Applied Mathematics Letters 73 (2017) 69–77
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Applied Mathematics Letters www.elsevier.com/locate/aml
Recovering the initial distribution for strongly damped wave equation Nguyen Huy Tuana , Doan Vuong Nguyenb,c , Vo Van Aud,e , Daniel Lesnicf, * a
Applied Analysis Research Group, Faculty of Mathematics and Statistics, Ton Duc Thang University, Ho Chi Minh City, Viet Nam b Faculty of Mathematics and Computer Science, University of Science, Vietnam National University, Ho Chi Minh City, Viet Nam c Faculty of Fundamental Science, Industrial University of Ho Chi Minh City, Viet Nam d Institute of Computational Science and Technology, Ho Chi Minh City, Viet Nam e Faculty of General Sciences, Can Tho University of Technology, Can Tho City, Viet Nam f Department of Applied Mathematics, University of Leeds, Leeds LS2 9JT, UK
article
info
Article history: Received 31 January 2017 Received in revised form 12 April 2017 Accepted 12 April 2017 Available online 19 April 2017
abstract We study for the first time the inverse backward problem for the strongly damped wave equation. First, we show that the problem is severely ill-posed in the sense of Hadamard. Then, under the a priori assumption on the exact solution belonging to a Gevrey space, we propose the Fourier truncation method for stabilizing the ill-posed problem. A stability estimate of logarithmic type is established. © 2017 Elsevier Ltd. All rights reserved.
Keywords: Fourier regularization method Final value problem Strongly damped wave equation
1. Introduction The strongly damped wave equation (SDWE), see (1.1), occurs in a wide range of applications modelling the motion of viscoelastic materials [1–4]. From both the theoretical and numerical point of view, the initial value problem for this equation has been extensively studied (e.g., [5–7]). However, to the best of our knowledge, the final value (backward) problem has not been solved yet (though it is worth mentioning that in [8], Lattes and Lions introduced the problem (1.1)–(1.2) but they did not regularize it). Our major objective is to provide a regularization method for solving the ill-posed nonlinear problem (1.1)–(1.2). Let T be a positive number and Ω be an open, bounded and connected domain in Rn , n ≥ 1 with a smooth boundary ∂Ω . We are interested in the following inverse backward problem: Find the initial data
*
Corresponding author. E-mail addresses:
[email protected] (N.H. Tuan),
[email protected] (D.V. Nguyen),
[email protected] (V.V. Au),
[email protected] (D. Lesnic). http://dx.doi.org/10.1016/j.aml.2017.04.014 0893-9659/© 2017 Elsevier Ltd. All rights reserved.
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u(x, 0) for x ∈ Ω , where u(x, t) satisfies the following semilinear SDWE: utt − αAut − Au = F (x, t, u(x, t)),
(x, t) ∈ Ω × (0, T ),
(1.1)
subject to the conditions ⎧ ⎨u = 0, (x, t) ∈ ∂Ω × (0, T ), u(x, T ) = g(x), (x, t) ∈ Ω × (0, T ), ⎩ ut (x, T ) = h(x), (x, t) ∈ Ω × (0, T ),
(1.2)
where α > 0 is a given damping constant, g(x) and h(x) are given functions, and the source function F will be defined later. Here, the operator −A : D(A) ⊂ L2 (Ω ) → L2 (Ω ) is a linear, positive( ) ∑d ∂ ∂ definite, self-adjoint operator with compact inverse in L2 (Ω ). For instance, A := i,j=1 ∂xi lij (x) ∂xj , x ∈ Ω is a linear second-order elliptic operator with smooth coefficients {lij }di,j=1 being symmetric and uniformly positive definite. Then, the Dirichlet eigenvalues (λj )j∈N∗ of −A satisfy, see Chapter 6.5 in [9], 0 < λ1 ≤ λ2 ≤ λ3 ≤ ... ≤ λj ≤ · · · and limj→∞ λj = ∞. In practice, the data (g, h) is obtained by measurement contaminated with noise. Hence, instead of (g, h), we have the observation data (g ϵ , hϵ ) ∈ L2 (Ω ) × L2 (Ω ) satisfying ∥g ϵ − g∥L2 (Ω) + ∥hϵ − h∥L2 (Ω) ≤ ϵ,
(1.3)
where the constant ϵ > 0 represents a bound on the measurement error. We will show that the inverse backward problem (1.1)–(1.2) is ill-posed in the sense of Hadamard in Section 2. In order to stabilize the solution, in Section 3 we develop a regularization method based on the truncated Fourier method for which a stability estimate of logarithmic type is established. 2. Ill-posedness of the inverse problem (1.1)–(1.2) Assume that the problem (1.1)–(1.2) has a unique solution in the series form u(x, t) =
∞ ∑
uj (t)ϕj (x),
(2.4)
j=1
where ϕj denotes the eigenfunction corresponding to the eigenvalue λj and { ′′ uj (t) + αλj u′j (t) + λj uj (t) = Fj (u)(t), t ∈ (0, T ), (2.5) uj (T ) = gj , u′j (T ) = hj , ∫ ∫ ∫ where Fj (u)(t) = Ω F (x, t, u(x, t))ϕj (x)dx, gj = Ω g(x)ϕj (x)dx and hj = Ω h(x)ϕj (x)dx. For given fixed damping α > 0, consider the decomposition ( { }) ( { }) ( { }) 4 4 4 N∗ = D1 = j ∈ N∗ |λj > 2 ∪ D2 = j ∈ N∗ |λj = 2 ∪ D3 = j ∈ N∗ |λj < 2 . α α α Then, the solution of (2.5) is given by: (i) if j ∈ D1 then uj (t) =
λ+ j e
(T −t)λ− j
∫ + t
− λ− e √ j Λj
T
e
(s−t)λ+ j
(T −t)λ+ j
−e √ Λj
gj −
e
(T −t)λ+ j
−e √ Λj
(T −t)λ− j
hj
(s−t)λ− j
Fj (u)(s)ds,
(2.6)
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(ii) if j ∈ D2 then [ ] ∫ T 2 2 2 2 uj (t) = e α (T −t) 1 − (T − t) gj − e α (T −t) (T − t)hj + (s − t)e α (s−t) Fj (u)(s)ds, α t (iii) if j ∈ D3 then αλj
2e 2 (T −t) uj (t) = √ −Λj
[√
αλj
2e 2 (T −t) − √ −Λj
(√ ) (√ )] −Λj −Λj −Λj αλj cos (T − t) + sin (T − t) gj 2 2 2 2 (√ ) (√ ) αλj ∫ T −Λj −Λj 2e 2 (s−t) √ sin sin (T − t) hj + (s − t) Fj (u)(s)ds, 2 2 −Λj t
where λ+ j =
αλj +
√ α2 λ2j − 4λj 2
λ− j =
,
αλj −
√ α2 λ2j − 4λj 2
,
Λj = α2 λ2j − 4λj .
(2.7)
Hence, the solution (2.4) can be represented as ∑ ∑ ∑ u(x, t) = uj (t)ϕj + uj (t)ϕj + uj (t)ϕj . j∈D1
j∈D2
j∈D3
∑
∑ From above observations, we can show that the term j∈D2 uj (t)ϕj + j∈D3 uj (t)ϕj containing sin and cos trigonometric functions is bounded and stable, and no regularization is needed for it. We only need to ∑ regularize the first term j∈D1 uj (t)ϕj which contains the exponential terms in (2.6). Alternatively, we can take D1 = D2 = ∅ by assuming that α2 λ1 > 4, as will be adopted in the remaining of the paper. Note that this also implies α2 λ2j − 4λj > 0 for all j ∈ N∗ , and hence the roots in (2.7) are real and distinct. From now on, we denote by ⟨·, ·⟩ and ∥ · ∥ the inner product and norm in L2 (Ω ), respectively. For φ ∈ L2 (Ω ), defining −
S(t)φ =
+
tλj tλj ∞ ∑ λ+ − λ− j e j e j=1
− λ+ j − λj
−
+
∞ tλ tλ ∑ e j −e j P(t)φ = − ⟨φ, ϕj ⟩ ϕj , λ+ j − λj j=1
⟨φ, ϕj ⟩ ϕj ,
(2.8)
we can recast (2.4) in the form ∫ u(x, t) = S(T − t)g(x) + P(T − t)h(x) −
T
P(s − t)F (u(s))ds.
(2.9)
t
Next, we give an example which shows that the solution um (x, t) (for any m ∈ N∗ ) of problem (1.1)–(1.2) (if it exists) is not stable. Let uT,m and vT,m and F0 be defined as follows: um (x, T ) = uT,m (x) = 0, F0 (w) =
∞ ∑ e−αT λj j=1
2T 2
ϕm (x) ∂t um (x, T ) = vT,m (x) = √ and λm
⟨w(x, t), ϕj ⟩ ϕj , ∀w ∈ L2 (Ω ).
(2.10)
Let um satisfy the integral equation ∫ um (x, t) = P(T − t)vT,m −
T
P(s − t)F0 (um (s))ds.
(2.11)
t
First, we show that (2.11) has a unique solution um ∈ C([0, T ]; L2 (Ω )). Indeed, we consider the function ∫ T H(w)(t) = P(T − t)vT,m (x) − P(s − t)F0 (w(s))ds. (2.12) t
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Then, for any v1 , v2 ∈ C([0, T ]; L2 (Ω )), we have ∫ T ∥H(v1 )(t) − H(v2 )(t)∥ ≤ P(s − t) (F0 (v1 (s)) − F0 (v2 (s))) ds t [ ]2 ∫ T ∞ (s−t)λ− (s−t)λ+ ∑ j −e j e 2 √ = ⟨F0 (v1 (s)) − F0 (v2 (s)), ϕj ⟩ ds − + − λ λ t j j j=1 [ ]2 − ∫ T ∞ (s−t)λj (s−t)λ+ ∑ j 1 e−2αT λj e −e 2 √ = ⟨v1 (s) − v2 (s), ϕj ⟩ ds. (2.13) − + 4 2 t T − λ λ j j j=1 Using the inequality |e−a − e−b | ≤ |a − b| for a, b > 0, we have [ (s−t)λ− ]2 [ −(s−t)λ+ ]2 (s−t)λ+ −(s−t)λ− j −e j −e j j e e−2αT λj e−2αT λj 2(s−t)(λ+ +λ− ) e j j = e − − T4 T4 λ+ λ+ j − λj j − λj ≤ (s − t)2 e2α(s−t)λj
e−2αT λj 1 ≤ 2. T4 T
From (2.13) and (2.14) we deduce that ∫ 1 T 1 1 ∥H(v1 )(t) − H(v2 )(t)∥ ≤ ∥v1 (s) − v2 (s)∥ds ≤ ∥v1 − v2 ∥C([0,T ];L2 (Ω)) , 2 t T 2
(2.14)
∀t ∈ [0, T ].
(2.15)
This implies that ∥H(v1 ) − H(v2 )∥C([0,T ];L2 (Ω)) ≤
1 ∥v1 − v2 ∥C([0,T ];L2 (Ω)) . 2
(2.16)
Hence, H is a contraction. Using the Banach fixed-point theorem, we conclude that H(w) = w has a unique solution um ∈ C([0, T ]; L2 (Ω )). It is easy to see that (here, noting that F0 (0) = 0) ∫ T 1 P(s − t)F0 (um (s))ds = ∥H(um )(t) − H(0)(t)∥ ≤ ∥um ∥C([0,T ];L2 (Ω)) . (2.17) 2 t Hence, ∫ T ∥um (t)∥ ≥ P(T − t)vT,m − P(s − t)F0 (um (s))ds t 1 ≥ P(T − t)vT,m − ∥um ∥C([0,T ];L2 (Ω)) . 2
(2.18)
This leads to ∥um ∥C([0,T ];L2 (Ω)) ≥
2 sup P(T − t)vT,m . 3 0≤t≤T
We continue to estimate the right hand side of this inequality. We have ( )2 √ 2 [ ] 2 2(T −t)λ+ + 2 m 1 − e−(T −t) α λm −4λm 2 e (T −t)λ− (T −t)λ m −e m 1 e = P(T − t)vT,m = ( ) − + − 2 λm λm − λm λm λ+ m − λm ( )2 √ 2 λ2 −4λ −(T −t) α 2(T −t)λ+ 1 1 m 1−e e ≥ . λm (α2 λ2m − 4λm )
(2.19)
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Since the function Φ(t) = e
(T −t)λ+ m
(
√
−(T −t)
1−e
(T −t)λ+ m
α2 λ2 −4λ1 1
(
)
√
−(T −t)
is a decreasing function, we deduce that α2 λ2 −4λ1 1
)
1−e √ λm (α2 λ2m − 4λm ) 0≤t≤T ( ) √ −T α2 λ2 −4λ1 1 eαT λm /2 1 − e √ ≥ . λm (α2 λ2m − 4λm )
e sup P(T − t)vT,m ≥ sup
0≤t≤T
73
e
T λ+ m
=
(
√
−T
1−e
√
α2 λ2 −4λ1 1
)
λm (α2 λ2m − 4λm ) (2.20)
Combining (2.19) and (2.20) yields
∥um ∥C([0,T ];L2 (Ω)) ≥
2 3
e
αT λm /2
√
( 1−e
−T
√
α2 λ2 −4λ1 1
) .
λm (α2 λ2m − 4λm )
(2.21)
As m → +∞, we see that lim (∥uT,m ∥ + ∥vT,m ∥) =
m→+∞
lim ∥um ∥C([0,T ];L2 (Ω)) ≥
m→+∞
lim
m→+∞
√
2 m→+∞ 3
1 = 0, λm
e
αT λm /2
lim
(
−T
√
α2 λ2 −4λ1 1
)
1−e √ λm (α2 λ2m − 4λm )
= +∞.
(2.22)
This shows that problem (1.1)–(1.2) is ill-posed in the sense of Hadamard in the L2 -norm. 3. Fourier’s truncation method For φ ∈ L2 (Ω ), let us define the truncated version of (2.8) as −
SN (t)φ =
+
tλj tλj N ∑ λ+ − λ− j e j e j=1
− λ+ j − λj
−
⟨φ, ϕj ⟩ ϕj ,
PN (t)φ =
+
N tλ tλ ∑ e j −e j − ⟨φ, ϕj ⟩ ϕj . λ+ j − λj j=1
Let us define the regularized solution by Fourier’s truncation method as follows: ∫ T uN,ϵ (x, t) = SN (T − t)g ϵ (x) + PN (T − t)hϵ (x) − PN (s − t)F (uN,ϵ )(x, s)ds,
(3.23)
(3.24)
t
where N is a parameter regularization to be prescribed. Lemma 3.1. The following estimates hold: √ 2α2 + 8T 2 αtλN e , ∥SN (t)∥L(L2 (Ω)) ≤ α2
∥PN (t)∥L(L2 (Ω)) ≤ T eαtλN ,
Proof . Let φ ∈ L2 (Ω ). From (3.23) we have [ + tλ− ] tλ+ 2 N j − λ− e j ∑ λ e j j 2 ∥SN (t)φ∥2 = ⟨φ, ϕj ⟩ − + λj − λj j=1 [ + −tλ+ ]2 −tλ− N j − λ− e j ∑ 2t(λ+ +λ− ) λj e j 2 j j = e ⟨φ, ϕj ⟩ . − + λ − λ j j j=1
∀t ∈ [0, T ].
(3.25)
(3.26)
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Using the inequality (a + b)2 ≤ 2a2 + 2b2 for a, b ∈ R and the inequality |e−a − e−b | ≤ |a − b| for a, b > 0, we obtain ⎞2 ]2 ⎛ [ + −tλ+ −tλ− −tλ+ −tλ− j j −e j + + λj e j − λ− e 2 2 −tλ j e − ⎠ ≤ 2e−2tλj + 2|λ− = ⎝e j + λj j | t − − + + λj − λj λj − λj ⎞2 ⎛ 2 2 2tλ j ⎠ ≤ 2α + 8T . √ (3.27) ≤2+⎝ α2 αλj + α2 λ2j − 4λj It follows from (3.26) and (3.27) that ∥SN (t)φ∥2 ≤ ≤
N N 2α2 + 8T 2 ∑ 2t(λ+ 2α2 + 8T 2 ∑ 2αtλj +λ− ) 2 j j ⟨φ, ϕ ⟩2 ≤ ⟨φ, ϕj ⟩ e e j 2 α2 α j=1 j=1
2α2 + 8T 2 2αtλN e ∥φ∥2 . α2
(3.28)
This completes the proof of the first part of (3.25). Also, using |e−a − e−b | ≤ |a − b| for a, b > 0, we obtain [ tλ− ] N tλ+ 2 ∑ e j −e j 2 2 ∥PN (t)φ∥ = ⟨φ, ϕj ⟩ − + λ − λ j j j=1 [ + − ]2 N N ∑ 2t(λ+ +λ− ) e−tλj − e−tλj ∑ 2t(λ+ +λ− ) 2 2 2 e j j = ⟨φ, ϕ ⟩ ≤ T e j j ⟨φ, ϕj ⟩ , (3.29) j − + λ − λ j j j=1 j=1 which completes the proof of the second part of (3.25). □ At this stage, in order to obtain the convergence rate (3.31) given in the following theorem, we introduce the abstract Gevrey class of functions of order γ > 0 and index σ > 0, e.g., [10], defined by { } ∞ ∑ 2 2γ 2σλj 2 Gγ,σ = v ∈ L (Ω ) : λj e |⟨v, ϕj (x)⟩| < ∞ , j=1
which is a Hilbert space equipped with the inner product ⟩ ⟨ √ √ ⟨v1 , v2 ⟩Gγ,σ := (−∆)γ eσ −∆ v1 , (−∆)γ eσ −∆ v2 ,
∀v1 , v2 ∈ Gγ,σ ,
√∑ ∞ 2 2γ 2σλj and the corresponding norm ∥v∥Gγ,σ = |⟨v, ϕj (x)⟩| < ∞. j=1 λj e We also assume that F is globally Lipschitz, i.e. there exists a constant K > 0 such that ∥F (x, t, u) − F (x, t, v)∥ ≤ K∥u − v∥,
∀u, v ∈ L2 (Ω ), ∀(x, t) ∈ Ω × (0, T ).
(3.30)
Theorem 3.1. The nonlinear integral equation (3.24) has a unique solution uN,ϵ ∈ C([0, T ]; L2 (Ω )). Assuming further that u ∈ L∞ (0, T ; Gγ,αT ) for some γ > 0, then we have the following estimate: ((√ ) ) 2α2 + 8T 2 −γ N,ϵ −αtλN KT 2 αT λN ∥u (·, t) − u(·, t)∥ ≤ e e +T e ϵ + λN ∥u∥L∞ (0,T ;Gγ,αT ) , α2 ∀t ∈ [0, T ].
(3.31)
( ) δ Remark 3.1. If we choose N = N (ϵ) such that λN ≤ αT ln 1ϵ for some δ ∈ (0, 1), then the error [ ]−γ ∥uN,ϵ (·, t) − u(·, t)∥ is of logarithmic order ln( 1ϵ ) . Also, if F (x, t, u) = F (x, t) does not depend on u
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then we do not need to employ the Gevrey space but only require that u ∈ L∞ (0, T ; L2 (Ω )). To remove the assumption on u belonging to L∞ (0, T ; Gγ,αT ), we can employ a new regularization method described in [11] for the Cauchy problem for semilinear elliptic equations, but its extension to the present damped semilinear wave equation (1.1) is deferred to a future work. Proof . Part 1. The existence and uniqueness solution of the nonlinear integral equation (3.24). For w ∈ C([0, T ]; L2 (Ω )), we define G(w)(x, t) = SN (T − t)g ϵ (x) + PN (T − t)hϵ (x) −
∫
T
PN (s − t)F (w(x, s))ds.
(3.32)
∀w1 , w2 ∈ C([0, T ]; L2 (Ω )).
(3.33)
t
We shall prove by induction that m G (w1 ) − G m (w2 ) C([0,T ];L2 (Ω)) ( )m KT eαT λN (T − t) ∥w1 − w2 ∥C([0,T ];L2 (Ω)) , ≤ m! For m = 1, we have
∫ T ∥G(w1 ) − G(w2 )∥ = PN (s − t) (F (w1 (s)) − F (w2 (s))) ds t ∫ T ≤ T eα(s−t)λN ∥F (w1 (s)) − F (w2 (s))∥ds t
≤ KT eαT λN (T − t)∥w1 − w2 ∥C([0,T ];L2 (Ω)) .
(3.34)
Assume that (3.33) holds for m = p. We show that (3.33) holds for m = p + 1. Indeed, we have ∥G
p+1
(w1 ) − G
p+1
(w2 )∥ = ≤ ≤
≤
≤
∫ T PN (s − t) (F (G p (w1 )(s)) − F (G p (w1 )(s))) ds t ∫ T T eα(s−t)λN F (G p (w1 )(s)) − F (G p (w1 )(s))ds t ∫ T αT λN KT e ∥G p (w1 (s)) − G p (w2 (s))∥ds t ( )p ∫ T KT eαT λN (T − s) ∥w1 − w2 ∥C([0,T ];L2 (Ω)) ds KT eαT λN p! t )p+1 ( KT eαT λN (T − t) ∥w1 − w2 ∥C([0,T ];L2 (Ω)) . (p + 1)! ( )m
(3.35)
KT 2 eαT λN
Therefore, by the induction principle, we have that (3.33) holds. Since limm→+∞ = 0 there m! exists a positive integer m0 such that G m0 is a contraction. It follows that the equation G m0 w = w has a unique solution uN,ϵ ∈ C([0, T ]; L2 (Ω )). We claim that G(uN,ϵ ) = uN,ϵ . Indeed, since G m0 (uN,ϵ ) = uN,ϵ , ( ) ( ) we know that G G m0 (uN,ϵ ) = G(uN,ϵ ). This is equivalent to G m0 G(uN,ϵ ) = G(uN,ϵ ). Hence, G(uN,ϵ ) is a fixed point of G m0 . Moreover, as noted above, uN,ϵ is a fixed point of G m0 . Part 2. Denote ∫ T U N (x, t) = SN (T − t)g(x) + PN (T − t)h(x) − PN (s − t)F (uN (x, s))ds. (3.36) t
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Step 1. Firstly, we estimate ∥uN,ϵ (·, t) − U N (·, t)∥. Using Lemma 3.1, we have ) ( ) ( ∥uN,ϵ (·, t) − U N (·, t)∥ ≤ SN (T − t) g − g ϵ + PN (T − t) h − hϵ ∫ T ( ) PN (s − t) F (uN,ϵ (·, s)) − F (U N (·, s)) ds + t √ 2α2 + 8T 2 α(T −t)λN ≤ e ∥g − g ϵ ∥ + T eα(T −t)λN ∥h − hϵ ∥ α2 ∫ T +T eα(s−t)λN F (uN,ϵ (·, s)) − F (U N (·, s))ds.
(3.37)
t
It follows from (1.3) and (3.30) that (√
N,ϵ
∥u
) 2α2 + 8T 2 + T eα(T −t)λN ϵ (·, t) − U (·, t)∥ ≤ α2 ∫ T + KT eα(s−t)λN ∥uN,ϵ (·, s) − U N (·, s)∥ds. N
(3.38)
t
Multiplying both sides by eαtλN and applying Gronwall’s inequality, we derive that (√ ) 2α2 + 8T 2 αtλN N,ϵ N e ∥u (·, t) − U (·, t)∥ ≤ + T eαT λN ϵeK(T −t)T . α2 Hence, (√ N,ϵ
∥u
N
(·, t) − U (·, t)∥ ≤ e
KT (T −t)
2α2 + 8T 2 +T α2
) eα(T −t)λN ϵ.
(3.39)
Step 2. Secondly, we estimate ∥u(·, t) − U N (·, t)∥. First, it is easy to see that ∫ T N ∑ uj (t)ϕj (x) = SN (T − t)g(x) + PN (T − t)h(x) − PN (s − t)F (u(x, s))ds. t
j=1
Using Lemma 3.1, we obtain N N ∑ ∑ ∥u(·, t) − U N (·, t)∥ ≤ u(·, t) − uj (t)ϕj (·) + uj (t)ϕj (·) − U N (·, t) j=1
j=1
∑ ∫ ∞ −2αtλN λ2γ e2αtλj |u (t)|2 + λ−2γ e ≤√ j j N
t
j=N +1 −αtλN ≤ λ−γ ∥u∥L∞ (0,T ;Gγ,αT ) + KT N e
∫
T
PN (s − t)F (u(·, s)) − F (U N (·, s))ds
T
eα(s−t)λN ∥u(·, s) − U N (·, s)∥ds.
(3.40)
t
Multiplying both sides by eαtλN and applying Gronwall’s inequality, we derive that −αtλN ∥u(·, t) − U N (·, t)∥ ≤ eKT (T −t) ∥u∥L∞ (0,T ;Gγ,αT ) λ−γ . N e
(3.41)
Combining (3.39) and (3.41) we deduce (3.31). □ Acknowledgement The first and the third author have been supported by the Institute for Computational Science and Technology, Ho Chi Minh City under the project entitled “Regularization of inverse problems for hyperbolic and pseudo-hyperbolic equations”.
N.H. Tuan et al. / Applied Mathematics Letters 73 (2017) 69–77
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