5.2. MONGE–AMPERE EQUATION 105
leading to the repeated root D x=y. us, we have only a single case to consider. From (5.74)
we have
yF
x
C ypF
u
yF
p
C xF
q
D 0; (5.106a)
yF
y
C yqF
u
C xF
p
yF
q
D 0: (5.106b)
Using the method of characteristic, we solve the first, giving
F D F
x C p; y; 2u C p
2
; 2yq x
2
: (5.107)
Substituting into the second gives
yF
y
C F
ˇ
C
F
˛
C 2˛F
ˇ
x F
ˇ
x
2
D 0; (5.108)
where ˛ D x C p, ˇ D u C
1
2
p
2
, and D
1
2
x
2
yq. With these variables, F is independent of
x; this leads to
F
ˇ
D 0; F
˛
C 2˛F
ˇ
D 0; yF
y
C F
ˇ
D 0; (5.109)
from which we obtain F D F ./, leading to the final form of first integral
F
2yq x
2
D 0: (5.110)
For example, if we choose
2yu
y
x
2
D 0 (5.111)
we integrate, giving
u D
1
2
x
2
ln jyj C C.x/; (5.112)
where C.x/ is an arbitrary function and substitution into (5.104) shows it is exactly satisfied.
Example 5.8 Consider
u
xx
C u
yy
C 2
u
xx
u
yy
u
2
xy
D 0: (5.113)
Here
A D 1; B D 0; C D 1; D D 2; E D 0; (5.114)
and from (5.72) we have
4
2
C 1 D 0; (5.115)
giving D ˙i=2. us, (5.74) and (5.75) become
2F
x
C 2pF
u
F
p
iF
q
D 0; (5.116a)
2F
y
C 2qF
u
C iF
p
F
q
D 0; (5.116b)
106 5. FIRST INTEGRALS
and
2F
x
C 2pF
u
F
p
C iF
q
D 0; (5.117a)
2F
y
C 2qF
u
iF
p
F
q
D 0: (5.117b)
e solutions of each are
F
1
.x C 2u
x
iy; u
x
C iu
y
/ D 0;
F
2
.x C 2u
x
C iy; u
x
iu
y
/ D 0:
(5.118)
Again, one will notice that the first integrals in (5.118) are complex. However, it is again possible
to construct real solutions to (5.113). For example, if we choose
x C 2u
x
iy C u
x
C iu
y
D 0; (5.119a)
x C 2u
x
C iy C u
x
iu
y
D 0; (5.119b)
then we can split into real and imaginary parts, giving
x C 3u
x
D 0; (5.120a)
y u
y
D 0; (5.120b)
which leads to the solution
u D
x
2
6
C
y
2
2
C c; (5.121)
where c is an arbitrary constant.
If we choose
.
x C 2u
x
iy
/
2
C u
x
C iu
y
D 0; (5.122a)
.
x C 2u
x
C iy
/
2
C u
x
iu
y
D 0; (5.122b)
then we can split into real and imaginary parts, giving
4yu
x
u
y
C 2xy D 0; (5.123a)
4u
2
x
C .4x C 1/u
x
C x
2
y
2
D 0; (5.123b)
which leads to the common solution
u D
2x
2
C 2y
2
C x
8
˙
.1 C 8x C 16y
2
/
3=2
96
C c; (5.124)
where c is an arbitrary constant. e reader can verify that (5.121) and (5.124) do indeed sat-
isfy (5.113).
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