Meaning of this definition of a system of ODE:s? (Multivariable ODE?)

Question

I always thought an ODE had one independent variable:

For a system of $n$ ODE:s, we have a vector function $f$ of one variable $t$, $f:\mathbb R\rightarrow \mathbb R^n$ and \begin{align} \dot x(t)=f(t,x(t)) \end{align}

However, in Ordinary Differential Equations by Logemann and Ryan, the defintion is

Our goal is the study of systems of ordinary differential equations of the form $$ \dot x(t)=f(t,x(t)) $$ Here $f:J\times G\rightarrow \mathbb R^N$ is a suitably regular function, $J\subset \mathbb R$ is an interval and $G$ is a non-empty open subset of $\mathbb R^N$.

From "$J\times G$", $f$ is now a function of $N+1$ variables: $f:\mathbb R^{N+1} \rightarrow \mathbb R^N$.

But hey, it is an ODE, $f$ should be function of one variable?

I'm stuck at this definition, I can't wrap my head around this!

I appreciate any clarification/reference. Thanks!

Answer 1

An ODE can be a function of as many variables as you like, the point is that there are only derivatives in one variable - this variable is often denoted $t$ (for time).

E.g. $t\mapsto x(t),$ $[0,1]\to \mathbb R^3$ can be thought of as a particle satisfying some ODE $$\frac{d}{dt}x(t)=f(t,x(t)),\quad x_0=x.$$ This is an ODE, as only a time-derivative occurs -- despite the fact that $f:[0,1]\times\mathbb R^3\to \mathbb R^3$ is a function of several variables. Indeed, $f$ needs to be a function of many variables as the solution to the ODE lives in $\mathbb R^3$ and we are composing $f$ with this solution!

Answer 2

Suppose you have a function

$$F(u,v) = [u^2 v, uv, 4v]$$

This is a vector-valued function of two variables: $F : \mathbb{R}^2 \rightarrow \mathbb{R}^3$.

Now, suppose you are interested in finding a function $y:\mathbb{R}\rightarrow \mathbb{R}^3$ such that: \begin{align*} \dot y(t) = [t^2 y(t),\, t\cdot y(t),\, 4y(t)] \end{align*}

You could express this problem by saying that you are looking for a function of one variable $y:\mathbb{R}\rightarrow \mathbb{R}^3$ such that for all real $t$ in an area of interest, the following equation is true:

$$\dot y (t) = F(t, y(t))$$

Evidently, both $\dot y$ and the composite function $h(t)\equiv F(t, y(t))$ are functions of a single variable. This is what makes the equation an ordinary differential equation.

$F$ has two independent variables. But the problem you've set up— find a function $y:\mathbb{R}\rightarrow \mathbb{R}^3$ such that $\dot y = F(t, y(t))$ has only single-valued functions on both sides.

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Maybe another way to put this is in terms of composition: given a function $y:\mathbb{R}\rightarrow \mathbb{R}$, you can explicitly mention a function $G_{[y]}:\mathbb{R}\rightarrow \mathbb{R}^2$ where $G_{[y]}(t) \equiv \langle t,\, y(t)\rangle$.

Then $F(t, y(t))$ can more clearly be written as a composition $F\circ G_{[y]}$ which sends $$t\mapsto \langle t,\,y(t)\rangle \mapsto F(t, y(t)).$$

$F$ is a function with two free variables (named $u$ and $v$, above). In contrast, $F\circ G_{[y]}$ is a function with one free variable (named $t$).

With an ordinary differential equation, you are searching for a function $y$ such that for all real $t$ of interest, $$\dot y = F \circ G_{[y]}$$

that is, such that $\dot y(t)$ and $F(t, y(t))$ are equal as functions of $t$.

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I write the function as $G_{[y]}$ as a reminder that $G$ was defined in terms of a particular function $y:\mathbb{R}\rightarrow \mathbb{R}$. In particular, $G_{[y]}$ has a geometric interpretation: it is the graph of $y:\mathbb{R}\rightarrow \mathbb{R}$.