124 5. FIRST INTEGRALS
We find the contact conditions are satisfied and lead to U D u C p
2
, and from
x D X 2U
X
; y D Y; u D U U
2
X
p D
1
U
X
; q D
U
Y
U
X
; (5.258)
the PDE (5.255) is transformed to U
XX
C U
Y Y
D 0.
Example 5.17 Consider
u
y
u
xx
u
yy
C
1 C u
y
u
xx
u
yy
u
2
xy
D u
y
: (5.259)
is PDE admits the first integrals
F .p x ˙ i q; y C q C ln q ˙ ix/ D 0: (5.260)
Here we try
X D p x; Y D q; P D x; Q D y C q C ln q: (5.261)
We find that the contact conditions (5.165) are satisfied, leading to
U D xp C yq u
1
2
q
2
C q ln q q: (5.262)
From (5.261) and (5.262),
x D U
X
; y D U
Y
Y ln Y; u D XU
X
C Y U
Y
U C
1
2
U
2
X
1
2
Y
2
Y (5.263)
transforms (5.259) to
U
XX
C U
Y Y
D 0: (5.264)
5.5 EXERCISES
5.1. Find a first integral for the following:
.i/ q
2
r 2pqs C p
2
t D 0;
.i i/ qr .p C q C 1/s C .p C 1/t D 0;
.i i i/ rt s
2
C 1 D 0;
.iv/ yr ps C t C y.rt s
2
/ C 1 D 0:
5.2. e equations that model one-dimensional gas flow are
t
C .u/
x
D 0; (5.265a)
u
t
C uu
x
D
P
x
; (5.265b)
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