Equations of the Form ∂ 2 w ∂ 2 w

7.2.1. Equations of the Form ∂ 2 w ∂ 2 w

F (x, y)

∂x 2 ∂y 2

◮ Suppose w(x, y) is a solution of the equation in question. Then the function

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

1. 2 = f (x)y 2 k .

∂x ∂y

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

1 k−2 w(x, C 1 y) + C 2 xy + C 3 x+C 4 y+C 5 ,

where C 1 , ...,C 5 are arbitrary constants, is also a solution of the equation.

2 ◦ . Generalized separable solutions:

Z x ( x − t)f (t)

w(x, y) = (C 1 x+C 2 ) y k+1 +

1 x+C 2 w(x, y) = (C x − t)f (t) ) y + dt + C 3 xy + C 4 x+C 5 y+C 6 ,

( k + 1)(k + 2) 0 ( C 1 t+C 2 ) where C 1 , ...,C 6 are arbitrary constants.

3 ◦ . Generalized separable solution:

k+2

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

where C 1 , ...,C 4 are arbitrary constants and the function ϕ = ϕ(x) is determined by the ordinary differential equation

k(k + 2)ϕϕ ′′ xx =4 f (x).

2. 2 2 = f (x)g(y).

∂x ∂y

1 ◦ . Additive separable solution:

w(x, y) = C 1 ( x − t)f (t) dt + C 2 x+

( y − τ )g(τ ) dτ + C 3 y+C 4 ,

0 C 1 0 where C 1 , ...,C 4 are arbitrary constants.

2 ◦ . Multiplicative separable solution:

w(x, y) = ϕ(x)ψ(y),

where the functions ϕ = ϕ(x) and ψ = ψ(y) are determined by the ordinary differential equations ( C 1 is an arbitrary constant)

ϕϕ ′′ xx = C 1 f (x),

ψψ yy ′′

= −1 C

1 g(y).

3. ∂x 2 ∂y 2

= f (ax + by).

Solutions:

1 p ( z − t)

w(x, y) = ✢

where C 1 , ...,C 4 are arbitrary constants.

= f (x)y + g(x)y + h(x)y .

4. 2 k

k–1

∂y Generalized separable solution:

∂x 2 2

w(x, y) = ϕ(x)y k+1 + ψ(x)y + χ(x),

where the functions ϕ = ϕ(x), ψ = ψ(x), and χ = χ(x) are determined by the system of ordinary differential equations

k(k + 1)ϕϕ ′′ xx = f (x), k(k + 1)ϕψ ′′ xx = g(x), k(k + 1)ϕχ ′′ xx = h(x).

5. ∂x 2 ∂y 2

= f (x)e λy .

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

w 1 = C 1 w x, y − ln | C 1 | + C 2 xy + C 3 x+C 4 y+C 5 ,

λ where C 1 , ...,C 5 are arbitrary constants, is also a solution of the equation.

2 ◦ . Generalized separable solution:

1 Z x ( x − t)f (t)

w(x, y) = (C 1 x+C 2 ) e λy + 2 dt + C 3 xy + C 4 x+C 5 y+C 6 ,

λ x 0 C 1 t+C 2

where C 1 , ...,C 6 are arbitrary constants, and x 0 is any number such that the integrand does not

have a singularity at x=x 0 .

3 ◦ . Generalized separable solution:

w(x, y) = ϕ(x)e λy/2 + C 1 xy + C 2 x+C 3 y+C 4 ,

where C 1 , ...,C 4 are arbitrary constants, and the function ϕ = ϕ(x) is determined by the ordinary

differential equation 2 λ ϕϕ ′′ xx =4 f (x).

6. = f (x)e 2 λy + g(x)e λy .

∂x 2 ∂y 2

Generalized separable solution:

w(x, y) = ϕ(x)e λy + 2 ( x − t)

g(t)

dt + C 1 xy + C 2 x+C 3 y+C 4 ,

λ x 0 ϕ(t)

where C 1 , ...,C 4 are arbitrary constants and the function ϕ = ϕ(x) is determined by the ordinary differential equation 2 λ ϕϕ ′′

xx = f (x).

∂ 2 7.2.2. Monge–Amp `ere equation 2 w

–∂ w ∂ 2 2 w 2 =

F (x, y)

∂x∂y

∂x ∂y

Preliminary remarks.

The Monge–Amp`ere equation is encountered in differential geometry, gas dynamics, and meteorol- ogy.

1 ◦ . Suppose w(x, y) is a solution of the Monge–Amp`ere equation. Then the function

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

where C 1 , C 2 , and C 3 are arbitrary constants, is also a solution of the equation.

2 ◦ . The transformation

2 x=a −2 ¯

1 x+b 1 y+c 1 , y=a ¯ 2 x+b 2 y+c 2 , w = kw + a ¯ 3 x+b 3 y+c 3 , F=k ¯ ( a 1 b 2 − a 2 b 1 ) F, where the a n , b n , c n , and k are arbitrary constants, takes the Monge–Amp`ere equation to an equation

of the same form.

3 ◦ . The transformation

x = x(1 + αx + βy) 4 ¯ ,¯ y = y(1 + αx + βy) , w = w(1 + αx + βy) ¯ , ¯ F = F (1 + αx + βy) , where α and β are arbitrary constants, takes the Monge–Amp`ere equation to an equation of the same

References : S. V. Khabirov (1990), N. H. Ibragimov (1994).

4 ◦ . In Lagrangian coordinates, the system of equations of the one-dimensional gas dynamics with plane waves is as follows:

where t is time, u the velocity, p the pressure, ξ the Lagrangian coordinate, and V the specific volume. The equation of state is assumed to have the form

V = V p, S(ξ) , where S = S(ξ) is a prescribed entropy profile. The Martin transformation

∂w

∂w

u(ξ, t) =

( x, y), t=

( x, y), x = ξ, y = p(ξ, t)

∂x

∂y

reduces the equations of the one-dimensional gas dynamics to the nonhomogeneous Monge–Amp `ere equation

F (x, y) = − ∂V ∂p p, S(ξ) .

References : M. N. Martin (1953), B. L. Rozhdestvenskii and N. N. Yanenko (1983).

Homogeneous Monge–Amp`ere equation .

1 ◦ . Suppose w(x, y) is a solution of the homogeneous Monge–Amp`ere equation. Then the functions w 1 = C 1 w(C 2 x+C 3 y+C 4 , C 5 x+C 6 y+C 7 )+ C 8 x+C 9 y+C 10 ,

w 2 = (1 + C 1 x+C 2 y)w

1+ C 1 x+C 2 y 1+ C 1 x+C 2 y

where C 1 , ...,C 10 are arbitrary constants, are also solutions of the equation.

where Φ 1 ( u, v) and Φ 2 ( u, z) are arbitrary functions of two arguments.

3 ◦ . General solution in parametric form:

w = tx + ϕ(t)y + ψ(t), x+ϕ ′ ( t)y + ψ ′ ( t) = 0,

where t is the parameter, and ϕ = ϕ(t) and ψ = ψ(t) are arbitrary functions.

4 ◦ . Solutions involving one arbitrary function:

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

w(x, y) = (C 1 x+C 2 y)ϕ

+ C 3 x+C 4 y+C 5 ,

1 2 3 C 4 x+C 5 y+C w(x, y) = (C 6 x+C y+C ) ϕ + C 7 x+C 8 y+C 9 ,

C 1 x+C 2 y+C 3

where C 1 , ...,C 9 are arbitrary constants and ϕ = ϕ(z) is an arbitrary function.

5 ◦ . Solutions involving arbitrary constants:

w(x, y) = C 1 y

C 2 2 + 2 xy + x + C 3 y+C 4 x+C 5 ,

w(x, y) =

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

w(x, y) = ✥

4 5 6 C x+C 8 y+C 9 ( , C x+C y+C )

y + b) 2 + C 5 x+C 6 y+C 7 , where the ✦✂✧

w(x, y) = ✥

x + a) 2 + C 2 ( x + a)(y + b) + C 3 (

a, b, and the C n are arbitrary constants.

References for equation 7.2.2.1: E. Goursat (1933), S. V. Khabirov (1990), N. H. Ibragimov (1994).

. First integrals for 2 A=a > 0:

where the Φ n ( u, v) are arbitrary functions of two arguments (n = 1, 2).

2 2 ◦ . General solution in parametric form for A=a > 0: β−λ

where β and λ are the parameters, ϕ = ϕ(β) and ψ = ψ(λ) are arbitrary functions.

w(x, y) = ★

x(C 1 x+C 2 y) + ϕ(C 1 x+C 2 y) + C 3 x+C 4 y,

w(x, y) = C 1 y + C 2 xy +

( C 2 − A)x + C 3 y+C 4 x+C 5 ,

w(x, y) =

w(x, y) = ★

( C 1 x−C 2 y + C 3 ) /2 + C 4 x+C 5 y+C 6 ,

where C 1 , ...,C 6 are arbitrary constants and ϕ = ϕ(z) is an arbitrary function. Another five solutions can be obtained:

(a) from the solution of equation 7.2.2.18 with α = 0 and f (u) = A, where β is an arbitrary constant; (b) from the solution of equation 7.2.2.20 with f (u) = A, where a, b, and c are arbitrary constants;

(c) from the solution of equation 7.2.2.21 with f (u) = A, where a, b, c, k, and s are arbitrary constants; (d) from the solution of equation 7.2.2.22 with α = 0 and f (u) = A, where β is an arbitrary constant; (e) from the solution of equation 7.2.2.24 with α = 0 and f (u) = A, where β is an arbitrary constant.

4 ◦ . The Legendre transformation

∂w

∂w

u = xξ + yη − w(x, y), ξ=

where u = u(ξ, η) is the new independent variable, and ξ and η are the new dependent variables, leads to an equation of the similar form

Reference : E. Goursat (1933).

= f (x).

∂x∂y

∂x 2 ∂y 2

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

w −1

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

where C 1 , ...,C 6 are arbitrary constants, are also solutions of the equation.

2 ◦ . Generalized separable solutions quadratic in y:

w(x, y) = C 1 y + C 2 xy +

( x − t)f (t) dt + C 3 y+C 4 x+C 5 ,

w(x, y) =

C 2 y + C 3 y+

( x − t)(t + C 1 ) f (t) dt + C 4 y+C 5 x+C 6 ,

x+C 1 4 C 2 2 C 2 0

where ✩✂✪ C 1 , ...,C 6 are arbitrary constants.

Reference : A. D. Polyanin and V. F. Zaitsev (2002).

3 ◦ . Generalized separable solutions for f (x) > 0:

Zp

w(x, y) = ★ y

f (x) dx + ϕ(x) + C 1 y,

where ✩✂✪ ϕ(x) is an arbitrary function.

References : M. N. Martin (1953), B. L. Rozhdestvenskii and N. N. Yanenko (1983).

D 2 x y(w x w yy − w y w xy + g y )− g−w x w y + D y y(w y w xx − w x w xy + g x )+( w x ) = 0,

where D x =

f (x) dx + ϕ(x) + ψ(y), and ϕ(x) and ψ(y) are arbitrary ∂x

Reference : S. V. Khabirov (1990).

5 ◦ . Let us consider some specific functions f = f (x). Solutions that can be obtained by the formulas of Items 1 ◦ and 2 ◦ are omitted.

5.1. Solutions with

f (x) = Ax k can be obtained:

(a) from the solution of equation 7.2.2.18 with f (u) = A and α = k/2, where β is an arbitrary constant; (b) from the solution of equation 7.2.2.24 with

f (u) = A and α = k/2, where β is an arbitrary constant. 5.2. Solutions for

f (x) = Ae λx :

w(x, y) = ✭ 2

A e λx/2 1 2 3 4 5 C 6

2 sin( C x+C y+C )+ C x+C y+C λ ,

w(x, y) =

e λx/2 sinh( C 1 x+C 2 y+C 3 )+ C 4 x+C 5 y+C 6 ,

w(x, y) =

y+C 6

e λx/2 cosh( C 1 x+C 2 y+C 3 )+ C 4 x+C 5 .

Another solution can be obtained from the solution of equation 7.2.2.22 with α = λ and f (u) = A, where β is an arbitrary constant.

= f (x)y.

∂x∂y

∂x 2 ∂y 2

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

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

where C 1 , ...,C 4 are arbitrary constants, are also solutions of the equation.

2 ◦ . Generalized separable solution quadratic in y:

w(x, y) = C 2

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

3 ◦ . Generalized separable solution quadratic in y:

w(x, y) = ϕ(x)y 2 + ψ(x)y + χ(x),

where

1 f (x) dx ϕ(x) =

1 ϕ(x)

f (x) dx

, ψ(x) = C 3 ϕ(x) + C 4 +

C x+C

C ϕ(x)]

C 1 [ ϕ(x)]

x − t) t ( ( t)] dt + C 5 x+C 6 .

χ(x) =

2 a ϕ(t)

4 ◦ . Generalized separable solutions cubic in y:

1 3 w(x, y) = C 1 y −

( x − t)f (t) dt + C 2 x+C 3 y+C 4 ,

1 w(x, y) = 2 2 − ( x − t)(C t+C 2 ) f (t) dt + C 3 x+C 4 y+C 5 ,

( C 1 x+C 2 )

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

5 ◦ . See solution of equation 7.2.2.7 in Item 3 ✫✂✬ ◦ with k = 1.

Reference : A. D. Polyanin and V. F. Zaitsev (2002).

= f (x)y 2 .

∂x∂y

∂x 2 ∂y 2

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

w −2 1 = ✯ C

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

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

2 ◦ . Generalized separable solution quadratic in y:

w(x, y) = ϕ(x)y 3 + C

1 ϕ ( x) dx + C 2 y+ C 1 ( x − t)ϕ ( t) dt + C 3 x+C 4 ,

where the function ϕ = ϕ(x) is determined by the ordinary differential equation

xx = 2( ϕ x ′ ) − 2 f (x).

3 ◦ . Generalized separable solutions in the form of polynomials of degree 4 in y:

w(x, y) = C 1 y −

( x − t)f (t) dt + C 2 x+C 3 y+C 4 ,

12 C 1 a

w(x, y) = 3

( x − t)(C 1 t+C 2 )

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

✰✂✱ 4 ◦ . See solution of equation 7.2.2.7 in Item 3 ◦ with k = 2.

Reference : A. D. Polyanin and V. F. Zaitsev (2002).

= f (x)y 2 + g(x)y + h(x).

∂x∂y

∂x 2 ∂y 2

Generalized separable solution quadratic in y:

w(x, y) = ϕ(x)y 2 + ψ(x)y + χ(x),

where the functions ϕ = ϕ(x), ψ = ψ(x), and χ = χ(x) are determined by the system of ordinary differential equations

xx = 2( ϕ x ) − 2 f (x),

xx ′′ =2 ϕ ′ x ψ ′ x − 2 g(x),

xx = 2 ( ψ ′ 2 x 1 ) − 2 h(x).

= f (x)y k .

∂x∂y

∂x 2 ∂y 2

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

1 k−2 w(x, C 1 y) + C 2 x+C 3 y+C 4 ,

where C 1 , ...,C 4 are arbitrary constants, are also solutions of the equation.

2 ◦ . Generalized separable solutions:

C 1 y k+2

w(x, y) =

( x − t)f (t) dt + C 2 x+C 3 y+C 4 ,

( k + 1)(k + 2) C 1 a

y k+2

w(x, y) =

( x − t)(C 1 t+C 2 ) k+1 f (t) dt + C 3 x+C 4 y+C 5 ,

( C 1 x+C 2 ) k+1

( k + 1)(k + 2) a

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

k+2

w(x, y) = ϕ(x)y 2 ,

where the function ϕ = ϕ(x) is determined by the ordinary differential equation

2 k(k + 2)ϕϕ 2 ′′

xx −( k + 2) ( ϕ ′ x ) +4 f (x) = 0.

4 ◦ . Let us consider the case where f is a power-law function of x, f (x) = Ax n , in more detail. Solutions:

C 1 x n+2

Ay k+2

w(x, y) =

+ C 2 y+C 3 x+C 4 ,

( n + 1)(n + 2) C 1 ( k + 1)(k + 2)

C 1 x n+2

Ay k+n+3

w(x, y) =

+ C 2 y+C 3 x+C 4 , ( n + 1)(n + 2)y

n+1 −

C 1 ( k + n + 2)(k + n + 3)

C 1 y k+2

Ax k+n+3

w(x, y) = ( 2 k + 1)(k + 2)x k+1 − +

C y+C x+C

( k + n + 2)(k + n + 3)

w(x, y) = (C − 1 x+C 2 ) k−1

y k+2 −

( x − t)t n ( C 1 t+C 2 ) k+1 dt + C 3 y+C 4 x,

( k + 1)(k + 2) a

w(x, y) = (C 1 y+C 2 ) n−1 x n+2 −

( y − t)t k ( C 1 t+C 2 ) n+1 dt + C 3 y+C 4 x,

( n + 1)(n + 2) a where C 1 , ...,C 4 are arbitrary constants.

There are also a multiplicative separable solution, see Item 3 ◦ with f (x) = Ax n , and a solution of the same type:

n+2

w(x, y) = ψ(y)x 2 ,

where the function ψ = ψ(y) is determined by the ordinary differential equation

yy −( n + 2) ( ψ y ) +4 Ay = 0.

n(n + 2)ψψ 2 ′′ ′ 2 k

The substitution − ψ=U n/2 brings it to the Emden–Fowler equation

yy ′′ = n 2 y U ( , n + 2)

n+1

whose exact solutions for various values of k and n can be found in the books by Polyanin and Zaitsev (1995, 2003).

Another exact solution for f (x) = Ax n can be obtained from the solution of equation 7.2.2.18 with f (u) = Au k ✲✂✳ and n = 2α + kβ, where α and β are arbitrary constants.

Reference : A. D. Polyanin and V. F. Zaitsev (2002).

= f (x)y 2 k+2 + g(x)y k .

∂x∂y

∂x 2 ∂y 2

Generalized separable solution:

k+2

g(t)

w(x, y) = ϕ(x)y

( x − t)

dt + C 1 x+C 2 y+C 3 ,

( k + 1)(k + 2) a ϕ(t)

where the function ϕ = ϕ(x) is determined by the ordinary differential equation

( 2 k + 1)(k + 2)ϕϕ ′′

xx −( k + 2) ( ϕ ′ 2 x ) + f (x) = 0.

Reference : A. D. Polyanin and V. F. Zaitsev (2002).

2 2 = f (x)e ∂x∂y . ∂x ∂y

λy

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

w 1 = ✴ C 1 w x, y −

ln | C 1 | + C 2 x+C 3 y+C 4 ,

λ where C 1 , ...,C 4 are arbitrary constants, are also solutions of the equation.

2 ◦ . Generalized separable solutions:

w(x, y) = C 1 ( x − t)f (t) dt + C 2 x−

e λy + C 3 y+C 4 ,

w(x, y) = C 1 e − βx+λy − 2 ( x − t)e βt f (t) dt + C 2 x+C 3 y+C 4 ,

C 1 λ a where C 1 , ...,C 4 and β are arbitrary constants.

3 ◦ . Multiplicative separable solution:

w(x, y) = ϕ(x) exp 1

2 λy ,

where the function ϕ = ϕ(x) is determined by the ordinary differential equation

Reference : A. D. Polyanin and V. F. Zaitsev (2002).

= f (x)e 2 2 λy 2 + g(x)e λy .

∂x∂y

∂x ∂y

Generalized separable solution:

w(x, y) = ϕ(x)e λy

λ a ϕ(t)

where the function ϕ = ϕ(x) is determined by the ordinary differential equation

Reference : A. D. Polyanin and V. F. Zaitsev (2002).

= f (x)g(y) + A 2

∂x∂y

∂x 2 ∂y 2

Generalized separable solutions:

w(x, y) = C 1 ( x − t)f (t) dt −

( y − ξ)g(ξ) dξ ✴ Axy + C 2 x+C 3 y+C 4 ,

a C 1 b where C 1 , ...,C 4 are arbitrary constants; a and b are any numbers.

= f (ax + by).

w(x, y) = ✴

f (z) dz + ϕ(z) + C 1 x+C 2 y,

z = ax + by,

b where C 1 and C 2 are arbitrary constants and ϕ(z) is an arbitrary function.

2 ◦ . The transformation

w = U (x, z), z = ax + by

leads to an equation of the form 7.2.2.3:

= −2 + b

f (z).

Here, x and z play the role of y and x in 7.2.2.3, respectively.

=x k

f (ax + by).

The transformation w = U (x, z), z = ax + by

leads to an equation of the form 7.2.2.7:

2 2 + b x k f (z).

∂x∂z

∂x ∂z

Here, x and z play the role of y and x in 7.2.2.7, respectively.

=x 2 2 k+2 2 f (ax + by) + x k g(ax + by). ∂x∂y

∂x ∂y

The transformation w = U (x, z), z = ax + by

leads to an equation of the form 7.2.2.8:

2 2 + b x k+2 f (z) + b x k ∂x∂z g(z). ∂x ∂z Here, x and z play the role of y and x in 7.2.2.8, respectively.

f (ax + by).

The transformation w = U (x, z), z = ax + by

leads to an equation of the form 7.2.2.9:

= −2 + b e λx f (z).

∂x∂z

∂x 2 ∂z 2

Here, x and z play the role of y and x in 7.2.2.9, respectively.

=e 2 2 λx 2 f (ax + by) + e λx g(ax + by). ∂x∂y

∂x ∂y

The transformation w = U (x, z), z = ax + by

leads to an equation of the form 7.2.2.10:

g(z). Here, x and z play the role of y and x in 7.2.2.10, respectively.

This is a special case of equation 7.2.2.18 with α = −2 and β = −1.

1 ◦ . Suppose w(x, y) is a solution of the equation in question. Then the functions w 1 = ✷ w(C 1 x, C 1 y) + C 2 x+C 3 y+C 4 , where C 1 , ...,C 4 are arbitrary constants, are also solutions of the equation.

x where

∂x

∂y

C is an arbitrary constant.

References : M. N. Martin (1953), B. L. Rozhdestvenskii and N. N. Yanenko (1983).

where ϕ(z) is an arbitrary function.

4 ◦ . Conservation law:

∂x ∂x∂y

where D x =

Reference : S. V. Khabirov (1990).

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

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

where C 1 , ...,C 4 are arbitrary constants, are also solutions of the equation.

2 ◦ . Self-similar solution:

w(x, y) = x α−β+1 u(z) z = x β y,

where the function u = u(z) is determined by the ordinary differential equation

2 [ 2 β(β + 1)zu ′ z +( α − β)(β − α − 1)u]u ′′ zz +( α + 1) ( u ′ z ) − f (z) = 0.

Reference : S. V. Khabirov (1990).

1 ◦ . Suppose w(x, y) is a solution of this equation. Then the functions w 1 = w(x + 2bC 1 y + abC 1 , y + aC 1 )+ C 2 x+C 3 y+C 4 ,

where C 1 , ...,C 4 are arbitrary constants, are also solutions of the equation.

2 ◦ . Solutions:

Zp

w(x, y) = 2 F (z) + C

1 dz + C 2 x+C 3 y+C 4 ,

F (z) = 2 f (z) dz, z = ax − by ,

a b where C 1 , ...,C 4 are arbitrary constants.

Reference : A. D. Polyanin and V. F. Zaitsev (2002).

Solution for 2 b ≠4 ac:

2 w(x, y) = u(z) z = ax 2 + bxy + cy ,

where the function u = u(z) is determined by the ordinary differential equation

z u ′′ zz + (4 ac − b )( u ′ 2 z ) + f (z) = 0. Integrating yields

u(z) = ✻

dz + C 1 ,

F (z) = 2 f (z) dz + C 2 ,

b −4 ac where C 1 and C ✼✂✽ 2 are arbitrary constants.

Reference : A. D. Polyanin and V. F. Zaitsev (2002).

2 w(x, y) = u(z), 2 z = ax + bxy + cy + kx + sy,

where the function u = u(z) is determined by the ordinary differential equation

2 2 2 2 2 2 (4 ac − b ) z + as + ck − bks u ′ z u ′′ zz + (4 ac − b )( u ′ z ) + f (z) = 0. ′ The substitution 2

V (z) = (u z ) leads to a first-order linear equation.

Reference : A. D. Polyanin and V. F. Zaitsev (2002).

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

1 w(x − 2 ln C 1 , C 1 y) + C 2 x+C 3 y+C 4 , where C 1 , ...,C 4 are arbitrary constants, are also solutions of the equation.

2 ◦ . Generalized self-similar solution:

2 α − β, where the function U = U (z) is determined by the ordinary differential equation

w(x, y) = e µx U (z), z=e βx

=e ky/x

f (x).

w(x, y) = exp

ϕ(x),

where the function ϕ = ϕ(x) is determined by the ordinary differential equation

2 x 2 ϕϕ

xx − x ( ϕ ′ 2 −2 x 4 ) +2 xϕϕ ′ x − ϕ +4 k x f (x) = 0.

w(x, y) = x α+2 u(z), z=x β e y/x ,

where the function u = u(z) is determined by the ordinary differential equation

2 z 2 βzu ′ +( α + 2)(α + 1)u u ′′

zz

+ z β − (α + 1) zu

z +( α + 2)(α + 1)u u z + f (z) = 0.

Reference : S. V. Khabirov (1990).

2 2 =y exp(2αy –1 )f (xy + βy ∂x∂y ). ∂x ∂y Solution:

w = y exp(αy −2 ) ϕ(z) + C

1 y+C 2 x+C 3 , z = xy + βy , where C 1 , C 2 , and C 3 are arbitrary constants, and the function ϕ = ϕ(z) is determined by the ordinary

differential equation

Reference : S. V. Khabirov (1990).

◮ For exact solutions of the nonhomogeneous Monge–Amp`ere equation for some specific

F = F (x, y) (without functional arbitrariness), see Khabirov (1990) and Ibragimov (1994). The Cauchy problem for the Monge–Amp`ere equation is discussed in Courant and Hilbert (1989).

–∂ w ∂ 2 7.2.3. Equations of the Form w

= F x, y, w, ∂w ∂x ∂y

, ∂w

∂x∂y

∂x 2

∂y 2

2 2 ∂x∂y = f (x)w + g(x). ∂x ∂y

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

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

2 ◦ . Generalized separable solution quadratic in y:

w = ϕ(x)y 2 + ψ(x)y + χ(x), where ϕ(x), ψ(x), and χ(x) are determined by the system of ordinary differential equations

xx + f (x)ϕ − 4(ϕ x ′ ) = 0,

2 ϕψ xx ′′ + f (x)ψ − 4ϕ ′ x ψ ′ x = 0,

xx + f (x)χ + g(x) − (ψ x ′ ) = 0.

Note that the second equation is linear in ψ and has a particular solution ψ = ϕ (hence, its general solution can be expressed via the particular solution of the first equation).

= f (x)w 2

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

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

where C 1 , C 2 , and C 3 are arbitrary constants, is also a solution of the equation.

w(x, y) = e λy u(x),

where λ is an arbitrary constant and the function u = u(x) is determined by the ordinary differential equation

f (x)u = 0.

= f (x)y n w k .

∂x∂y

∂x 2 ∂y 2

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

w 1 = C n+2 w(x, C k−2 y),

where

C is an arbitrary constant, is also a solution of the equation.

2 ◦ . Multiplicative separable solution with n ≠ −2 and k ≠ 2:

n+2

w(x, y) = y 2− k U (x),

where the function U (x) is determined by the ordinary differential equation

2 2 ( 2 n + 2)(n + k)U U

xx −( n + 2) ( U x ′ ) +( k − 2) f (x)U k = 0.

= f (x)e w .

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

C is an arbitrary constant, is also a solution of the equation.

2 ◦ . Multiplicative separable solution with k ≠ 2 and λ ≠ 0:

λy

w(x, y) = exp

U (x),

2− k

where the function U (x) is determined by the ordinary differential equation

′ 2 2 UU −2 xx −( U x ) +( k − 2) λ f (x)U k = 0. ∂ 2 w 2 ∂ 2 w ∂ 2 w

= f (w).

∂x∂y

∂x 2 ∂y 2

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

w 1 = w(A 1 x+B 1 y+C 1 , A 2 x+B 2 y+C 2 ),

| A 2 B 1 − A 1 B 2 | = 1, where C 1 , C 2 , and any three of the four constants A 1 , A 2 , B 1 , and B 2 are arbitrary, is also a solution

of the equation.

2 ◦ . Functional separable solution:

2 w(x, y) = u(z), 2 z = ax + bxy + cy + kx + sy,

where

a, b, c, k, and s are arbitrary constants and the function u = u(z) is determined by the ordinary differential equation

2 2 2 2 2 2 (4 ac − b ) z + as + ck − bks u z ′ u ′′ zz + (4 ac − b )( u ′ z ) + f (u) = 0.

w(x, y) = exp

u(x),

2− k

where the function u = u(x) is determined by the ordinary differential equation

x 2 uu ′′

xx −( xu ′

x − u) + λ x f (x)u k = 0.

This equation is used in meteorology for describing wind fields in near-equatorial regions.

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

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

2 ◦ . Solutions:

w = ϕ(x),

1 √ w= 2

a x + C) y + ϕ(x),

where ϕ(x) is an arbitrary function and C is an arbitrary constant.

4 aC 2 ( x+C 1 ) tanh( C 2 y+C 3 ),

w= 2

4 aC 2 ( x+C 1 ) coth( C 2 y+C 3 ),

w= 2

4 aC 2 ( x+C 1 ) tan( C 2 y+C 3 ),

where ϕ(x) is an arbitrary function and C 1 , C 2 , C 3 , and λ are arbitrary constants. The first solution is a solution in additive separable form and the other four are multiplicative separable solutions. ❁✂❂

Reference : E. R. Rozendorn (1984).

4 ◦ . Generalized separable solution quadratic in y:

w = F (x)y 2 + G(x)y + H(x),

where

F (x) =

, G(x) = −

H(x) =

( x − t) t

− aG(t)

dt + C 5 x+C 6 ,

2 0 F (t)

and C 1 , ...,C 6 are arbitrary constants.

5 ◦ . Generalized separable solution:

a w=C 1 exp( C 2 x+C 2 3 y) − x + C 4 x+C 5 .

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

w(x, y) = |x| k+2

U (z), z = y|x| − k ;

w(x, y) = e − kx V (ξ), ξ = ye kx ; w(x, y) = x 2 W (η),

η = y + k ln |x|;

where k is an arbitrary constant.

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

w −1

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

where C 1 , ...,C 5 are arbitrary constants, is also a solution of the equation.

2 ◦ . Solutions:

w = ϕ(x),

w=

f (x) dx + C y + ϕ(x),

w=C 1 exp( C 2 x+C 3 y) −

( x − t)f (t) dt + C 4 x+C 5 ,

where ϕ(x) is an arbitrary function and C is an arbitrary constant. For C 2 = 0, the last solution is an additive separable solution.

3 ◦ . Generalized separable solution quadratic in y:

w = ϕ(x)y 2 + ψ(x)y + χ(x),

where

C 3 ϕ(x) =

ψ 2 ′ ( t)] − f (t)ψ(t)

and C 1 , ...,C 6 are arbitrary constants. ∂ 2 w 2

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

where C 1 , ...,C 6 are arbitrary constants, are also solutions of the equation.

2 ◦ . Generalized separable solution linear in x:

w(x, y) = ϕ(y)x + ψ(y),

where ψ(y) is an arbitrary function and the function ϕ(y) is defined implicitly by

C is an arbitrary constant.

3 ◦ . Additive separable solution:

w(x, y) = C 2

1 y + C 2 y+C 3 + z(x),

where the function z(x) is determined by the autonomous ordinary differential equation 2C 1 z xx ′′ +

f (z ′ x ) = 0. Its general solution can be written out in parametric form as

dt

t dt

x = −2C 1

z = −2C 1

f (t)

f (t)

∂w

∂w

u = xξ + yη − w(x, y), ξ=

where u = u(ξ, η) is the new dependent variable and ξ and η are the new independent variables, leads to an equation of the form 7.2.2.3:

The Legendre transformation

∂w

∂w

u = xξ + yη − w(x, y), ξ=

where u = u(ξ, η) is the new dependent variable and ξ and η are the new independent variables, leads to the simpler equation

For exact solutions of this equation, see Subsection 7.2.2.

7.2.4. Equations of the Form

= f (x, y) ∂ w ∂ w

∂x 2 + ∂y g(x, y) 2

∂x∂y

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

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

2 ◦ . Degenerate solutions involving arbitrary functions:

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

where C 1 and C 2 are arbitrary constants and ϕ = ϕ(z) is an arbitrary function.

3 ◦ . Generalized separable solution quadratic in y:

[ ϕ ′ ( t)]

w(x, y) = ϕ(x)y +[ C 1 ϕ(x) + C 2 ] y+

( x − t) t

dt + C 3 x+C 4 ,

2 0 f (t)ϕ(t)

where the function ϕ = ϕ(x) is determined by the ordinary differential equation

f (x)ϕϕ ′′ xx − 2( ϕ ′ 2 x ) = 0.

4 ◦ . Generalized separable solution involving an arbitrary power of y:

w(x, y) = ϕ(x)y k + C 1 x+C 2 y+C 3

where the function ϕ = ϕ(x) is determined by the ordinary differential equation

( 2 k − 1)f (x)ϕϕ

xx ′′ − k(ϕ ′ x ) = 0.

5 ◦ . Generalized separable solution involving an exponential of y:

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

where C 1 , C 2 , C 3 , and λ are arbitrary constants and the function ϕ = ϕ(x) is determined by the ordinary differential equation

f (x)ϕϕ 2

xx ′′ −( ϕ ′ x ) = 0.

2. = f (x)

+ g(x).

∂x∂y

∂x 2 ∂y 2

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

w 1 = ❄ −1 C 1 w(x, C 1 y+C 2 )+ C 3 x+C 4 y+C 5 , where C 1 , ...,C 5 are arbitrary constants, are also solutions of the equation.

2 ◦ . Generalized separable solution linear in y:

Zp

w(x, y) = ❄ y

g(x) dx + ϕ(x) + C 1 y,

where ϕ(x) is an arbitrary function.

3 ◦ . Generalized separable solution quadratic in y:

w(x, y) = ϕ(x)y +[ C 1 ϕ(x) + C 1 2 ] y+ ( x − t) t

C [ ϕ ′ ( t)] − g(t)

dt + C 3 x+C 4 ,

2 0 f (t)ϕ(t)

where C 1 , ...,C 4 are arbitrary constants and the function ϕ = ϕ(x) is determined by the ordinary differential equation

f (x)ϕϕ ′′ xx − 2( ϕ ′ 2 x ) = 0, which has a particular solution ϕ=C 6 .

3. = f (x)

+ g(x)y.

∂x∂y

∂x 2 ∂y 2

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

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

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

2 ◦ . Generalized separable solution cubic in y:

g(t)

w(x, y) = C 1 y + C 2 y−

( x − t)

dt + C 3 x+C 4 ,

6 C 1 a f (t) where C 1 , ...,C 4 are arbitrary constants.

A more general solution is given by

g(t) dt

w(x, y) = ϕ(x)y + C 1 y−

( x − t)

+ C 2 x+C 3 ,

6 a f (t)ϕ(t)

where the function ϕ = ϕ(x) is determined by the ordinary differential equation

2 f (x)ϕϕ ′′ xx − 3( ϕ ′ 2 x ) = 0.

3 ◦ . For an exact solution quadratic in y, see equation 7.2.4.5 with g 2 = g 0 = 0.

4 ◦ . See the solution of equation 7.2.4.6 in Item 3 ◦ with k = 1.

4. = f (x)

+ g(x)y .

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

w −2

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

where C 1 , ...,C 4 are arbitrary constants, are also solutions of the equation.

1 4 2 w(x, y) = C g(t) y + C y− ( x − t) dt + C 3 x+C 4 ,

12 C 1 a f (t) where C 1 , ...,C 4 are arbitrary constants.

A more general solution is given by

g(t) dt

w(x, y) = ϕ(x)y + C 1 y−

( x − t)

+ C 2 x+C 3 ,

12 a f (t)ϕ(t)

where the function ϕ = ϕ(x) is determined by the ordinary differential equation

3 f (x)ϕϕ ′′ xx − 4( ϕ ′ 2 x ) = 0.

3 ◦ . For an exact solution quadratic in y, see equation 7.2.4.5 with g 1 = g 0 = 0.

4 ◦ . See the solution of equation 7.2.4.6 in Item 3 ◦ with k = 2.

5. = f (x)

+g 2 (x)y +g 1 (x)y + g 0 (x).

∂x∂y

∂x 2 ∂y 2

Generalized separable solution quadratic in y:

w(x, y) = ϕ(x)y 2 + ψ(x)y + χ(x),

where the functions ϕ = ϕ(x), ψ = ψ(x), and χ = χ(x) are determined by the system of ordinary differential equations

2 f (x)ϕϕ 1 ′′ = 2( ϕ ′

xx

x ) − 2 g 2 ( x),

f (x)ϕψ ′′ xx =2 ϕ ′ x ψ x ′ 1 − 2 g 1 ( x),

f (x)ϕχ ′′

xx = 2 ( ψ ′ x ) − 2 g 0 ( x).

6. = f (x)

+ g(x)y k .

∂x∂y

∂x 2 ∂y 2

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

1 k−2 w(x, C 1 y) + C 2 x+C 3 y+C 4 ,

where C 1 , ...,C 4 are arbitrary constants, are also solutions of the equation.

2 ◦ . Additive separable solution:

1 k+2

g(t)

w(x, y) =

( k + 1)(k + 2)

C 1 a f (t)

where C 1 , ...,C 4 are arbitrary constants.

3 ◦ . Multiplicative separable solution:

k+2

w(x, y) = ϕ(x)y 2 ,

where the function ϕ = ϕ(x) is determined by the ordinary differential equation

2 k(k + 2)f (x)ϕϕ 2 ′′

xx −( k + 2) ( ϕ x ′ ) +4 g(x) = 0.

4 ◦ . Generalized separable solution:

w(x, y) = ψ(x)y k+2 −

( k + 1)(k + 2) a f (t)ψ(t)

where the function ψ = ψ(x) is determined by the ordinary differential equation

( 2 k + 1)f (x)ψψ ′′

xx −( k + 2)(ψ x ′ ) = 0.

2 2 + g(x)y 2 k+2 + h(x)y k ∂x∂y . ∂x ∂y Generalized separable solution:

7. = f (x)

w(x, y) = ϕ(x)y k+2

h(t)

( x − t)

dt + C 1 x+C 2 y+C 3 , ( k + 1)(k + 2) a f (t)ϕ(t)

where the function ϕ = ϕ(x) is determined by the ordinary differential equation

x ) + g(x) = 0. ∂ 2 2 w

( 2 k + 1)(k + 2)f (x)ϕϕ ′′ −( k + 2) ( ϕ ′ 2

xx

8. = f (x)

+ g(x)e λy .

∂x∂y

∂x 2 ∂y 2

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

w 1 = C 1 w x, y − ln | C 1 | + C 2 x+C 3 y+C 4 ,

where C 1 , ...,C 4 are arbitrary constants, are also solutions of the equation.

2 ◦ . Additive separable solution:

g(t) w(x, y) = C

where C 1 , ...,C 4 are arbitrary constants.

3 ◦ . Multiplicative separable solution:

w(x, y) = ϕ(x) exp 1

2 λy ,

where the function ϕ = ϕ(x) is determined by the ordinary differential equation

2 f (x)ϕϕ −2 ′′

xx −( ϕ x ′ ) +4 λ g(x) = 0.

4 ◦ . Generalized separable solution:

w(x, y) = ψ(x)e λy

a f (t)ψ(t)

where the function ψ = ψ(x) is determined by the ordinary differential equation

f (x)ψψ 2

xx ′′ −( ψ x ′ ) = 0.

9. 2 = f (x) λy

+ g(x)e

+ h(x)e .

Generalized separable solution:

w(x, y) = ϕ(x)e λy

1 h(t)

− 2 ( x − t)

dt + C 1 x+C 2 y+C 3 ,

λ a f (t)ϕ(t) where the function ϕ = ϕ(x) is determined by the ordinary differential equation

f (x)ϕϕ ′′ xx −( ϕ ′ 2 x −2 ) + λ g(x) = 0.

2 2 +f 2 (x)g 2 ∂x∂y (y). ∂x ∂y

10. =f 1 (x)g 1 (y)

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

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

where C 1 , C 2 , and C 3 are arbitrary constants, is also a solution of the equation.

2 ◦ . Additive separable solution for f 1 g 1 ≠ 0:

f 2 ( t)

g 2 ( ξ)

w(x, y) = C 1 ( x − t)

where C 1 , ...,C 4 are arbitrary constants.

3 ◦ . Degenerate solutions for f 2 g 2 = 0: w(x, y) = ϕ(x) + C 1 y+C 2 ,

w(x, y) = ϕ(y) + C 1 x+C 2 ,

where C 1 and C 2 are arbitrary constants and ϕ = ϕ(z) is an arbitrary function.

4 ◦ . Generalized separable solution for f 2 g 2 = 0: w(x, y) = ϕ(x)ψ(y) + C 1 x+C 2 y+C 3 ,

where the functions ϕ = ϕ(x) and ψ = ψ(y) are determined by the ordinary differential equations

f 2 1 ( x)ϕϕ ′′

xx − C 4 ( ϕ ′ x ) = 0,

4 g 1 ( y)ψψ ′′ yy −( ψ y ′ ) = 0.

11. = f (ax + by)

+ g(ax + by).

2 w(x, y) = ϕ(z) + C 2

1 x + C 2 xy + C 3 y + C 4 x+C 5 y, z = ax + by, where C 1 , ...,C 5 are arbitrary constants and the function ϕ(z) is determined by the ordinary

differential equation

2 2 2 ( 2 abϕ zz + C ) = f (z)(a ϕ ′′ zz +2 C 1 )( b ϕ ′′ zz +2 C 3 )+ g(z), which is easy to integrate; to this end, the equation should first be solved for ϕ zz ′′ .