Machine-translated from Chinese.
非散度型方程
记$\Omega_T=\Omega\times(0,T],$
抛物边界$\partial_p\Omega=\overline\Omega_T\setminus \Omega_T.$
考虑方程$u_t+Lu=f,$ $Lu=-a^{ij}u_{ij}+b^iu_i+cu.$
要求$[a^{ij}]$在$\Omega_T$内半正定.
设$u\in C^{2,1}(\Omega_T):=\{u(x,t)\in C^2_x(\Omega),C^1_t\in (0,T]\}.$
定理 1 (弱极值原理). 设$\Omega$为有界开集, $a^{ij},b^i,c\in \Omega_T$满足适当光滑性条件, $u\in C^{2,1}(\Omega_T)\cap C(\overline\Omega_T),$ $c\ge 0,$ 则$u_t+Lu\le 0$可以推出$\max_{\overline\Omega_T}u\le \sup_{\partial_p\Omega_T}u^+.$ 取负可以得到最小值的控制.
反证法, 假设有
$$
\max_{\overline\Omega_T}=u(x_0,t_0)>\sup_{\partial_p\Omega_T}u^+\ge 0.
$$
取$v_\varepsilon=u-\varepsilon t,$ 只要$\varepsilon$充分小,
那么$v_\varepsilon$仍在内部取到非负最大值.
因此在最大值点$(x_\varepsilon,t_\varepsilon),$ $Dv_{\varepsilon}=0,$
$D^2v_\varepsilon\le 0,$ $v_{\varepsilon}'\ge 0.$
将$L$展开即可得到$v_{\varepsilon}'+Lv_\varepsilon\ge 0.$
(利用$c,v_\varepsilon\ge 0.$)
另一方面,
$$
0\le v_\epsilon'+Lv_\varepsilon=u_t+Lu-\varepsilon-c\varepsilon t\le -\varepsilon.
$$
矛盾, 因此$u$非负最大值在抛物边界上取到.
当$c\equiv 0$时, 可忽略正部的要求; 当$c\ge \mu$为常数时,
可取$v=e^{\mu t}u,$ 那么
$$
v_t+(L-\mu)v=e^{\mu t}(u_t+Lu).
$$
因此,
$$
u_t+Lu\le 0\Rightarrow v_t+\widehat Lv\le 0\Rightarrow \max_{\overline\Omega_T}e^{\mu t}u\le \sup_{\partial_p\Omega_T}(e^{\mu t}u)^{+}.
$$
定理 2. 设$\Omega\subset \mathbb{R}^n$为有界开集, $[a^{ij}]\ge 0,$ $c\ge \mu,$ 那么热方程初边值问题在$C(\overline\Omega_T)\cap C^{2,1}(\Omega_T)$上有唯一解.
最大模估计
考虑Cauchy问题:
$$
\begin{cases}
u_t+Lu=f&in\:Q_T:=\mathbb{R}^n\times (0,T]\\
u|_{t=0}=\psi&in\:\mathbb{R}^n
\end{cases}
$$
设系数满足: $\,\exists\,\alpha_1\in [0,2),$
$\alpha_2\in [0,1),$ $\mu_0\in \mathbb{R},$ $\mu_1\ge 0,$ 使得
$$
c\ge \mu_0,\quad\sum_{i,j=1}^n |a^{ij}|\le \mu_1(|x|^{\alpha_1}+1),\quad \sum_{i=1}^n |b^i|\le \mu_1(|x|^{\alpha_2}+1).
$$
定理 3. 设$u\in C(\overline Q_T)\cap C^{2,1}(Q_T)$为Cauchy问题解, $\lim\limits_{r\rightarrow\infty} \frac{M(r)}{r^2}=0,$ $M(r):=\sup\limits_{\begin{subarray}{c} |x|=r\\ 0<t<T \end{subarray} }|u(t)|,$ 则
$$
\sup_{Q_T}|u|\le e^{(-\mu_0)^+T}[T\Vert f\Vert_{L^\infty(Q_T)}+\Vert\psi\Vert_{L^\infty(\mathbb{R}^n)}].
$$
$\mu_0>0$
先来讨论$\mu_0>0$的情形, 此时只需证明
$$
\pm u+At+B\ge 0.
$$
取$v_r(x,t)=\frac{M(r)}{r^2}[|x|^2+mt],$ $m\ge 0$待定,
$w=\pm u+At+B+v_r.$ 我们希望通过证明$w\ge 0,$
令$r\rightarrow\infty$给出原不等式. 我们利用弱极值原理给出证明,
因此考虑$w_t+Lw\ge 0.$
$$
w_t+Lw=\pm f+A+c(At+B)+\frac{M(r)}{r^2}[m-2a^{ij}\delta_{ij}+2b^ix_i+c|x|^2+cmt]
$$
注意到$|x|^{\alpha_1}\le \varepsilon|x|^2+\frac{1}{4\varepsilon},$
那么由系数条件, 可以令$\varepsilon$充分小,
使得$a^{ij}$项被$c|x|^2,m$控制. $b^i$项同理.
因此当$m=C(\mu_0,\alpha_1,\alpha_2,\mu_1)>>1$时, $Q_T$上有$w_t+Lw\ge 0.$
由于$w|_{\partial B_r\times (0,t]}\ge (M(r)\pm u)|_{\partial B_r\times (0,t]}\ge 0,$
$w|_{t=0}\ge B\pm \psi\ge 0,$ 由弱极值原理我们有$w\ge 0,$
在$\overline B_r(0)\times [0,t]$上, 即
$$
|u(x,t)|\le At+B+v_r(x,t).
$$
那么任意固定一个$(x_0,t_0)\in Q_T,$ $\,\forall\,r>|x_0|,$ 有
$$
|u(x_0,t_0)|\le AT+B+v_r(x_0,t_0)\rightarrow AT+B.
$$
这样就给出了结论.
$\mu_0\le 0$
作$v(x,t)=u(x,t)e^{(\mu_0-\delta)t},$ $\delta>0.$
那么定义$\widehat L=L-(\mu_0-\delta),$ $\widehat c\ge \delta>0.$
我们有热方程
$$
v_t+\widehat Lv=\widehat f:=fe^{(\mu_0-\delta)t}.
$$
此时对$v$就可以应用得到的结果, 给出
$$
|v|\le \widehat AT+B,\quad |u|\le e^{-(\mu_0-\delta)t}[T\Vert e^{(\mu_0-\delta)t}f\Vert_{L^\infty(Q_T)}+\Vert\psi\Vert_{L^\infty(\mathbb{R}^n)}].
$$
令$\delta\rightarrow 0+$就给出了结论. 定理证毕.
一般的, 可以要求$M(r)=o(r^\alpha),$ 同时调整$\alpha_1,\alpha_2$的范围.
对于有界区域问题, 我们有
定理 4. 设$\Omega$为有界开集, $u\in C^{2,1}(\Omega_T)\cap C(\overline\Omega_T)$为热方程初边值问题的解, 若$[a^{ij}]$对称半正定, 则
$$
\sup_{\Omega_T}|u|\le e^{(-\mu_0)^+T}[T\Vert f\Vert_{L^\infty(\Omega_T)}+\Vert g\Vert_{L^\infty(\partial_p\Omega_T)}].
$$
由此立即得到热方程初边值问题解的唯一性.
Cauchy问题唯一性
Cauchy问题的唯一性要做很多要求, 如通过前面的估计, 我们有:
定理 5. 在最大模估计条件基础上, Cauchy方程至多只有一个由$r^\alpha$控制(多项式增长)的$C^{2,1}(\Omega_T)\cap C(\overline\Omega_T)$中的解.
接下来, 我们讨论在一种超指数增长控制下的解的唯一性. 设系数满足
$$
\sum|a^{ij}|+\sum |b^i|+|c|\le m,\quad \,\forall\,(x,t)\in Q_T,
$$
$$
E(Q_T):=\{u\in C(\overline Q_T):\,\exists\,k>1,\text{\,s.t.\,}|u|\le ke^{k|x|^2},\:\,\forall\,(x,t)\in Q_T\}.
$$
定理 6. 在$[a_{ij}]$半正定且系数满足如上条件下, Cauchy问题在$C^{2,1}(Q_T)\cap E(Q_T)$中的解唯一.
只需讨论$f=0,\psi=0$的情况. 我们希望构造$w_t+Lw\ge 0,$
然后令$v=\varepsilon w\pm u.$ 证明$v_t+Lv\ge 0$推出$v\ge 0.$
这样$|u|\le \varepsilon w,$ 令$\varepsilon\rightarrow 0$即可.
采用分层法解决. 断言存在$T_0=C(m,n,T)>0,$ 问题在$Q_{T_0}$内解唯一.
记$\widehat k=k+1.$
$w(x,t):=(\widehat k^2\mu t+\widehat k)e^{(\widehat k^2\mu t+\widehat k)|x|^2}.$
$\mu>0$待定. 计算:
$$
w_t=(\widehat k^2\mu+\widehat k^2\mu|x|^2 (\widehat k^2\mu t+\widehat k))e^{(\widehat k^2\mu t+\widehat k)|x|^2}
$$
$$
Lw=(\widehat k^2\mu t+\widehat k)e^{(\widehat k^2\mu t+\widehat k)|x|^2}(-2a_{ij}(\delta^{ij}(\widehat k^2\mu t+\widehat k)+x_ix_j(\widehat k^2\mu t+\widehat k)^2)+2b^ix_i(\widehat k^2\mu t+\widehat k)+c)
$$
取$T_0\le \min\{T,\frac{1}{\mu \widehat k}\},$
则$\,\forall\,t\in [0,T_0],$
$$
w_t\ge e^{(\widehat k^2\mu t+\widehat k)|x|^2}(\widehat k^2\mu +\widehat k^3\mu |x|^2),
$$
$$
|Lw|\le C(n,m)e^{(\widehat k^2\mu t+\widehat k)|x|^2}(\widehat k^2+\widehat k^3|x|^2).
$$
令$\mu\ge 2C$即有$w_t+Lw>0.$
令$v=\varepsilon w\pm u,$ 取$r$使得$\varepsilon e^{r^2}=1.$
当$\varepsilon\rightarrow 0$时,
$r_\varepsilon=\sqrt{|\log \varepsilon|}\rightarrow\infty.$ 记
$$
Q_{r,T_0}=\{|x|<r,0<t\le T_0\}.
$$
当$|x|=r$时,
$$
\varepsilon w\ge \varepsilon\widehat k e^{\widehat k r^2}\ge ke^{kr^2}\ge \pm u.
$$
因此由极值原理, $Q_{r,T_0}$上恒有$v\ge 0,$ 即$|u|\le \varepsilon w.$
$\,\forall\,(x_0,t_0)\in Q_{T_0},$ 总有充分小的$\varepsilon_0,$
使得$\,\forall\,\varepsilon\in (0,\varepsilon_0),$
$|x_0|<r_\varepsilon,$
因此$|u(x_0,t_0)|\le \varepsilon w(x_0,t_0)\rightarrow 0,$
由$(x_0,t_0)$任意性, $u\equiv 0.$
我们只证明了$Q_{T_0}$上的唯一性. 但从$t=T_0$平面出发,
我们可得到$Q_{2T_0}$上的唯一性. 逐层归纳,
我们便得到了$Q_T$上问题的唯一性.
定理 7. 在上述定理基础上, 有最大模估计
$$
\Vert u\Vert_{L^\infty(Q_T)}\le C(T,n,m,k)[\Vert f\Vert_{L^\infty(Q_T)}T+\Vert\psi\Vert_{L^\infty(\mathbb{R}^n)}].
$$
文章最后更新于 2022-12-24 09:06:43
Equations in Non-Divergence Form
Remember$\Omega_T=\Omega\times(0,T],$
Parabolic Boundary$\partial_p\Omega=\overline\Omega_T\setminus \Omega_T.$
Consider the equation $u_t+Lu=f,$ $Lu=-a^{ij}u_{ij}+b^iu_i+cu.$
It is required that $[a^{ij}]$ is positive semi-definite within $\Omega_T$.
Let $u\in C^{2,1}(\Omega_T):=\{u(x,t)\in C^2_x(\Omega),C^1_t\in (0,T]\}.$
Theorem 1 (Weak extreme value principle). Assume $\Omega$ is a bounded open set, $a^{ij},b^i,c\in \Omega_T$ satisfies appropriate smoothness conditions, $u\in C^{2,1}(\Omega_T)\cap C(\overline\Omega_T),$ $c\ge 0,$, then $u_t+Lu\le 0$ can be deduced, and $\max_{\overline\Omega_T}u\le \sup_{\partial_p\Omega_T}u^+.$ can be negative to obtain the control of the minimum value.
Proof by contradiction, assuming that there is
$$
\max_{\overline\Omega_T}=u(x_0,t_0)>\sup_{\partial_p\Omega_T}u^+\ge 0.
$$
Take $v_\varepsilon=u-\varepsilon t,$ as long as $\varepsilon$ is sufficiently small,
Then $v_\varepsilon$ still takes the non-negative maximum value internally.
Therefore at the maximum value point $(x_\varepsilon,t_\varepsilon),$ $Dv_{\varepsilon}=0,$
$D^2v_\varepsilon\le 0,$ $v_{\varepsilon}'\ge 0.$
Expand $L$ to get $v_{\varepsilon}'+Lv_\varepsilon\ge 0.$
(Use $c,v_\varepsilon\ge 0.$)
on the other hand,
$$
0\le v_\epsilon'+Lv_\varepsilon=u_t+Lu-\varepsilon-c\varepsilon t\le -\varepsilon.
$$
A contradiction, so the non-negative maximum value of $u$ is taken on the parabolic boundary.
When $c\equiv 0$, the requirement for the positive part can be ignored; when $c\ge \mu$ is a constant,
Desirable $v=e^{\mu t}u,$ Then
$$
v_t+(L-\mu)v=e^{\mu t}(u_t+Lu).
$$
Therefore,
$$
u_t+Lu\le 0\Rightarrow v_t+\widehat Lv\le 0\Rightarrow \max_{\overline\Omega_T}e^{\mu t}u\le \sup_{\partial_p\Omega_T}(e^{\mu t}u)^{+}.
$$
Theorem 2. Assume $\Omega\subset \mathbb{R}^n$ is a bounded open set, $[a^{ij}]\ge 0,$ $c\ge \mu,$, then the initial boundary value problem of the heat equation has a unique solution on $C(\overline\Omega_T)\cap C^{2,1}(\Omega_T)$.
maximum mode estimate
Consider the Cauchy problem:
$$
\begin{cases}
u_t+Lu=f&in\:Q_T:=\mathbb{R}^n\times (0,T]\\
u|_{t=0}=\psi&in\:\mathbb{R}^n
\end{cases}
$$
Assume that the coefficient satisfies: $\,\exists\,\alpha_1\in [0,2),$
$\alpha_2\in [0,1),$ $\mu_0\in \mathbb{R},$ $\mu_1\ge 0,$ makes
$$
c\ge \mu_0,\quad\sum_{i,j=1}^n |a^{ij}|\le \mu_1(|x|^{\alpha_1}+1),\quad \sum_{i=1}^n |b^i|\le \mu_1(|x|^{\alpha_2}+1).
$$
Theorem 3. Let $u\in C(\overline Q_T)\cap C^{2,1}(Q_T)$ be the solution of Cauchy problem, $\lim\limits_{r\rightarrow\infty} \frac{M(r)}{r^2}=0,$ $M(r):=\sup\limits_{\begin{subarray}{c} |x|=r\\ 0<t<T \end{subarray} }|u(t)|,$ then
$$
\sup_{Q_T}|u|\le e^{(-\mu_0)^+T}[T\Vert f\Vert_{L^\infty(Q_T)}+\Vert\psi\Vert_{L^\infty(\mathbb{R}^n)}].
$$
$\mu_0>0$
Let’s discuss the case $\mu_0>0$ first. At this time we only need to prove
$$
\pm u+At+B\ge 0.
$$
Take $v_r(x,t)=\frac{M(r)}{r^2}[|x|^2+mt],$ $m\ge 0$ to be determined,
$w=\pm u+At+B+v_r.$ We hope to prove $w\ge 0,$
Let $r\rightarrow\infty$ give the original inequality. We use the weak extreme value principle to give the proof,
Therefore consider $w_t+Lw\ge 0.$
$$
w_t+Lw=\pm f+A+c(At+B)+\frac{M(r)}{r^2}[m-2a^{ij}\delta_{ij}+2b^ix_i+c|x|^2+cmt]
$$
Notice$|x|^{\alpha_1}\le \varepsilon|x|^2+\frac{1}{4\varepsilon},$
Then according to the coefficient condition, $\varepsilon$ can be made sufficiently small,
Make the $a^{ij}$ item controlled by $c|x|^2,m$. The same goes for the $b^i$ item.
So when $m=C(\mu_0,\alpha_1,\alpha_2,\mu_1)>>1$, there is $w_t+Lw\ge 0.$ on $Q_T$
Due to $w|_{\partial B_r\times (0,t]}\ge (M(r)\pm u)|_{\partial B_r\times (0,t]}\ge 0,$
$w|_{t=0}\ge B\pm \psi\ge 0,$ From the weak extreme value principle we have $w\ge 0,$
On $\overline B_r(0)\times [0,t]$, that is
$$
|u(x,t)|\le At+B+v_r(x,t).
$$
Then arbitrarily fixing a $(x_0,t_0)\in Q_T,$ $\,\forall\,r>|x_0|,$ has
$$
|u(x_0,t_0)|\le AT+B+v_r(x_0,t_0)\rightarrow AT+B.
$$
This gives the conclusion.
$\mu_0\le 0$
Make $v(x,t)=u(x,t)e^{(\mu_0-\delta)t},$ $\delta>0.$
Then define $\widehat L=L-(\mu_0-\delta),$ $\widehat c\ge \delta>0.$
We have the heat equation
$$
v_t+\widehat Lv=\widehat f:=fe^{(\mu_0-\delta)t}.
$$
At this time, the obtained results can be applied to $v$, giving
$$
|v|\le \widehat AT+B,\quad |u|\le e^{-(\mu_0-\delta)t}[T\Vert e^{(\mu_0-\delta)t}f\Vert_{L^\infty(Q_T)}+\Vert\psi\Vert_{L^\infty(\mathbb{R}^n)}].
$$
Let $\delta\rightarrow 0+$ give the conclusion. The theorem is proved.
Generally, you can ask $M(r)=o(r^\alpha),$ to adjust the range of $\alpha_1,\alpha_2$ at the same time.
For bounded region problems, we have
Theorem 4. Assume $\Omega$ is a bounded open set, $u\in C^{2,1}(\Omega_T)\cap C(\overline\Omega_T)$ is the solution to the initial boundary value problem of the heat equation, if $[a^{ij}]$ is symmetric and positive semi-definite, then
$$
\sup_{\Omega_T}|u|\le e^{(-\mu_0)^+T}[T\Vert f\Vert_{L^\infty(\Omega_T)}+\Vert g\Vert_{L^\infty(\partial_p\Omega_T)}].
$$
From this, the uniqueness of the solution to the initial boundary value problem of the heat equation is immediately obtained.
Cauchy problem uniqueness
The uniqueness of the Cauchy problem requires many requirements. For example, through the previous estimation, we have:
Theorem 5. Based on the maximum modular estimation condition, the Cauchy equation has at most one solution in $C^{2,1}(\Omega_T)\cap C(\overline\Omega_T)$ controlled by $r^\alpha$ (polynomial growth).
Next, we discuss the uniqueness of the solution under a super-exponential growth control. Suppose the coefficient satisfies
$$
\sum|a^{ij}|+\sum |b^i|+|c|\le m,\quad \,\forall\,(x,t)\in Q_T,
$$
$$
E(Q_T):=\{u\in C(\overline Q_T):\,\exists\,k>1,\text{\,s.t.\,}|u|\le ke^{k|x|^2},\:\,\forall\,(x,t)\in Q_T\}.
$$
Theorem 6. When $[a_{ij}]$ is positive semi-definite and the coefficients meet the above conditions, the solution of the Cauchy problem in $C^{2,1}(Q_T)\cap E(Q_T)$ is unique.
Just discuss the case of $f=0,\psi=0$. We wish to construct $w_t+Lw\ge 0,$
Then let $v=\varepsilon w\pm u.$ prove $v_t+Lv\ge 0$ and deduce $v\ge 0.$
In this way, $|u|\le \varepsilon w,$ can be replaced by $\varepsilon\rightarrow 0$.
Use the hierarchical method to solve it. It is asserted that there is a problem in $T_0=C(m,n,T)>0,$ and the solution is unique in $Q_{T_0}$.
Remember$\widehat k=k+1.$
$w(x,t):=(\widehat k^2\mu t+\widehat k)e^{(\widehat k^2\mu t+\widehat k)|x|^2}.$
$\mu>0$ To be determined. Calculation:
$$
w_t=(\widehat k^2\mu+\widehat k^2\mu|x|^2 (\widehat k^2\mu t+\widehat k))e^{(\widehat k^2\mu t+\widehat k)|x|^2}
$$
$$
Lw=(\widehat k^2\mu t+\widehat k)e^{(\widehat k^2\mu t+\widehat k)|x|^2}(-2a_{ij}(\delta^{ij}(\widehat k^2\mu t+\widehat k)+x_ix_j(\widehat k^2\mu t+\widehat k)^2)+2b^ix_i(\widehat k^2\mu t+\widehat k)+c)
$$
Take $T_0\le \min\{T,\frac{1}{\mu \widehat k}\},$
Then $\,\forall\,t\in [0,T_0],$
$$
w_t\ge e^{(\widehat k^2\mu t+\widehat k)|x|^2}(\widehat k^2\mu +\widehat k^3\mu |x|^2),
$$
$$
|Lw|\le C(n,m)e^{(\widehat k^2\mu t+\widehat k)|x|^2}(\widehat k^2+\widehat k^3|x|^2).
$$
Let $\mu\ge 2C$ equal $w_t+Lw>0.$
Let $v=\varepsilon w\pm u,$ take $r$ so that $\varepsilon e^{r^2}=1.$
When $\varepsilon\rightarrow 0$,
$r_\varepsilon=\sqrt{|\log \varepsilon|}\rightarrow\infty.$ Note
$$
Q_{r,T_0}=\{|x|<r,0<t\le T_0\}.
$$
When $|x|=r$,
$$
\varepsilon w\ge \varepsilon\widehat k e^{\widehat k r^2}\ge ke^{kr^2}\ge \pm u.
$$
Therefore, according to the extreme value principle, there is always $v\ge 0,$ on $Q_{r,T_0}$, that is, $|u|\le \varepsilon w.$
$\,\forall\,(x_0,t_0)\in Q_{T_0},$ There is always enough small $\varepsilon_0,$
Make$\,\forall\,\varepsilon\in (0,\varepsilon_0),$
$|x_0|<r_\varepsilon,$
Therefore $|u(x_0,t_0)|\le \varepsilon w(x_0,t_0)\rightarrow 0,$
By $(x_0,t_0)$ arbitrariness, $u\equiv 0.$
We only proved the uniqueness on $Q_{T_0}$. But starting from the $t=T_0$ plane,
We can get the uniqueness on $Q_{2T_0}$. By induction layer by layer,
We then obtain the uniqueness of the problem on $Q_T$.
Theorem 7. Based on the above theorem, there is a maximum modular estimate
$$
\Vert u\Vert_{L^\infty(Q_T)}\le C(T,n,m,k)[\Vert f\Vert_{L^\infty(Q_T)}T+\Vert\psi\Vert_{L^\infty(\mathbb{R}^n)}].
$$
The article was last updated on 2022-12-24 09:06:43
