Singular Perturbation Theory: An Introduction with Examples, Slides of Mathematics

Download Singular Perturbation Theory: An Introduction with Examples and more Slides Mathematics in PDF only on Docsity! 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 Outline 1 Introduction 2 Perturbation Theory 3 Singular Perturbation Theory 4 Example Boundary Layer Outer Expansion Inner Expansion Matching Composite Approximation Analysis 5 Conclusion Erika May (Occidental College) Introduction to Singular Perturbation Theory February 25, 2016 2 / 24 Asymptotic Expansion Straightforward asymptotic expansion: (i) Assume solutions of given function can be asymptotically expanded in ε using power series: y = y0(x) + εy1(x) + ε2y2(x) + · · ·+ yn(x)εn +O(εn+1) (ii) Substitute expansion into original function (iii) Isolate zeroth order terms and solve Erika May (Occidental College) Introduction to Singular Perturbation Theory February 25, 2016 5 / 24 Singular Perturbation Theory Boundary layer problems Interval of rapid change Straightforward expansion using does not satisfy all boundary conditions Method of matched asymptotic expansions Construct separate asymptotic expansions for inside and outside of boundary layer and create composite approximation Erika May (Occidental College) Introduction to Singular Perturbation Theory February 25, 2016 6 / 24 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 Erika May (Occidental College) Introduction to Singular Perturbation Theory February 25, 2016 7 / 24 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) = y0(x) + εy1(x) +O(ε2) ⇓ ε(y ′′0 + εy ′′1 + . . . ) + 2(y ′0 + εy ′1 + . . . ) + (y0 + εy1 + . . . ) = 0 Since boundary layer is at x = 0 and we’re evaluating the outer region, impose boundary condition y(1) = 1 on expansion: 2y ′0 + y0 = 0 y0(1) = 1 Erika May (Occidental College) Introduction to Singular Perturbation Theory February 25, 2016 10 / 24 Example: Outer Expansion Solve linear first-order ODE: y0(x) = cesx 2s + 1 = 0 y0(x) = e 1 2 (1−x) Denote outer expansion as youter : youter = e 1 2 (1−x) Erika May (Occidental College) Introduction to Singular Perturbation Theory February 25, 2016 11 / 24 Example: Inner Expansion εy ′′ + 2y ′ + y = 0 y(0) = 0, y(1) = 1 To construct an inner expansion, rescale the narrow boundary layer using a stretching variable: X = x δ(ε) Seek inner solution: Y (X , ε) = y(x , ε) Chain rule gives us: y ′ = dy dx = dY dx = dY dX dX dx = 1 δ dY dX = 1 δ Y ′ ⇒ y ′′ = 1 δ2 Y ′′ Our original differential equation becomes: ε δ2 Y ′′ + 2 δ Y ′ + Y = 0 Erika May (Occidental College) Introduction to Singular Perturbation Theory February 25, 2016 12 / 24 Example: Inner Expansion From dominant balance, we can let: δ(ε) = ε, and so X = x ε New scaled differential equation: ε ε2 Y ′′ + 2 ε Y ′ + Y = 0 ⇓ Y ′′ + 2Y ′ + εY = 0 Y (0) = 0 Erika May (Occidental College) Introduction to Singular Perturbation Theory February 25, 2016 15 / 24 Example: Inner Expansion Construct expansion: Y (X , ε) = Y0(X ) + εY1(X ) +O(ε2) ⇓ (Y ′′0 + εY ′′1 + . . . ) + 2(Y ′0 + εY ′1 + . . . ) + ε(Y0 + εY1 + . . . ) = 0 Impose boundary condition at X = 0: Y ′′0 + 2Y ′0 = 0 Y0(0) = 0 Erika May (Occidental College) Introduction to Singular Perturbation Theory February 25, 2016 16 / 24 Example: Inner Expansion Solve second order ODE: Y0(X ) = c1e s1X + c2e s2X s2 + 2s = 0 Y0(X ) = c(1− e−2X ) Denote inner expansion as Yinner : Yinner = c(1− e−2X ) Erika May (Occidental College) Introduction to Singular Perturbation Theory February 25, 2016 17 / 24 Example: Composite Our composite approximation follows: ycomposite = yinner + youter − yoverlap Matching condition showed us yoverlap = e1/2, so: ycomposite = [e1/2(1− e−2x/ε)] + [e1/2(1−x)]− e1/2 ⇓ ycomposite = e 1−x 2 − e ε−2x 2ε Erika May (Occidental College) Introduction to Singular Perturbation Theory February 25, 2016 20 / 24 Example: Analysis Figure 3: Exact solution (red) and composite approximation (blue) at ε = 0.2 Figure 4: Exact solution (red) and composite approximation (blue) at ε = 0.01 Erika May (Occidental College) Introduction to Singular Perturbation Theory February 25, 2016 21 / 24 Conclusion To recap: Singular perturbation Boundary layer problems Method of matched asymptotic expansions Applications Navier-Stokes Equation: ∂u ∂t + u · ∇u = −1 ρ ∇p + ε∇2u ∇ · u = 0 ε = 1 Re Boundary layer theory Figure 5: Boundary layer flow Erika May (Occidental College) Introduction to Singular Perturbation Theory February 25, 2016 22 / 24