I want to consider whether the linearized equation of (nonlinear) uniformly elliptic equation maintains uniformly elliptic or not.
By definition, for the original uniformly elliptic equation:$$ F(D^2u(x))=f(x),\quad\in\mathbb{R}^n, $$ where F is smooth function and satisfies :$$ \lambda ||N||\leq F(M+N)-F(M)\leq\Lambda||N||,\ \forall N\geq0,\ M,N\in\mathtt{Sym}(n). $$ for some $0<\lambda\leq\Lambda<\infty$. Assume $F(I)=c$ for simplicity, where I stands for the unit metrix
Then we have two classical way to linearize it:
- Taking Derivatives $\partial_k$ to the equation: \begin{equation*} F_{M_{ij}}(D^2u)\cdot D_{ijk}u=\partial_kf(x). \end{equation*}
- Using Newton-Leibnitz formula to compare with $F(I)=c$. \begin{equation*} F(D^2u)-F(I)=f(x)-c \end{equation*} hence \begin{equation*} \int_0^1F_{M_{ij}}(I+t(D^2u(x)-I))\mathtt{d} t\cdot D_{ij}u=f(x)-c. \end{equation*}
Attempts: By definition of derivatives, the coefficients can be written as:$$ F_{M_{ij}}(D^2u(x)):=\lim_{h\rightarrow0}\dfrac{F(D^2u+h\delta_{ij})-F(D^2u)}{h} $$ where $\delta_{ij}$ is the matrix that only the $(i,j)$ position being 1. However this matrix is not symmetric, hence the condition of uniformly elliptic cannot be applied directly.
Is this result true or is this definition of uniformly elliptic equation not suitable ? Thanks for your kind help :)