Is this a typo?

From Ordinary Differential Equations and Dynamical Systems by Gerald Teschl (free copy), page 36, equation $2.10$ says:

Initial value problem \begin{align} \dot x &= f(t,x) \tag 1\\ x(t_0) &= x_0 \tag 2 \end{align} We suppose $f\in C(U,\mathbb R^n)$, where $U$ is an open subset of $\mathbb R^{n+1}$ and $(t_0,x_0)\in U$.


Isn't the order wrong in $(t_0,x_0)$ and $f(t,x)$ due to $\mathbb R^{n+1}$?

For simplicity I here use $\mathbb R^{n+1}$ instead of the subset $U$. So, from $(t_0,x_0)\in \mathbb R^{n+1}$ we have that $t_0 \in \mathbb R^n$ and $x_0 \in \mathbb R$, but in $(1)$ the variable is $t$? If $t\in \mathbb R$, isn't also $t_0 \in \mathbb R$?

I.e. shouldn't it be \begin{align} \dot x &= f(x,t) \tag 3\\ (x_0,t_0) &\in U \tag 4\\ U &\subset \mathbb R^{n+1} \tag 5 \end{align}

Alternatively $\mathbb R \times \mathbb R^n =\mathbb R^{1+n}$ for $(1)-(2)$.

  • 2
    $\begingroup$ But $\mathbb{R}^{1 + n}$ is the same as $\mathbb{R}^{n + 1}$. I think it is clear from the context that $x\in\mathbb{R}^n$ since he said $f\in C(U, \mathbb{R}^n)$. $\endgroup$
    – Chee Han
    Commented Dec 10, 2020 at 23:58

1 Answer 1


It sounds like you are viewing $\mathbb{R}^{n+1}$ as $\mathbb{R}^{n}\times\mathbb{R}$.

But $\mathbb{R}^{n+1}$ is also $\mathbb{R}\times \mathbb{R}^{n}$.

The integer $n+1$ is the same as the integer $1+n$.


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