A.1 The direct validation of the solution

To verify the solution (29) satisfies Eq. (19), let us substitute (29) in Eq. (19). Then we get from (19):

$$\begin{split} \int d^n \xi \left[ u\_{0i} \left( \overrightarrow{\xi} \right) \frac{\partial \det \hat{A}}{\partial t} \delta \left( \overrightarrow{\xi} - \overrightarrow{\varkappa} + t \overrightarrow{u} \overline{\phi} \left( \overrightarrow{\xi} \right) \right) - u\_{0i} u\_{0m} \det \hat{A} \frac{\partial \delta \left( \overrightarrow{\xi} - \overrightarrow{\varkappa} + t \overrightarrow{u} \overline{\phi} \left( \overrightarrow{\xi} \right) \right)}{\partial \textbf{x}\_m} \right] \\ + \int d^n \xi \int d^n \xi\_1 F = 0 \end{split} \tag{32}$$

where

To prove (23), it is necessary to use Taylor series decomposition wrt ξ

ð Þþ ξ<sup>m</sup> � ξ0<sup>m</sup> O ξ

! ξ � �!

! � ξ ! 0 � �<sup>2</sup>

Using variable substitution in the argument of the right-hand side of (25) (of the

This can be verified substituting (27) into (26) and taking into account that the

Let us use a known property of the delta function that for any smooth function

why, in the general case, it is possible to multiply both sides of (23) by det <sup>∂</sup>Φ<sup>k</sup> <sup>ξ</sup>

¼ f ! x ! 0 � �

det

� � � � � �

� x ! <sup>þ</sup> tu<sup>0</sup> ! ξ � � � �!

∂Φ<sup>k</sup> ξ � �!

∂ξ<sup>m</sup>

δ x ! � x<sup>0</sup> ! � �

> � � � � � �

> > det A,

. That solution of Eq. (19) is considered

det A^ > 0 (30)

!

and det <sup>∂</sup>Φ<sup>k</sup>

for the argument of the delta function Φ

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!

! and taking into account that d x!¼ d y

from (22); then Eq. (24) is reduced to the following equation:

ξ0 ! � x ! þ tu<sup>0</sup> ! ξ<sup>0</sup> � � !

ξ0 ! ¼ x ! � tu ! x !; t � �

general implicit solution of the equation (19) can be represented as

! x !� � δ x ! � x ! 0 � �

� ξ<sup>0</sup> � � !

that is used in (20).

¼ δ Φ ! ξ � � � �!

From (28) and (27), identical holding of the equality (21) follows.

initial velocity fields) solution of the Cauchy problem for Eq. (19) as

∂u0<sup>m</sup> ξ !� � ∂ξk � �

That is why, sign of det A^ is absent in (29). The condition (30) provides smoothness of the solution only on the finite-time interval defined above from (13).

Taking into account (21), from (20), we get an exact general (for any smooth

�ξ ! � x ! þ tu<sup>0</sup> ! ξ � �!

ξ ! ¼ ξ<sup>0</sup> !

δ Φ<sup>k</sup> ξ ! 0 � � þ

type A x ^!¼<sup>y</sup>

u ! x !; t � �

f ! x !� �

64

into account (24), we get

When in (23), Φ

¼ u<sup>0</sup> ! x ! � t u

getting the following:

∂Φ<sup>k</sup> ∂ξ<sup>m</sup> � �

ξ ! ¼ ξ ! 0

hand side of (25) the right-hand side of (23).

The solution of Eq. (26) is as follows:

!ð Þ x; t � �

δ ξ!

¼ ð dn ξu0<sup>i</sup> ξ � �! δ ξ!

, the following equality f

ui x !; t � �

where det <sup>A</sup>^ <sup>¼</sup> det <sup>δ</sup>mk <sup>þ</sup> <sup>t</sup>

under the following condition:

! ξ � �! ! near

taking

ð Þ ξ<sup>m</sup> � ξ0<sup>m</sup>

(25)

(27)

!� � ∂ξ<sup>m</sup>

(28)

� � � �

holds. That is

� � � �

^ (29)

! ! ξ<sup>0</sup> !

!

when in the limit ξ

∂Φ<sup>k</sup> ∂ξ<sup>m</sup> � �

ξ ! ¼ ξ ! 0

det <sup>A</sup>^ j j [48]), we get from the right-

<sup>∂</sup>ξ<sup>m</sup> ¼ det Akm where Akm is

¼ 0 (26)

¼ δ

$$F \equiv u\_{0m} \left(\overrightarrow{\xi}\_1\right) \det \hat{A} \left(\overrightarrow{\xi}\_1\right) \delta \left(\overrightarrow{\xi}\_1 - \overrightarrow{x} + t\overrightarrow{u\_0}\right) \overline{u}\_{0i} \left(\overrightarrow{\xi}\right) \det \hat{A} \left(\overrightarrow{\xi}\right) \frac{\delta \delta \left(\overrightarrow{\xi} - \overrightarrow{x} + t\overrightarrow{u\_0}\left(\overrightarrow{\xi}\right)\right)}{\delta x\_n}.$$
 
$$\text{To transform sub-integral expression in (33), the following identities shown.}$$

To transform sub-integral expression in (32), the following identities shall be used:

$$\frac{\partial \delta \left(\stackrel{\rightarrow}{\xi} - \stackrel{\rightarrow}{\varkappa} + t\iota\_0^{\cdots} \left(\stackrel{\rightarrow}{\xi}\right)\right)}{\partial \kappa\_m} = -A\_{km}^{-1} \frac{\partial \delta \left(\stackrel{\rightarrow}{\xi} - \stackrel{\rightarrow}{\varkappa} + t\iota\_0^{\cdots} \left(\stackrel{\rightarrow}{\xi}\right)\right)}{\partial \xi\_k} \tag{33}$$

$$\frac{\partial \det \hat{A}}{\partial t} \equiv \frac{\partial u\_{0m}}{\partial \xi\_k} A\_{km}^{-1} \det \hat{A} \tag{34}$$

$$\frac{\partial}{\partial \xi\_k} \left( A\_{km}^{-1} \det \hat{A} \right) \equiv \mathbf{0} \tag{35}$$

The identity (33) is obtained from the relationship (obtained by differentiating the delta function having argument as a given function of ξ ! ) ∂δ ξ! �x !þ tu<sup>0</sup> ! ξ ! � � � � <sup>∂</sup>ξ<sup>k</sup> ¼ � <sup>∂</sup>δ ξ! �x !þ tu<sup>0</sup> ! ξ ! � � � � <sup>∂</sup>xl Alk after multiplying it both sides by the inverse matrix A�<sup>1</sup> km (where AlkA�<sup>1</sup> km ¼ δlmи and δlm is the unity matrix or the Kronecker delta).

The validity of the identities (34) and (35) is proved by the direct checking. In the one-dimensional case, when <sup>A</sup>^ <sup>¼</sup> <sup>1</sup> <sup>þ</sup> <sup>t</sup> du<sup>01</sup> <sup>d</sup>ξ<sup>1</sup> <sup>¼</sup> det A; ^ <sup>A</sup>^ �<sup>1</sup> <sup>¼</sup> det <sup>A</sup>^ � ��<sup>1</sup> , it obviously follows directly from (34) and (35). Further, in Item 3, the proof of the identities (34) and (35) of the two- and three-dimensional cases is given.

Taking into account (33)–(35), from (32), we get

$$\int d^n \xi \delta \left( \stackrel{\rightarrow}{\xi} - \stackrel{\rightarrow}{\varkappa} + t \stackrel{\rightarrow}{u\_0} \left( \stackrel{\rightarrow}{\xi} \right) \right) A\_{km}^{-1} \det \hat{A} \left( u\_{0i} \frac{\partial u\_{0m}}{\partial \xi\_k} - \frac{\partial}{\partial \xi\_k} (u\_{0i} u\_{0m}) \right) + \int d^n \xi \left[ \stackrel{\leftarrow}{\xi} f\_1 \tilde{\xi}\_1 F\_1 = 0 \right] \tag{36}$$

A�<sup>1</sup>

<sup>11</sup> <sup>¼</sup> <sup>1</sup> <sup>þ</sup> <sup>t</sup>∂u02=∂ξ<sup>2</sup>

∂u0<sup>m</sup> ∂ξk

the two-dimensional case.

dimensional case.

A�<sup>1</sup> <sup>11</sup> <sup>¼</sup> <sup>1</sup>

A�<sup>1</sup> <sup>12</sup> <sup>¼</sup> <sup>1</sup>

A�<sup>1</sup> <sup>13</sup> <sup>¼</sup> <sup>1</sup>

A�<sup>1</sup> <sup>21</sup> <sup>¼</sup> <sup>1</sup>

A�<sup>1</sup> <sup>22</sup> <sup>¼</sup> <sup>1</sup>

A�<sup>1</sup> <sup>23</sup> <sup>¼</sup> <sup>1</sup>

A�<sup>1</sup> <sup>31</sup> <sup>¼</sup> <sup>1</sup>

A�<sup>1</sup> <sup>32</sup> <sup>¼</sup> <sup>1</sup>

A�<sup>1</sup> <sup>33</sup> <sup>¼</sup> <sup>1</sup>

67

A�<sup>1</sup>

Using (41), one gets from (44)

det <sup>A</sup>^ ; A�<sup>1</sup>

det <sup>A</sup>^ <sup>¼</sup> <sup>1</sup> <sup>þ</sup> <sup>t</sup>

DOI: http://dx.doi.org/10.5772/intechopen.80813

km det <sup>A</sup>^ <sup>¼</sup> <sup>∂</sup>u<sup>01</sup>

To prove the identity (35), let us introduce

<sup>B</sup><sup>1</sup> <sup>¼</sup> <sup>∂</sup> ∂ξ1

<sup>B</sup><sup>2</sup> <sup>¼</sup> <sup>∂</sup> ∂ξ1

det <sup>A</sup>^ <sup>1</sup> <sup>þ</sup> <sup>t</sup>

<sup>2</sup> ∂u<sup>01</sup> ∂ξ3

<sup>2</sup> ∂u<sup>01</sup> ∂ξ2

<sup>2</sup> ∂u<sup>02</sup> ∂ξ3

<sup>2</sup> ∂u<sup>01</sup> ∂ξ3

<sup>2</sup> ∂u<sup>02</sup> ∂ξ1

<sup>2</sup> ∂u<sup>01</sup> ∂ξ2

det <sup>A</sup>^ <sup>t</sup>

det <sup>A</sup>^ <sup>t</sup>

det <sup>A</sup>^ <sup>t</sup>

det <sup>A</sup>^ <sup>t</sup>

det <sup>A</sup>^ <sup>t</sup>

det <sup>A</sup>^ <sup>t</sup>

det <sup>A</sup>^ <sup>1</sup> <sup>þ</sup> <sup>t</sup>

det <sup>A</sup>^ <sup>1</sup> <sup>þ</sup> <sup>t</sup>

<sup>12</sup> ¼ � <sup>t</sup>∂u01=∂ξ<sup>2</sup>

∂u<sup>01</sup> ∂ξ1 þ ∂u<sup>02</sup> ∂ξ2

Here, (42) corresponds to the formula (13) for n = 2.

∂ξ1 þ ∂u<sup>02</sup> ∂ξ2

det <sup>A</sup>^ ; A�<sup>1</sup>

Bm <sup>¼</sup> <sup>∂</sup> ∂ξk

1 þ t

�t ∂u<sup>01</sup> ∂ξ2 

∂u<sup>02</sup> ∂ξ2 

need the following representation of the entries of the inverse matrix <sup>A</sup>^ �<sup>1</sup>

∂u<sup>02</sup> ∂ξ2 <sup>1</sup> <sup>þ</sup> <sup>t</sup>

> ∂u<sup>03</sup> ∂ξ2

> ∂u<sup>02</sup> ∂ξ3

∂u<sup>03</sup> ∂ξ1

∂u<sup>01</sup> ∂ξ1 <sup>1</sup> <sup>þ</sup> <sup>t</sup>

> ∂u<sup>02</sup> ∂ξ1

∂u<sup>03</sup> ∂ξ2

∂u<sup>03</sup> ∂ξ1

∂u<sup>01</sup> ∂ξ1 <sup>1</sup> <sup>þ</sup> <sup>t</sup>

þ ∂ ∂ξ2

The identities (45) and (46) confirm the truth of the identity (35) in the two-

Similarly, the identity (35) is proved in the three-dimensional case. For that, we

� t 1 þ t

� t 1 þ t

� t 1 þ t

� t 1 þ t

� t 1 þ t

� t 1 þ t

;

;

;

;

;

<sup>21</sup> ¼ � <sup>t</sup>∂u02=∂ξ<sup>1</sup>

<sup>2</sup> ∂u<sup>01</sup> ∂ξ1

∂u<sup>01</sup> ∂ξ1

∂u<sup>02</sup> ∂ξ2

� <sup>∂</sup>u<sup>01</sup> ∂ξ2

∂u<sup>02</sup> ∂ξ1 (43)

km det <sup>A</sup>^ (44)

þ t

Hydrodynamic Methods and Exact Solutions in Application to the Electromagnetic Field Theory…

Using (41), it is possible to show that the following equality holds (in the lefthand side of (43), summation is assumed on the repeating indices from 1 to 2):

þ 2t

From (42), it follows that the right-hand side of (43) exactly matches <sup>∂</sup> det <sup>A</sup>^

obtained when differentiating over time in (42). This proves the identity of (34) in

A�<sup>1</sup>

� ∂ ∂ξ2 t ∂u<sup>02</sup> ∂ξ1 

1 þ t

∂u<sup>03</sup> ∂ξ3 

> ∂u<sup>03</sup> ∂ξ3 ∂u<sup>01</sup>

> ∂u<sup>02</sup> ∂ξ2 ∂u<sup>01</sup>

> ∂u<sup>03</sup> ∂ξ3 ∂u<sup>02</sup>

> ∂u<sup>01</sup> ∂ξ1 ∂u<sup>02</sup>

∂u<sup>02</sup> ∂ξ2 ∂u<sup>03</sup>

∂u<sup>01</sup> ∂ξ1 ∂u<sup>03</sup>

∂u<sup>02</sup> ∂ξ2 

∂u<sup>03</sup> ∂ξ3 

;

;

∂u<sup>01</sup> ∂ξ1 

> � t <sup>2</sup> ∂u<sup>02</sup> ∂ξ3

> > ∂ξ2

∂ξ3

∂ξ1

∂ξ3

∂ξ1

∂ξ2

� t <sup>2</sup> ∂u<sup>01</sup> ∂ξ2

� t <sup>2</sup> ∂u<sup>01</sup> ∂ξ3

det <sup>A</sup>^ ; A�<sup>1</sup>

∂u<sup>02</sup> ∂ξ2

� <sup>∂</sup>u<sup>01</sup> ∂ξ2

<sup>22</sup> <sup>¼</sup> <sup>1</sup> <sup>þ</sup> <sup>t</sup>∂u01=∂ξ<sup>1</sup>

∂u<sup>02</sup> ∂ξ1 (42)

det <sup>A</sup>^ (41)

∂t

� 0 (45)

� 0 (46)

∂u<sup>03</sup> ∂ξ2

∂u<sup>03</sup> ∂ξ1

∂u<sup>02</sup> ∂ξ1

[49]:

(47)

where the sub-integral expression in the second term of the left-hand side of (36) is as follows:

$$F\_1 = u\_{0m} \left( \stackrel{\rightarrow}{\xi}\_1 \right) \frac{\partial u\_{0i} \left( \stackrel{\rightarrow}{\xi} \right)}{\partial \xi\_k} \det \hat{A} \left( \stackrel{\rightarrow}{\xi}\_1 \right) \det \hat{A} \left( \stackrel{\rightarrow}{\xi} \right) A\_{km}^{-1} \left( \stackrel{\rightarrow}{\xi} \right) \delta \left( \stackrel{\rightarrow}{\xi} - \stackrel{\rightarrow}{\mathsf{x}} + t \stackrel{\rightarrow}{u} \right) \delta \left( \stackrel{\rightarrow}{\xi}\_1 - \stackrel{\rightarrow}{\mathsf{x}} + t \stackrel{\rightarrow}{u} \right) \tag{37}$$

To transform (37), it is necessary to use the following identities:

$$\begin{split} \delta \left( \overrightarrow{\xi} - \overrightarrow{\mathbf{x}} + t\overrightarrow{u\_0} \left( \overrightarrow{\overline{\xi}} \right) \right) \delta \left( \overrightarrow{\xi\_1} - \overrightarrow{\mathbf{x}} + t\overrightarrow{u\_0} \left( \overrightarrow{\overline{\xi}\_1} \right) \right) \\ \equiv \delta \left( \overrightarrow{\xi} - \overrightarrow{\mathbf{x}} + t\overrightarrow{u\_0} \left( \overrightarrow{\overline{\xi}} \right) \right) \delta \left( \overrightarrow{\overline{\xi}\_1} - \overrightarrow{\overline{\xi}} + t \left( \overrightarrow{u\_0} \left( \overrightarrow{\overline{\xi}\_1} \right) - \overrightarrow{u\_0} \left( \overrightarrow{\overline{\xi}} \right) \right) \right) \\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \end{split} \tag{38}$$

$$\delta\left(\overrightarrow{\xi}\_1 - \overrightarrow{\xi} + t\left(\overrightarrow{u\_0}\left(\overrightarrow{\xi}\_1\right) - \overrightarrow{u\_0}\left(\overrightarrow{\xi}\right)\right)\right) \equiv \frac{\delta\left(\overrightarrow{\xi}\_1 - \overrightarrow{\xi}\right)}{\det \hat{A}}\tag{39}$$

In (39), as it is noted above, det A^ > 0, and that is why the sign is not used in the denominator of (39).

The identity (38) is a consequence of the noted above property of the delta function (see discussion before the formula (28)).

To infer the identity (39), it is necessary to consider in the argument of the delta function a Taylor series decomposition of the function

$$\begin{aligned} \mu\_{0k} \left( \stackrel{\rightarrow}{\xi}\_1 \right) &= \mu\_{0k} \left( \stackrel{\rightarrow}{\xi} \right) + \left( \partial \mu\_{0k} \left( \stackrel{\rightarrow}{\frac{\xi\_1}{\partial \xi\_{1n}}} \right)\_{\stackrel{\rightarrow}{\xi\_1 = \stackrel{\rightarrow}{\xi}}} (\xi\_{1m} - \xi\_m) + O \left( \stackrel{\rightarrow}{\xi}\_1 - \stackrel{\rightarrow}{\xi} \right)^2 \text{ near the point } \xi \\ &\rightarrow \ \cdots \end{aligned}$$

ξ1 ! ¼ξ ! . Then the left-hand side of (39) has the form δ A^ ξ<sup>1</sup> ! �ξ � � � �! similar to that of the right-hand side of (25), and according to (23), we get from here the identity (39).

After the application of the identity (39) to the expression (37), defining the form of the second term in (36), from (36), we get

$$\int d^n \xi \delta \left( \stackrel{\rightarrow}{\xi} - \stackrel{\rightarrow}{\varkappa} + t \stackrel{\rightarrow}{u\_0} \left( \stackrel{\rightarrow}{\xi} \right) \right) A\_{km}^{-1} \det \hat{A} \left[ u\_{0i} \frac{\partial u\_{0m}}{\partial \xi\_k} - \frac{\partial}{\partial \xi\_k} (u\_{0i} u\_{0m}) + u\_{0m} \frac{\partial u\_{0i}}{\partial \xi\_k} \right] = 0 \tag{40}$$

Equality (40) holds identically due to the identical equality to zero of the expression in the brackets in the sub-integral expression in (40).

Thus, we have proved that (29) exactly satisfies the Hopf equation (19) for any smooth initial velocity fields on the finite-time interval under condition det A^ > 0 in (13).
