# Find the phase flow of $\dot{x}= x-1$

definition.

The phase flow of the differential equation $$\dot{x}=\vec{v}\ (x)$$ is the one-parameter diffeomorphism group for which $$\vec{v}$$ is the phase velocity vector field, namely, $$\vec{v}=\frac{d}{dt} \Big|_{t=0} (g^tx)$$

In the book, for the problem to find the phase flow of $$\dot{x}=x-1$$, the provided answer, $$g^tx=(x-1)e^t+1$$, is easy to verify. However, I have not idea to solve this problem. Any helps?

• $\dot x=x-1$ makes $x$ look like a scalar. $\dot x=v(x)$ makes $x$ look like a vector. Can you clarify?
– robjohn
Aug 16, 2019 at 8:05
• @Mattos Yeah, since in the book it is defined as the $phase\ velocity\ vector\ field$, I use what the book indicates. Aug 16, 2019 at 13:03
• If $x$ is a vector, how can $\dot x=x-1$ since $1$ is not a vector?
– robjohn
Aug 16, 2019 at 15:32
• @robjohn It is just a notation in the definition, and the question is stated separately. You may assume it is a scalar and assume $v$ is also a scalar too. Aug 17, 2019 at 2:34

You seem to have a notation ambiguity. In $$g^tx$$ the $$g^t$$ is the group or semi-group action on the real line, not a power of some number $$g$$ with exponent $$t$$. What it means is $$g^t(x)$$ or $$g(t,x)$$. There is nothing more to do.

• Exactly, in the book, it says that it is a one-parameter group of transformation of a set, but how can I find this mapping tho? Aug 16, 2019 at 13:21

It is not that simple, $$x$$ is a function of $$t$$, so just integrating on both sides won't work. To solve $$\dot x=x-1$$ you could use separation of variables: $$\int \dot x/(x-1)dt=t$$ which, after substituting $$u$$ for $$x-1$$ becomes $$t=\int_{x_0}^{x(t)}\frac{1}{u}dt=\ln\left(x(t)-1\right)-\ln(x_0)$$ Therefore $$x(t)=x_0 e^t+1$$

• Why did you leave out the integration constant? It is quite important for a flow function to have the initial value as a variable. Aug 16, 2019 at 7:28
• I am sorry but your method for solving ordinary differential equation $\dot{x}-x=-1 (1)$ misses an arbitrary constant (moreover it does not necessitate integration).This ODE has associated homogeneous equation $\dot{x}-x=0$ whose evident general solution is $x=Ae^t$ ; therefore, as a particular solution of (1) is $1$, one adds it to the general solution found above, giving $x(t)=1+Ae^t$ Aug 16, 2019 at 7:37
• Thanks for answering, can you please indicate what I should do next to find the phase flow? Aug 16, 2019 at 7:50

The proceeding is the following (from Example 2, pag 65, Ordinary Differential Equations (3ed) - Arnol'd):

The phase velocity vector field of the ODE is: $$\vec{v}(x)= x-1$$, then the solution of the equation $$\dot{x} = \vec{v}(x)$$ with initial condition $$x_0$$ for $$t=0$$ can easily be found explicitly:

\begin{align} \frac{dx}{dt} &= x -1 \\ \frac{dx}{x-1}&= dt \\ \int \frac{1}{x-1} dx &= \int dt \\ \ln(x-1) &= t+ C \end{align}

We left the integration constant of both sides on the right side.

Then we know that for $$t=0$$ we have $$x(t=0)=x_0$$. Thus, evaluating this in the last expression, we obtain:

\begin{align} C = \ln(x_0 -1) \end{align}

Now we have:

\begin{align} \ln(x-1) &= t+ \ln(x_0 -1) \end{align}

rearranging

\begin{align} \ln(x-1)-\ln(x_0 -1) &= t \\ \ln\left(\frac{x-1}{x_0-1} \right) &= t\\ \frac{x-1}{x_0-1} &= e^t \\ x &= (x_0-1)e^t + 1 \end{align}

Now, the crucial final step is to identify correctly in the last expression the phase flow $$g^t$$. For this, we can think the right side as the motion under the action of the phase flow $$g^t$$ of a fixed point $$x_0$$ of the phase space, i.e, $$g^tx_0$$. Then¸ we can say that:

\begin{align} g^tx = (x-1)e^t + 1 \end{align}

(pag 64, Ordinary Differential Equations (3ed) - Arnol'd):

In other words, under the action of the phase flow ($$g^t$$) the phase point ($$x$$) moves so that its velocity vector at any instant equals the phase velocity vector at the point of the phase space at which the moving point is located.