I have a Cauchy problem
$$ \left\{ \begin{array}{l} \dot{x} = f(x) \\ x(0) = x_0 \end{array} \right. $$
with $x = x(t)$. Suppose to have $y(t) = a x\left(\frac{t}{b}\right)$, where $a$ and $b$ are real constants.
I would like to write another Cauchy problem for the variable $y$:
$$ \left\{ \begin{array}{l} \dot{y} = g(y, a, b) \\ y(0) = y_0(x_0, a, b) \end{array} \right. $$
It is easy to show that $y(0) = y_0(x_0, a, b) = ax_0$. What can I say about $g(y, a, b)$?
Best Answer
Finally I got it!!!
For a general ODE likes:
$$\frac{d}{dt}x(t) = f(x(t), t)$$
we have that:
$$\begin{array}{rl} \displaystyle\frac{d}{dt}y(t) & = \displaystyle\frac{d}{dt}a x\left(\frac{t}{b}\right) \Rightarrow \\ \displaystyle\frac{d}{dt}y(t) & = \displaystyle\frac{a}{b} f\left(x\left(\frac{t}{b}\right), \frac{t}{b}\right) = \frac{a}{b} f\left(\frac{y(t)}{a} , \frac{t}{b}\right) = g(y(t),t,a,b), \end{array}$$
and
$$y(0) = y_0(x_0, a, b) = ax_0$$