On the inverse variational problem for one class of quasilinear equations

Journal of Geometry and Physics 148 (2020) 103568

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On the inverse variational problem for one class of quasilinear equations D.V. Tunitsky Institute of Control Sciences of Russian Academy of Sciences, Moscow, Russia

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Article history: Received 11 June 2019 Received in revised form 21 November 2019 Accepted 25 November 2019 Available online 30 November 2019 MSC: 49N45 35A30 35L10

a b s t r a c t The paper concerns the inverse problem of calculus of variations for one class of elliptic and hyperbolic quasilinear second order equations with two independent variables. The equations of this class have a rather wide range of applications, among which are modeling of a two-conductor transmission line, motion of a hyperelastic homogeneous rod whose cross-sectional area varies along the rod, vibration of a string, wave propagation in a bar of elastic–plastic material, and isentropic flows of a compressible gas with plane symmetry. A constructive solution of the problem in hand is given and the corresponding Lagrangians are explicitly constructed. © 2019 Elsevier B.V. All rights reserved.

Keywords: Calculus of variations Inverse problem Quasilinear equations Elliptic equations Hyperbolic equations

1. Statement of the problem Consider a partial differential equation of the form h2 (zt ) ztt + ε g 2 (zx ) zxx + h (zt ) f1 (t ) − ε g (zx ) f2 (x) = 0,

(1.1)

where ε = −1 or ε = 1. It is a second order quasilinear equation with two independent variables t and x and unknown function z = z (t , x) .

(1.2)

The coefficients h (p), g (q), f2 (x), and f1 (t ) are given functions h: (α4 , β4 ) −→ R,

(1.3)

g: (α3 , β3 ) −→ R,

(1.4)

f2 : (α2 , β2 ) −→ R,

(1.5)

f1 : (α1 , β1 ) −→ R,

(1.6)

among which (1.3) and (1.4) are positive, h (p) > 0 and g (q) > 0. Here αk and βk are constant real values such that −∞ ≤ αk < βk ≤ +∞ for k = 1, 2, 3,4. E-mail address: [email protected]. https://doi.org/10.1016/j.geomphys.2019.103568 0393-0440/© 2019 Elsevier B.V. All rights reserved.

2

D.V. Tunitsky / Journal of Geometry and Physics 148 (2020) 103568

Since the coefficients h (1.3) and g (1.4) are positive, Eq. (1.1) is hyperbolic in case ε = −1 and elliptic in case ε = 1. The left hand side of this equation does not explicitly depend on the unknown function z and independent variables t, x; the first derivatives zt and zx are included symmetrically in it. Another distinctive feature of Eq. (1.1) is that it has intermediate integrals, cf. [5, section 475], of the form

√

√

−ε F2 (x) = const . √ Here j = 1, 2, −1 = i, and H, G, F2 , F1 are primitives of the coefficients (1.3)–(1.6), i.e., for fixed constants H (zt ) − (−1)j

γ4 ∈ (α4 , β4 ) ,

−ε G (zx ) + F1 (t ) + (−1)j

γ2 ∈ (α2 , β2 ) ,

γ3 ∈ (α3 , β3 ) ,

γ1 ∈ (α1 , β1 )

(1.7)

we put H (p) =

∫

p

γ4

h (τ ) dτ ,

G (q) =

∫

q

γ3

g (τ ) dτ ,

F2 (x) =

∫

x

γ2

f2 (τ ) dτ ,

F1 (t ) =

∫

t

γ1

f1 (τ ) dτ .

(1.8)

Equations of the type (1.1) have a rather wide range of applications. One example gives Eq. (1.1) with ε = −1, g (zx ) = 1, f1 (t ) = 1, and f2 (x) = 0. The application is a two-conductor transmission line with −zx as the current in the conductors, zt as the voltage between the conductors, h (zt ) as the leakage current per unit length, h2 (zt ) as the differential capacitance, x as a spatial variable, and t as time; see [8, ch. 1, § 1, sect. 1] and [1]. Another example is Eq. (1.1) with ε = −1, h (zt ) = 1, and f1 (t ) = 0. The application relates to the motion of a hyperelastic homogeneous rod whose cross-sectional area varies along the rod with zx as the displacement gradient related to the difference between spatial Eulerian and Lagrangian coordinates, zt as the velocity of a particle displaced by this difference, g (zx ) as essentially the stress-tensor such that g (zx ) = eµzx for some constant µ, f2 (x) as characterization of the cross-sectional area of the rod, and t as time; see [7,11], and [1]. Also the wave equation ztt − g 2 (zx ) zxx = 0,

(1.9)

which is a particular case of an equation (1.1) with ε = −1, h (p) = 1, f1 (t ) = 0, and f2 (x) = 0, is applied for description of a wide range of motions in one-dimensional homogeneous media. In particular, vibration of a string, see [15, ch. 2, § 1], [16, ch. 16, § 10], [17]; wave propagation in a bar of elastic–plastic material, see [3, sect. 98]; and isentropic flows of a compressible gas with plane symmetry, see [3, sect. 18], [13, ch. 1, § 5, sect. 2 and ch. 2, §2, sect. 8]. Let the function L = L (t , x, zt , zx , z )

(1.10)

be a Lagrangian of the variational problem

∫∫

L (t , x, zt (t , x) , zx (t , x) , z (t , x)) dtdx −→ min.

(1.11)

The Euler–Lagrange equation for this problem (1.11) is Lz −

∂ ∂ Lz − Lz = 0. ∂t t ∂x x

(1.12)

An inverse problem of calculus of variations has a long history that goes back to investigations by N. Ya. Sonin [14], H. Helmholtz [6], and A. Mayer [12]. An interesting case of this problem for systems of ordinary differential equations was solved by J. Douglas [4]. The range of its statements is quite wide and rather multifarious; for the inverse variational problem to be studied below it is the following. Find the necessary and sufficient conditions on that the left hand side of Eq. (1.1) is proportional to the left hand side of an Euler–Lagrange equation (1.12). The latter means that there exist functions L (1.10) and b = b (t , x, p, q, z ) in the domain M = (α1 , β1 ) × (α2 , β2 ) × (α3 , β3 ) × (α4 , β4 ) × R ⊆ R5

(1.13)

such that

(

h2 ztt + ε g 2 zxx + hf1 − ε gf2 = b Lz −

∂ ∂ Lz − Lz ∂t t ∂x x

) (1.14)

for all ztt , ztx , and zxx . If f1 (t ) = 0 and f2 (x) = 0, then (1.1) becomes a homogeneous quasilinear equation h2 (zt ) ztt + ε g 2 (zx ) zxx = 0,

(1.15)

of which the wave equation (1.9) is a particular case. For this equation (α1 , β1 ) = (α2 , β2 ) = (−∞, +∞) and the stated above problem is trivial. Indeed, Eq. (1.15) is an Euler–Lagrange equation for variation problem (1.11) with Lagrangian L (t , x, zt , zx , z ) = H (zt ) + ε G (zx ) ,

(1.16)

D.V. Tunitsky / Journal of Geometry and Physics 148 (2020) 103568

3

where H (zt ) =

zt

∫

∫

γ4

y

γ4

h2 (p) dp dy,

G (zx ) =

zx

∫

y

∫

γ3

γ3

g 2 (q) dq dy

(1.17)

and γ4 , γ3 are fixed numbers (1.7). Example 1.1. A particular case of a homogeneous equation (1.15) is zxx ztt + ε ( )2 = 0. 1 + zx 2

(1.18)

In this case the coefficients h, g, f2 , and f1 (1.3)–(1.6) of Eq. (1.1) and boundary points α1 , β1 , α2 , β2 , α3 , β3 , and α4 , β4 of their domains are h = 1,

g =

1 1 + q2

f1 (t ) = 0,

f2 (x) = 0,

,

α1 = α2 = α3 = α4 = −∞,

(1.19)

β1 = β2 = β3 = β4 = +∞.

So, we may put γ3 = γ4 = 0 and calculate the primitive G (zx ) (1.17) in Lagrangian (1.16): y

∫ 0

dq

(

1 + q2

)2 =

tan−1 y 2

+

y

(

2 1+

y2

zx

∫

),

0

y

∫ 0

dq

(

1 + q2

)2 dy =

zx tan−1 zx 2

.

Thus the Lagrangian L (1.16) becomes L (t , x, zt , zx , z ) =

(zt )2 2

+ε

zx tan−1 zx 2

.

A particular case of an equation (1.1) with f2 (x) ̸ = 0 is 1 zxx = 0. ztt + ε ( )2 − ε 2 1 + zx 2 1 + zx

(1.20)

In this case the coefficients h, g, f2 , and f1 (1.3)–(1.6) and boundary points α1 , β1 , α2 , β2 , α3 , β3 , α4 , β4 of their domains are the same as of (1.18)–(1.19) but f2 (x) = 1. Remarkably, there is no Lagrangian of the form (1.10) such that the left hand side of Eq. (1.20) could be proportional to the left hand side of an equation of the form (1.12), see Proposition 3.4. The solution of the problem in hand in general case of Eq. (1.1), when it may be both f1 (t ) ̸ = 0 and f2 (x) ̸ = 0, is given below in Section 4, see the theorem. 2. Exterior differential system Following Monge, we denote the derivatives of the unknown function z (1.2) as follows p = zt (t , x) ,

q = zx (t , x) .

(2.1)

The left hand side of Eq. (1.1) can be associated with the exterior differential 2-form

ω2 = h2 (p) dp ∧ dx + ε g 2 (q) dt ∧ dq + (h (p) f1 (t ) − ε g (q) f2 (x)) dt ∧ dx

(2.2)

and the equalities (2.1) and qt = px – with the differential forms

ω0 = dz − pdt − qdx,

ω1 = dt ∧ dp − dq ∧ dx,

for details see [9,10] Example 2.1. For Eq. (1.18) exterior differential 2-form ω2 (2.2) becomes

ω2 = dp ∧ dx + ε (

dt ∧ dq 1 + q2

)2

and for (1.20) –

ω2 = dp ∧ dx + ε (

dt ∧ dq

dt ∧ dx

1 + q2

1 + q2

)2 − ε

In both cases the domain M (1.13) is R5 .

.

(2.3)

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D.V. Tunitsky / Journal of Geometry and Physics 148 (2020) 103568

Example 2.2. The left hand side of Euler–Lagrange equation (1.12) is Lz −

∂ ∂ Lp − Lq = Lz − Lpt − pLpz − Lqx − qLqz − Lpp ztt − 2Lpq ztx − Lqq zxx . ∂t ∂x

(2.4)

In this case the corresponding exterior differential 2-form is

˜ ω2 = −Lpp(dp ∧ dx − Lpq (dt ∧ dp + dq ∧)dx) − Lqq dt ∧ dq + Lz − Lpt − pLpz − Lqx − qLqz dt ∧ dx.

(2.5)

Remind that a classical solution of Eq. (1.1) is a function z (1.2) that, being substituted in the left hand side of Eq. (1.1), turns it into a true equality. Let the map

σ : S −→ M ,

(2.6)

where M is the domain (1.13), be the graph (1.2), (2.1) of a classical solution z of Eq. (1.1). By definition of the pull-back σ ∗ of an exterior differential form, we have

) ( σ ∗ ω2 = h2 (p) pt + ε g 2 (q) qx + h (p) f1 (t ) − ε g (q) f2 (x) dt ∧ dx, σ ∗ ω0 = (zt − p) dt + (zx − q) dx. σ ∗ ω1 = (px − qt ) dt ∧ dx, Clearly, the function z (1.2) is a classical solution of Eq. (1.1) if and only if its graphic σ (2.6), (1.2), (2.1) satisfies the exterior differential equations

σ ∗ ω0 = 0,

σ ∗ ω1 = 0,

σ ∗ ω2 = 0.

(2.7)

An immersion (2.6) is a multivalued solution of Eq. (1.1) if it satisfies the system of external differential equations (2.7); cf. [9,10], and [2]. 3. Characteristic forms Consider the linear differential forms

√ √ j ωj,1 = h (p) dp − (−1)j −ε g (q) dq + f√ −ε f2 (x) dx, 1 (t ) dt + (−1) ωj,2 = h (p) dx − (−1)j −ε g (q) dt ,

(3.1)

which are complex-valued in case ε = 1. They are defined in the domain M (1.13) for j = 1, 2 and called characteristic forms. Obviously, we have the representation

ω2 −

√

−εh (p) g (q) ω1 = ω1,1 ∧ ω1,2 ,

for the 2-forms ω2 (2.2) and ω1 (2.3). characteristic forms (3.1), is through.

ω2 +

√ −ε h (p) g (q) ω1 = ω2,1 ∧ ω2,2

Therefore, the following statement, which explains the important role of

Lemma 3.1. (a) The 2-forms ω2 (2.2) and ω1 (2.3) and characteristic forms ωj,1 and ωj,2 (3.1) satisfy the equalities

ω2 =

1( 2

) ω1,1 ∧ ω1,2 + ω2,1 ∧ ω2,2 ,

ω1 = √

−1

2 −ε h (p) g (q)

) ( ω1,1 ∧ ω1,2 − ω2,1 ∧ ω2,2 .

(b) An immersion (2.6) satisfies the system of exterior differential equations (2.7) iff

σ ∗ ω0 = 0,

( ) σ ∗ ω1,1 ∧ ω1,2 = 0,

( ) σ ∗ ω2,1 ∧ ω2,2 = 0.

(3.2)

An immediate conclusion from this lemma and Example 2.2 is the following. Corollary 3.2. The left hand side of Eq. (1.1) is proportional to the left hand side of Euler–Lagrange equation (1.12), i.e., the equality (1.14) holds, if and only if there exists a differential 2-form ˜ ω2 of the type (2.5) such that the equalities

˜ ω2 ∧ ω1,k ∧ ω2,j = 0 hold for j, k = 1, 2. Direct calculations show that

( ) √ √ ˜ ω2 ∧ ω1,1 ∧ ω2,1 = ˜ ω2 ∧ ω1,1 ∧ ω1,1 − 2 −ε gdq + 2 −ε f2 (x) dx ( √ ) ( √ ) = −Lpp dp ∧ dx ∧ f1 dt ∧ − −ε 2gdq − Lqq dt ∧ dq ∧ hdp ∧ 2 −ε f2 (x) dx ( ) ( √ ) + Lz − Lpt − pLpz − Lqx − qLqz dt ∧ dx ∧ hdp ∧ −2 −ε gdq , ( 2 ) √ ˜ ω2 ∧ ω1,1 ∧ ω2,2 = h Lqq − 2 −εghLpq − ε g 2 Lpp dt ∧ dx ∧ dp ∧ dq, ( ) √ ˜ ω2 ∧ ω1,2 ∧ ω2,1 = −h2 Lqq − 2 −ε ghLpq + εg 2 Lpp dt ∧ dx ∧ dp ∧ dq, ˜ ω2 ∧ ω1,2 ∧ ω2,2 = 0.

D.V. Tunitsky / Journal of Geometry and Physics 148 (2020) 103568

5

Therefore, the conditions of Corollary 3.2 are equivalent to solvability of the system gf1 Lpp − hf2 Lqq + hg Lz − Lpt − pLpz − Lqx − qLqz = 0, √ √ − 2 −ε ghLpq − ε g 2 Lpp = 0, −h2 Lqq − 2 −ε ghLpq + ε g 2 Lpp = 0

(

h2 Lqq

)

(3.3)

with respect to the unknown function L (1.10). So, due to Corollary 3.2, analyzing the system (3.3) will bring us to solution of the problem in hand. Remark 3.3. Though rather simply derived, Eqs. (3.3) give a sufficiently effective tool for tackling the problem in hand, and, as far as the author can judge, this approach is an original one. Moreover, since the analog of Corollary 3.2 is through for general hyperbolic and elliptic Monge–Ampére equations, the approach under consideration is also applicable to all the hyperbolic and elliptic equations of the form A + Bztt + 2Cztx + Dzxx + E ztt zxx − ztx 2 = 0,

)

(

where the coefficients A, B, C , D, and E depend on t , x, z , zt , and zx . Proposition 3.4. The left hand side of Eq. (1.1) is proportional to the left hand side of Euler–Lagrange equation (1.12), i.e., the equality (1.14) holds, iff

(

1

)′′

h (p)

f1 (t ) +

(

1

)′′

g (q)

f2 (x) = 0

(3.4)

for all p ∈ (α4 , β4 ), q ∈ (α3 , β3 ), x ∈ (α2 , β2 ), t ∈ (α1 , β1 ). Proof. The last two equations (3.3) imply existence of three functions L1 (t , x, z , p), L2 (t , x, z , q), and L (t , x, z ) such that L (t , x, z , p, q) = L1 (t , x, z , p) + L2 (t , x, z , q) , L2qq (t , x, z , q) = L (t , x, z ) g 2 (q) , L1pp (t , x, z , p) = ε L (t , x, z ) h2 (p) .

Thenceforward, from the first equation (3.3) we get

(ε hf1 − gf2 ) L + L1z + L2z − L1tp − pL1zp − L2xq − qL2zq = 0. Differentiating the latter relation by p and q, we obtain the necessary and sufficient conditions for solvability of the first equation (3.3):

ε h′ f1 L − ε Lt h2 − ε pLz h2 = 0,

−g ′ f2 L − Lx g 2 − qLz g 2 = 0.

Thenceforth,

(ln L)z +

( )′′ 1 h

f1 = 0,

(ln L)z −

( )′′ 1

g

f2 = 0.

4. Main result The Proposition 3.4 allows to reveal all the possibilities for the left hand side of Eq. (1.1) being proportional to the left hand side of some Euler–Lagrange equation (1.12) and determine the corresponding Lagrangian (1.10). Namely, the following statement is true. Theorem. The left hand side of Eq. (1.1) is proportional to the left hand side of Euler–Lagrange equation (1.12), i.e., the equality (1.14) holds, iff one of the following five cases takes place. (a) f1 (t ) = 0 and f2 (x) = 0; Eq. (1.1) takes the form (1.15); the corresponding Lagrangian L (1.10) is given by the expressions (1.16), (1.17); ( the ) coefficient b in (1.14) is −1. (b) f1 (t ) = 0 and h2 (zt ) ztt + ε

1 g (q)

′′

= 0; Eq. (1.1) takes the form

zxx

(C0 + C1 zx )2

−ε

f2 (x) C0 + C1 zx

= 0,

where C0 and C1 are real constants such that C02 + C12 ̸ = 0; the corresponding Lagrangian (1.10) and coefficient b in (1.14) are L (t , x, zt , zx , z ) = eC1 F2 (x)

(

H (zt ) −

ε C12

)

ln |C0 + C1 zx | ,

b = −eC1 F2 (x)

6

D.V. Tunitsky / Journal of Geometry and Physics 148 (2020) 103568

if C1 ̸ = 0, and zx 2

L (t , x, zt , zx , z ) = H (zt ) + ε

−ε

2C02

x

∫

F2 (y) C0

γ2

dy,

b = −1

if C1 =(0. Here ) γ2 is taken from (1.7), F2 (x) given by (1.8), and H (zt ) – by (1.17). 1 h(p)

(c)

′′

= 0 and f2 (x) = 0; Eq. (1.1) takes the form

ztt

(C0 + C1 zt )

2

f1 (t )

+ ε g 2 (zx ) zxx +

C0 + C1 zt

= 0,

where C0 and C1 are real constants such that C02 + C12 ̸ = 0; the corresponding Lagrangian (1.10) and coefficient b in (1.14) are −C1 F1 (t )

L (t , x, zt , zx , z ) = e

(

) ln |C0 + C1 zt | + ε G (zx ) ,

−1 C12

b = −e−C1 F1 (t )

if C1 ̸ = 0, and L (t , x, zt , zx , z ) =

zt 2 2C02

t

∫

+ ε G (zx ) +

F1 (y) C0

γ1

dy,

b = −1

if C1 =(0. Here ) γ1 is taken ( from ) (1.7), F1 (t ) – from (1.8), and G (zx ) – from (1.17). 1 h(p)

(d)

′′

1 g (q)

= 0 and

ztt

(C0 + C1 zt )2

+ε

′′

= 0; Eq. (1.1) takes the form

zxx

(C2 + C3 zx )2

f1 (t )

+

C0 + C1 zt

−ε

f2 (x)

= 0,

C2 + C3 z x

where C0 , C1 , C2 , and C3 are real constants such that C02 + C12 ̸ = 0 and C22 + C32 ̸ = 0; the corresponding Lagrangian (1.10) and coefficient b in (1.14) are L (t , x, zt , zx , z ) = e−C1 F1 (t )+C3 F2 (x)

(

−1 C12

ε

ln |C0 + C1 zt | −

)

ln |C2 + C3 zx | ,

C32

−C1 F1 (t )+C3 F2 (x)

b = −e if C1 ̸ = 0 and C3 ̸ = 0, L (t , x, zt , zx , z ) = e−C1 F1 (t )

(

−1

ln |C0 + C1 zt | + ε

C12

zx 2

−ε

2C22

x

∫

F2 (y)

)

C2

γ2

b = −e−C1 F1 (t )

dy ,

if C1 ̸ = 0 and C3 = 0, C3 F2 (x)

L (t , x, zt , zx , z ) = e

zt 2

(

−

2C02

ε

∫

C32

ln |C2 + C3 zx | +

t

F1 (y) C0

γ1

)

b = −eC3 F2 (x)

dy ,

if C1 = 0 and C3 ̸ = 0, L (t , x, zt , zx , z ) =

zt 2 2C02

+ε

zx 2

∫

t

+

2C22

γ1

F1 (y) C0

dy − ε

∫

x

γ2

F2 (y) C2

dy,

b = −1

if C1 = 0 and C3 = 0. Here ( γ1)′′and γ2 are taken from (1.7), F1 (t ) and ( F)2 ′′(x) – from (1.8). 1 1 (e) f1 (t ) = C0 C1 , h(p) = C , f2 (x) = −C0 C2 , and g (1q) = C1 , where C0 , C1 , and C2 are non-zero real 1

2

constants; Eq. (1.1) takes the form ztt

(

C3 + C4 zt +

zt2 2C1

zxx

)2 + ε (

C5 + C6 zx +

zx2 2C2

)2 +

C0 C1 C3 + C4 zt +

zt2 2C1

C0 C2

+ε

C5 + C6 zx +

zx2 2C2

= 0,

where C3 , C4 , C5 , and C6 are real constants; the corresponding Lagrangian (1.10) and coefficient b in (1.14) are L (t , x, zt , zx , z ) =

⎛ e−C0 (C1 C4 t +C2 C6 x+z ) ⎝C −

zt

∫

y

∫

dp

⎜

γ

γ

(

C3 + C4 p +

p2 2C1

)2 dy − ε

b = e−C0 (C1 C4 t +C2 C6 x+z ) ,

⎞ zx

∫ γ

y

∫ γ

dq

(

C5 + C6 q +

q2 2C2

⎟ )2 dy⎠ ,

(4.1)

D.V. Tunitsky / Journal of Geometry and Physics 148 (2020) 103568 γ2

where γ is a real number such that C3 + C4 γ + C =

−C1 C3 + C4 γ +

−C2

+ε

γ2

2C1

C5 + C6 γ +

2C1

γ2

̸= 0, C5 + C6 γ +

7

̸= 0, and

2C2

.

γ2

(4.2)

2C2

Proof. By Proposition 3.4, the cases (a)–(e) cover all the possible logical outcomes. In the case (e) the left hand side of Euler–Lagrange equation (2.4) for Lagrangian (4.1) is Lz −

⎛

⎛

e−C0 (C1 C4 t +C2 C6 x+z ) ⎝−C0 ⎝C −

⎜

zt

∫

∂ ∂ Lzt − Lz = ∂t ∂x x

y

∫

dp

⎜

γ

(

γ

−C0 (C1 C4 + zt )

C3 + C4 p +

zt

∫ γ

C3 + C4 p +

)2 dy − ε

p2 2C1

dp

(

p2 2C1

⎞

)2 + (

zx

∫

y

∫

γ

dq

(

γ

C5 + C6 q +

q2 2C2

⎟ )2 dy⎠ .

ztt C3 + C4 p +

p2 2C1

)2 ⎞

−ε C0 (C2 C6 + zx )

zx

∫ γ

dq

(

C5 + C6 q +

q2 2C2

)2 + ε (

zxx C5 + C6 q +

q2 2C2

⎟ )2 ⎠ .

Put

⎛

⎞

W (zt , zx ) = −C0 ⎝C −

zt

∫

y

∫

dp

⎜

−C0 (C1 C4 + zt )

γ

γ

zt

∫

(

C3 + C4 p +

dp

(

γ

C3 + C4 p +

p2

p2 2C1

)2 dy − ε

zx

∫

y

∫ γ

γ

)2 − εC0 (C2 C6 + zx )

dq

(

C5 + C6 q +

zx

∫ γ

q2 2C2

dq

(

2C1

C5 + C6 q +

q2 2C2

⎟ )2 dy⎠ )2 .

Since

⎛ C0 C1 ∂W −C0 (C1 C4 + zt ) ∂ ⎝ = ( )2 = ∂ z ∂ zt zt2 t C3 + C4 z t + C3 + C4 zt + 2C 1

⎞ C0 C2

+ε

zt2 2C1

C5 + C6 zx +

zx2 2C2

⎠

and

⎛ ∂W C0 C1 C0 (C2 C6 + zx ) ∂ ⎝ = −ε ( )2 = 2 ∂ zx ∂ z z x C3 + C4 zt + C5 + C6 zx + 2Cx 2

⎞ zt2 2C1

+ε

C0 C2 C5 + C6 zx +

zx2 2C2

⎠,

then due to the choice (4.2) of the constant C W (zt , zx ) =

C0 C1 C3 + C4 zt +

zt2 2C1

+ε

C0 C2 C5 + C6 zx +

zx2 2C2

.

The cases (a)–(d) are obvious. Thus the theorem is proved. Acknowledgment The reported study was funded by RFBR and JSPS, project number 19-51-50005. References [1] [2] [3] [4] [5] [6] [7] [8] [9]

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