4

I've read some complex analysis texts and often there is some appeals to Green's theorems when proving facts about contour integrals of holomorphic functions yet there seems to be a lack of appeals to Green's functions (at least explicitly). We know that for a holomorphic function $f$, we have

$$\frac{\partial f}{\partial\bar{z}} = 0$$

and also that

$$f(z) = \frac{1}{2\pi i}\int_{\gamma} \frac{f(\zeta)}{\zeta-z}\,d\zeta,$$

where $\gamma$ is a circular contour around $z$ (with a winding number of one). Given the duality between partial differential operators and integral operators, this has made me a bit curious as to what the Green's function for $\dfrac{\partial}{\partial \bar{z}}$ is (when one restricts to holomorphic functions). Intuitively, I would hazard to guess that the Green's function is related to $\dfrac{1}{z}$ based on the above observations but I haven't figured out a way to show this. Solving

$$\frac{\partial}{\partial\bar{z}}G = \delta(z)$$

for $G$ seems to be a bit of a tall task. Distributions seem to be a little bit complicated on $\mathbb{C}$. Ultimately, I want to use this understanding to tackle some aspects of Green's functions in several complex variables. Any help would be greatly appreciated!

Cameron L. Williams
  • 30,485
  • You shouldn't have to do distributions on $\mathbb{C}$. Essentially all the arguments relating complex analysis with vector analysis use the interpretation of $\mathbb{C}$ as $\mathbb{R}^2$, so you are dealing with (in fact harmonic) functions from $\mathbb{R}^2$ to $\mathbb{R}^2$. – mlk Jun 13 '14 at 19:00
  • @CameronWilliams: obviously my comment is not considered to be offensive. I hope you gladly take my advice. apart from that "when one restricts to holomorphic functions" doesn't make sense, and what you are asking for is the fundamental solution of the C-R operator, which is well known and can be found in any PDE book (E.g. treves)...... –  Jun 21 '14 at 21:51
  • @Karl Your comment came off very rude and unkind (even if that was not your intention) but I see your point now. My phrasing was not the best since I was a bit frustrated and frantic but I am by no means a novice at complex analysis. I did extensive searches on this topic before asking here (probably spent a few hours). I was not using the right terms by which to search it seems. I apologize for getting upset but your first comment was horribly presumptuous. I hope you recognize that. – Cameron L. Williams Jun 21 '14 at 22:13
  • I can assure you that it wasn't my intention. you can read about this in e.g. Taylor, PDE I, chap. 3.4; or Treves, Basic linear PDE, chapter 1.5. (fundamental solutions of the cauchy-riemann operator). there some cool things are pointed out, such every distributional solution of $\bar{\partial}u=0$ is in fact an analytic function, derives cauchy's formula (from a distributional point of view), and also has many nice exercises (in case you want to prove some related results). Another is Krantz, Geometric function theory, chap. 7.2. I think it is exactly what you are looking for! –  Jun 21 '14 at 23:23
  • Thanks @Karl. Sorry for the misunderstanding. – Cameron L. Williams Jun 21 '14 at 23:27
  • Krantz's book discusses some things which you won't see in the usual more elementary complex analysis books. Also of course Lars Hörmanders introduction to complex analysis in several variables. by the way I would stop using "green's function" when you actually mean "fundamental solution", which is the modern term (the older has some weird connotations). apart from, ignoring the motivation in the first paragraph, finding the fundamental solution has little to do with complex analysis (if you look closely at daniels answer: he uses no complex analysis at all, just "complex notation"). –  Jun 21 '14 at 23:33
  • and for complex analysis in several variables it might well be that we have a different understanding what is meant by that. when I think of complex analysis in several variables then this is a highly nontrivial, sophisticated and very advanced subject, which uses lots of algebraic geometry, in particular sheaf theory. sorry for the spam, but I hopefully said something helpful. in hörmander's book chapter 1 contains the best and shortest presentation (about 20 pages) of complex analysis in one variable, just astounding –  Jun 21 '14 at 23:35
  • You definitely did. In our seminar we'll be avoiding algebraic geometry entirely since we're not trying to develop a full treatise on the subject (due to time constraints). We're working out of Range's text and he takes a mostly analytic approach to the book but that might be at the expense of making it a bit inelegant. Thanks for the clarification of fundamental solution versus Green's function. I've seen both plenty but it does seem that the term Green's function has been relegated to the likes of physics and mathematicians have been more keen on fundamental solution. (It even sounds better) – Cameron L. Williams Jun 21 '14 at 23:38

1 Answers1

7

Actually, it's not so hard to find a solution of $\frac{\partial}{\partial\overline{z}}G =\delta$. You have the right thing there already, modulo normalisation. We have, in the sense of distributions, for any $\varphi\in \mathscr{D}(\mathbb{C})$,

$$\begin{align} \left(\frac{\partial}{\partial\overline{z}}\frac{1}{z-w}\right)[\varphi] &= - \int_{\mathbb{C}} \frac{\partial\varphi}{\partial\overline{z}}(z)\frac{1}{z-w}\,d\lambda\\ &= -\lim_{\varepsilon\to 0} \int_{\varepsilon < \lvert z-w\rvert < R} \frac{\partial\varphi/\partial\overline{z}(z)}{z-w}\,d\lambda \end{align}$$

where $\lambda$ is the Lebesgue measure on $\mathbb{C}$ and $R$ is sufficiently large, so that the support of $\varphi$ is contained in $D_R(w)$. Now write the integral with differential forms, that is, replace $d\lambda$ with $dx\wedge dy$, and write the latter as $\frac{1}{2i}d\overline{z}\wedge dz$. We get

$$\begin{align} \left(\frac{\partial}{\partial\overline{z}}\frac{1}{z-w}\right)[\varphi] &= -\lim_{\varepsilon\to 0} \frac{1}{2i}\int_{\varepsilon < \lvert z-w\rvert < R} \frac{\partial\varphi/\partial\overline{z}(z)}{z-w}\,d\overline{z}\wedge dz\\ &= -\frac{1}{2i}\lim_{\varepsilon\to 0} \int_{\varepsilon < \lvert z-w\rvert < R} d\left(\frac{\varphi(z)}{z-w}\, dz\right) \tag{Stokes}\\ &= \frac{1}{2i}\lim_{\varepsilon\to 0} \int_{\lvert z-w\rvert = \varepsilon} \frac{\varphi(z)}{z-w}\,dz\\ &= \frac{1}{2i}\lim_{\varepsilon\to 0} \left(\int_{\lvert z-w\rvert = \varepsilon} \frac{\varphi(w)}{z-w}\,dz + \int_{\lvert z-w\rvert =\varepsilon} \frac{\varphi(z)-\varphi(w)}{z-w}\,dz \right)\\ &= \pi \varphi(w), \end{align}$$

since by the differentiability of $\varphi$ in $w$, the integrand of the second integral in the penultimate line remains bounded as $\varepsilon\to 0$, and so the second integral is $\leqslant C\cdot 2\pi\varepsilon$ in absolute value by the ML-inequality. That means

$$\frac{1}{\pi}\frac{\partial}{\partial\overline{z}} \frac{1}{z-w} = \delta_w.$$

Daniel Fischer
  • 211,575
  • Very nice Daniel! The problem I was having was exactly Stokes' theorem. I recognized that I would have $dxdy$ but did not think to write it in the language of differential forms. This is exactly what I was looking for. Thanks a bunch. – Cameron L. Williams Jun 13 '14 at 20:20
  • One concern that I have is that we say that $\varphi$ has compact support but then make use of the fact that it is holomorphic at the very end. Are these two ideas not at odds with each other? If $\varphi$ is holomorphic and has compact support, it must be identically zero. – Cameron L. Williams Jun 13 '14 at 20:26
  • 3
    No, we do not assume that $\varphi$ is holomorphic. (There is a certain scarcity of holomorphic test functions ;-) But since it is smooth, we can write $$\varphi(z) = \varphi(w) + \frac{\partial\varphi}{\partial z}(w)(z-w) + \frac{\partial\varphi}{\partial\overline{z}}(w)\overline{z-w} + O(\lvert z-w\rvert^2).$$ The constant term gives the $\varphi(w)$ (times $\pi$) independent of $\varepsilon$. The other terms are all bounded when divided by $z-w$, and so the integral of these terms is $\leqslant C\cdot 2\pi \varepsilon$ in absolute value. In the limit, only the constant term survives. – Daniel Fischer Jun 13 '14 at 20:34
  • Oh of course. That's very clever. Thanks again. – Cameron L. Williams Jun 13 '14 at 20:38