Wikipedia:Reference desk/Archives/Mathematics/2009 December 24

= December 24 =

Green's second identity and Green's functions for the Laplacian
Hi all,

I'm trying to prove that the for Green's function for the Laplacian, G(r;r0) in any arbitrary 3D domain, symmetry holds between r and r0; i.e. G(r;r0)=G(r0,r). My friend suggested I should try using the Second Green's identity (I sometimes wonder if my life would be a more interesting place if Green were never born!), but I can't seem to get anything out, perhaps I'm being slow this time of night.

Does anyone else have any luck using Green's 2nd identity? Thanks very much, Delaypoems101 (talk) 02:30, 24 December 2009 (UTC)

sequence space
I'm trying to prove that the sequence space of all complex sequences is a metric space with the metric $$d(x,y)=\sum_{j=1}^\infty \frac{1}{2^j}\frac{|x_j-y_j|}{1+|x_j-y_j|}$$. My questions are how can I show that this series is always convergent and why does d(x,y)=0 imply x=y. Thanks-Shahab (talk) 06:36, 24 December 2009 (UTC)
 * Doesn't really matter, d fails to satisfy the triangle inequality.--RDBury (talk) 07:18, 24 December 2009 (UTC)
 * No it satisfies the triangle inequality. I can reproduce the proof for that given in my book.-Shahab (talk) 07:26, 24 December 2009 (UTC)
 * My apologies, I got mixed up when I was checking it.--RDBury (talk) 12:13, 24 December 2009 (UTC)
 * (ec) Second question is easy: all fractions $$f_j = \frac{|x_j-y_j|}{1+|x_j-y_j|}$$ are non-negative, so d is a sum of non-negative terms, and thus can only be zero if all terms are zero, which implies all numerators are zero, so x=y. Now the first question gets easy: as all terms are non-negative AND less than 1 (because for $$f_j \neq 0$$ we have $$f_j = \frac 1 {\frac 1 {|x_j-y_j|} + 1}$$ which is a reciprocal of something greater than 1), the sum $$d(x,y)=\sum_{j=1}^\infty \frac {f_j} {2^j} \leq \sum \frac 1{2^j} = 1$$ --CiaPan (talk) 07:25, 24 December 2009 (UTC)
 * Thank you, it's clear. Instead of saying d is a sum of non-negative terms, and thus can only be zero if all terms are zero isn't it more appropriate to say that d(x,y)=0 is a limit of a monotonic increasing sequence of non-negative terms which is only possible if all terms are zero. I tend to think of series as sequences only.-Shahab (talk) 07:39, 24 December 2009 (UTC)
 * Both are valid arguments. CiaPan's argument is rooted on the assertion that if $$\sum_{i=1}^\infty a_i$$ is a convergent sum, with each term in the sum non-negative, $$a_j < \sum_{i=1}^\infty a_i$$ for all j. The argument you have suggested is rooted on the assertion that if $$S_j = \sum_{i=1}^j a_i$$ is the jth partial sum of a convergent series, $$0\leq S_k\leq S_{k+1}$$ for all k. Essentially, both arguments are correct (and similar in nature). However, you are correct to note that in a situation where basic intuition does not apply, it is often more appropriate to employ a formal argument. -- PS T  09:13, 24 December 2009 (UTC)
 * By the way, which book are you studying? [[Image:Smile.png]] -- PS T  09:14, 24 December 2009 (UTC)
 * Kreysig's. I found an online copy.-Shahab (talk) 09:40, 24 December 2009 (UTC)
 * Note that you can use other functions $$\scriptstyle\phi:[0,\infty)\to[0,\infty)$$ in place of t/(1+t) in the definition of d: precisely, any bounded continuous subadditive increasing function φ such that φ(0)=0 and φ(t)>0 if t>0 produces a distance $$\scriptstyle d(x,y)=\sum_{j=1}^\infty 2^{-j}\phi(|x_j-y_j|)$$ on the space of sequences. These are topologically equivalent, and induce the product topology. For instance, $$\phi(t):=\min(t,1)$$ is often used. --pm a (talk)  12:24, 24 December 2009 (UTC)
 * Thanks everyone and merry christmas-Shahab (talk) 04:34, 25 December 2009 (UTC)

Rolling sphere
An unconstrained sphere resting on the top of a fixed one is in unstable equilibrium. Suppose a minute disturbance (e.g. it's given an initial velocity of one millionth of the fixed sphere's radius per second) starts it rolling under gravity. Assuming no slipping, are there any circumstances which will make it leave the surface of the fixed one before the 90° point has been reached?→→86.155.184.27 (talk) 17:21, 24 December 2009 (UTC)


 * I suspect the answer might depend on whether "rolling" means it's not "slipping".
 * But then on another couple of seconds' thought (I haven't thought this one through) I would think it would have to leave the surface before reaching the 90&deg; point, because its motion has a horizontal component and there's inertia. Reaching the 90&deg; point would mean going straight down with no horizontal component to its motion. Michael Hardy (talk) 19:53, 24 December 2009 (UTC)
 * OK, now I see there's an explicit statement that it's not slipping. I don't know whether that actually matters. Michael Hardy (talk) 06:43, 25 December 2009 (UTC)
 * This problem is suited for Lagrangian mechanics. Bo Jacoby (talk) 23:09, 24 December 2009 (UTC).
 * I think in the general case of an object rolling off the fixed sphere (assuming the size of the rolling sphere is much smaller than that of the fixed one), it will depart at $$\cos \, \theta = \frac{2}{3 + \frac{I r^2}{m}}$$ (where $$I$$ is the moment of inertia of the sphere, $$m$$ its mass, and $$r$$ its radius). Note that this reduces to a constant in the special case of a particle sliding down the sphere ($$I = 0$$), giving $$\theta \approx 48.2^{\circ}$$.  Michael is completely right---since the object acquires some horizontal velocity, it has to leave before the 90&deg; point.  You can analyse this by working out the velocity as a function of angle (by conserving energy), then working out the angle at which the gravitational pull on the object (directed towards the sphere) is no longer enough to keep it in circular motion around it. — Zazou 00:40, 25 December 2009 (UTC)
 * I think in the general case of an object rolling off the fixed sphere (assuming the size of the rolling sphere is much smaller than that of the fixed one), it will depart at $$\cos \, \theta = \frac{2}{3 + \frac{I r^2}{m}}$$ (where $$I$$ is the moment of inertia of the sphere, $$m$$ its mass, and $$r$$ its radius). Note that this reduces to a constant in the special case of a particle sliding down the sphere ($$I = 0$$), giving $$\theta \approx 48.2^{\circ}$$.  Michael is completely right---since the object acquires some horizontal velocity, it has to leave before the 90&deg; point.  You can analyse this by working out the velocity as a function of angle (by conserving energy), then working out the angle at which the gravitational pull on the object (directed towards the sphere) is no longer enough to keep it in circular motion around it. — Zazou 00:40, 25 December 2009 (UTC)