Is the right-hand term of the dynamic system equivalent to the original system after being multiplied by a constant?

Question

Given two dynamical systems where $f$ is lipschitz for $x$ : $\begin{cases} x'(t)=af(x,t),\\ x(0)=x_0,\end{cases} t\in[0,\tau]$ and $\begin{cases} z'(t)=f(z,t),\\ z(0)=x_0,\end{cases} t\in[0,\tau']$, and their flowmaps $\phi_{af}^\tau(x),\phi_f^{\tau'}(x)$, if $\tau'=a\tau$, can we say $\phi_{af}^\tau(x_0)=\phi_f^{\tau'}(x_0)$ for any $x_0 \in \mathbb{R}$?

How about in a higher dimension? $\begin{cases} x'(t)=Af(x,t),\\ x(0)=x_0,\end{cases} t\in[0,\tau]$ and $\begin{cases}z'(t)=f(z,t),\\ z(0)=x_0,\end{cases} t\in[0,\tau']$ where $A\in\mathbb{R}^{n\times n}$ is diagonal matrix? what is the relationship between $\phi_{Af}^\tau(x_0)$ and $\phi_f^{\tau'}(x_0)$

Answer 1

Of course not. Take about any smooth $f(x,t)$ depending on $t$.

E.g., take $f(x,t)=x+t$, so that for $a\ne0$ $$\phi^t_{af}(x_0)=\frac{a x_0 e^{a t}-a t+e^{a t}-1}{a},$$ so that $$\phi^t_{af}(x_0)-\phi^{at}_f(x_0)=\frac{(a-1) \left(1+a t-e^{a t}\right)}{a}\ne0$$ if $a\notin\{0,1\}$ and $t\ne0$.

Answer 2

Regarding the second question: there is hardly any useful relationship between the flows. Such systems may not be even topologically equivalent even for the case of linear autonomous systems. For example: $$\tag{1} \left\{ \begin{array}{lrr} \dot x_1=&-x_1&-\frac12 x_2\\ \dot x_2=&3x_1&+x_2. \end{array} \right. $$ Since the eigenvalues of the matrix of the system are $\pm \frac1{\sqrt2}i$, the phase portrait of (1) is a center. If we multiply the right side of the system by the matrix $A=\left(\begin{array}{rr} 1&0\\ 0&2 \end{array}\right) $, we obtain $$\tag{2} \left\{ \begin{array}{lrr} \dot x_1=&-x_1&-\frac12 x_2\\ \dot x_2=&6x_1&+2x_2. \end{array} \right. $$ The eigenvalues of the matrix of the system are now $\frac12\pm \frac{\sqrt3}{2}i$, so the phase portrait of (2) is an unstable focus.