 1.首先我们证明只需要在\(\Omega\)的任何紧子集上证明即可.设\(\Omega=\bigcup_i\Omega_i\)其中\(\Omega_i\subset \Omega_{i+1}\)是\(\Omega\)的紧子集.对任意的\(x,y\in \Omega\),必然存在\(i\)使得\(x,y\in \Omega_i\),此时定义任何一个二阶导数我们都:
\[\begin{aligned}
\left(d_{x, y}^{(i)}\right)^{2+\alpha} \frac{\left|D^2 u(x)-D^2 u(y)\right|}{|x-y|^\alpha} & \leqslant[u]_{2, \alpha ; \Omega_i}^* \\
& \leqslant C\left(|u|_{0 ; \Omega_i}+|f|_{0, \alpha ; \Omega_i}^{(2)}\right) \text{如果结论成立}\\
& \leqslant C\left(|u|_{0 ; \Omega}+|f|_{0, \alpha ; \Omega}^{(2)}\right),
\end{aligned}
\]
其中\(d_{x,y}^{(i)}=\min\{\mathrm{dist}(x.\partial \Omega_i),\mathrm{dist}(y,\partial \Omega_i)\}\).令\(i\to \infty\),就得到了:
\[\left(d_{x, y}\right)^{2+\alpha} \frac{\left|D^2 u(x)-D^2 u(y)\right|}{|x-y|^\alpha}
\]
他们右者同一个无关\(i\)的上界.
2.下边我们就来证明\([u]^*_{2,\alpha;\Omega}\)是有界的.设\(x_0\ne y_0,d_{x_0}=d_{x_0,y_0}=\min \{d_{x_0},d_{y_0}\}\),设\(\mu\le \frac{1}{2}\)(\emph{待定}.)令\(d=\mu d_{x_0}\),\(B=B_d(x_0)\),下边我们冻结主部系数,将\(Lu=f\)改写为:
\[ \begin{aligned}
a^{i j}\left(x_0\right) D_{i j} u & =\left(a^{i j}\left(x_0\right)-a^{i j}(x)\right) D_{i j} u-b^i D_i u-c u+f \\
& \equiv F(x),
\end{aligned}
\]
在\(B\)中这里注意到我们在\(B\)中利用估计.,利用齐次方程的内估计,如果\(y\in B_{d/2}(x_0)\),则我们有:
\[\left(\frac{d}{2}\right)^{2+\alpha} \frac{\left|D^2 u\left(x_0\right)-D^2 u\left(y_0\right)\right|}{\left|x_0-y_0\right|^\alpha} \le d_{x,y}^{2+\alpha}\frac{\left|D^2 u\left(x_0\right)-D^2 u\left(y_0\right)\right|}{\left|x_0-y_0\right|^\alpha}\leqslant C\left(|u|_{0 ; B}+|F|_{0, \alpha ; B}^{(2)}\right)
\]
因此我们可以得到:
\[d_{x_0}^{2+\alpha} \frac{\left|D^2 u\left(x_0\right)-D^2 u\left(y_0\right)\right|}{\left|x_0-y_0\right|^\alpha} \leqslant \frac{C}{\mu^{2+\alpha}}\left(|u|_{0 ; B}+|F|_{0, \alpha ; B}^{(2)}\right) .
\]
而如果\(|x-y|\ge d/2\),则我们有:
\[d_{x_0}^{2+\alpha} \frac{\left|D^2 u\left(x_0\right)-D^2 u\left(y_0\right)\right|}{\left|x_0-y_0\right|^\alpha} \leqslant\left(\frac{2}{\mu}\right)^\alpha\left[d_{x_0}^2\left|D^2 u\left(x_0\right)\right|+d_{y_0}^2\left|D^2 u\left(y_0\right)\right|\right]
\leqslant \frac{4}{\mu^\alpha}[u]_{2 ; \Omega}^* .
\]
综上我们就得到了,对最初定义的\(x_0,y_0\in \Omega\),我们有
\[ d_{x_0}^{2+\alpha}\frac{\left|D^2 u\left(x_0\right)-D^2 u\left(y_0\right)\right|}{\left|x_0-y_0\right|^\alpha}\le \frac{C}{\mu+2}(|u|_{0;\Omega}+|F|^2_{0,\alpha;B})+\frac{4}{\mu^{\alpha}}[u]^*_{2;\Omega}
\]
3.下边我们估计\(|F|_{0,\alpha;B}^{(2)}\)
\[\begin{aligned}
|F|_{0, \alpha ; B}^{(2)} \leqslant \sum_{i, j}\left|\left(a^{i j}\left(x_0\right)-a^{i j}(x)\right) D_{i j} u\right|_{0, \alpha ; B}^{(2)} +\sum_i\left|b^i D_i u\right|_{0, \alpha ; B}^{(2)}+|c u|_{0, \alpha ; B}^{(2)}+\mid f_{0, \alpha ; B}\mid^{(2)} .
\end{aligned}
\]
由于出现了\(|\cdot|_{0,\alpha;B}\)而定理中要求的是\(|\cdot|_{\cdot,\Omega}\)为此我们需要研究这两者范数中间的关系.首先我们注意到如下关系对任意的\(x\in B,\)
\[d_x=\mathrm{dist}(x,\partial\Omega)>(1-\mu)d_{x_0}\ge \frac{1}{2}d_{x_0}
\]
因此我们有:
\[\begin{aligned}
|g|_{0, \alpha ; B}^{(2)} & \leqslant d^2|g|_{0 ; B}+d^{2+\alpha}[g]_{\alpha ; B} \\
& \leqslant \frac{\mu^2}{(1-\mu)^2}[g]_{0 ; \Omega}^{(2)}+\frac{\mu^{2+\alpha}}{(1-\mu)^{2+\alpha}}[g]_{0, \alpha ; \Omega}^{(2)} \\
& \leqslant 4 \mu^2[g]_{0 ; \Omega}^{(2)}+8 \mu^{2+\alpha}[g]_{0, \alpha ; \Omega}^{(2)} \leqslant 8 \mu^2|g|_{0, \alpha ; \Omega}^{(2)} .
\end{aligned}
\]
下边分别对:\(\left|\left(a^{i j}\left(x_0\right)-a^{i j}(x)\right) D_{i j} u\right|_{0, \alpha ; B}^{(2)} ,\left|b^i D_i u\right|_{0, \alpha ; B}^{(2)}+|c u|_{0, \alpha ; B}^{(2)},\mid f_{0, \alpha ; B}\mid^{(2)}\)进行估计.首先我们有:
对每一对 \(i, j\), 记 \(\left(a\left(x_0\right)-a(x)\right) D^2 u=\left(a^{i j}\left(x_0\right)-a^{i j}(x)\right) D_{i j} u\)(注意到下边用到了\([D^2u]_{0;\Omega}\le |u|_{2;\Omega}^*\)的事实.)
\[\begin{aligned}
\left|\left(a\left(x_0\right)-a(x)\right) D^2 u\right|_{0, \alpha ; B}^{(2)} & \leqslant\left|a\left(x_0\right)-a(x)\right|_{0, \alpha ; B}^{(0)}\left|D^2 u\right|_{0, \alpha ; B}^{(2)} \\
& \leqslant\left|a\left(x_0\right)-a(x)\right|_{0, \alpha ; B}^{(0)}\left(4 \mu^2[u]_{2 ; \Omega}^*+8 \mu^{2+\alpha}[u]_{2, \alpha ; \Omega}^*\right) .
\end{aligned}
\]
继续对\(|a(x)-a(x_0)|^{(0)}_{0,\alpha;B}\)进行估计:
\[\begin{aligned}
\left|a\left(x_0\right)-a(x)\right|_{0, \alpha ; B}^{(0)} & \leqslant \sup _{x \in B}\left|a\left(x_0\right)-a(x)\right|+d^\alpha[a]_{\alpha ; B} \\
& \leqslant 2 d^\alpha[a]_{\alpha ; B} \leqslant 2^{1+\alpha} \mu^\alpha[a]_{0, \alpha ; \Omega}^* \leqslant 4 \Lambda \mu^\alpha,
\end{aligned}
\]
因此我们有:
\[ \begin{aligned}
& \sum_{i, j}\left|\left(a^{i j}\left(x_0\right)-a^{i j}(x)\right) D_{i j} u\right|_{0, \alpha ; B}^{(2)} \\
\leqslant & 32 n^2 \Lambda \mu^{2+\alpha}\left([u]_{2 ; \Omega}^*+\mu^\alpha[u]_{2, \alpha ; \Omega}^*\right) \\
\leqslant & 32 n^2 \Lambda \mu^{2+\alpha}\left(C(\mu)|u|_{0 ; \Omega}+2 \mu^\alpha[u]_{2, \alpha ; \Omega}^*\right) .
\end{aligned}
\]
最后一步我们用到了内插定理中的
\[\boxed{[u]_{j, \beta ; \Omega}^* \leqslant C|u|_{0 ; \Omega}+\varepsilon[u]_{2, \alpha ; \Omega}^*}
\]
其中\(j=2,\beta=0,\varepsilon=\mu^{\alpha}\).
对每个\(i\),记\(b^iD_iu\)(这里就不求和了)为\(bDu\),因此我们有:
\[\begin{aligned}
|b D u|_{0, \alpha ; B}^{(2)} & \leqslant 8 \mu^2|b D u|_{0, \alpha ; \Omega}^2 \leqslant 8 \mu^2|b|_{0, \alpha ; \Omega}^{(1)}|D u|_{0, \alpha ; \Omega}^{(1)} \\
& \leqslant 8 \mu^2 \Lambda|u|_{1, \alpha ; \Omega}^* \leqslant 8 \mu^2 \Lambda\left(C(\mu)|u|_{0 ; \Omega}+\mu^{2 \alpha}[u]_{2, \alpha ; \Omega}^*\right) .
\end{aligned}
\]
于是得到了:
\[\begin{equation*}
\left|b^i D_i u\right|_{0, \alpha ; B}^{(2)} \leqslant 8 n \Lambda \mu^2\left(C(\mu)|u|_{0 ; \Omega}+\mu^{2 \alpha}[u]_{2, \alpha ; \Omega}^*\right) .
\end{equation*}
\]
其中用到了内插定理中的:
\[\boxed{|u|_{j, \beta ; \Omega}^* \leqslant C|u|_{0 ; \Omega}+\varepsilon[u]_{2, \alpha ; \Omega}^*}
\]
其中\(j=1,\beta=\alpha,\varepsilon=\mu^{2\alpha}\).
最后还有
\[ \begin{aligned}
|c u|_{0, \alpha ; B}^{(2)} & \leqslant 8 \mu^2|c|_{0, \alpha ; \Omega}^{(2)}|u|_{0, \alpha ; \Omega}^{(0)} \\
& \leqslant 8 \Lambda \mu^2\left(C(\mu)|u|_{0 ; \Omega}+\mu^{2 \alpha}[u]_{2, \alpha ; \Omega}^*\right) .
\end{aligned}
\]
\[\begin{equation*}
|f|_{0, \alpha ; B}^2 \leqslant 8 \mu^2|f|_{0, \alpha ; \Omega}^{(2)} .
\end{equation*}
\]
将上边的所有式子结合起来就得到了:
\[|F|_{0, \alpha ; B}^{(2)} \leqslant C \mu^{2+2 \alpha}[u]_{2, \alpha ; \Omega}^*+C(\mu)\left(|u|_{0 ; \Omega}+|f|_{0, \alpha ; \Omega}^{(2)}\right) .
\]
并且在
\[\boxed{[u]_{j, \beta ; \Omega}^* \leqslant C|u|_{0 ; \Omega}+\varepsilon[u]_{2, \alpha ; \Omega}^*}
\]
取\(j=2,\beta=0,\varepsilon=\mu^{2\alpha}\)来估计\([u]_{2;\Omega}^*\).就得到了
\[d_{x_0, y_0}^{2+\alpha} \frac{\left|D^2 u\left(x_0\right)-D^2 u\left(y_0\right)\right|}{\left|x_0-y_0\right|^\alpha}
\leqslant C \mu^\alpha[u]_{2, \alpha ; \Omega}^*+C(\mu)\left(|u|_{0 ; \Omega}+|f|_{0, \alpha ; \Omega}^{(2)}\right) .
\]
左边对\(x_0,y_0\)取上确界就得到了:
\[ [u]_{2, \alpha ; \Omega}^* \leqslant C \mu^\alpha[u]_{2, \alpha ; \Omega}^*+C(\mu)\left(|u|_{0 ; \Omega}+|f|_{0, \alpha ; \Omega}^{(2)}\right) .
\]
取\(C\mu^{\alpha}\le \frac{1}{2}\),就得到了:
\[[u]_{2, \alpha ; \Omega}^* \leqslant C\left(|u|_{0 ; \Omega}+|f|_{0, \alpha ; \Omega}^{(2)}\right) .
\]
再次利用:
\[\boxed{|u|_{j, \beta ; \Omega}^* \leqslant C|u|_{0 ; \Omega}+\varepsilon[u]_{2, \alpha ; \Omega}^*}
\]
取\(j=2,\beta=0\)就得到了结论.
这个结果的好处就是我们可以允许\(f\)和系数是无界的,下边是更常用的Schauder估计.
2.3:推论
推论设 \(u \in C^{2, \alpha}(\Omega), f \in C^\alpha(\bar{\Omega})\) 满足 \(L u=f\), 其中 \(L\) 在有界区域 \(\Omega\) 中满足 \((6.2)\), 并且 \(L\) 的系数在 \(C^\alpha(\bar{\Omega})\) 中. 那么如果 \(\Omega^{\prime} \subset \subset \Omega, \operatorname{dist}\left(\Omega^{\prime}, \partial \Omega\right) \geqslant d\), 就存在常数 \(C\), 使得
\[\begin{aligned}
& d|D u|_{0 ; \Omega^{\prime}}+d^2\left|D^2 u\right|_{0 ; \Omega^{\prime}}+d^{2+\alpha}\left[D^2 u\right]_{\alpha ; \Omega^{\prime}} \\
\leqslant & C\left(|u|_{0 ; \Omega}+|f|_{0, \alpha ; \Omega}^{(2)}\right),
\end{aligned}
\]
其中 \(C\) 仅依赖于椭圆性常数 \(\lambda\) 和 \(L\) 的系数的 \(C^\alpha(\bar{\Omega})\) 范数 (以及 \(n, \alpha\) 和 \(\Omega\) 的 直径).或者是:
\[|u|_{2,\alpha;\Omega'}\le C\left(|u|_{0;\Omega}+|f|_{0,\alpha;\Omega}\right)
\]
这里的\(C=C(n,\lambda,\Lambda,\alpha,\Omega',\Omega)\).