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Question: Let $\mathbb{X}$ be the vector field given by $\mathbb{X}(x,y)=(x,y)$

Compute its flow $\Phi(x,y)$

Attempt: We have $\dot{x}(t)=x\therefore$$$\int_{x_0}^{x(t)}dx'=\int_{0}^{t}x(t')dt'$$

$$\left.x'\right|_{x_0}^{x(t)}=\left.\pm e^{t'}\right|_0^t$$

$$x(t)-x_0=\pm e^t-1$$

$$x(t)=\pm e^t+x_0-1$$

And something similar for $y$

Giving flow:

$\Phi(x,y)=(\pm e^t+x-1,\pm e^t+y-1)$

Is this right? Any help or comments would be greatly appreciated

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  • $\begingroup$ Well, it doesn't look like a well-defined flow. It doesn't look like $\Phi^{t+s}(x,y) = \Phi^{t}(x,y) \circ \Phi^{s}(x,y)$ holds. $\endgroup$ – Evgeny Jan 12 '15 at 11:38
  • $\begingroup$ Yea, even setting $\pm$ to be $+$ doesn't work $\endgroup$ – Sam Houston Jan 12 '15 at 11:45
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Hint: vector field generates the system of differential equations:

$$ \dot{{\rm x}} = {\mathbb X}({\rm x}), $$ where ${\rm x} = (x,y)$. This system could be rewritten as

$$ \left \lbrace \begin{array}{ccc} \dot{x} &= x \\ \dot{y} &= y \end{array} \right . $$

Note on solving:

If you solve $\dot{x} = x $ by separation of variables, then you should obtain this

$$ \frac{d x}{dt} = x $$ $$ \frac{d x}{x} = dt $$ $$ \int\limits_{x_0}^{x(t^\ast)} \frac{d x}{x} = \int\limits_{0}^{t^\ast} dt $$ $$ \ln{\frac{x(t^\ast)}{x_0}} = t^\ast $$ $$ x(t^\ast) = x_0 \cdot e^{t^\ast} $$

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  • $\begingroup$ No, my "attempt" at continuing is above, I understand this system and this is what I have tried, the crux of my question is that I am unable to solve this. $\endgroup$ – Sam Houston Jan 12 '15 at 11:46
  • $\begingroup$ My "attempt" is literally trying and failing to solve this system $\endgroup$ – Sam Houston Jan 12 '15 at 11:47
  • $\begingroup$ The problem then is that general solution to $\dot{x} = x$ is $x(t) = x_0 \cdot e^t$, not as you've written it. $\endgroup$ – Evgeny Jan 12 '15 at 12:00
  • $\begingroup$ I fully understand this now thanks $\endgroup$ – Sam Houston Jan 12 '15 at 13:13
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The vector field $X(x,y):=(x,y)$ encodes the system of ODEs $$\dot x=x,\quad \dot y=y\ ,$$ which is separated as it stands. Given an initial point $(x_0,y_0)\in{\mathbb R}^2$ the solution is obviously given by $$x(t)=e^t x_0,\quad y(t)=e^t y_0\qquad(-\infty< t<\infty)\ .$$ In terms of "flow" this means that $$\Phi_t(x_0,y_0)=e^t(x_0,y_0)\ ,$$ where at the end we may drop the index ${}_0\,$: $$\Phi_t(x,y)=e^t(x,y)\qquad\bigl((x,y)\in{\mathbb R}^2\bigr)\ .$$ As expected, one has $$\Phi_{t+s}(x,y)=\Phi_s\circ \Phi_t(x,y)\ .$$

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  • $\begingroup$ At the end, why do we drop the index $0$? Is it because initial value changes with respect to $x$? i.e. it is a variable in $x$? $\endgroup$ – Serpenche May 17 at 15:54
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    $\begingroup$ @Serpenche : $(x_0,y_0)$ was considered "fixed" for the moment; but at the end the flow $\Phi_t(x,y)$ is a function of $t$ and the variable initial point $(x,y)$. $\endgroup$ – Christian Blatter May 17 at 17:53

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