I am confused about the following.
Suppose we wanted to find a solution $u(\textbf{x}, t): \mathbb{R}^n \times \mathbb{R} \rightarrow \mathbb{R}$ to the PDE
$$ u_t + \textbf{b(x)}\cdot \nabla u = 0, $$ with some initial data function $g(\textbf{x})$ on $\mathbb{R^n} \times \{t=0\}$.
By the method of characteristics, we have the following system of equations: $$ \begin{cases} \frac{d}{dt} \textbf{x}(t) = \textbf{b}(\textbf{x}(t)); \ \textbf{x}(0) = \textbf{x}_0\\ \frac{d}{dt}z(t) = 0 \end{cases} $$ Since we don't know the form of $\textbf{b}(\textbf{x}(t))$, we can't write an explicit solution to this ODE. My notes say that one can write
$$ \textbf{x}(t) = \int_0^t \textbf{b}(\textbf{x}(s)) \ ds + \textbf{x}_0 $$ as an implicit form of the solution to the characteristic ODE, but I don't see where this is possible. I get as far as
$$ x_i(t) = \int_0^t b_i(\textbf{x}(s)) \ ds + x_i^0 $$
where $b_i$ is the $i$'th component function of $\textbf{b}$, and $x_i$ is the $i$'th component function of $\textbf{x}$.
I'm really not too sure about what the notes are saying, because what I have doesn't seem to lead to the same thing as what's in the notes, and furthermore, the result of integrating is a vector, not a scalar, which I find confusing. Also, I am not sure if I should really be writing $$ x_i(t) = \int_0^t b_i(\textbf{x}(s)) \|\textbf{x}'(s)\| \ ds + x_i^0, $$ because after all aren't I integrating $b_i$ along a characteristic curve? I'm pretty confused here, and I really need some help.
Thanks!
$$\textbf{x}(t) = \int_0^t \textbf{b}(\textbf{x}(s)) \ ds + \textbf{x}_0$$ is meant to be understood as $n$ different identities
$$x_i(t) = \int_0^t b_i(\textbf{x}(s)) \ ds + x_i^0, \quad i=1,2,\dots,n$$ i.e. $$\begin{bmatrix}x_1\\ \vdots\\ x_n\end{bmatrix}= \int_0^t\begin{bmatrix}b_1(\mathbf x(s))\\ \vdots \\b_n(\mathbf x(s))\end{bmatrix}ds := \begin{bmatrix}\int_0^tb_1(\mathbf x(s))ds\\ \vdots \\\int_0^tb_n(\mathbf x(s))ds\end{bmatrix} $$ The integrals on the right are (assuming $\mathbf b$ is continuous, say) normal honest Riemann integrals; that is, one could say, $$ \int_0^tb_i(\mathbf x(s))ds \overset{\text{def}}= \lim_{|\Delta P|\to 0} \sum_{s_k\in P} b_i(\mathbf x(s_k)) (s_{k+1} - s_k)$$ where $P$ denotes some partition $s_0,\dots,s_n$ of $[0,T]$ and the limit sends the partition mesh size $\Delta P := \sup_k |s_{k+1} - s_k|$ to 0. Note $ds$ in this case does not indicate a line integral using arc length parameterisation.
The integral identity/identities came about by using the 1-D Fundamental Theorem of Calculus, $n$ times: if $f'=g$ (and $f,g$ are nice), then $f(t) = \int_0^t g(s)ds + f(0)$. You just need to apply this with $f=x_i, \ g=b_i\circ \mathbf x$. No mention of any geometry here, and no fussing with $\|x'(s)\|$ either.
You may actually check that the definition for a Riemann integral makes 100% perfect sense when the output is a vector, so long as you check all convergence results in the correct (i.e. vector) norm; so the above line (with the obvious modification of replacing $b_i$ with $\mathbf b$) is a perfectly good definition of $\int_0^t \mathbf b(\mathbf{x}(s)) \ ds$.