Second and Higher Order Differential Constraints

S.8.3. Second and Higher Order Differential Constraints

Constructing exact solutions of nonlinear partial differential equations with the help of second- and higher-order differential constraints requires finding exact solutions of these differential constraints. The latter is generally rather difficult or even impossible. For this reason, one employs some special differential constraints that involve derivatives with respect to only one variable. In practice, one considers second-order ordinary differential equations in, say, x and the other variable, t, is involved implicitly or is regarded as a parameter, so that integration constants depend on t.

The problem of compatibility of a second-order evolution equation

with a similar differential constraint

may be reduced to a problem with the first-order differential constraint considered in Subsec- tion S.8.2-1. To that end, one should first eliminate the second derivative w xx from the equations. Then, the resulting first-order equation is examined together with the original equation (or the original differential constraint).

Example 5. From the class of nonlinear heat equations with a source

one singles out equations that admit invariant manifolds of the form

The functions f 2 ( w), f 1 ( w), g 2 ( w), and g 1 ( w) are to be determined in the further analysis.

Eliminating the second derivative from (49) and (50), we obtain

∂w 2

∂w

ϕ(w)

+ ψ(w),

∂t

∂x

ϕ(w) = f 1 ( w)g 1 ( w) + f 1 ′ ( w),

ψ(w) = f 1 ( w)g 2 ( w) + f 2 ( w).

The condition of invariance of the manifold (50) under equation (49) is obtained by differentiating (50) with respect to t:

w 2 xxt =2 g 1 w x w xt + g

The derivatives w xxt , w xt , and w t should be eliminated from this relation with the help of equations (50) and (51) and those obtained by their differentiation. As a result, we get

(2 ϕg 2 1 +3 ϕ ′ g 1 + ϕg ′ 1 + ϕ ′′ ) w 4 x + (4 ϕg 1 g 2 +5 ϕ ′ g 2 + ϕg 2 ′ − g 1 ψ ′ − ψg ′ 1 + ψ ′′ ) w 2 x +2 ϕg 2 2 + ψ ′ g 2 − ψg ′ 2 = 0. Equating the coefficients of like powers of w x to zero, one obtains three equations, which, for convenience, may be written in the form

( ϕ ′ + ϕg 1 ) ′ +2 g 1 ( ϕ ′ + ϕg 1 ) = 0, 4 g 2 ( ϕ ′ + ϕg 1 )+( ϕg 2 − ψg 1 ) ′ + ψ ′′ = 0,

2 ( ψ/g 2 ) ′ . The first equation can be satisfied by taking ϕ ′ + ϕg 1 = 0. The corresponding particular solution of system (53) has the form

µ| µ ′ where µ = µ(w) is an arbitrary function. Taking into account (52), we find the functional coefficients of the original equation (49) and the invariant set (50):

w 1 f 2 =( µ−f 1 ) g 2 , g 1 =− ′ , g 2 = 2 C 1 + √ ′ . 55 |

µ| µ Equation (50), together with (55), admits the first integral

where σ(t) is an arbitrary function. Let us eliminate w 2 x from (51) by means of (56) and substitute the functions ϕ and ψ from (54) to obtain the equation

2 µ p w t =− C | µ| − σ t ′ ( t).

Let us dwell on the special case C 2 = C 3 = 0. Integrating equation (57) and taking into account that µ t = µ ′ w t yields

µ = −σ(t) + θ(x),

where θ(x) is an arbitrary function. Substituting (58) into (56) and taking into account the relation µ x = µ ′ w x , we obtain

θ 2 x −4 C 1 θ = 2σ t −4 C 1 σ.

Equating both sides of this equation to zero and integrating the resulting ordinary differential equations, we find the functions on the right-hand side of (58):

σ(t) = A exp(2C 1 t),

θ(x) = C 1 ( x + B) 2 ,

where A and B are arbitrary constants. Thus, an exact solution of equation (49) with the functions f 1 and f 2 from (55) can

be represented in implicit form for C 2 = C 3 = 0 as follows: µ(w) = −A exp(2C 2 1 t) + C 1 ( x + B) .

In the solution and the determining relations (55), the function µ(w) can be chosen arbitrary. Example 6. Consider the problem of finding nonlinear second-order equations

admitting invariant manifolds of the form

The compatibility analysis of these equations leads us to the following relations for the determining functions:

f 2 ( w) is an arbitrary function, f 1 ( w) = C 1 w+C 2 − (3 C 1 C 3 w+C 4 ) f 2 ( w),

2 3 w) = (−C 2

1 C 3 w − C 1 C 4 w + C 5 w+C 6 )[1 − C 3 f 2 ( w)],

g 1 ( w) = 3C 1 C 3 w+C 4 , g 0 ( w) = C 3 (− C 1 2 3 3 C 2 w − C 1 C 4 w + C 5 w+C 6 ),

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

Section S.8.4 contains examples of second- and third-order differential constraints that are essentially equivalent to most common structures of exact solutions. Note that third- or higher-order differential constraints are rarely used, since they lead to cum- bersome computations and rather complex equations (often, the original equations are simpler). ❈✂❉

References for Subsection S.8.3: A. F. Sidorov, V. P. Shapeev, and N. N. Yanenko (1984), V. A. Galaktionov (1994), V. K. Andreev, O. V. Kaptsov, V. V. Pukhnachov, and A. A. Rodionov (1999).

Second-order differential constraints corresponding to some classes of exact solutions representable in explicit form

No. Type of solution

Structure of solution

Differential constraints

1 Additive separable solution

w = ϕ(x)+ψ(y)

w xy =0

2 Multiplicative separable solution

w = ϕ(x)ψ(y)

ww xy − w x w y =0

3 Generalized separable solution

w = ϕ(x)y 2 + ψ(x)y+χ(x)

w yy − f (x) = 0 w − f (y)w

Generalized separable solution

yy

y =0

4 w = ϕ(x)ψ(y)+χ(x)

w xy − g(x)w y =0

5 Functional separable solution

w = f (z), z = ϕ(x)y+ψ(x)

w yy − g(w)w 2 y =0

6 Functional separable solution

w = f (z), z = ϕ(x)+ψ(y)

ww xy − g(w)w x w y =0

TABLE 21

Third-order differential constraints corresponding to some classes of exact solutions representable in explicit form

Type of solution

Structure of solution

Differential constraint

w yyy =0 Generalized separable

Generalized separable w = ϕ(x)y 2 + ψ(x)y+χ(x)

w y w xyy − w xy w yy =0 Functional separable

w = ϕ(x)ψ(y)+χ(x)

w y ( w x w yyy − w y w xyy )=2 w yy ( w x w yy − w y w xy ) Functional separable

w = f ϕ(x)y+ψ(x)

w = f ϕ(x)+ψ(y)

w x w y w xyy − w y w xxy = w xy ( w 2 w yy − w x 2 y w xx )