1.2 Statement of the problem

In this chapter we study the following problem:

$$L\_{\varepsilon}u\left(\left.\mathbf{x},t,\varepsilon\right) \equiv \partial\_{t}u - \varepsilon^{2}a(\mathbf{x})\partial\_{x}^{2}u - b(\mathbf{x},t)u = \sum\_{k=1}^{N} f\_{\mathbf{x}}(\mathbf{x},t) \exp\left(\frac{i\theta\_{\mathbf{x}}(t)}{\varepsilon}\right), (\mathbf{x},t) \in \Omega,\tag{1}$$

$$\mathbf{u}(\mathbf{x},\mathbf{t},\varepsilon)|\_{\mathbf{t}=0} = \mathbf{u}(\mathbf{x},\mathbf{t},\varepsilon)|\_{\mathbf{x}=0} = \mathbf{u}(\mathbf{x},\mathbf{t},\varepsilon)|\_{\mathbf{x}=1} = \mathbf{0}$$

∂2

L �

xu xð Þ� ;t; <sup>ε</sup> <sup>∂</sup><sup>2</sup>

Dξ,<sup>v</sup> � 2φ<sup>0</sup>

Dζ,<sup>v</sup> � 2φ<sup>0</sup>

¼ ∑ N k¼1

T1 � <sup>∂</sup><sup>η</sup> � <sup>∑</sup>

T2 � <sup>∂</sup><sup>t</sup> � <sup>∑</sup>

L<sup>ξ</sup> � a xð Þ ∑

L<sup>ζ</sup> � a xð Þ ∑

Lx � a xð Þ∂<sup>2</sup>

<sup>ε</sup>u Mð Þ� ; <sup>ε</sup> <sup>1</sup>

þ 1

<sup>x</sup> <sup>u</sup> <sup>ð</sup>M; <sup>ε</sup> � � <sup>þ</sup> <sup>∑</sup>

<sup>ε</sup><sup>2</sup> Dζ,vu Mð Þ ; ε

<sup>ν</sup>ð Þ <sup>x</sup> <sup>∂</sup><sup>2</sup>

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

<sup>ν</sup>ð Þ <sup>x</sup> <sup>∂</sup><sup>2</sup>

2 ν¼1 ∂2 ζv ,

2 ν¼1 ∂2

2 v¼1 Dξ,v,

2 v¼1 Dζ,v,

x:

1.4 Solution of iterative problems

T1u0ð Þ¼ M 0, T1u1ð Þ¼� M i ∑

N k¼1 θ0

N k¼1 θ0

T1u2ð Þ ¼¼ � M i ∑

T1uvð Þ¼� M i ∑

177

(

2 ν¼1

)!� � � � � γ¼p xð Þ ;t;ε ,

<sup>x</sup>, ξν þ φ<sup>00</sup>

<sup>x</sup>, ζν þ φ<sup>00</sup>

fкð Þ x;t exp rk þ

<sup>ε</sup><sup>2</sup> T0 u Mð Þþ ; <sup>ε</sup> <sup>∑</sup>

φ0 <sup>ν</sup>ð Þ x ε � �<sup>2</sup>

<sup>ν</sup> ð Þ <sup>x</sup> <sup>∂</sup>ξν ,

<sup>ν</sup> ð Þ <sup>x</sup> <sup>∂</sup>ζν :

iθ0 <sup>k</sup>ð Þt

iθкð Þ 0 ε � �

u Mð Þj ; <sup>ε</sup> <sup>t</sup>¼rk¼η¼<sup>0</sup> <sup>¼</sup> u Mð Þ ; <sup>ε</sup> <sup>x</sup>¼0, <sup>ξ</sup>1¼ζ1¼<sup>0</sup> <sup>¼</sup> u Mð Þ ; <sup>ε</sup> �

N k¼1

exp rð Þ<sup>k</sup> <sup>∂</sup><sup>σ</sup><sup>k</sup> ,

N k¼1

<sup>ξ</sup><sup>v</sup> � b xð Þþ ;t ∑

The problem (Eq. (5)) is regular in ε as ε ! 0:

L �

> N k¼1 θ0

<sup>ε</sup>u Mð Þ ; ε � ��

� �

u M ð Þ¼ ; ε ∑

<sup>k</sup>ð Þ<sup>t</sup> <sup>∂</sup>rku0ð Þ <sup>M</sup> ,

<sup>k</sup>ð Þ<sup>t</sup> <sup>∂</sup>rku1ð Þ� <sup>M</sup> T2u0ð Þþ <sup>M</sup> <sup>∑</sup>

The solution of problem (Eq. (5)) will be determined in the form of a series:

∞ v¼0

For the coefficients of this series, we obtain the following iterative problems:

N k¼1

The solution of this problem contains parabolic boundary layer functions; internal power boundary layer functions which are connected with a rapidly oscillating

<sup>k</sup>ð Þ<sup>t</sup> <sup>∂</sup>rkuv�<sup>1</sup>ð Þ� <sup>M</sup> T2uv�<sup>2</sup>ð Þþ <sup>M</sup> <sup>L</sup>ζuv�<sup>2</sup> <sup>þ</sup> <sup>L</sup>ξuv�<sup>3</sup>ð Þþ <sup>M</sup> Lxuv�<sup>4</sup>ð Þ <sup>M</sup> :

fкð Þ x;t exp rk þ

∂2 ξν u M ð Þþ ; <sup>ε</sup> <sup>φ</sup><sup>0</sup>

<sup>ε</sup> <sup>∂</sup>rku Mð Þþ ; <sup>ε</sup> T1 u Mð Þ ; <sup>ε</sup>

� �

<sup>ν</sup>ð Þ x ε2 � �<sup>2</sup>

<sup>þ</sup> <sup>L</sup><sup>ζ</sup> u Mð Þþ ; <sup>ε</sup> <sup>ε</sup>Lξu Mð Þþ ; <sup>ε</sup> <sup>ε</sup><sup>2</sup>

�

<sup>q</sup>¼q xð Þ ;t;<sup>ε</sup> � <sup>L</sup>εu x ð Þ ;t; <sup>ε</sup> : (6)

<sup>ε</sup>vuvð Þ <sup>M</sup> : (7)

iθкð Þ 0 ε � �

þ L<sup>ζ</sup> u0ð Þ M ,

(8)

<sup>x</sup>¼1, <sup>ξ</sup>2¼ζ2¼<sup>0</sup> <sup>¼</sup> <sup>0</sup>,

∂2 ζν u M ð Þþ ; <sup>ε</sup> <sup>1</sup>

<sup>ε</sup> <sup>D</sup>ξ,vu M ð Þ ; <sup>ε</sup>

Lxu Mð Þ ; ε

(5)

where ε . 0 is a small parameter and Ω = {(x, t): x∈ ð Þ 0; 1 ,t∈ð �g 0; T . The problem is solved under the following assumptions:

$$\mathbf{1}. \,\mathbf{a}(\mathbf{x}) \ge \mathbf{0}, \, a(\mathbf{x}) \in \mathbf{C}^{\infty}[0, 1], \, \mathbf{b}(\mathbf{x}, \mathbf{t}), \, \mathbf{f} \, (\mathbf{x}, \mathbf{t}) \in \mathbf{C}^{\infty}(\overline{\Omega}).$$

2. ∀x∈½ � 0; 1 function a xð Þ . 0.

3. θ<sup>0</sup> <sup>k</sup>ð Þt � � <sup>t</sup>¼<sup>0</sup> <sup>¼</sup> 0 is the phase function.

#### 1.3 Regularization of the problem

For the regularization of problem (Eq. (1)), we introduce regularizing independent variables using methods [5, 6]:

$$\begin{aligned} \boldsymbol{\eta} &= \frac{\mathbf{t}}{\mathbf{e}^2}, \mathbf{r}\_\mathbf{k} = \frac{\mathbf{i}\left[\boldsymbol{\theta}\_\mathbf{k}(\mathbf{t}) - \boldsymbol{\theta}\_\mathbf{k}(\mathbf{0})\right]}{\mathbf{e}}, \boldsymbol{\xi}\_\mathbf{v} = \frac{\boldsymbol{\upalpha}\_\boldsymbol{\nu}(\mathbf{x})}{\mathbf{e}}, \mathbf{i} = \sqrt{-1}, \\ \boldsymbol{\zeta}\_\mathbf{v} &= \frac{\boldsymbol{\upalpha}\_\boldsymbol{\nu}(\mathbf{x})}{\mathbf{e}^2}, \boldsymbol{\upalpha}\_\boldsymbol{\nu}(\mathbf{x}) = (-1)^{\nu-1} \int\_{\nu-1}^\mathbf{x} \frac{\mathbf{d}\mathbf{s}}{\sqrt{\mathbf{a}(\mathbf{s})}}, \quad \boldsymbol{\nu} = \mathbf{1}, \mathbf{2}, \\ \boldsymbol{\sigma}\_\mathbf{k} &= \int\_0^\mathbf{t} \exp\left(\frac{\mathbf{i}\left[\boldsymbol{\theta}\_\mathbf{k}(\mathbf{s}) - \boldsymbol{\uptheta}\_\mathbf{k}(\mathbf{0})\right]}{\mathbf{e}}\right) \mathbf{d}\mathbf{s} \equiv \mathbf{p}\_\mathbf{k}(\mathbf{t}, \mathbf{e}), \mathbf{l} = \overline{\mathbf{0}, \mathbf{r}}, \mathbf{j} = \overline{\mathbf{0}, \mathbf{k} - \mathbf{1}} \end{aligned} \tag{2}$$

Instead of the desired function u xð Þ ;t; ε , we will study the extended function

$$\mathbf{u}(\mathbf{M}, \mathbf{e}), \mathbf{M} = (\mathbf{x}, \mathbf{t}, \mathbf{r}, \eta, \sigma, \mathfrak{k}, \mathfrak{k}), \sigma = (\sigma\_1, \sigma\_2...\sigma\_N), \mathbf{r} = (\mathbf{r}\_1, \mathbf{r}\_2...\mathbf{r}\_N), \mathfrak{k} = (\mathfrak{k}\_1, \mathfrak{k}\_2), \mathfrak{k} = (\mathfrak{k}\_1, \mathfrak{k}\_2)$$

such that its restriction by regularizing variables coincides with the desired solution:

$$\begin{aligned} \left. \mathbf{\dot{u}}(\mathbf{M}, \mathbf{e}) \right|\_{\mathbf{y} = \mathbf{p}(\mathbf{x}, \mathbf{t}, \mathbf{e})} & \equiv \mathbf{u}(\mathbf{x}, \mathbf{t}, \mathbf{e}) \\\\ \mathbf{\dot{y}} &= (\mathbf{r}, \sigma, \eta, \xi, \xi) \end{aligned} \tag{3}$$

Taking into account (Eqs. (21)) and ((3)), we find the derivatives

On the basis of (Eqs. (1)–(4)) for the extended function u Mð Þ ; ε , we set the problem:

$$\partial\_{\mathbf{t}}\mathbf{u}(\mathbf{x},\mathbf{t},\varepsilon) \equiv \left(\partial\_{\mathbf{t}}\mathbf{\hat{u}}(\mathbf{M},\varepsilon) + \frac{1}{\varepsilon^{2}}\partial\_{\eta}\mathbf{\hat{u}}(\mathbf{M},\varepsilon) + \sum\_{\mathbf{k}=1}^{N} \left| \frac{\mathbf{i}\partial\_{\mathbf{k}}\mathbf{'}(\mathbf{t})}{\varepsilon} \partial\_{\mathbf{k}}\mathbf{\hat{u}}(\mathbf{M},\varepsilon) + \exp\left(\mathbf{r}\_{\mathbf{k}}\right) \partial\_{\mathbf{t}\_{\mathbf{k}}}\mathbf{\hat{u}}(\mathbf{M},\varepsilon)\right|\_{\mathbf{y}=\mathbf{p}(\mathbf{x},\mathbf{t},\varepsilon)},\tag{4.10}$$
 
$$\partial\_{\mathbf{x}}\mathbf{u}(\mathbf{x},\mathbf{t},\varepsilon) \equiv \left(\left(\partial\_{\mathbf{x}}\mathbf{\hat{u}}\left(\mathbf{M},\varepsilon\right) + \sum\_{\nu=1}^{2} \left\{\frac{\boldsymbol{\uprho}\_{\nu}^{\prime}(\mathbf{x})}{\varepsilon} \partial\_{\mathbf{t}\_{\nu}}\mathbf{\hat{u}}\left(\mathbf{M},\varepsilon\right) + \frac{\boldsymbol{\uprho}\_{\nu}^{\prime}(\mathbf{x})}{\varepsilon^{2}} \partial\_{\mathbf{t}\_{\nu}}\mathbf{\hat{u}}\left(\mathbf{M},\varepsilon\right)\right\}\right)\right|\_{\mathbf{y}=\mathbf{p}(\mathbf{x},\mathbf{t},\varepsilon)},\tag{4.10}$$

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

1.2 Statement of the problem

<sup>L</sup>εu xð Þ� ; <sup>t</sup>; <sup>ε</sup> <sup>∂</sup>tu � <sup>ε</sup><sup>2</sup>

2. ∀x∈½ � 0; 1 function a xð Þ . 0.

1.3 Regularization of the problem

dent variables using methods [5, 6]:

<sup>ζ</sup><sup>v</sup> <sup>¼</sup> φνð Þ <sup>x</sup>

<sup>η</sup> <sup>¼</sup> <sup>t</sup>

σ<sup>k</sup> ¼ ðt 0 exp

solution:

problem:

176

<sup>∂</sup>tu xð Þ�ð ;t; <sup>ε</sup> <sup>∂</sup>tu M ð Þþ ; <sup>ε</sup> <sup>1</sup>

<sup>∂</sup>xu xð Þ� ;t; <sup>ε</sup> ð Þþ <sup>∂</sup>xu <sup>ð</sup>M; <sup>ε</sup> <sup>∑</sup>

3. θ<sup>0</sup> <sup>k</sup>ð Þt � �

In this chapter we study the following problem:

Boundary Layer Flows - Theory, Applications and Numerical Methods

a xð Þ∂<sup>2</sup>

The problem is solved under the following assumptions:

1. a xð Þ . <sup>0</sup>, að Þ <sup>x</sup> <sup>∈</sup>С<sup>∞</sup>½ � <sup>0</sup>; <sup>1</sup> , b xð Þ ;<sup>t</sup> ,f xð Þ ;<sup>t</sup> <sup>∈</sup>С<sup>∞</sup> <sup>Ω</sup>

<sup>t</sup>¼<sup>0</sup> <sup>¼</sup> 0 is the phase function.

<sup>ε</sup><sup>2</sup> ,rk <sup>¼</sup> <sup>i</sup> ½ � <sup>θ</sup>kð Þ� <sup>t</sup> <sup>θ</sup>kð Þ <sup>0</sup>

i ½ � θкð Þ� s θкð Þ 0 ε � �

<sup>ε</sup><sup>2</sup> , φνð Þ¼ � <sup>x</sup> ð Þ<sup>1</sup> <sup>ν</sup>�<sup>1</sup>

xu � b xð Þ ; t u ¼ ∑

where ε . 0 is a small parameter and Ω = {(x, t): x∈ ð Þ 0; 1 ,t∈ð �g 0; T .

u xð Þj ;t; <sup>ε</sup> <sup>t</sup>¼<sup>0</sup> <sup>¼</sup> u xð Þ ;t; <sup>ε</sup> j j <sup>x</sup>¼<sup>0</sup> <sup>¼</sup> u xð Þ ;t; <sup>ε</sup> <sup>x</sup>¼<sup>1</sup> <sup>¼</sup> <sup>0</sup>

For the regularization of problem (Eq. (1)), we introduce regularizing indepen-

<sup>ε</sup> , ξν <sup>¼</sup> φνð Þ <sup>x</sup>

ðx

Instead of the desired function u xð Þ ;t; ε , we will study the extended function

u Mð Þ ; ε , M ¼ ð Þ x;t;r; η; σ; ξ; ζ , σ ¼ ð Þ σ1; σ2…σ<sup>N</sup> ,r ¼ ð Þ r1;r2…rN , ξ ¼ ξ1; ξ<sup>2</sup> ð Þ, ζ ¼ ζ1; ζ<sup>2</sup> ð Þ such that its restriction by regularizing variables coincides with the desired

u Mð Þj ; <sup>ε</sup> <sup>γ</sup>¼p xð Þ ;t;<sup>ε</sup> � u xð Þ ;t; <sup>ε</sup>

On the basis of (Eqs. (1)–(4)) for the extended function u Mð Þ ; ε , we set the

<sup>ε</sup> <sup>∂</sup>ξνu M ð Þþ ; <sup>ε</sup> <sup>φ</sup><sup>0</sup>

� � � � �

<sup>ν</sup>ð Þ x

<sup>ε</sup><sup>2</sup> <sup>∂</sup>ζνu M ð Þ ; <sup>ε</sup>

γ ¼ ð Þ r; σ; η; ξ; ζ

Taking into account (Eqs. (21)) and ((3)), we find the derivatives

N k¼1 ½ iθ<sup>k</sup> 0 ð Þt

<sup>ε</sup><sup>2</sup> <sup>∂</sup>ηu Mð Þþ ; <sup>ε</sup> <sup>∑</sup>

φ0 <sup>ν</sup>ð Þ x

2 ν¼1 ν�1

ds ffiffiffiffiffiffiffiffi a sð Þ <sup>p</sup> , <sup>ν</sup> <sup>¼</sup> <sup>1</sup>, <sup>2</sup>,

N k¼1

f <sup>к</sup>ð Þ x; t exp

� �:

<sup>ε</sup> , <sup>i</sup> <sup>¼</sup> ffiffiffiffiffiffi �<sup>1</sup> <sup>p</sup> ,

ds � pкð Þ t; ε , l ¼ 0,r, j ¼ 0, kl � 1

<sup>ε</sup> <sup>∂</sup>rku Mð Þþ ; <sup>ε</sup> exp rð Þ<sup>k</sup> <sup>∂</sup><sup>σ</sup>ku M ð ÞÞ ; <sup>ε</sup>

� � � γ¼p xð Þ ;t;ε ,

iθкð Þt ε � �

, xð Þ ; t ∈ Ω,

(1)

(2)

(3)

� � � � γ¼p xð Þ ;t;ε ,

(4)

∂2 xu xð Þ� ;t; <sup>ε</sup> <sup>∂</sup><sup>2</sup> <sup>x</sup> <sup>u</sup> <sup>ð</sup>M; <sup>ε</sup> � � <sup>þ</sup> <sup>∑</sup> 2 ν¼1 φ0 <sup>ν</sup>ð Þ x ε � �<sup>2</sup> ∂2 ξν u M ð Þþ ; <sup>ε</sup> <sup>φ</sup><sup>0</sup> <sup>ν</sup>ð Þ x ε2 � �<sup>2</sup> ∂2 ζν u M ð Þþ ; <sup>ε</sup> <sup>1</sup> <sup>ε</sup> <sup>D</sup>ξ,vu M ð Þ ; <sup>ε</sup> ( þ 1 <sup>ε</sup><sup>2</sup> Dζ,vu Mð Þ ; ε )!� � � � � γ¼p xð Þ ;t;ε , Dξ,<sup>v</sup> � 2φ<sup>0</sup> <sup>ν</sup>ð Þ <sup>x</sup> <sup>∂</sup><sup>2</sup> <sup>x</sup>, ξν þ φ<sup>00</sup> <sup>ν</sup> ð Þ <sup>x</sup> <sup>∂</sup>ξν , Dζ,<sup>v</sup> � 2φ<sup>0</sup> <sup>ν</sup>ð Þ <sup>x</sup> <sup>∂</sup><sup>2</sup> <sup>x</sup>, ζν þ φ<sup>00</sup> <sup>ν</sup> ð Þ <sup>x</sup> <sup>∂</sup>ζν : L � <sup>ε</sup>u Mð Þ� ; <sup>ε</sup> <sup>1</sup> <sup>ε</sup><sup>2</sup> T0 u Mð Þþ ; <sup>ε</sup> <sup>∑</sup> N k¼1 iθ0 <sup>k</sup>ð Þt <sup>ε</sup> <sup>∂</sup>rku Mð Þþ ; <sup>ε</sup> T1 u Mð Þ ; <sup>ε</sup> ¼ ∑ N k¼1 fкð Þ x;t exp rk þ iθкð Þ 0 ε � � <sup>þ</sup> <sup>L</sup><sup>ζ</sup> u Mð Þþ ; <sup>ε</sup> <sup>ε</sup>Lξu Mð Þþ ; <sup>ε</sup> <sup>ε</sup><sup>2</sup> Lxu Mð Þ ; ε u Mð Þj ; <sup>ε</sup> <sup>t</sup>¼rk¼η¼<sup>0</sup> <sup>¼</sup> u Mð Þ ; <sup>ε</sup> <sup>x</sup>¼0, <sup>ξ</sup>1¼ζ1¼<sup>0</sup> <sup>¼</sup> u Mð Þ ; <sup>ε</sup> � � � � <sup>x</sup>¼1, <sup>ξ</sup>2¼ζ2¼<sup>0</sup> <sup>¼</sup> <sup>0</sup>, T1 � <sup>∂</sup><sup>η</sup> � <sup>∑</sup> 2 ν¼1 ∂2 ζv , T2 � <sup>∂</sup><sup>t</sup> � <sup>∑</sup> 2 ν¼1 ∂2 <sup>ξ</sup><sup>v</sup> � b xð Þþ ;t ∑ N k¼1 exp rð Þ<sup>k</sup> <sup>∂</sup><sup>σ</sup><sup>k</sup> , L<sup>ξ</sup> � a xð Þ ∑ 2 v¼1 Dξ,v, L<sup>ζ</sup> � a xð Þ ∑ 2 v¼1 Dζ,v, Lx � a xð Þ∂<sup>2</sup> x: (5)

The problem (Eq. (5)) is regular in ε as ε ! 0:

$$\left. \left( \stackrel{\sim}{\mathbf{L}\_{\mathbf{e}}} \check{\mathbf{u}}(\mathbf{M}, \mathbf{e}) \right) \right|\_{\mathbf{q} = \mathbf{q}(\mathbf{x}, \mathbf{t}, \mathbf{e})} \equiv \mathbf{L}\_{\mathbf{e}} \check{\mathbf{u}}(\mathbf{x}, \mathbf{t}, \mathbf{e}). \tag{6}$$

#### 1.4 Solution of iterative problems

The solution of problem (Eq. (5)) will be determined in the form of a series:

$$\check{\mathbf{u}}(\mathbf{M}, \boldsymbol{\varepsilon}) = \sum\_{\mathbf{v}=\mathbf{0}}^{\infty} \boldsymbol{\varepsilon}^{\mathbf{v}} \mathbf{u}\_{\mathbf{v}}(\mathbf{M}).\tag{7}$$

For the coefficients of this series, we obtain the following iterative problems:

$$\begin{split} \mathbf{T}\_{1}\mathbf{u}\_{0}(\mathbf{M}) &= \mathbf{0}, \mathbf{T}\_{1}\mathbf{u}\_{1}(\mathbf{M}) = -\mathrm{i}\sum\_{\mathbf{k}=1}^{\mathsf{N}}\theta\_{\mathbf{k}}'(\mathbf{t})\partial\_{\mathbf{t}}\mathbf{u}\_{0}(\mathbf{M}), \\ \mathbf{T}\_{1}\mathbf{u}\_{2}(\mathbf{M}) &= -\mathrm{i}\sum\_{\mathbf{k}=1}^{\mathsf{N}}\theta\_{\mathbf{k}}'(\mathbf{t})\partial\_{\mathbf{t}}\mathbf{u}\_{1}(\mathbf{M}) - \mathbf{T}\_{2}\mathbf{u}\_{0}(\mathbf{M}) + \sum\_{\mathbf{k}=1}^{\mathsf{N}}\mathbf{f}\_{\mathbf{x}}(\mathbf{x},\mathbf{t})\exp\left(\mathbf{r}\_{\mathbf{k}} + \frac{\mathrm{i}\theta\_{\mathbf{x}}(\mathbf{0})}{\mathbf{c}}\right) + \mathbf{L}\_{\mathsf{f}}\,\mathbf{u}\_{0}(\mathbf{M}), \\ \mathbf{T}\_{1}\mathbf{u}\_{\mathbf{v}}(\mathbf{M}) &= -\mathrm{i}\sum\_{\mathbf{k}=1}^{\mathsf{N}}\theta\_{\mathbf{k}}'(\mathbf{t})\partial\_{\mathbf{t}}\mathbf{u}\_{\mathbf{v}-1}(\mathbf{M}) - \mathbf{T}\_{2}\mathbf{u}\_{\mathbf{v}-2}(\mathbf{M}) + \mathbf{L}\_{\mathsf{f}}\mathbf{u}\_{\mathbf{v}-2} + \mathbf{L}\_{\mathsf{f}}\mathbf{u}\_{\mathbf{v}-3}(\mathbf{M}) + \mathbf{L}\_{\mathsf{x}}\mathbf{u}\_{\mathbf{v}-4}(\mathbf{M}). \end{split} \tag{8}$$

The solution of this problem contains parabolic boundary layer functions; internal power boundary layer functions which are connected with a rapidly oscillating

free term in a phase which are vanished at t ¼ tl, l ¼ 0, 1, ……, n in addition; and the asymptotic also contain angular boundary layer functions. We introduce a class of functions in which the iterative problems will be solved:

Theorem 2. Suppose that the conditions of Theorem 1 are satisfied. Then, under

<sup>x</sup>¼l�1, <sup>ξ</sup>l¼0, <sup>ς</sup>l¼<sup>0</sup> <sup>¼</sup> <sup>0</sup>, <sup>l</sup> <sup>¼</sup> <sup>1</sup>, <sup>2</sup>:

�

Lςu Mð Þ¼ 0, Lξu Mð Þ¼ 0:

Proof. By Theorem 1 equation (Eq. (10)) has a solution that is representable in

ckð Þ <sup>x</sup>; <sup>0</sup> ; wl

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

�

<sup>2</sup> ffi t p

� � exp rð Þþ <sup>k</sup> <sup>½</sup>∂tvv�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> vv�<sup>1</sup>ðx;tÞ�

2 ffiffi t p � �

∂tY<sup>l</sup>

� � ��

<sup>3</sup> ð Þþ x;t ∑ 2 l¼1 h<sup>l</sup>,<sup>k</sup>

Condition (3) of the theorem will be ensured, if we choose arbitrarily (Eq. (9))

� � � � exp ð Þ <sup>τ</sup><sup>k</sup>

<sup>k</sup>,v�<sup>1</sup>ð Þ� Nl b xð Þ ;<sup>t</sup> <sup>Y</sup><sup>l</sup>

<sup>k</sup>,v�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> <sup>q</sup><sup>l</sup>

h ierfc <sup>ξ</sup><sup>l</sup>

<sup>k</sup>,v�<sup>1</sup>ð Þ Nl

<sup>k</sup>,v�<sup>1</sup>ð Þ <sup>x</sup>;<sup>t</sup>

2 ffiffi t p

<sup>1</sup>ð Þ <sup>x</sup>;<sup>t</sup> erfc <sup>ξ</sup><sup>l</sup>

2 l¼1 hl

<sup>3</sup> ð Þ <sup>x</sup>;<sup>t</sup> erfc <sup>ξ</sup><sup>l</sup>

� � � � <sup>σ</sup>k:

2 ffiffi t p

2 ffiffi t p � �

(13)

σk

2 l¼1

2 l¼1 ∂tql

2 ffiffi t p � � � � exp ð Þþ <sup>τ</sup><sup>k</sup> h0ð Þþ <sup>x</sup>;<sup>t</sup> <sup>∑</sup>

> N k¼1 hk

<sup>k</sup>ð Þ <sup>x</sup> , <sup>d</sup><sup>l</sup>

ð Þ x;t

� � � � t¼0

<sup>¼</sup> <sup>w</sup><sup>l</sup>

<sup>x</sup>¼l�<sup>1</sup> ¼ �zkð Þ <sup>l</sup> � <sup>1</sup>;<sup>t</sup> , <sup>l</sup> <sup>¼</sup> <sup>1</sup>, <sup>2</sup>:

� � is zero at <sup>θ</sup> <sup>¼</sup> 0, the values for

ð Þ x , (12)

<sup>k</sup>ð Þ<sup>t</sup> <sup>∂</sup>rkuvð Þþ <sup>M</sup> T2uv�1ð Þþ <sup>M</sup> h Mð Þ<sup>∈</sup> G3:

u Mð Þ <sup>t</sup>¼η¼<sup>0</sup> <sup>¼</sup> <sup>0</sup>; u Mð Þ � � �

the form (Eq. (9)). With satisfying condition (1), we obtain

N k¼1

v xð Þ ;t <sup>t</sup>¼<sup>0</sup> ¼ � ∑

ð Þ <sup>x</sup>;<sup>t</sup> <sup>x</sup>¼l�<sup>1</sup> ¼ �ckð Þ <sup>l</sup> � <sup>1</sup>;<sup>t</sup> ; <sup>q</sup><sup>l</sup>

Due to the fact that the function erfc <sup>θ</sup>

2 l¼1 Yl <sup>k</sup>,vð Þ Nl

<sup>v</sup>�<sup>1</sup>ð Þ <sup>x</sup>;<sup>t</sup> � �erfc <sup>ξ</sup><sup>l</sup>

<sup>∂</sup>tck,v�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> ck,v�<sup>1</sup>ð Þþ <sup>x</sup>;<sup>t</sup> <sup>∑</sup>

<sup>∂</sup>tzk,v�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> ð Þ <sup>x</sup>;<sup>t</sup> zk,v�<sup>1</sup>ð Þþ <sup>x</sup>;<sup>t</sup> <sup>∑</sup>

<sup>k</sup>,v�<sup>1</sup>ð Þ <sup>x</sup>;<sup>t</sup> erfc <sup>ξ</sup><sup>l</sup>

2 l¼1 ql

as the solutions of the following equations:

� <sup>t</sup>¼<sup>0</sup>=ql

<sup>k</sup>ð Þ <sup>x</sup>;<sup>t</sup> � � �

<sup>t</sup>¼<sup>0</sup> are chosen arbitrarily.

� � � �

i ∑ N k¼1 θ0

Eq. (10) is uniquely solvable.

<sup>k</sup>ð Þ <sup>x</sup>;<sup>t</sup> � � �

�

<sup>k</sup>ð Þ<sup>t</sup> <sup>∂</sup>rkuvð Þþ <sup>M</sup> T2uv�<sup>1</sup>ð Þþ <sup>M</sup> h Mð Þ

<sup>k</sup>ð Þt ck,vð Þþ x;t ∑

zk,v�<sup>1</sup>ð Þþ x;t ∑

<sup>2</sup> ð Þþ x;t ∑ 2 l¼1 h<sup>l</sup>,<sup>k</sup> <sup>2</sup> ð Þ x;t � � exp ð Þþ <sup>τ</sup><sup>k</sup> <sup>∑</sup>

<sup>v</sup>�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> wl

<sup>k</sup>ð Þ Nl <sup>t</sup>¼η¼<sup>0</sup> <sup>¼</sup> <sup>0</sup>; ql

<sup>k</sup>ð Þ <sup>x</sup>;<sup>t</sup> � � �

wl

ð Þ <sup>x</sup>;<sup>t</sup> <sup>t</sup>¼<sup>0</sup>; ql

∂twl

We calculate

additional conditions:

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

1.

2.

3.

Yl

wl

i ∑ N k¼1 θ0

> ¼ i ∑ N k¼1 θ0

þ ∑ 2 l¼1

þ ∑ N k¼1

þ ∑ N k¼1

þ ∑ N k¼1

þ ∑ N k¼1 hk

179

$$\mathbf{G}\_{0} \cong \mathbf{C}^{\mathbf{O}}(\overline{\mathbf{O}}), \ \mathbf{G}\_{1} = \left\{ \mathbf{u}(\mathbf{M}) : \mathbf{u}(\mathbf{M}) = \oplus\_{\mathbf{l}=1}^{2} \mathbf{G}\_{0} \otimes \mathbf{erfc}\left(\frac{\mathfrak{z}\_{\mathbf{l}}}{2\sqrt{\mathfrak{z}}}\right) \right\},$$

$$\mathbf{G}\_{2} = \left\{ \mathbf{u}(\mathbf{M}) : \mathbf{u}(\mathbf{M}) = \oplus\_{\mathbf{k}=1}^{N} \mathbf{G}\_{0} \otimes \exp\left(\mathfrak{n}\_{\mathbf{k}}\right) \right\},$$

$$\mathbf{G}\_{3} = \left\{ \mathbf{u}(\mathbf{M}) : \mathbf{u}(\mathbf{M}) = \oplus\_{\mathbf{k}=1}^{N} \oplus\_{\mathbf{l}=1}^{2} \mathbf{Y}\_{\mathbf{k}}(\mathbf{N}\_{\mathbf{l}}) \otimes \exp\left(\mathfrak{n}\_{\mathbf{k}}\right), \left|\left|\mathbf{Y}\_{\mathbf{k}}(\mathbf{N}\_{\mathbf{l}})\right|\right| < c \exp\left(-\frac{\mathfrak{z}\_{\mathbf{l}}^{2}}{8\mathfrak{n}}\right) \right\},$$

$$\mathbf{G}\_{4} = \left\{ \mathbf{u}(\mathbf{M}) : \mathbf{u}(\mathbf{M}) = \oplus\_{\mathbf{k}=1}^{N} \mathbf{G}\_{0} \left(\oplus\_{\mathbf{l}=1}^{2} \mathbf{G}\_{0} \otimes \mathbf{erfc}\left(\frac{\mathfrak{z}\_{\mathbf{l}}}{2\sqrt{\mathfrak{z}}}\right)\right) \sigma\_{\mathbf{k}}\right\}, \mathbf{N}\_{\mathbf{l}} = \left\langle \mathbf{x}, \mathbf{t}, \mathbf{n}, \varsigma\_{1}, \varsigma\_{2} \right\rangle.$$

From these spaces we construct a new space:

$$\mathbf{G} = \oplus\_{\mathbb{I}=0}^{4} \mathbf{G}\_{\mathbb{I}}.$$

The element u Mð ÞϵG has the form:

$$\begin{aligned} \mathbf{u}(\mathbf{M}) &= \mathbf{v}(\mathbf{x}, \mathbf{t}) + \sum\_{\mathbf{l}=1}^{2} \mathbf{w}^{\mathbf{l}}(\mathbf{x}, \mathbf{t}) \text{erfc}\left(\frac{\mathbf{\tilde{z}}\_{\mathbf{l}}}{2\sqrt{\mathbf{t}}}\right) \\ &+ \sum\_{\mathbf{k}=1}^{N} \left[ \mathbf{c}\_{\mathbf{k}}(\mathbf{x}, \mathbf{t}) + \sum\_{\mathbf{l}=1}^{2} \mathbf{Y}\_{\mathbf{k}}^{\mathbf{l}}(\mathbf{N}\_{\mathbf{l}}) \right] \mathbf{exp}\left(\mathbf{r}\_{\mathbf{k}}\right) \\ &+ \sum\_{\mathbf{k}=1}^{N} \left[ \mathbf{z}\_{\mathbf{k}}(\mathbf{x}, \mathbf{t}) + \sum\_{\mathbf{l}=1}^{2} \mathbf{q}\_{\mathbf{k}}^{\mathbf{l}}(\mathbf{x}, \mathbf{t}) \text{erfc}\left(\frac{\mathbf{\tilde{z}}\_{\mathbf{l}}^{2}}{2\sqrt{\mathbf{t}}}\right) \right] \sigma\_{\mathbf{k}}. \end{aligned} \tag{9}$$

#### 1.5 Solvability of intermediate tasks

The iterative problems (Eq. (9)) in general form will be written:

$$\mathbf{T}\_1 \mathbf{u}(\mathbf{M}) = \mathbf{H}(\mathbf{M}).\tag{10}$$

Theorem 1. Suppose that the conditions (1)–(3) and H Mð ÞϵG3 are satisfied. Then, equation (Eq. (10)) is solvable in G.

Proof. Let the free term H Mð ÞϵG3 be representable in the form:

$$\mathbf{H(M)} = \sum\_{\mathbf{k}=1}^{N} \sum\_{\mathbf{l}=1}^{2} \mathbf{H}\_{\mathbf{k}}^{\mathbf{l}}(\mathbf{N\_{l}}), \ \left\| \mathbf{H}\_{\mathbf{k}}^{\mathbf{l}}(\mathbf{N\_{l}}) \right\| \le c \ \exp\left(\frac{\mathfrak{s}\_{\mathbf{l}}^{2}}{8\eta}\right).$$

Then, by directly substituting function u Mð Þϵ G from (Eq. (9)) in (Eq. (10)), we see that this function is a solution if and only if the function Yl <sup>k</sup>ð Þ Nl will be a solution of equation:

$$
\partial\_{\boldsymbol{\eta}} \mathbf{Y}\_{\mathbf{k}}^{\mathrm{l}}(\mathbf{N}\_{\mathrm{l}}) = \partial\_{\boldsymbol{\xi}\boldsymbol{\eta}}^{2} \mathbf{Y}\_{\mathbf{k}}^{\mathrm{l}}(\mathbf{N}\_{\mathrm{l}}) + \mathbf{H}\_{\mathrm{k}}^{\mathrm{l}}(\mathbf{N}\_{\mathrm{l}}), \mathrm{l} = \mathbf{1}, \, \mathbf{2}, \, \mathrm{k} = \mathbf{1}, \, \mathbf{2}, \ldots, \mathrm{N}. \tag{11}
$$

With the corresponding boundary conditions, this equation has a solution which have the estimate:

$$\left\|\left|\mathbf{Y}\_{\mathbf{k}}^{\mathrm{l}}(\mathbf{N}\_{\mathrm{l}})\right|\right\| \leq c \exp\left(\frac{\mathfrak{s}\_{\mathrm{l}}^{2}}{8\eta}\right).$$

The theorem is proven.

Theorem 2. Suppose that the conditions of Theorem 1 are satisfied. Then, under additional conditions:

1.

free term in a phase which are vanished at t ¼ tl, l ¼ 0, 1, ……, n in addition; and the asymptotic also contain angular boundary layer functions. We introduce a class of

n o � �

� �,

� � � �

<sup>k</sup>ð Þ Nl <sup>⊗</sup> exp rð Þ<sup>k</sup> ; <sup>Y</sup><sup>l</sup>

<sup>l</sup>¼1G0⊗erfc <sup>ξ</sup><sup>l</sup>

<sup>l</sup>¼0Gl:

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

� � � �

� � � �

<sup>l</sup>¼<sup>1</sup>G0⊗erfc <sup>ξ</sup><sup>l</sup>

<sup>k</sup>¼1G0⊗ exp rð Þ<sup>k</sup>

� � �

2 ffiffi t p

2 ffiffi t p � �

<sup>k</sup>ð Þ <sup>x</sup>;<sup>t</sup> erfc <sup>ξ</sup><sup>2</sup>

exp rð Þ<sup>k</sup>

l 2 ffiffi t p

T1u Mð Þ¼ H Mð Þ: (10)

ς2 l 8η � � :

<sup>k</sup>ð Þ Nl , l ¼ 1, 2, k ¼ 1, 2, …, N: (11)

<sup>k</sup>ð Þ Nl will be a solution

� , c exp

σk:

2 ffi t p

<sup>k</sup>ð Þ Nl

σk

,

� , c exp � <sup>ς</sup><sup>2</sup>

, Nl ¼ x;t; η; ς1; ς<sup>2</sup> ð Þ:

l 8η

,

(9)

functions in which the iterative problems will be solved:

Boundary Layer Flows - Theory, Applications and Numerical Methods

� �, G1 <sup>¼</sup> u Mð Þ : u Mð Þ¼ <sup>⊕</sup><sup>2</sup>

G2 <sup>¼</sup> u Mð Þ : u Mð Þ¼ <sup>⊕</sup><sup>N</sup>

<sup>k</sup>¼1G0 <sup>⊕</sup><sup>2</sup>

� �

<sup>G</sup> <sup>¼</sup> <sup>⊕</sup><sup>4</sup>

2 l¼1 wl

ckð Þþ x;t ∑

zkð Þþ x;t ∑

The iterative problems (Eq. (9)) in general form will be written:

Proof. Let the free term H Mð ÞϵG3 be representable in the form:

<sup>k</sup>ð Þþ Nl Hl

2 l¼1 Yl <sup>k</sup>ð Þ Nl

2 l¼1 ql

Theorem 1. Suppose that the conditions (1)–(3) and H Mð ÞϵG3 are satisfied.

<sup>k</sup>ð Þ Nl , <sup>H</sup><sup>l</sup>

� � �

Then, by directly substituting function u Mð Þϵ G from (Eq. (9)) in (Eq. (10)), we

With the corresponding boundary conditions, this equation has a solution which

� , c exp

ς2 l 8η � � :

<sup>k</sup>ð Þ Nl

� �

<sup>k</sup>¼<sup>1</sup>⊕<sup>2</sup> <sup>l</sup>¼1Yl

From these spaces we construct a new space:

u Mð Þ¼ v xð Þþ ;t ∑

þ ∑ N k¼1

þ ∑ N k¼1

The element u Mð ÞϵG has the form:

1.5 Solvability of intermediate tasks

Then, equation (Eq. (10)) is solvable in G.

∂ηY<sup>l</sup>

The theorem is proven.

of equation:

178

have the estimate:

H Mð Þ¼ ∑

<sup>k</sup>ð Þ¼ Nl <sup>∂</sup><sup>2</sup>

N k¼1 ∑ 2 l¼1 Hl

> ςl Yl

see that this function is a solution if and only if the function Yl

Yl <sup>k</sup>ð Þ Nl

� � �

G0 ffi <sup>C</sup><sup>∞</sup> <sup>Ω</sup>

G3 <sup>¼</sup> u Mð Þ : u Mð Þ¼ <sup>⊕</sup><sup>N</sup>

G4 <sup>¼</sup> u Mð Þ : u Mð Þ¼ <sup>⊕</sup><sup>N</sup>

$$\mathbf{u(M)}|\_{\mathfrak{t}=\mathfrak{q}=\mathbf{0}} = \mathbf{0}, \mathbf{u(M)}|\_{\mathfrak{x}=\mathfrak{l}-\mathfrak{l}, \mathfrak{k}=\mathbf{0}, \mathfrak{z}=\mathbf{0}} = \mathbf{0}, \mathfrak{l} = \mathbf{1}, \mathbf{2}.$$

2.

$$\mathbf{L}\_{\xi}\mathbf{u}(\mathbf{M}) = \mathbf{0}, \mathbf{L}\_{\xi}\mathbf{u}(\mathbf{M}) = \mathbf{0}.$$

3.

$$\mathbf{i}\sum\_{\mathbf{k}=1}^{N}\boldsymbol{\Theta}\_{\mathbf{k}}^{\prime}(\mathbf{t})\boldsymbol{\partial}\_{\mathbf{r}\_{\mathbf{k}}}\mathbf{u}\_{\mathbf{v}}(\mathbf{M}) + \mathbf{T}\_{2}\mathbf{u}\_{\mathbf{v}-1}(\mathbf{M}) + \mathbf{h}(\mathbf{M}) \in \mathbf{G}\_{3}.$$

Eq. (10) is uniquely solvable.

Proof. By Theorem 1 equation (Eq. (10)) has a solution that is representable in the form (Eq. (9)). With satisfying condition (1), we obtain

$$\left. \mathbf{v}(\mathbf{x}, \mathbf{t}) \right|\_{\mathbf{t} = 0} = -\sum\_{\mathbf{k} = 1}^{\mathrm{N}} \mathbf{c}\_{\mathbf{k}}(\mathbf{x}, \mathbf{0}), \mathbf{w}^{\mathrm{l}}(\mathbf{x}, \mathbf{t}) \Big|\_{\mathbf{t} = \mathbf{0}} = \overline{\mathbf{w}}^{\mathrm{l}}(\mathbf{x}), \tag{12}$$
 
$$\left. \mathbf{Y}\_{\mathrm{k}}^{\mathrm{l}}(\mathbf{N}) \right|\_{\mathbf{t} = \mathbf{q} = \mathbf{0}} = \mathbf{0}, \mathbf{q}\_{\mathrm{k}}^{\mathrm{l}}(\mathbf{x}, \mathbf{t}) \Big|\_{\mathbf{t} = \mathbf{0}} = \overline{\mathbf{q}}\_{\mathrm{k}}^{\mathrm{l}}(\mathbf{x}), \ \mathbf{d}\_{\mathrm{k}}^{\mathrm{l}}(\mathbf{x}, \mathbf{t}) \Big|\_{\mathbf{t} = \mathbf{0}} = \overline{\mathbf{d}}\_{\mathrm{k}}^{\mathrm{l}}(\mathbf{x}),$$
 
$$\mathbf{w}^{\mathrm{l}}(\mathbf{x}, \mathbf{t})|\_{\mathbf{x} = \mathbf{l} - 1} = -\mathbf{c}\_{\mathrm{k}}(\mathbf{l} - \mathbf{1}, \mathbf{t}), \mathbf{q}\_{\mathrm{k}}^{\mathrm{l}}(\mathbf{x}, \mathbf{t})|\_{\mathbf{x} = \mathbf{l} - 1} = -\mathbf{z}\_{\mathrm{k}}(\mathbf{l} - \mathbf{1}, \mathbf{t}), \mathbf{l} = \mathbf{1}, 2.$$

Due to the fact that the function erfc <sup>θ</sup> <sup>2</sup> ffi t p � � is zero at <sup>θ</sup> <sup>¼</sup> 0, the values for wl ð Þ <sup>x</sup>;<sup>t</sup> <sup>t</sup>¼<sup>0</sup>; ql <sup>k</sup>ð Þ <sup>x</sup>;<sup>t</sup> � � � � <sup>t</sup>¼<sup>0</sup> are chosen arbitrarily.

We calculate

i ∑ N k¼1 θ0 <sup>k</sup>ð Þ<sup>t</sup> <sup>∂</sup>rkuvð Þþ <sup>M</sup> T2uv�<sup>1</sup>ð Þþ <sup>M</sup> h Mð Þ ¼ i ∑ N k¼1 θ0 <sup>k</sup>ð Þt ck,vð Þþ x;t ∑ 2 l¼1 Yl <sup>k</sup>,vð Þ Nl � � exp rð Þþ <sup>k</sup> <sup>½</sup>∂tvv�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> vv�<sup>1</sup>ðx;tÞ� þ ∑ 2 l¼1 ∂twl <sup>v</sup>�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> wl <sup>v</sup>�<sup>1</sup>ð Þ <sup>x</sup>;<sup>t</sup> � �erfc <sup>ξ</sup><sup>l</sup> 2 ffiffi t p � � þ ∑ N k¼1 <sup>∂</sup>tck,v�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> ck,v�<sup>1</sup>ð Þþ <sup>x</sup>;<sup>t</sup> <sup>∑</sup> 2 l¼1 ∂tY<sup>l</sup> <sup>k</sup>,v�<sup>1</sup>ð Þ� Nl b xð Þ ;<sup>t</sup> <sup>Y</sup><sup>l</sup> <sup>k</sup>,v�<sup>1</sup>ð Þ Nl � � � � exp ð Þ <sup>τ</sup><sup>k</sup> þ ∑ N k¼1 <sup>∂</sup>tzk,v�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> ð Þ <sup>x</sup>;<sup>t</sup> zk,v�<sup>1</sup>ð Þþ <sup>x</sup>;<sup>t</sup> <sup>∑</sup> 2 l¼1 ∂tql <sup>k</sup>,v�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> <sup>q</sup><sup>l</sup> <sup>k</sup>,v�<sup>1</sup>ð Þ <sup>x</sup>;<sup>t</sup> h ierfc <sup>ξ</sup><sup>l</sup> 2 ffiffi t p � � �� σk þ ∑ N k¼1 zk,v�<sup>1</sup>ð Þþ x;t ∑ 2 l¼1 ql <sup>k</sup>,v�<sup>1</sup>ð Þ <sup>x</sup>;<sup>t</sup> erfc <sup>ξ</sup><sup>l</sup> 2 ffiffi t p � � � � exp ð Þþ <sup>τ</sup><sup>k</sup> h0ð Þþ <sup>x</sup>;<sup>t</sup> <sup>∑</sup> 2 l¼1 hl <sup>1</sup>ð Þ <sup>x</sup>;<sup>t</sup> erfc <sup>ξ</sup><sup>l</sup> 2 ffiffi t p � � þ ∑ N k¼1 hk <sup>2</sup> ð Þþ x;t ∑ 2 l¼1 h<sup>l</sup>,<sup>k</sup> <sup>2</sup> ð Þ x;t � � exp ð Þþ <sup>τ</sup><sup>k</sup> <sup>∑</sup> N k¼1 hk <sup>3</sup> ð Þþ x;t ∑ 2 l¼1 h<sup>l</sup>,<sup>k</sup> <sup>3</sup> ð Þ <sup>x</sup>;<sup>t</sup> erfc <sup>ξ</sup><sup>l</sup> 2 ffiffi t p � � � � <sup>σ</sup>k: (13)

Condition (3) of the theorem will be ensured, if we choose arbitrarily (Eq. (9)) as the solutions of the following equations:

<sup>∂</sup>tvv�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> vv�<sup>1</sup>ð Þ¼� <sup>x</sup>;<sup>t</sup> h0ð Þ <sup>x</sup>;<sup>t</sup> , ∂twl <sup>v</sup>�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> <sup>w</sup><sup>l</sup> <sup>v</sup>�<sup>1</sup>ð Þ¼� <sup>x</sup>;<sup>t</sup> <sup>h</sup><sup>l</sup> <sup>1</sup>ð Þ x;t , ∂tYl <sup>k</sup>,v�<sup>1</sup>ð Þ� Nl b xð Þ ;<sup>t</sup> Yl <sup>k</sup>,v�<sup>1</sup>ð Þ¼� Nl <sup>h</sup><sup>l</sup>,<sup>k</sup> <sup>2</sup> ð Þþ <sup>x</sup>;<sup>t</sup> <sup>q</sup><sup>l</sup> <sup>k</sup>,v�<sup>1</sup>ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>ς</sup><sup>l</sup> 2 ffiffiffi η p � � � � , <sup>∂</sup>tzk,v�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> zk,v�<sup>1</sup>ð Þ¼� <sup>x</sup>;<sup>t</sup> <sup>h</sup><sup>k</sup> <sup>3</sup> ð Þ x;t , ∂tq<sup>l</sup> <sup>k</sup>,v�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> <sup>q</sup><sup>l</sup> <sup>k</sup>,v�<sup>1</sup>ð Þ¼� <sup>x</sup>;<sup>t</sup> <sup>h</sup><sup>l</sup>,<sup>k</sup> <sup>3</sup> ð Þ x;t , (14)

wl

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

ql

<sup>k</sup>,v�1ð Þ <sup>x</sup>;<sup>t</sup> erfc <sup>ξ</sup><sup>l</sup>

� � � �

<sup>1</sup>,v�1ð Þ <sup>x</sup>;<sup>t</sup> , <sup>H</sup><sup>l</sup>

1.6 Solution of iterative problems

2 l¼1 wl

<sup>k</sup>¼<sup>1</sup> ck,0ð Þþ x;t ∑

where H<sup>l</sup>

(Eq. (21)):

we define values of wl

L<sup>ξ</sup> wl

representable in the form:

u0ð Þ¼ M v0ð Þþ x;t ∑

<sup>þ</sup> <sup>∑</sup><sup>N</sup>

If the function Y<sup>l</sup>

which is satisfying that

Yl

d l

tion Yl

<sup>k</sup>ð Þ¼ 0 iθ<sup>0</sup>

(Eq. (14)))

<sup>k</sup>ð Þ<sup>t</sup> <sup>Y</sup><sup>l</sup>

Hl

Hl

181

<sup>k</sup>,0ð Þ¼ Nl <sup>d</sup><sup>l</sup>

takes the form:

Yl

<sup>k</sup>,v�1ð Þ¼ <sup>x</sup>;<sup>t</sup> wl

<sup>k</sup>,v�1ð Þ¼ <sup>x</sup>;<sup>t</sup> ql

<sup>k</sup>,v�1ð Þ <sup>x</sup> <sup>x</sup>¼l�1; ql

2 ffiffi t p

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

2 l¼1 Yl <sup>k</sup>,0ð Þ Nl � �erk <sup>þ</sup> zk,0ð Þþ <sup>x</sup>;<sup>t</sup> <sup>∑</sup>

<sup>k</sup>,0ð Þ Nl <sup>t</sup>¼η¼<sup>0</sup> <sup>¼</sup> <sup>0</sup>; Yl

2 ffiffiffi η p � �, dl

T1u1ð Þ¼� M i ∑

�

from the last problem, we define

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

2 ffiffi t p � �

� � � <sup>k</sup>,v�1ð Þ <sup>x</sup> B xð Þþ ;<sup>t</sup> Hl

<sup>k</sup>,v�1ð Þ <sup>x</sup> B xð Þþ ;<sup>t</sup> Hl

� � � x¼l�1

<sup>2</sup>,v�1ð Þ x;t - are known functions. With substituting (Eq. (16)) into the conditions under x ¼ l � 1 from (Eq. (12)),

<sup>k</sup>,v�1ð Þ <sup>x</sup>

<sup>¼</sup> <sup>0</sup>, <sup>L</sup><sup>ξ</sup> ql

Eq. (8) is homogeneous for k = 0; therefore, by Theorem 1, it has a solution in G,

<sup>k</sup>,0ð Þ Nl is the solution of the equation <sup>∂</sup>ηYl

�

<sup>k</sup>,0ð Þ x is the arbitrary function. In the next step, equation (Eq. (8)) for k = 1

According to Theorem 1, this equation is solvable in U, if ck,0ð Þ x;t =0; the func-

<sup>k</sup>ð Þt ck,0ð Þþ x;t ∑

<sup>k</sup>,0ð Þ Nl

� � �

� �

N k¼1 θ0

<sup>k</sup>,0ð Þ Nl is the solution of the differential equation <sup>∂</sup>ηYl

<sup>k</sup>,0ð Þ Nl , and its solution is representable in the form (Eq. (14)), where

� �

<sup>k</sup>,0ð Þ <sup>x</sup>; <sup>t</sup> <sup>x</sup>¼l�<sup>1</sup> ¼ �ck,0ð Þ <sup>l</sup> � <sup>1</sup>;<sup>t</sup> ; where dl

<sup>k</sup>,0ð Þ Nl : Satisfying condition (1)–(3) of Theorem 1, we obtain (see

2 l¼1 Yl <sup>k</sup>,0ð Þ Nl

� �erk :

solving differential equations which are obtained from the second condition of

Thus, function u Mð Þ is determined uniquely. The theorem is proven.

<sup>1</sup>,v�1ð Þ x;t ,

(16)

<sup>2</sup>,v�1ð Þ x;t

<sup>k</sup>,v�1ð Þ <sup>x</sup>;<sup>t</sup> erfc <sup>ξ</sup><sup>l</sup>

� � � �

2 l¼1 ql

<sup>x</sup>¼l�1, <sup>ς</sup>l¼<sup>0</sup> ¼ �ck,0ð Þ <sup>l</sup> � <sup>1</sup>;<sup>t</sup> :

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

<sup>k</sup>,0ð Þ¼ Nl <sup>∂</sup><sup>2</sup>

<sup>k</sup>,0ð Þ x;t

<sup>k</sup>,0ð Þ¼ Nl <sup>∂</sup><sup>2</sup>

ςl Yl

<sup>k</sup>,0ð Þþ Nl

� � � t¼0 <sup>¼</sup> <sup>d</sup><sup>l</sup> <sup>k</sup>,0ð Þ x :

� � � � <sup>σ</sup><sup>k</sup>

2 ffiffi t p

ςl Yl <sup>k</sup>,0ð Þ Nl

(17)

. These conditions are used in

2 ffiffi t p ¼ 0:

iθ0 <sup>k</sup>ð Þ<sup>t</sup> ck,vð Þ¼� <sup>x</sup>;<sup>t</sup> zk,v�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> <sup>½</sup>∂tck,v�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> ck,v�<sup>1</sup>ð Þ <sup>x</sup>;<sup>t</sup> � � hk <sup>2</sup> ð Þ x;t :

After this choice of arbitrariness, expression (Eq. (13)) is rewritten:

$$\mathrm{i}\sum\_{\mathbf{k}=1}^{N}\mathsf{G}\_{\mathbf{k}}^{\prime}(\mathbf{t})\partial\_{\mathbf{t}\mathbf{k}}\mathbf{u}\_{\mathbf{v}}(\mathbf{M}) + \mathsf{T}\_{2}\mathbf{u}\_{\mathbf{v}-1}(\mathbf{M}) + \mathrm{h}(\mathbf{M}) = \sum\_{\mathbf{k}=1}^{N}\sum\_{\mathbf{l}=1}^{2}\left[\mathrm{i}\theta\_{\mathbf{k}}^{\prime}(\mathbf{t})\mathbf{Y}\_{\mathbf{k},\mathbf{v}}^{\mathrm{l}}(\mathbf{N}\_{\mathbf{l}})\right] \exp\left(\mathsf{\tau}\_{\mathbf{k}}\right) \in \mathsf{G}\_{3}$$

In (Eq. (14)), transition was made from ξl=2 ffiffi <sup>t</sup> <sup>p</sup> to variable <sup>ς</sup>l=<sup>2</sup> ffiffiffi η p . The function Yl <sup>k</sup>ð Þ Nl is defined as the solution of equation (Eq. (30)) under the boundary conditions from (Eq. (12)) in the form:

$$\mathbf{Y}\_{\mathbf{k}}^{\mathrm{l}}(\mathbf{N}\_{\mathrm{l}}) = \mathbf{d}\_{\mathrm{k}}^{\mathrm{l}}(\mathbf{x}, \mathbf{t}) \mathrm{erfc} \left(\frac{\mathfrak{H}}{2\sqrt{\eta}}\right) + \frac{\mathbf{1}}{2\sqrt{\pi}} \int\_{0}^{\eta} \int\_{0}^{\infty} \frac{\mathbf{H}\_{\mathrm{k}}^{\mathrm{l}}(\cdot)}{\sqrt{\eta - \pi}} \left[ \exp\left(-\frac{\left(\mathfrak{g}\_{\mathrm{l}} - \mathbf{y}\right)^{2}}{4(\eta - \pi)}\right) - \exp\left(-\frac{\left(\mathfrak{g}\_{\mathrm{l}} + \mathbf{y}\right)^{2}}{4(\eta - \pi)}\right) \right] d\mathbf{y} d\mathbf{r} . \tag{15}$$

We substitute this function in the corresponding equation from (Eq. (14)); then with respect to d<sup>l</sup> <sup>k</sup>ð Þ x;t , we obtain a differential equation, which is solving under the initial condition d<sup>l</sup> <sup>k</sup>ð Þ x;t � � � <sup>t</sup>¼<sup>0</sup> <sup>¼</sup> <sup>d</sup> l <sup>k</sup>ð Þ x , and we find

$$\mathbf{d}\_{\mathbf{k}}^{\mathrm{l}}(\mathbf{x},\mathbf{t}) = \overline{\mathbf{d}}\_{\mathbf{k}}^{\mathrm{l}}(\mathbf{x},\mathbf{t})\mathbf{B}(\mathbf{x},\mathbf{t}) + \mathbf{P}\_{\mathbf{k}}^{\mathrm{l}}(\mathbf{x},\mathbf{t}),\\\mathbf{B}(\mathbf{x},\mathbf{t}) = \exp\left(\int\_{0}^{\mathrm{t}} \mathbf{b}(\mathbf{x},\mathbf{s})\mathbf{ds}\right),$$

where P<sup>l</sup> <sup>k</sup>ð Þ x;t is known as the function.

By substituting the obtained function into condition for d<sup>l</sup> <sup>k</sup>ð Þ x;t � � � x¼l�1 from (Eq. (12)), we define the value of d l <sup>k</sup>ð Þ x � � � x¼l�1 . The obtained value is used as an initial condition for a differential equation with respect to d l <sup>k</sup>ð Þ x , which is obtained after substitution d<sup>l</sup> <sup>k</sup>ð Þ x;t into the first condition of (2). With that we ensure fulfillment of this condition and uniqueness of the function Yl <sup>k</sup>ð Þ Nl : The last equation from (Eq. (14)) due to the fact that θ<sup>0</sup> <sup>k</sup>ð Þ¼ tk 0 is solvable if

$$\mathbf{z}\_{\mathbf{k}, \mathbf{v}-1}^{\mathrm{l}}(\mathbf{x}, \mathbf{0}) = -\mathbf{h}\_2^{\mathrm{k}}(\mathbf{x}, \mathbf{0}) - \left[\partial\_{\mathbf{t}} \mathbf{c}\_{\mathbf{k}, \mathbf{v}-1}(\mathbf{x}, \mathbf{t}) - \mathbf{b}(\mathbf{x}, \mathbf{t}) \mathbf{c}\_{\mathbf{k}, \mathbf{v}-1}(\mathbf{x}, \mathbf{t})\right] \Big|\_{\mathbf{t}=\mathbf{0}}.$$

The obtained ratio is used as the initial condition for the differential equation with respect to zl <sup>k</sup>,v�<sup>1</sup>ð Þ <sup>x</sup>;<sup>t</sup> from (Eq. (14)).

The equation with respect to vv�<sup>1</sup>ð Þ x;t under the initial condition from (12) determines this function uniquely. Equations with respect to wl <sup>k</sup>,v�<sup>1</sup>ð Þ <sup>x</sup>;<sup>t</sup> , ql <sup>k</sup>,v�<sup>1</sup>ð Þ <sup>x</sup>;<sup>t</sup> under the corresponding condition from (Eq. (12)) have solutions representable in the form:

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

<sup>∂</sup>tvv�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> vv�<sup>1</sup>ð Þ¼� <sup>x</sup>;<sup>t</sup> h0ð Þ <sup>x</sup>;<sup>t</sup> ,

<sup>v</sup>�<sup>1</sup>ð Þ¼� <sup>x</sup>;<sup>t</sup> <sup>h</sup><sup>l</sup>

<sup>2</sup> ð Þþ <sup>x</sup>;<sup>t</sup> <sup>q</sup><sup>l</sup>

<sup>k</sup>,v�<sup>1</sup>ð Þ¼� <sup>x</sup>;<sup>t</sup> <sup>h</sup><sup>l</sup>,<sup>k</sup>

N k¼1 ∑ 2 l¼1 iθ0 <sup>k</sup>ð Þ<sup>t</sup> <sup>Y</sup><sup>l</sup>

<sup>η</sup> � <sup>τ</sup> <sup>p</sup> exp � <sup>ς</sup><sup>l</sup> � <sup>y</sup> � �<sup>2</sup>

We substitute this function in the corresponding equation from (Eq. (14)); then

<sup>k</sup>ð Þ x , and we find

4ð Þ η � τ !

l

<sup>k</sup>ð Þ x;t , we obtain a differential equation, which is solving under the

<sup>k</sup>ð Þ x;t , B xð Þ¼ ;t exp

<sup>k</sup>ð Þ x;t into the first condition of (2). With that we ensure fulfillment

<sup>2</sup> ð Þ� <sup>x</sup>; <sup>0</sup> ½ � <sup>∂</sup>tck,v�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> ck,v�<sup>1</sup>ð Þ <sup>x</sup>;<sup>t</sup>

<sup>k</sup>,v�<sup>1</sup>ð Þ <sup>x</sup>;<sup>t</sup> under the corresponding condition from (Eq. (12)) have

<sup>k</sup>ð Þ¼ tk 0 is solvable if

The obtained ratio is used as the initial condition for the differential equation

The equation with respect to vv�<sup>1</sup>ð Þ x;t under the initial condition from (12)

<sup>k</sup>ð Þ Nl is defined as the solution of equation (Eq. (30)) under the boundary condi-

<sup>1</sup>ð Þ x;t ,

� � � �

<sup>3</sup> ð Þ x;t ,

<sup>3</sup> ð Þ x;t ,

<sup>k</sup>,v�<sup>1</sup>ð Þ <sup>x</sup>; <sup>t</sup> erfc <sup>ς</sup><sup>l</sup>

<sup>k</sup>,vð Þ Nl � � exp ð Þ <sup>τ</sup><sup>k</sup> <sup>∈</sup> G3

� exp � <sup>ς</sup><sup>l</sup> <sup>þ</sup> <sup>y</sup> � �<sup>2</sup>

b xð Þ ;s ds � �

> <sup>k</sup>ð Þ x;t � � � x¼l�1

. The obtained value is used as an initial

<sup>k</sup>ð Þ Nl : The last equation from

<sup>k</sup>ð Þ x , which is obtained after

� � � t¼0 :

4ð Þ η � τ

,

from

<sup>t</sup> <sup>p</sup> to variable <sup>ς</sup>l=<sup>2</sup> ffiffiffi

" # !

ðt 0 2 ffiffiffi η p ,

(14)

<sup>2</sup> ð Þ x;t :

η p . The function

dydτ:

(15)

<sup>v</sup>�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> <sup>w</sup><sup>l</sup>

Boundary Layer Flows - Theory, Applications and Numerical Methods

<sup>k</sup>,v�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> <sup>q</sup><sup>l</sup>

<sup>k</sup>ð Þ<sup>t</sup> <sup>∂</sup>rkuvð Þþ <sup>M</sup> T2uv�1ð Þþ <sup>M</sup> h Mð Þ¼ <sup>∑</sup>

In (Eq. (14)), transition was made from ξl=2 ffiffi

þ 1 2 ffiffiffi π p ðη 0 ð<sup>∞</sup> 0 Hl <sup>k</sup>ð Þ� ffiffiffiffiffiffiffiffiffiffiffi

<sup>k</sup>,v�<sup>1</sup>ð Þ¼� Nl <sup>h</sup><sup>l</sup>,<sup>k</sup>

<sup>∂</sup>tzk,v�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> zk,v�<sup>1</sup>ð Þ¼� <sup>x</sup>;<sup>t</sup> <sup>h</sup><sup>k</sup>

<sup>k</sup>ð Þ<sup>t</sup> ck,vð Þ¼� <sup>x</sup>;<sup>t</sup> zk,v�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> <sup>½</sup>∂tck,v�<sup>1</sup>ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> ck,v�<sup>1</sup>ð Þ <sup>x</sup>;<sup>t</sup> � � hk

After this choice of arbitrariness, expression (Eq. (13)) is rewritten:

∂twl

<sup>k</sup>,v�<sup>1</sup>ð Þ� Nl b xð Þ ;<sup>t</sup> Yl

∂tq<sup>l</sup>

tions from (Eq. (12)) in the form:

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

<sup>k</sup>ð Þ¼ x;t d

(Eq. (12)), we define the value of d

(Eq. (14)) due to the fact that θ<sup>0</sup>

<sup>k</sup>,v�<sup>1</sup>ð Þ¼� <sup>x</sup>; <sup>0</sup> <sup>h</sup><sup>k</sup>

solutions representable in the form:

2 ffiffiffi η p � �

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

l

<sup>k</sup>ð Þ <sup>x</sup>;<sup>t</sup> B xð Þþ ;<sup>t</sup> <sup>P</sup><sup>l</sup>

By substituting the obtained function into condition for d<sup>l</sup>

l <sup>k</sup>ð Þ x � � � x¼l�1

<sup>k</sup>ð Þ x;t is known as the function.

condition for a differential equation with respect to d

<sup>k</sup>,v�<sup>1</sup>ð Þ <sup>x</sup>;<sup>t</sup> from (Eq. (14)).

determines this function uniquely. Equations with respect to

of this condition and uniqueness of the function Yl

∂tYl

iθ0

i ∑ N k¼1 θ0

Yl

Yl

<sup>k</sup>ð Þ¼ Nl <sup>d</sup><sup>l</sup>

with respect to d<sup>l</sup>

initial condition d<sup>l</sup>

dl

where P<sup>l</sup>

substitution d<sup>l</sup>

zl

with respect to zl

<sup>k</sup>,v�<sup>1</sup>ð Þ <sup>x</sup>;<sup>t</sup> , ql

wl

180

$$\begin{aligned} \mathbf{w}\_{\mathbf{k}, \mathbf{v} - 1}^{\mathrm{l}}(\mathbf{x}, \mathbf{t}) &= \overline{\mathbf{w}}\_{\mathbf{k}, \mathbf{v} - 1}^{\mathrm{l}}(\mathbf{x}) \mathbf{B}(\mathbf{x}, \mathbf{t}) + \mathbf{H}\_{\mathbf{1}, \mathbf{v} - 1}^{\mathrm{l}}(\mathbf{x}, \mathbf{t}), \\ \mathbf{q}\_{\mathbf{k}, \mathbf{v} - 1}^{\mathrm{l}}(\mathbf{x}, \mathbf{t}) &= \overline{\mathbf{q}}\_{\mathbf{k}, \mathbf{v} - 1}^{\mathrm{l}}(\mathbf{x}) \mathbf{B}(\mathbf{x}, \mathbf{t}) + \mathbf{H}\_{\mathbf{2}, \mathbf{v} - 1}^{\mathrm{l}}(\mathbf{x}, \mathbf{t}) \end{aligned} \tag{16}$$

where H<sup>l</sup> <sup>1</sup>,v�1ð Þ <sup>x</sup>;<sup>t</sup> , <sup>H</sup><sup>l</sup> <sup>2</sup>,v�1ð Þ x;t - are known functions.

With substituting (Eq. (16)) into the conditions under x ¼ l � 1 from (Eq. (12)), we define values of wl <sup>k</sup>,v�1ð Þ <sup>x</sup> <sup>x</sup>¼l�1; ql <sup>k</sup>,v�1ð Þ <sup>x</sup> � � � � � � x¼l�1 . These conditions are used in solving differential equations which are obtained from the second condition of (Eq. (21)):

$$\mathbf{L}\_{\xi} \left( \mathbf{w}\_{\mathbf{k}, \mathbf{v} - 1}^{\mathsf{l}} (\mathbf{x}, \mathbf{t}) \text{erfc} \left( \frac{\xi\_{\mathsf{l}}}{2 \sqrt{\mathsf{t}}} \right) \right) = \mathbf{0}, \\ \mathbf{L}\_{\xi} \left( \mathbf{q}\_{\mathbf{k}, \mathbf{v} - 1}^{\mathsf{l}} (\mathbf{x}, \mathbf{t}) \text{erfc} \left( \frac{\xi\_{\mathsf{l}}}{2 \sqrt{\mathsf{t}}} \right) \right) = \mathbf{0}.$$

Thus, function u Mð Þ is determined uniquely. The theorem is proven.

#### 1.6 Solution of iterative problems

Eq. (8) is homogeneous for k = 0; therefore, by Theorem 1, it has a solution in G, representable in the form:

$$\begin{split} \mathbf{u}\_{0}(\mathbf{M}) &= \mathbf{v}\_{0}(\mathbf{x}, \mathbf{t}) + \sum\_{l=1}^{2} \mathbf{w}^{l}(\mathbf{x}, \mathbf{t}) \text{erfc}\left(\frac{\underline{\xi}\_{l}}{2\sqrt{\mathbf{t}}}\right) \\ &+ \sum\_{k=1}^{N} \left\{ \left( \mathbf{c}\_{\mathbf{k},0}(\mathbf{x}, \mathbf{t}) + \sum\_{l=1}^{2} \mathbf{Y}\_{\mathbf{k},0}^{l}(\mathbf{N}\_{l}) \right) \mathbf{e}^{\mathbf{x}\_{l}} + \left[ \mathbf{z}\_{\mathbf{k},0}(\mathbf{x}, \mathbf{t}) + \sum\_{l=1}^{2} \mathbf{q}\_{\mathbf{k},0}^{l}(\mathbf{x}, \mathbf{t}) \text{erfc}\left(\frac{\underline{\xi}\_{l}}{2\sqrt{\mathbf{t}}}\right) \right] \sigma\_{\mathbf{k}} \right\}, \end{split} \tag{17}$$

If the function Y<sup>l</sup> <sup>k</sup>,0ð Þ Nl is the solution of the equation <sup>∂</sup>ηYl <sup>k</sup>,0ð Þ¼ Nl <sup>∂</sup><sup>2</sup> ςl Yl <sup>k</sup>,0ð Þ Nl which is satisfying that

$$\left. \mathbf{Y}\_{\mathbf{k},0}^{\mathrm{l}}(\mathbf{N}\_{\mathrm{l}}) \right|\_{\mathbf{t}=\mathbf{q}=\mathbf{0}} = \mathbf{0}, \left. \mathbf{Y}\_{\mathbf{k},0}^{\mathrm{l}}(\mathbf{N}\_{\mathrm{l}}) \right|\_{\mathbf{x}=\mathbf{l}-\mathbf{1}, \mathbf{q}=\mathbf{0}} = -\mathbf{c}\_{\mathbf{k},0}(\mathbf{l}-\mathbf{1}, \mathbf{t}).$$

from the last problem, we define

$$\mathbf{Y}\_{\mathbf{k},0}^{\mathrm{l}}(\mathbf{N}\_{\mathrm{l}}) = \mathbf{d}\_{\mathrm{k},0}^{\mathrm{l}}(\mathbf{x},\mathbf{t}) \operatorname{erfc} \left(\frac{\mathfrak{H}}{2\sqrt{n}}\right), \mathbf{d}\_{\mathrm{k},0}^{\mathrm{l}}(\mathbf{x},\mathbf{t}) \Big|\_{\mathbf{x}=\mathbf{l}-\mathbf{1}} = -\mathbf{c}\_{\mathrm{k},0}(\mathbf{l}-\mathbf{1},\mathbf{t}), \text{where } \mathbf{d}\_{\mathrm{k},0}^{\mathrm{l}}(\mathbf{x},\mathbf{t}) \Big|\_{\mathbf{t}=\mathbf{0}} = \overline{\mathbf{d}}\_{\mathrm{k},0}^{\mathrm{l}}(\mathbf{x}).$$

d l <sup>k</sup>,0ð Þ x is the arbitrary function. In the next step, equation (Eq. (8)) for k = 1 takes the form:

$$\mathbf{T\_1u\_1(M) = -i\sum\_{\mathbf{k}=1}^N \theta\_\mathbf{k}'(\mathbf{t}) \left[ \mathbf{c\_{k,0}(x,\mathbf{t}) + \sum\_{\mathbf{l}=1}^2 Y\_{\mathbf{k,0}}^\mathbf{l}(\mathbf{N\_l}) \right] \mathbf{e^{r\_k}}} $$

According to Theorem 1, this equation is solvable in U, if ck,0ð Þ x;t =0; the function Yl <sup>k</sup>,0ð Þ Nl is the solution of the differential equation <sup>∂</sup>ηYl <sup>k</sup>,0ð Þ¼ Nl <sup>∂</sup><sup>2</sup> ςl Yl <sup>k</sup>,0ð Þþ Nl Hl <sup>k</sup>,0ð Þ Nl , and its solution is representable in the form (Eq. (14)), where Hl <sup>k</sup>ð Þ¼ 0 iθ<sup>0</sup> <sup>k</sup>ð Þ<sup>t</sup> <sup>Y</sup><sup>l</sup> <sup>k</sup>,0ð Þ Nl : Satisfying condition (1)–(3) of Theorem 1, we obtain (see (Eq. (14)))

<sup>∂</sup>tv0 � b xð Þ ;<sup>t</sup> v0ð Þ¼ <sup>x</sup>;<sup>t</sup> <sup>0</sup>, <sup>∂</sup>twl <sup>0</sup>ð Þ� <sup>x</sup>;<sup>t</sup> b xð Þ ;<sup>t</sup> wl <sup>0</sup>ð Þ¼ x;t 0, ∂td<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> <sup>k</sup>,0ð Þ¼� <sup>x</sup>;<sup>t</sup> ql <sup>k</sup>,0ð Þ x;t , <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>, ∂tql <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>,0ð Þ¼ x;t 0, iθ0 <sup>k</sup>ð Þt ck, <sup>1</sup>ð Þ¼� x;t zk,0ð Þþ x;t fkð Þ x;t exp iθкð Þ 0 ε � �, L<sup>ς</sup> d<sup>l</sup> <sup>k</sup>,0ð Þ <sup>x</sup>;<sup>t</sup> erfc <sup>ς</sup><sup>l</sup> 2 ffiffiffi η p � � � � ¼ 0: (18)

Theorem 3. Suppose that conditions (1)–(3) are satisfied. Then, the constructed

solution is an asymptotic solution of problem (Eq. (1)), i.e., ∀n ¼ 0, 1, 2, …; the

2. Two-dimensional parabolic problem with a rapidly oscillating

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

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

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

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.

Δau � b xð Þ ; t u ¼ f xð Þ ; t exp

<sup>l</sup>¼<sup>1</sup>alð Þ xl <sup>∂</sup><sup>2</sup>

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

xl . iθð Þt ε

<sup>u</sup>j j <sup>t</sup>¼<sup>0</sup> <sup>¼</sup> <sup>0</sup>; <sup>u</sup> <sup>∂</sup>Ω¼<sup>0</sup> <sup>¼</sup> <sup>0</sup>, (20)

, xð Þ ; <sup>t</sup> <sup>ϵ</sup>E,

estimate is fair (Eq. (18)).

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

free term

2.1 Introduction

layer.

2.2 Statement of the problem

Consider the problem:

2. b xð Þ ; <sup>t</sup> ,f xð Þ ; <sup>t</sup> <sup>∈</sup> <sup>С</sup><sup>∞</sup>½ � <sup>E</sup> .

2.3 Regularization of the problem

ð Þ¼ 0 0:

3. θ<sup>0</sup>

183

<sup>L</sup>εu xð Þ� ; <sup>t</sup>; <sup>ε</sup> <sup>∂</sup>tu � <sup>ε</sup><sup>2</sup>

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>

When the equation is obtained with respect to d<sup>l</sup> <sup>k</sup>,0ð Þ x;t in the ql <sup>k</sup>,0ð Þ <sup>x</sup>;<sup>t</sup> erfc <sup>ξ</sup><sup>l</sup> <sup>2</sup> ffi t p � �, a transition <sup>ξ</sup><sup>l</sup> <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)). Functions wl <sup>0</sup>ð Þ <sup>x</sup>;<sup>t</sup> , <sup>d</sup><sup>l</sup> <sup>k</sup>,0ð Þ <sup>x</sup>;<sup>t</sup> , ql <sup>k</sup>,0ð Þ x;t are expressed through arbitrary functions wl <sup>0</sup>ð Þ x , d l <sup>k</sup>,0ð Þ <sup>x</sup> , ql <sup>k</sup>,0ð Þ x . These arbitrary functions provide the condition:

$$
\mathbf{L}\_{\xi}\mathbf{u}\_{\mathbf{k}}(\mathbf{m}) = \mathbf{0}, \\
\mathbf{L}\_{\xi}\mathbf{u}\_{\mathbf{k}}(\mathbf{m}) = \mathbf{0}, \\
$$

ensuring the solvability of the equation with respect to cl <sup>k</sup>, <sup>1</sup>ð Þ x;t : Suppose that

$$\left. \mathbf{Z}\_{\mathbf{k}, \mathbf{0}}(\mathbf{x}, \mathbf{t}) \right|\_{\mathbf{t}=\mathbf{0}} = \mathbf{f}\_{\mathbf{k}}(\mathbf{x}, \mathbf{t}) \exp\left(\frac{\mathbf{i}\theta\_{\mathbf{k}}(\mathbf{0})}{\mathbf{e}}\right).$$

This relation is used by the initial condition for determining Zk,0ð Þ x;t from the equation entering into (Eq. (18)).

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

$$\mathbf{u}\_{\mathfrak{m}}(\mathbf{m}) = \sum\_{\mathbf{i}=\mathbf{0}}^{\mathbf{n}} \boldsymbol{\varepsilon}^{\mathbf{i}} \mathbf{u}\_{\mathbf{i}}(\mathbf{m}).$$

In each iteration with respect to við Þ <sup>x</sup>;<sup>t</sup> , <sup>w</sup><sup>l</sup> i ð Þ <sup>x</sup>;<sup>t</sup> , dl k,i ð Þ <sup>x</sup>;<sup>t</sup> , zk,ið Þ <sup>x</sup>;<sup>t</sup> , ql k,i ð Þ x;t , we obtain inhomogeneous equations.

#### 1.7 Assessment of the remainder term

For the remainder term

$$\mathbf{R}\_{\mathsf{en}}(\mathbf{x},\mathsf{t},\mathsf{e}) \equiv \mathbf{R}\_{\mathsf{en}}(\mathbf{m},\mathsf{e}) \Big|\_{\mathsf{Y}=\mathsf{p}(\mathbf{x},\mathsf{t},\mathsf{e})} = \mathbf{u}(\mathbf{x},\mathsf{t},\mathsf{e}) - \sum\_{i=0}^{\mathsf{n}} \mathsf{e}^{i} \mathbf{u}\_{i}(\mathbf{m}) \Big|\_{\mathsf{Y}=\mathsf{p}(\mathbf{x},\mathsf{t},\mathsf{e})},$$

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

$$\mathbf{L}\_{\boldsymbol{\mathfrak{e}}} \mathbf{R}\_{\boldsymbol{\mathfrak{e}}n}(\mathbf{x}, \mathbf{t}, \mathbf{e}) = \mathbf{e}^{\mathbf{n}+1} \mathbf{g}\_{\mathbf{n}}(\mathbf{x}, \mathbf{t}, \mathbf{e})$$

with homogeneous boundary conditions. Using the maximum principle, like work of [7], we get the estimate:

$$|\mathbf{R}\_{\rm{en}}(\mathbf{x}, \mathbf{t}, \mathbf{e})| \le \mathbf{c} \mathbf{e}^{\mathbf{n}+1}.\tag{19}$$

Theorem 3. Suppose that conditions (1)–(3) are satisfied. Then, the constructed solution is an asymptotic solution of problem (Eq. (1)), i.e., ∀n ¼ 0, 1, 2, …; the estimate is fair (Eq. (18)).
