International Journal of Differential Equations and Applications
Volume 14 No. 2 2015, 65-79
ISSN: 1311-2872
url: http://www.ijpam.eu
doi: http://dx.doi.org/10.12732/ijdea.v14i2.2091
AP
acadpubl.eu
A METHOD OF SOLVING LAGRANGE’S FIRST-ORDER
PARTIAL DIFFERENTIAL EQUATION WHOSE
COEFFICIENTS ARE LINEAR FUNCTIONS
Syed Md Himayetul Islam1 § , J. Das (Née Chaudhuri)2
1 Fatullapur
Adarsha High School
Vill.- Fatullapur, P.O.- Nimta
Dist.- North 24-Parganas, Kolkata, 700049, INDIA
@ Department of Pure Mathematics
Calcutta University
35, Ballygunge Circular Road, Kolkata, 700019
West Bengal, INDIA
Abstract: A method of solving Lagrange’s first-order partial differential equation of the form
P p + Qq = R,
where P , Q, R are linear functions of x, y, z, has been presented below.
AMS Subject Classification: 35F05, 35A99
Key Words: Lagrange’s first-order partial differential equation, linear functions, simultaneous ordinary differential equations, linear homogeneous algebraic equations
Received:
May 20, 2014
§ Correspondence
author
c 2015 Academic Publications, Ltd.
url: www.acadpubl.eu
66
S.M.H. Islam, J. Das (N. Chaudhuri)
1. Introduction
In the book [1] (page 46-47), the partial differential equation of the form
(a1 x + b1 y + c1 z + d1 )p + (a2 x + b2 y + c2 z + d2 )q = a3 x + b3 y + c3 z + d3 (1)
∂z
∂z
where p = ∂x
, q = ∂y
and ai , bi , ci , di (i = 1, 2, 3) are all real numbers, has
been discussed in brief. The present paper comprises a detailed discussion of
the same including the cases of failure of the method adopted there.
2. The Method
For determining the solutions of (1.1), it is required to consider the simultaneous
ordinary differential equations
dy
dz
dx
=
=
.
a1 x + b1 y + c1 z + d1
a2 x + b2 y + c2 z + d2
a3 x + b3 y + c3 z + d3
(2)
Suppose it is possible to find numbers λ, µ, ν, ρ such that each ratio of (2)
λdx + µdy + νdz
(a1 λ + a2 µ + a3 ν)x + (b1 λ + b2 µ + b3 ν)y + (c1 λ + c2 µ + c3 ν)z + (d1 λ + d2 µ + d3 ν)
=
λdx + µdy + νdz
.
ρ(λx + µy + νz)
(3)
Clearly, (3) holds if:
(a1 − ρ)λ + a2 µ + a3 ν =0,
b1 λ + (b2 − ρ)µ + b3 ν =0,
c1 λ + c2 µ + (c3 − ρ)ν =0,
(4)
d1 λ + d2 µ + d3 ν =0.
Considering the first three equations of (4) comprising a system of linear homogeneous algebraic equations in the three variables λ, µ, ν, for a non-trivial
solution (λ, µ, ν) 6= (0, 0, 0) of (4), one notes that the rank of the coefficient
matrix should not exceed two. So we should have
a1 − ρ
a
a
2
3
b1
b2 − ρ
b3 = 0.
(5)
c1
c2
c3 − ρ
A METHOD OF SOLVING LAGRANGE’S FIRST-ORDER...
67
This leads to the following cubic in ρ:
ρ3 − (a1 + b2 + c3 )ρ2 + (b2 c3 − b3 c2 + a2 b1 − a1 b2 + a3 c1 − a1 c3 )ρ
+(b2 c3 − b3 c2 )a1 + (b3 c1 − b1 c3 )a2 + (b1 c2 − b2 c1 )a3 = 0.
(6)
Let the roots of the equation (2.5) be ρ1 , ρ2 , ρ3 .
The following cases arise:
Case I: ρ1 , ρ2 , ρ3 are real and distinct,
Case II: One of ρ1 , ρ2 , ρ3 is real, the other two are complex,
Case III: ρ1 , ρ2 , ρ3 are real, but ρ1 = ρ2 6= ρ3 ,
Case IV: ρ1 , ρ2 , ρ3 are real, and ρ1 = ρ2 = ρ3 .
In the following sections the four cases I-IV have been discussed citing an
example in each case.
3. Case I: Roots of (6) are Real and Distinct
Let (λi , µi , νi ) be the solutions of the first three equations of the system (4)
corresponding to ρ = ρi (i = 1, 2, 3).
Then, from (3), we get
λ2 dx + µ2 dy + ν2 dz
λ3 dx + µ3 dy + ν3 dz
λ1 dx + µ1 dy + ν1 dz
=
=
.
ρ1 (λ1 x + µ1 y + ν1 z)
ρ2 (λ2 x + µ2 y + ν2 z)
ρ3 (λ3 x + µ3 y + ν3 z)
(7)
If each of (λi , µi , νi ) (i = 1, 2, 3) satisfies the last equation of (4), the above
relations lead to the required solution of the PDE (1.1) as
F (λ1 x+µ1 y+ν1 z)ρ2 (λ2 x+µ2 y+ν2 z)−ρ1 , (λ2 x+µ2 y+ν2 z)ρ3 (λ3 x+µ3 y+ν3 z)−ρ2
= 0,
where F is an arbitrary real-valued function of two real variables.
Example 1.
1
(y + 2z)p + (z + x)q = x − y.
2
For this PDE, the first three equations of (4) become
1
ρλ − µ − ν =0,
2
−λ + ρµ + ν =0,
−2λ − µ + ρν =0.
(8)
(9)
68
S.M.H. Islam, J. Das (N. Chaudhuri)
In this case the fourth equation is automatically satisfied and the equation (5)
becomes
ρ − 1 −1
2
−1 ρ
(10)
1 = 0,
−2 −1 ρ leading to the equation
3
ρ3 − ρ = 0.
2
Roots of the above equation are 0, ±
Using ρ = 0 in (9) we get
√
6
2 .
µ + 2ν = 0,
whence we have
Using ρ =
√
6
2
− λ + ν = 0,
µ
ν
λ
=
= .
1
−2
1
in (9) we get
√
6λ − µ − 2ν = 0,
− 2λ +
√
6µ + 2ν = 0.
It then follows that
λ
µ
ν
√
√ =
=
,
6
−
2
2( 6 − 1)
4−2 6
hence we have
Similarly, using ρ = −
√
6
2
µ
ν
λ
=√
= √
.
5
6−4
2( 6 + 1)
in (9) we get
λ
µ
ν
=√
= √
.
−5
6+4
2( 6 − 1)
So the relations (7) become
√
√
5dx + ( 6 − 4)dy + 2( 6 + 1)dz
dx − 2dy + dz
= √
√
√
6
0
2 (5x + ( 6 − 4)y + 2( 6 + 1)z)
√
√
−5dx + ( 6 + 4)dy + 2( 6 − 1)dz
= √
.
√
√
− 6
(−5x
+
(
6
+
4)y
+
2(
6
−
1)z)
2
(11)
A METHOD OF SOLVING LAGRANGE’S FIRST-ORDER...
69
From the above relations the required solution of the PDE (8) can be written
as
F x − 2y + z, 6(y + z)2 − (5x − 4y + 2z)2 = 0,
where F is an arbitrary real-valued function of two real variables.
4. Case II: ρ1 is real, ρ2 = ρ2 ′ + iρ2 ′′ and
ρ3 = ρ2 ′ − iρ2 ′′ , (ρ2 ′ , ρ2 ′′ ∈ R)
Let (λ1 , µ1 , ν1 ) be the solution of the first three equations of the system (4)
corresponding to the root ρ1 and (λ2 ′ + iλ2 ′′ , µ2 ′ + iµ2 ′′ , ν2 ′ + iν2 ′′ ), (λ2 ′ , λ2 ′′ ,
µ2 ′ , µ2 ′′ , ν2 ′ , ν2 ′′ ∈ R), be the solution of the above mentioned first three
equations of the system (4) corresponding to the root ρ2 ′ + iρ2 ′′ . Then the
solution of the system (4) corresponding to the root ρ2 ′ − iρ2 ′′ will be (λ2 ′ −
iλ2 ′′ , µ2 ′ − iµ2 ′′ , ν2 ′ − ν2 ′′ ).
Now, from (3), we get
λ1 dx + µ1 dy + ν1 dz
ρ1 (λ1 x + µ1 y + ν1 z)
(λ2 ′ + iλ2 ′′ )dx + (µ2 ′ + iµ2 ′′ )dy + (ν2 ′ + iν2 ′′ )dz
=
(ρ2 ′ + iρ2 ′′ ) (λ2 ′ + iλ2 ′′ )x + (µ2 ′ + iµ2 ′′ )y + (ν2 ′ + iν2 ′′ )z
=
(12)
(λ2 ′ − iλ2 ′′ )dx + (µ2 ′ − iµ2 ′′ )dy + (ν2 ′ − iν2 ′′ )dz
.
(ρ2 ′ − iρ2 ′′ ) (λ2 ′ − iλ2 ′′ )x + (µ2 ′ − iµ2 ′′ )y + (ν2 ′ − iν2 ′′ )z
It is assumed that each of (λ1 , µ1 , ν1 ), (λ2 ′ ± iλ2 ′′ , µ2 ′ ± iµ2 ′′ , ν2 ′ ± iν2 ′′ ) satisfies
the last equation of (4).
From the first equation of (12) we have a solution of the simultaneous
equations (2), say u1 , as
u1 = (λ1 x+ µ1 y + ν1 z)ρ2
′ +iρ ′′
2
(λ2 ′ + iλ2 ′′ )x+ (µ2 ′ + iµ2 ′′ )y + (ν2 ′ + iν2 ′′ )z
−ρ1
= constant = C1 , say.
This implies that
ln u1 = (ρ2 ′ + iρ2 ′′ ) ln(λ1 x + µ1 y + ν1 z)
− ρ1 ln (λ2 ′ + iλ2 ′′ )x + (µ2 ′ + iµ2 ′′ )y + (ν2 ′ + iν2 ′′ )z
70
S.M.H. Islam, J. Das (N. Chaudhuri)
= (ρ2 ′ + iρ2 ′′ ) ln(λ1 x + µ1 y + ν1 z)
− ρ1 ln (λ2 ′ x + µ2 ′ y + ν2 ′ z) + i(λ2 ′′ x + µ2 ′′ y + ν2 ′′ z)
= ln C1 .
Writing ln (λ2 ′ x + µ2 ′ y + ν2 ′ z) + i(λ2 ′′ x + µ2 ′′ y + ν2 ′′ z) = z1 + iz2
it is noted that
1 ′
z1 =
ln (λ2 x + µ2 ′ y + ν2 ′ z)2 + (λ2 ′′ x + µ2 ′′ y + ν2 ′′ z)2 ,
2
λ2 ′′ x + µ2 ′′ y + ν2 ′′ z
,
z2 = arctan
λ2 ′ x + µ 2 ′ y + ν 2 ′ z
(13)
(14)
and we get
ln u1 = (ρ2 ′ + iρ2 ′′ ) ln(λ1 x + µ1 y + ν1 z) − ρ1 (z1 + iz2 )
= ρ2 ′ ln(λ1 x + µ1 y + ν1 z) − ρ1 z1 + i ρ2 ′′ ln(λ1 x + µ1 y + ν1 z) − ρ1 z2
= ln C1 .
Hence
u1 = exp ρ2 ′ ln(λ1 x + µ1 y + ν1 z) − ρ1 z1
cos ρ2 ′′ ln(λ1 x + µ1 y + ν1 z) − ρ1 z2
+ i sin ρ2 ′′ ln(λ1 x + µ1 y + ν1 z) − ρ1 z2
!
′
= (λ1 x + µ1 y + ν1 z)ρ2 exp(−ρ1 z1 )
cos ρ2 ′′ ln(λ1 x + µ1 y + ν1 z) − ρ1 z2
+ i sin ρ2 ′′ ln(λ1 x + µ1 y + ν1 z) − ρ1 z2
!
=C1 .
So one solution of the simultaneous equations (12), say u, can be taken as
′
u = (λ1 x + µ1 y + ν1 z)ρ2 exp(−ρ1 z1 ) cos ρ2 ′′ ln(λ1 x + µ1 y + ν1 z) − ρ1 z2
A METHOD OF SOLVING LAGRANGE’S FIRST-ORDER...
71
= constant. (15)
From the last equation of (12) we have another solution of the simultaneous
equations (12), say v1 , as
ρ2 ′ −iρ2 ′′
v1 = (λ2 ′ x + µ2 ′ y + ν2 ′ z) + i(λ2 ′′ x + µ2 ′′ y + ν2 ′′ z)
−ρ2 ′ −iρ2 ′′
= constant
(λ2 ′ x + µ2 ′ y + ν2 ′ z) − i(λ2 ′′ x + µ2 ′′ y + ν2 ′′ z)
=C2 (say),
or,
ln v1 = (ρ2 ′ − iρ2 ′′ )(z1 + iz2 ) − (ρ2 ′ + iρ2 ′′ )(z1 − iz2 ) = ln C2 ,
where z1 , z2 are given by (13) and (14) respectively.
This shows that
ln v1 = 2i(ρ2 ′ z2 − ρ2 ′′ z1 ) = ln C2 .
Hence another solution of the simultaneous equations (12), say v, can be taken
as
v = ρ2 ′ z2 − ρ2 ′′ z1 = constant.
(16)
Therefore, in this case, the required solution of the PDE (1.1) is given by
F
′
(λ1 x + µ1 y + ν1 z)ρ2 exp(−ρ1 z1 ) cos ρ2 ′′ ln(λ1 x + µ1 y + ν1 z) − ρ1 z2 ,
ρ2 ′ z2 − ρ2 ′′ z1
!
= 0, (17)
where z1 , z2 are given by (13) and (14) respectively, and as in Case I and
Example 1, F is an arbitrary real-valued function of two real variables.
Example 2.
yp + zq = x.
(18)
The simultaneous ordinary differential equations corresponding to this PDE are
dx
dy
dz
=
=
.
y
z
x
It is noted that the fourth equation of (4) is automatically satisfied.
(19)
72
or,
S.M.H. Islam, J. Das (N. Chaudhuri)
In this case, the equation (5) becomes
−ρ 0
1
1 −ρ 0 = 0
0
1 −ρ
ρ3 − 1 = 0.
Roots of this equation are 1, ω, ω 2 ; where ω =
For ρ = 1,
(20)
√
−1±i 3
.
2
µ
ν
λ
= = .
1
1
1
For ρ = ω,
λ
µ
ν
= = 2.
ω
1
ω
For ρ = ω 2 ,
λ
µ
ν
= = .
ω2
1
ω
So each ratio of (19) is equal to
ωdx + dy + ω 2 dz
dx + dy + dz ωdx + dy + ω 2 dz
=
=
2
y+z+x
ωy + z + ω x
ω(ωx + y + ω 2 z)
ω 2 dx + dy + ωdz
ω 2 dx + dy + ωdz
=
=
.
ω 2 y + z + ωx
ω 2 (ω 2 x + y + ωz)
The required solutions of the PDE (18) are then obtained as
2
F (x + y + z)ω (ωx + y + ω 2 z)−1 , (x + y + z)ω (ω 2 x + y + ωz)−1 = 0,
where F is an arbitrary real-valued function of two real variables.
5. Case III: ρ1 = ρ2 6= ρ3
In this case, the method described above for the cases Case I and II does not
work. This is exhibited by citing the following example.
Example 3.
3
(y − z)p + (z + x)q = x + y.
4
(21)
A METHOD OF SOLVING LAGRANGE’S FIRST-ORDER...
73
The simultaneous ordinary differential equations corresponding to this PDE are
dx
dy
dz
.
=
=
(y − z)
(z + x)
x + 43 y
For this PDE the equation (5) becomes
−ρ 1
1 1 −ρ 3 = 0,
4 −1 1 −ρ
which leads to
(22)
(23)
(ρ − 1)(2ρ + 1)2 = 0.
Roots of this equation are 1, − 12 , − 12 .
For ρ = 1,
1
For ρ = − ,
2
µ
ν
λ
= = .
1
1
0
λ
µ
ν
= =
.
2
5
−6
So each ratio of (22) is equal to
2dx + 5dy − 6dz
dx + dy
= 1
.
y+x
− 2 (2x + 5y − 6z)
This relation gives us the single solution of the equation (2), in the present case,
as
u = (x + y)(2x + 5y − 6z)2 = constant.
Being unable to find another solution of the simultaneous equations (2), in the
present case, the method described above fails to derive solutions of the PDE
(21).
6. Case IV: ρ1 = ρ2 = ρ3
In this case, no solution of the simultaneous equations (2) can be derived and
hence the method presented fails to find any solution of the PDE under consideration. The following example exhibits the difficulty.
Example 4.
(x − 2y + z)p + (2x + y + z)q = 2x + 2y + z.
(24)
74
S.M.H. Islam, J. Das (N. Chaudhuri)
The simultaneous ordinary differential equations corresponding to this PDE are
dx
dy
dz
=
=
.
(x − 2y + z)
(2x + y + z)
2x + 2y + z
For this PDE the equation (5) becomes
1 − ρ
2
2 −2 1 − ρ
2 = 0,
1
1
1 − ρ
(25)
(26)
which leads to the equation
(1 − ρ)3 = 0.
Roots of this equation are 1, 1, 1.
µ
ν
λ
=
= .
1
−1
1
So, in this case, the equation (3) gives us the only ratio
For ρ = 1,
dx − dy + dz
.
x−y+z
(27)
Hence the method described above fails to give us the required solutions of the
PDE (24).
7. Alternative Approach to Solve the Problems
in Case III and Case IV
To find the solutions of the PDEs (21) and (24) the following approch has been
found to be suitable in case of the examples cited.
7.1. Solution of the PDE (21)
The simultaneous ordinary differential equations corresponding to the PDE (21)
are
dy
dz
dx
.
(28)
=
=
(y − z)
(z + x)
x + 43 y
Suppose it is possible to find numbers αi , βi , γi (i = 1, 2, 3) and ρ such that
each ratio of (28) is equal to
(α1 x + β1 y + γ1 z)dx + (α2 x + β2 y + γ2 z)dy + (α3 x + β3 y + γ3 z)dz
(α1 x + β1 y + γ1 z)(y − z) + (α2 x + β2 y + γ2 z)(z + x) + (α3 x + β3 y + γ3 z)(x + 34 y)
A METHOD OF SOLVING LAGRANGE’S FIRST-ORDER...
75
(α1 x + β1 y + γ1 z)dx + (α2 x + β2 y + γ2 z)dy + (α3 x + β3 y + γ3 z)dz
(α2 + α3 )x2 + (β1 + 34 β3 )y 2 + (−γ1 + γ2 )z 2 + (α1 + β2 + 43 α3 + β3 )xy
3
+ (−β1 + γ1 + β2 + γ3 )yz + (−α1 + α2 + γ2 + γ3 )zx
4
dD
=
,
(29)
ρd
=
where
3
D = (α2 + α3 )x2 + (β1 + β3 )y 2 + (−γ1 + γ2 )z 2
4
3
3
+ (α1 + β2 + α3 + β3 )xy + (−β1 + γ1 + β2 + γ3 )yz + (−α1 + α2 + γ2 + γ3 )zx,
4
4
and dD denotes the total derivative of D.
We see that (29) holds if
ρα1 = 2α2 + 2α3 ,
3
ρβ1 = α1 + β2 + α3 + β3 ,
4
ργ1 = −α1 + α2 + γ2 + γ3 ,
3
ρα2 = α1 + β2 + α3 + β3 ,
4
3
ρβ2 = 2β1 + β3 ,
2
3
ργ2 = −β1 + γ1 + β2 + γ3 ,
4
ρα3 = −α1 + α2 + γ2 + γ3 ,
3
ρβ3 = −β1 + γ1 + β2 + γ3 ,
4
ργ3 = −2γ1 + 2γ2 .
The above equations give us a linear homogeneous system of equations
−ρα1 + 2β1 + 2γ1
3
α1 − ρβ1 + β2 + γ1 + γ2
4
3
2β1 − ρβ2 + γ2
2
−α1 + β1 − ργ1 + γ2 + γ3
3
−β1 + β2 + γ1 − ργ2 + γ3
4
−2γ1 + 2γ2 − ργ3
= 0,
= 0,
= 0,
= 0,
= 0,
= 0,
(30)
76
S.M.H. Islam, J. Das (N. Chaudhuri)
where α2 = β1 , α3 = γ1 , β3 = γ2 . The linear homogeneous system (30) will
have a non-trivial solution for (α1 , β1 , β2 , γ1 , γ2 , γ3 ) if the determinant of the
coefficient matrix of the system (30) is zero, i.e.
−ρ 2
0
2
0
0 3
1 −ρ 1
1
0 4
3
0
2 −ρ 0
0 2
= 0.
−1 1
0 −ρ 1
1 0 −1 1
1 −ρ 34 0
0
0 −2 −2 −ρ
This leads to the equation
ρ6 −
1
15 4 7 3 9 2 3
ρ − ρ + ρ + ρ − = 0.
4
4
4
4
2
Roots of this equation are found to be 2, −1, −1, −1, 12 , 12 .
Using these values of ρ in (30) solutions for (α1 , β1 , β2 , γ1 , γ2 , γ3 ) are found
to be as follows:
For ρ = 2, α1 = β1 = β2 = 1, γ1 = γ2 = γ3 = 0.
For ρ = −1, α1 = −4, β1 = −10, β2 = −25,γ1 = 12, γ2 = 30, γ3 = −36.
For ρ = 21 , α1 = 4, β1 = 7, β2 = 10, γ1 = −6, γ2 = −6, γ3 = 0.
Again we have the relations α2 = β1 , α3 = γ1 , β3 = γ2 .
So using the above values of ρ, αi , βi , γi (i = 1, 2, 3) in (29) we have the
ratios:
2
2
d(2x2 + 25
d(x2 + y 2 + 2xy)
2 y + 18z + 10xy − 30yz − 12zx)
=
2
2
2(x2 + y 2 + 2xy)
−1(2x2 + 25
2 y + 18z + 10xy − 30yz − 12zx)
=
d(2x2 + 5y 2 + 7xy − 6yz − 6zx)
.
2(2x2 + 5y 2 + 7xy − 6yz − 6zx)
These ratios give us the required solution of the PDE (21) as
25
F (x + y)(2x2 + y 2 + 18z 2 + 10xy − 30yz − 12zx),
2
(2x2 + 5y 2 + 7xy − 6yz − 6zx)(x + y)2 = 0,
where F is an arbitrary real-valued function of two real variables.
A METHOD OF SOLVING LAGRANGE’S FIRST-ORDER...
77
7.2. Solution of the PDE (24)
The simultaneous ordinary differential equations corresponding to the PDE (24)
are
dx
dy
dz
=
=
.
(31)
x − 2y + z
2x + y + z
2x + 2y + z
Suppose it is possible to find numbers αi , βi , γi (i = 1, 2, 3) and ρ such that
each ratio of (31) is equal to
(α1 x + β1 y + γ1 z)dx + (α2 x + β2 y + γ2 z)dy + (α3 x + β3 y + γ3 z)dz
(α1 x + β1 y + γ1 z)(x − 2y + z) + (α2 x + β2 y + γ2 z)(2x + y + z) + (α3 x + β3 y + γ3 z)(2x + 2y + z)
=
(α1 x + β1 y + γ1 z)dx + (α2 x + β2 y + γ2 z)dy + (α3 x + β3 y + γ3 z)dz
(α1 + 2α2 + 2α3 )x2 + (−2β1 + β2 + 2β3 )y 2 + (γ1 + γ2 + γ3 )z 2 + (−2α1 + β1 + α2 + 2β2
+ 2α3 + 2β3 )xy + (β1 − 2γ1 + β2 + γ2 + β3 + 2γ3 )yz + (α1 + γ1 + α2 + 2γ2 + α3 + 2γ3 )zx
=
dD
ρd
(32)
,
where
D = (α1 + 2α2 + 2α3 )x2 + (−2β1 + β2 + 2β3 )y 2
+ (γ1 + γ2 + γ3 )z 2 + (−2α1 + β1 + α2 + 2β2 + 2α3 + 2β3 )xy
+ (β1 − 2γ1 + β2 + γ2 + β3 + 2γ3 )yz + (α1 + γ1 + α2 + 2γ2 + α3 + 2γ3 )zx,
and dD denotes the total derivative of D. We see that (32) holds if
ρα1 = 2α1 + 4α2 + 4α3 ,
ρβ1 = −2α1 + β1 + α2 + 2β2 + 2α3 + 2β3 ,
ργ1 = α1 + γ1 + α2 + 2γ2 + α3 + 2γ3 ,
ρα2 = −2α1 + β1 + α2 + 2β2 + 2α3 + 2β3 ,
ρβ2 = −4β1 + 2β2 + 4β3 ,
ργ2 = β1 − 2γ1 + β2 + γ2 + β3 + 2γ3 ,
ρα3 = α1 + γ1 + α2 + 2γ2 + α3 + 2γ3 ,
ρβ3 = β1 − 2γ1 + β2 + γ2 + β3 + 2γ3 ,
ργ3 = 2γ1 + 2γ2 + 2γ3 .
The above equations give us a linear homogeneous system of equations
(2 − ρ)α1 + 4β1 + 4γ1 = 0,
−2α1 + (2 − ρ)β1 + 2β2 + 2γ1 + 2γ2 = 0,
−4β1 + (2 − ρ)β2 + 4γ2 = 0,
α1 + β1 + (2 − ρ)γ1 + 2γ2 + 2γ3 = 0,
β1 + β2 − 2γ1 + (2 − ρ)γ2 + 2γ3 = 0,
2γ1 + 2γ2 + (2 − ρ)γ3 = 0,
(33)
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S.M.H. Islam, J. Das (N. Chaudhuri)
where α2 = β1 , α3 = γ1 , β3 = γ2 .
The linear homogeneous system (33) will have a non-trivial solution for
(α1 , β1 , β2 , γ1 , γ2 , γ3 ) if the determinant of the coefficient matrix of the system
(33) is zero, i.e.
2 − ρ
4
0
4
0
0 −2 2 − ρ
2
2
2
0 0
−4 2 − ρ
0
4
0 = 0.
1
1
0
2−ρ
2
2 0
1
1
−2 2 − ρ
2 0
0
0
2
2
2 − ρ
This leads to the equation
(2 − ρ)6 = 0.
So all the roots of this equation are equal to 2.
For ρ = 2 the system (33) has two linearly independent solutions, namely
(3, −1, 3, 1, −1, 0), (2, 0, 2, 0, 0, −1) for (α1 , β1 , β2 , γ1 , γ2 , γ3 ).
Using the values of ρ, αi , βi , γi (i = 1, 2, 3) in (32) we see that each of the
ratios in (32) is equal to
d(2x2 + 2y 2 − z 2 )
d(3x2 + 3y 2 − 2xy − 2yz + 2zx)
=
.
2(3x2 + 3y 2 − 2xy − 2yz + 2zx)
2(2x2 + 2y 2 − z 2 )
Again by (6.2) we have found that each ratios in (6.2) is equal to
d(x − y + z)
.
(x − y + z)
From the above three equal ratios we get the solution of the PDE (24) as
F
2x2 + 2y 2 − z 2
(x − y + z)2
,
3x2 + 3y 2 − 2xy − 2yz + 2zx = 0,
(x − y + z)2
where F is an arbitrary real-valued function of two real variables.
8. Remarks
The genesis of the method described in §§ 3, 4, 7, lies in finding an equivalent
ratio of the ratios in (2.1), in which the numerator is the total differential of the
denominetor. The idea may be employed to other types of equations. Further
works in this direction are under progress, to be reported in future.
A METHOD OF SOLVING LAGRANGE’S FIRST-ORDER...
79
References
[1] E.L. Ince, Ordinary Differential Equations, Dover Publications, New York,
1956.
80