Autonomous system

**alexmahone** · December 15th 2011, 04:18 AM

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?

**FernandoRevilla** · December 15th 2011, 09:25 AM

Originally Posted by alexmahone

How do I proceed?

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

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

**alexmahone** · December 15th 2011, 09:57 AM

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?