通过计算, 好像Yang–Mills方程即使在库伦规范下也不是严格椭圆的啊?
我记得Yang–Mills方程主项是dd^*A+d^*dA, 其中A=A_idx^i. 则其弱形式是
\int\langle d^*A,d^*B\rangle+\langle dA,dB\rangle=0,\quad\forall B=B_jdx^j\in C_0^\infty.
直接计算, 我们知道
\begin{align*}
dA&=d(A_idx^i)=\partial_jA_idx^j\wedge dx^i\\
dB&=\partial_k B_ld^k\wedge dx^l\\
d^*A&=-*d*A=-*d\left((-1)^{i-1}A_idx^1\wedge\cdots\wedge\widehat{dx^i}\wedge\cdots\wedge dx^n\right)\\
&=-*(\partial_iA_idx^1\wedge\cdots dx^n)\\
&=-\partial_iA_i\\
d^*B&=-\partial_kB_k.
\end{align*}
因此,
\begin{align*}
0&=\int\langle\partial_j A_i,\partial_kB_l\rangle\langle dx^j\wedge dx^i,dx^k\wedge dx^l\rangle+\langle\partial_kA_k,\partial_l B_l\rangle\\
&=\int\langle\partial_j A_i,\partial_kB_l\rangle(\delta_{jk}\delta_{il}-\delta_{jl}\delta_{ik})+\langle\partial_kA_k,\partial_l B_l\rangle\\
&=\int\langle\partial_j A_i,\partial_jB_i-\partial_iB_j\rangle+\langle\partial_kA_k,\partial_l B_l\rangle\\
&=\int\langle\partial_jA_i-\partial_iA_j,\partial_jB_i\rangle+\langle\partial_kA_k,\partial_l B_l\rangle\\
&=\int\langle\partial_jA_i-\partial_iA_j+\delta_{ij}\partial_mA_m,\partial_jB_i\rangle.
\end{align*}
这表明, 若记\xi_{j}^i=\partial_jB_i, u^i=A_i, p_j^i=\partial_jA_i以及q_i^j(x,u,p)=p_j^i-p_i^j+\delta_{ij}p_m^m, 则弱形式可改写为如下的标准形式(方便应用Morrey的定理1.1.11):
\int q_i^j(x,A,\nabla A)\xi_j^i=0,
此时, 方程的椭圆性可验证如下
\begin{align*}
\partial_{p_k^l}q_i^j\pi_k^l\pi_j^i&=\pi_l^k\pi_j^i(\delta_{jk}\delta_{il}-\delta_{jl}\delta_{ik}+\delta_{ij}\delta_{kl})\\
&=\pi_j^i\pi_j^i-\pi_j^i\pi_i^j+\pi_i^i\pi_j^j\\
&=\sum_{i<j}\pi_j^i\pi_j^i+\pi_i^j\pi_i^j-\pi_j^i\pi_i^j-\pi_i^j\pi_j^i+(\sum_i\pi_i^i)^2\\
&=\sum_{i<j}\pi_j^i(\pi_j^i-\pi_i^j)+\pi_i^j(\pi_i^j-\pi_j^i)+(\sum_i\pi_i^i)^2\\
&=\sum_{i<j}(\pi_j^i-\pi_i^j)^2+(\sum_i\pi_i^i)^2
\end{align*}
这表明它将在迹为零的对称矩阵上退化.