ε = 0.1 α = 1 β = 1
x(t)
y(t)
1
0.8
0.6
0.4
0.2
0
0
1
2
3
4
t
5
6
7
8
ε = 0.1
exact
approximate
2.5
2
1.5
1
0.5
0
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
x
ε = 0.05
exact
approximate
2.5
2
1.5
1
0.5
0
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
x
ε = 0.01
exact
approximate
2.5
2
1.5
1
0.5
0
−1
−0.8
−0.6
−0.4
−0.2
0
x
0.2
0.4
0.6
0.8
1
ε = 0.0001
exact
approximate
2.5
2
1.5
1
0.5
0
−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
x
Zoom in on the interior layer.
ε = 0.0001
2
1.9
exact
approximate
1.8
1.7
1.6
1.5
1.4
1.3
1.2
1.1
1
−0.03
−0.02
−0.01
0
x
0.01
0.02
0.03
Laplace Method Formulas
Laplace’s method is used to approximate integrals of the form
Z
b
I(x) =
f (t) exφ(t) dt as x → ∞.
a
The asymptotic approximation is determined at the point(s) where φ attains its absolute maximum on [a, b].
Case: φ0 (t) 6= 0 in [a, b]
In this case, φ has no local maximums in [a, b] so the absolute maximum must occur at an endpoint.
max at t = a:
xφ(a)
I(x) ∼ − xfφ(a)
0 (a) e
max at t = b:
I(x) ∼
f (b) xφ(b)
x φ0 (b) e
provided f (a) 6= 0
provided f (b) 6= 0
Case: φ0 (t) = 0 at a unique interior point, say at t = c ∈ (a, b)
In this case φ(c) is either a local maximum or a local minimum (or an inflection point).
local min: If φ(c) is a local minimum (or an inflection point), then the absolute maximum of φ(t)
occurs at an endpoint and the above formulas can be used.
• There is one exception: If φ(t) has a local minimum at an interior point, then it may be that
φ(a) = φ(b), in which case there is a significant contribution to the value of the integral at
each endpoint. The contributions given by the above formulas must be added.
local max: If φ00 (c) < 0, then φ(c) must be the absolute maximum of φ on [a, b]. In this case, the
dominant contribution to the integral occurs at t = c.
I(x) ∼
q
2π
−x φ00 (c)
f (c)exφ(c)
provided f (c) 6= 0
Case: φ0 (d) = 0 where d is an endpoint and the absolute maximum of φ
I(x) ∼
q
π
−2x φ00 (d)
f (d)exφ(d)
provided f (d) 6= 0
Case: Interior absolute max at t = c with φ0 (c)=φ00 (c)=· · · =φ(p−1) (c)=0
1/p
I(x) ∼
2 Γ( p1 )(p!)
p[
1/p
]
−x φ(p) (c)
f (c)exφ(c)
provided f (c) 6= 0
Case: The absolute maximum occurs at multiple points: t = c1 , c2 , . . . , cn
I(x) ∼ I1 (x) + I2 (x) + · · · + In (x),
where Ik (x) denotes the contribution to the value of the integral at t = ck . Ik (x) can be determined
from the above formulas. One example of this case is when φ is periodic.
ε = 0.02
2.5
exact
approximate
2
1.5
1
0.5
0
−0.5
−1
0
0.1
0.2
0.3
0.4
0.5
x
0.6
0.7
0.8
0.9
1
ε = 0.1
1.6
exact
approximate
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
0.1
0.2
0.3
0.4
0.5
x
0.6
0.7
0.8
0.9
1
ε = 0.01
1.6
exact
approximate
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0.03
0.035
0.04
0.045
0.05
x
Zoom in on the two-ply boundary layer.
ε = 0.01
1
exact
approximate
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.005
0.01
0.015
0.02
0.025
x
Trigonometric Identities
cos2A = 12 (1 + cos 2A)
sin2A = 12 (1 − cos 2A)
cosA cosB =
1
2
[cos(A + B) + cos(A − B)]
sinA sinB = 12 [cos(A − B) − cos(A + B)]
sinA cosB = 21 [sin(A + B) + sin(A − B)]
cos3A = 14 [3 cos A + cos 3A]
sin3A = 14 [3 sin A − sin 3A]
cos2A cosB = 41 [cos(2A + B) + cos(2A − B) + 2 cos B]
cos2A sinB = 14 [sin(2A + B) − sin(2A − B) + 2 sin B]
sin2A cosB = − 14 [cos(2A + B) + cos(2A − B) − 2 cos B]
sin2A sinB = − 14 [sin(2A + B) − sin(2A − B) − 2 sin B]