2.1 Introduction

<sup>∂</sup>tv0 � b xð Þ ;<sup>t</sup> v0ð Þ¼ <sup>x</sup>;<sup>t</sup> <sup>0</sup>, <sup>∂</sup>twl

∂tql

L<sup>ς</sup> d<sup>l</sup>

, a transition <sup>ξ</sup><sup>l</sup>

<sup>k</sup>,0ð Þ <sup>x</sup>;<sup>t</sup> , ql

In each iteration with respect to við Þ <sup>x</sup>;<sup>t</sup> , <sup>w</sup><sup>l</sup>

When the equation is obtained with respect to d<sup>l</sup>

<sup>k</sup>,0ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> <sup>d</sup><sup>l</sup>

Boundary Layer Flows - Theory, Applications and Numerical Methods

∂td<sup>l</sup>

iθ0

<sup>2</sup> ffi t p � �

<sup>0</sup>ð Þ <sup>x</sup>;<sup>t</sup> , <sup>d</sup><sup>l</sup>

equation entering into (Eq. (18)).

obtain inhomogeneous equations.

For the remainder term

work of [7], we get the estimate:

182

1.7 Assessment of the remainder term

ql

wl <sup>0</sup>ð Þ x , d l <sup>k</sup>,0ð Þ <sup>x</sup> , ql

<sup>k</sup>,0ð Þ <sup>x</sup>;<sup>t</sup> erfc <sup>ξ</sup><sup>l</sup>

Functions wl

the partial sum:

<sup>0</sup>ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> wl

<sup>k</sup>,0ð Þ¼ x;t 0,

¼ 0:

<sup>k</sup>,0ð Þ¼� <sup>x</sup>;<sup>t</sup> ql

2 ffiffiffi η p

<sup>k</sup>,0ð Þ x . These arbitrary functions provide the condition:

Lξukð Þ¼ m 0, Lςukð Þ¼ m 0,

This relation is used by the initial condition for determining Zk,0ð Þ x;t from the

Further repeating this process, we can determine all the coefficients of ukð Þ m of

n i¼0 εi uið Þ m :

> i ð Þ <sup>x</sup>;<sup>t</sup> , dl k,i

<sup>∂</sup>tzk,0ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> zk,0ð Þ¼ <sup>x</sup>;<sup>t</sup> <sup>0</sup>,

<sup>k</sup>,0ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> <sup>q</sup><sup>l</sup>

<sup>k</sup>ð Þt ck, <sup>1</sup>ð Þ¼� x;t zk,0ð Þþ x;t fkð Þ x;t exp

<sup>2</sup> ffi <sup>t</sup> <sup>p</sup> <sup>¼</sup> <sup>ς</sup><sup>l</sup> <sup>2</sup> ffiffi <sup>η</sup> <sup>p</sup> occurs: The initial conditions for equation (Eq. (18)) are determined from (Eq. (12)).

ensuring the solvability of the equation with respect to cl

Zk,0ð Þj <sup>x</sup>;<sup>t</sup> <sup>t</sup>¼<sup>0</sup> <sup>¼</sup> fkð Þ <sup>x</sup>;<sup>t</sup> exp

u<sup>ε</sup>nð Þ¼ m ∑

R<sup>ε</sup>nð Þ� x;t; ε R<sup>ε</sup>nð Þ m; ε <sup>γ</sup>¼ρð Þ <sup>x</sup>;t;<sup>ε</sup> ¼ u xð Þ� ;t; ε ∑

taking into account (Eqs. (3) and (6)), we obtain the equation

<sup>L</sup>εR<sup>ε</sup>nð Þ¼ <sup>x</sup>;t; <sup>ε</sup> <sup>ε</sup><sup>n</sup>þ<sup>1</sup>

with homogeneous boundary conditions. Using the maximum principle, like

j j <sup>R</sup><sup>ε</sup>nð Þ <sup>x</sup>;t; <sup>ε</sup> , <sup>c</sup>ε<sup>n</sup>þ<sup>1</sup>

� � � �

<sup>k</sup>,0ð Þ <sup>x</sup>;<sup>t</sup> erfc <sup>ς</sup><sup>l</sup>

� � � �

<sup>0</sup>ð Þ¼ x;t 0,

<sup>k</sup>,0ð Þ x;t ,

iθкð Þ 0 ε � �

<sup>k</sup>,0ð Þ x;t in the

<sup>k</sup>,0ð Þ x;t are expressed through arbitrary functions

iθkð Þ 0 ε � �

> n i¼0 εi uið Þ m

gnð Þ x;t; ε

:

ð Þ <sup>x</sup>;<sup>t</sup> , zk,ið Þ <sup>x</sup>;<sup>t</sup> , ql

� � � � γ¼ρð Þ x;t;ε

: (19)

k,i

,

ð Þ x;t , we

,

<sup>k</sup>, <sup>1</sup>ð Þ x;t : Suppose that

(18)

In the case when a small parameter is also included as a multiplier with a temporal derivative, the asymptotic of the solution acquires a complex structure.

Different classes of singularly perturbed parabolic equations are studied in [2]. There, regularized asymptotics of the solution of these equations are constructed, when a small parameter is in front of the time derivative and with one spatial derivative. It is shown that the constructed asymptotic contains exponential, parabolic, and angular products of exponential and parabolic boundary layer functions. The equations are studied when the limiting equation has a regular singularity. Such equations have a power boundary layer. If a small parameter is entering as the multiplier for all spatial derivatives, then the asymptotic solution contains a multidimensional parabolic boundary layer function. When entering into the equation, as free terms of rapidly oscillating functions, then the asymptotic of the solution additionally contains fast-oscillating boundary layer functions. If it is additionally assumed that the phase of this free term has a stationary point, in addition to the rapidly oscillating boundary layer function that arises as a power boundary layer.

This section is devoted to a two-dimensional equation of parabolic type.

#### 2.2 Statement of the problem

Consider the problem:

$$L\_{\varepsilon}u(\mathbf{x},t,\varepsilon) \equiv \partial\_{l}u - \varepsilon^{2}\Delta\_{d}u - b(\mathbf{x},t)u = f(\mathbf{x},t)\exp\left(\frac{i\theta(t)}{\varepsilon}\right), (\mathbf{x},t)\epsilon E,$$

$$u|\_{t=0} = \mathbf{0}, u|\_{\partial\varrho=0} = \mathbf{0},\tag{20}$$

where ε . 0 is the small parameter, x ¼ ð Þ x1; x<sup>2</sup> , Ω ¼ ð Þ 0 , x<sup>1</sup> , 1 x ð Þ <sup>0</sup> , <sup>x</sup><sup>2</sup> , <sup>1</sup> , E <sup>¼</sup> ð Þ <sup>0</sup> , <sup>t</sup> <sup>≤</sup><sup>T</sup> <sup>x</sup>Ω, <sup>Δ</sup><sup>a</sup> � <sup>∑</sup><sup>2</sup> <sup>l</sup>¼<sup>1</sup>alð Þ xl <sup>∂</sup><sup>2</sup> xl .

The problem is solved under the following assumptions:

1. <sup>∀</sup>xl <sup>∈</sup> ½ � <sup>0</sup>; <sup>1</sup> the function alð Þ xl <sup>∈</sup> <sup>С</sup><sup>∞</sup>½ � <sup>0</sup>; <sup>1</sup> , l <sup>¼</sup> <sup>1</sup>, 2.

$$2.b(\mathfrak{x},t), f(\mathfrak{x},t) \in \mathbb{C}^{\infty}[E].$$

$$3. \theta'(\mathbf{0}) = \mathbf{0}.$$

#### 2.3 Regularization of the problem

Following the method of regularization of singularly perturbed problems [1, 2], along with the independent variables xð Þ ;t , we introduce regularizing variables:

Boundary Layer Flows - Theory, Applications and Numerical Methods

$$\mu = \frac{t}{\varepsilon}, \xi\_l = \frac{(-1)^{l-1}}{\sqrt{\varepsilon^3}} \int\_{l-1}^{\chi\_1} \frac{ds}{\sqrt{a\_1(s)}}, \eta\_l = \frac{\varrho\_l(\chi\_1)}{\varepsilon^2}$$

$$\xi\_{l+2} = \frac{(-1)^{l-1}}{\sqrt{\varepsilon^3}} \int\_{l-1}^{\chi\_2} \frac{ds}{\sqrt{a\_2(s)}}, \eta\_{l+2} = \frac{\varrho\_{l+2}(\chi\_2)}{\varepsilon^3}$$

$$\sigma = \int\_0^t \varepsilon^{\frac{l[\theta(\gamma) - \theta(0)]}{\varepsilon}} ds, \tau\_2 = \frac{i[\theta(t) - \theta(0)]}{\varepsilon}, \tau\_1 = \frac{t}{\varepsilon^2},$$

$$\varrho\_l(\chi\_r) = (-1)^{l-1} \int\_{l-1}^{\chi\_r} \frac{ds}{\sqrt{a\_r(s)}},$$

<sup>L</sup>e<sup>ε</sup>e<sup>u</sup> � <sup>1</sup>

series

<sup>ε</sup><sup>2</sup> <sup>T</sup><sup>0</sup>e<sup>u</sup> <sup>þ</sup>

1 ε iθ0

Singularly Perturbed Parabolic Problems DOI: http://dx.doi.org/10.5772/intechopen.84339

eu|

ð Þ<sup>t</sup> <sup>∂</sup><sup>τ</sup>2e<sup>u</sup> <sup>þ</sup>

In this case, the identity is satisfied:

2.4 Solution of iterative problems

T0uvð Þ¼ M 0, v ¼ 0, 1,

T0u<sup>4</sup> ¼ f xð Þ ; t exp τ<sup>2</sup> þ

We introduce a class of functions:

T0uq ¼ �iθ<sup>0</sup>

T0ui ¼ �iθ<sup>0</sup>

U<sup>5</sup> ¼ V5ð Þ M : V2ð Þ¼ M ∑

185

<sup>t</sup>¼τ1¼τ2¼<sup>0</sup> <sup>¼</sup> <sup>0</sup>, <sup>e</sup>u<sup>|</sup>

1 ε

> Dr,l x, <sup>ξ</sup> � 2φ<sup>0</sup> l ð Þ xr <sup>∂</sup><sup>2</sup>

<sup>T</sup><sup>1</sup>e<sup>u</sup> <sup>þ</sup> <sup>D</sup><sup>σ</sup>e<sup>u</sup> � <sup>L</sup><sup>η</sup>e<sup>u</sup> � ffiffi

L<sup>η</sup> � ∑ 2 r¼1 ∑ 2r l¼2r�1

Δη � ∑ 4 k¼1 ∂2 ηk

<sup>L</sup>e<sup>ε</sup>e<sup>u</sup> � �� � �

ð Þ<sup>t</sup> <sup>∂</sup>τ2uq�<sup>2</sup> � <sup>T</sup>1uq�<sup>2</sup>, q <sup>¼</sup> <sup>2</sup>, <sup>3</sup>:

ui t¼τ¼<sup>0</sup> <sup>¼</sup> <sup>0</sup>; ui j jxl¼r�1, <sup>ξ</sup>k¼ηk¼<sup>0</sup> <sup>¼</sup> <sup>0</sup>, l, r <sup>¼</sup> <sup>1</sup>, <sup>2</sup>:<sup>k</sup> <sup>¼</sup> <sup>1</sup>, <sup>4</sup>

U<sup>4</sup> ¼ V4ð Þ M : V1ð Þ¼ M ∑

Any item u Mð Þ∈ U is representable in the form:

4 r,l¼<sup>1</sup> Y<sup>r</sup>þ2,l

iθð Þ 0 ε � �

<sup>T</sup><sup>0</sup> � <sup>∂</sup><sup>τ</sup><sup>1</sup> � Δη, T<sup>1</sup> � <sup>∂</sup><sup>μ</sup> <sup>þ</sup> Δξ, D<sup>σ</sup> � Dt <sup>þ</sup> exp ð Þ <sup>τ</sup><sup>2</sup> <sup>∂</sup>σ, Dt � <sup>∂</sup><sup>t</sup> <sup>þ</sup> b xð Þ ; <sup>t</sup> ,

For the solution of the extended function (Eq. (24)), we search in the form of

∞ i¼0 ε i

� T1u<sup>2</sup> � Dσu<sup>0</sup> þ Lηu0,

ð Þ Nl � � �

� , cexp � <sup>η</sup><sup>2</sup>

� , cexp � <sup>η</sup>r,l �

l 8τ<sup>1</sup>

> ; ηr,l � � � � ¼

� � � 2 8τ<sup>1</sup> !

ð Þ<sup>t</sup> <sup>∂</sup>τ2ui�<sup>2</sup> � <sup>T</sup>1ui�<sup>2</sup> � <sup>D</sup>σui�<sup>4</sup> <sup>þ</sup> <sup>L</sup>ηui�<sup>4</sup> <sup>þ</sup> <sup>L</sup>ξui�<sup>5</sup> <sup>þ</sup> <sup>Δ</sup>aui�<sup>8</sup>,

<sup>U</sup><sup>0</sup> <sup>¼</sup> <sup>V</sup>0ð Þ¼ <sup>N</sup> ½ � c xð Þþ ; <sup>t</sup> <sup>F</sup>1ð Þþ <sup>N</sup> <sup>F</sup>2ð Þ <sup>N</sup> exp ð Þ <sup>τ</sup><sup>2</sup> ; <sup>F</sup>1ð Þ <sup>N</sup> <sup>∈</sup> <sup>U</sup>4; <sup>F</sup>2ð Þ <sup>N</sup> <sup>∈</sup> <sup>U</sup>5;c xð Þ ; <sup>t</sup> <sup>∈</sup>C<sup>∞</sup> <sup>E</sup> � � ��

<sup>U</sup><sup>1</sup> <sup>¼</sup> <sup>V</sup>1ð Þ <sup>M</sup> : <sup>V</sup>1ð Þ¼ <sup>M</sup> v xð Þþ ; <sup>t</sup> <sup>F</sup>1ð Þþ <sup>M</sup> <sup>F</sup>2ð Þ <sup>M</sup> ; <sup>F</sup>1ð Þ <sup>M</sup> <sup>∈</sup> <sup>U</sup>4; <sup>F</sup>2ð Þ <sup>M</sup> <sup>∈</sup> <sup>U</sup>5; v xð Þ ; <sup>t</sup> <sup>∈</sup>C<sup>∞</sup> <sup>E</sup> � � ��

<sup>U</sup><sup>2</sup> <sup>¼</sup> <sup>V</sup>2ð Þ <sup>M</sup> : <sup>V</sup>2ð Þ¼ <sup>M</sup> ½ � z xð Þþ ; <sup>t</sup> <sup>F</sup>1ð Þþ <sup>M</sup> <sup>F</sup>2ð Þ <sup>M</sup> <sup>σ</sup>; <sup>F</sup>1ð Þ <sup>M</sup> <sup>∈</sup> <sup>U</sup>4; <sup>F</sup>2ð Þ <sup>M</sup> <sup>∈</sup> <sup>U</sup>5; z xð Þ ; <sup>t</sup> <sup>∈</sup>C<sup>∞</sup> <sup>E</sup> � � ��

� � � � ,

� �

U ¼ U<sup>0</sup> ⨁U<sup>1</sup> ⨁U2:

<sup>q</sup> ( )

ð Þ Nrþ2,l

4 l¼1 Yl ð Þ Nl ; <sup>Y</sup><sup>l</sup>

ð Þ Nrþ2,l ; <sup>Y</sup><sup>r</sup>þ2,l

From these classes we will construct a new one, as a direct sum:

�

<sup>e</sup>u Mð Þ¼ ; <sup>ε</sup> <sup>∑</sup>

Then, for the coefficients of this series, we get the following problems:

<sup>ε</sup> <sup>p</sup> <sup>L</sup><sup>ξ</sup>e<sup>u</sup> � <sup>ε</sup><sup>2</sup>

xl¼r�1, <sup>ξ</sup>k¼ηk¼<sup>0</sup> <sup>¼</sup> <sup>0</sup>, r <sup>¼</sup> <sup>1</sup>, <sup>2</sup>, l <sup>¼</sup> <sup>1</sup>, <sup>2</sup>, k <sup>¼</sup> <sup>1</sup>, <sup>4</sup>

<sup>l</sup> ð Þ xr <sup>∂</sup><sup>η</sup><sup>l</sup>

<sup>χ</sup>¼ψð Þ <sup>x</sup>;t;<sup>ε</sup> � <sup>L</sup>εu xð Þ ; <sup>t</sup>; <sup>ε</sup> : (25)

<sup>2</sup>uið Þ M : (26)

arð Þ xr <sup>D</sup>r,l x, <sup>η</sup>,

xrξ<sup>l</sup> þ φ<sup>00</sup>

, E<sup>1</sup> <sup>¼</sup> <sup>E</sup>хð Þ <sup>0</sup>; <sup>∞</sup> <sup>10</sup>

h i,

<sup>Δ</sup><sup>a</sup>e<sup>u</sup> <sup>¼</sup> f xð Þ ; <sup>t</sup> exp <sup>τ</sup><sup>2</sup> <sup>þ</sup>

iθð Þ 0 ε � �,

(24)

(27)

,

,

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi η2 <sup>r</sup> <sup>þ</sup> <sup>η</sup><sup>2</sup> l

,

:

For extended function <sup>e</sup>u Mð Þ ; <sup>ε</sup> , M <sup>¼</sup> ð Þ <sup>x</sup>; <sup>t</sup>; <sup>τ</sup>; <sup>ξ</sup>; <sup>η</sup> such that

<sup>e</sup>u Mð Þj ; <sup>ε</sup> <sup>μ</sup>¼ψð Þ <sup>x</sup>;t;<sup>ε</sup> � u xð Þ ; <sup>t</sup>; <sup>ε</sup> , χ ¼ ð Þ τ; ξ; η , τ ¼ ð Þ τ1; τ<sup>2</sup> , ξ ¼ ξ1; ξ2; ξ3; ξ<sup>4</sup> ð Þ, η ¼ η1; η2; η3; η<sup>4</sup> ð Þ, <sup>ψ</sup>ð Þ¼ <sup>x</sup>; <sup>t</sup>; <sup>ε</sup> <sup>t</sup> ε2 ; t ε ; i½ � θðÞ�t θð Þ 0 <sup>ε</sup> ; φð Þ x ε ; φð Þ x ε2 � �, φð Þ¼ x ð Þ φ1ð Þ x<sup>1</sup> ; φ2ð Þ x<sup>1</sup> ; φ3ð Þ x<sup>2</sup> ; φ4ð Þ x<sup>2</sup> <sup>e</sup>u Mð Þj ; <sup>ε</sup> <sup>μ</sup>¼ψð Þ <sup>x</sup>;t;<sup>ε</sup> � u xð Þ ; <sup>t</sup>; <sup>ε</sup> , <sup>χ</sup> <sup>¼</sup> ð Þ <sup>τ</sup>; <sup>ξ</sup>; <sup>η</sup> , τ ¼ ð Þ τ1; τ<sup>2</sup> , ξ ¼ ξ1; ξ2; ξ3; ξ<sup>4</sup> ð Þ, η ¼ η1; η2; η3; η<sup>4</sup> ð Þ, <sup>ψ</sup>ð Þ¼ <sup>x</sup>; <sup>t</sup>; <sup>ε</sup> <sup>t</sup> ε2 ; t ε ; i½ � θðÞ�t θð Þ 0 <sup>ε</sup> ; φð Þ x ε ; φð Þ x ε2 � �, (22)

Find from (Eq. (22)) the derivatives based on

<sup>∂</sup>tu � <sup>∂</sup><sup>t</sup>e<sup>u</sup> <sup>þ</sup> <sup>1</sup> <sup>ε</sup> <sup>∂</sup><sup>μ</sup>e<sup>u</sup> <sup>þ</sup> <sup>1</sup> <sup>ε</sup><sup>2</sup> <sup>∂</sup><sup>τ</sup>1e<sup>u</sup> <sup>þ</sup> <sup>i</sup>θ<sup>0</sup> ð Þt <sup>ε</sup> <sup>∂</sup><sup>τ</sup>2e<sup>u</sup> <sup>þ</sup> exp ð Þ <sup>τ</sup><sup>2</sup> <sup>∂</sup><sup>σ</sup>e<sup>u</sup> � �� � � χ¼ψð Þ x;t;ε , <sup>∂</sup>xru � <sup>∂</sup>xre<sup>u</sup> <sup>þ</sup> <sup>∑</sup> 2r l¼2r�1 φ0 l ð Þ xr ffiffiffiffi <sup>ε</sup><sup>3</sup> <sup>p</sup> <sup>∂</sup><sup>ξ</sup><sup>l</sup> <sup>e</sup><sup>u</sup> <sup>þ</sup> <sup>φ</sup><sup>0</sup> l ð Þ xr <sup>ε</sup><sup>2</sup> <sup>∂</sup><sup>ζ</sup><sup>l</sup> eu � � � � � � � � χ¼ψð Þ x;t;ε , ∂2 xr <sup>u</sup> � <sup>∂</sup><sup>2</sup> xr <sup>e</sup><sup>u</sup> <sup>þ</sup> <sup>∑</sup> 2r l¼2r�1 φ0 l 2ð Þ xr <sup>ε</sup><sup>3</sup> <sup>∂</sup><sup>2</sup> ξl <sup>e</sup><sup>u</sup> <sup>þ</sup> <sup>φ</sup><sup>0</sup> l 2ð Þ xr <sup>ε</sup><sup>4</sup> <sup>∂</sup><sup>2</sup> ζl eu � � � þ ∑ 2r l¼2r�1 2φ<sup>0</sup> l ð Þ xr ffiffiffiffi <sup>ε</sup><sup>3</sup> <sup>p</sup> <sup>∂</sup><sup>2</sup> xrξ<sup>l</sup> <sup>e</sup><sup>u</sup> <sup>þ</sup> <sup>φ</sup><sup>00</sup> l ð Þ xr ffiffiffiffi <sup>ε</sup><sup>3</sup> <sup>p</sup> <sup>∂</sup><sup>ξ</sup><sup>l</sup> <sup>e</sup><sup>u</sup> <sup>þ</sup> 1 <sup>ε</sup><sup>2</sup> <sup>φ</sup><sup>0</sup> l ð Þ xr <sup>∂</sup><sup>2</sup> xrη<sup>l</sup> <sup>e</sup><sup>u</sup> <sup>þ</sup> <sup>φ</sup><sup>00</sup> <sup>l</sup> ð Þ xr <sup>∂</sup><sup>η</sup><sup>l</sup> eu � � � ��� � � � χ¼ψð Þ x;t;ε (23)

φð Þ¼ x ð Þ φ1ð Þ x<sup>1</sup> ; φ2ð Þ x<sup>1</sup> ; φ3ð Þ x<sup>2</sup> ; φ4ð Þ x<sup>2</sup> :

Below, it is shown that the solution of the iterative problems does not contain terms depending on ξ1; ξ<sup>2</sup> ð Þ, ξ3; ξ<sup>4</sup> ð Þ, ζ1; ζ<sup>2</sup> ð Þ, ζ3; ζ<sup>4</sup> ð Þ, ξl; ζ<sup>k</sup> ð Þ, l, k ¼ 1, 2: Therefore, to simplify recording, the mixed derivatives of these variables are omitted. Based on (Eq. (20)), (Eq. (22)), and (Eq. (23)) for extended function u Með Þ ; <sup>ε</sup> , set the problem:

Singularly Perturbed Parabolic Problems DOI: http://dx.doi.org/10.5772/intechopen.84339

<sup>μ</sup> <sup>¼</sup> <sup>t</sup> ε

<sup>σ</sup> <sup>¼</sup> <sup>Ð</sup><sup>t</sup> 0 e i½ � θð Þ�s θð Þ 0

<sup>ξ</sup>lþ<sup>2</sup> <sup>¼</sup> ð Þ �<sup>1</sup> <sup>l</sup>�<sup>1</sup>

<sup>ψ</sup>ð Þ¼ <sup>x</sup>; <sup>t</sup>; <sup>ε</sup> <sup>t</sup>

<sup>ψ</sup>ð Þ¼ <sup>x</sup>; <sup>t</sup>; <sup>ε</sup> <sup>t</sup>

Find from (Eq. (22)) the derivatives based on

<sup>∂</sup>xru � <sup>∂</sup>xre<sup>u</sup> <sup>þ</sup> <sup>∑</sup>

<sup>u</sup> � <sup>∂</sup><sup>2</sup> xr

<sup>e</sup><sup>u</sup> <sup>þ</sup> <sup>φ</sup><sup>00</sup>

<sup>ε</sup> <sup>∂</sup><sup>μ</sup>e<sup>u</sup> <sup>þ</sup> <sup>1</sup>

2r l¼2r�1

<sup>e</sup><sup>u</sup> <sup>þ</sup> <sup>∑</sup> 2r l¼2r�1

l ð Þ xr ffiffiffiffi <sup>ε</sup><sup>3</sup> <sup>p</sup> <sup>∂</sup><sup>ξ</sup><sup>l</sup>

<sup>∂</sup>tu � <sup>∂</sup><sup>t</sup>e<sup>u</sup> <sup>þ</sup> <sup>1</sup>

∂2 xr

2φ<sup>0</sup> l ð Þ xr ffiffiffiffi <sup>ε</sup><sup>3</sup> <sup>p</sup> <sup>∂</sup><sup>2</sup> xrξ<sup>l</sup>

þ ∑ 2r l¼2r�1

problem:

184

, <sup>ξ</sup><sup>l</sup> <sup>¼</sup> ð Þ �<sup>1</sup> <sup>l</sup>�<sup>1</sup> ffiffiffiffi <sup>ε</sup><sup>3</sup> <sup>p</sup>

Boundary Layer Flows - Theory, Applications and Numerical Methods

ðx<sup>2</sup> l�1

<sup>φ</sup>lð Þ¼ � xr ð Þ<sup>1</sup> <sup>l</sup>�<sup>1</sup> <sup>Ð</sup> xr

ffiffiffiffi <sup>ε</sup><sup>3</sup> <sup>p</sup>

For extended function <sup>e</sup>u Mð Þ ; <sup>ε</sup> , M <sup>¼</sup> ð Þ <sup>x</sup>; <sup>t</sup>; <sup>τ</sup>; <sup>ξ</sup>; <sup>η</sup> such that

ε2 ; t ε ;

ε2 ; t ε ;

<sup>ε</sup><sup>2</sup> <sup>∂</sup><sup>τ</sup>1e<sup>u</sup> <sup>þ</sup> <sup>i</sup>θ<sup>0</sup>

φ0 l ð Þ xr ffiffiffiffi <sup>ε</sup><sup>3</sup> <sup>p</sup> <sup>∂</sup><sup>ξ</sup><sup>l</sup>

ðx<sup>1</sup> l�1

> ds ffiffiffiffiffiffiffiffiffiffi

<sup>ε</sup> ds, <sup>τ</sup><sup>2</sup> <sup>¼</sup> <sup>i</sup>½ � <sup>θ</sup>ðÞ�<sup>t</sup> <sup>θ</sup>ð Þ <sup>0</sup>

<sup>e</sup>u Mð Þj ; <sup>ε</sup> <sup>μ</sup>¼ψð Þ <sup>x</sup>;t;<sup>ε</sup> � u xð Þ ; <sup>t</sup>; <sup>ε</sup> , χ ¼ ð Þ τ; ξ; η , τ ¼ ð Þ τ1; τ<sup>2</sup> , ξ ¼ ξ1; ξ2; ξ3; ξ<sup>4</sup> ð Þ, η ¼ η1; η2; η3; η<sup>4</sup> ð Þ,

> i½ � θðÞ�t θð Þ 0 <sup>ε</sup> ;

φð Þ¼ x ð Þ φ1ð Þ x<sup>1</sup> ; φ2ð Þ x<sup>1</sup> ; φ3ð Þ x<sup>2</sup> ; φ4ð Þ x<sup>2</sup>

<sup>e</sup>u Mð Þj ; <sup>ε</sup> <sup>μ</sup>¼ψð Þ <sup>x</sup>;t;<sup>ε</sup> � u xð Þ ; <sup>t</sup>; <sup>ε</sup> , <sup>χ</sup> <sup>¼</sup> ð Þ <sup>τ</sup>; <sup>ξ</sup>; <sup>η</sup> ,

τ ¼ ð Þ τ1; τ<sup>2</sup> , ξ ¼ ξ1; ξ2; ξ3; ξ<sup>4</sup> ð Þ,

η ¼ η1; η2; η3; η<sup>4</sup> ð Þ,

φð Þ¼ x ð Þ φ1ð Þ x<sup>1</sup> ; φ2ð Þ x<sup>1</sup> ; φ3ð Þ x<sup>2</sup> ; φ4ð Þ x<sup>2</sup> :

ð Þt

<sup>e</sup><sup>u</sup> <sup>þ</sup> <sup>φ</sup><sup>0</sup> l ð Þ xr <sup>ε</sup><sup>2</sup> <sup>∂</sup><sup>ζ</sup><sup>l</sup> eu

� � �

� ��

� � � � �

φ0 l 2ð Þ xr <sup>ε</sup><sup>3</sup> <sup>∂</sup><sup>2</sup> ξl <sup>e</sup><sup>u</sup> <sup>þ</sup> <sup>φ</sup><sup>0</sup> l 2ð Þ xr <sup>ε</sup><sup>4</sup> <sup>∂</sup><sup>2</sup> ζl eu

� � � ���

Below, it is shown that the solution of the iterative problems does not contain terms depending on ξ1; ξ<sup>2</sup> ð Þ, ξ3; ξ<sup>4</sup> ð Þ, ζ1; ζ<sup>2</sup> ð Þ, ζ3; ζ<sup>4</sup> ð Þ, ξl; ζ<sup>k</sup> ð Þ, l, k ¼ 1, 2: Therefore, to simplify recording, the mixed derivatives of these variables are omitted. Based on (Eq. (20)), (Eq. (22)), and (Eq. (23)) for extended function u Með Þ ; <sup>ε</sup> , set the

<sup>e</sup><sup>u</sup> <sup>þ</sup> 1 <sup>ε</sup><sup>2</sup> <sup>φ</sup><sup>0</sup> l ð Þ xr <sup>∂</sup><sup>2</sup> xrη<sup>l</sup>

i½ � θðÞ�t θð Þ 0 <sup>ε</sup> ;

� �

� �

ds ffiffiffiffiffiffiffiffiffiffi

l�1

<sup>a</sup>1ð Þ<sup>s</sup> <sup>p</sup> , <sup>η</sup><sup>l</sup> <sup>¼</sup> <sup>φ</sup>lð Þ <sup>x</sup><sup>1</sup>

<sup>a</sup>2ð Þ<sup>s</sup> <sup>p</sup> , <sup>η</sup>lþ<sup>2</sup> <sup>¼</sup> <sup>φ</sup>lþ2ð Þ <sup>x</sup><sup>2</sup>

ds ffiffiffiffiffiffiffiffiffi arð Þ<sup>s</sup> <sup>p</sup> , ε2

ε2 ,

(21)

(22)

ε3

<sup>ε</sup> , <sup>τ</sup><sup>1</sup> <sup>¼</sup> <sup>t</sup>

φð Þ x ε ;

φð Þ x ε ;

<sup>ε</sup> <sup>∂</sup><sup>τ</sup>2e<sup>u</sup> <sup>þ</sup> exp ð Þ <sup>τ</sup><sup>2</sup> <sup>∂</sup><sup>σ</sup>e<sup>u</sup>

φð Þ x ε2

φð Þ x ε2

,

� � χ¼ψð Þ x;t;ε ,

,

� � � χ¼ψð Þ x;t;ε (23)

� � � χ¼ψð Þ x;t;ε

<sup>l</sup> ð Þ xr <sup>∂</sup><sup>η</sup><sup>l</sup> eu

<sup>e</sup><sup>u</sup> <sup>þ</sup> <sup>φ</sup><sup>00</sup>

,

$$\tilde{L}\_{\theta}\tilde{u} \equiv \frac{1}{\varepsilon^{2}}T\tilde{u}\tilde{u} + \frac{1}{\varepsilon}\dot{u}\theta(\vartheta)\partial\_{\tilde{x}}\tilde{u} + \frac{1}{\varepsilon}T\tilde{u} + D\_{\theta}\tilde{u} - L\_{\theta}\tilde{u} - \sqrt{\varepsilon}L\_{\varepsilon}\tilde{u} - \varepsilon^{2}\Delta\_{\theta}\tilde{u} = f(x,t)\exp\left(\tau\_{2} + \frac{i\theta(0)}{\varepsilon}\right),$$

$$\tilde{u}|\_{t=\tau\_{1}=\tau\_{2}=0} = 0,\ \tilde{u}|\_{\vert\eta-\tau-1,\ t\_{0}=\eta\_{0}=0} = 0,\ r=1,2,\ l=1,2,k=\overline{1,4}$$

$$T\_{0} \equiv \partial\_{t\_{1}} - \Delta\_{\theta},\ T\_{1} \equiv \partial\_{\mu} + \Delta\_{\theta}, D\_{\sigma} \equiv D\_{l} + \exp\left(\tau\_{2}\right)\partial\_{\sigma}, D\_{l} \equiv \partial\_{l} + b(\mathbf{x},t),$$

$$L\_{\eta} \equiv \sum\_{r=1}^{2}\sum\_{l=2r-1}^{2r} a\_{l}(\mathbf{x}\_{r})D^{r,l}\_{x\_{\eta},\eta}$$

$$D^{r,l}\_{x\_{\ast}\overline{t}} \equiv \left[2\rho^{\eta}\_{l}(\mathbf{x}\_{\ast})\partial^{2}\_{x\_{\ast}\overline{t}} + \rho^{\eta}\_{l}(\mathbf{x}\_{\ast})\partial\_{\eta}\right],$$

$$\Delta\_{\eta} \equiv \sum\_{k=1}^{4} \delta^{2}\_{\eta\_{k}},\ E\_{1} = E\mathbf{x}(0,\infty)^{10}$$

$$\text{(24)}$$

In this case, the identity is satisfied:

$$\left. \left( \widetilde{L}\_{\epsilon} \widetilde{\boldsymbol{u}} \right) \right|\_{\boldsymbol{\chi} = \boldsymbol{\psi}(\boldsymbol{x}, \boldsymbol{t}, \boldsymbol{\varepsilon})} \equiv L\_{\epsilon} \boldsymbol{u}(\boldsymbol{x}, \boldsymbol{t}, \boldsymbol{\varepsilon}). \tag{25}$$

#### 2.4 Solution of iterative problems

For the solution of the extended function (Eq. (24)), we search in the form of series

$$
\widetilde{u}(M,\varepsilon) = \sum\_{i=0}^{\infty} \varepsilon^{\dot{\xi}} u\_i(M). \tag{26}
$$

Then, for the coefficients of this series, we get the following problems:

$$\begin{aligned} T\_0 u\_\nu(M) &= 0, \nu = 0, 1, \\ T\_0 u\_q &= -i\theta'(t)\partial\_{\tau\_1} u\_{q-2} - T\_1 u\_{q-2}, q = 2, 3. \\ T\_0 u\_4 &= f(\mathbf{x}, t) \exp\left(\tau\_2 + \frac{i\theta(0)}{\varepsilon}\right) - T\_1 u\_2 - D\_\sigma u\_0 + L\_\eta u\_0, \\ T\_0 u\_i &= -i\theta'(t)\partial\_{\tau\_2} u\_{i-2} - T\_1 u\_{i-2} - D\_\sigma u\_{i-4} + L\_\eta u\_{i-4} + L\_\xi u\_{i-5} + \Delta\_d u\_{i-8}, \\ u\_i|\_{l=r=0} &= 0, u\_i|\_{\mathbf{x}=r-1, \xi\_k=\eta\_k=0} = 0, l, r = 1, 2. k = \overline{1, 4} \end{aligned} \tag{27}$$

We introduce a class of functions:

<sup>U</sup><sup>0</sup> <sup>¼</sup> <sup>V</sup>0ð Þ¼ <sup>N</sup> ½ � c xð Þþ ; <sup>t</sup> <sup>F</sup>1ð Þþ <sup>N</sup> <sup>F</sup>2ð Þ <sup>N</sup> exp ð Þ <sup>τ</sup><sup>2</sup> ; <sup>F</sup>1ð Þ <sup>N</sup> <sup>∈</sup> <sup>U</sup>4; <sup>F</sup>2ð Þ <sup>N</sup> <sup>∈</sup> <sup>U</sup>5;c xð Þ ; <sup>t</sup> <sup>∈</sup>C<sup>∞</sup> <sup>E</sup> � � �� , <sup>U</sup><sup>1</sup> <sup>¼</sup> <sup>V</sup>1ð Þ <sup>M</sup> : <sup>V</sup>1ð Þ¼ <sup>M</sup> v xð Þþ ; <sup>t</sup> <sup>F</sup>1ð Þþ <sup>M</sup> <sup>F</sup>2ð Þ <sup>M</sup> ; <sup>F</sup>1ð Þ <sup>M</sup> <sup>∈</sup> <sup>U</sup>4; <sup>F</sup>2ð Þ <sup>M</sup> <sup>∈</sup> <sup>U</sup>5; v xð Þ ; <sup>t</sup> <sup>∈</sup>C<sup>∞</sup> <sup>E</sup> � � �� , <sup>U</sup><sup>2</sup> <sup>¼</sup> <sup>V</sup>2ð Þ <sup>M</sup> : <sup>V</sup>2ð Þ¼ <sup>M</sup> ½ � z xð Þþ ; <sup>t</sup> <sup>F</sup>1ð Þþ <sup>M</sup> <sup>F</sup>2ð Þ <sup>M</sup> <sup>σ</sup>; <sup>F</sup>1ð Þ <sup>M</sup> <sup>∈</sup> <sup>U</sup>4; <sup>F</sup>2ð Þ <sup>M</sup> <sup>∈</sup> <sup>U</sup>5; z xð Þ ; <sup>t</sup> <sup>∈</sup>C<sup>∞</sup> <sup>E</sup> � � �� ,

$$U\_4 = \left\{ V\_4(\mathcal{M}) : V\_1(\mathcal{M}) = \sum\_{l=1}^4 Y^l(N\_l), \left| Y^l(N\_l) \right| < \exp\left( -\frac{\eta\_l^2}{8\tau\_1} \right) \right\},$$

$$U\_5 = \left\{ V\_5(\mathcal{M}) : V\_2(\mathcal{M}) = \sum\_{\nu\_l=1}^4 Y^{r+2,l}(N\_{r+2,l}), \left| Y^{r+2,l}(N\_{r+2,l}) \right| < \exp\left( -\frac{|\eta^{r,l}|^2}{8\tau\_1} \right), \left| \eta^{r,l} \right| = \sqrt{\eta\_r^2 + \eta\_l^2} \right\}.$$

From these classes we will construct a new one, as a direct sum:

$$U = U\_0 \oplus U\_1 \oplus U\_2.$$

Any item u Mð Þ∈ U is representable in the form:

Boundary Layer Flows - Theory, Applications and Numerical Methods

$$u(M) = v(\mathbf{x}, t) + c(\mathbf{x}, t) \exp\left(\tau\_2\right) + z(\mathbf{x}, t)\sigma + \left[\sum\_{l=1}^4 Y^l(N\_l) + \sum\_{r, l=1}^2 Y^{r+2, l}(N\_{r+2, l})\right] \exp\left(\tau\_2\right)$$

$$+ \sum\_{l=1}^4 w^l(\mathbf{x}, t) \text{erfc}\left(\frac{\tilde{\xi}\_l}{2\sqrt{\mu}}\right) + \sum\_{l, r=1}^2 w^{r+2, l}(M\_{r+2, l}) + \left[\sum\_{l=1}^4 q^l(\mathbf{x}, t) \text{erfc}\left(\frac{\tilde{\xi}\_l}{2\sqrt{\mu}}\right) + \sum\_{l, r=1}^2 Z^{r+2, l}(M\_{r+2, l})\right] \sigma,$$

$$N\_l = (\mathbf{x}, t, \tau\_1, \eta\_l), N\_{r+2, l} = (\mathbf{x}, t, \tau\_1, \eta\_l, \eta\_{r+2}),$$

$$M\_l = (\mathbf{x}, t, \mu, \xi\_l), M\_{r+2, l} = (\mathbf{x}, t, \mu, \xi\_l, \xi\_{r+2}). \tag{28}$$

Lηu ¼ ∑ 2 r¼1

Lξu ¼ ∑ 2 r¼1

þσ ∑ 2 r¼1

∑ 2r l¼2r�1

Singularly Perturbed Parabolic Problems DOI: http://dx.doi.org/10.5772/intechopen.84339

> ∑ 2r l¼2r�1

Dσu Mð Þ¼ Dtv xð Þþ ; t ∑

4 l¼1 ql

∑ 2r l¼2r�1

þ Dtc xð Þþ ; t ∑

þσ Dtz xð Þþ ; t ∑

þ z xð Þþ ; t ∑

T0Y<sup>l</sup>

ð Þ¼ Nrþ2,l <sup>Δ</sup>ηY<sup>r</sup>þ2,l

ð Þj Nrþ2,l <sup>η</sup>rþ2¼<sup>0</sup> ¼ �Y<sup>l</sup>

of the following equations:

ð Þ Nl <sup>τ</sup>1¼<sup>0</sup> <sup>¼</sup> <sup>0</sup>; <sup>Y</sup><sup>l</sup>

� �

Theorem 3. Let be <sup>H</sup>2ð Þ Nrþ2,l <sup>∈</sup> <sup>U</sup>5, Y<sup>l</sup>

2.5 The decision of the iterative problems

sentable in the form <sup>u</sup>0ð Þ <sup>M</sup> <sup>∈</sup> <sup>U</sup> if functions <sup>Y</sup><sup>l</sup>

Yl

T0Y<sup>l</sup>

<sup>v</sup>ð Þ¼ Nl dl

The proof of these theorems is given in [2].

�

ð Þ Nl ∈ U4.

<sup>H</sup>1ð Þ Nl , Y<sup>l</sup>

solution Y<sup>l</sup>

187

Dr,l x, <sup>η</sup>Y<sup>l</sup>

Dr,l x, <sup>ξ</sup>w<sup>l</sup>

Dr,l x, <sup>ξ</sup>q<sup>l</sup>

> 4 l¼1 DtY<sup>l</sup>

> > 4 l¼1 Dtq<sup>l</sup>

We write iterative equation (8) in the form:

Eq. (31) is solvable in U, if the equations are solvable:

ð Þ¼ Nl <sup>H</sup>1ð Þ Nl , l <sup>¼</sup> <sup>1</sup>, <sup>4</sup>, T0Y<sup>r</sup>þ2,l

ð Þ Nl

Theorem 2. Let be <sup>H</sup>1ð Þ Nl <sup>∈</sup> <sup>U</sup>4. Then, the problem <sup>∂</sup>τ1Y<sup>l</sup>

� <sup>η</sup>l¼<sup>0</sup> <sup>¼</sup> dl

ð Þþ Nrþ2,l <sup>H</sup>2ð Þ Nrþ2,l , Y<sup>r</sup>þ2,l

ð Þ Nl , r, l <sup>¼</sup> <sup>1</sup>, 2 has a solution <sup>Y</sup><sup>r</sup>þ2,l

Eq. (27) under v ¼ 0, 1 is homogeneous. By Theorem 1, it has a solution repre-

<sup>v</sup>ð Þ¼ Nl <sup>0</sup>, T0Y<sup>r</sup>þ2,l <sup>v</sup> ð Þ¼ Nrþ2,l <sup>0</sup>:

2 ffiffiffiffi τ1 p

Based on the boundary conditions from (Eq. (29)), the solution is written:

<sup>v</sup>ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>η</sup><sup>l</sup>

ð Þþ Nl ∑ 2 v¼1 ∑ 2 r,l¼<sup>1</sup>

ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>ξ</sup><sup>l</sup>

ð Þþ Nl ∑

" #

4 l¼1 Dtw<sup>l</sup>

ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>ξ</sup><sup>l</sup>

ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>ξ</sup><sup>l</sup>

Dv,l x, <sup>η</sup>Yrþ2,l

þ ∑ 2 v¼1 ∑ 2 r,l¼<sup>1</sup>

ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>ξ</sup><sup>l</sup>

DtY<sup>r</sup>þ2,l

2 ffiffiffi μ p � �

> þ ∑ 2 r,l¼<sup>1</sup>

" #

Theorem 1. Let be H Mð Þ∈ U<sup>4</sup> ⨁U<sup>5</sup> and condition (1) is satisfied. Then,

þ ∑ 2 v¼1 ∑ 2 r,l¼<sup>1</sup>

2 ffiffiffi μ p � �

> þ ∑ 2 r,l¼<sup>1</sup>

z<sup>r</sup>þ2,l

ð Þ Nrþ2,l

2 ffiffiffi μ p � �

" #

2 ffiffiffi μ p � �

> 2 r,l¼<sup>1</sup>

ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>ξ</sup><sup>l</sup>

2 ffiffiffi μ p � �

" #

ð Þ Nrþ2,l ,

Dv,l x, <sup>ξ</sup>wrþ2,l

þ ∑ 2 r,l¼<sup>1</sup>

Dtz<sup>r</sup>þ2,l

ð Þ Mrþ2,l

T0u Mð Þ¼ H Mð Þ: (31)

ð Þ x; t , l ¼ 1, 4 (Eq. (32)) has a

�

ð Þ Nl and <sup>Y</sup><sup>r</sup>þ2,l

ð Þ Nrþ2,l <sup>η</sup>l¼<sup>0</sup> ¼ �Y<sup>r</sup>þ<sup>2</sup>

ð Þ¼ Nrþ2,l H2ð Þ Nrþ2,l , r, l ¼ 1, 2:

ð Þ Nl <sup>∈</sup> <sup>U</sup>4, and then the problem <sup>∂</sup>τ1Y<sup>r</sup>þ2,l

ð Þ Nrþ2,l ∈ U5:

� �, l <sup>¼</sup> <sup>1</sup>, <sup>2</sup>, <sup>3</sup>, <sup>4</sup>: (32)

ð Þ¼ Nl <sup>Δ</sup>ηY<sup>l</sup>

ð Þ Nrþ<sup>2</sup> ; <sup>Y</sup><sup>r</sup>þ2,l �

ð Þ Nrþ2,l – are solutions

ð Þþ Nl

Dv,l x, <sup>ξ</sup> <sup>z</sup>rþ2,l ð Þ Mrþ2,l

,

ð Þ Mrþ2,l

ð Þ Mrþ2,l

Dtw<sup>r</sup>þ2,l

ð Þ Mrþ2,l

exp ð Þ τ<sup>2</sup>

exp ð Þ τ<sup>2</sup>

Let's satisfy this function to the boundary conditions:

v xð Þ¼� ; <sup>0</sup> c xð Þ ; <sup>0</sup> ; <sup>Y</sup><sup>l</sup> ð Þ Nl � � <sup>t</sup>¼τ1¼<sup>0</sup> <sup>¼</sup> <sup>0</sup> (29) Y<sup>r</sup>þ2,l ð Þ Nrþ2,l <sup>t</sup>¼τ1¼<sup>0</sup> <sup>¼</sup> <sup>0</sup>; <sup>w</sup><sup>l</sup> � � � � <sup>t</sup>¼<sup>0</sup> <sup>¼</sup> wl ð Þ x , ql � � <sup>t</sup>¼<sup>0</sup> <sup>¼</sup> <sup>q</sup><sup>l</sup> ð Þ x , w<sup>r</sup>þ2,l ð Þ Mrþ2,l � � <sup>t</sup>¼μ¼<sup>0</sup> <sup>¼</sup> <sup>0</sup>, z<sup>r</sup>þ2,l ð Þ Mrþ2,l � � <sup>t</sup>¼μ¼<sup>0</sup> <sup>¼</sup> <sup>0</sup>, wl ð Þ x; t � � <sup>x</sup>1¼l�<sup>1</sup> ¼ �v lð Þ � <sup>1</sup>; <sup>x</sup>2; <sup>t</sup> , ql ð Þ x; t � � <sup>x</sup>1¼l�<sup>1</sup> ¼ �z lð Þ � <sup>1</sup>; <sup>x</sup>2; <sup>t</sup> , Yl � � <sup>x</sup>1¼l�1, <sup>η</sup>l¼<sup>0</sup> ¼ �c lð Þ � <sup>1</sup>; <sup>x</sup>2; <sup>t</sup> , Y<sup>r</sup>þ2,l <sup>x</sup>1¼l�1, <sup>η</sup>l¼<sup>0</sup> ¼ �Y<sup>r</sup>þ2,l ð Þ Nrþ2,l � � � � x1¼l�1 , w<sup>r</sup>þ2,l � � <sup>x</sup>1¼l�1, <sup>ξ</sup>l¼<sup>0</sup> ¼ �w<sup>r</sup>þ<sup>2</sup> ð Þ <sup>l</sup> � <sup>1</sup>; <sup>x</sup>2; <sup>t</sup> erfc <sup>ξ</sup><sup>r</sup>þ<sup>2</sup> 2 ffiffi t p � �, z<sup>r</sup>þ2,l � � <sup>x</sup>1¼l�1, <sup>ξ</sup>l¼<sup>0</sup> ¼ �q<sup>r</sup>þ<sup>2</sup> ð Þ <sup>l</sup> � <sup>1</sup>; <sup>x</sup>2; <sup>t</sup> erfc <sup>ξ</sup><sup>r</sup>þ<sup>2</sup> 2 ffiffi t p � �, wl ð Þ <sup>x</sup>; <sup>t</sup> j j xr¼l�<sup>1</sup> ¼ �v xð Þ ; <sup>t</sup> xr¼l�<sup>1</sup>, ql ð Þ <sup>x</sup>; <sup>t</sup> j j xr¼l�<sup>1</sup> ¼ �z xð Þ ; <sup>t</sup> xr¼l�<sup>1</sup>, Y<sup>r</sup>þ<sup>2</sup> � � <sup>x</sup>2¼l�1, <sup>η</sup>rþ2¼<sup>0</sup> ¼ �c xð Þ <sup>1</sup>; <sup>l</sup> � <sup>1</sup>; <sup>t</sup> , Y<sup>r</sup>þ2,l <sup>x</sup>2¼l�1, <sup>η</sup>rþ2¼<sup>0</sup> ¼ �Y<sup>l</sup> � � � � x2¼l�1 , <sup>w</sup><sup>r</sup>þ2,l <sup>x</sup>2¼l�1, <sup>ξ</sup>rþ2¼<sup>0</sup> ¼ �w<sup>l</sup> � � � � x2¼l�1 erfc <sup>ξ</sup><sup>l</sup> 2 ffiffi t p � �, z<sup>r</sup>þ2,l <sup>x</sup>2¼l�1, <sup>ξ</sup>rþ2¼<sup>0</sup> ¼ �q<sup>l</sup> � � � � x2¼l�1 erfc <sup>ξ</sup><sup>l</sup> 2 ffiffi t p � �, l, r <sup>¼</sup> <sup>1</sup>, <sup>2</sup>:

We compute the action of the operators T0, T1, Lη, L<sup>ξ</sup> on function u Mð Þ∈ U, and we have

$$T\_1 u(M) = \sum\_{r,l=1}^2 \left\{ \partial\_\mu w^{r+2,l} - \Delta\_\xi w^{r+2,l} + \sigma \left[ \partial\_\mu z^{r+2,l} - \Delta\_\xi z^{r+2,l} \right] \right\},$$

Singularly Perturbed Parabolic Problems DOI: http://dx.doi.org/10.5772/intechopen.84339

u Mð Þ¼ v xð Þþ ; t c xð Þ ; t exp ð Þþ τ<sup>2</sup> z xð Þ ; t σ þ ∑

Boundary Layer Flows - Theory, Applications and Numerical Methods

þ ∑ 2 l,r¼<sup>1</sup>

w<sup>r</sup>þ2,l

Let's satisfy this function to the boundary conditions:

Y<sup>r</sup>þ2,l

v xð Þ¼� ; <sup>0</sup> c xð Þ ; <sup>0</sup> ; <sup>Y</sup><sup>l</sup>

ql � � <sup>t</sup>¼<sup>0</sup> <sup>¼</sup> <sup>q</sup><sup>l</sup>

w<sup>r</sup>þ2,l

z<sup>r</sup>þ2,l

<sup>x</sup>1¼l�1, <sup>η</sup>l¼<sup>0</sup> ¼ �c lð Þ � <sup>1</sup>; <sup>x</sup>2; <sup>t</sup> , Y<sup>r</sup>þ2,l <sup>x</sup>1¼l�1, <sup>η</sup>l¼<sup>0</sup> ¼ �Y<sup>r</sup>þ2,l

<sup>x</sup>1¼l�1, <sup>ξ</sup>l¼<sup>0</sup> ¼ �w<sup>r</sup>þ<sup>2</sup>

<sup>x</sup>1¼l�1, <sup>ξ</sup>l¼<sup>0</sup> ¼ �q<sup>r</sup>þ<sup>2</sup>

wl ð Þ x; t � �

ql ð Þ x; t � �

wl

ql

Y<sup>r</sup>þ2,l

<sup>x</sup>2¼l�1, <sup>ξ</sup>rþ2¼<sup>0</sup> ¼ �q<sup>l</sup> � � �

Y<sup>r</sup>þ<sup>2</sup> � �

w<sup>r</sup>þ2,l

2 r,l¼<sup>1</sup>

z<sup>r</sup>þ2,l

T1u Mð Þ¼ ∑

w<sup>r</sup>þ2,l � �

> z<sup>r</sup>þ2,l � �

ð Þ Nrþ2,l <sup>t</sup>¼τ1¼<sup>0</sup> <sup>¼</sup> <sup>0</sup>; <sup>w</sup><sup>l</sup> � � �

> ð Þ Mrþ2,l � �

ð Þ Mrþ2,l � �

�

ð Þ <sup>x</sup>; <sup>t</sup> j j xr¼l�<sup>1</sup> ¼ �v xð Þ ; <sup>t</sup> xr¼l�<sup>1</sup>,

ð Þ <sup>x</sup>; <sup>t</sup> j j xr¼l�<sup>1</sup> ¼ �z xð Þ ; <sup>t</sup> xr¼l�<sup>1</sup>,

<sup>x</sup>2¼l�1, <sup>η</sup>rþ2¼<sup>0</sup> ¼ �Y<sup>l</sup> � � �

> � x2¼l�1

<sup>x</sup>2¼l�1, <sup>ξ</sup>rþ2¼<sup>0</sup> ¼ �w<sup>l</sup> � � �

<sup>x</sup>2¼l�1, <sup>η</sup>rþ2¼<sup>0</sup> ¼ �c xð Þ <sup>1</sup>; <sup>l</sup> � <sup>1</sup>; <sup>t</sup> ,

� x2¼l�1

We compute the action of the operators T0, T1, Lη, L<sup>ξ</sup> on function u Mð Þ∈ U, and

þ∑ 4 l¼1 wl

> Yl � �

we have

186

ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>ξ</sup><sup>l</sup>

2 ffiffiffi μ p � �

4 l¼1 Yl

4 l¼1 ql

ð Þþ Mrþ2,l ∑

Nl ¼ x; t; τ1; η<sup>l</sup> ð Þ, Nrþ2,l ¼ x; t; τ1; ηl; ηrþ<sup>2</sup>

Ml ¼ x; t; μ; ξ<sup>l</sup> ð Þ, Mrþ2,l ¼ x; t; μ; ξl; ξrþ<sup>2</sup> ð Þ:

ð Þ Nl � �

ð Þ x ,

<sup>x</sup>1¼l�<sup>1</sup> ¼ �v lð Þ � <sup>1</sup>; <sup>x</sup>2; <sup>t</sup> ,

<sup>x</sup>1¼l�<sup>1</sup> ¼ �z lð Þ � <sup>1</sup>; <sup>x</sup>2; <sup>t</sup> ,

� <sup>t</sup>¼<sup>0</sup> <sup>¼</sup> wl

<sup>t</sup>¼μ¼<sup>0</sup> <sup>¼</sup> <sup>0</sup>,

<sup>t</sup>¼μ¼<sup>0</sup> <sup>¼</sup> <sup>0</sup>,

� �

ð Þ <sup>l</sup> � <sup>1</sup>; <sup>x</sup>2; <sup>t</sup> erfc <sup>ξ</sup><sup>r</sup>þ<sup>2</sup>

ð Þ <sup>l</sup> � <sup>1</sup>; <sup>x</sup>2; <sup>t</sup> erfc <sup>ξ</sup><sup>r</sup>þ<sup>2</sup>

� x2¼l�1 ,

erfc <sup>ξ</sup><sup>l</sup> 2 ffiffi t p � �

<sup>∂</sup>μw<sup>r</sup>þ2,l � <sup>Δ</sup>ξw<sup>r</sup>þ2,l <sup>þ</sup> <sup>σ</sup> <sup>∂</sup>μz<sup>r</sup>þ2,l � <sup>Δ</sup>ξz<sup>r</sup>þ2,l � � � � ,

erfc <sup>ξ</sup><sup>l</sup> 2 ffiffi t p � � ,

ð Þþ Nl ∑ 2 r,l¼<sup>1</sup>

ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>ξ</sup><sup>l</sup>

� �,

Y<sup>r</sup>þ2,l

" #

<sup>t</sup>¼τ1¼<sup>0</sup> <sup>¼</sup> <sup>0</sup> (29)

ð Þ Nrþ2,l

2 ffiffi t p � � ,

2 ffiffi t p � � ,

, l, r ¼ 1, 2:

� x1¼l�1 ,

ð Þ x ,

" #

2 ffiffiffi μ p � � ð Þ Nrþ2,l

zrþ2,l

þ ∑ 2 l,r¼<sup>1</sup> exp ð Þ τ<sup>2</sup>

ð Þ Mrþ2,l

(28)

σ,

Lηu ¼ ∑ 2 r¼1 ∑ 2r l¼2r�1 Dr,l x, <sup>η</sup>Y<sup>l</sup> ð Þþ Nl ∑ 2 v¼1 ∑ 2 r,l¼<sup>1</sup> Dv,l x, <sup>η</sup>Yrþ2,l ð Þ Nrþ2,l , Lξu ¼ ∑ 2 r¼1 ∑ 2r l¼2r�1 Dr,l x, <sup>ξ</sup>w<sup>l</sup> ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>ξ</sup><sup>l</sup> 2 ffiffiffi μ p � � þ ∑ 2 v¼1 ∑ 2 r,l¼<sup>1</sup> Dv,l x, <sup>ξ</sup>wrþ2,l ð Þ Mrþ2,l þσ ∑ 2 r¼1 ∑ 2r l¼2r�1 Dr,l x, <sup>ξ</sup>q<sup>l</sup> ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>ξ</sup><sup>l</sup> 2 ffiffiffi μ p � � þ ∑ 2 v¼1 ∑ 2 r,l¼<sup>1</sup> Dv,l x, <sup>ξ</sup> <sup>z</sup>rþ2,l ð Þ Mrþ2,l " #, Dσu Mð Þ¼ Dtv xð Þþ ; t ∑ 4 l¼1 Dtw<sup>l</sup> ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>ξ</sup><sup>l</sup> 2 ffiffiffi μ p � � þ ∑ 2 r,l¼<sup>1</sup> Dtw<sup>r</sup>þ2,l ð Þ Mrþ2,l þ Dtc xð Þþ ; t ∑ 4 l¼1 DtY<sup>l</sup> ð Þþ Nl ∑ 2 r,l¼<sup>1</sup> DtY<sup>r</sup>þ2,l ð Þ Nrþ2,l " # exp ð Þ τ<sup>2</sup> þσ Dtz xð Þþ ; t ∑ 4 l¼1 Dtq<sup>l</sup> ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>ξ</sup><sup>l</sup> 2 ffiffiffi μ p � � þ ∑ 2 r,l¼<sup>1</sup> Dtz<sup>r</sup>þ2,l ð Þ Mrþ2,l " # þ z xð Þþ ; t ∑ 4 l¼1 ql ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>ξ</sup><sup>l</sup> 2 ffiffiffi μ p � � þ ∑ 2 r,l¼<sup>1</sup> z<sup>r</sup>þ2,l ð Þ Mrþ2,l " # exp ð Þ τ<sup>2</sup>

We write iterative equation (8) in the form:

$$T\_0 u(\mathcal{M}) = H(\mathcal{M}).\tag{31}$$

Theorem 1. Let be H Mð Þ∈ U<sup>4</sup> ⨁U<sup>5</sup> and condition (1) is satisfied. Then, Eq. (31) is solvable in U, if the equations are solvable:

$$T\_0 Y^l(N\_l) = H\_1(N\_l), \\ l = \overline{1,4}, \\ T\_0 Y^{r+2,l}(N\_{r+2,l}) = H\_2(N\_{r+2,l}), \\ r, l = 1, 2.$$

Theorem 2. Let be <sup>H</sup>1ð Þ Nl <sup>∈</sup> <sup>U</sup>4. Then, the problem <sup>∂</sup>τ1Y<sup>l</sup> ð Þ¼ Nl <sup>Δ</sup>ηY<sup>l</sup> ð Þþ Nl <sup>H</sup>1ð Þ Nl , Y<sup>l</sup> ð Þ Nl <sup>τ</sup>1¼<sup>0</sup> <sup>¼</sup> <sup>0</sup>; <sup>Y</sup><sup>l</sup> ð Þ Nl � � � � <sup>η</sup>l¼<sup>0</sup> <sup>¼</sup> dl ð Þ x; t , l ¼ 1, 4 (Eq. (32)) has a solution Y<sup>l</sup> ð Þ Nl ∈ U4.

Theorem 3. Let be <sup>H</sup>2ð Þ Nrþ2,l <sup>∈</sup> <sup>U</sup>5, Y<sup>l</sup> ð Þ Nl <sup>∈</sup> <sup>U</sup>4, and then the problem <sup>∂</sup>τ1Y<sup>r</sup>þ2,l ð Þ¼ Nrþ2,l <sup>Δ</sup>ηY<sup>r</sup>þ2,l ð Þþ Nrþ2,l <sup>H</sup>2ð Þ Nrþ2,l , Y<sup>r</sup>þ2,l ð Þ Nrþ2,l <sup>η</sup>l¼<sup>0</sup> ¼ �Y<sup>r</sup>þ<sup>2</sup> ð Þ Nrþ<sup>2</sup> ; <sup>Y</sup><sup>r</sup>þ2,l � � ð Þj Nrþ2,l <sup>η</sup>rþ2¼<sup>0</sup> ¼ �Y<sup>l</sup> ð Þ Nl , r, l <sup>¼</sup> <sup>1</sup>, 2 has a solution <sup>Y</sup><sup>r</sup>þ2,l ð Þ Nrþ2,l ∈ U5: The proof of these theorems is given in [2].

2.5 The decision of the iterative problems

Eq. (27) under v ¼ 0, 1 is homogeneous. By Theorem 1, it has a solution representable in the form <sup>u</sup>0ð Þ <sup>M</sup> <sup>∈</sup> <sup>U</sup> if functions <sup>Y</sup><sup>l</sup> ð Þ Nl and <sup>Y</sup><sup>r</sup>þ2,l ð Þ Nrþ2,l – are solutions of the following equations:

$$T\_0 Y^l\_v(N\_l) = \mathbf{0}, \\ T\_0 Y^{r+2,l}\_v(N\_{r+2,l}) = \mathbf{0}.$$

Based on the boundary conditions from (Eq. (29)), the solution is written:

$$Y\_v^l(\mathbf{N}\_l) = d\_v^l(\mathbf{x}, t) \text{erfc}\left(\frac{\eta\_l}{2\sqrt{\tau\_1}}\right), l = 1, 2, 3, 4. \tag{32}$$

$$\begin{split} Y\_{v}^{r+2,l}(N\_{r+2,l}) &= -\int\_{0}^{\tau\_{1}} \int\_{0}^{\infty} Y\_{v}^{l}(\*) \left[ \frac{\partial}{\partial \xi} \, G(N\_{l}, \xi, \eta, \tau\_{1} - \tau) \right] \Big|\_{\xi=0} d\eta d\tau \\ &- \int\_{0}^{t} \int\_{0}^{\infty} Y\_{v}^{r+2}(\*) \left[ \frac{\partial}{\partial \eta} \, G(N\_{r+2,l}, \xi, \eta, \tau\_{1} - \tau) \right] \Big|\_{\eta=0} d\xi d\tau, \end{split}$$

�σ Dtz0ð Þþ x; t ∑

� z0ð Þþ x; t ∑

Singularly Perturbed Parabolic Problems DOI: http://dx.doi.org/10.5772/intechopen.84339

> ∑ 2r l¼2r�1

> > �iθ<sup>0</sup>

þ∑ 2 r¼1

then

4 l¼1 Dtq<sup>l</sup>

4 l¼1 ql

Dr,l x, <sup>ξ</sup>w<sup>p</sup>

DtY<sup>l</sup>

Dtw<sup>l</sup>

Dr,l x, <sup>ξ</sup>w<sup>l</sup>

F4ð Þ¼� M iθ<sup>0</sup>

� ∑ 4 l¼1 ql

Substituting the value Y<sup>l</sup>

arbitrary initial condition dl

determined from the relation:

� � �

following relations hold: Dtw<sup>r</sup>þ2,l

<sup>2</sup>ð Þ <sup>x</sup>; <sup>t</sup> , q<sup>l</sup>

The same is true for functions w<sup>r</sup>þ2,l

Solutions of equations with respect w<sup>r</sup>þ2,l

with respect to dl

condition L<sup>η</sup> Y<sup>l</sup>

dl

expresses through Y<sup>l</sup>

through w<sup>l</sup>

189

Dtq<sup>l</sup>

<sup>0</sup>ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>ξ</sup><sup>l</sup>

" #

2 ffiffiffi μ p � �

2 ffiffiffi μ p � �

By providing F4ð Þ M ∈ U<sup>4</sup> ⨁U<sup>5</sup> with regard to cvð Þ¼ x; t 0, v ¼ 0, 1, we set

Dtv0ð Þ¼ x; t 0, Dtz0ð Þ¼ x; t 0,

x, <sup>η</sup>Y<sup>r</sup>þ2,l

<sup>2</sup>ð Þþ Nl ∑

" #

which comes out from (Eq. (29)) and (Eq. (33)). The function Y<sup>r</sup>þ2,l

<sup>0</sup> <sup>¼</sup> <sup>0</sup>, Dv,l

<sup>0</sup> <sup>¼</sup> <sup>0</sup>, Dtzrþ2,l

conditions from (Eq. (29) are representable as (Eq. (33)), and they are expressed

<sup>2</sup>ð Þ x; t : The first equation (Eq. (36)) is solvable,

<sup>0</sup>ð Þ Nl therefore provided that

DtY<sup>r</sup>þ2,l

þ ∑ 2 r,l¼<sup>1</sup>

<sup>0</sup>ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>η</sup><sup>l</sup>

<sup>0</sup> ð Þ x; t

x, <sup>η</sup>Y<sup>r</sup>þ2,l <sup>0</sup> ¼ 0:

<sup>0</sup> <sup>¼</sup> <sup>0</sup>, Dv,l

x, <sup>ξ</sup>w<sup>r</sup>þ2,l

<sup>0</sup> ð Þ Mrþ2,l , z<sup>r</sup>þ2,l

<sup>0</sup> , zrþ2,l

2 r,l¼<sup>1</sup>

" #

<sup>2</sup> ¼ 0, T0z

<sup>0</sup> <sup>¼</sup> <sup>0</sup>, DtY<sup>r</sup>þ2,l

<sup>0</sup>ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>ξ</sup><sup>l</sup>

<sup>0</sup>ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>ξ</sup><sup>l</sup>

ð Þt c2ð Þþ x; t f xð Þ ; t exp

<sup>0</sup>ð Þ Nl , T0w<sup>r</sup>þ2,l

<sup>0</sup> <sup>¼</sup> <sup>0</sup>, Dtw<sup>r</sup>þ2,l

<sup>0</sup>ð Þ¼ <sup>x</sup>; <sup>t</sup> <sup>0</sup>, Dv,l

ð Þt ∑ 4 l¼1 Yl

In the last bracket, the transition is from the variables <sup>ξ</sup><sup>l</sup>

<sup>0</sup>ð Þ¼ Nl dl

<sup>0</sup>ð Þ <sup>x</sup>; <sup>t</sup> , we get the equation Dtdl

<sup>0</sup>ð Þ x; t � � � <sup>t</sup>¼<sup>0</sup> <sup>¼</sup> <sup>d</sup> l

<sup>0</sup> <sup>¼</sup> <sup>0</sup>;therefore, Dx, <sup>η</sup> <sup>Y</sup><sup>l</sup>

<sup>0</sup>ð Þ <sup>x</sup>; <sup>t</sup> <sup>x</sup>1¼l�<sup>1</sup> ¼ �c0ð Þ <sup>l</sup> � <sup>1</sup>; <sup>x</sup>2; <sup>t</sup> ; <sup>d</sup><sup>l</sup>þ<sup>2</sup>

2 ffiffiffiffi τ2 p � �

<sup>0</sup>ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>η</sup><sup>l</sup>

<sup>0</sup> ð Þ¼ <sup>x</sup>; <sup>t</sup> <sup>0</sup>, Dtz<sup>r</sup>þ2,l

2 ffiffiffi μ p � �

> þ ∑ 2 r,l¼<sup>1</sup>

þ ∑ 2 v¼1

iθð Þ 0 ε � �

" #

þ ∑ 2 r,l¼<sup>1</sup> Dtz rþ2,l <sup>0</sup> ð Þ Mrþ2,l

zrþ2,l <sup>0</sup> ð Þ Mrþ2,l

∑ 2 r,l¼<sup>1</sup>

> rþ2,l <sup>2</sup> ¼ 0,

<sup>0</sup> ð Þ¼ x; t 0,

<sup>0</sup> <sup>¼</sup> <sup>0</sup>, Dr,l

Y<sup>r</sup>þ2,l

z<sup>r</sup>þ2,l <sup>0</sup> ð Þ Nrþ2,l

> <sup>2</sup> ffiffiffi τ1 p

> > � � � x2¼l�1

<sup>0</sup> ¼ 0,

x, <sup>η</sup>Y<sup>l</sup>

<sup>2</sup> ð Þ Nrþ2,l

<sup>0</sup> ¼ 0,

exp ð Þ τ<sup>2</sup>

exp ð Þ τ<sup>2</sup> :

<sup>0</sup>ð Þ¼ x; t 0, which is solved under an

¼ �c0ð Þ x1; l � 1; t ,

<sup>0</sup> ð Þ Mrþ2,l ; in other words, the

<sup>0</sup> <sup>¼</sup> <sup>0</sup>, Dv,l

<sup>0</sup> under appropriate boundary

<sup>μ</sup> <sup>p</sup> to the variables <sup>η</sup><sup>l</sup>

<sup>2</sup> ffiffi

� � into equation DtY<sup>l</sup>

<sup>0</sup>ð Þ x . This arbitrary function provides the

<sup>0</sup> ¼ 0:The initial condition for this equation is

Dv,l x, <sup>η</sup>Yrþ2,l

� z0ð Þ¼ x; t 0,

exp ð Þ τ<sup>2</sup>

(35)

<sup>2</sup> ffiffiffi τ2 <sup>p</sup> .

<sup>0</sup>ð Þ¼ Nl 0,

<sup>0</sup> ð Þ Nrþ2,l

x, <sup>ξ</sup>zrþ2,l <sup>0</sup> ¼ 0:

<sup>0</sup> ð Þ Nrþ2,l :

where d<sup>l</sup> ð Þ x; t – is arbitrary function such as

$$\begin{split}d\_{v}^{p}(\mathbf{x},t)\Big|\_{t=0} &= -\overline{d}\_{v}^{p}(\mathbf{x}),d\_{v}^{l}(\mathbf{x},t)\Big|\_{\mathbf{x}\_{1}=l-1} = -c\_{v}(l-1,\mathbf{x}\_{2},t),\\G\big{(}\eta\_{l},\eta\_{r+2,l},\xi,\eta,\tau\_{1}\big) &= \frac{1}{4\pi\tau\_{1}}\Bigg{{}\exp\left(-\frac{\left(\eta\_{l}-\xi\right)^{2}}{4\tau\_{1}}\right)} - \exp\left(-\frac{\left(\eta\_{l}+\xi\right)^{2}}{4\tau\_{1}}\right)\Bigg{{}\\\-\left\{\exp\left(-\frac{\left(\eta\_{r+2}-\eta\right)^{2}}{4\tau\_{1}}\right)} - \exp\left(-\frac{\left(\eta\_{r+2}+\eta\right)^{2}}{4\tau\_{1}}\right)\right\}.\end{split}\tag{33}$$

Due to the fact that the function dl <sup>v</sup>ð Þ x; t при t ¼ τ<sup>1</sup> ¼ 0 multiplied by the function becomes as d<sup>l</sup> <sup>0</sup>ð Þ x; t � � � <sup>t</sup>¼<sup>0</sup> ¼ �<sup>d</sup> l <sup>0</sup>ð Þ x , an arbitrary function is accepted, and its values under x<sup>1</sup> ¼ l � 1 are determined from the second relation. According to Theorems 2 and 3, the functions found by the formula (Eq. (33)) satisfy the estimates:

$$\left|Y\_v^l(\mathbf{N}\_l)\right| \le c \exp\left(-\frac{\eta\_l^2}{8\tau\_1}\right), \left|Y\_v^{r+2,l}(\mathbf{N}\_{r+2,l})\right| \le c \exp\left(-\frac{\eta\_{r+2}^2 + \eta\_l^2}{8\tau\_1}\right), r, l = 1, 2. \tag{34}$$

Free member of equation (Eq. (27)) under v ¼ 2, 3 has a form

$$F\_{v-2}(\mathcal{M}) \equiv T\_1 \mathfrak{u}\_{v-2}(\mathcal{M}) + i\theta'(t)\partial\_{\sigma}\mathfrak{u}\_{v-2}(\mathcal{M}) = i\theta'(t) \left[ c\_{v-2}(\mathfrak{x}, t) + \sum\_{l=1}^{4} Y\_{v-2}^{l}(\mathcal{N}\_l) + \sum\_{r,l=1}^{2} Y\_{v-2}^{r+2, l}(\mathcal{N}\_{r+2, l}) \right],$$

$$\exp\left(\mathfrak{x}\right) + \sum\_{l\_l=1}^{2} \left\{ \partial\_{\mu} w\_{v-2}^{r+2, l} - \Delta\_l \mathfrak{u}\_{v-2}^{r+2, l} + \sigma \left[ \partial\_{\mu} \mathfrak{x}\_{v-2}^{r+2, l} - \Delta\_l \mathfrak{x}\_{v-2}^{r+2, l} \right] \right\},$$

so that equation (Eq. (27)), under v ¼ 2, 3, has a solution in U; we set

$$\mathcal{L}\_{v-2}(\mathbf{x}, t) = \mathbf{0}, \, T\_1 w\_{v-2}^{r+2, l} = \mathbf{0}, \, T\_1 z\_{v-2}^{r+2, l} = \mathbf{0}.$$

Solutions of the last equations under the boundary conditions from (Eq. (29)) have a form (Eq. (33)) for which estimates of the form (Eq. (35) are fair. Eq. (27), i=4, has a free term:

$$F\_{4}(M) = -i\theta'(t)\partial\_{\tau\_{2}} - T\_{1}u\_{2} + f(\mathbf{x},t)\exp\left(\frac{i\theta(0)}{\varepsilon}\right) - D\_{\sigma}u\_{0} + L\_{\eta}u\_{0}$$

$$= -i\theta'(t)\left[c\_{2}(\mathbf{x},t) + \frac{4}{l\_{1}}Y\_{2}^{l}(N\_{l}) + \sum\_{r,l=1}^{2}Y\_{2}^{r+2,l}(N\_{r+2,l})\right]\exp\left(\tau\_{2}\right)$$

$$-\sum\_{l\_{0}=1}^{2}\left[T\_{0}w\_{2}^{r+2,l}(M\_{r+2,l}) + \sigma T\_{0}z\_{2}^{r+2,l}\right] - D\_{l}v\_{0}(\mathbf{x},t) - \sum\_{l=1}^{4}D\_{l}w\_{0}^{l}(\mathbf{x},t)\text{erfc}\left(\frac{\xi\_{l}}{2\sqrt{\mu}}\right)$$

$$-\sum\_{l\_{0}=1}^{2}D\_{l}w\_{0}^{r+2,l}(\mathbf{x},t) - \exp\left(\tau\_{2}\right)\left[\partial\_{l}c\_{0}(\mathbf{x},t) + \sum\_{l=1}^{4}\partial\_{l}Y\_{0}^{l} + \sum\_{l\_{0}=1}^{2}D\_{l}Y\_{0}^{r+2,l}\right]$$

Singularly Perturbed Parabolic Problems DOI: http://dx.doi.org/10.5772/intechopen.84339

$$\begin{split} & -\sigma \left[ D\_{l} \boldsymbol{z}\_{0}(\boldsymbol{\kappa},t) + \sum\_{l=1}^{4} D\_{l} q\_{0}^{l}(\boldsymbol{\kappa},t) \text{erfc} \left( \frac{\xi\_{l}}{2\sqrt{\mu}} \right) + \sum\_{r\_{\boldsymbol{r}},l=1}^{2} D\_{l} \boldsymbol{z}\_{0}^{r+2,l}(\boldsymbol{M}\_{r+2,l}) \right] \\ & - \left[ \boldsymbol{z}\_{0}(\boldsymbol{\kappa},t) + \sum\_{l=1}^{4} q\_{0}^{l}(\boldsymbol{\kappa},t) \text{erfc} \left( \frac{\xi\_{l}}{2\sqrt{\mu}} \right) + \sum\_{r\_{\boldsymbol{r}},l=1}^{2} \boldsymbol{z}\_{0}^{r+2,l}(\boldsymbol{M}\_{r+2,l}) \right] \text{exp} \left( \boldsymbol{\tau}\_{2} \right) \\ & + \sum\_{r=1}^{2} \sum\_{l=2r-1}^{2r} D\_{\boldsymbol{x},\boldsymbol{\xi}}^{r,l} \boldsymbol{w}\_{0}^{p}(\boldsymbol{x},t) \text{erfc} \left( \frac{\xi\_{l}}{2\sqrt{\mu}} \right) + \sum\_{v=1}^{2} \sum\_{r\_{\boldsymbol{r}},l=1}^{2} D\_{\boldsymbol{x},\boldsymbol{\eta}}^{p,l} \boldsymbol{Y}\_{0}^{r+2,l}(\boldsymbol{N}\_{r+2,l}). \end{split}$$

By providing F4ð Þ M ∈ U<sup>4</sup> ⨁U<sup>5</sup> with regard to cvð Þ¼ x; t 0, v ¼ 0, 1, we set

$$\begin{aligned} -i\theta'(t)c\_2(\mathbf{x},t) + f(\mathbf{x},t)\exp\left(\frac{i\theta(0)}{\varepsilon}\right) - z\_0(\mathbf{x},t) &= \mathbf{0}, \\ D\_t v\_0(\mathbf{x},t) &= \mathbf{0}, D\_t \mathbf{z}\_0(\mathbf{x},t) = \mathbf{0}, \\ D\_t Y\_0^l(N\_l), T\_0 w\_2^{r+2,l} &= \mathbf{0}, T\_0 \mathbf{z}\_2^{r+2,l} = \mathbf{0}, \\ D\_t w\_0^l &= \mathbf{0}, D\_t w\_0^{r+2,l} = \mathbf{0}, D\_t Y\_0^{r+2,l} = \mathbf{0}, \\ D\_t q\_0^l \;(\mathbf{x},t) &= \mathbf{0}, D\_t \mathbf{z}\_0^{r+2,l} \;(\mathbf{x},t) = \mathbf{0}, \\ D\_{\mathbf{x},\xi}^{r,l} \underline{\boldsymbol{\mu}} \underline{\boldsymbol{\nu}}\_0(\mathbf{x},t) &= \mathbf{0}, D\_{\mathbf{x},\eta}^{r,l} Y\_0^{r+2,l} = \mathbf{0}, D\_{\mathbf{x},\eta}^{r,l} Y\_0^l = \mathbf{0}, \end{aligned} \tag{35}$$

then

<sup>Y</sup>rþ2,l <sup>v</sup> ð Þ¼� Nrþ2,l

p <sup>v</sup> ð Þ <sup>x</sup> ; <sup>d</sup><sup>l</sup>

where d<sup>l</sup>

<sup>v</sup> ð Þ x; t <sup>t</sup>¼<sup>0</sup> ¼ �d

G ηl; ηrþ2,l; ξ; η; τ<sup>1</sup> � � <sup>¼</sup> <sup>1</sup>

tion becomes as d<sup>l</sup>

mates:

� � �

Yl <sup>v</sup>ð Þ Nl

exp � <sup>η</sup>rþ<sup>2</sup> � <sup>η</sup> � �<sup>2</sup>

4τ<sup>1</sup> !

� , cexp � <sup>η</sup><sup>2</sup>

Fv�<sup>2</sup>ð Þ� M T1uv�<sup>2</sup>ð Þþ M iθ<sup>0</sup>

i=4, has a free term:

� ∑ 2 l,r¼<sup>1</sup>

188

F4ð Þ¼� M iθ<sup>0</sup>

¼ �iθ<sup>0</sup>

T0w<sup>r</sup>þ2,l

Dtw<sup>r</sup>þ2,l

� ∑ 2 l,r¼<sup>1</sup>

Due to the fact that the function dl

<sup>0</sup>ð Þ x; t � � �

l 8τ<sup>1</sup> � �

exp ð Þþ τ<sup>2</sup> ∑

�

2 l,r¼<sup>1</sup>

ð Þt c2ð Þþ x; t ∑

<sup>2</sup> ð Þþ Mrþ2,l σT0z

h i

� � �

dp

ðτ1 0

Boundary Layer Flows - Theory, Applications and Numerical Methods

� ðt 0

ð Þ x; t – is arbitrary function such as

<sup>v</sup>ð Þ x; t

4πτ<sup>1</sup>

( ) !

<sup>t</sup>¼<sup>0</sup> ¼ �<sup>d</sup>

l

, Y<sup>r</sup>þ2,l <sup>v</sup> ð Þ Nrþ2,l

Free member of equation (Eq. (27)) under v ¼ 2, 3 has a form

ð Þ<sup>t</sup> <sup>∂</sup>σuv�<sup>2</sup>ð Þ¼ <sup>M</sup> <sup>i</sup>θ<sup>0</sup>

∂μwrþ2,l

cv�<sup>2</sup>ð Þ¼ <sup>x</sup>; <sup>t</sup> <sup>0</sup>, T1w<sup>r</sup>þ2,l

<sup>v</sup>�<sup>2</sup> � <sup>Δ</sup>ξwrþ2,l

so that equation (Eq. (27)), under v ¼ 2, 3, has a solution in U; we set

ð Þ<sup>t</sup> <sup>∂</sup><sup>τ</sup><sup>2</sup> � <sup>T</sup>1u<sup>2</sup> <sup>þ</sup> f xð Þ ; <sup>t</sup> exp

rþ2,l 2

<sup>0</sup> ð Þ� <sup>x</sup>; <sup>t</sup> exp ð Þ <sup>τ</sup><sup>2</sup> <sup>∂</sup>tc0ð Þþ <sup>x</sup>; <sup>t</sup> <sup>∑</sup>

4 l¼1 Yl

Solutions of the last equations under the boundary conditions from (Eq. (29)) have a form (Eq. (33)) for which estimates of the form (Eq. (35) are fair. Eq. (27),

<sup>2</sup>ð Þþ Nl ∑

" #

2 r,l¼<sup>1</sup>

� �

� � � x1¼l�1

ð<sup>∞</sup> 0 Yl <sup>v</sup>ð Þ <sup>∗</sup> <sup>∂</sup> ∂ξ

ð<sup>∞</sup> 0

<sup>Y</sup>rþ<sup>2</sup> <sup>v</sup> ð Þ <sup>∗</sup> <sup>∂</sup>

exp � ð Þ <sup>η</sup><sup>l</sup> � <sup>ξ</sup> <sup>2</sup> 4τ<sup>1</sup> !

4τ<sup>1</sup>

� exp � <sup>η</sup>rþ<sup>2</sup> <sup>þ</sup> <sup>η</sup> � �<sup>2</sup>

values under x<sup>1</sup> ¼ l � 1 are determined from the second relation. According to Theorems 2 and 3, the functions found by the formula (Eq. (33)) satisfy the esti-

∂η

¼ �cvð Þ l � 1; x2; t ,

� , cexp � <sup>η</sup><sup>2</sup>

ð Þt cv�<sup>2</sup>ð Þþ x; t ∑

<sup>v</sup>�<sup>2</sup> <sup>þ</sup> <sup>σ</sup> <sup>∂</sup>μ<sup>z</sup>

n o h i

<sup>v</sup>�<sup>2</sup> <sup>¼</sup> <sup>0</sup>, T1z<sup>r</sup>þ2,l

( ) !

:

G Nl ð Þ ; ξ; η; τ<sup>1</sup> � τ � �

G Nrþ2,l ð Þ ; ξ; η; τ<sup>1</sup> � τ � �

� exp � ð Þ <sup>η</sup><sup>l</sup> <sup>þ</sup> <sup>ξ</sup> <sup>2</sup>

<sup>v</sup>ð Þ x; t при t ¼ τ<sup>1</sup> ¼ 0 multiplied by the func-

<sup>0</sup>ð Þ x , an arbitrary function is accepted, and its

<sup>r</sup>þ<sup>2</sup> <sup>þ</sup> <sup>η</sup><sup>2</sup> l 8τ<sup>1</sup> � �

> 4 l¼1 Yl

rþ2,l <sup>v</sup>�<sup>2</sup> � Δξz

<sup>v</sup>�<sup>2</sup> ¼ 0:

iθð Þ 0 ε � �

<sup>2</sup> ð Þ Nrþ2,l

4 l¼1 Dtw<sup>l</sup>

" #

<sup>0</sup> þ ∑ 2 l,r¼<sup>1</sup>

Y<sup>r</sup>þ2,l

4 l¼1 ∂tY<sup>l</sup>

� Dtv0ð Þ� x; t ∑

<sup>v</sup>�<sup>2</sup>ð Þþ Nl ∑

rþ2,l v�2

� Dσu<sup>0</sup> þ Lηu<sup>0</sup>

exp ð Þ τ<sup>2</sup>

<sup>0</sup>ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>ξ</sup><sup>l</sup>

DtY<sup>r</sup>þ2,l 0

2 ffiffiffi μ p � �

" #

2 r,l¼<sup>1</sup>

,

4τ<sup>1</sup>

<sup>ξ</sup>¼0dηd<sup>τ</sup> � �

<sup>j</sup><sup>η</sup>¼0dξdτ,

, r, l ¼ 1, 2: (34)

Y<sup>r</sup>þ2,l <sup>v</sup>�<sup>2</sup> ð Þ Nrþ2,l

(33)

$$\begin{split} F\_{4}(\mathcal{M}) &= -i\theta'(t) \left[ \sum\_{l=1}^{4} Y\_{2}^{l}(\mathcal{N}\_{l}) + \sum\_{r,l=1}^{2} Y\_{2}^{r+2,l}(\mathcal{N}\_{r+2,l}) \right] \exp\left(\tau\_{2}\right), \\ &- \left[ \sum\_{l=1}^{4} q\_{0}^{l}(\varkappa,t) \text{erfc}\left(\frac{\eta\_{l}}{2\sqrt{\tau\_{2}}}\right) + \sum\_{r,l=1}^{2} z\_{0}^{r+2,l}(\mathcal{N}\_{r+2,l}) \right] \exp\left(\tau\_{2}\right). \end{split}$$

In the last bracket, the transition is from the variables <sup>ξ</sup><sup>l</sup> <sup>2</sup> ffiffi <sup>μ</sup> <sup>p</sup> to the variables <sup>η</sup><sup>l</sup> <sup>2</sup> ffiffiffi τ2 <sup>p</sup> . Substituting the value Y<sup>l</sup> <sup>0</sup>ð Þ¼ Nl dl <sup>0</sup>ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>η</sup><sup>l</sup> <sup>2</sup> ffiffiffi τ1 p � � into equation DtY<sup>l</sup> <sup>0</sup>ð Þ¼ Nl 0, with respect to dl <sup>0</sup>ð Þ <sup>x</sup>; <sup>t</sup> , we get the equation Dtdl <sup>0</sup>ð Þ¼ x; t 0, which is solved under an arbitrary initial condition dl <sup>0</sup>ð Þ x; t � � � <sup>t</sup>¼<sup>0</sup> <sup>¼</sup> <sup>d</sup> l <sup>0</sup>ð Þ x . This arbitrary function provides the condition L<sup>η</sup> Y<sup>l</sup> <sup>0</sup> <sup>¼</sup> <sup>0</sup>;therefore, Dx, <sup>η</sup> <sup>Y</sup><sup>l</sup> <sup>0</sup> ¼ 0:The initial condition for this equation is determined from the relation:

$$\left.d\_0^l(\mathbf{x},t)\right|\_{\mathbf{x}\_1=l-1} = -c\_0(l-1,\mathbf{x}\_2,t), \\
\left.d\_0^{l+2}(\mathbf{x},t)\right|\_{\mathbf{x}\_2=l-1} = -c\_0(\mathbf{x}\_1,l-1,t),$$

which comes out from (Eq. (29)) and (Eq. (33)). The function Y<sup>r</sup>þ2,l <sup>0</sup> ð Þ Nrþ2,l expresses through Y<sup>l</sup> <sup>0</sup>ð Þ Nl therefore provided that

$$D\_t Y\_0^{r+2,l} = \mathbf{0}, \\ D\_{\mathbf{x}, \eta}^{\nu, l} Y\_0^{r+2,l} = \mathbf{0}.$$

The same is true for functions w<sup>r</sup>þ2,l <sup>0</sup> ð Þ Mrþ2,l , z<sup>r</sup>þ2,l <sup>0</sup> ð Þ Mrþ2,l ; in other words, the following relations hold: Dtw<sup>r</sup>þ2,l <sup>0</sup> <sup>¼</sup> <sup>0</sup>, Dtzrþ2,l <sup>0</sup> <sup>¼</sup> <sup>0</sup>, Dv,l x, <sup>ξ</sup>w<sup>r</sup>þ2,l <sup>0</sup> <sup>¼</sup> <sup>0</sup>, Dv,l x, <sup>ξ</sup>zrþ2,l <sup>0</sup> ¼ 0:

Solutions of equations with respect w<sup>r</sup>þ2,l <sup>0</sup> , zrþ2,l <sup>0</sup> under appropriate boundary conditions from (Eq. (29) are representable as (Eq. (33)), and they are expressed through w<sup>l</sup> <sup>2</sup>ð Þ <sup>x</sup>; <sup>t</sup> , q<sup>l</sup> <sup>2</sup>ð Þ x; t : The first equation (Eq. (36)) is solvable,

if <sup>z</sup>0ð Þj <sup>x</sup>; <sup>t</sup> <sup>t</sup>¼<sup>0</sup> <sup>¼</sup> f xð Þ ; <sup>0</sup> exp iθð Þ 0 ε � �: This ratio is used by the initial condition for the equation

j j R<sup>ε</sup>nð Þ x; t; ε , c:

Theorem 4. Suppose that the conditions (1)–(3) are satisfied. Then, using the above method for solving u xð Þ ; t; ε of the problem (Eq. (20)), a regularized series (Eq. (26)) such that ∀n ¼ 0, 1, 2, … can be constructed, and for small enough

j j u xð Þ� ; <sup>t</sup>; <sup>ε</sup> <sup>u</sup><sup>ε</sup>nð Þ <sup>x</sup>; <sup>t</sup>; <sup>ε</sup> <sup>¼</sup> j j <sup>R</sup><sup>ε</sup><sup>n</sup> ð Þ <sup>x</sup>; <sup>t</sup>; <sup>ε</sup> , <sup>c</sup>εnþ<sup>1</sup>

2,

Thus, we have proven the following:

Singularly Perturbed Parabolic Problems DOI: http://dx.doi.org/10.5772/intechopen.84339

ε . 0, inequality is fair:

Author details

191

Asan Omuraliev\* and Ella Abylaeva

provided the original work is properly cited.

Kyrgyz-Turkish Manas University, Bishkek, Kyrgyzstan

\*Address all correspondence to: asan.omuraliev@mail.ru

© 2019 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/ by/3.0), which permits unrestricted use, distribution, and reproduction in any medium,

where c is independent of ε.

Dtz0ð Þ¼ x; t 0. The remaining equations from (Eq. (36)) are solvable under the initial conditions from (Eq. (29)).

Thus, the main term of the asymptotics is uniquely determined. As can be seen from the representation (Eq. (28)) and the estimates (Eq. (35)), we note that the asymptotics of the solution have a complex structure. In addition to regular members, it contains various boundary layer functions. Parabolic boundary layer functions have an estimate:

$$\left|Y^{l}(N\_{l})\right| \leq c \exp\left(-\frac{\eta\_{l}^{2}}{8\tau\_{1}}\right), \left|w^{l}(\infty,t)\text{erfc}\left(\frac{\xi\_{l}}{2\sqrt{\mu}}\right)\right| \leq c \exp\left(-\frac{\xi\_{l}^{2}}{8\mu}\right).$$

Multidimensional and angular parabolic boundary layer functions have an estimate:

$$\begin{aligned} \left| Y^{r+2,l}(N\_{r+2,l}) \right| &< c \exp \left( -\frac{\eta\_{r+2}^2 + \eta\_l^2}{8\tau\_1} \right), \\ \left| w^{r+2,l}(M\_{r+2,l}) \right| &< c \exp \left( -\frac{\xi\_{r+2}^2 + \xi\_l^2}{8\mu} \right). \end{aligned}$$

The boundary layer functions with rapidly oscillating exponential and power type of change:

$$c(\varkappa, t) \exp\left(\tau\_2\right), \sigma = \int\_0^t e^{\frac{i[\vartheta(r) - \vartheta(0)]}{\tau}} ds.$$

In addition, the asymptotic contains the product of the abovementioned boundary layer functions.

Repeating the above process, we construct a partial sum:

$$
\widetilde{\boldsymbol{\mu}}\_{\boldsymbol{c}n}(\boldsymbol{M}) = \sum\_{i=0}^{n} \varepsilon^{i} \boldsymbol{\mu}\_{i}(\boldsymbol{M}).\tag{36}
$$

#### 2.6 Assessment of remainder term

Substituting the function <sup>e</sup>u Mð Þ¼ ; <sup>ε</sup> <sup>u</sup><sup>ε</sup>nð Þþ <sup>M</sup> <sup>ε</sup><sup>n</sup>þ<sup>1</sup> <sup>2</sup>R<sup>ε</sup>nð Þ M into problem (Eq. (24)), then taking into account the iterative tasks of (Eq. (27)) and (Eq. (29)), we obtain the following problem for the remainder term R<sup>ε</sup>nð Þ M :

$$\overline{L}\_r R\_m(M) = \mathbf{g}\_n(M, \varepsilon),\\ R\_m(M)|\_{t=0} = R\_m(M)|\_{\mathbf{x}\_l = r-1, \xi\_r = 0, \eta\_k} = \mathbf{0}, r = \mathbf{1}, \mathbf{2}; k = \overline{1, 4}, \tag{37}$$

where gnð Þ¼� M; ε iθ<sup>0</sup> ð Þ<sup>t</sup> <sup>∂</sup><sup>τ</sup>2un�<sup>1</sup> � <sup>ε</sup> 1 <sup>2</sup>iθ<sup>0</sup> ð Þ<sup>t</sup> <sup>∂</sup><sup>τ</sup>2unð Þ� <sup>M</sup> <sup>T</sup>1un�<sup>1</sup>ð Þ� <sup>M</sup> <sup>ε</sup> 1 <sup>2</sup>T1unð Þ� M D<sup>σ</sup> � L<sup>η</sup> � �∑<sup>3</sup> <sup>k</sup>¼<sup>0</sup><sup>ε</sup> k <sup>2</sup>un�3þ<sup>k</sup>ð Þþ <sup>M</sup> <sup>L</sup>η∑<sup>5</sup> <sup>k</sup>¼<sup>0</sup><sup>ε</sup> k <sup>2</sup>un�5þ<sup>k</sup>ð Þþ <sup>M</sup> <sup>Δ</sup>a∑<sup>7</sup> <sup>k</sup>¼<sup>0</sup><sup>ε</sup> k <sup>2</sup>un�7þ<sup>k</sup>ð Þ M :

We put in both parts (Eq. (38)) χ ¼ ψð Þ x; t; ε considering (Eq. (25)), with respect to

<sup>L</sup>εR<sup>ε</sup>nð Þ¼ <sup>x</sup>; <sup>t</sup>; <sup>ε</sup> <sup>g</sup><sup>ε</sup>nð Þ <sup>x</sup>; <sup>t</sup>; <sup>ε</sup> , R<sup>ε</sup>n tj j <sup>¼</sup><sup>0</sup> <sup>¼</sup> <sup>0</sup>; <sup>R</sup><sup>ε</sup><sup>n</sup> <sup>∂</sup>Ω¼<sup>0</sup>:

By virtue of the above constructions, the function is <sup>g</sup><sup>ε</sup>nð Þ <sup>x</sup>; <sup>t</sup>; <sup>ε</sup> � � � � , c, ∀ð Þ x; t ∈ ; therefore, applying the maximum principle, an estimate is established:

Singularly Perturbed Parabolic Problems DOI: http://dx.doi.org/10.5772/intechopen.84339

if <sup>z</sup>0ð Þj <sup>x</sup>; <sup>t</sup> <sup>t</sup>¼<sup>0</sup> <sup>¼</sup> f xð Þ ; <sup>0</sup> exp

tions have an estimate:

Yl ð Þ Nl

� � �

initial conditions from (Eq. (29)).

equation

estimate:

type of change:

ary layer functions.

2.6 Assessment of remainder term

where gnð Þ¼� M; ε iθ<sup>0</sup>

<sup>k</sup>¼<sup>0</sup><sup>ε</sup> k

D<sup>σ</sup> � L<sup>η</sup> � �∑<sup>3</sup>

respect to

190

iθð Þ 0 ε � �

Boundary Layer Flows - Theory, Applications and Numerical Methods

� , cexp � <sup>η</sup><sup>2</sup>

Y<sup>r</sup>þ2,l

w<sup>r</sup>þ2,l

�

�

l 8τ<sup>1</sup> � �

ð Þ Nrþ2,l

ð Þ Mrþ2,l

c xð Þ ; t exp ð Þ τ<sup>2</sup> , σ ¼

<sup>e</sup>uεnð Þ¼ <sup>M</sup> <sup>∑</sup>

Repeating the above process, we construct a partial sum:

Substituting the function <sup>e</sup>u Mð Þ¼ ; <sup>ε</sup> <sup>u</sup><sup>ε</sup>nð Þþ <sup>M</sup> <sup>ε</sup><sup>n</sup>þ<sup>1</sup>

we obtain the following problem for the remainder term R<sup>ε</sup>nð Þ M :

ð Þ<sup>t</sup> <sup>∂</sup><sup>τ</sup>2un�<sup>1</sup> � <sup>ε</sup>

<sup>2</sup>un�3þ<sup>k</sup>ð Þþ <sup>M</sup> <sup>L</sup>η∑<sup>5</sup>

<sup>L</sup>εR<sup>ε</sup>nð Þ¼ <sup>x</sup>; <sup>t</sup>; <sup>ε</sup> <sup>g</sup><sup>ε</sup>nð Þ <sup>x</sup>; <sup>t</sup>; <sup>ε</sup> , R<sup>ε</sup>n tj j <sup>¼</sup><sup>0</sup> <sup>¼</sup> <sup>0</sup>; <sup>R</sup><sup>ε</sup><sup>n</sup> <sup>∂</sup>Ω¼<sup>0</sup>:

� �

� �

, w<sup>l</sup>

Multidimensional and angular parabolic boundary layer functions have an

� �

Dtz0ð Þ¼ x; t 0. The remaining equations from (Eq. (36)) are solvable under the

Thus, the main term of the asymptotics is uniquely determined. As can be seen from the representation (Eq. (28)) and the estimates (Eq. (35)), we note that the asymptotics of the solution have a complex structure. In addition to regular members, it contains various boundary layer functions. Parabolic boundary layer func-

ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>ξ</sup><sup>l</sup>

� , cexp � <sup>η</sup><sup>2</sup>

� , cexp � <sup>ξ</sup><sup>2</sup>

ðt 0 e i½ � θð Þ�s θð Þ 0 <sup>ε</sup> ds:

In addition, the asymptotic contains the product of the abovementioned bound-

n i¼0 ε i

(Eq. (24)), then taking into account the iterative tasks of (Eq. (27)) and (Eq. (29)),

<sup>L</sup>eεR<sup>ε</sup>nð Þ¼ <sup>M</sup> gnð Þ <sup>M</sup>; <sup>ε</sup> , R<sup>ε</sup>nð Þ <sup>M</sup> j j <sup>t</sup>¼<sup>0</sup> <sup>¼</sup> <sup>R</sup><sup>ε</sup>nð Þ <sup>M</sup> xl¼r�1, <sup>ξ</sup>r¼0, <sup>η</sup><sup>k</sup> <sup>¼</sup> <sup>0</sup>, r <sup>¼</sup> <sup>1</sup>, <sup>2</sup>; k <sup>¼</sup> <sup>1</sup>, <sup>4</sup>,

<sup>2</sup>un�5þ<sup>k</sup>ð Þþ <sup>M</sup> <sup>Δ</sup>a∑<sup>7</sup>

1 <sup>2</sup>iθ<sup>0</sup>

We put in both parts (Eq. (38)) χ ¼ ψð Þ x; t; ε considering (Eq. (25)), with

therefore, applying the maximum principle, an estimate is established:

<sup>k</sup>¼<sup>0</sup><sup>ε</sup> k

By virtue of the above constructions, the function is <sup>g</sup><sup>ε</sup>nð Þ <sup>x</sup>; <sup>t</sup>; <sup>ε</sup> �

The boundary layer functions with rapidly oscillating exponential and power

� � � �

2 ffiffiffi μ p

<sup>r</sup>þ<sup>2</sup> <sup>þ</sup> <sup>η</sup><sup>2</sup> l 8τ<sup>1</sup> � �

<sup>r</sup>þ<sup>2</sup> <sup>þ</sup> <sup>ξ</sup><sup>2</sup> l 8μ � �

,

:

<sup>2</sup>uið Þ M : (36)

<sup>2</sup>R<sup>ε</sup>nð Þ M into problem

ð Þ<sup>t</sup> <sup>∂</sup><sup>τ</sup>2unð Þ� <sup>M</sup> <sup>T</sup>1un�<sup>1</sup>ð Þ� <sup>M</sup> <sup>ε</sup>

<sup>k</sup>¼<sup>0</sup><sup>ε</sup> k

� �

(37)

<sup>2</sup>T1unð Þ� M

� , c, ∀ð Þ x; t ∈ ;

1

<sup>2</sup>un�7þ<sup>k</sup>ð Þ M :

� � � � , cexp � <sup>ξ</sup><sup>2</sup>

l 8μ � �

:

: This ratio is used by the initial condition for the

$$|R\_{\epsilon n}(\mathfrak{x}, t, \mathfrak{e})| \le c.$$

Thus, we have proven the following:

Theorem 4. Suppose that the conditions (1)–(3) are satisfied. Then, using the above method for solving u xð Þ ; t; ε of the problem (Eq. (20)), a regularized series (Eq. (26)) such that ∀n ¼ 0, 1, 2, … can be constructed, and for small enough ε . 0, inequality is fair:

$$|u(\varkappa, t, \varepsilon) - u\_{\varepsilon n}(\varkappa, t, \varepsilon)| = |R\_{\varepsilon n}(\varkappa, t, \varepsilon)| \le c\varepsilon^{n + \frac{1}{2}},$$

where c is independent of ε.
