Asymptotics for solution of transport equation and characteristics

Question

Consider the transport equation $$u_t + v(t,x) \cdot \nabla u = 0.$$ Suppose that the solution of the characteristic equation decays to zero as $t \to \infty$. What happens to the solution of the PDE? Does it also decay to zero?

Answer 1

Let $X(t,y)$ be the flow defined by $X(0,y) = y$ and $\partial_t X(t,y) = v(t,X(t,y))$ and assume to simplify that the inverse flow $Y(t,x)$ is well defined (i.e. $X(t,Y(t,x))=x$). Let $u$ be the solution of $$ \partial_t u + v\cdot\nabla u = 0. $$ with initial condition $u(0,x) = f(x)$. Then $u(t,X(t,y))$ is constant by the equation, so $u(t,X(t,y)) = u(0,y) = f(y)$ and the solution is the function $$ u(t,x) = f(Y(t,x)). $$

-> Intuitively, since all the trajectories $X$ are converging to $0$, we see that the $Y$ will go in the other direction and go away from $0$, so the value of $u$ at $x$ will be the value of $f$ far away from $0$ (which is $0$ if $f$ is compactly supported).

-> To simplify, assume that $v(t,x) = -x$, so that $X(t,y)= y\,e^{-t} \underset{t\to \infty}{\longrightarrow} 0$. Then $Y(t,x) = x\,e^t$ and $$ u(t,x) = f(x\,e^t). $$

• If $f$ is compactly supported, then there is $R>0$ such that $f(y)=0$ when $|y|>R$, and for each point $x≠0$, when $t > \ln(|x|/R)$, then $|x\,e^t|>R$ so that $u(t,x) = 0$. Therefore, $f\to 0$ pointwise on $\mathbb{R}^d\setminus\{0\}$.
• With the same reasoning, if $f = \varphi + c$ with $\varphi$ compactly supported, then $f\to c$ pointwise on $\mathbb{R}^d\setminus\{0\}$
• And we can really get more general profiles by choosing the profile of $f$ at infinity. For example if $f(x) = g(x/|x|)$, then $u(t,x) = g(x/|x|)$ is constant and so converges towards $g(x/|x|)$ with $g$ any function defined on the sphere.

-> Remark in any case that if $f(0)=1$, $f$ is continuous and $f$ is compactly supported, even if $f(x)$ converges to $0$ almost everywhere, it does not converges to $0$ in $L^\infty$ (i.e. uniformly) since $\|u-0\|_{L^\infty} = \|f\|_{L^\infty} = 1$