Solution of Functional Differential Equations by Differentiation

S.4.3. Solution of Functional Differential Equations by Differentiation

S.4.3-1. Description of the method. Below we describe a procedure for constructing solutions to functional-differential equations. It

involves three successive stages.

results in a similar equation but with fewer terms: Φ e 1 ( Ψ 1 ( X) e Y)+e Φ 2 ( Ψ 2 ( X) e Y)+···+e Φ k−1 ( Ψ k−1 X) e ( Y ) = 0,

X) = Φ j ( X), Ψ e j ( Y ) = [Ψ j ( Y )/Ψ k ( Y )] ′ y .

We continue the above procedure until we obtain a separable two-term equation

(23) Three cases must be considered.

Φ b 1 ( Ψ 1 ( X) b Y)+b Φ 2 ( Ψ 2 ( X) b Y ) = 0.

0. Then equation (23) is equivalent to the ordinary differential equations

Nondegenerate case :|b Φ 1 ( X)| + |b Φ 2 ( X)| ✓

0 and | b Ψ 1 ( Y )| + | b Ψ 2 ( Y )| ✓

X) + C b Φ 2 (

X) = 0, Cb Ψ 1 ( Y)−b Ψ 2 ( Y ) = 0,

where

C is an arbitrary constant. The equations b Φ 2 = 0 and b Ψ 1 = 0 correspond to the limit case

C = ∞. Two degenerate cases :

X) ≡ 0, Φ b 2 (

X) ≡ 0 =⇒ b Ψ 1,2 ( Y ) are any; Ψ b 1 ( Y ) ≡ 0, Ψ b 2 ( Y)≡0 =⇒ b Φ 1,2 (

X) are any.

2 ◦ . The solutions of the two-term equation (23) should be substituted into the original functional- differential equation (21) to “remove” redundant constants of integration [these arise because equa- tion (23) is obtained from (21) by differentiation].

3 ◦ . The case Ψ k ≡ 0 should be treated separately (since we divided the equation by Ψ k at the first stage). Likewise, we have to study all other cases where the functionals by which the intermediate functional-differential equations were divided vanish.

Remark 1. The functional-differential equation (21) can happen to have no solutions. Remark 2. At each subsequent stage, the number of terms in the functional-differential equation

can be reduced by differentiation with respect to either y or x. For example, we can assume at the first stage that Φ k ✓

0. On dividing equation (21) by Φ k and differentiating with respect to x, we again obtain a similar equation that has fewer terms.

S.4.3-2. Examples of constructing exact generalized separable solutions. Below we consider specific examples illustrating the application of the above method to constructing

exact generalized separable solutions of nonlinear equations.

Example 7. Let us consider the nth-order nonlinear equation

∂y ∂x∂y

∂x ∂y 2 ∂y n

where f (x) is an arbitrary function. In the special case n = 3 and f (x) = const, it coincides with the equation of a steady boundary layer on a flat plate for the stream function (see Schlichting, 1981, and Loitsyanskiy, 1996). We look for generalized separable solutions to equation (24) in the form

w(x, y) = ϕ(x)ψ(y) + χ(x).

On substituting (25) into (24) and cancelling by ϕ, we arrive at the functional-differential equation

′ [( ′ ) ϕ 2 x ψ y − ψψ yy ′′ ]− χ ′ x ψ yy ′′ =

f (x)ψ ( y n) .

We divide (26) by f = f (x) and then differentiate with respect to x to obtain

( ϕ ′ x /f ) ′ x [( ψ y ′ ) 2 − ψψ yy ′′ ]−( χ ′ x /f ) x ′ ψ yy ′′ = 0.

( χ ′ x /f ) ′ x = C 1 ( ϕ ′ x /f ) ′ x , ( ψ ′ y ) 2 − ψψ ′′ yy − C 1 ψ yy ′′ = 0.

Integrating yields

ψ(y) = C 4 e λy − C 1 ,

ϕ(x) is any, χ(x) = C 1 ϕ(x) + C 2 f (x) dx + C 3 ,

where C 1 , ...,C 4 , and λ are constants of integration. On substituting (28) into (26), we establish the relationship between constants to obtain C 2 =− λ n−2 . Ultimately, taking into account the aforesaid and formulas (25) and (28), we arrive at a solution of equation (24) of the form (25):

w(x, y) = ϕ(x)e λy − λ n−2

f (x) dx + C,

where ϕ(x) is an arbitrary function, C and λ are arbitrary constants (C = C 3 , C 4 = 1).

Degenerate case. It follows from (27) that

( ϕ ′ x /f ) ′ x = 0,

( χ x ′ /f ) ′ x = 0,

ψ(y) is any.

Integrating the first two equations in (29) twice yields

ϕ(x) = C 1 f (x) dx + C 2 ,

χ(x) = C 3 f (x) dx + C 4 ,

where C 1 , ...,C 4 are arbitrary constants. Substituting (25) into (26) and taking into account (30), we arrive at an ordinary differential equation for ψ = ψ(y):

C 1 ( ψ y ′ ) 2 −( C 1 ψ+C 3 ) ψ ′′ yy = ψ ( n) y .

Formulas (25) and (30) together with equation (31) determine an exact solution of equation (24). Example 8. The two-dimensional stationary equations of motion of a viscous incompressible fluid are reduced to a

single fourth-order nonlinear equation for the stream function (see Loitsyanskiy, 1996):

(∆ w) = ν∆∆w,

We seek exact separable solutions of equation (32) in the form

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

Substituting (33) into (32) yields

g ′ y f xxx ′′′ − f x ′ g ′′′ yyy = νf xxxx ′′′′ + νg yyyy ′′′′ .

Differentiating (34) with respect to x and y, we obtain

g ′′ yy f xxxx ′′′′ − f xx ′′ g ′′′′ yyyy = 0.

Nondegenerate case. If f xx ′′ ✔

0 and g ′′ yy ✔

0, we separate the variables in (35) to obtain the ordinary differential equations

f xxxx ′′′′ = Cf xx ′′ ,

g ′′′′ yyyy = Cg yy ′′ ,

which have different solutions depending on the value of the integration constant C. 1 ◦ . Solutions of equations (36) and (37) for

g(y) = B 1 + B 2 y+B 3 y + B 4 3

where the A k and B k are arbitrary constants ( k = 1, 2, 3, 4). On substituting (38) into (34), we evaluate the integration constants. Three cases are possible:

A 4 = B 4 = 0, A n , B n are any numbers ( n = 1, 2, 3);

A k = 0,

B k are any numbers

( k = 1, 2, 3, 4);

B k = 0,

A k are any numbers

( k = 1, 2, 3, 4).

The first two sets of constants determine two simple solutions (33) of equation (32):

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

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

f (x) = A 1 + A 2 x+A 3 e λx + A 4 e − λx ,

g(y) = B 1 + B 2 y+B 3 −

e λy + B 4 e λy .

Substituting (39) into (34), dividing by λ 3 , and collecting terms, we obtain

νλ + B − 2 ) e λx + B 3 ( νλ + A 2 ) e λy + B 4 ( νλ − A 2 ) e λy = 0. Equating the coefficients of the exponentials to zero, we find

(The other constants are arbitrary.) These sets of constants determine three solutions (33) of equation (32):

w=C 1 e − λy + C 2 y+C 3 + νλx, w=C 1 e − λx + νλx + C 2 e − λy − νλy + C 3 , 1 w=C − e λx − νλx + C 2 e λy − νλy + C 3 ,

where C 1 , C 2 , C 3 , and λ are arbitrary constants. 3 ◦ . Solution of equations (36) and (37) for

C = −λ 2 < 0:

f (x) = A 1 + A 2 x+A 3 cos( λx) + A 4 sin( λx),

g(y) = B 1 +

B 2 y+B 3 cos( λy) + B 4 sin( λy).

Substituting (40) into (34) does not yield new real solutions.

Degenerate cases. If f xx ′′ ≡ 0 or g ′′ yy ≡ 0, equation (35) becomes an identity for any

g = g(y) or f = f (x), respectively. These cases should be treated separately from the nondegenerate case. For example, if f xx ′′ ≡ 0, we have

f (x) = Ax + B, where

A and B are arbitrary numbers. Substituting this f into (34), we arrive at the equation −Ag ′′′ yyy = νg ′′′′ yyyy . Its general solution is given by g(y) = C 1 exp(− Ay/ν) + C 2 y 2 + C 3 y+C 4 . Thus, we obtain another solution (33) of equation (32):

A = νλ, B = 0). Example 9. Consider the second-order nonlinear parabolic equation

We look for exact separable solutions of equation (41) in the form

w = ϕ(t) + ψ(t)θ(x).

Substituting (42) into (41) and collecting terms yields

ϕ 2 ′ t − c+ψ t ′ θ = aϕψθ xx ′′ + ψ aθθ ′′ xx + b(θ ′ 2 x ) .

On dividing this relation by ψ 2 and differentiating with respect to t and x, we obtain ( ψ t ′ /ψ 2 ) ′ t θ ′ x = a(ϕ/ψ) ′ t θ xxx ′′′ .

Separating the variables, we arrive at the ordinary differential equations

θ ′′′ xxx = Kθ ′ x ,

( ψ ′ t /ψ 2 ) ′ t = aK(ϕ/ψ) ′ t ,

where K is an arbitrary constant. The general solution of equation (44) is given by

A 1 sin( λx) + A 2 cos( λx) + A 3 if K = −λ 2

where A 1 , A 2 , and A 3 are arbitrary constants. Integrating (45) yields

ϕ(t) is any

t ′ , ψ(t) is any if K ≠ 0, aK ψ

ϕ = Bψ +

where B is an arbitrary constant. On substituting solutions (46) and (47) into (43), one can “remove” the redundant constants and define the functions ϕ and ψ. Below we summarize the results.

(corresponds to 2( a + b)

K = 0), where C 1 , C 2 , and C 3 are arbitrary constants.

( t+C 1 )+ C 2 ( t+C 1 ) a+2b −

2( a + 2b)(t + C 1 )

2 ◦ . Solution for b = −a:

w=

ψ(A − 1 e λx + A 2 e λx )

(corresponds to K=λ 2 > 0),

aλ ψ

where the function ψ = ψ(t) is determined from the autonomous ordinary differential equation

whose solution can be found in implicit form. In the special case A 1 = 0 or A 2 = 0, we have ψ=C 1 exp 1 2 2 2 acλ t + C 2 t . 3 ◦ . Solution for

b = −a: w=− 1 ψ ′ t

2 + ψ[A 1 sin( λx) + A 2 cos( λx)]

(corresponds to K = −λ 2 < 0),

aλ ψ

where the function ψ = ψ(t) is determined from the autonomous ordinary differential equation

2 2 4 2 2 tt 2 ′′ =− acλ + a λ ( A 1 + A 2 ) e Z ,

ψ=e Z ,

whose solution can be found in implicit form. Remark. The structure of solutions to equation (41) was obtained by Galaktionov (1995) by a different method (see

Subsection S.4.6, Example 14).

References for Subsection S.4.3: A. D. Polyanin (2002, Supplement B), A. D. Polyanin and V. F. Zaitsev (2002).