Wikipedia:Reference desk/Archives/Mathematics/2009 November 2

= November 2 =

Cauchy integral theorem
Dear collegues, article Cauchy's integral theorem says that "The Cauchy integral theorem is valid in a slightly stronger form than given above. Suppose U is an open simply connected subset of C whose boundary is the image of the rectifiable path &gamma;. If f is a function which is holomorphic on U and continuous on the closure of U, then &int;&gamma;f = 0".

I'm disappointed by that there is no any reference about it. Maybe the trick can be found in the proof of Goursat-lemma. If T is a triangle, int(T) &ne; 0 and f is regular at each point of int(T), then the sequence of the mid triangles (Tn) converges to a point of the int(T). Therefore, it is sufficient to require the differentiability of f at the points of int(T) (and of course the integrability of f along &part;T). Or, is it more difficult? Thanks, Mozó (talk) 11:56, 2 November 2009 (UTC).


 * To be precise it's not quite true if the hypotheses are stated that way. The obvious objection is that if U is an open simply connected subset of C, its closure $$\scriptstyle \overline {U}$$ need not  be simply connected. And the boundary of U may be the image of a smooth curve γ  that is not contractible in $$\scriptstyle \overline {U}$$. For example take $$U:=B(1,2)\setminus\overline {B(0,1)}$$ where $$B(x,r)$$ denotes the open disk of center x and radius r. There is a parametrization γ of the boundary of U which has index 2 to wrto 0, so f(z)=1/z has  &int;&gamma;f ≠ 0. Of course with the right hypotheses the generalization is true  and quite easy to prove. --pma (talk) 14:28, 2 November 2009 (UTC)
 * Wow, fascinating counterexample! Thanks! I think it's not about precisity, it's about true or false :) Then we should correct the article.
 * "quite easy to prove" typical mathematician. :) The generalization is not contained in basic complex analysis books. So, would you please offer us a reference? Or how can I extend my proof to non-triangular shapes (if it is correct)? Mozó (talk) 16:43, 2 November 2009 (UTC)
 * You're right about "easy to prove", sorry! But in fact what I meant is the trivial remark (and typical mathematician too)  that, under suitable hypotheses, every statement becomes true and easy to prove ;-) For instance, we may just change the assumption on γ in the original statement. Assume: (a): γ is a loop in $$\scriptstyle \overline {U}$$ which is limit in the  C1 sense (that is, uniformly with the derivative) of a sequence of  C1 loops γj in U . Then  0 = &int;&gamma; j f converges to &int;&gamma;f, and we are done. In fact (having defined conveniently the path integral for BV curves)  it would be sufficient: (b): γ is uniform limit of BV loops in U, with bounded length. However I understand that the spirit of that statement is giving a condition on U only. In this case, (a) is certainly true for any smooth loop in $$\scriptstyle \overline {U}$$ if $$\scriptstyle \overline {U}$$ is a  C 1 sub-manifold with boundary of R2. I am not sure at the moment  whether it is sufficient to assume a plain: "both U and $$\scriptstyle \overline {U}$$ are simply connected" though, or maybe just "$$\scriptstyle \overline {U}$$ is simply connected". Maybe there is a simple answer that I don't see now. I'll look for references, but I'm not optimistic. --pma (talk) 19:21, 2 November 2009 (UTC)
 * "spirit of that statement is giving a condition on U only" Well, complex analysts are esoteric guys, cause while we all know what is "sub-manifold with boundary", they don't use such clear concepts. They're always talking about "moving" contours, paths along the boundary, ... without any discussion on convergency or existence. By the way, Cauchy's integral formula is a highly surprising and inexplicable result, so they must be esoteric :) Mozó (talk) 21:02, 2 November 2009 (UTC)
 * Now, I finally see what you mean on "with the right hypotheses the generalization is true and quite easy to prove". The generalization is an immediate consequence of the Cauchy Theorem if we consider some well-known differential topological argument using the right hypotheses. Mozó (talk) 08:24, 3 November 2009 (UTC)

The answer is: Kodaira, Complex Analysis, Camb. Univ. Press 2007, Theorem 2.3. Since, he proved his celebrated vanishing theorem, I trust in him and indeed, he has nontrivial assumption about the boundary. Mozó (talk) 09:16, 3 November 2009 (UTC)


 * What I had in mind as easy case was essentially what I mentioned, an open simply connected domain U with a smooth boundary (or just Lipschitz). I do not see clear the full generality of "$$\scriptstyle \overline {U}$$ is simply connected", even if I'm possibly lacking something. Do you see a more general situation? What is the precise statement of Kodaira's result in the context we are talking of, that is U=a domain of C and its boundary? --pma (talki) 09:28, 3 November 2009 (UTC)
 * Kodaira's way: (1) If the boundary of a compact, connected set D consists of finite numbers of "disjoint", smooth Jordan curves then D has a cellular decomposition (the well-known "squares" in pictures of books) (2) He proves the generalization for rectangles by a construction of a uniformly convergent curve sequence, just like you. (3) He states the generalization for a &part;D in (1).
 * He does not require simply connectedness, since in his construction the integrals along the "inner" and "outer" loops balance each others. Mozó (talk) 18:52, 3 November 2009 (UTC)


 * Thanks, very interesting. In the meanwhile, I had some further thoughts that I hope may be of some interest for you:
 * 1. Let U be a bounded domain in C bounded by a simple closed rectifiable loop γ (here by rectifiable loop I mean: of bounded total variation, and of course continuous,  $$\textstyle \gamma:\partial D\to\C$$ )
 * A classical addendum to the Riemann mapping theorem says that in this situation the biholomorphic Riemann diffeomorphism $$\scriptstyle h:D\to U$$ extends to a continuous
 * $$h:\overline{D}\to\overline{U}$$
 * such that $$h_{|\partial{D}}$$ is a weakly monotone reparametrization of $$\gamma$$; moreover, h has finite energy (it's a minimizer for the energy indeed. It's a special case of Rado's solution of the Plateau's problem, i.e. for plane curves!).
 * 2. Actually we may think $$h_{|\partial{D}}=\gamma$$ with no loss of generality, because the path integral is certainly invariant by reparametrization. The preceding fact (1) then allows to conclude that &int;&gamma;f = 0 for any f holomorphic on U and continuous up to $$\scriptstyle\overline{U}$$, just because the analogous fact is true for the unit disk D, and you can change variable with h (the fact that h has finite energy is relevant to me, because it allows to approximate uniformly $$\gamma(\theta)=h(\theta, 1)$$ with a sequence $$\gamma_n(\theta):=h(\theta,\rho_n)$$ with $$\rho_n\to 1$$ (here I'm using polar coordinates) in such a way that $$\gamma_n$$ have bounded length. So one can pass to the limit in the sequence of the path integrals (which are all zero).
 * 3. Using the above fact (2), I think one could make a proof for this: assume U is any open set of C; that f is holomorphic on U and continuous up to $$\scriptstyle\overline{U}$$; that  γ is any rectifiable loop in $$\scriptstyle\partial{U}.$$ Assume further ind(γ,z)=0 for any z in U (this was exactly the missing condition in the counterexample). Then &int;&gamma;f = 0. The idea should be decomposing γ into a countable family of simple loops  γk that meet the conditions in (2), and in such a way that we can write &int;&gamma;f as a series  of the &int;&gamma; k f.
 * I think I see quite clearly the decomposition lemma, but it would be some effort to put it down here in all details.  I think one needs no finiteness assumptions however. I guess this  should be more or less equivalent to Kodaira's argument, with the difference that all the Jordan thing and the approximation scheme is already included in the known result (1).  So, what do you like more, mine or Kodaira's? forget about the fact that mine is only sketched ;-)   --pma (talk) 11:24, 4 November 2009 (UTC)

True or False? (Game Theory)
"In game theory, a player is supposed to choose only one strategy from her strategy set."

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==============True or False???

"To define a game, we must specify every player's payoff at every possible strategy profile."

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==============True or False???

"Among the games below, choose all games that have no pure-strategy Nash equilibrium." 1. Matching Penny; 2. Prisoners' dilemma; 3. Battle of sexes; 4. (Pure) Coordination games.

Thanks a lot! —Preceding unsigned comment added by Aaa50211 (talk • contribs) 14:06, 2 November 2009 (UTC)
 * You might find our article Game Theory to be helpful. --Sean 14:12, 2 November 2009 (UTC)


 * Have you tried out that search box at the top left of the page? just put in a term at a time like 'battle of sexes' and have a look at the results. Dmcq (talk) 19:33, 2 November 2009 (UTC)

Euler-Lagrange equations and (possibly!) Lagrange multipliers
Hi all,

I'm trying to solve the following problem:

Write down the Euler-Lagrange equation for the functional

$$I[u]=\int_{-\infty}^{+\infty}\frac{1}{2}(u')^2+(1-\cos u)\text{dx}=\int_{-\infty}^{+\infty}f$$

and find all solutions which satisfy $$\lim_{x\to -\infty}u(x)=0$$ and $$\lim_{x\to +\infty}u(x)=2\pi$$.

The first bit is just

$$\frac{\partial{f}}{\partial{u}}-\frac{d}{dx}\frac{\partial{f}}{\partial{u'}}=0=\sin u-u''$$, right?

Now how on earth do I solve that?

I have a feeling I may also be meant to incorporate the constraint "find all solutions which satisfy..." using Lagrange multipliers - the conditions imply $$\int_{-\infty}^{+\infty}u'=2\pi$$ so am I meant to say $$I[u]+\lambda \int_{-\infty}^{+\infty}u'= 0$$ for some $$\lambda$$? If so, I tried using the fact that the funtion f has no explicit dependence on x, i.e. $$f=f(u,u')$$, and then following up with the Beltrami Identity which gave me

$$k-\cos u=u'^2$$, if I did the math correctly. I'm not sure which route to pursue (they probably both lead the same way), or whether I'm doing the right thing: attempting a series solution seems like a terrible idea and other than that I'm out of suggestions!

Thanks for the help :) Spalton232 (talk) 21:56, 2 November 2009 (UTC)


 * Write $$\textstyle u'' = \frac{d(u')}{dx} = \frac{d(u')}{du} \frac{du}{dx} = \frac{1}{2} \frac{d \left(u'^2\right)}{du}$$ and separate the variables $$\textstyle u'^2$$ and $$\textstyle u$$. After integration, you'll have a first-order equation to separate and integrate again. The general solution is not an elementary function, but the demanded one may be (haven't checked).undefined&mdash;undefinedPt(T) 04:18, 3 November 2009 (UTC)


 * Few more words. Yours is the nonlinear simple pendulum equation. You'll find it more commonly written $$v=-\sin(v)$$ (corresponding to u=v-π in your equation). The limit conditions at $$\scriptstyle\pm\infty$$ describe a homoclinic orbit to the unstable equilibrium (the pendulum is vertical with the mass in the highest position at $$\scriptstyle -\infty$$, then falls in infinite time and reaches again the unstable equilibrium for $$\scriptstyle t\to+\infty.$$) The differential equation is solved by means of an elliptic integral and studied qualitatively, starting as shown by Pt(T), by looking at the phase space (u,u'). (according to the value of the energy, you have periodic or monotone solutions &c). In this context, the role of the action functional I(u) is classically bounded to describe some properties of the solutions (that you have found independently, by means of ODE techniques); so you do not have to bother too much about the domain of I(u) &c. Things are different if you consider more complicated (still variational) equations: then, the corresponding action functional becomes a key tool to prove existence and to get information on the solutions, which are found as critical points of the functional (there's a huge amount of calculus of variations and global analysis to do that). Note however that in your problem the limit conditions at $$\pm\infty$$ may be encoded in the domain of the functional, and you do not have to use Lagrange multipliers. Take as a domain the affine space of all u of the form $$u=u_0+w$$ where $$u_0$$ is a fixed smooth function with $$u_0(t)=0$$ for t<0 and $$u_0(t)=2\pi$$ for t>1, and where $$w$$ varies among all functions with $$\scriptstyle\int_\R(w'^2+w^2)dt<\infty$$ (say of class C1, but if you want a Hilbert space, the right choice is the Sobolev space $$\scriptstyle W^{1,2}(\R).$$ You may also check that if $$\scriptstyle u :\R\to\R$$ is any weakly differentiable function, and $$\scriptstyle I(u)<\infty$$, then u'' is a (Hölder) continuous function with limits at $$\scriptstyle \pm\infty$$ that are integer multiples of 2π).  --pma (talk) 08:14, 3 November 2009 (UTC)