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By differentiating (44) with respect to f and g, we obtain

(P + R)(h1 P ?1 ? h2 R?1 ) = 2(h1 ? h2 ),

?? ??

... ...

where h1 = P P ?1 and h2 = R R?1 .

? ?

Differentiation of the above equation with respect to f and g yields the following

relation:

(h1 P ?1 )· P ?1 = (h2 R?1 )· R?1 .

?? ??

? ? (45)

Since the functions P (f ), R(g) are independent, it follows from (45) that the

equalities

(h1 P ?1 )· = 12?P , (h2 R?1 )· = 12?R.

?? ??

? ? (46)

hold, where ? is an arbitrary real parameter.

Integration of equations (46) yields

?

P = ?P 4 + C1 P 3 + C2 P 2 + C3 P + C4 ,

?

R2 = ?R4 + D1 R3 + D2 R2 + D3 R + D4 ,

where C1 , . . . , C4 , D1 , . . . , D4 are arbitrary real constants. Substituting the above

result into the initial equation (44), we get restrictions on the choice of arbitrary

constants

C1 = ?D1 = ?, C3 = ?D3 = ?,

C2 = D2 = ?, C4 = D2 = ?.

Thus, we have obtained the potential listed in the lemma under number 3.

It is straightforward to verify that the equations obtained by the substitution of

functions V = m sin?2 H, V = m ch?2 with H = P (f ) + R(g) into (37) are reduced

to equation (44) by the following changes of variables:

P > arctg P, R > arctg R,

P > arth P, R > arth R,

P > arcth P, R > arcth R;

i.e., the potentials listed in the lemma under numbers 4–6 are obtained.

Equation (1) with the potential V = m exp H is reduced to the Klein–Gordon–Fock

equation (see case 4 and Note 2 below).

?

Case 3. Let V = 0 and assume, in addition, that equation (41) does not hold.

In this case, we can exclude from equations (37), (40) the third derivatives of the

function H

Hf f ? Hgg + A(H)(Hf ? Hg ) = 0,

2 2

(47a)

where

.... ... ...

A(H) = ( V ?2 V V V ?1 ? 4V 2 V + 5V V 2 V ?2 )(V ?4V V V ?1 + 3V 3 V ?2 )?1 .

? ? ?? ?? ?

Separation of variables in the two-dimensional wave equation with potential 303

It follows from (47a)

? ?

Hf f f = AHf (Hg ? Hf ) ? 2Hf f Hf A, Hggg = AHg (Hf ? Hg ) ? 2Hgg Hg A,

2 2 2 2

(we have used equation (38)).

By taking the first differential consequence of the above equations with account of

equation (38), we get

? ?2

2A(Hf f ? Hgg ) + A(Hf ? Hg ) = 0.

2

(48)

Clearly, equations (47a) and (48) are consistent iff the function A(H) satisfies the

following ordinary differential equation:

? ?

A = 2AA,

the general solution of which is given by one of the formulas (up to scaling H > CH).

A = C, A = tg (H + C), A = ?th (H + C),

A = ?cth (H + C), A = ?(H + C)?1 , C ? R1 .

Next, we consider the above cases separately.

Case 3.1. A(H) = C, C = 0. In this case, equation (47a) takes the form

Pf f ? Rgg + C(Pf ? Rg ) = 0

2 2

(47b)

or

? ? R1 .

2 2

Pf f + CPf = Rgg + CRg = ?,

Finally, we get

Pf f = ?CPf + ?, Rgg = ?CRg + ?.

2 2

(49)

Differentiating the first equation with respect to f , the second equation with

respect to g, and subtracting, we get

?1 ?1

Pf f f Pf ? Pggg Pg = ?2C(Pf f ? Rgg ). (50)

Substituting (49), (50) into equation (37), we come to the equation for V = V (H),

? ?

V ? 3C V + 2C 2 V = 0

the general solution of which reads

m2 , m2 ? R1 . (51)

V = m1 exp CH + m2 exp 2CH,

It is not difficult to check that function (51) satisfies equation (47b) provided that

A(H) = C. Consequently, if the potential is given by formula (51) (after rescaling

H > CH, we can choose C = 1), then the functions P (f ), R(g) are determined by

equations (35).

Case 3.2. A = tg(H + C). Multiplying equation (47) by ctg (H + C) and differen-

tiating the obtained expression with respect to f and g, we arrive at the equation

?1 ?1

(Pf f f Pf ? Pggg Pg ) ? 2ctg (H + C)(Pf f ? Rgg ) = 0. (52)

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

After excluding the function ctg (H + C) from (47) and (52), we get an equation

with separated variables

?1 ?1

(Pf f f Pf ? Pggg Pg ) + 2(Pf ? Rg ) = 0,

2 2

whence

?1 ?1

2 2

(53)

Pf f f Pf + 2Pf = ?, Rggg Rg + 2Rg = ?.

In (53), ? is an arbitrary real constant.

Substitution of formulas (52), (53) into equation (37) gives the equation for V =

V (H),

? ?

V ? 3 tg (H + C)V ? 2V = 0

the general solution of which has the form [17]

V = cos?2 (H + C)[m1 + m2 sin(H + C)]. (54)

As a direct check shows, the function V (H) (54) satisfies equation (47b) with

A = tg (H + C).

Integrating equations (53), we get

2 2

(55)

Pf = C1 sin 2P + C2 cos 2P + ?, Pg = D1 sin 2R + D2 cos 2R + ?,

where Ci , Di , and ? are arbitrary real constants.

Substitution of (55) into (47) with A = tg (H + C) yields the following restrictions

on the choice of the constants Ci , Di : C1 = D1 = ?, C2 = D2 = ?.

Thus, provided that the function V (H) is given by (44), the functions P (f ), R(g)

are determined by equations (32).

Case 3.3. A = ?th (H + C). In this case, one can obtain the following differential

consequence of equation (47):

?1 ?1

Pf f f Pf ? Pggg Pg = 2 cth (P + R + C)(Pf f ? Rgg ). (56)

Excluding the function ctg (H + C) from equations (47), (56), we get the equation

?1 ?1

Pf f f Pf ? Pggg Pg = 2(Pf ? Rg ),

2 2

whence

?1 ?1

Pf f f Pf ? 2Pf = ?, Rggg Rg ? 2Rg = ?.

2 2

(57)

In (57), ? is an arbitrary real constant.

Integration of equations (57) gives

2 2

(58)

Pf = C1 sh 2P + C2 ch 2P + ?, Rg = D1 sh 2R + D2 ch 2R + ?,

where Ci , Di , and ? are arbitrary real constants.

Substituting expressions (56), (57) into (37), we obtain an equation for V (H),

? ?

V + 3 th (H + C)V + 2V = 0,

the general solution of which has the form [17]

V = ch?2 (H + C)(m1 + m2 sh(H + C)), mi ? R1 . (59)

Separation of variables in the two-dimensional wave equation with potential 305

It is not difficult to become convinced of the fact that function (59) satisfies

equation (47b) with A = ?th (H + C).

At last, substituting (57) and (58) into (47), we get C1 = D1 = ?, C2 = ?D2 = ?.

Consequently, if the potential V (H) is given by formula (59), then functions P (f )

and R(g) are determined by equations (33).

Case 3.4. A = ?cth (H + C). In this case, one can obtain the following differential

consequence of equation (47):

?1 ?1

Pf f f Pf ? Rggg Rg = 2 th (P + R + C)(Pf f ? Rgg ). (60)

Using equations (37), (47), and (60), we get an equation for V (H),

? ?

V + 3 cth (H + C)V + 2V = 0,

the general solution of which has the form [17]

V = sh?2 (H + C)(m1 + m2 ch(H + C)), mi ? R1 . (61)

By direct computation, one can check that function (61) satisfies equation (47b)

with A = ?cth (H + C).

Next, by eliminating the function th (H + C) from equations (47) and (60), we get

an equation with separated variables

?1 ?1

Pf f f Pf ? Pggg Pg ? 2Pf + 2Rg = 0,

2 2

whence

?1 ?1

Pf f f Pf ? 2Pf = ?, Rggg Rg ? 2Rg = ?.

3 2

Here, ? is an arbitrary real constant.

Integration of the above ordinary differential equations shows that the functions

P (f ) and R(g) are determined by equations (58), where Ci , Di , and ? are arbitrary

real constants. Substituting (58) into equation (47), we have the following restrictions

on the choice of Ci , Di :

C1 = ?D1 = ?, C2 = D2 = ?.

Thus, if the function V (H) is given by (61), then functions P (f ) and R(g) are

determined by equations (34).

Case 3.5. A = ?(H + C)?1 . In this case, it follows from (47a) that the equality

?1 ?1

Pf f f Pf = Rggg Rg holds. Hence, we get equations for P (f ), R(g),

(62)

Pf f f = ?Pf , Rggg = ?Rg

? ?

with arbitrary ? ? R1 . Moreover, the equation for V (H) has the form V +3(H+C)V =

0, whence

V = m1 + m2 (H + C)?2 , mi ? R1 . (63)

It is not difficult to check that function (63) satisfies (47b) with A = ?(H + C)?1 .

Integration of equations (62) yields the following result:

Pf = ?P 2 + C1 P + C2 ,

2

Rg = ?R2 + D1 R + D2 ,

2

(64)

here ?, Ci , and Di are arbitrary real constants.

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

Next, substituting (64) into (47), we get C1 = ?D1 = ?, C2 = D2 = ?.

Thus, if the potential V is given by (63), then the functions P (f ), R(g) are

determined by equations (36).

?1

Case 4. V (H) = m = const. In this case, equation (37) reads Pf f f Pf =

?1

Rggg Rg , whence

(65)

Pf f f = ?Pf , Rggg = ?Rg ,

where ? ? R1 is an arbitrary constant.

Integrating (65), we get equations listed in the lemma under number 13.

??

Case 5. V (H) is an arbitrary function. In this case, the coefficients of V , V , V

in (37) must vanish. Consequently, the relations

?1 ?1 2 2

Hf f f Hf = Hggg Hg , Hgg = Hf f , Hf = Hg

hold. Hence, we have Hf = ?, Hg = ?, ? ? R1 . The lemma is proved.

Theorems 1, 2 are direct consequences of the above lemma. To prove Theorems

3–8, one has to integrate ordinary differential equations (30), (32)–(36) and substitute

the obtained expressions into (27),

?1 ? ?2

1 ?1 + ?2 1

(x + t) = P (f ) ? P (x ? t) = R(g) ? R

, ,

2 2 2 2

and (28).

Integration of equations (30), (32)–(36) is carried out in a standard way [17, 18],

the obtained result depends essentially on relations between parameters ?, ?, ?, ?, ?.

This procedure demands very cumbersome computations; this is why we omit details.

With the above remarks, the proof of Theorems 1–8 is completed.

4. Discussion. Let us say a few words about intrinsic characterization of SV in

equation (1). It is well known that the solution of a second-order partial differential

equation with separated variables is a joint eigenfunction of mutually commuting

second-order symmetry operators of the equation under study (for more details, see [6,

10, 14]). Below we construct, in an explicit form, a second-order symmetry operator

of equation (1) such that the solution with separated variables is its eigenfunction and

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