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In the answer to this question, a helpful Stack user said that the Poisson kernel does not necessarily give a unique harmonic function, given certain boundary data (in particular on the upper half-plane). I had learned that it did, based on the following argument: take the Green's identity $$\int_D u\Delta v - v\Delta u \,\,dV = \int_{\partial D} u \frac{\partial v}{\partial \nu} - v \frac{\partial u}{\partial \nu}d\sigma.$$ Now take $v(x,y)$ to be such that $\Delta_y v = -\delta_x$ and $v(x,y) = 0$ for $y \in \partial D$, and $\forall x \in D$. Suppose $u$ is a harmonic function with (integrable) boundary data $f$. Then the above gives $$u(x) = -\int_{\partial D}f\frac{\partial v}{\partial \nu}.$$ So $u$ is the unique such function, and altering $f$ on a set of measure zero should make no difference.

Is this argument correct?

This arose from trying to show that an analytic function defined on an infinite wedge in $\mathbb{C}$, for which we knew the real part on the boundary (except for two points), is unique up to an additive imaginary constant.

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  • $\begingroup$ The Poisson integral of the boundary values (if they are integrable) is of course unique. But not always is there only one harmonic function which has the given boundary values. Consider $u(x,y) = \sin x\cdot \sinh y$ on the upper half plane. Its boundary values are $0$, but $u \not\equiv 0$. $\endgroup$ Commented Dec 28, 2013 at 23:12
  • $\begingroup$ @DanielFischer Where does the above argument fail? (I.e. why do both functions not equal $$-\int_{\partial D}f\frac{\partial v}{\partial \nu}?)$$ Does it have to do with the Green's identity only holding provided everything is integrable? (So it should hold for nice boundary data on bounded domains?) $\endgroup$
    – Eric Auld
    Commented Dec 28, 2013 at 23:16
  • $\begingroup$ Yes. I don't remember exactly what regularity you need (hence not an answer), but if you transport the example to a disk, you get ugly behaviour near one boundary point (corresponding to the boundary point $\infty$ of the half plane in the sphere), that function is not the Poisson integral of a function (but I think, it is the Poisson integral of a measure). You need the function to have boundary values $f$ in a suitable sense to write it as a Poisson integral. $\endgroup$ Commented Dec 28, 2013 at 23:27
  • $\begingroup$ @DanielFischer OK, I am interested in what others have to say. It is surprising to me that there are two harmonic functions in the upper half-plane with zero boundary data, which you would think is the nicest boundary data there could be $\endgroup$
    – Eric Auld
    Commented Dec 29, 2013 at 5:02

1 Answer 1

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Philosophy

In a course in real analysis / measure theory one learns that sets of measure zero are negligible for integration. And so they are — for integration of functions. But when we integrate an expression involving derivatives of a function, something tricky can happen: if the function is not smooth enough, its derivative may become a distribution that is not represented by a function. Such a distribution may live on just one point (Dirac's delta) and yet contribute to the integral. I already wrote about this difference between straightforward integration and applications of some kind fo Fundamental theorem of Calculus. This is why careful statements of Green's identities (and other forms of the FTC / Stokes' theorem) impose smoothness up to and including the boundary. In specific cases, one may be able to use them for less smooth functions by exhausting the domain by compact subdomains and arguing that the relevant integrals converge; but this convergence is not automatic.


Concrete example

The Poisson kernel itself presents an example of how one boundary point can make a difference. The function $$ u(r,\theta) = \frac{1-r^2}{1-2r\cos\theta +r^2} \tag{1}$$ is harmonic in the unit disk $D$ and tends to zero at every boundary point except one. But it is not identically zero. It is not reproduced by the Poisson integral formula from its boundary values, understood as "$0$ a.e." And in particular,
$$ 0=\int_{D} \Delta u \ne \int_{\partial D} \frac{\partial u}{\partial n} =-2\pi \tag{2}$$ (taking exterior normal). In physical terms, the gradient field $\nabla u$ flows in across the whole boundary, and has no sinks inside of the domain. Where does it go? It squeezes out through the single point $1$.

What do we need to know about a harmonic function $h$ on $D$ to conclude that it's represented by a Poisson integral of an $L^1$ function? A necessary condition is $$ \sup_{0<r<1}\int_0^{2\pi} |h(r,\theta)|\,d\theta<\infty \tag{3}$$ The reason (3) is necessary is that convolution with Poisson kernel is a contraction on $L^1$ space; thus, the $L^1$ norm of $h$ on every concentric circle is at most the $L^1$ norm of its boundary values. But (3) is not sufficient for $h$ to be the Poisson integral of an $L^1$ function, because (1) satisfies it. It turns out that (3) is necessary and sufficient for $h$ to be the Poisson integral of a finite signed measure on $\partial D$. For the function (1), this measure is a point mass at $1$.

If a stronger condition holds: there is $p>1$ such that $$ \sup_{0<r<1}\int_0^{2\pi} |h(r,\theta)|^p\,d\theta <\infty \tag{4}$$ then $h$ is the Poisson integral of an $L^1$ (in fact $L^p$) function on the boundary. The boundary values can be understood in the sense of nontangential limits, or as a limit in $L^p$ of the restrictions of $h$ to concentric circles. In particular, every bounded harmonic function on $D$ is the Poisson integral of an $L^\infty$ function on the boundary. For example, if we prescribe $h$ to be $1$ on some part of the boundary, and $-2$ on another, and $\limsup_{z\to\zeta}|h(z)|\le 2$ when $\zeta$ belongs to the remaining part of boundary (of measure zero), then such a harmonic function is unique.


Remarks

The difference between $p=1$ and $p>1$ has to do with the fact that $L^1$ is not a dual space. For $p>1$, we can use the Banach-Alaoglu theorem to obtain a weak limit of restrictions of $h$ to circles $|z|=r$; this weak limit gives the notion of boundary values that's suitable for integration.

Condition (4) can be weakened to uniform integrability of the restrictions of $h$.

When instead of $D$ we look at the half-plane, we should realize that $\infty$ is a boundary point as well. A Möbius map makes this precise, sending $\infty$ into a boundary point of the disk.

Recommended reading: Potential theory in the complex plane by T. Ransford.

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  • $\begingroup$ Fantastic!----- $\endgroup$
    – Eric Auld
    Commented Dec 29, 2013 at 20:25
  • $\begingroup$ Is Wikipedia wrong here where they seem to claim that the function is unique provided the boundary data is in $L^1$? It seems they should need boundary data continuous. en.wikipedia.org/wiki/Poisson_kernel#On_the_unit_disc $\endgroup$
    – Eric Auld
    Commented Dec 31, 2013 at 0:04
  • $\begingroup$ Also, you said, "For example, if we prescribe h to be 1 on some part of the boundary, and −2 on another, and $\limsup_{z\to\zeta}|h(z)|\leq 2$ when $\zeta$ belongs to the remaining part of boundary (of measure zero), then such a harmonic function is unique." Can you offer a resource for that claim? (Ransford, perhaps?) Or sketch the proof? Thanks $\endgroup$
    – Eric Auld
    Commented Dec 31, 2013 at 0:09
  • $\begingroup$ Yes, the Wikipedia claim is incorrect. The maximum principle requires control at all boundary points, not "a.e." $\endgroup$ Commented Dec 31, 2013 at 2:50

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