From: Ferdinand Verhulst - Nonlinear Differential equations and dynamic systems - Page 1.

Let $\dot x = f(t,x)$.

Then apparently $f:G\to \Bbb R^n$ where $G$ is an open subset of $\Bbb R^{n+1}$ and so $x\in \Bbb R^n$.

Now $x\in \Bbb R^n$ where $x$ is really $x(t)$, a vector function with respect to time. So the derivative also lives in $\Bbb R^n$ right, also parametrised by time. Why does $f$ map from an open subset of one greater dimensional euclidean space?

My first thought was that we have points in $\Bbb R^n$ space, and then since we are parametrised by time, where $t\in \Bbb R$, we get $\Bbb R^{n+1}$, but the derivative is also parametrised by time.

  • $\begingroup$ If you write the ODE as $\dot x(t)=f(t,x(t))$, does that clarify the last point? $\endgroup$ – Dr. Lutz Lehmann Dec 26 '15 at 12:18

Only count the numbers.

$$\dot X_{n\times1} = f(X_{n\times1},t)$$

$\dot X_{n\times1}$ is made of $n$ numbers.

$(X_{n\times1},t)$ is made of $n+1$ numbers.

We do not care how much information they contain. We just care about how many number they are.


$$W_{n\times 1}(t)=h(U_{m\times 1},t)$$


$$h: \mathbb{R}^{m+1}\rightarrow\mathbb{R}^{n}$$

Let's make an example. Let's say at $t=2.5$,


How many numbers did $h$ received from us to give us the result? 5+1 (or m+1) And how many numbers did $h$ produced? 3 (or n)

It is true that the result is time dependent. However, we received only 3 numbers from the function $h$ and nothing more.

  • $\begingroup$ So really we have $\dot x_t$ where $\dot x_t$ is $\dot x(t)$ evaluated at $t$. Then this is just in $\Bbb R^n$ and $f$ takes a point in $\Bbb R^n$ as well as time for $\Bbb R^{n+1}$ $\endgroup$ – Thomas Dec 26 '15 at 12:34

The answer is that one does not want to restrict the class of problems covered by this theory if it can be avoided.

Differential equations $\dot x(t)=f(x(t))$ exist and are called autonomous. But they do not cover some of the most elementary applications. For instance a mechanical system with an external periodic force like $\ddot u+u=\sin 2t$ requires in a transcription for $x=(u,\dot u)$ the time $t$ as extra parameter, leading to $\dot x(t) = f(t,x(t))=(x_2(t),\sin 2t-x_1(t))$.


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