2.1. CHARPIT’S METHOD 13
Before trying to answer such questions, it is important to know that an actual solution exists.
Namely, does a solution exist that satisfies both the original PDE and second appended PDE?
So, in the first example, does a solution exist that satisfies both
xu
x
u
2
y
D 2u and u
y
D y‹ (2.38)
Here, we use the second in the first, and ask: does a solution exist to
u
x
D
2u
x
C
y
2
x
; u
y
D y‹ (2.39)
If so, then they certainly would be compatible, so
@u
x
@y
D
@u
y
@x
. Calculating these gives
2u
y
x
C
2y
x
?
D 0 (2.40)
and since u
u
D y then (2.40) is identically satisfied, so the two equations of (2.38) are com-
patible. For the second example we ask: are the following compatible?
xu
x
u
2
y
D 2u and u
x
D x: (2.41)
We substitute the second into the first and then seek compatibility of
x
2
u
2
y
D 2u and u
x
D x: (2.42)
To show compatibility, we differentiate the first of (2.42) with respect to x giving
2x 2u
y
u
xy
D 2u
x
(2.43)
and since
u
x
D x then (2.43) is identically satisfied then (2.41) are compatible. For the final
example, we ask: are these compatible?
xu
x
u
2
y
D 2u and u
x
C xu
y
D x xy: (2.44)
Definitely a harder problem to explicitly find u
x
and u
y
but is that really necessary? If they were
compatible, they would have the same second derivatives. If we calculate the x and y derivatives
of each we obtain
xu
xx
2u
y
u
xy
D u
x
; (2.45a)
xu
xy
2u
y
u
yy
D 2u
y
; (2.45b)
u
xx
C xu
xy
C u
y
D 1 y; (2.45c)
u
xy
C xu
yy
D x: (2.45d)
Solving (2.45c) and (2.45d) for u
xx
and u
xy
gives
u
xx
D x
2
u
yy
u
y
C x
2
y C1; u
xy
D xu
yy
x; (2.46)