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)