16 - Asymptotic Sequences

Asymptotic Sequence

Set of functions $\{{\varphi }_k(\xi ){\}}_{k=0,1,2,\cdots }$ is an asymptotic sequence as $ϵ\to 0$

If

$${\varphi }_{k+1}(ϵ)=o({\varphi }_k(ϵ))\text{ as }ϵ\to 0$$

Note that each term in the sequence is of smaller magnitude than the previous term

Examples of Asymptotic Sequences

1. $\{1,ϵ,{ϵ}^2,{ϵ}^3,\cdots \}\{1,ϵ,{ϵ}^2,{ϵ}^3,\cdots \}\{1,ϵ,{ϵ}^2,{ϵ}^3,\cdots \}\{1,ϵ,{ϵ}^2,{ϵ}^3,\cdots \}\{1,ϵ,{ϵ}^2,{ϵ}^3,\cdots \}$
2. $\{1,{ϵ}^{1/2},ϵ,{ϵ}^{3/2},\cdots \}$
3. $\{1,ϵ\log ϵ,ϵ,{ϵ}^2\log ϵ,{ϵ}^2,\cdots \}$

Asymptotic Expansion

Function $f(ϵ)$ has an asymptotic expansion of form

$$f(ϵ)\sim \sum _{k=1}^{\infty }{a}_k{\varphi }_k(ϵ)\text{ as }ϵ\to 0$$

If

1. Gauge Functions ${\varphi }_k$ form an asymptotic sequence

$${\varphi }_{k+1}(ϵ)\ll {\varphi }_k(ϵ)\text{ for all }k$$

$$f(ϵ)-\sum _{k=0}^N{a}_k{\varphi }_k(ϵ)\ll {\varphi }_N(ϵ)\text{ for all }N=0,1,\cdots$$

as $ϵ\to 0$

Note that $(1)$ ensures terms in expansion get successively smaller
Note that $(2)$ ensures approximation gets more accurate the more terms are incuded in expansion

Elementary Properties of Asymptotic Expansion

1. Coefficients $a_k$ are unique for a particular choice of Gauge Functions $\{{\varphi }_k\}$

2. Functions defines expansion but not vice-versa

Asymptotic Approach for Finding Asymptotic Expansions

Suppose find approximate solution $x$ of an algebraic equation of form

$$F(x;ϵ)=0$$

containing a small parameter $ϵ$

1. Scale variables to get a dominant balance
   So at least two of the terms balance in order and are much bigger than remaining terms

2. Plug in asymptotic expansion for $x$
   Usually form of expansion is clear from form of equation

3. Equate terms multiplying each power of $ϵ$ in equation to obtain coefficients in expansion

4. Repeat for any other possible dominant balances in equation
   In order to get approximation for other roots

$ϵx^2+x-1=0$ as $ϵ\to 0$

Set $ϵ=0$ to get leading-order solution
So $x-1=0$ hence

$$x\sim 1$$

Hence suppose $x$ can be expressed as a asymptotic expansion in powers of $ϵ$ so

$$x\sim 1+ϵx_1+{ϵ}^2{x}_2+{ϵ}^3{x}_3+\cdots \text{ as }ϵ\to 0$$

Substituting into original equation then

$$0\sim ϵ(1+ϵx_1+{ϵ}^2{x}_2+{ϵ}^3{x}_3+\cdots {)}^2+(1+ϵx_1+{ϵ}^2{x}_2+{ϵ}^3{x}_3+\cdots )-1$$

Grouping coefficients of each power of $ϵ$ then

$$\begin{array}{rlrlr}O(ϵ):& 1+x_1& & =0⟹x_1& =-1\\ O({ϵ}^2):& 2x_1+x_2& & =0⟹x_2& =2\\ O({ϵ}^3):& x_1^2+2x_2+x_3& & =0⟹x_3& =-5\end{array}$$

Since the equation is a quadratic then expecting two roots
First asymptotic expansion was found by balancing $x$ and $1$

Suppose dominant balancing $ϵx^2$ and $1$ so $x=O({ϵ}^{-1/2})$ but then $x$ dominates instead

$$\underset{O(1)}{\underbrace{ϵx^2}}+\underset{O({ϵ}^{-1/2})}{\underbrace{x}}-\underset{O(1)}{\underbrace{1}}=0$$

Balance $ϵx^2$ and $ϵ$ so $x=O({ϵ}^{-1})$ which then dominates

$$\underset{O({ϵ}^{-1})}{\underbrace{ϵx^2}}+\underset{O({ϵ}^{-1})}{\underbrace{x}}-\underset{O(1)}{\underbrace{1}}=0$$

Hence scale

$$x={ϵ}^{-1}y\text{ where }y=O(1)$$

Thus substituting then

$$\frac{y^2}{ϵ}+\frac{y}{ϵ}-1=0⟹y^2+y-ϵ=0$$

As $ϵ\to 0$ then suppose $y$ has asymptotic expansion of form

$$y\sim y_0+ϵy_1+{ϵ}^2{y}_2+{ϵ}^3{y}_3+\cdots \text{ as }ϵ\to 0$$

By setting $ϵ=0$ then $y_0=-1$ or $y_0=0$
However discard $y_0=0$ as it reproduces the first case found already

Hence substituting back in

$$0\sim (-1+ϵy_1+{ϵ}^2{y}_2+{ϵ}^3{y}_3+\cdots {)}^2+(y_0+ϵy_1+{ϵ}^2{y}_2+{ϵ}^3{y}_3+\cdots )-ϵ$$

Grouping coefficients of each power of $ϵ$ then

$$\begin{array}{rlrlr}O(ϵ):& -y_1-1& & =0⟹y_1& =-1\\ O({ϵ}^2):& y_1^2-y_2& & =0⟹y_2& =1\\ O({ϵ}^3):& 2y_1{y}_2-y_3& & =0⟹y_3& =-2\end{array}$$

Hence

$$y\sim -1-ϵ+{ϵ}^2-2{ϵ}^3+\cdots \text{ as }ϵ\to 0$$

Rescaling $x=\frac{y}{ϵ}$ then

$$x\sim -\frac{1}{ϵ}-1+ϵ-2{ϵ}^2+\cdots \text{ as }ϵ\to 0$$

Singular Pertubation

Suppose find approximate solution $x$ of an algebraic equation of form

$$F(x;ϵ)=0$$

Singular Perturbation is if $ϵ=0$ reduces degree of $F(x,ϵ)$