ñòð. 96 |

A simple method of finding solutions

of the nonlinear d’Alembert equation

P. BASARAB-HORWATH, W.I. FUSHCHYCH, M. SEROV

We consider the nonlinear d’Alembert equation

2u = F (u), (1)

where u = u(x) and x = (x0 , x1 , . . . , xn ) ? Rn ,

?2 ?2 ?2

2= ? 2 ? ··· ? 2

?x2 ?x1 ?xn

0

and F (u) is an arbitrary differentiable function. In equation (1) we make the local

change of variable

(2)

u = ?(w),

where w(x) is a new unknown function and ? is a function to be determined later.

On making this change, (1) becomes

? ?

?2w + ?wµ wµ = F (?), (3)

where

2 2 2

d? dw dw dw

? ? ? ··· ?

µ

?= , wµ w = .

dw dx0 dx1 dxn

Equation (3) is equivalent to the following equation

? ?

Pn ? Pn

? ? ?

? 2w ? ? + ?(wµ wµ ? ?) + ? ? + ? ? F (?) = 0, (4)

Pn Pn

where Pn (w) is an arbitrary polynomial of degree n in w, and ? = ?1, 0, 1. Choosing

? such that

?

? Pn

? (5)

? ?+? = F (?)

Pn

equation (4) becomes

?

Pn

? ?

? 2w ? ? + ?(wµ wµ ? ?) = 0. (6)

Pn

From this it is clear that a solution of the system

?

Pn

2w = ? wµ wµ = ? (7)

,

Pn

J. Phys. A: Math. Gen., 1992, 25, L871–L877.

408 P. Basarab-Horwath, W.I. Fushchych, M. Serov

is also a solution of (6), and in this way we obtain a solution of (1) provided ?

satisfies (5). There remains, of course, the problem of the existence of solutions

of (7). We have the following result.

Theorem 1. For n = 3 the system

2w = H(w), wµ wµ = ?

with ? = ?1, 0, 1 is compatible if and only if

?N

H(w) = ,

w+C

where N = 0, 1, 2, 3 and C is an arbitrary constant.

This result follows from theorem 2 of [2]. In theorem 1 of [3], it is further shown

that if the system in theorem 1 above is compatible, then it is necessarily of the type

given in equation (7). Moreover, as is mentioned in [3], the system (7) is always

compatible (for any n) if H(w) is as in theorem 1 above. Having discussed the

question of compatibility, we now turn to equation (5), which gives us the appropriate

choice of ?. We do this for several cases of the function F (u). Note that for the

remainder of this paper, we take n = 3.

Case 1. F (u) = uk with k = 1. If Pn = wm with m = 0, 1, 2, 3 then (5) becomes

? m?

? ? + ? = ?k . (8)

w

Assuming ? to be of the form

?(w) = ?w?

with ?, ? constants, we obtain

2/(1?k)

(1 ? k)w

(9)

?(w) =

(2?(1 + k + m ? km))1/2

as a solution of (8).

Case 2. F (u) = exp u. Again using Pn (w) = wm , m = 0, 1, 2, 3, equation (5)

becomes

? m? (10)

? ? + ? = exp ?.

w

and we seek solutions ? with the help of me ansatz

exp(?(w)) = ?w?

with ?, ? constants. We obtain

2?(m ? 1)

(11)

?(w) = log , m = 2, 3.

w2

? ?

Case 3. F (u) = ??(u)/?3 (u). Here we take ? to be an arbitrary differentiable

?

function such that ? = 0. If we take Pn (w) = ?0 = const, then (5) becomes

?

?(?)

?

?? = ? (12)

?

?3 (?)

A simple method of finding solutions of the nonlinear d’Alembert equation 409

and this gives us

v d?

(13)

? = w + c2 ,

?

(c1 + ??2 (?))1/2

where c1 , c2 are two constants of integration. On choosing these two constants of

integration to be zero, we obtain

v

w = ??(?)

as a solution of (12), and the change of variable (2) is then given by

v

(14)

w = ??(u).

In this case of F , we have replaced ? by another function ?; we now took at some

cases of ?.

Case 3(a). F (u) = ?1 sin u, where ?1 = const. On setting

?

?(u)

?1 sin u = ?

?

?3 (u)

we obtain

du

(15)

?(u) = + c2 .

(c1 ? 2?1 cos u)1/2

For c1 = 2?2 , c2 = 0, ?1 > 0 we find

1

?(u) = v log tan(u/4)

?1

and for c1 = ?2?1 , c2 = 0, ?1 < 0 one obtains

1

tanh?1 tan(u/4).

?(u) = v

??1

On putting ? = |?1 | in (14), the change of variable then takes on the form

v

4 arctan exp(w/ ?1 ) for ?1 > 0,

v

u=

4 arctan tanh(w/ ??1 ) for ?1 < 0.

Case 3(b). F (u) = sinh u. Integrating the equation

?

?(u)

sinh u = ?

?

?3 (u)

we obtain

du

(16)

?(u) = + c2 .

(c1 + 2 cosh u)1/2

For c1 = 2, c2 = 0 one finds

?(u) = 2 arctan tanh(u/4)

410 P. Basarab-Horwath, W.I. Fushchych, M. Serov

and for c1 = ?2, c2 = 0 one gets

?(u) = log tanh(u/4).

Then (14) gives, with ? > 0

v

u = 4 tanh?1 tan(w/ ?),

v

u = 4 tanh?1 exp(w/ ?).

Case 3(c). F (u) = sin u/ cos3 u. In this case, the equation

?

sin u ?(u)

=?

?

cos3 u ?3 (u)

yields

du

(17)

?(u) = + c2 .

(c1 + 2 tan2 u)1/2

Again, we choose values for the integration constants. For c1 = 1, c2 = 0 we find

?(u) = sin u

and for c1 = c2 = 0 we obtain

?(u) = log sin u.

Using (14), with ? > 0 the change of variable (2) is given by the equations

v

u = arcsin(w/ ?)

and

v

u = arcsin exp(w/ ?).

We present our results in table 1.

Table 1. Summary of results obtained in cases 1–3(c).

Solution of (1) System (7)

F (u)

2/(1?k)

(1?k)w

v

k

u ,k=1 u=

2?(1+k+m?km)

1 + k + m ? km = 0, m = 0, 1, 2, 3 2w = m?/w

2?(m?1)

wµ wµ = ?

, m = 2, 3

exp u u = log w2

??(u)/?3 (u)

? ? w

?(u) = v? , ? > 0

v

?1 sin u u = 4 arctan exp(w/v ?1 ), ?1 > 0

?? 2w = 0

u = 4 arctan tanh(w/ v 1 ), ?1 < 0

?1

wµ wµ = ?

sinh u u = 4 tanh tan(w/ v2?), ? > 0

u = 4 tanh?1 exp(w/ 2?), ? > 0

v

sin u/ cos3 u u = 4 arcsin(w/ ?), ? > 0

v

u = 4 arcsin exp(w/ ?), ? > 0

A simple method of finding solutions of the nonlinear d’Alembert equation 411

We now present some results from [1] concerning the general solutions of the

system

m?

2w = wµ wµ = ?, (18)

, n = 3.

w

Theorem 2. The general solution of the system (18) for m = 3, ? = 1 is given by

w2 = [xµ + Aµ (? )][xµ + Aµ (? )],

where the function ? (x) is defined implicitly by the equation

[xµ + Aµ (? )]B µ (? ) = 0

and Aµ , Bµ are arbitrary differentiable functions of one variable satisfying the

conditions

?

Aµ B µ = 0, Bµ B µ = 0.

Theorem 3. The general solution of the system (18) for m = 0, ? = ?1 is given by

w = Aµ (? )xµ + f1 (? ), (19)

where the function ? = ? (x) is implicitly defined by the equation

ñòð. 96 |