# Thread: Autonomous system

1. ## Autonomous system

Consider the autonomous system

$\frac{dx}{dt}=F(x, y),\ \frac{dy}{dt}=G(x, y)$

in a region where the functions $F$ and $G$ are continuously differentiable.

Suppose that $(x(t), y(t))$ is a solution of the system and that $\gamma\neq 0$. Define $\phi(t)=x(t+\gamma)$ and $\psi(t)=y(t+\gamma)$. Then show that $(\phi(t), \psi(t))$ is also a solution of the system.

My attempt:

$(x(t), y(t))$ is a solution of the autonomous system.

So, $x'(t)=F(x(t), y(t))$ and $y'(t)=G(x(t), y(t))$.

$\frac{d\phi(t)}{dt}=x'(t+\gamma)$

$\frac{d\psi(t)}{dt}=y'(t+\gamma)$

How do I proceed?

2. ## Re: Autonomous system

Originally Posted by alexmahone
How do I proceed?
$\frac{d\phi(t)}{dt}=x'(t+\gamma)=F(x(t+\gamma),y(t +\gamma))=F(\phi(t),\psi(t))$

$\frac{d\psi(t)}{dt}=y'(t+\gamma)=G(x(t+\gamma),y(t +\gamma))=G(\phi(t),\psi(t))$

3. ## Re: Autonomous system

Originally Posted by FernandoRevilla
$\frac{d\phi(t)}{dt}=x'(t+\gamma)=F(x(t+\gamma),y(t +\gamma))=F(\phi(t),\psi(t))$

$\frac{d\psi(t)}{dt}=y'(t+\gamma)=G(x(t+\gamma),y(t +\gamma))=G(\phi(t),\psi(t))$
I'm not sure that I completely understand how $x'(t+\gamma)=F(x(t+\gamma),y(t+\gamma))$ and $y'(t+\gamma)=G(x(t+\gamma),y(t+\gamma))$.

I know $x'(t)=F(x(t),y(t))$ and $y'(t)=G(x(t),y(t))$, but how do we know that substituting $t+\gamma$ for $t$ will lead to a valid equation?

4. ## Re: Autonomous system

For example, suppose $x'(t)=F(x(t),y(t))$ and $y'(t)=G(x(t),y(t))$ for all $t\in\mathbb{R}$ . What is $x'(u)$ and $y'(u)$ for all $u\in\mathbb{R}$ ?

5. ## Re: Autonomous system

Originally Posted by FernandoRevilla
For example, suppose $x'(t)=F(x(t),y(t))$ and $y'(t)=G(x(t),y(t))$ for all $t\in\mathbb{R}$ . What is $x'(u)$ and $y'(u)$ for all $u\in\mathbb{R}$ ?
$x'(u)=F(x(u),y(u))$ and $y'(u)=G(x(u),y(u))$.

6. ## Re: Autonomous system

Originally Posted by alexmahone
$x'(u)=F(x(u),y(u))$ and $y'(u)=G(x(u),y(u))$.
Right, now use the substitution $u=t+\gamma$ .

7. ## Re: Autonomous system

Originally Posted by FernandoRevilla
Right, now use the substitution $u=t+\gamma$ .
Hmm...makes sense. Thanks!