16 2. COMPATIBILITY
Denoting
G D u
x
u
y
1 D pq 1;
where p D u
x
and q D u
y
, then
G
x
D 0; G
y
D 0; G
u
D 0; G
p
D q; G
q
D p;
and the Charpit equations are
ˇ
ˇ
ˇ
ˇ
D
x
F F
p
0 q
ˇ
ˇ
ˇ
ˇ
C
ˇ
ˇ
ˇ
ˇ
D
y
F F
q
0 p
ˇ
ˇ
ˇ
ˇ
D 0;
or, after expansion,
qF
x
C pF
y
C 2pqF
u
D 0:
Note that the third term can be replaced by 2F
u
, due to the original equation. Solving this linear
PDE by the method of characteristics gives the solution as
F D F .uq 2x; up 2y; p; q/ (2.55)
or
F D F .uu
y
2x; uu
x
2y; u
x
; u
y
/: (2.56)
For example, if we choose
uu
y
2x D 0; (2.57)
then on integrating we obtain
u
2
D 4xy C f .x/ (2.58)
and substituting into the original PDE (2.54) gives
xf
0
D f (2.59)
and leads to the exact solution
u
2
D 4xy C cx: (2.60)
If we choose
uu
x
C u
y
2y D 0; (2.61)
then on integrating we obtain
u y
2
C f
yu x
2
3
y
3
D 0 (2.62)
and substituting into the original PDE (2.54) gives
ff
02
1 D 0 (2.63)
2.1. CHARPIT’S METHOD 17
and leads to the exact solution
u y
2
C
c ˙
3
2
yu x
2
3
y
3
2=3
D 0: (2.64)
Example 2.3 Consider
u
2
x
C u
2
y
D u
2
: (2.65)
Denoting p D u
x
and q D u
y
, then
G D u
2
x
C u
2
y
u
2
D p
2
C q
2
u
2
:
us,
G
x
D 0; G
y
D 0; G
u
D 2u; G
p
D 2p; G
q
D 2q;
and the Charpit equation’s are
ˇ
ˇ
ˇ
ˇ
D
x
F F
p
2pu 2p
ˇ
ˇ
ˇ
ˇ
C
ˇ
ˇ
ˇ
ˇ
D
y
F F
q
2qu 2q
ˇ
ˇ
ˇ
ˇ
D 0;
or, after expansion,
pF
x
C qF
y
C
p
2
C q
2
F
u
C puF
p
C quF
q
D 0: (2.66)
Note that the third term can be replaced by u
2
F
u
, due to the original equation. Solving (2.66),
a linear PDE, by the method of characteristics gives the solution as
F D F
x
p
u
ln u; y
q
u
ln u;
p
u
;
q
u
:
One particular example is
x
p
u
ln u C y
q
u
ln u D 0;
or
u
x
C u
y
D .x C y/
u
ln u
:
If we let u D e
p
v
, this becomes
v
x
C v
y
D 2.x C y/;
which, by the method of characteristics, has the solution
v D 2xy C f .x y/:
is, in turn, gives the solution for u as
u D e
p
2xyCf .xy/
: (2.67)
18 2. COMPATIBILITY
Substitution into the original equation (2.65) gives the following ODE:
f
02
2f
0
2f C 2
2
D 0;
where f D f ./ and D x y. If we let f D g C
1
2
2
then we obtain
g
02
2g D 0 (2.68)
whose solution is given by
g D
.r C c/
2
2
; g D 0; (2.69)
where c is an arbitrary constant of integration. is, in turn, gives
f D
2
C c C
1
2
c
2
; f D
1
2
2
(2.70)
and substitution into (2.67) gives
u D e
q
x
2
Cy
2
Cc.xy/C
1
2
c
2
; u D e
p
2xyC.xy/
2
=2
;
as exact solutions to the original PDE (2.65).
It is interesting to note that when we substitute the solution of the compatible equation
into the original it reduces to an ODE. A natural question is: does this always happen? is was
proven to be true in two independent variables by the author [38].
So now we ask, of the infinite possibilities of compatible equations, can we choose the
right one(s) to not only solve a given PDE but also satisfy the given boundary condition? e
following example illustrates.
Example 2.4 Solve
u
x
u
y
xu
x
yu
y
D 0 (2.71)
subject to the boundary conditions
.i/ u.x; 0/ D 0;
.i i/ u.x; 0/ D
1
2
x
2
;
.i i i/ u.x; x/ D 2x
2
:
(2.72)
Denoting
G D pq xp yq;
where p D u
x
and q D u
y
, then
G
x
D p; G
y
D q; G
u
D 0; G
p
D q x; G
q
D p y;
2.1. CHARPIT’S METHOD 19
and the Charpit equations are
ˇ
ˇ
ˇ
ˇ
D
x
F F
p
p q x
ˇ
ˇ
ˇ
ˇ
C
ˇ
ˇ
ˇ
ˇ
D
y
F F
q
q p y
ˇ
ˇ
ˇ
ˇ
D 0;
or, after expansion,
.q x/F
x
C .p y/F
y
C .2pq xp yq/F
u
C pF
p
C qF
q
D 0: (2.73)
Note that the third term can be replaced by pqF
u
, due to the original PDE. Solving (2.73) by
the method of characteristics gives the solution as
F D F .q
2
2xq; p
2
2yp; p=q; pq 2u/; (2.74)
or
F D F .u
2
y
2xu
y
; u
2
x
2yu
x
; u
x
=u
y
; u
x
u
y
2u/: (2.75)
So how do we incorporate the boundary conditions? We will look at each separately.
Boundary Condition (i) In this case u.x; 0/ D 0, and differentiating with respect to x gives
u
x
.x; 0/ D 0 on the boundary. From the original PDE (2.71) we then have u
y
.x; 0/ D 0. Sub-
stituting these into (2.75) gives
F D F .0; 0; ‹; 0/: (2.76)
Note that we have included a ? in the third argument of F since we have
0
0
. So how do we
now use (2.76)? If we choose any of the arguments in (2.76) that are zero, then the boundary
conditions are satisfied by that particular PDE. So, if we choose the first for example, we have
u
2
y
2xu
y
D 0 (2.77)
and we know that this is compatible with the original PDE (2.71) and satisfies the BC (2.72)
and will give rise to the desired solution. As there are two cases (a) u
y
D 0 and (b) u
y
2x D 0,
we consider each separately.
Case (a) If u
y
D 0 then u
x
D 0 from the original PDE, giving u D c and the BC
u.x; 0/ D 0 gives c D 0, so the solution is u 0.
Case (b) In the second case where u
y
2x D 0, integrating gives u D 2xy C g.x/ and
substituting into the original PDE gives xf
0
D 0 so f D c. us, we have the exact solution
u D 2xy C c. e BC u.x; 0/ D 0 gives that c D 0 and so the solution is u D 2xy:
Boundary Condition (ii) In this case u.x; 0/ D
1
2
x
2
, and differentiating with respect to x gives
u
x
.x; 0/ D x on the boundary. From the original PDE we then have xu
y
.x; 0/ x
2
D 0 or
u
y
.x; 0/ D x. Substituting these into (2.75) gives
F D F .x
2
; x
2
; 1; 0/: (2.78)
20 2. COMPATIBILITY
So how do we use (2.78)? Again, if we choose any combination of the arguments that is zero,
then the boundary conditions are satisfied by that particular PDE. us, if we choose the sum of
the first two arguments, i.e., u
2
y
2xu
y
C u
2
x
2yu
x
D 0, then the solution of this will satisfy
the BC u.x; y/ D
1
2
x
2
. However, this PDE is nonlinear. We would like to solve a linear problem
if we can so we will choose a different combination. Another choice would be choosing the third
argument in (2.78) equals 1, i.e., u
x
=u
y
D 1. So we are to solve
u
x
u
y
D 0: (2.79)
is is easily solved giving u D f .x C y/ and substitution into the original PDE gives
f
02
f
0
D 0 or f
0
D 0; f
0
D 0; (2.80)
where D x C y. Only the second gives rise to a correct solution and we find that f D
1
2
2
and leads to the exact solution u D
1
2
.x C y/
2
.
Boundary Condition (iii) In this case u.x; x/ D 2x
2
, and differentiating with respect
to x gives u
x
.x; x/ C u
y
.x; y/ D 4x on the boundary. From the original PDE we then
have u
x
.x; x/u
y
.x; x/ xu
x
.x; x/ y
y
.x; x/ D 0. Solving for u
x
.x; x/ and u
y
.x; x/ gives
u
x
.x; x/ D u
y
.x; x/ D 2x. Substituting these into (2.75) gives
F D F .0; 0; 1; 0/: (2.81)
So now we have lots of possibilities again: if we choose any combination of the arguments that
is zero, then the boundary conditions are satisfied by that particular PDE. For example,
.a/ u
2
y
2xu
y
D 0;
.b/ u
2
x
2yu
x
D 0;
.c/ u
x
=u
y
1 D 0;
(2.82)
will all leads to solutions that satisfy the BC. PDE (a) leads to u D 2xy, PDE (b) leads to the
same solution, and PDE (c) leads to u D
1
2
.x C y/
2
. As another possibility, consider
F .a; b; c; d / D
1
2
d
b
c
:
is leads to
yu
y
D u (2.83)
which is easily solved, giving u D yf .x/: Substitution into the original PDE (2.71) gives
ff
0
xf
0
f D 0: (2.84)
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