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Leading Order Analysis

A procedure for determining the behavior of an $n$th order Ordinary Differential Equation at a Removable Singularity without actually solving the equation. Consider

{d^n y\over dz^n} = F\left({{d^{n-1}y\over dz^{n-1}}, \ldots, {dy\over dx}, y, z}\right),
\end{displaymath} (1)

where $F$ is Analytic in $z$ and rational in its other arguments. Proceed by making the substitution
y(z)\equiv a(z-z_0)^\alpha
\end{displaymath} (2)

with $\alpha<1$. For example, in the equation
{d^2y\over dz^2} = 6y^2+Ay,
\end{displaymath} (3)

making the substitution gives
\end{displaymath} (4)

The most singular terms (those with the most Negative exponents) are called the ``dominant balance terms,'' and must balance exponents and Coefficients at the Singularity. Here, the first two terms are dominant, so
\alpha-2=2\alpha \Rightarrow \alpha=-2
\end{displaymath} (5)

6a=6a^2 \Rightarrow a=1,
\end{displaymath} (6)

and the solution behaves as $y(z)=(z-z_0)^{-2}$. The behavior in the Neighborhood of the Singularity is given by expansion in a Laurent Series, in this case,
y(z)=\sum_{j=0}^\infty a_j(z-z_0)^{j-2}.
\end{displaymath} (7)

Plugging this series in yields
$\sum_{j=0}^\infty a_j(j-2)(j-3)(z-z_0)^{j-4}$
$ = 6\sum_{j=0}^\infty\sum_{k=0}^\infty a_ja_k(z-z_0)^{j+k-4} + A\sum_{j=0}^\infty a_j(z-z_0)^{j-2}.\quad$ (8)
This gives Recurrence Relations, in this case with $a_6$ arbitrary, so the $(z-z_0)^6$ term is called the resonance or Kovalevskaya Exponent. At the resonances, the Coefficient will always be arbitrary. If no resonance term is present, the Pole present is not ordinary, and the solution must be investigated using a Psi Function.

See also Psi Function


Tabor, M. Chaos and Integrability in Nonlinear Dynamics: An Introduction. New York: Wiley, p. 330, 1989.

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© 1996-9 Eric W. Weisstein