# Method of dominant balance

In mathematics, the method of dominant balance approximates the solution to an equation by solving a simplified form of the equation containing 2 or more of the equation's terms that most influence (dominate) the solution and excluding terms contributing only small modifications to this approximate solution. Following an initial solution, iteration of the procedure may generate additional terms of an asymptotic expansion providing a more accurate solution.
An early example of the dominant balance method is the Newton polygon method. Newton developed this method to find an explicit approximation for an algebraic function. Newton expressed the function as proportional to the independent variable raised to a power, retained only the lowest-degree polynomial terms (dominant terms), and solved this simplified reduced equation to obtain an approximate solution. Dominant balance has a broad range of applications, solving differential equations arising in fluid mechanics, plasma physics, turbulence, combustion, nonlinear optics, geophysical fluid dynamics, and neuroscience.

## Asymptotic relations

The functions f(z){\textstyle f(z)} and $g(z)$ of parameter or independent variable z{\textstyle z} and the quotient f(z)/g(z){\textstyle f(z)/g(z)} have limits as z{\textstyle z} approaches the limit L{\textstyle L}.
The function f(z){\textstyle f(z)} is much less than g(z){\textstyle g(z)} as z{\textstyle z} approaches L{\textstyle L}, written as f(z)≪g(z) (z→L){\textstyle f(z)\ll g(z)\ (z\to L)}, if the limit of the quotient f(z)/g(z){\textstyle f(z)/g(z)} is zero as z{\textstyle z} approaches L{\textstyle L}.
The relation f(z){\textstyle f(z)} is lower order than g(z){\textstyle g(z)} as z{\textstyle z} approaches L{\textstyle L}, written using little-o notation f(z)=o(g(z)) (z→L){\textstyle f(z)=o(g(z))\ (z\to L)}, is identical to the f(z){\textstyle f(z)} is much less than g(z){\textstyle g(z)} as z{\textstyle z} approaches L{\textstyle L} relation.
The function f(z){\textstyle f(z)} is equivalent to g(z){\textstyle g(z)} as z{\textstyle z} approaches L{\textstyle L}, written as f(z)∼g(z) (z→L){\textstyle f(z)\sim g(z)\ (z\to L)}, if the limit of the quotient f(z)/g(z){\textstyle f(z)/g(z)} is 1 as z{\textstyle z} approaches L{\textstyle L}.
This result indicates that the zero function, f(z)=0{\textstyle f(z)=0} for all values of z{\textstyle z}, can never be equivalent to any other function.
Asymptotically equivalent functions remain asymptotically equivalent under integration if requirements related to convergence are met. There are more specific requirements for asymptotically equivalent functions to remain asymptotically equivalent under differentiation.

## Equation properties

An equation's approximate solution is s(z){\textstyle s(z)} as z{\textstyle z} approaches limit L{\textstyle L}. The equation's terms that may be constants or contain this solution are T$_0$(s), T$_1$(s), …, T$_n$(s){\textstyle T_{0}(s), T_{1}(s), \ldots , T_{n}(s)}. If the approximate solution is fully correct, the equation's terms sum to zero in this equation: $T_{0}(s)+T_{1}(s)+\ldots +T_{n}(s)=0.$ For distinct integer indices i, j{\textstyle i, j}, this equation is a sum of 2 terms and a remainder R$_{ij}$(s){\textstyle R_{ij}(s)} expressed as ${\begin{aligned}&T_{i}(s)+T_{j}(s)+R_{ij}(s)=0\\&R_{ij}(s)=\sum _{{k=0} \atop {k\neq i,k\neq j}}^{n}T_{k}(s).\end{aligned}}$ Balance equation terms T$_i$(s){\textstyle T_{i}(s)} and T$_j$(s){\textstyle T_{j}(s)} means make these terms equal and asymptotically equivalent by finding the function s(z){\textstyle s(z)} that solves the reduced equation T$_i$(s)+T$_j$(s)=0{\textstyle T_{i}(s)+T_{j}(s)=0} with T$_i$(s)≠0{\textstyle T_{i}(s)\neq 0} and T$_j$(s)≠0{\textstyle T_{j}(s)\neq 0}.
This solution s(z){\textstyle s(z)} is consistent if terms T$_i$(s){\textstyle T_{i}(s)} and T$_j$(s){\textstyle T_{j}(s)} are dominant; dominant means the remaining equation terms R$_{ij}$(s){\textstyle R_{ij}(s)} are much less than terms T$_i$(s){\textstyle T_{i}(s)} and T$_j$(s){\textstyle T_{j}(s)} as z{\textstyle z} approaches L{\textstyle L}. A consistent solution that balances two equation terms may generate an accurate approximation to the full equation's solution for z{\textstyle z} values approaching L{\textstyle L}. Approximate solutions arising from balancing different terms of an equation may generate distinct approximate solutions e.g. inner and outer layer solutions.
Substituting the scaled function s(z)=(z−L)$^p$s$^~$(z){\textstyle s(z)=(z-L)^{p}{\tilde {s}}(z)} into the equation and taking the limit as z{\textstyle z} approaches L{\textstyle L} may generate simplified reduced equations for distinct exponent values of p{\textstyle p}. These simplified equations are called distinguished limits and identify balanced dominant equation terms. The scale transformation generates the scaled functions. The dominant balance method applies scale transformations to balance equation terms whose factors contain distinct exponents. For example, T$_i$(s){\textstyle T_{i}(s)} contains factor (z−L)$^q${\textstyle (z-L)^{q}} and term T$_j$(s){\textstyle T_{j}(s)} contains factor (z−L)$^r${\textstyle (z-L)^{r}} with q≠r{\textstyle q\neq r}. Scaled functions are applied to differential equations when z{\textstyle z} is an equation parameter, not the differential equation´s independent variable. The Kruskal-Newton diagram facilitates identifying the required scaled functions needed for dominant balance of algebraic and differential equations.
For differential equation solutions containing an irregular singularity, the leading behavior is the first term of an asymptotic series solution that remains when the independent variable z{\textstyle z} approaches an irregular singularity L{\textstyle L}. The controlling factor is the fastest changing part of the leading behavior. It is advised to "show that the equation for the function obtained by factoring off the dominant balance solution from the exact solution itself has a solution that varies less rapidly than the dominant balance solution."

## Algorithm

The input is the set of equation terms and the limit L. The output is the set of approximate solutions. For each pair of distinct equation terms T$_i$(s), T$_j$(s){\textstyle T_{i}(s), T_{j}(s)} the algorithm applies a scale transformation if needed, balances the selected terms by finding a function that solves the reduced equation and then determines if this function is consistent. If the function balances the terms and is consistent, the algorithm adds the function to the set of approximate solutions, otherwise the algorithm rejects the function. The process is repeated for each pair of distinct equation terms.
    Inputs Set of equation terms {T$_0$(s), T$_1$(s), …, T$_n$(s)}{\textstyle \{T_{0}(s), T_{1}(s), \ldots , T_{n}(s)\}} and limit L{\textstyle L}
    Output Set of approximate solutions {s$_0$(z), s$_1$(z), …}{\textstyle \{s_{0}(z), s_{1}(z), \dots \}}
   * For each pair of distinct equation terms T$_i$(s), T$_j$(s){\textstyle T_{i}(s), T_{j}(s)} do:
   ** Apply a scale transformation if needed.
   ** Solve the reduced equation: T$_i$(s)+T$_j$(s)=0{\textstyle T_{i}(s)+T_{j}(s)=0} with T$_i$(s)≠0{\textstyle T_{i}(s)\neq 0} and T$_j$(s)≠0{\textstyle T_{j}(s)\neq 0}.
   ** Verify consistency: R$_{ij}$(s)≪T$_i$(s) (z→L){\textstyle R_{ij}(s)\ll T_{i}(s)\ (z\to L)} and R$_{ij}$(s)≪T$_j$(s) (z→L).{\textstyle R_{ij}(s)\ll T_{j}(s)\ (z\to L).}
   ** If function s(z){\textstyle s(z)} is consistent and solves the reduced equation, add this function to the set of approximate solutions, otherwise reject the function.

## Improved accuracy

The method may be iterated to generate additional terms of an asymptotic expansion to provide a more accurate solution. Iterative methods such as the Newton-Raphson method may generate a more accurate solution. A perturbation series, using the approximate solution as the first term, may also generate a more accurate solution.

## Examples
### Algebraic function

The dominant balance method will find an explicit approximate expression for the multi-valued function s=s(z){\textstyle s=s(z)} defined by the equation 1−16s+zs$^5$=0{\textstyle 1-16s+zs^{5}=0} as z{\textstyle z} approaches zero.

#### Input

The set of equation terms is {1, −16s, zs$^5$}{\textstyle \{1, -16s, zs^{5}\}} and the limit is zero.

#### First term pair

   * Select the terms 1{\textstyle 1} and −16s{\textstyle -16s}.
   * The scale transformation is not required.
   * Solve the reduced equation: $1-16s=0,s(z)={\tfrac {1}{16}}$.
   * Verify consistency: $zs^{5}\ll 1\ (z\to 0),\ zs^{5}\ll 16s\ (z\to 0)\$ for $s(z)={\tfrac {1}{16}}.$
   * Add this function to the set of approximate solutions: $s_{0}(z)={\tfrac {1}{16}}$.

#### Second term pair

   * Select the terms $-16s$ and $zs^{5}$.
   * Apply the scale transformation $s=z^{-1/4}{\tilde {s}}$. The transformed equation is $z^{1/4}-16{\tilde {s}}+{\tilde {s}}^{5}=0$.
   * Solve the reduced equation: $-16{\tilde {s}}+{\tilde {s}}^{5}=0,\ {\tilde {s}}=2,-2,2i,-2i$.
   * Verify consistency: $z^{1/4}\ll 16{\tilde {s}}\ (z\to 0),\ z^{1/4}\ll {\tilde {s}}^{5}\ (z\to 0)\$ for ${\tilde {s}}=2,-2,2i,-2i.$
   * Add these functions to the set of approximate solutions:
$s_{1}(z)={\frac {2}{z^{1/4}}},s_{2}(z)={\frac {-2}{z^{1/4}}},s_{3}(z)={\frac {2i}{z^{1/4}}},s_{4}(z)={\frac {-2i}{z^{1/4}}}.$

#### Third term pair

   * Select the terms $1$ and $zs^{5}$.
   * Apply the scale transformation $s=z^{-1/5}{\tilde {s}}$. The transformed equation is $1-16z^{-1/5}{\tilde {s}}+{\tilde {s}}^{5}=0.$
   * Solve the reduced equation: $1+{\tilde {s}}^{5}=0,\ {\tilde {s}}=(-1)^{1/5}.$
   * The function is not consistent: $-16z^{-1/5}{\tilde {s}}\gg 1\ (z\to 0),\ z^{-1/5}{\tilde {s}}\gg {\tilde {s}}^{5}\ (z\to 0)\$ for ${\tilde {s}}=(-1)^{1/5}.$
   * Reject this function: $s=z^{-1/5}(-1)^{1/5}.$

#### Output

The set of approximate solutions has 5 functions: $\left\{{\frac {1}{16}},{\frac {2}{z^{1/4}}},{\frac {-2}{z^{1/4}}},{\frac {2i}{z^{1/4}}},{\frac {-2i}{z^{1/4}}}\right\}.$

#### Perturbation series solution

The approximate solutions are the first terms in the perturbation series solutions.
    ${\begin{aligned}&s_{0}(z)={\frac {1}{16}}+{\frac {1}{16777216}}z^{1}+{\frac {5}{17592186044416}}z^{2}+\ldots ,\\&s_{1}(z)={\frac {2}{z^{1/4}}}-{\frac {1}{64}}-{\frac {5}{16384}}z^{\frac {1}{4}}-{\frac {5}{524288}}z^{\frac {1}{2}}-\ldots ,\\&s_{2}(z)=-{\frac {2}{z^{1/4}}}-{\frac {1}{64}}+{\frac {5}{16384}}z^{\frac {1}{4}}-{\frac {5}{524288}}z^{\frac {1}{2}}+\ldots ,\\&s_{3}(z)={\frac {2i}{z^{1/4}}}-{\frac {1}{64}}+{\frac {5i}{16384}}z^{\frac {1}{4}}+{\frac {5}{524288}}z^{\frac {1}{2}}-\ldots \\&s_{4}(z)=-{\frac {2i}{z^{1/4}}}-{\frac {1}{64}}-{\frac {5i}{16384}}z^{\frac {1}{4}}+{\frac {5}{524288}}z^{\frac {1}{2}}+\ldots ,\\\end{aligned}}$

### Differential equation

The differential equation z$^3$w$^{′′}$−w=0{\textstyle z^{3}w^{\prime \prime }-w=0} is known to have a solution with an exponential leading term. The transformation w(z)=e$^{s(z)}${\textstyle w(z)=e^{s(z)}} leads to the differential equation 1−z$^3$(s$^′$)$^2$−z$^3$s$^{′′}$=0{\textstyle 1-z^{3}(s^{\prime })^{2}-z^{3}s^{\prime \prime }=0}. The dominant balance method will find an approximate solution as z{\textstyle z} approaches zero. Scaled functions will not be used because z{\textstyle z} is the differential equation's independent variable, not a differential equation parameter.

#### Input

The set of equation terms is {1, −z$^3$(s$^′$)$^2$, −z$^3$s$^{′′}$}{\textstyle \{1, -z^{3}(s^{\prime })^{2}, -z^{3}s^{\prime \prime }\}} and the limit is zero.

##### First term pair

   * Select $1$ and $-z^{3}(s^{\prime })^{2}$.
   * The scale transformation is not required.
   * Solve the reduced equation: $1-z^{3}(s^{\prime })^{2}=0,\ s(z)=\pm 2z^{-1/2}$
   * Verify consistency: $z^{3}s^{\prime \prime }\ll 1\ (z\to 0),\ z^{3}s^{\prime \prime }\ll z^{3}(s^{\prime })^{2}\ (z\to 0)$ for $s(z)=\pm 2z^{-1/2}.$
   * Add these 2 functions to the set of approximate solutions: $s_{+}(z)=+2z^{-1/2},\ s_{-}(z)=-2z^{-1/2}.$

#### Second term pair

   * Select $1$ and $-z^{3}s^{\prime \prime }$
   * The scale transformation is not required.
   * Solve the reduced equation: $1-z^{3}s^{\prime \prime }=0,\ s(z)={\tfrac {1}{2}}z^{-1}$
   * The function is not consistent: $z^{3}(s^{\prime })^{2}\gg 1\ (z\to 0),\ z^{3}(s^{\prime })^{2}\gg z^{3}s^{\prime \prime }\ (z\to 0)$ for $s(z)={\tfrac {1}{2}}z^{-1}.$
   * Reject this function: $s(z)={\tfrac {1}{2}}z^{-1}.$.

#### Third term pair

   * Select $-z^{3}(s^{\prime })^{2}$ and $-z^{3}s^{\prime \prime }$.
   * The scale transformation is not required.
   * Solve the reduced equation: $z^{3}(s^{\prime })^{2}+z^{3}s^{\prime \prime }=0,\ s(z)=\ln z$.
   * The function is not consistent: $1\gg z^{3}(s^{\prime })^{2}\ (z\to 0)\$ and $\ 1\gg \ z^{3}s^{\prime \prime }\ (z\to 0)$ for $s(z)=\ln z.$
   * Reject this function: $s(z)=\ln z.$

#### Output

The set of approximate solutions has 2 functions: $\left\{+2z^{-1/2},-2z^{-1/2}\right\}.$

#### Find 2-term solutions

Using the 1-term solution, a 2-term solution is $s_{2\pm }(z)=\pm 2z^{-1/2}+s(z).$ Substitution of this 2-term solution into the original differential equation generates a new differential equation: ${\begin{aligned}1-z^{3}(s_{2\pm }^{\prime })^{2}-z^{3}s_{2\pm }^{\prime \prime }&=0\\\pm 1\mp {\frac {4}{3}}zs^{\prime }+{\frac {2}{3}}z^{5/2}(s^{\prime })^{2}+{\frac {2}{3}}z^{5/2}s^{\prime \prime }&=0.\end{aligned}}$

#### Input

The set of equation terms is {±1, ∓ 4/3 zs$^′$, 2/3 z$^{5/2}$(s$^′$)$^2$, 2/3 z$^{5/2}$s$^{′′}$}{\textstyle \{\pm 1, \mp {\frac {4}{3}}zs^{\prime }, {\frac {2}{3}}z^{5/2}(s^{\prime })^{2}, {\frac {2}{3}}z^{5/2}s^{\prime \prime }\}} and the limit is zero.

##### First term pair

    1. Select $1$ and $-{\tfrac {4}{3}}zs^{\prime }$.
    2. The scale transformation is not required.
    3. Solve the reduced equation: $1-{\tfrac {4}{3}}zs^{\prime }=0,\ s(z)={\tfrac {3}{4}}\ln z$.
    4. Verify consistency:
    ${\tfrac {2}{3}}z^{5/2}(s^{\prime })^{2}+{\tfrac {2}{3}}z^{5/2}s^{\prime \prime }\ll 1\ (z\to 0),{\text{for}}\ s(z)={\tfrac {3}{4}}\ln z$
    ${\tfrac {2}{3}}z^{5/2}(s^{\prime })^{2}+{\tfrac {2}{3}}z^{5/2}s^{\prime \prime }\ll {\tfrac {4}{3}}zs^{\prime }\ (z\to 0)\ {\text{for}}\ s(z)={\tfrac {3}{4}}\ln z.$
    5. Add these functions to the set of approximate solutions:
    s$_{2+}$(z)=+2z$^{−1/2}$+3/4 ln⁡z{\textstyle s_{2+}(z)=+2z^{-1/2}+{\tfrac {3}{4}}\ln z}
    s$_{2−}$(z)=−2z$^{−1/2}$+3/4 ln⁡z{\textstyle s_{2-}(z)=-2z^{-1/2}+{\tfrac {3}{4}}\ln z}.

#### Other term pairs

For other term pairs, the functions that solve the reduced equations are not consistent.

#### Output

The set of approximate solutions has 2 functions: $\left\{+2z^{-1/2}+{\tfrac {3}{4}}\ln z,-2z^{-1/2}+{\tfrac {3}{4}}\ln z\right\}.$

#### Asymptotic expansion

The next iteration generates a 3-term solution s$_{3±}$(z)=±2z$^{−1/2}$+3/4 ln⁡(z)+h(z){\textstyle s_{3\pm }(z)=\pm 2z^{-1/2}+{\tfrac {3}{4}}\operatorname {ln} (z)+h(z)} with h(z)≪1 (z→0){\textstyle h(z)\ll 1\ (z\to 0)} and this means that a power series expansion can represent the remainder of the solution.
The dominant balance method generates the leading term to this asymptotic expansion with constant A{\textstyle A} and expansion coefficients determined by substitution into the full differential equation:
    $w(z)=Az^{3/4}e^{\pm 2z^{-1/2}}\left(\sum _{n=0}^{m}\ a_{n}z^{n/2}\right)$
    $a_{n+1}=\pm {\frac {(n-1/2)(n+3/2)a_{n}}{4(n+1)}}.$
A partial sum of this non-convergent series generates an approximate solution. The leading term corresponds to the Liouville-Green (LG) or Wentzel–Kramers–Brillouin (WKB) approximation.

## Citations

* ↑ White 2010, p. 2.
* ↑ de Bruijn 1981, pp. 187–189.
* ↑ Christensen 1996.
* 1 2 White 2010, pp. 1–14.
* 1 2 3 4 5 Fishaleck & White 2008.
* ↑ Callaham et al. 2021.
* 1 2 3 4 Paulsen 2013, pp. 1–3, 7.
* ↑ Olver 1974, pp. 8, 9, 21.
* 1 2 Neu 2015, pp. 2–4, 14.
* 1 2 3 4 5 6 7 8 9 White 2010, pp. 49–51.
* 1 2 3 4 Bender & Orszag 1999, pp. 82–84.
* ↑ Kruskal 1962, p. 19.
* ↑ Hinch 1991, p. 62.
* 1 2 Rozman 2020.
* 1 2 Olver 1974, pp. 190–191.