Calculation of the flows of two vector fields

Question

Hi I have a doubt that I think is pretty stupid.

I am reading a book and at some point there is an example about computing a flow of 2 vector fields which says.

Set $\begin{equation}f_1(x)=\left(\begin{matrix} \exp(x_2) \\ 1 \end{matrix}\right)\end{equation}$ , $\begin{equation}f_2(x)=\left(\begin{matrix} 1 \\ 0 \end{matrix}\right)\end{equation}$.

The calculation of the flows of $f_1, f_2$ is easy.

For $f_1$ since:

$\dot{x}_1 = \exp(x_2) \\ \dot{x}_2 = 1$

is solved by

$x_1(t) = \exp(x_2^o)(\exp(t)-1) + x_1^o \\ x_2(t) = t+x_2^o$

(the question: How do I get this result? Just by integrating the $\dot{x_1}, \dot{x_2}$? I don't get it sorry)

from that we have

$\Phi_{z_1}^{f_1}(x) = \left(\begin{matrix} \exp(x_2)(\exp(z_1)-1)+x_1\\ z_1+x_2\end{matrix}\right)$

And this is clear to me, you just neet to put $(x_1^o,x_2^o)=(x_1,x_2)$ and $t=z_1$.

Then it computes, more or less in the same way the: $\Phi_{z_2}^{f_2}(x)$

Thanks a lot for the help.

Answer 1

The differential equation seems to be $$ \dot{x} = f(x) $$ for $x = (x_1, x_2)^T$.

Your example $$ \dot{x} = f_1(x) = \begin{pmatrix} \exp(x_2) \\ 1 \end{pmatrix} \quad (*) $$ for the second component reads $$ \dot{x}_2 = 1 $$ which can directly be integrated for both sides into \begin{align} \int\limits_0^t \dot{x}(\tau) d\tau &= \int_0^t\limits d\tau \iff \\ x_2(t) - x_2(0) &= t - 0 \iff \\ x_2(t) &= t + x_2(0) \end{align} Then the first component of the differential equation $(*)$ reads $$ \dot{x}_1 = \exp(x_2) = \exp(x_2(t)) = \exp(t + x_2(0)) $$ which after inserting the solution for $x_2(t)$ now only depends on $t$ and not on $x_1(t)$ or $x_2(t)$.

Again we integrate both sides over time and get: \begin{align} x_1(t) - x_1(0) &= \int\limits_0^t \exp(\tau + x_2(0)) \, d\tau \\ &= \int\limits_0^t \exp(\tau) \exp(x_2(0)) \, d\tau \\ &= \exp(x_2(0)) \int\limits_0^t \exp(\tau)\, d\tau \\ &= \exp(x_2(0)) \left[ \exp(\tau) \right]_{\tau=0}^{\tau=t} \\ &= \exp(x_2(0)) (\exp(t) - 1) \end{align} so we get \begin{align} x_1(t) &= \exp(x_2(0)) (\exp(t) - 1) + x_1(0) \\ x_2(t) &= t + x_2(0) \end{align} as solution.