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.^[1][2]

An early example of the dominant balance method is the Newton polygon method. Newton developed this method to find an explicit approximation for a function defined implicitly by an algebraic equation. He expressed the function as proportional to the independent variable raised to a power, retained only the lowest-degree polynomial terms (dominant terms) arising from this approximation, and solved this simplified reduced equation to obtain an approximate solution.^[3][4] 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.^[5][6]

Asymptotic relations

The functions ${\textstyle f(z)}$ and ${\displaystyle g(z)}$ of parameter or independent variable ${\textstyle z}$ and the quotient ${\textstyle f(z)/g(z)}$ have limits as ${\textstyle z}$ approaches ${\textstyle z_{0}}$.

The function ${\textstyle f(z)}$ is much less than ${\textstyle g(z)}$ as ${\textstyle z}$ approaches ${\textstyle z_{0}}$, written as ${\textstyle f(z)\ll g(z)\ (z\to z_{0})}$, if the limit of the quotient ${\textstyle f(z)/g(z)}$ is zero as ${\textstyle z}$ approaches ${\textstyle z_{0}}$ ^[7]

${\displaystyle \operatorname {Lim} \ {\frac {f(z)}{g(z)}}=0\ {\text{as}}\ z\to z_{0}}$.

The relation ${\textstyle f(z)}$ is lower order than ${\textstyle g(z)}$ as ${\textstyle z}$ approaches ${\textstyle z_{0}}$, written using little-o notation ${\textstyle f(z)=o(g(z))\ (z\to z_{0})}$, is identical to the ${\textstyle f(z)}$ is much less than ${\textstyle g(z)}$ as ${\textstyle z}$ approaches ${\textstyle z_{0}}$ relation.^[7]

The function ${\textstyle f(z)}$ is equivalent to ${\textstyle g(z)}$ as ${\textstyle z}$ approaches ${\textstyle z_{0}}$, written as ${\textstyle f(z)\sim g(z)\ (z\to z_{0})}$, if the limit of the quotient ${\textstyle f(z)/g(z)}$ is one as ${\textstyle z}$ approaches ${\textstyle z_{0}}$ ^[7]

${\displaystyle \operatorname {Lim} \ {\frac {f(z)}{g(z)}}=1\ {\text{as}}\ z\to z_{0}}$.

This result indicates that the zero function, ${\textstyle f(z)=0}$, can never be equivalent to any other function.^[7]

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.^[8]

Equation properties

Dominant balance applies to a minimum 3-term equation ${\textstyle A(f)+B(f)-C(f)=0}$ containing a function ${\textstyle f=f(z)}$.

Balance terms ${\textstyle B(f)}$ and ${\textstyle C(f)}$ means make these terms equal and asymptotically equivalent by finding the function ${\textstyle f(z)}$ that solves the reduced equation ${\textstyle B(f)-C(f)=0}$ with ${\textstyle B(f)\neq 0}$ and ${\textstyle C(f)\neq 0}$.^[9]

A solution ${\textstyle f(z)}$ is consistent if terms ${\textstyle B(f)}$ and ${\textstyle C(f)}$ are dominant; dominant means all other equation terms ${\textstyle A(f)}$ are much less than terms ${\textstyle B(f)}$ and ${\textstyle C(f)}$ as ${\textstyle z}$ approaches ${\textstyle z_{0}}$.^[10][11] A consistent solution that balances two equation terms may generate an accurate approximation to the full equation's solution for ${\textstyle z}$ values neighboring ${\textstyle z_{0}}$.^[11][12] Approximate solutions arising from balancing different terms of an equation may generate distinct approximate solutions e.g. inner and outer layer solutions.^[5]

Substituting the scaled function ${\textstyle f(z)=(z-z_{0})^{p}{\tilde {f}}(z)}$ into the equation and taking the limit as ${\textstyle z}$ approaches ${\textstyle z_{0}}$ may generate simplified reduced equations for distinct exponent values of ${\textstyle p}$.^[9] These simplified equations are called distinguished limits and identify balanced dominant equation terms.^[13] Scaled functions are often used when attempting to balance equation term ${\textstyle B(f)}$ containing factor ${\textstyle (z-z_{0})^{q}}$ and term ${\textstyle C(f)}$ containing factor ${\textstyle (z-z_{0})^{r}}$ with ${\textstyle q\neq r}$. Scaled functions are applied to differential equations when ${\textstyle z}$ is an equation parameter, not the differential equation´s independent variable.^[5] The Kruskal-Newton diagram facilitates identifying the required scaled functions needed for dominant balance of algebraic and differential equations.^[5]

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 ${\textstyle z}$ approaches an irregular singularity ${\textstyle z_{0}}$. 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."^[11]

Algorithm

• Select 2 equation terms to balance, ${\textstyle B(f)}$ and ${\textstyle C(f)}$, as ${\textstyle z}$ approaches ${\textstyle z_{0}}$.
• If ${\textstyle z}$ is not a differential equation's independent variable, apply a scaled function if the selected equation's terms contain factors ${\textstyle (z-z_{0})}$ with distinct exponents.
• Balance ${\textstyle B(f)}$ and ${\textstyle C(f)}$ by solving the reduced equation ${\textstyle B(f)-C(f)=0,\ B(f)\neq 0,\ C(f)\neq 0}$ for ${\textstyle f=f(z)}$.
• Verify that the solution ${\textstyle f(z)}$ is consistent.
• Accept the approximate solution ${\textstyle f(z)}$ if it is consistent and balances terms ${\textstyle B(f)}$ and ${\textstyle C(f)}$.

Improved accuracy

• The method may be iterated to generate additional terms of an asymptotic expansion to provide a more accurate solution.^[11]
• Iterative methods such as the Newton-Raphson method may generate a more accurate solution.^[4]
• A perturbation series, using the approximate solution as the first term, may also generate a more accurate solution.^[5]

Examples

Algebraic function

The dominant balance method will find an explicit approximate expression for the multi-valued function ${\textstyle f=f(z)}$ defined by the equation ${\textstyle 1-16f+zf^{5}=0}$ for ${\textstyle z}$ near zero.^[14]

Balance 1st and 2nd terms

• Select ${\displaystyle 1}$ and ${\displaystyle 16f}$.
• Scaled function is unnecessary.
• Solve reduced equation ${\displaystyle 1-16z=0,z={\tfrac {1}{16}}}$
• Verify consistency ${\displaystyle zf^{5}\ll 1\ (z\to 0),\ zf^{5}\ll 16f\ (z\to 0)\ }$ for ${\displaystyle f={\tfrac {1}{16}}}$
• Accept solution ${\displaystyle f_{1}(z)={\tfrac {1}{16}}}$

Balance 2nd and 3rd terms

• Select ${\displaystyle 16f}$ and ${\displaystyle zf^{5}}$.
• Apply scaled function ${\displaystyle f=z^{-1/4}{\tilde {f}}}$
• Transformed equation ${\displaystyle z^{1/4}-16{\tilde {f}}+{\tilde {f}}^{5}=0}$
• Solve reduced equation ${\displaystyle -16{\tilde {f}}+{\tilde {f}}^{5}=0,\ {\tilde {f}}=2,-2,2i,-2i}$
• Verify consistency ${\displaystyle z^{1/4}\ll 16{\tilde {f}}\ (z\to 0),\ z^{1/4}\ll {\tilde {f}}^{5}\ (z\to 0)\ }$ for ${\displaystyle {\tilde {f}}=2,-2,2i,-2i}$
• Accept solutions

${\displaystyle f_{2}(z)={\frac {2}{z^{1/4}}},f_{3}(z)={\frac {-2}{z^{1/4}}},f_{4}(z)={\frac {2i}{z^{1/4}}},f_{5}(z)={\frac {-2i}{z^{1/4}}}}$

Balance 1st and 3rd terms

The consistency condition fails for balance of 1st and 3rd terms.

Perturbation series solution

The approximate solutions are the first terms in the perturbation series solutions.^[14]

$${\displaystyle {\begin{aligned}&f_{1}(z)={\frac {1}{16}}+{\frac {1}{16777216}}z^{1}+{\frac {5}{17592186044416}}z^{2}+\ldots ,\\&f_{2}(z)={\frac {2}{z^{1/4}}}-{\frac {1}{64}}-{\frac {5}{16384}}z^{\frac {1}{4}}-{\frac {5}{524288}}z^{\frac {1}{2}}-\ldots ,\\&f_{3}(z)=-{\frac {2}{z^{1/4}}}-{\frac {1}{64}}+{\frac {5}{16384}}z^{\frac {1}{4}}-{\frac {5}{524288}}z^{\frac {1}{2}}+\ldots ,\\&f_{4}(z)={\frac {2i}{z^{1/4}}}-{\frac {1}{64}}+{\frac {5i}{16384}}z^{\frac {1}{4}}+{\frac {5}{524288}}z^{\frac {1}{2}}-\ldots \\&f_{5}(z)=-{\frac {2i}{z^{1/4}}}-{\frac {1}{64}}-{\frac {5i}{16384}}z^{\frac {1}{4}}+{\frac {5}{524288}}z^{\frac {1}{2}}+\ldots ,\\\end{aligned}}}$$

Differential equation

The differential equation ${\textstyle z^{3}w^{\prime \prime }-w=0}$ is known to have a solution with an exponential leading term.^[15] The transformation ${\textstyle w(z)=e^{f(z)}}$ leads to the differential equation ${\textstyle 1-z^{3}(f^{\prime })^{2}-z^{3}f^{\prime \prime }=0}$. The dominant balance method will find an approximate solution for ${\textstyle z}$ near 0. Scaled functions will not be used because ${\textstyle z}$ is the differential equation's independent variable, not a differential equation parameter.^[10]

Find 1-term solution

${\textstyle f_{1}(z)=f}$

${\textstyle 1-z^{3}(f_{1}^{\prime })^{2}-z^{3}f_{1}^{\prime \prime }=0}$.

${\textstyle 1-z^{3}(f^{\prime })^{2}-z^{3}f^{\prime \prime }=0}$.

Balance 1st and 2nd terms

• Select ${\displaystyle 1}$ and ${\displaystyle z^{3}(f^{\prime })^{2}}$.
• Solve reduced equation ${\displaystyle 1-z^{3}(f^{\prime })^{2}=0,\ f(z)=\pm 2z^{-1/2}}$
• Verify consistency ${\displaystyle z^{3}f^{\prime \prime }\ll 1\ (z\to 0),\ z^{3}f^{\prime \prime }\ll z^{3}(f^{\prime })^{2}\ (z\to 0)}$ for ${\displaystyle f(z)=\pm 2z^{-1/2}}$.
• Accept solution ${\displaystyle f_{1}(z)=\pm 2z^{-1/2}}$

Balance 1st and 3rd terms

• Select ${\displaystyle 1}$ and ${\displaystyle z^{3}f^{\prime \prime }}$
• Solve reduced equation ${\displaystyle 1-z^{3}f^{\prime \prime }=0,\ f(z)={\tfrac {1}{2}}z^{-1}}$
• Not consistent ${\displaystyle z^{3}(f^{\prime })^{2}\gg 1\ (z\to 0),\ z^{3}(f^{\prime })^{2}\gg z^{3}f^{\prime \prime }\ (z\to 0)}$ for ${\displaystyle f(z)={\tfrac {1}{2}}z^{-1}}$.
• Reject solution ${\displaystyle f(z)={\tfrac {1}{2}}z^{-1}}$.

Balance 2nd and 3rd terms

• Select ${\displaystyle z^{3}(f^{\prime })^{2}}$ and ${\displaystyle -z^{3}f^{\prime \prime }}$.
• Solve reduced equation ${\displaystyle z^{3}(f^{\prime })^{2}+z^{3}f^{\prime \prime }=0,\ f(z)=\operatorname {ln} (z)}$.
• Not consistent ${\displaystyle 1\gg z^{3}(f^{\prime })^{2}\ (z\to 0)\ }$ and ${\displaystyle \ 1\gg \ z^{3}f^{\prime \prime }\ (z\to 0)}$ for ${\displaystyle f(z)=\operatorname {ln} (z)}$.
• Reject solution ${\displaystyle f(z)=\operatorname {ln} (z)}$

Find 2-term solution

${\displaystyle f_{2+}(z)=+2z^{-1/2}+f}$

${\displaystyle 1-z^{3}(f_{2+}^{\prime })^{2}-z^{3}f_{2+}^{\prime \prime }=0}$.

${\displaystyle 1-{\tfrac {4}{3}}zf^{\prime }+{\tfrac {2}{3}}z^{5/2}(f^{\prime })^{2}+{\tfrac {2}{3}}z^{5/2}f^{\prime \prime }=0}$.

${\displaystyle f_{2-}(z)=-2z^{-1/2}+f}$

${\displaystyle 1-z^{3}(f_{2-}^{\prime })^{2}-z^{3}f_{2-}^{\prime \prime }=0}$.

${\displaystyle -1+{\tfrac {4}{3}}zf^{\prime }+{\tfrac {2}{3}}z^{5/2}(f^{\prime })^{2}+{\tfrac {2}{3}}z^{5/2}f^{\prime \prime }=0}$.

Balance 1st and 2nd terms

• Select ${\displaystyle 1}$ and ${\displaystyle {\tfrac {4}{3}}zf^{\prime }}$.
• Solve reduced equation ${\displaystyle 1-{\tfrac {4}{3}}zf^{\prime }=0,\ f(z)={\tfrac {3}{4}}\operatorname {ln} (z)}$.
• Verify consistency

${\displaystyle {\tfrac {2}{3}}z^{5/2}(f^{\prime })^{2}+{\tfrac {2}{3}}z^{5/2}f^{\prime \prime }\ll 1\ (z\to 0)}$ for ${\displaystyle f(z)={\tfrac {3}{4}}\operatorname {ln} (z)}$

${\displaystyle {\tfrac {2}{3}}z^{5/2}(f^{\prime })^{2}+{\tfrac {2}{3}}z^{5/2}f^{\prime \prime }\ll {\tfrac {4}{3}}zf^{\prime }\ (z\to 0)}$ for ${\displaystyle f(z)={\tfrac {3}{4}}\operatorname {ln} (z)}$

• Accept solutions^[10]

${\displaystyle f_{2+}(z)=+2z^{-1/2}+{\tfrac {3}{4}}\operatorname {ln} (z)}$

${\displaystyle f_{2-}(z)=-2z^{-1/2}+{\tfrac {3}{4}}\operatorname {ln} (z)}$

Balance other terms

The consistency condition fails for balance of other terms.^[10]

Asymptotic expansion

The next iteration generates a solution ${\textstyle f_{3}(z)=\pm 2z^{-1/2}+{\tfrac {3}{4}}\operatorname {ln} (z)+h(z)}$ with ${\textstyle h(z)\ll 1\ (z\to 0)}$ and this means that an asymptotic expansion can represent the remainder of the solution.^[10] The dominant balance method generates the leading term to this asymptotic expansion with constant ${\textstyle A}$ and expansion coefficients determined by substitution into the full equation.^[10]

${\displaystyle w(z)=Az^{3/4}e^{\pm 2z^{-1/2}}\left(\sum _{n=0}^{m}\ a_{n}z^{n/2}\right)}$

${\displaystyle a_{n+1}=\pm {\frac {(n-1/2)(n+3/2)a_{n}}{4(n+1)}}}$

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.^[15]

Citations

References

See also

• Asymptotic analysis