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June 10

Solution to a nonlinear ODE

The ODE f·f’’ - (f’)^2 = 1 has the simple solution f = cosh (x). Do deformations of the type n·f·f’’ - (f’)^2 = 1 with n an integer also have analytical solutions?XYZsquared (talk) 19:01, 10 June 2009 (UTC) ·[reply]

Yes; without entering general theorems, any solution is analytic in the open set where it is not 0. On the other hand, no solution is regular at any point where it vanishes, ~~even if~~ unless you allow complex valued solutions (in the equation, if you try equating the coefficients of the Taylor expansion of a solution

f

at a zero, you will immediately meet obstructions). Notice also that

\scriptstyle g:=f^{\frac {n-1}{n}}

puts your equation in normal form; computing you should get

\scriptstyle g\,^{''}={\frac {n-1}{n^{2}}}\,g^{-{\frac {n+1}{n-1}}}

.

--pma (talk) 19:57, 10 June 2009 (UTC)[reply]

thanks, the transformation to g is very useful for my purposes.XYZsquared (talk) 11:55, 11 June 2009 (UTC)[reply]

Topology question

Suppose we have two sets, $A_{1}$ and $A_{2}$ , both in the metric universe $\mathbb {R} ^{2}$ . Each of these sets consists of two discs touching each other at one single point. In $A_{1}$ , both discs are open. In $A_{2}$ , both are closed. Does the point where they are touching each other have any special property that any other point in their boundary (the distinction between a closed and an open disc) would not have? JIP | Talk 20:07, 10 June 2009 (UTC)[reply]

It has the property that deleting it from the boundary disconnects the boundary. Algebraist 20:10, 10 June 2009 (UTC)[reply]

Is it actually possible for two open discs to intersect at a single point as in A1?Rich (talk) 03:24, 11 June 2009 (UTC)[reply]

No; the intersection of two open sets is open, and an open subset of R^2 is either empty or has the cardinality of the continuum. In fairness to JIP, he didn't say that the point was in the intersection of the two discs. --Trovatore (talk) 03:53, 11 June 2009 (UTC)[reply]

thanks75.45.106.99 (talk) 05:53, 11 June 2009 (UTC)[reply]