9
C H A P T E R 2
Compatibility
We start our discussion by first solving the NPDE
xu
x
u
2
y
D 2u; (2.1)
subject to the boundary condition
u.x; x/ D 0: (2.2)
As with most introductory courses in partial differential equations (see, for example, [37]) we
use the method of characteristics. Here, we define F as
F D xp q
2
2u: (2.3)
e characteristic equations become
x
s
D F
p
D x; (2.4a)
y
s
D F
q
D 2q; (2.4b)
u
s
D pF
p
C qF
q
D xp 2q
2
; (2.4c)
p
s
D F
x
pF
u
D p; (2.4d)
q
s
D F
y
qF
u
D 2q: (2.4e)
In order to solve the PDE (2.1) we will need to solve the system (2.4). As (2.1) has a boundary
condition (BC), we will create BCs for the system (2.4). In the .x; y/ plane, the line y D x is
the boundary where u is defined. To this, we associate a boundary in the .r; s/ plane. Given the
flexibility, we can choose s D 0 and connect the two boundaries via x D r. erefore, we have
x D r; y D r; u D 0 when s D 0: (2.5)
To determine p and q on s D 0, it is necessary to consider the initial condition u.x; x/ D 0:
Differentiating with respect to x gives
u
x
.x; x/ C u
y
.x; x/ D 0: (2.6)
From the original PDE (2.1)
xu
x
.x; x/ u
2
y
.x; x/ D 0: (2.7)
If we denote p
0
D u
x
.x; x/ and q
0
D u
y
.x; x/, then (2.6) and (2.7) become
p
0
C q
0
D 0; rp
0
q
2
0
D 0: (2.8)
10 2. COMPATIBILITY
ese are easily solved giving
.i/ p
0
D 0; q
0
D 0 (2.9a)
.i i/ p
0
D r; q
0
D r: (2.9b)
We will only use case (ii) as case (i) gives rise to the trivial solution u D 0.
Solving (2.4a) gives
x D a.r/e
s
; (2.10)
where a is an arbitrary function. From the boundary condition (2.5), we find that a.r/ D r gives
x D re
s
: (2.11)
Solving (2.4d) gives
p D b.r/e
s
; (2.12)
where b is an arbitrary function. From the boundary condition (2.9b), we find that b.r/ D r
gives
p D re
s
: (2.13)
Solving (2.4e) gives
q D c.r/e
2s
; (2.14)
where c is an arbitrary function. From the boundary condition (2.9b), we find that c.r/ D r
gives
q D re
2s
: (2.15)
To solve (2.4b) we need to use q. us,
y
s
D 2q D 2re
2s
(2.16)
easily integrates, giving y D re
2s
C d.r/. Using the BC (2.5) shows that d D 0 gives
y D re
2s
: (2.17)
Finally, we see from (2.4b)
u
s
D xp 2q
2
D r
2
e
2s
2r
2
e
4s
(2.18)
which again integrates easily giving u D
1
2
.r
2
e
2s
r
2
e
4s
/ C e.r/, where e is an arbitrary func-
tion. From the boundary condition (2.5), we find that e.r/ D 0 gives
u D
r
2
e
2s
r
2
e
4s
2
: (2.19)
Eliminating r and s from (2.11), (2.17), and (2.19) gives
u D
x
2
y
2
2
: (2.20)
e reader can verify the solution (2.20) does indeed satisfy the PDE (2.1) and the BC (2.2).
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