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Consider the flow on a manifold $\mathcal{M}$ defined by $\dot{x} = f(x)$ with $x\in\mathcal{M}$ and $f : M\rightarrow TM$. Associated to this flow I can define the variational dynamics $\delta \dot{x} = \frac{\partial f(x)}{\partial x}\delta x$ with $\delta x \in T_xM$. Then clearly $\frac{\partial f(x)}{\partial x} : T_xM\rightarrow TTM$ or more generally $\frac{\partial f}{\partial x}\delta x : TM\rightarrow TTM$. I have two primary questions about these two dynamics.

The "second order" variational dynamics given by $\delta\delta\dot{x} = \frac{\partial \delta\dot{x}}{\partial \delta x}\delta\delta x = \frac{\partial f(x)}{\partial x}\delta\delta x$ obeys the same dynamics as the first-order variation. By analogy to above, it is natural to think that $\frac{\partial \delta\dot{x}}{\partial \delta x} : T_{(x, \delta x)}TM\rightarrow TTTM$ and more generally $\frac{\partial \delta\dot{x}}{\partial \delta x}\delta \delta x : TTM\rightarrow TTTM$. However, from above, we have also said that $\frac{\partial f(x)}{\partial x} : T_xM\rightarrow TTM$, and that $\frac{\partial \delta\dot{x}}{\partial \delta x} = \frac{\partial f}{\partial x}$. How can I resolve this apparent paradox?

My second question builds on the first. I will call a system exponentially incrementally stable with rate $\lambda$ if $\frac{d}{dt}\langle \delta x, M(x) \delta x\rangle \leq -2\lambda\langle \delta x, M(x)\delta x\rangle$ where $(x(t), \delta x(t))$ is an arbitrary trajectory on $TM$ induced by the flow and $M(x)$ is a Riemannian metric. Because the second-order variational dynamics is identical to the first-order variational dynamics, I would like to say that the $\delta\dot{x}$ system is also exponentially incrementally stable if the $\dot{x}$ system is.

However, it is not clear how to think of a trajectory $(\delta x(t), \delta\delta x(t))$. Where does this live, and what would be the corresponding metric? One could think of a flow on $TTM$ but this would require $(x, \delta x, \delta\delta x^h, \delta\delta x^v)$ where $\delta\delta x^h$ and $\delta\delta x^v$ are the horizontal and vertical projections of an element of $T_{(x, \delta x)}TM$ respectively.

I have a less rigorous proof of this in the case where everything is $\mathbb{R}^n$. Here, the condition for incremental exponential stability reduces to negative semi-definiteness of $\left(\frac{\partial f}{\partial x}\right)^TM + M\frac{\partial f}{\partial x} + \dot{M} + 2\lambda M$. We also have $x\in\mathbb{R}^n$, $\delta x\in\mathbb{R}^n$, and $\delta\delta x\in\mathbb{R}^n$, so that one can, for example, look directly at $\frac{d}{dt} \delta\delta x^T M(x) \delta\delta x$. Implicit in this I'm sure is some isomorphism that I hope can be applied in the general case.

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