Consider the $N$-dimensional autonomous system of ODEs
$$\dot{x}= f(x),$$
where a locally unique solution $x(t)$, starting from the initial condition $x$, is denoted as $x(t)=\phi(t,x)$. Show that$$\Big(\frac{\partial}{\partial{x}}\phi(t,x)\Big)f(x)=f(\phi(t,x))$$
Here is my proposed solution.
We are given that $\dot{x}= f(x),$ so we know that
$$\frac{dx(t)}{dt}=f(x(t))$$
If we integrate both sides with respect to $t$, we will get
$$\int_{t_0}^{t}\frac{dx(t)}{dt}dt =\int_{t_0}^{t} f(x(s)) ds$$
Therefore,
$$x(t) = x_0 + \int_{t_0}^{t} f(x(s)) ds$$
where the integration constant is chosen such that $x(t_0)=x_0$. As $x(t)=\phi(t,x)$, we have
$$\phi(t,x) = x_0 + \int_{t_0}^{t} f(x(s)) ds$$
Hence,
\begin{equation}
\begin{split}
\frac{\partial\phi}{\partial{t}}(t,x) & = 0 + \frac{\partial}{\partial{t}}\int_{t_0}^{t} f(x(s)) ds \\
& = f(x(t))\\
& = f(\phi(t,x))
\end{split}
\end{equation}
where the second equality follows from the fundamental theorem of calculus. Next, we are given that
- $\phi$ depends on $(t,x)$.
- $x$ depends on $t$.
Therefore, by the chain rule,
\begin{equation}
\begin{split}
\frac{\partial\phi}{\partial{t}}(t,x) & = \frac{\partial\phi}{\partial{x}}(t,x)\frac{dx}{dt} \\
& = \frac{\partial\phi}{\partial{x}}(t,x)f(x)\\
& = f(\phi(t,x))
\end{split}
\end{equation}
Thus, $\Big(\frac{\partial}{\partial{x}}\phi(t,x)\Big)f(x)=f(\phi(t,x))$.
It does seem rather strange that I concluded that
$$\frac{\partial\phi}{\partial{t}}(t,x) = \frac{\partial\phi}{\partial{x}}(t,x)\frac{dx}{dt}$$
This must be because $\frac{\partial\phi}{\partial{t}}(t,x) = \frac{\partial\phi}{\partial{x}}(t,x)\frac{dx}{dt}$ instead of $\frac{\partial\phi}{\partial{t}}(t,x) = \frac{\partial\phi}{\partial{t}}(t,x)\frac{dx}{dt}$ because of the chain rule.
Is this approach correct? Please let me know if there are any better alternatives.
Best Answer
If you make the flow definition more specific, such as not re-using the simple $x$ in two different meanings, you could define the flow as $$x(t)=ϕ(t;x_0) ~~ \text{ where } ~~ x(0)=ϕ(0;x_0)=x_0.$$
You need to invoke the autonomous nature of the ODE which has as consequence that $t\mapsto ϕ_t$ is a group action/representation of the additive group $(\Bbb R,+)$. This means that $ϕ_t\circ ϕ_s=ϕ_{t+s}$, or $$ ϕ(t+s;x_0)=ϕ(t;ϕ(s;x_0))\iff x(t+s)=ϕ(t;x(s)). $$ From your point-of-view the closest variant might be $x(t)= ϕ(t-s;x(s))$. Anyway, to get the $x$-derivative of $ϕ$ involved you need first some kind of explicit variability in the $x$-argument of $ϕ$ which is easiest done by moving the initial point along the solution curve.
Taking, in the first variant, the derivative for $s$ at $s=0$ then gives indeed by the chain rule $$ f(ϕ(t+s;x_0))=\frac{∂}{∂s}ϕ(t;ϕ(s;x_0))=\frac{∂ϕ}{∂x}(t;ϕ(s;x_0))\cdot f(ϕ(s;x_0))\\ \overset{s=0}\implies f(ϕ(t;x_0))=\frac{∂ϕ}{∂x}(t;x_0)f(x_0)\\ $$