Fourth Mississippi State Conference on Differential Equations and Computational Simulations, Electronic Journal of Differential Equations, Conference 03, 1999, pp 109–118. http://ejde.math.swt.edu or http://ejde.math.unt.edu ftp ejde.math.swt.edu or ejde.math.unt.edu (login: ftp)

On matched asymptotic analysis for laminar channel flow with a turning point ^{∗}

Chunqing Lu

Abstract

This paper presents a formal analysis of the asimptotic behaviour of solutions of type III for the Berman equation

ϵf^{iv} = ff′′′ − f′f′′, f(0) = f′′(0) = f′(1) = f(1) − 1 = 0 ,

where f describes a laminar flow in a channel with porous walls. A solution has a nonlinear turning point (1 − Δ), i.e. f(1 − Δ) = 0 for some Δ(ϵ). It is shown that

f(η) ∼ − \frac{1 − Δ}{πΔ} sin \frac{πη}{1 − Δ} ,

as ϵ → 0^{+}, for η ∈ [0, 1 − Δ) where Δ satisfies

\frac{Δ}{ϵ} e^{Δ/ϵ} ∼ \frac{1}{2eπ^{9}ϵ^{8}} .

1 Introduction

The laminar flow of a viscous fluid in a rectangular channel with porous walls is governed by the Berman equation [1]

f^{iv} = R(ff′′′ − f′f′′)  (1)

with boundary conditions

f(0) = f′′(0) = 0, f(1) = 1, f′(1) = 0  (2)

where f = f(η) is the unknown function related to the stream function of the flow, and R is the Reynold number of the flow. The case R > 0 corresponds to suction while R < 0 to injection. Setting ϵ = 1/R, one gets a singular perturbation problem for small |ϵ|:

ϵf^{iv} = ff′′′ − f′f′′  (3)

∗Mathematics Subject Classifications: 34B15, 34E05, 34E20.

Key words: singular perturbations, turning point, laminar flow, transcendental terms.

\begin{matrix}c \\ ◯ \end{matrix}2000 Southwest Texas State University and University of North Texas.

Published July 10, 2000.

subject to boundary conditions (2). The boundary value problem (3)-(2) has been investigated by many authors. In [2, 8, 9] it is shown that there exists a unique solution for each ϵ ≤ 0. When ϵ is positive and sufficiently small, the boundary value problem admits at least three solutions [7] which are classified as follows: Type I if the solution is increasing concave down, type II if the solution is increasing with an inflection point, and type III if the solution is non-monotone with a turning point.

A turning point, denoted by z_{ϵ} = 1 − Δ_{ϵ}, is defined as the value in (0, 1) at which the solution vanishes. Among the most interesting questions is the asymptotic behavior of the solutions as ϵ → 0, [2, 4, 10]. Hastings et. al. [2] showed that as ϵ → 0^{−}, f(η) → sin \frac{2}{π} η uniformly on [0, 1]. As ϵ → 0^{+}, however, type I and type II solutions approach the linear function in compact subsets of [0, 1) (see [4]), but type III solutions become unbounded on the left side of the turning point [5]. In addition, for type III solutions, the turning point z_{ϵ} → 1 or Δ_{ϵ} → 0, as ϵ → 0^{+}. Transcendentally small terms in the asymptotic expansion of the solutions of types I and II have been found by Robinson, Terrill and Lu et. al. Their validity has been confirmed by McLeod [6]. As for the type III solutions, the unboundedness and the existence of turning points cause the asymptotic analysis to be more difficult than for types I and II. Robinson [7] and Zaturska et. al. [11] used the method of matched asymptotic expansions and gave estimates on the asymptotic behavior for the type III solutions in the seventies and the eighties respectively. In 1994, MacGillivrary and Lu [5] considered the transcendental terms and formally obtained an asymptotic formula for type III solutions,

f ∼ −κ sin \frac{πη}{1 − Δ}

for η ∈ [0, 1 − Δ), where κ ∼ \frac{1−Δ}{πΔ} and Δ = Δ_{ϵ} satisfies the asymptotic formula

\frac{Δ}{ϵ}e^{\frac{Δ}{ϵ}} ∼ \frac{1}{2π^{9}ϵ^{8}}

for sufficiently small ϵ > 0. It was shown in [5], with numerical comparison, that this result is better than those given by Robinson and Zaturska et. al.. In 1997, Lu [3] rigorously proved that for the type III solutions, as ϵ → 0^{+},

f(η) ∼ −\frac{1 − Δ}{πΔ} sin \frac{πη}{1 − Δ}

uniformly on [0, 1 − Δ], and Δ satisfies

\frac{Δ}{ϵ}e^{\frac{Δ}{ϵ}} ∼ \frac{1}{2eπ^{9}ϵ^{8}} .

Interested readers may find that the formal result of [5] and the rigorous proof in [3] are of the same orders, but the coefficients are different. The purpose of this paper is to improve the formal matching method used in [5] to get the correct asymptotic result. It implies that the rigorous analysis is important for not only confirming but also improving the formal matching method, although sometimes it is hard and tedious. Also, the paper shows that the formal matching technique can exactly provide the asymptotic behavior of the solution when it is properly applied. The aim of the paper is to provide a general method of the formal asymptotic analysis to deal with the singular perturbation problems with boundary and internal layers. The paper is organized as follows. The outer solution that behaves like a sine function on the left side of the turning point is studied in §2. The inner solution near the turning point is given in §3 and §4. The Poincaré expansion is studied in §3, and the transcendentally small terms in §4. Finally, in §5, the relation between Δ and ϵ is determined by a formal matching process on the right side of the turning point.

Throughout the paper, the process ϵ → 0 means ϵ → 0^{+}, and the solution of the boundary value problem means the type III solution. Also, the crucial property f^{iv}(η) < 0 for η ∈ (0, 1], whose proof can be found in [2], is applied repeatedly in the paper.

2 Approximation on the left side of the turning point

Setting ϵ = 0 in (3), we obtain the reduced equation

ff′′′ − f′f′′ = 0 .  (4)

The initial condition f(0) = f′′(0) = 0 is then imposed. We see that equation (4) may have three possible solutions: Cη, C sin λη, and C sinh λη where C and λ are constants. Recall that the turning point is defined by f(1 − Δ) = 0. Thus, the outer solution on [0, 1 − Δ] that satisfies f(1 − Δ) = 0 must be neither the linear function nor the hyperbolic sine function. Thus, the solution of (4) with f(0) = f(1 − Δ) = 0 is C sin λη. Since f(η) ≤ 0 on [0, 1 − Δ], we can write

f(η) ∼ −κ sin \frac{πη}{1 − Δ},  (5)

where κ > 0 is a constant related to ϵ, and assume the convergence is valid uniformly on compact subintervals of [0, 1 − Δ). This is the outer solution on the left side of the turning point. It then remains to determine the values of κ and Δ in terms of ϵ as ϵ → 0.

3 The Poincaré expansion at the turning point

Noting that the turning point is moving toward the right end-point of the interval [0,1] as ϵ → 0, we introduce the interior-layer variable τ = \frac{η−1+Δ}{Δ}. Then, the turning point, which is τ = 0, becomes immobilized as ϵ → 0, and the interval [1 − Δ, 1] of η becomes the interval [0, 1] of τ.

Let f(η) = f(Δτ + 1 − Δ) = f̄(τ) and denote ϵ̄ = \frac{ϵ}{Δ}. The original equation (3 ) then takes the form

ϵ̄f̄^{iv} = ff̄′′′ − f̄′f̄′′,  (6)

with boundary conditions

f̄(0) = 0, f̄(1) = 1, f̄′(1) = 0.  (7)

As suggested in [5], assume ϵ̄ → 0 as ϵ → 0, which implies that (6)-(7) is again a singular perturbation problem.

Now we set ϵ̄ = 0 to get the reduced equation,

f̄f̄′′′ − f̄′f̄′′ = 0,  (8)

which must have a linear solution in a outer region [0, 1 − μ(\overline{ϵ})] of τ where μ(\overline{ϵ}) = o(1). This is seen from f̄(0) = 0, f̄′ > 0 for all τ ≥ 0, and f̄(1) = 1. In general, equation (8) would have three solutions: a linear function, sin \frac{π}{2} τ and \frac{sinh λτ}{sinh λ} where λ > 0 is a constant. The other two are excluded because they have positive fourth derivatives, which contradicts the property f^{iv} < 0 for η > 0. Thus, the outer solution, up to order O(1), for τ ∈ [0, 1) is a linear function, i.e.,

\overline{f}(τ) = λτ + …  (9)

where λ is to be determined.

The inner solution near the right end-point τ = 1 can be formally found by introducing the boundary layer variable x^{∗} defined by

ϵ̄x^{∗} = 1 − τ.  (10)

Let f̃(x^{∗}) = f̄(\frac{1−τ}{ϵ̄}). Then, the equation (6) becomes

f̃^{iv} = −f̃f̃′′′ + f̃′f̃′′.  (11)

At x^{∗} = 0, the initial conditions are f̃(0) = 1 and f̃′(0) = 0. Assume that f̃ ∼ 1 is valid for x^{∗} ∈ [0, A] where A is any positive constant. Then, we can set

f̃(x^{∗}) = 1 + ϵ̄f_{1}(x^{∗}) + ϵ̄^{2}f_{2}(x^{∗}) + ….  (12)

Substituting (12) into (11) and extract the powers of ϵ̄, we get the following system of equations:

f_{1}^{iv} + f′′′_{1} = 0, f_{1}(0) = f′_{1}(0) = 0,  (13)
f_{2}^{iv} + f′′′_{2} = −f_{1}f′′′_{1} + f′_{1}f′′_{1}, f_{2}(0) = f′_{2}(0) = 0,
f_{k}^{iv} + f′′′_{k} = − ∑_{i=1}^{k−1} f_{i}f′′′_{k−i} + ∑_{i=1}^{k−1} f′_{i}f′′_{k−i}, f_{k}(0) = f′_{k}(0) = 0, k = 3, 4, ….

The general solution of (13) is

f_{1}(x^{∗}) = a_{1}(−1 + x^{∗} + e^{−x^{∗}}) + \frac{1}{2}b_{1}x^{∗2}

where a_{1} and b_{1} are constant to be determined by matching with appropriate terms from the outer solution. So, the inner expansion is

f̃(x^{∗}) = 1 + ϵ̄[a_{1}(−1 + x^{∗} + e^{−x^{∗}}) + \frac{1}{2}b_{1}x^{∗2}] + o(ϵ̄).  (14)

Recall that the leading term of outer solution is given in (9 ). Thus, the Poincaré expansion of the outer solution can be written as

f̄(τ) = λτ + ∑_{i=1}^{∞} \overline{ϵ}^{i}g_{i}(τ).  (15)

If we match (14) with (15) to order O (1), then λ = 1. Substituting (15) with λ = 1 into (6) and comparing the powers of ϵ̄, we obtain

τg′′′_{1} − g′′_{1} = 0,  (16)

τg′′′_{i} − g′′_{i} = g_{i−1}^{iv} − ∑_{j=1}^{i−1} g_{j}g′′′_{i−j} + ∑_{j=1}^{i−1} g′_{j}g′′_{i−j},  (17)

for i = 2, 3, . . .. Since f(0) = 0, the boundary conditions

g_{i}(0) = 0, i = 1, 2, …  (18)

are imposed. We may also assume that all the functions g_{i}, i = 1, 2, …, are analytic at τ = 0. Thus, the function g_{1} can be solved from (16) and (18), which is

g_{1} = \frac{1}{6}A_{1}τ^{3} + B_{1}τ

for some constant A_{1} and B_{1}. Writing the outer expansion \overline{f}(τ) = τ +\overline{ϵ}g_{1}(τ)+… in terms of the variable x^{∗}, one obtains

f̄(τ) = (1 − ϵ̄x^{∗}) + ϵ̄[\frac{1}{6}A_{1}(1 − ϵ̄x^{∗})^{3} + B_{1}(1 − ϵ̄x^{∗})] + o(ϵ̄)
= 1 + ϵ̄[−x^{∗} + \frac{1}{6}A_{1} + B_{1}] + o(ϵ̄).  (19)

Now, matching the outer and inner expansions (14) and (19) to order O(ϵ̄), we find that

a_{1} = −1, b_{1} = 0, \frac{1}{6}A_{1} + B_{1} = 1.

Thus, at this stage, we have the following outer and inner approximations up to O(ϵ̄) :

outer: f̄ = τ + ϵ̄[\frac{1}{6}A_{1}τ^{3} + B_{1}τ] + o(ϵ̄),  (20)

and

inner: f̃ = 1 + ϵ̄[1 − x^{∗} − e^{−x^{∗}}] + o \binom{}{ϵ̄} .  (21)

Next, we want to argue A_{1} = 0, which implies that B_{1} = 1, hence, g_{1} is linear. To see this, we set i = 2 in (17) to obtain

τg′′′_{2} − g′′_{2} = \frac{1}{3}A_{1}^{2}τ^{3}.  (22)

Differentiating the both sides of (22) produces τg_{2}^{iv} = A_{1}^{2}τ^{2}. If A_{1} ≠ 0, the fourth derivative formally satisfies

f̄^{iv}(τ) = \overline{ϵ}^{2}A_{1}^{2}τ + o(\overline{ϵ}^{2}),

hence, f̄^{iv} > 0 for τ ≥ \frac{1}{2}, which contradicts the property f^{iv} < 0 for η > 0. Moreover, it can be argued that every function g_{i}, i = 1, 2, 3, ..., is linear. To see this, we use an argument by contradiction. Suppose not. Then, there would be the smallest integer greater than two, say k, such that g_{k} is the first nonlinear function, and g_{1}, g_{2}, ..., g_{k−1} are linear. Substitution of g_{1}, g_{2}, ..., g_{k−1} into (17) for i = k yields

τg′′′_{k} − g′′_{k} = 0,

which implies g_{k} = \frac{1}{6}A_{k}τ^{3} + B_{k}τ. If A_{k} ≠ 0, from (17), the function g_{k+1} satisfies

τg′′′_{k+1} − g′′_{k+1} = \frac{1}{3}A_{k}^{2}τ^{3},

which leads to τg_{k+1}^{iv} = A_{k}^{2}τ^{2}. Similar to the case k = 1 above, we would have f^{iv} > 0 for η > 0, again a contradiction. Thus, the Poincaré expansion is a linear function γ(\overline{ϵ})τ where

γ(\overline{ϵ}) = 1 + \overline{ϵ} + B_{2}\overline{ϵ} + B_{3}ϵ̄^{3} + ….  (23)

Following a similar procedure, one can determine the values of B_{2}, B_{3}, . . .. However, for the purpose of this paper, the result for B_{1}(= 1) is enough. In addition, we can assume that B_{n}, n = 2, 3, …, are all of order O(1) to guarantee the convergence γ(ϵ̄) → 1 as ϵ̄ → 0. From the inner expansion (21) we see that the exponentially small terms should be studied. This suggests that to have a precise outer expansion f̄(τ) for τ ∈ [0, 1), we may use a general form including both power terms of \overline{ϵ} and transcendentally small terms,

∑_{i=0}^{∞} \overline{ϵ}^{i}g_{i}(τ) + ∑_{i=1}^{∞} δ_{i}h_{i}(τ)  (24)

where the first series is the Poincaré expansion and the second are transcendental terms. It turns out, from the discussion above, that

f̄(τ) = γ(\overline{ϵ})τ + ∑_{i=1}^{∞} δ_{i}h_{i}(τ)  (25)

where δ_{1} = o(ϵ̄^{n}) for all positive integers n, δ_{i} = δ_{i}(ϵ̄), and δ_{i+1} ≪ δ_{i} for all i = 1, 2, 3, ..., are transcendentally small. The outer expansion (25) is valid in a neighborhood of the turning point, including [0, 1−ν(ϵ̄)] of τ, where ν(\overline{ϵ}) = o(1). It is interesting to point out here that the results in this section also show that f̄′′(0), f̄′′′(0), and f̄^{iv}(0) are all transcendentally small, which agrees to the rigorous proof in [3]. But there is no more information on those quantities that can be provided from the right side of the turning point at this stage. We must return to the left side of the turning point to determine the leading transcendental terms in the next section.

Remark The power expansion should be fully studied as above, but it was not completely given in [5].

4 The transcendental terms at the turning point

Note that the outer approximation on the right side of the turning point is the interior layer approximation which, from the previous section, is assumed as

f̄ = γτ + δ_{1}h_{1}(τ) + δ_{2}h_{2}(τ) + …  (26)

where γ = γ(ϵ̄) is given in (23), and δ_{1} = o(1), δ_{2} = o(δ_{1}), and h_{i}, i = 1, 2, are to be determined. Here o(1) = o(ϵ̄^{n}), for all positive integers n, is transcendentally small in terms of ϵ̄ (not the original ϵ). Since, from here on, the reasoning is similar to that in [5], only key steps of the formal analysis will be given in the remainder of the paper for completeness. The interior layer approximation (26) must be valid in the small neighborhood of η = 1 − Δ which should also include a small part of left side of the turning point, in terms of the original variable η. Thus, it may be assumed, for the interior variable τ, that the expression (26) is valid on any compact intervals of [−M, 1), where M > 0 is any constant, for sufficiently small ϵ̄ > 0. Substitution of (26) into (6) leads to

γ^{−1}ϵ̄h_{1}^{iv} = τh′′′_{1} − h′′_{1}  (27)

which has a solution (see [5] for details)

h_{1} = \frac{τ^{3}}{6} + r_{1}τ,  (28)

where r_{1} is a constant. Note that this h_{1} is valid for not only τ < 0 but also for τ ∈ [0, 1 − ν(ϵ̄)]. Replace h_{1} using (28), and substitute (26) into (6) again. We then balance the terms to get δ_{2} = δ_{1}^{2} and

γ^{−1}ϵ̄h_{2}^{iv} = τh′′′_{2} − h′′_{2} − γ^{−1} \frac{τ^{3}}{3} ,  (29)

from which one obtains

h_{2}^{iv} = −\frac{1}{ϵ̄}e^{\frac{γ}{2ϵ̄}τ^{2}} ∫_{0}^{τ} s^{2}e^{− \frac{γ}{2ϵ̄}s^{2}}ds − D_{2}e^{\frac{γ}{2ϵ̄}τ^{2}},  (30)

where D_{2} is a constant. To determine D_{2,} we choose τ < 0 bounded away from τ = 0 and apply integration by parts to evaluate the integral of (30). The result is

h_{2}^{iv} ∼ −\frac{1}{ϵ̄}e^{\frac{γ}{2ϵ̄}τ^{2}}\{−(\frac{2ϵ̄}{γ})^{3/2}[\frac{\sqrt{π}}{4} − (\frac{\sqrt{\frac{γ}{2ϵ̄}} |τ|}{2}e^{− \frac{γ}{2ϵ̄}τ^{2}} + \frac{1}{4|τ|\sqrt{γ/ (2ϵ̄)}}e^{− \frac{γ}{2ϵ̄}τ^{2}})]\}
−D_{2}e^{\frac{γ}{2ϵ̄}τ^{2}}.  (31)

To have δ_{2}h_{2}(τ) ≪ ϵ̄^{n} for all positive integers n, the first and last terms in the right hand side of (31) must be balanced each other, for otherwise, h_{2}(τ) would be exponentially large if τ is bounded away from zero. This may cause a contradiction. Thus, balancing the two terms, one obtains

D_{2} = \frac{\sqrt{2ϵ̄π}}{2γ\sqrt{γ}} .

Then, from (31), for τ < 0 bounded away from 0,

h_{2}^{iv} ∼ γ^{−1}τ

which gives that h_{2} is asymptotically equal to a polynomial of fifth degree, p(x) = γ^{−1}(\frac{τ^{5}}{5!} + …). Then, the inner expansion on the left side of the turning point is

f̄ = γτ + δ_{1}(\frac{τ^{3}}{6} + r_{1}τ) + δ_{1}^{2}γ^{−1}(\frac{τ^{5}}{5!} + …) + …  (32)

for τ < 0. Now, we match the expression (32) with the outer solution (5). Let η − (1 − Δ) = t. Then, t = τΔ. Expand the sine function at the turning point:

−κ sin \frac{πη}{1 − Δ} = κ\frac{πτΔ}{1 − Δ} − \frac{κ}{3!} (\frac{πτΔ}{1 − Δ})^{3} + \frac{κ}{5!} (\frac{πτΔ}{1 − Δ})^{5} + ….  (33)

Compare the corresponding terms in (32) and (33). We then see from the linear term that κ\frac{πΔ}{1−Δ} ∼ γ. Since γ ∼ 1, we have

κ ∼ \frac{1 − Δ}{πΔ}.  (34)

Then, from the cubic term,

δ_{1} ∼ − (\frac{1 − Δ}{πΔ})^{2} .

In closing this section, we write the interior layer approximation which is the out expansion of the solution on the right side of the turning point as follows:

f̄(τ) = γτ − (\frac{1 − Δ}{πΔ})^{2} \frac{τ^{3}}{6} + (\frac{1 − Δ}{πΔ})^{4} h_{2} (τ) + …,  (35)

where h_{2} satisfies (30).

5 On Δ and ϵ

The asymptotic relation between κ and Δ has been determined by (34). To complete the formal asymptotic analysis, our final task is to determine the asymptotic value of Δ in terms of ϵ. This can be done by matching (35) with the inner expansion (21). For τ ∼ 1, from §4,

h_{2}^{iv} = −\frac{1}{ϵ̄}e^{\frac{γ}{2ϵ̄}τ^{2}} ∫_{0}^{τ} s^{2}e^{− \frac{γ}{2ϵ̄}s^{2}}ds − \frac{\sqrt{2ϵ̄π}}{2γ\sqrt{γ}} e^{\frac{γ}{2ϵ̄}τ^{2}} = −\frac{\sqrt{2ϵ̄π}}{γ\sqrt{γ}} e^{\frac{γ}{2ϵ̄}τ^{2}} + τ + …. (36)

Differentiating (35) four times and applying (36), we find

\frac{d^{4}f̄}{dτ^{4}} = −π^{9/2} \frac{Δ^{7/2}}{(1 − Δ)^{4}} \sqrt{2ϵ̄}e^{\frac{γ}{2ϵ̄}τ^{2}} + (\frac{πΔ}{1 − Δ})^{4} τ + …  (37)

Writing this equation in terms of the boundary-layer variables x^{∗} by substituting (10) into (37), we obtain

\frac{1}{ϵ̄^{4}} \frac{d^{4}f̃}{dx^{∗4}} = −π^{9/2} \frac{Δ^{7/2}}{(1 − Δ)^{4}} \sqrt{2ϵ̄}e^{\frac{γ}{2ϵ̄}}e^{\frac{\overline{ϵ}γ}{2} x^{∗2}}e^{−γx^{∗}} +(\frac{πΔ}{1 − Δ})^{4} (1−ϵ̄x^{∗})+…. (38)

Then, from the inner expansion (21),

\frac{1}{ϵ̄^{4}} \frac{d^{4}f̃}{dx^{∗4}} = −\frac{1}{ϵ̄^{3}} e^{−x^{∗}} + o (\frac{1}{ϵ̄^{2}} )  (39)

Comparison of (38) and (39) shows the overlap domain must be such that

\frac{ϵ̄γ}{2} x^{∗2} ≪ 1and x^{∗} ≫ 1,

and hence

−π^{9/2} \frac{Δ^{7/2}}{(1 − Δ)^{4}} \sqrt{2ϵ̄}e^{\frac{γ}{2ϵ̄}} ∼ −\frac{1}{ϵ̄^{3}} .  (40)

Using the expression (23) and ϵ̄ = ϵ/Δ, and comparing the leading terms in (40), we finally obtain

\frac{Δ}{ϵ}e^{Δ/ϵ} = \frac{1}{2eπ^{9}ϵ^{8}} .

This result is the same as the rigorous result in [3].

6 Conclusion

The paper shows that when formal matching method is applied on the singular perturbation problem with a nonlinear turning point, like the Berman problem, an application of a full expansion of the asymptotic approximation is very important (see §2). This means that both the power terms and transcendental terms should involve. To determine the coefficients of their leading terms, the formal matching, in usual, must travel around the turning point several times. In other words, we have to match the inner and outer solutions repeatedly. Of course, it makes the computation tedious, but is necessary and important.

Acknowledgment. The author wants to express his gratitude to Professor A. Dean MacGillivray for his suggestions.

References

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[3] C. Lu, On the asymptotic solution of laminar channel flow with large suction, SIAM J. Math. Anal., vol. 28, No. 5, pp. 1113-1134, 1997.

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Chunqing Lu Department of Mathematics and Statistics Southern Illinois University at Edwardsville Edwardsville, IL 62026 USA e-mail: clu@siue.edu