Suppose I have the system $x'=F(x)$ with $F:\mathbb{R}^n\rightarrow\mathbb{R}^n$. I denote by $J(x)$ the Jacobian matrix, that is, $J_{ij}(x)=\partial F_i/\partial x_j (x)$. Suppose I know that for all $x\in\mathbb{R}^n$ the eigenvalues of $J(x)$ have negative real part, so local stability holds. Now I want to understand under which conditions, if any, global stability follows. To this end I define the function $$V(x)=\frac{1}{2}\sum_{i=1}^nF_i(x)^2,$$ wihch I hope could be a Lyapunov function. It is always non-negative and $$\nabla V(x)\cdot F(x)=\Big(\sum_{i=1}^nF_i(x)\nabla F_i(x)\Big)\cdot F(x)=\sum_{i=1}^nF_i(x)\sum_{j=1}^n\frac{\partial F_i(x)}{\partial x_j}F_j(x)=F(x)^TJ(x)F(x).$$ As I understand, if this expression is always negative then I have global stability. If $J(x)$ is always symmetric (not sure if this is at all possible in the non linear case) then the condition assumed above that the real parts of the eigenvalues are negative means that $J(x)$ is always negative definite and I should be done. If $J(x)$ is not symmetric, then I don't think that the sign of this quadratic form and the sign of the eigenvalues are necessarily related. Am I wrong? Can I be educated?
I know that without loss of generality I can take the symmetric part of $J(x)$ and have $$\nabla V(x)\cdot F(x)=J(x)^T \Big(\frac{J(x)+J(x)^T}{2}\Big)J(x)$$ and so the sign of this quadratic form is determined by the sign of the eigenvalues of the symmetric part of $J(x)$. But I don't know if I can say anything about the sign of these eigenvalues even if I know the sign of the eigenvalues of $J(x)$ everywhere. Or can I? Perhaps for some special class of matrices $J(x)$?
Thanks!!!!!