Looking for a first order perturbation of the Laplacian having 0 in its spectrum

Question

I would like to find an explicit example of a linear elliptic operator having the following form: $$Lu=-\Delta u +b(x)\cdot \nabla u, $$ where $b\colon \mathbb{R}^n \to \mathbb{R}^n$, and such that there exists a non trivial (weak) solution of the Dirichlet problem $$ \begin{cases} Lu = 0 & \text{in}\ \Omega \\ u=0 & \text{on}\ \partial \Omega\end{cases}.$$ Here $\Omega$ can be any bounded and smooth domain of $\mathbb{R}^n$.

In functional terms, I am trying to convince myself that a first-order perturbation of the Laplacian can alter the spectrum to the point that $0$ becomes an eigenvalue.

Can somebody provide me with such an example? I have tried looking at the one-dimensional case, when $\Omega$ reduces to an interval, but I could not find one.

Thank you for reading.

Answer 1

Let $\phi\colon[\,a,b\,]\to\mathbb{R}$ be a $C^1$ function such that $\int_a^b\phi(t)\,dt=0$. Then $u(x)=\int_a^x\phi(t)\,dt$ is a $C^2$ function such that $u(a)=u(b)=0$ and $$ u''=b\,u'\quad\text{with}\quad b(x)=\frac{\phi'(x)}{\phi(x)}. $$ All explicit examples I have tried give $b$'s that are not in any $L^p$, $1\le p\le\infty$, so that $u$ might not be a weak solution.

On the other hand, it is easy to see that if $b$ is $C^1$, then no regular solution exits. From $u(a)=u(b)$ it follows that $u'(c)=0$ at some $c\in(a,b)$. Since $b$ is regular this implies that $u''(c)=0$. A uniqueness argument implies that $u$ must be constant.

Answer 2

I am starting to think that such an example is not that easy to find. Indeed, it won't exist if $b$ is the gradient of a scalar field $B$. I propose a proof of this: please, let me know if you find it wrong. Thank you.

Claim Suppose that $B\colon \mathbb{R}^n\to \mathbb{R}$ is differentiable. Then the only weak solution to the problem $$\tag{1}\begin{cases} -\Delta u + \nabla B\cdot \nabla u =0 & \Omega \\ u=0 & \partial \Omega\end{cases}$$ is the trivial one.

Proof As you can see in this answer, the problem (1) is the Euler-Lagrange equation of the following functional: $$S(u)=\int_{\Omega} e^{-B(x)}\frac{\lvert \nabla u\rvert^2}{2}\, dx,$$ so that (1) is equivalent to the relation $$dS(u)v=0, \quad \forall v \in H^1_0(\Omega).$$ We compute $dS(u)v$ and discover that $$dS(u)v=\int_\Omega e^{-B(x)}\nabla u\cdot \nabla v\, dx.$$ So if $u$ satisfies (1) we have, setting $v=u$ above, $$\int_\Omega \lvert \nabla u\rvert^2e^{-B(x)}\, dx=0,$$ from which we infer that $\nabla u =0 $ on $\Omega$.