Math gifts

- Art Gallery -

In mathematics, the method of dominant balance is used to determine the asymptotic behavior of solutions to an ordinary differential equation without fully solving the equation. The process is iterative, in that the result obtained by performing the method once can be used as input when the method is repeated, to obtain as many terms in the asymptotic expansion as desired.[1]

The process goes as follows:

Assume that the asymptotic behavior has the form

\( {\displaystyle y(x)\sim e^{S(x)}.} \)

Make an informed guess as to which terms in the ODE might be negligible in the limit of interest.
Drop these terms and solve the resulting simpler ODE.
Check that the solution is consistent with step 2. If this is the case, then one has the controlling factor of the asymptotic behavior; otherwise, one needs try dropping different terms in step 2, instead.
Repeat the process to higher orders, relying on the above result as the leading term in the solution.


For arbitrary constants c and a, consider

\( {\displaystyle xy''+(c-x)y'-ay=0.} \)

This differential equation cannot be solved exactly. However, it is useful to consider how the solutions behave for large x: it turns out that y behaves like \( {\displaystyle e^{x}\,} \) as x → ∞ .

More rigorously, we will have \( \log(y)\sim {x} \), not \( y\sim e^{x} \). Since we are interested in the behavior of y in the large x limit, we change variables to y = exp(S(x)), and re-express the ODE in terms of S(x),

x \( {\displaystyle xS''+xS'^{2}+(c-x)S'-a=0,\,} \)


\( S''+S'^2+\left(\frac{c}{x}-1\right)S'-\frac{a}{x}=0\, \)

where we have used the product rule and chain rule to evaluate the derivatives of y.

Now suppose first that a solution to this ODE satisfies

\( {\displaystyle S'^{2}\sim S',} \)

as x → ∞, so that

\( S'',~\frac{c}{x}S',~\frac{a}{x}=o(S'^2),~o(S')\, \)

as x → ∞. Obtain then the dominant asymptotic behaviour by setting

\( {\displaystyle S_{0}'^{2}=S_{0}'.} \)

If \( S_{0} \) satisfies the above asymptotic conditions, then the above assumption is consistent. The terms we dropped will have been negligible with respect to the ones we kept.

\( S_{0} \) is not a solution to the ODE for S, but it represents the dominant asymptotic behavior, which is what we are interested in. Check that this choice for \( S_{0} \) is consistent,

\( {\displaystyle {\begin{aligned}S_{0}'&=1\\S_{0}'^{2}&=1\\S_{0}''&=0=o(S_{0}')\\{\frac {c}{x}}S_{0}'&={\frac {c}{x}}=o(S_{0}')\\{\frac {a}{x}}&=o(S_{0}')\end{aligned}}} \)

Everything is indeed consistent.

Thus the dominant asymptotic behaviour of a solution to our ODE has been found,

\( {\displaystyle {\begin{aligned}S_{0}&\sim x\\\log(y)&\sim x.\end{aligned}}} \)

By convention, the full asymptotic series is written as

\( {\displaystyle y\sim Ax^{p}e^{\lambda x^{r}}\left(1+{\frac {u_{1}}{x}}+{\frac {u_{2}}{x^{2}}}+\cdots +{\frac {u_{k}}{x^{k}}}+o\left({\frac {1}{x^{k}}}\right)\right),} \)

so to get at least the first term of this series we have to take a further step to see if there is a power of x out the front.

Proceed by introducing a new subleading dependent variable,

\( S(x)\equiv S_0(x)+C(x)\,

and then seek asymptotic solutions for C(x). Substituting into the above ODE for S(x) we find

\( {\displaystyle C''+C'^{2}+C'+{\frac {c}{x}}C'+{\frac {c-a}{x}}=0.} \)

Repeating the same process as before, we keep C' and (c − a)/x to find that

\( {\displaystyle C_{0}=\log x^{a-c}.} \)

The leading asymptotic behaviour is then

\( {\displaystyle y\sim x^{a-c}e^{x}.} \)

See also

Asymptotic analysis


Bender, C.M.; Orszag, S.A. (1999). Advanced Mathematical Methods for Scientists and Engineers. Springer. pp. 549–568. ISBN 0-387-98931-5.

Undergraduate Texts in Mathematics

Graduate Texts in Mathematics

Graduate Studies in Mathematics

Mathematics Encyclopedia



Hellenica World - Scientific Library

Retrieved from ""
All text is available under the terms of the GNU Free Documentation License