0
$\begingroup$

Consider the transport equation $$u_t(t,x) + v(t,x) \cdot \nabla u(t,x) = 0.$$ Suppose that the solution of the characteristic equation $$\dot X(t) = v(t,X(t)) $$ decays to zero as $t \to \infty$. What happens to the solution $u$ of the PDE as $t \to \infty$? Does it also decay to zero or to the Dirac delta as the weak solution formula $$\int_{\mathbb R^N} \phi u(t,x)dx = \int_{\mathbb R^N} \phi(X(t,x))u_0(x)dx \qquad \phi \in C^\infty_c$$ suggests?

$\endgroup$
  • $\begingroup$ V is invompressible if its divergence is 0 everywhere. Such flows cannot have sinks as required by your 2nd eq. $\endgroup$ – Piyush Grover Jun 28 '20 at 16:34
  • $\begingroup$ @PiyushGrover Thanks. Can you show why if $\mathrm{div} v \neq 0$ then the solution of the ODE cannot decay to zero? $\endgroup$ – Zac Jun 28 '20 at 16:39
  • $\begingroup$ I am saying the opposite, that is if div v=0, then ODE cannot decay to 0 for all initial conditions.. Just take tiny circle around origin and apply divergence thm. Since all traj. are going into that circle, the line integral will be non-zero, but the area integral is 0 if div.v=0. $\endgroup$ – Piyush Grover Jun 28 '20 at 16:44
  • $\begingroup$ @PiyushGrover This counterexample is not clear to me: where are you applying the divergence theorem? Let's start over: if div v = 0, is it possible to prove that $X(t) > c \ge 0$ for every $t>0$? $\endgroup$ – Zac Jun 28 '20 at 16:55
  • $\begingroup$ Take a 2D example with 0 divergence. $\dot{x}=x$,$\dot{y}=-y$. See what you get. $\endgroup$ – Piyush Grover Jun 28 '20 at 21:20
1
$\begingroup$

Let me change slightly your notations with the flow $\psi (t,y)$ defined by $$ \dot \psi(t,y)=v(t, \psi(t,y)),\quad \psi(0,y)=y. $$ The solution $u$ is constant along the integral curves of the vector field, i.e. $$ u(t,\psi(t,y))=u(0, y). $$ Using the inverse function theorem you can introduce $\phi(t,x)$ to be a first integral defined by $$ x=\psi(t,y)\Longleftrightarrow y=\phi(t,x). $$ It is possible locally and let us assume that we can do that globally. Then we have $$ u(t,x)=u(t=0, \phi(t,x))=u_{0}(\phi(t,x)). $$ Assuming for instance that the initial datum $u_{0}$ is compactly supported or decays at infinity, you will get decay for the solution $u$ whenever $\phi(t,x)$ goes to infinity when $t\rightarrow+\infty$. The natural condition for decay of $u$ whenever the Cauchy datum $u_{0}$ is say compactly supported is that the first integral (which is the inverse function of the flow) goes to infinity with $t$.

$\endgroup$
  • $\begingroup$ Thanks! I have some additional questions: 1. Does $\psi \to 0$ imply $\phi \to \infty$ as $t \to \infty$? 2. What happens in general, for example for measure initial data? Do we have convergence to the Dirac delta? $\endgroup$ – Zac Jun 28 '20 at 20:42
  • $\begingroup$ @Zac Since $x=\psi(t,y)$ is equivalent to $y=\phi(t,x)$, we have $x=\psi(t,\phi(t,x))$. Assuming $x\not=0$ we must have that $\phi(t,x)$ goes to infinity, otherwise under a mild continuity assumption, $x=\psi (+\infty,0)=0$. $\endgroup$ – Bazin Jun 29 '20 at 10:20
  • $\begingroup$ Thank you! What about the second question on the convergence to the Dirac delta in case of measure initial data? $\endgroup$ – Zac Jun 29 '20 at 10:45
  • $\begingroup$ @Zac Well, under some conditions of regularity and behavior at infinity, the solution of the transport equation is given via the first integral above with the formula $u=u_0(\phi(t,x))$, where $u_0$ is the Cauchy datum. To prove convergence to the Dirac mass at $x=0$, you take $u_0=\delta_0$ which is indeed well localized; I guess that the arguments sketched above show that $\phi(t,x)$ goes to infinity for $x\not=0$, so that $u=u_0(\phi(t,x))=0$, proving that the limit distribution is supported at 0. $\endgroup$ – Bazin Jun 29 '20 at 16:28
  • $\begingroup$ It seems that the weak formulation implies that the limit distribution is a measure, thus proportional to the Dirac mass at 0. $\endgroup$ – Bazin Jun 29 '20 at 16:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.