Introduction to Singular Perturbation Theory
Erika May
Department of Mathematics
Occidental College
February 25, 2016
Erika May (Occidental College) Introduction to Singular Perturbation Theory February 25, 2016 1 / 24
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Singular Perturbation Theory: An Introduction with Examples, Slides of Mathematics
An introduction to Singular Perturbation Theory, a mathematical method used to solve problems where the parameter is small but nonzero and the problem's nature changes qualitatively. the concept of perturbation theory, the difference between regular and singular perturbations, and the use of asymptotic expansion, boundary layer theory, and matched asymptotic expansions to find solutions. An example of a second-order, linear, constant coefficient ODE is used to illustrate the concepts.
What you will learn
- What is the role of Boundary Layer Theory in Singular Perturbation Theory?
- How is the exact solution of a singular perturbation problem found?
- What is the significance of the matching condition in Singular Perturbation Theory?
- What is Singular Perturbation Theory and how does it differ from Regular Perturbation Theory?
- How is the Method of Matched Asymptotic Expansions used to solve Singular Perturbation Problems?
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Introduction to Singular Perturbation Theory
Erika May
Department of Mathematics Occidental College
February 25, 2016
Outline
1 Introduction
(^2) Perturbation Theory
(^3) Singular Perturbation Theory
(^4) Example Boundary Layer Outer Expansion Inner Expansion Matching Composite Approximation Analysis (^5) Conclusion
Perturbation Theory
Regular perturbation example: x^2 − x + ε = 0 Exact solution: x =^1 ±
1 − 4 ε 2 Let ε = 0: x^2 − x = 0 ⇒ x = 0, 1 Singular perturbation example: εx^2 + 2x + 1 = 0 Exact solution: x = −^2 ±
4 − 4 ε 2 ε Let ε = 0: 2 x + 1 = 0 ⇒ x = −
Asymptotic Expansion
Straightforward asymptotic expansion: (i) Assume solutions of given function can be asymptotically expanded in ε using power series: y = y 0 (x) + εy 1 (x) + ε^2 y 2 (x) + · · · + yn(x)εn^ + O(εn+1) (ii) Substitute expansion into original function (iii) Isolate zeroth order terms and solve
Example
Motivating example: boundary value problem of second-order, linear, constant coefficient ODE
εy ′′^ + 2y ′^ + y = 0, x ∈ (0, 1)
y (0) = 0, y (1) = 1 ⇒ This is a singular perturbation problem
Example: Exact Solution
εy ′′^ + 2y ′^ + y = 0 y (0) = 0, y (1) = 1 Characteristic polynomial: εs^2 + 2s + 1 = 0
s 1 = −1 +^
1 − ε ε
, s 2 = −^1 −
1 − ε ε Thus, the general solution will be: y (x, ε) = c 1 es^1 x^ + c 2 es^2 x where c 1 , c 2 are arbitrary constants. Imposing the boundary conditions at x = 0 and x = 1, the solution is:
y (x, ε) = e
s 1 x (^) − es 2 x es^1 − es^2
Example: Outer Expansion
εy ′′^ + 2y ′^ + y = 0 y (0) = 0, y (1) = 1 Outer region varies slowly (unperturbed), so proceed with straightforward expansion: y (x, ε) = y 0 (x) + εy 1 (x) + O(ε^2 ) ⇓ ε(y 0 ′′ + εy 1 ′′ +... ) + 2(y 0 ′ + εy 1 ′ +... ) + (y 0 + εy 1 +... ) = 0 Since boundary layer is at x = 0 and we’re evaluating the outer region, impose boundary condition y (1) = 1 on expansion:
2 y 0 ′ + y 0 = 0
y 0 (1) = 1
Example: Outer Expansion
Solve linear first-order ODE:
y 0 (x) = cesx
2 s + 1 = 0
y 0 (x) = e
(^12) (1−x)
Denote outer expansion as youter :
youter = e
(^12) (1−x)
Example: Inner Expansion
ε δ^2 Y^
′′ +^2
δ Y^
′ + Y = 0
After rescaling, we must determine correct two-term dominant balancing of terms. We have three coefficients: ε δ^2
δ
Two options: (a)
ε δ^2 and 1 are of the same magnitude and dominant over
δ (b)
ε δ^2 and
δ are of the same magnitude and dominant over 1
Example: Inner Expansion
ε δ^2
δ
Outcomes: (a) (^) δε 2 ∼ 1 implies δ(ε) = O(√ε):
1 √^2
ε
Since ε is small, so no dominant balance (b) (^) δε 2 ∼ (^2) δ implies δ(ε) = O(ε):
1 ε
ε 1
Example: Inner Expansion
Construct expansion:
Y (X , ε) = Y 0 (X ) + εY 1 (X ) + O(ε^2 )
⇓ (Y 0 ′′ + εY 1 ′′ +... ) + 2(Y 0 ′ + εY 1 ′ +... ) + ε(Y 0 + εY 1 +... ) = 0 Impose boundary condition at X = 0:
Y 0 ′′ + 2Y 0 ′ = 0
Y 0 (0) = 0
Example: Inner Expansion
Solve second order ODE:
Y 0 (X ) = c 1 es^1 X^ + c 2 es^2 X
s^2 + 2s = 0 Y 0 (X ) = c(1 − e−^2 X^ ) Denote inner expansion as Yinner :
Yinner = c(1 − e−^2 X^ )
Example: Matching
Determine unknown constant c by ”matching” inner and outer solution, given by the matching condition:
lim X →∞
Yinner (X ) = lim x→ 0 +^
youter (x)
Xlim →∞ c(1^ −^ e−^2 X^ ) =^ xlim→ 0 +^ e^
(^12) (1−x)
Which implies: c = e^1 /^2 = yoverlap
Final inner expansion, with y (x, ε) = Y (X , ε) and X = x ε
yinner = e^1 /^2 (1 − e−^2 x/ε)
Example: Composite
Our composite approximation follows:
ycomposite = yinner + youter − yoverlap
Matching condition showed us yoverlap = e^1 /^2 , so:
ycomposite = [e^1 /^2 (1 − e−^2 x/ε)] + [e^1 /2(1−x)] − e^1 /^2
⇓ ycomposite = e 1 −^2 x− e ε−^2 ε^2 x