Analytical solutions of a differential equation (from Archimedes' Spiral) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T16:45:35Z http://mathoverflow.net/feeds/question/54393 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54393/analytical-solutions-of-a-differential-equation-from-archimedes-spiral Analytical solutions of a differential equation (from Archimedes' Spiral) Mikhail Gaichenkov 2011-02-05T10:22:15Z 2011-11-03T20:44:07Z <p>There is a differential equation in polar coordinates: $r'^2+r^2=(kt)^2$, $r(t=0)=0$, k- Const.</p> <p>I've found that a) if $\phi \in (0,t)$, t is quite small, then $r(\phi) \approx k/2 *\phi^2$ b) if $\phi \in (c, + \infty)$, c is quite large positive number, then $r(\phi)=k\phi+o(\phi)$, as $\phi \to \infty$.</p> <p>I am looking for approximation between a &amp; b. How the Archimedes' Spiral transforms into other curve. Any ideas are highly welcomed.</p> http://mathoverflow.net/questions/54393/analytical-solutions-of-a-differential-equation-from-archimedes-spiral/54428#54428 Answer by Julián Aguirre for Analytical solutions of a differential equation (from Archimedes' Spiral) Julián Aguirre 2011-02-05T18:31:38Z 2011-02-06T01:09:24Z <p>By a change of time scale we can assume that $k=1$. Rewrite the equation as $$r'=\sqrt{t^2-r^2},\quad r(0)=0.$$ The solution must be increasing, hence positive, and bounded by $t$, which by the way is a supersolution. It is also clear that $0\le r(t)\le t^2/2$. From this, it follows that $$r'\ge t \sqrt{1-t^2/4}\ ,\quad 0\le t\le 2.$$ Integrating we obtain $4/3\le r(2)&lt;2$. The function $t-4/(3t)$ is a subsolution on $[2,\infty)$ and its value at $t=2$ is $4/3$. It follows that $$t-\frac{4}{3t}\le r(t)\le t\quad\forall t\ge2.$$</p> http://mathoverflow.net/questions/54393/analytical-solutions-of-a-differential-equation-from-archimedes-spiral/54439#54439 Answer by Dick Palais for Analytical solutions of a differential equation (from Archimedes' Spiral) Dick Palais 2011-02-05T18:49:37Z 2011-02-05T18:49:37Z <p>If you go to the Archimedean Spiral page of the Virtual Math Museum:</p> <p><a href="http://virtualmathmuseum.org/Curves/archimedean_spiral/archimedean_spiral.html" rel="nofollow">http://virtualmathmuseum.org/Curves/archimedean_spiral/archimedean_spiral.html</a></p> <p>you will find a lot of material about Archimede's Spiral and the more general class of Archimedean spirals to which it belongs, and these may be of use to you. In particular, try experimenting with the 3D-XplorMath-j applet that is there, and download the file Archimedean_Spiral.pdf.</p> http://mathoverflow.net/questions/54393/analytical-solutions-of-a-differential-equation-from-archimedes-spiral/79737#79737 Answer by Aleksey Pichugin for Analytical solutions of a differential equation (from Archimedes' Spiral) Aleksey Pichugin 2011-11-01T17:56:46Z 2011-11-03T20:44:07Z <p><strong>[Updated 3/11/2011]</strong></p> <p>Your approximate solutions are the leading order terms of the asymptotic expansions of $r(t)$ for $t\ll1$ and $t\gg1$. More terms can be computed using perturbation techniques: $$r(t)/k = \frac{1}{2}t^2-\frac{1}{32}t^4+\frac{1}{768}t^6+O(t^8),\quad t\ll1,\quad (1)$$ and $$r(t)/k = t-\frac{1}{2}t^{-1}-\frac{5}{8}t^{-3}+O(t^{-5}),\quad t\gg1.\quad (2)$$ The expansion for small $t$ reproduces the first few terms of the Maclaurin series for $r(t)$. The full series can be obtained by substituting the formal power series expansion into your equation and matching the terms at equal powers of $t$. Coefficients of the resulting power series $$r(t)=\sum_{n=1}^{\infty} R_{2n}t^{2n}\quad (3)$$ can be computed using the recurrent formulae $$R_2=k/2,\quad R_4=-k/32,$$ $$R_{2n}=-\frac{R_{2n-2}}{8n} -\frac{1}{4kn}\sum_{i=1}^{n-2} \left( R_{2i}+4(i+1)(n-i)R_{2i+2} \right)R_{2n-2i} ,\quad n=3,4,\dots$$</p> <p>You said that you were interested in an approximate solution valid for $t=O(1)$; neither expansion (1) nor expansion (2) is valid there. It would not be possible to match (1) and (2) without having to derive an intermediate asymptotic, because their present regions of validity do not overlap. This means that series (3) is likely to be your best bet at the direct analytic computation of $r(t)$. I was unable to compute the radius of its convergence analytically. At the same time, coefficients $R_{2n}$ appear to decay in a rapid fashion, so the radius is not likely to be small. If unsure, you can always estimate the value of the radius numerically using the Domb-Sykes plot.</p> <p>In situations like yours people sometimes construct two-point Pade approximants $$r(t)/k \approx \sum_{k=2}^Na_kt^k/\left(1+\sum_{k=1}^{N-1}b_kt^k\right),$$ i.e. ratios of two polynomials that, when expanded into a power series at $t=0$ or at the $t=\infty$, match the first few terms of the appropriate power series expansions, see G.A. Baker &amp; P. Graves-Morris "Pade Approximants. Part II: Extensions and Applications", Addison-Wesley (1981) for more details. For example, for $N=2$, $$r(t)/k \approx t^2/(2+t)$$ matches the leading order behaviour of both expansions given above. It is actually a very poor approximation, because it has an $O(1)$ absolute error for large $t$. Higher-order approximants get progressively better (and uglier); e.g. for $N=5$ one obtains $$r(t)/k \approx \frac{2t^2(772+432t+184t^2+65t^3)}{(3088+1728t+929t^2+368t^3+130t^4)}.$$ This expression recovers the first couple of terms in expansions (1) and (2) and its relative error in the intermediate region is below $5\%$.</p> <p>It is worth noting that two-point Pade approximants are not always well-behaved in the intermediate region. In this particular case they seem to be converging to the exact solution reasonably well.</p>