ñòð. 73 |

Here, 2 ? ?t ? ?x .

2 2

Thus, to separate variables in the linear differential equation (1) one has to con-

struct a general solution of system of nonlinear partial differential equations (15)–

(18). The same assertion holds true for a general linear differential equation, i.e., the

problem of SV is essentially nonlinear. This is the reason why, even for the classical

d’Alembert equation 24 u ? utt ? ?3 u = 0, there is no complete description of all

coordinate systems that provide its separability [6].

It is not difficult to become convinced of that from (18). Since the functions ?1 ,

?2 are real, we have

(?1t ? ?1x )(?2t ? ?2x ) = 0.

2 2 2 2

(19)

Differentiating equations (15), (16) with respect to ? and using (19), we get A1? =

A2? = 0.

Consequently, the relation B1? B2? = 0 holds. Differentiating with respect to ? we

have

B1? (?1t ? ?1x ) + B2? (?2t ? ?2x ) = 0

2 2 2 2

or B1? /B2? = ?(?2t ? ?2x )/(?1t ? ?1x ). Differentiation of the above equality with

2 2 2 2

respect to ? yields B1?? /B1? = B2?? /B2? . But the functions B1 = B1 (?1 ), B2 =

B2 (?2 ) are independent, whence it follows that there exists a function such that

Bi?? = K(?)Bi? , i = 1, 2.

Integrating the above differential equation with respect to ?, we get

Bi (?i ) = ?(?)fi (?i ) + gi (?i ), i = 1, 2,

where fi , gi are arbitrary smooth functions.

On redefining the parameter ? > ?(?), we have

(20)

Bi (?i ) = ?fi (?i ) + gi (?i ), i = 1, 2,

Substitution of (20) into (17) with a subsequent splitting with respect to ? yields

the following equations:

2A + A[g1 (?1 )(?1t ? ?1x ) + g2 (?2 )(?2t ? ?2x ) + V (x)A = 0,

2 2 2 2

(21)

f1 (?1 )(?1t ? ?1x ) + f2 (?2 )(?2t ? ?2x ) = 0.

2 2 2 2

(22)

Separation of variables in the two-dimensional wave equation with potential 299

Thus, system (15)–(18) is equivalent to the system of equations (15), (16), (20)–

(22). Before integrating it, we make a remark. It is evident that the structure of

ansatz (2) is not changed by the transformation

A > A = Ah1 (?1 )h2 (?2 ),

(23)

?i > ?i = Ri (?i ), i = 1, 2,

where hi , Ri are some smooth functions.

This is why solutions of the system under the study, connected by relations (23),

are considered as equivalent ones.

By a proper choice of the functions hi , we can put Ri , f1 = f2 = 1 and A1 = A2 =

0 in equations (15), (16), (22).

Consequently, the functions ?1 , ?2 satisfy equations of the form

?1t ?2t ? ?1x ?2x = 0, ?1t ? ?1x + ?2t ? ?2x = 0,

2 2 2 2

whence (?1 ± ?2 )2 ? (?1 ± ?2 )2 = 0. Integrating the above equations, we get

t x

?2 = f (?) ? g(?), (24)

?1 = f (?) + g(?),

where f, g ? C 2 (R1 , R1 ) are arbitrary functions, ? = (x + t)/2, ? = (x ? t)/2.

Substitution of (24) into equations (15), (16) with A1 = A2 = 0 yields the following

equations for a function A = A(t, x): (ln A)t = 0, (ln A)x = 0, whence A = 1.

At last, substituting the obtained results into equation (21), we have

df dg

V (x) = [g1 (f + g) ? g2 (f ? g)] (25)

.

d? d?

Thus, the problem of integration of the overdetermined system of nonlinear diffe-

rential equations (15)–(18) is reduced to the integration of the functional-differential

equation (25).

Let us sum up the obtained results. The general form of the solution of equation (1)

with separated variables is as follows:

u1 = ?1 (f (?) + g(?))?2 (f (?) ? g(?)); (26)

here, ?i are arbitrary solutions of equations (8) and the functions f (?), g(?), g1 (f +g),

g2 (f ? g), V (x) are determined by (25).

To integrate equation (25) we make the hodograph transformation

(27)

? = P (f ), ? = R(g),

? ?

where P ? 0, R ? 0.

After making transformation (27), we get

? ?

g1 (f + g) ? g2 (f ? g) = P (f )R(g)V (P + R). (28)

Evidently, equation (28) is equivalent to the equation

? ?

(?f ? ?g )[P (f )R(g)V (P + R)] = 0

2 2

or

... ...

(P P ?1 ? R R?1 )V + 3(P ? R)V + (P 2 ? R2 )V = 0.

? ? ? ?? ? ?? (29)

300 R.Z. Zhdanov, I.V. Revenko, W.I. Fushchych

Thus, to integrate equation (25) it suffices to construct all functions P (f ), R(g),

V (P + R) satisfying (29) and substitute them into equation (28).

Let us prove the following assertion.

Lemma. The general solution of equation (29), determined up to transforma-

tions (4), is given by the one of the following formulas:

? ?

V = V (x) is an arbitrary function,

(1) P = ?, R = ?;

? ?

P 2 = ?P + ?, R2 = ?R + ?; (30)

(2) V = mx,

V = mx?2 , P = F (f ), R = G(g),

(3)

?

F 2 = ?F 4 + ?F 3 + ?F 2 + ?F + ?, (31)

?

G2 = ?G4 ? ?G3 + ?G2 ? ?G + ?;

V = m sin?2 x, P = arctg F (f ), R = arctg G(g),

(4)

and F, G are determined by (31);

V = m sh?2 x, P = arth F (f ), R = arth G(g)

(5)

and F, G are determined by (31);

V = m ch?2 x, P = arcth F (f ), R = arcth G(g)

(6)

and F, G are determined by (31);

(7) V = m exp x,

? ?

P 2 = ? exp 2P + ? exp P + ?, R2 = ? exp 2R + ? exp R + ?;

V = cos?2 x(m1 + m2 sin x),

(8)

(32)

? ?

P 2 = ? sin 2P + ? cos 2P + ?, R2 = ? sin 2R + ? cos 2R + ?;

V = ch?2 x(m1 + m2 sh x),

(9)

(33)

? ?

R2 = ? sh 2R ? ?ch 2R + ?;

P 2 = ? sh 2P + ?ch 2P + ?,

V = sh?2 x(m1 + m2 chx),

(10)

(34)

? ?

R2 = ?? sh 2R + ?ch 2R + ?;

P 2 = ? sh 2P + ?ch 2P + ?,

(11) V = (m1 + m2 exp x) exp x,

(35)

? ? ? ?

P = ?P 2 + ?, R = ?P 2 + ?;

V = m1 + m2 x?2 ,

(12)

(36)

? ?

R2 = ?R2 ? ?R + ?,

P 2 = ?P 2 + ?P + ?,

(13) V = m,

? ?

P 2 = ?P 2 + ?P + ?, R2 = ?R2 + ?R + ?.

Here ?, ?, ?, ?, ?, m1 , m2 , m are arbitrary real parameters; x = ? + ? = P + R.

Proof. Since the functions P , R in (29) are arbitrary, equation (29) is equivalent to

the following system of equations:

?1 ?1 ? 2?

(Hf f f f Hf ? Hggg Hg )V (H) + 3(Hf f ? Hgg )V (H) + (Hf ? Hg )V (H) = 0,(37)

2

Separation of variables in the two-dimensional wave equation with potential 301

(38)

Hf g = 0;

here, H = P (f ) + R(g).

Taking differential consequences of equation (37), we have

?1

Hf f f f = Hf f f Hf f Hf + V V ?1 (Hggg Hg Hf ? 4Hf f f Hf ) +

?1 2

?

...

+ V V ?1 (3Hgg H 2 ? 5Hf f H 2 ) + V V ?1 (Hg H 2 ? H 4 ),

? 2

f f f f

(39)

?1 2

Hggg Hgg Hg + V V ?1 (Hf f f Hf Hg ? 4Hggg Hg ) +

?1 ?

Hgggg =

...

+ V V ?1 (3Hf f Hg ? 5Hgg Hg ) + V V ?1 (Hf Hg ? Hg ).

? 2 2 22 4

For system (39) to be compatible, it is necessary that relations Hf f f f g = Hggggf =

0 hold. Differentiating the first equation in (39) with respect to g and taking into

account relations (39), we get

?1

(Hf f f Hf ? Hggg Hg )(5V 2 V ?2 ? 4V V ?1 ) + (Hf f ? Hgg ) ?

?1 ? ?

(40)

... ... ...

? (8V V V ?2 ? 5 V V ?1 ) + (Hf ? Hg )(V V V ?2 ? (V V ?1 )· ) = 0.

?? ?

2 2

Since equation (40) is a necessary compatibility condition for a system (39), one

has to supplement the system under study (equations (37), (38)) by equation (40). To

investigate the system of equations (37), (38), (40) it is necessary to consider several

inequivalent cases.

? ?

Case 1. Let V = 0, V = 0. Then equalities Hf f = Hgg = 2?, ? = const hold.

Hence, we have

V = m(H + C) ? m(x + C),

?, ? ? R1 ,

P (f ) = ?f 2 + ?, R(g) = ?g 2 + ?,

i.e., we obtain the potential listed in the lemma under number 2.

?

Case 2. Let V = 0 and let equation (40) be a consequence of equation (37). In this

??

case, the coefficients of V , V , V must be proportional

...

(5V 2 V ?2 ? 4V V ?1 ) = (8V V V ?2 ? 5 V V ?1 )(3V )?1 =

? ? ??

... ....

= (2 V V V ?2 ? V V ?1 )(V )?1 .

? ?

From the above equalities, we get a system of two ordinary differential equations

for a function V = V (H)

...

? ?1 ? 3V 3 V ?2 ,

? (41)

V = 4V V

.... ...

? ?1 ? 4V 2 V ?1 + 5V 2 V V ?2 .

? ?? (42)

V = 2V V V

But equation (42) is the differential consequence of equation (41). The general

solution of equation (41), determined up to equivalence relations (4), is given by one

of the following formulas [17]:

V2 = m sin?2 H,

V1 = mH ?2 ,

(43)

V3 = m sh?2 H, V4 = m ch?2 H, V5 = m exp H;

i.e., we obtain potentials listed in the lemma under numbers 3–7.

302 R.Z. Zhdanov, I.V. Revenko, W.I. Fushchych

By substituting V = V1 = mH ?2 into (37) and replacing H by P (f ) + R(g), we

get

... ...

(P + R)2 (P P ?1 ? R R?1 ) ? 6(P + R)(P ? R) + 6(P 2 ? R2 ) = 0.

ñòð. 73 |