ñòð. 78 |

y1 + ty2 y1

?1/2

? ? ? arctan t ?

2

v = t +1 exp

t2 + 1

4t 4t

y1 + ty2

? Dv? ±(1 + i)

t2 + 1

and the real and imaginary parts of this function give us exact solutions of the heat

equation (9).

3.1.10.

(? ? 0, ? ? R),

J12 + 2?M, S + T + 2?M

2 2

t y1 + y2

y1

?1/2

exp ?? arctan t + ? arctan ?

2

v = t +1 ?(?),

4 (t2 + 1)

y2

2 2

?2

y1 + y 2 1 1 ?

?= 2 , ?+ ?+

? ? + + ? = 0.

16 4? 4? 2

t +1 ?

The solutions of this equation can be given in terms of Whittaker functions [11],

and we obtain the following exact solutions of the heat equation as a result:

2 2

y1 t(y1 + y2 )

2 ?1/2

exp ?? arctan t + ? arctan ? ?

2

v= y1 + y2

4(t2 + 1)

y2

2 2

i? i? i y1 + y2

?W ;; .

8 2 2 (t2 + 1)

In the above cases we have been able to describe exact solutions of (8) in terms

of elementary functions or confluent hypergeometric functions. Using the notation

introduced in equations (8) and (7), we are thus able to construct strikingly new

exact solutions of the linear wave equation (1).

3.2. Reduction to partial differential equations by one-dimensional subal-

gebras. Here we list the subalgebras, the relevant parameters, ansatzes and reduced

equations, without constructing their exact solutions. We use ?1 to denote the partial

derivative with respect to ?1 , and ?22 means the second derivative with respect to ?2 ,

and so on.

320 P. Basarab-Horwath, L. Barannyk, W.I. Fushchych

3.2.1.

P2 : v = ?(?1 , ?2 ), ?1 = t, ?2 = y 1 , ?1 = ?22 .

This is the heat equation in 1 + 1 spacetime dimensions. The symmetries and condi-

tional symmetries of the heat equation are well known. A discussion of these can be

found in [6] and in appendix 7 of [2].

3.2.2.

2

y1

v = exp ?

G1 + P2 : ?(?1 , ?2 ), ?1 = t, ?2 = y1 + ty2 ,

4t

?2 1

(1 + ?1 )?22 ? ?1 ? ?2 ?

2

? = 0.

?1 2?1

3.2.3.

?t

(? = 0, ±1) : v = exp ?

T + ?M ?(?1 , ?2 ), ?1 = y 1 , ?2 = y 2 ,

2

1

?11 + ?22 + ?? = 0.

2

This equation is the Laplace equation for ? = 0. Solutions can be obtained by using

separation of variables.

3.2.4.

t3 y1 t

? ?1 = t2 ? 2y1 ,

T + G1 : v = exp ?(?1 , ?2 ), ?2 = y 2 ,

6 2

1

4?11 + ?22 ? ?1 ? = 0.

4

3.2.5.

(? ? 0),

J12 + 2?M

y1 2 2

v = exp ? arctan ?(?1 , ?2 ), ?1 = y1 + y2 , ?2 = t,

y2

4?1 ?11 + 4?1 ?1 + ?1 ?2 + ?2 ? = 0.

2

3.2.6.

(? ? R),

J12 + T + 2?M

y1

2 2

v = exp(??t)?(?1 , ?2 ), ?1 = y 1 + y 2 , ?2 = t + arctan ,

y2

4?1 ?11 + ?22 + 4?1 ?1 ? ?1 ?2 + ??1 ? = 0.

2

3.2.7.

?

D + ?(2? ? 1)M (? ? 0, ? ? 1/2),

J12 +

2

y 2 + y2

2

y1

, ?2 = 1

v = t? ?(?1 , ?2 ), ?1 = log t + ? arctan ,

y2 t

?2 ?11 + 4?2 ?22 ? ?2 ?1 + (4?2 + ?2 )?2 + ??2 ? = 0.

2 2

New solutions of the wave equation 321

3.2.8.

2 2

y1 y2

v = t?? ?(?1 , ?2 ),

D + (4? ? 2)M (? ? 1/2) : ?1 = , ?2 = ,

t t

4?1 ?11 + 4?2 ?22 + (2 + ?1 )?1 + (2 + ?2 )?2 + ?? = 0.

3.2.9.

(? > 0, ? ? R),

S + T + ?J12 + 2?M

2 2

t(y1 + y2 )

?1/2

exp ?? arctan t ?

v = t2 + 1 ?(?1 , ?2 )

4 (t2 + 1)

2 2

y1 + y2 y1

?1 = 2 , ?2 = arctan + ? arctan t,

t +1 y2

1 ?1

?22 + 4?1 ? ??2 + ? +

4?1 ?11 + ? = 0.

?1 4

3.2.10.

(? ? R),

S + T + 2?M

2 2

t(y1 + y2 )

?1/2

exp ?? arctan t ?

2

v = t +1 ?(?1 , ?2 ),

4(t2 + 1)

y2 y2

? 1 = 2 1 , ?2 = 2 2 ,

t +1 t +1

?1 + ?2

4?1 ?11 + 4?2 ?22 + 2?1 + 2?2 + ? + ? = 0.

4

3.2.11.

S + T + J12 + ?(G1 + P2 ) (? > 0),

(1 ? t2 )(y1 + ty2 )2 y2

v = (t2 + 1)?1/2 exp ? 1 ?(?1 , ?2 ),

4t(t2 + 1)2 4t

ty1 ? y2

y1 + ty2

?1 = 2 , ?2 = 2 = ? arctan t,

t +1 t +1

?11 + ?22 ? (2?1 ? ?)?2 + ?1 ? = 0.

2

4. Some conditional symmetries of the 2 + 1 heat equation

In this section we give the conditional symmetries of equation (8). The defining

equations are nonlinear coupled partial differential equations, which we do not solve,

except in one case, leaving the others for consideration in a later publication. We have

the following result.

Proposition 3. Equation (8) is conditionally invariant under

? ? ? ?

X = ?0 + ?1 + ?2 +?

?t ?y1 ?y2 ?v

when the coefficients satisfy the following conditions:

?y2 = ??y1 ,

(i) ? 0 = 1 : 1 2 1 2

?y1 = ?y2 , ? = Av + B,

where ? 1 , ? 2 , A, B are functions of t, y1 , y2 and satisfy the system

?t + 2? 1 ?y1 + 2Ay1 = 0,

1 1

?t + 2? 2 ?y2 + 2Ay2 = 0,

2 2

At = Ay1 y1 + Ay2 y2 ? 2A?y2 , Bt = By1 y1 + By2 y2 ? 2B?y2 .

2 2

322 P. Basarab-Horwath, L. Barannyk, W.I. Fushchych

(ii) ? 0 = 0, ? 1 = 1 : ?y2 = ? 2 ?y1 ,

2 2

? = Av + B,

where ? 2 , A, B are functions of t, y1 , y2 and satisfy the system

?t ? ?y1 y1 ? ?y2 y2 + 2?y1 ?y2 ? 2? 2 Ay1 ? 2A?y1 = 0,

2 2 2 22 2

At = Ay1 y1 + Ay2 y2 + 2AAy1 ? 2Ay2 ?y1 ,

2

Bt = By1 y1 + By2 y2 + 2BAy1 ? 2By2 ?y1 .

2

(iii) ? 0 = ? 1 = 0, ? 2 = 1 : ? = Av + B,

where A is a function of t, y2 only, and B is a function of t, y1 , y2 and satisfy the

equations

At = Ay2 y2 + 2AAy2 , Bt = By1 y1 + By2 y2 + 2BAy2 .

As is clear in the above three cases, the systems of equations involved are highly

nonlinear, and cannot be solved in general. However, the equation for the function A

in case (iii) is recognized to be the Burgers equation. This equation can be linearized

by the Hopf–Cole transformation A = wy2 /w, where w is a solution of the heat

equation wt = wy2 y2 (see for example [2]). The solutions obtained in this way can

then be used to build ansatzes first for the 2 + 1 heat equation (8) and then, in turn,

the linear wave equation (1), using the ansatz (7).

Ansatzes can also be obtained from the symmetry algebra of the Burgers equation.

Indeed, the symmetry algebra of the equation

(10)

At = Ay2 y2 + 2AAy2

is generated by the operators

?t , ?y2 , 2t?y2 ? ?A , 2t?t + y2 ?y2 ? A?A ,

(11)

y2

t2 ?t + ty2 ?y2 ? tA + ?A .

2

The operator (11) gives the ansatz

y2 1 y2

A=? (12)

+?

2t t t

which gives, on substituting into (10), the equation

? ?

? + 2? ? = 0

for ?, where the dot denotes differentiation with respect to the variable ? = y2 /t.

This equation readily integrates to

?

? + ? 2 = c,

where c is a constant. This gives us three cases:

(13)

c=0: ? = t/(kt + y2 ),

where k is a constant;

2ay2 2ay2

?1

c = a2 , a > 0 : (14)

? = a t exp t exp +1

t t

New solutions of the wave equation 323

with l = 0 a constant;

ay2

c = ?a2 , a > 0 : ? = ?a tan a2 + (15)

.

ñòð. 78 |