Machine-translated from Chinese.
考虑自共轭算子$S:H_0^1(\Omega)\rightarrow H^{-1}(\Omega),$
$$
Su:=-(a^{ij}u_i+b^ju)_j+b^ju_j+cu.
$$
系数均在$L^\infty(\Omega)$中,
$a^{ij}\xi_i\xi_j\ge \mu|\xi|^2.$ 我们有双线性有界泛函:
$$
B_S[u,v]=\int_\Omega a^{ij}u_iv_j+b^i(uv_j+vu_j)+cuv=\left<{}Su,v\right>=\left<{}u,Sv\right>=B_S[v,u].
$$
满足能量估计:
$$
B_S[u,u]\ge \frac{\mu}{2}\Vert Du\Vert_{L^2}^2-C(\mu,S)\Vert u\Vert_{L^2}^2.
$$
若存在$(u,\lambda),$ 使得在$H^{-1}(\Omega)$中有$Su=\lambda u,$
则称其为特征对, $\lambda$为特征值, $u$为特征函数.
一个自然的问题是如何找到所有的$\lambda.$
若$\,\exists\,$特征对$(u,\lambda),$ 则$\,\forall\,v\in H_0^1(\Omega),$
$$
B_S[u,v]=\left<{}\lambda u,v\right>=\lambda\int_\Omega uvdx.
$$
此时,
$$
\lambda=\frac{B_S[u,u]}{\int_\Omega u^2dx}\ge -C(\mu,S).
$$
由此启发, 我们定义
$$
\lambda_1:=\inf_{u\in H_0^1(\Omega)\setminus \{0\} }\frac{B_S[u,u]}{\Vert u\Vert_{L^2}^2}
$$
我们来证明$\lambda_1$为第一特征值. 首先证明它是可以达到的.
命题 1. $\,\exists\,w\in H_0^1(\Omega),$ $B_S[w,w]=\lambda_1$
$\,\exists\,w_k,$ 满足$\Vert w_k\Vert_{L^2}=1,$ 使得
$$
\lambda_1\le B_S[w_k,w_k]\le \lambda_1+\frac{1}{k}.
$$
由后面的不等号,
有
$$
\frac{\mu}{2}\Vert Dw_k\Vert_{L^2}^2\le \lambda_1+1+C(\mu,S).
$$
因此$\{w_k\}$为$H_0^1$中的有界列, 有子列在$L^2$内收敛, $H_0^1$中弱收敛.
不妨设子列为自身, 即$w_k\xrightarrow{L^2}w,$
$w_k\rightharpoonup w\in H_0^1.$
希望证明$B_S[w_k,w_k]\rightarrow B_S[w,w].$ 为此逐项考虑.
$$
|\left<{}w_k,w_k\right>-\left<{}w,w\right>|\le |\left<{}w_k,w_k-w\right>|+|\left<{}w_k-w,w\right>|\le 2\Vert w_k-w\Vert_{L^2}\rightarrow 0
$$
$$
|\left<{}\partial_iw_{k},w_k\right>-\left<{}\partial_i w,w\right>|\le |\left<{}\partial_iw_{k},w_k-w\right>|+|\left<{}\partial_i (w_k-w),w\right>|\le (\Vert Dw_k\Vert_{L^2}+\Vert Dw\Vert_{L^2})\Vert w_k-w\Vert_{L^2}
$$
但是$\left<{}\partial_iw_k,\partial_iw_k\right>$项并不好证明收敛性,
我们转而考虑证明
$$
\varliminf_{k\rightarrow 0}\int a^{ij}\partial_iw_k\partial_jw_k\ge \int a^{ij}w_iw_j.
$$
这样结合前面的收敛就有,
$$
\lambda_1\le B_S[w,w]\le \varliminf_{k\rightarrow 0}B_S[w_k,w_k]=\lambda_1\quad\Rightarrow\quad B_S[w,w]=\lambda_1.
$$
由于$a^{ij}$一致椭圆,
$\int_\Omega a^{ij}u_iv_j$定义了$H_0^1(\Omega)$上的内积$B_a[u,v],$
诱导了范数$\Vert\cdot\Vert_{a}.$ 这时发现想证明的等式恰恰是
$$
\varliminf_{k\rightarrow 0} \Vert w_k\Vert_a\ge \Vert w\Vert_a
$$
那么由弱Sharp连续性, 我们就得到了结论.
命题 2. $Sw=\lambda_1 w\in H^{-1}(\Omega).$
定义
$$
f(t)=B_S[w+t\eta,w+t\eta]-\lambda_1\Vert w+t\eta\Vert_{L^2}^2,\quad \,\forall\,\eta\in C_0^\infty(\Omega).
$$
那么$f(t)\ge 0,$ $f(0)=0,$ 在$t=0$处达到极小, 因此$f'(t)=0,$
$$
f'(t)=B_S[w,\eta]-\lambda_1\left<{}w,\eta\right>=0,\quad \left<{}Sw,\eta\right>=\left<{}\lambda_1w,\eta\right>.
$$
由稠密性, 这就证明了命题.
归纳定义特征子空间$(\lambda_m,V_m):$
$$
V_1:=\{w\in H_0^1(\Omega)|Sw=\lambda_1w\},
$$
$$
\lambda_2=\inf_{\substack{u\in H_0^1\setminus\{0\}\\u\perp_{L^2} V_1} }\frac{B[u,u]}{\Vert u\Vert_{L^2}^2},\quad V_2:=\{\cdots\}
$$
定理 3. $\Omega\subset \mathbb{R}^n$为有界开集, $S$系数满足前述条件, 则$S$有无穷多个特征值, 趋于$+\infty.$ 不同特征值对应特征向量$\{w_k\}$在$L^2$中正交, 构成标准正交基. 若$B_S$满足强制性条件, 则$\{\frac{w_k}{\sqrt{\lambda_k} }\}\in H_0^1$在内积$B_S[u,v]$下也构成标准正交基. 此时$u=\sum d_kw_k\in L^2(\Omega),$ 在$H_0^1$中也是.
由$S$的对称性, 由谱理论前两个性质正确. 当$B_S$满足强制性条件时,
只需验证若$B_S[u,\frac{w_k}{\sqrt{w_k} }]=0,$
$\,\forall\,k\ge 1$推出$u=0.$
而这可根据$\{w_k\}$是$L^2$下的标准正交基得到. 再后一结论显然.
取$S=-\Delta,$ 特征问题变为
$$
-\Delta w_k-\lambda w_k=0.
$$
这时由第四章的理论, 可要求
$$
\{w_k\}\subset C^\infty(\Omega)\cap H^2(\Omega)\cap H_0^1(\Omega)\subset H^{m+2}(\Omega).
$$
最后一个包含关系要求$\partial\Omega\in C^m.$
文章最后更新于 2022-12-02 16:40:24
Consider the self-conjugate operator $S:H_0^1(\Omega)\rightarrow H^{-1}(\Omega),$
$$
Su:=-(a^{ij}u_i+b^ju)_j+b^ju_j+cu.
$$
The coefficients are all in $L^\infty(\Omega)$,
$a^{ij}\xi_i\xi_j\ge \mu|\xi|^2.$ We have bilinear bounded functionals:
$$
B_S[u,v]=\int_\Omega a^{ij}u_iv_j+b^i(uv_j+vu_j)+cuv=\left<{}Su,v\right>=\left<{}u,Sv\right>=B_S[v,u].
$$
Satisfies the energy estimate:
$$
B_S[u,u]\ge \frac{\mu}{2}\Vert Du\Vert_{L^2}^2-C(\mu,S)\Vert u\Vert_{L^2}^2.
$$
If $(u,\lambda),$ exists, then there is $Su=\lambda u,$ in $H^{-1}(\Omega)$
It is called a eigenpair, $\lambda$ is the eigenvalue, and $u$ is the eigenfunction.
A natural question is how to find all $\lambda.$
If $\,\exists\,$ feature pair $(u,\lambda),$ then $\,\forall\,v\in H_0^1(\Omega),$
$$
B_S[u,v]=\left<{}\lambda u,v\right>=\lambda\int_\Omega uvdx.
$$
At this time,
$$
\lambda=\frac{B_S[u,u]}{\int_\Omega u^2dx}\ge -C(\mu,S).
$$
Inspired by this, we define
$$
\lambda_1:=\inf_{u\in H_0^1(\Omega)\setminus \{0\} }\frac{B_S[u,u]}{\Vert u\Vert_{L^2}^2}
$$
Let's prove that $\lambda_1$ is the first eigenvalue. First prove that it is achievable.
Proposition 1. $\,\exists\,w\in H_0^1(\Omega),$ $B_S[w,w]=\lambda_1$
$\,\exists\,w_k,$ satisfies $\Vert w_k\Vert_{L^2}=1,$ such that
$$
\lambda_1\le B_S[w_k,w_k]\le \lambda_1+\frac{1}{k}.
$$
From the following inequality sign,
Yes
$$
\frac{\mu}{2}\Vert Dw_k\Vert_{L^2}^2\le \lambda_1+1+C(\mu,S).
$$
Therefore, $\{w_k\}$ is a bounded column in $H_0^1$, there are sub-columns that converge in $L^2$, and weak convergence in $H_0^1$.
Let’s assume that the sub-column is itself, that is $w_k\xrightarrow{L^2}w,$
$w_k\rightharpoonup w\in H_0^1.$
It is hoped that the proof $B_S[w_k,w_k]\rightarrow B_S[w,w].$ will be considered item by item for this purpose.
$$
|\left<{}w_k,w_k\right>-\left<{}w,w\right>|\le |\left<{}w_k,w_k-w\right>|+|\left<{}w_k-w,w\right>|\le 2\Vert w_k-w\Vert_{L^2}\rightarrow 0
$$
$$
|\left<{}\partial_iw_{k},w_k\right>-\left<{}\partial_i w,w\right>|\le |\left<{}\partial_iw_{k},w_k-w\right>|+|\left<{}\partial_i (w_k-w),w\right>|\le (\Vert Dw_k\Vert_{L^2}+\Vert Dw\Vert_{L^2})\Vert w_k-w\Vert_{L^2}
$$
However, the $\left<{}\partial_iw_k,\partial_iw_k\right>$ term is not easy to prove convergence.
Let us turn to proving
$$
\varliminf_{k\rightarrow 0}\int a^{ij}\partial_iw_k\partial_jw_k\ge \int a^{ij}w_iw_j.
$$
In this way, combined with the previous convergence, we have,
$$
\lambda_1\le B_S[w,w]\le \varliminf_{k\rightarrow 0}B_S[w_k,w_k]=\lambda_1\quad\Rightarrow\quad B_S[w,w]=\lambda_1.
$$
Since $a^{ij}$ is uniformly elliptic,
$\int_\Omega a^{ij}u_iv_j$ defines the inner product $B_a[u,v],$ on $H_0^1(\Omega)$
The norm $\Vert\cdot\Vert_{a}.$ was induced. At this time, it was found that the equation wanted to prove was exactly
$$
\varliminf_{k\rightarrow 0} \Vert w_k\Vert_a\ge \Vert w\Vert_a
$$
Then based on weak Sharp continuity, we get the conclusion.
Proposition 2. $Sw=\lambda_1 w\in H^{-1}(\Omega).$
definition
$$
f(t)=B_S[w+t\eta,w+t\eta]-\lambda_1\Vert w+t\eta\Vert_{L^2}^2,\quad \,\forall\,\eta\in C_0^\infty(\Omega).
$$
Then $f(t)\ge 0,$ $f(0)=0,$ reaches minimum at $t=0$, so $f'(t)=0,$
$$
f'(t)=B_S[w,\eta]-\lambda_1\left<{}w,\eta\right>=0,\quad \left<{}Sw,\eta\right>=\left<{}\lambda_1w,\eta\right>.
$$
By density, this proves the proposition.
Inductively define the feature subspace $(\lambda_m,V_m):$
$$
V_1:=\{w\in H_0^1(\Omega)|Sw=\lambda_1w\},
$$
$$
\lambda_2=\inf_{\substack{u\in H_0^1\setminus\{0\}\\u\perp_{L^2} V_1} }\frac{B[u,u]}{\Vert u\Vert_{L^2}^2},\quad V_2:=\{\cdots\}
$$
Theorem 3. $\Omega\subset \mathbb{R}^n$ is a bounded open set, and the $S$ coefficient satisfies the aforementioned conditions, then $S$ has infinite eigenvalues, tending to $+\infty.$. The eigenvectors $\{w_k\}$ corresponding to different eigenvalues are orthogonal in $L^2$, forming a standard orthonormal basis. If $B_S$ meets the mandatory conditions, then $\{\frac{w_k}{\sqrt{\lambda_k} }\}\in H_0^1$ also forms a standard orthonormal basis under the inner product $B_S[u,v]$. At this time, $u=\sum d_kw_k\in L^2(\Omega),$ is also in $H_0^1$.
Due to the symmetry of $S$, the first two properties of the spectral theory are correct. When $B_S$ satisfies the mandatory conditions,
Just verify if $B_S[u,\frac{w_k}{\sqrt{w_k} }]=0,$
$\,\forall\,k\ge 1$ Launched $u=0.$
And this can be obtained based on the fact that $\{w_k\}$ is the standard orthonormal basis under $L^2$. The latter conclusion is obvious.
Taking $S=-\Delta,$, the characteristic problem becomes
$$
-\Delta w_k-\lambda w_k=0.
$$
At this time, based on the theory of Chapter 4, it can be required
$$
\{w_k\}\subset C^\infty(\Omega)\cap H^2(\Omega)\cap H_0^1(\Omega)\subset H^{m+2}(\Omega).
$$
The last inclusion relation requirement $\partial\Omega\in C^m.$
The article was last updated on 2022-12-02 16:40:24
