2.1. CHARPIT’S METHOD 11
2.1 CHARPIT’S METHOD
Obtaining exact solutions to NLPDEs such as (2.1) can be a difficult task as we are required
to solve equations such as (2.4)! e difficulty is not so much in solving these characteristic
equations but in eliminating the five arbitrary functions that appear upon integration. So we
ask, is it possible to obtain exact solutions another way?
Consider the PDE
u
y
D y: (2.21)
is integrates to give
u D
y
2
2
C f .x/ (2.22)
and substitution into the original PDE (2.1) gives
xf
0
D 2f: (2.23)
is ODE is solved giving
f D cx
2
(2.24)
which from (2.22) leads to
u D cx
2
y
2
2
(2.25)
and the initial condition (2.2) gives c D 1=2, and leads to the solution given in (2.20).
Consider the PDE
u
x
D x: (2.26)
is integrates to give
u D
x
2
2
C g.y/ (2.27)
and substitution into the original PDE (2.1) gives
g
02
D 2g: (2.28)
is ODE is solved giving (we will omit the trivial solution g D 0)
g D
y
2
2
C cy
c
2
2
(2.29)
which from (2.27) and the initial condition (2.2) leads again to the solution (2.20).
As for the final example, consider the PDE
u
x
C xu
y
D x xy (2.30)
is integrates to give
u D y
1
2
y
2
C h
y
1
2
x
2
(2.31)
12 2. COMPATIBILITY
and substitution into the original PDE (2.1) gives
2. 1/h
0
h
02
1 D 2h; (2.32)
where h D h./ and D y
1
2
x
2
. is ODE actually has two solutions
h D c
1
2
.c C 1/
2
; h D
1
2
2
; (2.33)
and leads to the exact solutions
u D y
1
2
2
C c
y
1
2
x
2
1
2
.c C 1/
2
;
u D
1
2
.x
2
y
2
/ C
1
2
y
1
2
x
2
2
:
(2.34)
However, only the first solution will give rise to (2.20). A question we ask: is it really necessary
to solve the reduced ODEs given in (2.23), (2.28), and (2.32)? Consider the solution forms that
lead to these ODEs
u D
y
2
2
C f .x/
u D
x
2
2
C g.y/
u D y
1
2
y
2
C h
y
1
2
x
2
:
(2.35)
If we impose the initial condition u.x; x/ D 0 in each solution form (2.35), we obtain the so-
lution u D
1
2
.x
2
y
2
/, but we certainly would have missed obtaining the second exact solution
in (2.34). So are there others? For example, both
u
y
D x
2
;
xu
x
C 2yu
y
D 2u y
2
(2.36)
will lead to exact solutions of the given PDE (2.1), however, the boundary conditions might not
be satisfied. In fact any PDE of the form
F
u
x
x
;
u
2
x
u
y
; u
y
C y; xu
x
u
2
y
2u
D 0 (2.37)
will give rise to an exact solution to (2.1). A number of questions arise.
1. Where did these associated PDEs come from?
2. How do we know that they will lead to a solution that also satisfies the BC?
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)
14 2. COMPATIBILITY
and the two remaining equations in (2.45) become
x.2u
y
C x
2
/u
yy
u
x
C xu
y
C x
3
xy C x D 0; (2.47a)
.2u
y
C x
2
/.u
yy
C 1/ D 0: (2.47b)
If (2.44) were to be compatible, then (2.47) would be identically satisfied. So we see two cases
emerge:
.i/ 2u
y
C x
2
D 0;
.ii/ 2u
y
C x
2
¤ 0:
In the first case where u
y
D
1
2
x
2
, (2.47a) reduces to
2u
x
1
3
x
3
x C xy D 0;
which is compatible with u
y
D
1
2
x
2
and and so (2.44) are compatible. In the second case where
u
yy
D 1, (2.47a) reduces to
u
x
C xu
y
C x.y 1/ D 0
which is identically satisfied by virtue of (2.44) and again, (2.44) are compatible.
So now we know how to determine when two first-order PDEs are compatible. Our next
step is to determine how they come about.
Consider the compatibility of the following first-order PDEs
F .x; y; u; p; q/ D 0;
G.x; y; u; p; q/ D 0;
(2.48)
where p D u
x
and q D u
y
. Calculating the x and y derivatives of (2.48) gives
F
x
C pF
u
C u
xx
F
p
C u
xy
F
q
D 0;
F
y
C qF
u
C u
xy
F
p
C u
yy
F
q
D 0;
G
x
C pG
u
C u
xx
G
p
C u
xy
G
q
D 0;
G
y
C qG
u
C u
xy
G
p
C u
yy
G
q
D 0:
(2.49)
Solving the first three (
2.49) for u
xx
, u
xy
, and u
yy
gives
u
xx
D
F
x
G
q
p F
u
G
q
C F
q
G
x
C p F
q
G
u
F
p
G
q
F
q
G
p
;
u
xy
D
F
p
G
x
p F
p
G
u
C F
x
G
p
C p F
u
G
p
F
p
G
q
F
q
G
p
;
u
yy
D
F
2
p
G
x
C p F
2
p
G
u
F
y
F
p
G
q
q F
u
F
p
G
q
Cq F
u
F
q
G
p
F
x
F
p
G
p
p F
u
F
p
G
p
C F
y
F
q
G
p
.F
p
G
q
F
q
G
p
/F
q
:
2.1. CHARPIT’S METHOD 15
Substitution into the last of (2.49) gives
F
p
G
x
C F
q
G
y
C .p F
p
C q F
q
/G
u
.F
x
C p F
u
/G
p
.F
y
C q F
u
/G
q
D 0;
conveniently written as
ˇ
ˇ
ˇ
ˇ
D
x
F F
p
D
x
G G
p
ˇ
ˇ
ˇ
ˇ
C
ˇ
ˇ
ˇ
ˇ
D
y
F F
q
D
y
G G
q
ˇ
ˇ
ˇ
ˇ
D 0; (2.51)
where D
x
F D F
x
C p F
u
, D
y
F D F
y
C q F
u
and j j is the usual determinant. ese are
known as the Charpit equations. We also assumed that F
p
G
q
F
q
G
p
¤ 0 and F
q
¤ 0. ese
cases would need to be considered separately.
Example 2.1 Consider
xu
x
u
2
y
D 2u: (2.52)
is is the example we considered already; now we will determine all classes of equations that
are compatible with this one. Denoting
G
D
xu
x
u
2
y
2u
D xp q
2
2u;
where p D u
x
and q D u
y
, then
G
x
D p; G
y
D 0; G
u
D 2; G
p
D x; G
q
D 2q;
and the Charpit equations are
ˇ
ˇ
ˇ
ˇ
D
x
F F
p
p x
ˇ
ˇ
ˇ
ˇ
C
ˇ
ˇ
ˇ
ˇ
D
y
F F
q
2q 2q
ˇ
ˇ
ˇ
ˇ
D 0;
or, after expansion,
xF
x
2qF
y
C
xp 2q
2
F
u
C pF
p
C 2qF
q
D 0:
Solving this linear PDE by the method of characteristics gives the solution as
F
p
x
;
p
2
q
; q C y; xp q
2
2u
D 0 (2.53)
which is exactly the one given in (2.37)!
Example 2.2 Consider
u
x
u
y
D 1: (2.54)
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