A simple ordinary differential equation - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T15:29:21Z http://mathoverflow.net/feeds/question/46104 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46104/a-simple-ordinary-differential-equation A simple ordinary differential equation Marc Palm 2010-11-15T08:21:20Z 2010-11-23T11:03:31Z <p>Consider an entire function $f : \mathbb{C} \rightarrow \mathbb{C}$! We search the function $$g: (a,b) \rightarrow \mathbb{C},$$ which solves the following equation locally: $g'(t)=f(g(t))$ and $g(0)=f(x_0)$.</p> <p>I can compute the inverse $G$ of $g$, if $f(x_0) \neq 0$, i.e. $$G(y) = \int\limits_{f(x_0)}^y \frac{d s}{f(s)}.$$</p> <p>I also known how to compute the Taylor expansion recursively, whose radius of convergence is positive (see below). Also we can give suitable approximations of the solution in terms of Picard iterations. I am not interested in such a solution!</p> <p>Is there an alternative to this integral expression?</p> http://mathoverflow.net/questions/46104/a-simple-ordinary-differential-equation/46109#46109 Answer by Florian for A simple ordinary differential equation Florian 2010-11-15T10:41:39Z 2010-11-15T10:41:39Z <p>The power series for $g$ has a positive radius of convergence; this is a consequence of the Cauchy-Kovalevskaya theorem (which is a statement about PDEs, but an ODE is just a PDE with one variable).</p> http://mathoverflow.net/questions/46104/a-simple-ordinary-differential-equation/46114#46114 Answer by Michael Renardy for A simple ordinary differential equation Michael Renardy 2010-11-15T13:00:32Z 2010-11-15T13:00:32Z <p>You don't need the Cauchy-Kovalevskaya theorem. Just the analytic inverse function theorem.</p> http://mathoverflow.net/questions/46104/a-simple-ordinary-differential-equation/46119#46119 Answer by Dick Palais for A simple ordinary differential equation Dick Palais 2010-11-15T15:41:41Z 2010-11-15T15:41:41Z <p>It is hard to guess what you are looking for. Take the apparently simpler case where $f$ is a polynomial, say of degree $d$. If $d = 1$ you have an explicit solution in terms of the exponential function (because your $G$ is logarithmic). If $d = 2$ the solution can be written in terms of trigonometric functions. If $d = 3$ you need elliptic functions to express the solution explicitly. As soon as $d$ is greater than $3$, I don't know of any standard naming for the functions you get or any interesting theory of these functions. </p>