Question

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.

Answer 1

I love this problem and have spent half the evening thinking about it.

Here is a rough sketch of an idea that could possibly work. Making it rigorous might be a bit of a chore due to the unboundedness of $\mathbb{R}^n$, etc, but my guess is that it is doable.

Let $L$ denote the elliptic operator \[ Lu : = -\Delta u - \mathrm{div}( f u ). \]

Suppose the operator $L$ has a principal eigenvalue $\lambda_1$, which is the smallest number $\lambda$ for which there exists a nonzero solution of the equation $Lu = \lambda u$ in $\mathbb{R}^n$. Then $\lambda_1$ should be simple (!) and have a principal eigenfunction which does not change sign. Let $\varphi \in L^1(\mathbb{R}^n)$ denote the principal eigenfunction, which we normalize to be positive.

Assume for now that $\varphi$ and its derivatives tend to zero at infinity, and $f$ and its derivatives stay bounded. Then we may simply integrate the equation $L\varphi = \lambda_1\varphi$ to get $\lambda_1 \int_{\mathbb{R}^n} \varphi dx = 0$. Well, that means that $\lambda_1 = 0$.

Recalling the simplicity of $\lambda_1=0$, we see that the equation $Lu = 0$ not only has a positive solution, its set of solutions is precisely $\{ c \varphi : c\in \mathbb{R} \}$. This implies the result.

Now, you may have already noticed that there aren't really such principal eigenvalues in general, for example when $f=0$. But I think it is possible that the idea can still be made into a rigorous proof.

If we look at a really large domain, say the ball $B(0,R)$ with $R> 0$ very large, there is a principal eigenvalue $\lambda_{1,R}$ of $L$ on $B(0,R)$ and it is going to be close to zero. This can be shown by considering the adjoint operator $L^*$, which has the same principal eigenvalue $\lambda_{1,R}$, and it is easy to show that $0 < \lambda_{1,R} \leq CR^{-2}$, due to boundedness of $f$. This is where we use the special form of the equation.

If we have a solution to $Lu = 0$ on the whole space $\mathbb{R}^n$, with $u(0)>0$, I believe it should be possible to prove that if we choose the normalization $\varphi_{1,R}(0) = u(0)$, then $\varphi_{1,R}$ converges to $u$ (at least locally uniformly) as $R\to \infty$. In particular, $u$ must be positive everywhere.

There are obviously some details left to work out. It is very possible I am making a silly error and it doesn't work at all.