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Existence and uniqueness of a weak solution to a non-automonous time-fractional di ff usion equation (of distributed order)

Karel Van Bockstal

Research Group NaM2, Department of Electronics and Information systems, Ghent University, Krijgslaan 281, 9000 Ghent, Belgium

Abstract

In this contribution, we investigate an initial-boundary value problem for a fractional diffusion equation of distributed order where the coefficients of the elliptic operator are dependent on spatial and time variables.

We consider a homogeneous Dirichlet boundary condition. Using a classical variational approach, we establish the existence of a unique weak solution to the problem in C

[0,T],H10(Ω)

∩L

(0,T),H10(Ω) if the initial data belongs to H10(Ω). The same result is also valid for the fractional diffusion equation with Caputo derivative.

Keywords: time-fractional diffusion equation, distributed order, non-autonomous, time discretization, existence, uniqueness

2010 MSC: 35A15, 35R11, 47G20, 65M12

1. Introduction

LetΩ⊂Rd (d∈N) be a bounded domain with a Lipschitz continuous boundary∂Ω. The final time is denoted byT,QT := Ω×(0,T] andΣT :=∂Ω×(0,T]. Consider a general second-order linear differential operator given by

L(x,t)u(x,t)=−∇ ·(A(x,t)∇u(x,t)+b(x,t)u(x,t))+c(x,t)u(x,t), (1) where ((x,t)∈QT)

A(x,t)=

ai,j(x,t)

i,j=1,...,d, b(x,t)=(b1(x,t),b2(x,t), . . . ,bd(x,t)).

The goal of this contribution is to show the existence and uniqueness ofufor given f andu0such that









D(µ)t u

(x,t)+(Lu) (x,t) = f(x,t) (x,t)∈QT, u(x,t) =0 (x,t)∈ΣT, u(x,0) =u0(x) x∈Ω.

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Here,D(µ)t udenotes the distributed order fractional derivative defined by D(µ)t u

(x,t)=Z 1 0

βtu

(x,t)µ(β) dβ, (x,t)∈QT, (3)

with weight functionµ: [0,1]→Rsatisfying

µ∈L1(0,1), µ>0, µ.0, and with∂βtuthe Caputo derivative of orderβ∈(0,1) defined by

βtu

(x,t)=Z t 0

(t−s)−β

Γ(1−β)∂su(x,s) ds, (x,t)∈QT,

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whereΓdenotes the Gamma function. The equation in (2) is used to model ultra slow diffusion processes with a logarithmic growth of the mean square displacement, see e.g. [1, 2, 3, 4].

We only mention the most relevant results available in the literature concerning the subject of this paper.

First, we discuss the papers [5, 6, 7, 8, 9]. In these papers, the considered operatorLis time-independent (autonomous) and symmetric (i.e. AT=Aandb=0). In [5], an appropriate maximum principle (and as a consequence also uniqueness of a solution) for problem (2) is proved. In [6], a strong solution to problem (2) is provided whenµ∈C1and f =0. The asymptotic behavior of the solutions to initial-boundary-value problems for distributed order time-fractional diffusion equations (without source) is studied in [7]. The authors of [8] used the method of the eigenfunctions expansion in combination with the Laplace transform to prove the uniqueness and existence of a solution to (2) (and its analyticity in time) when f =u0 =0.

In [9], the authors investigated the existence and uniqueness of a weak solution to problem (2). They characterize the weak solution to (2) as the original of the solution to the Laplace transform (of tempered distributions) of (2) with respect to the time variable.

Finally, we mention also the paper [10]. The authors showed the existence of a weak solution to (2) by an approach based on Galerkin method and energy estimates obtained for a special approximating sequence (the approximate solution is separated in the space and time variable). However, their approach (based on Banach fixed-point theorem with the solution space depending onµ) does not seem to be suitable for performing computations and the obtained continuity of the solution in the time variable is under the additional assumption thatR1

1/2µ(β) dβ >0, see [10, Theorem 2 and 4]. Note also that in the papers above, the domain has either a smooth or C1,1boundary, whilst in this paper a Lipschitz domain is considered.

The approach followed in this contribution illustrates that the standard procedure (time-discretization) for showing the existence of a weak solution to classical parabolic problems also can be applied on problem (2), which is a new result and also includes a scheme for computations. The weak formulation of the problem and assumptions on the data are discussed in Section 2 and the existence of a unique solution is shown in Section 3.

2. Weak formulation

We can rewrite the distributed order fractional derivative defined in Eq. (3) as follows D(µ)t u

(x,t)=(k∗∂tu) (x,t), (x,t)∈QT, where k(t)=Z 1 0

t−β

Γ(1−β)µ(β) dβ.

Note that the kernelksatisfiesk(t)>0 fort>0 and it is singular att=0. The convolution kernelkbelongs to L1(0,T) and it is a decreasing function in time. Now, we state the assumptions on the data functions. The matrix A=

ai j(x,t)

is ad×dmatrix-valued function such that A∈ L

QT

d×d

is uniformly elliptic, i.e. there exists a constantα >0 such that

d

X

i,j=1

ai j(x,t)ξiξj>α|ξ|2, for a.a. (x,t)∈QTand for allξ∈Rd.

Moreover, we suppose thatc∈L QT

. For the vector functionb, we assume that

b∈L QT

s.t. c(x,t)− kbk2

L QT

2α >0, (x,t)∈QT. We associate a bilinear formLwith the differential operatorLdefined in (1) as follows

L(t) (u(t), ϕ) :=(Lu, ϕ)=(A(t)∇u(t)+b(t)u(t),∇ϕ)+(c(t)u(t), ϕ), withu(t), ϕ∈H10(Ω). Using the properties above, we obtain that

L(t) (u, ϕ)6CkukH1

0()kϕkH1

0(), ∀u, ϕ∈H10(Ω), and L(t) (ϕ, ϕ)> α

2 k∇ϕk2, ∀ϕ∈H10(Ω), (4)

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i.e. the bilinear form L is continuous and H10(Ω)-elliptic due to the Friedrichs inequality. Next, the variational formulation of problem (2) can be defined as follows: search u ∈ L2

(0,T),H10(Ω) with k∗∂tu∈L2

(0,T),H10(Ω)

such that for a.a.t∈(0,T) it holds that h(k∗∂tu) (t), ϕiH1

0()×H10()+L(t) (u(t), ϕ)=hf(t), ϕiH1

0()×H10(), ∀ϕ∈H10(Ω). (5) 3. Existence of a solution

The existence of a solution is shown by the aid of Rothe’s method, which is a time-discretization method. The time interval [0,T] is divided inton∈Nequidistant subintervals [ti−1,ti] with lengthτ=Tn <

1. The approximation of a functionzat timet=ti, 06i6n, is denoted byzi. The same notation is also used for a given function. Moreover, we approximate∂tz(ti), 16i6n, by the backward Euler difference δzi=zi−zτi−1. Finally, the time discrete convolution is defined as follows

(k∗z)(ti)≈(k∗z)i:=

i

X

l=1

ki+1−lzlτ.

Note that (k∗z)0 :=0. From [11, Lemma 3.2], using the evolution triple H10(Ω)⊂L2(Ω)(L2(Ω)) ⊂ H10(Ω), it follows that for a sequence (vi)i∈Nin H10(Ω) it holds that

j

X

i=1

hδ(k∗v)i,viiH1

0()×H10()τ=

j

X

i=1

(δ(k∗v)i,vi)τ>1 2

k∗ kvk2

j+1 2

j

X

i=1

kikvik2τ, j∈N, (6) with

k∗ kvk2

j:=

j

X

l=1

kj+1−lkvlk2τ.

where (·,·) denotes the standard inner product in L2(Ω) andk·kits induced norm. We approximate prob- lem (5) at timet=tiby: Findui∈H10(Ω),i=1,2, . . . ,n, such that

h(k∗δu)i, ϕiH1

0()×H10()+Li(ui, ϕ)=hfi, ϕiH1

0()×H10(), ∀ϕ∈H10(Ω). (7) Using the time discrete convolution (3), the discrete problem can be equivalently written as

ai(ui, ϕ) :=k(τ) (ui, ϕ)+Li(ui, ϕ)=hfi, ϕi+k(τ) (ui−1, ϕ)−

i−1

X

l=1

ki+1−l(ul−ul−1, ϕ)=:hFi, ϕi, ∀ϕ∈H10(Ω).

(8) The existence of a unique solution on every time step follows from the Lax-Milgram lemma.

Lemma 3.1. Suppose that u0 ∈ L2(Ω)and f ∈ L2

(0,T),H10(Ω)

. Then, for any i = 1,2, . . . ,n, there exists a unique ui∈H10(Ω)solving(7).

The following two lemmas are required to ensure the existence of a solution to (5) and to prove the convergence of approximations towards this solution.

Lemma 3.2. Let the assumptions of Lemma 3.1 be fulfilled. Then, there exist positive constants C such that for every j=1,2, . . . ,n, the following relations hold

k∗ kuk2

j+

j

X

i=1

kikuik2τ+

j

X

i=1

k∇uik2τ6C and

j

X

i=1

k(k∗δu)ik2

H10()τ6C.

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Proof. We setϕ=uiτin (7) and sum the result up fori=1, . . . ,jwith 16 j6n. Using the relation

δ(k∗u)i=kiu0+(k∗δu)i, (9)

we obtain that

j

X

i=1

(δ(k∗u)i,ui)τ+

j

X

i=1

Li(ui,ui)τ=

j

X

i=1

hfi,uiiτ+

j

X

i=1

ki(u0,ui)τ.

Employing Eq. (4) and Eq. (6) gives the first estimate. The second estimate follows from k(k∗δu)ikH1

0() = sup

kϕkH1

0(Ω)=1

h(k∗δu)i, ϕiH1

0()×H10()

= sup

kϕkH1

0(Ω)=1

|hfi, ϕi − Li(ui, ϕ)|6kfikH1

0()+Ck∇uik, and the result of the first estimate.

The L-bound obtained in the next two lemmas is necessary to obtain the continuity in time of the solution. Here, we assume thatb=0and we consider two different condition on∂tf.

Lemma 3.3. Let the assumptions of Lemma 3.1 be fulfilled. Moreover, assume that u0∈H10(Ω),

tf ∈ L2

(0,T),H10(Ω)

, b = 0,∂tA ∈ L

QT

d×d

and∂tc∈ L QT

. Then, there exists a positive constant C such that for every j=1,2, . . . ,n, the following relation holds

∇uj

2+

j

X

i=1

k∇ui− ∇ui−1k26C.

Proof. We setϕ=δuiτin (7) and sum the result up fori=1, . . . ,jwith 16 j6n. We obtain that

j

X

i=1

h(k∗δu)i, δuiiH1

0()×H10()τ+

j

X

i=1

Li(ui, δui)τ=

j

X

i=1

hfi, δuiiH1

0()×H10()τ.

The positivity of the first term on the left-hand side follows from [12, Eq. 3.2]. The term on the right-hand side can be estimated by using the partial summation rulePj

i=1hfi, δuiiτ=hfj,uji−hf0,u0i−Pj

i=1hδfi,ui−1iτ and Lemma 3.2.

Lemma 3.4. Let the assumptions of Lemma 3.1 be fulfilled. Moreover, assume that u0∈H10(Ω),k∂tf(t)kH1 0()6 Ct−αfor all t∈(0,T]and some constantα∈(0,1),b=0,∂tA∈

L QT

d×d

and∂tc∈L QT

. Then, there exists aτ0>0such that the estimate from Lemma 3.3 also holds true forτ < τ0.

Proof. The estimate follows from takingτsufficiently small (note thatτ−ατ→0 asτ→0) and the discrete Gr¨onwall lemma [13, Corollary 15.5] after considering that

j

X

i=1

hδfi,ui−1iH1

0()×H10()τ

6C







 1+

j

X

i=1

t−αi kui−1k2

H10()τ







6C







 1+

j

X

i=1

t−αi k∇uik2τ+τ−ατ

j

X

i=1

k∇ui− ∇ui−1k2







 .

Now, we define the functionszn,zn: [0,T]→L2(Ω) by zn(t)=





z0 t=0

zi−1+(t−ti−1)δzi t∈(ti−1,ti], 16i6n, and zn(t)=





z0 t=0

zi t∈(ti−1,ti], 16i6n.

Using these so-called Rothe’s functions and Eq. (9), Eq. (7) can be rewritten on the whole time frame as h∂t(k∗u)n(t)−kn(t)u0, ϕiH1

0()×H10()+Ln(t)(un(t), ϕ)= fn(t), ϕ

, ∀ϕ∈H10(Ω), (10) where

L (t) (u (t), ϕ)=

A (t)∇u (t)+b (t)u (t),∇ϕ

+(c (t)u (t), ϕ).

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Theorem 3.1 (Existence). Suppose that the conditions of Lemma 3.3 or 3.4 are fulfilled. Then there exists a unique weak solution u to the problem(5)with u ∈ C

[0,T],H10(Ω)

∩L

(0,T),H10(Ω) and k∗∂tu∈L2

(0,T),H10(Ω) .

Proof. From the Rellich-Kondrachov theorem [14, Theorem 6.6-3], we have that H10(Ω) ,→,→ L2(Ω).

Lemma 3.2 gives that (un)n∈Nis bounded in L2

(0,T),H10(Ω)

. Thus we have the existence of an element u∈L2

(0,T),L2(Ω)

and a subsequence (unl)l∈Nof (un)n∈Nsuch that unl →uin L2

(0,T),L2(Ω)

asl→ ∞.

From Lemma 3.2 and the reflexivity of the space L2

(0,T),H10(Ω)

, we have the existence of a subsequence (indexed bynlagain) such that

unl *uin L2

(0,T),H10(Ω)

asl→ ∞. (11)

Now, we integrate Eq. (10) in time over (0, η)⊂(0,T) for the resulting subsequence to get that h(k∗u)nl(η), ϕiH1

0()×H10()− Z η

0

hu0knl(t), ϕiH1

0()×H10()dt+Z η 0

Lnl(t)(unl(t), ϕ) dt=Z η 0

hfnl(t), ϕidt. (12) We only point out the limit transition in the first term by showing that

l→∞lim

Z T

0

h(k∗u)nl(t), ϕiH1

0()×H10()dt− Z T

0

h(k∗unl)(t), ϕiH1

0()×H10()dt

=0. (13) This limit transition can be done in two steps

(i) lim

l→∞

Z T

0

h(k∗u)nl(t) dt− Z T

0

(k∗u)nl(t), ϕiH1

0()×H10()dt

=0;

(ii) lim

l→∞

Z T

0

h(k∗u)nl(t) dt− Z T

0

(k∗unl)(t), ϕiH1

0()×H10()dt

=0.

The limit transition (i) follows from Lemma 3.2 and the Lebesgue dominated theorem as follows Z T

0

h(k∗u)nl(t)−(k∗u)nl(t), ϕiH1

0()×H10()

dt=

nl

X

i=1

Z ti

ti1

h(t−ti)δ(k∗u)i, ϕiH1

0()×H10()

dt

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6

nl

X

i=1

τ2nl

h(k∗δu)i−kiu0, ϕiH1

0()×H10()

6Cτnl+C Z 1

0

τ1−βnl µ(β) dβl→∞−→0.

For the limit transition (ii), we deduce that (dteτ=tiwhent∈(ti−1,ti]) Z T

0

h(k∗u)nl(t)−(k∗unl)(t), ϕiH1

0()×H10()

dt=Z T 0

h(knl∗unl)(dteτ)−(k∗unl)(t), ϕiH1

0()×H10()

dt 6

Z T

0

h

Z t

0

knl(dteτ−s)−k(t−s)

unl(s) ds, ϕiH1

0()×H10()

+ h

Z dteτ

t

knl(dteτ−s)unl(s) ds, ϕiH1

0()×H10()

! dt→0 asl→ ∞by applying H¨older’s inequality twice on both terms on the right-hand side, since from Lemma 3.2,

it follows that (fort∈(ti−1,ti]) Z dteτ

t

knl(dteτ−s) unl(s)

2

H10() ds6C Z ti

0

knl(ti−s) unl(s)

2 ds=C

k∗ kuk2

i6C,

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and Z T

0

Z t

0

knl(dteτ−s)−k(t−s) unl(s)

2 H10() ds

! dt

(?)

6kkkL1(0,T)

unl

L2((0,T),H10())+Z T 0

Z t

0

knl(dteτ−s) unl(s)

2 H10() ds

! dt6C, where we used Young’s inequality for convolutions at position (?); and since for t ∈ (ti−1,ti], by the Lebesgue dominated theorem, it holds that

Z dteτ

t

knl(dteτ−s) ds=Z ti−t 0





 Z 1

0

τ−βnl

Γ(1−β)µ(β) dβ





dξ6 Z 1

0

τ1−βnl µ(β) dβ→0 asτnl →0, and

Z t

0

knl(dteτ−s)−k(t−s)

ds6C Z 1

0

τ1−βnl µ(β) dβ→0 asτnl →0.

Now, we integrate the equation (12) again in time overη ∈(0, ξ)⊂(0,T). Then, using Eq. (11) and (13), we are allowed to pass to the limit forl→ ∞. We get that (knl→kpointwise in (0,T))

Z ξ 0

h(k∗u)(η), ϕiH1

0()×H10()dη− Z ξ

0

Z η 0

hu0k(t), ϕiH1

0()×H10()dtdη +Z ξ

0

Z η 0

L(t)(u(t), ϕ) dtdη=Z ξ 0

Z η 0

(f(t), ϕ) dtdη.

Differentiating this relation with respect toξleads to h(k∗u)(ξ), ϕiH1

0()×H10()− Z ξ

0

hu0k(t), ϕiH1

0()×H10()dt+Z ξ 0

L(t)(u(t), ϕ) dt=Z ξ 0

(f(t), ϕ) dt. (14) i.e. (k∗u)(t) ∈ H10(Ω) for allt ∈ (0,T). The reasoning up to now is valid whenu0 ∈ L2(Ω). From Lemma 3.3 or 3.4, we obtain thatu∈L

(0,T),H10(Ω)

. Then, from Eq. (14), we have that

ξ→0lim+h(k∗u)(ξ), ϕiH1

0()×H10() =0 ⇒ (k∗u)(0)=0 in H10(Ω). Moreover, differentiating (14) with respect toξ(and replacingξbyt) gives

h∂t(k∗u)(t)−k(t)u0, ϕiH1

0()×H10()+L(t) (u(t), ϕ)=hf(t), ϕiH1

0()×H10(), ∀ϕ∈H10(Ω). (15) Thusuis satisfying (5) since∂t(k∗u)−ku0=k∗∂tu∈L2

(0,T),H10(Ω)

asuturns out to be absolutely continuous. Since∂t(k∗u)(t)−k(t)u0=∂t(k∗(u−u0)) (t) and (k∗(u−u0)) (0)=0 in H10(Ω), integrating (15) in time gives that (k∗(u−u0))(t) is absolutely continuous with values in H10(Ω). Following [10, Proposition 6], there exists a non-negative kernelg∈L1(0,T) such thatg∗k=1 and thus

(g∗∂t(k∗(u−u0)))(t)=∂t(g∗k∗(u−u0)) (t)=u(t)−u(0) in H10(Ω).

Therefore, applying this convolution operation on (15) gives thatu is absolutely continuous in the time variable, i.e. u ∈C

[0,T],H10(Ω)

. We conclude the proof by showing the uniqueness of a solution. We putϕ=u(t) in (5) withu0= f =0 and integrate over the time interval (0,T). We obtain that

Z T

0

h∂t(k∗u) (t),u(t)iH1

0()×H10()dt+Z T 0

L(t) (u(t),u(t)) dt=0.

From [10, Corollary 16], it follows that k(T)

2 Z T

0

ku(t)k2dt+α 2

Z T

0

k∇u(t)k2dt=0, i.e.u=0 a.e. inQT.

Remark 3.1. The result of Theorem 3.1 is also valid for the fractional diffusion equation (i.e. when k(t)= t−β ), which gives an improvement of the result obtained in [15, Theorem 2 and 4].

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Acknowledgement. The author is supported by a postdoctoral fellowship of the Research Foundation - Flanders (106016/12P2919N).

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Recall that a solution to a partial differential equation is usually referred to as a singular solution if it satisfies (1.2), the main examples being the so-called

Parabolic equation, fractional Laplacian, Riemann-Liouville fractional integrals and derivatives, local existence, critical exponent, blow-up