Open access peer-reviewed chapter

# Perturbation Expansion to the Solution of Differential Equations

By Jugal Mohapatra

Submitted: July 21st 2020Reviewed: September 24th 2020Published: October 29th 2020

DOI: 10.5772/intechopen.94173

## Abstract

The main purpose of this chapter is to describe the application of perturbation expansion techniques to the solution of differential equations. Approximate expressions are generated in the form of asymptotic series. These may not and often do not converge but in a truncated form of only two or three terms, provide a useful approximation to the original problem. These analytical techniques provide an alternative to the direct computer solution. Before attempting to solve these problems numerically, one should have an awareness of the perturbation approach.

### Keywords

• perturbation methods
• asymptotic expansion
• boundary layer
• principle of least degeneracy
• inner and outer expansion

## 1. Introduction

The governing equations of physical, biological and economical models often involve features which make it impossible to obtain their exact solution. For instance, problems where we observe “a complicated algebraic equations”, “the occurrence of a complicated integral”, in case of differential equations (DE), “a varying coefficients or nonlinear term” sometimes problems with an awkwardly shaped boundary are tough to solve with the limited methods for finding analytical solutions. The main purpose of this chapter is to describe the application of perturbation expansion techniques to the solution of DE. Approximate expressions are generated in the form of asymptotic series. These may not and often do not converge but in a truncated form of only two or three terms, provide a useful approximation to the original problem. These analytical techniques provide an alternative to the direct computer solution. Before attempting to solve these DE numerically, one should have an awareness of the perturbation approach. An example of this occurs in boundary layer problems where there are regions of rapid change of quantities such as fluid velocity, temperature or concentration. Appropriate scaling of the boundary layer dimension is required before a numerical solution can be generated which will capture the behavior in the rapidly changing region.

When a large or small parameter occurs in a mathematical model of a process there are various methods of constructing perturbation expansions for the solution of the governing equations. Often the terms in the perturbation expansions are governed by simpler equations for which the exact solution techniques are available. Even if exact solutions cannot be obtained, the numerical methods used to solve the perturbation equations approximately are often easier to construct than the numerical approximation for the original governing equation.

First, we consider a model problem for which an exact solution is available against which the perturbation expansion can be compared. A feature of the perturbation expansions is that they often form divergence series. The concept of an asymptotic expansion will be introduced and the value of a truncated divergent series will be demonstrated.

## 2. Projectile motion

This example studies the effect of small damping on the motion of a particle. Consider a particle of mass Mwhich is projected vertically upward with an initial speed U0. Let Udenote the speed at some general time T. If air resistance is neglected then the only force acting on the particle is gravity, Mg, where gis the acceleration due to gravity and the minus sign occurs because the upward direction is chosen to be the positive direction. Newton’s second law governs the motion of the projectile, i.e.,

MdUdT=Mg.E1

Integrating (1), we obtain the solution U=CgT. The constant of integration is determined from the initial condition U0=U0, so that

U=U0gT.E2

On defining the non-dimensional velocity v, and time t, by v=U/U0and t=gT/U0, the governing equation becomes

dvdt=1,v0=1,E3

with the solution vt=1t.

Taking account of the air resistance, and is included in the Newton’s second law as a force dependent on the velocity in a linear way, we obtain the following linear equation

MdUdT=MgKU,E4

where the drag constant Kis the dimensions of masa/time. In the non-dimensional variables, it becomes

dvdt=1KU0Mgv.E5

Let us denote the dimensionless drag constant by ε, then the governing equation is

dvdt=1εv,v0=1,E6

where ε>0is a “small” parameter and the disturbances are very “small”. The damping constant Kin (4) is small, since Khas the dimensions of mass/time and a small quantity in units of kilograms per second.

### 2.1 Perturbation expansion

It is possible to solve (6) exactly since it is of variables separable form. Here, we solve by an iterative process, known as perturbation expansion for the solution.

Let videnotes the ith iterate, which is obtained from the equation

dvidt=1εvi1,vi0=1,E7

The justification for this iterative scheme is that the term εvinvolves the small multiplying coefficient ε, and so the term itself may be expected to be small. Thus, the term εviwhich should appear on the RHS of (7) to make it exact, may be replaced by εvi1with an error which is expected to be small.

The first iterate is obtained by neglecting the perturbation, thus

dv0dt=1,v0=1.

This is known as the unperturbed problem, and direct integration yields

v0=1t.

The next iterate v1, satisfies

dv1dt=1ε1t,v1=1.

and integration yields

v1=1t1+ε+12εt2

Similarly, v2satisfies

dv2dt=1ε1t1+ε+12εt2,v2=1.

Direct integration yields the solution

v2=1t1+ε+ε1+εt2216ε2t3.

Rearranging the terms in these iterates in ascending powers of ε, we obtain

v0=1t,v1=1t+εt22t,v2=1t+εt22t+ε2t22t36.E8

Clearly as the iteration proceeds the expressions are refined by terms which involve increasing powers of ε. These terms become progressively smaller since εis a small parameter. This is an example of a perturbation expansion. It will often be the case that perturbation expansions involve ascending integer powers of the small parameter, i.e., ε0ε1ε2. Such a sequence is called an asymptotic sequence. Although this is the most common sequence which we shall meet, it is by no means unique. Examples of other asymptotic sequences are ε1/2εε3/2ε2and ε0ε2ε4. In each case the essential feature is that subsequent terms tend to zero faster than previous terms as ε0.

An alternative procedure to that of developing the expansion by iteration is to assume the form of the expansion at the outset. Thus, if we assume that the perturbation expansion involves the standard asymptotic sequence ε0ε1ε2, then the solution v, which depends on the variable t, and the parameter ε, is expressed in the form

vtε=ε0v0t+ε1v1t+ε2v2t+.E9

The coefficients v0t,v1t,of powers of εare functions of tonly. Substituting expansion (9) in the governing Eq. (6) yields the following

dv0dt+εdv1dt+ε2dv2dt+=1εv0ε2v1v00+εv10+ε2v20+=1.E10

Thus, the coefficients of powers of εcan be equated on the left– and right–hand sides of (10):

ε0:dv0dt=1,v00=1,ε1:dv1dt=v0,v10=0ε2:dv2dt=v1,v20=0,etc.E11

The proof of validity of this fundamental procedure can be developed by first setting ε=0in (10) which yields the first equation of (11). This result allows the first member of the left– and right–hand side of Eq. (10) to be removed. Then, after dividing the remaining terms by εwe obtain the equation

dv1dt+εdv2dt+=v0εv1

This is valid for all nonzero values of εso that on taking the limit as ε0we obtain the second equation of (11). Repeating the procedure, we obtain the other equations.

Integrating the equations in (11), we obtain

v0=1t,v1=t2/2t,v2=t2/2/t3/6.

Using these values in (9), we obtain that

vtε=1t+εt2t+ε2t2/2t3/6+E12

This is the same as the expansion (8) which is generated by iteration.

The IVP (6) can be solved exactly as

vt=1+εeεt1ε1.

The perturbation expansion can be obtained from (12) by replacing the exponential function by its Maclaurin expansion, i.e.,

vt=1ε1εt+ε2t22ε3t36++εε2t+ε3t22+1E13
=1t+εt22t+ε2t22t36+E14

This is the same as the expansion (12). Thus, the perturbation expansion approach is justified in this case. One can refer the books [1, 2].

## 3. Asymptotics

The letters Oand oare order symbols. They are used to describe the rate at which functions approach limit values. We will consider the types of limit values, namely zero, a finite number but nonzero and infinite.

If a function fxapproaches a limiting value at the same rate of another function gxas xx0, then we write

fx=Ogx,asxx0E15

The functions are said to be of the same order as xx0. The test for this is the limit of the ratio. Thus, if limxx0fxgx=C,where Cis finite, then we say (15) holds.

For example, we have the following functions:

x2=Ox,x<2,sinx=Ox,x0,sinx=Ox,<x<.

The expression

fx=ogx,asxx0E16

means that limxx0fxgx=0.This is a stronger assertion that the corresponding O–formula. The relation (16) implies the relation (15), as convergence implies boundedness from a certain point onwards.

We have the following functions satisfy the o–relation:

cosx=1+ox,x<2,ex=1+ox,x0n!=ennn2πn1+o1,n.

### 3.1 Asymptotic expansions

Consider the expansion

fx=a0+a1x+a2x2++aNxN+RN,E17

is an asymptotic expansion as x, if, for any N,

RN=O1xN+1,asxE18

The following expansion is used when (17) and (18) hold,

fxn=0anxn,asxE19

Here, limnRN=0, for any value of N.

The sequence 11/x1/x2is an asymptotic sequenceas x. The characteristic feature of such sequences is that each member is dominated by the previous member. In constructing examples it is easier to deal with the limit zero than any other. Thus, for the case x, we let ε=1/x, which for xx0, we let ε=xx0so that without loss of generality we may confirm our attention to the limit ε0. The standard asymptotic sequence is 1εε2as ε0. If we let δnεrepresent members of an asymptotic sequence δ0εδ1εas ε0, then the following condition must hold

δn+1ε=oδnε,asε0.

Some examples of asymptotic sequences are

1. 1sinεsinε2sinε3, here we have

limε0δn+1δn=limε0sinε=0.

• 1ln1+εln1+ε2ln1+ε3, with δ0=1,δn=ln1+εnn1, we have

• limε0δ1δ0=limε0ln1+ε=0,limε0δn+1δn=limε0ln1+εn+1ln1+εn=limε0εn+1+Oε2n+2εn+Oε2n=0.

The general expression for an asymptotic expansion of a function fε, in terms of an asymptotic sequence δnεis

fxn=0anδnε,asε0,E20

where the coefficients anare independent of ε. The expression (20) involving the symbol , means that for all N,

fx=n=0Nanδnε+RN,E21

where

RN=OδN+1ε,asε0,E22
an=limε0fεn=0N1anδnεδNε.E23

If a function possesses an asymptotic expansion involving the sequence δ0εδ1εthen the coefficients anof the expansion (21) given by the expression (24) are unique. However, another function may share the same set of coefficients. Thus, while functions have unique expansions, an expansion does not correspond to a unique function.

Consider a function fxε, which depends on both an independent variable x, and a small parameter ε. Suppose that fxεis expanded using an asymptotic sequence δnε,

fxε=n=0Nanxδnε+RNxε.E24

The coefficients of the gauge functions δnεare functions of x, and the remainder after Nterms is a function of both xand ε. For this to be an asymptotic expansion, we require

RNxε=OδN+1ε,asε0.E25

Refer [3, 4] for more details. For (24) to be a uniform asymptotic expansion the ultimate proportionality between RNand δN+1must be bounded by a number independent of x, i.e.,

RNxεKδN+1ε,E26

for εin the neighborhood near zero, where Kis a fixed constant.

An example of a uniform asymptotic expansion is fxε=11εsinx.

An example of a nonuniform expansion is

fxεn=0Nxnεn+RNxε,asε0.E27

Here, one cannot find a fixed Kwhich satisfy RNKεN+1,because for any choice of K, xcan be chosen so that xN+1exceeds this value.

### 3.2 Nonuniformity

The expansion (27) becomes nonuniform when subsequent terms are no longer small corrections to previous terms. This occurs when subsequent terms are of the same order or of dominant order than previous terms. Subsequent terms dominate previous terms for larger x, for example, when x=O1/ε2. The expansion is valid for x=O1since then subsequent terms decrease by a factor of ε. The expansion remains valid for large x, provided xis not as large as 1/ε. For instance, the expansion is valid for x=O1/ε, as ε0.

The critical case is such that subsequent terms are of the same order. This determines the region of nonuniformity. In (27), the region of nonuniformity occurs when εx=O1, i.e., x=Oε1, as ε0.

#### 3.2.1 Sources of nonuniformity

There are two common reasons for nonuniformities in asymptotic expansions, they are

1. Infinite domains which allow long-term effects of small perturbations to accumulate.

2. Singularities in governing equations which lead to localized regions of rapid change.

Consider the nonlinear Duffing equation

d2udt2+u+εu3=0,t0u0=a,dudt0=0.E28

Suppose the solution may be expanded using the standard asymptotic sequence

utεu0t+εu1t+ε2u2t+.E29

On substituting this in (28) and in the initial conditions, we get

d2u0dt2+εd2u1dt2++u0+εu1++εu03+0,u00+εu10+=a+0ε+,du0dt0+εdu1dt0+=0+0ε+.

Equating like of powers of εon both sides, we get

O1:d2u0dt2+u0=0,u00=a,du0dt0=0,E30

and

Oε:d2u1dt2+u1=u03,u10=0,du1dt0=0.E31

Solving Eqs. (30) and (31), we obtain

uacost+εa332cos3tcost3a38tsint+.E32

The term tsintin the expansion (32) is called a secular term. It is an oscillating term of growing amplitude. All other terms are oscillating of fixed amplitude. The secular term leads to a nonuniformity for large t. The region of nonuniformity is obtained by equating the order of the first and second terms,

cost=Oεtsint,asε0.

The trigonometric functions are treated as O1terms. Thus, the region of nonuniformity is t=O1/ε, as ε0.

The second common source of nonuniformities is associated with the presence of singularities. Consider, the following initial-value problem:

εdydx+y=ex,x>0y0=2,E33

where ε>0is a small parameter. Suppose yhas the expansion

yy0x+εy1x+ε2y2x+.E34

Substituting (34) in (33), we have

εdy0dx+εdy1dx++y0+εy1+=ex,y00+εy10+=2.E35

Equating coefficients of like powers of εon both sides, we get

O1:y0=ex,y00=2,Oε:y1=dy0dx=ex,y10=0,Oε2:y2=dy1dx=ex,y20=0.

Clearly, y0cannot satisfy the boundary condition y00=2as no constant of integration is available because the equation determining y0is an algebraic equation not a differential equation, and no additional conditions are required. Thus, we have obtain the expression

yex+εex+ε2ex+,E36

but the initial condition y0=2has not been satisfied.

The unperturbed problem, obtained by setting ε=0is not a DE, but an algebraic equation y=ex. This cannot satisfy an arbitrarily imposed condition at x=0. For any nonzero value of ε, (33) becomes a first-order DE which can satisfy an initial condition. This is an example of a singular perturbation problem (SPP), where the behavior of the perturbed problem is very different from that of the unperturbed problem.

Thus, the perturbation expansion (36) is a good approximation of the exact solution away from the region x=0. To see this, let us compare (36) with the following exact solution:

yex=12ε1εex/ε+ex1ε=1εε2ex/ε+1+ε+ε2+ex.E37

The perturbation expansion (36) generates the second member of (37), but not the first member. The coefficient ex/ε)is a rapidly varying function which takes the value of unity at x=0, and rapidly decays to zero for x>0. Clearly, y0provides a good approximation away from the region x=0. The region near x=0is called the boundary layer. These regions usually occur when the highest order derivative of a DE is multiplied by a small parameter. The unperturbed problem, obtained by setting ε=0is of lower order and consequently cannot satisfy all the boundary conditions. This leads to boundary layer regions where the solution varies rapidly in order to satisfy the boundary condition.

Boundary layers are regions of nonuniformity in perturbation expansions of the form (36).

## 4. Boundary layer

Boundary layers are regions in which a rapid change occurs in the value of a variable. Some physical examples include “the fluid velocity near a solid wall”, “the velocity at the edge of a jet of fluid”, “the temperature of a fluid near a solid wall.” Ludwig Prandtl pioneered the subject of boundary layer theory in his explanation of how a quantity as small as the viscosity of common fluids such as water and air could nevertheless play a crucial role in determining their flow. The viscosity of many fluids is very small and yet taking account of this small quantity is vital. The essential point is that the viscous term involves higher order derivatives so that its omission necessitates the loss of a boundary condition. The ideal flow solution allow slip to occur between a solid and fluid. In reality the tangential velocity of a fluid relative to a solid is zero. The fluid is brought to rest by the action of a tangential stress resulting from the viscous force.

Mathematically the occurrence of boundary layers is associated with the presence of a small parameter multiplying the highest derivative in the governing equation of a process. A straightforward perturbation expansion using an asymptotic sequence in the small parameter leads to differential equations of lower order than the original governing equation. In consequence not all of the boundary and initial conditions can be satisfied by the perturbation expansion. This is an example of what is commonly referred to as a singular perturbation problem. The technique for overcoming the difficulty is to combine the straightforward expansion, which is valid away from the layer adjacent to the boundary. The straightforward expansion is referred to as the outer expansion. The inner expansionassociated with the boundary layer region is expressed in terms of a stretched variable, rather than the original independent variable, which takes due account of the scale of certain derivative terms. The inner and outer expansions are matched over a region located at the edge of the boundary layer. The technique is called the method of matched asymptotic expansions.

Consider the following two-point boundary value problem:

εd2udx2+dudx=2x+1,x01u0=1,u1=4,E38

where ε>0is a small parameter. If we assume that upossesses a straightforward expansion in powers of ε,

uxεu0x+εu1x+ε2u2x+,E39

then the equations associated with powers of εleads to

O1:du0dx=2x+1,E40
Oεn:dundx=d2un1dx2,forn=1,2,3,.E41

and the boundary conditions require

u00+εu10+1+ε0+,u01+εu11+4+ε0+,

u00=1,u01=4,un0=0,un1=0,forn=1,2,.E42

Equation (42) require that each unxsatisfy two boundary conditions. This is in general impossible since Eqs. (41) and (42) governing each unare of first-order. Now the question is which one of the boundary condition has to be taken into account. We will find out that the boundary condition at x=0must be abandoned and consequently the expansion (39) is invalid near x=0.

The general solution of (42) is u0x=x2+x+C, using the boundary condition u01=4, we obtain

u0x=x2+x+2.

From (42), we obtain the equations

du1dx=2,u11=0,du2dx=0,u21=0,

and its solutions are

u1x=2x1,unx=0,n2.

Therefore, the outer expansion is

uoutxε=x2+x+2+ε21x,E43

where ‘out’ label is used to indicate that the solution is valid away from the region near x=0. Clearly uoutfails to satisfy the boundary condition at x=0. The reason why the outer solution is of use is that it closely follows the exact solution of the problem except in a narrow region near x=0, where the exact solution changes rapidly in order to satisfy the boundary condition.

The exact solution of the BVP (38) can be obtained as

ux=A+Bex/ε+x2+x12ε.E44

The constants Aand Bare determined from the boundary conditions:

A+B=1,A+Be1/ε+22ε=4E45

We know that e1/ε=oεN, as ε0, for all N. This means that the exponential term tends to zero faster than any power of ε, as ε0. It is called a transcendentally small term(T.S.T.) and can always be neglected since its contribution is asymptotically always less than any power of ε. Thus, (45) gives

A=21+ε,B=1+2ε,

and the exact solution is

uexx=21+ε1+2εex/ε+x2+x12ε,E46

after rearranging the terms in asymptotic order, we obtain

uexx=x2+x+2ex/ε+ε21x2ex/ε.E47

Comparing the exact solution with the outer expansion shows that the terms involving ex/εare absent. The effect of these terms is negligible when x=O1. But, when x=Oε, then e=O1. It is clear that as ε0the region in which the outer solution departs from the exact solution becomes arbitrarily close to x=0with a thickness Oε. This region is called the boundary layer.

The behavior of the exact solution and the zeroth-order term of the outer expansion are plotted in Figure 1 for various values of ε.

By differentiating the leading order term u0ex, of the exact solution, we have

u0ex=x2+x+2ex/εdu0exdx=2x+1+1εex/εd2u0exdx2=21ε2ex/ε

Outside the boundary layer, i.e., for x=O1, we have ex/ε=oεN,N, so ε1ex/εand ε2ex/εare also transcendentally small. Within the boundary layer when x=Oε, we have ex/ε=O1. The order of u0exand its derivatives are given below:

OutsideBLInsideBLu0exO1O1du0exdxO1O1εd2u0exdx2O1O1ε2

This indicates that xis the appropriate independent variable outside the boundary layer where u0exand its derivatives are of O1quantities. However, within the boundary layer the appropriately scaled independent variable is s=x/ε, then

dudx=ε1dvds,d2udx2=ε2d2vds2,

so that within the boundary layer

dudx=O1,andd2udx2=O1.

The variable s=x/εis called a stretched variable. The differential equations becomes

d2vds2+dvds=ε+2ε2s.E48

We assume a boundary layer expansion, called the inner expansionof the form

vsεv0s+εv1s+.E49

The inner expansion will satisfy the boundary condition at x=s=0namely v0s=0=1giving v00=1, and vn0=0,n=1,2,. Substituting (49) into the DE (48), we obtain the following set of equations:

O1:d2v0ds2+dv0ds=0,v00=1Oε:d2v1ds2+dv1ds=1,v10=0Oε2:d2v2ds2+dv2ds=2s,v10=0Oεn:d2vnds2+dvnds=0,vn0=0,n=3,4,E50

with solutions

v0=A+1Aesv1=BBes+sv2=CCes+s22svn=DnDnes,n=3,4,E51

The boundary condition at x=1cannot be used to determine the constants appearing in these solutions because the DEs (50) are only valid in the boundary layer. The constants in (51) are determined by matching the inner and outer expansions. We shall first restrict our attention to matching the leading order expansions u0and v0. The method which we shall apply is Prandtl’s matching condition.

The leading order terms in the ‘inner’ and ‘outer’ expansions are to be matched at the ‘edge of the boundary layer’. Of course there is no precise edge of the boundary layer, we simply know that it has thickness of order Oε. A plausible matching procedure would be to equate u0and v0at a value of xsuch that the region of rapid change has passed. We might choose to equate the terms at the point x=5ε. The leading order expansions are

u0=x2+x+2v0=A+1Aes.

Equating at x=5εgives the following:

A=2+5ε+25ε2e51e5.

If, instead we choose to match at x=6ε, then we obtain

A=2+6ε+36ε2e61e6.

These two expressions differ in the argument of the exponential and differ algebraically with 5εreplaced by 6ε. The exponential functions are approaching transcendentally small values so that their contribution can be neglected. The algebraic difference is of Oε. Thus, the arbitrariness in the decision of the point at which we choose to equate the expansions leads to a difference of Oε. But we are only dealing with leading order expansions anyway. The difference between the exact solution and the leading order expansions will of Oεso that an arbitrariness in v0and u0of Oεis immaterial. Rather than choose between, for example, 5εand 6εas the value of xto evaluate u0we may take the value at x=0, since

u0x=Oε=u00+Oε,

where the remainder is uniformly Oεsince the gradient of u0is O1. For the inner expansion we are to ensure that the rapidly varying function has achieved its asymptotic value at the edge of the boundary layer. This means that the term ex/εshould be replaced by zero. This can be achieved by taking the limit s. Thus, rather than choosing a specific point to equate the inner and outer terms er are led to the following Prandtl’s matching condition:

limx0u0x=limsv0s.E52

The limit smay appear rather dangerous since although it certainly removes the exponential term it could lead to an algebraically unbounded term. For example, if v0=As+1Aes, then the first member would be unbounded as s. This possibility can be eliminated since the inner expansion must be of a form which varies rapidly for x=Oεbut not for x=O1, i.e., not for s. In practice, if the boundary layer has been properly located and the correct inner variable is used then Prandtl’s matching condition is valid and elegantly avoids the need to choose an arbitrary ‘edge’ of the boundary layer.

Applying these conditions to the current example leads to

limx0x2+x+2=limsA+1Aes,

which yields A=2. Thus the leading order terms in the expansion solutions are

Outerregion:u0=x2+x+2,forx=O1Innerregion:v0=2ex/ε,forx=Oε

To prove that these are valid leading terms we consider uex:

Ifx=O1,thenu0ex=x2+x+2+T.S.T.Ifx=Oε,thenu0ex=2ex/ε+Oε

We conclude that the matching condition has correctly predicted the leading order terms.

### 4.1 Composite expansion

As single composite expression for these leading order terms can be constructed using the combination

u0comp=u0+v0u0match,E53

where u0matchis given by (52). Then,

forx=O1,v0=u0match+T.S.T.,sothatu0comp=u0match+T.S.T.forx=Oε,u0=u0match+Oε,sothatu0comp=v0+Oε

For the current example, u0match=2, so the composite expansion is

u0match=x2+x+2ex/ε.E54

Prandtl’s matching condition can only be used for the leading order terms in the asymptotic expansions.

The outer, inner and composite expansions of the BVP (38) are presented in Figures 2 and 3 for different values of ε. From these figures, one can easily identify the need and efficiency of the composite expansion.

### 4.2 Boundary layer location

Consider the following linear DE

εd2udx2+axdudx+bxu=cx,xx1x2.E55

The following general statements can be made about the boundary layer location and the nature of the inner expansion.

Case I.If ax>0throughout x1x2, then the boundary layer will occur at x=x1. The stretching transformation will be s=xx1/ε, and the one-term inner expansion will satisfy

d2v0ds2+ax1dv0ds=0.

The solution of this equation is

v0=A+Beax1xx1/ε,

where A+B=ux=x1. The other condition to determine the constants Aand Bis obtained by matching with the value of the outer expansion at x=x1.

Case II.If ax<0throughout x1x2, then the boundary layer will occur at x=x2. The stretching transformation will be s=x2x/ε, and the one-term inner expansion will involve the rapidly decaying function eax2x2x/ε.

Case III.If axchanges sign in the interval x1<x<x2, then a boundary layer occurs at an interior point x0, where ax0=0and boundary layers may also occur at both ends x1and x2.

### 4.3 Boundary layer thickness and the principle of least degeneracy

The boundary layers which we have met so far have all had thickness Oε. By this we mean that a variation of Oεin the independent variable will encompass the region of rapid change in the dependent variable. The associated stretched independent variable s, appropriate for the boundary layer is related to xby a linear transformation involving division by ε.

There are practical situations where the boundary layer thickness will be of Oεp. This means that if the boundary layer is located at x=x0, then the appropriate stretching transformation is s=xx0/εp. More generally, the choice of the function δεto use in the stretching transformation s=xx0/δεis determined by the need to represent the region of rapid change correctly. We must ensure that the boundary layer solution contains rapidly varying functions. The form of the governing equation in the boundary layer region must have sufficient structure to allow such solutions.

Consider the example

εd2udx2+dudx+u=x,01u0=1,u1=2.E56

Since the signs of the first and second derivatives are the same, and the boundary layer will occur at x=0. We are not going to assume at the outset that the boundary layer thickness is Oε. Our intension is to deduce that the appropriate stretching variable is s=x/ε.

The one-term outer expansion u0satisfies du0dx+u0=x,u01=2The solution is

u0x=2e1x+x1.E57

To determine the inner expansion we first wrongly assume that the boundary layer thickness is Oε1/2. The stretching transformation s=x/ε1/2changes the original DE (56) into the following one:

d2vds2+1ε1/2dvds+v=ε1/2sE58

If the appropriate stretching transformation has been used for the boundary layer then dv/dsand d2v/ds2will be of O1within it. The leading order expansion v0will satisfy the dominant part of (58), i.e., the component of Oε1/2

dv0ds=0,v00=1.E59

The solution is v0s=1. This of course does not have the rapidly varying behavior which we anticipate in the boundary layer. Prandtl’s matching condition cannot be satisfied since

limx02e1x+x1=2e1limsv0s=1.

Thus, we reject the assumption of a boundary layer of thickness Oε1/2.

Next, suppose that the boundary layer thickness is Oε2and again we will discover that this is incorrect because the corresponding inner expansion cannot be matched to the outer expansion. Proceeding with the analysis we introduce the stretching transformation s=x/ε2which leads to the equation

1ε3d2vds2+1ε2dvds+v=ε2s.

Again we argue that if the appropriate stretching has been used then all derivatives are of O1so that the governing equation for the leading term in Oε3, namely

d2v0ds2=0,v00=1.E60

The solution is v0s=1+As, where the constant Ais to be determined from matching. This solution is rapidly varying but the rapidity does not decay at the edge of the boundary layer (i.e., as s). Indeed, we cannot match v0to the outer expansion because the term Asbecomes arbitrarily large as s.

The correct choice of stretching transformation is s=x/εshowing that the boundary layer thickness is Oε. The boundary layer equation becomes

1εd2vds2+1εdvds+v=εs.

The dominant equation satisfied by v0is O1/ε, namely

d2v0ds2+dv0ds=0,v00=1.E61

The solution is v0s=1A+Aes. The last member provides the necessary rapid decay away from the point x=s=0. Prandtl’s matching condition requirest

limx02e1x+x1=lims1A+Aes,

v0x=2e1+21eex/ε.

The one-term composite expansion is

ucomp=2e1x+x1+2e1+21eex/ε2e1.E62

The leading order boundary layer equation associated with the stretching transformation s=x/ε, (61) involves more terms than (59), associated with s=x/ε1/2, and (60) associated with s=x/ε2. The extra term in (61) allows sufficient structure in the solution to produce the required boundary layer behavior. An aid for choosing the boundary layer thickness is to seek a stretching transformation which retains the largest number of terms in the dominant equation governing v0. This referred to as the principle of least degeneracyby Van Dyke.

The composite expansion (62) can be verified by comparing with the exact solution of (56). The general solution of (56) is

uex=C1em1x+C2em2x+x1,

where

m1=1+14ε2ε,m2=114ε2ε.

We expand 14εusing the binomial series, 14ε=12ε+Oε2, then

m1=1+Oε,andm2=1ε+1+Oε,

so that

uex=C1ex+C2ex/εex+x1+Oε.E63

Using the boundary conditions and by neglecting the transcendentally small term e1/ε, we have C1=2e,C2=21e. Then, (63) becomes

uex=2e1x+21eex/εex+x1+Oε.E64

There is an apparent discrepancy between (64) and the composite expansion (62) in the coefficient of the ex/εterm. There is an extra term only contributes in the boundary layer where x=Oεso that the coefficient exmay to leading order, be replaced by unity. Thus, the leading order composite expansion and the leading order term in the exact solution are in complete agreement.

### 4.4 Boundary layer of thickness of Oε

Consider the following two-point BVP:

εd2udx2+x2dudxu=0,01u0=1,u1=2.E65

We seek a one-term composite expansion for the above BVP. We will tentatively assume that a boundary layer occurs at x=0although the vanishing of the coefficient of the first derivative suggests the possibility of nonstandard behavior.

The one term outer expansion satisfies

x2du0dxu0=0,u01=2.

Its exact solution is u0x=2e11/x.

Let us assume that the boundary layer thickness is of Oεp, where pis to be determined from the principle of least degeneracy. The stretched variable is s=x/εp, and (65) becomes

ε12pd2vds2+εps2dvdsv=0.

The second-term is always dominated by the third, so the principle of degeneracy requires the first term to be of the same order as the third term (i.e., O1). Thus, p=1/2, and the one-term inner expansion satisfies

d2v0ds2+dv0ds=0,v00=1.

The solution of the above problem is v0s=Aes+1Aes. Prandtl’s matching condition requires

limx02e11/x=limsAes+1Aes

which yields A=0. This example is rather special in that Awill be zero for all boundary conditions.

The on-term composite expansion is

u0comp=2e1x+ex/ε.

We conclude this example with the observation that a choice for the value of the index pother than p=1/2leads to boundary layer equations with insufficient structure to generate the required rapidly decaying behavior.

Thus, if p>1/2, the dominant equation becomes

d2v0ds2=0,v00=1,

which gives v0s=1+As. It is obvious that Prandtl’s matching condition cannot be used to determine A. Whereas, if p<1/2the dominant equation degenerates to v0s=0which does not satisfy the boundary condition at s=0.

### 4.5 Interior layer

Consider the BVP:

εd2udx2+xdudx+xu=0,11u1=e,u1=2e1.E66

The coefficient of the first derivative (convective term) is positive in 01which indicates the occurrence of a boundary layer at the left hand limit x=0. While the corresponding coefficient is negative in the range 1<x<0indicates a boundary layer located at the right-hand limit which is again is x=0. Thus, we are led to expr = ect two outer expansions for positive and negative xrespectively and an inner expansion in the interior layer located at x=0. We denote the leading term in the outer expansion for positive xby u0+, it satisfies

du0+dx+u0+=0,u0+1=2e1E67

with the solution u0+x=2ex.

The outer expansion for negative x, u0satisfies

du0dx+u0=0,u0+1=eE68

with the solution u0x=ex.

We suppose the boundary layer at x=0has thickness Oεpand determine the index pusing the principle of least degeneracy. Let s=x/εpso that the DE becomes

ε12pd2vds2+sdvds+εpsv=0.

The third term is dominated by the second term. The first term has the same order as the second term if p=1/2. For this choice of pthe leading term of the inner expansion v0satisfies

d2v0ds2+sdv0ds=0.

Its solution can be given by

v0s=Berfs/2+v00,

Prandtl;s matching condition applied to the region x>0is

lims+v0s=limx0+u0+x

and corresponding for x<0, we have

limsv0s=limx0u0x

Using the limiting values erf±=±1yields v00=1.5and B=0.5. The leading order terms over the whole region are

u0+x=2ex,x>Oεv0=0.5erfx/2ε+1.5,x=Oεu0x=ex,x<Oε

A composite expansion cannot be formed in the standard way when there is more than one outer solution. However, the behavior of v0for x>Oεis as follows:

v0x>Oε=0.5+1.5+T.S.Tv0x<Oε=0.5+1.5+T.S.T

Utilizing this enables a uniformly valid one-term composite expansion to be constructed which yields the correct coefficient of exoutside the boundary layer and the correct leading order behavior within the boundary layer. It is

u0comp=0.5erfx/2ε+1.5ex.

### 4.6 Nonlinear differential equation

Consider the following semilinear

εd2udx2+dudx+u2=0,01u0=2,u1=1/2.E69

The coefficient of the first and second order derivatives have the same sign, so the boundary layer will occur at the left boundary x=0. The one-term outer expansion satisfies

du0dx+u02=0,u01=1/2,

and the solution is u0x=1/1+x. The stretching transformation for the inner region will be s=x/εand therefore, the inner expansion satisfies

d2vds2+dvds+εv2=0,v0=2.

The one-term inner expansion v0satisfies the dominant part of this equation, i.e.,

d2v0ds2+dv0ds=0,v00=2,

which gives v0s=A+2Aes. Prandtl’s matching condition yields A=1, and the composite one-term uniformly valid expansion is

u0comp=11+x+ex/ε.

Next, consider the quasilinear problem

εd2udx2+2ududx4u=0,01u0=0,u1=4.E70

The nonlinearity is associated with the first derivative term. The location of the boundary layer depends on the relative sign of the first and second derivative coefficients. If we assume that the dependent variable is nonnegative throughout the interval 0<x<1, then the boundary layer will occur at x=0. The one-term outer expansion satisfies

2u0du0dx4u0=0,u01=4,

with the solution u0x=2x+2.

Assuming that the boundary layer thickness is Oε, therefore, the dominant-order equation for the one-term inner expansion becomes

d2v0ds2+2v0dv0ds=0,v00=0.

Its solution is v0s=atanhas. Prandtl’s matching condition yields a=2. Thus, v0s=2tanh2s, and the uniformly valid one-term composite expansion is

u0comp=2x+2+2tanh2s2.

Application of perturbation techniques to partial differential equations, and other types of problems can be seen in the books [5, 6].

## Conflict of interest

The authors declare no conflict of interest.

## Notes/thanks/other declarations

I owe a great debt to my mentor Prof. S Natesan, Department of Mathematics, IIT Guwahati, who introduced me to this topic. The chapter was discussed during my stay at IIT Guwahati.

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Jugal Mohapatra (October 29th 2020). Perturbation Expansion to the Solution of Differential Equations, A Collection of Papers on Chaos Theory and Its Applications, Paul Bracken and Dimo I. Uzunov, IntechOpen, DOI: 10.5772/intechopen.94173. Available from:

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