Solvability in Holder Space of an Initial Boundary Value Problem for the Time-Fractional Diffusion
In this paper we consider an initial boundary value problem for the timefractional diffusion equation with mixed boundary conditions. We prove a theorem of existence and uniqueness of solution to this problem in Hölder spaces.
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| Цитувати: | Solvability in Holder Space of an Initial Boundary Value Problem for the Time-Fractional Diffusion / M.V. Krasnoschok // Журнал математической физики, анализа, геометрии. — 2016. — Т. 12, № 1. — С. 48-77. — Бібліогр.: 24 назв. — англ. . |
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irk-123456789-1405472018-07-11T01:23:01Z Solvability in Holder Space of an Initial Boundary Value Problem for the Time-Fractional Diffusion Krasnoschok, M.V. In this paper we consider an initial boundary value problem for the timefractional diffusion equation with mixed boundary conditions. We prove a theorem of existence and uniqueness of solution to this problem in Hölder spaces. Рассмотрена начально-краевая задача для уравнения диффузии с дробной производной по времени со смешанными граничными условиями. Доказана теорема существования и единственности решения этой задачи в пространствах Гёльдера. 2016 Article Solvability in Holder Space of an Initial Boundary Value Problem for the Time-Fractional Diffusion / M.V. Krasnoschok // Журнал математической физики, анализа, геометрии. — 2016. — Т. 12, № 1. — С. 48-77. — Бібліогр.: 24 назв. — англ. . 1812-9471 DOI: doi.org/10.15407/mag12.01.048 Mathematics Subject Classification 2000: 35B65 (primary); 26A33 (secondary) http://dspace.nbuv.gov.ua/handle/123456789/140547 en Журнал математической физики, анализа, геометрии Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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In this paper we consider an initial boundary value problem for the timefractional diffusion equation with mixed boundary conditions. We prove a theorem of existence and uniqueness of solution to this problem in Hölder spaces. |
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Article |
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Krasnoschok, M.V. |
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Krasnoschok, M.V. Solvability in Holder Space of an Initial Boundary Value Problem for the Time-Fractional Diffusion Журнал математической физики, анализа, геометрии |
| author_facet |
Krasnoschok, M.V. |
| author_sort |
Krasnoschok, M.V. |
| title |
Solvability in Holder Space of an Initial Boundary Value Problem for the Time-Fractional Diffusion |
| title_short |
Solvability in Holder Space of an Initial Boundary Value Problem for the Time-Fractional Diffusion |
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Solvability in Holder Space of an Initial Boundary Value Problem for the Time-Fractional Diffusion |
| title_fullStr |
Solvability in Holder Space of an Initial Boundary Value Problem for the Time-Fractional Diffusion |
| title_full_unstemmed |
Solvability in Holder Space of an Initial Boundary Value Problem for the Time-Fractional Diffusion |
| title_sort |
solvability in holder space of an initial boundary value problem for the time-fractional diffusion |
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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2016 |
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http://dspace.nbuv.gov.ua/handle/123456789/140547 |
| citation_txt |
Solvability in Holder Space of an Initial Boundary Value Problem for the Time-Fractional Diffusion / M.V. Krasnoschok // Журнал математической физики, анализа, геометрии. — 2016. — Т. 12, № 1. — С. 48-77. — Бібліогр.: 24 назв. — англ.
. |
| series |
Журнал математической физики, анализа, геометрии |
| work_keys_str_mv |
AT krasnoschokmv solvabilityinholderspaceofaninitialboundaryvalueproblemforthetimefractionaldiffusion |
| first_indexed |
2025-07-10T10:42:09Z |
| last_indexed |
2025-07-10T10:42:09Z |
| _version_ |
1837256286684577792 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2016, vol. 12, No. 1, pp. 48–77
Solvability in Hölder Space of an Initial Boundary
Value Problem for the Time-Fractional Diffusion
Equation
M.V. Krasnoschok
Institute of Applied Mathematics and Mechanics
National Academy of Sciences of Ukraine
1 Dobrovol’skogo Str., Slov’yansk 84100, Ukraine
E-mail: iamm012@ukr.net
Received October 13, 2014, revised April 20, 2015,
published online December 4, 2015
In this paper we consider an initial boundary value problem for the time-
fractional diffusion equation with mixed boundary conditions. We prove a
theorem of existence and uniqueness of solution to this problem in Hölder
spaces.
Key words: fractional-order diffusion, Hölder spaces, Caputo derivative.
Mathematics Subject Classification 2010: 35B65 (primary); 26A33 (se-
condary).
1. Introduction
Let ∂αt (0 < α < 1) be a regularized fractional derivative (the Caputo deriva-
tive), that is,
∂αt u(x, t) =
1
Γ(1− α)
∂
∂t
t∫
0
(t− τ)−αu(x, τ)dτ − t−αu(x, 0)
. (1.1)
Let Ω be a doubly connected bounded domain in Rn with the boundary
∂Ω = Σ1∪Σ2 (Σ1∩Σ2 = ∅). Denote ΩT = Ω× (0, T ), Σi
T = Σi× (0, T ), i = 1, 2,
T > 0.
We need to find the function u(x, t) satisfying the equation
L
(
x, t,
∂
∂x
, ∂αt
)
u = f(x, t), in ΩT , (1.2)
c© M.V. Krasnoschok, 2016
Solvability in Hölder Space of an Initial Boundary Value Problem
with the initial and boundary conditions
u|t=0 = u0(x), x ∈ Ω, (1.3)
u|Σ1 = ψ1(x, t), B
(
x, t,
∂
∂x
)
u|Σ2 = ψ2(x, t), t ∈ (0, T ), (1.4)
where
L
(
x, t,
∂
∂x
, ∂αt
)
= ∂αt −
n∑
i,j=1
aij(x, t)
∂2
∂xi∂xj
−
n∑
i=1
ai(x, t)
∂
∂xi
− a0(x, t), (1.5)
B
(
x, t,
∂
∂x
)
=
n∑
i=1
bi(x, t)
∂
∂xi
+ b0(x, t).
We assume
νξ2 ≤
n∑
i,j=1
aij(x, t)ξiξj ≤ µξ2 for all (x, t) ∈ ΩT , (1.6)
n∑
i=1
bi(x, t)ni(x) ≤ −δ < 0 for all (x, t) ∈ Σ2
T , (1.7)
where n(x) is the unit outward normal to Σ2 at the point x.
Fractional diffusion equations were applied to numerous problems in biology,
chemistry, hydrology, etc. (see, for example, [1, 14, 23], Chapter 10 in [17] and
Chapter 5 in [16]).
In [9], Kochubei found the expression for the fundamental solution Γα(x, t)
to the Cauchy problem
∂αt u−∆u = f(x, t), (x, t) ∈ Rn × (0, T ), (1.8)
u(x, 0) = u0(x), x ∈ Rn, (1.9)
in terms of Fox’s H-functions (see also [22]). In [19], Pskhu represented Γα(x, t)
using Wright functions
φ(−α, δ; z) =
∞∑
k=0
zk
k!Γ(δ − αk)
, β ∈ (0, 1). (1.10)
Besides, some estimates of Γα(x, t) and its derivatives are derived in [9, 19].
The Cauchy problem for the operator L
(
x, t, ∂∂x , ∂
α
t
)
(see (1.5)) and fractional
order parabolic systems were studied in [3, 10].
Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1 49
M.V. Krasnoschok
The papers [4, 15, 21, 24] give the results on solvability of some initial bound-
ary value problems for linear and quasilinear fractional diffusion equations in
Sobolev spaces.
H. Lopushans’ka and A. Lopushans’kyi [13] proved the theorem on the exis-
tence and uniqueness of the solution to the space-time fractional Cauchy problem
in spaces of generalized functions.
Kemppainen used in [6] the boundary integral approach to prove the existence
of the solution for a time-fractional diffusion equation with Robin boundary con-
dition in the space of continuous up to the boundary functions.
Clément, Londen, Simonett obtained in [2] the existence, uniqueness, and
continuaty on quasilinear parabolic equation with time-fractional derivative. They
considered this derivative in spaces of continuous functions having a prescribed
singularity as t→ 0.
In [18], R. Ponce described the well-posedness (or maximal regularity) of the
abstract fractional differential equation (β > 0),
∂βt u(t) = Au(t) +
t∫
−∞
a(t− s)Au(s)ds+ f(t), t ∈ R,
in Cγ(R;X), γ ∈ (0, 1), where A is a closed linear operator defined on a Banach
space X.
The aim of this paper is to extend classical theory [12] to the case of the
problem of fractional order and to prove solvability of problem (1.2)–(1.4) in
Hölder spaces. We consider the case n ≥ 2. The one-dimensional case was
studied in [11].
The paper is organized as follows. In Sec. 2, we define the functional spaces
and state the main result. The estimates of solutions of model problems are given
in Sec. 3. In Sec. 4 we prove the main result. Appendix contains the proof of
some auxiliary assertions used in Sec. 3.
2. Hölder Spaces and the Main Result
Let us define the Hölder spaces used in this paper. Let Q be a domain in Rn,
QT = Q× (0, T ), θ ∈ (0, 1).
By Cθα(QT ), we mean the set of functions defined in QT and having a finite
norm
|w|(θ)α,QT = |w|QT + 〈w〉(θ)α,QT ,
where
|w|QT = sup
(x,t)∈QT
|w(x, t)|,
〈w〉(θ)α,QT = 〈w〉(θ)x,QT + 〈w〉(
θα
2
)
t,QT
,
50 Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1
Solvability in Hölder Space of an Initial Boundary Value Problem
〈w〉(θ)x,QT = sup
x,y∈Q
sup
t∈(0,T )
|w(x, t)− w(y, t)||x− y|−θ,
〈w〉(θ)t,QT = sup
x∈Q
sup
t,τ∈(0,T )
|w(x, t)− w(x, τ)||t− τ |−θ.
Let Dl
x = ∂|l|
∂x
l1
1 ...∂x
ln
n
, l = (l1, . . . , ln), |l| = l1 + . . . + ln. By Ck+θ
α (QT ),
(k = 1, 2), we mean the set of functions w(x, t) with a finite norm (m = 0, 1),
|w|(k+θ)
α,QT
=
∑
|l|+2m≤k
|∂αmt Dl
xw|QT + 〈w〉(k+θ)
α,QT
,
〈w〉(k+θ)
α,QT
=
∑
|l|+2m=k
〈∂αmt Dl
xw〉θα,QT +
∑
|l|=k−1
〈Dl
xw〉
k−1+θ
2
α
t,QT
.
By the symbol Ck+θ
α,0 (QT ), we denote the subspace of Ck+θ
α (QT ), (k = 0, 1, 2)
consisting of functions w(x, t) such that ∂αmt u|t=0 = 0, where m = 0 for k = 0, 1
and m = 0, 1 for k = 2. With the help of local coordinates and partition of unity,
all these spaces can be introduced on manifolds Σ1
T , Σ2
T .
If α = 1, then the spaces Ck+θ
α (QT ) coincide with the classical Hölder spaces
Hk+θ, k+θ
2 (QT ) (see Chapter I in [12]).
Now we assume that the compatibility conditions
ψ1(x, 0) = u0(x), ∂αt ψ1(x, 0) =
n∑
i,j=1
aij(x, 0)
∂2u0(x)
∂xi∂xj
+
n∑
i=1
ai(x, 0)
∂u0(x)
∂xi
+a0(x, 0)u0(x), x ∈ Σ1, B
(
x, 0,
∂
∂x
)
u0(x) = ψ2(x, 0), x ∈ Σ2 (2.1)
hold.
The main result of this paper is the following.
Theorem 2.1. Suppose that
Σ1, Σ2 ∈ C2+θ, aij , ai, a0 ∈ Cθα(ΩT ), bi, b0 ∈ C1+θ
α (Σ1
T ) i, j = 1, . . . , n, (2.2)
and assumptions (1.6), (1.7), (2.1) hold. Then for every function u0 ∈ C2+θ(Ω),
f ∈ Cθα(ΩT ), ψ1 ∈ C2+θ
α (Σ1
T ), ψ2 ∈ C1+θ
α (Σ2
T ), problem (1.2)–(1.4) has a unique
solution u ∈ C2+θ
α (ΩT ) satisfying the estimate
|u|(2+θ)
α,ΩT
≤ C
(
|u0|(2+θ)
Ω + |f |(θ)α,QT + |ψ1|(2+θ)
α,Σ1
T
+ |ψ2|(1+θ)
α,Σ2
T
)
. (2.3)
Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1 51
M.V. Krasnoschok
Further, let us recall that
Iγt v(x, t) =
1
Γ(γ)
t∫
0
v(x, τ)
(t− τ)1−γ dτ (2.4)
is the Riemann–Liouville fractional integral of order γ > 0, and
Dγ
t v(x, t) =
∂k
∂tk
Ik−γt v(x, t), k − 1 < γ ≤ k, k ∈ N, (2.5)
is the Riemann–Liouville fractional derivative of order γ (see Chapter 2 in [8]).
We put
I0
t v(x, t) = v(x, t), Dζ
t v(x, t) = I−ζt v(x, t) if ζ < 0. (2.6)
It can easily be checked that
∂αt v(x, t) = Dα
t (v(x, t)− v(x, 0)). (2.7)
The properties of Wright functions φ(−α, δ, z) are described in detail, for
example, in [19, 20, 7]. We put conventionally in (1.10) that 1
Γ(−k) = 0 if k ∈
N
⋃
0.
Lemma 3.3.1 in [20] implies
φ(−α, δ,−z) > 0 if z > 0, δ ≥ 0. (2.8)
To this end, we prove the following auxiliary result.
Lemma 2.1. Let δ ∈ R. The Wright function φ(−α, δ,−z) with z > 0 satisfies
the inequalities
|φ(−α,−δ,−z)| ≤ c exp(−σz
1
1−α )
1 , if δ /∈ N
⋃
0,
z, if δ ∈ N
⋃
0.
(2.9)
P r o o f. The estimate
|φ(−α, ν,−z)| ≤ c exp(−σz
1
1−α ) for all ν ∈ R (2.10)
follows immediately from Lemmas 2, 3 in [19], and thus inequality (2.9) for δ /∈
N
⋃
0 is proved.
For δ ∈ N
⋃
0, we use the formula (see (2.2.5) in [20])
φ(−α,−k,−z) = αzφ(−α,−k + 1− α,−z)− kφ(−α,−k + 1,−z). (2.11)
By using this formula and the method of induction, one can easily prove
φ(−α,−k,−z) = αz
k∑
m=0
(−1)k−m
k!
m!
φ(−α,−m+ 1− α,−z). (2.12)
Combining (2.10), (2.12), we obtain (2.9) for δ ∈ N
⋃
0. The proof is finished.
52 Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1
Solvability in Hölder Space of an Initial Boundary Value Problem
3. Model Problems
An important role in the proof of Theorem 2.1 is played by the estimates of
the solutions to the following three model problems. Firstly, we determine the
function u(x, t) as the solution to the Cauchy problem (1.8), (1.9). Secondly, we
find the function v(x, t) as the solution of the first boundary value problem in
the half-space Rn
+ = {x ∈ Rn : xn > 0}:
∂αt v −∆v = 0, (x, t) ∈ Rn
+,T = Rn
+ × (0, T ),
v(x, 0) = 0, x ∈ Rn, v → 0 with |x| → ∞,
v(x′, 0, t) = Φ(x′, t), x′ = (x1, · · · , xn−1) ∈ Rn−1. (3.1)
Thirdly, we find the function w(x, t) as the solution of the oblique derivative
boundary value problem
∂αt w −∆w = 0, (x, t) ∈ Rn
+,T = Rn
+ × (0, T ),
w(x, 0) = 0, x ∈ Rn, w → 0 with |x| → ∞,
n∑
i=1
hi
∂w
xi
(x′, 0, t) = Ψ(x′, t), x′ = (x1, · · · , xn−1) ∈ Rn−1. (3.2)
Here h = (h1, . . . , hn) is a constant vector.
We assume that
u0 ∈ C2+θ(Rn), f ∈ Cθα(Rn
T );
u0(x), f(x, t) = 0 for all |x| > R > 0, (3.3)
Φ ∈ C2+θ
α,0 (Rn−1
T ), Ψ ∈ C1+θ
α,0 (Rn−1
T );
Φ(x′, t) = 0, Ψ(x′, t) = 0 for all |x′| > R > 0. (3.4)
Owing to (3.4), we can take
Φ(x′, t) = 0, Ψ(x′, t) = 0, t ≤ 0. (3.5)
Lemma 3.1. Let assumptions (3.3) hold. Then the solution of (1.8), (1.9) is
represented as
u(x, t) = (Γα ∗ f) + (Dα−1
t Γα ∗1 u0)
≡
t∫
0
∫
Rn
Γα(x− y, t− τ)f(y, τ)dy +
∫
Rn
Dα−1
t Γα(x− y, t)u0(y)dy,
where Γα(x, t) = (4π)−
n
2
∞∫
0
λ−
n
2 exp(−|x|
2
4λ
)t−1φ(−α, 0,−λt−α)dλ. (3.6)
Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1 53
M.V. Krasnoschok
P r o o f. We take the Fourier transform in the spatial variables x and the
Laplace transform in t,
F [u] =
∫
Rn
v(x, t) exp(−ix · ξ)dx, ξ = (ξ1, . . . , ξn),
L[u] =
∞∫
0
u(x, t) exp(−pt)dt. (3.7)
Problem (1.8), (1.9) then reduces to the relation (see (2.253) in [17]):
pαũ− pα−1F [u0] + |ξ|2ũ = f̃ , (3.8)
here ũ = F [L[u]]. By (3.8), we have
ũ =
f̃
pα + |ξ|2
+
pα−1
pα + |ξ|2
F [u0]
=
∞∫
0
exp(−(pα + |ξ|2)λ)dλ f̃ +
∞∫
0
pα−1 exp(−(pα + |ξ|2)λ)dλ F [u0]. (3.9)
Combining
F−1[exp(−|ξ|2λ)] = (4πλ)−
n
2 exp(−|x|
2
4λ
), (3.10)
L−1[p−µ exp(−(pαλ)] = tµ−1φ(−α, µ,−λt−α) (3.11)
(see (3.2.7) in [20]),
Dν
t (tµ−1φ(−α, µ,−λt−α)) = tµ−ν−1φ(−α, µ− ν,−λt−α) (3.12)
(see (8) in [19]), and the convolution formula, we deduce (3.6) from (3.9). The
lemma is proved.
R e m a r k. Let us recall the fundamental solution to the Cauchy problem
for the heat equation
Γ1(x, t) = (4πt)−n/2 exp(−|x|
2
4t
). (3.13)
We can rewrite Γα in a more convenient form
Γα(x, t) =
∞∫
0
Γ1(x, λ) t−1φ(−α, 0,−λt−α)d λ (3.14)
54 Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1
Solvability in Hölder Space of an Initial Boundary Value Problem
and use below the well-known estimate
|Dl
xΓ1(x, λ)| ≤ cλ−
n+|l|
2 exp(−C |x|
2
λ
) (3.15)
in analysis of Γα(x, t).
We assume that
hn ≤ −δ0 < 0, |h′| ≤M0. (3.16)
Lemma 3.2. Let assumptions (3.4), (3.16) hold. Then the solutions of (3.1)
and (3.2) are represented as
v(x, t) = −2(
∂Γα
∂xn
∗2 Φ) ≡ −2
t∫
−∞
∫
Rn
∂Γα
∂xn
(x′ − y′, xn, t− τ)Φ(y′, τ)dy′, (3.17)
w(x, t) = (G ∗2 Ψ), where G(x, t) = −2
∞∫
0
∂Γα
∂xn
(x− hλ, t)dλ. (3.18)
The proof of this Lemma is given in Appendix 5.1. In our analysis we follow
very closely the classical approach of V.A. Solonnikov (see Chapter IV in [12]).
First, we derive the estimates of Dν
tD
l
xΓα(x, t) (ν ∈ R, |l| ≥ 0). Then we establish
some integral estimates of Γα, G and their derivatives. Finally, we estimate Hölder
constants with respect to t and x of the potentials (Γα∗f), (Γα∗1 Φ) and (G∗2 Φ).
Lemma 3.3. Let ν ∈ R, |l| ≥ 0. The function Γα(x, t), defined by (3.6),
satisfies the following estimates (β = α
2 ):
|Dν
tD
l
xΓα(x, t)| ≤ ctβ(2−n−|l|)−ν−1γp(ν,l)(|x|t−β) exp(−σ(|x|t−β)
1
1−β ), (3.19)
here (z > 0),
γm(z) =
1, if m ≤ 3,
| log z|+ 1, if m = 4,
z4−n, if m ≥ 5,
p(ν, l) =
n+ |l|, if ν ∈ N ∪ {0},
n+ |l|+ 2, if ν /∈ N ∪ {0}.
The proof of this lemma is given in Appendix 5.2. Estimates (3.19) are estab-
lished in [19] for the case 0 ≤ |l| ≤ 2.
Estimates (3.19) allow us to deduce the integral estimates of Γα, G.
Lemma 3.4. The following estimate holds:
∞∫
0
|Dl
xΓα(x, t)|dt ≤ C|x|−n+2−|l|, n+ |l| ≥ 3. (3.20)
Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1 55
M.V. Krasnoschok
Lemma 3.5. The function G, defined by formula (3.18), satisfies the following
estimates:
∞∫
0
|Dl
xG(x, t)|dt ≤ C|x|−n+2−|l|, 0 ≤ |l| ≤ 2, (3.21)
∞∫
0
tβ/2|Gxi(x, t)|dt ≤ C|x|−n+3/2, i = 1, . . . , n, (3.22)
∞∫
0
tδβ|Dα
t D
l
xG(x, t)|dt ≤ C|x|δ−n−|l|, |l| = 0, 1, δ > 0, (3.23)
∫
Rn−1
|Dα−1+ζ
t G(x, t)|dx′ ≤ Ct−β−ζ , ζ ∈ R, (3.24)
∫
Rn−1
|Dk
tGxi(x, t)|dx′ ≤ Ct−1−k, k = 0, 1, i = 1, . . . , n, (3.25)
∫
Rn−1
|Dl
xG(x, t)|dx′ ≤ Ctβ−1−β|l|, 0 ≤ |l| ≤ 2. (3.26)
P r o o f of Lemma 3.4. By (3.19), we get
|Dl
xΓα(x, t)| ≤ C
∞∫
0
tβ(2−n−|l|)−1γn+|l|(|x|t−β) exp(−σ(|x|t−β)
1
1−β )dt.
By the change of variable t→ λ = |x|t−β, we obtain
|Dl
xΓα(x, t)| ≤ c|x|−n+2−|l|
∞∫
0
λn+|l|−3γn+|l|(λ) exp(−σλ
1
1−β )dλ ≤ c|x|−n+2−|l|,
where we used the definition of γm and the assumption n+ |l| ≥ 3. The proof is
finished.
P r o o f of Lemma 3.5. Begin with considering the expression
∑n
i=1(xi −
hiλ)2. We claim that
n∑
i=1
(xi − hiλ)2 ≥ C2
0 (
n−1∑
i=1
x2
i + λ2) + x2
n, C2
0 =
δ2
0
2
min
{
1,
1
δ2
0 +M2
0
}
. (3.27)
56 Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1
Solvability in Hölder Space of an Initial Boundary Value Problem
It follows from (3.16) that for any a ∈ (0, 1),
n∑
i=1
(xi − hiλ)2 ≥
n−1∑
i=1
(x2
i − 2axi
hi
a
λ+ h2
iλ
2) + x2
n + h2
nλ
2
≥ (1− a2)
n−1∑
i=1
x2
i −
(
1
a2
− 1
)
M2
0λ
2 + δ2
0λ
2 + x2
n.
Putting a2 = 1
2
(
1 + M2
0
δ20+M2
0
)
, we derive (3.27).
Let us establish (3.22), (3.26). The rest of estimates can be proved by the
same arguments.
Apply (3.18), (3.19) to see
J =
∞∫
0
tβ/2|Gxi(x, t)|dt
≤ C
∞∫
0
dt
∞∫
0
tβ/2−βn−1γn+2(|x− hλ|t−β) exp(−σ(|x− dλ|t−β)
1
1−β )dλ.
Consider the cases n = 2 and n ≥ 3 separately.
If n = 2, then from the inequality
(| log z|+ 1) exp(−σ|z|
1
1−β ) ≤ C|z|−1/4 exp(−σ
2
|z|
1
1−β ) (3.28)
and (3.27) it follows that
J ≤ C
∞∫
0
dt
∞∫
0
tβ/2−2β−1(
|x− hλ|
tβ
)−1/4 exp(−σ
2
(
|x− hλ|
tβ
)
1
1−β )dλ
≤ C(C0)
∞∫
0
dt
∞∫
0
tβ/2−2β−1(
|x′|
tβ
)−
1
4 exp(−σ0
4
(
|x′|
tβ
)
1
1−β ) exp(−σ0
4
(
λ
tβ
)
1
1−β )dλ,
here and below σ0 depends on C0. By the change of variable t→ s = |x′|t−β, we
deduce
J ≤ C(C0)
∞∫
0
(
|x′|
s
)−
1
2 s−
1
4 exp(−σ0
4
s
1
1−β )
ds
s
≤ C(C0)|x′|−
1
2
∞∫
0
s−
3
4 exp(−σ0
4
s
1
1−β )ds ≤ C(C0)|x′|−
1
2 . (3.29)
Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1 57
M.V. Krasnoschok
Analogously, for n ≥ 3, we get
J ≤ C
∞∫
0
dt
∞∫
0
tβ/2−βn−1(
|x− hλ|
tβ
)2−n exp(−σ(
|x− hλ|
tβ
)
1
1−β )dλ
≤ C(C0)
∞∫
0
dt
∞∫
0
tβ/2−βn−1(
|x′|
tβ
)2−n exp(−σ0
2
(
|x′|
tβ
)
1
1−β ) exp(−σ0
2
(
λ
tβ
)
1
1−β )dλ
≤ C(C0)
∞∫
0
tβ/2−β(n−1)−1(
|x′|
tβ
)2−n exp(−σ0
2
(
|x′|
tβ
)
1
1−β )dt
≤ C(C0)
∞∫
0
(
|x′|
s
)
1
2
−(n−1)s2−n exp(−σ0
2
s
1
1−β )
ds
s
≤ C(C0)|x′|
3
2
−n
∞∫
s−
1
2 exp(−σ0
2
s
1
1−β )ds ≤ C(C0)|x′|
3
2
−n.
Use this inequality and (3.29) to obtain (3.22).
By (3.18), (3.13), (3.14), we have
Dl
xG(x, t) = −2
∞∫
0
Dl
x
∂Γα
∂xn
(x− hµ, t)dµ
= −2
∞∫
0
dλ
∞∫
0
Dl
x
∂Γ1
∂xn
(x− hµ, t)t−1φ(−α, 0,−λt−α)dµ.
Using (3.15), (2.8) and the change of variable y′ = x′−hµ
λ1/2 , we obtain
J =
∫
Rn−1
|Dl
xG(x, t)|dx′
≤ C
∞∫
0
dλ
∞∫
0
λ−
|l|
2
−1 exp(−C (xn − hnµ)2
λ
t−1φ(−α, 0,−λt−α)dµ.
The change of variable ζ = µ√
λ
and assumption (3.16) give
J ≤ C
∞∫
0
λ−
|l|+1
2 t−1φ(−α, 0,−λt−α)dλ
∞∫
0
exp(−Cδ0ζ
2)dζ.
58 Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1
Solvability in Hölder Space of an Initial Boundary Value Problem
Estimate (2.9), formula (2.11) and the change of variable λ→ η = λ
tβ
imply
J ≤ Ctβ−1−β|l|
∞∫
0
η
1−|l|
2 exp(−ση
1
1−αdη ≤ Ctβ−1−β|l| if 0 ≤ |l| ≤ 2.
The proof of (3.26) is finished.
Now we estimate Hölder constants of the potentials derived in (3.6), (3.17),
(3.18).
Lemma 3.6. The following estimates hold:〈
∂2(Γα ∗ f)
∂xi∂xj
〉(θ)
α,RnT
≤ C〈f〉(θ)α,RnT , (3.30)
〈
∂(Γα ∗ f)
∂xi
〉( 1+θ
2
α)
t,RnT
≤ C〈f〉(θ)α,RnT , (3.31)
〈
(Dα−1
t Γα ∗1 u0)
〉(θ)
x,RnT
≤ C〈u0〉(θ)x,Rn , (3.32)〈
(Dα−1
t Γα ∗1 u0)
〉( k+θ
2
α)
t,RnT
≤ C〈u0〉(k+θ)
x,Rn , k = 0, 1. (3.33)
Lemma 3.7. The following estimates hold:
〈(Gxi ∗2 ψ)〉(θ)x,Rn+,T ≤ C〈ψ〉
(θ)
α,Rn−1
T
, (3.34)
〈∂αt (G ∗2 ψ)〉(
θ
2
α)
t,Rn+,T
≤ C〈ψ〉(
1+θ
2
α)
t,Rn−1
T
, (3.35)
〈∂αt (G ∗2 Ψ)〉(θ)x,Rn+,T ≤ C〈ψ〉
( 1+θ
2
α)
t,Rn−1
T
, (3.36)
〈(Gxi ∗2 ψ)〉(θα)
t,Rn+,T
≤ C〈ψ〉(2θ)
α,Rn−1
T
, (3.37)
〈(G ∗2 ψ)〉(
1+θ
2
α)
t,Rn+,T
≤ C〈ψ〉(θ)
α,Rn−1
T
. (3.38)
The proof of Lemma 3.6 follows the same line as the proof of (2.1), (2.2) in
§2, Chapter IV [12] and is based on estimates (3.19), (3.20). Thus it is omitted.
P r o o f of Lemma 3.7. Similarly to Chapter IV [12], we write the potential
(Gxi ∗2 ψ) in the form
(Gxi ∗2 ψ)(x, t) =
∞∫
0
dτ
∫
Rn−1
Gxi(x
′ − y′, xn, τ)(ψ(y′, t− τ)− ψ(y′, t))dy′
Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1 59
M.V. Krasnoschok
+
∞∫
0
dτ
∫
Rn−1
Gxi(x
′ − y′, xn, τ)(ψ(y′, t)− ψ(x′, t))dy′
+ψ(x′, t)
∞∫
0
dτ
∫
Rn−1
Gxi(x
′ − y′, xn, τ)dy′. (3.39)
Now we represent the difference (Gxi ∗2 ψ)(x, t) − (Gxi ∗2 ψ)(x, t) as the sum of
seven integrals
(Gxi ∗2 ψ)(x, t)− (Gxi ∗2 ψ)(x, t)
=
|x−x|1/β∫
0
dτ
∫
Rn−1
Gxi(x
′ − y′, xn, τ)(ψ(y′, t− τ)− ψ(y′, t))dy′
−
|x−x|1/β∫
0
dτ
∫
Rn−1
Gxi(x
′ − y′, xn, τ)(ψ(y′, t− τ)− ψ(y′, t))dy′
+
∞∫
|x−x|1/β
dτ
∫
Rn−1
(Gxi(x
′− y′, xn, τ)−Gxi(x′− y′, xn, τ)(ψ(y′, t− τ)−ψ(y′, t))dy′
+
∫
K
(ψ(y′, t)− ψ(x′, t))dy′
∞∫
0
Gxi(x
′ − y′, xn, τ)dτ
−
∫
K
(ψ(y′, t)− ψ(x′, t))dy′
∞∫
0
Gxi(x
′ − y′, xn, τ)dτ
+
∫
Rn−1\K
dy′
∞∫
0
(Gxi(x
′ − y′, xn, τ)−Gxi(x′ − y′, xn, τ))(ψ(y′, t)− ψ(x′, t))dτ
−(ψ(x′, t)− ψ(x′, t)
∞∫
0
dτ
∫
K
Gxi(x
′ − y′, xn, τ)dx′ =
7∑
i=1
Ii, (3.40)
here we use the notation
K = {y′ ∈ Rn−1 : |x′ − y′| ≤ 2|x− x|}.
60 Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1
Solvability in Hölder Space of an Initial Boundary Value Problem
Applying (3.25) to I1, I2, we obtain
|I1|+ |I2| ≤ c〈ψ〉
( θα
2
)
t,Rn−1
T
|x−x|1/β∫
0
τ
θα
2
−1dτ ≤ c〈ψ〉(
θα
2
)
t,Rn−1
T
|x− x|θ. (3.41)
As for I3, we represent it as
I3 =
n∑
k=1
∞∫
|x−x|1/β
dτ
1∫
0
dλ
∫
Rn−1
Gηiηk(η′ − y′, ηn, τ)(xk − xk)
×(ψ(y′, t− τ)− ψ(y′, t))dy′,
where η = x+ λ(x− x). By (3.26), we get
|I3| ≤ c〈ψ〉
( θα
2
)
t,Rn−1
T
|x− x|
∞∫
|x−x|1/β
τ
θα
2
−1−βdτ ≤ c〈ψ〉(
θα
2
)
t,Rn−1
T
|x− x|θ. (3.42)
With the help of (3.21) we derive
|I4|+ |I5| ≤ c〈ψ〉θx,Rn−1
T
(∫
K
|x′ − y′|θ−(n−1)dy′ +
∫
|x′−y′|≤3|x−x|
|x− y′|θ−(n−1)dy′
)
≤ c〈ψ〉(θ)
x,Rn−1
T
|x− x|θ. (3.43)
Now we have
I6 =
n∑
k=1,n
∫
Rn−1\K
dy′
1∫
0
dλ
∞∫
0
Gηiηk(η′ − y′, ηn, τ)(xk − xk)(ψ(y′, t)− ψ(x′, t))dτ,
where η = x+ λ(x− x). One can easily check that
|x− x| ≤ |η′ − y′|, |x′ − y′| ≤ 2|η′ − y′|
for any η = x+ λ(x− x), y′ ∈ K, λ ∈ (0, 1). From this fact and estimate (3.21)
it follows that
|I6| ≤ c〈ψ〉(θ)x,Rn−1
T
|x− x|
1∫
0
dλ
∫
|η′−y′|≥|x−x|
|η′ − y′|θ−ndy′
Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1 61
M.V. Krasnoschok
≤ c〈ψ〉(θ)
x,Rn−1
T
|x− x|θ. (3.44)
Then we decompose I7 into the sum
I7 = (ψ(x′, t)− ψ(x′, t))
( ∞∫
|x−x|1/β
dτ
∫
K
Gxi(x
′ − y′, xn, τ)dy′
+
|x−x|1/β∫
0
dτ
∫
K
Gxi(x
′ − y′, xn, τ)dy′
)
= (ψ(x′, t)− ψ(x′, t)(J ′ + J ′′). (3.45)
By (3.22), we obtain
|J ′| ≤ |x− x|−1/2
∞∫
|x−x|1/β
dτ
∫
K
τβ/2|Gxi(x′ − y′, xn, τ)|dy′
≤ c|x− x|−1/2
∫
K
dy′
|x′ − y′|n−3/2
≤ c. (3.46)
Before proceeding to prove
J ′′ ≤ c, (3.47)
we establish the identity
G(x, t) = − 2
hn
Γα(x, t) +
2
hn
n−1∑
i=1
∞∫
0
hi
∂Γα
∂xi
(x− hλ, t)dλ
= G1(x, t) +G2(x, t). (3.48)
Since
d
dλ
Γα(x− hλ, t) = −
n−1∑
i=1
hi
∂Γα
∂xi
(x− hλ, t)− hn
∂Γα
∂xi
(x− hλ, t),
we obtain
−Γα(x, t) = −
n−1∑
i=1
∞∫
0
hi
∂Γα
∂xi
(x− hλ, t)dλ− hn
∞∫
0
∂Γα
∂xn
(x− hλ, t)dλ.
Hence (3.48) is true.
62 Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1
Solvability in Hölder Space of an Initial Boundary Value Problem
Now we consider the cases a) i 6= n and b) i = n. For the case a), we get
J ′′ =
|x−x|1/β∫
0
dτ
∫
Σ
G(x′ − y′, xn, τ)νidSy′
=
|x−x|1/β∫
0
dτ
∫
Σ
G2(x′ − y′, xn, τ)νidSy′ ,
here Σ = {y′ : |x′ − y′| = 2|x − x|} and ν is the unit outward normal to Σ in
Rn−1. By (3.19), we obtain
|J ′′| ≤
|x−x|1/β∫
0
dτ
∫
Σ
|G2(x′ − y′, xn, τ)|dSy′
≤ c
|x−x|1/β∫
0
dτ
∞∫
0
dλ
∫
Σ
τβ(1−n)−1γn+2(ξ) exp(−σξ
1
1−β )dSy′ , (3.49)
where ξ = |x′ − y′ − hλ|τ−β.
If n = 2, we use (3.28), (3.27) to get
|J ′′| ≤ c
|x−x|1/β∫
0
dτ
τ
∞∫
0
z−
1
4 exp(−σ0
4
z
1
1−β )
∣∣∣∣
z=
|x−x|
τβ
exp(−σ0
4
(
λ
τβ
)
1
1−β )
dλ
τβ
≤ c
|x−x|1/β∫
0
z−
1
4 exp(−σ0
4
z
1
1−β )
∣∣∣∣
z=
|x−x|
τβ
dτ
τ
.
By the change of variable z = |x− x|τ−β, we get
|J ′′| ≤ c
∞∫
1
z−5/4 exp(−σ0
4
z
1
1−β )dz ≤ c, n = 2, i 6= n. (3.50)
In a similar way, we get (see (3.49)) for n ≥ 3 and i 6= n,
|J ′′| ≤ c|x− x|n−2
∞∫
1
(
z
|x− x|
)n−2z2−n exp(−σ0
4
z
1
1−β )
dz
z
≤ c. (3.51)
Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1 63
M.V. Krasnoschok
Here, let us recall that meas Σ = 2 π
n−1
2
Γ(n−1)(2|x− x|)n−2.
In the case b), by (3.48), (3.18) we have
∂G
∂xn
= − 2
hn
∂Γα
∂xn
(x, t)− 1
hn
n−1∑
i=1
hi
∂G
∂xi
(x, t). (3.52)
The sum in the right-hand side of (3.52) is already estimated.
In remains to prove
I ′′ =
|x−x|1/β∫
0
dτ
∫
K
|∂Γα
∂xn
(x′ − y′, xn, τ)|dy′ ≤ c. (3.53)
By (3.14), we get
−2
∂Γα
∂xn
=
1
(4π)n/2
∞∫
0
xn
λ
λ−n/2 exp(−|x|
2
4λ
)t−1φ(−α, 0,−λt−α)dλ. (3.54)
In view of (2.8), we have
−2
∂Γα
∂xn
(x, t) ≥ 0.
We establish the identity
I ≡ −2
∞∫
0
dt
∫
Rn−1
∂Γα
∂xn
(x′, xn, t)dx′ = 1. (3.55)
Then (3.53) follows immediately from
I ′′ ≤ c
∞∫
0
dτ
∫
Rn−1
|∂Γα
∂xn
(x′, xn, τ)|dx′ = c
2
.
First, let us recall the identity (see (10) in [19]):
∞∫
0
φ(−α, 1− α,−z)dz = 1. (3.56)
We use (3.54) and (2.11) to rewrite the integral I in the form
I =
1
(4π)n/2
∞∫
0
exp
(
−x
2
n
4λ
)
xndλ
λ3/2
∞∫
0
φ(−α, 1−α,−λt−α)
αλdt
t1+α
∫
Rn−1
exp(−|x
′|2
4λ
)
dx′
λ
n−1
2
.
64 Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1
Solvability in Hölder Space of an Initial Boundary Value Problem
Then we change the variables x′ → y′ = x′√
λ
, t→ η = λt−α, λ→ µ = xn√
λ
to get
I =
2
(4π)n/2
∞∫
0
exp
(
−µ
2
4λ
)
dµ
∞∫
0
φ(−α, 1− α,−η)dη
∫
Rn−1
exp(−|y
′|2
4λ
)dy′.
We apply (3.56) and
+∞∫
−∞
exp(−z
2
4
)dz = 2
√
π
to obtain (3.55).
Thus the proof of (3.47) is finished.
Using (3.41), (3.42), (3.43), (3.44), (3.45), (3.46), we obtain (3.34).
In Appendix 5.3, we derive
∂αt (G ∗2 ψ)(x, t) =
∞∫
0
dτ
∫
Rn−1
Dα
t G(x′ − y′, xn, τ)
×(ψ(y′, t− τ)− ψ(y′, τ))dy′. (3.57)
In order to prove (3.35), we estimate the following difference (t > t):
∂αt (G ∗2 ψ)(x, t)− ∂αt (G ∗2 ψ)(x, t) =
t∫
2t−t
dτ
∫
Rn−1
Dα
t G(x′ − y′, xn, t− τ)(ψ(y′, τ)− ψ(y′, t))dy′
−
t∫
2t−t
dτ
∫
Rn−1
Dα
t G(x′ − y′, xn, t− τ)(ψ(y′, τ)− ψ(y′, t))dy′
+
2t−t∫
−∞
dτ
∫
Rn−1
(Dα
t G(x′ − y′, xn, t− τ)−Dα
t G(x′ − y′, xn, t− τ))
×(ψ(y′, τ)− ψ(y′, t))dy′
+
2t−t∫
−∞
dτ
∫
Rn−1
(ψ(y′, t)− ψ(y′, t))Dα
t G(x′ − y′, xn, t− τ)dy′ =
4∑
i=1
Ii.
Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1 65
M.V. Krasnoschok
By (3.24), we obtain
|I1|+ |I2| ≤ c〈ψ〉
( 1+θ
2
α)
t,Rn−1
T
t∫
2t−t
(t− τ)
1+θ
2
α−β−1dτ
+
t∫
2t−t
(t− τ)
1+θ
2
α−β−1dτ
≤ c〈ψ〉( 1+θ
2
α)
t,Rn−1
T
|t− t|
θ
2
α (3.58)
and
|I3| ≤ c〈ψ〉
( 1+θ
2
α)
t,Rn−1
T
2t−t∫
−∞
dτ
t∫
t
∫
Rn−1
Dα+1
s G(x′ − y′, xn, s− τ)(t− τ)
1+θ
2
αdy′
≤ c〈ψ〉(
1+θ
2
α)
t,Rn−1
T
2t−t∫
−∞
dτ
t∫
t
(s− τ)β−2(t− τ)
1+θ
2
αds
≤ c〈ψ〉(
1+θ
2
α)
t,Rn−1
T
t∫
t
ds
2t−t∫
−∞
(t− τ)
θ
2
α−2dτ ≤ c〈ψ〉(
1+θ
2
α)
t,Rn−1
T
|t− t|
θ
2
α. (3.59)
To this end, we obtain from (3.24), integrating with respect to τ ,
|I4| =
∣∣∣∣∣∣
∫
Rn−1
(ψ(y′, t)− ψ(y′, t))Dα−1
t G(x′ − y′, xn, 2(t− t))dy′
∣∣∣∣∣∣
≤ c〈ψ〉(
1+θ
2
α)
t,Rn−1
T
|t− t|
1+θ
2
α−β = c〈ψ〉(
1+θ
2
α)
t,Rn−1
T
|t− t|
θ
2
α. (3.60)
Inequalities (3.58)–(3.60) yield estimate (3.35).
Estimates (3.36), (3.37), (3.38) are proved in a similar manner with the help
of classical methods [12]. The proof of Lemma (3.5) is finished.
Arguing as in §2, Chapter IV [12], from the results of Lemmas 3.4, 3.5 we can
deduce the assertions of the following existence lemmas.
Lemma 3.8. Assume that assumptions (3.3) hold. Then problem (1.8), (1.9)
has a unique solution u ∈ C2+θ
α (Rn
T ), and
〈u〉2+θ
α,RnT
≤ C
(
〈f〉θα,RnT + 〈u0〉2+θ
Rn
)
. (3.61)
66 Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1
Solvability in Hölder Space of an Initial Boundary Value Problem
Lemma 3.9. Assume that assumptions (3.4) hold. Then problem (3.1) has a
unique solution v ∈ C2+θ
α,0 (Rn
+,T ), and
〈v〉2+θ
α,RnT
≤ C〈Φ〉2+θ
α,Rn−1
T
. (3.62)
Lemma 3.10. Assume that assumptions (3.4), (3.16) hold. Then problem
(3.2) has a unique solution w ∈ C2+θ
α,0 (Rn
+,T ), and
〈w〉2+θ
α,RnT
≤ C(δ0,M0)〈Ψ〉1+θ
α,Rn−1
T
. (3.63)
4. The Proof of Theorem 2.1
First we consider the case of zero initial data:
Theorem 4.1. Let assumptions (1.6), (1.7), (2.2) hold. Suppose
u0 = 0, f ∈ Cθα,0(ΩT ), ψ1 ∈ C2+θ
α,0 (Σ1
T ), ψ2 ∈ C1+θ
α,0 (Σ2
T ). (4.1)
Then problem (1.2)–(1.4) for sufficiently small τ ∈ (0, T ) has a unique solution
u ∈ C2+θ
α,0 (Ωτ ) satisfying the estimate
|u|(2+θ)
α,Ωτ
≤ c(|f |(θ)α,Ωτ + |ψ1|(2+θ)
α,Σ1
τ
+ |ψ2|(1+θ)
α,Σ2
τ
). (4.2)
We prove this theorem by constructing a regularizer. Since it is a standard
procedure (see Chapter IV [12]), we give only a sketch of the proof.
We cover the domain Ω with the balls K(k)
λ and K
(k)
2λ of radii λ and 2λ,
respectively, with a common center ξ(k) for sufficiently small λ > 0.
The index k belongs to one of the sets: k ∈ N(i) if K(k)
λ ∩Σi 6= ∅, i = 1, 2 and
k ∈M if K(k)
λ ∩ ∂Ω = ∅.
We take
τ = κλ
2
α , where κ < 1. (4.3)
Let ζ(k), η(k) be the sets of smooth functions subordinated to the indicated
overlapping of Ω such that∑
k
ζ(k)(x)η(k)(x) = 1, x ∈ Ω,
|Dl
xζ
(k)|+ |Dl
xη
(k)| ≤ cλ−|l|, |l| ≥ 0.
We suppose that the boundary Σi ∩K(k)
2λ (i = 1, 2) can be given by the equation
yn = F (y′)
Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1 67
M.V. Krasnoschok
in some local coordinates {y} with origin at the point ξ(k), where the axis yn is
oriented at the direction of the outward normal vector n(ξ(k)) to the surface Σi.
We ”straighten” the boundary by the formulae
z′ = y′, zn = yn − F (y′).
Let z = Zk(x) be the transformation of the coordinates {x} into {z}.
We denote
L0(x, t,
∂
∂x
, ∂αt ) = ∂αt −
n∑
i,j=1
aij(x, t)
∂2
∂xi∂xj
,
B0(x, t,
∂
∂x
) =
n∑
i
bi(x, t)
∂
∂xi
.
Let L(k)
0 , B(k)
0 be the operators L0, B0 in local coordinates {y} at the point
(ξ(k), 0):
L(k)
0 (ξ(k), 0,
∂
∂y
, ∂αt ) = ∂αt −
n∑
i,j=1
a
(k)
ij
∂2
∂yi∂yj
,
B(k)
0 (ξ(k), 0,
∂
∂y
) =
n∑
i
b
(k)
i
∂
∂yi
.
We set g = (f, ψ1, ψ2) and define a regularizer R by the formulae
Rg =
∑
k
η(k)(x)uk(x, t),
where the functions uk are found as follows.
If k ∈M, we find the function uk(x, t) as the solution to the Cauchy problem
L(k)
0 (ξ(k), 0,
∂
∂y
, ∂αt )uk(x, t) = fk(x, t), (x, t) ∈ Rn
T ,
uk(x, 0) = 0, x ∈ Rn, (4.4)
here fk(x, t) = ζk(x)f(x, t).
If k ∈ N(1) ∪N(2), we set
f ′k(z, t) = ζk(x)f(x, t)|x=Zk(z), ψ
′
i,k(z, t) = ζk(x)ψi(x, t)|x=Zk(z), i = 1, 2.
If k ∈ N(1), we find the function u′k(x, t) as the solution to the first boundary
value problem
L(k)
0 (ξ(k), 0,
∂
∂z
, ∂αt )u′k(z, t) = f ′k(z, t), (x, t) ∈ Rn
+,T ,
68 Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1
Solvability in Hölder Space of an Initial Boundary Value Problem
u′k(z
′, 0, t) = ψ′1,k(z
′, t), (z′, t) ∈ Rn−1
T , u′k(z, 0) = 0, z ∈ Rn
+. (4.5)
If k ∈ N(2), we find the function u′k(x, t) as the solution to the oblique bound-
ary value problem
L(k)
0 (ξ(k), 0,
∂
∂z
, ∂αt )u′k(z, t) = f ′k(z, t), (x, t) ∈ Rn
+,T ,
B(k)
0 (ξ(k), 0,
∂
∂z
)u′k(z
′, 0, t) = ψ′2,k(z
′, t), (z′, t) ∈ Rn−1
T ,
u′k(z, 0) = 0, z ∈ Rn
+. (4.6)
Now we set
uk(x, t) = u′k(z, t)|z=Z−1
k (x), k ∈ N(1) ∪N(2).
So we define uk(x, t) for all k.
Due to § 6, Chapter IV in [12], we can reduce problems (4.4), (4.5), (4.6) to
the case a(k)
ij = δij . Moreover, we can repeat routine calculations of § 6, 7 Chapter
IV in [12] in order to prove that the parameter δ0 in (3.16) depends only on δ
and µ from (1.6), (1.7).
For the sake of convenience, problem (1.2)–(1.4) (with zero initial data) can
be conventionally written in the operator form
Au = g,
where Au is a linear operator determined by the expressions in the left-hand sides
of (1.2), (1.4). Moreover, A : C2+θ
α,0 (ΩT ) → H(ΩT ), where H(ΩT ) = Cθα,0(ΩT ) ×
C2+θ
α,0 (Σ1
T )× C1+θ
α,0 (Σ2
T ) represents the space of functions g = (f, ψ1, ψ2) with the
norm
|g|H(ΩT ) = |f |(θ)α,ΩT + |ψ1|(2+θ)
α,Σ1
T
+ |ψ2|(1+θ)
α,Σ2
T
.
On the base of estimates (3.61)–(3.63), we obtain
|Rg|(2+θ)
α,Ωτ
≤ c|g|H(Ωτ ), (4.7)
where c does not depend on λ and τ , and for any h ∈ H(Ωτ ), u ∈ C(2+θ)
α (ΩT ),
|RAu− u|2+θ
α,ΩT ) ≤
1
2
|u|(2+θ)
α,ΩT ),
|ARg − g|H(ΩT ) ≤
1
2
|g|H(Ωτ ) (4.8)
if τ is sufficiently small and (4.3) is in force. The calculations are simple but
tedious. Inequalities (4.7), (4.8) yield the assertion of Theorem 4.1. It should
Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1 69
M.V. Krasnoschok
be emphasized that in our analysis we follow very closely the standard technique
found in Chapter IV [12].
To remove the restrictions imposed on the initial data in (4.1) and then ex-
tend the solution from [0, τ ] to [0, T ], we reduce problem (1.2)–(1.4) to the new
unknowns with zero initial data a) at t = 0, b) at t = τ .
In the case a), we set
u1(x) = f(x, 0) +
n∑
i,j=1
aij(x, 0)
∂2u0(x)
∂xi∂xj
+
n∑
i=1
ai(x, 0)
∂u0(x)
∂xi
+ a0(x, 0)u0(x),
f1(x) = u1(x)−∆u0(x), x ∈ Ω.
We observe that u1, f1 ∈ Cθ(Ω). By Hesten’s lemma, we construct û0 ∈
C2+θ(Rn), f̂1 ∈ Cθ(Rn
T ) such that
û0(x) = u0(x), f̂1(x) = f(x), x ∈ Ω.
Then we determine the auxiliary function u(0)(x, t) under conditions
u(0)(x, 0) = u0(x), ∂αt u
(0)(x, 0) = u1(x), x ∈ Ω, (4.9)
as a solution to the Cauchy problem
∂αt u
(0)(x, t)−∆u(0)(x, t) = f̂(x), (x, t) ∈ Rn
T ,
u(0)(x, 0) = û0(x), x ∈ Rn. (4.10)
We employ Lemma 3.8 to deduce u(0) ∈ C2+θ
α (Rn
T ) and
|u(0)|(2+θ)
α,RnT
≤ c(|u0|(2+θ)
Rn + |f(·, 0)|(θ)Rn).
Now we look for the solution of (1.2)–(1.4) as u(x, t) = v(x, t) + u(0)(x, t), where
v is found from the problem of the form (1.2)–(1.4) with zero initial data and
recalculated right-hand terms.
In the case b), we have to find the function u(τ)(x, t) ∈ C2+θ
α (Ω2τ ) satisfying
the condition
u(τ)(x, t) = u(x, t), x ∈ Ω, t ∈ [0, τ ], (4.11)
where u(x, t) is a local solution of problem (1.2)–(1.4) on [0, τ ]. We remark that
(4.11) implies
∂αt u
(τ)(x, t) = ∂αt u(x, t), x ∈ Ω, t ∈ [0, τ ].
It should be noticed that we need to know the function u(τ)(x, t) on the whole seg-
ment [0, τ ]. This is connected with nonlocal property of the fractional derivative
∂αt .
70 Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1
Solvability in Hölder Space of an Initial Boundary Value Problem
By Hesten’s lemma, again we can continue u(x, t) to the set Rn
τ :
û ∈ C2+θ
α (Rn
τ ), |û|(2+θ)
α,Rnτ ≤ c|u|
(2+θ)
α,Ωτ
, û(x, t) = u(x, t), (x, t) ∈ Ωτ .
Then we put
ω(x, t) =
∂αt û(x, t)− û(x, t), x ∈ Rn, t ∈ [0, τ ],
∂αt û(x, τ)− û(x, τ), x ∈ Rn, t ∈ [τ, 2τ ].
Obviously, ω ∈ Cθα(Rn
2τ ). We are now in position to determine u(τ) as a solution
of the Cauchy problem
∂αt u
(τ)(x, t)−∆u(τ)(x, t) = ω(x, t), (x, t) ∈ Rn
2τ ,
u(τ)(x, 0) = û0(x), x ∈ Rn. (4.12)
After that we look for the solution of problem (1.2)–(1.4) in the form
u(x, t) = v(x, t) + u(τ)(x, t),
where v(x, t) satisfies (1.2), (1.4), and
v(x, t) = 0, x ∈ Ω, t ∈ [0, τ ].
The boundary value problem of the form (1.2)–(1.4) for v is obtained by transla-
tion in time: t→ t− τ (see also [11]).
We can repeat this procedure in order to prove the solvability of problem
(1.2)–(1.4) on segment [0, T ].
5. Appendix
5.1. The proof of Lemma 3.2
We take the Fourier transform F ′ in the tangent space variables x′
F ′[v] =
∫
Rn−1
v(x, t) exp(−ix′ · ξ)dx′, ξ = (ξ1, . . . , ξn−1)
and the Laplace transform in t (3.7). Problem (3.1) then reduces to the ordinary
differential equation (ṽ = F [L[v]])
pαṽ(ξ, xn, p) + |ξ|2ṽ(ξ, xn, p)− ṽxnxn(ξ, xn, p) = 0 (5.1)
with the boundary conditions
ṽ → 0, xn →∞, ṽ(ξ, 0, p) = Φ̃(ξ, p). (5.2)
Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1 71
M.V. Krasnoschok
We solve problem (5.1), (5.2) to obtain
ṽ(ξ, xn, p) = exp(−
√
pα + |ξ|2xn)Φ̃(ξ, p). (5.3)
We need to prove that
LF ′
[
−2
∂Γα
∂xn
]
= exp(−
√
pα + |ξ|2xn). (5.4)
Then identity (3.17) follows from (5.4) and the convolution formula.
Since
F ′[exp(−|x
′|2
4λ
] = exp(−|ξ|2λ)(4
√
πλ)
n−1
2
and
L[t−1φ(−α, 0,−λt−α)] = exp(−pαλ)
(see (3.10), (3.11)), we obtain from (3.54)
LF ′
[
−2
∂Γα
∂xn
]
=
1
2
√
π
∞∫
0
xnλ
−3/2 exp(−x
2
n
4λ
− (pα + |ξ|2)λ)dλ.
The change of variable λ→ z = xn√
λ
and formula (3.325) in [5] give
LF ′
[
−2
∂Γα
∂xn
]
=
1√
π
∞∫
0
exp(−z
2
4
− (pα + |ξ|2)x2
n
z2
)dλ = exp(−
√
(pα + |ξ|2)x2
n).
Thus identity (5.4) is proved.
Then we apply the Laplace transform in t and the Fourier transform in x′ to
problem (3.2) to obtain
pαw̃(ξ, xn, p) + |ξ|2w̃(ξ, xn, p)− w̃xnxn(ξ, xn, p) = 0,
(hnw̃xn + ih′ · ξw̃)|xn=0 = Ψ̃(ξ, p), w̃ → 0, xn →∞.
Assumption (3.16) allows us to write
w̃(ξ, xn, p) =
exp(−
√
pα + |ξ|2xn)
−hn
√
pα + |ξ|2 + ih′ · ξ
Ψ̃(ξ, p)
=
∞∫
0
exp(−
√
pα + |ξ|2(xn − hnλ)− ih′ · ξλ)dλ Ψ̃(ξ, p) ≡ G̃(ξ, xn, p)Ψ̃(ξ, p).
Identity (5.4) implies L−1F ′−1[G̃] = −2
∫∞
0
∂Γα
∂xn
(x− hλ, t)dλ = G(x, t).
72 Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1
Solvability in Hölder Space of an Initial Boundary Value Problem
To this end, we need to prove that potentials (3.17), (3.18) satisfy the bound-
ary conditions in (3.1), (3.2).
As for (3.1), similarly to the classical case, we need the identity
−2
∞∫
0
dt
∫
Rn−1
∂Γα
∂xn
(x′, xn, t)dx′ = 1, (5.5)
derived above in Lemma 3.7 (see (3.55)).
As for problem (3.3), we see
n∑
i=1
hi
∂G
∂xi
(x, t) = −2
∞∫
0
n∑
i=1
∂2Γα
∂xn∂xi
(x− hλ, t)dλ
= 2
∞∫
0
d
dλ
∂Γα
∂xn
(x− hλ, t)dλ = −2
∂Γα
∂xn
(x, t).
This fact implies that
∑n
i=1 hi
∂
∂xi
(G ∗2 Ψ)|xn=0 = Ψ.
5.2. The proof of Lemma 3.3
By (3.14), (3.12), we obtain
Dl
xD
ν
t Γα(x, t) =
∞∫
0
Dl
xΓ1(x, λ) t−ν−1φ(−α,−ν,−λt−α)dλ.
We prove (3.19) for ν ∈ N∪0. If ν /∈ N∪0, the proof is the same. Inequalities
(3.15) and (2.9) give
|Dl
xD
ν
t Γα(x, t)| ≤ c
∞∫
0
λ−
n+|l|
2 exp(−C |x|
2
4λ
)λ t−α−ν−1 exp(−σ(λt−α)
1
1−α )dλ.
By the change of variable λ→ η = λt−α, we get
|Dl
xD
ν
t Γα(x, t)| ≤ ctβ(2−n−|l|)−ν−1
∞∫
0
η1−n+|l|
2 exp
(
−A
(
|x|2
tα
1
η
+ η
1
1−α
))
dη.
It is not hard to prove (z ≥ 0)
z
η
+ η
1
1−α ≥ σ∗(α)z
1
2−α for all η ∈ (0,∞).
Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1 73
M.V. Krasnoschok
Then it follows that
|Dl
xD
ν
t Γα(x, t)| ≤ ctβ(2−n−|l|)−ν−1 exp(−σ∗
A
2
(
|x|
tβ
)
1
1−β )
×
∞∫
0
η1−n+|l|
2 exp
(
−A
2
(
|x|2
tα
1
η
+ η
1
1−α
))
dη. (5.6)
We set (B > 0, z > 0),
Jm(z) =
∞∫
0
η1−m
2 exp
(
−B
(
z
η
+ η
1
1−α
))
dη
and claim that
Jm(z) ≤ c
1, m = 1, 2, 3;
| log z|+ 1, m = 4;
z2−m
2 , m ≥ 5.
(5.7)
If m = 1, 2, 3, one can easily see
Jm(z) ≤ c
∞∫
0
η1−m
2 exp
(
−Bη
1
1−α
)
dη ≤ c. (5.8)
If m = 4, we get
J4,1(z) =
z∫
0
exp
(
−Bz
η
)
dη
η
=
∞∫
1
exp (−Bs) ds
s
≤ c,
J4,2(z) =
∞∫
z
exp
(
−Bη
1
1−α
) dη
η
= log η exp
(
−Bη
1
1−α
) ∣∣∣∣∞
z
+
B
1− α
∞∫
z
log η η
α
1−α exp
(
−Bη
1
1−α
)
dη
≤ c
| log z|+
∞∫
0
log η η
α
1−α exp
(
−Bη
1
1−α
)
dη
≤ c(| log z|+ 1).
Hence, it follows that
J4(z) ≤ J4,1(z) + J4,2(z) ≤ c(| log z|+ 1). (5.9)
74 Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1
Solvability in Hölder Space of an Initial Boundary Value Problem
If m ≥ 5, we put ζ = η
z and deduce
Jm(z) =
∞∫
0
(ζz)1−m
2 exp
(
−B
(
1
ζ
+ (zζ)
1
1−α
))
zdζ
≤ cz2−m
2
∞∫
0
ζ1−m
2 exp
(
−B 1
ζ
)
dζ ≤ cz2−m
2 . (5.10)
Using (5.8), (5.9), (5.10), we get (5.7). Substituting (5.7) and z = |x|2
tα in
(5.6), we obtain (3.19).
5.3. The development of formula (3.57)
Estimate (3.24) implies
|(G ∗2 Ψ)(x, t)| ≤ c|Ψ|Rn−1
T
t∫
0
τβ−1dτ ≤ c|Ψ|Rn−1
T
tβ,
so
(G ∗2 Ψ)(x, t)|t=0 = 0.
By definitions (2.5), (2.7) and assumption (3.5), we see
∂αt (G ∗2 Ψ) = Dα
t (G ∗2 Ψ) =
∂
∂t
(
Iα−1
t (G ∗2 Ψ)
)
.
Now we introduce a sequence
uε(x, t) =
t−ε∫
0
dτ
∫
Rn−1
Iα−1
t G(x′ − y′, xn, t− τ)Ψ(y′, τ)dy′.
We need to prove that lim
ε→0
∂
∂tuε is equal to the right-hand side of (3.57).
First we obtain
∂
∂t
uε(x, t) =
t−ε∫
−∞
dτ
∫
Rn−1
Dα
t G(x′ − y′, xn, t− τ)Ψ(y′, τ)dy′
+
∫
Rn−1
Iα−1
t G(x′ − y′, xn, ε)Ψ(y′, τ − ε)dy′
Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 1 75
M.V. Krasnoschok
=
∞∫
ε
dτ
∫
Rn−1
Dα
t G(x′ − y′, xn, τ)(Ψ(y′, t− τ)−Ψ(y′, t− ε))dy′.
Then we consider the difference between ∂
∂tuε and the right-hand side of (3.57)
δε =
∂
∂t
uε(x, t)−
∞∫
0
dτ
∫
Rn−1
Dα
t G(x′ − y′, xn, τ)(Ψ(y′, t− τ)−Ψ(y′, t))dy′
=
∞∫
ε
dτ
∫
Rn−1
Dα
t G(x′ − y′, xn, τ)(Ψ(y′, t)−Ψ(y′, t− ε))dy′
+
ε∫
0
dτ
∫
Rn−1
Dα
t G(x′ − y′, xn, τ)(Ψ(y′, t− τ)−Ψ(y′, t))dy′.
We claim that
lim
ε→0
δε = 0.
Indeed, by (3.24), we get
|δε| ≤ c〈Ψ〉
( 1+θ
2
α)
t,Rn−1
T
ε 1+θ
2
α
∞∫
ε
τ−1−βdτ +
ε∫
0
τ
θ
2
α−1dτ
≤ c〈Ψ〉( 1+θ
2
α)
t,Rn−1
T
ε
θ
2
α.
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