An approximate solution for a nonlinear biharmonic equation with discrete random data

Journal Pre-proof An approximate solution for a nonlinear biharmonic equation with discrete random data Nguyen Huy Tuan, Yong Zhou, Tran Ngoc Thach, Nguyen Huu Can

PII: DOI: Reference:

S0377-0427(20)30002-9 https://doi.org/10.1016/j.cam.2020.112711 CAM 112711

To appear in:

Journal of Computational and Applied Mathematics

Received date : 23 March 2019 Revised date : 24 November 2019 Please cite this article as: N.H. Tuan, Y. Zhou, T.N. Thach et al., An approximate solution for a nonlinear biharmonic equation with discrete random data, Journal of Computational and Applied Mathematics (2020), doi: https://doi.org/10.1016/j.cam.2020.112711. This is a PDF file of an article that has undergone enhancements after acceptance, such as the addition of a cover page and metadata, and formatting for readability, but it is not yet the definitive version of record. This version will undergo additional copyediting, typesetting and review before it is published in its final form, but we are providing this version to give early visibility of the article. Please note that, during the production process, errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.

© 2020 Elsevier B.V. All rights reserved.

Journal Pre-proof

An approximate solution for a nonlinear biharmonic equation with discrete random data Nguyen Huy Tuan1 , Yong Zhou2,3 , Tran Ngoc Thach4 , Nguyen Huu Can4∗, 1

of

Institute of Research and Development, Duy Tan University, Da Nang 550000, Vietnam Faculty of Information Technology, Macau University of Science and Technology, Macau 999078, China 3 Faculty of Mathematics and Computational Science, Xiangtan University, Hunan 411105, China 4 Applied Analysis Research Group, Faculty of Mathematics and Statistics, Ton Duc Thang University, Ho Chi Minh City, Vietnam

p ro

2

January 3, 2020

Pr e-

Abstract

In this paper, we study a problem of finding the solution for the nonlinear biharmonic equation ∆2 u = f (x, t, u(x, t)) from the final data. By using a simple example, the ill-posedness of the present problem with random noise is demonstrated. The Fourier method is conducted in order to establish an estimator for the mild solution (called regularized solution) and the convergence results in some different cases are proposed. Finally, numerical experiments are presented for showing that this regularization method is flexible and stable.

1

Introduction

urn

al

In this work, the problem of finding a function u = u(x, t) which satisfies the following nonlinear biharmonic equation   (x, t) ∈ Ω × (0, T ), ∆2 u = f (x, t, u(x, t)),    u(0, t) = u(π, t) = ∆u(0, t) = ∆u(π, t) = 0, t ∈ (0, T ), (1) u(x, T ) = ϕ1 (x), ∆u(x, T ) = ϕ2 (x), x ∈ Ω,       ∂u (x, T ) = 0, ∂∆u (x, T ) = 0, x ∈ Ω, ∂t ∂t

Jo

is considered. Here, Ω = (0, π), T > 0 is a fixed value, ∆2 u = uxxxx + 2uxxtt + utttt is the biharmonic operator and ϕ1 , ϕ2 are given final data. The biharmonic equation have many applications in physical practical for example, in plate theory (Chapter 5, [8]), the motion of fluids, free boundary problems and non-linear elasticity [9, 10]. The biharmonic equation appears in many various applications in science and engineering [34, 30, 31]. For example, the equation describing the displacement vector ~u in elastodynamics is given by [30, 31] (λ + µ) ∇ (∇ · ~u) + µ∇2 ~u + F~ = 0,

(2)

where λ and µ are the Lam´e coefficients, and F~ is the body force acting on the object. By decomposing ~ Eq. (2) gives the displacement vector ~u = ∇φ + ∇ × ψ, ∇2 ∇2 φ = ∇4 φ = ∆2 φ = − ∗

1 ~ = ∇4 ψ ~ = 1 ∇ × F~ , ∇ · F~ , ∇2 ∇2 ψ λ+µ µ

Corresponding author: [email protected] (Nguyen Huu Can)

1

(3)

Journal Pre-proof

~ are the inhomogeneous scalar and vector biharmonic equations [30]. that is, the equations for φ and ψ Some related models of elastic bodies have been studied by using some mathematical methods which are studied by [32, 33]. The biharmonic equation have many applications in gravitational theories. For example, let us take the gravitational field of Dirac δ-type mass distribution with the form ρ = 4πGmδ (~r), where G is gravitational constant, m the mass, and δ (~r) is the Dirac delta function. Then the gravitational potential Φ satisfies the Poisson equation [35], (4)

of

∆Φ = 4πGmδ (~r) ,

Jo

urn

al

Pr e-

p ro

with the radial solution given by Φ(r) = −Gm/r. As we know before, this potential is singular at r = 0, giving rise to infinite tidal forces. However, a modified Poisson equation is given as follows [35]
∆ 1 + M −2 ∆ Φ = 4πGmδ (~r) , (5)
where M is a constant, gives the solution Φ(r) = −Gm 1 − e−M r /r, which is nonsingular at r = 0, and tends towards the Newtonian potential when M → ∞ (we can see more details in [36].) The numerical solution for Cauchy problem for biharmonic equation was studied by many authors, such as [11, 13]. Some other works on nonlinear biharmonic equation can be found in [23]. From the study of P. Schafer [20, 21, 22], we know that Cauchy problem for biharmonic equation are not well-posed problem ( called improperly posed problem or ill-posed problem). For a history of the biharmonic problem and the relation with elasticity from an engineering viewpoint see the survey of Meleshko [38]. Since the biharmonic equation is a higher-order partial differential equation, some Cauchy data on all of the boundary of the solution domain is not enough to give a well-posed problem. If a part of the boundary is over-specified and the remaining part is under-specified then the problem is known as a Cauchy problem (see [37]). Such incomplete data imply that the biharmonic equation is an ill-posed problem. Such a problem (1) is ill-posed in the sense of Hadamard, that a small pertubation of the given data (ϕ1 , ϕ2 ) may effect a very large error on the solution. It is difficult to do a numerical calculation. Hence, a regularization method is needed to give a stable approximation solution of the exact solution. Since the input data is based on the observations, there must be measurement errors in reality. If the quantities causing these errors is under our control, the mathematical model is often deterministic. By contrast with that case, there are many practical situations in which the sources generating these errors are not controllable - wind, rain are some understandable examples. For that reason, the regularization and convergence estimate should be investigated in statistics setting. It is the fact that a small perturbation in the input data can make a large deviation in the solution, see [1] for more details on the influence of the errors. In the case of deterministic errors, Problem (1) with homogeneous case i.e, f = 0, has been studied in [7, 12] with some well-known regularization method. Since Problem (1) is nonlinear, then it is much more complicated than some linear case. Until now, there are not any results on the ill-posedness and regularization results for nonlinear biharmonic equation. In the real-world problem, the data is defined by experiments. Hence, we only know its values inexactly. When we measure or observe the data, we thus only know a finite number of discrete data which is closer to practice. In this work, we choose the fixed points 0 < χ1 < χ2 < ... < χn < π as follows χj =

π(2j − 1) , 2n

j = 1, 2, ..., n,

and assume that the measured data at points χj ∈ Ω are contaminated by n observed values (φej , ψej ) as in two models φej = ϕ1 (χj ) + αj Yj ,

ψej = ϕ2 (χj ) + βj Wj .

Here, Yj , Wj denote the mutually independent standard normal variables, Yj ∼ N (0, 1), Wj ∼ N (0, 1), αj , βj > 0, j = 1, 2, ..., n, are bounded by the positive constants α∗ , β∗ respectively. The statistical 2

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2

p ro

of

inverse problems for linear problem has a long history. Some theoretical examples in the case of random noise can be found in [2]. In [3], we can refer some methods (spectral cut-off and Tikhonov) for dealing with linear statistical inverse problems. Until now, we have not found any article in the literature studied the regularization and convergence results for nonlinear biharmonic equation with perturbed random noise. Our paper may be the first work considering the nonlinear model such as (1) with random case. In our model, the reader can realized that the random case is more difficulty than the deterministic case. Our aim of this paper is to focus on the ill-posedness and regularization results by using a Fourier truncated expansion method, and then establish an estimator for the solution. Our random model is introduced in some recent paper, [15, 24]. We list the content of the present article as follows. In Section 2, we give some auxiliaries which are necessary for stating our results. Some representations and the instability of the mild solution are reviewed in Section 3. In Section 4, we construct an approximate regularized solution for the sought solution and then show the convergence estimate in two different cases under some priori assumptions. We finally test some numerical examples in the last Section.

Some auxiliaries

Pr e-

The following notations and spaces are used through out the present article. For m ∈ N∗ , we set p Xm = 2/π sin mx. Then, {Xm }m∈N∗ is a completely orthonormal in L2 (Ω), Xm = 0 on ∂Ω and the problem −∆Xm (x) = λm Xm (x), x ∈ Ω, admits the eigenvalues λm = m2 , m ∈ N∗ . Moreover, Xm are the corresponding eigenfunctions. For σ > 1, we recall the Hilbert space ( ) ∞ X 2 σ 2 σ H (Ω) = g ∈ L (Ω) : λm |hg, Xm i| < ∞. m=1

The norm in H σ (Ω) is given by

al

kgkH σ (Ω)

v u ∞ uX σ =t λm |hg, Xm i|2 . m=1

urn

Recall that Lp (Ω; X ), where X is a Banach space, p > 2, to represent the Bochner space (see p.3197, [17])   Z   Lp (Ω; X ) = Lp ((Ω, F, P) ; X ) = v : E kvkpX = kv(ω)kpX dP(ω) < ∞, ω ∈ Ω .   Ω

Here, Lp (Ω, X ) is equipped with the following norm

Jo

kvkLp (Ω;X ) = E kvkpX


1/p

,

v ∈ Lp (Ω; X ).

Let us denote WT space of all L2 -valued predictable processes w (see [18]) ( ) WT =

w : sup kw(·, t)kL2 (Ω;L2 (Ω)) < ∞ , t∈[0,T ]

with the following WT -norm kwkWT = sup kw(·, t)kL2 (Ω;L2 (Ω)) = sup t∈[0,T ]

t∈[0,T ]

3



E kw(·, t)k2L2 (Ω)

1/2

.

Journal Pre-proof

3

The mild solution and the instability

3.1

The mild solution

Throughout this article, we use the following notations to represent the m-th Fourier coeffcients of u, f, ϕ1 , ϕ2 um (t) := hu(·, t), Xm i , fm (u)(t) := hf (·, t, u(·, t)) , Xm i , ϕk;m := hϕk , Xm i , k = 1, 2.

t

By setting


√ t sinh λm t √ Hm (t) = , 2 λm

for t ∈ [0, T ], the solution u can be rewritten as follows u(x, t) =

∞ X

m=1

"



√ √ t cosh λm t sinh λm t p Km (t) = , + 2λm 2 λ3m

Pr e-

p  Gm (t) = cosh λm t ,

p ro

of

Using the Fourier method, we can obtain a formula for the solution u as follows "
√ ∞ p  X (T − t) sinh λm (T − t) √ u(x, t) = cosh λm (T − t) ϕ1;m + ϕ2;m 2 λm m=1 #

! √ √ ZT (τ − t) cosh λm (τ − t) sinh λm (τ − t) p fm (u)(τ)dτ Xm (x). + + 2λm 2 λ3m

Gm (T − t)ϕ1;m + Hm (T − t)ϕ2;m +

ZT t

#

Km (τ − t)fm (u)(τ)dτ Xm (x).

(6)

(7)

al

Our next purpose is to find a representation for u expressed based on the discrete values ϕ1 (χj ), ϕ2 (χj ), j = 1, 2, . . . , n. The following forms of ϕ1 , ϕ2 are useful to do this end:
Lemma 3.1. Let m be a positive integer less than n and suppose that ϕk ∈ C 1 Ω , k = 1, 2. Then ∞

j=1

i=1

urn

ϕk;m

n

X πX = ϕk (χj )Xm (χj ) − (−1)i ϕk;2in±m , n

in which ϕk;2in±m := ϕk;2in+m − ϕk;2in−m . Proof. For k = 1, 2 and m ∈ N∗ , by using Lemma 3.5 of [19] and ϕk = n−1

n X

l=1

Jo

j=1

ϕk (χj )Xm (χj ) = n−1

∞ X

ϕk;l

n X

"

∞ X

ϕk;l Xl , we can check that

l=1

Xl (χj )Xm (χj ) = π −1 ϕk;m +

j=1

∞ X i=1

#

(−1)i ϕk;2in±m .

This completes the proof.

P i By the next lemma, it can be seen that the terms ∞ i=1 (−1) ϕk;2in±m are very small under the σ priori assumption: ϕk belongs to H (Ω) with σ > 1, for k = 1, 2. Lemma 3.2. Let m be a positive integer less than n and ϕ1 , ϕ2 belongs to H σ (Ω) with σ > 1. Then, ∞ X for k = 1, 2, the absolute of (−1)i ϕk;2in±m is of order n−σ . i=1

4

Journal Pre-proof

Proof. The assumption ϕk ∈ H σ (Ω) yields that |ϕk;m | ≤

kϕk kH σ (Ω) mσ

,

for any m < n.

In view of the above, we can verify that ∞ ∞ X X   (2in + m)−σ + (2in − m)−σ kϕk kH σ (Ω) (−1)i ϕk;2in±m ≤ i=1



−σ

≤ (2n)

∞ X

i=1

i−σ kϕk kH σ (Ω) .

(8)

i−σ , the inequality (8) can be rewritten as follows

i=1

∞ X Cσ kϕk k σ H (Ω) i . (−1) ϕ k;2in±m ≤ nσ i=1

Pr e-

Here, Cσ < ∞.

∞ X

p ro

By setting Cσ := (2−σ + 1)

+n

−σ

of

i=1

From Lemma 3.1, we get a representation of the solution u which is needed for constructing its estimator: Theorem 3.1. Let N be a positive integer number less than n and assume that ϕ1 , ϕ2 belongs to H σ (Ω). Then, we have

i=1

al

u(x, t) " n ! ZT N X πX = Gm (T − t)ϕ1 (χj ) + Hm (T − t)ϕ2 (χj ) Xm (χj ) + Km (τ − t)fm (u)(τ)dτ n m=1 j=1 t # ∞ ∞ ∞ X X X i i − Gm (T − t) (−1) ϕ1;2in±m − Hm (T − t) (−1) ϕ2;2in±m Xm (x) + um (t)Xm (x), i=1

(9)

m=N+1

urn

where Gm (t), Hm (t), Km (t) and ϕk;2in±m are defined in (6) and Lemma 3.1 respectively. In the next subsection, we will show the instability of the solution u, which leads to the ill-posedness of the problem (1). The following inequalities are introduced throughout this subsection:

and

Jo

Lemma 3.3. For m ∈ N∗ and t ∈ [0, T ], we have
√ p  exp λm t ≤ Gm (t) ≤ exp λm t , 2

Km (t) ≤

t 1 + p 2λm 2 λ3m


√ t exp λm t √ Hm (t) ≤ , 2 λm !

exp

p  λm t ,

where Gm (t), Hm (t), Km (t) are defined in (6).

Proof. It is easy to prove this Lemma by using the inequalities (a + b)2 ≤ 2(a2 + b2 ), for a, b ≥ 0. 5

ea 2

≤ cosh a ≤ ea , sinh a ≤ ea and

Journal Pre-proof

3.2

The instability property of the solution

Now, we give an example which demonstrates that the solution of (1) is instable. Let the exact data be as follows ϕ1 (x) = ϕ2 (x) = 0,

f (x, t, u) =

∞ X

m=1

1 um (t)Xm (x). 2T (T + 1) exp (T m)

(10)

of

1 and the observations be φej = √ Yj and ψej = exp (−(n − 1)T ) Wj . Then, we have the following result: n

Lemma 3.4. The problem (1) with respect to the data (10) has a unique solution u ≡ 0. Proof. From (10), we have 1 um (t), 2T (T + 1) exp (T m)

It can be seen from (7) that

m=1

It follows from Lemma 3.3 that

 

1 2T (T + 1) exp (T m)

ZT t

(11)



Km (τ − t)um (τ)dτ Xm (x).

Pr e-

u(x, t) =

∞ X

for m = 1, 2, ...

p ro

fm (u)(t) =

Km (τ − t) ≤ (T + 1) exp (T m) . Applying the Parseval’s identity and the H¨older inequality, we get 2  ZT ∞ X 1  Km (τ − t)um (τ)dτ ku(·, t)k2L2 (Ω) = 2T (T + 1) exp (T m) m=1

m=1

This leads to

t

1 4T 2 (T + 1)2 exp (2T m)

al

≤

∞ X

(12)

urn

ku(·, t)k2L2 (Ω)

1 ≤ 4T

ZT t

ZT

2

(T + 1) exp (2mT ) dτ

t

ZT t

|um (τ)|2 dτ.

ku(·, τ)k2L2 (Ω) dτ.

By the Gronwall’s inequality, we conclude that ku(·, t)k2L2 (Ω) = 0. This shows that u(x, t) = 0.

Jo

The couple of following lemmas will prove that the solution does not depend continuously on the input data, i.e, if there are small errors in the input data ϕ1 , ϕ2 then we can get a large error in the solution u. Lemma 3.5. Let us define the approximate functions for ϕ1 , ϕ2 as follows   n−1 n X πX  ϕ1 (x) ≈ φej Xm (χj ) Xm (x) =: ϕ bn1 (x), n m=1 j=1   n−1 n X πX  ψej Xm (χj ) Xm (x) =: ϕ bn2 (x). ϕ2 (x) ≈ n m=1

j=1

Then, for k = 1, 2, the mean of kϕ bnk − ϕk k2L2 (Ω) tends to zero as n → ∞. 6

(13)

(14)

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Proof. By the Parseval’s identity, it can be seen that E kϕ bn1 − ϕ1 k2L2 (Ω) = E kϕ bn1 k2L2 (Ω) =

m=1

π2 n3



E

Using the properties of random variables Yj , we have (
1, if j = j 0 , E Yj Yj 0 = 0, if j 6= j 0 . Lemma 3.5 of [19] and (15) give us

n X j=1

2

Yj Xm (χj ) .

(15)

of

n−1 X

p ro

 2 n n X X 2   = E Yj Xm (χj ) Xm (χj ) = nπ −1 . j=1

This leads to

(16)

j=1

π(n − 1) . n2

Pr e-

E kϕ bn1 − ϕ1 k2L2 (Ω) =

By a similarly way as above, we can obtain E kϕ bn2 − ϕ2 k2L2 (Ω) =

n−1 X

m=1

 2 n X π2 π(n − 1) exp (−2(n − 1)T )E  Wj Xm (χj ) = exp (−2(n − 1)T ) . 2 n n j=1

Hence, for k = 1, 2, the mean of kϕ bnk − ϕk k2L2 (Ω) tends to zero as n → ∞.

Lemma 3.6. The problem (1) with respect to the data ϕ bn1 , ϕ bn2 defined in (13)-(14) has a unique solution 2 u bn ∈ WT and kb un − ukWT tends to infinity as n tends to infinity.

al

Proof. Noting that the solution u bn (x, ·) is a trigonometric polynomial with order less than n, we can see from the representation of u in Theorem 3.1 that

urn

u bn (x, t) " # ZT n−1 n n X π X X π n e e = Gm (T − t) φj Xm (χj ) + Hm (T − t) ψj Xm (χj ) + Km (τ − t)fm (b u )(τ)dτ Xm (x). n n m=1

j=1

Let us set

j=1

t

(17)

m=1

Jo

F (u(x, t)) " # ZT n n n−1 X X X π π e e Gm (T − t) φj Xm (χj ) + Hm (T − t) ψj Xm (χj ) + Km (τ − t)fm (u)(τ)dτ Xm (x). = n n j=1

j=1

t

Then, we have

n

n

kF (b u (·, t)) − F (b v

(·, t))k2L2 (Ω)

=

n−1 X

m=1

ZT t

7

n

n

!2

Km (τ − t) [fm (b u )(τ) − fm (b v )(τ)] dτ

.

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Using (11)-(12), we obtain kF (b un (·, t)) − F

(b v n (·, t))k2L2 (Ω)

≤ (T + 1)2

n−1 X

exp (2T m)

m=1

" ZT t

#2 hb un (·, τ), Xm i − hb v n (·, τ), Xm i dτ . 2T (T + 1) exp (T m)

The H¨ older inequality yields that Z n−1 Z 1 X ≤ dτ |hb un (·, τ), Xm i − hb v n (·, τ), Xm i|2 dτ 2 4T

kF (b u (·, t)) − F (b v

(·, t))k2L2 (Ω)

≤

1 4T

m=1 t ZT t

It follows that E kF (b un (·, t)) − F (b v n (·, t))k2L2 (Ω) ≤

t

kb un (·, τ) − vbn (·, τ)k2L2 (Ω) dτ.

1 sup E kb un (·, t) − vbn (·, t)k2L2 (Ω) , 4 t∈[0,T ]

∀t ∈ [0, T ].

Pr e-

Hence

T

p ro

n

of

T

n

kF(b un ) − F(b v n )kWT ≤

1 n kb u − vbn kWT . 2

Since F is a contraction with respect to the norm k·kWT , we can conclude by using the Banach fixedpoint theorem that there exists a unique function u bn ∈ WT such that F(b un ) = u bn almost every. Next, 2 we demonstrate that the mean of kb un − ukWT is very large. It follows from (17) that kb un (·, t)k2L2 (Ω) " #2 ZT n−1 n n X π X X π = Gm (T − t) φej Xm (χj ) + Hm (T − t) ψej Xm (χj ) + Km (τ − t)fm (b un )(τ)dτ . n n j=1

Thus

j=1

al

m=1

t

urn

kb un (·, t)k2L2 (Ω) #2 " ZT n n X X π π φej Xm (χj ) + Hm (T − t) ψej Xm (χj ) + Km (τ − t)fm (b un )(τ)dτ , ≥ Gm (T − t) n n j=1

j=1

t

with m is a fixed positive integer number less than n. It follows that

Jo

2 kb un (·, t)k2L2 (Ω)  2  2 n n X X π π ≥  Gm (T − t) φej Xm (χj ) − 4  Hm (T − t) ψej Xm (χj ) n n j=1

j=1

 T 2 Z −4  Km (τ − t)fm (b un )(τ)dτ t

=: P1 − P2 − P3 .

(18)

8

Journal Pre-proof

Now, we estimate the mean of each of three terms above. From Lemma 3.3 and (16), we first have  2
√ n   2 X p π exp 2 λm (T − t) π . (19) EP1 ≥ 3 exp 2 λm (T − t) E  Yj Xm (χj ) = 4n 4n2 j=1

Similarly, we get


√ πT 2 exp 2 λm T EP2 ≤ exp (−2(n − 1)T ). nλm

EP3 ≤ 4(T + 1) exp (2mT )

ZT t

This gives us 2

t

E |fm (b un )(τ)|2 dτ.

t

E kb un (·, τ)k2L2 (Ω)

4T 2 (T + 1)2 exp (2T m)

dτ.

Pr e-

EP3 ≤ 4T (T + 1) exp (2mT )

ZT

dτ

ZT

p ro

2

of

Using H¨ older inequality, we also obtain

(20)

In the inequality above, we use the fact that |hb un (·, τ), Xm i|2 ≤ kb un (·, τ)k2L2 (Ω) . Hence EP3 ≤ sup E kb un (·, t)k2L2 (Ω) .

(21)

t∈[0,T ]

By choosing m = n − 1, we have from (18)-(21) 2E kb un (·, t)k2L2 (Ω) ≥

π exp (2(n − 1)(T − t)) πT 2 − − sup E kb un (·, t)k2L2 (Ω) . 4n2 n(n − 1)2 t∈[0,T ]

Therefore

t∈[0,T ]

urn

This implies that

πT 2 π exp (2(n − 1)(T − t)) − . 4n2 n(n − 1)2

al

3 sup E kb un (·, t)k2L2 (Ω) ≥

lim kb un − uk2WT = lim kb un k2WT = ∞.

n→∞

n→∞

Jo

Remark 3.1. 1. The backward uniqueness (BU) problem is: if u vanishes at , that is u(x, T ) = 0, x ∈ Ω does u vanish identically in Ω × (0, T ) ? The uniqueness of backward PDEs has attracted the attention of many authors; see, for example, [25, 26, 27]. It is also a challenging and open problem, and should be the topic for another paper. 2. Since our model is nonlinear, we need to assume that the sought solution u satsfies (39). The backward uniqueness in C([0, T ]; L2 (Ω)) is open problem, however, the backward uniqueness in C([0, T ]; Eθγ (Ω)) is showed by as follows. The Gevrey class of functions of order γ ≥ 0 and index θ ≥ 0, see e.g. [29], defined by the spectrum of the Laplacian as ( )
√ ∞ X exp 2 λ γ m hf, Xm (x)i2L2 (Ω) ≤ ∞ , (22) Eθγ (Ω) := f ∈ L2 (Ω) : λ2θ m m=1

9

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is equipped with the norm defined by

kf kEθγ (Ω)

 1
p 2 ∞ X exp 2 λj γ 2 = hf, `j (xiL2 (Ω)  < ∞. λ2θ j j=1

(23)

→0

p ro

By a similar explanation as above, if v ∈ C([0, T ]; Eθγ (Ω)) then

of

Let assume Problem (1) have two solution u, v ∈ C([0, T ]; Eθγ (Ω)). Let us recall the regularized solution Uβ defined in (3.1) of the paper [28]. From Theorem 3.2, we know that if u ∈ C([0, T ]; Eθγ (Ω)) then

lim Uβ (., t) − u(., t) L2 (Ω) = 0. (24)

lim Uβ (., t) − v(., t) L2 (Ω) = 0.

→0

(25)

Then from two above limitation, we get the following conclusion

lim ku(., t) − v(., t)kL2 (Ω) = 0.

→0

4

Pr e-

which implies the desired result.

The main results

4.1

(26)

Regularized solution and convergence rate in L2 (Ω)


Theorem 4.1. Let N ∈ N∗ such that N < n.
Assume that ϕ1 , ϕ2 ∈ C 1 Ω ∩ H σ , σ > 1, the problem (1) has a unique solution u ∈ C [0, T ]; L2 (Ω) and f satisfies the global Lipschitz property with respect to the variable u, i.e. there exists a positive constant K such that (27)

al

kf (·, t, u(·, t)) − f (·, t, v(·, t))kL2 (Ω) ≤ K ku(·, t) − v(·, t)kL2 (Ω) . Let an estimator for the solution u (called regularized solution) be established as follows

urn

uN (x, t) # " ZT n n N X X X π π Gm (T − t) φej Xm (χj ) + Hm (T − t) ψej Xm (χj ) + Km (τ − t)fm (uN )(τ)dτ Xm (x). = n n m=1

j=1

j=1

t

where Gm (t), Hm (t), Km (t)are defined in (6). If there exists γ > 0 and a positive constant Λ1 such that sup

∞ X

Jo

0≤t≤T i=1

2 λ2γ i exp (2λi t) | hu(·, t), Xi i | ≤ Λ1 ,

t ∈ [0, T ],

(28)

then we have

2

N

E u (·, t) − u(·, t) 2

L (Ω)

" #  ΓN exp 2√λ T
p
N −2γ ≤ exp −2 λN t + λN Λ1 exp 3K 2 T 2 (T + 1)2 , n 

where

  3T 2   Γ := 6 πα∗2 + Cσ2 kϕ1 k2H σ (Ω) + πβ∗2 + Cσ2 kϕ2 k2H σ (Ω) . 2 10

(29)

Journal Pre-proof

Remark 4.1. For 0 < µ < 12 , let Nn be a positive integer satisfying µ Nn ≤ log n. 2T

2

Then, E uNn (·, t) − u(·, t) L2 (Ω) is of order n o max n2µ−1 , (log nµ )−2γ .

m=1

Pr e-

p ro

of

Proof of the Theorem 4.1: The proof of the main result above is divided into two parts. Part A (Construct an estimator for the sought solution u). Let N ∈ N∗ satisfies N < n. In order to get a estimator for u (called regularized solution), we consider the finite system as follows   (x, t) ∈ Ω × (0, T ), ∆2 uN = f (x, t, uN (x, t)),     uN (0, t) = uN (π, t) = ∆uN (0, t) = ∆uN (π, t) = 0, t ∈ (0, T ), (30) uN (x, T ) = ϕN ∆uN (x, T ) = ϕN x ∈ Ω,  1 (x), 2 (x),   N N    ∂u (x, T ) = 0, ∂∆u (x, T ) = 0, x ∈ Ω, ∂t ∂t in which     N N n n X X X X π π ϕN ϕN φej Xm (χj ) Xm , ψej Xm (χj ) Xm . 1 = 2 = n n m=1

j=1

j=1

 By a similar argument as in the subsection 3.1, we obtain the solution uN ∈ span Xl : l = 1, N (the regularized solution) as follows " # ZT N X N Gm (T − t)ϕN Km (τ − t)fm (uN )(τ)dτ Xm (x), (31) uN (x, t) = 1;m + Hm (T − t)ϕ2;m + m=1

t

in which

n D E πX N ϕN := ϕ , X = ψej Xm (χj ), m 2;m 2 n

al

n D E πX N ϕN := ϕ , X = φej Xm (χj ), m 1;m 1 n j=1

j=1

urn

for m = 1, 2, . . . , N. Under the priori assumption (27), we now prove that the nonlinear integral equation (31) has a unique solution uN ∈ WT . For u ∈ WT , let us set # " ZT N X N N F (u(x, t)) = Gm (T − t)ϕ1;m + Hm (T − t)ϕ2;m + Km (τ − t)fm (u)(τ)dτ Xm (x). m=1

t

Jo

Then, we have the following inequality for u, v ∈ WT  k 1/2 2 T 2 (T + 1)2 exp (2T N)

K

k

  ≤

F (u) − Fk (v)  ku − vkWT , k! WT

k ∈ N∗ .

Indeed, if k = 1, we have kF (u(·, t)) −

F (v(·, t))k2L2 (Ω)

=

N X

m=1

" ZT t

≤ (T + 1)2

#2

Km (τ − t) [fm (u)(τ) − fm (v)(τ)] dτ N X

m=1

 T 2 Z exp (2T m)  [fm (u)(τ) − fm (v)(τ)] dτ .

11

t

(32)

Journal Pre-proof

By the H¨ older inequality, it can be seen that N Z X

T

2

≤ (T + 1) exp (2T N)

m=1 t ZT

≤ T (T + 1)2 exp (2T N)

t

dτ

ZT t

kf (·, τ, u(·, τ)) − f (·, τ, v(·, τ))k2L2 (Ω) dτ.

By using the assumption (27), we get 2

2

≤ K T (T + 1) exp (2T N)

It follows that kF(u) − F(v)kWT

ZT

ku(·, τ) − v(·, τ)k2L2 (Ω) dτ.

p ro

kF (u(·, t)) −

F (v(·, t))k2L2 (Ω)

|fm (u)(τ) − fm (v)(τ)|2 dτ

of

kF (u(·, t)) −

F (v(·, t))k2L2 (Ω)



t

K 2 T 2 (T + 1)2 exp (2T N) ≤ 1!

1/2

ku − vkWT .



Pr e-

By similarly argument as in above, we can check that (32) holds for every k ∈ N∗ . Since k 1/2 2 T 2 (T + 1)2 exp (2T N) K   lim   = 0, k→∞ k!

there exists k0 ∈ N∗ such that Fk0 is a contraction with respect to the norm k·kWT . Hence, there exist a unique function uN ∈ WT such that Fk0 (uN ) = uN ,

almost every.

urn

that

al

It means that the equation Fk0 (u) = u has a unique solution uN ∈ WT . It follows from   F Fk0 (uN ) = F(uN ), almost every,   Fk0 F(uN ) = F(uN ),

almost every.

Jo

Thus, F(uN ) is a fixed point of Fk0 . We can now conclude that the equation F(uN ) = uN has a unique solution uN ∈ WT . Part B (Convergence rate of the regularized solution). In order to obtain the convergence result presented in Theorem 4.1, we need to undergo some following steps. Step 1 (Construct a formula for the error with respect to k·k2L2 (Ω) ). We can deduce from (7) and (31) that " N     X N N u (x, t) − u(x, t) = Gm (T − t) ϕN − ϕ + H (T − t) ϕ − ϕ 1;m m 2;m 1;m 2;m m=1

+

ZT t



N



#

Km (τ − t) fm (u )(τ) − f (u)(τ) dτ Xm (x) −

12

∞ X

um (t)Xm (x).

m=N+1

(33)

Journal Pre-proof

It can be seen by using the Parseval’s identity that " N

2     X

N

N ≤ Gm (T − t) ϕN − ϕ + H (T − t) ϕ − ϕ

u (x, t) − u(x, t) 2 1;m m 2;m 1;m 2;m L (Ω)

m=1

+

ZT t

#2 ∞  X Km (τ − t) fm (u )(τ) − f (u)(τ) dτ + |um (t)|2 . 

N

m=N+1

L (Ω)

≤3 +3

N X

m=1 N X

m=1

N   2 2 X 2 N + 3 |H (T − t)| |Gm (T − t)|2 ϕN − ϕ ϕ − ϕ m 1;m 2;m 1;m 2;m

 

m=1

ZT t

2 ∞  X N  Km (τ − t) fm (u )(τ) − f (u)(τ) dτ + |um (t)|2 Xm (x) 

:= Q1 + Q2 + Q3 +

∞ X

m=N+1

m=N+1

|um (t)|2 .

(34)

Pr e-

The first term can be estimated as follows

p ro

2

N

u (x, t) − u(x, t) 2

of

Step 2 (Estimate the mean of the errors). From step 1, we can easily see that

" N #  2  p X N ϕ1;m − ϕ1;m . EQ1 ≤ 3 exp 2 λN (T − t) E m=1

al

For m ≤ N, we use the fact that (see Lemma 3.1) n n ∞ π X X X π N i e φj Xm (χj ) − ϕ1 (χj )Xm (χj ) + (−1) ϕ1;2in±m ϕ1;m − ϕ1;m = n n j=1 j=1 i=1 n X ∞ π X i ≤ αj Yj Xm (χj ) + (−1) ϕ1;2in±m . n j=1

Hence

N  X

m=1

2

#

 2 ∞ 2 N n N X X X 2π 2 X i  +2 E α Y X (χ ) ≤ (−1) ϕ . j j m j 1;2in±m n2

urn

E

"

ϕN 1;m − ϕ1;m

i=1

m=1

m=1 i=1

j=1

By the property (15) and under the assumption ϕ1 ∈ H σ (Ω) with σ > 1 as in Lemma 3.2, we get  2 # " N N n 2 2 2 X X X 2Cσ2 kϕ1 k2H σ (Ω) 2π α∗   ϕN − ϕ ≤ E Y X (χ ) + N. E 1;m j m j 1;m n2 n2σ

Jo

m=1

m=1

j=1

Using the property (16), we get # " N 2 X E ϕN ≤2 1;m − ϕ1;m m=1

2 2 πα∗2 Cσ kϕ1 kH σ (Ω) + n n2σ

!

N.

2 2 πα∗2 Cσ kϕ1 kH σ (Ω) + n n2σ

!

N.

Hence  p  EQ1 ≤ 6 exp 2 λN (T − t) 13

(35)

Journal Pre-proof

The second term can be estimated by similar method " N #  2  p X 3T 2 N ϕ2;m − ϕ2;m EQ2 ≤ exp 2 λN (T − t) E 4λN m=1 !  πβ 2 Cσ2 kϕ2 k2 σ  p 3T 2 H (Ω) ∗ . ≤ exp 2 λN (T − t) + 2N n n2σ

(36)

m=1

of

Now, we estimate the next term under the assumption (27).  T 2 Z N p   X  exp Q3 ≤ 3(T + 1)2 λm (τ − t) fm (uN )(τ) − fm (u)(τ) dτ . The H¨ older inequality yields that

N Z X

T

Q3 ≤ 3(T + 1)

2

dτ

m=1 t

ZT t

p ro

t

2   p exp 2 λm (τ − t) fm (uN )(τ) − fm (u)(τ) dτ.

L (Ω)

we get

2

Q3 ≤ 3K T (T + 1)

2   p

exp 2 λN (τ − t) uN (·, τ) − u(·, τ) 2

ZT

2   p

exp 2 λN (τ − t) E uN (·, τ) − u(·, τ) 2

t

t

L (Ω)

,

dτ.

L (Ω)

al

EQ3 ≤ 3K T (T + 1)

2

L (Ω)

ZT

2

This leads to 2

2

≤ K u (·, t) − u(·, t) 2 2 N

Pr e-

Using the fact that

2

N

f (·, t, u (·, t)) − f (·, t, u(·, t)) 2

dτ.

(37)

Combining (34)-(37) and using the fact that

 p  −2γ |um (t)|2 ≤ λN exp −2 λN t Λ1 ,

urn

∞ X

m=N+1

which follows from the assumption (28), we obtain

2  p 

exp 2 λN t E uN (·, t) − u(·, t) 2 L (Ω)

√




Jo

ΓN exp 2 λN T 2 2 ≤ + λ−2γ N Λ1 + 3K T (T + 1) n

ZT t

2  p 

exp 2 λN τ E uN (·, τ) − u(·, τ) 2

L (Ω)

dτ,

(38)

in which Γ is defined as in (29). By using the Gronwall’s inequality, we have " #
√

2  p 
ΓN exp 2 λN T

N

−2γ exp 2 λN t E u (·, t) − u(·, t) 2 ≤ + λN Λ1 exp 3K 2 T 2 (T + 1)2 . n L (Ω)

This completes the proof. In the next subsection, we will give the convergence rate of the regularized solution uN in another space by similar argument and under stronger assumption. 14

Journal Pre-proof

4.2

Convergence rate in H σ (Ω)

Theorem 4.2. Let N ∈ N∗ such that N < n. Assume that ϕ1 , ϕ2 , f fulfill Theorem 4.1 and the problem
2 (1) has a unique solution u ∈ C [0, T ]; L (Ω) . For δ > 0, if there exists a positive constant Λ2 such that sup

∞ X

0≤t≤T i=1

exp (2λi (t + δ)) | hu(·, t), Xi i |2 ≤ Λ2 ,

(39)

of

then we have

2

E uN (·, t) − u(·, t)

t ∈ [0, T ],

H σ (Ω)

" #  ΓN exp 2√λ T
p
N ≤ λσN exp −2 λN (t) + exp (−2λN δ) Λ2 exp 3K 2 T 2 (T + 1)2 λσN . n

p ro



in which Γ is defined as in (29). Remark 4.2. Setting

Π := T + 3K 2 T 2 (T + 1)2 ,

Nn ≤

Pr e-

and letting Nn be a positive integer satisfying ν

log n



1 2σ

,

Π

2

N 5 Then, E u n (·, t) − u(·, t) H σ (Ω) is of order n 2 ν−1 .

2 for 0 < ν < . 5

Proof. It follows from the representation (33) that " N

2     X

N

N = λσm Gm (T − t) ϕN

u (·, t) − u(·, t) σ 1;m − ϕ1;m + Hm (T − t) ϕ2;m − ϕ2;m H (Ω)

m=1

+

urn

t

Hence

#2 ∞  X Km (τ − t) fm (uN )(τ) − f (u)(τ) dτ + λσm |um (t)|2 . 

al

ZT

2

N

u (·, t) − u(·, t)

H σ (Ω)

m=N+1

≤ λσN (Q1 + Q2 + Q3 ) +

∞ X

m=N+1

λσm |um (t)|2 ,

(40)

where Q1 , Q2 , Q3 are defined in (34). Using (35)-(37), we have

Jo

E (Q1 + Q2 + Q3 )
√ ZT

2  p  ΓN exp 2 λN (T − t)

2 2 ≤ + 3K T (T + 1) exp 2 λN (τ − t) E uN (·, τ) − u(·, τ) σ dτ. (41) n H (Ω) t

Next, we consider the last term of the right-hand side of (40). Setting h(y) = y σ exp (−cy) ,

y > 0,

with c is a positive constant. We can check that h(y) ≤ h(z),

for y ≥ z ≥ σ/c. 15

(42)

Journal Pre-proof

Since λσm = m2σ tends to infinity as m → ∞, for N large enough, we have λN ≥ inequality (42) with c = 2(t + δ), we get λσm exp (−2λm (t + δ)) ≤ λσN exp (−2λN (t + δ)) ,

σ 2(t+δ) .

Applying the

for m ≥ N.

Hence

m=N+1

λσm |um (t)|2 = ≤

∞ X

λσm exp (−2λm (t + δ)) exp (2λm (t + δ)) |um (t)|2

m=N+1 λσN exp (−2λN (t

+ δ)) Λ2 .

2

N u (·, t) − u(·, t) λ−σ E

N

p ro

Combining (40), (41), (43), we get

H σ (Ω)

√




ΓN exp 2 λN (T − t) ≤ + 3K 2 T (T + 1)2 n

ZT t

+ exp (−2λN (t + δ)) Λ2 .

of

∞ X

(43)

2  p 

exp 2 λN (τ − t) E uN (·, τ) − u(·, τ)

H σ (Ω)

dτ

2

E uN (·, t) − u(·, t)

H σ (Ω)

≤

λσN

Pr e-


√ Multiplying both sides of the above by λσN exp 2 λN (t) and using the Gronwall’s inequality, we obtain " #  ΓN exp 2√λ T
p
N exp −2 λN (t) + exp (−2λN δ) Λ2 exp 3K 2 T 2 (T + 1)2 λσN . n 

This completes the proof.

5

urn

al

Remark 4.3. Our problem is restricted to a rectangular geometry for which the eigenvalues and eigenfunctions of the Laplacian are readily available. The analysis here comes from the trigonemetric functions ( sine, the cosine function) of eigenfunctions. Lemma 3.1 give the representations of the exact solution which is given by trigonemetric functions. However, if we let an arbitrary domain Ω with a C 2 -boundary, the analysis in this paper is not applied and such problem is more difficult. This challenge and open problem may be addressed in future works.

Numerical example

Jo

To study the performance our regularized method, we propose some examples for the problem of finding a function u := u(x, t) that satisfies the biharmonic equation as follows ∂4u ∂4u ∂4u + 2 + = f (x, t, u), ∂x4 ∂x2 ∂t2 ∂t4

(44)

where (x, t) ∈ (0, π) × (0, 1), with x identified as the space variable, t as the time variable, and the boundary conditions on the interval [0, π] take the form ( u(0, t) = u(π, t) = 0, t ∈ (0, 1), (45) ∆u(0, t) = ∆u(π, t) = 0, t ∈ (0, 1), 16

Journal Pre-proof

and the final value conditions  u(x, 1) = ϕ1 (x), ∆u(x, 1) = ϕ2 (x), x ∈ (0, π),  ∂u (x, 1) = 0, ∂∆u (x, 1) = 0, x ∈ (0, π), ∂t ∂t

(46)

of

The eigenvalues are λm = m2 , with m = 1, 2, ..., and the corresponding normalized eigenfunctions r 2 Xm (x) = sin(mx) π form a complete orthonormal set in L2 (0, π). For any w ∈ L2 (0, π), the inner product is defined by

p ro

wm

r Z π 2 := hw, Xm i = w(x) sin(mx)dx, and kwkL2 (0,π) = π 0

∞ X

2 wm

m=1

!1 2

.

The inner product in L2 (0, π) can be approximated by the Composite Simpson rule of numerical integration as

0

π

G(x)dx ≈

π 3(N + 1)

(N+1)/2

X

[G(x2k−2 ) + 4G(x2k−1 ) + G(x2k ))] ,

Pr e-

Z

k=1

kπ , x0 = 0, xN+1 = π. N+1 We recall the fixed points χj as follows where xk =

χj =

π(2j − 1) ∈ (0, π), 2n

and assume that the observations (φej , ψej ) as follows

ψej = ϕ2 (χj ) + βj Wj ,

al

φej = ϕ1 (χj ) + αj Yj ,

j = 1, 2, ..., n,

i.i.d

with Yj , Wj ∼ N (0, 1),

urn

where αj , βj are positive numbers given. In Matlab software, the function randn(argument,arraytype) may be used to simulate a random number drawn from the N (0, 1). As an example, we use the function randn([m n],’distributed’) to generate a fixed set of random numbers then we get a matrix m × n with the average of the elements is zero (see Figure (1)). At the discretization, a uniform grid of mesh point (xj , tk ) is used to discretize the space and time intervals to perform the solutions as follows xj = (j − 1)∆x, tk = (k − 1)∆t,

Jo

where

∆x =

π 1 , ∆t = , j = 1, Nx + 1, k = 1, Nt + 1. Nx Nt

The the errors are esimated by 

1/2 NX x +1  2 1 ξ(t) =  uN (xj , t) − u(xj , t)  , Nx + 1 j=1

where u, uN are the exact and regularized solutions, respectively. In order to simulate our results, we consider the input data as follows 17

(47)

Journal Pre-proof

3

1

of

0

-1

-2

p ro

Value of the function randn

2

-3

-4 0

50

100

150

200

Pr e-

m = 1, n = 200

Figure 1: The function randn([1 200],’distributed’)

5.1

Case 1: The linear source input data

In this subsection, we consider the following input data r r r 2 2 2 ϕ1 (x) = sin(x), ϕ2 (x) = sin(2x), f (x, t) = t sin(3x). π π π

(48)

al

Instead of giving the input exact data (48), we get the observation values of φej (for example, see Figure 2-a) and ψej (for example, see Figure 2-b) follow the random models input data (48) as follows

0.7 0.6 0.5 0.4

0.2 0.1

Jo

0.3

ϕ1 = φe

0 -0.1 -0.2 0

0.5

1

1.5

q

2 π

i.i.d

with Yj , Wj ∼ N (0, 1).

0.8

ϕ2 = ψe

0.6

Value of the functions

0.8

Value of the functions

ψej = ϕ2 (χj ) + 0.08Wj ,

urn

φej = ϕ1 (χj ) + 0.1Yj ,

sin(x)

0.4

q

2 π

sin(2x)

0.2 0 -0.2 -0.4 -0.6 -0.8

2

2.5

3

3.5

0

x ∈ [0, π]

0.5

1

1.5

2

2.5

x ∈ [0, π]

e (a) The functions ϕ1 and φ

(b) The functions ϕ2 and ψe

Figure 2: The input data and its approximations for j = 1, 2, ..., 100

18

3

3.5

Journal Pre-proof

The exact solution of the problem (44) with the conditions (48) are given by

of

uex := u(x, t) " √ (1 − t) sinh(2 − 2t) sin(2x) sin(3x)  π2 cosh(1 − t) sin(x) + = + (−9t − 3 3 + 11) sinh(3 − 3t) 4 4 486 #  √ √ + (2 − 3 3) sinh(3t) + 3(t + 3 3 − 2) cosh(3 − 3t) − 3t cosh(3t) . The regularized solution of the problem (49) as follows

m=1

j=1

p ro

ureg := uN (x, t) " # Z1 N n n X X X π π = Gm (1 − t) φej Xm (χj ) + Hm (1 − t) ψej Xm (χj ) + Km (τ − t)fm (τ)dτ Xm (x). n n j=1

in which Gm (t), Hm (t) and Km (t) are defined in (6).

t

Pr e-

The results of this case are summarized in the following Table 5.1 and Figures 3, 4 and 5, i.e., in Table 5.1, we show the error estimations between the exact solution and its the solution at t ∈ {0.1, , 0.5, 0.7} for n ∈ {50, 100, 150}, respectively. In Figures 3, 4 and 5, we show the graph of the solutions t = 0.5 for n ∈ {50, 100, 150}, respectively. From observing the above results, we can see that the larger the number of points of observation, the better the convergence results. Error ξ(0.1)

n = 50 0.166156740422917

n = 100 0.097637257438274

n = 150 0.080541876791263

ξ(0.5)

0.123179586324836

0.093045699811714

0.063109664468757

ξ(0.7)

0.115213324990308

0.100200674101592

0.083129814384141

al

Table 1: The error between the exact and regularized solutions at t ∈ {0.1, 0.5, 0.7}

urn

3

1.5

1

0.5

Error 0.6

Errors between uex and ureg

2

Jo

Value of the solutions

2.5

0.7

uex ureg

0

0.5

1

1.5

0.4

0.3

0.2

0.1

-0.5

0

0.5

0 2

2.5

3

3.5

0

0.5

1

1.5

2

2.5

3

3.5

x ∈ [0, π]

x ∈ [0, π]

(a) The solutions at t = 0.5, n = 50

(b) The error at t = 0.5 , n = 50

Figure 3: The exact solution uex , the regularized solution ureg and the error at t = 0.5, n = 50

19

Journal Pre-proof

0.4

3

0.35

2

1.5

1

0.5

0

0.3 0.25 0.2 0.15 0.1 0.05

-0.5

0 0

0.5

1

1.5

2

2.5

3

3.5

0

0.5

of

Errors between uex and ureg

2.5

Value of the solutions

Error

uex ureg

1

1.5

2

2.5

3

3.5

x ∈ [0, π]

x ∈ [0, π]

(b) The error at t = 0.5 , n = 100

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(a) The solutions at t = 0.5, n = 100

Figure 4: The exact solution uex , the regularized solution ureg and the error at t = 0.5, n = 100 3

0.35

uex ureg

Errors between uex and ureg

2

1.5

1

0.5

0

Error

0.3

Pr e-

Value of the solutions

2.5

0.25

0.2

0.15

0.1

0.05

-0.5

0

0

0.5

1

1.5

2

x ∈ [0, π]

2.5

3

3.5

0

1

1.5

2

2.5

3

3.5

x ∈ [0, π]

(b) The error at t = 0.5 , n = 150

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(a) The solutions at t = 0.5, n = 150

0.5

Figure 5: The exact solution uex , the regularized solution ureg and the error at t = 0.5, n = 150

Case 2: The nonlinear source input data

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5.2

where

the nonlinear source input data as follows (x, t) ∈ (0, π) × (0, 1), x ∈ (0, π), x ∈ (0, π),

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In the second case, we consider the problem (44) for   f (u) = g(u) + h(x, t), ϕ1 (x) = − sin(x),   ϕ2 (x) = (π 2 + 1) sin(x), 

(49)

u3 , |u| ≤ 210 0, |u| > 210   1 h(x, t) = cos(πt) sin(x) cos(2πt − 2x) + cos(2πt + 2x) − 2 cos(2πt) + 2 cos(2x) + 8π 2 + 16π + 6 . 8

g(u) =

The sought solution of the problem (49) is given by

uex := u(x, t) = cos(πt) sin(x).

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Journal Pre-proof

The regularized solution of the problem (49) as follows " N n n X X π π(1 − t) sinh (m(1 − t)) X e N e ureg := u (x, t) = cosh (m(1 − t)) ψj Xm (χj ) φj Xm (χj ) + n 2nm m=1

j=1

j=1

Z1 



2 π

0

The results are summarized in the following table and figures.

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(τ − t) cosh (m(τ − t)) sinh (m(τ − t)) √ + 2m2 2m m t #  Z π (u3 (x, τ) + h(x, τ)) sin(mx)dx dτ sin(mx). ×

+

n = 50 0.040942137054414

n = 100 0.023170315514223

n = 150 0.022577794149926

ξ(0.5)

0.024383141486002

0.023882414992822

0.022087168979915

ξ(0.7)

0.024435068363780

0.023840300324347

0.023804411781118

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Error ξ(0.1)

Pr e-

Table 2: The error between the exact and regularized solutions at t ∈ {0.1, 0.5, 0.7}

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urn

al

In the first row in Table 2 indicate number of the fixed points, the remaining rows represent the results of the estimates at t ∈ {0.1, 0.5, 0.7}, respectively. It is clear that the error between the sought solution and the regularized solution is acceptable. Moreover, we also show the solutions 2D graphs Figure 6 (a)-(b)-(c) and 3D graphs in Figure 7 (a)-(b)-(c)-(d) at t = 0.1 and n ∈ {50, 100, 150}, respectively.

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Journal Pre-proof

1.2

1

uex ureg

1

uex ureg

0.9

0.6

0.4

0.2

0.7 0.6 0.5 0.4 0.3 0.2

0 0.1 -0.2

0 0

0.5

1

1.5

2

2.5

3

3.5

0

0.5

1

1.5

2

2.5

3

3.5

x

p ro

x

of

The solution u at t = 0.1

The solution u at t = 0.1

0.8 0.8

(a) The regularized solution ureg at t = 0.1, n = 50

(b) The regularized solution ureg at t = 0.1, n = 100

1

uex ureg

0.9

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

Pr e-

The solution u at t = 0.1

0.8

0.5

1

1.5

2

2.5

3

3.5

x

(c) The regularized solution ureg at t = 0.1, n = 150

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urn

al

Figure 6: The exact solution and the regularized solution at t = 0.1, n ∈ {50, 100, 150}

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Journal Pre-proof

1 1 0.8 0.8

1

0.2 0

-0.5

0.5

0.4

0

0.2 0

-0.5

-0.2

-0.2

-1

-1 -0.4

-0.4

-1.5 3

-0.6

1

2 0.5

1

-0.8

2 1

-1

x

x 0

(a) The regularized solution ureg at t = 0.1, n = 50

0.8

0

0.6 0.4 0.2

0

0

-0.5

-0.2

-1

0.8

uex

Value of u

0.2

-0.5

1

0.5

Pr e-

ureg

0.4

0

-1

t

1

0.6

0.5

0

(b) The regularized solution ureg at t = 0.1, n = 100

1

1

-0.8

0.5

0

t

-0.6

1

p ro

0

of

-1.5 3

-0.2

-0.4

-1.5 3

1

0.5

1 0

t

-0.8

0.5

1

-1

0

-0.6

1

2

-0.8

x

-0.4

-1 3

-0.6

2

Value of u

0

ureg

0.4 Value of u

ureg

0.6

0.6

0.5

x

(c) The regularized solution ureg at t = 0.1, n = 150

Value of u

1

-1

0

0

t

(d) The exact solution uex at t = 0.1, n = 50

Conclusion

urn

6

al

Figure 7: The exact solution and the regularized solution at t = 0.1, n ∈ {50, 100, 150}

In this paper we propose a Fourier truncation method for Cauchy problem for nonlinear biharmonic equation. The ill-posedness of our problem is discussed. Based on an a priori assumption for the exact solution, the error estimates are obtained under an a priori regularization parameter choice rule. Numerical tests are given which shows the effectiveness of our method. In the future work, we try to consider ill-posed problem with the case of an arbitrary domain Ω with a C 2 -boundary.

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Acknowledgement

This research was supported by the Macau Science and Technology Development Fund (Grant No.0074/2019/A2) from the Macau Special Administrative Region of the P.R. China.

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