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I have just started studying partial differential equations from PDE by Lawrence.C Evans.
While describing the transport equation, it is mentioned that,
$u_t+b\cdot Du=0$ in $\mathbb{R}^n\times(0,\infty)$, where
$b$ is a fixed vector in $\mathbb{R}^n$,$b=(b_1,\dots,b_n)$
and $u:\mathbb{R}\times[0,\infty)\to \mathbb{R}$ is the unknown $u=u(x,t)$.

What I don't understand is $\mathbb{R}^n\times(0,\infty)$? I know its the domain of $u$ but I don't know how to understand Cartesian product as a domain.

1500kook12
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    The cartesian product is just another set. The elements of $\Bbb{R}^n \times (0,\infty)$ are ordered pairs $(x,t)$ with $x\in \Bbb{R}^n$ and $t\in (0,\infty)$. – peek-a-boo Jul 28 '20 at 17:12
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    See my answer to this question here: https://math.stackexchange.com/questions/3770733/what-does-x-y-in-10-10-times-0-0-mean-multiple-of-two-domains/3771468#3771468 – Taylor Rendon Jul 29 '20 at 03:25
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    Are you sure you did read correctly? Shouldn't it be $u = u(x,t)$ and $u : \mathbb R^n \times (0,\infty) \to \mathbb R^n$? – LL 3.14 Jul 30 '20 at 10:14

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Given two sets $A$ and $B$, their Cartesian Product $A \times B$ is a new set, whose elements are ordered pairs. These ordered pairs have an element of $A$ in the first coordinate, and an element of $B$ in the second coordinate. $A \times B$ is the set that contains all possible ordered pairs. In formal language, $$ A \times B = \{ (a,b) : a \in A , b \in B \} $$ But you are trying to understand the cartesian product as a domain. To do this, I think it's best to consider a familiar example, the Cartesian Plane a.k.a the x-y plane. Recall that this is the 2-dimensional plane is defined by points $$(x,y)$$ where $x \in \mathbb{R}$ and $y \in \mathbb{R}$. This can be written as $$ \mbox{Cartesian Plane } = \mathbb{R} \times \mathbb{R} $$ and any real-valued function with this domain, $f:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$, can be seen as a 3-dimensional surface over the plane.

The idea is similar for $\mathbb{R}^n \times (0,\infty)$, although it may not be easy to imagine. For the case $n=1$, we have $\mathbb{R} \times (0,\infty)$ which is the upper half plane. So any real-valued function can be pictured as a surface over the upper half plane

NazimJ
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