Escher J., Guidotti P., Hieber M. et al. (editors) Parabolic Problems: The Herbert Amann Festschrift
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68 G. Bizhanova
by (3.17), (3.29). So we determine the functions W
j
(x, t) as solutions of the first
boundary value problems
∂
t
W
j
− aΔW
j
= −z
(1)
j
(x
,t)erfc
x
n
2
√
at
in D
T
,
W
j
"
"
t=0
=0 inD, W
j
"
"
x
n
=0
=0onR
T
,j=0, 1,
(3.30)
then the sum Z
j
(x, t)=V
j
(x, t)+W
j
(x, t) will satisfy the homogeneous equation
∂
t
Z
j
−aΔZ
j
=0,j =0, 1, i.e., (3.22).
We can see that in problem (3.30) the compatibility conditions of zero order
is fulfilled
(3.30) is the problem (A.1). In Theorem 3.3 it was proved that the function
z
(1)
j
, j =0, 1, belongs to C
α
s−2
(R
T
), but then Theorem A.1 is valid for prob-
lem (3.30). In accordance to this theorem, problem (3.30) has a unique solution
W
j
(x, t) ∈ C
2
s,δ
0
(D
T
), δ
0
=
1
8a
, and is subjected to an estimate
|W
j
|
(2)
s,δ
0
,D
T
≤ c
18
|z
(1)
j
|
(α)
s−2,R
T
≤ c
19
|A
j
|
(s−2j)
R
,j=0, 1,
due to (A.3) and (3.20).
By direct substitution of the function Z
j
(x, t)=z
j
(x
,t)erfc
x
n
2
√
at
+ W
j
(x, t),
j =0, 1, into initial and boundary conditions (3.23), we are convinced that it
satisfies these conditions. Thus Z
j
(x, t) are the unique solutions of problem (3.22),
(3.23).
Now we study the second boundary value problem with unknown function
Z
2
(x, t),
∂
t
Z
2
− aΔZ
2
=0 in D
T
, (3.31)
Z
2
|
t=0
=0 inD, ∂
x
n
Z
2
|
x
n
=0
= z
2
(x
,t)onR
T
, (3.32)
where a function z
2
(x
,t) was constructed in Theorems 3.3, 3.4 as a solution of a
Cauchy problem (3.21) with an initial data z
2
|
t=0
= B
0
(x
)inR,whereB
0
(x
)=
ψ(x
, 0) − ∂
x
n
u
0
(x)|
x
n
=0
= 0. We can see that in this problem the compatibility
condition of zero order is not fulfilled for s ∈ [1, 2+α]. We recall also that
ierfc ζ =
∞
ζ
erfc ξdξ, erfc ζ =
2
√
π
∞
ζ
e
−ζ
2
dζ.
Theorem 3.7. Let B
0
(x
) ∈ C
s−1
(R), s ∈ [1, 2+α], α ∈ (0, 1). Then problem
(3.31), (3.32) has the unique solution
V
2
(x, t)=−2
√
at z
2
(x
,t)ierfs
x
n
2
√
at
, (3.33)
where z
2
(x
,t) is defined in Theorem 3.3: z
2
∈C
2+α
s−1
(R
T
), |z
2
|
(2+α)
s−1,R
T
≤C
9
|B
0
|
(s−1)
R
.
Theorem 3.8. Let B
0
(x
) ∈ C
1+α
(R), α ∈ (0, 1). Then problem (3.31), (3.32)
has the unique solution V
2
(x, t) defined by formula (3.33),where z
2
(x
,t) is
determined in Theorem 3.4: z
2
∈ C
2+α
1+α
(R
T
), |z
2
|
(2+α)
1+α, R
T
≤ C
12
|B
0
|
(1+α)
R
.
This theorem follows from Theorem 3.7 for s =2+α.
On the Classical Solvability 69
Proof of Theorem 3.7. The solution of problem (3.31), (3.32) may be written in
the explicit form
Z
2
(x, t)=−2a
t
0
dτ
R
n−1
z
2
(y
,τ)Γ(x
− y
,x
n
,τ)dy
(3.34)
= V
2
(x, t) − 2a
t
0
dτ
R
n−1
z
2
(y
,τ) − z
2
(x
,t)
Γ(x
− y
,x
n
,τ)dy
,
where
V
2
(x,t)=−2az
2
(x
,t)
t
0
dτ
R
n−1
Γ(x
−y
,x
n
,τ)dy
=−2
√
atz
2
(x
,t)ierfs
x
n
2
√
at
.
The function
√
t ierfs
x
n
2
√
at
satisfies an equation
∂
t
− a∂
2
x
n
x
n
√
t ierfs
x
n
2
√
at
=0.
This is confirmed by direct computations with the help of formulas (1.13), (1.14)
for the iterated integrals of the probability. The function z
2
is a solution of a
Cauchy problem (3.21) with an initial condition z
2
|
t=0
= B
0
(x
). But then the
function V
2
(x, t) is a solution of an equation (3.31),
∂
t
V
2
−aΔV
2
= − 2
√
at ierfc
x
n
2
√
at
∂
t
z
2
− aΔ
z
2
− 2
√
az
2
(x
,t)
∂
t
− a∂
2
x
n
x
n
√
t ierfc
x
n
2
√
at
=0.
Moreover, by the relations ierfc ∞ =0, erfc0=1 wehave
V
2
"
"
t=0
=0,∂
x
n
V
2
"
"
x
n
=0
= z
2
(x
,t)erfc
x
n
2
√
at
"
"
x
n
=0
= z
2
(x
,t).
We remark that the last potential in (3.34) satisfies an equation (3.31) and
the homogeneous conditions (3.32).
Thus, we have shown that the function V
2
(x, t) is a unique solution of problem
(3.31), (3.32).
4. Problem 1
Consider the first boundary value problem (1.1)–(1.3) – Problem 1.
Proof of Theorem 2.1. It is known [1, 2, 8] that the solution of Problem 1 belongs
to C
2+α
s
(R
T
), if there are fulfilled the compatibility conditions of zero order for
s ∈ (α, 2) : A
0
(x
)=0onR, and of zero and first orders for s ∈ [2, 2+α]:
A
0
(x
)=0,A
1
(x
)=0onR,where
A
0
(x
):=ϕ(x
, 0) − u
0
(x)
"
"
x
n
=0
on R, (4.1)
A
1
(x
):=∂
t
ϕ(x
,t)
"
"
t=0
− (a Δu
0
(x)+f(x, 0)
"
"
x
n
=0
on R. (4.2)
We study the problem under the conditions A
0
(x
) =0,A
1
(x
) =0on R.
70 G. Bizhanova
In Theorems 3.3, 3.4 we have extended the functions A
j
(x)intoR
T
by the
functions z
j
(x
,t), which satisfy conditions (3.8)–(3.10) and in Theorems 3.5, 3.6
we have continued z
j
(x
,t)intoD
T
by the functions Z
j
(x, t), j =0, 1, as the
solutions of problems (3.22), (3.23):
∂
t
Z
j
−a Δ Z
j
=0inD
T
,Z
j
|
t=0
=0 in D, Z
j
|
x
n
=0
= z
j
(x
,t)onR
T
, (4.3)
and have represented them in the form (3.24), (3.25),
Z
0
(x, t)=V
0
(x, t)+
0,s∈ (α, 2),
W
0
(x, t),s∈ [2, 2+α],
(4.4)
Z
1
(x, t)=V
1
(x, t)+W
1
(x, t),s∈ [2, 2+α], (4.5)
V
j
(x, t)=z
j
(x
,t)erfc
x
n
2
√
at
,j=0, 1. (4.6)
With the help of the functions Z
j
(x, t), j =0, 1, we reduce the original
Problem 1 to a problem with the fulfilled compatibility conditions of the necessary
orders.
We make the substitution
u(x, t)=Z
0
(x, t)+v
1
(x, t),s∈ (α, 2),
u(x, t)=Z
0
(x, t)+Z
1
(x, t)+v
2
(x, t),s∈ [2, 2+α],
(4.7)
in Problem 1 (1.1)–(1.3), where v
1
,v
2
are the new unknowns. For these functions
due to (4.3) we shall have the problems
∂
t
v
i
− a Δ v
i
= f(x, t)inD
T
,i=1, 2, (4.8)
v
1
|
t=0
= u
0
(x)inD, v
1
|
x
n
=0
= ϕ(x
,t) − z
0
(x
,t)onR
T
, (4.9)
and
v
2
|
t=0
= u
0
(x)inD, v
2
|
x
n
=0
= ϕ(x
,t) −
z
0
(x
,t)+z
1
(x
,t)
on R
T
. (4.10)
In problems (4.8), i = 1, (4.9) and (4.8), i = 2, (4.10) the compatibility condi-
tions of zero and of zero and first orders are fulfilled respectively. Really, taking into
account the initial conditions (3.8), (3.9), (3.10) for z
j
(x
,t), ∂
t
z
j
(x
,t), j =0, 1,
and formulas (4.1), (4.2) we derive the identities (compatibility conditions) on the
boundary x
n
=0att = 0 for problem (4.8), i = 1, (4.9)
v
1
"
"
x
n
=0,
t=0
≡ u
0
(x
, 0) = ϕ(x
, 0) − z
0
(x
, 0) = ϕ(x
, 0) − A
0
(x
) ≡ u
0
(x
, 0)
and for problem (4.8), i = 2, (4.10),
v
2
"
"
x
n
=0,
t=0
≡ u
0
(x
, 0) = ϕ(x
, 0) −
z
0
(x,
t)+z
1
(x,
t)
"
"
t=0
= ϕ(x
, 0) − A
0
(x
) ≡ u
0
(x
, 0),
∂
t
v
2
"
"
x
n
=0,
t=0
≡
a Δu
0
(x)+f(x, 0)
"
"
x
n
=0
= ∂
t
ϕ(x
,t)
"
"
t=0
−
∂
t
z
0
(x
,t)+∂
t
z
1
(x
,t)
"
"
t=0
= ∂
t
ϕ(x
,t)
"
"
t=0
− A
1
(x
) ≡
a Δu
0
(x)+f(x, 0)
"
"
x
n
=0
.
On the Classical Solvability 71
Then each of problems (4.8), i = 1, (4.9) and (4.8), i = 2, (4.10) has a unique
solution v
i
∈ C
2+α
s
(D
T
), i =1, 2, and for v
i
the following estimates are valid
[1,2,6,8]:
|v
1
|
(2+α)
s,D
T
≤ c
1
|u
0
|
(s)
D
+ |f|
(α)
s−2,D
T
+ |ϕ|
(2+α)
s,R
T
+ |z
0
|
(2+α)
s,D
T
,
|v
2
|
(2+α)
s,D
T
≤ c
2
|u
0
|
(s)
D
+ |f|
(α)
s−2,D
T
+ |ϕ|
(2+α)
s,R
T
+ |z
0
|
(2+α)
s,D
T
+ |z
1
|
(2+α)
s,D
T
.
(4.11)
Here the functions z
j
(x
,t), j =0, 1, satisfy the estimates (3.12), (3.13):
|z
j
|
(2+α)
s,D
T
≤ c
3
|A
j
|
(s−2j)
R
, and the norms of A
0
and A
1
(see (4.1), (4.2)) are evaluated
by the norms of the given functions u
0
,f,ϕrespectively. Thus, (4.11) leads to
the estimate (2.2). The functions Z
0
, Z
1
in the solution (4.7) of Problem 1 are
expressed via the functions z
j
(x
,t), W
j
(x, t), j =0, 1, satisfying the estimates
(3.12), (3.13), (3.26), i.e., (2.1).
The solution (4.7), (4.4)–(4.6) of Problem 1 contains the function
h
0
(x
n
,t):=erfc
x
n
2
√
at
=
2
√
π
∞
x
n
2
√
at
e
−ξ
2
dξ.
Consider this function and its derivatives
∂
x
n
h
0
(x
n
,t)=−
1
√
aπt
e
−
x
2
n
4at
,∂
t
h
0
= a∂
2
x
n
x
n
h
0
=
x
n
2
√
aπt
3/2
e
−
x
2
n
4at
.
The function h
0
and its derivatives ∂
x
n
h
0
, ∂
t
h
0
, a∂
2
x
n
x
n
h
0
have different limits
at the point ((x
, 0), 0) depending on the approach of the point (x, t)tothispoint
((x
, 0), 0). Really,
lim
t→0
lim
x
n
→0
h
0
(x
n
,t)=1, lim
x
n
→0
lim
t→0
h
0
(x
n
,t)=0, (4.12)
lim
t→0
lim
x
n
→0
∂
x
n
h
0
(x
n
,t)=−∞, lim
x
n
→0
lim
t→0
∂
x
n
h
0
(x
n
,t)=0. (4.13)
Let the point x =(x
1
,...,x
n
) tend to the boundary x
n
= 0 of a domain as
(x
,l
0
t
β
), t → 0, β>0, l
0
=const> 0, then
h
0
(l
0
t
β
,t)=erfs
l
0
2
√
a
t
β−1/2
,
∂
x
n
h
0
(x
n
,t)
"
"
x
n
=l
0
t
β
= −
1
√
aπt
e
−
l
2
0
4a
t
2β−1
,
∂
t
h
0
(x
n
,t)
"
"
x
n
=l
0
t
β
= a∂
2
x
n
x
n
h
0
(x
n
,t)
"
"
x
n
=l
0
t
β
=
l
0
2
√
aπ
t
β−3/2
e
−
l
2
0
4a
t
2β−1
and
lim
t→0
h
0
(l
0
t
β
,t)=
⎧
⎪
⎨
⎪
⎩
0, 0 <β<1/2,
erfc
l
0
2
√
a
,β=1/2,
1,β>1/2,
(4.14)
lim
t→0
∂
x
n
h
0
(x
n
,t)
"
"
x
n
=l
0
t
β
=
0, 0 <β<1/2,
−∞,β≥ 1/2,
(4.15)
72 G. Bizhanova
lim
t→0
∂
t
h
0
(x
n
,t)
"
"
x
n
=l
0
t
β
= lim
t→0
a∂
2
x
n
x
n
h
0
(x
n
,t)
"
"
x
n
=l
0
t
β
=
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
0, 0 <β<1/2,
∞, 1/2 ≤ β<3/2,
l
0
2
√
aπ
,β=3/2,
0,β>3/2.
(4.16)
In particular, for x
n
= l
0
√
t, l
0
=const> 0, we have
h
0
(l
0
√
t, t)=erfc
l
0
2
√
a
=const, (4.17)
∂
x
n
h
0
(x
n
,t)
"
"
x
n
=l
0
√
t
= −
1
√
aπt
e
−
l
2
0
4a
, (4.18)
∂
t
h
0
(x
n
,t)
"
"
x
n
=l
0
√
t
= a∂
2
x
n
x
n
h
0
(x
n
,t)
"
"
x
n
=l
0
√
t
=
l
0
2
√
aπ
1
t
e
−
l
2
0
4a
. (4.19)
From here it is seen that the derivatives ∂
x
n
h
0
"
"
x
n
=l
0
√
t
and ∂
t
h
0
"
"
x
n
=l
0
√
t
,
∂
2
x
n
x
n
h
0
"
"
x
n
=l
0
√
t
tend to −∞ and ∞ as −1/
√
t and 1/t respectively, when t → 0
(x|
x
n
=l
0
√
t
,t) → ((x
, 0), 0
.
Let x
n
≥ r
0
=const> 0, then applying the estimates |ξ|
β
≤ c
β
e
−ξ
2
/2
, β ≥ 0,
erfc ζ ≤
√
2erfs
ζ
√
2
e
−ζ
2
/2
≤
√
2 e
−ζ
2
/2
we shall have
h
0
(x
n
,t) ≤ c
4
e
−
x
2
n
8at
,
|∂
x
n
h
0
(x
n
,t)|≤c
5
x
n
√
t
1
x
n
e
−
x
2
n
4at
≤ c
6
1
r
0
e
−
x
2
n
8at
,
|∂
t
h
0
(x
n
,t)|, |∂
2
x
n
x
n
h
0
(x
n
,t)|≤c
7
1
r
2
0
e
−
x
2
n
8at
,x
n
≥ r
0
.
(4.20)
These inequalities show that the function h
0
=erfc
x
n
2
√
at
and its derivatives ∂
x
n
h
0
,
∂
t
h
0
, ∂
2
x
n
x
n
h
0
tend to zero exponentially as t → 0 in the interior of the domain
x
n
≥ r
0
.
We see that nonfulfillment of the compatibility conditions of zero (A
0
=
0) and first (A
1
= 0) orders in Problem 1 leads to appearance of the singular
functions Z
0
(x, t)andZ
1
(x, t) in the solution respectively (see (4.7), (4.4)–(4.6)),
the principle parts of which are
V
j
(x, t)=z
j
(x
,t)erfc
x
n
2
√
at
,j=0, 1,
where z
j
(x
,t) ∈ C
2+α
s
(R
T
), |z
0
|
(2+α)
s,R
T
≤ c
8
|A
0
|
(s)
R
,s∈ (α, 2+α], |z
1
|
(2+α)
s,R
T
≤
c
9
|A
1
|
(s−2)
R
,s∈ [2, 2+α].
On the Classical Solvability 73
Consider the functions V
j
(x, t), j =0, 1, and its derivatives:
∂
x
V
j
(x, t)=erfc
x
n
2
√
at
∂
x
z
j
(x
,t),
∂
2
x
x
V
j
(x, t)=erfc
x
n
2
√
at
∂
2
x
x
z
j
(x
,t),
(4.21)
∂
x
n
V
j
(x, t)=−
1
√
at
e
−
x
2
n
4at
z
j
(x
,t),
∂
2
x
x
n
V
j
(x, t)=−
1
√
aπt
e
−
x
2
n
4at
∂
x
z
j
(x
,t),
(4.22)
∂
2
x
n
x
n
V
j
(x, t)=
x
n
2a
√
aπt
3/2
e
−
x
2
n
4at
z
j
(x
,t),
∂
t
V
j
(x, t)=erfc
x
n
2
√
at
∂
t
z
j
(x
,t)+a∂
2
x
n
x
n
V
j
(x, t),j=0, 1,
(4.23)
where z
0
(x
, 0) = A(x
) =0,z
1
(x
, 0) = 0 on R.
Due to the function h
0
(x
n
,t):=erfc
x
n
2
√
at
satisfying the estimates (4.20) for
x
n
≥ r
0
V
0
, ∂
x
V
0
, ∂
2
xx
V
0
, ∂
t
V
0
go to zero exponentially as t → 0intheinterior
of the domain. V
0
is bounded, but discontinuous at the point ((x
, 0), 0), as it was
shown (see (4.12), (4.14), (4.17)). The derivatives ∂
x
n
V
0
and ∂
2
x
n
x
n
V
0
containing
∂
x
n
h
0
and ∂
2
x
n
x
n
h
0
respectively have finite or infinite limits as (x, t) → ((x
, 0), 0)
in accordance with (4.13), (4.15), (4.18) and (4.16), (4.19). We can see also that
the derivatives ∂
2
x
n
x
n
V
0
, ∂
t
V
0
are integrable functions with respect to t ∈ (0,T),
T>0. The derivatives ∂
x
V
0
, ∂
2
x
x
V
0
, ∂
t
V
0
and erfc
x
n
2
√
aπt
∂
t
z
0
(x
,t) in (4.23) are
subjected to the estimates
|∂
x
V
0
|≤c
10
t
s−1
2
e
−
x
2
n
4at
,s∈ (α, 1),
|∂
xx
V
0
|, erfc
x
n
2
√
aπt
|∂
t
z
0
(x
,t)|≤c
11
t
s−2
2
e
−
x
2
n
8at
,s∈ (α, 2),
the derivatives ∂
x
z
0
for s ∈ [1, 2+α]and∂
2
x
x
z
0
, ∂
t
z
0
for s ∈ [2, 2+α], in formulas
(4.21), (4.22), (4.23) are bounded functions.
Thanks to the function z
1
(x
,t) satisfying an estimate (3.14): |z
1
|≤c
12
|A
1
|
(s−2)
R
t in R
T
, s ∈ [2, 2+α], the functions V
1
(x, t)and∂
x
V
1
(x, t)arecontinu-
ous, the higher derivatives ∂
t
V
1
(x, t), ∂
2
xx
V
1
(x, t) are bounded, but discontinuous
in the vicinity of a boundary x
n
= 0 of a domain as t → 0.
If A
0
(x
) = 0, but A
1
(x
)=0onR,thenZ
1
(x, t)=0inD
T
and the solution
of Problem 1 takes the form
u(x, t)=Z
0
(x, t)+
v
1
(x, t),s∈ (α, 2),
v
2
(x, t),s∈ [2, 2+α].
If A
0
(x
) = 0, but A
1
(x
) =0onR,thenZ
0
(x, t)=0inD
T
and
u(x, t)=
v
1
(x, t),s∈ (α, 2),
Z
1
(x, t)+v
2
(x, t),s∈ [2, 2+α].
74 G. Bizhanova
For A
0
(x
)=0, A
1
(x
)=0onR we have
u(x, t)=
v
1
(x, t),s∈ (α, 2),
v
2
(x, t),s∈ [2, 2+α].
The derivatives ∂
x
v
1
; ∂
t
v
1
, ∂
2
xx
v
1
and their H¨older constants have singular-
ities of orders t
s−1
2
, s ∈ (α, 1); t
s−2
2
, s ∈ (α, 2), and t
s−2−α
2
, s ∈ (α, 2+α),
respectively in D for t → 0. The function v
2
(x, t) possesses the bounded derivatives
∂
t
v
2
(x, t), ∂
2
x
v
2
(x, t), and their H¨older constants are unbounded of t
s−2−α
2
order,
s ∈ [2, 2+α) and bounded for s =2+α as t → 0.
We can see that the character of the singularities of the functions Z
0
(x, t),
Z
1
(x, t)andv
1
(x, t), v
2
(x, t) is different. Nonfulfillment of the compatibility condi-
tions of the given functions on the boundary x
n
=0att = 0 leads to appearance
of the functions Z
0
(x, t), Z
1
(x, t), which are singular only in the vicinity of a
boundary x
n
= 0 of a domain as t → 0. The derivatives of the function v
1
(x, t)
and the H¨older constants of the higher derivatives of v
1
(x, t)andv
2
(x, t)maybe
singular as t → 0 in the closure of a domain D and their singularities depend on
an initial function u
0
(x) of Problem 1 belonging to C
s
(D).
5. Problem 2
Consider the second boundary value problem (1.1), (1.2), (1.4) – Problem 2.
Proof of Theorem 2.3. 1. For s ∈ (α, 1) the derivatives ∂
x
u have singularities of
order t
s−1
2
as t → 0, so the compatibility condition can not be fulfilled and it is
not required. In this case Problem 2 has a unique solution u(x, t):=v
1
(x, t) ∈
C
2+α
s
(D
T
), and an estimate (2.3), i = 1, for it is valid [1, 2, 8]. The higher deriva-
tives ∂
t
v
1
(x, t), ∂
2
x
v
1
(x, t)andtheirH¨older constants have in D singularities of
orders t
s−2
2
and t
s−2−α
2
respectively as t → 0.
2. Let s ∈ [1, 2+α]. This case requires the fulfillment of the zero-order compati-
bility condition: B
0
(x
)=0onR,where
B
0
(x
):=ψ(x
, 0) − ∂
x
n
u
0
(x)
"
"
x
n
=0
∈ C
s−1
(R
T
)onR (5.1)
for the solution of Problem 2 to belong to C
2+α
s
(D
T
) [1, 2, 6, 8].
We study the problem under the condition B
0
(x
) =0onR.InTheorem
3.3 we have extended the function B
0
(x
)intoR
T
by the function z
2
(x
,t)asa
solution of the Cauchy problem (3.21) satisfying an initial condition
z
2
|
t=0
= B
0
(x
)onR, (5.2)
then we have extended the function z
2
(x
,t)intoD
T
by the function
V
2
(x, t)=−2
√
atz
2
(x
,t)ierfs
x
n
2
√
at
(5.3)
On the Classical Solvability 75
which is a solution of problem (3.31), (3.32),
∂
t
V
2
−aΔV
2
=0 in D
T
,V
2
|
t=0
=0 inD, ∂
x
n
V
2
|
x
n
=0
= z
2
(x
,t)onR
T
. (5.4)
We recall that
ierfs ζ =
∞
ζ
erfs ξdξ, erfs ζ =
2
√
π
∞
ζ
e
−ξ
2
dξ.
After the substitution
u(x, t)=V
2
(x, t)+v
2
(x, t)
in Problem 2 (1.1), (1.2), (1.4) and taking into account that V
2
(x, t)isasolution
of a problem (5.4) we obtain the following problem for the new unknown function
v
2
(x, t),
∂
t
v
2
− a Δ v
2
= f(x, t)inD
T
,v
2
"
"
t=0
= u
0
(x)inD,
∂
x
n
v
2
"
"
x
n
=0
= ψ(x
,t) − ∂
x
n
V
2
"
"
x
n
=0
≡ ψ(x
,t) − z
2
(x
,t)erfs
x
n
2
√
at
"
"
x
n
=0
≡ ψ(x
,t) − z
2
(x
,t).
(5.5)
Due to (5.2), (5.1) in this problem the compatibility condition on the bound-
ary x
n
=0att = 0 is fulfilled:
∂
x
n
v
2
"
"
x
n
=0,
t=0
≡ ∂
x
n
u
0
(x)
"
"
x
n
=0
= ψ(x
, 0) − B
0
(x
) ≡ ∂
x
n
u
0
(x)
"
"
x
n
=0
.
As it was shown in Theorem 3.3, z
2
(x
,t) ∈ C
2+α
s−1
(R
T
) ⊂ C
1+α
s−1
(R
T
)andan
estimate (3.15) for it holds: |z
2
|
(2+α)
s−1,R
T
≤ c
9
|B
0
|
(s−1)
R
,hereB
0
(x
)isevaluatedby
the norms of the functions u
0
,ψ. But then problem (5.5) has a unique solution
v
2
(x, t) ∈ C
2+α
s
(R
T
) and it satisfies an estimate (2.3) [1, 2, 6, 8].
Consider the function V
2
(x, t) (see (5.3)) and its derivatives. V
2
is a continuous
function in D
T
due to the cofactor
√
t. In the vicinity of a boundary x
n
=0the
derivative
∂
x
n
V
2
(x, t)=z
2
(x
t)erfs
x
n
2
√
at
is bounded, but discontinuous as (x, t) → ((x
, 0), 0) (see (4.12), (4.14), (4.17)),
and the second derivative
∂
2
x
n
x
n
V
2
(x, t)=−
1
√
aπt
z
2
(x
t) e
−
x
2
n
4at
is singular as (x, t) → ((x
, 0), 0) (see (4.13), (4.15), (4.18)).
In formula (5.3) the function z
2
(x
,t) belongs to C
2+α
s−1
(R
T
), but thanks to
a cofactor
√
t the orders of the singularities with respect to t as t → 0ofthe
derivatives ∂
x
V
2
, ∂
2
x
x
V
2
,
√
t ierfs
x
n
2
√
at
∂
t
z
2
(x
,t) decrease.
In the interior of a domain x
n
≥ r
0
=const > 0 the function V
2
and its
derivatives ∂
x
V
2
, ∂
2
xx
V
2
, ∂
t
V
2
tend to zero as t → 0 exponentially due to the
estimates erfs ζ ≤
√
2erfs
ζ
√
2
e
−ζ
2
/2
≤
√
2 e
−ζ
2
/2
,ierfsζ ≤ 2ierfs
ζ
√
2
e
−ζ
2
/2
≤
√
πe
−ζ
2
/2
and (4.20).
76 G. Bizhanova
Consider the function v
2
(x, t). The higher derivatives ∂
t
v
2
, ∂
2
xx
v
2
and their
H¨older constants have in D the singularities of orders t
s−2
2
for s ∈ [1, 2) and t
s−2−α
2
,
s ∈ [1, 2+α) respectively as t → 0. These singularities are caused by the initial
function u
0
(x) belonging to C
s
(D). Nonfulfillment of the boundary and initial
functions leads to appearance of the function V
2
(x, t) in the solution of Problem
2, which is singular only in the vicinity of a boundary x
n
=0ast → 0.
Appendix
We have let D := R
n
+
, n ≥ 2, R := {x =(x
,x
n
) |x
∈ R
n−1
,x
n
=0}, x
=
(x
1
,...,x
n−1
), D
T
= D × (0,T), R
T
= R × [0,T].
In Theorems 3.5, 3.6 for s ∈ [2, 2+α] we have constructed the singular
functions Z
j
(x, t)=V
j
(x, t)+W
j
(x, t), V
j
(x, t)=z
j
(x
,t)erfc
x
n
2
√
at
, j =0, 1, (see
(3.24), (3.25)), which appear due to the incompatible boundary and initial data in
Problem 1. The functions V
j
(x, t), j =0, 1, are the principle parts of the singular
solutions Z
j
(x, t), j =0, 1. The functions W
j
(x, t), j =0, 1, are those that remain
after extracting from the singular solutions their principle parts V
j
(x, t), j =0, 1,
and they are the solutions of problems (3.30).
Consider this problem with unknown function W (x, t):
∂
t
W − a Δ W = g(x
,t)erfs
x
n
2
√
at
in D
T
,
W
"
"
t=0
=0 in D, W
"
"
x
n
=0
=0 on R
T
,
(A.1)
where
erfs ζ =
2
√
π
∞
ζ
e
−ξ
2
dξ ≤
√
2erfs
ζ
√
2
e
−ζ
2
/2
≤
√
2 e
−ζ
2
/2
,ζ≥ 0. (A.2)
We can see that in problem (A.1) the compatibility condition of zero order
is fulfilled.
Theorem A.1. Let s ∈ [2, 2+α], α ∈ (0, 1). For every function g(x
,t) ∈ C
α
s−2
(R
T
)
the problem (A.1) has a unique solution W (x, t) ∈ C
2
s,δ
0
(D
T
), δ
0
=
1
8a
,which
satisfies an estimate
|W |
(2)
s,δ
0
,D
T
≤ c
1
|g|
(α)
s−2,R
T
. (A.3)
From this theorem for s =2+α we have the following one.
Theorem A.2. For every function g(x
,t) ∈ C
α, α/2
x
t
(R
T
), α ∈ (0, 1), the problem
(A.1) has a unique solution W (x, t) ∈ C
2
2+α,δ
0
(D
T
), δ
0
=
1
8a
, which satisfies an
estimate
|W |
(2)
2+α,δ
0
,D
T
≤ c
2
|g|
(α)
R
T
.
Here the norms |g|
(α)
s−2,R
T
, |g|
(α)
R
T
and |W |
(2)
s,δ
0
,D
T
, |W |
(2)
2+α,δ
0
,D
T
are defined
by formulas (1.5), (1.9) and (1.10), (1.11) respectively.
On the Classical Solvability 77
Proof of Theorem A.1. The solution of problem (A.1) may be written in the ex-
plicit form
W (x, t)=
t
0
dτ
R
n
+
g(y
,τ)erfc
y
n
2
√
aτ
Γ
n
(x −y,t −τ) − Γ
n
(x −y
∗
,t− τ)
dy
(A.4)
≡
t
0
dτ
R
n−1
g(y
,τ)Γ
n−1
(x
− y
,t− τ) dy
×
∞
0
erfc
y
n
2
√
aτ
Γ
1
(x
n
− y
n
,t− τ) − Γ
1
(x
n
+ y
n
,t− τ)
dy
n
;
here y
∗
=(y
1
,...,y
n−1
, −y
n
),
Γ
n
(x, t)=
1
(2
√
aπt)
n
e
−
x
2
4at
(A.5)
is a fundamental solution of a heat equation satisfying an estimate
|∂
k
t
∂
m
x
Γ
n
(x, t)|≤c
3
1
t
n+2k+|m|
2
e
−
x
2
8at
. (A.6)
We evaluate the norm (1.10) of the potential (A.4). First, with the help of a
tabular formula
∞
−∞
e
−A(x−y)
2
−B(y−z)
2
dy =
√
π
√
A + B
e
−
AB(x−z)
2
A+B
,A,B>0,
we compute an integral
J
1
(x
n
,t−τ, τ; k)=
∞
0
e
−
(x
n
−y
n
)
2
ka(t−τ)
+ e
−
(x
n
+y
n
)
2
ka(t−τ)
e
−
y
2
n
kaτ
dy
n
=
∞
−∞
e
−
(x
n
−y
n
)
2
ka(t−τ)
e
−
y
2
n
kaτ
dy
n
=
√
πak
τ(t − τ)
√
t
e
−
x
2
n
kat
,τ∈ (0,t),k>0,
(A.7)
and estimate it as
J
1
(x
n
,t− τ, τ; k) ≤
√
πak
√
t − τe
−
x
2
n
kat
. (A.8)
Let R
t
= R × [t/2,t],
M
1
:= sup
t≤T
|g(x
,t)|
R
, (A.9)
M
2
:= sup
t≤T
t
2+α−s
2
sup
(x
,t),(z
,t)∈R
t
"
"
g(x
,t) − g(z
,t)
"
"
|x
− z
|
−α
, (A.10)
M
3
:= sup
t≤T
t
2+α−s
2
sup
(x
,t),(x
,t
1
)∈R
t
"
"
g(x
,t) − g(x
,t
1
)
"
"
|t − t
1
|
−α/2
. (A.11)