Given the equation [here](https://mathoverflow.net/questions/397816/linear-transport-equation-with-unbounded-coefficients), I would like to ask the following relaxed question:


Consider the PDE


$$\partial\_t f(x,t) = \langle q(x), \nabla \rangle f(t,x) + p(x),$$


with Schwartz initial data $f(0,x) = f\_0(x) \in \mathscr S(\mathbb R^n).$


I am wondering then if $q$ is Lipschitz and $p$ is Schwartz as $f\_0$, does there exist a solution $(t,x)\mapsto f(t,x)$ for this equation that decays faster than any polynomial in the space variable $x$ for any fixed time $t>0$?



Linear transport equation with Lipschitz conditions
The anwer is **yes**.


**Preliminary observations:**


Let us first consider the case $p=0$.


Since $q$ is globally Lipschitz on $\mathbb{R}^n$, say with Lipschitz constant $L$, all solutions of the ODE $\dot x = q(x)$ on $\mathbb{R}^n$ exist globally and, if $(\varphi\_t)\_{t \in \mathbb{R}}$ denotes the induced flow in $\mathbb{R}^n$, each function $\varphi\_t: \mathbb{R}^n \to \mathbb{R}^n$ is Lipschitz continuous with constant $e^{|t|L}$ (this follows from Grönwall's lemma).


Since we assumed $p=0$, the solution of your PDE is given by
$$
 f(t,x) = f\_0(\varphi\_t(x)) \tag{\*}\label{1}
$$
for each $t$ and $x$.


Hence, if $f\_0$ decays faster than every polynomial, so does $f(t, \cdot)$ for each fixed time $t$.


Let us now phrase this in a more functional analytic language: For each integer $k \ge 0$, consinder the weight function $w\_k: \mathbb{R}^n \to [1,\infty)$ given by $w\_k(x) =1 + \lvert x \rvert^k$, and consider the Banach space $E\_k$ of all continuous functions $f: \mathbb{R}^n \to \mathbb{R}$ which satisfies $w\_k f \in C\_0(\mathbb{R}^n; \mathbb{R})$; we endow the space $E\_k$ with the weighted supremum norm $\|\cdot\|\_{w\_k}$ given by
$$
 \|f\|\_{w\_k} = \|w\_k f\|\_\infty.
$$


It follows from $\eqref{1}$ that the *Koopman group* $(T\_k(t))\_{t \in \mathbb{R}}$ given by
$$
 T\_k(t)f = f \circ \varphi\_t
$$
for each $f \in E\_k$ is a well-defined one-parameter group of bounded linear operators on $E\_k$. Moreover, the orbits of the group are continuous with respect to the weak topology on $E\_k$ (use that (i) $E\_k$ is isomorphic to $C\_0(\mathbb{R}^n; \mathbb{R})$ via multiplication with $w\_k$, that (ii) for each $x \in \mathbb{R}^n$, the mapping $t \mapsto \varphi\_t(x)$ is continuous, and (iii) the Riesz representation theorem for measures on $C\_0(\mathbb{R}^n; \mathbb{R})$).
Hence, by a standard result from $C\_0$-semigroup theory, the the group $(T\_k(t))\_{t \in \mathbb{R}}$ is even continuous with respect to the strong operator topology, i.e., it is a $C\_0$-group on $E\_k$.


It is trivial, but important to observe, that the groups $(T\_k(t))\_{t \in \mathbb{R}}$ act consistently on the spaces $E\_k$.


Using these preliminary observations, we can show:


**Theorem.** Let $p$ be continuous and decay faster than every polynomial.


(i) For each $f\_0 \in C\_0(\mathbb{R}^n; \mathbb{R})$ there exists a unique mild solution $f: \mathbb{R} \to C\_0(\mathbb{R}^n; \mathbb{R})$ of the Cauchy problem
$$
 \dot f = \langle q, \nabla \rangle f(t) + p, \qquad f(0) = f\_0.
$$


(ii) If $f\_0$ decays faster than every polynomial and $f: \mathbb{R} \to C\_0(\mathbb{R}^n; \mathbb{R})$ denotes the mild solution from (i), then for each fixed time $t \in \mathbb{R}$, the function $f(t)$ also decays faster (in space) than every polynomial.


*Proof.*
Fix $k \ge 0$. Since $(T\_k(t))$ is a $C\_0$-group on $E\_k$ and since $p \in E\_k$, it follows that, if $f\_0 \in E\_k$, the PDE has a unique mild solution $f: [0,\infty) \to E\_k$ given by
$$
 f(t) = T\_k(t)f\_0 + \int\_0^t T\_k(s) p \, ds,
$$
where the integral is a Bochner integral in $E\_k$. Applying this to $k=0$, we obtain (i).


Moreover, these solutions are consistent on the $E\_k$-spaces (i.e., if $f\_0 \in E\_{k+1} \subseteq E\_k$, then the mild solutions in $E\_k$ and $E\_{k+1}$ are the same). If $f\_0$ decays faster than every polynomial, then we have $f\_0 \in E\_k$ for each $k$. Hence, for each time $t$ and each $k$, we have $f(t) \in E\_k$. Thus, for each time $t$, the function $f(t)$ decays faster in space than any polynomial. $\square$


