Hua C., Wong R. Differential Equations and Asymptotic Theory in Mathematical Physics
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352
Proof.
By the assumption (4.5) we can choose
b
E
CC:
such that
dP
a2
P
off
-(O,O,
b)
=
ab
and
-(0,0,b)
#
[0,
co)
U
{
2,
3,.
.
.}
.
8X
&dX
(4.6)
Since
(dP/dz)(O,y,z)
=
0
is assumed in Case
(3),
the function
P(O,y,z)
is independent of
z
and
so
we can write
P(O,y,z)
=
q(y)
=
y2qo(y);
this
means that
P(x,
y,
z)
is written in the form
P(X,
Yl
2)
=
Y240(Y)
+
xR(z,
Yl.>.
(4.7)
Under the expression (4.7) we have
(dP/dx)(O,
y,
z)
=
R(0,
y,
z)
and
so
by
(4.6) we have
dR
aa
dz
R(O,O,b)
=
ab
and
-(0,0,b)
#
[0,
co)
U
{
2,~'.
.
.}
For simplicity we set
(Y
=
(dR/dz)(O,O,b);
we have
a
#
0,
a
#
[O,co)
and
(Y
#
{a/2,
a/3,.
. .
}.
Since the degree of each term of
R(O,O,
z)
in
z
is
greater than or equal to 2, the condition
(dR/dz)(O,
0,
b)
#
0
implies
b
# 0.
The condition
(dR/dz)(O,
0,
b)
#
0
is
also
used to solve the following
functional equation with respect to
z:
xY2qo(xy)+R(x,~y,y+z)
=ay
at
(X,Y,Z)
=
(O,b,O).
(4.8)
z
=
H(x1
Y),
(4.9)
By the implicit function theorem, the equation (4.8) is reduced to the form
where
H(x,
y)
is a holomorphic function in
a
neighborhood of
(z,
y)
=
(0,
b)
satisfying
H(0,
b)
=
0.
Moreover we see
by the assumption
a
ordinary differential equation
{a/2,
a/3,.
. .
}
we have
(dH/dy)(O,
b)
#
{1,2,.
.
.}.
Now we note that
n
=
1
is assumed and then let us consider the following
By (4.7) this is equivalent to
If
we set
~(z)
=
z$(x),
this is rewritten into the form
(4.10)
353
moreover, if we consider (4.10) under the condition
$(O)
=
b,
by (4.9) we
see that (4.10) is reduced to
d$
2-
=
H(x,$),
$(O)
=
b.
dx
(4.11)
Since
H(0,
b)
=
0
holds, (4.11) is nothing but a Briot-Bouquet’s ordinary
differential equation and its characteristic exponent is
(dH/dy)(O,
b).
Since
(dH/dy)(O,
b)
#
{1,2,.
.
.}
is known, we can conclude that (4.11) has a
unique holomorphic solution
$(x)
in a neighborhood of
x
=
0.
Thus, to complete the proof of Theorem 4.4 it is sufficient to show that
the above
p(x)
=
x$(x)
satisfies the conditions 1)-4)
of
(C,) in Theorem
3.2. We will do this now.
Since
$(O)
=
b
#
0,
this implies
p(x)
$
0;
this is
1).
2)
is clear from
the construction of
p(x).
Note that the vector field
X(p)
is given by
therefore the condition
3)
is clear. Since
(dR/dz)(O,
p(O),
(dp/dx)(O))
=
(dR/dz)(O,
0,
b)
=
a
holds, the condition 4) is equivalent to the condition
that the vector field
d d
t-
-
ffx-
at
dX
satisfies the Poincar6 condition which is also equivalent to
a
#
[0,
m).
Thus,
0
the condition 4) is also verified.
Example
(3).
Let
(t,x)
E
C2
and let us consider
dU
du
2
-
at
=
4.)
u2
+
x
(-)
dX
+
c(t, x),
where
u(x)
and
c(t,x)
are holomorphic functions.
Then
0
=
-1,
P
=
a(x)y2
+
xz2
and
so
the conditions in Case (3) are satisfied. We have
C
=
{a
E
@;
a2
=
-a}
=
(0,
-1)
and
(d2P/dzdx)(0,
0,
C)
=
(0,
-2}.
Since
-2
$i
[O,oo)u{-1/2, -1/3,.
.
.}
we have the condition (4.5). Thus, we
can apply Theorem 4.4 to this case and obtain the following: this equation
has a solution with singularities only on
{t
=
0)
of order
1tI-l.
References
1.
H.
Chen and
H.
Tahara, On the totally characteristic type non-linear partial
differential equations in the complex domain,
Publ.
RIMS,
Kyoto
Univ.,
35
(1999),
621-636.
354
2.
H.
Chen,
Z.
Luo and H. Tahara, Formal solutions of nonlinear first
or-
der totally characteristic type
PDE
with irregular singularities,
Ann. Inst.
FourierC Grenoble,
51
(2001), 1599-1620.
3.
R.
Gkrard and
H.
Tahara, Holomorphic and singular solutions of nonlinear
singular first order partial differential equations,
Publ. RIMS, Kyoto Univ.,
26
(1990), 979-1000.
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R.
Gkrard and H. Tahara, Singular nonlinear partial differential equations,
Aspects
of
Mathematics,
E
28,
Vieweg-Verlag,
1996.
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T.
Kobayashi, Singular solutions and prolongation of holomorphic solutions
to nonlinear differential equations,
Publ. RIMS. Kyoto Univ.,
34
(1998), 43-
63.
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J.E.C.
Lope and
H.
Tahara, On the analytic continuation
of
solutions to
nonlinear partial differential equations,
J.
Math. Pures Appl.,
81
(2002),
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H.
Tahara, On the singularities of solutions of nonlinear partial differen-
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273-283,
World Sci.,
2002.
8.
H. Tahara, Solvability of nonlinear totally characteristic type partial differ-
ential equations with resonances,
J.
Math. SOC. Japan,
55
(2003), 1095-1113.
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Y.
Tsuno, On the prolongation of local holomorphic solutions of nonlinear
partial differential equations,
J.
Math. SOC. Japan,
27
(1975), 454-466.
10.
H.
Yamazawa, Singular solutions of the Briot-Bouquet type partial differ-
ential equations,
J.
Math. SOC. Japan,
55
(2003), 617-632.
11.
M.
Zerner, Domaines d’holomorphie des fonctions vkrifiant une kquation
aux derivkes partielles,
C.
R.
Acad. Sci. Paris SLr.
I.
Math.,
272
(1971),
811-826.
1646-1648.
IDENTIFYING CORROSION BOUNDARY BY
PERTURBATION
METHOD
YONGJI TAN*AND XINXING CHEN
Fudan University
E-mad:
yjtan @fudan. edu. en,
012018041
@fudan.edu.cn
In this paper we investigate an inverse boundary value problem to determine the
boundary corrosion
of
a
tube, which is formulated to
a
problem to determine
a
free
boundary for
a
heat initial-boundary value problem
.
By perturbation method it
can be reduced to inverse problems with fixed boundaries. We develops
a
method
to solve these problems and give several numerical examples to show that this
method works quite well.
1.
Introduction
We consider an insulated tube one end of which is eroded for some physical
and chemical reasons.
To
determine the corrosion, some sensors are pre-
installed on the wall of the tube to measure the temperature, since the
corrosion can not be measured directly. This problem can be formulated to
an inverse problem
to
determine
a
free boundary for the initial-boundary
problem of an 1-D heat equation which is also a model problem for some
complicated situations e.g. the bottom corrosion of the blast furnace[l].
Since the length of corrosion is relatively small, we denote it by
Eg(t)
with a small parameter
E
.
Let
u
=
u(x,
t)
be the temperature in the tube
located
x
at time
t
.
By
some scaling the mathematical model can be
written as:
To
determine
u(x,
t),g(t)
with
g(0)
=
0,
g'(t)
2
0
such that
2
-
u(2,O)
=
cp(x)
u(0,t)
=
0
=
f(2,t)
t
>
0
,
0
<
2
<
1
+Eg(t)
0
I
Ic
I
1
8U
I
I"=l+Eg(t)
=
0
'Project
1017-1020
supported by
NSFC
355
356
and
where
21
and
$11
are the location and the measuring temperature of the
1-th sensor respectively
.
It is noticed that this free boundary problem is
different from Stefen problem
[2]
.
In Stefen problem, one extra Stefen
condition is given in the free boundary, nevertheless, in this problem the
extra conditions are given inside the domain
.
The plan of this paper is as follows
.
In section
2
we solve direct problem
by perturbation method
.
In section
3
we give the method of reconstruction
of
the free boundary
.
In section
4
we show some numerical results
.
In
section
5
we describe some theoretical results.
2.
Perturbation Solution for Direct Problem
To
solve the inverse problem of identifying corrosion boundary we have to
solve many direct problems in which the boundaries are constantly chang-
ing. It causes some inconvenience in numerical solution.
To
overcome this
difficulty, we use perturbation method[3]. The problems can be changed
into initial-boundary value problems with fixed boundaries. This idea will
be more useful for high dimensional problem.
If
g(t)
is known,the direct problem is:
&-&f(,,t)t>O
at
,
O<X<l+Eg(t)
u(z,O)
=
cp(z)
u(0,
t)
=
0
ax
Iz=l+Eg(t)
=
0
0
5
z
5
1
au
-
Formally asymptotic expand
~(z,
t)
about
E
as
u(z,
t)
=
uo(2,
t)
+
EZLl(Z,
t)
+
. . . .
Expanding the boundary condition at the right end into Taylor series about
x
near x=l, we have
Substituting them into problem(2) and comparing the coefficients of the
same power of
E,
we obtain two problems satisfied by
UO(X,~),ZL~(Z,~)
as
357
follows:
t>O
,O<x<l
Olzll
(4)
Ul(0,t)
=
0
It is noticed that they are both initial-boundary value problems with fixed
boundaries. However, in
(4)
the boundary condition at the right end de-
pends on
ug(z,t).
By separation variable technique we solve problem(3) and obtain
:
Differentiate
u~(z,t)
twice about x. Under some smooth and compatible
conditions on boundary data, the order
of
differentiation and integration
can be exchanged. Then
358
Substituting
x=l
into the expression, we obtain
1
exp(-7r2(k
+
2)2t)
Now we are going
to
solve problem
(4).
Let
Transform the unknown function by
and then problem
(4)
turns into
aw
a2
at
-_
=
--Zp/(t)
t
>
0
,
0
<
5
<
1
w(5,O)
=
0
0<2<1
w(0,
t)
=
0
zlr=1
=
0
aW
Similarly, by separation of variable we obtain
(5)
M
where
and
359
Therefore
and the first order formal asymptotic solution for the original problem is
a(z,
t)
=
zlo(z;
t)
+
&Ul(Z,
t)
where
=
2g(t){
&)kr."(k
+
-)2
1
x
2
k=O
It
is easy to see that this asymptotic solution depends
on
g(t) linearly.
360
3.
Reconstruct
Free
Boundary
The original problem is to find
u(z,
t),
g(t)
with
g(0)
=
0
such that
au
a%
--
at
=
f(x,t)
t
>
0
9
0
<
<
1
+
Eg(t)
u(z,O)
=
p(z)
u(0,t)
=
0
(6)
0
I
Ic
5
1
au
Iz=l+q(t)
=
0
u(x1,t)
=
+l(t)
,
(1
51
I
N,0
<
t
ST)
(6)'
1
and
While g(t) is given, we use the asymptotic solution as the approximate
solution of problem and denote it by
u(z,t;g(t)).
We relax the condition
(6)'
into
The problem of reconstructing free boundary is reduced to a variational
problem: Find g(t) such that
N
..r
ii(q,
t;
g(t))
-
q!~l(t))~dt
=
min
EL-(
1=1
where
ii(zl,t;g(t))
is the first order asymptotic solution
of
problem
(2),
provided that g(t) is given.
We discrete the problem by using polynomial to approximate g(t)
n
j=1
and taking the measurements at the discrete time points
tS(1
I
s
I
L)
Denoting
G(Q,
t;
al,
a2,.
.
+
,
a,)
=
G(q,
t;
g(t)),
the variational problem
turns into a function minimization
:
LN
j=11=1
Noticing that the asymptotic solution depends on g(t) linearly, this prob-
lem is a typical least square problem. We can easily solve it and obtain
al,
a2,
. . .
,
a,,
and solve the inverse problem
.
36
1
4.
Numerical Examples
Example
1
+g+=f(x,t)t>o
,
O<x<l+J&$
u(x,O)
=
0
OLzll
u(0,t)
=
0
I
&l+
$,t)
=
0
(7)
+
2tx
t
+
t2
(1
+
2t)tz
t+t2
x3
f(z,
t)
=
(1
+
-)2x
-
-
3-
(1
+
-)
50
100
100 3
The exact solution for this problem is
t
+
t2
23
u(2,
t)
=
t((1
+
-)%
-
-)
100
3
Taking
E
=
0.01,
solving this problem by perturbation method and compar-
ing the result with exact solution in
9
x
25
net points, we find the maximal
relative error is about
3%.
It
is acceptable.
Then by taking the values at
x=0.3,0.5,0.7
of the exact solution as
measurements and using above method to solve the inverse problem we
obtain
S(t)
=
c,”=,
ajtj
as
the approximation
of
g(t)
.
The graph for both of exact boundary(’-’) and reconstructed
boundary(‘*‘) is shown in figure
1
and the biggest error happens
at
the
end
of
time interval where the value of relative error is about
6%.
Figure
1.
g(t)
=
t
+
t2