# Displacement of differential equation

In [Forni and Sepulchre, arXiv:1305.3456] the authors state that given the differential equation \begin{equation} \dot{x}=f(x,u),\quad (1) \end{equation} where $f:\mathbb{R}^{n\times m}\to\mathbb{R}^n$ and $u:\mathbb{R}_+\to\mathbb{R}^m$ are $\mathcal{C}^2$-continuous function, one can always associate the exact differential: \begin{equation*} \delta\dot{x}=\dfrac{\partial f}{\partial x}(x,u)\delta x+\dfrac{\partial f}{\partial u}(x,u)\delta u, \end{equation*} for all $(x,u)\in\mathbb{R}^{n+m}$, where $\delta x$ stands for the notion of differential length.

I can proof that this is true, whenever $u\equiv0$ as follows. Consider any pair of initial conditions $x_0,x_1\in\mathbb{R}^n$ for Eq. (1). Let $\psi:[0,1]\to\mathbb{R}^n$ be any smooth curve such that $\psi(0)=x_0$ and $\psi(1)=x_1$. Define $$\delta x(t,r)=\dfrac{\partial }{\partial r}\phi(t,\psi(r)),$$ where $\phi$ is the solution of Eq. (1) with $u\equiv0$.

Now, \begin{equation*} \dfrac{d}{dt}\delta x(t,r)=\dfrac{d}{dt}\dfrac{\partial }{\partial r}\phi(t,\psi(r))=\dfrac{\partial }{\partial r}\dfrac{d}{dt}\phi(t,\psi(r))=\dfrac{\partial }{\partial r}f(\phi(t,\psi(r)),0)=\dfrac{\partial f}{\partial \phi}(\phi(t,\psi(r)),0)\dfrac{\partial}{\partial r}\phi(t,\psi(r))=\dfrac{\partial f}{\partial \phi}(\phi(t,\psi(r)),0)\delta x(t,r). \end{equation*} Since the this holds for every two initial conditions, and for every smooth curve connecting them, \begin{equation*} \delta\dot{x}=\dfrac{\partial f}{\partial x}(x,0)\delta x \end{equation*}

The question is, how can I generalize the previous proof for the case in which $u\neq0$?

## 1 Answer

I realize how the authors managed to solve the problem. consider two inputs $p,q\in\mathcal{C}^2(\mathbb{R},\mathbb{R}^m)$, two initial conditions $y,z\in\mathbb{R}^n$, and the solutions $\phi(\cdot,y,p)$ and $\phi(\cdot,z,q)$ of the initial value problems associated with (1).

Consider the smooth curve [\ \begin{array}{rrcl} \psi:&[0,1]&\to&\mathbb{R}^{n+m} \\ &r&\mapsto&(\psi_x(r),\psi_u(r)) \end{array} ]

satisfying $\psi(0)=(y,v)$ and $\psi(1)=(z,w)$. For every $r\in[0,1]$, let $\chi(\cdot,r)=\phi(\cdot,\psi_x(r),\psi_u(r))$, and define $\delta x(\cdot,r)=\partial \chi(\cdot,r)/\partial r$ and $\delta u(\cdot,r)=\partial \psi_u(\cdot,r)/\partial r$. It follows that, for every $t\geq0$, and for every $r\in[0,1]$ \begin{equation*} \begin{array}{rcl} \dfrac{d \delta x}{d t}(t,r)&=&\dfrac{d }{d t}\left[\dfrac{\partial \chi}{\partial r}(t,r)\right]=\dfrac{\partial }{\partial r}\left[\dfrac{d \chi}{d t}(t,r)\right]=\dfrac{\partial}{\partial r}\left[f(\chi(t,r),\psi_u(t,r))\right]\\ \\ &=&\dfrac{\partial f}{\partial \chi}(\chi(t,r),\psi_u(t,r))\dfrac{\partial\chi}{\partial r}(t,r)+\dfrac{\partial f}{\partial \psi_u}(\psi_u(t,r),\psi_u(t,r))\dfrac{\partial\psi_u}{\partial r}(t,r)\\ \\ &=&\dfrac{\partial f}{\partial \chi}(\chi(t,r),\psi_u(t,r))\delta x(t,r)+\dfrac{\partial f}{\partial \psi_u}(\psi_u(t,r),\psi_u(t,r))\delta u(t,r)\\ \end{array} \end{equation*} From the generality of the initial conditions, inputs, and curves, the previous equation can be rewritten as \begin{equation*} \dot{\delta x}=F(x,u)\delta x + G(x,u)\delta u, \end{equation*}