Inverse source problem of heat conduction equation with time dependent diffusivity on a spherical symmetric domain

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Inverse Problems in Science and Engineering

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Inverse source problem of heat conduction

equation with time-dependent diffusivity on a

spherical symmetric domain

Xiaoxiao Geng, Hao Cheng & Mian Liu

To cite this article: Xiaoxiao Geng, Hao Cheng & Mian Liu (2021): Inverse source problem of heat conduction equation with time-dependent diffusivity on a spherical symmetric domain, Inverse Problems in Science and Engineering, DOI: 10.1080/17415977.2021.1899172

To link to this article: https://doi.org/10.1080/17415977.2021.1899172

Published online: 13 Mar 2021.

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Inverse source problem of heat conduction equation with time-dependent diffusivity on a spherical symmetric domain

Xiaoxiao Geng, Hao Cheng and Mian Liu

School of Science, Jiangnan University, Wuxi, Jiangsu, People’s Republic of China

ABSTRACT

In this paper, we consider the inverse source problem of heat

con-duction equation with time-dependent diffusivity on a spherical

symmetric domain This problem is ill-posed, i.e the solution of the

problem does not depend continuously on the measured data To

solve this problem, we propose an iterative regularization method

and obtain the Hölder type error estimates Numerical examples are

presented to demonstrate the effectiveness of the proposed method.

ARTICLE HISTORY

Received 8 August 2020 Accepted 24 February 2021

KEYWORDS

Inverse source problem; heat conduction equation; time-dependent diffusivity; iterative regularization method; error estimates

2010 MATHEMATICS SUBJECT

CLASSIFICATIONS

35R25; 47A52

1 Introduction

In recent years, the inverse source problem has played an important role in dealing with inverse problems for partial differential equations Meanwhile, the inverse source problem often occurs in the migration of groundwater, the identification of radiogenic source for helium diffusion, the identification and control of pollution sources and so forth [1,2] The purpose of the inverse source problem is to determine the unknown source term from the final temperature distribution or the boundary temperature distribution

In [3], the authors reconstructed the spatially variable heat source from the measured data by applying the dual least squares method Wei and Zhang [4] used a boundary element method combined with a generalized Tikhonov regularization to determine the time-dependent source In [5], the authors proved the existence, uniqueness and con-tinuously dependence on data of heat source depending on time variable by using the generalized Fourier method Ma et al [6] applied a variational method to reconstruct the unknown source term depending on both time and space variables and so on [7 13] How-ever, most of these works were only devoted to the heat conduction equation with constant coefficient

In practical engineering, the variable case of diffusivity has more applications than the constant one In [14], the authors considered the model of helium production and diffusion

CONTACT Hao Cheng chenghao@jiangnan.edu.cn School of Science, Jiangnan University, Wuxi, 214122 Jiangsu, People’s Republic of China

in a spherical diffusion geometry:

4He= 8238

U(t)(e λ238t − 1) + 7235

U(t)(e λ235t − 1) + 6232Th(t)(e λ232t − 1),

∂4He(r, t)

D(t)

a2



∂2 4He(r, t)

∂r2 +2

r

∂4He(r, t)

∂r



+ 8λ238

238U(t) + 7λ235

235U(t) + 6λ232

232Th(t),

where 4He(t), U(t) and Th(t) are amounts present at time t, λ’s are the decay constants, a and r are the radius and radial position of spherical diffusion domain, respectively, D (t) is

the time-dependent diffusion coefficient obeying an Arrhenius relationship such that:

D(t)

a2 = D0

a2exp



RT(t)

 ,

where D0is the diffusivity at infinite temperature and Eathe activation energy measured

in laboratory experiments, R is the constant, and T (t) is an arbitrary thermal history.

In this paper, we consider the following heat conduction equation with time-dependent diffusivity on a spherical symmetric domain:

⎧

⎪

⎪

⎪

⎪

⎪

⎪

u t (r, t) = a(t)



u rr (r, t) +2

r u r (r, t)



+ f (r), 0 < r < R0, 0< t < T, u(r, 0) = φ(r), 0 ≤ r ≤ R0,

u(R0, t ) = 0, 0 ≤ t ≤ T,

lim

r→0u(r, t) bounded, 0 ≤ t ≤ T,

(1)

where a (t) is a time-dependent diffusion coefficient, assuming that it has a lower limit a0

and an upper limit a1, i.e

Here we want to identify the unknown source term f (r) from the additional final value

data

For the uniqueness results, we refer to the papers [15–18] and the references cited therein Since the measurement is noise-contaminated inevitably, we denote the noise

measure-ment of g (r) as g δ (r) which satisfies

where ·  is the L2([0, R0]) norm and δ > 0 is the noise level.

For the diffusion coefficient a (t) = constant, Cheng and Fu [19] and Yang et al [20]

used a spectral method and an a posteriori truncation regularization method to solve this problem, respectively For the diffusion coefficient a (t) = function of t, Bao et al [18] and Zhang and Yan [21] used Tikhonov regularization and spectral truncation regularization to solve this problem, respectively Inspired by Wang and Ren [22], we will adopt an iterative regularization method to solve this inverse problem Compared with the above methods,

the iterative regularization method seems more simple and direct Meanwhile, the iterative regularization method may be used to solve some nonlinear inverse problems

The structure of this paper is as follows In Section2, we provide some preliminary knowledge In Section3, we analyse the ill-posedness of the inverse problem and con-ditional stability An iterative regularization method is introduced in Section4, and the error estimates under two parameter choice rules are obtained In Section5, three differ-ent numerical examples are presdiffer-ented to verify the effectiveness of our method Finally, a brief conclusion is given in Section6

2 Preliminaries

Lemma 2.1: Assuming λ n = ( nπ R0 )2, for any n ≥ 1, there holds

C

n2 ≤

T

0

e −λ nT

τ a(s) dsdτ ≤ C

where C= (1−e −a1λ1T a1π2 )R2and C= R2

a0π2.

Proof: For n ≥ 1, λ n ≥ · · · ≥ λ1> 0 and according to (2), we have

T

0

e −λ nT

τ a(s) dsdτ ≥

T

0

e −a1λn (T−τ)dτ = 1− e −a1λn T

a1λ n ≥ 1− e −a1λ1T

a1λ n

= (1 − e −a1λ1T)R20

a1π2n2 = C

n2, T

0

e −λ nT

τ a(s) dsdτ ≤

T

0

e −a0λn (T−τ)dτ = 1− e −a0λn T

a0λ n ≤ 1

a0λ n = R20

a0π2n2 = C

n2



Lemma 2.2 ([ 23]): For 0 < λ < 1 and k ≥ 1, define p k (λ) = k−1

i=0(1 − λ) i and r k (λ) =

1− λpk (λ) = (1 − λ) k Then,

where

θ ν =



1, 0≤ ν ≤ 1,

ν ν, ν > 1.

3 Ill-posedness and conditional stability

Let

ω n (r) :=

√

2n π



R30

sin(nπr/R0) (nπr/R0) .

The eigenfunctionsω n (r) are standard orthogonal function systems with weight r2in the

[0, R0], and are complete in the class of square integrable functions on [0, R0] [24] The

solution u (r, t) and source term f (r) can be represented in the following form:

u(r, t) =∞

n=1



R30

√

f (r) =∞

n=1

where

f n=

R0

0 r2f (r)ω n (r) dr, n = 1, 2,

Substituting (8) and (9) into the equation and initial condition in (1), and denotingφ n =

(φ(r), ω n (r)), we can obtain

⎧

⎪

⎪

⎪

⎪

⎪

⎪

Tn (t) + λ n a(t)T n (t) =

√

2n π



R30

f n,

T n (0) =

√

2n π



R30

φ n,

whereλ n = ( nπ R0 )2for n∈N∗ The solution of this problem is

T n (t) =

√

2n π



R30



φ n e −λ nt

0a(s) ds + fn

t

0

e −λ nt

τ a(s) dsdτ

 , n = 1, 2, (10)

Thus, we have

u(r, t) =∞

n=1



φ n e −λ nt

0a(s) ds + fn

t

0

e −λ nt

τ a(s) dsdτ



ω n (r). (11)

Using the final value data u (r, T) = g(r), it can be obtained that

g n =g(r), ω n (r)= φn e −λ nT

0 a(s) ds + fn

T

0

e −λ nT

Denote hn = gn − φn e −λ nT

0 a(s) ds and define the linear operator K : f (·) −→ h(·),

Kf (r) =∞

n=1

T

0

e −λ nT

τ a (s) dsdτf (r), ω n (r)ω n (r) = h(r).

It is easy to show that the operator K is a linear self-adjoint operator with singular

values T

0 e −λ nT

τ a (s) dsdτ and eigenfunctions ω n (r) According to the property of the

eigenfunctionsω n (r), we have

f (r) =

∞



n=1

h n

T

0 e −λ nT

From Lemma 2.1, we find

1 T

0 e −λ nT

τ a(s) dsdτ ≥

n2

Therefore, it can be seen from (13) that small error in the measured data will lead to great

changes in the source term f (r), so the inverse problem is ill-posed.

In order to ensure the stability of the solution, we assume that an a priori bound of the source term f (r) is known as

where the normf (·)pis defined as

f (·) p= ∞

n=1

(1 + n2) p2(f (·), ω n (·))ω n (·)

Theorem 3.1: If f (r) satisfies f (r) p ≤ E, then we have

where C1= C−p+2p is a constant.

Proof: From Hölder inequality and (13), we have

f (r) 2=

∞



n=1

f n2=

∞



n=1

h2n (T

0 e −λ nT

τ a(s) dsdτ)2 =

∞



n=1

h

4 +2

n (T

0 e −λ nT

τ a(s) dsdτ)2h

2p

p+2

n

≤

⎛

⎝∞

n=1

h2n (T

0 e −λ nT

τ a(s) dsdτ) p+2

⎞

⎠

2 +2∞



n=1

h2n

 p

p+2

Applying Lemma 2.1, we obtain

∞



n=1

h2n (T

0 e −λ nT

τ a(s) dsdτ) p+2 ≤∞

n=1

h2n (T

0 e −λ nT

τ a(s) dsdτ)2

n 2p

C p

≤

∞



n=1

f n2(1 + n2) p C −p = C −p f (r) 2p ≤ C −p E2 (17)

Combining (16) and (17), it is easy to get that

f (r) 2≤ C−p 2p+2E +24 h(r) p 2p+2



Remark 3.1: According to the proof of Theorem 3.1, we can also obtain

f (r) C1 f (r) +22

which means that we can easily get the bound as the L2-norm of f (r) by estimating f (r) p and the L2-norm of Kf (r).

4 An iterative regularization method and error estimates

In this section, we will obtain the regularized approximate solution of the source term f (r)

by using an iterative regularization method, and give the error estimates between the exact

solution and its approximation under the a priori and a posteriori parameter choice rules,

respectively

We solve the inverse problem by constructing the following direct problem

⎧

⎪

⎪

⎪

⎨

⎪

⎪

⎪

u k t (r, t) = a(t)



u k rr (r, t) +2

r u

k

r (r, t)



+ f k, δ (r), 0 < r < R0, 0< t < T,

u k (r, 0) = φ(r), 0 ≤ r ≤ R0,

u k (R0, t ) = 0, 0 ≤ t ≤ T,

lim

r→0u

k (r, t) bounded, 0 ≤ t ≤ T,

(19)

where the source term f k, δ (r) is given by the following iteration

f0,δ (r) = 0, f k, δ (r) = f k −1,δ (r) − su k−1(r, T) − g δ (r), k = 1, 2, 3, , (20)

where k is the number of iterations, equivalent to the regularization parameter, and s is an

acceleration factor satisfying 0< sT

0 e −λ nT

τ a(s) dsdτ < 1 for all n ∈N

Denoting f n k, δ = (f k, δ (r), ω n (r)), we can easily obtain that

u k (r, t) =∞

n=1



φ n e −λ nt

0a(s) ds + f n k,δ

t

0 e −λ nt

τ a(s) dsdτ



ω n (r).

For simplicity, we denote σ n = sT

0 e −λ nT

τ a (s) dsdτ, and the iteration scheme (20)

becomes

f k, δ (r) =∞

n=1



f n k −1,δ − s



φ n e −λ nT

0 a (s) ds + f k −1,δ

n

T

0

e −λ nT

τ a (s) dsdτ − g δ

n



ω n (r)

=

∞



n=1

(1 − σ n ) f k −1,δ

n + sg n δ − φn e −λ nT

0 a (s) ds

ω n (r)

n=1

(1 − σ n ) f k −1,δ

n + sh δ nω n (r)

=∞

n=1



(1 − σ n ) k f n0,δ+

k−1



i=0

(1 − σ n ) i sh δ n



ω n (r)

=∞

n=1

k−1



i=0

(1 − σ n ) i sh δ n



4.1 Error estimate under the a priori regularization parameter choice rule

Theorem 4.1: Let f (r) be the exact solution and f k, δ (r) be the regularized approximate solu-tion given by (21) Assuming that the a priori bound (14) and the noise assumpsolu-tion (4) are both true, if we choose

k=

 1

s



E δ

 2 +2

we can obtain the following error estimate

k,δ (r) − f (r) 1+ θ p

2C−p2

E +22 δ p+2p , (23)

where [x] represents a maximum integer not exceeding x.

Proof: Applying the triangle inequality, we have

k,δ (r) − f (r) k,δ (r) − f k (r) k (r) − f (r) (24) According to (21), (4) and (6) (takingμ = 0), it can be obtained that

k,δ (r) − f k (r) ∞

n=1

p k (σ n )s(h δ n − hn )ω n (r) ∞

n=1

p k (σ n )s(g n δ − gn )ω n (r)

≤ sup

n∈Np k (σ n )s g δ (r) − g(r) sup

n∈Np k (σ n )s ≤ skδ. (25)

From the a priori bound (14), Lemma 2.1 and (7) (letting v= p

2), there holds

k (r) − f (r) ∞

n=1

p k (σ n )sh n ω n (r) −∞

n=1

h n

T

0 e −λ nT

τ a(s) dsdτ ω n (r)

=

∞



n=1

(−r k (σ n ))T h n

0 e −λ nT

τ a(s) dsdτ ω n (r)

=

∞



n=1

r k (σ n )f n ω n (r)



n=1

r2k (σ n )f2

n

1

=



n=1

r2k (σ n )(1 + n2) −p (1 + n2) p f n2

1

≤ sup

n∈N



r k (σ n )(1 + n2)−p2

 ∞

n=1

(1 + n2) p f n2

1 2

≤ E sup

n∈N



r k (σ n )n −p

≤ E sup

n∈N

⎛

⎝r k (σ n )C−p2

 T

0

e −λ nT

τ a(s) dsdτ

p

2

⎞

⎠

= EsC−p

2sup

n∈N



r k (σ n )σ p2

n



≤ θ p

2



sC−p

2E (k + 1)−p2 (26)

Combining (25) and (26) and choosing k= [1

s ( E

4.2 Error estimate under the a posteriori regularization parameter choice rule

In Section4.1, the regularization parameter k is chosen by (22), but in fact the a priori bound E is usually unknown or cannot be exactly obtained Therefore, we consider the a posteriori rule that does not depend on the a priori information to choose the regularization

parameter

Applying the discrepancy principle, a posteriori regularization parameter k is chosen to

satisfy

k, δ (r) − h δ (r) k −1,δ (r) − h δ (r) (27) whereτ > 1 is a constant.

Theorem 4.2: Let f (r) be the exact solution and f k,δ (r) be the regularized approximate solu-tion given by (21) Assume that the a priori bound (14) and the noise assumpsolu-tion (4) are both valid, and the regularization parameter k is chosen in (27) Then,

k, δ (r) − f (r)

⎛

⎜

⎝C1(τ + 1) p+2p +

⎛

⎝C−

p

2θ p+2 2

τ − 1

⎞

⎠

2 +2⎞

⎟

⎠ E +22 δ p+2p (28)

Proof: Applying the triangle inequality, it can be obtained that

k,δ (r) − f (r) k,δ (r) − f k (r) k (r) − f (r) (29) From (25), we know

According to (27), Lemma 2.1 and (4), there holds

τδ < k −1,δ (r) − h δ (r) ∞

n=1

r k−1(σ n )h δ n ω n (r)

n=1

r k−1(σ n )h δ n − hnω n (r) ∞

n=1

r k−1(σ n )h n ω n (r)

n=1

r k−1(σ n )f n

T

0

e −λ nT

τ a(s) dsdτω n (r)

≤ δ +

∞



n=1

r k−1(σ n )

T

0

e −λ nT

τ a(s) dsdτn −p (1 + n2) p2f n ω n (r)

≤ δ + E sup

n∈N



r k−1(σ n )

T

0

e −λ nT

τ a (s) dsdτn −p

≤ δ + E sup

n∈N

⎛

⎝r k−1(σ n )C−p

2

 T

0

e −λ nT

τ a (s) dsdτ

p+2 2

⎞

⎠

≤ δ + C−p2θ p+2

(sk) p+22

From (31), we have

sk≤

⎛

⎝C

−p

2θ p+2 2

τ − 1

⎞

⎠

2 +2

E δ

 2 +2

Substituting (32) into (30), we obtain

k, δ (r) − f k (r)

⎛

⎝C−

p

2θ p+2 2

τ − 1

⎞

⎠

2 +2

E p+22 δ p+2p (33)

On the other hand,



f k (r) − f (r)

∞



n=1

r k (σ n )h n ω n (r)

n=1

r k (σ n )h n − h δ nω n (r) + ∞

n=1

r k (σ n )h δ n ω n (r)

According to the a priori bound of f (r) and the definition of norm f (·) p, we know

k (r) − f (r)

n=1

r k (σ n ) h n

T

0 e −λ nT

τ a (s) dsdτ (1 + n

2) p2ω n (r)

≤ ∞

n=1

f n (1 + n2) p2ω n (r) = f (r) p ≤ E. (35)

So from Remark 3.1, we deduce that

k (r) − f (r) 1(τ + 1) p+2p E +22 δ p+2p (36)

5 Numerical implementation and numerical examples

In this part, we provide the numerical experiments of the above iterative regularization method Considering the difficulty in getting the exact solution of the direct problem,

we use the finite difference method to solve the direct problem According to the given

functions a (t), φ(r) and f (r), the final value data g(r) can be obtained In the numeri-cal experiments, the numbers of discrete points in time and space are N + 1 and M + 1,

respectively, and the corresponding step sizes are τ = T

N and h= R0

M Denote tn = nτ (n = 0, 1, , N) and r i = ih (i = 0, 1, , M), and the values of each grid point are u n

u(r i , t n ).

We first present the numerical solution of the direct problem

The derivatives of time and space terms are approximated by the following schemes, respectively:

u t (r i, tn ) ≈ 1

τ



u n i − u n−1

i

 ,

u rr (r i, tn ) ≈ 1

h2



u n i+1− 2u n

i + u n

i−1

 , u r (r i, tn ) ≈ 1

2h



u n i+1− u n

i−1



In this way, the discrete format of equation ut (r, t) = a(t)[u rr (r, t) +2

r u r (r, t)] + f (r) can

be obtained as follows

1

τ



u n i − u n−1

i



= a(tn )

 1

h2



u n i+1− 2u n

i + u n

i−1



r i h



u n i+1− u n

i−1



+ f (ri ), (37)

where i = 1, 2, , M − 1, n = 1, 2, , N.

Therefore, the matrix equation corresponding to the discrete format is

AU n= h2

where vector

F=f (r1), f (r2), , f (r M−1)T

,

B=



−a(tn )



h

r1 − 1



u n0, 0, , 0, a(t n )



r M−1



u n M

T ,

U n = (u n

1, u n2, , u n

M−1)T, n = 1, 2, , N,

acceleration factor satisfying 0< sT

0... R0], and are complete in the class of square integrable functions on [0, R0] [24] The