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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 $g(0)=f(x_0)$.

I can compute the inverse $G$ of $g$, if $f(x_0) \neq 0$, i.e. $$ G(y) = \int\limits_{f(x_0)}^y \frac{d s}{f(s)}.$$

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!

Is there an alternative to this integral expression?

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4  
Though f may be entire, the solution won't necessarily be so; take $f(u)=1+u^2$ for instance. –  J. M. Nov 15 '10 at 8:54
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If $f$ is real valued on the reals and $x_0$ is real your equation reduces to a standard autonomous equation $g'=f(g)$ on $R$. Why do you expect anything better than the standard representation of the solution, which is given precisely by your last formula? –  Piero D'Ancona Nov 15 '10 at 10:13
    
@J.M: That is definitely true, I am not hoping for something entire. This example related to the tangens is a good illustration. @Piero D'Ancona: I hope that $f$ entire, does allow for a better description of the solutions. –  plusepsilon.de Nov 15 '10 at 10:25

3 Answers 3

up vote 2 down vote accepted

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.

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I don't think you need elliptic functions yet for $d=3$; you may have been thinking of the DE for Weierstrass: ${y^{\prime}}^2=4y^3-a y-b$ where the derivative is squared. On the other hand, $y^{\prime}=4y^3-a y-b$ requires the solution of a nasty-looking transcendental equation involving sums of logarithms. –  J. M. Nov 15 '10 at 16:20
    
Thanks, you're right, J.M., I did forget the square! :-( –  Dick Palais Nov 15 '10 at 16:56
    
I choose this as the correct answer since it comes closest to what I wanted. In the case of an elliptic equation, we get Jacobi's theory of elliptic function. This theory is well understood. If there a nicer expression I will probably find them here. The function $\sqrt{x}$ is "almost entire", hence I accept this argument as an indication! Thanks for this illustrating examples. –  plusepsilon.de Nov 17 '10 at 11:10

You don't need the Cauchy-Kovalevskaya theorem. Just the analytic inverse function theorem.

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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).

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@Florian and Michael Renardy: The problem with the radius of convergence is now solved. Since power series are hard to analyse, I am searching for another expression. My problem is not as well posed as I have thought. –  plusepsilon.de Nov 15 '10 at 14:37

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