100 5. FIRST INTEGRALS
which simplifies (5.60b) to
uF
p
F
x
C uF
y
C
.
p Cuq
/
F
u
D 0 (5.62)
or
F
x
C uF
y
C
.
p Cuq
/
F
u
D 0: (5.63)
Note that we have excluded the case where F
p
D 0, as this would gives F
q
D 0. e solution
of (5.61) is
F D G
.
x; y; u; p C uq
/
: (5.64)
Substituting this into (5.63) and letting D p C uq gives
G
x
C uG
y
C G
u
C qG
D 0: (5.65)
Since G is independent of q, this gives
G
D 0: (5.66)
Normally we would say that G
D 0 which would then give that F is independent of p and q,
i.e., no first integral. But we also have the choice that D 0, and one can show that, in fact,
D p C uq D 0 is a first integral.
5.2 MONGE–AMPERE EQUATION
In this section we consider Monge–Ampere equations. ese are second-order PDEs in two
independent variables in the form
Ar C Bs C C t C D
rt s
2
D E; (5.67)
where in addition to the forms of A; B; C; and E in the previous section, we also assume that
D ¤ 0 and is a function of x; y; u; u
x
; and u
y
. We again assume a first integral of the form (5.8)
and on isolating the derivatives r and t and substituting into (5.67) and isolating coefficients
with respect to s gives
AF
2
q
BF
p
F
q
C CF
2
p
D D
F
x
F
p
C F
y
F
q
C pF
u
F
p
C qF
u
F
q
;
(5.68a)
AF
q
.
F
x
C pF
u
/
C CF
p
F
y
C qF
u
C EF
p
F
q
D D
F
x
F
y
C pF
y
F
u
C qF
x
F
u
C pqF
2
u
:
(5.68b)
In an attempt to solve (5.68), we assume there exists a .x; y; u; p; q/, such that
AF
2
q
BF
p
F
q
C CF
2
p
D
F
x
F
p
C F
y
F
q
C pF
u
F
p
C qF
u
F
q
CAF
q
.
F
x
C pF
u
/
C CF
p
F
y
C qF
u
C EF
p
F
q
D
F
x
F
y
C pF
y
F
u
C qF
x
F
u
C pqF
2
u
D
A
1
F
x
C B
1
F
u
C C
1
F
p
C D
1
F
q
A
2
F
y
C B
2
F
u
C C
2
F
p
C D
2
F
q
:
(5.69)
5.2. MONGE–AMPERE EQUATION 101
Expanding (5.69) and equating coefficients of the product of F
x
; F
y
; F
u
; F
p
; F
q
gives
A
1
A
2
D D; A
2
B
1
D pD; A
1
B
2
D qD;
B
1
B
2
D pqD; A
1
C
2
D D; D
1
A
2
D D;
A
2
C
1
D C; A
1
D
2
D A; C
1
C
2
D C; D
1
D
2
D A; (5.70)
B
1
D
2
C D
1
B
2
D pA qD; B
1
C
2
C C
1
B
2
D qC pD;
C
1
D
2
C D
1
C
2
D B CE;
from which we find the following:
B
1
D pA
1
; C
1
D
A
1
C
D
; D
1
D A
1
;
A
2
D
D
A
1
; B
2
D
qD
A
1
; C
2
D
D
A
1
; D
2
D
A
A
1
; (5.71)
where satisfies
D
2
2
BD C AC CDE D 0: (5.72)
Solving (5.72) for gives rise to the following possibilities:
(i) two real distinct roots,
(ii) two real repeated roots, and
(iii) two complex roots.
In the case of two distinct roots, say
1
and
2
, (5.69) can be factored as
DF
x
C pDF
u
CF
p
C
1
DF
q
DF
y
C qDF
u
C
1
DF
p
AF
q
D 0; (5.73a)
DF
x
C pDF
u
CF
p
C
2
DF
q
DF
y
C qDF
u
C
2
DF
p
AF
q
D 0: (5.73b)
ere are four systems to be considered, however, two of these give
1
D
2
; as we have assumed
that
1
¤
2
, these are inadmissible. us, we shall only consider the systems
D
.
F
x
C pF
u
/
CF
p
C
1
DF
q
D 0; (5.74a)
D
F
y
C qF
u
C
2
DF
p
AF
q
D 0; (5.74b)
and
D
.
F
x
C pF
u
/
CF
p
C
2
DF
q
D 0; (5.75a)
D
F
y
C qF
u
C
1
DF
p
AF
q
D 0: (5.75b)
In the case of a repeated root, we only have one set of equations for F to consider, as (5.74)
and (5.75) coalesce into one. e following examples illustrate.
102 5. FIRST INTEGRALS
Example 5.5 Obtain first integral to
r C 3s C t C rt s
2
D 1: (5.76)
Identifying that A D 1, B D 2, C D 1, D D 1, and E D 1, then (5.72) gives
2
3 C 2 D 0; (5.77)
leading to D 1; 2. us, we have two cases to consider.
Case 1: From (5.74) we have
F
x
C pF
u
F
p
C F
q
D 0; (5.78a)
F
y
C qF
u
C 2F
p
F
q
D 0: (5.78b)
Using the method of characteristic, we solve the first, giving
F D F
x C p; y; u C
1
2
p
2
; x q
: (5.79)
Substituting into the second gives
F
y
C .x /F
ˇ
C 2
F
˛
C .˛ x/F
ˇ
C F
D 0; (5.80)
where ˛ D x C p, ˇ D u C
1
2
p
2
, and D x q. With these variables, F is independent of x;
this leads to
F
ˇ
D 0; F
y
C 2F
˛
C F
D 0; (5.81)
from which we obtain the final form of first integral
F .x 2y Cp; y x C q/ D 0: (5.82)
Case 2: From (5.75) we have
F
x
C pF
u
F
p
C 2F
q
D 0; (5.83a)
F
y
C qF
u
C F
p
F
q
D 0: (5.83b)
Using the method of characteristic, we solve the first, giving
F D F
x C p; y; u C
1
2
p
2
; 2x q
: (5.84)
Substituting into the second gives
F
y
C .2x /F
ˇ
C F
˛
C .˛ x/F
ˇ
C F
D 0; (5.85)
5.2. MONGE–AMPERE EQUATION 103
where ˛ D x C p, ˇ D u C
1
2
p
2
, and D 2x q. With these variables, F is independent of x;
this leads to
F
ˇ
D 0; F
y
C F
˛
C F
D 0; (5.86)
from which we obtain the final form of first integral
F .x y Cp; y 2x C q/ D 0: (5.87)
Any choice of F in (5.82) or (5.87) will lead to an exact solution of the original PDE. For
example, from (5.87), choosing
a.x y C p/ C b.y 2x C q/ D 0; (5.88)
where a and b are arbitrary constants and f is an arbitrary function, gives
u D
b
2
ab a
2
2a
2
x
2
C
2a b
a
xy Cf .bx ay/; (5.89)
an exact solution to (5.76).
Example 5.6 Obtain first integral to
xqr .x Cy/s Cypt C xy
rt s
2
D 1 pq: (5.90)
Identifying that A D xq, B D .x C y/, C D yp, D D xy, and E D 1 pq, then from (5.72)
gives
x
2
y
2
2
C xy.x C y/ C xypq C xy.1 pq/ D 0; (5.91)
leading to D
1
x
;
1
y
. us, we have two cases to consider.
Case 1: From (5.74) we have
xF
x
C xpF
u
pF
p
F
q
D 0; (5.92a)
yF
y
C yqF
u
F
p
qF
q
D 0: (5.92b)
Using the method of characteristic, we solve the first, giving
F D F
.
xp; y; u xp ln x; ln x Cq
/
: (5.93)
Substituting into the second gives
F
y
C y. ln x/F
ˇ
xF
˛
C x ln xF
ˇ
C
.
ln x
/
F
D 0; (5.94)
or, re-arranging gives
F
y
C yF
ˇ
F
xF
˛
C ln x
F
yF
ˇ
C x ln xF
ˇ
D 0; (5.95)
104 5. FIRST INTEGRALS
where ˛ D x Cp, ˇ D u C
1
2
p
2
, and D x q. With these variables, F is independent of x
and thus leads to
F
y
D 0; F
˛
D 0; F
ˇ
D 0; F
D 0; (5.96)
gives that F is constant. So in this case, there is no first integral.
Case 2: From (5.75) we have
xyF
x
C xypF
u
ypF
p
xF
q
D 0; (5.97a)
xyF
y
C xyqF
u
yF
p
xqF
q
D 0: (5.97b)
Using the method of characteristic, we solve the first, giving
F D F
.
xp; y; u C xp ln x; x C yq
/
: (5.98)
Substituting (5.98) into (5.97b) and re-arranging gives
F
y
yF
˛
F
ˇ
x C F
ˇ
x
2
xy ln xF
ˇ
D 0; (5.99)
where ˛ D xp, ˇ D u C xp ln x, and D x C yq. With these variables, F is independent of x;
this leads to
F
ˇ
D 0; F
y
F
˛
D 0; (5.100)
from which we obtain F D F .; y C ˛/ and the final form of first integral
F .xp C y; yq Cx/ D 0: (5.101)
Any choice of F in (5.101) will lead to an exact solution of the original PDE. For example,
choosing
xp C yq C x C y D 0 (5.102)
leads to the exact solution
u D .x Cy/ C f .x=y/; (5.103)
where f is arbitrary as an exact solution to (5.90).
Example 5.7 Obtain first integral to
yqr C 2xys C y
2
t C y
2
rt s
2
D x
2
yq: (5.104)
Identifying that A D yq, B D 2xy, C D y
2
, D D y
2
, and E D x
2
yq, from (5.72) we get
y
4
2
2xy
3
C x
2
y
2
D 0; (5.105)
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