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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

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