An introduction to perturbation methods applied to industrial mathematics

Transcription

1 An introduction to perturbation methods applied to industrial mathematics Tim Myers Centre de Recerca Matemàtica, Bellaterra, Barcelona, Spain December 10th 2014

2 Perturbation methods Begin with a simple example... Consider the quadratic equation x 2 2ǫx 1 = 0 ǫ 1 Exact solution Since ǫ 1 x = 2ǫ± 4ǫ = ǫ± 1+ǫ 2 ǫ22 ǫ48 x ǫ± (1+ + ) = ±1+ǫ± ǫ2 2 ǫ4 8 +O(ǫ6 )

3 Perturbation methods x ±1+ǫ± ǫ2 2 ǫ4 8 +O(ǫ6 ) Take ǫ = 0.1 and the positive root we find x 0 = 1 x 1 = 1.1 x 2 = x 3 = Exact x = , i.e. in 3 terms we have 3 decimal place accuracy. This improves as ǫ 0. What if we try a series solution from the start?

4 Perturbation methods Let x = x 0 +ǫx 1 +ǫ 2 x 2 +O(ǫ 3 ) Substitute into the original equation (x 0 +ǫx 1 +ǫ 2 x 2 ) 2 2ǫ(x 0 +ǫx 1 +ǫ 2 x 2 ) 1 = 0 Gather together terms of the same order x ǫ(2x 0x 1 2x 0 )+ǫ 2 (x x 0x 2 2x 1 ) 2 +O(ǫ 3 ) = 0 Solve at each order of ǫ Hence ǫ 0 : x = 0 x 0 = ±1 ǫ 1 : 2x 0 x 1 2x 0 = 0 x 1 = 1 ǫ 2 : x x 0x 2 2x 1 = 0 x 2 = ±1/2 x = ±1+ǫ± ǫ2 2 +O(ǫ3 ) = expansion of the exact solution

5 What is perturbation theory? In general we would not use perturbation techniques to solve a quadratic equation. Wikipedia: Perturbation Theory Perturbation theory comprises mathematical methods that are used to find an approximate solution to a problem which cannot be solved exactly, by starting from the exact solution of a related problem (in the previous example we first solved x 2 = 1). Perturbation theory is applicable if the problem at hand can be formulated by adding a small term to the mathematical description of the exactly solvable problem. Note, the terms perturbation and asymptotic expansion are often used interchangeably (see Howison P177)

6 What is perturbation theory? Bender & Orszag Perturbation theory is a large collection of iterative methods for obtaining approximate solutions to problems involving a small parameter ǫ. These methods are so powerful that sometimes it is advisable to introduce a small parameter temporarily into a difficult problem... finally to set ǫ = 1 to recover the original problem The thematic approach of perturbation theory is to decompose a tough problem into an infinite number of relatively easy ones. So, once we understand how to perturb a simple problem, we may start to apply the technique to more difficult ones...

7 Order notation The notation O() may (amongst other things) describe the error term in an approximation to a mathematical function. The most significant terms are written explicitly, and then the least-significant terms are summarized in a single O() term. For example, Indicates that exp(x) = 1+x + x2 2 +O(x3 ) asx 0 exp(x) (1+x + x2 2 ) < C x3 as x 0

8 A simple ODE example Consider the differential equation with a small parameter Exact solution du dt u = +ǫu = t u(0) = 1. (1+ 1ǫ 2 ) e ǫt + t ǫ 1 ǫ 2. Expand exponential for small ǫt u (1+ 1ǫ )(1 ǫt 2 +ǫ 2t2 = 1+ t2 2 ǫ 2 ǫ3t3 6 ) (t + t3 +O(ǫ 2 ). 6 ) + t ǫ 1 ǫ 2

9 ODE example II Alternatively, look for series solution u = u 0 +ǫu 1 + d dt (u 0 +ǫu 1 + )+ǫ(u 0 +ǫu 1 + ) = t. Boundary condition becomes u 0 (0)+ǫu 1 (0)+ = 1 Equating coefficients of ǫ u 0 (0) = 1 u 1 (0) = u n (0) = 0 Equating coefficients in governing equation and applying BCs etc. ǫ 0 : ǫ 1 : du 0 dt = t u 0 = t du 1 dt +u 0 = 0 u 1 = t t3 6

10 ODE example III As expected asymptotic solution = expansion of exact solution ( ) u = 1+ t2 2 ǫ t + t3 +O(ǫ 2 ) u(t) ǫ = 0.1 Leading order Exact 1.2 First order t Since ǫ = 0.1 expect errors of 10% at leading order. At first order we neglect terms of O(ǫ 2 ) and therefore expect errors of 1%. From the figure it is clear that the first order solution is more accurate than the leading order, and the accuracy could be improved by looking for higher order corrections. Accuracy will improve as ǫ decreases.

11 Small parameter? In practice a problem is seldom presented with a nice small parameter. We usually have to nondimensionalise or examine the physics Hence we now divert to nondimensionalisation...

12 Nondimensionalising birds Consider a population initially of 1000 birds on a remote island, the maximum number of birds the resources can sustain (the carrying capacity) is After one month the population has increased to 1125 birds. Population modelled by a logistic equation dx dt = r(1 x/k)x where x is the number of birds, r the growth rate and K the carrying capacity Introduce dimensionless variables ˆx = x/x, ˆt = t/t x and t are (constant) typical values of x and t Choose x = K d(x ˆx) d(tˆt) = r ( 1 xˆx K dˆx dˆt = rt(1 ˆx)ˆx ) xˆx Choose t = 1/r Having scaled x and t with typical values expect ˆx and its derivatives to have size around unity (O(1)) Initial condition x(0) = 1000 ˆx(0) = ˆx 0 = 1000/6000 = 1/6 Problem reduced from one depending on three parameters to depending on one parameter

13 Nondimensionalising birds Solution is ˆx = ˆx 0 ˆx 0 +(1 ˆx 0 )e ˆt Dimensional result replace ˆt with t/t = rt and ˆx with x/x = x/k x K = x 0 x 0 +(K x 0 )e rt. Value of r = 0.14 found by setting x = 1125 at t = x(t) x0 = x0 = x0 = 0.2 t Non-dimensional bird population for ˆx 0 = 0.2,0.8,1.5

14 Nondimensionalising birds Another possible scaling based on initial condition Could scale x with x 0 ˆx = x K or ˆx = x x 0 Scaling not always unique (and not always obvious) First scaling if interest lies in long time behaviour Second for small times, when x is close to x 0 Setting x = x 0 ( dˆx dˆt = 1 ˆx ) ˆx α where α = K/x 0 The governing equation has a parameter α, but the initial condition is simply ˆx(0) = 1 Now we know how to nondimensionalise, lets apply it to a real problem

15 Football motion z r(t) x y Newton s 2nd law F = mẍ where F = mg+f d +F l and F d = 1 2 ρa v 2 C d v F l = 1 2 ρa v 2 C l σ v m is the mass of the ball, A is its cross-sectional area, ρ is the density of air, C are drag coefficients, v = v/ v v = (ẋ 2 +ẏ 2 +ż 2 )

16 Two-dimensional equations of motion Focus on 2D in x y plane (to cut down on algebra) ẍ = v { k d ẋ k l sinγẏ } ÿ = v { k d ẏ +k l sinγẋ } k d = ρac d /2m, k l = ρac l /2m, γ is angle of spin axis Initial conditions x(0) = y(0) = 0 ẋ(0) = 0, ẏ(0) = v

17 Nondimensionalisation ẍ = v { k d ẋ k l sinγẏ } Scale with typical value ˆx = x/l 1,ŷ = y/l 2,ˆt = t/τ L 2 distance of the free kick - 20m (also possible L 2 = 1/k d 100m) τ L 2 /v time taken for the ball to travel the distance L 2 1s L 1 unknown L 2 1 ˆv = ˆx 2 τ 2 + L2 2 ŷ 2 τ 2 = L 2 ŷ 1+ L2 1 ˆx 2 τ L 2 2 ŷ 2 ˆx = ŷ 1+ L2 1 L 2 2 ˆx 2 ŷ 2 ( ) k d L 2 ˆx k l sinγ L2 2 ŷ L 1 Indicates L 1 = k l sinγl 2 2 taking k l = 0.013,γ = π/2,l 2 = 20 L 1 = 5.2m Denote ǫ = k d L 2 ( 0.3 for the given parameter values) L 1 /L 2 = 5.2/20 = 0.26 L 1 /L 2 = aǫ

18 Perturbation solution Governing equations ˆx = ŷ ŷ = ŷ 1+a 2 ǫ ˆx 2 2 ( ǫ ˆx ŷ ŷ ) 2 1+a 2 ǫ 2 ˆx 2 ŷ 2 ( ǫ ŷ +bǫ 2 ˆx ) Note, RHS of y equation is O(ǫ) Now drop hat notation Initial conditions become x(0) = y(0) = 0, ẋ(0) = 0, ẏ(0) = 1 Look for series solution x = x 0 +ǫx 1 +ǫ 2 x 2 +ǫ 2 x 3 +, y = y 0 +ǫy 1 +ǫ 2 y 2 +ǫ 3 y 3 +

19 Perturbation 2 Things get messy... ẍ 0 + ǫẍ 1 + = (ẏ 0 + ǫẏ 1 + ) Expand and collect terms 1+a 2 ǫ 2(ẋ 0 +ǫẋ 1 + ) 2 (ẏ 0 +ǫẏ 1 + ) 2 (ǫ(ẋ 0 +ǫẋ 1 + ) (ẏ 0 +ǫẏ 1 + )) O(ǫ 0 ) : ẍ 0 = ẏ 2 0 O(ǫ) : ẍ 1 = ẏ 0 (ẋ 0 ẏ 1 ẏ 0 ẏ 1 ) need y solution Recall RHS of y equation was O(ǫ) O(ǫ 0 ) : ÿ 0 = 0 O(ǫ) : ÿ 1 = ẏ 2 0 Leads to y = t ǫ t2 2 +ǫ2 (2 b) t3 6 ǫ3 (6+a 2 7b) t4 24 +O(ǫ4 ) x = t2 2 ǫt3 2 +ǫ2 (11 +a 2 2b) t4 24 ǫ3 (50 +15a 2 25b) t O(ǫ4 )

20 Perturbation versus numerics X Y Z T Non-dimensional three-dimensional trajectories of X, Y and Z against T: full numerical solution (solid line), O(ǫ 0 ) solution (dotted line), O(ǫ) solution (dot-dashed line) and O(ǫ 2 ) solution (dashed line). Parameter values are C d = 0.3, C s = 0.25 and C l = 0.1.

21 Comparison of results Physics often clearer in dimensional form x x/l 1 = x/(k l L 2 2 sinγ), y y/l 2, t tv/l 2 x = k l sinγ(vt) 2 2 [ { kd vt y = vt 1 2 [ 1 {k d vt g2 t 2 }] +O(ǫ 2 ) Analytical solution makes the important factors clear First term is most important So, how to get swerve 6v 2 }] +O(ǫ 2 ) large spin ( make γ high) kick ball hard ( make v high) do it from far away ( allow t to get high - but beware ẏ then decreases)... swerve goals Increase k l = ρac l /2m Note, first order must be leading order. Here this imposes a time restriction. For y equation t 2/(k d v). To carry solution further may require Method of Multiple Scales

22 Perturbation versus experiment 10 8 y (m) t (s) z (m) t (s) Comparison of experimental data of Carre et al 2002 for a two-dimensional kick in y z plane (no swerve) (asterisks) and perturbation (solid line) with θ 20, (ẏ(0),ż(0)) (17.59,6.29), k d = 0.008, k l =

23 So, was it useful? Drag coefficients k proportional to air density, so Coastal teams get high drag and expect swerve, high altitude teams do not High altitude team choose ball that swerves the least (i.e. smooth), low altitude team choose ball that swerves the most (rough) to confuse opposition This information was given to a South African premiership team, Bidvest Wits, who play at high altitude Before 2nd Feb 2011 Wits had gone eight games without a win Recent results in South African premiership (since meeting): 6 Feb Wits 6-0 Vasco da Gama; 19 Feb, Wits 3-0 Mpumalanga; 16 Feb, Wits 3-1 Santos; 22 Feb, Free State 2-2 Wits; 26 Feb, Sundowns 2-0 Wits; 2 March, Wits 2-0 Ajax Cape Town; 5 March, AmaZulu 1-1 Wits; 16 March Wits 3-2 Golden Arrows; Did not lose a home game all season after the meeting (away games not so good) Moved from 13th to 6th

24 Laser drilling A one-dimensional model for laser drilling (see Andrews and Athey 1975, Crank 1984, Fulford 2002, Mitchell & Myers 2008) The laser heats the material up to such a high temperature that it turns from a solid to a vapour (sublimation). Equivalent to melting of heat shields on space vehicles

25 Mathematical model One-phase Stefan problem T t = T α 2 x 2 ds ρl s dt = k T x + W x=s A, where... Subject to T(s,t) = T s T(x,0) = T(,t) = T s(0) = 0

26 Nondimensionalisation T T T = T T T s T T s s L t t τ L,τ unknown T t = ατ 2 T L 2 x 2 ds dt = kτ T T ρl sl 2 x + Wτ x=s AρL sl Heat equation suggests τ = L 2 /α Sublimation driven by the laser so L = αaρl s/w T t = 2 T x 2 ds dt = ǫ T x +1 x=s where the ǫ = c T/L s T(s,t) = 1 T(x,0) = T(,t) = 0 s(0) = 0 Typical values for metals ǫ 0.2

27 Laser perturbation Take T = T 0 +ǫt 1 + s = s 0 +ǫs 1 + Take care with BC at x = s(t) T(s,t) = T 0 (s 0 +ǫs 1 +,t)+ǫt 1 (s 0 +ǫs 1 +,t)+ Equations to first order = T 0 (s 0,t)+ǫs 1 T 0 (s 0,t) x +ǫt 1 (s 0,t)+ = 1, T 0 t +ǫ T 1 t ds 0 dt +ǫds 1 dt = 2 T 0 T 1 x 2 +ǫ 2 x 2 = ǫ T 0 x +1 x=s0 Note s 0 = t gives position of the boundary (to leading order).

28 Laser perturbation Now focus on melt front change the co-ordinate system We know front approximately at x = s 0 = t, so take ξ = x s 0 (t) = x t T 0 (x,t) = θ 0 (ξ(x,t),t) The leading order heat equation becomes T 0 t = 2 T 0 θ 0 x 2 t θ 0 ξ = 2 θ 0 ξ 2 with θ 0 (0,t) = 1 θ 0 (,t) = θ 0 (ξ,0) = 0 We may solve this problem exactly using Laplace transforms or get a simpler solution by looking at large times By large times we mean t = r/ǫ hence ǫθ 0r θ 0ξ = θ 0ξξ In the limit ǫ 0 we obtain the steady state

29 Laser perturbation Steady-state problem 2 θ 0 ξ 2 + θ 0 ξ = 0 θ 0 = A+Be ξ Boundary conditions θ 0 = 1,0 at ξ = 0, gives θ 0 = e ξ or T 0 = e (x t) Now use this to correct position of moving boundary Hence, to first order ds 1 dt = T 0 x = e (x t) x=s0 = 1 x=s0 ( s = (1 ǫ)t Dimensional s = 1 ) c(ts T ) Wt L s AρL s Leading order perturbation indicates energy from the laser is nearly all used to change phase First order some heat conducted away from moving front, leaving less heat for phase change and slowing down the process Sublimation rate power W/A and physical parameters

30 Singular perturbation Of course things don t always work out so well... Recall original example which gave x 2 2ǫx 1 = 0 ǫ 1 Now consider x = ±1+ǫ± ǫ2 2 ǫ4 8 +O(ǫ6 ) ǫx 2 2x 1 = 0 ǫ 1 and let x = x 0 +ǫx 1 +

31 Singular perturbation Now consider ǫ(x 0 +ǫx 1 + ) 2 2(x 0 +ǫx 1 + ) 1 = 0 ǫ 1 Giving the series of problems ǫ 0 : 2x 0 1 = 0 x 0 = 1/2 ǫ 1 : x0 2 2x 1 = 0 x 1 = 1/8 etc x = 1/2+ǫ/8+O(ǫ 2 ) i.e. we only have a single solution to a quadratic The exact solution is x = 1± 1+ǫ ǫ 8, 2 ǫ ǫ 8

32 Singular perturbation Why only picking up first solution? Second solution is x = O(1/ǫ), so our assumption x = x 0 +ǫx 1 + that indicates x 0 = O(1) is incorrect Rescale x = X/ǫ ǫx 2 2x 1 = 0 X = X 0 +ǫx 1 +O(ǫ 2 ) and look at initial terms ǫ X2 ǫ 2 2X ǫ 1 = 0 (X 0 +ǫx 1 ) 2 2(X 0 +ǫx 1 ) ǫ = 0 X = ǫ/2+o(ǫ 2 ), 2+ǫ/2+O(ǫ 2 ) x = 1/2+O(ǫ), 2/ǫ+1/2 +O(ǫ) ǫ 0 : X 2 0 2X 0 = 0 X 0 = 0,2 ǫ 1 : 2X 1 (X 0 1) = 1 X 1 = 1/2,1/2 Often get singular perturbation when small parameter multiplies highest order/derivative term

33 Conclusion Perturbation methods provide powerful technique for solving problems with no exact solution Analytical solutions make physics clear in a way numerics cannot Applicability increases when used in conjunction with other techniques Nondimensionalisation Change of co-ordinate system Rescaling near boundary layers And when things get tricky... Method of multiple scales Matched asymptotic expansions Singular perturbation theory

34 Further reading The pull-off test for viscoelastic soft solids for Unilever 2009 (Atomic force microscopy) Dynamical Models of Extreme Rolling of Vessels in Head Waves for MARIN 2009 (Ship movement) Homogenization of the Equations Governing the Flow Between a Slider and a Rough Spinning Disk for Hitachi 2009 (Modelling hard drives) Spin-coating on nanoscale topography and phase separation of diblock copolymers for CRANN 2008 (Nanofabrication) Sensitivity of Markov chains for wireless protocols for BT 2007 (Wireless networks) Myers & Mitchell, A mathematical analysis of the motion of an in-flight soccer ball. Sports Engng 2013.

35 Further reading S. Howison, Practical Applied Mathematics, Cambridge University Press, Cambridge, 2005 C.M. Bender & S.A. Orszag Advanced Mathematical Methods for Scientists and Engineers: Asymptotic Methods and Perturbation Theory. Springer 1978 R.S. Johnson Singular Perturbation Theory. Springer E.J. Hinch Perturbation Methods. Cambridge 1991.