# Uniqueness for linear elliptic PDE given existence

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 $$C^1$$.

• I agree with all your observations. You might be better off treating this as an eigenvalue-type problem, that is, trying to prove non-uniqueness by showing there exists an $f$ with a non-zero solution $u$ Jul 22, 2023 at 6:10
• @JackT thank you for the comment. So it seems I should head for a counterexample. I'd also be interested in that! Jul 22, 2023 at 6:58

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 $$u$$.

• Thank you for the answer. Very nice construction. Is it problematic that $a$ depends on $\lambda$? I think there should be a fix given the weighted eigenvalue problem has a non-empty set of eigenvalues. But I don't see how. Is this a legit issue? Jul 22, 2023 at 10:12
• Yeah, good question. I actually hadn’t thought about it. It probably is an issue, but I’m not 100% Jul 22, 2023 at 10:41

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 :-)

• Thank you very much for the answer. Let me know if I understood your argument correctly. You calculated the second derivative of $I$ to show convexity, right? At some point, you say integrate over $I$, but I think you mean $J$? Also very nice where the term $(fu)_x$ is used. Finally, I think this argument also applies with $v\cdot \nabla (fu)$ by considering an orthogonal coordinate transformation which keeps the Laplacian invariant and such that $v\cdot\nabla$ is the derivative in $x_1$ in the new coordinates (tho domain changed). I need some time to fully digest your answer! Very neat one! Jul 27, 2023 at 17:33
• @Shashi Yes, the integrations should be over $J$. I edited. As to the second derivative, the answer is "sort of" because the smoothness properties of $I$ are not so clear even at the level of the first derivative, so the formal double differentiation under the integral sign is a bit ugly business. What I did was just to show that for every $\varepsilon>0$, we have $I(y+t)+I(y-t)-2I(t)\ge -\varepsilon t^2$ for all $y$ and small enough $t$, which in the class of continuous functions implies convexity. And yes, the domain doesn't matter too much as long as the boundary is reasonable. Jul 27, 2023 at 18:04
• Thank you for the clarification, indeed I wasn't aware of that characterization of convex functions! Jul 27, 2023 at 22:44
• Hi sorry one more question, is this a well known technique? Like is there similar techniques in some textbooks? Jul 30, 2023 at 18:48
• @Shashi That is something I'm a wrong person to ask about: I have my toolbox, but I don't keep the origin labels on my tools :-) It is certainly not unknown but whether this particular variation of the general maximum principle idea appears in some textbook is hard to tell: it is more reminiscent of some tricks used for parabolic PDE than for elliptic ones, so I would look there first if you need a reference. Jul 30, 2023 at 19:03