A.2 The validation of identities (34) and (35)

In the two-dimensional case, the elements of the inverse matrix A�<sup>1</sup> km and the determinant of the matrix A^ are

Hydrodynamic Methods and Exact Solutions in Application to the Electromagnetic Field Theory… DOI: http://dx.doi.org/10.5772/intechopen.80813

$$A\_{11}^{-1} = \frac{\mathbf{1} + t \vartheta u\_{02}/\vartheta \xi\_2}{\det \hat{A}};\\ A\_{12}^{-1} = -\frac{t \vartheta u\_{01}/\vartheta \xi\_2}{\det \hat{A}};\\ A\_{21}^{-1} = -\frac{t \vartheta u\_{02}/\vartheta \xi\_1}{\det \hat{A}};\\ A\_{22}^{-1} = \frac{\mathbf{1} + t \vartheta u\_{01}/\vartheta \xi\_1}{\det \hat{A}} \tag{41}$$

$$\det \hat{A} = \mathbf{1} + t \left( \frac{\partial u\_{01}}{\partial \xi\_1} + \frac{\partial u\_{02}}{\partial \xi\_2} \right) + t^2 \left( \frac{\partial u\_{01}}{\partial \xi\_1} \frac{\partial u\_{02}}{\partial \xi\_2} - \frac{\partial u\_{01}}{\partial \xi\_2} \frac{\partial u\_{02}}{\partial \xi\_1} \right) \tag{42}$$

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

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):

$$\frac{\partial u\_{0m}}{\partial \xi\_k} A\_{km}^{-1} \det \hat{A} = \frac{\partial u\_{01}}{\partial \xi\_1} + \frac{\partial u\_{02}}{\partial \xi\_2} + 2t \left( \frac{\partial u\_{01}}{\partial \xi\_1} \frac{\partial u\_{02}}{\partial \xi\_2} - \frac{\partial u\_{01}}{\partial \xi\_2} \frac{\partial u\_{02}}{\partial \xi\_1} \right) \tag{43}$$

From (42), it follows that the right-hand side of (43) exactly matches <sup>∂</sup> det <sup>A</sup>^ ∂t obtained when differentiating over time in (42). This proves the identity of (34) in the two-dimensional case.

To prove the identity (35), let us introduce

$$B\_m = \frac{\partial}{\partial \xi\_k} \left( A\_{km}^{-1} \det \hat{A} \right) \tag{44}$$

Using (41), one gets from (44)

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

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A�<sup>1</sup>

km det A u ^ <sup>0</sup><sup>i</sup>

det A^ ξ � �! A�<sup>1</sup> km ξ � �! δ ξ! � x ! þ tu<sup>0</sup> ! ξ � � � �!

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

δ ξ<sup>1</sup> ! � ξ !

� � � � � �!

δ ξ<sup>1</sup> ! � x ! þ tu<sup>0</sup> ! ξ<sup>1</sup>

þ t u<sup>0</sup> ! ξ<sup>1</sup> � � ! ∂u0<sup>m</sup> ∂ξk

where the sub-integral expression in the second term of the left-hand side of

� � � � !

� u<sup>0</sup> ! ξ

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

To infer the identity (39), it is necessary to consider in the argument of the delta

!ð Þþ <sup>ξ</sup>1<sup>m</sup> � <sup>ξ</sup><sup>m</sup> <sup>O</sup> <sup>ξ</sup>

The identity (38) is a consequence of the noted above property of the delta

the right-hand side of (25), and according to (23), we get from here the identity

After the application of the identity (39) to the expression (37), defining the

∂u0<sup>m</sup> ∂ξk

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

Equality (40) holds identically due to the identical equality to zero of the

In the two-dimensional case, the elements of the inverse matrix A�<sup>1</sup>

� ∂ ∂ξk

þ t u<sup>0</sup> ! ξ<sup>1</sup> � � !

� ∂ ∂ξk

� �

ð Þ u0iu0<sup>m</sup>

� u<sup>0</sup> ! ξ � � � � � �! (38)

> δ ξ<sup>1</sup> ! � ξ � �!

�

! <sup>1</sup>� ξ � �! <sup>2</sup>

! �ξ � � � �!

ð Þþ u0iu0<sup>m</sup> u0<sup>m</sup>

� �

þ ð dn ξ ð dn

> δ ξ<sup>1</sup> ! � x ! þ tu<sup>0</sup> ! ξ<sup>1</sup> ! Þ

det <sup>A</sup>^ (39)

near the point

∂u0<sup>i</sup> ∂ξk

similar to that of

¼ 0 (40)

km and the

ξ1F<sup>1</sup> ¼ 0

(36)

(37)

� � �

ð dn ξδ ξ!

� x ! þ tu<sup>0</sup> ! ξ � � � �!

� � ! <sup>∂</sup>u0<sup>i</sup> <sup>ξ</sup>

δ ξ!

denominator of (39).

¼ u0<sup>k</sup> ξ � �!

u0<sup>k</sup> ξ ! 1 � �

ξ1 ! ¼ξ !

(39).

ð dn ξδ ξ! � x ! <sup>þ</sup> tu<sup>0</sup> ! ξ � � � �!

in (13).

66

� δ ξ!

� �!

det A^ ξ<sup>1</sup> � � !

∂ξk

� x ! þ tu<sup>0</sup> ! ξ � � � �!

> � x ! þ tu<sup>0</sup> ! ξ � � � �!

> > δ ξ<sup>1</sup> ! � ξ !

function (see discussion before the formula (28)).

<sup>þ</sup> <sup>∂</sup>u0<sup>k</sup>

form of the second term in (36), from (36), we get

A�<sup>1</sup>

A.2 The validation of identities (34) and (35)

determinant of the matrix A^ are

function a Taylor series decomposition of the function

ξ1 ! Þ ∂ξ1<sup>m</sup> � �

. Then the left-hand side of (39) has the form δ A^ ξ<sup>1</sup>

km det A u ^ <sup>0</sup><sup>i</sup>

expression in the brackets in the sub-integral expression in (40).

ξ1 ! ¼ξ

(36) is as follows:

F<sup>1</sup> ¼ u0<sup>m</sup> ξ<sup>1</sup>

$$B\_1 = \frac{\partial}{\partial \xi\_1} \left( \mathbf{1} + t \frac{\partial u\_{02}}{\partial \xi\_2} \right) - \frac{\partial}{\partial \xi\_2} \left( t \frac{\partial u\_{02}}{\partial \xi\_1} \right) \equiv \mathbf{0} \tag{45}$$

$$B\_2 = \frac{\partial}{\partial \xi\_1} \left( -t \frac{\partial u\_{01}}{\partial \xi\_2} \right) + \frac{\partial}{\partial \xi\_2} \left( \mathbf{1} + t \frac{\partial u\_{01}}{\partial \xi\_1} \right) \equiv \mathbf{0} \tag{46}$$

The identities (45) and (46) confirm the truth of the identity (35) in the twodimensional case.

Similarly, the identity (35) is proved in the three-dimensional case. For that, we need the following representation of the entries of the inverse matrix <sup>A</sup>^ �<sup>1</sup> [49]:

A�<sup>1</sup> <sup>11</sup> <sup>¼</sup> <sup>1</sup> det <sup>A</sup>^ <sup>1</sup> <sup>þ</sup> <sup>t</sup> ∂u<sup>02</sup> ∂ξ2 <sup>1</sup> <sup>þ</sup> <sup>t</sup> ∂u<sup>03</sup> ∂ξ3 � t <sup>2</sup> ∂u<sup>02</sup> ∂ξ3 ∂u<sup>03</sup> ∂ξ2 ; A�<sup>1</sup> <sup>12</sup> <sup>¼</sup> <sup>1</sup> det <sup>A</sup>^ <sup>t</sup> <sup>2</sup> ∂u<sup>01</sup> ∂ξ3 ∂u<sup>03</sup> ∂ξ2 � t 1 þ t ∂u<sup>03</sup> ∂ξ3 ∂u<sup>01</sup> ∂ξ2 ; A�<sup>1</sup> <sup>13</sup> <sup>¼</sup> <sup>1</sup> det <sup>A</sup>^ <sup>t</sup> <sup>2</sup> ∂u<sup>01</sup> ∂ξ2 ∂u<sup>02</sup> ∂ξ3 � t 1 þ t ∂u<sup>02</sup> ∂ξ2 ∂u<sup>01</sup> ∂ξ3 ; A�<sup>1</sup> <sup>21</sup> <sup>¼</sup> <sup>1</sup> det <sup>A</sup>^ <sup>t</sup> <sup>2</sup> ∂u<sup>02</sup> ∂ξ3 ∂u<sup>03</sup> ∂ξ1 � t 1 þ t ∂u<sup>03</sup> ∂ξ3 ∂u<sup>02</sup> ∂ξ1 ; A�<sup>1</sup> <sup>22</sup> <sup>¼</sup> <sup>1</sup> det <sup>A</sup>^ <sup>1</sup> <sup>þ</sup> <sup>t</sup> ∂u<sup>01</sup> ∂ξ1 <sup>1</sup> <sup>þ</sup> <sup>t</sup> ∂u<sup>03</sup> ∂ξ3 � t <sup>2</sup> ∂u<sup>01</sup> ∂ξ3 ∂u<sup>03</sup> ∂ξ1 ; A�<sup>1</sup> <sup>23</sup> <sup>¼</sup> <sup>1</sup> det <sup>A</sup>^ <sup>t</sup> <sup>2</sup> ∂u<sup>01</sup> ∂ξ3 ∂u<sup>02</sup> ∂ξ1 � t 1 þ t ∂u<sup>01</sup> ∂ξ1 ∂u<sup>02</sup> ∂ξ3 ; A�<sup>1</sup> <sup>31</sup> <sup>¼</sup> <sup>1</sup> det <sup>A</sup>^ <sup>t</sup> <sup>2</sup> ∂u<sup>02</sup> ∂ξ1 ∂u<sup>03</sup> ∂ξ2 � t 1 þ t ∂u<sup>02</sup> ∂ξ2 ∂u<sup>03</sup> ∂ξ1 ; A�<sup>1</sup> <sup>32</sup> <sup>¼</sup> <sup>1</sup> det <sup>A</sup>^ <sup>t</sup> <sup>2</sup> ∂u<sup>01</sup> ∂ξ2 ∂u<sup>03</sup> ∂ξ1 � t 1 þ t ∂u<sup>01</sup> ∂ξ1 ∂u<sup>03</sup> ∂ξ2 A�<sup>1</sup> <sup>33</sup> <sup>¼</sup> <sup>1</sup> det <sup>A</sup>^ <sup>1</sup> <sup>þ</sup> <sup>t</sup> ∂u<sup>01</sup> ∂ξ1 <sup>1</sup> <sup>þ</sup> <sup>t</sup> ∂u<sup>02</sup> ∂ξ2 � t <sup>2</sup> ∂u<sup>01</sup> ∂ξ2 ∂u<sup>02</sup> ∂ξ1 (47)

From (44), in the three-dimensional case, we get on the basis of (47) that all three components of the vector Bm � 0. For each m ¼ 1, 2, 3, we get identical zeroing separately for the sum of terms proportional to t and separately for the sum of the terms proportional to t 2.

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For example, in the expression for B<sup>1</sup> the sum of terms proportional to the first degree of time has the form t <sup>∂</sup> ∂ξ1 ∂u<sup>02</sup> <sup>∂</sup>ξ<sup>2</sup> <sup>þ</sup> <sup>∂</sup>u<sup>03</sup> ∂ξ3 � � � <sup>∂</sup>2u<sup>02</sup> <sup>∂</sup>ξ2∂ξ<sup>1</sup> � <sup>∂</sup>2u<sup>03</sup> ∂ξ3∂ξ<sup>1</sup> h i � 0, and similarly we can show the vanishing of the sum of twelve terms proportional to the square of time. Thus, the identity (35) is also proved in the three-dimensional case.

Proof of the identity (34) also is possible in the 3D case on the basis of (47) and (13) but is related to the cumbersome transformations.
