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?

(10)

vµ v µ = .

f?2 (v)

The system (9), (10) is obviously compatible since it is the local transformation of

an already compatible system. However, it should be noted that this does not mean

that the exact solutions we obtain by using (9), (10) are equivalent to those obtained

from (8), since we have introduced some extra freedom via the function f .

We now equate the right-hand sides of (6), (7) with the right-hand sides of (9),

(10), respectively. A little algebraic manipulation gives us

1/2

f?(v)

(11)

g(v) = ? ,

f N (v)

where ? is an arbitrary non-zero constant. Thus we have a differential relation between

f and g which we can integrate. For N = 1 we obtain

v

1

g 2 (?)d? (12)

f (v) = C exp

?2

and g has to satisfy the integro-differential equation

v

C2 6 1

g? ? 2g ? g ?

2 2

g 2 (?)d? F (g) = 0. (13)

g ? g exp

?? 4 ?2

For N = 1 we find

1/(1?N )

1?N v

2

(14)

f (v) = g (?)d? + C ,

?2

C being an arbitrary real constant, and with the following condition on g:

2N/(1?N )

1?N v

16

g? ? 2g ? g ?

2 2 2

(15)

g ? g g (?)d? + C F (g) = 0.

?? 4 ?2

Solutions of nonlinear d’Alembert equations 329

Our result is summarized in the following:

Result 2. (i) The function

?(x) = g(v(x)) exp[iv(x)]

is a solution of (1) whenever g is a solution of (13) and w(x) = f (v(x)) is a solution

of

?

2w = , wµ wµ = ?

w

with f given by (12).

(ii) The function

?(x) = g(v(x)) exp[iv(x)]

is a solution of (1) whenever g is a solution of (15) and w(x) = f (v(x)) is a solution

of

?N

2w = , wµ wµ = ?

w

with f given by (14) for N = 1.

One may treat (13) and (15) in two ways: consider F as given, and then attempt

to solve for g, or make an assumption about g and then find the corresponding F .

We take this second approach, and in doing so, we determine the function f which

appears in (12) and (14), which also relates (4) to the system (6), (7).

This is illustrated in the following example, where we take g as g(v) = v ? . Then

we obtain after some elementary manipulation

2

w = f (v) = Cv 1/?

when N = 1, ? = ? 1 In this case we find the corresponding nonlinear version of (1)

2

and an exact solution:

?? 4 1 4/?2 2 2

2? + 2 |?| ? |?|4(1?? )/? ? = 0,

C 4

?? 2 /2 ?2

1 1

?(x) = w(x) exp i w(x) ,

C C

where w is a solution of

?

2w = , wµ wµ = ?.

w

The solutions of this system are given in table 1. We can choose the nonlinearity

in the above wave equation by choosing ?. For instance, for ? 2 = 2 we obtain the

3

equation

2

2 ? 1

2? ? |?|2 ? |?|6 ? = 0. (16)

C2

3 4

Equation (16) is of the type considered by Grundland and Tuczynski [12].

Table 1. Solutions for the system 2n w = ?N/w, (?n w)2 = ?. Inner products are with respect to the Minkowski metric.

330

? N Solutions w Conditions on aµ ? Rn , bµ ? Rn

1/2

±1 N ? {1, 2, . . . , n ? 1} (a0 · x)2 ± (a1 · x)2 ± · · · ± (aN · x)2 a0 · a0 = 1, a0 · aj = 1, aj · ak = ±?jk ,

j, k = 1, 2, . . . , N

1/2

1 N ? {1, 2, . . . , n ? 1} aµ · aµ = ?1, aµ · a? = 0,,

(a0 · x)2 ± (a1 · x)2 ± · · · ± (aN · x)2

?, µ = 0, 1, . . . , N

?

2

k

v

1 N ? {1, 2, . . . , n ? 3} +

? b1 · x + h1 ka0 · x + a0 · a0 = 1

ai · x

i=1

2

k

v

+ + ···

b2 · x + h2 ka0 · x + ai · x ai · ai = ?1

i=1

? 1/2

2

k

v

?

··· + ka0 · x + ai · x bj · bj = ?1

bN +1 · x + hN +1

i=1

h1 , h2 . . . , hN +1 are arbitrary real functions a0 · ai = 0, a0 · bj = 0, ai · bj = 0, bj · bl = 0

i = 1, . . . , k, (k ? n ? 1), j = l = 1, . . . , N + 1

k

v

±1 N =0 +

b1 · x cos h1 ka0 · x + ai · x a0 · a0 = ±1, ai · ai = ?1, bj · bj = ?1

i=1

k k

v v

+ b2 · x sin h1 ka0 · x + + h2 ka0 · x +

ai · x ai · x a0 · ai = 0, a0 · bj = 0, ai · bj = 0, bj · bl = 0

i=1 i=1

i = 1, . . . , k, (k ? n ? 1), j = l = 1, 2

h1 and h2 are arbitrary real functions

k

v

±1 N =0 +

b1 · x cos h1 ka0 · x + ai · x a0 · a0 = 1, ai · ai = ?1

i=1

k k

v v

+ b2 · x sin h1 ka0 · x + + h2 ka0 · x + b1 · b1 = ±1, b2 · b2 = ?1

ai · x ai · x

i=1 i=1

h1 and h2 are arbitrary real functions a0 · ai = 0, a0 · bj = 0, ai · bj = 0, bj · bl = 0

i = 1, . . . , k, (k ? n ? 1), j = l = 1, 2

k

v

0 N =0 h1 ka0 · x + ai · x a0 · a0 = 1, ai · ai = ?1, a0 · ai = 0

i=1

i = 1, . . . , k (k ? n ? 1)

h1 is an arbitrary real function

P. Basarab-Horwath, N. Euler, M. Euler, W.I. Fushchych

Solutions of nonlinear d’Alembert equations 331

For N = 2, ? = ?1 we obtain the following wave equation and exact solution:

1 1

2? + |?|2 + |?|6 ? = 0, (17)

?8 4

??

1

?(x) = (C ? w(x))? 2 exp i (18)

,

(C ? w(x))? 2

where w is a solution of the compatible system

2?

2w = , wµ wµ = ?

w

and exact solutions of this system are given in table 1. Equation (17) is also of a type

considered by Grundland and Tuczynski [12]. Our exact solutions are new.

2.2. ? = g(u). We now assume that the phase is a function of the amplitude:

v = g(u). On substituting this in equations (2), (3), we obtain

(u2 g + 2ug)F (u)

? ?

2u = ? F1 (u), (19)

u? + 2g + u2 g 3

g ? ?

?

?u2 F (u)

? F2 (u). (20)

uµ uµ =

u? + 2g + u2 g 3

g ? ?

Here g = dg/du.

?

We perform the same analysis as before. First, letting F1 (u) = ?N/u, F2 (u) = ?,

? = 0, we find (after some computation)

?uN +1 ?N ??1

g(u) = ? ? 2(N +2) .

+ ?1 , F (u) =

u2

N +1 u

Having determined g and the nonlinearity of the wave equation (19), we have the

following:

Result 3. An exact solution of (1) with nonlinearity

F (|?|) = ?N |?|?2 ? ??1 |?|?2(N +2)

is given by

?i?

?(x) = Cu(x) exp ,

(N + 1)u(x)(N +1)

where ? = 0 and C = 0 is an arbitrary real constant, and where u(x) is a solution of

the system (4).

Another way of dealing with (19), (20) is to transform (4) locally using the

transformation w = f (u) with f? = 0, which gives us

?

?N ?f (u)

2u = ? (21)

,

f (u)f?(u) f?3 (u)

?

(22)

uµ uµ = .

f?2 (u)

332 P. Basarab-Horwath, N. Euler, M. Euler, W.I. Fushchych

Then, equating the right-hand sides of (21), (22) with the right-hand sides of (19),

(20), we find (for ? = 0, as before) that

f?(u)

u2 g(u) = ?

? ,

f N (u)

where ? = 0 is an arbitrary real constant. Again we see that there are two cases to

consider: N = 1 and N = 1.

For N = 1 we obtain

u

1 dg(?)

?2 (23)

f (u) = C exp d?

? d?

with C an arbitrary real constant. The condition on g is

u

u4 C 2 3 1 dg(?)

u? + 2g + u2 g 3 + ?2 (24)

g ? ? g exp

? d? F (u) = 0.

?? 2 ? d?

When N = 1, f is given by

1/(1?N )

N ?1 u

2 dg(?)

C? (25)

f (u) = ? d?

? d?

with C an arbitrary real constant and with the following condition on g:

2N/(1?N )

N ?1 u

u4 2 dg(?)

C?

23

u? + 2g + u g +

g ? ? ? d? F (u) = 0.

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