Applied Mathematics E-Notes, 14(2014), 161-172 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
Well-Posedness Results For A Third Boundary Value
Problem For The Heat Equation In A Disc
Arezki Khelou…y
Received 10 March 2014
Abstract
In this work we prove well-posedness results for the following one space linear
second order parabolic equation @t u @x2 u = f , set in a domain
= (t; x) 2 R2 :
r < t < r; '1 (t) < x < '2 (t)
1
of R2 , where 'i (t) = ( 1)i r2 t2 2 ; i = 1; 2 and with lateral boundary conditions of Robin type. The right-hand side f of the equation is taken in L2 ( ).
The method used is based on the approximation of the domain by a sequence
of subdomains ( n )n which can be transformed into regular domains.
1
Introduction
Let = D (0; r) be the open disc centred at the origin of R2 and with radius r > 0,
characterized by = (t; x) 2 R2 : r < t < r; '1 (t) < x < '2 (t) ; where '1 and '2
1
k
are de…ned on [ r; r] by 'k (t) = ( 1) r2 t2 2 ; k = 1; 2. The lateral boundary of
is de…ned by k = (t; 'k (t)) 2 R2 : r < t < r , k = 1, 2. In , we consider the
Robin type boundary value problem
@t u @x2 u = f in ;
@x u + k uj k = 0, k = 1, 2,
(1)
where the coe¢ cients k , k = 1,2 are real numbers satisfying non-degeneracy assumptions (to be made more precise later) and the right-hand side term f of the equation
lies in L2 ( ), the space of square-integrable functions on with the measure dtdx.
The main di¢ culty related to this kind of problems is due to the fact that '1
coincides with '2 for t = r and for t = r, which prevents the domain
to be
transformed into a regular domain by means of a smooth transformation.
The case k = 1, k = 1, 2, corresponding to Dirichlet boundary conditions is
considered in [19]. We can …nd in [6] a study of the case k = 0, k = 1, 2, corresponding to Neumann boundary conditions and in [23] an abstract study in the case
Mathematics Subject Classi…cations: 35K05, 35K20.
des Mathématiques Appliquées, Faculté des Sciences Exactes, Université de Bejaia,
6000, Béjaia, Algeria. Lab. E.D.P.N.L., Ecole Normale Supérieure, 16050-Kouba, Algiers, Algeria
y Laboratoire
161
162
Well-Posedness Results for the Heat Equation
( 1 ; 2 ) = (1; 0), corresponding to mixed (Dirichlet-Neumann) lateral boundary conditions. However, the boundary assumptions dealt with by the authors exclude our
domain. Further references on the analysis of parabolic problems in non-cylindrical
domains are: Labbas et al. [13, 14, 15], Khelou… et al. [8, 9, 10, 12], Degtyarev [5],
Aref’ev and Bagirov [3, 4], Sadallah [20, 21, 22], Alkhutov [1, 2] and Paronetto [17].
In this work, we consider the case of Robin type boundary condition, namely, the
case where k 6= 0, k = 1,2, and we look for su¢ cient conditions (as weak as possible)
on the lateral boundary of the domain and on the coe¢ cients k , k = 1, 2 in order
to obtain the maximal regularity of the solution in an anisotropic Hilbertian Sobolev
space.
In previous works (see [7, 11]), we have studied the case where
= (t; x) 2 R2 : 0 < t < T ;
1
(t) < x <
2
(t)
with the fundamental hypothesis 1 (0) = 2 (0) and we have proved that the solution
u of Problem (1) is unique and has the optimal regularity, that is a solution u belonging
to the anisotropic Sobolev space
H 1;2 ( ) := u 2 H 1;2 ( ) : @x u +
k uj
k
= 0, k = 1,2
with
H 1;2 ( ) = u 2 L2 ( ) : @t u; @x u; @x2 u 2 L2 ( ) ;
under su¢ cient conditions on
0
k
(t) (
2
(t)
k;
1
k = 1; 2, that are
(t))
!
0
as t ! 0, k = 1; 2.
1
k
Examples of functions satisfying this last condition are k (t) = ( 1) r2 t2 2+ ; k =
1; 2 for all < 0. However, the above condition is false in the case = 0 corresponding
to the class of domains considered in this article. So, the well-posedeness result which
we will prove here can not be derived from [7] and [11]. In order to overcome this
di¢ culty, we impose su¢ cient conditions on the lateral boundary of the domain and
on the coe¢ cients k , k = 1, 2; that are,
1
( 1)
k
k
t
p
2
2 r
< 0;
t2
2
> 0,
(2)
0 a.e. t 2 ] r; r[ , k = 1; 2;
(3)
and
1
(16 + 4
2
1
+4
2
2 )r
+ (4 j
1j
+ 4j
2 j) r
2
+ (8 + 4
2
1
+4
2 3
2 )r
> 0:
(4)
Then, our main result is following:
THEOREM 1. Under the hypothesis (2), (3) and (4), the heat operator L = @t
is an isomorphism from H 1;2 ( ) into L2 ( ).
@x2
A. Khelou…
163
It is not di¢ cult to prove the injectivity of the operator L. Indeed, If u is a solution
of Problem (1) with a null right-hand side, the calculations show that the inner product
hLu; ui in L2 ( ) gives
0=
2 Z
X
k=1
( 1)
k
p
k
t
2 r2
k
u2 (t; 'k (t)) dt +
t2
Z
2
(@x u) dtdx.
The hypothesis (3) implies that @x u = 0 and consequently @x2 u = 0. Then, the equation
of Problem (1) gives @t u = 0. Thus, u is constant. The boundary conditions and the
fact that k 6= 0, k = 1, 2 imply that u = 0 in . So, in the sequel, we will be interested
only by the question of the surjectivity of the operator L.
The method used here is the domain decomposition method. More precisely, we
divide into two parts
1
= f(t; x) 2
:
r < t < 0g and
So, we obtain two solutions uk 2 H 1;2 (
function u de…ned by
u :=
k)
in
u1 in
u2 in
2
= f(t; x) 2
k,
: 0 < t < rg .
k = 1, 2. Finally, we prove that the
1;
2;
is the solution of problem (1) and has the optimal regularity, that is u 2 H 1;2 ( ).
The plan of this paper is as follows. In Section 2, we prove that Problem (1) admits
a (unique) solution in the case of a "truncated" domain. Then, in Section 3, we
approximate
by a sequence ( n ) of such truncated domains and we establish an
energy estimate which will allow us to pass to the limit and complete the proof of our
main result.
2
Resolution of Problem (1) in a Truncated Disc
n
For each n 2 N , we de…ne
n
:=
(t; x) 2 R2 :
THEOREM 2. Assume that
and let fn = f j n and
n;k
=
r<t<r
k
1
; ' (t) < x < '2 (t) :
n 1
and 'k , k = 1; 2 verify assumptions (2) and (3)
(t; 'k (t)) 2 R2 :
r<t<r
1
n
for k = 1; 2:
Then, for each n 2 N , the problem
@t un @x2 un = fn 2 L2 ( n ) ;
@x un + k un j n;k = 0, k = 1, 2,
admits a (unique) solution un 2 H 1;2 (
n ).
(5)
164
Well-Posedness Results for the Heat Equation
PROOF. We divide
n,
= f(t; x) 2
So, we have
n
=
:
[
LEMMA 1. Let f
k
n 2 N into two parts
+
n
r < t < 0g and
+
n
[ (f0g
= fj
=
(t; x) 2
:0<t<r
1
n
.
]'1 (0) ; '2 (0)[) .
and
= (t; 'k (t)) 2 R2 :
r<t<0
for k = 1; 2:
Then, the problem
@t u
@x2 u = f 2 L2 ( ) ;
@x u + k u j = 0, k = 1,2,
k
admits a (unique) solution u 2 H
1;2
(
).
PROOF. Since '1 is a decreasing function on ] r; 0[ and '2 is an increasing function
on ] r; 0[, then the result follows from [18].
Hereafter, we denote the trace u jf0g ]'1 (0);'2 (0)[ by , which is in the Sobolev
space H 1 (f0g ]'1 (0) ; '2 (0)[) because u 2 H 1;2 ( ) (see [16]). Now, consider the
following problem on +
n, n 2 N
8
2
+
+
2 +
+
>
< @t un @x un = fn 2 L ( n ) ;
+
un jf0g ]'1 (0);'2 (0)[ = 2 H 1 (f0g ]'1 (0) ; '2 (0)[) ;
(6)
>
+
: @x u+
n + k un j + = 0, k = 1, 2,
n;k
1
2
where +
n;k = (t; 'k (t)) 2 R : 0 < t < r
n , k = 1, 2.
We use the following result, which is a consequence of Theorem 4.3 in [16] to solve
Problem (6).
PROPOSITION 1. Let Q be the rectangle ]0; T [ ]0; 1[, f 2 L2 (Q) and
with 0 = f0g ]0; 1[. Then, the problem
8
2
2
< @t u @x u = f 2 L (Q) ;
uj 0 = ;
:
@x u + k uj k = 0, k = 1, 2,
where
1
= ]0; T [
f0g and
2
= ]0; T [
2 H1 (
0)
f1g admits a (unique) solution u 2 H 1;2 (Q).
REMARK 1. We have
lies in H 1 (f0g ]'1 (0) ; '2 (0)[), then @x is (only)
2
in L (f0g ]'1 (0) ; '2 (0)[) and its pointwise values should not make sense. So in
the application of [[16] Theorem 4.3, Vol. 2], there are no compatibility conditions to
satisfy.
Thanks to the transformation (t; x) 7! (t; y) = (t; ('2 (t)
deduce the following result:
'1 (t)) x + '1 (t)) ; we
A. Khelou…
165
PROPOSITION 2. For each n 2 N , Problem (6) admits a unique solution u+
n 2
H ( +
n ).
1;2
So, the function un 2 H 1;2 (
n ),
n 2 N de…ned by
un :=
u in
u+
n in
;
+
n;
is the (unique) solution of Problem (5). This completes the proof of Theorem 2.
3
Resolution of Problem (1) in the Half Disc
+
In this section, we de…ne
+
:= (t; x) 2 R2 : 0 < t < r; '1 (t) < x < '2 (t)
and consider the following problem in +
8
+
@x2 u+ = f + 2 L2 ( + ) ,
< @t u
+
u jf0g ]'1 (0);'2 (0)[ = 0,
:
@x u+ + k u+ j + = 0, k = 1, 2,
(7)
k
where f
+
= fj
+
and
+
k
= (t; 'k (t)) 2 R2 : 0 < t < r
for k = 1; 2:
We assume that k and 'k , k = 1; 2 verify assumptions (2), (3) and (4) and we denote
+
1;2
( +
and u+
fn+ = f + j +
n . Such a solution
n ) the solution of Problem (7) in
n 2 H
n
exists by Proposition 2.
PROPOSITION 3. There exists a constant K > 0 independent of n such that
u+
n
H 1;2 (
+
n
K f+
K fn+
L2 (
+
n
)
+
) + @t un
2
L2 (
+
n
+
) + @x un
)
L2 (
+)
,
where
u+
n H 1;2 (
+
n
)=
r
u+
n
2
L2 (
+
n
2
L2 (
+
n
2 +
) + @x un
2
L2 (
+
n
In order to prove Proposition 3, we need the following result
LEMMA 2. We have the following estimations
(i) j'0k (t)j ('2 (t) '1 (t)) 2r for t 2 ] r; r[ and k = 1, 2.
R ' (t)
2 R ' (t)
2
2
(ii) ' 2(t) [@xj u+
['2 (t) '1 (t)] ' 2(t) [@xj+1 u+
n (s; x)] ds
n (s; x)] ds for j = 0; 1.
1
1
).
166
Well-Posedness Results for the Heat Equation
2
(iii) k@x u+
n kL2 (
+
n
2
L2 (
4r2 @x2 u+
n
)
+
n
).
PROOF OF PROPOSITION 3. We have
fn+
2
L2 (
+
n
)
= h@t u+
n
+
@x2 u+
n ; @t un
2
@t u+
n L2 (
=
Let us consider the term
R
2
+
n
+
n
)+
@x2 u+
ni
2
@x2 u+
n L2 (
2
Z
+
n
2 +
@t u+
n :@x un dtdx
=
=
2
)
2 +
@t u+
n :@x un dtdx.
+
n
2 +
@t u+
n :@x un dtdx. We have
2 +
+
+
@t u+
n :@x un = @x @t un :@x un
Then
+
n
Z
2
Z
@
Z
+
n
+
n
h
1
@t @x u+
n
2
+
@x @t u+
n @x un
@x u+
n
2
t
dt dx +
2
.
Z
+
2@t u+
n @x un
+
n
i
x
@t @x u+
n
2
dt dx
d ,
with t , x are the components of the unit outward normal vector at @ +
n . We shall
rewrite the boundary integral making use of the boundary conditions. On the part
+
+
of the boundary of +
n where t = 0, we have un = 0 and consequently @x un = 0.
The corresponding boundary integral vanishes. On the part of the boundary where
1
t = r
n , we have x = 0 and t = 1. Accordingly the corresponding boundary
R '2 (r n1 )
2
integral ' r 1 (@x u+
n ) dx is nonnegative. On the parts of the boundary where
1(
n)
x = 'k (t), k = 1, 2, we have
k
x
( 1)
=q
,
2
1 + ('0k ) (t)
k+1
t
( 1)
'0k (t)
=q
and @x u+
n (t; 'k (t)) +
2
0
1 + ('k ) (t)
+
k un
(t; 'k (t)) = 0:
Consequently, the corresponding boundary integrals In;k and Jn;k , k = 1; 2 are the
following:
Z r n1
2
k+1
In;k = ( 1)
'0k (t) @x u+
n (t; 'k (t)) dt, k = 1, 2,
Jn;k
k
=
( 1) 2
Z
0
r
1
n
+
k @t un
(t; 'k (t)) :u+
n (t; 'k (t)) dt, k = 1, 2.
0
We have
2
Z
+
n
2 +
@t u+
n :@x un dtdx
jIn;1 j
jIn;2 j
jJn;1 j
jJn;2 j .
It is the reason for which we look for an estimate of the type
jIn;1 j + jIn;2 j + jJn;1 j + jJn;2 j
@x2 u+
n
2
L2 (
+
n
);
(8)
A. Khelou…
167
where is a positive constant independent of n belonging to the interval ]0; 1[. By
2 (t) x
introducing the function (t; x) = ' '(t)
' (t) like in [18], we write for In;1
2
Z
jIn;1 j =
Z
1
n
r
(Z
'2 (t)
'01 (t) @x
'1 (t)
0
+
n
'01 (t) ('2 (t)
2
2r @x2 u+
n
+
(t; x) @x u+
n (t; x)
Z
2
dtdx
+
2
'1 (t)) @x2 u+
n
Z
1
2
2
+
@x2 u+
j'01 j @x u+
n
n
)
2
dx dt
+
n
2
2 +
j'01 j @x u+
n @x un dtdx
dtdx
+
n
2r + +
4r2
2
@x2 u+
n L2 (
7r
1
@x2 u+
n
+
n
2
L2 (
+
n
)
).
The last inequality is obtained by choosing
jIn;2 j
= r. Similarly, we have
7r @x2 u+
n
2
L2 (
+
n
).
2
Let us now consider the terms Jn;k , k = 1, 2. By setting h (t) = (u+
n ) (t; 'k (t)) , we
obtain
Jn;k
=
k
( 1)
Z
r
1
n
k:
0
=
( 1)
k
r
k :h (t)
h
h0 (t)
1
n
'0k (t) @x u+
n
k+1
+ ( 1)
0
Z
r
1
n
2
i
(t; 'k (t)) dt
0
k :'k
(t) @x u+
n
2
(t; 'k (t)) dt.
0
2
r
k
1
n
Condition (2) and the fact that (u+
0: In
k :h (t)
n ) (0; 'k (0)) = 0 give ( 1)
0
the sequel, we estimate the last boundary integral in the expression of Jn;k , namely
Ln;k = ( 1)
k+1
Z
r
1
n
0
k :'k
(t) @x u+
n
2
(t; 'k (t)) dt.
0
We have
@x u+
n
2
(t; '1 (t))
x=' (t)
=
=
=
Z
2
'2 (t) x
2
@x u+
(t; x)
n
'2 (t) '1 (t)
x='1 (t)
Z '2 (t)
'2 (t) x
2
@x
@x u+
(t; x) dx
n
'2 (t) '1 (t)
'1 (t)
'2 (t)
'1 (t)
1
'2 (t)
'1 (t)
@x u+
n
2
(t; x)
'2 (t) x
@ 2 u+
'2 (t) '1 (t) x n
2
(t; x) dx.
168
Well-Posedness Results for the Heat Equation
So,
Ln;1 =
Z
0
1 :'1
+
n
'2 (t)
(t)
@x u+
n
'1 (t)
2
'2 (t) x
'2 (t) '1 (t)
(t; x)
0
1 :'1
(t) @x2 u+
n
2
(t; x) dtdx.
By using the equalities
2
@x u+
n
and
2
@x2 u+
n
+
(t; x) = 2@x u+
n (t; x) un (t; x)
+
+
(t; x) = 2@x2 u+
n (t; x) un (t; x) + 2 @x un (t; x)
2
;
we obtain
Ln;1
=
=
Z
2 1 :'01 (t)
+
@x u+
n (t; x) un (t; x) dtdx
+ ' (t)
'
(t)
2
1
Zn
'2 (t) x
+
2 1 :'01 (t) @x2 u+
n (t; x) un (t; x) dtdx
+ ' (t)
'
(t)
2
1
Z n
'2 (t) x
2
2 1 :'01 (t) @x u+
n (t; x) dtdx
+ ' (t)
'
(t)
2
1
n
An;1 + Bn;1 + Cn;1 :
Estimation of An;1 , Bn;1 and Cn;1
a) We have
Z
2 1 :'01 (t)
+
@x u+
An;1 =
n (t; x) un (t; x) dtdx;
+ ' (t)
'
(t)
2
1
n
then
jAn;1 j
Z
+
Z
+
1
[
+
n
Z
2
1
2
(t)]
@x u+
n (t; x)
1
+
n
1
[
+
n
Z
0
1 :'1
['2 (t)
0
1 :'1
(t)] ['2 (t)
2
['2 (t)
'1 (t)]
4r2 + 4r2
@x2 u+
n
+
n
dtdx
u+
n (t; x)
2
'1 (t)]
2
2
2
'1 (t)]
2
@x2 u+
n (t; x)
@x2 u+
n (t; x)
2
L2 (
+
n
)
dtdx
2
2
dtdx
dtdx
4r3 + 4
2
1r
@x2 u+
n
2
L2 (
The last inequality is obtained by choosing = r.
b) We have
Z
'2 (t) x
+
2 :'0 (t) @x2 u+
Bn;1 =
n (t; x) un (t; x) dtdx;
+ ' (t)
'1 (t) 1 1
2
n
+
n
).
A. Khelou…
169
then
2
1
jBn;1 j
2
1
Z
2
+
n
j'01 (t)j
2
L2 (
2 4
1r
+
n
2
L2 (
+
n
The last inequality is obtained by choosing
c) We have
Z
Cn;1 =
+
n
4
@x2 u+
n
'1 (t)]
2
L2 (
+
n
2
L2 (
)
+
n
)
)
@x2 u+
n
+
@x2 u+
n
dtdx +
2
@x2 u+
n
4
2
j'01 (t)j ['2 (t)
sup
t2[0;r]
+
u+
n (t; x)
4
)
2 3
1r
+r
2
L2 (
@x2 u+
n
+
n
):
= r.
'2 (t) x
2
'2 (t) '1 (t)
0
1 :'1
(t) @x u+
n (t; x)
2
dtdx
then
jCn;1 j
Z
2j
1j
4j
1j r
2
+
n
j'01 (t)j j'2 (t)
@x2 u+
n
2
L2 (
+
n
'1 (t)j
2
@x2 u+
n (t; x)
2
dtdx
).
Consequently,
jLn;1 j
(4 + 4
2 3
1 )r
+ 4j
1j r
2
+ (1 + 4
2
1 )r
@x2 u+
n
2
L2 (
+
n
).
2 3
2 )r
+ 4j
2j r
2
+ (1 + 4
2
2 )r
@x2 u+
n
2
L2 (
+
n
).
Similarly, we can obtain
jLn;2 j
(4 + 4
Summing up the above estimates, we obtain
fn+
2
L2 (
+
n
)
2
L2 (
2
2 +
jIn;1 j jIn;2 j jLn;1 j jLn;2 j
) + @x un L2 ( +
n)
n
h
2
@x2 u+
1
(16 + 4 21 + 4 22 )r + (4 j 1 j + 4 j 2 j) r2
n L2 ( +
n)
io
2
+(8 + 4 21 + 4 22 )r3 + @t u+
n L2 ( + ) :
@t u+
n
+
n
n
2
Using the condition (4) and since kfn+ kL2 (
proved.
2
+
n
)
kf + kL2 (
+)
, then Proposition 3 is
THEOREM 3. Problem (7) admits a (unique) solution u+ 2 H 1;2 (
+
).
170
Well-Posedness Results for the Heat Equation
PROOF. The estimation of Proposition 3 shows that
+
uf
n
L2 (
+)
+
+ @]
t un
L2 (
+)
+
2
X
i +
@]
x un
i=1
L2 (
+)
C f+
L2 (
+)
;
+ ]
+ ]
+
i +
where e: denotes the 0 extension of u+
. This means that uf
n , @t un ; @x un , i = 1; 2
n to
2
+
are bounded functions in L ( ). The following compactness result is well known: A
bounded sequence in a re‡exive Banach space (and in particular in a Hilbert space) is
weakly convergent. So for a suitable increasing sequence of integers nk , k = 1; 2; :::,
there exists functions u+ ; v + ; vi+ ; i = 1; 2 in L2 ( + ) such that
g
+
+
+
+ ^
+ ^
i +
u
nk * u ; @t unk * v ; @x unk * vi ; i = 1; 2
weakly in L2 ( + ) as k ! 1: Clearly, v + = @t u+ ; vi+ = @xi u+ ; i = 1; 2 in the sense of
distributions in + and so in L2 ( + ). Finally, u+ 2 H 1;2 ( + ) and
@t u+
@x2 u+ = f in
+
:
On the other hand, the solution u+ satis…es the boundary conditions, since
8n 2 N ; u+
+
n
= u+
n.
REMARK 2. The function u 2 H 1;2 ( ) de…ned by
u :=
u in
u+ in
+
;
,
is the (unique) solution of Problem (1).
Acknowledgment. I want to thank the anonymous referee for a careful reading
of the manuscript and for his/her helpful suggestions.
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