77
Internat. J. Math. & Math. Sci.
VOL. 16 NO. 4 (1993) 775-782
SOLUTION TO A PARABOLIC EQUATION
WITH INTEGRAL TYPE BOUNDARY CONDITION
IGNACIO BARRADAS
Centro de Investigaci6n en Matemgticas, A.C.
A partado Postal 402
36000 Guanajuato Gto. M6.xico
SALVADOR PEREZ-ESTEVA
Instituto de Matemgtica.s,
Universidad Nacional Aut6noma de M6xico,
Ciudad Universitaria,
04510, M6xico, D.F., Mxico.
(Received September 28, 1992 and in revlsed form January 21, 1993)
ABSTRACT In this paper we study the existence, and continuous dependence of the solution 0
0(z,t) on a H61der space Ua’l+’d2(’,.)(-, [0,1] [0, T], 0 < 7 < 1) of a linear parabolic
equation, prescribing 0(z,0) f(z),O=(1, r) g(r) the integral type condition f O(x, ’)dx E(r).
0
KEY WORD AND PHRASES. Parabolic equation, integral
1991 AMS SUBJECT CLASSIFICATION CODE: 35K20.
Ioundary condition.
1. INTRODUCTION.
Consider the problem of finding 0
O(:, ’) such that
0
<
-
,
(1.1)
(1.2)
(1.3)
(1.4)
E(O) 0f f(z)dz, for b fixed with 0 < b < and r(z, ’) >_ r0 > 0 on [0,1] x [0, T].
In Cannon, Yanpin Lin [1] it is proved a result on existence, uniqueness and continuous dependence for this problem. In this paper we give conditions for which the solution of (1.1)-(1.4) belongs
with
to a Hfilder space and we prove that this solution depends continuously upon the data with respect
to the corresponding Hfilder norms. Similar problems are considered in [2,3,5,6,8,9,10].
Notice that function O satisfies
(1.1)-(1.4)if and only if u(z,t)
O(z -), with
f
0
.
76
BARRADAS AND S. PEREZ-ESTEVA
satisfies
u(x,O)
f(x),
0
< z_< l,
(1.7)
/ u(x,t)dx
(t),
0
<_ <_ T,
(1.8)
0
T
do
r(x, r), T
(--{.,, and (0) E(O) f f(x)dx.
(A) and (B) will denote problems (1.1)-(1.4) and (1.5)-(1.8), respectively. The results
where
(t) E(r), (t)
g(r), a(x,t)
of
on
existence, uniqueness and continuous dependence will be based on a standard fixed point argument
for a contraction defined on a subset of an appropriate functional space. We shall follow Ladyzenskaja
et al. [11] to define the spaces of HSlder continuous functions:
(0, 1) x (0, T), Q--T
[0,1] x [0, T]. For M > 0, k 0,1, 2 and 0 < 7 < 1, H+[0, M]
shall denote the spaces of functions h h(t)in [0,MI, with Ilhll+’) < oo; where
Let QT
k
Ilhll+’)
II’OIIM / IIh’Oll),
Ilhll=
sup
Ih(t)l,
I(t) (t’)l
Ilhll -I(O)1 +
where h(") denotes the derivative of h of order n.
For u Q T "-* R, let
..
lu(:, t)- t,(=’, t)l
sup
Ix
t*.,
telo,rl
Ha,,..,(u)
I,,(:, t) ,(:, t’)l
sup
-t*.,
t,t’e[o33
I1,,11o,
Then
H"/(r) and H+’a+’d(-T
It
sup
will denote the space of all functions
that
I1,,117,’’/
Ilull,,. + uL,(u) + H,r..,/(,,) < o,:,
u"
r "* R such
PARABOLIC EQUATION WITH INTEGRAL TYPE BOUNDARY CONDITION
777
and
IIll/"’’+"/
I111, + I1,,.11, + I1=11, + I1,11,
respectively.
K
and 0
.
K(z, ) will denote the fundental lution to the heat equation
0(z, ) shall be the Theta hnction
EXISTENCE, UNIQUgNgSS AND CONTINUOUS DPENDENCE.
DEFINITION. A function u(z, ) on
is called a solution of problem (B), if
1) u d u are continuous in Q,
2) u is unded in
3)
u satisfi
(1.5)-(1.8).
We notice that if
solution of problem
(B)
u is such that
u is
continuous in
Q and satises (1.)-(1.?), then
u is a
if and only if
=(a, )g’()
=(, ).(a, ) =(0, )(0, )
(.)
or
E’(’) r(b, ’)O.(b, ’) r(O, ’)O.(O, ’),
(2.2)
" < 7", 0 _< < T, provided E is ditferentiable.
We shall assume the following compatibility hypothesis:
for 0 <
H1) .6(0)
f(1),
H2) a(b, 0)g’(0)
R1) /
a(b, O)f(b) a(0,0)f’(0), and the regularity conditions:
H’+(-)[O,T, 6 H[O,T],f
H +’,
T
R2) a, a., a.. H"/a(-T) and I’,s(at)
< oo for some 5 > 0.
Let Vr { E H(v)[0,T] o(0) f’(0)}. We define a nonlinear operator
V- V- as
follows: For So E VT, let u be the unique solution in H+’’1+"/(T) of (1.5)-(1.7), with u=(O,t)
(t), (cf [11], Theorem 5.3 p. 320). Then we define
"
:(t)
of
a(b,t)
(u=(b,t)
a(O, t)
Since u
H+’,’+’da(r) and (H2) holds, we have ’o
then u is a solution of problem (B) and conversely.
Vr, furthermore, if p is a fixed point
I. BARRADAS AND S. PEREZ-ESTEVA
778
LEMMA 2.1. There exists e > 0 not depending on f,,
PROOF. Let T" < T,
(, )
and b in Vr-, h
-2
/ o(,
o
K(z + 2m, t), then
,
)()d +
,
and w
such that if 0 < T < e then
u*
u *. Then
PARABOLIC EQUATION WITH INTEGRAL TYPE BOUNDARY CONDITION
can be estimated just as
779
I, to obtain
IJl _< Ca’llhllr. for _< T.
To estimate J
Since
K(z, ) at (0, 0).
< C’l- bl, then integrating by parts as before, we have
we have to take case of the singularity of
’0-0’’)o0,0
1") l(b 1")
IJl _<
(b,i
0
)(o, )1
+
o o
< 6’s(T + rln)llhllr..
-< T’, where C depends on T, b and function
Hence Iffi(b, )1 < IIl+l&l+l&l-< 6’Tffinllll,
z, t). From this (a) follows immediately.
Now we estimate
For < s we have
+
O.(+,)(F(,.-)-(,-,.)),.
+
+
+
.(-,,.)(F(,.-)-F(,-,’))dd;
.(
,
)F(,
)d
L+L+L+L+Ls+Ls+L+Ls.
We claim that
,=
IL, <
MT’llbllr)l,-tl
IL, <
MT"/’IIIIr’.":PI
ILsl <
MsT’llllls-tl
tl
"
(2.6)
where Mi depends on T, b and function a(x,t),
1, ...,8.
The proof of (2.4) follows as the proof of part (a). For (2.5) we let c(x,t)
L.
] K.,(b, 1")(c(0,
o
s
(2.4)
,...,6,
1")w.(O, s 1") c(O,
1")wffi(O,
*(,O--(b.qo(b,0
rI)dr
then
I. BARRADAS AND S. PEREZ-ESTEVA
780
J+J+Ja.
Since c(, t).=
o(1
Sl), we obtain
Ig, _< K,"
IIllT)lt-l
(2.8)
n y (V),
I( blSr..(
J
(, )w.((,
)
o o
0% I(
bl
Hence
II. I+’Y,I +/’I I.
,"- 1.
IJl < KaT’t/’II’IIT"
We obtain (2.5) from (2.7), (2.8), (2.9) and the fact that
(2.9)
Ilwll+."’’+’n _< MIIhll’.-P), where M
does not depend on T* (ee [] Theorem 5.4, p. 322). With a similar axgumnt we obtain (2.6),
and the proof of the Lemma follows from (2.4) (2.5) and (2.6).
REMARK. Notice that Lemma 2.1(a) holds for any two functions 0, /, for which u u axe
well defined, u= u= axe continuous in
and u.,
are b6unded in
THEOREM 2.2. Assume that H, Ha, R, R hold. Then there exists a unique solution
u u(x,) of Problem (B). This solution belongs to H+"’+"n(-T and satisfies
,
.
r-
II,ll+’’’+’n _<
PROOF. Let
>0
a
o,r i q0. ’1
,’
T"
C(T){llll’+(’,- + IIll"-’ + IIfll:+’}.
in I.,emrna 2.1 and T"
’.-V o
,io.
< ," then if we define the equence (a:)
,o.
Furthermore
,11.’) + t1,11
< 211,
_< c,
Then for u "QT"
R defin by u
..,+,,/,
lull.,
S C,
{llll+."’-’ + IIll’ + IIfll:+’}
u
,
we ha
{ll#llt’ + I111 + IlYll?+"}.
Hene u is solution o he 1o problem. Since
u csn be obtned by
&
ad
dend on
stdsrd step by step construction, d
+/
"
only,
loba lufion
PARABOLIC EQUATION WITH INTEGRAL TYPE BOUNDARY CONDITION
781
Finally, the remark after Lemma 2.1 implies that any solution of (B) in QT has to be u.
[2]
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