Asymptotic and Perturbation Methods: Lecture Notes.

Transcription

1 Asymptotic and Perturbation Methods: Lecture Notes. Blerta Shtylla University of Utah Mathematics Department Fall

2 Lecture 1 and Lecture 2 Week of August 26, 2008 Introduction: For many problems it is more advantageous to build approximations to solutions rather than solving the exact problem. This is especially advantageous in applied problems where certain approximations allow for solutions that are easier to interpret physically. In math biology we encounter many problems in which we might want to ignore some effects. We need to make sure that if we have to throw away terms the effect is not significant. Perturbation methods give us a way to study how certain approximations affect our models. All the problems we consider in this course have a small parameter. There are two classes of problems we will consider in this course: 1. Regular Perturbation Problems 2. Singular Perturbation Problems (nonuniformities) (a) Boundary Value Problems (b) Initial Value Problems Ex: dx dt = y x dy dt = y x(0) = 1, y(0) = 1 Notice that the solution x tracks y but it is delayed. So there is a difference in time scales. If we take = 0 the problem is inconsistent! Need a way to approximate the solution. 2

3 (c) Oscillatory Processes with multiple time scales (d) Microscale Problems (homogenization) Ex: Diffusion through media with a lot of micro-structure. Let us now consider a typical perturbation problem from physics. Example 1: Projectile Motion Consider an object projected radially upward from the surface of the Earth with initial velocity v 0. From our intro to physics course we usually write the following equation for the position of the object at a given time t: m d2 x dt 2 = f = mg (1) x(0) = 0 (2) x (0) = v 0 (3) with g the gravitational constant and m the mass of the object. Such a system can be easily solved to obtain, x(t) = 1 2 gt2 + v 0 t. (4) However these equations assume that the distance travelled from the object is much smaller than the radius of the Earth. A more precise statement of our problem reads: m d2 x dt 2 = GM em (x + R) 2 (5) m d2 x = mgr2 dt 2 (x + R) 2 (6) d 2 x = gr2 dt 2 (x + R). 2 (7) Notice that this problem reduces to the first one if x is much smaller than R however if that is not the case we are left with a nonlinear ode. In order to solve this problem we first rescale the variables. Let x = αy and t = στ, with y, τ dimensionless. Pick α = R and substitution yields, 3

4 If we choose σ = R g Rd 2 y = gr2 σ 2 dτ 2 (R + Ry) 2 (8) d 2 y = gσ2 /R dτ 2 (1 + y) 2 (9) y(0) = 0 (10) y (0) = σ R v 0. (11) the problem reduces to : d 2 y 1 = dτ 2 (1 + y) 2 (12) y(0) = 0 (13) y (0) = v 0. Rg (14) If we pick = v 0 Rg as our small parameter we are in trouble. This is because in the case that = 0 the solutions to the problem can take negative values which is not physically acceptable(negative height!). Let us try another choice of scales. We start by substituting x = αy and t = στ directly into the equation, α d 2 y g = σ 2 dτ 2 (1 + αy R )2 (15) d 2 y gσ 2 = dτ 2 α(1 + αy R )2 (16) y(0) = 0 (17) y (0) = σv 0 α. (18) We now pick the scales by setting gσ2 = 1 and σv 0 = 1 which gives σ = v 0 α α g 4

5 and α = v2 0 g. Substitution transforms our original equations into: d 2 y 1 = (19) dτ 2 (1 + y) 2 y(0) = 0 (20) y (0) = 1 (21) with = v2 0 Rg dimensionless. For R = 4000mi then v 0 s 2 ft 2, thus if v 0 is smaller than 10 3 then is small and the problem can be reduced to a linear ode. This would imply that for small the solution x(t) = 1 2 gt2 + v 0 t is a reasonable approximation to the solution of the projectile motion. We will later be able to estimate just how well our approximate solution is for the nonlinear ode. Approximations The typical approximation for any function is a power series f(x) f n (x) = f(0) + f (0)x + f (0) x R 2 n (x n ). How close is this expansion to the actual function? There are two ways of measuring close 1. Convergence: We say that f(x) converges to f n (x) if lim n f(x) f n (x) = 0 for all x with x > R. Notice that in order to get a good approximation here we might need to include a lot of terms. 2. Asymptotic: We say that f(x) is asymptotic to f n (x) if lim x x n (f(x) f n (x)) = 0 for n fixed. Notice that such an approximation does not ask for convergence thus there is no need to require a lot of terms. In fact many asymptotic series are divergent. Order notation: Let there be two functions f(x) and φ(x)(gauge function) defined in some interval Ω of the real numbers then, a) We say that f(x) is of the order of φ(x) as x x 0 denoted f(x) = O(φ(x)) if there is a positive constant A > 0 and a neighborhood U of x 0 so that f(x) < A φ(x) for all x U Ω. Example: Consider f(x) = x 2 + x 3 + x 4. Then f(x) = O(x 2 ) for all x < 1 since x 2 (1 + x + x 2 ) < 3x 2. 5

6 b) We say that f(x) = o(φ(x)) if for any > 0 there is a neighborhood U of x 0 so that f(x) φ(x) for all x U Ω. Example: Consider f(x) = x 2 ln(x) then f(x) = o(x) since f(x) x = x ln(x) 0 for small enough x. Order Notation Examples: 1. e 1 = O( n ) for all n i.e it is transcedentally small (take using L Hopital lim 0 e 1 = lim 0 e Let f(x) = x + e x then f(x) x = O( n ) for x 0. Notice that in this case lim 0 (lim x 0 f(x, )) lim x 0 (lim 0 f(x, )) so this is essentially a singular function. Definition(Asymptotic Sequences): A sequence of gauge functions {φ j (x)} j=0 is an asymptotic sequence as x x 0 if φ j+1 = o(φ j ) as x x 0 for j = 0, 1, 2... Examples of Asymptotic sequences: 1. φ(x) = x : 1, x, x 2,...x n as x φ(x) = 1 x : 1, 1/x, 1/x2,...1/x n+1 as x. 3. φ(x) = e n x : 1, e 1 x...e n x as x , x ln(x), x, x 2 ln(x)... as x 0. Definition(Asymptotic Representations): A function f(x) has an asymptotic representation f(x) n k=0 a kφ k (x) = f(x) as x x 0 if for each fixed n, lim f(x) f n (x) x x 0 φ n (x) = 0. (22) Perturbation Approach to Estimating Solutions of Algebraic Equations The simplest examples of perturbation methods when approximating solutions are the ones related to algebraic equations. We will start off by looking to approximate solutions to simple equations. The advantage to such examples is that for the simple cases we will know the exact solutions so we can get a feel for how well perturbation methods can do. Let us start with a few examples. 6

7 Example: 1. Find an approximate solution to the equation x = 0. Recall here that we want to approximate the solutions. Perturbation methods always involve problems with a small parameter. So let us state our problem in a way that would more naturally lead to perturbation methods. Let us consider instead x 2 (1 + ) = 0. Notice here that we know exactly how small (our small parameter) is. In order to tackle this problem we can think of the variable x as a function of so that x = x() so that around 0 one can taylor expand x() = x(0) + x (0) +... or in the context of our definitions x() = a 0 + a 1 + a Substituting into the original expression the expansion for x we obtain, (a 0 + a 1 + a ) 2 1 = 0 (23) a a 0 a 1 + a a 0 a = 0 (24) Equating same order terms we obtain: a = 0 (25) 2a 0 a 1 1 = 0 (26) 2a 0 a 2 + a 2 1 = 0. (27) It is now easy to see that the terms in the expansion for x are a 0 = ±1, a 1 = ± 1 2 which produces the solution x = ±1 ± 2. Recall however that we chose =.01 thus the approximate solution is x = ± Compare this with the exact solution, we did pretty well! Notice that we were able to solve this problem by using a taylor expansion for x. We call such problems regular perturbation expansions (it is regular in ). Example: Find the solution of x x 1 = 0 using a regular perturbation expansion as above. The above examples could be easily solved by thinking of the variable as a function of the small parameter and taylor expanding. Nevertheless, such techniques do not work for all problems. Notice that the 7

8 small parameter in the above equations was not coupled with the highest order terms, thus sending 0 did not change the number of solutions for the problem. There are plenty of problems where sending the small parameter to 0 severely affects our approximations. In such scenarios taylor expansions do not work ( as 0 convergence fails). Such problems are known as singular perturbation problems. Let us consider a few examples: Example: Find an approximate solution using perturbation methods for the equation x 2 + x 1 = 0. Solution: Notice that for = 0 the problem reduces to x 1 = 0 and we only have one root. Let us think of x as a function of again, but a regular taylor expansion in this case does no good. Instead we try x() = α y() = α (x 0 + x 1 +..) where α > 0 is some constant. Our goal will be to find the appropriate value of α. Substituting we have, 1+2α y() 2 + α y() 0 = 0. (28) Notice that the powers in the exponents of decide how fast each term changes as 0. Since we want to recover two roots from this equation we will try and have different terms of the equation change at the same rate(i.e the terms of the equation must balance in order to produce 0). This implies that the powers of these terms will have to be matched. Let us consider a few combinations and what they produce: α = α so α = 1 and we use x = y. With this scaling the first two terms in the equation blow up together as 0 whereas the last one is left constant. This means that the two solutions can be obtained from the first two terms for this value of α and we have: 1 ( (x0 + x ) 2 + x 0 + x 1 ) = 0 (29) x x 0 x x 0 + x 1 = 0 (30) Such an equation is solved by solving for all the different orders of terms. Therefore we obtain the following set of equations: O(1) : x x 0 = 0 x 0 = 0, 2 (31) O() : 1 2x 0 x 1 + x 1 1 = 0 x 1 = (1 + 2x 0 ). (32) 8

9 Therefore we obtain two approximations for the roots of this equation, namely x = 1( 2 ) and x = 1 (0 + ). Notice that by 3 matching the first two terms we were able to recover two roots for the equation. Let s see what the other matchings give us α = 0 so α = 1 2. Using our scaling we get x = y.such a scaling substituted into the original equation gives that the first and third term are O(1) but the second term is O( 1/2 ) which means that this term blows up as 0(also the first and third term are O(1) whereas the second term is O( 1 ) but we need the second term to be higher order). Therefore this case is not possible and we set it aside. 3. α = 0 requires that we use a regular perturbation method which we already know cannot give us two roots so such a value is not useful. Example: Find the approximate solutions using singular perturbations methods for the equation x x 3 + x + = 0. Solution: Like we did before we try x = α y() and substitute 4α+1 y() 4 + 3α+2 y() 3 + α y() + = 0. (33) Observe that this equation has four roots which we will find using power matching. Before we start approximating a quick look at the equation tells us that x = is one root. Matching powers implies that we will be looking at the behavior of the terms of the equation as 0. In this case matching will be successful if it can give us the remaining three roots for the equation. We create a table with all the possible matchings for the powers. 4α + 1 3α + 2 α 1 α = α = α = α = Let us consider each row of the above table. 1. For the first row of the table we notice that the last two terms dominate as 0 this implies that we use the substitution x = y(). 9

10 This produces y + = 0 which does not give us the needed three extra roots. 2. For the second row we see that the first and third term dominate yielding x = 1 y(). Going back to the original equation we see that 1/3 the dominating terms are x 4 + x 0 so that x = 0. We can immediately see that these two terms can produce three roots. We can easily find them by substituting x = y() into the roots of x = 0. 1/3 3. For the third row the O(1) term dominates but that is not enough to give us three roots. 4. Again this matching does not yield any roots. Example: Approximate the roots using singular perturbation methods for the following equation: 2 x 4 + x 3 + x + = 0. Solution: Using singular perturbation we try x = α y(). We match the exponents just like before and we summarize all the possible matchings in a table. 4α + 2 3α + 2 α 1 α = α = α = α = α = The first two terms dominate in the limit 0 which leaves us with 2 x 4 + x 3 0 so that x which only produces one root (but 2 we need four!). 2. This matching does not produce any roots since the O( 1 ) overpowers all the other terms in the limit. 3. This matching also produces no roots since the O( 1/4 ) overpowers all the other terms in the limit. 4. The second and third terms here dominate in the limit 0, so in order to find the roots we have to consider x 3 + x 0 which gives x So this matching produces two solutions. 10

11 5. Only the last two terms dominate here which gives us x + 0 so that we get only one root x = which for x = y has y = 1. Example:Find the approximate solutions for the equation x 2 + e x = 5. Solution:For this transcedental equation we notice that sending 0 does not affect the number of solutions to the problem (hint graph the function and vary ). So we try a regular expansion with x() = x 0 + x Substitution yields, (x 0 + x ) 2 + e (x 0+x ) = 5. (34) In order to solve here we try a Taylor expansion for exp(x) so we have, (x 0 + x ) (x 0 + x 1 +..) + 2 (x 0 + x 1 +..) 2 We solve by picking same order terms: 2 = 5. (35) with solution x = ± O(2 ). O(1) : x 2 0 = 4 (36) O() : 2x 0 x 1 + x 0 = 0 (37) Example: Approximate the solutions to x sech( x ) = 0. Solution: Let us start with a guess of the solution x() = 1 + µ()(notice that the guess here does not have to be some Taylor series expansion). Substitute into the original equation to obtain, 1 + µ() µ() + sech( ) = 0 (38) 2 µ() + = 0 (39) e µ 1 + e 1 µ 2 µ() + e 1 + e 1 = 0 (40) µ() 2e 1 = 0. (41) Immediately this gives us the solution x = 1 2e 1 a Taylor expansion. which clearly is not 11

12 Let us now return to the first example we examined, namely the gravitational problem. Recall that we rescaled the problem to, d 2 y dt 2 = 1 (1 + y) 2 (42) with y(0) = 0 and y (0) = 1. We try a regular expansion on y() and also Taylor expand 1 (1+y) 2. Substituting into the equation we have, y () = 1(1 2y +...) (43) y 0 + y 1 = 1(1 2y ). (44) Equating same order terms we have, O(1) : y 0 = 1 (45) O() : y 1 = 2y 0, (46) with y 0 (0) = 0 (47) y 0(0) = 1. (48) The above equations have solutions, y 0 (τ) = τ τ 2 2 y 1 (τ) = τ 3 3 τ 4 12 (49) (50) which gives y(τ) = τ τ 2 + ( τ 3 τ 4 ). Now let us see how our approximate solution does. Let us calculate the landing time using the approximate solution we just calculated. The landing time is found by solving for the roots of, 0 = τ τ (τ 3 τ 4 ). (51) 12 Let us assume that the root we are looking for is of the form τ = 2 + a α. Substitution gives, 0 = (2 + a α ) (2 + aα ) 2 + ( (2 + aα ) 3 3 (2 + aα ) 4 ) (52) = 2 + a α 2a α + ( ) (53) 12

13 so that α = 1 and α = 4 3 and τ = α. Putting back the dimensional v 2 0 scales we get t 0 = v 0 g (2 + 4 ). Notice that this landing time is a little 3 Rg longer than the time we would obtain when = 0. Thus when is not very small the landing time we calculate for this problem is shorter than the actual value. Notice that in the above example we used regular perturbation methods on a differential equation. In order to better understand the validity of such an approximation for ODE s we need to refer to the Implicit Function Theorem. Implicit Function Theorem: Let f = (f 1, f 2...f n ) be a continuously differentiable, vector-valued function mapping an open set E R n+m into R n. Let (a, b) = (a 1,..., a n, b 1,..., b m ) be a point in E for which f(a,b) = 0 and such that the n n determinant D j f i (a,b) 0 (54) for i = 1,..., n, where D j f i = f i x j. Then there exists an m-dimensional neighbourhood W of b and a unique continuously differentiable function g : W R n such that g(b)=a and for all t W. f(g(t),t) = 0 (55) In our context we start with the map f : R 2 R so that f(x, ) evaluated at (x 0, 0) is 0 with R a small parameter. Then there exist a unique continuously differentiable function x() such that x(0) = x 0 with f(x(), ) = 0 provided f x (x 0,0) 0. This means that the regular perturbation methods can be applied if f x (x 0,0) 0. Let us consider a few examples. Examples: 1. Consider x + = 0. Then we have f(x, ) = x +, with f(x, 0) = 0 and f (x 0, 0) = 1 0 so the implicit function theorem can be applied for this problem. 2. x 2 + = 0, with f(x, ) = x 2 + and f(x, 0) = x 2. We calculate f (x 0, 0) = 2x 0 so f (0, 0) = 0 so the Implicit function theorem doesn t apply here. 13

14 3. Finally, lets go back to a familiar problem For = 0 the map reduces to, f(y, ) = y 1 (x) + (1 + y) 2 (56) y(0) = 0 (57) y (0) = 1. (58) f(y, ) = y (x) + 1 (59) y(0) = 0 (60) y (0) = 1. (61) Here we are concerned with the derivative of the map. More specifically we need to examine f(y + δz) f(y) = y + δz y 1 +. Substituting into (1 + (y + δz)) 2 the derivative expresion we have, f(y + δz) f(y) = δz + (1 + y)2 1 + (y + δz) (1 + y) 2 (1 + (y + δz)) 2 (62) = δz 2δz(1 + y) ( )( ) + O(δ2 ) (63) = δz So we have reduced our problem to 2δz(1 + y) (1 + y) 4 + O(δ 2 ) (64) = δ(z 2z (1 + y) 3 ) + O(δ2 ) (65) Lz = z (66) z(0) = 0 (67) z (0) = 0 (68) with operator L = d2 2. Clearly we can find an approximate (1+y) solution to this problem if the operator L is invertible which requires dx 2 14

15 for us to evoke the Fredholm alternative (i.e the nullspace of the operator needs to only contain 0). We will discuss in more detail the application of the Fredholm alternative to perturbation problems. Nevertheless it should be clear that checking the conditions of the Fredholm alternative on the operator is equivalent to checking the applicability of the Implicit function theorem on a map. 15