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vspace{0.5cm} The flowmap denoted $phi_{t,t_{0}}in Diff(R^{N})$ where $Diff(R^{N})$ denotes the group of diffeomorphisms of $R^{N}$ is defined as a map that takes the initial solution, that is the solution at $t_{0}$ to the solution at any time $t$, this can be expressed mathematically by egin{center} $phi_{t,t_{0}}: Y_{t_{0}}longmapsto Y_{t}$ end{center} That is to say, given any initial data $Y_{0}in R^{N}$, the solution $ Y_{t}$ at any later time can be easily specified. This can be done by applying the action of the flowmap to the initial data $Y_{0}$ in order to $Y_{t}=phi_{t,t_{0}} circ Y_{t_{0}}$. Consider a function $fin Diff(R^{N})$, by using the chain rule, we obtain $ frac{d}{dt}(f(Y))=V(Y).partial_{Y} f(Y)$ This means that action of the vector fields on $Diff(R^{N})$ is as first order partial differential operators since $V(Y).partial_{Y} f(Y)$ is considered as first order partial differential operators. vspace{0.5cm} The evolution of $fcircphi_{t}$ is given by egin{center} $ frac{d}{dt}(fcircphi_{t})=Vcirc fcircphi_{t}$ end{center} The above equation is an autonomous linear functional differential equation for $fcircphi_{t}$. Such equation has a solution $fcircphi_{t}=exp(tV)circ fcircphi_{t}$ as $phi_{0}=id$ Setting $f=id$ provides the representation of the flowmap as follows egin{center} $phi_{t}=exp(tV)$ end{center} Hence, in this considered case, the flowmap is the exponential of the vector field. By compositing the above equation

vspace{0.5cm} The flowmap denoted $phi_{t,t_{0}}in Diff(R^{N})$ where $Diff(R^{N})$ denotes the group of diffeomorphisms of $R^{N}$ is defined as a map that takes the initial solution, that is the solution at $t_{0}$ to the solution at any time $t$, this can be expressed mathematically by egin{center} $phi_{t,t_{0}}: Y_{t_{0}}longmapsto Y_{t}$ end{center} That is to say, given any initial data $Y_{0}in R^{N}$, the solution $ Y_{t}$ at any later time can be easily specified. This can be done by applying the action of the flowmap to the initial data $Y_{0}$ in order to $Y_{t}=phi_{t,t_{0}} circ Y_{t_{0}}$. Consider a function $fin Diff(R^{N})$, by using the chain rule, we obtain $ frac{d}{dt}(f(Y))=V(Y).partial_{Y} f(Y)$ This means that action of the vector fields on $Diff(R^{N})$ is as first order partial differential operators since $V(Y).partial_{Y} f(Y)$ is considered as first order partial differential operators. vspace{0.5cm} The evolution of $fcircphi_{t}$ is given by egin{center} $ frac{d}{dt}(fcircphi_{t})=Vcirc fcircphi_{t}$ end{center} The above equation is an autonomous linear functional differential equation for $fcircphi_{t}$. Such equation has a solution $fcircphi_{t}=exp(tV)circ fcircphi_{t}$ as $phi_{0}=id$ Setting $f=id$ provides the representation of the flowmap as follows egin{center} $phi_{t}=exp(tV)$ end{center} Hence, in this considered case, the flowmap is the exponential of the vector field. By compositing the above equation

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