Siberian Mathematical Journal, Vol. 49, No. 1, pp. 123–129, 2008 c 2008 Mamontov E. V. Original Russian Text Copyright 

EQUIVALENCE TRANSFORMATIONS OF THE CLEBSCH EQUATIONS E. V. Mamontov

UDC 533;517.958

Abstract: We consider the Clebsch equations describing the motion of an ideal incompressible fluid in the absence of mass forces. These equations contain an arbitrary element, namely a function of three variables. We find the equivalence transformations of the variables which act on the arbitrary element and preserve the structure of the equations. The equivalence transformation is pointed out that takes the arbitrary element to zero. Some examples are given. Keywords: ideal incompressible fluid, Clebsch equations, arbitrary element, equivalence transformation

1. Preliminaries The barotropic motions of an ideal fluid are described by the system of equations [1] 1 Du + ∇p = F, ρ

Dρ + ρ div u = 0,

(1)

where u = (u, v, w), F = (F1 , F2 , F3 ), p = p(ρ), D = ∂t + u · ∇, and ∇ = (∂x , ∂y , ∂z ). The independent variables are t and x = (x, y, z). The sought functions are u and ρ. The functions F and p(ρ) are given. To analyze the solutions to (1), Clebsch used the following representation for the velocity vector [2, 3]: u = ∇ϕ + λ∇μ with ϕ, λ, and μ functions of (t, x). Ovsyannikov proposed [4, 5] to use the representation 1 u = ∇ϕ + (λ∇μ − μ∇λ) 2

(2)

with ϕ, λ, and μ functions of (t, x). The advantage of this representation is its symmetry. Note that rot u = ∇λ × ∇μ

(3)

and thereby the motion is a vortex motion in general. Suppose that the external force is potential; i.e., F = −∇Ω. Then the acceleration field is potential too:   dp a = Du = −∇ +Ω . (4) ρ Put

1 1 Q = ϕt + (λμt − μλt ) + |u|2 + 2 2



dp + Ω. ρ

(5)

The author was partially supported by the Russian Foundation for Basic Research (Grant 05–01–00080), the State Maintenance Program for the Leading Scientific Schools of the Russian Federation (Grant NSh–5245.2006.1), and the Integration Grant of the Siberian Division of the Russian Academy of Sciences (No. 2.15). Novosibirsk. Translated from Sibirski˘ı Matematicheski˘ı Zhurnal, Vol. 49, No. 1, pp. 153–160, January–February, 2008. Original article submitted October 25, 2006. c 2008 Springer Science+Business Media, Inc. 0037-4466/08/4901–0123 

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Lemma (Ovsyannikov). The Clebsch identity is valid: ∇Q = (Dμ)∇λ − (Dλ)∇μ.

(6)

Proof. Indeed,

  1 1 dp ∇ϕt + ∇(λμt − μλt ) = ut + μt ∇λ − λt ∇μ, ∇ + Ω = ∇p − F, 2 ρ ρ   1 2 |u| = (u · ∇)u + u × rot u = (u · ∇)u + u × (∇λ × ∇μ) ∇ 2 = (u · ∇)u + (u · ∇μ)∇λ − (u · ∇λ)∇μ.

Summing these equalities and recalling the first equation of (1) yields the sought identity.  If we define λ and μ as solutions to the system Dλ = fμ ,

Dμ = −fλ

(7)

with some function f = f (t, λ, μ) then (6) takes the form ∇(Q + f ) = 0. Hence, Q + f = 0.

(8)

An inessential function of the time can be included into ϕ. The continuity equation, together with (7), (8), and the second equation of (1), constitutes a system of equations in ρ, λ, μ, and ϕ; u is determined from (2) and the pressure p is a known function of ρ, while the motion is barotropic. Ovsyannikov noted [4, 5] that in the case of an ideal incompressible fluid in the absence of mass forces we can obtain a closed system of three equations in λ, μ, and ϕ equivalent to the initial system. In this case the motion of a fluid is described by the system of equations Du + ∇p = 0,

div u = 0

(9)

and

1 1 Q = ϕt + (λμt − μλt ) + |u|2 + p. 2 2 The representation for p in terms of ϕ, λ, and μ has the form 1 1 p = −ϕt + (μλt − λμt ) − |u|2 − f. 2 2 Moreover, by (2), the equation div u = 0 takes form (with Δ the Laplacian) 2Δϕ + λΔμ − μΔλ = 0.

(10)

(11)

The object of our further study is the Clebsch system Dλ = fμ , in more detail,

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Dμ = −fλ ,

2Δϕ + λΔμ − μΔλ = 0;

    1 1 λt + ϕx + (λμx − μλx ) λx + ϕy + (λμy − μλy ) λy 2 2   1 + ϕz + (λμz − μλz ) λz = fμ , 2     1 1 μt + ϕx + (λμx − μλx ) μx + ϕy + (λμy − μλy ) μy 2 2   1 + ϕz + (λμz − μλz ) μz = −fλ , 2 λ μ ϕxx + ϕyy + ϕzz + (μxx + μyy + μzz ) − (λxx + λyy + λzz ) = 0. 2 2

(12)

2. Equivalence Transformations The kernel of admissible algebras is generated by the operators Z1 = y∂x − x∂y ,

Z2 = z∂x − x∂z ,

Z5 = b2 (t)∂y + b2 (t)y∂ϕ ,

Z4 = b1 (t)∂x + b1 (t)x∂ϕ ,

Z3 = z∂y − y∂z ,

Z6 = b3 (t)∂z + b3 (t)z∂ϕ ,

Z7 = h(t)∂ϕ .

In the case f = 0 the following operators are admissible (Ovsyannikov): X1 = ∂t ,

X2 = 2t∂t + x∂x + y∂y + z∂z ,

X3 = x∂x + y∂y + z∂z + λ∂λ + μ∂μ + 2ϕ∂ϕ , X5 = x∂z − z∂x ,

X6 = y∂x − x∂y ,

X8 = b2 (t)∂y + b2 (t)y∂ϕ ,

X4 = z∂y − y∂z ,

X7 = b1 (t)∂x + b1 (t)x∂ϕ ,

X9 = b3 (t)∂z + b3 (t)z∂ϕ ,

1 (λFλ + μFμ − 2F ) ∂ϕ , X11 = h(t)∂ϕ , 2 bi (t), h(t), and F (λ, μ) are arbitrary functions and the prime stands for differentiation. The function f = f (t, λ, μ) is an arbitrary element. The problem is to find the groups of equivalence transformations; i.e., the group of transformations acting on the arbitrary element and preserving the structure of (12) (cp. [6]). Supplement (12) with the equations X10 = Fμ ∂λ − Fλ ∂μ +

fx = 0,

fy = 0,

fz = 0,

fϕ = 0.

(13)

Find the equivalence transformations of (12) and (13). The corresponding operator has the form X e = α t ∂t + α x ∂x + α y ∂y + α z ∂z + α λ ∂λ + α μ ∂μ + α ϕ ∂ϕ + α f ∂f . All coefficients are functions of t, x, y, z, λ, μ, ϕ, and f . Inspection of the defining equations demonstrates that they are x-autonomous and αt = αt (t),

αx = αx (t, x, y, z),

αλ = αλ (t, λ, μ),

αμ = αμ (t, λ, μ),

αy = αy (t, x, y, z),

αz = αz (t, x, y, z),

αϕ = αϕ (t, x, y, z, λ, μ, ϕ), αf = αf (t, λ, μ, f ).

Consequently, αt = 2(C1 − C5 )t + C6 ,

αy = −C2 x + C1 y + C4 z + b2 (t),

αx = C1 x + C2 y + C3 z + b1 (t),

αz = −C3 x − C4 y + C1 z + b3 (t),

αλ = C5 λ + ψμ ,

λψλ + μψμ − ψ + b1 (t)x + b1 (t)y + b3 (t)z, 2 where ψ(t, λ, μ) is an arbitrary function. The basis operators have the form αϕ = 2C5 ϕ +

X1e = ∂t ,

αμ = C5 μ − ψλ ,

αf = (4C5 − 2C1 )f + ψt + τ (t),

X2e = 2t∂t + x∂x + y∂y + z∂z − 2f ∂f ,

X3e = x∂x + y∂y + z∂z + λ∂λ + μ∂μ + 2ϕ∂ϕ + 2f ∂f , X4e = z∂y − y∂z , X7e = b1 (t)∂x + b1 (t)x∂ϕ ,

X5e = x∂z − z∂x ,

X6e = y∂x − x∂y ,

X8e = b2 (t)∂y + b2 (t)y∂ϕ ,

X9e = b3 (t)∂z + b3 (t)x∂ϕ ,

e = 2ψμ ∂λ − 2ψλ ∂μ + (λψλ + μψμ − 2ψ)∂ϕ + 2ψt ∂f , X10

e X11 = τ (t)∂f .

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e , and X e . The operators X e and X e The arbitrary element f is transformed only by X2e , X3e , X10 11 2 3 e dilate f and the action of X11 reduces to addition of an arbitrary function of t. The most interesting e . transformations are those connected with X10 e The operator X10 generates the following finite transformations: t = t and dλ dμ dϕ df  = 2ψμ , = −2ψλ , = (λ ψλ + μ ψμ − 2ψ), = 2ψt , (14) da da da da     λ (0) = λ, μ (0) = μ, ϕ (0) = ϕ, f (0) = f. The equations dμ dλ = 2ψμ , = −2ψλ , λ (0) = λ, μ (0) = μ (15) da da is a Hamilton subsystem and ψ(t, λ, μ) is its first integral; i.e., ψ(t, λ (a), μ (a)) is independent of a: ψ(t, λ , μ ) = ψ(t, λ, μ). Then the fourth equation of (14) can be integrated: (16) f  = f (t, λ, μ) + 2aψt (t, λ, μ).    Let λ = χ1 (t, λ, μ; a), μ = χ2 (t, λ, μ; a) be a solution to system (15). The function ϕ is found by quadrature. Consequently, we obtain (17) λ = χ1 (t, λ, μ; a), μ = χ2 (t, λ, μ; a), ϕ = ϕ + χ3 (t, λ, μ; a), f  = f (t, λ, μ) + 2aψt (t, λ, μ). (18) Put 1 1 1 u(a) = ∇ϕ + (λ ∇μ − μ ∇λ ), p(a) = −ϕt + (μ λt − λ μt ) − |u(a)|2 − f  . 2 2 2 d d u(a) = 0 and da p(a) = 0. Assertion. da

d d u(a), da p(a), using (14) and the equalities ∇ψλ = ψλ λ ∇λ + ψλ μ ∇μ and Proof. Calculate da   ∇ψμ = ψλ μ ∇λ + ψμ μ ∇μ . 

Theorem. Suppose that a smooth function f (t, λ, μ) is defined in a domain Ω ⊂ R3 . For every point (t0 , λ0 , μ0 ) ∈ Ω, there exist numbers T > 0 and δ > 0 and an equivalence transformation t = t, λ = λ (t, λ, μ), μ = μ (t, λ, μ), ϕ = ϕ (t, λ, μ, ϕ), defined in the parallelepiped Π = {|t − t0 | ≤ T, |λ − λ0 | ≤ δ, |μ − μ0 | ≤ δ} (for an arbitrary ϕ) and taking f (t, λ, μ) to f  (t, λ , μ ) ≡ 0. The inverse transformation is also defined. Proof. Fix an arbitrary point (t0 , λ0 , μ0 ) ∈ Ω and take the function t 1 1 f (τ, λ, μ)dτ =⇒ ψt = − f. ψ(t, λ, μ) = − 2 2 t0

If t = t0 then ψ = 0 and problem (15) dμ dλ = 0, = 0, λ (0) = λ0 , μ (0) = μ0 da da has a unique inextensible solution λ ≡ λ0 , μ ≡ μ0 defined on the whole real axis −∞ < a < ∞. Then the theorem on the continuous dependence on parameters implies existence of T > 0 and δ > 0 such that the solution to (15) with |t − t0 | ≤ T , |λ − λ0 | ≤ δ, and |μ − μ0 | ≤ δ exists for −1 ≤ a ≤ 1. Putting a = 1 in (17), we obtain the equivalence transformation λ = χ1 (t, λ, μ; 1), μ = χ2 (t, λ, μ; 1), ϕ = ϕ + χ3 (t, λ, μ; 1) (19) which takes f (t, λ, μ) to zero. Note that ∂(λ , μ ) ≡ 1. j= ∂(λ, μ) The inverse transformation is defined by the equalities λ = χ1 (t, λ , μ ; −1), μ = χ2 (t, λ , μ ; −1), ϕ = ϕ + χ3 (t, λ , μ ; −1).  126

3. Examples Example 1. Assume that f (t, λ, μ) = q(t)λμ. dQ Put ψ(t, λ, μ) = − Q(t) 2 λμ and dt = q(t). System (14) has the form dμ dϕ dλ = −Q(t)λ , = Q(t)μ , = 0, da da da λ (0) = λ, μ (0) = μ, ϕ (0) = ϕ. Thereby

λ = λe−aQ(t) ,

μ = μeaQ(t) ,

ϕ = ϕ.

The sought equivalence transformation (a = 1) is as follows: λ = λe−Q(t) ,

μ = μeQ(t) ,

ϕ = ϕ.

Verify this straightforwardly. Consider, for example, the second equation of (12). We have μt = μt e−Q − qμ e−Q , μx = μx e−Q ,     1 1 ϕx + (λμx − μλx ) μx = ϕx + (λ μx − μ λx ) μx e−Q , . . . , −fλ = −qμ = −qμ e−Q . 2 2 The equation takes the form μt e−Q −qμ e−Q

  1  1     ϕ + (λ μx − μ λx ) μx e−Q + · · · = −qμ e−Q . + 2 x 2

Using the new variables, we thus obtain an equation with the zero right-hand side. Example 2. Assume that f (t, λ) = q(t)r(λ). dQ Put ψ(t, λ) = − Q(t) 2 r(λ) and dt = q(t). System (14) has the form   dμ λ  dϕ  = Q(t)r (λ), = Q(t) r(λ) − r (λ) , da da 2    λ (0) = λ, μ (0) = μ, ϕ (0) = ϕ.

dλ = 0, da Thereby 

λ = λ,





μ = μ + aQ(t)r (λ),

 λ  ϕ = ϕ + aQ(t) r(λ) − r (λ) . 2

The sought equivalence transformation (a = 1) is as follows: 

λ = λ,





μ = μ + Q(t)r (λ),





 λ  ϕ = ϕ + Q(t) r(λ) − r (λ) . 2 



Verify this straightforwardly. We have λt = λt ,

ϕxx

λx = λx ,

μt = μt − q(t)r (λ ) − Q(t)r (λ )λt ,  2 μx = μx − Q(t)r (λ )λx , μxx = μxx − Q(t)r (λ ) λx − Q(t)r (λ )λxx ,

1 ϕx = ϕx + Q(t) λ r (λ )λx − r (λ )λx , 2 1 1 ϕx + (λμx − μλx ) = ϕx + (λ μx − μ λx ), 2 2 1 = ϕxx + Q(t)[r (λ )(λx )2 + λ r (λ )(λx )2 + λ r (λ )λxx − r (λ )λxx − r (λ )(λx )2 ], 2 127

whence λt

  1      + ϕx + (λ μx − μ λx ) λx + · · · = 0, 2

and the first equation holds. Now,     1 1      ϕx + (λμx − μλx ) μx = ϕx + (λ μx − μ λx ) (μx − Q(t)r (λ )λx ), 2 2   1 μt + ϕx + (λ μx − μ λx ) μx − q(t)r (λ ) − Q(t)r (λ )λt 2   1        −Q(t)r (λ ) ϕx + (λ μx − μ λx ) λx + · · · = −q(t)r (λ ), 2 and the second equation holds. The third equation holds as well: λ μ λ μ ϕxx + μxx − λxx + · · · = ϕxx + μxx − λxx + . . . . 2 2 2 2 Example 3. Assume that f (t, λ, μ) = λ(1 + k 2 μ2 ), where k > 0 is a parameter. Put ψ(t, λ, μ) = − 2t λ(1 + k 2 μ2 ). System (14) has the form t dμ dϕ dλ = −2k 2 tλ μ , = t(1 + k 2 μ2 ), = λ (1 − k 2 μ2 ), da da da 2 λ (0) = λ, μ (0) = μ, ϕ (0) = ϕ. Thereby 1 kμ + tan(kta) , k 1 − kμ tan(kta)   kλμ2 λ λμ λμ  + cos(2kta) + − sin(2kta). ϕ =ϕ− 2 2 4k 4 The sought equivalence transformation (a = 1) is as follows: 1 kμ + tan(kt) λ = (cos(kt) − kμ sin(kt))2 λ, μ = , k 1 − kμ tan(kt)   λ kλμ2 λμ λμ  + cos(2kt) + − sin(2kt). ϕ =ϕ− 2 2 4k 4 λ = (cos(kta) − kμ sin(kta))2 λ,

μ =

1 , j = 0, ±1, ±2, . . . , and μ = k tan(kt) . This transformation is defined provided that t = (2j+1)π 2k Straightforward verification (for the first and second equations): It is convenient to start with the second equation. After the change of variables this equation takes the form   1       (20) μt + ϕx + (λ μx − μ λx ) μx + · · · = 0, 2 i.e., becomes homogeneous. The first equation takes the form 2k sin(kt)λ μ + · · · = 0. (21) λt + cos(kt) + k sin(kt)μ t 

2k sin(kt)λ Subtracting from (21) the equation (20) multiplied by cos(kt)+k sin(kt)μ , we find that the first equation takes the sought form:   1       (22) λt + ϕx + (λ μx − μ λx ) λx + · · · = 0. 2 The author expresses his gratitude to L. V. Ovsyannikov and the participants of the SUBMODELS program: S. V. Khabirov, A. P. Chupakhin, S. V. Golovin, and A. A. Cherevko for the useful discussions of this article.

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References 1. Serrin J. B., Mathematical Principles of Classical Fluid Mechanics [Russian translation], Izdat. Inostr. Lit., Moscow (1963). ¨ 2. Clebsch A., “Uber eine allgemeine Transformation der hydrodynamischen Gleichungen,” Crelle LIV (1857), LVI (1859). 3. Lamb H., Hydrodynamics, Cambridge University Press, Cambridge (1975). 4. Ovsyannikov L. V., “The Clebsch equations and new models of vortex flows of a fluid,” in: Abstracts: XVI All-Russia Conference “Analytic Methods in Gas Dynamics SAMGAD 2006,” St. Petersburg, 5–10 July 2006, p. 64. 5. Ovsyannikov L. V., “The Clebsch equations and new models of vortex flows of a fluid,” in: Abstracts: IX All-Russia Congress on Theoretic and Applied Mechanics, Nizhni˘ı Novgorod, 22–28 August 2006, Vol. II, p. 140. 6. Ovsyannikov L. V., Group Analysis of Differential Equations, Academic Press, New York (1982). E. V. Mamontov Lavrent ev Institute of Hydrodynamics Novosibirsk, Russia E-mail address: [email protected]

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