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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?

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  • $\begingroup$ V is invompressible if its divergence is 0 everywhere. Such flows cannot have sinks as required by your 2nd eq. $\endgroup$ Jun 28, 2020 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, 2020 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. $\endgroup$ Jun 28, 2020 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, 2020 at 16:55
  • $\begingroup$ Take a 2D example with 0 divergence. $\dot{x}=x$,$\dot{y}=-y$. See what you get. $\endgroup$ Jun 28, 2020 at 21:20

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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 $t$.

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  • $\begingroup$ 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, 2020 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, 2020 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, 2020 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, 2020 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, 2020 at 16:29

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