Hindawi Publishing Corporation
Boundary Value Problems
Volume 2010, Article ID 471491, 12 pages
doi:10.1155/2010/471491
Research Article
Slowly Oscillating Solutions of a Parabolic Inverse
Problem: Boundary Value Problems
Fenglin Yang and Chuanyi Zhang
Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China
Correspondence should be addressed to Fenglin Yang, fengliny@hit.edu.cn
Received 11 October 2010; Accepted 20 December 2010
Academic Editor: Colin Rogers
Copyright q 2010 F. Yang and C. Zhang. This is an open access article distributed under the
Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
The existence and uniqueness of a slowly oscillating solution to parabolic inverse problems for a
type of boundary value problem are established. Stability of the solution is discussed.
1. Introduction
It is well known that the space APR of almost periodic functions and some of its
generalizations have many applications e.g., 1–13 and references therein. However, little
has been done for APR to inverse problems except for our work in 14–16. Sarason in 17
studied the space SOR of slowly oscillating functions. This is a C∗ -subalgebra of CR, the
space of bounded, continuous, complex-valued functions f on R with the supremum norm
f sup{|fx| : x ∈ R}. Compared with APR, SOR is a quite large space see 17–20.
What we are interested in SOR is based on the belief that SOR certainly has a variety of
applications in many mathematical areas too. In 15, we studied slowly oscillating solutions
of a parabolic inverse problem for Cauchy problems. In this paper, we devote such solutions
for a type of boundary value problem.
Set J ∈ {R, Rn }. Let CJ resp., CJ × Ω, where Ω ⊂ Rm denote the C∗ -algebra of
bounded continuous complex-valued functions on J resp., J × Ω with the supremum norm.
For f ∈ CJ resp., CJ × Ω and s ∈ J, the translate of f by s is the function Rs ft ft s
resp., Rs ft, Z ft s, Z, t, Z ∈ J × Ω.
Definition 1.1. 1 A function f ∈ CJ is called slowly oscillating if for every τ ∈ J, Rτ f − f ∈
C0 J, the space of the functions vanishing at infinity. Denote by SOJ the set of all such
functions.
2 A function f ∈ CJ × Ω is said to be slowly oscillating in t ∈ J and uniform on
compact subsets of Ω if f·, Z ∈ SOJ for each Z ∈ Ω and is uniformly continuous on
2
Boundary Value Problems
J × K for any compact subset K ⊂ Ω. Denote by SOJ × Ω the set of all such functions. For
convenience, such functions are also called uniformly slowly oscillating functions.
3 Let X be a Banach space, and let CJ, X be the space of bounded continuous
functions from J to X. If we replace CJ in 1 by CJ, X, then we get the definition of
SOJ, X.
As in 17, we always assume that f ∈ SOJ is uniformly continuous.
The following two propositions come from 15, Section 1.
Proposition 1.2. Let f ∈ SOJ SOJ × Ω be such that ∂f/∂xi is uniformly continuous on J.
Then ∂f/∂xi ∈ SOJ SOJ × Ω.
For H h1 , h2 , . . . , hn ∈ CRn , suppose that Ht ∈ Ω for all t ∈ R. Define H × ι →
Ω × R by
H × ιt h1 t, h2 t, . . . , hn t, t
t ∈ R.
1.1
The following proposition shows that the composite is also slowly oscillating.
Proposition 1.3. Let f ∈ SOR × Ω. If H ∈ SORn and Ht ∈ Ω for all t ∈ R, then f ◦ H × ι ∈
SOR.
m
In the sequel, we will use the notations: Rm
T R × 0, T , FT sup{|Fx, t| : x ∈
m
n
1
2
R , 0 ≤ t ≤ T }. F ∈ SOR × RT means that Fx , x , t is slowly oscillating in x1 ∈ Rn
n
m
1
2
and uniformly on x2 , t ∈ Rm
T ; F ∈ SOR × R means that Fx , x is slowly oscillating
1
n
2
m
in x ∈ R and uniformly on x ∈ R .
Let
n
1
Zx, t; ξ, s nm
2 πt − s
xi − ξi 2
exp −
x, ξ ∈ Rnm 4t − s
1.2
be the fundamental solution of the heat equation 21.
2. A Type of Boundary Value Problem
We will keep the notation in Section 1 and at the same time introduce the following new
notation:
x x1 , x2 , . . . , xn−1 ,
X x, xn ,
ζ ξ, ξn ,
ξ ξ1 , ξ2 , . . . , ξn−1 ,
Dn {X ∈ Rn : xn > 0}.
2.1
In this section, we always assume the following: f, fxn xn ∈ SORn−1 × DT0 , hx, t ≥
n−1
×D, ϕ ∈ C3 Rn−1 ×D, and g, Δg−gt ∈
const > 0, h, Δh−ht ∈ SORn−1
T0 , ϕ, ϕxn xn ∈ SOR
SORn−1
T0 .
Boundary Value Problems
3
Let
GX, t; ζ, τ ZX, t; ξ, ξn , τ ZX, t; ξ, −ξn , τ
2.2
be Green’s function for the boundary value problems 22, 23.
The following estimates are easily obtained:
t GX, t; ζ, sdζ ≤ m1 T ,
ds
0
n
D
t ZX, t; ξ, 0, sdξ ≤ m2 T ,
ds
0
Rn−1
2.3
t ∂ZX, t; ζ, s dζ ≤ m3 T ,
ds
0
∂x
n
n
R
where mi T i 1, 2, 3 are positive and increasing for T ≥ 0 and mi T → 0 as T → 0.
To show the main results of this section, the following lemmas are needed. The first
lemma is Lemma 3.1 on page 15 in 24.
Lemma 2.1. Let ϕ, φ, and χ be real, continuous functions on 0, T with χ ≥ 0. If
ϕt ≤ φt t
χsϕsds
t ∈ 0, T ,
2.4
0
then
ϕt ≤ φt t
t
χsφs exp
0
χ ρ dρ ds
t ∈ 0, T .
2.5
s
Lemma 2.2. Let ϕ be a continuous function on 0, T . If φ, χ1 , and χ2 are nondecreasing and
nonnegative on 0, T and
ϕt ≤ φt χ1 t
t
0
ϕsds χ2 t
t
0
ϕs
ds t ∈ 0, T ,
√
t−s
2.6
then
√
ϕt ≤ φt 1 tχ1 t 2 tχ2 t etχt ,
2.7
√
χt tχ21 t 4 tχ1 tχ2 t πχ22 t.
2.8
where
4
Boundary Value Problems
Proof. Replacing ϕs in the two integrals of 2.6 by the expression on the right hand
side in 2.6, changing the integral order of the resulting inequality and making use of the
monotonicity of φ, χ1 and χ2 , one gets
t
√
√
2
2
ϕt ≤ φt 1 tχ1 t 2 tχ2 t tχ1 t 4 tχ1 tχ2 t πχ2 t
ϕsds.
2.9
0
Apply Lemma 2.1 to get the conclusion.
n
Lemma 2.3. Let FX, t ∈ SODTn , φx, t, qx, t ∈ SORn−1
T , and ϕ ∈ SOD . Then the
problem
ut − Δu qu FX, t,
uX, 0 ϕX,
X, t ∈ DTn ,
X ∈ Dn ,
uxn x, 0, t φx, t,
2.10
x, t ∈ Rn−1
T
has a unique solution u, and u is in SODTn and satisfies
uT ≤ KT T FT ϕ T
φ ,
T
2
√
2.11
where KT 21 T qT eT qT .
One sees that KT depends on qT only and is bounded near zero.
Proof. The existence and uniqueness of the solution comes from Theorem 5.3 on page 320 in
25.
As in 22, 23, the solution u can be written as
uX, t ϕζGX, t; ζ, 0dζ Dn
−
ds
0
t
Fζ, sGX, t; ζ, sdζ
Dn
qξ, suζ, sGX, t; ζ, sdζ − 2
ds
0
t
Dn
vx, t −
ds
0
t
Rn−1
φξ, sZX, t; ξ, 0, sdξ
2.12
ds
0
t
qξ, suζ, sGX, t; ζ, sdζ.
Dn
So,
ut ≤ 2ϕ 2
t t
φ
Fs ds 2 √ s ds 2 qs us ds.
t−s
0
0
0
t
By Lemma 2.1, one gets the desired inequality.
2.13
Boundary Value Problems
5
Now we show that u ∈ SODTn . As in the proofs of Lemmas 2.1 and 2.3 in 15, one
gets v ∈ SODTn . For x, τ ∈ Rn−1 with |x| ≥ A > 0,
ux τ, xn , t − ux, xn , t
vx τ, xn , t − vx, xn , t −
t
qξ, suζ, sGx τ, xn , t; ζ, s − Gx, xn , t; ζ, sdζ
ds
0
Dn
vx τ, xn , t − vx, xn , t
t − ds
qx τ ξ, sux τ ξ, xn ξn , s − qx ξ, sux ξ, xn ξn , s Gθ, t; ζ, sdζ
Dn
0
vx τ, xn , t − vx, xn , t
t − ds
qx τ ξ, s − qx ξ, s ux τ ξ, xn ξn , sGθ, t; ζ, sdζ
0
−
t
Dn
ds
0
Dn
ux τ ξ, xn ξn , s − ux ξ, xn ξn , sqx ξ, sGθ, t; ζ, sdζ.
2.14
Note that
t qx τ ξ, s − qx ξ, s ux τ ξ, xn ξn , sGθ, t; ζ, sdζ ≤ B · distA Rτ q − q t
ds
0
n
D
≤ B q ,
qξ,
sGθ,
t;
ζ,
sdζ
s
Dn
2.15
where B is a constant and
distA Rτ q, q t sup
qx τ, s − qx, s.
2.16
s∈0,t,|x|≥A
So,
distA Rτ u, ut ≤ distA Rτ v, vt B · distA Rτ q, q
t
B
t
0
distA Rτ u, us qs ds.
2.17
By Lemma 2.1, one has
distA Rτ u, ut ≤ m distA Rτ v, vt B · distA Rτ q, q t ,
2.18
where m is a constant. Since v and q are slowly oscillating, the right-hand sides of the
inequality above approaches zero as A → ∞. This means that u ∈ SODTn . The proof is
complete.
6
Boundary Value Problems
Consider the following problem.
Problem 1. Find functions u ∈ SORn−1 × DT and q ∈ SORn−1
T such that
ut − Δu qx, tu fX, t,
2.19
X ∈ Dn ,
2.20
x, t ∈ Rn−1
T ,
2.21
x, t ∈ Rn−1
T , a ∈ 0, ∞.
2.22
uX, 0 ϕX,
uxn x, 0, t gx, t,
ux, a, t hx, t,
X, t ∈ DTn ,
One sees that
2.23
hx, 0 ϕx, a, ϕxn x, 0 gx, 0, x ∈ Rn−1 ,
ht x, 0 ut |xn a,t0 Δu − qu fX, t xn a,t0 ΔϕXxn a − qx, 0ϕx, a fx, a, 0,
gt x, 0 utxn |xn 0,t0 Δϕxn Xxn 0 − qx, 0ϕxn x, 0 fxn x, 0, 0.
2.24
It follows from 2.24 that
ϕxn x, 0ΔϕXxn a fx, a, 0ϕxn x, 0 − ht x, 0ϕxn x, 0
ϕx, aΔϕxn Xxn 0 fxn x, 0, 0ϕx, a − gt x, 0ϕx, a.
2.25
Let V X, t uxn X, t, and let WX, t Vxn X, t. We have the following two
additional problems for V and W, respectively.
Problem 2. Find functions V ∈ SORn−1 × DT and q ∈ SORn−1
T such that
Vt − ΔV qx, tV fxn X, t,
2.26
X ∈ Dn ,
2.27
x, t ∈ Rn−1
T ,
2.28
V X, 0 ϕxn X,
V x, 0, t gx, t,
X, t ∈ DTn ,
Vxn x, a, t ht − Δh qh − fx, a, t,
x, t ∈ Rn−1
T .
2.29
Problem 3. Find functions W ∈ SORn−1 × DT and q ∈ SORn−1
T such that
Wt − ΔW qx, tW fxn xn X, t,
WX, 0 ϕxn xn X,
X, t ∈ DTn ,
X ∈ Dn ,
Wxn x, 0, t gt − Δg qg − fxn x, 0, t,
Wx, a, t ht − Δh hq − fx, a, t,
2.30
2.31
,
x, t ∈ Rn−1
T
2.32
x, t ∈ Rn−1
T .
2.33
Boundary Value Problems
7
Lemma 2.4. Problems 1, 2, and 3 are equivalent to each other.
Proof. The existence and uniqueness of the solution V, q of Problem 2 can be easily obtained
from that of the solution u, q of Problem 1. Conversely, let V, q be the solution of Problem
2. We show that Problem 1 has a unique solution u, q. The uniqueness comes from the
uniqueness of 2.19–2.21. For the existence, let
uX, t xn
V x, y, t dy hx, t.
2.34
a
Obviously, uX, t ∈ SORn−1 × DT and satisfies 2.22. Also u satisfies 2.21 because
uxn x, 0, t V x, 0, t gx, t. By 2.23 and 2.27, one sees that 2.20 is true. Finally, we
show that u satisfies 2.19 and therefore, along with q, constitutes a solution of Problem 1. In
fact,
ut − Δu qu ht − Δh qh xn
Vt x, y, t − ΔV x, y, t qV x, y, t dy
a
xn
∂
∂2
V
x,
y,
t
dy
−
∂xn2
∂y2
2
a
xn
V x, y, t dy
a
2.35
ht − Δh qh fX, t − fx, a, t Vxn X, t − Vxn x, a, t − Vxn X, t
fX, t. by 2.29
Thus, we have shown the equivalence of Problems 1 and 2. Replacing 2.34 by the function
V X, t xn
W x, y, t dy gx, t,
2.36
a
the equivalence of Problems 2 and 3 can be proved similarly. The proof is complete.
By Lemma 2.4, to solve Problem 1, we only need to solve Problem 3. By 2.30–2.32,
we have the integral equation about W:
WX, t Dn
−
ϕξn ξn ζGX, t; ζ, 0dζ ds
fξn ξn ζ, sGX, t; ζ, sdζ
qξ, sWζ, sGX, t; ζ, sdζ
t
ds
0
Dn
2.37
Dn
0
−2
ds
0
t
t
Rn−1
gs − Δg qg − fξn ξ, 0, s ZX, t; ξ, 0, sdξ.
Rewrite 2.33 as
q Lq h−1 x, t Δh − ht fx, a, t Wx, a, t ,
where W is determined by 2.37.
2.38
8
Boundary Value Problems
One can directly test that Problem 3 is equivalent to 2.37-2.38.
Note that for a given qx, t ∈ SORn−1
T , Lemma 2.3 shows that 2.30–2.32 or
equivalently, 2.37 have a unique solution W ∈ SORn−1 × DT . Thus, 2.38 does define
an operator L. Therefore, we only need to show that the integral 2.38 has a unique solution
n−1
q and q ∈ SORn−1
T . That is, L has a fixed point in SORT . Let
t
Δh − ht fx, a, t 2ϕξ ξ ds
f
sGx,
a,
t;
ζ,
sdζ
ζ,
ξ
ξ
n
n
n
n
T0
0
n
D
T0
t M
−1 .
2 ds
Δg − gs fξn ξ, 0, s Zx, a, t; ξ, 0, sdξ
h 0
T
2
n−1
0
R
2.39
T0
Set BM, T {q ∈ SORn−1
T : qT ≤ M}, where T ≤ T0 . If q ∈ BM, t, then, by Lemma 2.3,
WX, t is in SORn−1 × DT , and so, by 2.38, Lq is in SORn−1
T with
Lq ≤ M h−1 2m2 T g T0 m1 T WT M.
T
T0
2
2.40
Equation 2.37 gives the estimate
WT ≤ 2ϕξn ξn 2m2 T0 gt − Δg − fxn x, 0, tT0 2Mm2 T0 g T0
m1 T0 fxn xn T0 Mm1 T WT .
2.41
Choose t0 < T0 such that when T ≤ t0 , one has 1 < 21 − Mm1 T . It follows that
WT ≤ 2 2ϕxn xn 2m2 T0 gt − Δg − fxn x, 0, tT0 2Mm2 T0 g T0 m1 T0 fxn xn T0 .
2.42
Choose T1 ≤ t0 such that when T ≤ T1 , one has
2h−1 m2 T g T0 m1 T T0
1
× 2ϕxn xn 2m2 T0 gt − Δg − fxn x, 0, tT0 2Mm2 T0 g T0 m1 T0 fxn xn < ,
2
2.43
and therefore, LqT ≤ M.
Boundary Value Problems
9
Let q1 ,q2 ∈ BM, T . By 2.38, Lq1 − Lq2 T ≤ h−1 T W1 − W2 T . Note that the
function W W1 − W2 is the solution of the problem
Wt − ΔW qW W2 q2 − q1 ,
X, t ∈ DTn ,
WX, 0 0, X ∈ Dn ,
Wxn x, 0, t q2 − q1 gx, t, x, t ∈ Rn−1
T .
2.44
So, by Lemma 2.3, one has
√
T
q1 − q2 g T q1 − q2 W2 T
T
T
T
2
WT ≤ KT 2.45
.
Choose T2 < t0 such that for T ≤ T2 , h−1 T0 W1 − W2 T ≤ 1/2q1 − q2 T . Now, set
T ≤ min{T1 , T2 }. Then L is a contraction from BM, T into itself, and therefore, has a unique
fixed point. Thus, we have shown.
Theorem 2.5. Let functions f, g, h, and ϕ be as above. Then, for small T , Problem 3 has a unique
solution ( W, q) in RnT with W ∈ SORn−1 × DT and q ∈ SORn−1
T .
Let W i , qi i 1.2 be the solutions of Problem 3 in DTn for the functions f i , g i , hi , and
ϕ . Set h0 h1 − h2 , f 0 f 1 − f 2 , ϕ0 ϕ1 − ϕ2 , and g 0 g 1 − g 2 . For the stability of the solution,
we have the following.
i
Theorem 2.6. For 0 ≤ t ≤ T , one has
q1 − q2 ≤ c1 h0 c2 g 0 c3 fx0n xn c4 ϕ0xn xn c5 h0t − Δh0 − f 0 x, a, t
t
t
t
c6 gt0 − Δg 0 − fx0n x, 0, t ,
t
t
2.46
t
1
1
1
where ci 1 ≤ i ≤ 6 depends on t, h−1
1 t , g t , fxn xn t , ϕxn xn , q1 t , q2 t , and
1
1
1
gt − Δg − fxn x, 0, tt .
Proof. By 2.33,
−1 q1 − q2 h1
Δh0 − h0t f 0 x, a, t − q2 h0 W1 − W2 .
2.47
So,
−1 0
0
0
0
q1 − q2 ≤ h1 Δh − ht f x, a, t q2 t h W1 − W2 t .
t
t
t
t
2.48
10
Boundary Value Problems
Note that the function W W1 − W2 is the solution of the problem
Wt − ΔW q2 W fx0n xn − W1 q1 − q2 ,
X, t ∈ DTn ,
2.49
WX, 0 ϕ0xn xn X,
X ∈ Dn ,
Wxn x, 0, t gt0 − Δg 0 q2 g 0 − fx0n x, 0, t q1 − q2 g 1 ,
x, t ∈ Rn−1
T .
Using a formula similar to 2.37 and Lemma 2.2 for the function W, one gets
Wt ≤
t
t 0
0
0 0 q2 t g 2
tfxn xn ϕxn xn 2
gt − Δg 0 − fx0n x, 0, t
t
t
t
π
π
1 t
t
g t q1 − q2 s
t
q1 − q2 s ds √
q2 ρ dρ ds .
exp
W1 t
π 0
t − s
0
0
2.50
Applying Lemma 2.2 and 2.48, one gets the desired conclusion with
1 −1 q2 ,
c1 φt
h
t
t
t
−1 t
1
h
q2 s ds ,
c2 2φt
q2 t exp
π
t
0
t
1 −1 q2 s ds ,
c3 tφt h
exp
t
0
t
1 −1 q2 s ds ,
c4 φt h
exp
t
2.51
0
1 −1 ,
c5 φt
h
t
t
−1 t
1
h
q2 s ds ,
c6 2φt
exp
π
t
0
where
√
φt 1 tχ1 t 2 tχ2 t etχt ,
√
χt tχ21 t 4 tχ1 tχ2 t πχ22 t,
t
1 −1 q2 s ds ,
χ1 t h
Φt exp
t
χ2 t π
0
t
1 −1 1 q2 s ds
h
g t exp
−1/2 t
0
2.52
Boundary Value Problems
11
and Φt is majorant of W1 t . One can specially assume that
t
t g 1 − Δg 1 − f 1 x, 0, s
s
xn
1 1 q2 s ds .
ds exp
ϕxn xn tfxn xn t
πt − s
0
s
Φt 2.53
The proof is complete.
Corollary 2.7. Under the conditions in Theorem 2.6, the solution of Problem 3 is unique.
Acknowledgment
The research is supported by the NSF of China no. 11071048.
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Boundary Value Problems
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