Handbook of mathematics for engineers and scienteists part 107 pot

710 NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS
Equating the functional coefficient of u

to zero and taking into account (15.7.1.12), we obtain
g =–
1
ϕ
2
(ϕ

t
x + ψ

t
)
2
. (15.7.1.13)
Substituting the expressions (15.7.1.12) and (15.7.1.13) into (15.7.1.11), we arrive at the relation
ϕ
6
(au

+ uu

+ u
2
)+ϕ
2
(xϕ
tt
+ ψ

tt
)u

+ 2ϕϕ
tt
u –

ϕ
–2
(ϕ
t
x + ψ
t
)
2

tt
+ 6ϕ
–4
ϕ
2
t
(ϕ
t
x + ψ
t
)
2
= 0.
Let us perform the double differentiation of the expression in square brackets and then divide all terms by ϕ

6
.
Excluding x with the help of the relation x =(z – ψ)/ϕ,weget
au

+ uu

+(u

)
2
+ ϕ
–5
(ϕ
tt
z + ϕψ
tt
– ψϕ
tt
)u

+ 2ϕ
–5
ϕ
tt
u + ···= 0. (15.7.1.14)
Let us require that the functional coefficient of u

be a function of only one variable, z, i.e.,
ϕ

–5
(ϕ
tt
z + ϕψ
tt
– ψϕ
tt
)=ϕ
–5
ϕ
tt
z + ϕ
–5
(ϕψ
tt
– ψϕ
tt
) ≡ Az + B,
where A and B are arbitrary constants. Hence, we obtain the following system of ordinary differential equations
for the functions ϕ and ψ:
ϕ
tt
= Aϕ
5
,
ψ
tt
=(Aψ + B)ϕ
4
.

(15.7.1.15)
Let us eliminate the second and the third derivatives of the functions ϕ and ψ from (15.7.1.14). As a result, we
arrive at the following ordinary differential equation for the function u(z):
au

+ uu

+(u

)
2
+(Az + B)u

+ 2Au – 2(Az + B)
2
= 0. (15.7.1.16)
Formulas (15.7.1.1), (15.7.1.12), and (15.7.1.13), together with equations (15.7.1.15)–(15.7.1.16), describe
an exact solution of the Boussinesq equation (15.7.1.10).
15.7.1-2. Description of the Clarkson–Kruskal direct method.
1
◦
. The basic idea of the method is the following: for an equation with the unknown
function w = w(x, t), an exact solution is sought in the form
w(x, t)=f(x, t)u(z)+g(x, t), z = z(x, t). (15.7.1.17)
The functions f(x, t), g(x, t), and z(x, t) are determined in the subsequent analysis, so that
ultimately one obtains a single ordinary differential equation for the function u(z).
2
◦
. Inserting (15.7.1.17) into a nonlinear partial differential equation with a quadratic or a
power nonlinearity, we obtain

Φ
1
(x, t)Π
1
[u]+Φ
2
(x, t)Π
2
[u]+···+ Φ
m
(x, t)Π
m
[u]=0.(15.7.1.18)
Here, Π
k
[u] are differential forms that are the products of nonnegative integer powers of
the function u and its derivatives u

z
, u

zz
,etc.,andΦ
k
(x, t) depend on the functions f(x, t),
g(x, t), and z(x, t) and their partial derivatives with respect to x and t. Suppose that the
differential form Π
1
[u] contains the highest-order derivative with respect to z. Then the
function Φ

1
(x, t) is used as a normalizing factor. This means that the following relations
should hold:
Φ
k
(x, t)=Γ
k
(z)Φ
1
(x, t), k = 1, , m,(15.7.1.19)
where Γ
k
(z) are functions to be determined.
15.7. DIRECT METHOD OF SYMMETRY REDUCTIONS OF NONLINEAR EQUATIONS 711
3
◦
. The representation of a solution in the form (15.7.1.17) has “redundant” generality and
the functions f , g, u,andz are ambiguously determined. In order to remove the ambiguity,
we use the following three degrees of freedom in the determination of the above functions:
(a)iff =f(x, t)hastheformf =f
0
(x, t)Ω(z), then we can take Ω ≡ 1, which corresponds
to the replacement u(z) → u(z)/Ω(z);
(b)ifg = g(x, t)hastheformg = g
0
(x, t)+f(x, t)Ω(z), then we can take Ω ≡ 0,which
corresponds to the replacement u(z) → u(z)–Ω(z);
(c)ifz = z(x, t) is determined by an equation of the form Ω(z)=h(x, y), where Ω(z)is
any invertible function, then we can take Ω(z)=z, which corresponds to the replacement
z → Ω

–1
(z).
4
◦
. Having determined the functions Γ
k
(z), we substitute (15.7.1.19) into (15.7.1.18) to
obtain an ordinary differential equation for u(z),
Π
1
[u]+Γ
2
(z)Π
2
[u]+···+ Γ
m
(z)Π
m
[u]=0.(15.7.1.20)
Below we illustrate the main points of the Clarkson–Kruskal direct method by an
example.
Example 3. We seek a solution of the Boussinesq equation (15.7.1.10) in the form (15.7.1.17). We have
afz
4
x
u

+ a(6fz
2
x

z
xx
+ 4f
x
z
3
x
)u

+ f
2
z
2
x
uu

+ ···= 0. (15.7.1.21)
Here, we have written out only the first three terms and have omitted the arguments of the functions f and z.
The functional coefficients of u

and uu

should satisfy the condition [see (15.7.1.19)]:
f
2
z
2
x
= afz
4

x
Γ
3
(z),
where Γ
3
(z) is a function to be determined. Hence, using the degree of freedom mentioned in Item 3
◦
(a), we
choose
f = z
2
x
, Γ
3
(z)=1/a. (15.7.1.22)
Similarly, the functional coefficients of u

and u

must satisfy the condition
6fz
2
x
z
xx
+ 4f
x
z
3

x
= fz
4
x
Γ
2
(z), (15.7.1.23)
where Γ
2
(z) is another function to be determined. Hence, with (15.7.1.22), we find
14 z
xx
/z
x
= Γ
2
(z)z
x
.
Integrating with respect to x yields
ln z
x
= I(z)+lnϕ(t), I(z)=
1
14

Γ
2
(z) dz,
where ϕ(t) is an arbitrary function. Integrate again to obtain


e
–I(z)
dz = ϕ(t)x +

ψ(t),
where

ψ(t) is another arbitrary function. We have a function of z on the left and, therefore, using the degree of
freedom mentioned in Item 3
◦
(c), we obtain
z = xϕ(t)+ψ(t), (15.7.1.24)
where ϕ(t)andψ(t) are to be determined.
From formulas (15.7.1.22)–(15.7.1.24) it follows that
f = ϕ
2
(t), Γ
2
(z)=0. (15.7.1.25)
Substituting (15.7.1.24) and (15.7.1.25) into (15.7.1.17), we obtain a solution of the form (15.7.1.1) with
the function f defined by (15.7.1.12). Thus, the general approach based on the representation of a solution
in the form (15.7.1.17) ultimately leads us to the same result as the approach based on the more simple
formula (15.7.1.1).
712 NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS
Remark 1. In a similar way, it can be shown that formulas (15.7.1.1) and (15.7.1.17) used for the
construction of an exact solution of the generalized Burgers–Korteweg–de Vries equation (15.7.1.2) lead us to
the same result.
Remark 2. The above examples clearly show that it is more reasonable to perform the initial analysis of
specific equations on the basis of the simpler formula (15.7.1.1) rather than the general formula (15.7.1.17).

Remark 3. A more general scheme of the Clarkson–Kruskal direct method is as follows: for an equation
with the unknown function w = w(x, t), an exact solution is sought in the form
w(x, t)=F

x, t, u(z)

, z = z(x, t). (15.7.1.26)
The functions F (x, t, u)andz(x, t) should be chosen so as to obtain ultimately a single ordinary differential
equation for u(z). Unlike formulas (15.7.1.1) and (15.7.1.17), the relationship between the functions w and u
in (15.7.1.26) can be nonlinear.
15.7.2. Some Modifications and Generalizations
15.7.2-1. Symmetry reductions based on the generalized separation of variables.
1
◦
. The Clarkson–Kruskal direct method based on the representation of solutions in the
forms (15.7.1.17) and (15.7.1.26) attaches particular significance to the function u = u(z),
because the choice of the other functions is meant to ensure a single ordinary differential
equation for u(z). However, in some cases it is reasonable to combine these methods with
the ideas of the generalized and functional separation of variables, with all determining
functions being regarded as equally important. Then the function u(z) is described by an
overdetermined system of equations.
2
◦
. Exact solutions of nonlinear partial differential equations with quadratic or power
nonlinearities may be sought in the form (15.7.1.1) with g(x, t)=g
1
(t)x+g
0
(t). Substituting
(15.7.1.1) into an equation under consideration, we replace x by x =[z – ψ(t)]/ϕ(t). As

a result, we obtain a functional differential equation with two arguments, t and z.Its
solution can sometimes be obtained by the differentiation and splitting methods outlined in
Subsections 15.5.3 and 15.5.4.
Example 1. Consider the equation of an axisymmetric steady hydrodynamic boundary layer
∂w
∂y
∂
2
w
∂x∂y
–
∂w
∂x
∂
2
w
∂y
2
=
∂
∂y

y
∂
2
w
∂y
2

+ F(x), (15.7.2.1)

where w is the stream function, y =
1
4
r
2
,andx and r are axial and radial coordinates.
Its solution is sought in the form
w(x, y)=f(x)u(z)+g(x), z = ϕ(x)y + ψ(x). (15.7.2.2)
Let us substitute this expression into equation (15.7.2.1) and eliminate y using the relation y =[z –ψ(x)]/ϕ(x).
After the division by ϕ
2
f, we arrive at the functional differential equation
(zu

zz
)

z
– ψu

zzz
+ f

x
uu

zz
+ g

x

u

zz
–
(fϕ)

x
ϕ
(u

z
)
2
+
F
fϕ
2
= 0.(15.7.2.3)
General methods for solving such equations are outlined in Section 15.5. Here we use a simplified scheme
for the construction of exact solutions. Assume that the functional coefficients of uu

zz
, u

zz
,(u

z
)
2

,and1 are
linear combinations of the coefficients 1 and ψ of the highest-order terms (zu

zz
)

z
and u

zzz
, respectively. We
have
f

x
= A
1
+ B
1
ψ,
g

x
= A
2
+ B
2
ψ,
–(fϕ)


x
/ϕ = A
3
+ B
3
ψ,
F/(fϕ
2
)=A
4
+ B
4
ψ,
(15.7.2.4)
15.7. DIRECT METHOD OF SYMMETRY REDUCTIONS OF NONLINEAR EQUATIONS 713
where A
k
and B
k
are arbitrary constants. Let us substitute the expressions of (15.7.2.4) into (15.7.2.3) and
sum up the terms proportional to ψ (it is assumed that ψ ≠ const). Equating the functional coefficient of ψ to
zero, we obtain the following overdetermined system:
(zu

zz
)

z
+ A
1

uu

zz
+ A
2
u

zz
+ A
3
(u

z
)
2
+ A
4
= 0,(15.7.2.5)
–u

zzz
+ B
1
uu

zz
+ B
2
u


zz
+ B
3
(u

z
)
2
+ B
4
= 0.(15.7.2.6)
Case 1.Let
A
1
= A
3
= A
4
= 0, A
2
=–n.(15.7.2.7)
Then the solution of equation (15.7.2.5) has the form
u(z)=
C
1
n(n + 1)
z
n+1
+ C
2

z + C
3
,(15.7.2.8)
where C
1
, C
2
,andC
3
are integration constants. The solution (15.7.2.8) of equation (15.7.2.5) can be a solution
of equation (15.7.2.6) only if the following conditions are satisfied:
n =–2, B
1
= B
3
, C
1
=–4/B
1
, C
2
2
=–B
4
/B
1
, C
3
=–B
2

/B
1
.(15.7.2.9)
Let us insert the coefficients (15.7.2.7), (15.7.2.9) into system (15.7.2.4). Integrating yields
g(x)=2x – C
3
f, ϕ =
C
4
f
2
, ψ =–
C
1
4
f

x
, F =–(C
2
C
4
)
2
f

x
f
3
, (15.7.2.10)

where f = f(x) is an arbitrary function.
Formulas (15.7.2.2), (15.7.2.8), (15.7.2.10) define an exact solution of the axisymmetric boundary layer
equation (15.7.2.1).
Case 2.For
B
1
= B
3
= B
4
= 0, B
2
=–λ, A
2
= 0, A
3
=–A
1
, A
4
= λ
2
/A
1
(15.7.2.11)
a common solution of system (15.7.2.5), (15.7.2.6) can be written in the form
u(z)=
1
A
1

(C
1
e
–λz
+ λz – 3). (15.7.2.12)
A solution of system (15.7.2.4) with coefficients (15.7.2.11) is described by the formulas
f = A
1
x + C
2
, ϕ = C
3
, ψ =–
1
λ
g

x
, F =
(C
3
λ)
2
A
1
(A
1
x + C
2
), (15.7.2.13)

where C
1
, C
2
,andC
3
are arbitrary constants and g = g(x) is an arbitrary function.
Formulas (15.7.2.2), (15.7.2.12), (15.7.2.13) define an exact solution of the axisymmetric boundary layer
equation (15.7.2.1).
Case 3. System (15.7.2.5)–(15.7.2.6) also admits solutions of the form
u(z)=C
1
z
2
+ C
2
z + C
3
,
with constants C
1
, C
2
,andC
3
related to A
n
and B
n
. The corresponding solution is easier to obtain directly

from the original equation (15.7.2.1) by substituting w = ϕ
2
(x)y
2
+ ϕ
1
(x)y + ϕ
0
(x) into it, which corresponds
to the method of generalized separation of variables. This results in the solution
w(x, y)=C
1
y
2
+ ϕ(x)y +
1
4C
1
ϕ
2
(x)–
1
2C
1

F(x) dx – x + C
3
,
where ϕ(x) is an arbitrary function and C
1

and C
3
are arbitrary constants.
Example 2. Consider the equation with a cubic nonlinearity
∂w
∂t
+ σw
∂w
∂x
= a
∂
2
w
∂x
2
+ b
3
w
3
+ b
2
w
2
+ b
1
w + b
0
. (15.7.2.14)
Let us seek its solution in the form
w(x, t)=f(x, t)u(z)+λ, z = z(x, t), (15.7.2.15)

714 NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS
where the functions f = f(x, t), z = z(x, t), and u = u(z), as well as the constant λ, are to be determined.
Substituting (15.7.2.15) into equation (15.7.2.14), we obtain
afz
2
x
u

– σf
2
z
x
uu

+(afz
xx
+ 2af
x
z
x
– σλfz
x
– fz
t
)u

+ b
3
f
3

u
3
+(3b
3
λf
2
+ b
2
f
2
– σff
x
)u
2
+(3b
3
λ
2
f + 2b
2
λf + b
1
f + af
xx
– σλf
x
– f
t
)u
+ b

3
λ
3
+ b
2
λ
2
+ b
1
λ + b
0
= 0.
(15.7.2.16)
From the overdetermined system of ordinary differential equations resulting from the condition of proportion-
ality of the three functions u

, uu

,andu
3
and that of the two functions u

and u
2
, it follows that
u(z)=1/z, (15.7.2.17)
where the constant factor is taken equal to unity [this factor can be included in f , since formula (15.7.2.15)
contains the product of u and f]. Let us substitute (15.7.2.17) into (15.7.2.16) and represent the resulting
expression as a finite expansion in negative powers of z. Equating the functional coefficient of z
–3

to zero, we
obtain
f = βz
x
, (15.7.2.18)
where β is a root of the quadratic equation
b
3
β
2
+ σβ + 2a = 0. (15.7.2.19)
Equating the functional coefficients of the other powers of z to zero and taking into account (15.7.2.18), we
find that
z
t
–(3a + βσ)z
xx
+(σλ + βb
2
+ 3βb
3
λ)z
x
= 0 (coefficient of z
–2
),
z
xt
– az
xxx

+ σλz
xx
–(b
1
+ 2λb
2
+ 3b
3
λ
2
)z
x
= 0 (coefficient of z
–1
),
b
3
λ
3
+ b
2
λ
2
+ b
1
λ + b
0
= 0 (coefficient of z
0
).

(15.7.2.20)
Here, the first two linear partial differential equations form an overdetermined system for the function z(x, t),
while the last cubic equation serves for the determination of the constant λ.
Using (15.7.2.15), (15.7.2.17), and (15.7.2.18), we can write out a solution of equation (15.7.2.14) in the
form
w(x, t)=
β
z
∂z
∂x
+ λ. (15.7.2.21)
Let β be a root of the quadratic equation (15.7.2.19), and λ be a root of the last (cubic) equation in
(15.7.2.20). According to the value of the constant b
3
, one should consider two cases.
1
◦
. Case b
3
≠ 0. From the first two equations in (15.7.2.20), one obtains
z
t
+ p
1
z
xx
+ p
2
z
x

= 0,
z
xxx
+ q
1
z
xx
+ q
2
z
x
= 0,
where
p
1
=–βσ – 3a, p
2
= λσ + βb
2
+ 3βλb
3
, q
1
=–
βb
2
+ 3βλb
3
βσ + 2a
, q

2
=–
3b
3
λ
2
+ 2b
2
λ + b
1
βσ + 2a
.
Four situations are possible.
1.1. For q
2
≠ 0 and q
2
1
≠ 4q
2
,wehave
z(x, t)=C
1
exp(k
1
x + s
1
t)+C
2
exp(k

2
x + s
2
t)+C
3
,
k
n
=–
1
2
q
1
1
2

q
2
1
– 4q
2
, s
n
=–k
2
n
p
1
– k
n

p
2
,
where C
1
, C
2
,andC
3
are arbitrary constants; n = 1, 2.
1.2. For q
2
≠ 0 and q
2
1
= 4q
2
,
z(x, t)=C
1
exp(kx + s
1
t)+C
2
(kx + s
2
t)exp(kx + s
1
t)+C
3

,
k =–
1
2
q
1
, s
1
=–
1
4
p
1
q
2
1
+
1
2
p
2
q
1
, s
2
=–
1
2
p
1

q
2
1
+
1
2
p
2
q
1
.
1.3. For q
2
= 0 and q
1
≠ 0,
z(x, t)=C
1
(x – p
2
t)+C
2
exp[–q
1
x + q
1
(p
2
– p
1

q
1
)t]+C
3
.
1.4. For q
2
= q
1
= 0,
z(x, t)=C
1
(x – p
2
t)
2
+ C
2
(x – p
2
t)–2C
1
p
1
t + C
3
.
2
◦
. Case b

3
= 0, b
2
≠ 0. The solutions are determined by (15.7.2.21), where
β =–
2a
σ
, z(x, t)=C
1
+ C
2
exp

Ax + A

b
1
σ
2b
2
+
2ab
2
σ

t

, A =
σ(b
1

+ 2b
2
λ)
2ab
2
,
and λ = λ
1,2
are roots of the quadratic equation b
2
λ
2
+ b
1
λ + b
0
= 0.
15.7. DIRECT METHOD OF SYMMETRY REDUCTIONS OF NONLINEAR EQUATIONS 715
15.7.2-2. Similarity reductions in equations with three or more independent variables.
The procedure of the construction of exact solutions to nonlinear equations with three or
more independent variables sometimes involves (at intermediate stages) the solution of
functional differential equations considered in Subsections 15.5.3 and 15.5.4.
Example 3. Consider the nonlinear nonstationary wave equation anisotropic in one of the directions
∂
2
w
∂t
2
= a
∂

2
w
∂x
2
+
∂
∂y

(bw + c)
∂w
∂y

. (15.7.2.22)
Let us seek its solution in the form
w = U(z)+f(x, t), z = y + g(x, t). (15.7.2.23)
Substituting (15.7.2.23) into equation (15.7.2.22), we get
[(bU + ag
2
x
– g
2
t
+ bf + c)U

z
]

z
+(ag
xx

– g
tt
)U

z
+ af
xx
– f
tt
= 0.
Suppose that the functions f and g satisfy the following overdetermined system of equations:
af
xx
– f
tt
= C
1
, (15.7.2.24)
ag
xx
– g
tt
= C
2
, (15.7.2.25)
ag
2
x
– g
2

t
+ bf = C
3
, (15.7.2.26)
where C
1
, C
2
,andC
3
are arbitrary constants. Then the function U(z) is determined by the autonomous ordinary
differential equation
[(bU + c + C
3
)U

z
]

z
+ C
2
U

z
+ C
1
= 0. (15.7.2.27)
The general solutions of equations (15.7.2.24)–(15.7.2.25) are expressed as
f = ϕ

1
(ξ)+ψ
1
(η)–
1
2
C
1
t
2
,
g = ϕ
2
(ξ)+ψ
2
(η)–
1
2
C
2
t
2
,
ξ = x + t
√
a, η = x – t
√
a.
Let us insert these expressions into equation (15.7.2.26) and then eliminate t with the help of the formula
t =

ξ – η
2
√
a
. After simple transformations, we obtain a functional differential equation with two arguments,
bϕ
1
(ξ)+C
2
ξϕ

2
(ξ)–kξ
2
–C
3
+bψ
1
(η)+C
2
ηψ

2
(η)–kη
2
+ψ

2
(η)[4aϕ


2
(ξ)–C
2
ξ]+η[2kξ–C
2
ϕ

2
(ξ)] =0, (15.7.2.28)
where
k =
1
8a
(bC
1
+ 2C
2
2
).
Equation (15.7.2.28) can be solved by the splitting method described in Section 15.5. According to the
simplified scheme, set
bϕ
1
(ξ)+C
2
ξϕ

2
(ξ)–kξ
2

– C
3
= A
1
,
4aϕ

2
(ξ)–C
2
ξ = A
2
,
2kξ – C
2
ϕ

2
(ξ)=A
3
,
(15.7.2.29)
where A
1
, A
2
,andA
3
are constants. The common solution of system (15.7.2.29) has the form
ϕ

1
(ξ)=–
C
2
2
8ab
ξ
2
–
BC
2
b
ξ +
A
1
+ C
3
b
, ϕ
2
(ξ)=
C
2
8a
ξ
2
+ Bξ (15.7.2.30)
and corresponds to the following values of the constants:
A
1

is arbitrary, A
2
= 4aB, A
3
=–BC
2
, B is arbitrary, C
1
=–
C
2
2
b
, C
2
and C
3
are arbitrary, k =
C
2
2
8a
.
(15.7.2.31)
From (15.7.2.28) and (15.7.2.29) we obtain an equation that establishes a relation between the functions ψ
1
and ψ
2
,
A

1
+ bψ
1
(η)+C
2
ηψ

2
(η)–kη
2
+ A
2
ψ

2
(η)+A
3
η = 0. (15.7.2.32)
716 NONLINEAR PARTIAL DIFFERENTIAL EQUATIONS
Hence, taking into account (15.7.2.31), we get
ψ
1
(η)=–
1
b
(C
2
η + 4aB)ψ

2

(η)+
1
b

C
2
2
8a
η
2
+ BC
2
η – A
1

, ψ
2
(η) is an arbitrary function.
Ultimately, we find the functions that determine solution (15.7.2.23):
f(x, t)=–
C
2
2
2
√
ab
xt +
C
2
2

2b
t
2
–
2
√
aBC
2
b
t +
C
3
b
–
1
b
(C
2
η + 4aB)ψ

2
(η),
g(x, t)=
C
2
8a

x
2
+ 2

√
axt– 3at
2

+ B(x +
√
at)+ψ
2
(η), η = x – t
√
a.
Remark. In the special case of a = 1, b < 0,andc > 0, equation (15.7.2.22) describes spatial transonic
flows of an ideal polytropic gas.
15.8. Classical Method of Studying Symmetries of
Differential Equations
Preliminary remarks. The classical method of studying symmetries of differential equa-
tions* presents a routine procedure that allows to obtain the following:
(i) transformations under which equations are invariant (such transformations bring the
given equation to itself);
(ii) new variables (dependent and independent) in which equations become considerably
simpler.
The transformations of (i) convert a solution of an equation to the same or another
solution of this equation. In the former case, we have an invariant solution, which can be
found by symmetry reduction, rewriting the equation in new, fewer variables. In the latter
case, we have noninvariant solutions, which may be “multiplied” to a family of solutions.
Remark 1. In a sense, the classical method of symmetry analysis of differential equations may be treated
as significant extension of the similarity method, outlined in Subsection 15.3.3.
Remark 2. Subsections 15.8.1–15.8.3 give a description of the classical method in a nontraditional way,
with minimal use of the special (group) terminology, for the reader’s easier understanding. Subsection 15.8.4
will explain the origin of the term “Lie group analysis.”

15.8.1. One-Parameter Transformations and Their Local Properties
15.8.1-1. One-parameter transformations. Infinitesimal operator.
We will consider invertible transformations of the form
¯x = ϕ
1
(x, y, w, ε), ¯x|
ε=0
= x,
¯y = ϕ
2
(x, y, w, ε), ¯y|
ε=0
= y,
¯w = ψ(x, y, w, ε), ¯w|
ε=0
= w,
(15.8.1.1)
where ϕ
1
, ϕ
2
,andψ are sufficiently smooth functions of their arguments, and ε is a real pa-
rameter. It is assumes that the successive application (composition) of two transformations
of the form (15.8.1.1) with parameters ε and ¯ε is equivalent to a single transformation of the
form with parameter ε + ¯ε (this means that such transformations have the group property).
* It is also known as the Lie group analysis of differential equations or the classical method of symmetry
reductions.