Uniqueness for linear elliptic PDE given existence

Question

I am having an issue showing that zero is the only solution for the following PDE. Let $M=[-1,1]^2$ be a two dimensional square. And let $f\in W^{k,p}(M)$ for $k$ and $p$ arbitrarily high (or $C^\infty$ and Lipschitz up to the boundary etc, or any assumption with smoothness that is required) but without any given sign (e.g. we do not know $\partial_1 f>0$ etc). Consider the PDE

$$ \Delta u=\partial_1(fu), \ \ \ u|_{\partial M}=0. $$ I would love to have a prove or a counterexample for:

Problem. Does the PDE above has uniqueness (in the sense of weak solutions)? i.e., is $u\equiv 0$ the only solution?

The actual problem that I have is slightly different, but I thought the one above is easier to handle.

Alternative reformulation of problem. Let $G$ be the Green operator for the Laplacian with zero Dirichlet boundary condition. Is it true that any $u$ satisfying $$ u=G(\partial_1 (fu))$$ satisfies $u\equiv 0$?

Attempt. I had several attempts, first one is to expand via product rule $$ \Delta u=\partial_1 fu+f\partial_1u,$$ which by standard assumptions a solution exists (via Lax-Milgram). This requires suitable smallness of $\partial_1 f$, which I do not have.

Another attempt is to use Maximum principle, but in there I need a sign condition of $\partial_1 f$ which I do not have.

My feeling is that the smoothness of $f$ might play a role as I'm suspecting a counterexample exists when $f$ is not so regular.

Finally, to give a bit of context, this problem arises in some work I am currently doing where I need to identify two solutions and reduced the problem to the PDE above. Also one can a second derivative as well $\partial_2(gu)$ for example, but I do not think that makes it any different. I may be missing something, so I would appreciate any lead. I'm also happy with a counterexample. I just couldn't find any.

Addendum. Actually, if one is attempting for a counterexample, I would love to see any equation of the form $$ \Delta u=v\cdot \nabla(fu), \ \ \ u|_{\partial M}=0, $$ with non-zero solution $u$, where $v$ is a constant vector and $f$ is at least $C^1$.

Answer 1

Let $\Omega \subset \mathbb R^n$ be an open bounded set and $f\in C^1(\Omega)$. Consider the boundary value problem $$\left \{ \begin{split} \Delta u &= \partial_1(fu), \qquad&\text{in } \Omega\\ u&=0 \qquad &\text{on }\partial \Omega . \end{split} \right . \tag{$\ast$}$$ Let $a\in C^1(\Omega)$ be a function to be chosen later. Then, using that $u$ satisfies $(\ast)$, we have that \begin{align*}\operatorname{div} (a(x)\nabla u) &= a\Delta u + \nabla a \cdot \nabla u \\ &= af\partial_1u + \nabla a \cdot \nabla u+a (\partial_1 f)u. \end{align*} Next, assume that $a$ depends only $x_1$, so $$\operatorname{div} (a(x)\nabla u) = (af+ \partial_1a)\partial_1 u + a(\partial_1f) u . $$ Now, consider $f(x) = -\lambda x_1$ for some $\lambda >0$. Moreover, let $a(x) = \exp \big( \frac12 \lambda x_1^2 \big )$, so $\partial_1a = \lambda x_1 \exp \big( \frac12 \lambda x_1^2 \big )=-f(x)a(x).$ Thus, for this choice of $f$ and $a$, we have that $$\left \{ \begin{split} -\operatorname{div}(a\nabla u ) &= \lambda a u, \qquad &\text{in } \Omega\\ u&=0 \qquad &\text{on }\partial \Omega . \end{split} \right . $$ This is a weighted eigenvalue problem, so, particularly since $a$ is positive, it should be possible to choose $\lambda$ in such a way that there is a non-trivial solution $u$.

Answer 2

You have uniqueness at least for the classical solution (assuming $u\in C^2$ up to the boundary and $f\in C^1$, say. That can be relaxed quite a bit but you'd better tell us exactly how "weak" your weak solution is before we try to generalize).

The idea is very simple. Consider $U=\max(u,0)$ and $I(y)=\int_{-1}^1 U(x,y)\,dx$. We will show that $I$ is a convex function of $y$ and, since it is non-negative and tends to $0$ as $y\to\pm 1$, we'll have $U\equiv 0$.

For each $y\in(0,1)$, we define $G_y=\{x:u(x,y)>0\}$ and $E_y=[-1,1]\setminus G_y$. Fix $\varepsilon>0$ and take a very small $t$. Write $$ I(y+t)+I(y-t)-2I(t)= \\ \int_{G_y}[U(x,y+t)+U(x,y-t)-2U(x,y)]\,dx+\int_{E_y}[U(x,y+t)+U(x,y-t)-2U(x,y)]\,dx\,. $$ The second integral is trivially non-negative since $U(x,y)=0$ for $x\in E_y$ and $U\ge 0$ everywhere. In the first one $U(x,y)=u(x,y)$ and $U\ge u$ everywhere, so we can bound it from below by $$ \int_{G_y}[u(x,y+t)+u(x,y-t)-2u(x,y)]\,dx\ge t^2\int_{G_y}u_{yy}(x,y)\,dx-\varepsilon t^2 $$ if $t$ is small enough (I formally used the uniform continuity of $u_{yy}$, though you need it only in the integral sense).

Now $G_y$ consists of (possibly countably many) open intervals $J$. On each of them we have $u>0$ inside and $u$ vanishing at the endpoints plus our equation $u_{xx}+u_{yy}+(fu)_x=0$. Integrating that equation over $J$, we conclude that $\int_J [u_{xx}(x,y)+u_{yy}(x,y)]\,dx=0$ ($(fu)_x$ integrates to $0$). Also $u_{xx}$ integrates to the difference of $u_x$ at the endpoints and, since we ascend from $0$ at the left end and descend to $0$ at the right end of $J$, this difference is non-positive. Thus the integral of $u_{yy}$ over $J$, and, thus, over $G_y$, is nonnegative, and we have shown that $I(y)+\frac\varepsilon 2 y^2$ is convex. Since $\varepsilon>0$ was arbitrary, we are done.

Note that the cancellation coming from the possibility to integrate $(fu)_x$ to $0$ between two points where $u$ vanishes (i.e., the full derivative form of the lower order term) is crucial here.

I hope this clarifies the issue a bit :-)