In this paper, we consider an inverse source problem with nonlocal boundary conditions for the heat equation involving multi-term time-fractional derivatives. We determine a source term independent of the space variable, and the temperature distribution from the energy measurement. We reduce the solution of the inverse problem to finding solutions to two problems. The well-posedness of each problem is shown using the generalized Fourier method.
Citation: Bauyrzhan Derbissaly, Makhmud Sadybekov. Inverse source problem for multi-term time-fractional diffusion equation with nonlocal boundary conditions[J]. AIMS Mathematics, 2024, 9(4): 9969-9988. doi: 10.3934/math.2024488
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In this paper, we consider an inverse source problem with nonlocal boundary conditions for the heat equation involving multi-term time-fractional derivatives. We determine a source term independent of the space variable, and the temperature distribution from the energy measurement. We reduce the solution of the inverse problem to finding solutions to two problems. The well-posedness of each problem is shown using the generalized Fourier method.
For a fixed positive integer m, let αi and qi, (i=1,...,m) be positive constants such that 1>α1>⋅⋅⋅>αm>0. We assume q1=1 without loss of generality. Consider the inverse problem of finding a pair of functions {r,u} such that it satisfies the equation
m∑i=1qi∂αitu=uxx+r(t)f(x,t), (x,t)∈DT, | (1.1) |
the initial condition
u(x,0)=φ(x), x∈[0,1], | (1.2) |
the nonlocal boundary conditions
u(0,t)+u(1,t)=0, t∈[0,T],a1ux(0,t)+b1ux(1,t)+a0u(0,t)=0, t∈[0,T], | (1.3) |
and the overdetermination condition
1∫0u(x,t)dx=E(t), t∈[0,T], | (1.4) |
where T>0; DT={(x,t): 0<x<1, 0<t≤T}; f, φ and E are given functions; a0, a1 and b1 are real constants such that |a1|+|b1|>0, a1+b1≠0; ∂αit is the Caputo time-fractional derivative defined in [1] by
∂αitu=1Γ(1−αi)t∫0∂u(x,s)∂sds(t−s)αi, |
and Γ(⋅) is the Gamma function. Note that, in the case |a1|+|b1|>0, a1+b1=0, problem (1.1)–(1.4) will immediately be incorrect.
For (1.1)–(1.3), the direct problem is the determination of u in ¯DT such that u(⋅,t)∈C2[0,1] and ∂αitu(x,⋅)∈C(0,T] when the initial temperature φ and the source term rf are given and continuous.
If the function r(t), t∈[0,T] is unknown, the inverse problem is formulated as the problem of finding a pair of functions {r,u} which satisfy (1.1)–(1.4) with r∈C[0,T], u(⋅,t)∈C2[0,1], ∂αitu(x,⋅)∈C(0,T].
The integral condition (1.4) arises when the data on the boundary cannot be measured directly, but only the average value of the solution can be measured along the boundary [2].
Fractional calculus is used in quantum mechanics [3], biophysics [4], control theory [5], viscoelasticity [6], signal processing [7], biological sciences [8], and many other disciplines.
In [9], the inverse problem for the classical heat equation satisfying the boundary conditions
a1ux(0,t)+b1ux(1,t)+a0u(0,t)+b0u(1,t)=0, t∈[0,T],c1ux(0,t)+d1ux(1,t)+c0u(0,t)+d0u(1,t)=0, t∈[0,T], |
was considered. The boundary conditions (1.3) are the particular case of the boundary conditions that considered in [9] when b0=0, c0=d0=1 and c1=d1=0.
The practical applications of nonlocal boundary value problems span across diverse fields, encompassing chemical diffusion [10], thermal conductivity [11], biological processes [12], and others. Particularly in scenarios like multiphase flows involving liquids, solids, and gases, heat flow is often proportional to the variations in boundary temperatures among distinct phases, alongside the parameters a0,a1,b1 outlined in the nonlocal boundary conditions (1.3). This elucidates the growing importance and extensive utilization of inverse problems featuring nonlocal boundary conditions across various disciplines.
We note several papers devoted to the study the inverse problem for a time-fractional diffusion equation with nonlocal boundary conditions. In [13], the inverse source problem for the time-fractional diffusion equation in two dimensions was considered. [14] focused on determining a time-dependent factor of an unknown source under certain sub-boundary conditions for the time-fractional diffusion equation using nonlocal measurement data. The paper established the existence and uniqueness of the solution to the inverse source problem by applying Lax-Milgram's lemma in appropriate Sobolev spaces. In [15], the inverse source problem for the time-fractional diffusion equation in two dimensional space was considered, where the time fractional derivative is the Hilfer derivative. In [16], the authors considered the problem of determining the distribution and the source term for the time-fractional diffusion equation. Two inverse problems for the time-fractional diffusion equation with a family of nonlocal boundary conditions were discussed in [17].
We also note some papers related to the study of impulsive differential equations and the inverse source problem for a time-fractional diffusion equation. In [18], the study focused on multipoint BVPs concerning a generalized class of impulsive fractional-order nonlinear differential equations. [19] addressed the study of a class of impulsive integro-differential equations, where impulses are not instantaneous. In [20], the estimation of an unknown source term in the time-fractional diffusion equation from measurement data was explored using the alternating direction method of multipliers. The work presented in [21] involved the mathematical analysis of an inverse source problem governed by a time-fractional diffusion equation. The objectives of this research included identifying the source function using additional data via a regularized optimal control approach and determining the regularization parameters through bi-level optimization.
This paper is an extension of the problem considered in [22].
It is well known that the boundary conditions (1.3) are not strongly regular boundary conditions [23]. For this reason, when we solve the problem (1.1)–(1.4) by the Fourier method, then the system of eigenfunctions of the auxiliary spectral problem does not form a basis. Therefore, the application of the Fourier method is impossible and additional research is needed.
The multinomial Mittag–Leffler function is defined as [24]
E(β1,...,βm),β0(z1,...,zm)=∞∑k=0∑k1+⋅⋅⋅+km=k(k;k1,...,km)m∏i=1zkiiΓ(β0+m∑i=1βiki), |
where 0<β0<2, 0<βi<1 (i=1,...,m), and zi∈C (i=1,...,m). Here, (k;k1,...,km) denotes the multinomial coefficient
(k;k1,...,km):=k!k1!⋅⋅⋅km! where k=m∑i=1ki, |
and ki (1≤i≤m) are non-negative integers.
Lemma 2.1. [25] Let 0<β0<2, 0<βi<1 (i=1,...,m), and zi∈C (i=1,...,m) be fixed. Then,
1Γ(β0)+m∑i=1ziE(β1,...,βm),β0+βi(z1,...,zm)=E(β1,...,βm),β0(z1,...,zm). |
Lemma 2.2. [25] Let 0<β<2 and 1>α1>⋅⋅⋅>αm>0 be given. Assume that α1π/2<μ<α1π, μ≤|arg(z1)|≤π, and there exists K>0 such that −K≤zi<0, (i=2,...,m). Then, there exists a constant c>0 depending on μ, K, αi (i=1,...,m), and β such that
|E(α1,α1−α2,...,α1−αm),β(z1,...,zm)|≤c1+|z1|≤c. |
Let us denote
E(⋅),α1(t):=E(α1,α1−α2,...,α1−αm),α1(−l1tα1,−l2tα1−α2,...,−lmtα1−αm), t>0, |
where l1,...,lm are some positive constants.
Lemma 2.3. [25] Let 1>α1>⋅⋅⋅>αm>0. Then,
ddt(tα1E(⋅),1+α1(t))=tα1−1E(⋅),α1(t), t>0. |
Lemma 2.4. [26] Let fn be a sequence of functions defined on (a,b] for each n∈N, such that:
(1). For some α>0, ∂αtfn(t) exists for all n∈N, t∈(a,b];
(2). Both the series ∑∞n=1fn(t) and ∑∞n=1∂αtfn(t) are uniformly convergent on the interval [a+ε,b] for any ε>0.
Then, ∑∞n=1fn(t) is α>0 differentiable, where ∑∞n=1fn(t) is a series of functions that must satisfy
∂αt∞∑n=1fn(t)=∞∑n=1∂αtfn(t). |
Consider the Volterra integral equation
g0(t)=t∫0Q(t,t1)g0(t1)dt1+g1(t), 0≤t≤1. | (2.1) |
Denote Δ:={(t,t1):0≤t1<t≤1} and introduce the class Sα of kernels Q(t,t1) that are defined and continuous on Δ and for (t,t1)∈Δ satisfy the inequality
|Q(t,t1)|≤c0(t−t1)−α, 0<α<1, c0=const>0. |
Lemma 2.5. [27] Let g1∈C[0,1] and Q(t,t1)∈Sα with 0≤α<1. Then, (2.1) has a unique solution g0∈C[0,1].
The application of the Fourier method to solve problem (1.1)–(1.4) leads to a spectral problem
{−z″(x)=λz(x), x∈(0,1),z(0)+z(1)=0,a1z′(0)+b1z′(1)+a0z(0)=0, | (3.1) |
where |a1|+|b1|>0, a1+b1≠0. The system of eigenfunctions of problem (3.1) does not form a basis in L2(0,1) [28]. For this reason, we cannot solve problem (1.1)–(1.4) by the Fourier method. Therefore, using the method from [29], we introduce even u1 and odd u2 with respect to the variable x parts of the function u as
u(x,t)=u1(x,t)+u2(x,t), |
where
2u1(x,t)=u(x,t)+u(1−x,t), 2u2(x,t)=u(x,t)−u(1−x,t). |
At the same time, for all (x,t)∈DT, the relations
u2(x,t)=−u2(1−x,t), u2x(x,t)=u2x(1−x,t),u1(x,t)=u1(1−x,t), u1x(x,t)=−u1x(1−x,t), | (3.2) |
hold. Equality (3.2) implies boundary relations
u2(0,t)=−u2(1,t), u2x(0,t)=u2x(1,t),u1(0,t)=u1(1,t), u1x(0,t)=−u1x(1,t). | (3.3) |
Substituting u1 and u2 into boundary conditions (1.3) and using (3.3), we obtain
u1(0,t)=0, (a1+b1)u2x(0,t)+a0u2(0,t)=(b1−a1)u1x(0,t). | (3.4) |
Also, substituting u1 and u2 into (1.1), (1.2) and (1.4) and using (3.3) and (3.4), we have the first problem of finding a pair of functions {r,u1} in the form
m∑i=1qi∂αitu1=u1xx+r(t)f1(x,t), (x,t)∈DT, | (3.5) |
u1(x,0)=φ1(x), x∈[0,1], | (3.6) |
u1(0,t)=0, u1(1,t)=0, t∈[0,T], | (3.7) |
1∫0u1(x,t)dx=E(t), t∈[0,T], | (3.8) |
where
2f1(x,t)=f(x,t)+f(1−x,t), 2φ1(x)=φ(x)+φ(1−x). |
And, for the function u2, we have the second problem in the form
m∑i=1qi∂αitu2=u2xx+r(t)f2(x,t), (x,t)∈DT, | (3.9) |
u2(x,0)=φ2(x), x∈[0,1], | (3.10) |
(a1+b1)u2x(0,t)+a0u2(0,t)=(b1−a1)u1x(0,t), t∈[0,T],(a1+b1)u2x(1,t)−a0u2(1,t)=(b1−a1)u1x(0,t), t∈[0,T], | (3.11) |
where
2f2(x,t)=f(x,t)−f(1−x,t), 2φ2(x)=φ(x)−φ(1−x). |
Consider the spectral problem
{−y″(x)=λy(x), x∈(0,1),y(0)=0, y(1)=0. | (4.1) |
The spectral problem (4.1) has only eigenfunctions
yk(x)=√2sinkπx, k=1,2,..., |
and the eigenvalues are defined by
λk=(kπ)2, k=1,2,.... |
Since (4.1) is the self-adjoint problem, the system of eigenfunctions {yk(x)}, (k=1,2,...) forms an orthonormal basis in L2(0,1).
Lemma 4.1. Let φ1(x)∈C4[0,1] be a function satisfying the conditions
φ1(0)=φ″1(0)=0, φ1(1)=φ″1(1)=0. | (4.2) |
Then, the following inequality
∞∑k=1|λkφ1k|≤ˆc1‖φ1‖C4[0,1]≤c1 |
holds, where c1 is a constant, φ1k=(φ1,yk), (k=1,2,...).
Proof. By using (4.2), integration by parts four times and the Schwarz and Bessel inequalities, we obtain
∞∑k=1|λkφ1k|=∞∑k=1|λkλkλkφ1k|≤(∞∑k=11|λk|2)12(∞∑k=1|λ2kφ1k|2)12≤ˆc1‖φ(4)1‖L2[0,1]≤ˆc1‖φ(4)1‖C[0,1]. |
Theorem 4.2. Let the following conditions be satisfied:
(A1) φ1∈C4[0,1], φ1(0)=φ″1(0)=0, φ1(1)=φ″1(1)=0;
(A2) f1∈C(¯DT), f1(⋅,t)∈C4[0,1], f1(0,t)=f1xx(0,t)=0, f1(1,t)=f1xx(1,t)=0, 0<1m0≤min0≤t≤T|∫10f1(x,t)dx|;
(A3) E(t)∈C1[0,T], E(0)=1∫0φ1(x)dx,
where m0 is a constant. Then, the inverse problem (3.5)–(3.8) has a unique classical solution.
Proof. (Existence of the solution of the first problem) To construct a formal solution of problem (3.5)–(3.8), we will use the Fourier method. Following this method, we seek the solution of (3.5)–(3.8) in a Fourier series as
u1(x,t)=∞∑k=1u1k(t)yk(x), |
where u1k(t)=(u1(⋅,t),yk), (k=1,2,...). For the functions u1k we obtain the Cauchy problem
m∑i=1qi∂αitu1k(t)+λku1k(t)=r(t)f1k(t),u1k(0)=φ1k, |
where f1k(t)=(f1(⋅,t),yk), φ1k=(φ1,yk), (k=1,2,...). The solution of this Cauchy problem is given in [24] by
u1k(t)=(tα1−1E(⋅),α1(t))∗r(t)f1k(t)+φ1kˆu1k(t), | (4.3) |
where
E(⋅),α1(t)=E(α1,α1−α2,...,α1−αm),α1(−λktα1,−q2tα1−α2,...,−qmtα1−αm), |
ˆu1k(t)=1−λktα1E(⋅),1+α1(t)−q2tα1−α2E(⋅),1+α1−α2(t)−⋅⋅⋅−qmtα1−αmE(⋅),1+α1−αm(t), |
(tα1−1E(⋅),α1(t))∗r(t)f1k(t)=t∫0(t−τ)α1−1E(⋅),α1(t−τ)r(τ)f1k(τ)dτ. |
According to Lemma 2.1, we rewrite (4.3) as
u1k(t)=(tα1−1E(⋅),α1(t))∗r(t)f1k(t)+φ1kE(⋅),1(t). | (4.4) |
Hence, the formal solution of problem (3.5)–(3.8) is expressed via the series
u1(x,t)=∞∑k=1[(tα1−1E(⋅),α1(t))∗r(t)f1k(t)+φ1kE(⋅),1(t)]yk(x). | (4.5) |
Now, we get the expression of the term r. Integrating Eq (3.5) between 0 and 1, we obtain
1∫0[m∑i=1qi∂αitu1(x,t)]dx=1∫0[u1xx(x,t)+r(t)f1(x,t)]dx. |
By using the overdetermination conditions (3.8) and (3.3), it is easy to deduce that
r(t)=h(t)(m∑i=1qi∂αitE(t)+2u1x(0,t)), | (4.6) |
where
u1x(0,t)=√2∞∑k=1kπ((tα1−1E(⋅),α1(t))∗r(t)f1k(t)+φ1kE(⋅),1(t)).h(t)=(1∫0f1(x,t)dx)−1. |
Let us denote
L(t):=h(t)(m∑i=1qi∂αitE(t)+√2∞∑k=1kπφ1kE(⋅),1(t)(1−(−1)k)),K(t,τ):=√2(t−τ)α1−1∞∑k=1kπf1k(τ)E(⋅),α1(t−τ)(1−(−1)k). |
Then, we obtain the Volterra integral equation of the second kind with respect to r in the form
r(t)=h(t)t∫0K(t,τ)r(τ)dτ+L(t). | (4.7) |
Before we proceed further, notice that, under assumption (A2), the series
∞∑k=1|λkf1k(t)|≤ˆc2max0≤t≤T‖f1(⋅,t)‖C4[0,1]≤c2, |
is uniformly convergent, and for E(t)∈C1[0,T] the term ∑mi=1qi∂αitE(t) is continuous, being the difference of m pieces of continuous functions. According to Lemma 2.2, we estimate the kernel of (4.7) and L(t) in the form
|L|≤m0(m1+cc1):=m2, |K(t,τ)|≤m0cc2(t−τ)α1−1:=c0(t−τ)α1−1, | (4.8) |
where m1 is a bound of ∑mi=1qi∂αitE(t) and c,c1 are constants defined in Lemmas 2.2 and 4.1, respectively. Therefore, the kernel of Eq (4.7) is weakly singular and, by Lemma 2.5, there exists a unique solution r∈C[0,T].
Due to assumptions (A1) and (A2) in Theorem 4.2, we have
φ1k=√2(kπ)41∫0φ(4)1(x)sinkπxdx:=1(kπ)4φ(4)1k,f1k(t)=√2(kπ)41∫0f(4)1x(x,t)sinkπxdx:=1(kπ)4f(4)1k(t). |
Since the solution u1 is formally given by series (4.5), we need to show that the series corresponding to u1, u1xx, and ∑mi=1qi∂αitu1 converge. Under assumptions (A1)–(A3), for all (x,t)∈¯DT, the series corresponding to u1,u1xx are bounded from above by the series
|u1|≤c∞∑k=11(kπ)4[m3‖f(4)1k‖C[0,T]Tα1α1+|φ(4)1k|],|u1xx|≤c∞∑k=11(kπ)2[m3‖f(4)1k‖C[0,T]Tα1α1+|φ(4)1k|], |
where m3:=‖r‖C[0,T]. Obviously, these majorizing series are convergent.
Now we show that ∑mi=1qi∂αitu1(x,t) are continuous functions on DT. For this, we first calculate u′1k(t), then we estimate ∂α1tu1k(t) on [ε,T] for all ε>0. The estimates for ∂α2tu1k(t),...,∂αmtu1k(t) on [ε,T] are obtained in a similar way.
First, let us get an estimate for u′1k. According to Lemma 2.3, we obtain the estimate
|u′1k(t)|≤c(kπ)4|t∫0(t−τ)α1−1|m3‖f(4)1k‖C[0,T]+c(kπ)4|φ(4)1k||λktα1−1+q2tα1−α2−1+⋅⋅⋅+qmtα1−αm−1|=c(kπ)4tα1−11−α1m3‖f(4)1k‖C[0,T]+c(kπ)4|φ(4)1k||λktα1−1+q2tα1−α2−1+⋅⋅⋅+qmtα1−αm−1|. |
Then, for ∂α1tu1k, we have the estimate
|∂α1tu1k(t)|≤1Γ(1−α1)t∫0|u′1k(τ)|(t−τ)α1dτ≤m3m4‖f(4)1k‖C[0,T]Γ(1−α1)(1−α1)t∫0τα1−1(t−τ)α1dτ+m4|φ(4)1k|Γ(1−α1)t∫01(t−τ)α1|λkτα1−1|dτ+m4|φ(4)1k|Γ(1−α1)t∫01(t−τ)α1|q2τα1−α2−1|dτ+⋅⋅⋅+m4|φ(4)1k|Γ(1−α1)t∫01(t−τ)α1|qmτα1−αm−1|dτ, | (4.9) |
where m4:=c/(kπ)4. By applying the change of variable s=τ/t in (4.9), we have
|∂α1tu1k(t)|≤m3m4‖f(4)1k‖C[0,T]Γ(1−α1)(1−α1)1∫0sα1−1(1−s)−α1ds+λkm4|φ(4)1k|Γ(1−α1)1∫0(1−s)−α1sα1−1ds+q2m4t−α2|φ(4)1k|Γ(1−α1)1∫0(1−s)−α1sα1−α2−1ds+⋅⋅⋅+m4qmt−αm|φ(4)1k|Γ(1−α1)1∫0(1−s)−α1sα1−αm−1ds=m3m4‖f(4)1k‖C[0,T]Γ(1−α1)(1−α1)B(α1,1−α1)+λkm4|φ(4)1k|Γ(1−α1)B(α1,1−α1)+m4q2t−α2|φ(4)1k|Γ(1−α1)B(α1−α2,1−α1)+⋅⋅⋅+m4qmt−αm|φ(4)1k|Γ(1−α1)B(α1−αm,1−α1), | (4.10) |
where
B(z1,z2)=1∫0(1−s)z2−1sz1−1ds. |
Consequently, by (4.10), the functions ∑mi=1qi∂αitu1(x,t) are continuous on DT.
Proof. (Uniqueness of the solution of the first problem) Let us show that the solution of problem (3.5)–(3.8) is unique. Suppose that there are two pairs of solutions {ˆr,ˆu1} and {˜r,˜u1} of the inverse problem (3.5)–(3.8). Then from (4.5) and (4.7), we have
ˆu1(x,t)−˜u1(x,t)=∞∑k=1[(tα1−1E(⋅),α1(t))∗(ˆr(t)−˜r(t))f1k(t)]yk(x), | (4.11) |
and
ˆr(t)−˜r(t)=h(t)(t∫0K(t,τ)(ˆr(τ)−˜r(τ))dτ). | (4.12) |
Then, (4.12) yields ˆr=˜r. After substituting ˆr=˜r in (4.11), we have ˆu1=˜u1.
Theorem 4.3. Let F be the set of triples {φ1,f1,E} where the functions φ1,f1 and E satisfy the assumptions of Theorem 4.2, and
‖φ1‖C4[0,1]≤M0, ‖f1‖C4,0(¯DT)≤M1, ‖E‖C1[0,T]≤M2, |
for some positive constants M0, M1 and M2. Then, the solution (r,u1) of the inverse problem (3.5)–(3.8) depends continuously upon the data on F.
Proof. Let F={φ1,f1,E} and ¯F={¯φ1,¯f1,¯E} be two sets of data, and ‖F‖=‖f1‖C4,0(¯DT)+‖φ1‖C4[0,1]+‖E‖C1[0,T], (r,u1) and (¯r,¯u1) be the solutions of the inverse problem (3.5)–(3.8) corresponding to the data F and ¯F, respectively.
Let us denote
¯h(t):=(1∫0¯f1(x,t)dx)−1, ψk:=√2kπ(1−(−1)k). |
For the difference E−¯E we have the estimate
m∑i=1qi∂αit|E−¯E|≤M3‖E−¯E‖C1[0,T], |h−¯h|≤m20‖h−¯h‖C[0,T], | (4.13) |
where
M3=q1Γ(1−α1)T1−α11−α1+⋅⋅⋅+qmΓ(1−αm)T1−αm1−αm. |
First, we write the function L, the difference r−¯r, and the integral from 0 to t of the kernel of Eq (4.7) as
t∫0(K(t,τ)−¯K(t,τ))dτ=t∫0∞∑k=1(t−τ)α1−1ψk(f1k−¯f1k)(τ)E(⋅),α1(t−τ)dτ, | (4.14) |
L(t)−¯L(t)=h(t)(m∑i=1qi∂αit(E−¯E)(t)+∞∑k=1ψk(φ1k−¯φ1k)E(⋅),1(t))+(h(t)−¯h(t))(m∑i=1qi∂αit¯E(t)+∞∑k=1ψk¯φ1kE(⋅),1(t)), | (4.15) |
r(t)−¯r(t)=L(t)−¯L(t)+¯h(t)t∫0(K(t,τ)−¯K(t,τ))r(τ)dτ+¯h(t)t∫0¯K(t,τ)(r(τ)−¯r(τ))dτ+(h(t)−¯h(t))t∫0K(t,τ)r(τ)dτ. | (4.16) |
Then, from Lemma 2.2 and equality (4.15) we obtain
‖L−¯L‖C[0,T]≤M4‖φ1−¯φ1‖C4[0,1]+M5‖f1−¯f1‖C4,0(¯DT)+M6‖E−¯E‖C1[0,T], | (4.17) |
where M4=m20(m1+cc1), M5=m0cˆc1,and M6=m0M3.
By Lemma 2.2, equality (4.14) we arrive at
‖K−¯K‖C[0,T]×C[0,T]≤M7‖f1−¯f1‖C4,0(¯DT), | (4.18) |
where M7=m0cTα1/α1(ˆc2+m0c2).
According to inequalities (4.17) and (4.18) from (4.16), we have
‖r−¯r‖C[0,T]≤M9(‖φ1−¯φ1‖C4[0,1]+‖f1−¯f1‖C4,0(¯DT)+‖E−¯E‖C1[0,T]), |
where
M9=max(M4+m3TM7M8, M5M8, M6M8), M8=1−c0Tα1α1. |
From (4.5), a similar estimate can be obtained for u1−¯u1.
To begin with, let us rewrite problem (3.9)–(3.11) as
m∑i=1qi∂αitu2=u2xx+r(t)f2(x,t), (x,t)∈DT, | (5.1) |
u2(x,0)=φ2(x), x∈[0,1], | (5.2) |
u2x(0,t)+αu2(0,t)=γ(t), t∈[0,T],u2x(1,t)−αu2(1,t)=γ(t), t∈[0,T], | (5.3) |
where
γ(t)=√2b1−a1b1+a1∞∑k=1kπ[tα1−1E(⋅),α1(t)∗r(t)f1k(t)+φ1kE(⋅),1(t)],α=a0b1+a1. |
We search for the solution to problem (5.1)–(5.3) in the form
u2(x,t)=u0(x,t)+a(x)γ(t), |
where a(x)=20x7−70x6+84x5−35x4+x. Consequently, for the unknown function u0 we have the problem
m∑i=1qi∂αitu0=u0xx+f0(x,t), (x,t)∈DT, | (5.4) |
u0(x,0)=φ0(x), x∈[0,1], | (5.5) |
u0x(0,t)+αu0(0,t)=0, t∈[0,T],u0x(1,t)−αu0(1,t)=0, t∈[0,T], | (5.6) |
where
f0(x,t)=f2(x,t)r(t)−a(x)b(t)+a″(x)γ(t), φ0(x)=φ2(x)−a(x)γ(0),b(t)=m∑i=1qi∂αitγ(t). |
The auxiliary spectral problem for the considered direct problem (5.4)–(5.6) is
{μ″(x)+λμ(x)=0, x∈(0,1),μ′(0)+αμ(0)=0,μ′(1)−αμ(1)=0. | (5.7) |
The spectral problem (5.7) has only eigenfunctions
μk(x)=√2(cos(√λkx)−α√λksin(√λkx)), k=1,2,..., |
and the eigenvalues are defined by
tan√λk=2α√λkα2−λk, k=1,2,.... |
Since problem (5.7) is self-adjoint, the system of eigenfunctions {μk(x)}, (k=1,2,...) forms an orthonormal basis in L2(0,1). We consider only the case α>0. The case α<0 will be similar.
For sufficiently large k, the asymptotic representation of eigenvalues of problem (5.7) has the form
√λk=kπ+O(1k). |
Lemma 5.1. Let φ0∈C4[0,1] be a function satisfying the conditions
φ′0(0)+αφ0(0)=0, φ‴0(0)+αφ″0(0)=0,φ′0(1)−αφ0(1)=0, φ‴0(1)−αφ″0(1)=0. | (5.8) |
Then, the inequality
∞∑k=1|λkφ0k|≤ˆc3‖φ0‖C4[0,1]≤c3 |
holds, where c3 is a constant and φ0k=(φ0,μk), (k=1,2,...).
Proof. By using (5.8), integration by parts four times, and the Schwarz and Bessel inequalities, we obtain
∞∑k=1|λkφ0k|=∞∑k=1|λkλkλkφ0k|≤(∞∑k=11|λk|2)12(∞∑k=1|λ2kφ0k|2)12≤ˆc3‖φ(4)0‖L2[0,1]≤ˆc3‖φ(4)0‖C[0,1]. |
The class of functions which satisfy the conditions of Lemma 5.1 will be denoted by
Φ≡{φ0∈C4[0,1]:φ′0(0)+αφ0(0)=0, φ‴0(0)+αφ″0(0)=0,φ′0(1)−αφ0(1)=0, φ‴0(1)−αφ″0(1)=0}. |
Similarly, as for the inverse problem (3.5)–(3.8), we search for a solution to problem (5.4)–(5.6) by the Fourier method. Then, we have
u0(x,t)=∞∑k=1[(tα1−1E(⋅),α1(t))∗f0k(t)+φ0kE(⋅),1(t)]μk(x), | (5.9) |
where f0k(t)=(f0(⋅,t),μk), φ0k=(φ0,μk), (k=1,2,...).
Theorem 5.2. Suppose φ0∈Φ, f0∈C(¯DT) and f0(⋅,t)∈Φ for every t∈[0,T]. Then, (5.9) gives a classical solution u0 to (5.4)–(5.6) and u0(⋅,t)∈C2[0,1], ∂αitu0(x,⋅)∈C(0,T], (i=1,...,m).
Proof. Let us denote
φ0k=√2λ2k1∫0φ(4)0(x)μk(x)dx:=1λ2kφ(4)0k,f0k(t)=√2λ2k1∫0f(4)0x(x,t)μk(x)dx:=1λ2kf(4)0k(t). |
As in the previous section, we need to show that the series corresponding to u0, u0xx, and ∑mi=1qi∂αitu0 converge. The series corresponding to u0 and u0xx are bounded from above by the series
|u0|≤c∞∑k=11λ2k[‖f(4)0k‖C[0,T]Tα1α1+|φ(4)0k|],|u0xx|≤c∞∑k=11λk[‖f(4)0k‖C[0,T]Tα1α1+|φ(4)0k|]. |
According to Lemma 2.3, we obtain the estimate of ∂α1tu0k(t) on [ε,T] for all ε>0 as
|∂α1tu0k(t)|≤c‖f(4)0k‖C[0,T]λ2kΓ(1−α1)(1−α1)B(α1,1−α1)+c|φ(4)0k|λkΓ(1−α1)B(α1,1−α1)+cq2t−α2|φ(4)0k|λ2kΓ(1−α1)B(α1−α2,1−α1)+⋅⋅⋅+cqmt−αm|φ(4)0k|λ2kΓ(1−α1)B(α1−αm,1−α1). |
Therefore, ∂α1tu0k(t) represent the continuous function on [ε,T] for all ε>0.
Remark 5.3. The uniqueness of the solution of problem (5.4)–(5.6), under the conditions of Theorem 5.2 is obtained from the uniqueness of the representation (5.9).
Lemma 5.4. [30] Let u0 satisfy Eq (5.4) in ¯DT. If f0(x,t)≤0 in ¯DT, then
u0(x,t)≤max{0, max0≤x≤1u0(x,0), max0≤t≤Tu0(0,t), max0≤t≤Tu0(1,t)}. |
If f0(x,t)≥0 in ¯DT, then
u0(x,t)≥min{0, min0≤x≤1u0(x,0), min0≤t≤Tu0(0,t), min0≤t≤Tu0(1,t)}. |
Theorem 5.5. The classical solution of problem (5.4)–(5.6) depends continuously on φ0∈C[0,1], f0∈C(¯DT) in the sense that
‖u0−¯u0‖C(¯DT)≤‖φ0−¯φ0‖C[0,1]]+(α+1)‖f0−¯f0‖C(¯DT), | (5.10) |
where u0 and ¯u0 are classical solutions of (5.4)–(5.6) with the data f0, φ0, and ¯f0, ¯φ0, respectively.
Proof. Let u0 be a classical solution of problem (5.4)–(5.6). We introduce the following:
R=‖f0‖C(¯DT), M=‖φ0‖C[0,1]. |
Let us construct the function
ω(x,t)=u0(x,t)−Rx22. |
The function ω is the classical solution of problem
m∑i=1qi∂αitω−ωxx=f0(x,t)−R,ω(x,0)=φ0(x)−Rx22,ωx(0,t)+αω(0,t)=0,ωx(1,t)−αω(1,t)=R2(α−1). |
Using the maximum principle, we obtain the estimates
ω(x,t)≤max{0, φ0(x)−Rx22, R2(α−1)} |
and
u0(x,t)≤max{0, φ0(x)−Rx22, R2(α−1)}+R≤φ0(x)−Rx22+R2(α−1)+R≤(α+1)R+M. |
Similarly, if we introduce the function
h(x,t)=u0(x,t)+Rx22 |
and using the minimum principle we arrive at the opposite estimate
u0(x,t)≥−(α+1)R−M. |
Hence, if u0 is the classical solution of problem (5.4)–(5.6), we have the estimate
‖u0‖C(¯DT)≤‖φ0‖C[0,1]]+(α+1)‖f0‖C(¯DT). | (5.11) |
To prove the continuous dependence on the data, we study the difference g(x,t)=u0(x,t)−¯u0(x,t). This function is the classical solution of (5.4)–(5.6) with f0−¯f0 and φ0−¯φ0 replaced by f0 and φ0, respectively. Applying inequality (5.11) to g, we arrive at estimate (5.10).
As a consequence of Theorem 5.2, we obtain the following main theorem for problem (5.1)–(5.3):
Theorem 5.6. Let φ2∈C4[0,1], f2∈C(¯DT), f2(⋅,t)∈C4[0,1] for every t∈[0,T], and the following conditions
φ′2(0)+αφ2(0)=γ(0), φ‴2(0)+αφ″2(0)=0,φ′2(1)−αφ2(1)=γ(0), φ‴2(1)−αφ″2(1)=0,f2x(0,t)+αf2(0,t)=b(t)r(t), f2xxx(0,t)+αf2xx(0,t)=0,f2x(1,t)−αf2(1,t)=b(t)r(t), f2xxx(1,t)−αf2xx(1,t)=0 | (5.12) |
be satisfied. Then, the classical solution u2 of problem (5.1)–(5.3) exists, is unique and u2(⋅,t)∈C2[0,1], ∂αitu2(x,⋅)∈C(0,T], (i=1,...,m).
Proof. The solution to problem (5.1)–(5.3) has the form
u2(x,t)=a(x)γ(t)+∞∑k=1[(tα1−1E(⋅),α1(t))∗f0k(t)+φ0kE(⋅),1(t)]μk(x). | (5.13) |
It can be seen from (5.13) that the majorizing series for (5.13) and (5.9) are the same and these series converge when conditions (5.12) are met. The uniqueness comes from the fact that the homogeneous problem (5.1)–(5.3) (that is, when φ2, γ, rf2≡0) has only a trivial solution.
Theorem 6.1. Let the following conditions be satisfied:
(H1) φ∈C1[0,1], φ(0)+φ(1)=0, a1φ′(0)+b1φ′(1)+a0φ(0)=0;
(H2) f∈C(¯DT), f(⋅,t)∈C1[0,1], f(0,t)+f(1,t)=0, fx(0,t)+fx(1,t)+α(f(0,t)−f(1,t))=0;
(H3) E∈C1[0,T], E(0)=1∫0φ(x)dx.
Then the classical solution u of problem (1.1)–(1.4) exists, is unique and u(⋅,t)∈C2[0,1], ∂αitu(x,⋅)∈C(0,T], (i=1,...,m).
Proof. The smoothness conditions specified in assumptions (H1)–(H3) are derived from Theorems 4.2 and 5.6. To prove the consistency conditions in Theorem 6.1, it is enough for us to prove the following lemma.
Lemma 6.2. Let assumptions (H1)–(H3) be fulfilled. Then, the conditions
φ1(0)=φ1(1)=0, φ′2(0)+αφ2(0)=γ(0), φ′2(1)−αφ2(1)=γ(0),f1(0,t)=f1(1,t)=0, f2x(0,t)+αf2(0,t)=b(t)r(t), f2x(1,t)−αf2(1,t)=b(t)r(t),E(0)=1∫0φ2(x)dx |
are satisfied.
Proof. It is easy to see from the representation γ that
γ(0)=b1−a1b1+a1φ′1(0). | (6.1) |
Substituting the expression φ(x)=φ1(x)+φ2(x) into φ(0)+φ(1)=0, we have
φ1(0)+φ1(1)+φ2(0)+φ2(1)=0. | (6.2) |
Using the conditions φ2(0)+φ2(1)=0, φ1(0)=φ1(1) from (6.2), we obtain φ1(0)=φ1(1)=0. Also, substituting the expression φ(x)=φ1(x)+φ2(x) into a1φ′(0)+b1φ′(1)+a0φ(0)=0, we have
a1(φ′1(0)+φ′2(0))+b1(φ′1(1)+φ′2(1))+a0(φ1(0)+φ2(0))=0. |
The conditions φ′1(0)=−φ1(1), φ1(0)=0, φ′2(0)=φ′2(1) imply that
φ′2(0)+a0b1+a1φ2(0)=b1−a1b1+a1φ′1(0), |
which, from (6.1), becomes φ′2(0)+αφ2(0)=γ(0).
Similarly, we get φ′2(1)−αφ2(1)=γ(0).
Substituting the expression f(x,t)=f1(x,t)+f2(x,t) into f(0,t)+f(1,t)=0, we have
f1(0,t)+f2(0,t)+f1(1,t)+f2(1,t)=0. | (6.3) |
Then using conditions f1(0,t)=f1(1,t), f2(0,t)+f2(1,t)=0 from (6.3), we arrive at f1(0,t)=f1(1,t)=0.
Applying f(x,t)=f1(x,t)+f2(x,t) and the conditions f1(0,t)=f1(1,t), f2(0,t)+f2(1,t)=0, f1x(0,t)+f1x(1,t)=0, f2x(0,t)=f2x(1,t) from fx(0,t)+fx(1,t)+α(f(0,t)−f(1,t))=0, we obtain f2x(0,t)+αf2(0,t)=b(t)r(t).
Similarly, we have f2x(1,t)−αf2(1,t)=b(t)r(t).
Substituting the expression φ(x)=φ1(x)+φ2(x) into (H3), we obtain E(0)=1∫0φ2(x)dx.
The uniqueness of the solution of problem (1.1)–(1.4) comes from the fact that u1 and u2 are unique. This completes the proof.
Remark 6.3. If we consider the direct problem (1.1)–(1.3), it is easy to see that, for the direct problem (1.1)–(1.3), we have a theorem of existence and uniqueness, which is similar to Theorem 6.1 without assumption (H3).
In this paper, we considered the inverse source problem with the nonlocal boundary conditions for the heat equation involving multi-term time-fractional derivatives. Since the eigenvalues of the auxiliary spectral problem do not form a basis, we have divided the problem into two sub-problems, one of which is the inverse problem, and the second the direct problem. The well-posedness of the inverse and direct problems are shown by Fourier expansion in terms of eigenfunctions of the corresponding spectral problems. Also, for the well-posedness of the inverse problem, the properties of the Volterra integral equation of the second kind were used. The continuous dependence on the data of the solutions of the inverse and direct problems was proved.
Since a feature of this paper is the nonlocal boundary condition (1.3), performing some numerical tests is more difficult even without the fractional derivative (see for example [31]). This is due to the fact that the sufficient conditions for the existence of a solution are not satisfied using standard methods. It is also difficult to show the convergence of the series that arises when solving (1.1)–(1.4) using the Fourier method.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
The authors declare no conflict of interest.
The authors are grateful to the anonymous reviewers for their careful reading of the article, as well as for many useful comments and suggestions that helped improve the presentation of the article.
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