Methods of Applied Mathematics Notes

Brian Bockelman

October 18, 2006

1 Scaling and Dimensions

1.1 Linear Algebra Background

Monday, August 28, 2006:

Definition 1.1.1. Let A be n × m. So we can think of A as a linear map-ping from Rm → Rn. The kernel K(A) is x ∈ Rm : Ax = 0. K(A) isa subspace of Rm whose dimension is the nullity of A. The range R(A) isy ∈ Rn|Ax = y for some x ∈ Rm. The range is a subspace of Rn, the dimen-sion of which is the rank of A.

Theorem 1.1.2. The rank of A equals:

(i) The number of linearly independent columns of A,

(ii) The number of linearly independent rows of A.

Proof. (i) Let

A =

a11 . . . a1m

...an1 . . . anm

=

[A1|A2| . . . |Am

]Let x = (x1, . . . , xm) ∈ Rm. Then, Ax = x1A1 + · · · + xmAm. Thus,every vector in the range is a lienar combination of the column vectorsA1, . . . , Am. Thus, a maximal linearly independent subset of the vectorsA1, . . . , Am forms a basis of the range.

Example 1.1.3. Let

A =[1 0 −1 12 1 2 0

]Since A is 2×4, we think of it as a linear mapping: R4 → R2. Since a (maximal)linear independent set of columns has two elements, the rank of A is 2.

1

What is the kernel of A? We have to determine which vectors

α =

α1

...α4

satisfy the homogeneous system Aα = 0. Term by term, this system is:

α1 −α3 +α4 = 02α1 +α2 +2α3 = 0.

Use Gaussian elimination: [1 0 −1 1 02 1 2 0 0

][1 0 −1 1 00 1 4 −2 0

]Thus,

α1 −α3 +α4 = 0α2 +4α3 −2α4 = 0

So, the solutions to the system Aα = 0 are vectors in R4 of the form:

α =

α3 − α4

−4α3 + 2α4

α3

α4

Since the solution α depends only on α3 and α4, dimK(A) = 2.

We can now write down a basis of K(A); for one vector, use α3 = 0 andα4 = 1. For the other vector, use α3 = 1 and α4 = 0.

Theorem 1.1.4. Let A be an n×m matrix with rank r and nullity k. Then,

r + k = m

Thus, to find k, just compute m− r.

1.2 Dimensional Analysis

Consider a mathematical model involving physical quantities q1, . . . , qm. LetL1, . . . ,Ln be the fundamental dimensions of the model. Let the dimension ofqj be

[qj ] = La1j

1 La2j

2 · · · Lanjn , j = 1, . . . ,m

Let A be the dimension matrix [aij ].

2

Consider the projectile problem:d2x∗

dt∗2 = −gR2

(x∗+R)2

x∗(0) = 0x∗′ = V.

Our physical quantities are x∗, t∗, R, g, and V with dimensions:

[x∗] = [R] = L, [t∗] = T , [g] = LT −2, [V ] = LT −1

Thus, the fundamental dimensions are L and T . The dimension matrix is:

A =[1 0 1 10 1 −2 −1

]Exercise 1.2.1.

A =

1 0 0 0 −1 00 1 0 0 2 −30 0 1 0 0 −10 0 0 1 0 1

1. Determine the rank.

2. Determine the nullity.

3. Find a basis for K(A).

Definition 1.2.2. If a physical quantity q has dimension

[q] = Li

for some i = 1, . . . , n, then q is called a primary quantity. Quantities obtainedby multiplying powers of primary quantities are called secondary.

In the projectile problem, x∗, t∗, and R were primary; g and V are secondary.

Let q be a physical quantity of the form

q = qα11 · · · qαm

m .

Thus, q has the dimensions

[q] = [q1]α1 · · · [qm]αm

= (La111 · · · Lan1

n )α1 · · · (La1m1 · · · Lanm

n )αm

= La11α1+a12α2+···+a1mαm1 · · ·Lan1α1+···+anmαm

n

Therefore, q is dimensionless iff

Aα = 0

3

where α = (α1, . . . , αm)T . Thus, q is dimensionless iff α ∈ K(A).

Let

αi =

αi1...αi

m

; i = 1, . . . ,K,

form a basis for the kernel K(A). Let

πi = qαi

11 · · · qαi

mm ; i = 1, . . . ,K.

Since the vectors αi lie in K(A), π1, . . . , πk are dimensionless.

Theorem 1.2.3. Letπ = qv1

1 · · · qvmm

be dimensionless. Then there are numbers c1, . . . , cK such that

π = πc11 · · ·πcK

K

Thus, if we have π1, . . . , πk, we have all the dimensionless quantities.

Proof. Since π = qv11 · · · qvm

m , we know that the vector v = (v1, . . . , vm)T belongsto the kernel K(A). Thus, there are scalars c1, . . . , ck such that

v =

v1...vm

= c1α1 + · · ·+ ckα

k

= c1

α11...α1

m

+ · · ·+ ck

αk1...αk

m

=

c1α11 + · · ·+ ckα

k1

...c1α

1m + · · ·+ ckα

km

Therefore,

vi = c1α1i + · · ·+ ckα

ki ; i = 1, . . .m

Consequently,

π = qv11 + ·s+ qvm

m = qc1α1

1+···+ckαk1

1 · · · qc1α1m+···+ckαk

mm

=(q

α11

1 qα1

22 · · · qα1

mm

)c1

· · ·(q

αk1

1 qαk

22 · · · qαk

mm

)ck

= πc11 π

c22 · · ·πck

k

Note: since α1, . . . , αk are linearly independent, π1, . . . , πk are called inde-pendent dimensionless quantities.

4

1.3 Conversion Factors, Unit-Free Laws

We have L1, . . . ,Ln, measured in some system of units. Let L1, . . . , Ln be thefundamental dimensions, given in some other system of units. Let λ1, . . . , λn bethe conversion factors:

Li = λiLi, i = 1, . . . , n

Let q1, . . . , qm be the physical quantities measured in the new unit system.Since,

[qj ] = La1j

1 · · · Lanjn ,

the dimension of qj is

[qj ] = La1j

1 · · · Lanjn = (λ1L1)a1j · · · (λnLn)anj

= λa1j

1 · · ·λanjn Laj

1 · · · Lanjn

= λa1j

1 · · ·λanjn [qj ].

Therefore,qj = λ

α1j

1 · · ·λanjn qj , j = 1, . . . ,m

Suppose now that we have a physical law relating q1, . . . , qm:

f(q1, . . . , qm) = 0

Hence, we should havef(q1, . . . , qm) = 0

for all choices of positive conversion factors λ1, . . . , λn.

Definition 1.3.1. A physical law is unit-free if

f(q1, . . . , qm) = 0

holds ifff(q1, . . . , qm) = 0

holds, for all choices of positive conversions factors λ1, . . . , λn.

Example:Let e,m, and c be energy, mass, and the speed of light, all in some system ofunits.

[e] = ML2T−2

[m] = M

[c] = LT−1

The dimension matrix is:

A = [aij ] =

2 0 1−2 0 −11 1 0

5

Let λ1, λ2, and λ3 be conversion factors for L, T , and M :

e = λ21λ−22 λ3e

m = λ3m

c = λ1λ−12 c.

Is f(e,m, c) = e−mc2 = 0 unit free?

f(e, m, c) = e− mc2

= λ21λ−22 λ3e− (λ3m)λ2

1λ−22 c2

= λ21λ−22 λ3(e−mc2) = 0

So, f(e,m, c) = 0 is unit-free.

1.4 Buckingham Pi Theorem

Theorem 1.4.1. Buckingham Pi Theorem Consider a unit-free law

g(q1, . . . , qm) = 0

relating physical quantities q1, . . . , qm with fundamental dimensions L1, . . . ,Ln.Suppose further that the dimension matrix A has rank r. Then, there are k =m− r dimensionless quantities π1, . . . , πk and there is a law of the form

G(π1, . . . , πk) = 0

equivalent to g.

Proof. We’ve already proven the existence of the π1, . . . , πk. Rather than es-tablish the second part, we’ll consider an example.

Example: Consider the simplified projectile problem:d2xdt2 = −g,x∗(0) = 0,

(x∗)′(0) = v

Or, in implicit form,f(x∗, t∗, V, g) = 0

wheref(x∗, t∗, V, g) = x∗ − 1

2g(t∗)2 − V t∗

We claim the implicit form is unit free. Since,

[x∗] = L, [t∗] = T , [V ] = LT −1, and [g] = LT −2

6

we’ll take L and T as fundamental dimensions and obtain the dimension matrix:

A =[1 0 1 10 1 −1 −2

]Since rank(A) = 2, then we have 4−2 = 2 independent dimensionless quantities.Let α = (α1, α2, α3, α4)T . We need to find basis vectors for K(A); in this case,we have:

α1 =

−1110

, α2 =

−1201

Thus, we have independent, dimensionless quantities (x∗)−1t∗V and (x∗)−1(t∗)2g.Us the reciprocal

π1 =x∗

V t∗

π2 =x∗

g(t∗)2

0 = x∗ − 12g(t∗)2 − V t

=x∗

V t∗− 1

2g(t∗)2

V t∗− 1

=x∗

V t∗− 1

2g(t∗)2

x∗x∗

V t∗− 1

= π1 −12π1

π2− 1 = F (π1, π2)

Hence, we have the dimensionless law,

F (π1, π2) = 0

equivalent to the original f(x∗, t∗, V, g). This can be taken a step further andwritten in explicit form as

π2 = G(π1)

Example: Obtain information about the radius of the blast wave from theatomic bomb. How rapidly does the blast wave move, etc?

Assume that the energy is released instantaneously in a region small enoughto be considered a point. Assume a unit-free law of the form

f(t, r, ρ, e) = 0

where t is the time, r is blast radius, ρ is the initial air density, and e the energyyield;

[r] = L1, [t] = T , [ρ] = ML−3, [e] = µL2, T −2

7

Our dimension matrix is:

A =

1 0 0 −20 1 −3 20 0 1 1

The rank of this matrix is 3, so the nullity is 1. A vector in the kernel is:

α =

−251−1

Thus, our dimensionless quantity is

π =r5ρ

t2e

Our original law is equivalent to a dimensionless law of the form

F (π) = 0

Hence, for a constant c, π = c, so that

r =(Ct2e

ρ

) 15

Thus,r ∼ t

25

Wednesday, September 6, 2006:

2 Perturbation Theory

The (Hardy) Order symbols. These are defined carefully in appendix 3.1, pages112 - 113.

We say that φ(x) = O(ψ(x)) as x → x0 if there is a positive constant Msuch that ∣∣∣∣φ(x)

ψ(x)

∣∣∣∣ ≤M

for x “near” x0.We sometimes say that φ(x) is order ψ(x) as x→ x0.We say that φ(x) = o(ψ(x)) if ∣∣∣∣φ(x)

ψ(x)

∣∣∣∣ → 0

8

as x→ x0.

If φ(x) = O(ψ(x)) as x→ x0, then φ(x) can’t be “much larger” than ψ(x).If φ(x) = o(ψ(x)) as x → x0, then φ(x) → c as x → x0 faster than ψ(x) → 0(or φ(x) → ±∞ less rapidly than φ(x).

Example: sin θ = O(θ) as θ → 0.Since

limθ→0

sin θθ

= 1,

sin(θ) = O(θ) as θ → 0.

Example: 1− cos(x) = o(x) as x→ 0.

limx→0

1− cos(x)x

= limx→0

sin(x)1

= 0

Example: 1− cos(x) = O(x2) as x→ 0:

limx→0

1− cos(x)x2

= limx→0

sin(x)2x

=12

Example: e−1ε = o(ε) as ε→ 0+. Set ε = 1

η ; then,

limε→0

e−1ε

ε= lim

η→∞ηe−η = 0

ε usually denotes a small positive quantity (ε 1). So, if φ(ε) = O(ε) asε→ 0, then φ(ε) is “small.” On the other hand, if φ(ε) = O(ε−1) as ε→ 0, thenφ(ε) is large. Finally, if φ(ε) = O(1), then φ(x) is of “moderate” size.

When we scale a problem, we obtain dimensionless terms. In a well-scaledproblem, the dimensionless variables should be O(1) so that the relative mag-nitude of the term can be determined from the dimensionless coefficients alone.

2.1 Perturbation Theory

Suppose you’ve scaled a second-order ODE and obtained a dimensionless versionthat depends on a small parameter ε.

F (x, u, u′, u′′, ε) = 0.

You could look for a solution u as a perturbation expansion:

u = u0 + εu1 + εu22 + · · ·

We plug the expansion into the ODE, and then solve iteratively for u0(x), u1(x), . . . , un(x), . . ..We then have an approximate solution.

9

Friday, September 8, 2006:Example of Scaling:Consider the motion of a body of mass m in a medium that resists with force:

FR = −av + bv2

for velocity v in a certain range. We regard bv2 as a (small) nonlinear pertur-bation. Thus, the velocaity as a function of time τ satisfies the ODE

mdv

dτ= −av + bv2

Let v(0) = v0. We’ll scale to obtain a dimensionless version of the IVP.:m dv

dτ = −av + bv2

v(0) = v0(1)

The physical quantities have dimensions:

[τ ] = T , [v] = [v0] = LT−1

[a] = MT −1, [b] = ML−1

[m] = M

Choose V0 and m/a as our velocity and time scales. Why are these good scales?To begin with, we expect v

v0to be O(1).

If bv2 is in some sense “small,” we could drop it and consider the linear problem:m dv

dτ = −av + bv2

v(0) = v0(2)

The solution isv(τ) = v0e

− am t (3)

Suppose that a velocity drop to 12v0 is “significant”. When does this occur?

v(τ) =12v0 = v0e

− am t

τ =m

aln 2

So, ma is a good characteristic scale for the time τ .

We thus set

t =τ

m/a, u(t) =

v(τ)v0

The dimensionless problem is then:mdu

dt = −v + εv2

v(0) = 1(4)

10

whereε =

bv0a 1

To find an approximate solution to (P2), expand u:

u(T ) = u0(t) + εu1(t) + ε2u2(t) + · · ·

Plug this into the ODE and match powers of ε.

(u′0 + εu′1 + ε2u′2 + · · · ) = −(u0 + εu1 + ε2u2 + · · · ) + ε(u0 + εu1 + · · · )2u′0 = −u0

u0(0) = 1

u0 = e−tu′1 = −u1 + e−2t

u1(0) = 0

u1 = e−t − e−2tu′2 = −u2 + 2u0u1

u2(0) = 0

We thus have our approximate solution:

uA = u0 + εu1 + ε2u2

uA = e−t + ε(e−t − e−2t

)+ ε2

(e−t − 2e−2t + e−3t

)For this to be a good approximation, we should have

u(t) = uA(t) +O(ε3)

We can actually solve (4) exactly. The ODE is a Bernoulli equation.

du

dt= −u+ εu2

Set u = 1x . Plug in x, get a linear equation, recover the solution in terms of u:

u =e−t

1− e(1− e−t); t ≥ 0

Recall that for |r| < 1,

11− r

= 1 + r + r2 + r3 + · · ·

and thus1

1− r= 1 + r + r2 + · · ·+ rn +O(rn+1)

11

Since ε 1, |ε(1 − e−t)| 1 for t ≥ 0. We thus have a geometric seriesrepresentation for u(t), valid for all t ≥ 0:

u(t) = e−t(1 + ε(1− e−t) + ε2(1− e−t)2

)If we expand this, we end up with our previous singular-perturbation solution.

= uA(t) +O(ε3)

Monday, September 11, 2006:

2.2 Simple Pendulum Problem:

Suppose we have a pendulum consisting of a massless rod of length L and a bobof mass m. We would like to derive an initial value problem for θ∗ as a functionof the time t∗. By Newton’s second law,

mLd2θ∗

dt∗2+mg sin θ∗ = 0

d2θ∗

dt∗2+ ω2 sin θ∗ = 0,

where ω =√

gL . Note that [w] = T −1.

Our initial value problem is thus:d2θ∗

dt∗2 + ω2 sin θ∗ = 0θ∗(0) = 0θ∗′(0) = Ω.

For convenience, take a > 0 and Ω = 0. To scale the problem, set θ = 1aθ∗. For

time, assuming θ∗ is small, then sin θ∗ ≈ θ∗. We thus obtain a linear problem:d2θ∗

dt∗2 + ω2θ∗ = 0θ∗(0) = 0θ∗′(0) = Ω.

The solution to this problem is θ∗ = a cos(ωt∗). Thus ω appears as the frequencyof the solution to the linearized problem. Hence, the period is 2π

ω . This suggeststhat 1

ω is a good characteristic time. We therefore set t = t∗ω. So, our finalscaling is:

t = t∗ω

θ(t) =1aω∗(t∗).

The scaled nonlinear problem is then:d2θdt2 + sin(aθ)

a = 0θ(0) = 1θ′(0) = 0.

12

We’ll derive an approximate solution using a perturbation series θ = θ0 + aθ1 +a2θ2:

0 = θ′′ + a−1 sin(aθ)

= θ′′ + a−1

(aθ − 1

3!(aθ)3 +

15!

(aθ)3 − · · ·)

= θ′′0+aθ′′1+a2θ′′2+· · ·+1a

((aθ0 + a2θ1 + a3θ2 + · · · )− 1

3!(aθ0 + a2θ1 + a3θ2 + · · · )3 + · · ·

)For O(1),

θ′′0 + θ0 = 0θ0(0) = 1θ′0(0) = 0.

θ0 = cos t

For O(a), θ′′1 + θ1 = 0θ1(0) = 0,θ′1(0) = 0.

θ1(t) = 0.

For O(a2), θ′′2 + θ2 − 1

3!θ30 = 0,

θ2(0) = 0,θ′2(0) = 0

where θ0 = cos t. The solution to this is:

θ2 =1

192[cos t− cos 3t] +

116t sin t

We thus obtain our approximate solution:

θA(t) = cos t+ a2

1

192(cos t− cos 3t) +

116t sin t

We hope that θ(t) and θA(t) differ by at most an O(a3) term, at least over someshort time interval.

2.3 Scaled Projectile Problemx′′ = −(1 + εx)−2, ε 1x(0) = 0,x′(0) = 1.

13

Assume a perturbation expansion of the forom

x = x(t, ε) = x0(t) + εx1(t) + ε2x2(t) + · · ·

Plug the perturbation series into the ODE, and once again use a Taylor Seriesto expand the nonlinear term. In this case, it is easier to start from a binomialseries than to compute the Taylor series directly:

− 1(1 + r)2

=d

dr

11 + r

= −1 + 2r − 3r2 + 4r3 − · · ·

x′′0 + εx′′1 + ε2x′′2 + · · · ) = −1 + 2ε(x0 + εx1 + · · · )− 3ε2(x0 + εx1 + · · · )2 + · · ·

Homework:Solve the initial value problem (1)Solve:

θ′′ + ω2θ = cosωt, ω 6= 0θ(0) = 0θ′(0) = 0

Wednesday, September 13, 2006:Friday, September 15, 2006:

3 Method of Successive Approximations

3.1 Successive Approximation for a Polynomial

We used successive approximation to derive the approximate solution

x2(ε) = 2 + 8ε+ 96ε2 +O(ε3)

for a root of the polynomial

εz3 − z + 2 = 0.

(ε small). If ε = .0005, and we drop the O(ε3) terms, we get x2 = 2.042. Theactual solution is 2.04261152.

3.2 Successive Approximation for an IVP

We wish to solve the problem:F (t, x, x′, x′′) = G(t, x, x′, x′′)x(0) = A

x′(0) = B

14

We make an initial guess: this should be the solution x0 to the problemF (t, x, x′, x′′) = 0x(0) = A

x′(0) = B

With x0, we next solveF (t, x, x′, x′′) = G(t, x0, x

′0, x

′′0)

x1(0) = A

x′1(0) = B

Repeat to obtain successive approximations, x0, x1, . . ..

Consider the projectile problem:x′′ = −(1 + εx)−2,

x(0) = 0,x′(0) = 1

Taking ε 1, using series, we derived an approximate solution:

x(t, ε) = t− 12t2 + ε

(13t3 − 1

12t4

)+ ε2

(−1

4t4 +

1160t5 − 11

360t6

)We obtain x0 by solving F = 0. It is natural to take x0 to be the solution withε = 0:

x′′ = −1

This tell us that we should take:

F = F (x′′) = x′′ + 1.

So, x0 is the solution to x′′ + 1 = 0,x(0) = 0,x′(0) = 1

and thus, x0(t) = t− t2

2 .We next solve F = G0(t, x0, x

′0, x

′′0) for x1. Since F (x′′) = x′′ + 1, we should

take:G = G(x) = −(1 + εx)−2 + 1.

Therefore, our problem for x1 is:

F (x1) = G(x0)

x′′1 + 1 = −(1 + εx0)−2 + 1

15

with initial valuesx1(0) = 0, x′1(0) = 1.

x1 satisfies the equation:

x′′1 + 1 = 1− 1(1 + εx0)2

=ε(2x0 + εx2

0)(1 + εx0)2

.

Recall that:

1(1 + εx0)2

= 1− 2εx0 + 3ε2x20 = 1− 2εx0 +O(ε2)

Hence, the problem becomes:

x′′1 + 1 = ε(2x0 + εx20)(1− 2εx0 +O(ε2))

= ε(2t− t2) +O(ε2)

Since we want the two-term approximation, drop O(ε2). Thus,

x1 = 1− t2

2+ ε

(t3

3− t4

12

),

which agrees up to O(ε) with xA obtained using series.

How do we judge the accuracy and validity of these approximate solutions?Suppose that xA(t, ε) is an approximation to x(t, ε) for ε 1, and t in someinterval I. We say that xA(t, ε) is asymptotically valid as ε→ 0 uniformly on I.

For the simple pendulum, we derived the approximate solution:

θA(t, a) = cos t+ a2

(1

192cos t− 1

192cos 3t+

116t sin t

)The presence of the secular term t sin t suggests that θA(t, a) is an asymptoti-cally valid approximation to the true solution θ(t, a) as a→ 0 over a finite timeinterval.

Read, in 2.2 (53, 54) Poincare’s perturbation method.Wednesday, September 20, 2006:

4 Singular Perturbation

Let p(m, ε) = εm2 + 2m+ 1, where ε 1. We are to approximate the roots ofp(m, ε).

16

Since ε 1, we start by taking εm2 = 0. We get:

2m+ 1 = 0

m = −12

for ε 1. We can improve this estimate by setting:

m = −12

+ εm1 + ε2m2 + · · ·

Plug this expansion into our equation:

p(m, ε) = ε(−12

+ εm1 + · · · )2 + 2(−12

+ εm1 + · · · ) + 1 = 0

For O(ε),14

+ 2m1 = 0, m1 = −18.

For O(ε2),

−m1 + 2m2 = 0, m2 = − 116

Thus, we have the perturbation expansion. Note that we lose a solution - whenwe took ε → 0, we lost a degree. Hence, we are only able to approximate onesolution. This characterizes a singular perturbation problem. The actual rootsare given by:

m =−1 +

√1− ε

ε=−1 + 1− 1

2ε

ε= −1

2for ε small. The other root is:

m =−1−

√1− ε

ε= O(ε−1), ε→ 0

Therefore, the leading term in the polynomial is:

εm2 = O(ε−1), ε→ 0

Therefore, we cannot drop εm2, even though ε is small.

Idea: If all three terms, εm2, 2m and 1 are all of the same order, we can’tsimplify the equation by dropping one of them. We look for a two-term dom-inant balance (find two terms that have the greatest magnitude, and have asimilar magnitude). When we find the balance, we simply drop the others andsolve the simplified equation. Use that solution to start a series or successiveapproximation solution.

For our problem, we have the following option:

17

(a) The dominate pair is εm2 and 1. Assuming this true, we consider 2m tobe negligible 2m:

εm2 + 1 ≈ 0, m = ± i√ε

= O(ε−12 )

The balance of these are:

εm2 = εO(ε−12 ) = O(1) = 1

The remaining term is:2m = O(ε−

12 )

So, this term is much larger, and these two terms are not dominant.

(b) The pair 2m and 1 give the dominant balance. However, they lead to thesolution we have already found.

(c) The pair εm2 and 2m give the dominant balance. We thus take 1 to benegligible, and get:

εm2 + 2m ≈ 0

m = 0 cannot be order ε−1, so it is not the solution we are looking for.So,

m =2ε

= O(ε−1)

Since m = O(ε−1) and εm2 = O(ε−1), we have a balance.This is also clearly dominant since 1 = O(1) O(ε−1) as ε→ 0.

We would like to improve the m estimate by successive approximations.m0 = − 2

ε is the solution to

F (m) := εm+ 2 = 0

So, we take

G(m) = − 1m

This implies that

m1 =12− 2ε

Monday, September 25, 2006:

4.1 Singular Perturbation and ODEsεy′′ + 2y′ + y = 0, 0 < x < 1, ε 1y(0) = 0,y(1) = 1

(5)

18

The order of the equation changes from 2 to 1 when ε → 0+. Thus, thisproblem might be a candidate for singular perturbation.

Try setting ε = 0. You get the problem:2y′ + y = 0,y(0) = 0,y(1) = 1

The general solution is y = Kex/2; note we cannot chose K to satisfy bothboundary conditions.

If we choose the condition y(1) = 1, we get the solution

y = e12 (x−1)

which doesn’t satisfy y(0) = 0. We’ll see that the above solution is in somesense a good approximation to the solution near x = 1. We’ll use singular per-turbation to find a good approximation near x = 0.

We will then match (in some sense) these approximations in some interme-diate region (IR). The correct solution to (5) is:

y(x) =em1x − em2x

em1 − em2

where m1 and m2 are the roots of

εm2 + 2m+ 1 = 0

Use the approximations m1 = − 12 , m2 = − 2

ε ; we get the approximate solution

yA(x) =e−

12 x − e−

2ε x

e−12 − e−

2ε

For x = O(1), ε 1, yA(x) = O(1):

y′A(x) =23e− 2

3x− 12e− x

2

e−12 − e−

2ε

= O(1)

And y′′A(x) = O(1) for x = O(1).

For the O(ε) terms, y′A(x) = O(ε−1) for x = O(ε). The interval, or zonex = O(ε), where the function changes rapidly is called the boundary layer, inthis case of width O(ε).

If we take yA to be a good enough approximation to y, then the fact thatyA(x), y′A(x), and y′′A(x) are O(1) for x = O(1) suggets that the problem is

19

well-scaled in the outer region, x = O(1).

Thus, for x = O(1), εy′′ is O(ε), and may be neglected. We end up with:2y′ + y = 0y(1) = 1.

Thus, we solve for the “outer solution,” y0(x) = e12 (x−1). How good is this

approximation? You can show that

1. For 0 < x ≤ 1, fixed,lim

ε→0+y(x) = y0(x).

2. For any fixed a ∈ (0, 1), the convergence is uniform on [a, 1].

Homework: 9.1 (1a, 1c, 2a, 5a, 5c). Note in the book, they use:

yA(x) =ε−

12 x − e−

2ε x

e−12

Wednesday, September 27, 2006:

Last time, we dropped εy′′ and the boundary condition y(0) = 0 and solvedto get the “outer equation” for the “outer approximation” or “outer solution”

y0(x) = e12 (1−x)

We’ll next obtain an approximate solution that is valid in the boundary layerx = O(ε). Recall the approximate solution:

yA =e−

12 x − e−

2ε x

e−12 − e−

2ε

If we just let ε → 0+, we have to let x → 0+, since x = O(ε) in the boundarylayer. This leaves us with the “poor” approximation

y ≈ y(0) = 0

in the boundary layer. To get around this difficulty, define a variable ξ that isO(1) in the boundary layer.

Define Y (ξ, ε) = y(x, ε) = y(εξ, ε). Now let ε → 0+, bearing in mind that ξremains O(1). Let ε→ 0+ and obtain the inner approximation:

yI = limε→0+

Y (ξ, ε) = limε→0+

yA(εξ, ε) = e12 (1− e2ξ)

valid for ξ = O(1).We obtained inner and outer approximations by using the exact solution.

We were able to show that:

20

(a) For x = O(1), y′′, y′, and y were O(1). We were thus able to drop εy′′ andthe condition y(0) = 0. We got y0.

(b) We show that there was a boundary layer at x = 0.

(c) We showed that the boundary had thickness O(ε). With that, we wereable to define our O(1) boundary layer variable, and thus obtain the innerapproximation yI .

How do we get any of this information without having the approximatesolution? For now, we take points (a) and (b) as a given. Since y(x) is notchanging rapidly for x = O(1), we do as before (drop εy′′ and y(0) = 0, etc) toget the outer solution:

yo(x) = e12 (1−x)

We’ll next find an inner solution, using the balancing procedure to guess thethickness of the boundary layer. When we have the inner solution yI , valid inthe boundary, and the outer solution y0, valid for x = O(1), we will still haveto paste the solutions together in an intermediate region. Since we’ll need yI

to satisfy yI(0) = 0 as well as a matching condition in the intermediate region,the ODE defining YI must be of the second order.

We’ll derive an inner equation for yI . We’ll scale the problem in the boundarylayer by setting

ξ =x

δ(ε)where δ(ε) is the width of the boundary layer. Let Y (ξ) = y(x). The scaledproblem fro Y (ξ) is

εδ2Y

′′ + 2δY

′ + Y = 0, ξ = O(1)Y (0) = 0

Assuming this to be well-scaled, we tell the magnitudes of the terms from the

coefficientsε

δ2,1ε, and 1. Look for a two-term dominant balance:

(a) 1δ , 1: Can’t give the correct dominant balance. It is balanced, and theyare dominant, but they lead to the outer solution.

(b) εδ2 , 1: This is balanced, but not dominant.

(c) εδ2 ,

1δ . This gives the dominant balance. This also shows that the bound-

ary layer width is O(ε).

We have a boundary layer thickness δ(ε) = O(ε). For scaling purposes,set δ(ε) = ε. Thus, our “inner equation” is:

1εy′′ +

2εy′ + y = 0

21

y′′ + 2y′ + εy = 0

in the boundary layer. Now that the ODE is correctly scaled, we can doregular perturbation through normal means. We get

y0(ξ) = c(1− e−2ξ)

In terms of x, this is:yI(x) = c(1− e−2x/ε)

Now, we must do matching for the inner and outer solutions. As x = O(1)in the outer region, and ξ = O(1) in the inner region, we’ll define an in-termediate value which will be O(1) in the IR.

Suppose that in the IR, x = O(θ), where θ = θ(ε). We stipulate that:

limε→0+

θ

ε= ∞

limε→0+

θ = 0

Let’s pick θ =√ε (of course θ(ε) = εb for any b ∈ (0, 1) will do). Thus,

our intermediate variable isη =

x

θ(ε)

Think the IR as the left extremity of the outer region, and the rightextremity of the boundary layer. Thus, YI and YO should agree as ε→ 0+

in the IR. We thus have our matching condition:

limε→0+

yI(ξ)|ξ= ηθδ

= limε→0+

yO(x)|x=ηθ

Remember, when taking limits, that θδ →∞ and θ → 0 as ε→ 0+.

So, we evaluate the limit. On the left,

limε→0+

c(1− e−2 ηθ

δ

)= c

On the right,limε→0

yO(x)|x=ηθ = limε→0+

e12 (1−ηθ) = e

12

Hence, we pick c = e12 . We can then add the two solutions together, and

simplify:yU (x) = e

12

(e−

12 x − e−

23 x

)

22

Monday, October 2, 2006:

Let’s consider the example:εy′′ + 2y′ + ey = 0, 0 < x < 1, ε 1y(0) = 0y(1) = 0.

Assume that we can do regular perturbation for x = O(1), to obtain anouter approximation; it is easy to drop εy′′ and solve the resulting first-order,separable problem:

2y′ + ey = 0, x = O(1)y(1) = 0.

We thus obtain the outer solution

y0 = ln(

2x+ 1

),

valid in the outer region x = O(1).There is a boundary layer at x = 0. Let δ(ε) be its width. Our boundary layervariable is ξ = x

δ , which is O(1) in the boundary layer. Set Y (ξ) = y(x). Theequation for Y is:

ε

δ2Y ′′ +

2δY ′ + eY = 0

along with the initial condition Y (0) = 0.

Here, the dominant balance is given byε

δ2and

1δ. Hence, δ(ε) = O(ε). We just

set δ(ε) = ε. We now have the inner problem:Y ′′ + 2Y ′ + εeY = 0, ξ = O(1),Y (0) = 0

Plug in the regular perturbation Y = Y0 + εY1 + · · · , save the O(1) problem:Y ′′0 + 2Y ′0 = 0Y0(0) = 0.

Our inner solution is:Y0(ξ) = C(1 + e−2ξ)

oryI(x) = C

(1− e−

2xδ

),

valid for ξ = O(1).Let the IR be characterized by x = O(θ), where θ = θ(ε). We should have

limε→0+

θ

δ= ∞

23

limε→0+

θ = 0.

The intermediate variable is η = xθ . Thus, η = O(1) in the IR, x = ηθ, and

ξ = ηθδ .

To match yI and yO in the intermediate zone, we stipulate that

limε→0+

yI(ε)|ξ= ηθδ

= limε→0+

yO(x)|x=ηθ

Thus,lim

ε→0+yI(ξ)|ξ= ηθ

δ= lim

ε→0+C

(1− e−2 ηθ

δ

)= C

= limε→0

yO(x)|x=ηθ = limε→0

ln(

2ηθ + 1

)= ln 2

Thus, our uniform approximation yU to the true solution y is:

yU (x) = yI(x) + yO(x)− limε→0

yO(x)|x=ηθ

= ln(2)(1− e−

2xε

)+ ln

(2

x+ 1

)− ln 2

= (ln 2)(1− e−

xε

)− ln(x+ 1)

Theorem 4.1.1. Let p(t) and q(t) be continuous on [0, 1] with p(t) > 0. Then,for the boundary value problem,

εy′′ + p(t)y′ + q(t)y = 0, 0 < t < 1, ε 1y(0) = a,

y(1) = b

there exists a boundary layer of width O(ε) at x = 0.

Example from chemical kinetics, in chapter 10. Skipped from taking notes.

5 Asymptotic Expansions:

Definition 5.0.2. Let gn(t, ε) be a sequence of functions defined on a t-intervalI, for values of a parameter ε in some relevant range. The (formal) series∞∑

n=0

gn(t, ε) is called an asymptotic expansion for g(t, ε) on I as ε→ ε0 if

g(t, ε)−N∑

n=0

gn(t, ε) = o(gN (t, ε))

as ε → ε0, at each t ∈ I, for each N . The asymptotic expansion∑∞

n=0 gn(t, ε)is uniform if the convergence is uniform.

24

We often need an asymptotic expansion of an expression g(ε). In this case,∑∞n=0 gn(ε) is an asymptotic expansion for g(ε) if

g(ε)−N∑

n=0

gn(ε) = o(gN (ε))

as ε→ ε0 for each N . IfN∑

n=0

gn(ε) is an expansion for g(ε) as ε→ ε0, we write:

g(ε) ∼∞∑

n=0

gn(ε)

as ε→ ε0.

We usually approximate g(ε) with the first few terms of an asymptotic ex-pansions. We still write:

g(ε) ∼N∑

n=0

gn(ε) as ε→ ε0

Asymptotic expansions are very important in approximate solutions to differ-ential equations given as parameter independent integrals:

I(λ) =∫ b

a

f(t, λ)dt,

orI(λ) =

∫C

f(z, λ)dz

where C is a contour in the complex plane. We then use asymptotic methodsto obtain expansions for I(λ) as λ → λ0 for some relevant value λ0 - usually0,±λ.

There are a few common methods - integration by parts, Laplace’s method,Watson’s Lemma, Stationary phase (method of steepest descent).

Ex:

I(x) =∫ ∞

0

e−t

1 + xt, x ≥ 0.

We’ll obtain an asymptotic expansion for I(x) as x→ 0:

I(x) =∫ ∞

0

e−t

1 + xtdt.

(Integrate by parts with dv = e−tdt, u =1

1 + xt. So, v = e−t and du =

− x(1+xt)2 dt ):

=−e−t

1 + xt

∣∣∣∣∞t=0

−∫ ∞

0

xe−t

(1 + xt)2dt = 1− x

∫ ∞

0

e−t

(1 + xt)3dt.

25

Doing this twice again, we get:

I(x) = 1− x+ 2x2 − 6x3

∫ ∞

0

e−t

1 + xt)4dt.

Can we write I(x) ∼ 1−x+2x2 as x→ 0? We can if I(x)−[1−x+2x2] = o(2x2)as x→ 0. We need only show that:

I(x)− [1− x+ 2x2] = −6x3

∫ ∞

0

e−t

(1 + xt)4dt

is O(2x2) as x→ 0, i.e. that,

limx→∞

x3∫∞0

e−t

(1+xt)4 dt

x2= 0

This is true. Hence, I(x) ∼ 1− x+ 2x2.Doing this N times gives us:

I(x) =N∑

k=0

(−1)kk!xk + (−1)N+1(N + 1)!xN+1

∫ ∞

0

e−t

(1 + xt)N+2dt.

It is easily checked that

I(x)−N∑

n=0

(−1)nn!xn = o(xN )

as x→ 0. Thus,

I(x) ∼∞∑

n=0

(−1)nn!xn,

as x→ 0. Why bother with the “∼” notation? Can’t we let N →∞ and write:

I(x) =∞∑

n=0

(−1)nn!xn

For which values of x does the asymptotic expansion actually converge? Usethe ratio test:

limn→∞

∣∣∣∣ (n+ 1)!xn+1

n!xn

∣∣∣∣ = limn→∞

(n+ 1)x =

0, x = 0∞, x 6= 0

Thus, the asymptotic expansion actually diverges for x 6= 0. Thus,

I(x) =∞∑

n=0

(−1)nn!xn

26

is false for all x 6= 0, and not a power series representations. When we write

I(x) ∼∞∑

n=0

(−1)nn!xn

we mean only that

I(x)−N∑

n=0

(−1)nn!xn = o(xN )

for each N as x→ 0. We cannot improve the approximation by fixing x near 0and letting N →∞. But, the larger N is, the more rapidly the approximationwill improve.In maple, we get I(.03) = .9716549. WithN = 4, the approximation is .9716574.

Laplace’s Method: We consider integrals of the form:

I(λ) =∫ b

a

f(xx)e−λφ(x)dx

where a can be −∞ and b can be ∞. We want the leading order term.

We first consider the case in which φ(x) has a unique minimum at a pointc ∈ (a, b). Specifically:

• φ ∈ C2(a, b)

• φ(c) < φ(x) for a ≤ x ≤ b, x 6= c.

• φ′(c) = 0, φ′′(c) > 0.

Rewrite I(λ):

I(λ) =∫ b

a

f(x)e−λ[φ(x)−φ(c)]dx · e−λφ(c)

Let 0 < δ 1 Break I(λ) up into three integrals:

I(λ) =∫ c−δ

a

f(x)e−λ[φ(x)−φ(c)]dx·e−λφ(c)+∫ b

c+δ

(· · · )dx·e−λφ(c)+∫ c+δ

c−δ

(· · · )dx·e−λφ(c)

Note that φ(x)− φ(c) > 0 for a ≤ x ≤ c− δ, φ(x)− φ(c) > 0 for c+ δ ≤ x ≤ b.Thus, as λ→∞, the “main contribution” to I(λ) comes from a “small” intervalaround x = c. So, since we only want the leading order behavior, we drop I1and I2. So,

I(λ) = I3(λ) + (HOTS)

To the leading order,

I(λ) ∼ I3(λ) =∫ c+δ

c−δ

e−λ[φ(x)−φ(c)]f(x)dx · e−λφ(c)

27

as λ → ∞. As it is, this is not very helpful. We use the Taylor approximationfro φ(x)− φ(c), valid on the “small” interval (c− δ, c+ δ):

φ(x)− φ(c) ≈ φ′(c)(x− c) +12φ′′(c)(x− c)2 =

12φ′′(c)(x− c)2

Moreover, f(x) ≈ f(c) for c− δ < x < c+ δ, δ small. Then, we get:

I(λ) ∼∫ c+δ

c−δ

e−λφ′′(c)

2 (x−c)2dx · f(c)e−λφ(c)

as λ→∞.A Gaussian function is one of the form

g(x) = e−a(x−λ)2

where a > 0. Thus, as a→∞, ∫ µ+δ

µ−δ

makes the main contribution to the integral over the whole real line.So, returning to our original problem, to facilitate calculation, we extend the

interval of integration to all (−∞,∞). We are only adding integrals that decaymuch faster than the one we are considering. Thus, to leading order,

I(λ) ∼∫ ∞

−∞e−λφ′′(c)

2 (x−c)2dx · e−λφ(c)f(c)

By Liouville, we know: ∫ ∞

−∞e−ax2

dx =

√φ

a, a > 0

So,

I(λ) ∼ f(c)e−λφ(c)

√2π

λφ′′(c),

as λ→∞.Example: Determine the leading order behavior of

I(λ) =∫ 3π

2

π2

= e−λ cos xx4dx,

as λ→∞. Here, f(x) = x4, φ(x) = cosx, φ(c) = −1, φ′′(c) = 1, c = π:

I(λ) ∼ π4eλ

√2πλ

as λ→∞.

28

Wednesday, October 18, 2006:

Last time, we considered the case where φ(t) has a unique global minimumat c ∈ (a, b), and derived the leading asymptotic behavior of I(λ) as λ → ∞.We also assumed that f(c) 6= 0 (but, in the case of f(c) = 0, we can use ahigher-order Taylor expansion).

The main idea: the primary contribution to I(λ) comes from a “small” in-terval [c− δ, c+ δ], 0 < δ 1.

Now, let’s assume that φ has its minimum over [a, b] at t = a. Assume thatφ′(a) = 0, φ′′(a) > 0, and φ(a) < φ(t) for a < t ≤ b. We’ll assume (always)that we can approximate f and φ by Taylor polynomials( of appropriate degree)near φ’s minumum, and that f and φ are such that I(λ) is convergent for λ large.

I(λ) =∫ b

a

f(t)e−λφ(t)dt ≈∫ a+δ

a

f(t)e−λφ(t)dt

=∫ a+δ

a

f(t)e−λ[φ(t)−φ(a)]dt · e−λphi(a) =∫ a+δ

a

f(a)e−λ2 φ′′(a)(t−a)2dt · e−λφ(a)

≈∫ ∞

a

e−λ2 φ′′(a)(t−a)2dt · f(a)e−λφ(a)

= f(a)e−λφ(a)

√π

2λφ′′(a)

We have found the leading order term in an asymptotic expansion of I(λ)as λ→∞. Thus,

I(λ) ∼ f(a)e−λφ(a)

√π

2λφ′′(a)

as λ→∞.

Note: The derivation we used to obtain the leading order terms for I(λ)is valid on infinite ((−∞,∞)) and semi-infinite ((−∞, a], [b,∞)) intervals also.You only need to be a little more careful about the convergence of I(λ).

5.1 The Gamma Function Γ(x)

For x > 0, the gamma function is

Γ(x) =∫ ∞

0

tx−1e−tdt.

Γ(12) =

∫ ∞

0

e−t

√tdt

29

Set u =√t, du = 1

2√

tdt,

= 2∫ ∞

0

e−u2du = 2

12√π =

√π

Γ(1) =∫ ∞

0

e−tdt = 1.

The factorial property of Γ(x). Let x > 0;

Γ(x+ 1) =∫ ∞

0

txe−tdt

Let tx = u, dv = e−tdt, du = xtx−1dt, v = −e−t. Then,

−txe−t∣∣t=∞t=0

+ x

∫ ∞

0

e−ttx−1dt.

30