Machine-translated from Chinese.
记号
略. 参考辜联崑《二阶抛物型偏微分方程》.
抛物方程在仿射变换$x=Ay$下还是抛物方程.
弱极值原理
设$u\in C_p^2,$ $\Delta_x u-u_t>0.$ 假设$u$在$P(D)$内取极大值,
那么在极值点$\Delta_x u(x_0,t_0)\le 0,$
$\frac{\partial {}u}{\partial {}t}\ge 0,$ 矛盾.
此时$u$只能在边界取到极大值.
对于一般的$L,$ 我们有定理:
定理 1. 若$c\le 0,$ $Lu>0,$ 则$u$不能在$P(D)$内达到非负极大值; $Lu<0,$ 则$u$不能在$P(D)$内达到非正极小值.
不失一般性, 设$Hess=\operatorname{diag}\{\Gamma_1,\cdots,\Gamma_n\}.$
假设有这样的极值, 在极值点$\frac{\partial {}u}{\partial {}x_i}=0,$
$\frac{\partial {}u}{\partial {}t}\ge 0,$ $a^{ij}u_{ij}\le 0,$
$cu\le 0,$ 因此$0<Lu\le 0.$ 矛盾.
对于$c=0$的情形, $u$的正负性限制可以取消. 因为此时$Lu=L(u+C),$
总能将极值改为正负性适当.
若$Lu\ge 0,$ $c\le 0,$ $u\in C_p^2(D)\cap C(\overline D),$
那么考虑$v=u-\varepsilon t$即可(更严谨的, $t-t_0$).
此时$Lv=Lu-c\varepsilon t+\varepsilon>0.$
那么$u\le \max_{\partial_pD}(u-\varepsilon t)^++\varepsilon t\le \max_{\partial_pD}u^++\varepsilon T.$
令$\varepsilon\rightarrow 0$即可. 注意此时的极值原理是弱的.
若$Lu\ge 0,$ $c\le C_0,$ 可考虑$v=ue^{-(C_0+1)t}$满足的方程. 此时
$$
\widetilde Lv:=a_{ij}v_{ij}+b_iv_i+(c-(C_0+1))v-v_t=e^{-(C_0+1)t}Lu\ge 0,
$$
$\widetilde L$中常数项系数非正, 因此可用前面的结论.
不过此时只能推断出$u\le e^{(C_0+1)t}\max_{\partial_pD}(ue^{-(C_0+1)t})^+.$
常用的结论是若$u|_{\partial_pD}\le 0,$ 则$u\le 0.$
强极值原理
定理 2. 考虑$Q=Q(x_0,r,t_1,t_0).$ 若$Lu\ge 0,$ $c\le 0,$ $u\in C_p^2(P(Q))\cap C(\overline Q),$ $u(x_0,t_0)=\max_{P(Q)}u> 0,$ 则$u$在$\overline{Q(x_0,r,t_1,t_0)}$上为常数.
首先说明$u(x_0,t_0)=u(x_0,t),$ $\,\forall\,t<t_0.$
不然不妨假设$0<u(x_0,t_1)<u(x_0,t_0).$
取$\delta$使得在$B_n^\delta(x_0)\times\{t_1\}$处,
$u(x)<u(x_0,t_0)-\varepsilon_1.$ 希望找到函数$v$使得$Lv>0,$
同时其在$\widetilde Q:=Q(x_0,\delta,t_1,t_0)$内部达到极值. 考虑
$$
v=\varepsilon(\delta^2-r^2)^2e^{-a(t-t_0)}+u.
$$
固定$a,$ 存在这样的$\varepsilon,$
使得$v|_{B_n^\delta(x_0)\times \{t_1\} }<u(x_0,t_0).$
在$S(\widetilde Q)$上, $v=u\le u(x_0,t_0).$ 结合来看,
在抛物边界$\partial_p\widetilde Q$上, $v\le u(x_0,t_0).$
因此其在$P(\widetilde Q)$内部达到极值.
只需验证$Lv>0,$
$$
\begin{aligned}
Lv=Lu+\varepsilon e^{-a(t-t_0)}&\left\{8a_{ij}(x_i-x_{0,i})(x_j-x_{0,j})-4a_{ii}(\delta^2-r^2)\right.\\
&\left.-4b_i(\delta^2-r^2)(x_i-x_{0,i})+(a+c)(\delta^2-r^2)^2\right\}
\end{aligned}
$$
只需说明大括号内可以取正即可.
在$r$靠近$\delta$时, 利用第一项. 由$a_{ij}$的一致椭圆性,
$\,\exists\,\delta_0,$ 当$\delta_0<r<\delta$时, 括号内取正;
当$\delta<\delta_0$时, 利用最后一项. 令$a$充分大, 即可令括号内取正.
这样与$v$的极值原理矛盾, 从而$u(x_0,t)=u(x_0,t_0),$ $\,\forall\,t<t_0.$
接下来, 由仿射变换, $u(x_0,t_0)=u(k(x_0',t_0')+(1-k)(x_0,t_0)),$
$\,\forall\,k\in (0,1),$ $(x_0',t_0')\in Q.$ 于是我们得到了结论.
对一般的区域$D$, 利用多个小柱体可以得到更一般的结论.
$u(x_0,t_0)>0$若是极大值, 则可以控制沿时间单调递减能到达的区域,
其上函数值均为常数.
若$c=0,$ 同理可去掉正负性的要求.
先验估计
基本思路是利用弱极值原理加加减减.
定理 3. 若$c\le C_0,$ $u\in C_p^2(P(D))\cap C(\overline D).$ 当$Lu\ge f$时, $u\le e^{C_0^+t}(\max_{\partial_pD}u^++t\max_{\overline D}f^-);$ 若$Lu\le f,$ 则$u\ge -e^{C_0^+t}(\max_{\partial_pD}u^-+t\max_{\overline D} f^+).$
令$v=u-e^{C_0^+t}(\max_{\partial_pD}u^++t\max_{\overline D} f^-),$
那么$v|_{\partial_pD}\le 0,$
$$
Lv=Lu+e^{C_0^+t}(C_0^+-c)(\max_{\partial_pD}u^++t\max_{\overline D} f^-)+e^{C_0^+t}\max_{\overline D} f^-\ge f+\max_{\overline D} f^-\ge 0.
$$
这就说明了$v\le 0,$
从而$u\le e^{C_0^+}(\max_{\partial_pD}u^++t\max_{\overline D}f^-).$
若$Lu\le f,$ 则$L(-u)\ge -f,$
$$
u\ge -e^{C_0^+t}(\max_{\partial_pD}(-u)^++t\max_{\overline D}(-f)^-)=-e^{C_0^+t}(\max_{\partial_pD}u^-+t\max_{\overline D} f^+).
$$
推论 4. $Lu=f,$ $c\le C_0,$ 那么$\Vert u\Vert_{C_0(\overline D)}\le e^{C_0^+t}(\Vert u\Vert_{C_0(\partial_pD)}+t\Vert f\Vert_{C_0(\overline D)}).$
推论 5. 若$Lu_1=Lu_2=f,$ $c\le C_0,$ 在$\partial_pD$上$u_1=u_2.$ 那么在整个$\overline D$上, 也有$u_1\equiv u_2.$
推论 6. 若$c=0,$ $Lu=0,$ 则$\min_{\partial_pD}u\le u\le \max_{\partial_pD}u.$
文章最后更新于 2023-03-06 13:01:03
mark
Briefly. Please refer to Gu Liankun's "Second Order Parabolic Partial Differential Equations".
A parabolic equation is still a parabolic equation under affine transformation $x=Ay$.
weak extreme value principle
Assume $u\in C_p^2,$ $\Delta_x u-u_t>0.$ and assume $u$ takes the maximum value within $P(D)$,
Then at the extreme point $\Delta_x u(x_0,t_0)\le 0,$
$\frac{\partial {}u}{\partial {}t}\ge 0,$ Contradiction.
At this time $u$ can only take the maximum value at the boundary.
For general $L,$ we have theorem:
Theorem 1. If $c\le 0,$ $Lu>0,$ then $u$ cannot reach the non-negative maximum value within $P(D)$; $Lu<0,$ then $u$ cannot reach the non-positive minimum value within $P(D)$.
Without loss of generality, let $Hess=\operatorname{diag}\{\Gamma_1,\cdots,\Gamma_n\}.$
Suppose there is such an extreme value, at the extreme point $\frac{\partial {}u}{\partial {}x_i}=0,$
$\frac{\partial {}u}{\partial {}t}\ge 0,$ $a^{ij}u_{ij}\le 0,$
$cu\le 0,$ Therefore $0<Lu\le 0.$ is contradictory.
For the case of $c=0$, the positivity restriction of $u$ can be canceled. Because at this time $Lu=L(u+C),$
The extreme values can always be changed to be positive or negative appropriately.
If $Lu\ge 0,$ $c\le 0,$ $u\in C_p^2(D)\cap C(\overline D),$
Then consider $v=u-\varepsilon t$ (more rigorously, $t-t_0$).
At this time $Lv=Lu-c\varepsilon t+\varepsilon>0.$
Then $u\le \max_{\partial_pD}(u-\varepsilon t)^++\varepsilon t\le \max_{\partial_pD}u^++\varepsilon T.$
Just let $\varepsilon\rightarrow 0$. Note that the extreme value principle at this time is weak.
If $Lu\ge 0,$ $c\le C_0,$ can consider the equation that $v=ue^{-(C_0+1)t}$ satisfies. At this time
$$
\widetilde Lv:=a_{ij}v_{ij}+b_iv_i+(c-(C_0+1))v-v_t=e^{-(C_0+1)t}Lu\ge 0,
$$
The coefficient of the constant term in $\widetilde L$ is not positive, so the previous conclusion can be used.
However, we can only infer $u\le e^{(C_0+1)t}\max_{\partial_pD}(ue^{-(C_0+1)t})^+.$ at this time
The commonly used conclusion is that if $u|_{\partial_pD}\le 0,$ then $u\le 0.$
strong extreme value principle
Theorem 2. Consider $Q=Q(x_0,r,t_1,t_0).$ if $Lu\ge 0,$ $c\le 0,$ $u\in C_p^2(P(Q))\cap C(\overline Q),$ $u(x_0,t_0)=\max_{P(Q)}u> 0,$ then $u$ is constant on $\overline{Q(x_0,r,t_1,t_0)}$.
First explain $u(x_0,t_0)=u(x_0,t),$ $\,\forall\,t<t_0.$
Otherwise, we might as well assume $0<u(x_0,t_1)<u(x_0,t_0).$
Take $\delta$ so that at $B_n^\delta(x_0)\times\{t_1\}$,
$u(x)<u(x_0,t_0)-\varepsilon_1.$ hopes to find the function $v$ such that $Lv>0,$
At the same time, it reaches the extreme value inside $\widetilde Q:=Q(x_0,\delta,t_1,t_0)$. Consider
$$
v=\varepsilon(\delta^2-r^2)^2e^{-a(t-t_0)}+u.
$$
Fixed $a,$ exists such that $\varepsilon,$
Make$v|_{B_n^\delta(x_0)\times \{t_1\} }<u(x_0,t_0).$
On $S(\widetilde Q)$, combined with $v=u\le u(x_0,t_0).$,
On the parabolic boundary $\partial_p\widetilde Q$, $v\le u(x_0,t_0).$
Therefore it reaches its extreme value inside $P(\widetilde Q)$.
Just verify $Lv>0,$
$$
\begin{aligned}
Lv=Lu+\varepsilon e^{-a(t-t_0)}&\left\{8a_{ij}(x_i-x_{0,i})(x_j-x_{0,j})-4a_{ii}(\delta^2-r^2)\right.\\
&\left.-4b_i(\delta^2-r^2)(x_i-x_{0,i})+(a+c)(\delta^2-r^2)^2\right\}
\end{aligned}
$$
Just explain that the curly brackets can be corrected.
When $r$ is close to $\delta$, the first term is used. Due to the consistent ellipticity of $a_{ij}$,
$\,\exists\,\delta_0,$ When $\delta_0<r<\delta$, take the positive value in the brackets;
When $\delta<\delta_0$, use the last term. Make $a$ sufficiently large to make the values in the brackets positive.
This contradicts the extreme value principle of $v$, so $u(x_0,t)=u(x_0,t_0),$ $\,\forall\,t<t_0.$
Next, by affine transformation, $u(x_0,t_0)=u(k(x_0',t_0')+(1-k)(x_0,t_0)),$
$\,\forall\,k\in (0,1),$ $(x_0',t_0')\in Q.$ So we get the conclusion.
For the general area $D$, more general conclusions can be obtained by using multiple small cylinders.
If $u(x_0,t_0)>0$ is a maximum value, you can control the area that can be reached by monotonically decreasing along time,
The function values above are all constants.
If $c=0,$, the requirement of positivity and negativity can be removed in the same way.
a priori estimate
The basic idea is to use the weak extreme value principle to add, add, subtract and subtract.
Theorem 3. If $c\le C_0,$ $u\in C_p^2(P(D))\cap C(\overline D).$ When $Lu\ge f$, $u\le e^{C_0^+t}(\max_{\partial_pD}u^++t\max_{\overline D}f^-);$ If $Lu\le f,$ then $u\ge -e^{C_0^+t}(\max_{\partial_pD}u^-+t\max_{\overline D} f^+).$
Let$v=u-e^{C_0^+t}(\max_{\partial_pD}u^++t\max_{\overline D} f^-),$
Then $v|_{\partial_pD}\le 0,$
$$
Lv=Lu+e^{C_0^+t}(C_0^+-c)(\max_{\partial_pD}u^++t\max_{\overline D} f^-)+e^{C_0^+t}\max_{\overline D} f^-\ge f+\max_{\overline D} f^-\ge 0.
$$
This explains $v\le 0,$
Thus $u\le e^{C_0^+}(\max_{\partial_pD}u^++t\max_{\overline D}f^-).$
If $Lu\le f,$ then $L(-u)\ge -f,$
$$
u\ge -e^{C_0^+t}(\max_{\partial_pD}(-u)^++t\max_{\overline D}(-f)^-)=-e^{C_0^+t}(\max_{\partial_pD}u^-+t\max_{\overline D} f^+).
$$
Corollary 4. $Lu=f,$ $c\le C_0,$ Then $\Vert u\Vert_{C_0(\overline D)}\le e^{C_0^+t}(\Vert u\Vert_{C_0(\partial_pD)}+t\Vert f\Vert_{C_0(\overline D)}).$
Corollary 5. If $Lu_1=Lu_2=f,$ $c\le C_0,$ is $u_1=u_2.$ on $\partial_pD$, then on the whole $\overline D$, there is also $u_1\equiv u_2.$
Corollary 6. If $c=0,$ $Lu=0,$ then $\min_{\partial_pD}u\le u\le \max_{\partial_pD}u.$
The article was last updated on 2023-03-06 13:01:03
