Escher J., Guidotti P., Hieber M. et al. (editors) Parabolic Problems: The Herbert Amann Festschrift

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58 G. Bizhanova

go continuously. If we choose an initial moment T

∗

in the problem, then the

compatibility conditions will be fulﬁlled.

If we study the problem since t = T

∗

or since the moment of a jump of

all characteristics of the process (given functions, coeﬃcients, parameters in the

problem), then, in general, the compatibility conditions are not fulﬁlled, but the

physical process continues, and the problem can also have a solution.

To study the ﬁrst and second boundary value problems for the parabolic

equations in the classes C

2, 1

(Ω

) ∩C(Ω

)andC

2, 1

(Ω

) ∩C

1, 0

(Ω

) respectively,

we assume that the compatibility conditions of zero order are fulﬁlled in these

problems, because we look for, in the closure of a domain Ω

, a continuous solution

of the ﬁrst boundary value problem and a continuous solution together with all

its derivatives of ﬁrst order with respect to the spatial variables of the second

boundary value problem.

Solutions of boundary value problems in a weighted H¨older space C

(Ω

s ≤ l, introduced by V.S. Belonosov, permits us to get rid of one compatibility

condition [1, 2, 8]. Considering the ﬁrst boundary value problem in this class we

must require fulﬁllment of the compatibility condition of zero order, but the ﬁrst-

order compatibility condition can not take place. Y. Martel and Ph. Souplet in

[7] proved that the solution of the ﬁrst boundary value problem for the parabolic

equation with incompatible data is not continuous in the closure of a domain.

One-dimensional boundary value problems with incompatible data were studied

in [3, 4].

We study the ﬁrst and second boundary value problems for heat equations

in the half-space R

, n ≥ 2, with incompatible initial and boundary data on

the boundary x

= 0 of a domain at t = 0. The existence, uniqueness and esti-

mates of the solutions are proved in H¨older and weighted spaces. Nonfulﬁllment of

the compatibility conditions of initial and boundary data in the ﬁrst and second

boundary value problems leads to appearance of the functions z



,t)erfc

√

(x, t), j =0, 1, (see Theorems 2.1, 2.2) and −2

√

at z



,t)ierfs

√

(see The-

orems 2.3, 2.4) in the solutions of these problems respectively, which are singular

in the vicinity of a boundary of a domain as t → 0. These functions permit us to

reduce the original problems to problems with a fulﬁlled compatibility conditions

of all necessary orders.

In Chapter 1 the H¨older and weighted spaces are determined, the deﬁnition of

the special functions – repeated integrals of the probability and the compatibility

conditions for the considered problems are given. The main results of the paper

are formulated in Chapter 2. In Chapter 3 there are constructed and studied the

singular solutions of the auxiliary problems. In Chapters 4 and 5 with the help

of these singular solutions the original ﬁrst and second boundary value problems

are reduced to problems that have unique solutions in the weighted and classical

H¨older spaces. In the Appendix the auxiliary ﬁrst boundary value problem is

studied.

On the Classical Solvability 59

Let

D := R

= {x =(x



) |x



∈ R

n−1

> 0)},n≥ 2,

R := {x |x



∈ R

n−1

=0},x



=(x

,...,x

n−1

= D × (0,T),R

= R × [0,T].

We consider two problems. We are required to ﬁnd the solution u(x, t)ofthe

ﬁrst boundary value problem – Problem 1

∂

u −a Δ u = f(x, t)inD

, (1.1)

t=0

= u

(x)inD, (1.2)

= ϕ(x



,t)onR

, (1.3)

and the solution u(x, t) of the second boundary value problem – Problem 2, which

satisﬁes an equation (1.1), initial condition (1.2) and the boundary condition

∂

= ψ(x



,t)onR

. (1.4)

Here a =const> 0, ∂

= ∂/∂t, ∂

= ∂/∂x

,Δ=∂

+···+∂

.Byc

,...

we shall denote positive constants.

Determine the weighted and classical H¨older spaces.

Let l be a positive non-integer. By C

), s ≤ l, we shall denote a weighted

H¨older space deﬁned by V.S. Belonosov with the norm [1, 2, 8],

|u|

(l)

s,D

=sup

t≤T

l−s

[u]

(l)



s<2k+|m|<l

sup

t≤T

2k+|m|−s

|∂

∂





|u|

(s)

,s≥ 0,

0,s<0,

(1.5)

where D



= D × [t/2,t], |v |

=sup

(x,t)∈D

|v(x, t) |,

[u]

(l)

2k+|m|=[l]

[∂

∂

(l−[l])

x,D

0<l−2k−|m|<2

[∂

∂



l−2k−|m|



t,D

, (1.6)

[v]

(α)

x,D

=sup

(x,t),(z,t)∈D

v(x, t) − v(z,t)

|x −z|

−α

, (1.7)

[v]

(α)

t,D

=sup

(x,t),(x,t

)∈D

v(x, t) −v(x, t

)

|t − t

−α

,α∈ (0, 1), (1.8)

|u|

(s)

is the norm of the classical H¨older space C

s, s/2

)[6],

|u|

(s)

2k+|m|≤[s]

|∂

∂



0,san integer,

[u]

(s)

,snot an integer,

(1.9)

where [u]

(s)

is determined by (1.6)–(1.8).

For s = l, C

)isthespaceC

l, l/2

From the norm (1.5) we can see that for s<lthe function u(x, t)has

a singularity with respect to t as t → 0 in the whole domain D including its

60 G. Bizhanova

boundary. For instance, if an initial function u

(x) in the Cauchy problem for the

parabolic equation is from the H¨older space C

(D), s<l, then the solution of this

problem belongs to C

We introduce a weighted space C

s, δ

), δ

> 0, 2 ≤ s ≤ 2+α,ofthe

functions u(x, t) with the norm

|u|

(2)

s,δ

:= sup

t≤T

−1



n−1

μ=1

sup

t≤T

1−s

∂



+sup

t≤T

−1/2

∂



i=1

n−1

μ=1

sup

t≤T

2−s

∂



∂

, (1.10)

and, in particular, for s =2+α,

|u|

(2)

2+α,δ

:= sup

t≤T

−1



n−1

μ=1

sup

t≤T

−

1+α

∂



+sup

t≤T

−1/2

∂



i=1

n−1

μ=1

sup

t≤T

−

∂



∂

. (1.11)

We write |u|

(2)

s, δ

:= M . From (1.10), (1.11) we shall have

∂

≤ Me

−δ

,i=1,...,n, μ =1,...,n− 1, for s =2,

∂

u(x, t)

∂

u(x, t)

≤ Me

−δ

for 2 ≤ s ≤ 2+α.

(1.12)

Let the point x be in the interior of the domain: x

≥ r

=const> 0,

then from (1.12) we obtain that the derivatives ∂

u for s =2, and ∂

u(x, t),

∂

u(x, t)for2≤ s ≤ 2+α tend to zero exponentially as t → 0andonthe

boundary x

=0ofD they are bounded and can not be equal to zero at t =0,

i.e., they can be discontinuous in D

The negative powers of the t weights in the norms (1.10), (1.11) mean that

the function and its derivatives with respect to x with such weights tend to zero

as t → 0 on the boundary x

= 0 of a domain and exponentially in the interior of

a domain due to the weight e

We point out also that x

in exponential power is the distance between a

point x =(x

,...,x

) and a boundary of a domain x

=0.

An example of a function from C

s, δ

) is a solution of problem (A.1) in

the Appendix (see Theorems A.1 and A.2).

On the Classical Solvability 61

We do not include the H¨older constants of the derivatives into the norms

(1.10), (1.11). But from these norms we see the behavior of the function and its

derivatives in the domain and on its boundary.

We deﬁne the compatibility conditions of the boundary and initial functions

for Problems 1 and 2.

Let



):=ϕ(x



, 0) − u

(x)



):=∂

ϕ(x



,t)

t=0

−(a Δu

(x)+f(x, 0)





):=ψ(x



, 0) − ∂

(x)

The compatibility conditions of zero and ﬁrst orders on the boundary x

for Problem 1 (1.1), (1.2), (1.3) are A



)=0,A



)=0onR and of zero order

for Problem 2 (1.1), (1.2), (1.4) – B



)=0onR. Evidently, nonfulﬁllment of

the compatibility conditions of zero and ﬁrst orders means A



) =0,A



) =0,



) =0onR.

Later on we shall apply special functions – iterated integrals of the probability

erfc ζ; they are determined by the formulas [5], Ch. 7.2,

erfc ζ :=



∞

n−1

erfc ξdξ, n=0, 1, 2,...,

−1

erfc ζ :=

√

−ζ

, i

erfc ζ := erfc ζ =

√



∞

−ξ

dξ, i

erfc ζ := ierfc ζ.

The following relations for them hold:

dζ

erfc ζ = −i

n−1

erfc ζ, n =0, 1, 2,..., (1.13)

erfc 0 =

Γ(n/2+1)

,n= −1, 0, 1,...,

where Γ(·) – Euler gamma – function,

erfc ζ =

n−2

erfc ζ −

n−1

erfc ζ, n =1, 2,.... (1.14)

erfc ζ ≤ i

erfc 0,ζ≥ 0,n= −1, 0, 1,....

2. Main results

We formulate the main results for Problems 1, 2.

Theorem 2.1. Let α ∈ (0, 1), s ∈ (α, 2+α]. For all functions u

(x) ∈C

(D),

f(x,t) ∈C

s−2

), ϕ(x



,t)∈C

2+α

) that do not satisfy, on the boundary x

of the domain D, compatibility conditions of zero order for s ∈ (α, 2) (A



):=

ϕ(x



, 0) − u

(x)

=0on R) and of zero and ﬁrst orders for s ∈ [2, 2+α]



) =0, A



):=∂

ϕ(x



,t)

t=0

− (a Δu

(x)+f(x, 0)



=0on R),

62 G. Bizhanova

Problem 1 (1.1), (1.2), (1.3) has a unique solution u(x, t)=V

(x, t)+v

(x, t)

for s ∈ (α, 2) and u(x, t)=V

(x, t)+W

(x, t)+V

(x, t)+W

(x, t)+v

(x, t) for

s ∈ [2, 2+α],whereV

(x, t)=z



,t)erfc

√

, z

∈ C

2+α

), W

∈ C

s,δ

j =0, 1, δ

, v

∈ C

2+α

), i =1, 2, and the following estimates for them

hold:

(2+α)

s,R

≤ c

(s−2j)

, |W

(2)

s,δ

≤ c

(s−2j)

,j=0, 1, (2.1)

(2+α)

s,D

≤ c



(s)

+ |f|

(α)

s−2,D

+ |ϕ|

(2+α)

s,R



,i=1, 2. (2.2)

From this theorem for s =2+α we obtain the following one.

Theorem 2.2. For all functions u

(x) ∈ C

2+α

(D), f (x, t) ∈ C

α,α/2

), ϕ(x



,t) ∈

2+α,1+α/2



), α ∈ (0, 1) that do not satisfy, on the boundary x

=0of the

domain D, compatibility conditions of zero and ﬁrst orders (A



) =0, A



) =

0 on R), Problem 1 (1.1), (1.2), (1.3) has a unique solution u(x, t)=V

(x, t)+

(x, t)+V

(x, t)+W

(x, t)+v

(x, t) where V

(x, t)=z



,t)erfc

√

, z

∈

2+α,1+α/2



), W

∈ C

2+α,δ

), j =0, 1, δ

, v

∈ C

2+α,1+α/2

and the following estimates for them hold:

(2+α)

≤ c

(2+α−2j)

, |W

(2)

2+α,δ

≤ c

(2+α−2j)

,j=0, 1,

(2+α)

≤ c



(2+α)

+ |f|

(α)

+ |ϕ|

(2+α)



Theorem 2.3. Let α ∈ (0, 1), s ∈ (α, 2+α]. For all functions u

(x) ∈C

(D),

f(x,t) ∈C

s−2

), ψ(x



,t) ∈C

1+α

s−1

), s ∈ (α, 1),andifs ∈ [1, 2+α], then for

all functions u

,f,ψthat do not satisfy, on the boundary x

=0of the domain D,

the compatibility condition of zero order (B



):=ψ(x



, 0) − ∂

(x)

=0

on R), Problem 2 (1.1), (1.2), (1.4) has a unique solution u(x, t)=v

(x, t) for

s ∈ (α, 1) and u(x, t)=−2

√

at z



,t)ierfs

√

+ v

(x, t) for s ∈ [1, 2+α, 2],

where z

∈ C

2+α

s−1

), v

∈ C

2+α

), i =1, 2, and the following estimates for

them hold:

(2+α)

s−1,R

≤ c

(s−1)

(2+α)

s,D

≤ c



(s)

+ |f|

(α)

s−2,D

+ |ψ|

(1+α)

s−1,R



,i=1, 2.

(2.3)

From this theorem for s =2+α we obtain the following one.

Theorem 2.4. For al l functions u

(x) ∈ C

2+α

(D), f(x, t) ∈ C

α,α/2

), ψ(x



,t) ∈

1+α,

1+α



) that do not satisfy, on the boundary x

=0of the domain D,the

compatibility condition of zero order (B



) =0on R), Problem 2 (1.1), (1.2),

(1.4) has a unique solution u(x, t)=−2

√

at z



,t)ierfs

√

+ v

(x, t),where

On the Classical Solvability 63

∈ C

2+α

1+α

), v

∈ C

2+α, 1+α/2

) and the following estimates for them hold:

(2+α)

1+α,R

≤ c

(1+α)

(2+α)

s,D

≤ c



(2+α)

+ |f|

(α)

+ |ψ|

(1+α)



For Theorems 2.1–2.4, the functions z



,t), z



,t)andW

(x, t), j =0, 1,

are deﬁned in Theorems 3.3, 3.4 and 3.5, 3.6, A.1, A.2 respectively.

Remark 2.5. If in Problem 1 the compatibility condition of zero (ﬁrst) order is

fulﬁlled, i.e., A



)=0 (A



) = 0) on R,theninTheorems2.1,2.2V

(x, t)=0,

(x, t)=0 (V

(x, t)=0,W

(x, t)=0) inD

.IfA



)=0, A



)=0onR,

then V

(x, t)=0,W

(x, t)=0 inD

, j =0, 1.

If in Problem 2 the compatibility condition of zero order is fulﬁlled, i.e.,



)=0 onR,theninTheorems2.3,2.4z



,t)=0 onR

3. Auxiliary problems

We recall that D := R

, n ≥ 2, R := {x =(x



) |x



∈ R

n−1

=0},

= D × (0,T), R

= R × [0,T], Δ



:= ∂

+ ···+ ∂

n−1

First, we construct a function z(x



,t) satisfying the conditions

t=0

= μ



),∂

t=0

= μ



)inR. (3.1)

Lemma 3.1. [6] Let μ



) ∈ C

(R), μ



) ∈ C

s−2

(R), s ∈ [2, 2+α], α ∈ (0, 1).

Then there exists a unique function z(x



,t) ∈ C

2+α

), for which the following

estimate holds:

|z|

(2+α)

s,R

≤ c



|μ

(s)

+ |μ

(s−2)



. (3.2)

Lemma 3.2. [6] Let μ



) ∈ C

2+α

(R), μ



) ∈ C

(R), α ∈ (0, 1). Then there ex-

ists a unique function z(x



,t) ∈ C

2+α,1+α/2



), for which the following estimate

holds:

|z|

(2+α)

≤ c



|μ

(2+α)

+ |μ

(α)



This lemma follows from Lemma 3.1 for s =2+α.

Proof of Lemma 3.1. Consider the Cauchy problem

∂

z − a Δ



z = z

(1)



,t)inR

, (3.3)

t=0

= μ



)inR, (3.4)

where z

(1)



,t) is a solution of the Cauchy problem

∂

(1)

− a Δ



(1)

=0 in R

, (3.5)

(1)

t=0

= μ



) − aΔ



) ∈ C

s−2

(R)inR. (3.6)

Problem (3.5), (3.6) has a unique solution [1, 2, 6, 8] z

(1)



,t) ∈ C

2+α

s−2

), such

that

(1)

(2+α)

s−2,R

≤ c



|μ

(s)

+ |μ

(s−2)



64 G. Bizhanova

Due to embedding C

2+α

s−2

) ⊂ C

s−2

)weobtain

(1)

(α)

s−2,R

≤ c



|μ

(s)

+ |μ

(s−2)



. (3.7)

Then problem (3.3), (3.4) has a unique solution z ∈ C

2+α

), for which a valid

estimate is

|z|

(2+α)

s,R

≤ c



(1)

(α)

s−2,R

+ |μ

(s)



From here by (3.7) we shall have an estimate (3.2). Moreover, z(x



,t) satisﬁes the

conditions (3.1). Really, the ﬁrst condition in (3.1) is fulﬁlled due to (3.4) and from

(3.3) and (3.6) we obtain

∂

t=0

= aΔ



)+z

(1)

t=0

= μ



i.e., the second condition in (3.1) also holds. 

We recall that the compatibility conditions of zero and ﬁrst orders for Prob-

lem 1 and of zero order for Problem 2 are not fulﬁlled, i.e., A



):=ϕ(x



, 0) −

(x)

=0,A



):=∂

ϕ(x



,t)

t=0

−(a Δu

(x)+f(x, 0)



=0onR and



):=ψ(x



, 0) − ∂

(x)

=0onR.

To extract from the solutions of Problems 1 and 2 the singular parts, which

appear due to a nonfulﬁllment of the compatibility conditions on R,weextend

the functions A



), j =0, 1, B



)intoR

and then into D

. For that, ﬁrst,

we construct the functions z



,t), j =0, 1, and z



,t) under the conditions

t=0

= A



)inR, s ∈ (α, 2), (3.8)

t=0

= A



),∂

t=0

=0inR, s ∈ [2, 2+α], (3.9)

t=0

=0,∂

t=0

= A



)inR, s ∈ [2, 2+α] (3.10)

for Problem 1 (1.1)–(1.3) and

t=0

= B



)inR, s ∈ [1, 2+α] (3.11)

for Problem 2 (1.1), (1.2), (1.4).

Theorem 3.3. Let A



) ∈ C

(R), s ∈ (α, 2+α], A



) ∈ C

s−2

(R), s ∈ [2, 2+α],



) ∈ C

s−1

(R), s ∈ [1, 2+α], α ∈ (0, 1). Then there exist unique functions



,t) ∈C

2+α

), j =0,1, z



,t) ∈ C

2+α

s−1

), which satisfy the conditions

(3.8)–(3.11) respectively, and the estimates for them hold:

(2+α)

s,R

≤ c

(s)

,s∈ (α, 2+α], (3.12)

(2+α)

s,R

≤ c

(s−2)

,s∈ [2, 2+α], (3.13)

|≤c

(s−2)

t in R

,s∈ [2, 2+α], (3.14)

(2+α)

s−1,R

≤ c

(s−1)

,s∈ [1, 2+α]. (3.15)

Theorem 3.4. Let A



) ∈ C

2+α

(R), A



) ∈ C

(R), B



) ∈ C

1+α

(R),

α ∈ (0, 1). Then there exist unique functions z



,t) ∈ C

2+α,1+α/2



), j =0, 1,

On the Classical Solvability 65



,t) ∈ C

2+α

1+α

), which satisfy the conditions (3.8)–(3.11) respectively, and

the estimates for them hold:

(2+α)

≤ c

(2+α)

≤ c

(α)

, |z

|≤c

(α)

t in R

(2+α)

1+α,R

≤ c

(1+α)

This theorem follows from Theorem 3.3 for s =2+α.

Proof of Theorem 3.3. 1. For s ∈ (α, 2) we can take the function z



,t)asa

solution of the Cauchy problem in R

:= R

n−1

×[0,T] for the equation

∂

− a Δ



=0 inR

(3.16)

with an initial data (3.8). This solution belongs to C

2+α

), and an estimate

(3.12) for it holds [1, 2, 8].

2. Let s ∈ [2, 2+α]. With the help of Lemma 3.1 we construct the functions



,t), z



,t) as solutions of the problems for the equations

∂

− a Δ



= z

(1)



,t)inR

,j=0, 1, (3.17)

with the initial conditions (3.9) for j = 0 and (3.10) for j = 1. Here the functions

(1)



,t) are the solutions of the following Cauchy problems:

∂

(1)

−a Δ



(1)

=0 inR

,j=0, 1, (3.18)

(1)

t=0

= −aΔ



)inR; z

(1)

t=0

= A



)inR. (3.19)

The solutions z

(1)

of problems (3.18), (3.19) exist, belong to C

2+α

s−2

)⊂C

s−2

)

and the estimates (3.7) for them are fulﬁlled with μ

= A

, μ

=0forz

(1)

and

with μ

=0,μ

= A

for z

(1)

, i.e.,

(1)

(α)

s−2,R

≤ c

(s−2j)

,j=0, 1. (3.20)

By Lemma 3.1 the functions z



,t), j =0, 1, belong to C

2+α

)andthe

estimates (3.12), (3.13) for them are valid.

Due to (3.13) and the ﬁrst condition (3.10) we shall have an estimate (3.14)

for z

, really,



,t)| =



∂



,τ) dτ

≤ c

(s−2)

t, s ∈ [2, 2+α].

3. Let s ∈ [1, 2+α]. We determine a function z



,t) as a solution of the Cauchy

problem

∂

− a Δ



=0 inR

t=0

= B



)inR. (3.21)

By [1, 2, 8] this problem has a unique solution z



,t) ∈ C

2+α

s−1

) and it satisﬁes

an estimate (3.15). 

66 G. Bizhanova

Thus, we have extended the functions A



), j =0, 1, and B



)intoR

the functions z



,t), j =0, 1, and z



,t) respectively. Now we extend z



,t),

j =0, 1, and z



,t)intoD

:= R

× (0,T).

Consider two of the ﬁrst boundary value problems with unknown functions

(x, t), j =0, 1,

∂

− a Δ Z

=0 in D

, (3.22)

t=0

=0 in D, Z

= z



,t)onR

, (3.23)

where the function z

is the solution of the problems (3.16), (3.8) for s ∈ (α, 2)

and (3.17), j = 0, (3.9) for s ∈ [2, 2+α], and z

is the solution of the problem

(3.17), j = 1, (3.10).

In the problem for Z

(x, t) the compatibility condition of zero order is not

fulﬁlled and the ﬁrst-order compatibility condition holds by (3.8), (3.9) and for

(x, t) inversely, the compatibility condition of zero order is fulﬁlled and of the

ﬁrst order is not, by (3.10).

We remind that erfs ζ =

√



∞

−ξ

dξ and the norms |u|

(2)

s,δ

, |u|

(2)

2+α,δ

are determined by (1.10), (1.11).

Theorem 3.5. Let A



) ∈ C

(R), s ∈ (α, 2+α], A



) ∈ C

s−2

(R), s ∈ [2, 2+α].

Then each of the problems (3.22), (3.23) has the unique solution

(x, t)=V

(x, t)+



0,s∈ (α, 2),

(x, t),s∈ [2, 2+α],

(3.24)

(x, t)=V

(x, t)+W

(x, t),s∈ [2, 2+α], (3.25)

(x, t)=z



,t)erfc

√

,j=0, 1,

where the functions z



,t) are deﬁned in Theorem 3.3: z



,t) ∈ C

2+α

) and

(2+α)

s,D

≤ c

(s−2j)

,j=0, 1,

(x, t) ∈ C

2+α

s,δ

), δ

,and

(2)

s,δ

≤ c

(s−2j)

,j=0, 1. (3.26)

Theorem 3.6. Let A



) ∈ C

2+α

(R), A



) ∈ C

(R), α ∈ (0, 1). Then each of

problems (3.22), (3.23) has the unique solution Z

(x, t)=V

(x, t)+W

(x, t), j =

0, 1,whereV

= z



,t)erfc

√

, the functions z



,t) are deﬁned in Theorem

3.4: z



,t) ∈ C

2+α,1+α/2



) and

(2+α)

≤ c

(2+α−2j)

,j=0, 1,

(x, t) ∈ C

2+α,δ

), δ

,and

(2)

2+α,δ

≤ c

(2+α−2j)

,j=0, 1.

This theorem follows from Theorem 3.5 for s =2+α.

On the Classical Solvability 67

Proof of Theorem 3.5. The solutions of problems (3.22), (3.23) may be represented

in the explicit form

(x, t)=−2a



dτ



n−1



,τ)∂

Γ(x



−y



,τ)dy



, (3.27)

where Γ(x, t) is a fundamental solution of the heat equation (3.22),

Γ(x, t)=

√

aπ)

−

4at

We construct the solutions of problems (3.22), (3.23) in more suitable forms

than (3.27) to see the character of their singularity. For this, ﬁrst, we write the

potential (3.27) as follows:

(x, t)=V

(x, t) − 2a



dτ



n−1





,τ) − z



,t)



∂

Γ(x



−y



,τ)dy



(3.28)

(x, t):= −2az



,t)



dτ



n−1

∂

Γ(x



− y



,τ)dy



= z



,t)erfc

√

,j=0, 1.

1. Let s ∈ (α, 2). Consider the function V

(x, t)=z



,t)erfc

√

.Itsatis-

ﬁes the homogenous heat equation. Really, z

was constructed in Theorem 3.3 as

a solution of a Cauchy problem for equation (3.16) with an initial condition (3.8):

t=0

= A



). By direct computation with the help of formula (1.13) we can see

that

∂

erfc

√

−a∂

erfc

√

=0, (3.29)

but then

∂

− aΔV

=erfc

√



∂

− aΔ





+ z



,t)



∂

−a∂



erfc

√

=0.

Moreover, due to the relations erfc ∞ = 0, erfc 0 = 1 the function V

(x, t)satisﬁes

the conditions (3.23),

t=0

= z



,t)erfc

√

t=0

=0 in D, V

= z



,t)onR

Thus, the function V

(x, t) is a solution of problem (3.22), (3.23), j =0.We

point out that the last potential in (3.28) satisﬁes equation (3.22) and initial and

boundary conditions (3.23) with zero in the right-hand sides.

2. Let s ∈ [2, 2+α]. The functions V

(x, t)=z



,t)erfc

√

, j =0, 1, are

solutions of the nonhomogeneous equations

∂

−a Δ V

= z

(1)



,t)erfc

√