2
$\begingroup$

OK, then I read Frobenius method in mathworld (I learned when I took ODE 2): http://mathworld.wolfram.com/FrobeniusMethod.html

My question is: Are there any ODEs where the solution is given by full Laurent series?, i.e its negative indexed coeffecients are not zero starting from some negative integer.

Thanks in advance.

Improve this question
$\endgroup$

3 Answers 3

Reset to default
1
$\begingroup$

Every analytic function in a ring has a Laurent expansion. Thus if your differential equation has a solution analytic in a ring it has a Laurent expansion. See the examples given in other answers.

The difference between a full Laurent series (I understand that "full" means infinitely many coefficients in both directions) and a Frobenius series (which is infinite in only one direction) is that you cannot manipulate with a full Laurent series formally. In particular, you cannot multiply two such series: the coefficient of the product is a series, rather than a finite sum.

For this reason, the use of full Laurent series is limited. The advantage of the one-suded series is that you can substitute a formal one-sided series to the differential equation and effectively determine its coefficients (like in Frobenius method). You cannot do this with a full Laurent series.

Thus a full Laurent series solution (when exists) is usually impossible to find explicitly.

Improve this answer
$\endgroup$
2
$\begingroup$

Take any two linearly independent functions analytic in an annulus and you can find a second-order linear DE that has these as a fundamental set of solutions.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Ok, what I meant is that if I have some nonlinear ode, like: $$y''+(y')^2+y^3=0$$ How do I know if I need to guess a Laurent series or any other solution? I mean before imposing initial conditions. $\endgroup$
    – Alan
    Commented Feb 26, 2013 at 3:19
  • 1
    $\begingroup$ You might be interested in the Painlevé property (= no mobile essential singularity) mathworld.wolfram.com/PainleveProperty.html $\endgroup$
    – Loïc Teyssier
    Commented Feb 26, 2013 at 9:57
0
$\begingroup$

There are two related notions for linear ODEs with meromorphic coefficients that have a pole at say $z=0$. The notion of regular singular point (which means that any solution grows no faster at $0$ than $|z|^{-d}$ for some $d$) and the notion of the Fuchs condition (which means that the orders of the poles at $0$ of the meromorphic coefficients satisfy some inequalities). The growth condition implies that the Laurent expansion of any solution necessarily starts at some finite negative order (determined by $d$). According to the second link, these two conditions are equivalent. Unfortunately, I don't have a handy reference for where this is discussed in more detail.

For a non-linear equation, it's probably more difficult to figure out when a singular point is regular in the above sense.

Here's an example of an ODE with an irregular singular point at $z=0$, where the solution has a non-terminating Laurent series: $z^2 \frac{du}{dz}=u$, $u(z) = C \exp(-1/z)$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .