hyunjoong kim 1 & chee han tan 1 January 3, 2018 contents list of figures abstract

M ATC H E D A S Y M P TOT I C E X PA N S I O N S hyunjoong kim 1 & chee han tan 1 January 3, 218 contents 1 Introductory example 2 1.1 Outer solution by regular perturbation.................. 2 1.2 Boundary layer................................ 3 1.3 Matching.................................... 3 1.4 Composite expression............................ 3 2 Extensions: multiple boundary layers, etc. 5 2.1 Multiple boundary layers.......................... 5 2.2 Interior layers................................. 5 3 Partial differential equations 6 4 Strongly localized perturbation theory 8 4.1 Eigenvalue asymptotics in 3D........................ 9 4.2 Eigenvalue asymptotics in 2D........................ 1 4.3 Summing all logarithmic terms....................... 11 5 Exercises 12 list of figures Figure 1 Two choices of boundary layers.................. 2 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 email hkim@math.utah.edu or tan@math.utah.edu. 1 Department of Mathematics, University of Utah, Salt Lake City, UT 1

introductory example 2 (a) Possible boundary layer at x = 1 (b) Possible boundary layer at x = Figure 1: Two choices of boundary layers. It can be chosen by investigating the sign of y near the boundary layer by looking at the concavity of the function. introduction For most of singular perturbation problem of differential equations, the solution has extreme changes because a singular problem converges to a differential equation with different order or behavior as ɛ. If we apply the regular asymptotic expansion, it fails to represent such drastic change and to match all boundary condition. To resolve the problem, we introduce matched asymptotic expansion which approximates the exact solution by zooming in the extreme changing zones, such as inner or boundary layers, together with the regular expansion for outer region. 1 introductory example Consider a singular problem { ɛy + 2y + 2y =, < x < 1 y() = y(1) = 1 (1.1) If ɛ =, then we have a first order ODE. It only needs one boundary condition. It yields to have drastic dynamics on boundary layer. Remark that boundary layer could be interior, not only near boundary of domain. 1.1 Outer solution by regular perturbation Set y(x) y (x) + ɛy 1 (x) +. Substitute into equation (1.1) and we have ɛ(y (x) + ɛy 1 (x) + ) + 2(y (x) + ɛy 1 (x) + ) + 2(y (x) + ɛy 1 (x) + ) =. Balance O(1) and it provides y + y = = y (x) = ae x. It leads to dilemma that the solution has only one arbitrary constant but we have two boundary conditions. It is over-determined. Moreover, the outer solution cannot describe solution over the whole domain [, 1]. The following question is which boundary layer would we use?

introductory example 3 1.2 Boundary layer Assume that boundary layer is at x =. Introduce the stretched coordinate x = x/ɛ α, α >. Treat x as fixed when ɛ is reduced. Setting Y( x) = y(x) yields Try a solution of a form ɛ 1 2α d2 Y dy + 2ɛ α + 2Y =, Y() =. (1.2) d x 2 d x Y( x) Y ( x) + ɛ γ Y 1 ( x) +, γ >. Substitution into inner equation provides 1 2α d2 ɛ d x 2 (Y + ɛ γ Y 1 + ) + 2ɛ α d }{{} d x (Y + ɛ γ Y 1 + ) + 2(Y + ɛ γ Y 1 + ) =. }{{}}{{} (3) (1) (2) One need to determined correct balance condition: Balance (1) and (3) with taking (2) is higher order. Then it requires α = 1/2. Then (1), (3) = O(1), but (3) = Oɛ 1/2. ( ) Balancing (2) and (3) gives outer solution. ( ) Balance (1) and (2) with taking (3) is higher order. Then it requires α = 1. Then (1), (2) = O(ɛ 1 ) and (3) = O(1). (Yay!) Choosing the last balance, one can obtain an equation from O(ɛ 1 ) terms Y + 2Y =, < x <. One can get inner solution Y ( x) = A(1 e 2 x ), where A is unknown constant. 1.3 Matching It remains to determine the constant A. The inner and outer solutions are both approximations of the same function. Hence they sould agree in the transition zone between inner and outer layers. Thus lim Y ( x) = lim y (x), (1.3) x x + and yields A = e. Therefore, Y (x) = e(1 e 1 2x/ɛ ). 1.4 Composite expression So far, we have a solution in two pieces, neither is uniformly valid in x [, 1]. We would like to construct a composite solution that holds everywhere. One way is subtracting constant to match each one Y(x) y (x) + Y(x/ɛ) y (). (1.4) Near x =, y (x) is canceled out with the constant and vice versa. The matching condition Y (+ ) = y ( + ) may not work in general. First, the limits might not exist. Second, complication may arise when constructing second order terms. A more general approach is to explicitly introduce an intermediate region between inner and outer domain. Introduce an intermediate variable x η = x/η(ɛ) with ɛ η 1. The inner and outer solution should give same result when expression in terms of x η. Then

introductory example 4 1. change from x to x η in outer expansion y outer (x η ). Assume there is η 1 (ɛ) such that y outer is valid for η 1 (ɛ) η(ɛ) 1. 2. Change variable x to x η in inner expansion to obtain y inner (x η ). Assume there is η 2 (ɛ) such that inner is valid for ɛ η(ɛ) η 2 (ɛ). 3. If η 1 η 2, then domain of validity overlap (because inner expansion valid on x η 2 and outer expansion valid on x η 2 ) and we require y outer y innter in the overlap region. Return to our particular example. Let x η = x/ɛ β with < β < 1. Then and y inner A(1 e 2x η/ɛ 1 β ) A + O(ɛ β 1 ), y outer e 1 x ηɛ β e + O(ɛ β ). These are hard to match so we consider higher-order term. Find the second balance equation y 1 + y 1 = 1 2 y, y 1(1) = = y 1 (x) = 1 (1 x)e1 x 2 from O(ɛ) terms of outer expansion and Y 1 + 2Y 1 = 2Y, Y 1 () = = Y 1 ( x) = B(1 e 2 x ) xe(1 + e 2 x ) from O(1) terms of inner expansion. Determine B by matching on intermediate zone y outer e 1 x ηɛ β + ɛ 2 (1 x ηɛ β )e 1 x ηɛ β e 1 e x η ɛ β + ɛ 2 e 1 + e 1 2 x2 ηɛ 2β + ( y inner e(1 e ξ ) + ɛ B(1 e ξ ) x ) η ɛ 1 β e(1 + eξ ), ξ = 2x η /ɛ 1 β e ɛ β x η e + ɛ B +, and yields B = e/2. Therefore, the composite solution is y(x) y (x) + ɛy 1 (x) + Y (x/ɛ) + ɛy 1 (x/ɛ) e x η ɛ b e + e }{{} 2 ɛ. =x (1.5) Remark 1. Things to look for in more general problems on [, 1] 1. The boundary layer could be at x = 1 or there could be boundary layers at both ends. At x = 1, the stretched coordinate is x = (x 1)/ɛ α. 2. There is an interior layers at some x (ɛ) x = x x ɛ α. 3. ɛ-dependence could be funky, e.g. ν = 1/ log ɛ. 4. The solution odes not have layered structure.

extensions: multiple boundary layers, etc. 5 2 extensions: multiple boundary layers, etc. 2.1 Multiple boundary layers Consider a boundary value problem ɛ 2 y + ɛxy y = e x, with y() = 2, and y(1) = 1, (2.1) which is singular and non-linear. Note that in case when ɛ =, we get y = e x and it does not match any boundary conditions. This solution is the first term in the outer solution, i.e. y (x) = e x. Start to find inner solution at x =. Set x = x/ɛ α and Y( x) = y(x). Then we have 2 2α d2 ɛ d x 2 Y + ɛ x d d x Y }{{} (2) } {{ } (1) Y }{{} (3) = e xɛα = (1 + xɛ α + ). }{{} (4) In order to balance (1),(3) and (4), we require α = 1. Then taking Y Y + yields the following balance equation for O(1) Its general solution is Y Y = 1, Y () = 2. Y ( x) = 1 + Ae x + (1 A)e x, < x <. To achieve A, matching Y (+ ) = y () implies A = 1. At x = 1, setting x = (x 1)/ɛ β and Y( x) = y(x) provides that 2 2β d2 ɛ d x 2 Y + (1 + ɛβ x)ɛ 1 β d d x Y Y = e 1+ɛβ x. Achieve balance for β = 1 and we obtain Its general solution is Y + Y Y = e, < x <, with Y () = 1. Y ( x) = e + Be r + x + (1 e B)e r x, where r ± = ( 1 ± 5)/2. Matching Y ( ) = y (1) provides B = 1 e. Therefore, its composite solution is [ ( x ) ] [ ( x ) ] y y (x) + Y Y (+ ) + Y Y ( ) ɛ ɛ e x + e x/ɛ + (1 e)e r +(1 x)/ɛ. 2.2 Interior layers It is also possible for a boundary layer to occur in the interior of the domain rather than at a physical boundary matching now has to determined the location at the interior layer. Consider a boundary value problem ɛy = y(y 1), < x < 1 (2.2) with y() = 1 and y(1) = 1. For its outer equation, setting y y + yields y (y 1) = = y = or y (x) = x + a for some constant a. Since the outer equation does not satisfy both boundary condition at once, we need to find a boundary layer to fit boundary conditions.

partial differential equations 6 Assume that boundary layer is at x =. In the boundary layer, y > and y <. Since y can be positive, we cannot match signs of differential equation everywhere. If boundary layer is at x = 1, then y < and y 1 <. Since y can be negative, it cannot match signs everywhere in boundary layer. What if it has interior layer at x = x? For x x =, we have y <, y 1 >, but y >. For x x = +, we have y >, y 1 <, but y <. Thus interior layer can match the signs. From the argument of interior layer argument, find inner solution by setting x = (x x )/ɛ α, < x < 1. Then we have two outer regions x < x and x < x 1. The inner equation is ɛ 1 2α Y = ɛ α YY Y, and one can balance if α = 1. Setting Y( x) Y ( x) + gives Y = Y Y = Y = 1 2 Y2 + A. It has three general solution depending on sign of A: [ ] 1. Y = B 1 De B x if A >, 1+De B x [ ] 2. Y = B tan C B 2 x if A <, 3. Y = C 2 x if A =. Three forms (rather than a single general solution) reflects non-linearity. Next, match the inner solution with outer solution { x + 1, x < x y (x) = x 2, x < x. Only inner solution 1. can match these outer solutions. Without loss of generality, assume B > and we get B = Y (+ ) = y (x + ) = x 2 and B = Y ( ) = y (x ) = x + 1. It yields x = 1/2 and B = 3/2. What about D? Remember that y(x ) =. This implies that Y (x ) = = 3 2 1 D 1 + D = D = 1. Therefore, Y ( x) 3 2 1 e3 x/2 1 + e 3 x/2. Finally, the composite solution can be constructed in the two domains [, x ) and (x, 1] x + 1 + 2 3 y(x) 1 e3(2x 1)/4ɛ 1+e 3(2x 1)/4ɛ 2 3, x < x x 2 + 2 3 1 e3(2x 1)/4ɛ + 3 1+e 3(2x 1)/4ɛ 2, x < x 1. 3 partial differential equations Consider Burger s equation u t + u u x = ɛu xx, < x <, t > (3.1) u(x, ) = φ(x) (3.2)

partial differential equations 7 Notice that this perturbation problem is singular because type of solution is changed from parabolic to hyperbolic when ɛ > to ɛ =. Assume that φ(x) is smooth and bounded except for a jump continuity at x = with φ( ) > φ( + ) and φ. For concreteness, set { 1, x < u(x, ) =, < x. This is an example of a Riemann problem evolves into a traveling front that sharpens as ɛ. We can handle it in similar way for boundary layer problem. For outer solution, expanding u(x, t) u (x, t) + gives balance equation for O(1) terms t u + u x u =. Solve it using the method of characteristics dt dτ = 1, dx dτ = u and and it yields characteristic straight lines x = x + φ(x )t. du dτ =, Characteristic into set at the shock x = s(t) with determined using the Rankine- Hugoniot equation ṡ = 1 2 [φ(x+ )]2 [φ(x )]2 φ(x + ) φ(x ) = 1 2 [φ(x+ ) + φ(x )]. (3.3) We will derive an equation for s(t) using match asymptotics. Introduce a moving inner layer around s(t) x = x s(t) ɛ α. The inner PDE for U( x, t) = u(x, t) t U ɛ α s (t) x U + ɛ α U x U = ɛ 1 2α 2 xu. In order to balance terms, require α = 1 and U U + s (t) x U + U x U = 2 xu. Integrating with respect to x gives x U = 1 2 U2 s (t)u + A(t). Its matching conditions are lim U = u x and lim U = u + x + where u ± = lim x s(t) ± u (x, t). Since U ( x, t) is a constant for x ±, we have x U as x ±. Then we have = 1 2 [u ]2 s (t)u + A(t), = 1 2 [u+ ]2 s (t)u + + A(t).

strongly localized perturbation theory 8 Subtracting part of equations yields s (t) = 1 2 [φ(x+ )]2 [φ(x )]2 φ(x + ) φ(x ) = 1 2 [φ(x+ ) + φ(x )]. Hence A(t) = 1 2 u+ u. We now note that the inner equation can be rewritten as x U = 1 2 (U u + )(U u ), with u ± = u± (t). Then one can achieve the following equations [ ] 1 1 du U u + U u = 1 d x(u + 2 u ) U u + = log U u = 1 2 (u+ u ) x + C(t) = U u + u U where B(t) = e C(t). Therefore, = b( x, t) = B(t)e (u+ u ) x/2 U ( x, t) = u+ + b( x, t)u 1 + b( x, t). In order to determine B(t), we have to go to next order [See Holmes for more details]. You may find, in the end, B(t) = 1 + tφ (x + ) 1 + tφ (x ). 4 strongly localized perturbation theory This work is mainly done by Michael J. Ward. Consider a diffusion equation with small holes. Before we start to apply perturbation theory on the problem, recall Green s function in two and three dimensional space. Green s function is solution with single input data, especially in case of Laplace operator, u = δ(x x ), in R n, n = 2, 3. Then u 1/4π x x as x x in 3D and u log x x /2π as x x in 2D. In case of 3D, the Laplace operator in spherical coordinate with angular symmetry is u = u r r + 2 r u r, for x x >, and its solution is u(r) = B/r for some constant r. By taking integral in Ω ɛ, ball centered at x with radius ɛ, we get udx = u nds = 4πr 2 u r = 4πB = δ(x x )dx = 1. Ω ɛ Ω ɛ Ω ɛ It yields that u(r) = 1/4πr, which is the Green s function in 3D.

strongly localized perturbation theory 9 4.1 Eigenvalue asymptotics in 3D Let Ω be a 3D bounded domain with a hole of radius O(ɛ), denoted by Ω ɛ, removed from Ω. Consider an eigenvalue problem in Ω\Ω ɛ as follows: u + λu =, in Ω\Ω ɛ u =, on Ω. (4.1) u =, on Ω ɛ Ω\Ω ɛ u 2 dx = 1 We assume that Ω ɛ shrinks to a point x as ɛ. For example, we could assume Ω ɛ to be the sphere x x ɛ. The unperturbed problem is φ + λφ =, in Ω φ =, on Ω. (4.2) Ω\Ω ɛ φ 2 dx = 1 Assume this has eigenpair φ j (x) and µ j for j =, 1, with Ω φ jφ k dx = if j = k and φ (x) > for x Ω. We look for perturbed eigenpair near the φ (x) and µ. Expand λ µ + ν(ɛ)λ 1 + where (ν(ɛ) as ɛ.) In the outer region away from the hole, we take u φ (x) + ν(ɛ)u 1 (x) +. Since Ω ɛ {x } as ɛ, we have the following u 1 + µ u 1 = λ 1 φ, in Ω\{x } u 1 =, on Ω. (4.3) Ω 2u 1φ dx = Construct the inner solution near the hole. Let y = (x x )/ɛ and set V(x; ɛ) = u(x + ɛy). Then we find that V satisfies y V + λɛ 2 V =, outside of Ω = Ω ɛ /ɛ. Take V V + ν(ɛ)v 1 + and get y V =, outside Ω V =, on Ω. V φ (x ) as y Try a solution of it by V = φ (x )(1 V c (y)). Then V c satisfies y V c =, outside Ω V c = 1, on Ω. V c as y A classical result from PDE theory is V c C/ y as y where C is electrostatic capacitance of Ω, determined by shape and size of Ω. We now have [ V (x) φ (x ) 1 ɛc ]. x x It has to match φ (x ) + ν(ɛ)u 1 as x x. This yields that ν(ɛ) = ɛ and u ( x) φ (x )C/ x x as x x. To evaluate perturbed eigenvalue λ 1, return to equation (4.3). Since u 1 4πφ (x )C ( 1/4π x x ) as x x, then we have the modified problem { Lu 1 u 1 + µu 1 = λ 1 φ + 4πCφ (x )δ(x x ), in Ω u 1 =, on Ω. (4.4)

strongly localized perturbation theory 1 Use Green s identity φ Lu 1 u 1 Lφ dx = Ω Ω φ n u 1 u 1 n φ ds. Since φ = u 1 = on Ω and Lφ =, we have = φ Lu 1 dx = φ [ λ 1 φ + 4πCφ (x )δ(x x )]dx, and it yields Ω λ 1 = 4πCφ2 (x ) Ω φ2 dx. Ω Therefore, λ µ + ɛλ 1. Remark 2. 1. Let us assume that u = on Ω is replaced by the no-flux condition on Ω. Then ɛ = problem becomes φ + µφ =, n φ =, Ω φ2 dx = 1 Ω. The principal eigenvalues µ = and φ (x) = 1/ ω 1/2. In this case, λ 1 4πCɛ/ Ω (to leading order it is independent of location x.) 2. For multiple holes Ω ɛj for j = 1, n and well-separated, its eigenvalue expansion is λ µ + 4πɛ j c j [φ (x j )] 2 / Ω φ2 dx. 4.2 Eigenvalue asymptotics in 2D In the same fashion with 3D case, we want to find an asymptotic expansion of the same problem (4.1), but 2D. Let µ and φ be principal eigenpair of unperturbed problem (4.2). Set λ µ + ν(ɛ)λ 1 + for eigenvalue and u φ + ν(ɛ)u 1 + in outer region, where ν(ɛ) as ɛ. Then the equation for second term of outer expansion is (4.3). In the inner region, set y = (x x )/ɛ and take u(x) = ν(ɛ)v (y) where y V =. We want V (y) A log y as y. To do so, setting V = A V c where { y V c =, y Ω V c =, y on Ω (4.5) gives that V c log y log d + O(1/ y ), as y where d is logarithmic capacitance determined by shape of Ω. It is interesting enough to notice the logarithmic capacitance of simple objects in the table. Then write inner solution in outer variable u(x) ν(ɛ)a log y d ν(ɛ)a [ log(ɛd) + log x x ]. Matching solution yields that φ (x ) + ν(ɛ)u 1 (x) log(ɛd)a ν(ɛ) + A ν(ɛ) log x x, as x x. In order to match the conditions, set ν(ɛ) = 1/ log(ɛd). Then unknown constant A = φ (x ). Thus, u 1 (x) φ (x ) log x x,

strongly localized perturbation theory 11 Ω Geometric info Capacitance d Circle radius a a Ellipse radius a, b (a + b)/2 Triangle side h 3[Γ(1/3)] 3 h/(8π 2 ) Rectangle side h [Γ(1/4)] 2 h/(4π 3/2 ) Table 1: The logarithmic capacitance in 2D for simple geometric figures. as x x. Hence, by the same procedure in 3D case by using Green s identity, one can find eigenvalue expansion λ µ + 2π ν(ɛ) [φ (x )] 2 Ω φ2 dx. Remark 3. Further terms in expansion yields λ µ + A 1 ν + A 2 ν 2 + A 3 ν 3 +. Its potential problem is that the log decreases very slowly as ɛ decreases. Then the remaining term is quite large and break the asymptotic expansions. By summing the log series, one can solve the problem. 4.3 Summing all logarithmic terms Consider Poisson s equation in a domain with one small hole given w = B, in Ω\Ω ɛ w =, on Ω. (4.6) w =, on Ω ɛ In the outer region, set w(x; ɛ) = w (x; ν(ɛ)) + σ(ɛ)w 1 (x; ν(ɛ)) + where ν(ɛ) = 1/ log(ɛd) and σ ν k for any k >. It gives the outer equation w = B, in Ω\{x } w =, on Ω. (4.7) w is singular, as x x In the inner region, set y = (x x )/ɛ and V(y; ɛ) = w(x + ɛy; ɛ). Expand V(y; ɛ) = V (y; ν(ɛ)) + µ (ɛ)v 1 (y; ν(ɛ)) + where µ ν k for all k >. Then V satisfies { y V =, outside Ω. (4.8) V =, on Ω The leading order matching condition is lim w (x; ν) lim V (y; ν). x x y Introduce an unknown function γ = γ(ν) with γ() = 1 and let V (y; ν) = νγv c (y). Then it follows that y V c =, outside Ω V c =, on Ω. V c log y, as y

exercises 12 Thus, V c (y) log y log d + O(1/ y ) for 1 y. In original coordinate, V (y; ν) γ + νγ log x x. Matching condition gives w νγ log x x + γ as x x. So outer problem is w = B, in Ω\{x } w =, on Ω. (4.9) w γ + νγ log x x as x x Introduce w OH (x) and G(x; x with { woh = B, in Ω w OH =, on Ω, { G = δ(x x ), in Ω G =, on Ω. One can find G(x; x ) = 1 2π log x x + R (x; x ) where R is the regular prt of Green s function which converges as x x. Then we can write down the solution w (x; ν) = w OH (x) + 2πγνG(x; x ). As x x, we obtain the asymptotic condition [ ] 1 w OH (x ) + 2πγν 2π log x x + R(x; x ) = γ + γν log x x, and it yields γ(ν) = w OH (x ) 1 2πνR (x ; x ). Therefore, the final expansion of the Poisson equation is w(x) w OH (x) + ν(ɛ) 1 2πR (x ; x )ν(ɛ) 2πw OH(x )G(x; x ). 5 exercises Problem 1. Find a composite expansion of the solution to the following problems on x [, 1] with a boundary layer at the end x = : 1. ɛy + 2y + y 3 =, y() =, y(1) = 1/2. Proof. To find outer expansion y, set y y + and balance O(1) terms + 2y + y3 =. It yields a general solution y 2 = x + D for some constant D. Since the boundary layer is at x =, boundary condition at x = 1 determines integrating constant D and yields y (x) = 1 x + 3. Now, construct inner expansion near x =. Setting x = x/ɛ α and Y( x) = y(x) yields ODE for inner solution ɛ 1 2α Y + 2ɛ α Y + Y 3 =.

exercises 13 In order to balance terms, require α = 1 and Y Y +. Then we achieve Y + 2Y =. A general solution of Y is Y ( x) = C(1 e 2 x ) with boundary condition Y () =. Matching condition lim Y ( x) = lim y (x) x x + yields C = 1/ 3. Therefore, the composite expansion of the solution is ( x ) y(x) y (x) + Y 1 = ɛ 3 1 1 e 2x/ɛ. x + 3 3 2. ɛy + (1 + 2x)y 2y =, y() = ɛ, y(1) = sin(ɛ). Proof. To find outer expansion y, set y y + ɛy 1 + and balance O(1) terms + (1 + 2x)y 2y =, y () = y (1) =, and its general solution is y (x) = C(2x + 1). Since we know that it has a boundary layer at x =, match boundary condition and get C =. Thus y (x) =. Balancing O(ɛ) terms yields that y + (1 + 2x)y 1 2y 1, y 1() = y 1 (1) = 1, and its solution with boundary condition at x = 1 is y 1 (x) = (2x + 1)/3. It follows that the outer expansion is y(x) ɛ (2x + 1) +. 3 Now, consider inner expansion near x =. Setting x = x/ɛ α and Y( x) = y(x) yields ODE for inner solution ɛ 1 2α Y + ɛ α Y + 2 xy 2Y =. To balance the equation, it requires that α = 1 by setting Y Y. Then we have Y + Y =, Y () = = Y ( x) = D(1 e x ). Matching condition gives lim Y( x) = D = lim y(x) = ɛ x x 3. Therefore, the composite expansion of the solution is y(x) ɛy 1 (x) + (Y(x/ɛ) ɛ/3) = ɛ 3 (2x + 1 e x/ɛ ).

exercises 14 Problem 2. Consider the integral equation x ɛy(x) = q(x) [y(s) f (s)]sds, x 1, where f (x) is positive and smooth. 1. Taking q(x) = 1 find a composite expansion of the solution y(x). [Hint: convert to an ODE.] Proof. Observe that ɛy() = = y() =. Taking derivative on given integral equation gives ɛy (x) + xy(x) = x f (x). One can get the outer expansion by setting y(x) y (x) + xy (x) = x f (x) = y (x) = f (x), < x 1. Since f is positive function lim x y (x) = f () >, which does not match boundary condition. It implies that the expansion has boundary layer at x =. Scale near x = by taking a new coordinate x = x/ɛ α and Y( x) = y(x). In this coordinate, the smooth function f (x) can be count as a constant f (x) f (). It follows the ODE for inner expansion ɛ 1 α Y + ɛ α xy = ɛ α x f (). To balance the equation, it requires 1 α = α, i.e. α = 1/2 and its general solution with boundary condition Y() = yields the first term inner expansion ( ) Y( x) = f () 1 e x2 /2, and it matches with outer solution lim Y( x) = f () = lim f (x) = lim y (x). x x x Therefore, the composite expansion of the integral equation is y(x) f (x) f ()e x2 /2ɛ. 2. Generalize to the case that q(x) is a positive smooth function. Proof. It is also true that y() = because f is positive function. Taking derivative and substituting integral term gives ɛy = ɛ q y q(y f )x, q and one can rewrite it as ( ) y ( ) y ɛ + xq = x f q q = ɛz + xqz = x f

exercises 15 by setting z = y/q. In the same fashion in part 1., obtain outer expansion by balancing O(1) z(x) z (x) = f (x) q(x) = y(x) y (x) = f (x). Since f, q are positive, then z has boundary layer at x =. With the same argument in part 1., one can get the ODE for inner expansion Z( x) Z + xq()z = x f () = Z( x) = f () q() ( 1 e q() x2 /2 ), ( ) that is Y( x) f () 1 e q() x2 /2. Therefore, its composite expansion is y(x) f (x) f ()e q(x)x2 /2ɛ. 3. Show that solution of part 2. still holds if q(x) is continuous but not differentiable everywhere on [, 1]. Proof. The basic idea showing the claim is to derive all the expansions from integral equation. For the outer expansion, setting y y gives x = q(x) (y (s) f (s))sds = x (y (s) f (s))sds = because q(x) is positive. Without worrying about differentiability of q, take derivative on the equation and get same outer expansion y (x) = f (x). In the similar way, to find the inner expansion, set the new coordinate x = x/ɛ α and Y( x) = y(x). By approximating continuous function q(x) = q() and f (x) = f () in the boundary layer, it follows that ɛα ɛ 1 α x Y = q() (Y(s/ɛ α ) f ())sds. Now, one can take derivative and get the same differential equation for inner expansion. Therefore, one can achieve the same composite expansion. Problem 3. (Boundary layer at both ends) Find a composite expansion of the following problem on [, 1] and sketch the solution: ɛy + ɛ(x + 1) 2 y y = x 1, y() =, y(1) = 1. Proof. To find outer expansion y, set y y + and balance O(1) terms = y + x 1 = y (x) = 1 x. It does not satisfy neither boundary conditions. Hence there are two boundary layer at x = and x = 1. First, consider boundary layer at x = by setting x = x/ɛ α for α > and U( x) = y(x). It follows ODE for U ɛ 1 2α U + (ɛ 1+α x 2 + ɛ 2 x + ɛ 1 α )U = U 1 + ɛ α x.

exercises 16 Since α >, the smallest order of LHS is O(1 2α) and RHS is O(1). balance them, require α = 1/2 and setting U U provides To U = U 1 = U( x) = Ae x + Be x + 1. By boundary condition at x =, y() =, we achieve A + B + 1 =. Matching condition yields lim x U( x) = lim y(x) = 1 = A =, x + and U( x) = 1 e x. Similarly, to find inner expansion at x = 1, set ξ = (x 1)/ɛ β and V(ξ) = y(x). It provides ODE for V ɛ 1 2β V + (ɛ 1+β x 2 + ɛ 4 x + ɛ 1 β 4)V = Vɛ β x. Since β >, the smallest order of LHS is O(1 2α) and RHS is O(1). balance them, require β = 1/2 and setting U U provides To V = V = V(ξ) = Ce ξ + De ξ. By boundary condition at x = 1, that is y(1) = 1, we obtain C + D = 1. Matching outer and inner layer near x = 1 gives that lim ξ V(ξ) = lim y(x) = = D =, x 1 and V(ξ) = e ξ. Therefore, the composite expansion of the solution is and it follows that [ ( x ) ] [ y(x) y (x) + U ɛ 1/2 1 + U y(x) 1 x e x/ɛ1/2 e x 1/ɛ1/2. ( ) ] x 1 ɛ 1/2, Problem 4. (Matched asymptotics can also be used in the time domain) The Michaelis- Menten reaction scheme for an enzyme catalyzed reaction is ds = s + (µ + s)c, dt ɛ dc = s (κ + s)c, dt where s() = 1, c() =. Here s(t) is the concentration of substrate, c(t) is the concentration of the catalyzed chemical product, and µ, κ are positive constants with µ < κ. Find the first term in the expansions in the outer layer, the initial layer around t =, and the composite expansion. Proof. Find the expansions in the outer layer by setting s s + and c c + and balancing O(1) terms ds dt = s + (µ + s )c, = s (κ + s )c.

exercises 17 It yields that s (t) 1 + κ log s (t) = (µ κ)t, c (t) = s (t) s (t) + κ. Notice that s is implicitly determined. One can observe that c has a layer near t =. Setting t = t/ɛ α, S( t) = s(t) and C( t) = c(t) gives the system of ODE ɛ α ds ɛ d t 1 α dc d t = S + (µ + S)C, = S (κ + S)C It requires that α = 1 to balance equation for C not same with outer expansion. By setting S S and C C +, it follows that ds =, d t dc = S (κ + S )C. d t First equation with initial condition s() = 1 gives that S ( t) = 1. Hence we write ODE for C as dc d t = 1 (κ + 1)C = C ( t) = 1 ( ) 1 e (κ+1) t. κ + 1 Fortunately, this solution satisfies matching condition lim C ( t) = 1 t κ + 1 = s () s () + κ = lim c (t). t Therefore, the composite solution of perturbation equation is c(t) c (t) 1 κ + 1 e (κ+1)t/ɛ where c is implicitly determined by s. Problem 5. (Implicit inner solution) A classical model in gas lubrication theory is the Reynolds equation ɛ d ( H 3 yy ) = d (Hy), < x < 1, dx dx where y() = y(1) = 1. Here H(x) is a known, smooth, positive function with H() = H(1). 1. Suppose that there is a boundary layer at x = 1. Construct the first terms of the outer and inner solutions. Note that the leading order term Y of the inner solution is defined implicitly according to (x 1)/ɛ = F(Y ). Calculate the function F. Proof. Setting y y and balancing O(1) terms yields outer solution equation = d dx (Hy ) = y (x) = C H(x) where C is constant. Since we have boundary layer at x = 1, then applying boundary condition at x = to outer solution gives y (x) = H()/H(x).

exercises 18 In the inner layer, setting x = (x 1)/ɛ α and Y( x) = y(x) provides the ODE for inner solution ɛ 1 2α d ( H 3 YY ) = ɛ α d d x d x (HY). Since the inner layer near x = 1, then continuous function H can be approximated as H(x) H(1). It follows that ɛ 1 2α H 3 (1) d d x (Y Y ) = ɛ α H(1) d d x Y. for first expansion term Y of Y. To balance the equation, it requires α = 1 and now get d d x (Y Y ) = 1 d H 2 (1) d x Y. The general solution of ODE is given by Y ( x) 1 C log 1 + Y C = x H 2 (1) with boundary condition Y () = 1. Matching condition gives lim Y ( x) = lim y (x) = H() x x 1 H(1). It determines C = H()/H(1). Thus we have F(Y ) = H 2 (1)(Y 1) + H()H(1) log 1 H(1)Y H() = x. 2. Use matching to construct the composite solution. Proof. By the result from part 1., one can write the composite solution as y(x) H() [ ( ) x 1 H(x) + F 1 ɛ H() ]. H(1) 3. Show that if the boundary layer was assumed to be at x =, then the inner and outer solutions would not match. Proof. It follows the same procedure in part 2., but achieve different F F(Y ) = H 2 ()(Y 1) + H()H(1) log 1 H()Y H(1) = x. However, as x, the RHS tends to negative infinity. It does not match the conditions.

exercises 19 Problem 6. (Boundary layer at both ends) In a one-dimensional bounded domain, the potential φ(x) of an ionized gas satisfies with boundary conditions d2 φ + h(φ/ɛ) = α, < x < 1, dx2 φ () = γ, φ (1) = γ. Charge conservation requires 1 h(φ(x)/ɛ)dx = β. The function h(s) is smooth and strictly increasing with h() =. The positive constants α and β are known (and independent of ɛ), and the constant γ is determined from the conservation equation. 1. Calculate γ in terms of α and β. Proof. Integration on given differential equation over [, 1] gives 1 [φ (1) φ ()] + h(φ(x)/ɛ)dx = α 1 = 2γ + β = α, and it yields γ = (β α)/2. 2. Find the exact solution for the potential when h(s) = s. Sketch the solution for γ < and small ɛ, and describe the boundary layers that are present. Proof. With h(s) = s, we have d2 φ dx 2 + φ ɛ = α, and its general solution is ( ) ( ) x x φ(x) = A sinh + B cosh + ɛα, ɛ ɛ where A, B are constants. Then one can obtain it derivative φ (x) = 1 ( ) ( )] x x [A cosh + B sinh ɛ ɛ ɛ To determine A and B, imposing boundary conditions to the general solution we have φ () = A ɛ = γ, and φ (1) = 1 ( ) ( )] 1 1 [A cosh + B sinh = γ ɛ ɛ ɛ Solving them for A, B yields A = γ ɛ, B = γ ɛ 1 + cosh(1/ ɛ) sinh(1/. ɛ)

exercises 2 Then it follows that φ (x) = [ ( ) ( )] γ x 1 x sinh(1/ sinh + sinh. ɛ) ɛ ɛ For x =, 1, then φ (x) decays to zero as ɛ. Since φ () and φ () are nonzero, then it implies that φ has boundary layers at x =, 1. 3. Suppose that h(s) = s 2k+1, where k is a positive integer, and assume β < α. Find the first term in the inner and outer expansions of the solution. Proof. With h(s) = s 2k+1, we have ɛ 2k+1 d2 φ dx 2 + φ2k+1 = ɛ 2k+1 α, (5.1) with same boundary conditions. For ɛ =, φ has a trivial solution. Thus we expand φ as and its derivatives are φ ɛ p (φ + ɛ q φ 1 + ), φ ɛ p (φ + ɛq φ 1 + ), φ ɛ p (φ + ɛq φ 1 + ). First, consider the boundary layer at x =. Rescale as x = x/ɛ r and set Φ( x) = φ(x). Then we have d dx d d x d dx = d ɛ r d x. It allows the governing equation in the boundary layer at x = to be ɛ 2k+1 2r Φ + Φ 2k+1 = ɛ 2k+1 α, (5.2) with the boundary condition ɛ r Φ () = γ, (5.3) and it requires r = p and gives Φ () = γ. Then (5.2) turns out to be ɛ 2k+1 p (Φ + ɛq Φ 1 + )+ ɛ (2k+1)p (Φ + ɛφ 1 + ) 2k+1 = ɛ 2k+1 α. To construct a boundary layer at x =, the only remaining case is to balancing O(ɛ 2k+1 p ) and O(ɛ (2k+1)p ) and it requires p = (2k + 1)/(2k + 2). Then we have a differential equation for boundary layer at x = Φ + Φ2k+1 =. (5.4) Multiplying Φ and perform integration gives 1 2 (Φ )2 + Φ2k+2 2k + 2 = C, for some constant C. As x, Φ matches with the outer solution φ(x) = for < x < 1. It implies that C =. Then we have its general solutions [ k 1/k Φ ( x) = (± x D)], k + 1

exercises 21 for some constant D. Its derivative becomes Φ ( x) = 1 [ k (k+1)/k ( (± x D)] ± k ). (5.5) k k + 1 k + 1 Imposing boundary condition at x = gives 1 [ kd ] (k+1)/k k k + 1 ( ± k k + 1 ) = γ. Since γ = (β α)/2 <, then we choose the negative sign and determine D such that kd = ( γ k + 1) k/(k+1) := λ. k + 1 Setting κ = k/ k + 1 gives Φ ( x) = (λ κ x) 1/k. (5.6) Since the boundary layer at x = 1 satisfies the same differential equation (5.4), then one can derive the lowest order boundary layer Ψ with rescaling ˆx = (1 x)/ɛ r Ψ ( ˆx) = (λ + κ ˆx) 1/k. (5.7) Therefore, the match asymptotic expansion of the differential equation is y(x) [ λ κx ] [ 1/k + ɛ r λ + ] κ(1 x) 1/k ɛ r, (5.8) where r = (2k + 1)/(2k + 2). 4. Can one construct a composite solution using the first terms? Proof. Not exactly :) Problem 7. (Internal boundary layer) Consider the problem ɛy + y(1 y)y xy =, < x < 1, with y() = 2 and y(1) = 2. A numerical solution for small ɛ shows that there is a boundary later at x = 1 and an internal layer at some x, where y. 1. Find the first term in the expansion of the outer solution. Assume that this function satisfies the boundary condition at x =. Proof. Setting y y + and balancing O(1) yields the following equation y (1 y )y xy =. Since we assume that it satisfies y() = 2, then y (x) =. Then it follows that (1 y )y = x = y (x) = 1 + 1 x 2.

exercises 22 2. Explain why there cannot be a boundary layer at x = 1, which links the boundary condition at x = 1 with the outer solution of part 1. evaluated at x = 1. Proof. If it has a boundary layer at near x = 1, then the solution connects lim t 1 y (x) = 1 and boundary condition y(1) = 2. Then there is x in the boundary layer such that < y < 1. Since y, y < in the layer, then one can conclude ɛy + y(1 y)y xy <, that is such expansion cannot satisfy IVP. 3. Assume that there is an interior layer at some point x, which links the outer solution calculated in (a) for x < x with the outer solution y for x < x < 1. From the matching show that x = 3/2. Note that there will be an undetermined constant. Proof. In the interior layer, scale the coordinate as x = (x x )/α and set Y( x) = y(x). Then one can achieve equation for the interior solution ɛ 1 2α Y + ɛ α Y(1 Y)Y (ɛ α x + x )Y =. Expanding the interior solution as Y Y + and setting α = 1 yields the balance equation for O(ɛ 1 ) terms Y + Y(1 Y)Y =. Taking integration on both sides, Y + 1 2 Y2 1 3 Y3 = C, where C is constant. From the right matching condition, lim x Y( x) = lim x Y ( x) =. Hence, C =. Invoking partial fraction to the separable ODE gives the general solution 1 6 x + D = 2 9 log Y 1 3Y + 2 log(3 2Y). 9 The left matching condition yields that lim Y( x) = 3 x 2 = lim y(x) = 1 + 1 x x x 2, and it implies that x = 3/2. With the undetermined constant D, the interior expansion Y( x) = G 1 ( x) where [ G(Y) = 6 2 9 log Y 1 3Y + 2 ] log(3 2Y) D. 9 4. Given the interior layer at x, construct the first term in the expansion of the inner solution at x = 1.

references 23 Proof. In the similar fashion, setting ξ = (x 1)/ɛ β and V(ξ) = y(x). Then we get the same ODE with the left matching condition V + V(1 V)V =. Then it follows the general solution 1 6 ξ + E = 2 9 log( V) 1 3V + 2 log(3 2V), 9 where E is a constant. Since V() = 2, then we have E = 2 9 log 2 + 1 6 + 2 9 log 7 = 1 6 + 2 9 log ( 7 2 ). Therefore, the expansion in the inner layer at x = 1 is V(ξ) = H 1 (ξ) where [ H(V) = 6 2 ( ) V 9 log 1 2 3V + 2 ( ) 3 2V 9 log 1 ]. 7 6 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