14
$\begingroup$

Let $f\colon\mathbb{R}^n\to\mathbb{R}^n$ be a smooth, bounded vector field. Further, let $u\colon\mathbb{R}^n\to\mathbb{R}$ satisfy $$-\Delta u=\operatorname{div}(fu).$$ If $u\in L^1(\mathbb{R}^n)$, then $u$ has one sign, i.e., either $u>0$ everywhere, or $u<0$ everywhere, or $u=0$ everywhere.

I have a direct proof of this for $n=1$. For $n>1$, I have a proof using the theory of parabolic equations (see below). My question: Is there a direct proof using only the theory of elliptic PDEs?

(Edited to assume $f$ bounded and fix the case $n=1$ below.)

For $n=1$, my proof goes as follows. The equation is $-u''=(fu)'$, which integrates to $u'+fu=A$ for some constant $A$. If $u$ changes sign then we may without loss of generality take $u(0)=0$. Thus $$u(x)=A\int_0^x e^{F(t)-F(x)}\,dt$$ where $F'=f$. If $|f|\le c$ then $F(t)-F(x)\ge c(t-x)$ for $t<x$, so $u(x)\ge Ac^{-1}(1-e^{-cx})$ when $x>0$, and hence $u\notin L^1$ (unless $A=0$).

For $n>1$, my only proof is much more involved. Here is a brief outline. Assume the conclusion is wrong, so we can write $u=u_+-u_-$ with $u_\pm\ge0$ everywhere and neither identically zero, and $u_+u_-=0$ everywhere.

Now let $v_\pm$ solve $$\frac{\partial v_\pm}{\partial t}=\Delta v_\pm+\operatorname{div}(fv_\pm)$$ for $t>0$, with initial conditions $v_\pm(0,x)=u_\pm(x)$. By uniqueness for this equation (with suitable growth conditions at infinity), $v_+(t,x)-v_-(t,x)=u(x)$ for $t>0$ and $x\in\mathbb{R}^n$. Also, for $t>0$ we find $v_\pm>0$ everywhere, and also $$\int_{\mathbb{R}^n} v_\pm(t,x)\,dx=\int_{\mathbb{R}^n} u_\pm(x)\,dx$$ since the equation is on divergence form. We conclude $$\int_{\mathbb{R}^n}|u(x)|\,dx=\int_{\mathbb{R}^n}(u_+(x)+u_-(x))\,dx=\int_{\mathbb{R}^n}(v_+(t,x)+v_-(t,x))\,dx>\int_{\mathbb{R}^n}|v_+(t,x)-v_-(t,x)|\,dx,$$ which is a contradiction.

Improve this question
$\endgroup$
3
  • $\begingroup$ I might add a bit of intuition for why this is true: The laplacian makes stuff diffuse, while the divergence term transports stuff along the vector field while preserving the total amount of stuff. If a solution has two signs, the diffusion will mix the positive and negative parts together, making them shrink. This intuition is of course formalized in the parabolic proof. $\endgroup$
    – Harald Hanche-Olsen
    Commented Feb 9, 2010 at 0:52
  • $\begingroup$ Regarding my edit: It may be that it is sufficient to assume $f$ grows at most linearly. Some such restriction is necessary, though: I can make a counterexample for $n=1$ with $f$ odd, $f(x)=ax^{a-1}$ for $x>0$ where $a>1$. The odd solution $u$ with $A=1$ in the notation of the question will belong to $L^1$. We find $\int_0^\infty u=\int_0^\infty \int_0^x e^{y^a-x^a}\,dy\,dx$. The part where $y<1$ is no problem, and the other part is handled by noting $y^a-x^a<ay^{a-1}(y-x)$ when $y<x$ and integrating. $\endgroup$
    – Harald Hanche-Olsen
    Commented Feb 9, 2010 at 3:31
  • $\begingroup$ For a purely radial vector field, there is a purely radial solution, easily found, and with $g=0$ in your notation. For a purely rotational field, I don't know of any solution. You clearly can't get away with $g=0$ in that case. $\endgroup$
    – Harald Hanche-Olsen
    Commented Feb 9, 2010 at 21:15

1 Answer 1

Reset to default
1
$\begingroup$

A formal argument is to integrate both sides over the region where u>0. The integral of the right hand side is zero, while the integral of the left hand side yields something strictly positive if there is a boundary. This argument is not rigorous because we do not a priori know if the integrals converge. But you could try to make it rigorous by multiplying with a factor like $\exp(-\epsilon |x|^2)$ and then letting $\epsilon\to 0$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .