93
C H A P T E R 5
First Integrals
In an introductory course in PDEs, the wave equation
u
tt
D c
2
u
xx
; (5.1)
where c > 0 is a constant wave speed, is introduced. e general solution of (5.1) is
u D F .x ct/ C G.x Cct /; (5.2)
where F and G are arbitrary functions of their arguments. In this solution, there are two waves;
one is traveling right (F .x ct/) and one is traveling left (G.x C ct /). Each of these solutions
can be obtained from the following first-order PDEs
u
t
C cu
x
D 0; (5.3a)
u
t
cu
x
D 0: (5.3b)
PDE (5.3a) gives rise to u D F .x ct /, whereas PDE (5.3b) gives rise to u D G.x C ct/.
We question whether it is possible to show this directly—whether the solutions of (5.3a)
and/or (5.3b) will give rise to solutions of (5.1). For example, differentiating (5.3a) with re-
spect to t and x gives
u
t t
C cu
tx
D 0;
u
tx
C cu
xx
D 0:
(5.4)
It is a simple matter to show that upon elimination of u
tx
in (5.4) we obtain (5.1). e same
follows from (5.3b). In this example, we considered the linear wave equation, where the solution
of a lower-order PDE gave rise to solutions of a higher order PDE. Does this idea apply for
NLPDEs? e following example illustrates the concept.
Consider the following pair of PDEs:
u
x
u
y
D 1; (5.5a)
u
xx
u
4
y
u
yy
D 0: (5.5b)
We ask, will solutions of (5.5a) give rise to solutions of (5.5b)? For example, one exact solution
of (5.5a) is u D 2
p
xy. Calculating all necessary derivatives gives
u
x
D
p
y
p
x
; u
y
D
p
x
p
y
; u
xx
D
p
y
2x
p
x
; u
yy
D
p
x
2y
p
y
; (5.6)
94 5. FIRST INTEGRALS
and substituting into (5.5b) shows it is identically satisfied. e reader can also verify that u D
x C y also satisfies both PDEs. We naturally ask, will all of the solutions of (5.5a) give rise to
solutions (5.5b)? It’s a simple matter to calculate the derivatives of (5.5a)
u
x
D
1
u
y
u
xx
D
u
xy
u
2
y
u
xy
D
u
yy
u
2
y
(5.7)
and eliminating u
xy
from (5.7) gives (5.5b). If a PDE of lower-order
F .x; y; u; u
x
; u
y
/ D 0 (5.8)
exists such that differential consequences of (5.8) gives rise to the original (higher-order) PDE,
then the lower-order PDE is called a first integral. We see that (5.5a) is one first integral
of (5.5b). Are there others and, if so, how do we find them? If we assume a general form as
in (5.8), then differential consequences give
F
x
C F
u
u
x
C F
p
u
xx
C F
q
u
xy
D 0; (5.9a)
F
y
C F
u
u
y
C F
p
u
xy
C F
q
u
yy
D 0; (5.9b)
where p D u
x
and q D u
y
. Eliminating u
xy
from (5.9) gives
F
2
p
u
xx
F
2
q
u
yy
C F
q
.
F
x
C pF
u
/
F
p
F
y
C qF
u
: (5.10)
In order to obtain our targeted PDE (5.5b), we need to solve the following system of PDE for
F
F
2
p
q
4
F
2
q
D 0; (5.11a)
F
q
.
F
x
C pF
u
/
F
p
F
y
C qF
u
D 0: (5.11b)
Since (5.11a) factors
F
p
q
2
F
q
F
p
C q
2
F
q
D 0, there are two cases to consider. We consider
the first case here and leave it to the reader to consider the second case. Solving
F
p
q
2
F
q
D 0 (5.12)
gives
F D G
x; y; u; p
1
q
; (5.13)
where G is an arbitrary function of its arguments. Substituting (5.13) into (5.11b) and letting
D p 1=q gives
.
G
x
C G
u
/
q
2
G
y
D 0: (5.14)
Since G D G.x; y; u; / and is now independent of q, then from (5.14) we have both
G
x
C G
u
D 0;
G
y
D 0;
(5.15)
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