Equations Involving Third-Order Mixed Derivatives

9.5.2. Equations Involving Third-Order Mixed Derivatives

1. 2 = ae λw .

∂x ∂y

1 ◦ . Suppose w(x, y) is a solution of this equation. Then the function

w 1 = wC 1 x+C 2 , C 3 y+C 4 + ln( C 2 1 C 3 ),

λ where C 1 , ...,C 4 are arbitrary constants, is also a solution of the equation.

3 1 Z dy

w(x, y) = − ln z,

z = f (y)x − aλf (y)

, ( y)

where f (y) is an arbitrary function. ∂w

BBM equation (Benjamin–Bona–Mahony equation). It describes long waves in dispersive systems.

1 ◦ . Suppose w(x, t) is a solution of the equation in question. Then the functions

w 1 = C 1 w( t x+C 2 , t C 1 t+C 3 ),

where C 1 , C 2 , and C 3 are arbitrary constants, are also solutions of the equation (either plus or minus signs are taken).

2 ◦ . Traveling-wave solution:

x − at

w(x, t) = −a + ℘

2 3 aβ p

a, C 1 , C 2 , and C 3 are arbitrary constants. See also equation 9.5.2.3, Item 2 ◦ with a = −1, b = β, and k = 1.

where ℘(z, C 2 , C 3 ) is the Weierstrass elliptic function ℘ ′ z = 4 ℘ 3 − C 2 ℘−C 3 ;

3 ◦ . Multiplicative separable solution:

w(x, t) = u(x)/t,

where the function u = u(x) is determined by the autonomous ordinary differential equation βu ′′ xx − uu ′ x − u = 0. Its solution can be written out in parametric form

w(x, t) = U (ξ)/t,

ξ = x − a ln |t|,

where the function U = U (ξ) is determined by the autonomous ordinary differential equation β(aU ξξξ ′′′ + U ξξ ′′ )−( U + a)U ξ ′ − U = 0.

5 ◦ . Conservation laws for β = 1:

✉✂✈ where D x = ∂ ∂x and D t = ∂ ∂t .

References : D. N. Peregrine (1966), T. B. Benjamin, J. L. Bona, and J. J. Mahony (1972), P. O. Olver (1979), N. H. Ibragimov (1994).

1 ◦ . Suppose w(x, t) is a solution of the equation in question. Then the functions

w 1 = C 1 w( t x+C 2 , t C k 1 t+C 3 ),

where C 1 , C 2 , and C 3 are arbitrary constants, are also solutions of the equation (either plus or minus signs are taken).

2 ◦ . Traveling-wave solution (soliton):

1 C /k

w(x, t) = cosh −2 √ ( x−C 1 t+C 2 ) ,

1 ( k + 1)(k + 2)

✉✂✈ where C 1 and C 2 are arbitrary constants.

Private communications : W. E. Schiesser (2003), S. Hamdi, W. H. Enright, W. E. Schiesser, and J. J. Gottlieb (2003).

3 ◦ . There is a multiplicative separable solution of the form w(x, t) = t −1 /k θ(x).

∂x∂t 2

1 ◦ . Suppose w(x, t) is a solution of the equation in question. Then the functions

w 1 = C 2 w( ✇ C − 1 k 1 x+C 2 , ✇ C 1 k t+C 3 ),

where C 1 , C 2 , and C 3 are arbitrary constants, are also solutions of the equation (either plus or minus signs are taken).

2 ◦ . Traveling-wave solution (soliton):

k + 1)(k + 2) /k w(x, t) =

①✂② where C 1 and C 2 are arbitrary constants.

Private communication : W. E. Schiesser (2003).

3 ◦ . There is a self-similar solution of the form w(x, t) = x 2 /k U (z), where z = xt.

4 ◦ . Generalized separable solution for k = 1:

x+C 2 2 ab

w(x, t) =

This equation is encountered at the interface between projective geometry and gravitational theory.

1 ◦ . Suppose w(x, t) is a solution of the equation in question. Then the function

w 1 = C 1 wC 2 x+C 2 kϕ(t), C 1 C 2 t+C 3 + ϕ ′ t ( t),

where C 1 , C 2 , and C 3 are arbitrary constants and ϕ(t) is an arbitrary function, is also a solution of the equation.

2 ◦ . Degenerate solution:

w(x, t) = Cx 2 + ϕ(t)x + ψ(t),

where ϕ(t) and ψ(t) are arbitrary functions and C is an arbitrary constant.

3 ◦ . Self-similar solution:

w(x, t) = t − α−1 U (z),

z=t α x,

where α is an arbitrary constant and the function U (z) is determined by the ordinary differential equation ( α − 1)U zz ′′ + αzU zzz ′′′ = kU U zzz ′′′ .

4 ◦ . Multiplicative separable solution:

w(x, t) = (Akt + B) −1 u(x),

A and B are arbitrary constants, and the function u(x) is determined by the autonomous ordinary differential equation uu ′′′ xxx + Au ′′ xx = 0.

where

5 ◦ . There is a first integral:

where ψ(t) is an arbitrary function. For ψ = 0, the substitution u = kw leads to an equation of the

form 7.1.1.2 with a=− 1 ①✂② 2 .

References : V. S. Dryuma (2000), M. V. Pavlov (2001).

6. 2 = f (t)w

+ g(x, t).

∂x ∂t

∂x 3

There is a first integral:

∂ 2 w 2 ∂ 2 w 1 ∂w

= f (t)w

f (t)

g(x, t) dx + ϕ(t),

where ϕ(t) is an arbitrary function. ∂w ∂ 3 w

∂x ∂x∂y 2

1 ◦ . Suppose w(x, y) is a solution of this equation. Then the function

w 1 = C 1 w(C 2 x+C 3 , C 4 y+C 5 )+ C 6 , where C 1 , ...,C 6 are arbitrary constants, is also a solution of the equation.

2 ◦ . Solutions: w(x, y) = axy + f (x) + g(y);

aλ Z

w(x, y) =

f (x) + g(y) − ln

exp f (x) dx +

exp g(y) dy

w(x, y) = ϕ(z),

z = ax + by;

w(x, y) = ψ(ξ),

ξ = xy;

where f = f (y), g = g(y), ϕ(z), and ψ(ξ) are arbitrary functions; a, b, and λ are arbitrary constants.

3 ◦ . There are exact solutions of the following forms:

w(x, y) = |x| a F (r), r = y|x| b ;

w(x, y) = e ax G(η),

η = bx + cy;

w(x, y) = e ax H(ζ),

ζ = ye bx ;

w(x, y) = |x| a U (ρ),

ρ = y + b ln |x|;

w(x, y) = V (r) + a ln |x|,

r = y|x| b ;

w(x, y) = W (ρ) + a ln |x|,

ρ = y + b ln |x|;

a, b, and c are arbitrary constants. Another set of solutions can be obtained by swapping x and y in the above formulas.

where

4 ◦ . The left-hand side of the original equation represents the Jacobian of w and v = w xy . The fact that the Jacobian of two quantities is zero means that these are functionally dependent, i.e., v can be treated as a function of w:

where Φ( w) is an arbitrary function. Any solution of the second-order equation (1) with arbitrary Φ( w) will be a solution of the original equation.

1 ◦ . Suppose w(x, y) is a solution of this equation. Then the functions

w 1 = C 1 w(C 2 x+C 3 , C 4 y+C 5 )+ C 6 , w 2 = w x + ϕ(y), y ,

where C 1 , ...,C 6 are arbitrary constants and ϕ(y) is an arbitrary function, are also solutions of the equation.

w(x, y) = Ax 2 + f x + g; w(x, y) = f exp(Ax) + g exp(−Ax); w(x, y) = f sin(Ax) + g cos(Ax);

w(x, y) = A ln ( x+f) 2 + B; w(x, y) = A ln sin 2 ( f x + g) + B; w(x, y) = A ln 2 sinh ( f x + g) + B; w(x, y) = A ln cosh 2 ( f x + g) + B;

w(x, y) = ϕ(z),

z = Ax + By;

where f = f (y), g = g(y), and ϕ(z) are arbitrary functions; A and B are arbitrary constants.

3 ◦ . The left-hand side of the original equation represents the Jacobian of w and v = w xx . The fact that the Jacobian of two quantities is zero means that these are functionally dependent, i.e., v can be treated as a function of w:

where ϕ(w) is an arbitrary function. Any solution of the second-order equation (1) with arbitrary ϕ(w) will be a solution of the original equation.

Integrating (1) yields the general solution of the original equation in implicit form:

f (y) + 2 ϕ(w) dw

dw = g(y) ③ x,

where f = f (y), g = g(y), and ϕ(w) are arbitrary functions. ∂w ∂ 3 w

1 ◦ . Suppose w(x, y) is a solution of this equation. Then the functions

w 1 = C 1 −2 w(C 1 x+C 2 , y) + C 3 , w 2 = w x + ϕ(y), y ,

where C 1 , C 2 , and C 3 are arbitrary constants and ϕ(y) is an arbitrary function, are also solutions of the equation.

2 ◦ . Generalized separable solution quadratic in x:

w(x, y) = − x 2 f (y) dy + C + xϕ(y) + ψ(y),

where ϕ(y) and ψ(y) are arbitrary functions and C is an arbitrary constant.

3 ◦ . Additive separable solutions:

w(x, y) = C 1 e kx +

2 e kx + C 3 +

f (y) dy,

f (y) dy, where C 1 , C 2 , C 3 , and k are arbitrary constants.

w(x, y) = C 1 cos( kx) + C 2 sin( kx) + C 3 −

4 ◦ . The original equation can be rewritten as the relation where the Jacobian of two functions, Z w and v=w xx +

f (y) dy, is equal to zero. It follows that w and v are functionally dependent, i.e., v can be treated as a function of w:

∂ 2 w ∂x 2

(1) where ϕ(w) is an arbitrary function. Any solution of the second-order equation (1) with arbitrary

f (y) dy = ϕ(w),

ϕ(w) will be a solution of the original equation.

parameter y. Integrating yields the general solution of (1) in implicit form:

dw = ψ 2 ( y) ④ x, where ψ 1 ( y), ψ 2 ( y), and ϕ(w) are arbitrary functions.

ψ 1 ( y) − 2w

f (y) dy + 2 ϕ(w) dw

+ g(x)

First integral:

= ϕ(w) +

g(x) dx −

f (y) dy,

∂x 2

where ϕ(w) is an arbitrary function. This equation can be treated as a second-order ordinary differential equation with independent variable x and parameter y.

∂y ∂x∂y General solution:

∂y ∂x ∂y

∂y ∂x∂y

∂y

∂x ∂y

w = f ϕ(x)y + ψ(x) ,

where ϕ(x), ψ(x), and f (z) are arbitrary functions. Remark. The equation in question can be represented as the equality of the Jacobian of two

functions, w and v, to zero:

∂x ∂x∂y 2 ∂y ∂x 2 ∂y

Two forms of representation of the general solution:

w = f ϕ(x) + ψ(y) , w=¯ f¯ ϕ(x) ¯ ψ(y) ,

where ϕ(x), ψ(y), ¯ ϕ(x), ¯ ψ(y), f (z 1 ), and ¯ f(z 2 ) are arbitrary functions.

Remark. The equation in question can be represented as the equality of the Jacobian of two functions to zero:

w x v y − w y v x = 0,

where v=w xy /(w x w y ).

9.5.3. Equations Involving ∂ w

and ∂ w

∂x 3 ∂y 3

1. a + b =( ay 3 + bx 3 ) f (w).

∂x 3 ∂y 3

Solution:

w = w(z),

z = xy,

where the function w(w) is determined by the autonomous ordinary differential equation

w ′′′ zzz = f (w).

Remark. The above remains true if the constants a and b in the original equation are replaced by arbitrary functions a = a(x, y, w, w x , w y , . . . ) and b = b(x, y, w, w x , w y , . . . ).

1 ◦ . Suppose w(x, y) is a solution of this equation. Then the function

w 1 = w(C 1 x+C 2 , C 1 y+C 3 )+ C 4 , where C 1 , ...,C 4 are arbitrary constants, is also a solution of the equation.

2 ◦ . Additive separable solutions:

w(x, y) = cλx + C 1 e aλy + C 2 y+C 3 , w(x, y) = C 1 e λx + C 2 x + bλy + C 3 , w(x, y) = C 1 e − aλx cλ + x+C 2 e λy − abλy + C 3 ,

where C 1 , C 2 , C 3 , and λ are arbitrary constants.

3 ◦ . Solution:

w = u(z) + C 3 x,

z=C 1 x+C 2 y,

where C 1 , C 2 and C 3 are arbitrary constants. The function u(z) is determined by the second-order

autonomous ordinary differential equation ( C 4 is an arbitrary constant)

C 1 C 2 ( C 1 + aC 2 )( u ′ z ) 2 +2 aC 2 2 C 3 u ′ z = 2( bC 1 3 + cC 2 3 ) u ′′ zz + C 4 .

F (u) = (u 2 z ′ ) leads to a first-order linear equation.

To C 3 = 0 there corresponds a traveling-wave solution. In this case, the substitution

4 ◦ . There is a self-similar solution of the form w = w(y/x). ∂ 2 w ∂ 2 w

3. = a + b . ∂x 2 ∂y 2 ∂x 3 ∂y 3

1 ◦ . Suppose w(x, y) is a solution of this equation. Then the function w 1 = C −1 1 w(C 1 x+C 2 , C 1 y+C 3 )+ C 4 xy + C 5 x+C 6 y+C 7 ,

where C 1 , ...,C 7 are arbitrary constants, is also a solution of the equation.

2 ◦ . Traveling-wave solution:

aC 3

1 + bC 3 2

w(x, y) = −

2 2 z(ln |z| − 1),

z=C 1 x+C 2 y+C 3 .

3 ◦ . Additive separable solutions: w(x, y) = 1 2 2 bC 1 x + C 2 x+C 3 exp( C 1 y) + C 4 y+C 5 ,

w(x, y) = 1 2 aC 1 y 2 + C 2 y+C 3 exp( C 1 x) + C 4 x+C 5 ,

where C 1 , ...,C 5 are arbitrary constants.

where the function U (ζ) is determined by the autonomous ordinary differential equation ( 2 ′′ +2 )( 2 ′′ +2 )=( 3 C 3 1 U ζζ C 3 C 2 U ζζ C 4 aC 1 + bC 2 ) U ζζζ ′′′ , which can be integrated with the substitution

F (ζ) = U ζζ ′′ .

5 ◦ . There is a self-similar solution of the form w = xu(y/x).

Chapter 10

Fourth›Order Equations