# Existence and Uniqueness of the Solutions for Some Initial-Boundary Value Problems with the Fractional Dynamic Boundary Condition

Authors: Mykola Krasnoschok; Nataliya Vasylyeva
Journal: International Journal of Partial Differential Equations
 (2013)

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## Abstract

In this paper, we analyze some initial-boundary value problems for the subdiffusion equation with a fractional dynamic boundary condition in a one-dimensional bounded domain. First, we establish the unique solvability in the Hölder space of the initial-boundary value problems for the equation∂tαu(x,t)=Lu(x,t)+f0(x,t), α∈(0,1), where L is a uniformly elliptic operator with smooth coefficients with the fractional dynamic boundary condition. Second, we apply the contraction theorem to prove the existence and uniqueness locally in time in the Hölder classes of the solution to the corresponding nonlinear problems.


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## Body

## 1. Introduction

Leta and b be any numbers from R1 and let Ω=(a,b), ΩT=Ω×(0,T); Γ1T={a}×[0,T]; Γ2T={b}×[0,T]; T>0 be a fixed value. In this paper, we consider a partial differential equation with the fractional derivative in time t as follows:
(1)∂tαu(x,t)-a0(x,t)∂2u(x,t)∂x2+a1(x,t)∂u(x,t)∂x+a2(x,t)u(x,t)=f0(x,t),(x,t)-a0(x,t)∂2u(x,t)∂x2(x,t)∈ΩT,α∈(0,1].
Here, ∂tα denotes the Caputo fractional derivative with respect to t and is defined by (see, e.g., (2.4.6) in [1]),
(2)∂tαg=1Γ(1-α)ddt∫0tg(τ)dτ(t-τ)α-g(0)Γ(1-α)tα,α∈(0,1),
where Γ is the gamma function, ai(x,t), i=0,2¯, are the given functions, and a0(x,t) is a positive. Note that if α=1, then (1) represents a parabolic equation. As we are interested in the fractional cases, we restrict the order α to the case 0<α<1.We will solve (1) satisfying the following conditions:
(3)u(x,0)=u0(x)inΩ(initialcondition),
the fractional dynamic boundary condition on Γ2T:
(4)∂tαu(x,t)+b0(t)∂u(x,t)∂x+b1(t)u(x,t)=f1(t)onΓ2T
and one of the following conditions on Γ1T: the Dirichlet boundary condition:
(5)u(x,t)=f2(t)onΓ1T,
or the Neumann boundary condition:
(6)∂u(x,t)∂x=f3(t)onΓ1T,
or the fractional dynamic boundary condition:
(7)∂tαu(x,t)-b2(t)∂u(x,t)∂x+b3(t)u(x,t)=f4(t)onΓ1T.Here,b0(t) and b2(t) are given positive functions, and b1(t), b3(t), and fi(t), i=2,4¯, are given functions.Concerning problems (1)–(6), they have the following features. First of all, (4) is a fractional dynamic boundary condition; next, these problems are formulated for the subdiffusion linear equation.Note that ifα=1, conditions (4) and (6) are called normal dynamic boundary conditions. These conditions are very natural in many mathematical models, including heat transfer in a solid in contact with a moving fluid, thermoelasticity, diffusion phenomena, and problems in fluid dynamics, and in the Stefan problem, (see [2–4] and the references therein). At the present moment, there are a lot of works concerning linear and nonlinear problems with dynamic boundary conditions. Here we make no pretence to provide a complete survey on the results related to problems of the type (1)-(6), if α=1, and present only some of them. The initial-boundary value problems for the heat equation in the certain shape of domains with linear dynamic boundary condition have been solved with the separation variables method or with the Laplace transformation in [3]. In the case of smooth domains, these problems have been researched with the approaches of the general theory for evolution equations in Hilbert and Banach spaces, and the weak solutions of the above mentioned problems have been obtained in [5–7]. Using the Schauder method, Grigor’eva and Mogilevskii [8] have got the coercive estimates of the solution in the anisotropic Sobolev spaces. The one-to-one solvability in the case of the linear parabolic equation with variable coefficients has been proved by Bazaliy [4] in the Hölder spaces and by Bizhanova and Solonnikov [9] in the weighted Hölder classes. The global and local existence for the solution to initial-boundary value problem for linear and quasilinear equations with nonlinear dynamic boundary conditions has been discussed in [10–12] (see also references there).Over the past few decades, an intensive effort has been put into developing theoretical models for systems with diffusive motion that cannot be modelled as the standard Brownian motion [13, 14]. The signature of this anomalous diffusion is that the mean square displacement of the diffusing species 〈(Δx)2〉 scales as a nonlinear power law in time, that is, 〈(Δx)2〉~tα. If α∈(0,1), this is referred to as subdiffusion. In recent years, the additional motivation for these studies has been stimulated by experimental measurements of subdiffusion in porous media [15], glass forming materials [16], and biological media [17]. The review paper by Klafter et al. [18] provides numerous references to physical phenomena in which anomalous diffusion occurs.Here we refer to several works on the mathematical treatments for linear equation (1). Kochubei [19, 20], and Pskhu [21, 22] constructed the fundamental solution in Rd and proved the maximum principle for the Cauchy problem. Gejji and Jafari [23] solved a nonhomogeneous fractional diffusion-wave equation in a one-dimensional bounded domain. Metzler and Klafter [14], using the method of images and the Fourier-Laplace transformation technique, obtained the solutions of different boundary value problems for the homogenous fractional diffusion equation in a half-space and in a box. Agrawal [24] constructed a solution of a fractional diffusion equation using a finite transform technique and presented numerical results in a one-dimensional bounded domain. Mophou and N’Guérékata [25] and Sakamoto and Yamamoto [26] proved the one-valued solvability of the initial-boundary value problem for the fractional diffusion equation with variable coefficients which is t-independent with the homogenous Dirichlet conditions in the Sobolev space. Note that, in [26], the authors obtained the certain regularities of the solution given by the eigenfunction expansions and established several results of uniqueness for related inverse problems.As source books related with fractional derivatives, see the work of Samko et al. [27] which is an encyclopedic treatment of the fractional calculus and also Kilbas et al. [1], Mainardi [28], Podlubny [29], and Pskhu [21].As for the quasilinear equation like (1), Clément et al. [30] analyzed the abstract fractional parabolic quasilinear equations. Via maximal regularity results in the corresponding linear equation, they arrived to results on existence (locally in time), uniqueness, and continuation on the quasilinear equation in the BUC classes with a weight. As for investigation of the problem with fractional dynamic boundary conditions, Kirane and Tatar [31] have analyzed the issue of nonexistence of local and global solutions for elliptic systems with nonlinear fractional dynamic boundary conditions.To the authors’ best knowledge, there are no works published concerning the solvability of problems (1)–(6) in the Hölder classes. The first purpose of this paper is to prove the well-posedness and the regularity of the solutions to problems (1)–(6) in the smooth classes. Second, we obtain a local in time solvability in the smooth classes of the corresponding nonlinear problems. This paper is organized as follows. In the second section, we state the main results, Theorems 3–5, and define the functional spaces. In Section 3, we establish the one-valued solvability of certain model problems in RT+. The principal results of this section are given in Theorems 9 and 13. In Section 4, we prove the main results of this paper. To this end, we will combine ideas from [32] with coercive estimates of the solutions to the corresponding model problems (Section 3). In Section 5, we address the corresponding nonlinear problems. We first reduce them to a form 𝔄w=𝔗(w), where 𝔗(w) is a nonlinear function of w and 𝔄 is the linear operator derived in Section 4; that is, 𝔄-1𝔗 is the solution of the model problem for data 𝔗. Setting 𝔖=𝔄-1𝔗, we will then prove that the mapping w→𝔖(w), where 𝔖(w)=𝔄-1𝔗w, is a contraction, so that it has a unique fixed point. The principal results of this section are formulated in Theorem 18 and Remarks 19 and 20. The Appendix contains the proofs of some auxiliary assertions which are applied in Section 3.


## 2. The Functional Spaces and the Main Results

Let us introduce the functional spaces. LetΩ be a bounded or an unbounded domain in R1, x¯, x∈Ω¯; t,τ∈[0,T]; θ,γ∈(0,1). Denote
(8)〈v〉x,ΩT(θ)=sup(x,t),(x¯,t)∈ΩT,x≠x¯|v(x,t)-v(x¯,t)||x-x¯|θ;〈v〉t,ΩT(γ)=sup(x,t),(x,τ)∈ΩT,t≠τ|v(x,t)-v(x,τ)||t-τ|γ.Definition 1.
We will say that functionsv(x,t)∈Cθ,(θ/2)α(Ω¯T) and that w(x,t)∈C2+θ,((2+θ)/2)α(Ω¯T) if and only if the functions v(x,t) and ∂kw(x,t)/∂xk, k=0,2¯, ∂tαw(x,t) are continuous and the following norms are finite:
(9)∥v∥Cθ,(θ/2)α(Ω¯T)=supΩ¯T|v(x,t)|+〈v〉x,ΩT(θ)+〈v〉t,ΩT((θ/2)α),(10)∥w∥C2+θ,((2+θ)/2)α(Ω¯T)=∑k=02‍supΩ¯T|∂kw∂xk|+supΩ¯T|∂tαw|+〈∂2w∂x2〉x,ΩT(θ)+〈∂tαw〉x,ΩT(θ)+∑k=12‍〈∂kw∂xk〉t,ΩT(((2+θ-k)/2)α)+〈∂tαw〉t,ΩT((θ/2)α).Note that ifα=1, the spaces Cl+θ,((l+θ)/2)α(Ω¯T), l=0,2, coincide with the ordinary Hölder spaces (see (1.10)–(1.12) in Chapter 1 in [32]). Further, we also use the Banach spaces Cγα([0,T]) and C(1+γ)α([0,T]) of the functions V(t) and W(t) with the finite norms
(11)∥V∥Cγα([0,T])=sup[0,T]|V(t)|+〈V〉t,[0,T](γα),(12)∥W∥C(1+γ)α([0,T])=sup[0,T]|W(t)|+sup[0,T]|∂tαW|∥W∥C(1+γ)α([0,T])=+〈∂tαW〉t,[0,T](γα).
In a similar way, we introduce the spaces C(k+γ)α(∂ΩT), k=0,1.Definition 2.
We will say that functionsv1(x,t)∈C0θ,(θ/2)α(Ω¯T) and that v2(x,t)∈C02+θ,((2+θ)/2)α(Ω¯T) if and only if v1(x,t)∈Cθ,(θ/2)α(Ω¯T) and v1(x,0)=0 and v2(x,t)∈C2+θ,((2+θ)/2)α(Ω¯T) and v2(x,0)=∂tαv2(x,0)=0.
Ifu1(t)∈Cγα([0,T]),u1(0)=0(u2(t)∈C(1+γ)α([0,T]), and u2(0)=∂tαu2(0)=0), then u1(t)∈C0γα([0,T])(u2(t)∈C0(1+γ)α([0,T])).We introduce the spaces ofC0(k+γ)α(∂ΩT), k=0,1, with the same way.Letδ1 and δ2 be some positive numbers. We assume that the given functions ai(x,t), i=0,2¯, and bj(t), j=0,2¯, in (1), (4), and (7) are subject to the following conditions:
(13)(i)a0(x,t)≥δ1>0inΩ¯T;b0(t),b2(t)≥δ2>0a0(x,t)≥δ1>0inΩ¯T;b0(t)b0(t)∀t∈[0,T];(14)(ii)ai(x,t)∈Cθ,(θα/2)(Ω¯T),bj(t)∈C((1+θ)/2)α([0,T]),ai(x,t)∈Cθ,(θα/2)(Ω¯T),11bj(t)∈i=0,2¯,j=0,3¯;(15)(iii)f1(0)-b0(0)∂u0(b)∂x-b1(0)u0(b)(iii)=a0(b,0)∂2u0(b)∂x2-a1(b,0)∂u0(b)∂x-a2(b,0)u0(b)+f0(b,0);and one of the following:(16)f2(0)=u0(a),∂tαf2(0)=a0(a,0)∂2u0(a)∂x2-a1(a,0)∂u0(a)∂x∂tαf2(0)=-a2(a,0)u0(a)+f0(a,0)(inthecaseofDirichletcondition(5));or(17)f3(0)=∂u0(a)∂x(inthecaseofNeumanncondition(6));or(18)f4(0)+b2(0)∂u0(a)∂x-b3(0)u0(a)=a0(a,0)∂2u0(a)∂x2-a1(a,0)∂u0(a)∂x=-a2(a,0)u0(a)+f0(a,0)(inthecaseoffractionaldynamicboundarycondition(7)).Note that requirements (15)–(18) are called the consistency conditions.The main results of our paper are the following:Theorem 3.
Letθ,α∈(0,1), and conditions (13)–(16) hold, and f0(x,t)∈Cθ,(θ/2)α(Ω¯T), u0(x)∈C2+θ(Ω¯), f1(t)∈C((1+θ)/2)α([0,T]), f2(t)∈C((2+θ)/2)α([0,T]) for any positive number T. Then there exists a unique solution u(x,t) of problem (1)–(5): u(x,t)∈C2+θ,((2+θ)/2)α(Ω¯T), ∂tαu(x,t)∈C((1+θ)/2)α(Γ2T), and
(19)∥u∥C2+θ,((2+θ)/2)α(Ω¯T)+∥∂tαu∥C((1+θ)/2)α(Γ2T)≤C[∥f0∥Cθ,(θ/2)α(Ω¯T)+∥u0∥C2+θ(Ω¯)+∥f1∥C((1+θ)/2)α([0,T])+∥f2∥C((2+θ)/2)α([0,T])],
where a positive constant C depends only on the measure of Ω and ∥ai∥Cθ,(θ/2)α(Ω¯T), i=0,2¯, ∥bj∥C((1+θ)/2)α([0,T]), j=0,1.Theorem 4.
Letθ,α∈(0,1), and conditions (13)–(15), and (17) hold, and let f0(x,t)∈Cθ,(θ/2)α(Ω¯T), u0(x)∈C2+θ(Ω¯), fi(t)∈C((1+θ)/2)α([0,T]), i=1,3 for any positive number T. Then there exists a unique solution u(x,t) of problem (1)–(4), (6) as follows: u(x,t)∈C2+θ,((2+θ)/2)α(Ω¯T), ∂tαu(x,t)∈C((1+θ)/2)α(Γ2T), and
(20)∥u∥C2+θ,((2+θ)/2)α(Ω¯T)+∥∂tαu∥C((1+θ)/2)α(Γ2T)≤C[∥f0∥Cθ,(θ/2)α(Ω¯T)+∥u0∥C2+θ(Ω¯)+∥f1∥C((1+θ)/2)α([0,T])+∥f3∥C((1+θ)/2)α([0,T])],
where C is a positive constant and depends only on ∥bj∥C((1+θ)/2)α([0,T]), j=0,1, ∥ai∥Cθ,(θ/2)α(Ω¯T), i=0,2¯, and the measure of Ω.Theorem 5.
Letθ,α∈(0,1), and conditions (13)–(15) and (18) hold, and f0(x,t)∈Cθ,(θ/2)α(Ω¯T), u0(x)∈C2+θ(Ω¯), fi(t)∈C((1+θ)/2)α([0,T]), and i=1,4 for any positive number T. Then there exists a unique solution u(x,t) of problem (1)–(4), (7) as follows: u(x,t)∈C2+θ,((2+θ)/2)α(Ω¯T), ∂tαu(x,t)∈C((1+θ)/2)α(ΓiT), i=1,2, and
(21)∥u∥C2+θ,((2+θ)/2)α(Ω¯T)+∑i=12‍∥∂tαu∥C((1+θ)/2)α(ΓiT)≤C[∥f0∥Cθ,(θ/2)α(Ω¯T)+∥u0∥C2+θ(Ω¯)+∥f1∥C((1+θ)/2)α([0,T])+∥f4∥C((1+θ)/2)α([0,T])],
where C is a positive constant and depends only on ∥bj∥C((1+θ)/2)α([0,T]),j=0,3¯, ∥ai∥Cθ,(θ/2)α(Ω¯T),  i=0,2¯, and the measure of Ω.


## 3. Model Problems

LetD+=(0,+∞), and DT+=D+×(0,T), and a0 and b0 be some positive numbers. Here we will discuss the first initial-boundary value problem for the fractional diffusion equation in DT+ and the initial-boundary value problem with the fractional boundary condition in DT+.


### 3.1. The Solvability of the First Initial-Boundary Value Problem for the Subdiffusion Equation

We look for the functionv(x,t) by the following conditions:
(22)∂tαv(x,t)-a0∂2v(x,t)∂x2=g0(x,t)inDT+,α∈(0,1);(23)v(x,0)=v0(x),x∈D+;v(x,t)⟶0ifx⟶+∞,∀t∈[0,T];(24)v(0,t)=g1(t),t∈[0,T],
where g0(x,t) and g1(t),v0(x) are some given functions.We assume that the following conditions hold:(25)(i)v0(0)=g1(0);∂tαg1(0)=a0∂2v0∂x2+g0(x,0);(26)(ii)v0(x),g0(x,t)≡0,if|x|<R0∀t∈[0,T],
for some positive number R0.First, we will study problem (22)–(24) under restriction
(27)v0(x),g0(x,t)≡0.
We will search a solution of (22)–(24) under restriction (27) in the class C2+θ,((2+θ)/2)α(D¯T+), if g1(t)∈C((2+θ)/2)α([0,T]) for θ∈(0,1).Note that conditions (25) and (26) together with restriction (27) allow us to apply the Laplace transformation in t to the right hand sides of (22)–(24). Indeed, conditions (25)–(27) mean that the right hand sides in (22)–(24) except g1(t) equal zero and g1(0)=0. Thus, we can extend the right hand sides in (22)–(24) by 0 for t≤0 and save, for simplicity, the same notation for the extension of the function g1(t). Therefore, we can apply, at least formally, the Laplace transformation in t to (22)–(24) in the case of (25)–(27) hold.Denote byw^(x,p) the Laplace transformation of the function w(x,t); that is,
(28)w^(x,p)=∫0+∞e-ptw(x,t)dt.
The Laplace transformation in (22)–(24) leads to the problem
(29)pαv^(x,p)-a0∂2v^(x,p)∂x2=0,x∈D+;v^(x,p)⟶0ifx⟶+∞;v^(0,p)=g^1(p).
Here we used the following formula from [33]:
(30)∂tαw^=pαw^(x,p)-pα-1w(x,0).
One can easily check that the following function solves the equations in (29):
(31)v^(x,p)=exp(-xpα/2a0)g1(p).
Due to formula (2.30) in [34] and the inverse Laplace transformation, we get the integral representations of v(x,t) as follows:
(32)v(x,t)=∫0tG(x,t-τ)g1(τ)dτ,
where
(33)G(x,t)=t-1W(-xt-α/2a0;-α2,0).
Here W(z;β,γ) is the Wright function, which is defined for z,β,γ∈ℂ as (see formula (1.8.1(27)) in v.3 [35])
(34)W(z;β,γ)=∑k=0∞‍zkk!Γ(βk+γ).
The main properties of the Wright functions are described in Chapters 4.1, v.1 and 18.1 v3 in [35], Chapter 1.11 in [1], Chapter 1.3 in [34], and Chapter 2 in [21, 36].In Lemma6, we describe the properties of the kernel G(x,t) which will be necessary to estimate the function v(x,t). Its proof is represented in Appendix A.Lemma 6.
Letκ, δ, and θ¯ be some positive constants, κ=κ(α), θ¯∈(0,1], x∈D+, t∈[0,T]. Then one has the following.
(35)(i)ThekernelG(x,t)ispositiveforx>0.(36)(ii)∫0+∞G(x,t)dt=1.(37)(iii)tG(x,t)≤Cexp(-κ(xt-α/2)1/(1-α/2));|∂G(x,t)∂t|≤Ct-21+(xt-α/2)4/α;|∂G(x,t)∂x|≤Ct-1-α/21+(xt-α/2)1+2/α.(38)(iv)∫0tG(x,τ)τ(θ¯/2)αdτ≤Ct(θ¯/2)α.(39)(v)∫δ+∞|∂G(x,τ)∂x|τ(θ¯/2)αdτ≤Cδ-(1-θ¯)α/2.(40)(vi)∫δ+∞|∂G(x,τ)∂τ|τθ¯α/2dτ≤Cδ-1+θ¯α/2.(41)(vii)∫0t(t-τ)-αG(x,τ)dτ≤Ct-αexp(-κ(xt-(α/2))1/(1-α/2)).(42)(viii)DtαG(x,t)=a0∂2G(x,t)∂x2inDT+,whereDtα is the Riemann-Liouville fractional derivative, and its definition is in (2.1.8) in [1].Lemma 7.
Letθ∈(0,1), and conditions (25)–(27) hold, and g1(t)∈C((2+θ)/2)α([0,T]). Then the function v(x,t) represented by formula (32) satisfies the following inequalities:
(43)supD¯T+|∂tαv|≤C(sup[0,T]|∂tαg1|+〈∂tαg1〉t,[0,T]((θ/2)α));(44)〈∂tαv〉t,DT+((θ/2)α)≤C∥∂tαg1∥C(θ/2)α([0,T]);(45)〈∂tαv〉x,DT+(θ)≤C∥∂tαg1∥C(θ/2)α([0,T]).Proof.
First, we obtain the representation of∂tαv. To this end, we need the following properties of the fractional derivative (see Lemma 2.10 and formula (2.4.10) in [1]): (i)
(46)∂tαρ(t)=Dtαρ(t),ifρ(0)=0,where (see (2.1.8) in [1])
(47)Dtαρ(t)=1Γ(1-α)ddt∫0tρ(τ)(t-τ)αdτ,α∈(0,1).(ii)
If the functionsρi(t), i=1,2, and Dtαρ1(t) are bounded in [0,T], then
(48)Dtα∫0tρ1(t-τ)ρ2(τ)dτ=∫0tρ2(t-τ)Dταρ1(τ)dτ+ρ2(t)limz→0Iz1-αρ1(z),=∫0tρ2(t-τ)Dταρ1(τ)dτ+ρ2(t)limz→0∀t∈[0,T],whereItν is the Riemann-Liouville fractional integral of order ν (see, e.g. (2.1.1) in [1])
(49)Itνρ(t)=1Γ(ν)∫0t‍ρ(τ)(t-τ)1-νdτ,Reν>0.One can easily see that ∂tαg1(t)∈C(θ/2)α([0,T]) and ∂tαg1(0)=0 (see (25) and (27)). Then, using properties (46) and (48) and equality (36), we represent the function ∂tαv(x,t) as
(50)∂tαv(x,t)=∫0tG(x,t-τ)[∂tαg1(τ)-∂tαg1(t)]dτ+∂tαg1(t).
Namely, this representation will be useful below. To prove inequality (43), we will use this representation, statement (i) from Lemma 6 and the first estimate in (37) as follows:
(51)|∂tαv|≤〈∂tαg1〉t,[0,T]((θ/2)α)|∂tαv|≤×∫0tτ(θ/2)α-1exp(-κ(xτ-α/2)(1/(1-α/2)))dτ|∂tαv|≤+max[0,T]|∂tαg1||∂tαv|≤C∥∂tαg1∥C(θ/2)α([0,T]).
In view of ∂tαg1(t)=0, if t≤0, we can rewrite representation (50) as
(52)∂tαv(x,t)=∫-∞tG(x,t-τ)[∂tαg1(τ)-∂tαg1(t)]dτ+∂tαg1(t).
Letx1, x2∈D+ and x2>x1. Denote
(53)Δx≔x2-x1,Δx∂tαv≔∂tαv(x2,t)-∂tαv(x1,t).
Then,
(54)Δx∂tαv=∫0(Δx)2/αG(x2,τ)[∂ταg1(t-τ)-∂ταg1(t)]dτΔx∂tαv=+∫0(Δx)2/αG(x1,τ)[∂ταg1(t)-∂ταg1(t-τ)]dτΔx∂tαv=+∫(Δx)2/α+∞[G(x2,τ)-G(x1,τ)]o×[∂ταg1(t-τ)-∂ταg1(t)]dτ≡∑i=13‍ℑi.
Using inequality (38) with t≔(Δx)2/α, θ¯≔θ, we obtain
(55)|ℑ1|+|ℑ2|≤const.〈∂tαg1〉t,[0,T]((θ/2)α)(x2-x1)θ.
As for the term ℑ3, we apply the mean-value theorem to the difference [G(x2,τ)-G(x1,τ)] together with inequality (39) (where δ≔(Δx)2/α, θ¯≔θ) and deduce
(56)|ℑ3|≤const.〈∂tαg1〉t,[0,T]((θ/2)α)(Δx)1+(2/α)(-(α/2)+θ(α/2)).
Thus, representation (54) together with inequalities (55) and (56) prove the correctness of (44).
To complete the proof of Lemma7, we need to obtain inequality (45). Let t1,t2∈[0,T] and t2>t1. Denote
(57)Δt≔t2-t1,Δt∂tαv≔∂tαv(x,t2)-∂tαv(x,t1).
We analyze the difference
(58)Δt∂tαv=∫2t1-t2t1G(x,t1-τ)[∂tαg1(t1)-∂tαg1(τ)]dτΔt∂tαv=+∫2t1-t2t2G(x,t2-τ)[∂tαg1(τ)-∂tαg1(t2)]dτΔt∂tαv=+∫-∞2t1-t2G(x,t1-τ)dτ[∂tαg1(t1)-∂tαg1(t2)]Δt∂tαv=+∫-∞2t1-t2[∂tαg1(τ)-∂tαg1(t2)]Δt∂tαv=Δt∂tαv=×[G(x,t2-τ)-G(x,t1-τ)]dτΔt∂tαv=+[∂tαg1(t2)-∂tαg1(t1)]≡∑i=15‍𝔍i.
As for the last term in this sum, it is estimated by (Δt)θα/2〈∂tαg1〉t,[0,T](θ/2)α. We change the variable τ=t1-z in the term 𝔍1 and apply estimate (38) with t≔Δt, θ¯≔θ. Thus, we have
(59)|𝔍1|≤const.(Δt)θα/2〈∂tαg1〉t,[0,T]((θ/2)α).
In the same way, we evaluate the function 𝔍2. The estimate of the term 𝔍3 follows from the properties of the function ∂tαg1 and inequality (36). At last, the mean-value theorem together with estimate (40), where δ≔Δt, θ¯≔θ, lead to
(60)|𝔍4|≤C(Δt)θα/2〈∂tαg1〉t,[0,T]((θ/2)α).
Therefore, inequality (45) is deduced from (58)–(60).Now, based on the results of Lemmas6 and 7, we can infer the next assertion.Lemma 8.
Let conditions of Lemma7 hold. Then there exists a unique solution v(x,t)∈C2+θ,((2+θ)/2)α(D¯T+) of problem (22)–(24), which is represented with (32) and
(61)∥v∥C2+θ,((2+θ)/2)α(D¯T+)≤C∥g1∥C((2+θ)/2)α([0,T]).Proof.
First of all we obtain estimate (61). One can get the following inequality using the results of Lemma 7 and (22), where g0≡0:(62)∥∂2v∂x2∥Cθ,(θ/2)α(D¯T+)≤const.∥∂tαv∥Cθ,(θ/2)α(D¯T+)∥∂2v∂x2∥Cθ,(θ/2)α(D¯T+)≤C∥g1∥C((2+θ)/2)α([0,T]).
Next, we use formula (3.5.4) from [1] as follows:
(63)w(t)-w(0)=1Γ(α)∫0t∂ταw(τ)(t-τ)1-αdτ,
to evaluate the maximum of |v(x,t)|. Hence, (43) and (63) lead to inequality
(64)supD¯T+|v|≤const.supD¯T+|∂tαv|≤C∥g1∥C((2+θ)/2)α([0,T]);
here we use the fact that v(x,0)=0.
After that, the minor seminorms of the functionv(x,t) are estimated with the interpolation inequalities from Section 8.8 [37] and (43)–(45), (62), and (64). Therefore, the arguments above prove inequality (61) and the embedding v(x,t)∈C2+θ,((2+θ)/2)α(D¯T+).
Next, we show that the functionv(x,t) given by (32) satisfies (22). To this end, we use equality (46) and (48) and represent ∂tαv(x,t) as
(65)∂tαv(x,t)=∫0tg1(t-τ)DtαG(x,τ)dτ+g1(t)limτ→0Iτ1-αG(x,τ).
Then (41) leads to
(66)∂tαv(x,t)=∫0tg1(t-τ)DtαG(x,τ)dτ.
Next, due to property (42) of the function G(x,t), one can check that
(67)∂tαv(x,t)=a0∂2v∂x2.
As it follows from the first inequality in (37), the function v(x,t) represented by (32) satisfies the following conditions:
(68)limt→0v(x,t)=0;v(x,t)⟶0ifx⟶+∞.
Finally, it is necessary to show that the functionv(x,t) meets boundary condition (24). To this end, we observe the next difference (here we will essentially use statement (35))
(69)|v(x,t)-g1(t)|≤|∫x+∞G(x,τ)[g1(t-τ)-g1(t)]dτ|≤+|∫0xG(x,τ)[g1(t-τ)-g1(t)]dτ|≤2sup[0,T]|g1|∫x+∞G(x,τ)dτ+〈g1〉t,[0,T]((θ/2)α)≤×∫0xG(x,τ)ταθ/2dτ≡A1+A2.
Applying inequality (38) with t≔x, θ¯≔θ to the term A2, we get
(70)|A2|≤const.∥g1∥C((2+θ)/2)α([0,T])xαθ/2.
To estimate the term A1 in (69), we use the Wright formula (see [38] or (2.2.5) in [21]):
(71)-βzW(z;β,1-β)=W(z,β,0),
and rewrite the function G(x,t) as
(72)G(x,t)=α2a0xt-1-(α/2)W(-xa0t-α/2;-α2;1-α2).
After that, due to Lemmas 2.2.4 and 2.2.7 in [21] and representation (72), we have
(73)0<G(x,t)<const.xt-1-α/2.
Thus,
(74)|A1|≤const.sup[0,T]|g1|∫x+∞xτ-1-α/2dτ|A1|≤const.sup[0,T]|g1|x1-α/2.
Then, we joint estimates (69)–(74) and obtain
(75)|v(x,t)-g1(t)|≤const.∥g1∥C((θ/2)α)α([0,T])(x1-α/2+xαθ/2)⟶0≤const.11111i11111∥g1∥C((θ/2)α)α([0,T])ifx⟶0,
which means that limx→0v(x,t)=g1(t).
Therefore, as it was written above, the functionv(x,t) given by (32) is a solution of (22)–(24) in the case of (27). The uniqueness of this solution is proved like the arguments of Theorem 3.2 from [39].Now we remove restriction (27). To this end, it is enough to consider the Cauchy problem:
(76)∂tαv¯(x,t)-a0∂2v¯(x,t)∂x2=g¯0(x,t)∂tαv¯(x,t)inR1×(0,T),α∈(0,1);v¯(x,0)=v¯0(x),x∈R1;v¯(x,t)⟶0,if|x|⟶+∞,∀t≥0.
Here, g¯0(x,t) and v0(x) are extensions of the functions g0(x,t) and v0(x), correspondingly, onto x<0. These functions together with their corresponding derivatives have finite supports and
(77)∥g¯0∥Cθ,(θ/2)α(R1×[0,T])≤C∥g0∥Cθ,(θ/2)α(D¯T+),∥v¯0∥C2+θ(R1)≤C∥v0∥C2+θ(D¯+).
The results of Theorem 3.2 from [39] give the one-valued solvability of (76) and
(78)∥v¯∥C2+θ,((2+θ)/2)α(R1×[0,T])≤C(∥g¯0∥Cθ,(θ/2)α(D¯T+)+∥v¯0∥C2+θ(D¯+)),
or, due to inequalities (77),
(79)∥v¯∥C2+θ,((2+θ)/2)α(R1×[0,T])≤C(∥g0∥Cθ,(θ/2)α(D¯T+)+∥v0∥C2+θ(D¯+)).
After that, we will look for the solution v(x,t) of problem (22)–(24) as
(80)v(x,t)=v¯(x,t)+v¯¯(x,t),(x,t)∈D¯T+,
where v¯¯(x,t) satisfies conditions (22)–(24) with the new right-hand sides which meet requirements of Lemma 8. Hence, we can apply the results of Lemma 8 to the function v¯¯(x,t). This fact and the properties (see (77)) of the function v¯(x,t) allow us to obtain the next results.Theorem 9.
Letα,θ∈(0,1), and conditions (25) and (26) hold, and g0∈Cθ,(θ/2)α(D¯T+), v0∈C2+θ(D¯+), g1(t)∈C((2+θ)/2)α([0,T]) for any positive number T. Then there exists a unique solution v∈C2+θ,((2+θ)/2)α(D¯T+) of problem (22)–(24) and
(81)∥v∥C2+θ,((2+θ)/2)α(D¯T+)≤C(∥g1∥C((2+θ)/2)α([0,T])+∥g0∥Cθ,(θ/2)α(D¯T+)+∥v0∥C2+θ(D¯+)∥g1∥C((2+θ)/2)α([0,T])+∥g0∥Cθ,(θ/2)α(D¯T+)).


### 3.2. The Model Initial-Boundary Value Problem with a Fractional Dynamic Boundary Condition

Here we study the following problem: it is necessary to find the functionw(x,t) by the following conditions:
(82)∂tαw(x,t)-a0∂2w(x,t)∂x2=Φ0(x,t)inDT+,α∈(0,1),w(x,0)=w0(x),x∈D+;w(x,t)⟶0ifx⟶+∞,∀t∈[0,T],∂tαw(0,t)-b0∂w(0,t)∂x=Φ1(t),t∈[0,T],
where Φi,i=0,1,w0 are the given functions. Let the following conditions hold:
(83)b0∂w0(0)∂x+Φ1(0)=a0∂2w0(0)∂2x+Φ0(0,0);w0(x),Φ0(x,t)≡0,if|x|<R0∀t∈[0,T];
for some positive number R0.At the beginning, we assume that(84)w0(x),Φ0(x,t)≡0
and search a solution of (82) under this restriction in the class C2+θ,((2+θ)/2)α(D¯T+), if Φ1(t)∈C((1+θ)/2)α([0,T]), θ∈(0,1).Note that conditions (83) and (84) allow us to extend the right-hand sides in (82) by 0 for t≤0. We save, for simplicity, the same notation for the extended function Φ1(t).After the application of the Laplace transformation in timet to problem (82), we have the following:
(85)pαw^(x,p)-a0∂2w^(x,p)∂x2=0,x∈D+;w^(x,p)⟶0ifx⟶+∞;pαw^(0,p)-b0∂w^(0,p)∂x=Φ^1(p).
Here, we used again formula (30). Some simple calculations lead to the function
(86)w^(x,p)=Φ^1(p)pα/2(pα/2+b0a0-1/2)exp(-xa0-1/2pα/2)w^(x,p)≡F^(p)exp(-xa0-1/2pα/2),
which is the solution of problem (85). Due to formulas (2.30) in [34] and (1.80) in [29], we obtain, after applying of the inverse Laplace transformation to (86), that
(87)w(x,t)=∫0tG(x,t-τ)F(τ)dτ,
where the kernel G(x,t) is given by (33) and
(88)F(t)=∫0tG2(τ)Φ1(t-τ)dτ,G2(t)=tα-1Eα/2,α(-b0a0-1/2tα/2).
Here, Eγ,β(z) is the function of the Mittag-Leffler type, which is defined by the series expansion (see, e.g., (1.56) in [29] or (1.8.17) in [1])
(89)Eγ,β(z)=∑k=0∞‍zkΓ(γk+β),Reγ>0,β,z∈ℂ.
Note that this two-parameter function of the Mittag-Leffler type was in fact introduced by Agarwal [40].The functionG(x,t) has been studied in Section 3.1 (see Lemma 6). Thus, to describe the properties of the function w(x,t), we have to observe the function F(t). To this end, we will use the following properties of the kernel G2(t), which are proved in Appendix B.Lemma 10.
LetN, δ, and θ be some positive constants, θ∈(0,1); Itν and Dtν be the fractional Riemann-Liouville integral and derivative, correspondingly (their definitions are given in (49) and (47)). Then the following is true.
(90)(i)0≤G2(t)≤Ctα-11+tα/2,∫0tG2(τ)dτ≤Ctα.(91)(ii)limt→0It1-αG2(t)=1.(92)(iii)∫0+∞DταG2(τ)dτ=-1.(93)(iv)∫0t|DταG2(τ)|τ(1+θ)α/2dτ≤Ct(1+θ)α/2.(94)(v)∫N+∞|DταG2(τ)|dτ≤C.(95)(vi)|∫δ+∞DταG2(τ)dτ|≤1+C1+δα/2.(96)(vii)∫δ+∞|ddτDταG2(τ)|τ(1+θ)α/2dτ≤Cδ-1+α(θ+1)/2.Lemma 11.
LetΦ1(t) meet requirement (83) and let Φ1∈C((1+θ)/2)α([0,T]). Then the function F(t) represented by (88) belongs to C((3+θ)/2)α([0,T]), F(0)=0, and the following estimate holds:
(97)∥F∥C((3+θ)/2)α([0,T])≤C∥Φ1∥C((1+θ)/2)α([0,T]).Proof.
First of all, we evaluate the value ofmax[0,T]|F(t)|. To this end, we use the first inequality in (90). Thus, one has
(98)max[0,T]|F(t)|≤max[0,T]|Φ1(t)|∫0t|G2(τ)|dτ≤Cmax[0,T]|Φ1(t)|tα.
This inequality gives that
(99)F(0)=0.
Next, we obtain the representation of ∂tαF(t). Due to equality (99) and properties (46) and (48), we conclude that
(100)∂tαF(t)=DtαF(t)∂tαF(t)=Φ1(t)limt→0It1-αG2(t)+∫0tΦ1(t-τ)DταG2(τ)dτ.
Since Φ1(0)=0, we can rewrite the last equality as
(101)∂tαF(t)=Φ1(t)limt→0It1-αG2(t)+∫0+∞Φ1(t-τ)DταG2(τ)dτ,
or, applying (91) and (92), we have
(102)∂tαF(t)=∫0+∞[Φ1(t-τ)-Φ1(t)]DταG2(τ)dτ.
To estimate max[0,T]|∂tαF(t)|, we use representation (102) and get
(103)|∂tαF(t)|≤∫0Nτ(1+θ)α/2|DταG2(τ)|dτ〈Φ1〉t,[0,T]((1+θ/2)α)|∂tαF(t)|≤+2max[0,T]|Φ1(t)|∫N+∞|DταG2(τ)|dτ.
After that, we apply inequality (93) with t≔δ to the first term in (103) and (94) to the second and get
(104)max[0,T]|∂tαF(t)|≤C∥Φ1∥C((1+θ)/2)α([0,T]).
Finally, to complete the proof of Lemma 11 is necessary to estimate of 〈∂tαF〉t,[0,T](((1+θ)/2)α). Let t2,t1∈[0,T] and t2>t1. Denote
(105)Δt≔t2-t1,Δt∂tαF=∂tαF(t2)-∂tαF(t1).
Using formula (102), we represent the difference Δt∂tαF as
(106)Δt∂tαF=∫t1-Δtt2‍DταG2(t2-τ)[Φ1(τ)-Φ1(t2)]dτΔt∂tαF=+∫t1-Δtt1‍DταG2(t1-τ)[Φ1(t1)-Φ1(τ)]dτΔt∂tαF=+∫-∞t1-Δt‍[DταG2(t2-τ)-DταG2(t1-τ)]×[Φ1(τ)-Φ1(t2)]dτΔt∂tαF=+∫-∞t1-Δt‍DταG2(t1-τ)dτ[Φ1(t1)-Φ1(t2)]Δt∂tαF≡∑i=14‍𝔅i.
Changing the variable t2-τ=z in 𝔅1 and t1-τ=z in 𝔅2, we get
(107)𝔅i=∫02Δt‍DταG2(z)[Φ1(ti-z)-Φ1(ti)]dz,i=1,2.
Then, the property of the function Φ1(t) together with inequality (93) (where t≔2Δt) lead to
(108)|𝔅1|+|𝔅2|≤C∥Φ1∥C((1+θ)/2)α([0,T])(Δt)(1+θ)α/2.
To get the same estimate for the term 𝔅3, we apply the mean-value theorem to the difference [DταG2(t2-τ)-DταG2(t1-τ)] and inequality (96) and have after some simple calculations
(109)|𝔅3|≤C∥Φ1∥C((1+θ)/2)α([0,T])(Δt)(1+θ)α/2.
As for the estimate of 𝔅4, one follows immediately from (95) where δ≔Δt and properties of the function Φ1 are
(110)|𝔅4|≤const.(Δt)(1+θ)α/2|∫Δt+∞‍DταG2(τ)dτ||𝔅4|≤C∥Φ1∥C((1+θ)/2)α([0,T])(Δt)(1+θ)α/2.
Hence, inequalities (106)–(110) lead to estimate
(111)〈∂tαF〉t,[0,T](((1+θ)/2)α)≤C∥Φ1∥C((1+θ)/2)α([0,T])(Δt)(1+θ)α/2,
which completes the proof of Lemma 11.Due to results of this lemma and arguments like (70)–(74), we can infer that
(112)w(x,t)|∂DT+=F(t),∂tαw(x,t)|∂DT+=∂tαF(t),w(x,t)∈C((3+θ)/2)α(∂DT+).
Based on the results of Lemma 11, properties of the function G(x,t) (see Lemmas 7 and 8), and (112), we can get the next assertion.Lemma 12.
Let conditions (83) and (84) hold and let Φ1(t) meet the requirements of Lemma 11. Then the function w(x,t) represented by (87) satisfies (112) and
(113)∥w∥C((3+θ)/2)α(∂DT+)+∥∂tαw∥Cθ,(θ/2)α(D¯T+)≤C∥Φ1∥C((1+θ)/2)α([0,T]).Now, we can prove the solvability of problem (82).Theorem 13.
Letθ,α∈(0,1), and conditions (83) hold, and Φ0(x,t)∈Cθ,(θ/2)α(D¯T+), w0(x)∈C2+θ(D¯+), Φ1(t)∈C((1+θ)/2)α([0,T]) for any positive number T. Then there exists a unique solution w(x,t)∈C2+θ,((2+θ)/2)α(D¯T+) of problem (82) and
(114)∥w∥C2+θ,((2+θ)/2)α(D¯T+)+∥∂tαw∥C((1+θ)/2)α(∂DT+)≤C[∥Φ0∥Cθ,(θ/2)α(D¯T+)+∥Φ1∥C((1+θ)/2)α([0,T])+∥w0∥C2+θ(D¯+)∥Φ0∥Cθ,(θ/2)α(D¯T+)+∥Φ1∥C((1+θ)/2)α([0,T])].Proof.
To prove this theorem in the case of (84) it is enough to consider the following Dirichlet problem:
(115)∂tαw(x,t)-a0∂2w(x,t)∂x2=0inDT+,α∈(0,1);w(x,0)=0,x∈D+;w(x,t)⟶0ifx⟶+∞,∀t∈[0,T];w(0,t)=F(t),t∈[0,T],
and to apply the results of Theorem 9 and Lemma 12.
To remove restriction (84), we will look for the solution of problem (82) in the following form:
(116)w(x,t)=w1(x,t)+w2(x,t),
where w1(x,t) is the solution of the following Dirichlet problem:
(117)∂tαw1(x,t)-a0∂2w1(x,t)∂x2=Φ0(x,t)inDT+,α∈(0,1);w1(x,0)=w0(x),x∈D+;w1(x,t)⟶0ifx→+∞,∀t∈[0,T];w1(0,t)=0,t∈[0,T]
and the function w2(x,t) solves the following problem:
(118)∂tαw2(x,t)-a0∂2w2(x,t)∂x2=0inDT+,α∈(0,1);w2(x,t)⟶x→+∞0∀t∈[0,T]w2(x,0)=0,x∈D+;∂tαw2(0,t)-b0∂w2(0,t)∂x=Φ1(t)+b0∂w1(0,t)∂x≡Φ¯1(t),-b0∂w2(0,t)∂x=Φ1(t)+1b0∂w1(0,t)∂x≡Φ¯1(t),t∈[0,T].
The one-valued solvability of problem (117) follows from Theorem 9, which gives
(119)∥w1∥C2+θ,((2+θ)/2)α(D¯T+)+∥∂tαw1∥C((1+θ)/2)α(∂DT+)≤C[∥Φ0∥Cθ,(θ/2)α(D¯T+)+∥w0∥C2+θ(D¯+)],∂tαw1(0,t)=0,t∈[0,T].
These properties of the function w1(x,t) allow us to conclude that Φ¯1(t)∈C((1+θ)/2)([0,T]) and Φ¯1(0)=0; that is, Φ¯1(t) satisfies the conditions of Lemma 12, and the right-hand sides of (118) meet requirement (84). It means that there exists a unique solution w2∈C2+θ,((2+θ)/2)α(D¯T+) of (118) which satisfies to inequality (119) with Φ0(x,t)=w0(x)≡0.
Finally, returning to representation (116) for w(x,t) and using the described properties of wi(t,x), i=1,2, we complete the proof of Theorem 13.


## 3.1. The Solvability of the First Initial-Boundary Value Problem for the Subdiffusion Equation

We look for the functionv(x,t) by the following conditions:
(22)∂tαv(x,t)-a0∂2v(x,t)∂x2=g0(x,t)inDT+,α∈(0,1);(23)v(x,0)=v0(x),x∈D+;v(x,t)⟶0ifx⟶+∞,∀t∈[0,T];(24)v(0,t)=g1(t),t∈[0,T],
where g0(x,t) and g1(t),v0(x) are some given functions.We assume that the following conditions hold:(25)(i)v0(0)=g1(0);∂tαg1(0)=a0∂2v0∂x2+g0(x,0);(26)(ii)v0(x),g0(x,t)≡0,if|x|<R0∀t∈[0,T],
for some positive number R0.First, we will study problem (22)–(24) under restriction
(27)v0(x),g0(x,t)≡0.
We will search a solution of (22)–(24) under restriction (27) in the class C2+θ,((2+θ)/2)α(D¯T+), if g1(t)∈C((2+θ)/2)α([0,T]) for θ∈(0,1).Note that conditions (25) and (26) together with restriction (27) allow us to apply the Laplace transformation in t to the right hand sides of (22)–(24). Indeed, conditions (25)–(27) mean that the right hand sides in (22)–(24) except g1(t) equal zero and g1(0)=0. Thus, we can extend the right hand sides in (22)–(24) by 0 for t≤0 and save, for simplicity, the same notation for the extension of the function g1(t). Therefore, we can apply, at least formally, the Laplace transformation in t to (22)–(24) in the case of (25)–(27) hold.Denote byw^(x,p) the Laplace transformation of the function w(x,t); that is,
(28)w^(x,p)=∫0+∞e-ptw(x,t)dt.
The Laplace transformation in (22)–(24) leads to the problem
(29)pαv^(x,p)-a0∂2v^(x,p)∂x2=0,x∈D+;v^(x,p)⟶0ifx⟶+∞;v^(0,p)=g^1(p).
Here we used the following formula from [33]:
(30)∂tαw^=pαw^(x,p)-pα-1w(x,0).
One can easily check that the following function solves the equations in (29):
(31)v^(x,p)=exp(-xpα/2a0)g1(p).
Due to formula (2.30) in [34] and the inverse Laplace transformation, we get the integral representations of v(x,t) as follows:
(32)v(x,t)=∫0tG(x,t-τ)g1(τ)dτ,
where
(33)G(x,t)=t-1W(-xt-α/2a0;-α2,0).
Here W(z;β,γ) is the Wright function, which is defined for z,β,γ∈ℂ as (see formula (1.8.1(27)) in v.3 [35])
(34)W(z;β,γ)=∑k=0∞‍zkk!Γ(βk+γ).
The main properties of the Wright functions are described in Chapters 4.1, v.1 and 18.1 v3 in [35], Chapter 1.11 in [1], Chapter 1.3 in [34], and Chapter 2 in [21, 36].In Lemma6, we describe the properties of the kernel G(x,t) which will be necessary to estimate the function v(x,t). Its proof is represented in Appendix A.Lemma 6.
Letκ, δ, and θ¯ be some positive constants, κ=κ(α), θ¯∈(0,1], x∈D+, t∈[0,T]. Then one has the following.
(35)(i)ThekernelG(x,t)ispositiveforx>0.(36)(ii)∫0+∞G(x,t)dt=1.(37)(iii)tG(x,t)≤Cexp(-κ(xt-α/2)1/(1-α/2));|∂G(x,t)∂t|≤Ct-21+(xt-α/2)4/α;|∂G(x,t)∂x|≤Ct-1-α/21+(xt-α/2)1+2/α.(38)(iv)∫0tG(x,τ)τ(θ¯/2)αdτ≤Ct(θ¯/2)α.(39)(v)∫δ+∞|∂G(x,τ)∂x|τ(θ¯/2)αdτ≤Cδ-(1-θ¯)α/2.(40)(vi)∫δ+∞|∂G(x,τ)∂τ|τθ¯α/2dτ≤Cδ-1+θ¯α/2.(41)(vii)∫0t(t-τ)-αG(x,τ)dτ≤Ct-αexp(-κ(xt-(α/2))1/(1-α/2)).(42)(viii)DtαG(x,t)=a0∂2G(x,t)∂x2inDT+,whereDtα is the Riemann-Liouville fractional derivative, and its definition is in (2.1.8) in [1].Lemma 7.
Letθ∈(0,1), and conditions (25)–(27) hold, and g1(t)∈C((2+θ)/2)α([0,T]). Then the function v(x,t) represented by formula (32) satisfies the following inequalities:
(43)supD¯T+|∂tαv|≤C(sup[0,T]|∂tαg1|+〈∂tαg1〉t,[0,T]((θ/2)α));(44)〈∂tαv〉t,DT+((θ/2)α)≤C∥∂tαg1∥C(θ/2)α([0,T]);(45)〈∂tαv〉x,DT+(θ)≤C∥∂tαg1∥C(θ/2)α([0,T]).Proof.
First, we obtain the representation of∂tαv. To this end, we need the following properties of the fractional derivative (see Lemma 2.10 and formula (2.4.10) in [1]): (i)
(46)∂tαρ(t)=Dtαρ(t),ifρ(0)=0,where (see (2.1.8) in [1])
(47)Dtαρ(t)=1Γ(1-α)ddt∫0tρ(τ)(t-τ)αdτ,α∈(0,1).(ii)
If the functionsρi(t), i=1,2, and Dtαρ1(t) are bounded in [0,T], then
(48)Dtα∫0tρ1(t-τ)ρ2(τ)dτ=∫0tρ2(t-τ)Dταρ1(τ)dτ+ρ2(t)limz→0Iz1-αρ1(z),=∫0tρ2(t-τ)Dταρ1(τ)dτ+ρ2(t)limz→0∀t∈[0,T],whereItν is the Riemann-Liouville fractional integral of order ν (see, e.g. (2.1.1) in [1])
(49)Itνρ(t)=1Γ(ν)∫0t‍ρ(τ)(t-τ)1-νdτ,Reν>0.One can easily see that ∂tαg1(t)∈C(θ/2)α([0,T]) and ∂tαg1(0)=0 (see (25) and (27)). Then, using properties (46) and (48) and equality (36), we represent the function ∂tαv(x,t) as
(50)∂tαv(x,t)=∫0tG(x,t-τ)[∂tαg1(τ)-∂tαg1(t)]dτ+∂tαg1(t).
Namely, this representation will be useful below. To prove inequality (43), we will use this representation, statement (i) from Lemma 6 and the first estimate in (37) as follows:
(51)|∂tαv|≤〈∂tαg1〉t,[0,T]((θ/2)α)|∂tαv|≤×∫0tτ(θ/2)α-1exp(-κ(xτ-α/2)(1/(1-α/2)))dτ|∂tαv|≤+max[0,T]|∂tαg1||∂tαv|≤C∥∂tαg1∥C(θ/2)α([0,T]).
In view of ∂tαg1(t)=0, if t≤0, we can rewrite representation (50) as
(52)∂tαv(x,t)=∫-∞tG(x,t-τ)[∂tαg1(τ)-∂tαg1(t)]dτ+∂tαg1(t).
Letx1, x2∈D+ and x2>x1. Denote
(53)Δx≔x2-x1,Δx∂tαv≔∂tαv(x2,t)-∂tαv(x1,t).
Then,
(54)Δx∂tαv=∫0(Δx)2/αG(x2,τ)[∂ταg1(t-τ)-∂ταg1(t)]dτΔx∂tαv=+∫0(Δx)2/αG(x1,τ)[∂ταg1(t)-∂ταg1(t-τ)]dτΔx∂tαv=+∫(Δx)2/α+∞[G(x2,τ)-G(x1,τ)]o×[∂ταg1(t-τ)-∂ταg1(t)]dτ≡∑i=13‍ℑi.
Using inequality (38) with t≔(Δx)2/α, θ¯≔θ, we obtain
(55)|ℑ1|+|ℑ2|≤const.〈∂tαg1〉t,[0,T]((θ/2)α)(x2-x1)θ.
As for the term ℑ3, we apply the mean-value theorem to the difference [G(x2,τ)-G(x1,τ)] together with inequality (39) (where δ≔(Δx)2/α, θ¯≔θ) and deduce
(56)|ℑ3|≤const.〈∂tαg1〉t,[0,T]((θ/2)α)(Δx)1+(2/α)(-(α/2)+θ(α/2)).
Thus, representation (54) together with inequalities (55) and (56) prove the correctness of (44).
To complete the proof of Lemma7, we need to obtain inequality (45). Let t1,t2∈[0,T] and t2>t1. Denote
(57)Δt≔t2-t1,Δt∂tαv≔∂tαv(x,t2)-∂tαv(x,t1).
We analyze the difference
(58)Δt∂tαv=∫2t1-t2t1G(x,t1-τ)[∂tαg1(t1)-∂tαg1(τ)]dτΔt∂tαv=+∫2t1-t2t2G(x,t2-τ)[∂tαg1(τ)-∂tαg1(t2)]dτΔt∂tαv=+∫-∞2t1-t2G(x,t1-τ)dτ[∂tαg1(t1)-∂tαg1(t2)]Δt∂tαv=+∫-∞2t1-t2[∂tαg1(τ)-∂tαg1(t2)]Δt∂tαv=Δt∂tαv=×[G(x,t2-τ)-G(x,t1-τ)]dτΔt∂tαv=+[∂tαg1(t2)-∂tαg1(t1)]≡∑i=15‍𝔍i.
As for the last term in this sum, it is estimated by (Δt)θα/2〈∂tαg1〉t,[0,T](θ/2)α. We change the variable τ=t1-z in the term 𝔍1 and apply estimate (38) with t≔Δt, θ¯≔θ. Thus, we have
(59)|𝔍1|≤const.(Δt)θα/2〈∂tαg1〉t,[0,T]((θ/2)α).
In the same way, we evaluate the function 𝔍2. The estimate of the term 𝔍3 follows from the properties of the function ∂tαg1 and inequality (36). At last, the mean-value theorem together with estimate (40), where δ≔Δt, θ¯≔θ, lead to
(60)|𝔍4|≤C(Δt)θα/2〈∂tαg1〉t,[0,T]((θ/2)α).
Therefore, inequality (45) is deduced from (58)–(60).Now, based on the results of Lemmas6 and 7, we can infer the next assertion.Lemma 8.
Let conditions of Lemma7 hold. Then there exists a unique solution v(x,t)∈C2+θ,((2+θ)/2)α(D¯T+) of problem (22)–(24), which is represented with (32) and
(61)∥v∥C2+θ,((2+θ)/2)α(D¯T+)≤C∥g1∥C((2+θ)/2)α([0,T]).Proof.
First of all we obtain estimate (61). One can get the following inequality using the results of Lemma 7 and (22), where g0≡0:(62)∥∂2v∂x2∥Cθ,(θ/2)α(D¯T+)≤const.∥∂tαv∥Cθ,(θ/2)α(D¯T+)∥∂2v∂x2∥Cθ,(θ/2)α(D¯T+)≤C∥g1∥C((2+θ)/2)α([0,T]).
Next, we use formula (3.5.4) from [1] as follows:
(63)w(t)-w(0)=1Γ(α)∫0t∂ταw(τ)(t-τ)1-αdτ,
to evaluate the maximum of |v(x,t)|. Hence, (43) and (63) lead to inequality
(64)supD¯T+|v|≤const.supD¯T+|∂tαv|≤C∥g1∥C((2+θ)/2)α([0,T]);
here we use the fact that v(x,0)=0.
After that, the minor seminorms of the functionv(x,t) are estimated with the interpolation inequalities from Section 8.8 [37] and (43)–(45), (62), and (64). Therefore, the arguments above prove inequality (61) and the embedding v(x,t)∈C2+θ,((2+θ)/2)α(D¯T+).
Next, we show that the functionv(x,t) given by (32) satisfies (22). To this end, we use equality (46) and (48) and represent ∂tαv(x,t) as
(65)∂tαv(x,t)=∫0tg1(t-τ)DtαG(x,τ)dτ+g1(t)limτ→0Iτ1-αG(x,τ).
Then (41) leads to
(66)∂tαv(x,t)=∫0tg1(t-τ)DtαG(x,τ)dτ.
Next, due to property (42) of the function G(x,t), one can check that
(67)∂tαv(x,t)=a0∂2v∂x2.
As it follows from the first inequality in (37), the function v(x,t) represented by (32) satisfies the following conditions:
(68)limt→0v(x,t)=0;v(x,t)⟶0ifx⟶+∞.
Finally, it is necessary to show that the functionv(x,t) meets boundary condition (24). To this end, we observe the next difference (here we will essentially use statement (35))
(69)|v(x,t)-g1(t)|≤|∫x+∞G(x,τ)[g1(t-τ)-g1(t)]dτ|≤+|∫0xG(x,τ)[g1(t-τ)-g1(t)]dτ|≤2sup[0,T]|g1|∫x+∞G(x,τ)dτ+〈g1〉t,[0,T]((θ/2)α)≤×∫0xG(x,τ)ταθ/2dτ≡A1+A2.
Applying inequality (38) with t≔x, θ¯≔θ to the term A2, we get
(70)|A2|≤const.∥g1∥C((2+θ)/2)α([0,T])xαθ/2.
To estimate the term A1 in (69), we use the Wright formula (see [38] or (2.2.5) in [21]):
(71)-βzW(z;β,1-β)=W(z,β,0),
and rewrite the function G(x,t) as
(72)G(x,t)=α2a0xt-1-(α/2)W(-xa0t-α/2;-α2;1-α2).
After that, due to Lemmas 2.2.4 and 2.2.7 in [21] and representation (72), we have
(73)0<G(x,t)<const.xt-1-α/2.
Thus,
(74)|A1|≤const.sup[0,T]|g1|∫x+∞xτ-1-α/2dτ|A1|≤const.sup[0,T]|g1|x1-α/2.
Then, we joint estimates (69)–(74) and obtain
(75)|v(x,t)-g1(t)|≤const.∥g1∥C((θ/2)α)α([0,T])(x1-α/2+xαθ/2)⟶0≤const.11111i11111∥g1∥C((θ/2)α)α([0,T])ifx⟶0,
which means that limx→0v(x,t)=g1(t).
Therefore, as it was written above, the functionv(x,t) given by (32) is a solution of (22)–(24) in the case of (27). The uniqueness of this solution is proved like the arguments of Theorem 3.2 from [39].Now we remove restriction (27). To this end, it is enough to consider the Cauchy problem:
(76)∂tαv¯(x,t)-a0∂2v¯(x,t)∂x2=g¯0(x,t)∂tαv¯(x,t)inR1×(0,T),α∈(0,1);v¯(x,0)=v¯0(x),x∈R1;v¯(x,t)⟶0,if|x|⟶+∞,∀t≥0.
Here, g¯0(x,t) and v0(x) are extensions of the functions g0(x,t) and v0(x), correspondingly, onto x<0. These functions together with their corresponding derivatives have finite supports and
(77)∥g¯0∥Cθ,(θ/2)α(R1×[0,T])≤C∥g0∥Cθ,(θ/2)α(D¯T+),∥v¯0∥C2+θ(R1)≤C∥v0∥C2+θ(D¯+).
The results of Theorem 3.2 from [39] give the one-valued solvability of (76) and
(78)∥v¯∥C2+θ,((2+θ)/2)α(R1×[0,T])≤C(∥g¯0∥Cθ,(θ/2)α(D¯T+)+∥v¯0∥C2+θ(D¯+)),
or, due to inequalities (77),
(79)∥v¯∥C2+θ,((2+θ)/2)α(R1×[0,T])≤C(∥g0∥Cθ,(θ/2)α(D¯T+)+∥v0∥C2+θ(D¯+)).
After that, we will look for the solution v(x,t) of problem (22)–(24) as
(80)v(x,t)=v¯(x,t)+v¯¯(x,t),(x,t)∈D¯T+,
where v¯¯(x,t) satisfies conditions (22)–(24) with the new right-hand sides which meet requirements of Lemma 8. Hence, we can apply the results of Lemma 8 to the function v¯¯(x,t). This fact and the properties (see (77)) of the function v¯(x,t) allow us to obtain the next results.Theorem 9.
Letα,θ∈(0,1), and conditions (25) and (26) hold, and g0∈Cθ,(θ/2)α(D¯T+), v0∈C2+θ(D¯+), g1(t)∈C((2+θ)/2)α([0,T]) for any positive number T. Then there exists a unique solution v∈C2+θ,((2+θ)/2)α(D¯T+) of problem (22)–(24) and
(81)∥v∥C2+θ,((2+θ)/2)α(D¯T+)≤C(∥g1∥C((2+θ)/2)α([0,T])+∥g0∥Cθ,(θ/2)α(D¯T+)+∥v0∥C2+θ(D¯+)∥g1∥C((2+θ)/2)α([0,T])+∥g0∥Cθ,(θ/2)α(D¯T+)).


## 3.2. The Model Initial-Boundary Value Problem with a Fractional Dynamic Boundary Condition

Here we study the following problem: it is necessary to find the functionw(x,t) by the following conditions:
(82)∂tαw(x,t)-a0∂2w(x,t)∂x2=Φ0(x,t)inDT+,α∈(0,1),w(x,0)=w0(x),x∈D+;w(x,t)⟶0ifx⟶+∞,∀t∈[0,T],∂tαw(0,t)-b0∂w(0,t)∂x=Φ1(t),t∈[0,T],
where Φi,i=0,1,w0 are the given functions. Let the following conditions hold:
(83)b0∂w0(0)∂x+Φ1(0)=a0∂2w0(0)∂2x+Φ0(0,0);w0(x),Φ0(x,t)≡0,if|x|<R0∀t∈[0,T];
for some positive number R0.At the beginning, we assume that(84)w0(x),Φ0(x,t)≡0
and search a solution of (82) under this restriction in the class C2+θ,((2+θ)/2)α(D¯T+), if Φ1(t)∈C((1+θ)/2)α([0,T]), θ∈(0,1).Note that conditions (83) and (84) allow us to extend the right-hand sides in (82) by 0 for t≤0. We save, for simplicity, the same notation for the extended function Φ1(t).After the application of the Laplace transformation in timet to problem (82), we have the following:
(85)pαw^(x,p)-a0∂2w^(x,p)∂x2=0,x∈D+;w^(x,p)⟶0ifx⟶+∞;pαw^(0,p)-b0∂w^(0,p)∂x=Φ^1(p).
Here, we used again formula (30). Some simple calculations lead to the function
(86)w^(x,p)=Φ^1(p)pα/2(pα/2+b0a0-1/2)exp(-xa0-1/2pα/2)w^(x,p)≡F^(p)exp(-xa0-1/2pα/2),
which is the solution of problem (85). Due to formulas (2.30) in [34] and (1.80) in [29], we obtain, after applying of the inverse Laplace transformation to (86), that
(87)w(x,t)=∫0tG(x,t-τ)F(τ)dτ,
where the kernel G(x,t) is given by (33) and
(88)F(t)=∫0tG2(τ)Φ1(t-τ)dτ,G2(t)=tα-1Eα/2,α(-b0a0-1/2tα/2).
Here, Eγ,β(z) is the function of the Mittag-Leffler type, which is defined by the series expansion (see, e.g., (1.56) in [29] or (1.8.17) in [1])
(89)Eγ,β(z)=∑k=0∞‍zkΓ(γk+β),Reγ>0,β,z∈ℂ.
Note that this two-parameter function of the Mittag-Leffler type was in fact introduced by Agarwal [40].The functionG(x,t) has been studied in Section 3.1 (see Lemma 6). Thus, to describe the properties of the function w(x,t), we have to observe the function F(t). To this end, we will use the following properties of the kernel G2(t), which are proved in Appendix B.Lemma 10.
LetN, δ, and θ be some positive constants, θ∈(0,1); Itν and Dtν be the fractional Riemann-Liouville integral and derivative, correspondingly (their definitions are given in (49) and (47)). Then the following is true.
(90)(i)0≤G2(t)≤Ctα-11+tα/2,∫0tG2(τ)dτ≤Ctα.(91)(ii)limt→0It1-αG2(t)=1.(92)(iii)∫0+∞DταG2(τ)dτ=-1.(93)(iv)∫0t|DταG2(τ)|τ(1+θ)α/2dτ≤Ct(1+θ)α/2.(94)(v)∫N+∞|DταG2(τ)|dτ≤C.(95)(vi)|∫δ+∞DταG2(τ)dτ|≤1+C1+δα/2.(96)(vii)∫δ+∞|ddτDταG2(τ)|τ(1+θ)α/2dτ≤Cδ-1+α(θ+1)/2.Lemma 11.
LetΦ1(t) meet requirement (83) and let Φ1∈C((1+θ)/2)α([0,T]). Then the function F(t) represented by (88) belongs to C((3+θ)/2)α([0,T]), F(0)=0, and the following estimate holds:
(97)∥F∥C((3+θ)/2)α([0,T])≤C∥Φ1∥C((1+θ)/2)α([0,T]).Proof.
First of all, we evaluate the value ofmax[0,T]|F(t)|. To this end, we use the first inequality in (90). Thus, one has
(98)max[0,T]|F(t)|≤max[0,T]|Φ1(t)|∫0t|G2(τ)|dτ≤Cmax[0,T]|Φ1(t)|tα.
This inequality gives that
(99)F(0)=0.
Next, we obtain the representation of ∂tαF(t). Due to equality (99) and properties (46) and (48), we conclude that
(100)∂tαF(t)=DtαF(t)∂tαF(t)=Φ1(t)limt→0It1-αG2(t)+∫0tΦ1(t-τ)DταG2(τ)dτ.
Since Φ1(0)=0, we can rewrite the last equality as
(101)∂tαF(t)=Φ1(t)limt→0It1-αG2(t)+∫0+∞Φ1(t-τ)DταG2(τ)dτ,
or, applying (91) and (92), we have
(102)∂tαF(t)=∫0+∞[Φ1(t-τ)-Φ1(t)]DταG2(τ)dτ.
To estimate max[0,T]|∂tαF(t)|, we use representation (102) and get
(103)|∂tαF(t)|≤∫0Nτ(1+θ)α/2|DταG2(τ)|dτ〈Φ1〉t,[0,T]((1+θ/2)α)|∂tαF(t)|≤+2max[0,T]|Φ1(t)|∫N+∞|DταG2(τ)|dτ.
After that, we apply inequality (93) with t≔δ to the first term in (103) and (94) to the second and get
(104)max[0,T]|∂tαF(t)|≤C∥Φ1∥C((1+θ)/2)α([0,T]).
Finally, to complete the proof of Lemma 11 is necessary to estimate of 〈∂tαF〉t,[0,T](((1+θ)/2)α). Let t2,t1∈[0,T] and t2>t1. Denote
(105)Δt≔t2-t1,Δt∂tαF=∂tαF(t2)-∂tαF(t1).
Using formula (102), we represent the difference Δt∂tαF as
(106)Δt∂tαF=∫t1-Δtt2‍DταG2(t2-τ)[Φ1(τ)-Φ1(t2)]dτΔt∂tαF=+∫t1-Δtt1‍DταG2(t1-τ)[Φ1(t1)-Φ1(τ)]dτΔt∂tαF=+∫-∞t1-Δt‍[DταG2(t2-τ)-DταG2(t1-τ)]×[Φ1(τ)-Φ1(t2)]dτΔt∂tαF=+∫-∞t1-Δt‍DταG2(t1-τ)dτ[Φ1(t1)-Φ1(t2)]Δt∂tαF≡∑i=14‍𝔅i.
Changing the variable t2-τ=z in 𝔅1 and t1-τ=z in 𝔅2, we get
(107)𝔅i=∫02Δt‍DταG2(z)[Φ1(ti-z)-Φ1(ti)]dz,i=1,2.
Then, the property of the function Φ1(t) together with inequality (93) (where t≔2Δt) lead to
(108)|𝔅1|+|𝔅2|≤C∥Φ1∥C((1+θ)/2)α([0,T])(Δt)(1+θ)α/2.
To get the same estimate for the term 𝔅3, we apply the mean-value theorem to the difference [DταG2(t2-τ)-DταG2(t1-τ)] and inequality (96) and have after some simple calculations
(109)|𝔅3|≤C∥Φ1∥C((1+θ)/2)α([0,T])(Δt)(1+θ)α/2.
As for the estimate of 𝔅4, one follows immediately from (95) where δ≔Δt and properties of the function Φ1 are
(110)|𝔅4|≤const.(Δt)(1+θ)α/2|∫Δt+∞‍DταG2(τ)dτ||𝔅4|≤C∥Φ1∥C((1+θ)/2)α([0,T])(Δt)(1+θ)α/2.
Hence, inequalities (106)–(110) lead to estimate
(111)〈∂tαF〉t,[0,T](((1+θ)/2)α)≤C∥Φ1∥C((1+θ)/2)α([0,T])(Δt)(1+θ)α/2,
which completes the proof of Lemma 11.Due to results of this lemma and arguments like (70)–(74), we can infer that
(112)w(x,t)|∂DT+=F(t),∂tαw(x,t)|∂DT+=∂tαF(t),w(x,t)∈C((3+θ)/2)α(∂DT+).
Based on the results of Lemma 11, properties of the function G(x,t) (see Lemmas 7 and 8), and (112), we can get the next assertion.Lemma 12.
Let conditions (83) and (84) hold and let Φ1(t) meet the requirements of Lemma 11. Then the function w(x,t) represented by (87) satisfies (112) and
(113)∥w∥C((3+θ)/2)α(∂DT+)+∥∂tαw∥Cθ,(θ/2)α(D¯T+)≤C∥Φ1∥C((1+θ)/2)α([0,T]).Now, we can prove the solvability of problem (82).Theorem 13.
Letθ,α∈(0,1), and conditions (83) hold, and Φ0(x,t)∈Cθ,(θ/2)α(D¯T+), w0(x)∈C2+θ(D¯+), Φ1(t)∈C((1+θ)/2)α([0,T]) for any positive number T. Then there exists a unique solution w(x,t)∈C2+θ,((2+θ)/2)α(D¯T+) of problem (82) and
(114)∥w∥C2+θ,((2+θ)/2)α(D¯T+)+∥∂tαw∥C((1+θ)/2)α(∂DT+)≤C[∥Φ0∥Cθ,(θ/2)α(D¯T+)+∥Φ1∥C((1+θ)/2)α([0,T])+∥w0∥C2+θ(D¯+)∥Φ0∥Cθ,(θ/2)α(D¯T+)+∥Φ1∥C((1+θ)/2)α([0,T])].Proof.
To prove this theorem in the case of (84) it is enough to consider the following Dirichlet problem:
(115)∂tαw(x,t)-a0∂2w(x,t)∂x2=0inDT+,α∈(0,1);w(x,0)=0,x∈D+;w(x,t)⟶0ifx⟶+∞,∀t∈[0,T];w(0,t)=F(t),t∈[0,T],
and to apply the results of Theorem 9 and Lemma 12.
To remove restriction (84), we will look for the solution of problem (82) in the following form:
(116)w(x,t)=w1(x,t)+w2(x,t),
where w1(x,t) is the solution of the following Dirichlet problem:
(117)∂tαw1(x,t)-a0∂2w1(x,t)∂x2=Φ0(x,t)inDT+,α∈(0,1);w1(x,0)=w0(x),x∈D+;w1(x,t)⟶0ifx→+∞,∀t∈[0,T];w1(0,t)=0,t∈[0,T]
and the function w2(x,t) solves the following problem:
(118)∂tαw2(x,t)-a0∂2w2(x,t)∂x2=0inDT+,α∈(0,1);w2(x,t)⟶x→+∞0∀t∈[0,T]w2(x,0)=0,x∈D+;∂tαw2(0,t)-b0∂w2(0,t)∂x=Φ1(t)+b0∂w1(0,t)∂x≡Φ¯1(t),-b0∂w2(0,t)∂x=Φ1(t)+1b0∂w1(0,t)∂x≡Φ¯1(t),t∈[0,T].
The one-valued solvability of problem (117) follows from Theorem 9, which gives
(119)∥w1∥C2+θ,((2+θ)/2)α(D¯T+)+∥∂tαw1∥C((1+θ)/2)α(∂DT+)≤C[∥Φ0∥Cθ,(θ/2)α(D¯T+)+∥w0∥C2+θ(D¯+)],∂tαw1(0,t)=0,t∈[0,T].
These properties of the function w1(x,t) allow us to conclude that Φ¯1(t)∈C((1+θ)/2)([0,T]) and Φ¯1(0)=0; that is, Φ¯1(t) satisfies the conditions of Lemma 12, and the right-hand sides of (118) meet requirement (84). It means that there exists a unique solution w2∈C2+θ,((2+θ)/2)α(D¯T+) of (118) which satisfies to inequality (119) with Φ0(x,t)=w0(x)≡0.
Finally, returning to representation (116) for w(x,t) and using the described properties of wi(t,x), i=1,2, we complete the proof of Theorem 13.


## 4. The Proofs of Theorems3–5

Note that the proofs of Theorems4 and 5 are analogous to the one of Theorem 3 and use the technique from Chapter 4 [32] together with the results of the solvability to the model problems from Section 3. That is why, we represent here only the proof of Theorem 3. In this route, we will need the solvability in C2+θ,((2+θ)/2)α(Ω¯T) of the next initial-boundary value problem:
(120)∂tαU(x,t)-a0(x,t)∂2U(x,t)∂x2+a1(x,t)∂U(x,t)∂x∂tαU(x,t)+a2(x,t)U(x,t)=g0(x,t),∂tαU(x,t)a0(x,t)∂2U(x,t)∂x2(x,t)∈ΩT,α∈(0,1];U(x,t)=gi(t)onΓiT,i=1,2;U(x,0)=U0(x)inΩ.
Reaching this goal is enough to repeat the arguments of Section 3.3 from [39] and apply the results of Theorem 9. Thus, we can assert the following.Theorem 14.
Letθ,α∈(0,1) and the following conditions
(121)g1(0)=U0(a);∂tαg1(0)=a0(a,0)∂2U0(a)∂x2-a1(a,0)∂U0(a)∂x∂tαg1(0)=-a2(a,0)U0(a)+g0(a,0);g2(0)=U0(b);∂tαg2(0)=a0(b,0)∂2U0(b)∂x2-a1(b,0)∂U0(b)∂x∂tαg2(0)=-a2(b,0)U0(b)+g0(b,0)
hold. Let and g0(x,t)∈Cθ,(θ/2)α(Ω¯T), U0(x)∈C2+θ(Ω¯), gi(t)∈C((2+θ)/2)α([0,T]), i=1,2, for any positive number T. Then there exists a unique solution u(x,t) of problem (120): U(x,t)∈C2+θ,((2+θ)/2)α(Ω¯T), and
(122)∥U∥C2+θ,((2+θ)/2)α(Ω¯T)≤C[∑i=12‍∥g0∥Cθ,(θ/2)α(Ω¯T)+∥U0∥C2+θ(Ω¯)+∑i=12‍∥gi∥C((2+θ)/2)α([0,T])],
where a positive constant C depends only on the measure of Ω and ∥ai∥Cθ,(θ/2)α(Ω¯T), i=0,2¯.At the beginning, we prove Theorem9 in the case of t∈[0,τ] for some small τ<τ0. Then, we will show how the solution u(x,t) is extended from [0,τ] into [0,T], where T=τ+τ0, for all τ0>0.Lemma 15.
Let the conditions of Theorem3 hold. Then there exists a unique solution u(x,t)∈C2+θ,((2+θ)/2)α(Ω¯τ) of problem (1)–(5), for any τ∈(0,τ0], and
(123)∥u∥C2+θ,((2+θ)/2)α(Ω¯τ)+∥∂tαu∥C((1+θ)/2)α(Γ2τ)≤C[∥f0∥Cθ,(θ/2)α(Ω¯τ)+∥u0∥C2+θ(Ω¯)+∥f1∥C((1+θ)/2)α(Γ2τ)+∥f2∥C((2+θ)/2)α(Γ1τ)],
where the positive constant C depends on ∥ai∥Cθ,(θ/2)α(Ω¯τ), i=0,2¯, ∥bj∥C((1+θ)/2)α(Γ2τ), j=0,1, and the measure of Ω.Proof.
If the right-hand sides of (1)–(5) meet the following requirements:
(124)f0(x,t)∈C0θ,(θ/2)α(Ω¯τ),f1(t)∈C0((1+θ)/2)α(Γ2τ),f2(t)∈C0((2+θ)/2)α(Γ1τ),u0(x)≡0;
then by repeating the arguments of §4–§7 from Chapter 4 [32] and using the results of Theorems 9 and 13 and Theorem 3.2 from [39], we have proved the assertion of Lemma 15.
To remove restriction (124), we look for the solution of problem (1)–(5) as
(125)u(x,t)=U(x,t)+V(x,t),
where the function U(x,t) is the solution of problem (120) with
(126)g0(x,t)≔f0(x,t);U0(x)≔u0(x);g1(t)≔u0(a)+U1(a)tααΓ(α);g2(t)≔u0(b)+U1(b)tααΓ(α);U1(x)=f0(x,0)+a0(x,0)∂2u0(x)∂x2U1(x)=-a1(x,0)∂u0(x)∂x-a2(x,0)u0(x)
and the unknown function V(x,t) is the solution of the problem
(127)∂tαV(x,t)-a0(x,t)∂2V(x,t)∂x2+a1(x,t)∂V(x,t)∂x+a2(x,t)V(x,t)=0inΩτ;V(x,0)=0inΩ;V(x,t)=f2(t)-g1(t)V(x,t)≡f2*(t)onΓ1τ;∂tαV(x,t)+b0(t)∂V(x,t)∂x+b1(t)V(x,t)=f1(t)-∂tαg2(t)-b0(t)∂U(b,t)∂x-b1(t)g2(t)≡f1*(t)onΓ2τ.
Our further arguments are divided into two parts. In the first step, we will show that the right-hand sides of (126) meet the requirements of Theorem 14, which will ensure the existence of the unique solution U(x,t)∈C2+θ,((2+θ)/2)α(Ω¯τ). The next step deals with the proving of the following equalities:
(128)fi*(0)=0,i=1,2.
We obtain after simple calculations the following properties for the functionsgi(t), for all t∈[0,τ]:
(129)g1(0)=u0(a),g2(0)=u0(b);∂tαg1(t)=U1(a),∂tαg2(t)=U2(b);gi(t)∈C((2+θ)/2)α([0,τ]),i=1,2.
After that, using (129), one can easily check that the right-hand sides in (126) satisfy conditions of Theorem 14. That is why, there exists a unique solution U(x,t)∈C2+θ,((2+θ)/2)α(Ω¯τ) as follows as follows:
(130)∥U∥C2+θ,((2+θ)/2)α(Ω¯τ)≤C[∥f0∥Cθ,(θ/2)α(Ω¯τ)+∥u0∥C2+θ(Ω¯)],(131)∂U(b,0)∂x=∂u0(b)∂x,
where the positive constant C depends on ∥ai∥Cθ,(θ/2)α(Ω¯τ), i=0,2¯, and the measure of Ω.
Then we return to problem (127). Using properties (129) and (131), we get that
(132)f2*(0)=f2(0)-u0(a);f1*(0)=f1(0)-b0(0)∂u0(b)∂x-b1(0)u0(b)-f0(b,0)f1*(0)=+a2(b,0)u0(b)-a0(b,0)∂2u0(b)∂x2f1*(0)=+a1(b,0)∂u0(b)∂x.
After that, to get statement (128) is enough to apply condition (15) to the right-hand sides of equalities (132).
Finally, taking into account (129)–(131), we obtain the following statements:
(133)∥f1*∥C((1+θ)/2)α([0,τ])+∥f2*∥C((2+θ)/2)α([0,τ])≤C[∥f0∥Cθ,(θ/2)α(Ω¯τ)+∥u0∥C2+θ(Ω¯)+∥f1∥C((1+θ)/2)α([0,τ])+∥f2∥C((2+θ)/2)α([0,τ])],
where the positive constant C depends on ∥ai∥Cθ,(θ/2)α(Ω¯τ), i=0,2¯, ∥bj∥C((1+θ)/2)α([0,τ]), j=0,1, and the measure of Ω.
Therefore, the right-hand sides of (127) meet requirements (124). It means that there exists a unique solution V(x,t) of problem (127) and that V(x,t) satisfies to inequality (123) with the corresponding right-hand sides. This fact together with the properties of the function U(x,t) (see (130)) and representation (125) completes the proof of Lemma 15.Now we will extend the obtained solutionu(x,t),t∈[0,τ], in Lemma 15, into [τ,T]. In other words, we construct the function U¯(x,t)∈C2+θ,((2+θ)/2)α(Ω¯T), ∂tαU¯(x,t)∈C((1+θ)/2)α(Γ2T) as follows:
(134)∂tαU¯(x,t)-a0(x,t)∂2U¯(x,t)∂x2+a1(x,t)∂U¯(x,t)∂x+a2(x,t)U¯(x,t)=f0(x,t)inΩT;U¯(x,t)=f2(t)onΓ1T;∂tαU¯(x,t)+b0(t)∂U¯(x,t)∂x+b1(t)U¯(x,t)=f1(t)onΓ2T;U¯(x,t)=u(x,t),x∈Ω¯,t∈[0,τ];∂tαU¯(b,t)=∂tαu(b,t),t∈[0,τ],
where the function u(x,t) is constructed in Lemma 15; that is, the function u(x,t): u(x,t)∈C2+θ,((2+θ)/2)α(Ω¯τ), ∂tαu(x,t)∈C((1+θ)/2)α(Γ2τ), is the unique solution of (1)–(5) for t∈[0,τ], τ<τ0.First, we assume thatT=2τ and consider problem (120) with the following right-hand sides:
(135)g0(x,t)≔f0(x,t);U0(x)≔u0(x);g1(t)≔f2(t);g2(t)≔u0(b)+1Γ(α)∫0t‍τα-1g(t-τ)dτ,
where
(136)g(t)={∂tαu(b,t),ift∈[0,τ];∂tαu(b,τ),ift>τ.
After some calculations, we can infer that
(137)∂tαg2(t)=g(t);g2(0)=u0(b);∂tαg2(0)=f0(b,0)+a0(b,0)∂2u0(b)∂x2∂tαg2(0)=-a1(b,0)∂u0(b)∂x-a2(b,0)u0(b);∥g2∥C((3+θ)/2)α([0,2τ])≤C[∥∂tαu∥C((1+θ)/2)α(Γ2τ)+∥u0∥C2+θ(Ω¯)].
Due to the fact the functions f0,f2, and u0 satisfy the consistency conditions, properties (137) allow us to conclude that the right-hand sides (135) meet requirements of Theorem 14. This means that there exists a unique solution U(x,t)∈C2+θ,((2+θ)/2)α(Ω¯T), T=2τ, and that
(138)∥U∥C2+θ,((2+θ)/2)α(Ω¯T)≤C[∥f0∥Cθ,(θ/2)α(Ω¯T)+∥u0∥C2+θ(Ω¯)+∥f2∥C((2+θ)/2)α([0,T])+∥∂tαu∥C((1+θ)/2)α(Γ2τ)],∂tαU=g(t)onΓ2T;∂tαU|Γ2T∈C((1+θ)/2)α([0,T]);U(x,t)=u(x,t)inΩ¯τ.
After that, we search the solution of problem (134) in the following form:
(139)U¯(x,t)=U(x,t)+Θ(x,t),
where the new unknown function Θ(x,t) satisfies the following conditions:
(140)∂tαΘ(x,t)-a0(x,t)∂2Θ(x,t)∂x2+a1(x,t)∂Θ(x,t)∂x+a2(x,t)Θ(x,t)=0inΩ2τ;∂tαΘ(b,t)+b0(t)∂Θ(b,t)∂x+b1(t)Θ(b,t)=f1(t)-g(t)-b0(t)∂U(b,t)∂x-b1(t)U(b,t)≡Φ2(t),Θ(a,t)=0,t∈[0,2τ];Θ(x,t)=u(x,t)-U(x,t)≡Φ3(x,t)inΩ¯τ;∂tαΘ(b,t)=∂tαu(b,t)-g(t)≡Φ4(t),t∈[0,τ].
As it follows from (136) and (138), if t∈[0,τ] then
(141)Φ3(x,t)=0,x∈Ω¯,Φ2(t)=Φ4(t)=0,∥Φ2∥C((1+θ)/2)α([0,2τ])≤C[∥f0∥Cθ,(θ/2)α(Ω¯2τ)+∥u0∥C2+θ(Ω¯)+∥f2∥C((2+θ)/2)α([0,2τ])+∥f1∥C((1+θ)/2)α([0,2τ])].In problem (140), we introduce the new variable: σ=t-τ. Let
(142)Θ¯(x,σ)≔Θ(x,σ+τ);a¯i(x,σ)≔ai(x,σ+τ),i=0,2¯b¯j(x,σ)≔bj(x,σ+τ),j=0,1;Φ¯2(σ)≔Φ2(σ+τ),
where σ∈[-τ,τ] and ai¯, b¯j satisfy the conditions of Theorem 3. Due to (141), we have
(143)Φ¯2(σ)=0,ifσ∈[-τ,0],(144)Φ¯2(σ)∈C((1+θ)/2)α([-τ,τ]),
and therefore
(145)Θ¯(x,σ)=0,ifσ∈[-τ,0].To rewrite problem (140) in the new variable σ, we use formula (3.110) from [39] as follows:
(146)∂tαΘ(x,t)=∂σαΘ¯(x,σ).
Thus, problem (140) in the new variable can be rewritten as
(147)∂σαΘ¯(x,σ)-a¯0(x,σ)∂2Θ¯(x,σ)∂x2+a¯1(x,σ)∂Θ¯(x,σ)∂x+a¯2(x,σ)Θ¯(x,σ)=0inΩτ;∂σαΘ¯(b,σ)+b¯0(σ)∂Θ¯(b,σ)∂x+b¯1(σ)Θ¯(b,σ)=Φ¯2(σ),σ∈[0,τ];Θ¯(a,σ)=0,σ∈[0,τ];Θ¯(x,0)=∂σαΘ(b,0)=0inΩ¯.
Then we apply Lemma 15 to problem (147) and get the one-to-one solvability in C2+Θ,((2+Θ)/2)α(Ω¯τ), and the estimate like (123) holds. Returning to the old variable t=σ+τ, we obtain the unique solution of problem (140), Θ(x,t)∈C2+θ,((2+θ)/2)α(Ω¯2τ), and Θ(x,t) satisfies inequality (123), where τ≔2τ.Thus, representation (139) together with the properties (138) for the functions U(x,t) and the corresponding properties of Θ(x,t) allow us to construct the solution of problem (1)–(5) for t∈[τ,2τ], if there is the solution for t∈[0,τ]. Repeating this procedure, we prove the existence of the solution for t∈[0,T] and obtain estimate (19). As for the uniqueness of the solution, it follows from inequality (19). This ends the proof of Theorem 3.


## 5. The Local Solvability of the Nonlinear Problem with a Fractional Dynamic Boundary Condition

In this section, we indicate how our results may be applied to the nonlinear problem:(148)∂tαv-a0(v,vx)vxx=a1(v,vx),∂tαv1-(x,t)∈ΩT,α∈(0,1);v(x,0)=u0(x)inΩ;∂tαv-b0i(v)vx=b1j(v)onΓjT,j=1,2.
We require the following conditions on the functions ai,  bkj, k, i=0,1; j=1,2:  (i)
There exist some positive constantsδl,l=0,7¯ such that
(149)0<δ0<a0(v,vx)<δ1;0<δ2<b01(v)<δ3;-δ4<b02(v)<-δ5<0;|b1j|<δ6,j=1,2;|a1(v,vx)|≤δ7.(ii)
Consistency conditions hold:(150)a0(u0,u0x)u0xx+a1(u0,u0x)=b0j(u0)u0x+b1j(u0)onΓjT,j=1,2.(iii)
There exist∂ai(v,vx)/∂v, ∂ai(v,vx)/∂vx, ∂bkj(v)/∂v, k,i=0,1, j=1,2, such that for some positive constant L(151)∑i=01‍[|∂ai(v,vx)∂v-∂ai(u,ux)∂u|+|∂ai(v,vx)∂vx-∂ai(u,ux)∂ux|]≤L[|v-u|+|vx-ux|];∑j=12‍|∂b0j(v)∂v-∂b0j(u)∂u|+|∂b1j(v)∂v-∂b1j(u)∂u|≤L|v-u|,for any bounded functionsu,v together with their derivatives.(iv)
There are positive constantsAi,Bj,  i,j=0,3¯ as follows:
(152)|∂a0(v,vx)∂v|≤A0;|∂a1(v,vx)∂v|≤A1;|∂a0(v,vx)∂vx|≤A2;|∂a1(v,vx)∂vx|≤A3;|∂b01(v)∂v|≤B0;|∂b02(v)∂v|≤B1;|∂b11(v)∂v|≤B2;|∂b12(v)∂v|≤B3;(v)
(153)u0(x)∈C2+θ(Ω¯).We introduce the functional spaces ℍ and ℍ0 as follows:
(154)ℍ=C2+θ,((2+θ)/2)α(Ω¯T)×C((1+θ)/2)α(Γ1T)×C((1+θ)/2)α(Γ2T)ℍ0=Cθ,(θ/2)α(Ω¯T)×C((1+θ)/2)α(Γ1T)×C((1+θ)/2)α(Γ2T),
and for any elements z=(v,∂tαv∣∂ΩT)∈ℍ, and 𝔉(z)=(Φ0,Φ1,Φ2)∈ℍ0, we denote
(155)∥z∥ℍ≔∥v∥C2+θ,((2+θ)/2)α(Ω¯T)+∥∂tαv∥C((1+θ)/2)α(Γ1T)∥z∥ℍ≔+∥∂tαv∥C((1+θ)/2)α(Γ2T);∥𝔉(z)∥ℍ0≔∥Φ0∥Cθ,(θ/2)α(Ω¯T)+∥Φ1∥C((1+θ)/2)α(Γ1T)∥𝔉(z)∥ℍ0≔+∥Φ2∥C((1+θ)/2)α(Γ2T).LetBr(v)={v∈ℍ:∥v∥ℍ≤r} and Br0(v)={v∈ℍ:∥v∥ℍ≤r,v(x,0)=0}, r<r0, be balls of radius r in the space ℍ, centered at the origin, for some positive r0 to be determined later on.The simple calculations lead to the following assertion.Proposition 16.
Let conditions (149)–(153) hold and the function u∈Br(u); then ai(u,ux), ∂ai(u,ux)/∂u, ∂ai(u,ux)/∂ux∈Cθ,(θ/2)(Ω¯T); bkj(u), ∂bkj/∂u∈C((1+θ)/2)(ΓjT); i,k=0,1, j=1,2, and
(156)∥a0∥Cθ,(θ/2)α(Ω¯T)≤max{δ0,δ1}+[A0+A2]∥u∥ℍ[1+Tα/2+T((2-θ)/2)α];∥a1∥Cθ,(θ/2)α(Ω¯T)≤δ7+[A1+A3]∥u∥ℍ[1+Tα/2+T((2-θ)/2)α];∑j=12‍∥b0j∥C((1+θ)/2)α(ΓjT)≤max{δ2,δ3}+max{δ4,δ5}≤+[B0+B1]∥u∥ℍ[Tα+T((1+θ)/2)α];∥b11∥C((1+θ)/2)α(Γ1T)+∥b12∥C((1+θ)/2)α(Γ2T)≤δ6+[B3+B2]∥u∥ℍ[Tα+T((1+θ)/2)α];∥∂a0∂u∥Cθ,(θ/2)α(Ω¯T)+∥∂a0∂ux∥Cθ,(θ/2)α(Ω¯T)≤2L∥u∥ℍ[1+T((2-θ)/2)α+Tα/2]+A0+A2;∥∂a1∂u∥Cθ,(θ/2)α(Ω¯T)+∥∂a1∂ux∥Cθ,(θ/2)α(Ω¯T)≤2L∥u∥ℍ[1+T((2-θ)/2)α+Tα/2]+A1+A3;∥∂b01∂u∥C((1+θ)/2)α(Γ1T)+∥∂b02∂u∥C((1+θ)/2)α(Γ2T)≤2L∥u∥ℍ[1+T((1-θ)/2)α]+B0+B1;∥∂b11∂u∥C((1+θ)/2)α(Γ1T)+∥∂b12∂u∥C((1+θ)/2)α(Γ2T)≤2L∥u∥ℍ[1+T((1-θ)/2)α]+B3+B2.
Moreover, for the functions pσ(u,v)≔σu+(1-σ)v and qσ(u,v)≔σux+(1-σ)vx, for all σ∈[0,1], u,v∈Br(u), the following inequalities hold:
(157)|∂ai(pσ(u,v),vx)∂pσ(u,v)|≤LmaxΩ¯T|u|+maxx,t,v,vx|∂ai(v,vx)∂v|;|∂ai(v,qσ(u,v))∂qσ(u,v)|≤LmaxΩ¯T|ux|+maxx,t,v,vx|∂ai(v,vx)∂vx|;|∂bij(pσ(u,v))∂pσ(u,v)|≤LmaxΩ¯T|u|+maxx,t,v|∂bij(v)∂v|,≤LmaxΩ¯T|u|+maxx,t,v|∂bij(v)∂v|,i=0,1;j=1,2.First, we linearize problem (148) on the initial data and represent one as a system 𝔄z=𝔗(z), where 𝔄 is a linear operator and 𝔗(z) is a nonlinear perturbation. To this end, we introduce the new unknown function as follows:
(158)w(x,t)=u(x,t)-V0(x,t),
where the function V0(x,t) is a solution of the following problem:
(159)∂tαV0-a0(u0,u0x)V0xx=a1(u0,u0x),i(x,t)∈ΩT,α∈(0,1);V(x,0)=u0(x)inΩ;∂tαV0-b0i(u0)V0x=b1j(u0)onΓjT,j=1,2.
By Theorem 5 and Proposition 16, there exists a unique solution V0(x,t) of the problem (159) and
(160)∥V0∥ℍ≤C0∥u0∥C2+θ(Ω¯),
where the positive constant C0 depends on δi, i=0,7¯, Bk,Aj, k,j=0,3¯, L.Next, we rewrite problem (148) in terms of the function w(x,t) and after some calculations get the problem in the form:
(161)∂tαw-a0(V0,V0x)wxx=Φ0(w,wx,wxx),∂tαw-a0(V0,V0x)(x,t)∈ΩT,α∈(0,1);w(x,0)=0inΩ;∂tαw-b0i(V0)wx=Φ1j(w,wx)onΓjT,j=1,2,
where
(162)Φ0(w,wx,wxx)=f0(x,t)+[a0(V0+w,V0x+wx)-a0(V0,V0x)]=×[wxx+V0xx]=+[a1(V0+w,V0x+wx)-a1(V0,V0x)],(163)f0(x,t)=[a0(V0,V0x)-a0(u0,u0x)]V0xxf0(x,t)=+[a1(V0,V0x)-a1(u0,u0x)],(164)Φ1j(w,wx)=f1j(x,t)Φ1j(w,wx)=+[b0j(V0+w)-b0j(V0)][wx+V0x]Φ1j(w,wx)=+[b1j(V0+w)-b1j(V0)],(165)f1j(x,t)=[b0j(V0)-b0j(u0)]V0x+[b1j(V0)-b1j(u0)].Thus, we represent nonlinear problem (148) as
(166)𝔄w=𝔗(w).
Note that if we froze the functional arguments in the functions Φ0(w,wx,wxx) and Φ1j(w,wx), j=1,2, then problem (161) will be a linear problem with variable coefficients, which has been studied in detail in Sections 2–4. By Theorem 5, 𝔄 has a bounded inverse 𝔄-1, so that
(167)w=𝔄-1𝔗(w)≡𝔖(w),
and 𝔖 is a nonlinear operator. We will show that 𝔖 is a contraction operator.Lemma 17.
The following inequalities hold for the right-hand sides of problem (167):
(168)∥𝔗(0)∥ℍ0≤C1(T),(169)∥𝔗(w)-𝔗(v)∥ℍ0≤C2(T,r)∥w-v∥ℍ,≤C2(T,r)1∥w-v∥ℍℍ0,∀w,v,∈Br0(u),
with C1(T)→0C2(T,r)→0 as r,T→0.Proof.
First we prove (168). Note that
(170)∥𝔗(0)∥ℍ0=∥f0∥Cθ,(θ/2)α(Ω¯T)+∑j=12‍∥f1j∥C((1+θ)/2)α(ΓjT).
Let
(171)pσ≔pσ(z1,z2)=σz1(x,t)+(1-σ)z2(x,t);qσ≔qσ(z1,z2)qσ:=σz1x(x,t)+(1-σ)z2x(x,t);∀zi(x,t)∈Br(u),i=1,2,σ∈[0,1].
Using definitions (163), (165) and notations (171) with z1≔V0(x,t) and z2≔u0(x), we have
(172)f0(x,t)=V0xx∫01‍∂a0(pσ,V0x)∂pσdσ[V0(x,t)-u0(x)]f0(x,t)=+V0xx∫01‍∂a0(u0,qσ)∂qσdσ[V0x(x,t)-u0x(x)]f0(x,t)=+∫01‍∂a1(pσ,V0x)∂pσdσ[V0(x,t)-u0(x)]f0(x,t)=+∫01‍∂a1(u0,qσ)∂qσdσ[V0x(x,t)-u0x(x)];f1j(x,t)=V0x∫01‍∂b0j(pσ)∂pσdσ[V0(x,t)-u0(x)]f1j(x,t)=+∫01‍∂b1j(pσ)∂pσdσ[V0(x,t)-u0(x)].
It is easy to examine that the following are true:
(173)|V0(x,t)-u0(x)|≤Tα∥V0∥ℍ;|V0x(x,t)-u0x(x)|≤T((1+θ)/2)α∥V0∥ℍ;〈V0-u0〉x,ΩT(θ)≤T((1+θ)/2)α∥V0∥ℍ;〈V0x-u0x〉x,ΩT(θ)≤T(θ/2)α∥V0∥ℍ;〈V0-u0〉t,ΩT((θ/2)α)≤T((2-θ)/2)α∥V0∥ℍ;〈V0x-u0x〉t,ΩT((θ/2)α)≤Tα/2∥V0∥ℍ.
Hence, using inequalities (173), representation (172), and results of Proposition 16, we get after some tedious calculations
(174)∥f0∥Cθ,(θ/2)α(Ω¯T)≤C1Tθα/2[1+∑i=03‍Ai]≤×[1+L∥u0∥C2+θ(Ω¯)(1+Tα/2+T(2-θ)α/2)]2;∑j=12‍∥f1j∥C((1+θ)/2)α(ΓjT)≤C1max{Tθ/α2,T(1-θ)α/2}[1+∑i=03‍Bi]≤×[1+L∥u0∥C2+θ(Ω¯)(1+T(1-θ)α/2)]2,C1=max{1,C0},
where the constant C0 is from (160).
Let(175)C1(T)≔C1max{Tθ/α2,T(1-θ)α/2}C1(T)≔×[1+∑i=03‍(Bi+Ai)]C1(T)≔×[1+L∥u0∥C2+θ(Ω¯)(1+Tθ/α2+T(2-θ)α/2)]2.
It is easy to see that C1(T)→0, as T→0. Thus inequalities (174) together with representation (170) lead to estimate (168), where C1(T) is given by (175).
Next, we will obtain inequality (169). Note that, as it follows from definitions (162) and (164), one has
(176)∥𝔗(w)-𝔗(v)∥ℍ0=∥[Φ0(w,wx,wxx)-f0]=i-[Φ0(v,vx,vxx)-f0]∥Cθ,(θ/2)α(Ω¯T)=+∑j=12‍∥[Φ1j(w,wx)-f1j]=iiiiiiiii-[Φ1j(v,vx)-f1j]∥C((1+θ)/2)α(ΓjT).
Denote(177)lσ=σwxx(x,t)+(1-σ)vxx(x,t),σ∈[0,1];Q(z,zx,zxx)≔[a0(V0+z,V0x+zx)-a0(V0,V0x)][zxx+V0xx];Q1j(z,zx)≔[b0j(V0+z)-b0j(V0)][zx+V0x],∀z∈Br0(u).
Then, we have the following:(178)[Φ0(w,wx,wxx)-f0]-[Φ0(v,vx,vxx)-f0]=∫01‍∂Q(pσ,wx,wxx)∂pσdσ(w-v)=+∫01‍∂Q(v,qσ,wxx)∂qσdσ(wx-vx)=+∫01‍∂Q(v,vx,lσ)∂lσdσ(wxx-vxx)+(w-v)=×∫01‍∂a1(pσ,V0x+wx)∂pσdσ=+∫01‍∂a1(V0+v,qσ)∂qσdσ(wx-vx);[Φ1j(w,wx)-f1j]-[Φ1j(v,vx)-f1j]=∫01‍∂Q1j(pσ,wx)∂pσdσ(w-v)=+∫01‍∂Q1j(v,qσ)∂qσdσ(wx-vx)=+∫01‍∂b1j(pσ)∂pσdσ(w-v).
After that, the simple calculations together with the results of Proposition16 allow us to get the following for any function z(x,t)∈Br0(u). 							(179)(i)∥∂Q(z,zx,zxx)∂z∥Cθ,(θ/2)α(Ω¯T)+∥∂Q(z,zx,zxx)∂zx∥Cθ,(θ/2)α(Ω¯T)+∥∂Q11(z,zx)∂z∥C((1+θ)/2)α(Γ1T)+∥∂Q12(z,zx)∂z∥C((1+θ)/2)α(Γ2T)≤[∥z∥ℍ+C0∥u0∥C2+θ(Ω¯)]≤×[2L∥z∥ℍ+∑j=01‍Bj+A0+A2].(180)(ii)maxΩ¯T|∂Q(z,zx,zxx)∂zxx|≤T(1+θ)α/2∥z∥ℍ[2L∥z∥ℍ+A0+A2][1+Tα/2];〈∂Q(z,zx,zxx)∂zxx〉t,ΩT((θ/2)α)≤Tα/2∥z∥ℍ[2L∥z∥ℍ+A0+A2]≤×[1+Tθα/2+T(1-θ)α/2];〈∂Q(z,zx,zxx)∂zxx〉x,ΩT(θ)≤T(1+θ)α/2∥z∥ℍ[2L∥z∥ℍ+A0+A2]≤×[1+T(1-θ)α/2].(181)(iii)maxΩ¯T|∂Q1j(z,zx)∂zx|≤Tα∥z∥ℍ[L∥z∥ℍ+∑j=01‍Bj];〈∂Q1j(z,zx)∂zx〉t,ΩT(((1+θ)/2)α)≤T(1-θ)α/2[1+T(1+θ)α/2][L∥z∥ℍ+∑j=01‍Bj];〈∂Q1j(z,zx)∂zx〉x,ΩT(θ)≤T(1+θ)α/2∥z∥ℍ[L∥z∥ℍ+∑j=01‍Bj][1+T(1-θ)α/2].
Moreover, due tov,w∈Br0(u), we have the following:
(182)T-αmaxΩ¯T|v-w|+T-(1+θ)α/2maxΩ¯T|vx-wx|+T-θα/2maxΩ¯T|vxx-wxx|≤∥w-v∥ℍ;T-(1+θ)α/2〈v-w〉x,ΩT(θ)+T-θα/2〈vx-wx〉x,ΩT(θ)≤∥w-v∥ℍ;T-((2-θ)/2)α〈v-w〉t,ΩT((θ/2)α)+T-α/2〈vx-wx〉t,ΩT((θ/2)α)≤∥w-v∥ℍ.
Thus, applying inequalities from Proposition 16 and (179)–(182) to the right-hand sides in (178), we get (169), where
(183)C2(T,r)=C1max{Tθα/2,T(1-θ)α/2}C2(T,r)=×[Lr+∑i=03‍(Bi+Ai)][1+r+∥u0∥C2+θ(Ω¯)]C2(T,r)=×[1+T3α/2+T(3+θ)α/2+Tα+αθ].
It is obviously C2(T,r)→0 as r,T→0.As it follows from Lemma17, for sufficiently small T and r0, the nonlinear operator 𝔖 satisfies the conditions of the fixed point theorem for a contraction operator. Hence, we have proved the following theorem.Theorem 18.
Let conditions (149)–(153) hold. Then there exists a unique solution of problem (148) for small interval 0<t<T, such that v∈C2+θ,((2+θ)/2)α(Ω¯T) and that ∂tαv∈C((1+θ)/2)α(ΓjT), j=1,2.Using the analogous arguments, it is possible to assert the following results.Remark 19.
If we exchange the dynamic boundary condition onΓ1T by either (5) or (6) and assume that the corresponding consistency conditions hold, then the results of Theorem 18 save.Remark 20.
The results of Theorem18 hold if conditions (152) are changed by
(184)|∂a0(v,vx)∂v|≤A0[1+|v|+|vx|];|∂a1(v,vx)∂v|≤A1[1+|v|+|vx|];|∂a0(v,vx)∂vx|≤A2[1+|v|+|vx|];|∂a1(v,vx)∂vx|≤A3[1+|v|+|vx|];|∂b01(v)∂v|≤B0[1+|v|];|∂b02(v)∂v|≤B1[1+|v|];|∂b11(v)∂v|≤B2[1+|v|];|∂b12(v)∂v|≤B3[1+|v|]
and constants δ6 and δ7 in requirements (149) are replaced by δ6[1+|v|] and δ7[1+|v|+|vx|], correspondingly.



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*Source: 796430-2013-11-07.xml*