Giải phương trình nhiệt ngược bằng phương pháp tựa biên

1. c
de Gruyter 2009
J. Inv. Ill-Posed Problems 17 (2009), 913–932 DOI 10.1515 / JIIP.2009.053
A new version of quasi-boundary value method
for a 1-D nonlinear ill-posed heat problem
P. H. Quan, D. D. Trong, and N. H. Tuan
Abstract. In this paper, a simple and convenient new regularization method which is called modified
quasi-boundary value method for solving nonlinear backward heat equation is given. Some new
quite sharp error estimates between the approximate solution are provided and generalize the results
in our paper [17, 19, 20]. The approximation solution is calculated by the contraction principle. A
numerical experiment is given.
Key words. Backward heat problem, nonlinearly ill-posed problem, quasi-boundary value methods,
quasi-reversibility methods, contraction principle.
AMS classification. 35K05, 35K99, 47J06, 47H10.
1. Introduction
Let T be a positive number. We consider the problem of finding the temperature u(x, t),
(x, t) ∈ (0, π) × [0, T] such that
ut − uxx = f(x, t, u(x, t)), (x, t) ∈ (0, π) × (0, T), (1.1)
u(0, t) = u(π, t) = 0, t ∈ (0, T), (1.2)
u(x, T) = g(x), x ∈ (0, π), (1.3)
where g(x), f(x, t, z) are given. The problem is called the backward heat problem, the
backward Cauchy problem or the final value problem.
As is known, the nonlinear problem is severely ill-posed, i.e., solutions do not al-
ways exist, and in the case of existence, these do not depend continuously on the given
data. In fact, from small noise contaminated physical measurements, the correspond-
ing solutions have large errors. This makes it difficult to find numerical calculations.
Hence, a regularization is in order. In the mathematical literature various methods
have been proposed for solving backward Cauchy problems. We can notably men-
tion the method of quasi-reversibility (QR method) of Lattes and Lions [2, 3, 7], the
quasi-boundary value method (Q.B.V method) [4] and the C-regularized semigroups
technique [11].
In the method of quasi-reversibility, the main idea is to replace A by Aε = gε(A). In
the original method in [7], Lattes and Lions have proposed gε(A) = A−εA2, to obtain
well-posed problem in the backward direction. Then, using the information from the
solution of the perturbed problem and solving the original problem, we get another
All authors were supported by the Council for Natural Sciences of Vietnam.
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2. 914 P. H. Quan, D. D. Trong, and N. H. Tuan
well-posed problem and this solution sometimes can be taken to be the approximate
solution of the ill-posed problem.
Difficulties may arise when using the method quasi-reversibility discussed above.
The essential difficulty is that the order of the operator is replaced by an operator of
second order, which produces serious difficulties on the numerical implementation,
in addition, the error c(ε) introduced by small change in the final value g is of the
order eT/ε
.
In 1983, Showalter, in [15, 16], presented a different method called the quasi-
boundary value (QBV) method to regularize that linear homogeneous problem which
gave a stability estimate better than the one of the disscused method. The main idea of
the method is to add an appropriate “corrector” into the final data. Using the method,
Clark and Oppenheimer, in [4], and Denche-Bessila, very recently in [5], regularized
the backward problem by replacing the final condition by
u(T) + εu(0) = g (1.4)
and
u(T) − εu0
(0) = g, (1.5)
respectively.
To the authors’ knowledge, so far there are many papers on the linear homogeneous
case of the backward problem, but we only find a few papers on the nonhomogeneous
case, and especially, the nonlinear case is very scarce.
For recent articles considering the nonlinear backward-parabolic heat, we refer the
reader to [14,17,20]. In [17], the authors used the quasi-boundary value method to reg-
ularize the latter problem. Thus, they have established, under the hypothesis that f is a
global Lipschitzian function, the existence of a unique solution for some approximated
well-posed problem as follows:
V ε
(x, t) =
∞
X
p=1
e−tp2
ε + e−T p2 ϕp −
Z T
t
e−tp2
εs/T + e−sp2 fp(V ε
)(s) ds
sin px, (1.6)
where ε is a positive parameter such that ε 0 and
fp(u)(t) =
2
π
hf(x, t, u(x, t)), sin (px)i =
2
π
Z π
0
f(x, t, u(x, t)) sin (px) dx, (1.7)
ϕp =
2
π
hϕ(x), sin (px)i =
2
π
Z π
0
ϕ(x) sin (px) dx (1.8)
and h · , · i is the inner product in L2(0, π).
Let ϕ and ϕε denote the exact and measured data at t = T, respectively, which
satisfy
kϕ − ϕεk ≤ ε.
Under a strong smoothness assumption on the original solution namely
Z T
0
∞
X
p=1
e2sp2
f2
p (u)(s) ds ∞, (1.9)
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3. Quasi-boundary value method 915
we obtain an error estimate
ku( · , t) − V ε
( · , t)k ≤
p
M1 exp
3k2T(T − t)
2
εt/T
, (1.10)
where
M1 = 3ku( · , 0)k2
+ 6π
Z T
0
∞
X
p=1
e2sp2
f2
p (u)(s) ds.
And in [19], the error is also given as the similar form
ku(t) − uε
(t)k ≤ Mβ(ε)t/T
.
It is easy to see that the two above errors do not tend to zero, if ε is fixed and t tends to
zero . Hence, the convergence of the approximate solution is very slow when t is in a
neighborhood of zero. Moreover, the error in case t = 0 is not given in here. These are
some disadvantages of the above methods.
In the present paper, we will improve the latter results by using the method given
in [17]. The nonlinear backward problem is approximated by the following one dimen-
sional problem:
uε
t − uε
xx =
∞
X
k=1
e−T k2
εk2 + e−T k2 fk(uε
)(t) sin (kx), (x, t) ∈ (0, π) × (0, T), (1.11)
uε
(0, t) = uε
(π, t) = 0, t ∈ [0, T], (1.12)
uε
(x, T) =
∞
X
k=1
e−T k2
εk2 + e−T k2 gk sin (kx), x ∈ [0, π], (1.13)
where ε ∈ (0, e T), and
gk =
2
π
hg(x), sin kxi =
2
π
Z π
0
g(x) sin (kx) dx,
fk(u)(t) =
2
π
hf(x, t, u(x, t)), sin kxi =
2
π
Z π
0
f(x, t, u(x, t)) sin kx dx
and h · , · i is the inner product in L2(0, π).
The paper is organized as follows. In Theorem 2.1 and 2.2, we will show that (1.11)–
(1.13) is well-posed and that the unique solution uε
(x, t) of it is given by
uε
(x, t) =
∞
X
k=1
εk2
+ e−T k2 −1
e−tk2
gk −
Z T
t
e(s−t−T )k2
fk(uε
)(s) ds
sin kx.
(1.14)
Then, in Theorem 2.4 and 2.6, we estimate the error between an exact solution u of
problem (1.1)–(1.3) and the approximation solution uε
of (1.14). Under some addi-
tional conditions on the smoothness of the exact solution u, we also show that
kuε
( · , t) − u( · , t)k ≤ Cε(t+m)/(T +m)
, (1.15)
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4. 916 P. H. Quan, D. D. Trong, and N. H. Tuan
where m 0, C depends on u and k · k is the norm of L2(0, π). Note that (1.15) is an
improved result of many recent papers, such as [17,19,20]. Oncemore, the convergence
of the approximate solution in t = 0 is also proved. The notations about the usefulness
and advantages of this method can be found in Remarks 2.3, 2.5. Finally, a numerical
experiment will be given in Section 4, which proves the efficiency of our method.
2. The main results
From now on, for clarity, we denote the solution of (1.1)–(1.3) by u(x, t), and the
solution of the problem (1.11)–(1.13) by uε
(x, t). Let ε be a positive number such that
0 ε e T.
The function f is called global Lipchitz function if f ∈ L∞
([0, π] × [0, T] × R)
satisfies
|f(x, y, w) − f(x, y, v)| ≤ L|w − v| (H1)
for a constant L 0 independent of x, y, w, v.
Theorem 2.1 (Existence and uniqueness of the regularized problem). Let (H1) hold.
Then problem (1.11)–(1.13) has a unique weak solution
uε
∈ W = C([0, T]; L2
(0, π)) ∩ L2
(0, T; H1
0 (0, π)) ∩ C1
(0, T; H1
0 (0, π)) (2.1)
satisfying (1.14).
Theorem 2.2 (Stability of the regularized solution). Let u and v be two solutions of
(1.11)–(1.13) corresponding to the final values g and h in L2(0, π). Then we have the
inequality
ku( · , t) − v( · , t)k ≤ εt/T −1
T
1 + ln (T/ε)
1−t/T
exp (L2
(T − t)2
) kg − hk.
Remark 2.3. In [18], the stability of magnitude is eT/ε
. And in [4, 5, 14, 17], the
stability estimate is of order εt/T −1, which is better than the previous results.
In this paper, we give different estimation of the stability is order of
Cεt/T −1
T
1 + ln (T/ε)
1−t/T
. (2.2)
It is clear to see that the order of stability (2.2) is less than the orders given in the above
results. This is one of the advantages of our method.
Despite the uniqueness, problem (1.1)–(1.3) is still ill-posed. Hence, we have to
resort to a regularization. We have the following result.
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5. Quasi-boundary value method 917
Theorem 2.4. Let (H1) hold.
a) If u(x, t) ∈ W is the solution of the problem (1.1)–(1.3) such that
Z T
0
∞
X
k=1
k4
e2sk2
f2
k(u)(s) ds ∞ (2.3)
and kuxx( · , 0)k ∞, then
ku( · , t) − uε
( · , t)k ≤ CM εt/T
T
1 + ln (T/ε)
1−t/T
. (2.4)
b) If u(x, t) satisfies
Q = sup
0≤t≤T
∞
X
k=1
k4
e2tk2
|hu(x, t), sin kxi|2
∞, (2.5)
then
ku( · , t) − uε
( · , t)k ≤ CQεt/T
T
1 + ln (T/ε)
1−t/T
,
for every t ∈ [0, T], where
M = 3kuxx(0)k2
+
3π
2
T
Z T
0
∞
X
k=1
k4
e2sk2
f2
k(u)(s) ds, (2.6)
CM =
p
πM e3L2T (T −t), (2.7)
CQ =
q
πQ e3L2T (T −t). (2.8)
c) If there exists a positive number m 0 such that
Qm = sup
0≤t≤T
∞
X
k=1
k4
e(2t+2m)k2
|hu(x, t), sin kxi|2
∞, (2.9)
then
ku( · , t) − Uε
( · , t)k ≤ ε(t+m)/(T +m)
p
Qm eL2
(T +1)(T −t)
(2.10)
for every t ∈ [0, T], where Uε
is the unique solution of the problem
Uε
(x, t) =
∞
X
k=1
εk2
+ e−(T +m)k2 −1
×
e−(t+m)k2
gk −
Z T
t
e(s−t−T −m)k2
fk(Uε
)(s) ds
sin kx. (2.11)
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6. 918 P. H. Quan, D. D. Trong, and N. H. Tuan
Remark 2.5. 1. In [17, p. 241] and [20] the error estimate between the exact solution
and the approximate solution is U(ε, t) = Cεt/T
. So, if the time t is close to the
original time t = 0, the convergence rates here are very slow. This implies that the
methods studied in [31, 33] are not useful to derive the error estimations in the case t
is near zero. To improve this, we give the convergence rate in the present theorem in
slightly different form, defined by V (ε, t) = Dεt/T
[T/(1 + ln (T/ε)]1−t/T
. We note
that limε→0 V (ε, t)/U(ε, t) = 0. Moreover, we also have limε→0(limt→0 U(ε, t)) = C
and limε→0(limt→0 V (ε, t)) = limε→0[DT/(1 + ln (T/ε))] = 0. This also proves
that our method gives a better approximation than the previous case which we know.
Comparing (1.12) with the previous results obtained in [17, 20], we realize that this
estimate is sharp and the best known estimate.
2. One superficial advantage of this method is that there is an error estimation for
the time t = 0. We have the following logarithmic estimate:
ku( · , 0) − uε
( · , 0)k ≤
p
M e3L2T 2 T
1 + ln (T/ε)
. (2.12)
These estimates, as noted above, are very seldom in the theory of ill-posed problems.
3. In the linear nonhomogeneous case f(x, t, u) = f(x, t), the error estimates were
given in [19]. And the assumption of f in (2.3) is not used, only in L2(0, T; L2(0, π)).
4. In Theorem 2.4a) we ask for a condition on the expansion coefficient fk. This
condition on u is severe and requires more than analiticy of x → f(x, t, u(x, t)) for all
t ∈ [0, T]. We note that the solution u depends on the nonlinear term f and therefore fk
and fk(u) are very difficult to be valued. Such an obscurity makes this theorem hard
to be used for numerical computations. To improve this, in Theorem 2.6b) we only
require the assumption on u, depending not on the function f(u). Once more, we note
that in the simple case of the right hand side f(u) = 0, the term Q becomes
∞
X
k=1
k4
e2tk2
|hu(x, t), sin kxi|2
= kuxx( · , 0)k.
So, the condition (2.5) is nature and acceptable.
5. In Theorem 2.4c) we give the error which is the Hölder form on the whole t ∈
[0, T]. Comparing (2.10) with (2.12), we can see that (2.10) is the optimal order.
In the case of nonexact data, we have the following theorem.
Theorem 2.6. Let the exact solution u of (1.1)–(1.3) correspond to g. Let gε be a
measured data such that
kgε − gk ≤ ε.
Then there exists a function wε
satisfying
a) kwε
( · , t) − u( · , t)k ≤ (1 +
√
M ) exp
3L2
T (T −t)
2
εt/T
T
1+ln (T/ε)
1−t/T
,
for every t ∈ [0, T], where u is defined in Theorem 2.4a).
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7. Quasi-boundary value method 919
b) kwε
( · , t) − u( · , t)k ≤ (1 +
√
πQ ) exp
3L2
T (T −t)
2
εt/T
T
1+ln (T/ε)
1−t/T
,
for every t ∈ [0, T], where u is defined in Theorem 2.4a), and M, Q are defined in
Theorem 2.4b).
3. Proofs of the main theorems
We give some lemmas which will be useful for proving the main theorems.
Lemma 3.1. Denote h(x) = 1/(εx + e−xT
) and Mε = T(ε + ε ln (T/ε))−1 then
h(x) ≤ Mε =
T
ε(1 + ln (T/ε))
. (3.1)
The proof of this lemma can be found in [19].
For 0 ≤ t ≤ s ≤ T, denote
Gε(s, t, k) =
e(s−t−T )k2
εk2 + e−T k2 . (3.2)
It is easy to see that
Gε(T, t, k) =
e−tk2
εk2 + e−T k2 . (3.3)
Lemma 3.2.
Gε(s, t, k) ≤ M(s−t)/T
ε . (3.4)
Proof. We have
Gε(s, t, k) =
e(s−t−T )k2
εk2 + e−T k2 =
e(s−t−T )k2
(εk2 + e−T k2
)(s−t)/T (εk2 + e−T k2
)(T +t−s)/T
≤
e(s−t−T )k2
(e−T k2
)(T +t−s)/T
1
(εk2 + e−T k2
)s/T −t/T
≤
T
ε(1 + ln (T/ε))
s/T −t/T
= εt/T −s/T
T
1 + ln (T/ε)
s/T −t/T
= M(s−t)/T
ε . (3.5)
2
Lemma 3.3.
Gε(T, t, k) ≤ M(T −t)/T
ε . (3.6)
Proof. Let s = T in Lemma 3.2. Then we obtain Lemma 3.3. 2
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8. 920 P. H. Quan, D. D. Trong, and N. H. Tuan
Proof of Theorem 2.1. The proof consists of Steps 1–3.
Step 1. Existence and uniqueness of the solution of the integral equation (1.14).
Put
F(w)(x, t) = P(x, t) −
∞
X
k=1
Z T
t
Gε(s, t, k)fk(w)(s) ds sin (kx)
for w ∈ C([0, T]; L2(0, π)), where
P(x, t) =
∞
X
k=1
Gε(T, t, k) hg(x), sin kxi sin kx.
We claim that, for every w, v ∈ C([0, T]; L2(0, π)), p ≥ 1, we have
kFp
(w)( · , t) − Fp
(v)( · , t)k2
≤
LT
ε(1 + ln (T/ε))
2p (T − t)p
Cp
p!
k|w − v|k2
, (3.7)
where C = max {T, 1}, and k| · |k is the sup norm on C([0, T]; L2(0, π)).
We shall prove the latter inequality by induction. For p = 1, using (3.2) and
Lemma 3.2, we find that
kF(w)( · , t) − F(v)( · , t)k2
=
π
2
∞
X
k=1
h Z T
t
Gε(s, t, k)(fk(w)(s) − fk(v)(s)) ds
i2
≤
π
2
∞
X
k=1
Z T
t
e(s−t−T )k2
εk2 + e−T k2
2
ds
Z T
t
(fk(w)(s) − fk(v)(s))2
ds
≤
π
2
∞
X
k=1
M2
ε (T − t)
Z T
t
(fk(w)(s) − fk(v)(s))2
ds
=
π
2
M2
ε (T − t)
Z T
t
∞
X
k=1
(fk(w)(s) − fk(v)(s))2
ds
= M2
ε (T − t)
Z T
t
Z π
0
(f(x, s, w(x, s)) − f(x, s, v(x, s)))2
dx ds
≤ L2
M2
ε (T − t)
Z T
t
Z π
0
|w(x, s) − v(x, s)|2
dx ds
= CL2
M2
ε (T − t) k|w − v|k2
.
Thus (3.7) holds.
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9. Quasi-boundary value method 921
Suppose that (3.7) holds for p = m. We prove that (3.7) holds for p = m + 1. We
have
kFm+1
(w)( · , t) − Fm+1
(v)( · , t)k2
=
π
2
∞
X
k=1
h Z T
t
Gε(s, t, k)(fk(Fm
(w))(s) − fk(Fm
(v))(s)) ds
i2
≤
π
2
M2
ε
∞
X
k=1
h Z T
t
|fk(Fm
(w))(s) − fk(Fm
(v))(s)| ds
i2
≤
π
2
M2
ε (T − t)
Z T
t
∞
X
k=1
|fk(Fm
(w))(s) − fk(Fm
(v))(s)|2
ds
≤ M2
ε (T − t)
Z T
t
kf( · , s, Fm
(w)( · , s)) − f( · , s, Fm
(v)( · , s))k2
ds
≤ M2
ε (T − t)L2
Z T
t
kFm
(w)( · , s) − Fm
(v)( · , s)k2
ds
≤ M2
ε (T − t)L2m+2
M2m
ε
Z T
t
(T − s)m
m!
dsCm
k|w − v|k2
≤ (LMε)2m+2 (T − t)m+1
(m + 1)!
Cm+1
k|w − v|k2
.
Therefore, by the induction principle, we have
k|Fp
(w) − Fp
(v)|k ≤
LT
ε(1 + ln (T/ε))
p Tp/2
√
p!
Cp/2
k|w − v|k
for all w, v ∈ C([0, T]; L2(0, π)).
We consider F : C([0, T]; L2(0, π)) → C([0, T]; L2(0, π)). Since
lim
p→∞
LT
ε(1 + ln (T/ε))
p Tp/2Cp/2
√
p!
= 0,
there exists a positive integer p0 such that
LT
ε(1 + ln (T/ε))
p0 Tp0/2Cp0/2
p
(p0)!
1
and Fp0 is a contraction. It follows that the equation Fp0 (w) = w has a unique solution
uε
∈ C([0, T]; L2(0, π)).
We claim that F(uε
) = uε
. In fact, we have F(Fp0 (uε
)) = F(uε
). Hence it
follows Fp0 (F(uε
)) = F(uε
). By the uniqueness of the fixed point of Fp0 , we
get that F(uε
) = uε
, that is, the equation F(w) = w has a unique solution uε
∈
C([0, T]; L2(0, π)).
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10. 922 P. H. Quan, D. D. Trong, and N. H. Tuan
Step 2. If uε
∈ W satisfies (1.14) then uε
is solution of (1.11)–(1.13). We have
uε
(x, t) =
∞
X
k=1
(εk2
+ e−T k2
)−1
e−tk2
gk −
Z T
t
e(s−t−T )k2
fk(uε
)(s) ds
sin kx,
0 ≤ t ≤ T. (3.8)
We can verify directly that uε
∈ C([0, T]; L2(0, π) ∩ C1((0, T); H1
0 (0, π)) ∩
L2(0, T; H1
0 (0, π))). In fact, uε
∈ C∞
((0, T]; H1
0 (0, π))). Moreover, by direct compu-
tation, one has
uε
t (x, t) =
∞
X
k=1
−k2
(εk2
+ e−T k2
)−1
e−tk2
gk −
Z T
t
e(s−t−T )k2
fk(uε
)(s) ds
sin kx
+
∞
X
k=1
e−T k2
(εk2
+ e−T k2
)−1
fk(uε
)(t) sin kx
= −
2
π
∞
X
k=1
k2
huε
(x, t), sin kxi sin kx
+
∞
X
k=1
e−T k2
(εk2
+ e−T k2
)−1
fk(uε
)(t) sin kx
= uε
xx(x, t) +
∞
X
k=1
e−T k2
(εk2
+ e−T k2
)−1
fk(uε
)(t) sin kx (3.9)
and
uε
(x, T) =
∞
X
k=1
e−T k2
(εk2
+ e−T k2
)−1
gk sin (kx). (3.10)
So uε
is the solution of (1.11)–(1.13).
Step 3. The problem (1.11)–(1.13) has at most one (weak) solution uε
∈ W.
Let uε
and vε
be two solutions of problem (1.11)–(1.13) such that uε
, vε
∈ W.
Putting wε
(x, t) = uε
(x, t) − vε
(x, t), then wε
satisfies the equation
wε
t − wε
xx =
∞
X
k=1
e−T k2
(εk2
+ e−T k2
)−1
(fk(uε
)(t) − fk(vε
)(t)) sin (kx). (3.11)
Noticing that e−T k2
(εk2 + e−T k2
)−1 ≤ 1/ε, we have
kwε
t − wε
xxk2
≤
1
ε2
∞
X
k=1
(fk(uε
)(t) − fk(vε
)(t))2
≤ kf( · , t, uε
( · , t)) − f( · , t, vε
( · , t))k2
/ε2
≤ M2
kuε
( · , t) − vε
( · , t)k2
/ε2
= M2
kwε
( · , t)k2
/ε2
.
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11. Quasi-boundary value method 923
Using the Lees–Protter in [29], we get wε
( · , t) = 0. This completes the proof of
Step 3.
The three steps complete the proof of the theorem. 2
Proof of Theorem 2.2. Using (1.14) and the inequality (a+b)2 ≤ 2(a2 +b2), we obtain
ku( · , t) − v( · , t)k2
=
π
2
∞
X
k=1

14. Gε(T, t, k)(gk − hk) −
Z T
t
Gε(s, t, k)(fk(u)(s) − fk(v)(s)) ds

17. 2
≤ π
∞
X
k=1
(Gε(T, t, k) |gk − hk|)2
+ π
∞
X
k=1
Z T
t
Gε(s, t, k) |fk(u)(s) − fk(v)(s)| ds
2
. (3.12)
Using Lemma 3.2, Lemma 3.3 and (3.12), we get
ku( · , t) − v( · , t)k2
≤ M2−2t/T
ε kg − hk2
+ 2L2
(T − t)M−2t/T
ε
Z T
t
M2s/T
ε ku( · , s) − v( · , s)k2
ds.
(3.13)
It follows from (3.13) that
M2t/T
ε ku( · , t) − v( · , t)k2
≤ M−2
ε kg − hk2
+ 2L2
(T − t)
Z T
t
M2s/T
ε ku( · , s) − v( · , s)k2
ds.
Using Gronwall’s inequality we have
M2t/T
ε ku( · , t) − v( · , t)k2
≤ M−2
ε exp (2L2
(T − t)2
) kg − hk2
.
Thus
ku( · , t) − v( · , t)k ≤ εt/T −1
T
1 + ln (T/ε)
1−t/T
exp (L2
(T − t)2
) kg − hk.
This completes the proof of the theorem. 2
Proof of Theorem 2.4a). Suppose the problem (1.1)–(1.3) has an exact solution u,
then u is given by
u(x, t) =
∞
X
k=1
e−(t−T )k2
gk −
Z T
t
e−(t−s)k2
fk(u)(s) ds
sin kx. (3.14)
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18. 924 P. H. Quan, D. D. Trong, and N. H. Tuan
Since
uk(0) = eT k2
gk −
Z T
0
esk2
fk(u)(s) ds
implies
gk = e−T k2
uk(0) +
Z T
0
e(s−T )k2
fk(u)(s) ds,
we get
u(x, T) =
∞
X
k=1
gk sin kx =
∞
X
k=1
e−T k2
uk(0) +
Z T
0
e−(T −s)k2
fk(u)(s) ds
sin kx.
Since (1.15) and (3.14), we have
uε
k(t) = (εk2
+ e−T k2
)−1
e−tk2
gk −
Z T
t
e(s−t−T )k2
fk(uε
)(s) ds
, (3.15)
uk(t) = eT k2
e−tk2
gk −
Z T
t
e(s−t−T )k2
fk(u)(s) ds
. (3.16)
With (3.3), (3.15) and (3.16) we have
uk(t) − uε
k(t) =
eT k2
−
1
εk2 + e−T k2
e−tk2
gk −
Z T
t
e(s−t−T )k2
fk(u)(s) ds
+
Z T
t
e(s−t−T )k2
εk2 + e−T k2 (fk(uε
)(s) − fk(u)(s)) ds
=
εk2 e−tk2
εk2 + e−T k2
eT k2
gk −
Z T
0
esk2
fk(u)(s) ds +
Z t
0
esk2
fk(u)(s) ds
+
Z T
t
Gε(s, t, k)(fk(uε
)(s) − fk(u)(s)) ds. (3.17)
Since (3.6) and
eT k2
gk −
Z T
0
esk2
fk(u)(s) ds = uk(0),
we have
|uk(t) − uε
k(t)| ≤ εGε(T, t, k)

21. k2
eT k2
gk −
Z T
0
k2
esk2
fk(u)(s) ds

24. + εGε(T, t, k)

27. Z t
0
k2
esk2
fk(u)(s) ds

30. +
Z T
t
Gε(s, t, k)|fk(u)(s) − fk(uε
)(s)| ds
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31. Quasi-boundary value method 925
≤ ε · M(T −t)/T
ε
|k2
uk(0)| +
Z t
0
|k2
esk2
fk(u)(s)| ds
+
Z T
t
M(s−t)/T
ε |fk(u)(s) − fk(uε
)(s)| ds
= ε · M(T −t)/T
ε
|k2
uk(0)| +
Z T
0
|k2
esk2
fk(u)(s)| ds
+
Z T
t
1
ε
Ms/T −1
ε |fk(u)(s) − fk(uε
)(s)| ds
.
Applying the inequality (a + b + c)2 ≤ 3(a2 + b2 + c2), we get
ku( · , t) − uε
( · , t)k2
=
π
2
∞
X
k=1
|uk(t) − uε
k(t)|2
≤
3π
2
ε2
· M(2T −2t)/T
ε
∞
X
k=1
h
|k2
uk(0)|2
+
Z T
0
|k2
esk2
fk(u)(s)| ds
2
+
Z T
t
1
ε
Ms/T −1
ε |fk(u)(s) − fk(uε
)(s)| ds
2i
= ε2
· M(2T −2t)/T
ε (I1 + I2 + I3),
where
I1 =
3π
2
∞
X
k=1
|k2
uk(0)|2
, I2 =
3π
2
∞
X
k=1
Z T
0
|k2
esk2
fk(u)(s)| ds
2
,
I3 =
3π
2
∞
X
k=1
Z T
t
1
ε
Ms/T −1
ε |fk(u)(s) − fk(uε
)(s)| ds
2
.
The terms I1, I2, I3 can be estimated as
I1 = 3ε2
M(2T −2t)/T
ε kuxx(0)k2
.
Using the Hölder inequality, we estimate the term I2 as follows:
I2 ≤
3π
2
T
Z T
0
∞
X
k=1
(k2
esk2
fk(u)(s))2
ds ≤
3π
2
T
Z T
0
∞
X
k=1
k4
e2sk2
f2
k(u)(s) ds
≤
3π
2
T
Z T
0
∞
X
k=1
k4
e2sk2
f2
k(u)(s) ds. (3.18)
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32. 926 P. H. Quan, D. D. Trong, and N. H. Tuan
Using condition (H1) from page 916, we get
I3 ≤
3π
2
(T − t)
Z T
t
1
ε2
M2s/T −2
ε
∞
X
k=1
(fk(u)(s) − fk(uε
)(s))2
ds
≤ 3(T − t)
Z T
t
1
ε2
M2s/T −2
ε kf( · , s, u( · , s)) − f( · , s, uε
( · , s))k2
ds
≤ 3L2
T
Z T
t
1
ε2
M2s/T −2
ε ku( · , s) − uε
( · , s)k2
ds. (3.19)
Combining (3.18), (3.19), we obtain
ku( · , t) − uε
( · , t)k2
≤ ε2
· M(2T −2t)/T
ε
h
3kuxx(0)k2
+
3π
2
T
Z T
0
∞
X
k=1
k4
e2sk2
f2
k(u)(s) ds
+ 3L2
T
Z T
t
1
ε2
M2s/T −2
ε ku( · , s) − uε
( · , s)k2
ds
i
.
It follows that
1
ε2
· M(2t−2T )/T
ε ku( · , t) − uε
( · , t)k2
≤ M + 3L2
T
Z T
t
1
ε2
M2s/T −2
ε ku( · , s) − uε
( · , s)k2
ds.
Using Gronwall’s inequality, we get
ε−2t/T
T
1 + ln (T/ε)
2t/T −2
ku( · , t) − uε
( · , t)k2
≤ M e3L2
T (T −t)
.
Hence
ku( · , t) − uε
( · , t)k2
≤ M e3L2
T (T −t)
ε2t/T
T
1 + ln (T/ε)
2−2t/T
.
This completes the proof of Theorem 2.4a). 2
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33. Quasi-boundary value method 927
Proof of Theorem 2.4b). Since (3.17), we prove in a similar manner to that in the above
part, that
|uk(t) − uε
k(t)|
≤

36. eT k2
−
1
εk2 + e−T k2
e−tk2
gk −
Z T
t
e(s−t−T )k2
fk(u)(s) ds

39. +

42. Z T
t
Gε(s, t, k)(fk(uε
)(s) − fk(u)(s)) ds

45. ≤

48. εk2 e−tk2
εk2 + e−T k2
eT k2
gk −
Z T
t
esk2
fk(u)(s) ds

51. +
Z T
t
Gε(s, t, k)|fk(uε
)(s) − fk(u)(s)| ds
≤
ε e−tk2
εk2 + e−T k2 k2
etk2
|uk(t)| +
Z T
t
Gε(s, t, k)|fk(u)(s) − fk(uε
)(s)| ds (3.20)
≤ ε · M(T −t)/T
ε |k2
etk2
uk(t)| +
Z T
t
M(s−t)/T
ε |fk(u)(s) − fk(uε
)(s)| ds. (3.21)
This implies that
ku( · , t) − uε
( · , t)k2
=
π
2
∞
X
k=1
|uk(t) − uε
k(t)|2
≤ π
∞
X
k=1
ε2
· M(2T −2t)/T
ε |k2
etk2
uk(t)|2
+ π
∞
X
k=1
ε2
· M(2T −2t)/T
ε
×
Z T
t
ε−s/T
T
1 + ln (T/ε)
s/T −1
|fk(u)(s) − fk(uε
)(s)| ds
2
.
Since Mε = T(ε + ε ln (T/ε))−1 implies that
ε2
· M(2T −2t)/T
ε = ε2t/T
T
1 + ln (T/ε)
2−2t/T
,
we obtain
ku( · , t) − uε
( · , t)k2
≤ ε2t/T
T
1 + ln (T/ε)
2−2t/T
π
∞
X
k=1
k4
e2tk2
u2
k(t)
+ 2L2
Tε2t/T
T
1 + ln (T/ε)
2−2t/T
Z T
t
ε−2s/T
T
1 + ln (T/ε)
2s/T −2
× ku( · , s) − uε
( · , s)k2
ds.
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52. 928 P. H. Quan, D. D. Trong, and N. H. Tuan
Using again Gronwall’s inequality, we get
ε−2t/T
T
1 + ln (T/ε)
2t/T −2
ku( · , t) − uε
( · , t)k2
≤ Q e2L2
T (T −t)
.
This completes the proof of Theorem 2.4b. 2
Proof of Theorem 2.4c). We have
|uk(t) − Uε
k (t)|
≤

55. eT k2
−
e−mk2
εk2 + e−(T +m)k2
e−tk2
gk −
Z T
t
e(s−t−T )k2
fk(u)(s) ds

58. +

61. Z T
t
e(s−t−T −m)k2
εk2 + e−(T +m)k2 (fk(Uε
)(s) − fk(u)(s)) ds

64. ≤

67. εk2 e−tk2
εk2 + e−(T +m)k2
eT k2
gk −
Z T
t
esk2
fk(u)(s) ds

70. +
Z T
t
e(s−t−T −m)k2
εk2 + e−(T +m)k2 |fk(Uε
)(s) − fk(u)(s)| ds
≤
ε e−(t+m)k2
εk2 + e−(T +m)k2 k2
e(t+m)k2
|uk(t)| +
Z T
t
Gε(s, t, k)|fk(u)(s) − fk(uε
)(s)| ds
≤ ε(t+m)/(T +m)
k2
e(t+m)k2
|uk(t)| +
Z T
t
M(s−t)/(T +m)
ε |fk(u)(s) − fk(Uε
)(s)| ds
≤ ε(t+m)/(T +m)
k2
e(t+m)k2
|uk(t)| +
Z T
t
ε
T
(s−t)/(T +m)
|fk(u)(s) − fk(Uε
)(s)| ds.
(3.22)
Then, we obtain
ku( · , t) − Uε
( · , t)k2
=
π
2
∞
X
k=1
|uk(t) − Uε
k (t)|2
≤ ε2(t+m)/(T +m)
∞
X
k=1
k4
e2(t+m)k2
u2
k(t)
+ 2L2
(T + 1)ε2t/(T +m)
Z T
t
ε−2s/(T +m)
ku( · , s) − Uε
( · , s)k2
ds.
Using again Gronwall’s inequality, we get
ε−2t/T +m
ku( · , t) − Uε
( · , t)k2
≤ ε2m/(T +m)
Qm e2L2
(T +1)(T −t)
.
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71. Quasi-boundary value method 929
Hence, we obtain
ku( · , t) − Uε
( · , t)k ≤ ε(t+m)/(T +m)
p
Qm eL2
(T +1)(T −t)
. (3.23)
2
Proof of Theorem 2.6. Let uε
be the solution of problem (1.11)–(1.13) corresponding
to g and let wε
be the solution of problem (1.11)–(1.13) corresponding to gε.
Using Theorem 2.2 and 2.4a), we get
kwε
( · , t) − u( · , t)k ≤ kwε
( · , t) − uε
( · , t)k + kuε
( · , t) − u( · , t)k
≤ εt/T −1
T
1 + ln (T/ε)
1−t/T
exp (L2
(T − t)2
)kgε − gk
+
p
M e3L2T (T −t) εt/T
T
1 + ln (T/ε)
1−t/T
≤ (1 +
√
M ) exp
3L2T(T − t)
2
εt/T
T
1 + ln (T/ε)
1−t/T
for every t ∈ [0, T].
Theorem 2.6b) is similarly proved as Theorem 2.6a). 2
4. A numerical experiment
We consider the equation
−uxx + ut = f(u) + g(x, t),
where
f(u) = u4
, g(x, t) = 2 et
sin x − e4t
sin4
x,
and
u(x, 1) = ϕ0(x) ≡ e sin x.
The exact solution of the equation is
u(x, t) = et
sin x.
Especially
u(x, 99/100) ≡ u(x) = exp (99/100) sin x.
Let ϕε(x) ≡ ϕ(x) = (ε + 1) e sin x. We have
kϕε − ϕk2 =
sZ π
0
ε2 e2 sin2
x dx = ε e
r
π
2
.
We find the regularized solution uε(x, 99/100) ≡ uε(x) having the form
uε(x) = vm(x) = w1,m sin x + w2,m sin 2x + w3,m sin 3x,
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72. 930 P. H. Quan, D. D. Trong, and N. H. Tuan
where
v1(x) = (ε + 1) e sin x, w1,1 = (ε + 1) e, w2,1 = 0, w3,1 = 0
and
a = 1/10000, tm = 1 − am, m = 1, 2, . . . , 100,
wi,m+1 =
e−tm+1i2
εi2 + e−tmi2 wi,m
−
2
π
Z tm
tm+1
e−tm+1i2
εi2 + e−tmi2 e(s−tm)i2
Z π
0
(v4
m(x) + g(x, s)) sin ix dx
ds,
i = 1, 2, 3.
Let aε = kuε − uk be the error between the regularization solution uε and the exact
solution u. Letting ε = ε1 = 10−5, ε = ε2 = 10−7, ε = ε3 = 10−11, we have
ε uε aε
ε1 = 10−5
2.685490624 sin (x) − 0.00009487155350 sin (3x) 0.005744631447
ε2 = 10−7
2.691122866 sin (x) + 0.00001413193606 sin (3x) 0.0001124971593
ε3 = 10−11
2.691180223 sin (x) + 0.00002138991088 sin (3x) 0.00005831365439
We have, in view of the error table in [17, p. 214],
ε uε aε
ε1 = 10−5
2.430605996 sin (x) − 0.0001718460902 sin (3x) 0.3266494251
ε2 = 10−7
2.646937077 sin (x) − 0.002178680692 sin (3x) 0.05558566020
ε3 = 10−11
2.649052245 sin (x) − 0.004495263004 sin (3x) 0.05316693437
By applying the stabilized quasi-reversibility method from [20], we have the ap-
proximate solution uε(x, 99/100) ≡ uε(x) having the form
uε(x) = vm(x) = w1,m sin x + w6,m sin 6x,
where v1(x) = (ε + 1) e sin x, w1,1 = (ε + 1) e, w6,1 = 0, and a = 1/10000, tm =
1 − am for m = 1, 2, . . . , 100, and
wi,m+1 = (ε + e−tmi2
)(tm+1−tm)/tm
wi,m
−
2
π
Z tm
tm+1
e(s−tm+1)i2
Z π
0
(v4
m(x) + g(x, s)) sin ix dx
ds
for i = 1, 6. The following table shows the approximation error in this case.
ε uε kuε − uk
ε1 = 10−5
2.690989330 sin(x) − 0.06078794774 sin(6x) 0.003940316590
ε2 = 10−7
2.691002638 sin(x) − 0.05797060493 sin(6x) 0.003592425036
ε3 = 10−11
2.691023938 sin(x) − 0.05663820292 sin(6x) 0.003418420030
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73. Quasi-boundary value method 931
Looking at three above tables with comparison between other methods, we can see
that the error results of the second and third tables are smaller than those in the first
table. This shows that our approach has a nice regularizing effect and gives a better
approximation compared to the previous method in [17,20].
Acknowledgments. The authors would like to thank the anonymous referees for their
careful reading and valuable comments and suggestions.
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Received April 14, 2009
Author information
P. H. Quan, Department of Mathematics, Sai Gon University, 273 An Duong Vuong, Q.5, Ho Chi
Minh city, Vietnam.
D. D. Trong, Department of Mathematics, HoChiMinh City National University, 227 Nguyen Van
Cu, Q. 5, Ho Chi Minh city, Vietnam.
N. H. Tuan, Department of Mathematics, Sai Gon University, 273 An Duong Vuong, Q.5, Ho Chi
Minh city, Vietnam.
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