Asymptotics and Integral Transforms

Example of asymptotics

• f(x) = 1/x sinx - (sin ε)/ε ∼ 1 − ε²/6 + ε⁴/120

• Stirling’s formula: n! ∼ √(2πn)nⁿe⁻ⁿ

• Prime Number Theorem. If π(n) is the number of primes less than or equal to n, then π(n) ∼ n/ log n for large n

• Much of applied maths research

Asymptotics

____ is the study of mathematical objects (e.g. roots of equations, solutions to PDEs, distribution of primes, etc) as a parameter ε gets small. Or, as a parameter gets large: if X → ∞ then ε = 1/X → 0.

Asymptotics can

• Give often accurate solutions with very little computational effort.

• Often make sense of why things happen, what is important, and the mechanism behind them.

• Give analytic approximate solutions where no exact analytic solution exists

• Works when numerical solutions don’t.

Big O Notation

For functions f(x) and g(x), we say that f(x) = O(g(x)) as x → a

⇔ |f(x)/g(x)| is bounded as x → a

⇔ ∃ M, δ s.t. |f(x)/g(x)| < M ∀ |x − a| < δ

⇔ lim sup_x→a |f(x)/g(x)| < ∞

Little o Notation

For functions f(x) and g(x), we say that f(x) = o(g(x)) as x → a ⇔ |f(x)/g(x)| → 0 as x → a. Think of this as “f is smaller than order g”

Big Θ notation

For functions f(x) and g(x), we say that f(x) = Θ(g(x)) as x → a ⇔ f(x) = O(g(x)) and g(x) = O(f(x)) as x → a.

Asymptotic Series

The function f(x) is said to have an _____ as x → a where f(x) ∼ N∑j=1 fj(x) as x → a ⇔ ∀ finite M ≤ N, f(x) − M∑ j=1 fj(x) = o(fM(x)) ⇔ | (f(x) − M∑j=1 fj (x))/ fM(x)| → 0 as x → a

The Error Function

Erf(z) = 2/√π z∫0 e^{−t^2} dt

Asymptotic Sequences

We want fn(x) to be of a particular form so we write f(x) ∼ N∑n=1 a_nα_n(x) as x → a, where αn is an ____

An ____ ⇐⇒ αn+1 = o(αn) as x → a for all n

Uniqueness

Given an asymptotic sequence α_n(x), and an asymptotic series f(x) ∼ N∑n=1 a_nα_n(x) as x → a, then the an coefficients are unique.

Indeed, by definition, an = limx→a (f(x) − n−1∑j=1 ajαj(x))/α_n(x)

Iteration

• No guarantees

• Rearranging, x to get x on its own and εx on other side

• Try x0 = 1 and x1 = εxo eqn

• Repeat for x2, x3, …

Successive Approximation

• The slow and steady method.

• Try for if ε = 0, and find value of x (x0)

• Then try x = x0 + e1 (e1 is previous error)

• Put this into original eqn (neglect terms that are smaller than the ones we have included)

• Set x = x0 +e1 + e2 and repeat

Series Expansion

• The inspired guess method

• Guess the series expansion x+ = a0 + εa1 + ε²a2 + ε³a3 + Θ(ε⁴) and substitute it in.

• Expand and collect powers of ε to find values of a0,a1,…

• Place values into guess

Singular problem

Setting ε=0 makes a solution “disappear” / changes solution a lot.

Solving singular perturbations

Identify the scaling needed by balancing terms. At least two of the terms must be the same order to cancel, and the remaining terms must be smaller

Poincaré asymptotic series

• f(t, ε) ∼ N∑n=1 a_n(t)α_n(ε)

If this is an asymptotic series for each fixed t as ε → 0

Uniformly Valid

Rescale the problem to find inner and outer expansions.

Then try a composite expansion which recovers both expansions when needed.

Integrals with small parameter

Approximate by asymptotic series using successive approximation

The Exponential integral

for z > 0, E₁(z) = ∞∫z e^−x/x dx

Splitting

Split at x=a and use either expansion in its valid area. If correct, all the a’s will cancel

Watson’s Lemma

If f(x) ∼ N∑n=0 anx^αn as x → 0 with α₀ > −1, and if f(x) is bounded for x ∈ [ε, a] for all ε > 0, then

a∫0 f(x)e^−λxdx ∼ N∑n=0 a_nΓ(α_n + 1)/λ^αn+1 as λ → ∞

Use of Watson’s Lemma

Says what to do when an exponentially weighted integral is dominated by one end of the integration range.

Laplace’s Method

If g(x) is twice continuously differentiable on [a, b], and ∃ x0 ∈ (a, b) such that g(x0) = max_x∈[a,b] g(x) and g’’(x0) < 0, then b∫a e^λg(x)dx ∼ e^λg(x0) √(2π/ −λg’’(x0)) as λ → ∞.

Stirling’s Formula

Γ(λ + 1) = ∞∫0 x^λ e^−x dx

Nondimensionalisation

Change variables to some which are dimensionless.

dy/st scales like y/t

Damping of R

• Small R means little damping.

• Large R means damping dominates

• R = Θ(1) means all are important.

Regular ODE problem

Use techniques from algebraic equations such as series expansion

Rescaling and Balancing

• Try the change of variable t = aτ

• Match different terms s.t. third term is smaller

• Ensure fits boundary conditions

• Use successive approximation

Singular differential equation

A ____ is one where standard methods don’t work. Often, we lose the highest derivative. In these cases, we must rescale to find solutions valid in particular regions (e.g. for t = ετ and t = T /ε).

Outer/ inner solution

• Suppose our solution f(x) has an _____ which satisfies, as ε → 0 with x fixed, f(x) ∼ P∑n=0 εⁿf_n(x) = E_Pf where E_P means taking the first (P + 1) terms of the outer asymptotic series.

• Suppose for x = εξ, then as ε → 0 with ξ fixed we obtain the ____ f(εξ) ∼ Q∑n=0 εⁿg_n(ξ) = H_Qf where H_Q means taking the first (Q+ 1) terms of the inner asymptotic series.

Van Dyke’s matching rule

The two asymptotic series should match according to the matching rule E_PH_Qf = H_QE_Pf for “suitable” values of P and Q; i.e. expanding the inner solution in the outer variable gives the same as expanding the outer solution in the inner variable.

Composite Expansions

C(x; ε) = E_Pf + H_Qf − E_PH_Qf

• valid as ε → 0 for fixed x or fixed ξ

• not a Poincaré series, and so is not necessarily unique

Multiplicative composite expansion

M(x; ε) = (E_Pf)(H_Qf) / E_PH_Qf

Rescaling for nonlinear ODE

• Rescale

• Match powers ε

• Makes a full nonlinear problem with no small parameters

Integral Transform

Given a function f(x), an ___ of f yields another function g(y), where g(y) = ∫_C f(x)K(x, y) dx

Kernel

K(x, y) is called the _______ of the integral transform.

Contour integral

Let f(z) be a function D → C defined on an open set D ⊆ C, and let q(t) be a piecewise differentiable function [a, b] → D.

The ______ of f along the curve C = {q(t) : a ≤ t ≤ b} is ∫_C f(z) dz = ^b∫_a f(q(t)) dq/dt dt = ^b∫_a Re(f(q(t)) dq/dt) dt + i^b∫_a Im(f(q(t)) dq/dt) dt

Bounding contour integral

|∫_C f(z) dz| ≤ max_z∈C |f(z)| ∫^b_a |dq/dt| dt = max_z∈C |f(z)|L where L is the length of the curve C

Anti-derivative

If f(z) has an ____, meaning f(z) = F‘(z) for some function F(z), then ∫_C f(z) dz = ^b∫_a F’(q(t)) dq/dt dt = ^b∫_a d/dt(F(q(t))) dt = F(q(b)) − F(q(a))

Cauchy’s Theorem

The domain D is simply connected if any path in D can be smoothly deformed into any other path in D between the same end points without moving the end points

If f(z) is differentiable everywhere in D and D is simply connected, then the integral of f(z) is independent of the path.

Removable singularity

If lim_z→z0 f(z) exists, then z0 is a ____

Simple pole

If lim_z→z0 (f(z) − a1/(z − z0)) exists for some a1 ≠ 0, then f has a ____at z0, with residue Res(f, z0) = a1

Pole of order n

If the limit lim_z→z0 (f(z) − a1/(z − z0) − a2/(z − z0)² − · · · − an/(z − z0)ⁿ ) exists for some n and some a1, . . . , an with an ≠ 0, then f has a _____ at z0.

The residue is still Res(f, z0) = a1 (note, not an).

Essential singularity

Otherwise, f has an ____ at z0.

Residue theorem

∫_C f(z) dz → 2π_ia₁ as ε →0 = 2π_iRes(f,z0)

Singularity at the origin

• Bound the integral along ε+ as R → ∞

• Integrate along ε_ε and take the limit that ε → 0

Fourier Transform

The ___ of a function f(t) is ˜f(ω) = (Ff)(ω) = ∫^∞_−∞ f(t)e^−iωt dt where-ever the integral exists.

Inverse Fourier Transform

of a function ˜f(ω) is f(t) = (F⁻¹˜f)(t) = 1/2π ∫^∞_−∞ ˜f(ω)e^+iωt dω

Inversion

Provided f is “well behaved”, F⁻¹(Ff) = f that is, the inverse transform is actual the inverse of the forward transform.

More specifically, if on any finite interval, f(t) has finitely many minima, maxima, and finite discontinuities, and no infinite discontinuities, and if in addition ^∞∫_−∞ |f(t)| dt < ∞ (absolutely integrable), then 1/2π ∫^∞_−∞ e^iωτ(∫^∞_−∞ f(t)e^−iωt dt)dω = 1/2(lim_t→τf(t) + lim_t→τf(t)) i.e. we recover f(t), except we average at discontinuities.

Linearity

h(t) = af(t) + bg(t) ⇒ h˜(ω) = a ˜f(ω) + bg˜(ω)

Shift Property

• If g(t) = f(t − a), then ˜g(ω) = e^−iωa ˜f(ω)

• If g(t) = e^iΩtf(t), then ˜g(ω) = ˜f(ω − Ω)

Scaling

If g(t) = f(at), then ˜g(ω) = 1/|a|˜f(ω/a)

Differentiation

• If g(t) = f’(t), then ˜g(ω) = iω ˜f(ω)

• If g(t) = tf(t), then ˜g(ω) = i ˜f’(ω)

Multiple Derivatives

g(t) = dⁿf/dtⁿ then ^~g(w) = (iw)^n~f(w)

Convolution

Given f(t) and g(t), we define the ___ h = f∗g by h(t) = ∫^∞_−∞ f(τ)g(t − τ) dτ = ∫^∞_−∞ f(t − τ)g(τ) dτ

• If h = f∗g then h˜(ω) = ˜f(ω)˜g(ω)

• If h(t) = f(t)g(t), then h˜ = 1/2π ˜f∗g˜

Solution of Integral equation

Consider finding the function f(t) such that λf(t) + ^∞∫_−∞ f(t − τ)g(τ) dτ = u(t), with g(t), u(t) and λ known.

Fourier transforming gives a solution: λ˜f + ˜fg˜ = ˜u ⇒ ˜f = u˜/(λ + ˜g)

Parseval’s Theorem

• Given two functions f(t) and g(t), and using x + iy = x − iy to denote the complex conjugate, 1/2π ∫^∞_−∞ ˜f(ω)g˜(ω) dω = ∫^∞_−∞ f(t)g(t) dt.

Plancherel’s theorem

If f(t) = g(t), then 1/2π ∫^∞_−∞ |˜f(ω)|² dω = ∫^∞_−∞|f(t)|² dt

Multi-dimensional Fourier transforms

• Take multiple Fourier transforms, one in each scalar variable

• Often written in vector notation. ^~f(k) = ∫R2 f(x)e^−ik·x d²x. Fourier inversion is similar: f(x) = 1/(2π)² ∫_R2 ˜f(k)e^ik·x d²k

Properties of multi-dimensional transform

• g(x) = ∇f ⇒ g˜(k) = ik ˜f

• h = ∇²f = ∇·(∇f) ⇒ h˜ = ik·ik ˜f.

Limitation of Fourier Transform

They only exist for functions f : R → C, and f also needs to be “well behaved” at ±∞. However, we often only know/want f on a domain D ⊆ R.

Zero-extending

Extend f : [0, ∞] → C to g : R → C by setting g(t) = ( f(t) t ≥ 0, 0 t < 0

Laplace Transform

f(s) = ˜g(−is) = ∫^∞₀ f(t)e^−st dt

Differentiation - Laplace

• If g(t) = tf(t), then ˆg(s) = − dfˆ/ds (s)

• If g(t) = df/dt (t) = ˙f(t), then gˆ(s) = −f(0) + sˆf(s)

• If g(t) = ¨f(t), then ˆg(s) = −˙f(0) − sf(0) + s²ˆf(s).

Inverse of Laplace issues

• ˆf(s) is only defined for Re(s) sufficiently positive

• f(t) ≡ 0 for t < 0

Inverse of Laplace

Take f(t) = 1/2πi ∫^i∞+α _−i∞+α ˆf(s)e^st ds with α > Re(sj ) for all singularities sj of ˆf(s)

Convolution (Laplace)

If f(t) = g(t) = 0 for t < 0, then h(t) = (f ∗ g)(t) = ∫^t₀ f(τ)g(t − τ) dτ and Laplace transforming gives hˆ(s) = ˆf(s)ˆg(s)

Solving PDEs

• the Laplace transform is used for the time variable

• Form an ODE in x

• Solve using integrating factors

• Invert the Laplace transform

Limits (Abelian & Tauberian theorems)

sˆf − f(0) = ^˙f = ∫^∞₀ ˙f(t)e^−st dt and ^˙f(0) = ∫^∞₀ ˙f(t) dt = f(∞) − f(0)