2.2. SECOND-ORDER PDES 21
Now (2.84) is nonlinear and can be integrated. Sometimes the solution form and boundary
condition give us a hint of what solution to look for. Since u.x; x/ D 2x
2
, using u D yf .x/
gives xf .x/ D 2x
2
or f .x/ D 2x, which in fact does satisfy the ODE (2.84)!
Up to this point, we have considered the compatibility of two first-order PDEs. We now
extend this technique to higher-order PDEs.
2.2 SECOND-ORDER PDES
Consider the following pairs of PDEs
u
t
C u
x
D x; u
t
D u
xx
(2.85)
and
u
t
C u
x
D t; u
t
D u
xx
: (2.86)
We ask: does a common solution exist for each pair, (2.85) and (2.86)? We certainly could solve
each of the first-order PDEs in (2.85) and (2.86) and substitute the solutions into the second
equation of (2.85) and (2.86) and determine whether solutions exist, but instead we will ask
if they are compatible. Compatible in the sense that all higher order derivatives would be the
same.
Let us consider the pair (2.85). If we isolate the derivatives u
t
and u
xx
, then
u
t
D x u
x
; u
xx
D x u
x
: (2.87)
If these are to be compatible, then .u
t
/
xx
D .u
xx
/
t
. Using (2.87), this compatibility restraint
becomes
.
x u
x
/
xx
D
.
x u
x
/
t
or u
xxx
D u
tx
(2.88)
and, by virtue of the diffusion equation in (2.85), (2.88) is identically satisfied and, thus, the pair
(2.85) are compatible.
For the second pair, (2.86), isolating the derivatives u
t
and u
xx
gives
u
t
D t u
x
; u
xx
D t u
x
: (2.89)
If these are to be compatible, then .u
t
/
xx
D .u
xx
/
t
. erefore,
.
t u
x
/
xx
D
.
t u
x
/
t
or u
xxx
D 1 u
tx
; (2.90)
and, by virtue of the diffusion equation in (2.86), gives 0 D 1 which clearly is not true, so (2.86)
are not compatible. As a matter of exercise, let us determine functions f .t; x/ such that the
following are compatible
u
t
C u
x
D f .t; x/; u
t
D u
xx
: (2.91)
As we did previously, we isolate the derivatives u
t
and u
xx
. is gives
u
t
D f .t; x/ u
x
; u
xx
D f .t; x/ u
x
: (2.92)
22 2. COMPATIBILITY
If these are to be compatible, then .u
t
/
xx
D .u
xx
/
t
. is gives
.
f .t; x/ u
x
/
xx
D
.
f .t; x/ u
x
/
t
or f
xx
u
xxx
D f
t
u
tx
(2.93)
and, by virtue of the diffusion equation in (2.91), gives f
xx
D f
t
. us, if f is any solution of
the diffusion equation, then (2.91) will be compatible. So how complicated can this get? e
next example illustrates.
Determine functions f and g such that the following are compatible:
u
t
C f .t; x/u
x
D g.t; x/u; u
t
D u
xx
: (2.94)
We first isolate the derivatives u
t
and u
xx
. is gives
u
t
D g.t; x/u f .t; x/u
x
; u
xx
D g.t; x/u f .t; x/u
x
: (2.95)
Imposing .u
t
/
xx
D .u
xx
/
t
gives
g
xx
u C 2g
x
u
x
C gu
xx
f
xx
u
x
2f
x
u
xx
f u
xxx
D g
t
u C gu
t
f
t
u
x
f u
tx
: (2.96)
Using (2.95) to eliminate the derivatives u
t
; u
tx
; u
xx
and u
xxx
gives
.
g
xx
2gf
x
g
t
/
u C
.
2g
x
f
xx
C 2ff
x
C f
t
/
u
x
D 0: (2.97)
For (2.94) to be compatible, then (2.97) this would have to be identically zero and, since u can
vary, then the coefficients of u and u
x
in (2.97) must be zero. is means that f and g must
satisfy
g
t
C 2gf
x
g
xx
D 0; (2.98a)
f
t
C 2ff
x
C 2g
x
f
xx
D 0: (2.98b)
So, the problem of determining compatibility for the pair (2.94) turns to a very complicated
problem, as we now have to solve a coupled nonlinear system of PDEs for f and g. e good
news is, we can actually solve this system (see Section 3.5.1).
In light of our discussion so far, we have been seeking compatible equations with the
diffusion equation that are linear. Must we assume this on the onset? e answer is no, as the
next example illustrates.
Determine functions f .u/ such that
u
t
D u
xx
and u
t
D f .u/u
2
x
(2.99)
are compatible. As we have done previously, we rewrite (2.99) as
u
t
D f .u/u
2
x
; u
xx
D f .u/u
2
x
; (2.100)
2.2. SECOND-ORDER PDES 23
and the compatibility of (2.100) requires that
.
u
t
/
xx
.
u
xx
/
t
D 0: It follows from the first of
(2.100) that
u
tx
D 2f u
x
u
xx
C f
0
u
3
x
D 2f
2
u
3
x
C f
0
u
3
x
D
f
0
C 2f
2
u
3
x
: (2.101)
Note that the second of (2.100) has been used. Further,
u
txx
D
f
00
C 4ff
0
u
4
x
C 3
f
0
C 2f
2
u
2
x
u
xx
D
f
00
C 4ff
0
u
4
x
C 3
f
0
C 2f
2
f u
4
x
D
f
00
C 7ff
0
C 6f
3
u
4
x
: (2.102)
Similarly,
u
xxt
D 2f u
x
u
xt
C f
0
u
t
u
2
x
D 2f u
x
f
0
C
2f
2
u
3
x
C
f
0
f u
4
x
D
3ff
0
C 4f
3
u
4
x
: (2.103)
erefore, compatibility gives rise to
f
00
C 7ff
0
C 6f
3
u
4
x
3ff
0
C 4f
3
u
4
x
D 0
or
f
00
C 4ff
0
C 2f
3
u
4
x
D 0
and since u
x
¤ 0, then
f
00
C 4ff
0
C 2f
3
D 0: (2.104)
e solution of (2.104) is rather complicated but it does admit the simple solution f .u/ D
1
u
and, thus,
u
t
D
u
2
x
u
; u
t
D u
xx
(2.105)
are compatible.
Up to this point, we have been determining compatible equations for the diffusion
equation—a linear second-order PDE. We would really like to harness the power of this tech-
nique to hopefully gain some insight into NLPDEs. e next example illustrates this very nicely.
Example 2.5 Seek compatibility with the nonlinear diffusion equation
u
t
D
.
u
m
u
x
/
x
(2.106)
.m ¤ 0/ and the linear first-order PDE
u
t
C A.t; x/u
x
D B.t; x/u: (2.107)
24 2. COMPATIBILITY
Isolating u
t
and u
xx
from (2.106) and (2.107) gives
u
t
D B.t; x/u A.t; x/u
x
; u
xx
D
B.t; x/u A.t; x/u
x
u
m
mu
2
x
u
; (2.108)
and requiring compatibility
.
u
xx
/
t
D
.
u
t
/
xx
gives (using (2.108) where appropriate)
.
A
t
C 2AA
x
mAB
/
u
x
u
m
C
.
2.m C 1/B
x
A
xx
/
u
x
CB
xx
u C
mB
2
B
t
2BA
x
u
m1
D 0:
Since A and B are independent of u and u
x
, we obtain
A
t
C 2AA
x
mAB D 0; (2.109a)
A
xx
2.m C 1/B
x
D 0; (2.109b)
B
t
C 2BA
x
mB
2
D 0; (2.109c)
B
xx
D 0:
(2.109d)
At this point we impose compatibility for A and B, i.e.,
.
A
xx
/
t
D
.
A
t
/
xx
and
.
B
xx
/
t
D
.
B
t
/
xx
.
is gives
.3m C 4/B
x
.
.m C 1/B A
x
/
D 0; (2.110a)
.3m C 4/B
2
x
D 0; (2.110b)
and from (2.110b) we see two cases emerge:
(i) m D 4=3;
(ii) m ¤ 4=3 .so B
x
D 0/:
Case (i) m D 4=3
In this case we solve (2.109b) and (2.109d) for A and B giving
A D
1
3
a.t /x
2
C c.t/x Cd.t/; B D a.t /x C b.t/; (2.111)
where a d are arbitrary functions of integration. Further, substitution into the remaining equa-
tions of (2.109) gives
1
3
Pa C 2ac C
4
3
ab
x
2
C
Pc C 2c
2
C
4
3
bc
x C
P
d C 2cd C
4
3
bd D 0; (2.112a)
Pa C 2ac C
4
3
ab
x C
P
b C 2bc C
4
3
b
2
D 0; (2.112b)
and since a d are independent of x, this leads to
P
a
C
2ac
C
4
3
ab
D
0;
P
b C 2bc C
4
3
b
2
D
0;
(2.113a)
Pc C 2c
2
C
4
3
bc D 0;
P
d C 2cd C
4
3
cd D 0: (2.113b)
2.2. SECOND-ORDER PDES 25
We will assume that a ¤ 0 as this is contained in case (ii). Eliminating the nonlinear terms in
(2.113) gives
b Pa a
P
b D 0; c Pa a Pc D 0; d Pa a
P
d D 0; (2.114)
and are easily solved giving
b D k
1
a; c D k
2
a; d D k
3
a; (2.115)
where k
1
; k
2
and k
3
are arbitrary constants. With these assignments, (2.113a) becomes
Pa C
2k
2
C
4
3
k
1
a
2
D 0; (2.116)
and is easily solved giving
a D
1
2k
2
C
4
3
k
1
t C k
4
; (2.117)
where k
4
is an additional constant. us, from (
2.115) and (2.117) we obtain the forms for a; b; c
and d , and through (2.111), the forms of A and B. us, (2.107) is compatible with equations
of the form
u
t
C
1
3
x
2
C k
2
x C k
3
2k
2
C
4
3
k
1
t C k
4
u
x
D
x C k
2
2k
2
C
4
3
k
1
t C k
4
u: (2.118)
Case (ii) m ¤ 4=3
In this case, we have B
x
D 0. We solve (2.109b) and (2.109d) giving
A D c.t/x C d.t/; B D b.t/; (2.119)
where b; c; and d are arbitrary functions of integration. Further, substitution into the remaining
equations of (2.109) gives
Pc C 2c
2
mbc
x C
P
d C 2cd mbd D 0; (2.120a)
P
b C 2bc mb
2
D 0; (2.120b)
and since b; c; and d are independent of x, this leads to
P
b C 2bc mb
2
D 0; Pc C 2c
2
mbc D 0;
P
d C 2cd mbc D 0: (2.121a)
ese can be solved as in the previous case. eir solution is:
If b D 0 then
c D
c
0
2c
0
t C k
; d D
d
0
2c
0
t C k
; (2.122)
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