Machine-translated from Chinese.
动机
考虑问题
$$
\left\{\begin{aligned}
u_t-\Delta u=0&\quad in\:\Omega_T\\
u=0&\quad on\:\partial\Omega\\
u|_{t=0}=h&\quad in\:\Omega
\end{aligned}\right.
$$
一种很经典的处理方法是分离变量法, 如果有$u(x,t)=w(x)H(t),$ 那么
$$
H_tw-\Delta wH=0,\quad w|_{\partial\Omega}=0.
$$
进而,
$$
\frac{\Delta w}{w}=\frac{H_t}{H}=-\lambda,
$$
此时$w$为$\Delta$的特征函数.
取$\lambda_1\le \lambda_2\le\cdots\rightarrow \infty,$
那么$\{w_k\}_{k=1}^\infty$构成一组正交基.
我们希望通过$u_m=\sum_{k=1}^m C_m^k(t) w_k\rightarrow u$给出弱解.
Galerkin逼近
可以取到$\{w_k\}\subset C^\infty(\Omega)\cap H_0^1(\Omega)\cap H^2(\Omega),$
使得$\{w_k\}$为$L^2(\Omega)$标准正交基, $H_0^1(\Omega)$正交基.
考虑条件
$$
a^{ij},b^i,c\in L^\infty(\Omega_T),\quad f\in L^2(\Omega_T),\quad h\in L^2(\Omega).
$$
$\,\forall\,m\ge 1,$ 取$u_m=\sum_{k=1}^m C_m^k(t)w_k(x),$
使得$\,\forall\,v\in L(w_1,\cdots,w_m),$
$$
\begin{cases}
\left<{}u_m'(t),v\right>+B_L[u_m,v,t]=(f,v),&a.e.\,t\in (0,T)\\
u_m|_{t=0}=h_m:=\sum_{k=1}^m (h,w_k)w_k(x)
\end{cases}
$$
$h_m\rightarrow h=\sum_{k=1}^\infty (h,w_k)w_k.$
将$u_m$展开, 记$e_l^k=B[w_l,w_k;t],$ 我们只需解ODE方程组
$$
\begin{cases}
{C_m^k}'(t)+C_m^le_l^k=(f,w_k)\\
C_m^k(0)=(h,w_k)
\end{cases}
$$
注意到$(f,w_k)(t)=\int_\Omega f(x,t)w_k(x)dx\in L^2(0,T),$
$e_l^k(t)\in L^\infty(0,T),$ 由ODE理论, 我们有如下引理:
引理 1. 上述ODE方程组在$W^{1,2}(0,T)$中解$\{C_m^k\}$存在唯一. 因此原方程有$W^{1,2}(0,T;H_0^1(\Omega))$中的解$u_m.$
在$f$关于时间$L^\infty$有界时, 方程在$W^{1,\infty}(0,\infty)$中也有解,
此时$u_m$还可以取在 $W^{1,\infty}(0,T;H_0^1(\Omega))$ 中.
能量估计
零阶项估计
为了说明收敛性, 我们需要给出能量的有界性. 在方程组中取$v=u_m,$ 那么
$$
\frac{1}{2}\frac{d {} }{d {}t}\Vert u_m(t)\Vert_{L^2(\Omega)}^2+B[u_m,u_m,t]=(f,u_m)\le \frac{\Vert f(t)\Vert_{L^2(\Omega)}^2+\Vert u_m(t)\Vert_{L^2(\Omega)}^2}{2}.
$$
由椭圆型方程能量估计,
$$
B[u_m,u_m;t]\ge \frac{\lambda}{2}\Vert Du_m\Vert_{L^2}^2-\mu(L,\Omega)\Vert u_m\Vert_{L^2}^2.
$$
因此,
$$
\frac{d {} }{d {}t}\Vert u_m\Vert_{L^2}^2+\lambda\Vert u_m\Vert_{H^1}^2\le C(\Vert f\Vert_{L^2}^2+\Vert u_m\Vert_{L^2}^2).
$$
对于
$$
\frac{d {}F}{d {}t}\le CF(t)+g(t),\quad \frac{d {} }{d {}t}(e^{-Ct}F(t))\le e^{-Ct}g(t),
$$
$$
e^{-Ct}F(t)\le F(0)+\int_0^te^{-cs}g(s)ds,\quad F(t)\le e^{Ct}F(0)+\int_0^te^{C(t-s)}g(s)ds.
$$
这样我们就得到了Gronwall不等式. 进一步有
$$
F(t)\le C(T)\left(F(0)+\int_0^Tg(s)ds\right).
$$
若$g$非负非减,
我们也能给出更精确的估计.
于是由原不等式, 我们得到
$$
\Vert u_m(t)\Vert_{L^2(\Omega)}^2\le C(T)\left(\Vert\varphi_m\Vert_{L^2(\Omega)}^2+\Vert f\Vert_{L^2(\Omega_T)}^2\right).
$$
进一步有
$$
\Vert u_m(t)\Vert_{L^2(\Omega)}^2+\int_0^t\Vert u_m(s)\Vert_{H^1}^2ds\le C(T)\left(\Vert\varphi_m\Vert_{L^2(\Omega)}^2+\Vert f\Vert_{L^2(\Omega_T)}^2\right).
$$
于是$u_m\in L^2(0,T;H_0^1(\Omega)),$
事实上也在$L^\infty(0,T;H_0^1(\Omega))$中.
接下来需要说明的是$u_m'\in L^2(0,T;H^{-1}(\Omega)).$
导数项估计
$\,\forall\,v\in H_0^1(\Omega),$ 取$v_m=\sum_{k=1}^m (v,w_k)w_k.$
对几乎处处的$t\in (0,T),$ 有
$$
\begin{aligned}
|\left<{}u_m',v\right>|&=|(u_m',v_m)|=|(f,v_m)-B[u_m,v_m;t]|\\
&\le \Vert f\Vert_{L^2}\Vert v_m\Vert_{L^2}+C\Vert u_m\Vert_{H^1}\Vert v_m\Vert_{H^1}\le (\Vert f\Vert_{L^2}+C\Vert u_m\Vert_{H^1})\Vert v\Vert_{H^1}
\end{aligned}
$$
因此,
$$
\Vert u_m'(t)\Vert_{H^{-1}(\Omega)}\le \Vert f\Vert_{L^2}+C\Vert u_m\Vert_{H^1}
$$
定理 2. 在合适的系数条件下, 逼近解$u_m\in C([0,T];L^2(\Omega))$满足
$$
\max_{[0,T]}\Vert u_m\Vert_{L^2}+\Vert u_m\Vert_{L^2(0,T;H_0^1)}+\Vert u_m'\Vert_{L^2(0,T;H^{-1})}\le C(T,L,\Omega,n)\left(\Vert f\Vert_{L^2(\Omega_T)}+\Vert\varphi_m\Vert_{L^2(\Omega)}\right).
$$
当然可将$\varphi_m$换为$\varphi$给出一致的上界.
弱解存在性
分为三步进行:
弱收敛
由于$\Vert u_m\Vert_{L^2(0,T;H_0^1)},$
$\Vert u_m'\Vert_{L^2(0,T;H^{-1})}$一致有界, 我们可以取到弱收敛子列,
分别在各自空间中弱收敛到$u,v.$ 下证$u'=v\in H^{-1},$ $a.e.$
$t\in [0,T].$
结论等价于
$$
\frac{d {} }{d {}t}\left<{}u(t),\varphi\right>=\left<{}u'(t),\varphi\right>=\left<{}v,\varphi\right>,\quad \,\forall\,\varphi\in H_0^1(\Omega),\quad a.e.\:t\in [0,T].
$$
由关于时间弱导数定义, 这又等价于$\,\forall\,\eta\in C_0^\infty(0,T),$
$$
-\int_0^T\left<{}u(t),\varphi\right>\eta'=\int_0^T\left<{}v,\varphi\right>\eta.
$$
而$\,\forall\,\varphi\in H_0^1(\Omega),$
$$
\int_0^T\left<{}v,\varphi\right>\eta dt=\lim_{m\rightarrow\infty}\int_0^T(u_m',\varphi)\eta dt=-\lim_{m\rightarrow\infty}\int_0^T(u_m,\varphi)\eta' dt=-\int_0^T(u,\varphi)\eta'dt.
$$
这就给出了$u'=v$.
热方程弱解
接下来验证$u$确实是热方程的弱解. 取$v_N:=\sum_{k=1}^N(v,w_k)w_k.$
若$m>N,$ 则有
$$
\int_0^T(u_m',v_N)\eta dt=\int_0^T\left((f,v_N)-B[u_m,v_N;t]\right)\eta dt.
$$
令$m\rightarrow\infty,$ 再令$N\rightarrow\infty,$ 即有
$$
\int_0^T(u',v)\eta dt=\int_0^T\left((f,v)-B[u,v;t]\right)\eta dt,\quad \,\forall\,\eta\in C[0,T].
$$
这就说明了$u$为热方程弱解.
初值条件
最后还需验证满足初值条件. 由于取$u_m$时考虑了初值条件,
取$\eta\in C^1[0,T],$ $\eta(T)=0,$ $\eta(0)=1.$ 对于$m>N,$ 我们有
$$
\int_0^T(u_m',v_N)\eta dt=(u_m,v_N)\eta|^{T}_{t=0}-\int_0^T(u_m,v_N)\eta'dt=-(\varphi_m,v_N)-\int_0^T(u_m,v_N)\eta'dt.
$$
令$m\rightarrow\infty,$ 再令$N\rightarrow\infty,$ 我们有:
$$
\int_0^T(u',v)\eta dt=-(\varphi,v)-\int_0^T(u,v)\eta'dt.
$$
另一方面, 由Lions定理, $u\in C([0,T],L^2),$
在$L^2$意义下$u(0)=\lim_{t\rightarrow 0+}u(t).$ 因此和前面一样的有
$$
\int_0^T(u',v)\eta dt=(u,v)\eta|^{T}_{t=0}-\int_0^T(u,v)\eta'dt=-(u(0),v)-\int_0^T(u,v)\eta'dt
$$
于是$u$也满足初值条件. 由范数的弱sharp连续性, 我们有:
定理 3. 在热方程系数满足适当条件时, 热方程存在弱解, 且满足能量估计:
$$
\max_{[0,T]}\Vert u\Vert_{L^2}+\Vert u\Vert_{L^2(0,T;H_0^1)}+\Vert u'\Vert_{L^2(0,T;H^{-1})}\le C(T,L,\Omega,n)\left(\Vert f\Vert_{L^2(\Omega_T)}+\Vert\varphi\Vert_{L^2(\Omega)}\right).
$$
弱解唯一性
前面的能量估计只能说明通过Galerkin方法得到的弱解唯一,
但不见得所有弱解都是通过Galerkin方法得到的, 因此我们仍需寻求另外的做法.
另外证明了唯一性之后,
我们便说明了唯一的弱解一定可以通过Galerkin方法逼近得到.
这样任意的弱解也就都适用前面得到的能量估计了.
只需考虑$f\equiv 0,$ $\varphi\equiv 0$的情形. $u(t)\in H_0^1(\Omega),$
$a.e.$ $t.$ 只需考虑如下方程是否只有零解.
$$
\left<{}u_t,u\right>+B[u,u;t]=0.
$$
和前面一样的, 我们有
$$
\frac{d {} }{d {}t}\Vert u(t)\Vert^2_{L^2}=\left<{}u_t,u\right>,\quad B[u,u;t]\ge \frac{\lambda}{2}\Vert Du\Vert_{L^2}^2-\mu\Vert u\Vert_{L^2}^2.
$$
因此
$$
\frac{d {} }{d {}t}\Vert u(t)\Vert_{L^2}^2=-B\le \mu\Vert u\Vert_{L^2}^2.
$$
这样由基本的ODE知识, 就有
$$
\Vert u(0)\Vert_{L^2}^2=0\quad\Rightarrow\quad \Vert u(t)\Vert_{L^2}^2\equiv 0\quad\Rightarrow\quad u\equiv 0\quad in\:\Omega_T.
$$
这样我们就证明了弱解的唯一性, 也证明了弱解总可由Galerkin逼近得到.
文章最后更新于 2022-12-08 10:24:09
motivation
Consider the problem
$$
\left\{\begin{aligned}
u_t-\Delta u=0&\quad in\:\Omega_T\\
u=0&\quad on\:\partial\Omega\\
u|_{t=0}=h&\quad in\:\Omega
\end{aligned}\right.
$$
A very classic processing method is the separation of variables method. If there is $u(x,t)=w(x)H(t),$ then
$$
H_tw-\Delta wH=0,\quad w|_{\partial\Omega}=0.
$$
Furthermore,
$$
\frac{\Delta w}{w}=\frac{H_t}{H}=-\lambda,
$$
At this time $w$ is the characteristic function of $\Delta$.
Take $\lambda_1\le \lambda_2\le\cdots\rightarrow \infty,$
Then $\{w_k\}_{k=1}^\infty$ forms a set of orthogonal basis.
We hope to give a weak solution through $u_m=\sum_{k=1}^m C_m^k(t) w_k\rightarrow u$.
Galerkin approaches
Can get $\{w_k\}\subset C^\infty(\Omega)\cap H_0^1(\Omega)\cap H^2(\Omega),$
Let $\{w_k\}$ be $L^2(\Omega)$ orthonormal basis and $H_0^1(\Omega)$ orthonormal basis.
Consider conditions
$$
a^{ij},b^i,c\in L^\infty(\Omega_T),\quad f\in L^2(\Omega_T),\quad h\in L^2(\Omega).
$$
$\,\forall\,m\ge 1,$ takes $u_m=\sum_{k=1}^m C_m^k(t)w_k(x),$
Make$\,\forall\,v\in L(w_1,\cdots,w_m),$
$$
\begin{cases}
\left<{}u_m'(t),v\right>+B_L[u_m,v,t]=(f,v),&a.e.\,t\in (0,T)\\
u_m|_{t=0}=h_m:=\sum_{k=1}^m (h,w_k)w_k(x)
\end{cases}
$$
$h_m\rightarrow h=\sum_{k=1}^\infty (h,w_k)w_k.$
Expand $u_m$ and write it as $e_l^k=B[w_l,w_k;t],$. We only need to solve the ODE equations
$$
\begin{cases}
{C_m^k}'(t)+C_m^le_l^k=(f,w_k)\\
C_m^k(0)=(h,w_k)
\end{cases}
$$
Notice$(f,w_k)(t)=\int_\Omega f(x,t)w_k(x)dx\in L^2(0,T),$
$e_l^k(t)\in L^\infty(0,T),$ From ODE theory, we have the following lemma:
Lemma 1. The above ODE system of equations has a unique solution $\{C_m^k\}$ in $W^{1,2}(0,T)$. Therefore, the original equation has a solution $u_m.$ in $W^{1,2}(0,T;H_0^1(\Omega))$
When $f$ is bounded with respect to time $L^\infty$, the equation also has a solution in $W^{1,\infty}(0,\infty)$,
At this time $u_m$ can also be taken from $W^{1,\infty}(0,T;H_0^1(\Omega))$.
energy estimate
Zero-order term estimate
In order to illustrate the convergence, we need to give the boundedness of the energy. Taking $v=u_m,$ in the system of equations then
$$
\frac{1}{2}\frac{d {} }{d {}t}\Vert u_m(t)\Vert_{L^2(\Omega)}^2+B[u_m,u_m,t]=(f,u_m)\le \frac{\Vert f(t)\Vert_{L^2(\Omega)}^2+\Vert u_m(t)\Vert_{L^2(\Omega)}^2}{2}.
$$
Estimated from the elliptic equation energy,
$$
B[u_m,u_m;t]\ge \frac{\lambda}{2}\Vert Du_m\Vert_{L^2}^2-\mu(L,\Omega)\Vert u_m\Vert_{L^2}^2.
$$
Therefore,
$$
\frac{d {} }{d {}t}\Vert u_m\Vert_{L^2}^2+\lambda\Vert u_m\Vert_{H^1}^2\le C(\Vert f\Vert_{L^2}^2+\Vert u_m\Vert_{L^2}^2).
$$
for
$$
\frac{d {}F}{d {}t}\le CF(t)+g(t),\quad \frac{d {} }{d {}t}(e^{-Ct}F(t))\le e^{-Ct}g(t),
$$
$$
e^{-Ct}F(t)\le F(0)+\int_0^te^{-cs}g(s)ds,\quad F(t)\le e^{Ct}F(0)+\int_0^te^{C(t-s)}g(s)ds.
$$
In this way we get Gronwall's inequality. Further we have
$$
F(t)\le C(T)\left(F(0)+\int_0^Tg(s)ds\right).
$$
If $g$ is non-negative and non-decreasing,
We can also give more precise estimates.
So from the original inequality, we get
$$
\Vert u_m(t)\Vert_{L^2(\Omega)}^2\le C(T)\left(\Vert\varphi_m\Vert_{L^2(\Omega)}^2+\Vert f\Vert_{L^2(\Omega_T)}^2\right).
$$
There are further
$$
\Vert u_m(t)\Vert_{L^2(\Omega)}^2+\int_0^t\Vert u_m(s)\Vert_{H^1}^2ds\le C(T)\left(\Vert\varphi_m\Vert_{L^2(\Omega)}^2+\Vert f\Vert_{L^2(\Omega_T)}^2\right).
$$
So$u_m\in L^2(0,T;H_0^1(\Omega)),$
In fact it's also in $L^\infty(0,T;H_0^1(\Omega))$.
What needs to be explained next is $u_m'\in L^2(0,T;H^{-1}(\Omega)).$
Derivative term estimation
$\,\forall\,v\in H_0^1(\Omega),$ takes $v_m=\sum_{k=1}^m (v,w_k)w_k.$
Yes for $t\in (0,T),$ almost everywhere
$$
\begin{aligned}
|\left<{}u_m',v\right>|&=|(u_m',v_m)|=|(f,v_m)-B[u_m,v_m;t]|\\
&\le \Vert f\Vert_{L^2}\Vert v_m\Vert_{L^2}+C\Vert u_m\Vert_{H^1}\Vert v_m\Vert_{H^1}\le (\Vert f\Vert_{L^2}+C\Vert u_m\Vert_{H^1})\Vert v\Vert_{H^1}
\end{aligned}
$$
Therefore,
$$
\Vert u_m'(t)\Vert_{H^{-1}(\Omega)}\le \Vert f\Vert_{L^2}+C\Vert u_m\Vert_{H^1}
$$
Theorem 2. Under appropriate coefficient conditions, the approximate solution $u_m\in C([0,T];L^2(\Omega))$ satisfies
$$
\max_{[0,T]}\Vert u_m\Vert_{L^2}+\Vert u_m\Vert_{L^2(0,T;H_0^1)}+\Vert u_m'\Vert_{L^2(0,T;H^{-1})}\le C(T,L,\Omega,n)\left(\Vert f\Vert_{L^2(\Omega_T)}+\Vert\varphi_m\Vert_{L^2(\Omega)}\right).
$$
Of course, $\varphi_m$ can be replaced by $\varphi$ to give a consistent upper bound.
existence of weak solution
It is divided into three steps:
weak convergence
Due to $\Vert u_m\Vert_{L^2(0,T;H_0^1)},$
$\Vert u_m'\Vert_{L^2(0,T;H^{-1})}$ is uniformly bounded, and we can obtain weakly convergent subsequences,
Weakly converge to $u,v.$ in their respective spaces and prove $u'=v\in H^{-1},$ $a.e.$
$t\in [0,T].$
The conclusion is equivalent to
$$
\frac{d {} }{d {}t}\left<{}u(t),\varphi\right>=\left<{}u'(t),\varphi\right>=\left<{}v,\varphi\right>,\quad \,\forall\,\varphi\in H_0^1(\Omega),\quad a.e.\:t\in [0,T].
$$
Defined by weak derivatives with respect to time, this is equivalent to $\,\forall\,\eta\in C_0^\infty(0,T),$
$$
-\int_0^T\left<{}u(t),\varphi\right>\eta'=\int_0^T\left<{}v,\varphi\right>\eta.
$$
And $\,\forall\,\varphi\in H_0^1(\Omega),$
$$
\int_0^T\left<{}v,\varphi\right>\eta dt=\lim_{m\rightarrow\infty}\int_0^T(u_m',\varphi)\eta dt=-\lim_{m\rightarrow\infty}\int_0^T(u_m,\varphi)\eta' dt=-\int_0^T(u,\varphi)\eta'dt.
$$
This gives $u'=v$.
Weak solution to the heat equation
Next, verify that $u$ is indeed a weak solution to the heat equation. Take $v_N:=\sum_{k=1}^N(v,w_k)w_k.$
If $m>N,$ then there is
$$
\int_0^T(u_m',v_N)\eta dt=\int_0^T\left((f,v_N)-B[u_m,v_N;t]\right)\eta dt.
$$
Let $m\rightarrow\infty,$ and then $N\rightarrow\infty,$ then we have
$$
\int_0^T(u',v)\eta dt=\int_0^T\left((f,v)-B[u,v;t]\right)\eta dt,\quad \,\forall\,\eta\in C[0,T].
$$
This shows that $u$ is a weak solution to the heat equation.
initial value condition
Finally, it is necessary to verify that the initial value condition is met. Since the initial value condition is taken into account when taking $u_m$,
Taking $\eta\in C^1[0,T],$ $\eta(T)=0,$ $\eta(0)=1.$ for $m>N,$ we have
$$
\int_0^T(u_m',v_N)\eta dt=(u_m,v_N)\eta|^{T}_{t=0}-\int_0^T(u_m,v_N)\eta'dt=-(\varphi_m,v_N)-\int_0^T(u_m,v_N)\eta'dt.
$$
Let $m\rightarrow\infty,$ and then $N\rightarrow\infty,$ we have:
$$
\int_0^T(u',v)\eta dt=-(\varphi,v)-\int_0^T(u,v)\eta'dt.
$$
On the other hand, according to Lions’ theorem, $u\in C([0,T],L^2),$
In the sense of $L^2$, $u(0)=\lim_{t\rightarrow 0+}u(t).$ therefore has the same as before
$$
\int_0^T(u',v)\eta dt=(u,v)\eta|^{T}_{t=0}-\int_0^T(u,v)\eta'dt=-(u(0),v)-\int_0^T(u,v)\eta'dt
$$
So $u$ also satisfies the initial value condition. From the weak sharp continuity of the norm, we have:
Theorem 3. When the coefficients of the heat equation meet appropriate conditions, there is a weak solution to the heat equation and the energy estimation is satisfied:
$$
\max_{[0,T]}\Vert u\Vert_{L^2}+\Vert u\Vert_{L^2(0,T;H_0^1)}+\Vert u'\Vert_{L^2(0,T;H^{-1})}\le C(T,L,\Omega,n)\left(\Vert f\Vert_{L^2(\Omega_T)}+\Vert\varphi\Vert_{L^2(\Omega)}\right).
$$
Weak solution uniqueness
The previous energy estimation can only show that the weak solution obtained by the Galerkin method is unique,
But not all weak solutions are obtained through the Galerkin method, so we still need to find another method.
In addition, after proving the uniqueness,
We have shown that the only weak solution can be approximated by the Galerkin method.
In this way, any weak solution can be applied to the energy estimate obtained previously.
Just consider the case $f\equiv 0,$ $\varphi\equiv 0$. $u(t)\in H_0^1(\Omega),$
$a.e.$ $t.$ Just consider whether the following equation has zero solution.
$$
\left<{}u_t,u\right>+B[u,u;t]=0.
$$
As before, we have
$$
\frac{d {} }{d {}t}\Vert u(t)\Vert^2_{L^2}=\left<{}u_t,u\right>,\quad B[u,u;t]\ge \frac{\lambda}{2}\Vert Du\Vert_{L^2}^2-\mu\Vert u\Vert_{L^2}^2.
$$
Therefore
$$
\frac{d {} }{d {}t}\Vert u(t)\Vert_{L^2}^2=-B\le \mu\Vert u\Vert_{L^2}^2.
$$
In this way, based on basic ODE knowledge, we have
$$
\Vert u(0)\Vert_{L^2}^2=0\quad\Rightarrow\quad \Vert u(t)\Vert_{L^2}^2\equiv 0\quad\Rightarrow\quad u\equiv 0\quad in\:\Omega_T.
$$
In this way, we have proved the uniqueness of the weak solution and also proved that the weak solution can always be obtained by Galerkin approximation.
The article was last updated on 2022-12-08 10:24:09
