**4.3 Application of TPPAs in problems of incompressible liquid and gas mechanics**

Consider the Blasius equation (45), which describes laminar boundary layers on a flat plate:

$$\begin{aligned} \rho^{''} + \rho \rho^{''} &= 0; \\ \rho(\mathbf{0}) = \rho'(\mathbf{0}) &= \mathbf{0}; \; \rho'(\boldsymbol{\Leftrightarrow}) = \mathbf{2} \end{aligned} \tag{25}$$

where *φ ζ*ð Þ¼ *<sup>ψ</sup><sup>=</sup>* ffiffiffi *<sup>x</sup>* <sup>p</sup> , *<sup>ψ</sup>*ð Þ*<sup>y</sup>* is the stream function, *<sup>ζ</sup>* <sup>¼</sup> *<sup>y</sup>* 2 ffiffiffiffiffi Re *x* q is the automodel variable, and *x* and *y* are the Cartesian coordinates such that the axis *x* is directed along the flow. The interior asymptotic (ζ ! 0) has the form:

$$
\rho = a\_2 \zeta^2 - \frac{a\_2^2}{30} \zeta^5 + O\left(\zeta^8\right) \tag{26}
$$

ð ∞

*Uds* ¼ 1 (21)

, *U*<sup>00</sup> … *,* we get after integration

) *UU*0<sup>2</sup>

� � � �

*ds* ¼ *a*

*ds* ¼ *a* (22)

� � � �

3 5 ) ∞ð

*sU*<sup>2</sup> *ds*

0

2, … increases the

, *U*<sup>00</sup> is

*U*<sup>0</sup> ¼ *t*, *dt* ¼ *U*00*ds U*00*ds* ¼ *dV*, *V* ¼ *U*<sup>0</sup>

> ð ∞

0 *sU*0 *Uds*

∞

0 � ∞ð

0

*ds* <sup>¼</sup> *<sup>a</sup>*<sup>2</sup> (23)

*U*2 2 *ds*

2 � � � �

2 4 can be obtained by

� � � ∞ <sup>0</sup> � ∞ð

*U*0<sup>2</sup> *ds*

0

0 *s* 2

> � � � �

*sU*<sup>2</sup> *ds* )

This is the first integral relation for the second method of producing it:

*<sup>U</sup>*0<sup>2</sup> <sup>þ</sup> *sU*<sup>2</sup> � �

*Uds* )

And this is the next integral relation for the second method of producing it:

Using Eq. (19), from Eqs. (22) and (23), we calculate *а* = 0.7277, and *b* = 0.7966. From the given example, it follows that the features of the asymptotic connection method are the ambiguity of the algorithm, the freedom to choose both the form of TPPAs, integral relations, and methods for calculating the parameters of the TPPAs. The question of choosing integral relations is, in fact, a question of controlling the asymptotic approximation using weights selected to obtain integral relations. Choosing the weight allows you to achieve acceptable accuracy in a particular area of the boundary layer: a weight equal to 1 means that the uniform influence of

using quadrature integration formulas *a* = 0.7287 and *b* = 0.7922. In the same manner, integral relations with weights *U, U'*

*Mathematical Theorems - Boundary Value Problems and Approximations*

part integration. Multiplying Eq. (18) by *U*, *U*<sup>0</sup>

∞ð

*sU*<sup>2</sup> *ds* )

0

ð ∞

0

ð ∞

0 *sU*0

∞ð

0 *sU*0

� � � � � �

2

∞ð

0 *U*2

the entire layer is taken into account; a weight equal to 1, *s*, *s*

influence of the outer region of the layer; and if the desired solution *U*, *U*<sup>0</sup>

*U*00*ds* ¼

*U*00*ds* ¼

*Uds* <sup>¼</sup> *dV*, *<sup>V</sup>* <sup>¼</sup> *<sup>U</sup>*<sup>2</sup>

0

*U*0<sup>2</sup> *ds* ¼ ð ∞

*UU*00*ds* ¼

∞ð

0

ð ∞

0

*ds* ) *a* �

*U* )

∞ð

0 *U*0

∞ð

0 *U*0

*s* ¼ *t*, *dt* ¼ *ds*

*<sup>U</sup>*<sup>2</sup> � �*ds* <sup>¼</sup> *<sup>a</sup>*<sup>2</sup>

from 0 to ∞,

*UU*<sup>00</sup> <sup>¼</sup> *sU*<sup>2</sup> )

¼ ð ∞

*U*0

**84**

*U*<sup>00</sup> ¼ *sU*<sup>0</sup>

)

) ∞ð

0

) *<sup>U</sup>*0<sup>2</sup> � � � ∞ <sup>0</sup> �

> � � � � � �

*U*0

0

*sU*<sup>2</sup>

Substituting in Eqs. (20) and (21) instead of *U* (4) interpolation *Ua* (7), calculate

*U* ¼ *t*, *dt* ¼ *dU U*00*ds* ¼ *dV*, *V* ¼ *U*<sup>0</sup>

*<sup>U</sup>*0<sup>2</sup> <sup>þ</sup> *sU*<sup>2</sup> � �

∞ð

0

� � � �

*Uds* ) �*a*<sup>2</sup> <sup>¼</sup> <sup>2</sup>

) �*a*<sup>2</sup> <sup>¼</sup> <sup>2</sup> *sU*<sup>2</sup>

The procedure for obtaining external asymptotics is nontrivial due to the presence of logarithmic components in the main elements. We describe in detail the mechanism for obtaining and evaluating both primary and secondary members of asymptotic. From Eq. (25) follows:

$$\frac{\rho^{\prime}}{\rho^{\prime\prime}} = \rho \tag{27}$$

In the resulting equation, the first compound is the principal member of the external asymptotics. To obtain the following members of the asymptotic, we will

*φ* ¼ 2*ζ* � *c* þ *z*

where *z* ! 0, if *ζ* ! ∞. Given the last expression of the function *φ*, Eq. (29) is

0 @ ð *ζ*

ð Þ 2*ζ*<sup>1</sup> � *c* þ *z dζ*<sup>1</sup>

ð *ζ*

0 @

0

1 A

ð Þ *φ* � 2*ζ*<sup>1</sup> þ *c dζ*<sup>1</sup>

1 *ζ*2

*<sup>a</sup>* 2 � *φ*<sup>0</sup> ð Þ*dζ* (31)

2 � *φ*<sup>0</sup> ð Þ*dζ* (32)

A (33)

(34)

1

1 A

� � (30)

0

present the function as

obtained as follows:

*φ*0

asymptotic:

*φ*0

*DOI: http://dx.doi.org/10.5772/intechopen.93084*

If *z* ¼ *φ* � 2*ζ* þ *c*, then

ð Þ¼ *ζ* 2 þ

where *<sup>D</sup>* <sup>¼</sup> exp � <sup>Ð</sup>

and using weight equal to 1:

Therefore,

*φ*0

**87**

and

ð Þ¼ *ζ* 2 þ

2*a*<sup>2</sup>

ð Þ¼ *ζ* 2 þ

*φ*0

At that, in external domain, *ζ* ! ∞

∞ 0

2*a*<sup>2</sup>

*Padé Approximation to Solve the Problems of Aerodynamics and Heat Transfer…*

<sup>2</sup>*<sup>ζ</sup>* � *<sup>c</sup>* <sup>þ</sup> *<sup>z</sup>* exp �

<sup>2</sup>*<sup>ζ</sup>* � *<sup>c</sup>* <sup>þ</sup> *<sup>z</sup>* exp �*ζ*<sup>2</sup> <sup>þ</sup> <sup>c</sup>*<sup>ζ</sup>* � � exp �

2*a*2*D*

ð Þ *φ* � 2*ζ* þ *c dζ* � �.

> *<sup>a</sup>*<sup>2</sup> <sup>¼</sup> <sup>1</sup> 2 ∞ð

> > *c* ¼ ∞ð

> > > 0 @

*D* ¼ exp �

Type of generalized and normalized TPPA of order (4,4):

0

∞ð

ð Þ *φ* � 2*ζ* þ *c dζ*

<sup>1</sup> <sup>þ</sup> *<sup>β</sup>*1*<sup>ζ</sup>* <sup>þ</sup> *<sup>β</sup>*2*ζ*<sup>2</sup> <sup>þ</sup> *<sup>β</sup>*3*ζ*<sup>3</sup> <sup>þ</sup> *<sup>β</sup>*4*ζ*<sup>4</sup>

" #

0

*<sup>a</sup>*ð Þ¼ *<sup>ζ</sup>* 2 1 � <sup>1</sup> <sup>þ</sup> *<sup>α</sup>*1*<sup>ζ</sup>* <sup>þ</sup> *<sup>α</sup>*2*ζ*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*3*ζ*<sup>3</sup> <sup>þ</sup> *<sup>α</sup>*4*ζ*<sup>4</sup> � � exp �*ζ*<sup>2</sup> <sup>þ</sup> *<sup>c</sup><sup>ζ</sup>* � �

0 *φ*0

*φ* ¼ 2*ζ* � *c*, ð Þ *z* ! 0

In the external domain, where *ζ* ! ∞ and *z* ! 0, let us receive an exterior

<sup>2</sup>*<sup>ζ</sup>* � *<sup>c</sup>* exp �*ζ*<sup>2</sup> <sup>þ</sup> <sup>c</sup>*<sup>ζ</sup>* � � <sup>þ</sup> *<sup>o</sup>*

To calculate parameter *a*2, use the procedure of Section 4.1 [see formula (20)],

After integration of Eq. (27) by the coordinate *ζ* follows:

$$\begin{aligned} \ln \left[ \rho''(\zeta) \right] \Big|\_{0}^{\zeta} &= - \int\_{0}^{\zeta} \rho d\zeta \Rightarrow \ln \left[ \rho''(\zeta) \right] - \ln \left( 2a\_{2} \right) = - \int\_{0}^{\zeta} \rho d\zeta \Rightarrow \rho''(\zeta) \\\ &= 2a\_{2} \exp \left( - \int\_{0}^{\zeta} \rho d\zeta \right) \end{aligned} \tag{28}$$

After reintegration of Eq. (28) by the coordinate

$$\int\_0^{\zeta} \rho''(\zeta\_1) d\zeta\_1 = \int\_0^{\zeta} 2a\_2 \exp\left(-\int\_0^{\zeta\_4} \rho d\zeta\_2\right) d\zeta\_1$$

follows:

$$|\rho'(\zeta\_1)|\_0^\zeta = \int\_0^\zeta 2a\_2 \exp\left(-\int\_0^\zeta \rho d\zeta\_2\right) d\zeta\_1$$

subject to boundary conditions

$$\begin{aligned} \rho'(\zeta\_1) &= \int\_0^{\zeta\_1} 2a\_2 \exp\left(-\int\_0^{\zeta\_1} \rho d\zeta\_2\right) d\zeta\_1 \Rightarrow \rho'(\zeta\_1) = \int\_0^{\zeta} 2a\_2 \frac{\rho(\zeta\_1) \exp\left(-\int\_0^{\zeta\_1} \rho(\zeta\_2) d\zeta\_2\right)}{\rho(\zeta\_1)} d\zeta\_1 \\ &\Rightarrow \rho'(\zeta\_1) = 2a\_2 \left[\frac{1}{\rho(\zeta\_1)} d\left(\exp\left(-\int\_0^{\zeta\_1} \rho(\zeta\_2) d\zeta\_2\right)\right)\right] \end{aligned}$$

Let us make a limit transition *ζ* ! ∞ in the last equation and represent the integration interval as

½ Þ¼ 0, ∞ ½ � 0, *ζ* ∪½ Þ *ζ*, ∞ follows:

$$\rho^{\prime}(\zeta) = 2 + 2a\_2 \int\_{\zeta}^{\infty} \frac{1}{\rho(\zeta\_1)} d\left( \exp\left( - \int\_0^{\zeta\_1} \rho(\zeta\_2) d\zeta\_2 \right) \right).$$

We use the mean theorem in the last equation

$$\rho'(\zeta) = 2 + 2a\_2 \frac{1}{\rho(\zeta)} \exp\left(-\int\_0^{\zeta\_1} \rho(\zeta\_2) d\zeta\_2\right) \tag{29}$$

*Padé Approximation to Solve the Problems of Aerodynamics and Heat Transfer… DOI: http://dx.doi.org/10.5772/intechopen.93084*

In the resulting equation, the first compound is the principal member of the external asymptotics. To obtain the following members of the asymptotic, we will present the function as

$$
\varphi = \mathfrak{L}\zeta - c + x
$$

where *z* ! 0, if *ζ* ! ∞. Given the last expression of the function *φ*, Eq. (29) is obtained as follows:

$$\rho'(\zeta) = 2 + \frac{2a\_2}{2\zeta - c + z} \exp\left(-\int\_0^{\zeta} (2\zeta\_1 - c + z) d\zeta\_1\right).$$

If *z* ¼ *φ* � 2*ζ* þ *c*, then

The procedure for obtaining external asymptotics is nontrivial due to the presence of logarithmic components in the main elements. We describe in detail the mechanism for obtaining and evaluating both primary and secondary members of

*<sup>φ</sup>*<sup>00</sup> <sup>¼</sup> *<sup>φ</sup>* (27)

*φdζ* ) *φ*00ð Þ*ζ*

ð *ζ*

0

1 A*dζ*<sup>1</sup>

A (28)

*φ*‴

*φdζ* ) ln *φ*<sup>00</sup> ½ �� ð Þ*ζ* ln 2ð Þ¼� *a*<sup>2</sup>

1

After integration of Eq. (27) by the coordinate *ζ* follows:

*Mathematical Theorems - Boundary Value Problems and Approximations*

ð *ζ*

0 @

After reintegration of Eq. (28) by the coordinate

*<sup>φ</sup>*<sup>0</sup> *<sup>ζ</sup>*<sup>1</sup> ð Þj*<sup>ζ</sup>* <sup>0</sup> ¼ ð *ζ*

ð *ζ*1

*φdζ*<sup>2</sup>

1

1

*φ ζ*<sup>1</sup> ð Þ *<sup>d</sup>* exp �

∞ð

1

1

*ζ*

0 @

0

ð *ζ*

0

ð Þ¼ *ζ* 2 þ 2*a*<sup>2</sup>

We use the mean theorem in the last equation

ð Þ¼ *ζ* 2 þ 2*a*<sup>2</sup>

*φ*0

*φ*<sup>00</sup> *ζ*<sup>1</sup> ð Þ*dζ*<sup>1</sup> ¼

*φdζ*

ð *ζ*

2*a*<sup>2</sup> exp �

2*a*<sup>2</sup> exp �

0 @

0 @ ð *ζ*4

*φdζ*<sup>2</sup>

0

ð *ζ*4

*φdζ*<sup>2</sup>

1 A*dζ*<sup>1</sup>

*φ ζ*<sup>1</sup> ð Þ exp � <sup>Ð</sup>

*ζ*1 0

*φ ζ*<sup>2</sup> ð Þ*dζ*<sup>2</sup> !

*φ ζ*<sup>1</sup> ð Þ *<sup>d</sup>ζ*<sup>1</sup>

0

ð *ζ*

0 2*a*<sup>2</sup>

> 1 A

ð *ζ*1

0 @

*φ ζ*<sup>2</sup> ð Þ*dζ*<sup>2</sup>

*φ ζ*<sup>2</sup> ð Þ*dζ*<sup>2</sup>

1 A

1

1 A

A (29)

0

ð *ζ*1

0 @

0

1 A

*φ ζ*<sup>2</sup> ð Þ*dζ*<sup>2</sup>

0

0

A*dζ*<sup>1</sup> ) *φ*<sup>0</sup> *ζ*<sup>1</sup> ð Þ¼

0 @ ð *ζ*1

0

*φ ζ*<sup>1</sup> ð Þ *<sup>d</sup>* exp �

*φ ζ*ð Þ exp �

0 @

Let us make a limit transition *ζ* ! ∞ in the last equation and represent the

0

asymptotic. From Eq. (25) follows:

0 ¼ �

ð *ζ*

0

ð *ζ*

0

subject to boundary conditions

0 @

½ Þ¼ 0, ∞ ½ � 0, *ζ* ∪½ Þ *ζ*, ∞ follows:

*φ*0

2*a*<sup>2</sup> exp �

) *φ*<sup>0</sup> *ζ*<sup>1</sup> ð Þ¼ 2*a*<sup>2</sup>

¼ 2*a*<sup>2</sup> exp �

ln *<sup>φ</sup>*<sup>00</sup> ½ �j ð Þ*<sup>ζ</sup> <sup>ζ</sup>*

follows:

*φ*<sup>0</sup> *ζ*<sup>1</sup> ð Þ¼

**86**

ð *ζ*

0

integration interval as

$$\rho'(\zeta) = 2 + \frac{2a\_2}{2\zeta - c + z} \exp\left(-\zeta^2 + c\zeta\right) \exp\left(-\int\_0^\zeta (\rho - 2\zeta\_1 + c)d\zeta\_1\right).$$

In the external domain, where *ζ* ! ∞ and *z* ! 0, let us receive an exterior asymptotic:

$$\log'(\zeta) = 2 + \frac{2a\_2D}{2\zeta - c} \exp\left(-\zeta^2 + c\zeta\right) + o\left(\frac{1}{\zeta^2}\right) \tag{30}$$

where *<sup>D</sup>* <sup>¼</sup> exp � <sup>Ð</sup> ∞ 0 ð Þ *φ* � 2*ζ* þ *c dζ* � �.

To calculate parameter *a*2, use the procedure of Section 4.1 [see formula (20)], and using weight equal to 1:

$$a\_2 = \frac{1}{2} \int\_0^\infty \rho\_a'(2 - \rho') d\zeta \tag{31}$$

At that, in external domain, *ζ* ! ∞

$$
\varphi = 2\zeta - c, \ (z \to 0),
$$

Therefore,

$$\mathcal{L} = \bigcap\_{0}^{\infty} (2 - \rho') d\zeta \tag{32}$$

and

$$D = \exp\left(-\int\_0^\infty (\rho - 2\zeta + c)d\zeta\right) \tag{33}$$

Type of generalized and normalized TPPA of order (4,4):

$$\rho\_a'(\zeta) = 2 \left[ \mathbf{1} - \frac{\left( \mathbf{1} + a\mathbf{1}\zeta + a\mathbf{2}\zeta^2 + a\mathbf{3}\zeta^3 + a\mathbf{4}\zeta^4 \right) \exp\left( -\zeta^2 + c\zeta \right)}{\mathbf{1} + \beta\_1\zeta + \beta\_2\zeta^2 + \beta\_3\zeta^3 + \beta\_4\zeta^4} \right] \tag{34}$$

Species of TPPA taking into account four nontrivial parameters:

$$a\_3, \ \beta\_1, \ \beta\_2, \ \beta\_4$$

Therefore,

$$\rho\_a'(\zeta) = 2 \left[ 1 - \frac{\left( 1 + a\_3 \zeta^3 \right) \exp \left( -\zeta^2 + c\zeta \right)}{1 + \beta\_1 \zeta + \beta\_2 \zeta^2 + \beta\_4 \zeta^4} \right] \tag{35}$$

are discussed. The equations of laminar boundary layer near a semi-infinite plate in the supersonic flow of viscous perfect gas, as it is known [2, 7], can be reduced to

<sup>þ</sup> *σφT*<sup>0</sup> <sup>þ</sup> *<sup>a</sup><sup>σ</sup> <sup>μ</sup>*

2 ffiffiffi *<sup>x</sup>* <sup>p</sup> , *<sup>η</sup>* <sup>¼</sup>

*M* is the Mach number, *σ* is the Prandtl number**,** *κ* is the adiabatic index, *ψ* is the stream function, *T* is the temperature, *μ* is the viscosity coefficient, and *x* and *y* are

> *T*1 *Ts*

*<sup>ζ</sup>*<sup>2</sup> <sup>þ</sup> *<sup>O</sup> <sup>ζ</sup>*<sup>3</sup> � �,

*T*2 1 *Ts* � �*ζ*<sup>2</sup> <sup>þ</sup> *<sup>O</sup> <sup>ζ</sup>*<sup>3</sup> � �

ln �*T*<sup>0</sup> ð Þ¼�*σζ*<sup>2</sup> <sup>þ</sup> *<sup>σ</sup>c<sup>ζ</sup>* <sup>þ</sup> ln *<sup>B</sup>* <sup>þ</sup> *<sup>o</sup>*ð Þ<sup>1</sup> (45)

*<sup>β</sup>*<sup>0</sup> <sup>þ</sup> *<sup>β</sup>*1*<sup>ζ</sup>* exp *<sup>σ</sup>* �*ζ*<sup>2</sup> <sup>þ</sup> *<sup>c</sup><sup>ζ</sup>* � � � � (47)

∞ð

0 *T*0

<sup>2</sup> <sup>þ</sup> ð Þ *<sup>n</sup>* � <sup>1</sup> 2

<sup>2</sup> <sup>þ</sup> *<sup>c</sup><sup>ζ</sup>* <sup>þ</sup> ln *<sup>A</sup>* <sup>þ</sup> *<sup>o</sup>*ð Þ<sup>1</sup> ,

We solve boundary problems (40) and (41) approximately by connecting

*<sup>a</sup>*ð Þ¼ *<sup>ζ</sup>* 2 1 � <sup>1</sup> <sup>þ</sup> *<sup>A</sup>ζ*<sup>3</sup> � � exp �*ζ*<sup>2</sup> <sup>þ</sup> *<sup>c</sup><sup>ζ</sup>* � �

*<sup>a</sup>*ð Þ*ζ dζ*, *Ta*ð Þ¼ *ζ Ts* þ

<sup>1</sup> <sup>þ</sup> *<sup>α</sup>*1*<sup>ζ</sup>* <sup>þ</sup> *<sup>α</sup>*2*ζ*<sup>2</sup> <sup>þ</sup> *<sup>α</sup>*4*ζ*<sup>4</sup>

" #

ð *y*

*dy <sup>T</sup>* , *<sup>a</sup>* <sup>¼</sup> <sup>1</sup>

ð Þ¼ 0 0, *T*ð Þ¼ 0 *Ts* (42)

ð Þ¼ ∞ 2, *T*ð Þ¼ ∞ 1*:* (43)

0

þ *φφ*<sup>00</sup> ¼ 0, (40)

*<sup>T</sup> <sup>φ</sup>*00<sup>2</sup> <sup>¼</sup> <sup>0</sup> (41)

4 *M*<sup>2</sup>

ð Þ *κ* � 1

(44)

(46)

*<sup>a</sup>*ð Þ*ζ dζ* (48)

*<sup>φ</sup>*<sup>00</sup> *<sup>μ</sup> T* � �0

*Padé Approximation to Solve the Problems of Aerodynamics and Heat Transfer…*

*μ T*0 *T* � �0

*DOI: http://dx.doi.org/10.5772/intechopen.93084*

*<sup>x</sup>* <sup>p</sup> <sup>¼</sup> *φ ζ*ð Þ, *<sup>T</sup>* <sup>¼</sup> *<sup>T</sup>*ð Þ*<sup>ζ</sup>* , *<sup>ζ</sup>* <sup>¼</sup> *<sup>η</sup>*

*φ*ð Þ¼ 0 *φ*<sup>0</sup>

*φ*0

Interior asymptotic expansions are for *<sup>μ</sup>* <sup>¼</sup> *<sup>T</sup><sup>n</sup>*

*φ*<sup>0</sup> ¼ 2*a*2*ζ* � ð Þ *n* � 1 *a*<sup>2</sup>

*<sup>T</sup>* <sup>¼</sup> *Ts* <sup>þ</sup> *<sup>T</sup>*1*<sup>ζ</sup>* � <sup>2</sup>*aσa*<sup>2</sup>

where two constants *a*<sup>2</sup> and *T*<sup>1</sup> remain undefined.

ln *φ*<sup>00</sup> ¼ *c*

where three constants are unknown: *c*, *A*, and *B.*

The boundary conditions at the wall are

At external boundary of layer is

Exterior asymptotics for *ς* ! ∞

asymptotics (44) and (45) TPPA

**89**

*φ*0

*T*0

*φa*ð Þ¼ *ζ*

*<sup>a</sup>*ð Þ¼ *<sup>ζ</sup> <sup>ζ</sup><sup>m</sup>* � *<sup>ζ</sup>*

Boundary conditions (45) and (46) are satisfied if to put

∞ð

0 *φ*0

the form:

where

*<sup>φ</sup>* <sup>¼</sup> *<sup>ψ</sup>* ffiffiffi

the Cartesian coordinates.

Parameter values are determined using local asymptotic and TPPA in the respective domain. Taking into account the decomposition of the exponent in the internal domain, we will write down the local equality:

$$2a\_2\zeta - \frac{a\_2^2}{6}\zeta^4 = 2\left[\mathbf{1} - \frac{\left(\mathbf{1} + a\_3\zeta^3\right)\left(\mathbf{1} + c\zeta - \zeta^2 + \frac{\left(c\zeta - \zeta^2\right)^2}{2} + \dots\right)}{\mathbf{1} + \beta\_1\zeta + \beta\_2\zeta^2 + \beta\_4\zeta^4}\right] \tag{36}$$

Taking into account Eq. (33) in the external domain, we will write down the second local equality:

$$2 - \frac{2a\_2 D}{2\zeta - c} \exp\left(-\zeta^2 + c\zeta\right) = 2\left[1 - \frac{\left(1 + a\_3 \zeta^3\right) \exp\left(-\zeta^2 + c\zeta\right)}{1 + \beta\_1 \zeta + \beta\_2 \zeta^2 + \beta\_4 \zeta^4}\right] \tag{37}$$

$$\Rightarrow a\_2 D \left(1 + \beta\_1 \zeta + \beta\_2 \zeta^2 + \beta\_4 \zeta^4\right) = \left(2\zeta - c\right)\left(1 + a\_3 \zeta^3\right) \tag{37}$$

Equalizing the coefficients in Eqs. (36) and (37) at the same degrees *ζ*, we get

$$a\_3 = a\_2 D, \ \beta\_1 = a\_2 + c, \ \beta\_2 = -\frac{c^2}{2} + a\_2(a\_2 + c), \ \beta\_4 = 2.1$$

Therefore, the TPPA has the form:

$$\rho\_a'(\zeta) = 2 \left[ 1 - \frac{\left( 1 + a\_2 D \zeta^3 \right) \exp\left( -\zeta^2 + c\zeta \right)}{1 + (a\_2 + c)\zeta + \left( a\_2^2 + a\_2 c + \frac{c^2}{2} - 1 \right) \zeta^2 + 2\zeta^4} \right] \tag{38}$$

After systems (31)–(33) are solved, we will obtain

$$a\_2 = 0.6641,$$

$$c = 1.7308,$$

$$D = 0.3357$$

By substituting (39) in (38), we get an explicit expression for the TPPA.
