hyunjoong kim 1 & chee han tan 1 January 2, 2018 contents list of figures abstract

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1 I N T R O D U C T I O N TO A S Y M P TOT I C A P P R O X I M AT I O N hyunjoong kim 1 & chee han tan 1 January 2, 218 contents 1 Asymptotic expansion Accuracy and convergence Manipulating asymptotic expansion Asymptotic solution of algebraic and transcendental equations Example: algebraic equation Example: transcendental equation Differential equations - regular perturbations Projectile problem Non-linear potential problem Fredholm alternative Exercises 1 list of figures Figure 1 Solution of ɛx 2 = 1 2x Figure 2 Solution of e ɛx = 5 x Figure 3 First approximation of projectile problem abstract These notes are largely based on Math 673: Asymptotic and Perturbation Methods course, taught by Paul Bressloff in Fall 217, at the University of Utah. Additional examples or remarks or results from other sources are added as we see fit, mainly to facilitate our understanding. These notes are by no means accurate or applicable, and any mistakes here are of course our own. Please report any typographical errors or mathematical fallacy to us by hkim@math.utah.edu or tan@math.utah.edu. 1 Department of Mathematics, University of Utah, Salt Lake City, UT 1

2 asymptotic expansion 2 introduction Our main goal is to construct approximate solutions of differential equations to gain insight of the problem, since they are nearly impossible to solve analytically in general due to the nonlinear nature of the problem. Among the most important machinery in approximating functions in some small neighbourhood is the Taylor s theorem. It says that given f C N+1) B δ x )), we can write f x) as f x) = N k= where R N+1 x) is the remainder term f k) x ) x x ) k + R k! N+1 x), R N+1 x) = f N+1) ξ) N + 1)! x x ) N+1, for some ξ between x and x, provided x B δ x ). Taylor s theorem can be used to solve the following problem: Given a fixed ɛ = x x >, how many terms should we include in the Taylor polynomial to achieve a certain accuracy? Asymptotic approximation concerns about a slightly different problem: Given a fixed number of terms N, how accurate is the asymptotic approximation as ɛ? We want to avoid from including as many terms as possible as ɛ and in contrast to Taylor s theorem, we do not care about convergence of the asymptotic approximation. In fact, most asymptotic approximations diverge as N for a fixed ɛ. Remark 1. If the given function is sufficiently differentiable, then Taylor s theorem offers a reasonable approximation and we can easily analyse the error as well. 1 asymptotic expansion We begin the section with a motivating example. Suppose that we want to evaluate the integral f ɛ) = e t dt, ɛ >. 1 + ɛt We can develop an approximation of f ɛ) for small ɛ by repeatedly integration by parts where e f ɛ) = 1 t ɛ 1 + ɛt) 2 dt = = N k= 1) k k!ɛ k + R N ɛ), R N ɛ) = 1) N+1 N + 1)!ɛ N+1 e t dt. 1 + ɛt) N+2

3 asymptotic expansion 3 In practice, it is hard to estimate a bound for the remaining term. In this case, since then e t 1 + ɛt) N+2 dt e t dt = 1, R N ɛ) 1) N+1 N + 1)!ɛ N+1 1) N N!ɛ N. For fixed N, f ɛ) k= N lim a kɛ k ɛ =. That is, f ɛ) = ɛ N N a k ɛ k + Oɛ N+1 ). k= The formal series N k= a kɛ k is said to be an asymptotic expansion of f t) such that for fixed How to choose N in practice? A thumb rule is beginning at N = 2 and increasing to see physical phenomena) N it provides a good approximation to f ɛ) as ɛ. Note that asymptotic expansion in general is not convergent for fixed ɛ since 1) N N!ɛ N, N, that is the correction term actually blows up! Remark 2. Observe that for sufficiently small ɛ >, R N ɛ) 1) N N!ɛ N, which means that the remainder R N ɛ) is dominated by the N + 1th term of the approximation. In other words the error is of higher-order of the approximating function. This property is something that we would want to impose on the asymptotic expansion, and this idea can be made precise using the Landau symbols. Definition 1. a) f ɛ) = Ogɛ)) as ɛ means that there exists a finite M for which f ɛ) M gɛ). b) f ɛ) = ogɛ)) as ɛ means lim ɛ f ɛ)/gɛ) =. c) The ordered sequence of function {φ k ɛ)}, k =, 1, is called an asymptotic sequence as ɛ if φ k+1 ɛ) = oφ k ɛ)) as ɛ. d) Let f ɛ) be a continuous function of ɛ and let {φ k ɛ)} be an asymptotic sequence. The formal series expansion N a k φ k ɛ) 1.1) k= is called an asymptotic expansion a.e.) valid to order φ N ɛ) if for all N f ɛ) k= N lim a kφ k ɛ) =. 1.2) ɛ φ N ɛ) One typically write f ɛ) N k= a kφ k ɛ) as ɛ.

4 asymptotic expansion 4 In solution to ODEs we will need to consider time-dependent asymptotic expansions. Suppose x t) = f x, ɛ) for x R n where ɛ is a parameter of vector field. Then xt, ɛ) N a k t)φ k ɛ) k= In case of PDE, ux, t, ɛ) k= N a kx, t)φ k ɛ).) which will tend to be valid over some range of times t. It is often useful to characterize the time interval over which a.e. exists, i.e. say that estimate is valid on a time scale 1/ ˆδɛ) if lim ɛ where C is independent of ɛ. xt, ɛ) N k= a kt)φ k ɛ) φ N ɛ) = for t C/ ˆδɛ). 1.1 Accuracy and convergence In the case of a Taylor series expansion, one can increase accuracy for fixed ɛ) by including more terms assuming convergence). This is not usually the case for an a.e. a.e. concerns limit ɛ whereas the number of terms concerns N for fixed ɛ. 1.2 Manipulating asymptotic expansion Two asymptotic expansions can be added together term by term, assuming both involve the same basis functions {φ k ɛ)}. Multiplication can also be carried out provided the asymptotic sequences can be ordered in a particular way. What about differentiation? Suppose f x, ɛ) φ 1 x, ɛ) + φ 2 x, ɛ), ɛ. It is not necessarily the case d d f x, ɛ) dx dx φ 1x, ɛ) + d dx φ 2x, ɛ), ɛ. There are two possible scenarios: Example 1. Consider f x, ɛ) = e x/ɛ sine x/ɛ ). Note that f is fast oscillation with different scale. Observe that for x > lim f x, ɛ) ɛ ɛ n = for all finite n, which means that as ɛ. However, f x, ɛ) + ɛ + ɛ 2 +, d dx f x, ɛ) = 1 ɛ e x/ɛ sine x/ɛ ) + 1 ɛ cosex/ɛ ), ɛ, that is derivative cannot be expanded using ɛ, ɛ 2,. Example 2. Even if {φ k x, ɛ)} is an ordered asymptotic sequence, its derivative {φ k x, ɛ)} need not be. Consider φ 1x, ɛ) = 1 + x and φ 2 x, ɛ) = ɛ sinx/ɛ) for x [, 1]. Then φ 1 x, ɛ) = 1, φ which are not ordered. 2 x, ɛ) = cosx/ɛ)

5 asymptotic solution of algebraic and transcendental equations 5 y y = 1 2x y = ɛx 2 1/2,) x Figure 1: Solution of ɛx 2 = 1 2x. and In practice, if f x, ɛ) a 1 x)φ 1 ɛ) + a 2 x)φ 2 ɛ), ɛ, 1.3) d x f x, ɛ) b 1 x)φ 1 ɛ) + b 2 x)φ 2 ɛ), ɛ, 1.4) then b k x) = d x a k x). Throughout this course, we assume that it holds whenever we are given 1.3) which almost holds in practice. Integration, on the other hand, is less problematic. If 1.3) holds on x [a, b] and all functions are integrable, then b a b f x, ɛ)dx a ) b ) a 1 x)dx φ 1 ɛ) + a 2 x)dx φ 2 ɛ), ɛ. a 1.5) 2 asymptotic solution of algebraic and transcendental equations We study three examples where approximate solutions are found using asymptotic expansions, but each uses different method. They serve to illustrate the important point that instead of performing the routine procedure with standard asymptotic sequence, we should taylor our asymptotic expansion to extract the physical property or behavior of our problem. 2.1 Example: algebraic equation Consider the quadratic equation ɛx 2 + 2x 1 =. 2.1) This is singular problem since order of polynomial changes when ɛ =, see Fig. 1. In this case, the unique solution is x = 1/2. Therefore, try x x 1ɛ α +. Substitute it into the equation and we have ) ) 1 1 ɛ 4 + x2 1 ɛ2α + x 1 ɛ α x 1ɛ α + 1 =.

6 asymptotic solution of algebraic and transcendental equations 6 y y = 5 x 2 y = e ɛx - 5,) 5,) x Figure 2: Solution of e ɛx = 5 x 2. The O1) terms cancel as we anticipate. To cancel Oɛ) terms, set α = 1 and obtain x 1 = = x 1 = 1 8 = x ɛ. One method to generate other root is to consider x x 1 )x x 2 ) =. More systematic method, applicable to ODEs, is to avoid the O1) solution. Take x ɛ γ x + x 1 ɛ α + ). Now we have ɛ 1+2γ x 2 + 2x x 1 ɛ α + ) + 2ɛ γ x + x 1 ɛ α + ) + 1) =. }{{}}{{}}{{} a) b) c) Terms on LHS must balance and there are three possibilities: 1. Set γ = and we recover the previous solution balancing a) and c). 2. Balance a) c) and make b) to be higher-order. Require 1 + 2γ = = γ = 1/2 such that a) c) for O1) with b) = Oɛ 1/2 ), not higher-order. Thus, this is not a good approximation. 3. Balance a) b) and make c) to be higher order. Require 1 + 2γ = γ, i.e. γ = 1. Then a), b) = Oɛ 1 ) and c) = O1). Setting γ = 1 gives x 2 + 2x x 1 ɛ α + ) + 2x + x 1 ɛ α + ) ɛ =. Balancing O1) gives x 2 + 2x = = x =, 2. If x =, then it recovers the original solution. Now root given by taking x = 2. To balance ɛ in the last term take α = 1 and balance Oɛ). Then we have 2x x 1 + 2x 1 1 = impliesx 1 = 1/2. Therefore, x 1 ɛ 2 ɛ ) Example: transcendental equation For transcendental equations, one needs to determine how many solution exist graphically. Consider x 2 + e ɛx = 5, 2.2) and one can figure out that it has two solution for ɛ >, see Fig. 2 and this is not singular because ɛ = provides x = ±2). Applying same arguement in example of algebraic equation, assume asymptotic expansion x x + x 1 ɛ α + and Taylor expand the exponent and have x 2 + 2x x 1 ɛ α + ) ɛx + ) = 5.

7 differential equations - regular perturbations 7 Balance O1) provides x = 4 = x = ±2. Take α = 1 and balance Oɛ) follows that x 1 = 1/2. Therefore, x ±2 ɛ 2. Consider one more transcendental equation x ) x ɛ sech =. 2.3) ɛ One can determine it has one solution graphically. If you try with polynomials x x + x 1 ɛ α +, then one cannot balance terms. In fact, it suffices to find an asymptotic sequece {φ 1 ɛ), φ 2 ɛ), } such that φ 1 ɛ) 1, φ 2 ɛ) φ 1 ɛ) and so on. Therefore, try x x + µɛ) with µɛ) 1 where ɛ 1. Substituting into original equation gives x x + µɛ) ɛ sech ɛ + µɛ) ) =. ɛ Balance O1) yields x = 1. Then we have µɛ) + ɛ sech 1 ɛ + µɛ) ) =. ɛ Since µɛ) 1, then one can approximate sech 1 ɛ + µɛ) ) sech 1 ) 2 = ɛ ɛ e 1/ɛ + e 1/ɛ 2e 1/ɛ. Therefore, it requires µɛ) = 2ɛe 1/ɛ. To proceed further, take with νɛ) ɛe 1/ɛ. x 1 2ɛe 1/ɛ + νɛ) 3 differential equations - regular perturbations Roughly speaking, the regular perturbation theory is a variant of Talyor s theorem, in the sense that we seek power series solution in ɛ. More precisely, we assume that the solution takes the form x x + x 1 ɛ + x 2 ɛ 2 +, as ɛ. In this chapter, we apply this idea to some differential equations. 3.1 Projectile problem Consider the motion of a gerbil projected radially upward from the surface of the earth. Let xt) be the height of the gerbil from the surface of the Earth. Newton s law of motion asserts that d 2 dt 2 x = g R 2 x + R) 2 3.1) where R is radius of the earth and g is the gravitational constant. If x R, then to a first approximation d 2 dt 2 x = g = x t) = 1 2 gt2 + v t,

8 differential equations - regular perturbations 8 xt) ) v g, v2 2g 2v g, ) t Figure 3: First approximation of projectile problem. where v is given initial velocity, see Fig. 3. It has characteristic values that t c = v /g is time reaching highest height and x c = v 2 /g is highest reached height. Before perturbing the ODE, non-dimensionalize by τ = t/t c and y = x/x c. Then it provides that d 2 dτ 2 y = ɛy) 2 with y) = and y ) = ) Consider an asymptotic expansion with power basis yτ) y τ) + ɛ α y 1 τ) +. Assume we can differentiate term by term. Then by Taylor expanding to RHS with initial condition y + ɛα y 1 + = ɛy + )) 2 = 1 + 2ɛy + y ) + ɛ α y 1 ) + = and y ) + ɛα y 1 ) + = 1. Balance O1) with y ) = and y ) = 1 and it yields y = 1 and y τ) = 1 2 τ2 + τ. To balance the next term α = 1, determined by RHS. Balance Oɛ), then one can obtain y 1 = 2y with y 1 ) = y 1 ) = and y 1 τ) = 1 3 τ τ2. Therefore, the solution of asymptotic expansion would be yτ) τ 1 12 ) τ ɛτ3 1 τ ). 3.3) Non-linear potential problem Let us consider diffusion of ions through a solution containing charged molecule. Electrostatic potential φx) satisfies Poisson-Boltzmann equation 2 φ = k i=1 α i z i e z iφ, x Ω, 3.4) where α i is constant and z i is valence of ith ionic species, with electro-neutrality and boundary conditions k α i z i = and n φ = n φ = ɛ, Ω. 3.5) i=1

9 differential equations - regular perturbations 9 Until now, there is no known exact solution. Try φ ɛφ x) + ɛφ 1 x) + ) because one can get trivial solution as ɛ. Then we have ɛ 2 φ + ɛ 2 φ 1 + ) = α i z i e ɛz iφ +ɛφ 1 + ). i Expanding the right-hand side by Taylor series ɛ 2 φ + ɛ 2 φ 1 + ) α i z i 1 ɛz i φ + ɛφ 1 + ) + 1 ) 2 ɛ2 z 2 i φ + ) 2 +, i and rearranging it gives ɛ 2 φ + ɛ 2 φ 1 + ) ɛ i Set κ 2 = i α i z 2 i and it follows an equation of balancing Oɛ) { 2 κ 2 )φ =, x Ω n φ = 1, x Ω. Balance Oɛ 2 ) and we get { 2 κ 2 )φ 1 = λφ, x Ω n φ 1 =, x Ω. α i z 2 i φ + ɛ 2 α i z 2 i φ 1 1 ) 2 z iφ 2 +. i where λ = 2 1 i α i z 3 i. The solutions are called basis of classical Debye-Huckel theory in electrochemistry. Specifically, take Ω to be the domain outside of the unit sphere. It follows that solutions have radial symmetry. By the radial Laplace operator, we obtain { ) 1 d r 2 dr r 2 dφ dr κ 2 φ =, 1 < r < φ 1) = 1. It yields φ r) = 1 1+κ)r eκ1 r). In the same fashion, one can get { ) 1 d r 2 dr r 2 dφ dr κ 2 φ = λ e 2κ1 r), 1 < r < 1+κ) 2 r 2 φ 1 1) =, and it yields the solution φ 1 r) = α r e κr + γ κr [ e κr E 1 3κr) e κr E 1 κr) ] where γ = λe 2κ /2κ and E 1 is the exponential integral E 1 z) = x e t t 1 dt. 3.3 Fredholm alternative Let L and L 1 be differential operators on L 2 R with inner product f, g = f x)gx)dx. Consider eigenvalue problem L + ɛl 1 )φ = λφ. Suppose that ɛ = and equation has a unique solution L φ = λ φ

10 exercises 1 with λ is non-degenerate. Take L to be self-adjoint operator. Introduce asymptotic expansion and it yields φ φ + ɛφ 1 + ɛ 2 φ 2 and λ λ + ɛλ 1 + ɛ 2 λ 2, L + ɛl 1 )φ + ɛφ 1 + ɛ 2 φ 2 + ) = λ + ɛλ 1 + ɛ 2 λ 2 )φ + ɛφ 1 + ɛ 2 φ 2 + ). Balance O1) and we have L φ = λ φ, i.e. L λ I)φ =. Balance Oɛ) and get equation for φ 1 L λ I)φ 1 = λ 1 φ L 1 φ. Use the fact that L λ I is self-adjoint and φ kerl λ I). Then we get = λ 1 φ, φ + φ, L 1 φ = λ 1 = φ, L 1 φ φ, φ. Therefore, λ λ + ɛλ 1 and solve for φ 1. With the same analogy, one can find λ n and φ n. 4 exercises Problem 1. Consider the transcendental equation 1 + x 2 + ɛ = e x. 4.1) Explain why there is only one small root for small ɛ. Find the three term expansion of the root: x x + x 1 ɛ α + x 2 ɛ β, β > α >. Proof. Consider two graph f x) = x 2 + ɛ and gx) = e x 1. If x <, then f x) > > gx). It means that there is no solution in negative region. If x >, then f x) x as x starting its curve from f ) = ɛ. One can draw graph of f x) and gx) on x >, then it yields there is only one solution. To obtain first expansion, set ɛ =. Then we get 1 + x = e x = x =. Since there is only on solution for all ɛ >, then one can set x = and expand x further at this point. To do so, rewrite the equation 4.1) as x 2 + ɛ = e x 1) 2 and x 1 as ɛ 1, it is reasonable to expand RHS as Taylor series. Then one can obtain x 2 + ɛ = x + 1 2! x2 + 1 ) 2 3! x3 + = x 2 + x x ) Before we balance both sides, consider the leading order of both sides. Without doubt, the leading order is ɛ 2α with coefficient x1 2. For LHS, we have three cases

11 exercises If 2α > 1, then the leading order is ɛ with coefficient 1. It leads to a contradiction when balancing both sides because 2α = 1. ) 2. If 2α = 1, the balancing equation yields x = x2 1 and it does not make sense. ) 3. Thus, the only case is 2α < 1. Then one can rewrite equation 4.2) as ɛ = x x4 + = x 3 1 ɛ3α + 3x 2 1 x 2ɛ 2α+β + ) x4 1 ɛ4α + ). Since the leading order of RHS is ɛ 3α, it provides that 1 = 3α and 1 = x 3 1. Thus, α = 1/3 and x 1 = 1. The next leading term is ɛ 4α. Since there is no remaining term on LHS, then balance RHS as 2α + β = 4α and = 3x 2 1 x x4 1, yields β = 2α = 2/3 and x 2 = 7/36. Therefore, the three term expansion of root is x + ɛ 1/ ɛ2/3. 4.3) Problem 2. A classical eigenvalue problem is the transcendental equation λ = tanλ). 1. After sketching the two functions in the equation, establish that there is an infinite number of solutions, and for sufficiently large λ take the form with x n small. λ = πn + π 2 x n Proof. Tangent is π-periodic function with asymptotic line λ n = πn + π/2. tanλ) as λ λ + n. Since f λ) = λ passes through all the asymptotic line and tangent function is close to the asymptotic line, then for sufficiently large n, λ takes the form λ = λ n x n where x n is a small number and tends to zero as n. 2. Find an asymptotic expansion of the large solutions of the form λ ɛ α λ + ɛ β λ 1 ) and determine ɛ, α, β, λ, λ 1. Proof. Set λ = 1/ɛ and see the asymptotic behavior of x n. Then one can figure out an asymptotic expansion of λ. For convenience, set x n = x. Then one can get ) 1 1 ɛ x = tan ɛ x = cotx)

12 exercises 12 because tan1/ɛ) =. By multiplying ɛ tanx) on both sides and we get tanx) ɛx tanx) = ɛ. Sicne we know that x as ɛ, take x x ɛ θ, θ >. It follows that x ɛ θ + ) ɛx ɛ θ + )x ɛ θ + ) = ɛ Since θ >, the leading order of LHS is ɛ θ. To balance both sides with Oɛ), set θ = 1 and get x = 1. Therefore, an asymptotic expansion of λ is λ = 1 ɛ x 1 ɛ ɛ = ɛ ɛ 2 1)). It follows that α = 1, β = 2, λ = 1 and λ 1 = 1. Problem 3. In the study of porous media one is interested in determining the permeability ks) = F cs)), where F 1 c ɛr)dr = s and F 1 c) F 1 c ɛ) = β, and β is a given positive constant. The functions Fc) and constant c both depends on ɛ, whereas s and β are independent of ɛ. Find the first term in the expansion of the permeability for small ɛ. [Hint: consider an asymptotic expansion of c and use the fact that s is independent of ɛ.] Proof. Take c c + c 1 ɛ +. Substituting it into given integral equation and expanding F 1 as Taylor series centered at c = c yields F 1 c ) + ɛc 1 r) df 1 dc c ) + Oɛ 2 )dr = F 1 c ) + ɛ Since s is independent of ɛ, then it gives us that c 1 1 ) df 1 2 dc c ) + Oɛ 2 ) = s. F 1 c ) = s and c 1 2 = = c = Fs) and c 1 = 1 2. From the second condition, expanding F 1 as Taylor series follows provides that It follows that F 1 c ) + ɛc 1 df 1 dc c ) F 1 c ) ɛc 1 1) df 1 dc c ) + Oɛ 2 ) = β. 1 ɛ F s) + Oɛ2 ) = β = ks) = F s) ɛ β. Problem 4. Let A and D be real n n matrices. 1. Suppose A is symmetric and has n distinct eigenvalues. Find a two-term expansion of the eigenvalues of the perturbed matrix A + ɛd, where D is positive definite.

13 exercises 13 Proof. Take λ λ + λ 1 ɛ and its corresponding eigenvector x x + x 1 ɛ. It follows that A + ɛd)x + x 1 ɛ + ) = λ + λ 1 ɛ + )x + x 1 ɛ + ). Balance for O1) and we have Ax = λ x. Since A has n distinct eigenvalues, then λ can be one of them and x is corresponding eigenvector. Similarly, balance for Oɛ) and it yields Ax 1 + Dx = λ x 1 + λ 1 x = A λ I)x 1 = Dx + λ 1 x Take inner product with x and by symmetry of A λ I, one can obtain x, Dx + λ 1 x, x = x, A λ I)x 1 = A λ I)x, x 1 =. Thus λ 1 = x, Dx / x, x. It is well-defined because x = = x, x =. One can find x 1 by solving A λ I)x 1 = Dx + λ 1 x. 2. Consider the matrices [ ] 1 A =, D = [ ]. 1 Use this example to show that the Oɛ) perturbation of a matrix need not result in a Oɛ) perturbation of the eigenvalues, nor that the perturbation is smooth at ɛ = ). Proof. Its characteristic equation is pλ) = det A + ɛd λi) = λ 2 ɛ and it provides two eigenvalues λ = ± d ɛ. At ɛ, dɛ λ and it means that the perturbation is not smooth at ɛ =. Problem 5. The eigenvalue problem for the vertical displacement yx) of an elastic string with variable density is y + λ 2 ρx, ɛ)y =, < x < 1, where y) = y1) =. For small ɛ, assume ρ 1 + ɛµx), where µx) is positive and continuous. Consider the asymptotic expansions yx) y x) + ɛy 1 x) and λ λ + ɛλ Find y, λ and λ 1. The latter will involve an integral expression.) Proof. Substituting all asymptotic expansions into given differential equation leads that y + y 1 ɛ + )+ λ + λ 1 ɛ + ) µɛ + )y + y 1 ɛ + ) =.

14 exercises 14 Balancing O1) terms yields y x) + λ2 y x) =. The solution of equation is known as y x) = sinnπx) with λ = nπ, for n = 1, 2,. Similarly, balance Oɛ) terms and get that is y 1 x) + 2λ λ 1 y x) + λ 2 µx)y x) + λ 2 y 1x) =, y 1 x) + λ2 y 1x) = λ 2 µx)y x) 2λ λ 1 y x). Taking integral with y x) and integration by parts yields that Therefore, = λ 2 µs)y 2 s)ds 2λ λ 1 y 2 s)ds. λ 1 = λ µs)y 2 s)ds = nπ µs) sin 2 nπs)ds. 2. Using the equation for y 1, explain why the asymptotic expansion can break down when λ is large. Proof. From the previous results, one can find the equation for y 1 as ) y 1 x) + λ2 y 1x) = λ 2 µx)y x) + 2 µs)y 2 s)ds. Notice that the RHS proportional to λ 2 and it follows that the particular solution of y 1 is proportional to λ 2, then it implies that y 1. This can break down the expansion mixed with ɛ. Problem 6. Consider the following eigenvalue problem: a Kx, s)ys)ds = λyx), < x < a. This is a Fredholm integral equation, where the Kernel Kx, d) is known and is asumed to be smooth and positive. The eigenfunction yx) is taken to be positive and normalized so that a y 2 ds = a. Both yx) and λ depends on the parameter a, which is assumed to be small. 1. Find the first two terms in the expansion of λ and yx) for small a.

15 exercises 15 Proof. Since the LHS of Fredholm integral equation is proportional to a, because integral contains a, the leading order of eigenvalue λ is Oa). So, take λ λ a + λ 1 a 2 and yx) y x) + y 1 x)a. Expand Kx, s) and ys) as Taylor series in terms of s centered at s = because < s < a is also small. Then we get a Kx, ) + K d x, )s + )y) + y )s + )ds = λyx), and it follows that akx, )y) + a2 2 Kx, )y ) + K d x, )y)) + = λyx). Balance Oa) terms and one can obtain Kx, )y ) = λ y x). 4.4) Balance Oa 2 ) terms and one can find Kx, )y 1 ) Kx, )y ) + K dx, )y )) = λ 1 y x) + λ y 1 x). 4.5) In the same fashion, find one more asymptotic equation from given normalization equation a y) + y )s + ) 2 ds = a and it follows that a y)) 2 + a2 2 2y)y ) + = a. Balance Oa) terms and one can obtain y )) 2 = ) Balance Oa 2 ) terms and one can find 2y )y 1 ) y )y ) =. 4.7) From equation 4.4,4.6), one can find This implies that y ) = 1 and λ = K, ). y x) = Similarly, one can find and Kx, ) K, ). λ 1 = 1 2 K, )y ) + K d, )) = 1 2 K x, ) + K d, )) y 1 x) = [ 1 Kx, )y 1 ) + 1 Kx, )Kx x, ) 2 K, ) λ ) ] + K d x, ) λ 1 y x),

16 exercises 16 and it follow that y 1 x) = [ ] 1 Kx, ) 2λ K, ) [K xx, ) K x, )] + K d x, ) 2λ 1 y x). 2. By changing variables, transform the integral equation into Kaξ, ar)φr)dr = λ φξ), < ξ < 1. a Write down the normalization condition for φ. Proof. Substituting x = aξ and s = ar into the Fredholm integral equation yields Kaξ, ar)yar)adr = λyar), and set φr) = yar). It follows that Kaξ, ar)φr)dr = λ a φr). In the same fashion, consider the normalization equation a y 2 s)ds = a = φ 2 r)adr = a = φ 2 r)dr = From part b) find the two-term expansion for λ and φξ) for small a. Proof. Take λ aλ + a 2 λ 1 and φ φ + aφ 1. In the same fashion we did in part a), expand K inside of integral centered at zero K, ) + K x, )aξ + K d, )ar + )φr)dr = K, ) ak d, ) φr)dr + aξk x, ) rφr)dr +. φr)dr+ Then balance O1) terms in both sides of the equations and we get { K, ) φ r)dr = λ φ ξ) φ r)) 2. dr = 1 It implies that φ is constant, and it yields that φ ξ) = 1 and λ = K, ). 4.8) Similarly, balance Oa) terms of the equations and one can obtain K, ) φ 1r)dr + ξk x, ) φ r)dr + K d, ) rφ r)dr = λ φ 1 ξ) + λ 1 φ ξ) φ r)φ 1 r)dr =

17 exercises 17 It follows that φ 1r)dr = and one can have ξk x, ) K d, ) = λ φ 1 ξ) + λ 1. Integrate both side with respect to ξ on [, 1] and get λ 1 = 1 2 K x, ) K d, ). 4.9) This eigenvalue yields that φ 1 ξ) = K x, ) λ ξ 1 ). 4.1) 2 4. Explain why expansion in part a) and c) are the same for λ but not the eigenfunction. Proof. The eigenvalue is coordinate invariant, so it is not affected by change of variables. However, the eigenfunctions are. Problem 7. In quantum mechanics, the perturbation theory for bound states involves the time-independent Schrodinger equation ψ [V x) + ɛv 1 x)]ψ = Eψ, < x <, where ψ) = ψ ) =. In this problem, the eigenvalue E represents energy and V 1 is a perturbing potential. Assume that the unperturbed ɛ = ) is nonzero and nondegenerate. 1. Assuming ψx) ψ x) + ɛψ 1 x) + ɛ 2 ψ 2 x) and E E + ɛe 1 + ɛ 2 E 2, write down the equation for ψ x) and E. We will assume in the following that ψ 2 dx = 1 and V 1 x) dx <. Proof. Substituting expansion of ψ and E into the Schrodinger equation, one can balance O1) terms and get ψ x) V x)ψ x) = Eψ x). 2. Substituting ψx) = expφx)) into the Schrodinger equation, derive the equation for φx). Proof. Take derivative twice to ψ and it yields that ψ x) = φ x)e φx) and ψ x) = φ x) + φ x)) 2 )e φx). Plug them into the Schrodinger equation and drop common term e φx). Then it follows that φ x) + φ x)) 2 V x) + ɛv 1 x)) = E.

18 references By expanding φx) for small ɛ, determine E 1 and E 2 in terms of ψ and V 1. Proof. Assume that φx) φ x) + ɛφ 1 x) + ɛ 2 φ 2 x). Substituting it into the new Schrodinger equation and balance Oɛ) terms. Then one can obtain φ 1 + 2φ φ 1 = V 1 E 1. Define an differential operator L = d 2 /dx 2 + 2φ d/dx. Notice that for sufficiently smooth f, ψ 2, L f = Performing integration by parts ψ 2, L f = e 2φ x) f x) + 2φ x) f x))dx. e 2φx) f x)dx + e 2φx) f x) e 2φx) f x)dx, and it follows that ψ 2, L f =. From the observation, take inner product with ψ 2 to the first order balance equation and get Therefore, ψ 2, Lφ 1 = = ψ 2, V 1 E 1 ψ 2, 1 = ψ2, V 1 E 1. E 1 = ψ 2, V 1 = x V 1 x)[ψ x)] 2 dx. 4.11) To find φ 1, solve the first order inhomogeneous ODE of φ 1 by integrating factor, or observe that x x ) ψ 2 Lφ d 1dy = ψ 2 dφ 1 dy = ψ 2 dy dy x)φ 1 x) and it yields that = φ 1 x) = 1 ψ 2 x) x ψ 2 V 1 E 1 )dy ψ 2 V 1 E 1 )dy. Similarly, find the second order balance equation φ 2 + 2φ φ 2 + φ 1 )2 = E 2 = Lφ 2 = φ 1 )2 E 2. Therefore, E 2 = ψ 2, φ 1 )2. references [1] Bressloff P C 214 Waves in Neural Media Springer [2] Holmes M H 213 Introduction to Perturbation Methods Springer 2/e [3] Pikovsky A, Rosenblum M and Kurths J 23 Synchronization: A Universal Concept in Nonlinear Sciences Cambridge University Press