8
$\begingroup$

Let $u(x, y)$ be a harmonic function on the upper half-plane $\mathbb{R}\times \mathbb{R}^+$, that is, $$\partial_x^2 u(x, y) + \partial_y^2 u(x, y) = 0$$ for $x \in \mathbb{R}, y>0$. Assume $u(x, 0)$ is smooth for $x \in \mathbb{R}$. In addition, we assume that both $u(x, 0)$ and $\partial_yu(x, 0)$ vanish at $x = 0$ to infinite order, i.e., for every $k \in \mathbb{Z}^+$, $$\lim_{x \to 0}\frac{u(x, 0)}{x^k} = \lim_{x\to 0}\frac{\partial_yu(x,0)}{x^k} = 0.$$

More explanation: As pointed out by Alexandre Eremenko, here we assume that $u(x, y)$ is continuous up to the boundary $\mathbb{R}\times \{0\}$, and we treat $-\partial_yu(x, 0)$ as the outer normal derivative of $u$ at the boundary point $(x, 0)$.

Question: can we conclude that $u(x, y)$ vanish to infinite order at the origin with respect to interior points? In other words, does the following limit hold for every $k \in \mathbb{Z}^+$? $$\lim_{\substack{(x, y) \to (0, 0)\\ x\in \mathbb{R}, y>0}}\frac{u(x, y)}{(|x|+|y|)^k} = 0.$$ If not, is there a counter-example?

$\endgroup$
6
  • $\begingroup$ (I think you mean "the upper half-plane $\mathbb R\times \mathbb R_+$" ) $\endgroup$ Aug 5 at 12:17
  • $\begingroup$ The physical intuition already suggests it can't be true. Imagine a soap film bounded by circular wire of radius 1, with an obstacle in the middle, given by a unit wire straight segment, located at some distance from the disk on a plane parallel to it. We expect the shape of the film not to be smooth near the segment, but more similar to a Canadian tent. $\endgroup$ Aug 5 at 12:26
  • $\begingroup$ You should add some conditions, explaining what $u(x,0)$ and $u_y(x,0)$ exactly are. Your function is defined only in the upper half-plane. Is it assumed to be continuous in the closed half-plane? $u_y$ is continuous in the closed half-plane? What about $u_x$? $\endgroup$ Aug 5 at 13:47
  • $\begingroup$ I just misses to see the condition on $u_y$ .. $\endgroup$ Aug 5 at 15:21
  • $\begingroup$ @PietroMajer As a Canadian, what is a Canadian tent? $\endgroup$ Aug 5 at 20:40
5
$\begingroup$

There is a counterexample. Consider the harmonic function $$u(x,y) = Re\left(e^{-1/z^2}\right) = e^{-\frac{x^2-y^2}{r^4}}\cos\left(\frac{2xy}{r^4}\right),$$ where $r^2 = x^2 + y^2$. We have that $$u(x,\,0) = e^{-1/x^2}$$ vanishes to infinite order in $x$, and that $$u_y(x,\,0) \equiv 0$$ since $u$ is even in $y$. However, $$u(x,\,x) = \cos(2/x^2)$$ does not vanish to infinite order.

$\endgroup$
2
  • $\begingroup$ This $u$ is not even bounded near the origin: $u(0, y) = e^{1/y^2}$, so what would $u_y(0, 0) = 0$ mean? $\endgroup$ Aug 5 at 14:59
  • $\begingroup$ @MateuszKwaśnicki: That is a fair point. As Alexandre Eremenko points out in the comments, it could be helpful for the question-asker to clarify what is meant by $u,\,\nabla u$ on $\{y = 0\}$. $\endgroup$ Aug 5 at 15:50
4
$\begingroup$

Assuming $u$ is smooth enough, the answer seems to be affirmative.

The $k$-th term in the power series of $u$ near the origin must be a solid harmonic polynomial $P_k$ of degree $k$, satisfying two independent conditions: $\partial_x^k P_k = 0$ and $\partial_x^{k-1} \partial_y P_k = 0$. The space of harmonic polynomials of degree $k$ is two-dimensional, so this essentially tells us that $P_k = 0$. Consequently, the power series of $u$ near $(0, 0)$ is zero, and hence all partial derivatives of $u$ vanish at the origin.


Edit: Here are some additional details. Suppose that $u$ is the Poisson integral of the boundary data $f$ (so, for example, it suffices to know that $u$ is bounded, or non-negative — this is a rather mild condition). Suppose, furthermore, that $f$ is infinitely smooth in a neighbourhood of $0$. Then it is an easy exercise to see that $u$ is infinitely smooth in a neighbourhood of $(0,0)$ (intersected with $\mathbb R \times [0, \infty)$, of course).

In particular, we can develop $u$ into the power series at $(0, 0)$. Of course, this power series need not be convergent, it is just a convenient formal way to speak about the partial derivatives of $u$. The $k$-th term of this power series, call it $P_k$, is a homogeneous polynomial of degree $k$, and using smoothness of $u$ it is easy to check that $P_k$ is a harmonic polynomial.

The remaining part of the argument is given in the original answer.

$\endgroup$
4
  • $\begingroup$ If we want to expand $u$ as harmonic polynomials near the origin, do we need $u(x, y)$ to be harmonic on the whole plane $\mathbb{R}^2$? I am not sure whether this would work when the origin lies on the boundary. $\endgroup$
    – Jacob Lu
    Aug 6 at 4:23
  • $\begingroup$ I added some details. Of course the power series need not converge to $u$ (consider $u(x,y)=\Re(\exp(-1/\sqrt{y-ix}))$), but it allows one to easily handle the partial derivatives of $u$. $\endgroup$ Aug 6 at 7:15
  • $\begingroup$ I see. This makes sense. If $u$ is harmonic on the upper half-plane and smooth up to the boundary, and vanishes at $(0, 0)$ up to infinity order from inside, do you think now $u$ is identically zero? $\endgroup$
    – Jacob Lu
    Aug 13 at 4:44
  • $\begingroup$ Yes, I think this is what I tried to write. $\endgroup$ Aug 15 at 22:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.