Stepwise regularization method for a nonlinear Riesz–Feller space-fractional backward diffusion problem

Dang Duc Trong; Dinh Nguyen Duy Hai; Nguyen Dang Minh

doi:10.1515/jiip-2018-0033

Article

Stepwise regularization method for a nonlinear Riesz–Feller space-fractional backward diffusion problem

Dang Duc Trong , Dinh Nguyen Duy Hai and Nguyen Dang Minh

Published/Copyright: April 6, 2019

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From the journal Journal of Inverse and Ill-posed Problems Volume 27 Issue 6

Abstract

In this paper, we consider the backward diffusion problem for a space-fractional diffusion equation (SFDE) with a nonlinear source, that is, to determine the initial data from a noisy final data. Very recently, some papers propose new modified regularization solutions to solve this problem. To get a convergence estimate, they required some strongly smooth conditions on the exact solution. In this paper, we shall release the strongly smooth conditions and introduce a stepwise regularization method to solve the backward diffusion problem. A numerical example is presented to illustrate our theoretical result.

Keywords: Space-fractional backward diffusion problem; ill-posed problem; regularization; convergence estimate

MSC 2010: 35R30; 47A52; 65M30; 65M32

A Appendix

Proof of Lemma 2.4.

Denote

∥h∥*:-supT+≤t≤T++⁡λ++⁢(β)-tT++⁢∥h⁢(⋅,t)∥

for all h∈C⁢([T+,T++];L2⁢(ℝ)). It is easy to show that ∥⋅∥* is a norm of C⁢([T+,T++];L2⁢(ℝ)). We shall prove that problem (2.8) exists a unique solution by proving that LT+,T++,χλ++⁢(β) has unique fixed point in C⁢([T+,T++];L2⁢(ℝ)). To begin with, for every ϖ1,ϖ2∈C⁢([T+,T++];L2⁢(ℝ)), we prove the estimate

(A.1)∥LT+,T++,gλ++⁢(β)⁢(ϖ1)-LT+,T++,gλ++⁢(β)⁢(ϖ2)∥*≤K⁢(T++-T+)⁢∥ϖ1-ϖ2∥*.

In fact, for every ϖ1,ϖ2∈C⁢([T+,T++];L2⁢(ℝ)), we have from (2.9) that

L^T+,T++,gλ++⁢(β)⁢(ϖ1)⁢(ω,t)-L^T+,T++,gλ++⁢(β)⁢(ϖ2)⁢(ω,t)=Fβϵ⁢(ω)⁢∫tT++Φ⁢(ψαθ⁢(ω),s-t)⁢(f^ϖ1⁢(ω,s)-f^ϖ2⁢(ω,s))⁢d⁢s+(1-Fβϵ⁢(ω))⁢∫T+tΦ⁢(ψαθ⁢(ω),s-t)⁢(f^ϖ1⁢(ω,s)-f^ϖ2⁢(ω,s))⁢d⁢s.

Thus, using the inequality (a+b)2≤(1+m)⁢a2+(1+m-1)⁢b2 for all real numbers a and b and m>0, we obtain the following estimate

(A.2)∥L^T+,T++,gλ++⁢(β)⁢(ϖ1)⁢(⋅,t)-L^T+,T++,gλ++⁢(β)⁢(ϖ2)⁢(⋅,t)∥2=∫ℝ|L^T+,T++,gλ++⁢(β)⁢(ϖ1)⁢(ω,t)-L^T+,T++,gλ++⁢(β)⁢(ϖ2)⁢(ω,t)|2⁢d⁢ω≤(1+m)⁢J1+(1+m-1)⁢J2,

where

J1=∫ℝ|Fβϵ⁢(ω)⁢∫tT++Φ⁢(ψαθ⁢(ω),s-t)⁢(f^ϖ1⁢(ω,s)-f^ϖ2⁢(ω,s))⁢d⁢s|2⁢d⁢ω,

J2=∫ℝ|(1-Fβϵ⁢(ω))⁢∫T+tΦ⁢(ψαθ⁢(ω),s-t)⁢(f^ϖ1⁢(ω,s)-f^ϖ2⁢(ω,s))⁢d⁢s|2⁢d⁢ω.

We first estimate J1. By using the Hölder inequality, we have that

J1≤(T++-t)⁢∫ℝ∫tT++|Fβϵ⁢(ω)⁢Φ⁢(ψαθ⁢(ω),s-t)|2⁢|f^ϖ1⁢(ω,s)-f^ϖ2⁢(ω,s)|2⁢d⁢s⁢d⁢ω.

Therefore, we conclude from condition (F1) and the Lipschitz condition (1.1) that

(A.3)J1≤K2⁢(T++-t)⁢∫tT++λ++⁢(β)2⁢(t-s)T++⁢∥ϖ1⁢(⋅,s)-ϖ2⁢(⋅,s)∥2⁢d⁢s≤λ++⁢(β)2⁢tT++⁢K2⁢(T++-t)2⁢supT+≤s≤T++⁡λ++⁢(β)-2⁢sT++⁢∥ϖ1⁢(⋅,s)-ϖ2⁢(⋅,s)∥2=λ++⁢(β)2⁢tT++⁢K2⁢(T++-t)2⁢∥ϖ1-ϖ2∥*2.

Now we estimate J2. Applying again the Hölder inequality, we arrive at

J2≤(t-T+)⁢∫ℝ∫T+t|(1-Fβϵ⁢(ω))⁢Φ⁢(ψαθ⁢(ω),s-t)|2⁢|f^ϖ1⁢(ω,s)-f^ϖ2⁢(ω,s)|2⁢d⁢s⁢d⁢ω.

At this time, using condition (F2) and the Lipschitz condition (1.1) yield

(A.4)J2≤K2⁢(t-T+)⁢∫T+tλ++⁢(β)2⁢(t-s)T++⁢∥ϖ1⁢(⋅,s)-ϖ2⁢(⋅,s)∥2⁢d⁢s≤λ++⁢(β)2⁢tT++⁢K2⁢(t-T+)2⁢supT+≤s≤T++⁡λ++⁢(β)-2⁢sT++⁢∥ϖ1⁢(⋅,s)-ϖ2⁢(⋅,s)∥2≤λ++⁢(β)2⁢tT++⁢K2⁢(t-T+)2⁢∥ϖ1-ϖ2∥*2.

Substituting (A.3) and (A.4) into (A.2), we obtain for all t∈(T+,T++) that

∥L^T+,T++,gλ++⁢(β)⁢(ϖ1)⁢(⋅,t)-L^T+,T++,gλ++⁢(β)⁢(ϖ2)⁢(⋅,t)∥2≤(1+m)⁢λ++⁢(β)2⁢tT++⁢K2⁢(t-T+)2⁢∥ϖ1-ϖ2∥*2+(1+m-1)⁢λ++⁢(β)2⁢tT++⁢K2⁢(T++-t)2⁢∥ϖ1-ϖ2∥*2.

By choosing m=T++-tt-T+, we have the following estimate for all t∈(T+,T++):

(A.5)λ++⁢(β)-2⁢tT++⁢∥L^T+,T++,gλ++⁢(β)⁢(ϖ1)⁢(⋅,t)-L^T+,T++,gλ++⁢(β)⁢(ϖ2)⁢(⋅,t)∥2≤K2⁢(T++-T+)2⁢∥ϖ1-ϖ2∥*2.

On other hand, letting t=T+ in (A.3), we get that

(A.6)λ++⁢(β)-2⁢T+T++⁢∥L^T+,T++,gλ++⁢(β)⁢(ϖ1)⁢(⋅,T+)-L^T+,T++,gλ++⁢(β)⁢(ϖ2)⁢(⋅,T+)∥2≤K2⁢(T++-T+)2⁢∥ϖ1-ϖ2∥*2.

Replacing t=T++ in (A.4) yields

(A.7)λ++⁢(β)-2⁢∥L^T+,T++,gλ++⁢(β)⁢(ϖ1)⁢(⋅,T++)-L^T+,T++,gλ++⁢(β)⁢(ϖ2)⁢(⋅,T++)∥2≤K2⁢(T++-T+)2⁢∥ϖ1-ϖ2∥*2.

Combining (A.5), (A.6) and (A.7), we obtain the following estimate for all [T+,T++]:

λ++⁢(β)-tT++⁢∥L^T+,T++,gλ++⁢(β)⁢(ϖ1)⁢(⋅,t)-L^T+,T++,gλ++⁢(β)⁢(ϖ2)⁢(⋅,t)∥≤K⁢(T++-T+)⁢∥ϖ1-ϖ2∥*,

which leads to (A.1) by the Parseval formula. Since K⁢(T++-T+)<1, LT+,T++,gλ++⁢(β) is a contraction. Therefore, by the Banach fixed point theorem, the equation LT+,T++,gλ++⁢(β)⁢(ϖ)=ϖ has a unique solution uβϵ∈C⁢([T+,T++];L2⁢(ℝ)). This completes the proof of the lemma. ∎

Proof of Lemma 2.5.

By (2.9), we have for all t∈[T+,T++] that

L^T+,T++,gλ++⁢(β)⁢(ϖ)⁢(ω,t)=Fβϵ⁢(ω)⁢Φ⁢(ψαθ⁢(ω),T++-t)⁢(g⁢(ω)-∫tT++Φ⁢(ψαθ⁢(ω),s-T++)⁢f^ϖ⁢(ω,s)⁢d⁢s)+(1-Fβϵ⁢(ω))⁢∫T+tΦ⁢(ψαθ⁢(ω),s-t)⁢f^ϖ⁢(ω,s)⁢d⁢s.

Thus, for all t∈[T+,T++], using (2.7), one has after some rearrangements that

where conditions (F1) and (F2) have been applied. At this time, applying the inequality

(a+b+c)2≤2⁢(1+m-1)⁢a2+2⁢(1+m-1)⁢b2+(1+m)⁢c2

for any real numbers a,b,c and m>0, we have

Thus the Parseval formula, the Hölder inequality and the Lipschitz condition (1.1) imply that

∥LT+,T++,gλ++⁢(β)⁢(ϖ)⁢(⋅,t)-u⁢(⋅,t)∥2=∥L^T+,T++,gλ++⁢(β)⁢(ϖ)⁢(⋅,t)-u^⁢(⋅,t)∥2≤2⁢(1+m-1)⁢λ++⁢(β)2⁢(t-T++)T++⁢∥g⁢(⋅)-u^⁢(⋅,T++)∥2+2⁢(1+m-1)⁢λ++⁢(β)2⁢(t-T+)T++⁢∥u⁢(⋅,T+)∥2+K2⁢(1+m)⁢(T++-T+)⁢∫T+T++λ++⁢(β)2⁢(t-s)T++⁢∥LT+,T++,gλ++⁢(β)⁢(ϖ)⁢(⋅,s)-u⁢(⋅,s)∥2⁢d⁢s,

where we have used that ϖ=LT+,T++,gλ++⁢(β)⁢(ϖ) on t∈[T+,T++]. Multiplying λ++⁢(β)2⁢(T+-t)T++ by both sides, we thus obtain

(A.8)λ++⁢(β)2⁢(T+-t)T++⁢∥LT+,T++,gλ++⁢(β)⁢(ϖ)⁢(⋅,t)-u⁢(⋅,t)∥2≤2⁢(1+m-1)⁢λ++⁢(β)2⁢(T+-T++)T++⁢∥g⁢(⋅)-u^⁢(⋅,T++)∥2+2⁢(1+m-1)⁢∥u⁢(⋅,T+)∥2+K2⁢(1+m)⁢(T++-T+)⁢∫T+T++λ++⁢(β)2⁢(T+-s)T++⁢∥LT+,T++,gλ++⁢(β)⁢(ϖ)⁢(⋅,s)-u⁢(⋅,s)∥2⁢d⁢s.

Put

Ω⁢(t)=λ++⁢(β)2⁢(T+-t)T++⁢∥LT+,T++,gλ++⁢(β)⁢(ϖ)⁢(⋅,t)-u⁢(⋅,t)∥2 for all⁢t∈[T+,T++].

Since LT+,T++,gλ++⁢(β)⁢(ϖ), u∈C⁢([T+,T++];L2⁢(ℝ)), we see that the function Ω is continuous on [T+,T++] and attains over there its maximum I at some t0∈[T+,T++]. It follows that

(A.9)∫T+T++λ++⁢(β)2⁢(T+-s)T++⁢∥LT+,T++,gλ++⁢(β)⁢(ϖ)⁢(⋅,s)-u⁢(⋅,s)∥2⁢d⁢s≤∫T+T++supT+≤s≤T++⁡λ++⁢(β)2⁢(T+-s)T⁢∥LT+,T++,gλ++⁢(β)⁢(ϖ)⁢(⋅,s)-u⁢(⋅,s)∥2⁢d⁢s≤I⁢(T++-T+).

Substituting (A.9) into (A.8), we get

λ++⁢(β)2⁢(T+-t)T++⁢∥LT+,T++,gλ++⁢(β)⁢(ϖ)⁢(⋅,t)-u⁢(⋅,t)∥2≤2⁢(1+m-1)⁢λ++⁢(β)2⁢(T+-T++)T++⁢∥g⁢(⋅)-u^⁢(⋅,T++)∥2+2⁢(1+m-1)⁢∥u⁢(⋅,T+)∥2+K2⁢(1+m)⁢(T++-T+)2⁢I.

Choosing t=t0 on the left-hand side of the latter inequality, we obtain

[1-K2⁢(1+m)⁢(T++-T+)2]⁢I≤2⁢(1+m-1)⁢λ++⁢(β)2⁢(T+-T++)T++⁢∥g⁢(⋅)-u^⁢(⋅,T++)∥2+2⁢(1+m-1)⁢∥u⁢(⋅,T+)∥2.

Since 0<m<1K2⁢(T++-T+)2-1, it follows that the left hand-side bracket is positive. This implies for all t∈[T+,T++] that

λ++⁢(β)2⁢(T+-t)T++⁢∥LT+,T++,gλ++⁢(β)⁢(ϖ)⁢(⋅,t)-u⁢(⋅,t)∥2≤2⁢(1+m-1)⁢λ++⁢(β)2⁢(T+-T++)T++⁢∥g⁢(⋅)-u^⁢(⋅,T++)∥2+2⁢(1+m-1)⁢∥u⁢(⋅,T+)∥21-K2⁢(1+m)⁢(T++-T+)2,

which leads to

∥LT+,T++,gλ++⁢(β)⁢(ϖ)⁢(⋅,t)-u⁢(⋅,t)∥2≤2⁢(1+m-1)⁢[λ++⁢(β)2⁢(T+-T++)T++⁢∥g⁢(⋅)-u^⁢(⋅,T++)∥2+∥u⁢(⋅,T+)∥2]⁢λ++⁢(β)2⁢(t-T+)T++1-K2⁢(1+m)⁢(T++-T+)2.

Therefore, we obtain for all t∈[T+,T++] that

∥LT+,T++,gλ++⁢(β)⁢(ϖ)⁢(⋅,t)-u⁢(⋅,t)∥≤2⁢(1+m-1)1-K⁢1+m⁢(T++-T+)⁢(λ++⁢(β)T+-T++T++⁢∥g⁢(⋅)-u^⁢(⋅,T++)∥+∥u⁢(⋅,T+)∥)⁢λ++⁢(β)t-T+T++,

which is the desired estimate. This completes the proof of the lemma. ∎

Acknowledgements

The authors wish to thank the handling editor as well as anonymous referees for their most helpful comments on this paper.

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Received: 2018-04-18

Revised: 2019-02-18

Accepted: 2019-02-27

Published Online: 2019-04-06

Published in Print: 2019-12-01

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Articles in the same Issue

https://doi.org/10.1515/jiip-2018-0033

Keywords for this article

Space-fractional backward diffusion problem; ill-posed problem; regularization; convergence estimate