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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.

Giuseppe Negro
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  • What are the regularity conditions on $b$? – Julián Aguirre May 02 '13 at 13:20
  • @JuliánAguirre: Just $b\in L^\infty(\Omega)$. If you can provide an example with unbounded $b$ it's fine, too. – Giuseppe Negro May 02 '13 at 13:22
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    I think there is no such example, i.e., the weak maximum principle holds for this equation. But I'd like to understand the precise notion of weak solution used here, given the equation is in the non-divergence form. How do you test the validity of equation on test functions? – 75064 May 03 '13 at 18:46
  • @user75064: I mean that a weak solution is a function $u\in H^1_0$ such that $$\int \nabla u\cdot\nabla v+b(x)\cdot \nabla u v = 0$$ for all $v\in H^1_0$. – Giuseppe Negro May 03 '13 at 23:17

2 Answers2

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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 $\Omega$.

Community
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Giuseppe Negro
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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.

Julián Aguirre
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  • Thank you, this is a nice and useful observation for me. Moreover, the nonexistence of non trivial solutions when $b$ is regular points in the same direction as my other intervention below (the CW one). – Giuseppe Negro May 02 '13 at 14:20