© Pleiades Publishing, Ltd., 2009.Published in Russian in Doklady Akademii Nauk, 2009, Vol.. In this article a semilinear boundary value problem is studied for a degenerate parabolic pse
Trang 1ISSN 1064–5624, Doklady Mathematics, 2009, Vol 80, No 1, pp 482–486 © Pleiades Publishing, Ltd., 2009.
Published in Russian in Doklady Akademii Nauk, 2009, Vol 427, No 2, pp 155–159.
In this article a semilinear boundary value problem
is studied for a degenerate parabolic pseudodifferential
equation The main result generalizes the famous
theo-rem of Agranovich and Vishik (see [1]) We prove the
existence of a solution using the Rothe theorem on a
fixed point (see [2])
1. Let p∈ (1, ∞), l∈+, q = µ + iτ∈, µ≥ 0, d≥ 1,
d∈ , δ = (δ1, δ2, …, δn – 1, 0), δi≥ 0, i = 1, 2, …, n – 1
For each ξ' = (ξ1, ξ2, …, ξn – 1) ∈n – 1, x n ∈ +, set
Define the Laplace transform
ξ' q,
( )δ,d ξi
p/ 1( + δi)
x n pδi
i= 1
n 1
,
=
ξ' x, ,n q
p
x n pδi
i= 1
n 1
,
=
n 1u(ξ' xn, ,q)
2π ( )(n 1 ) /2 - e–i x'〈 ,ξ'〉u x' x( , ,n q)d x',
n∫1
n 1v(ξ' q, )
2π ( )(n 1 ) /2 - e–i x'〈 ,ξ'〉v(x' q, )d x'.
n 1
∫
The space H(l, p, δ)(n – 1) is defined as the comple-tion of the space (n – 1) with the norm
The space H (l, p, δ)( ) (l ∈ +) is defined as the
completion of the space P( ) = {u ∈ (n ): suppu ∈
} using the norm
The space P l, p (d, µ, δ, × (0, +∞)) is defined
as the completion of the space P(n ) = {u ∈ (n ×
(–∞, +∞)): suppu ⊂ × (0, +∞)} with the norm
2π ( )1/2 - e–qt u x t( , )d t.
0
+ ∞
∫
=
C0∞
v
l p dδ n 1 , , , ,
= 1( + (ξ' q, )δ,d)lp
n 1v(ξ' q, ) p
ξ'
d
n 1
∫
+
n
+
n
C0∞
+
n
u l p d, , , ,δ +
= 1( + (ξ' q, )δ,d+ (ξ' x, ,n q)δ,d)(l–j)p
0
+ ∞
∫
n∫1
j= 0
l
∑
⎝
⎜
⎛
∫ × D n
j
n 1u(ξ' xn, ,q) p
x n dξ'
d
⎠
⎟
⎞ -1p
+
n
C0∞
+
n
u
P l p, (d, , , µ δ + × ( 0 + , ∞ ) )
= 1( + (ξ' q, )δ,d+ (ξ' xn, ,q)δ,d)p l( –j)
0
+ ∞
∫
ξn' , τ
∫
j= 0
l
∑
⎝
⎜
⎜
⎛
∫ × D n
j
n u e( –µt u x t( , )) ξ( ' xn, ,τ) p
x n dξ'dτ
d
⎠
⎟
⎞1 -p
MATHEMATICS
On a Semilinear Boundary Value Problem
Yu V Egorova, Nguyen Minh Chuongb, and Dang Anh Tuanc
Presented by Academician V.S Vladimirov February 18, 2009
Received March 6, 2009
DOI: 10.1134/S1064562409040085
a Université Paul Sabatier, Toulouse, France
b Institute of Mathematics, Hanoi, Vietnam
c Hanoi National University, Hanoi, Vietnam
1 The article was translated by the authors.
Trang 2The space P l, p (d, µ, δ, n – 1× (0, +∞)) is defined as
the completion of the space P∞(n – 1) = {v∈ (n – 1×
(–∞, +∞)): suppv⊂n – 1× (0, +∞)} with the norm
The space E l, p (d, µ, δ, n – 1) is defined as the
com-pletion of the space P(n – 1) with the norm
The space E l, p (d, µ, δ, ) is defined as the
comple-tion of the space P∞(n) with the norm
The spaces P l, p (d, µ, δ, ∂Ω× (0, +∞)), P l, p (d, µ, δ,
Ω × (0, +∞)), E l, p (d, µ, δ, ∂Ω) and E l, p (d, µ, δ, Ω),
where Ω is a bounded set in n with smooth boundary
∂Ω, are defined in the standard way using a unity parti-tion
2 Consider a pseudodifferential operator L(x, D, q, y)
with a symbol
where m ∈+, γi = 1 + δi , i = 1, 2, …, n, lα≥ 0,
,
supp (·, ξ') ⊂ Kα, where Kα are compact sets inde-pendent of ξ'
Let us introduce the operator
Let us now set
where
supp (·, ξ') ⊂ K j , K j are compact sets independent
of ξ',
A pseudodifferential operator b(x', D', q, d) with a symbol b(x', ξ', q, d) is denned as usually:
3 Consider the problem
(2)
where
C0∞
v
P l p, dµ δ n 1 , , , × 0 + , ∞ )
= 1 (ξ' q, )δ,d)pl
n[e–µt v(x' t, )] ξ( ',τ) p dξ'dτ +
ξn' , τ
∫
v
E l p, dµ δ n 1
, , ,
( ) v l p q dδ n 1
, , , , ,
p
τ
d
τ
∫
=
+
n
v
E l p, (d, , , µ δ + ) v l p q d, , , , , δ +
τ
∫
-
=
L x( , , ,ξ q d) aα(x,ξ')x n lαξn
α
qβ,
α +dβ ≤2m
γ α ,
〈 〉 +dβ –lα =2m
∑
=
aα(x,ξ') = aα( )0 ( )ξ' +aα( )1 (x,ξ'),
aα( )0 ( )· C n 1
\ 0{ } ( ), aα( )1(x ·, ) C n 1
\ 0{ }
x
∀ ,∀α
aα( )1 (·,ξ') C0∞( )n , ∀ξ' n 1
\ 0{ },
aα( )1
aα( )0 (λγ1ξ1,λγ2ξ2, ,… λγn 1ξn 1) = aα( )0 ( )ξ' ,
ξ'
∀ n 1
\ 0{ }, ∀λ>0,
∈
aα( )1(x,λγ1ξ1,λγ2ξ2, ,… λγn 1ξn 1) = aα( )1 (x,ξ'),
x
, ∀ξ' n 1
\ 0{ }, ∀λ>0
L x D q d( , , , )u x q( , ) 1
2π ( )(n 1 ) /2
-α +dβ ≤2m
γ α ,
〈 〉+dβ–lα=2m
∑
= e–i x'〈 ,ξ'〉aα(x,ξ')x n lα( )ξ' α'
qβD nαnn 1u(ξ' x, ,n q)dξ'
∫n 1
b x'( , , ,ξ' q d) (b( )j0 ( )ξ' +b( )j1 (x',ξ'))q j,
j= 0
k
∑
=
b( )j0 C n 1
\ 0{ } ( ), b( )j1 (x' ·, ) C n 1
\ 0{ }
x'
∀ n 1
, j∀ ,
∈
b j
1
( )
·,ξ'
( ) C0∞( )n , ∀ξ' n 1
\ 0{ }, j,∀
b( )j1
b j
0
( )(λγ1ξ1,λγ2ξ2, ,… λγn 1ξn 1) λk–dj
b j
0
( )( )ξ' ,
= λ
∀ 0, ∀ξ' n 1
\ 0{ },
∈
>
b( )j1 (x',λγ1ξ1,λγ2ξ2, ,… λγn 1ξn 1) λk–dj
b( )j1 (x,ξ'),
= λ
∀ 0, ∀ξ' n 1
\ 0{ }, x'∀ n 1
>
b x' D' q d( , , , )v(x' q, ) 1
2π ( )(n 1 ) /2 - e i x'〈 ,ξ'〉
n∫1
=
×b x'( , , ,ξ' q d)n 1v(ξ' q, )dξ'
L x( n, , ,D q d)U x q( , ) F x q( , ), x +
n
∈
=
B j(D q d, , )U x q( , ) x= 0 = G j(x' q, ),
j = 1 2, , ,… m,
Trang 3Set l0 = max{2m, m j + 1}, (x n , D, q, d)u = (L(x n , D, q,
For l ≥ l0, we set
Proposition 1 The operator (x n , D, q, d) is
bounded from the space E l, p (d, µ, δ, ) to the space
E l, p (d, µ, δ, , n – 1 ).
The operator (x n , D, q, d) and problem (1), (2) are
elliptic, if
(i) L0(x n, ξ, q, d) = (ξ') ξαqβ≠ 0,
∀ξ∈n , q ∈, |ξ| + |q|≠ 0, ∀x n > 0;
(ii) the equation L0(x n, ξ', ξn , q, d) = 0 has m
solu-tions ξn with positive imaginary parts for x n > 0, |ξ'| +
|q|≠ 0;
(iii) the Cauchy problem
has a unique solution, tending to 0 as x n→ +∞ for any
h j∈
Proposition 2 Suppose that L0(x n, ξ, q, d) ≠ 0, ∀q ∈
, |ξ| + |q| ≠ 0, ∀x n > 0, and the equation L0(x, ξ', ξn,
q, d) = 0 has m solutions ξn with positive Imξn for |ξ'| +
|q|≠ 0, x n > 0 Then the space Nξ', q of solutions to the
equation L(x n, ξ', D n , q, d) v (x n , q) = 0 has dimension m,
with the base Φ1(ξ', x n , q), …, Φm(ξ', x n , q) and Φk(ξ',
L x( n, , ,D q d) aα( )ξ' x n
lαξα
qβ,
α +dβ ≤2m
γ α ,
〈 〉+dβ–lα=2m
∑
=
\ 0{ } ( ), aα(λγ1ξ1,λγ2ξ2, ,… λγn 1ξn 1)
∈
= aα( )ξ' , ∀λ 0, ∀ξ' n 1
\ 0{ },
∈
>
B j(ξ' q d, , ) b jkl( )ξξ' n
k
q l,
k+dl=m j
∑
=
\ 0{ }
∈
b jkm(λγ1ξ1,λγ2ξ2, ,… λγn 1ξn 1) = b jkm( )ξ' ,
λ
∀ 0, ∀ξ' n 1
\ 0{ }
∈
>
u x
n= 0
E l p, d µ δ +
n
n 1
, , , ,
= E l–2m,p (d µ δ +
n
l–m j 1 p
-– (d, , ,µ δ n 1)
j= 0
m
∏
×
+
n
+
n
aα
α +dβ ≤2m
δ α ,
〈 〉 +dβ –lα =2m
L0(x n, ,ξ' D n, ,q d)v(x n,q) = 0, x n>0;
B j(ξ' Dn, , ,q d)v(0 q, ) = h j, j = 1 2, , ,… m
D n j
x n , q), 1 ≤ k ≤ m, 0 ≤ j ≤ 2m, are continuous in (ξ', x n , q)
Moreover,
Suppose that
Theorem 1 Let be an ellipic operator If
E l, p (d, µ, δ, α, , n – 1 ) = {(F, G) ∈ E l, p (d, µ, δ, α, , n – 1)|F(x', x n ) = 0 for x n > α}, then there exists a
bounded operator R0 from (d, µ, δ, α, , n – 1)
to E l, p (d, µ, δ, , n – 1 ) such that
(i) for any α > 0: R0 is bounded from E l, p (d, µ, δ, α)
to E l, p (d, µ, δ);
(ii) for any α > 0, (F, G) ∈ E l, p (d, µ, δ, α), 0R0(F, G) = (F, G) + Q(F, G), where Q is bounded from E l, p (d,
µ, δ, α) to (d, µ, δ, α);
(iii) U ∈ E l, p (d, µ, δ, and U(x', x n) = 0 for x n≥α
with some α > 0, then R00U = U.
Proof Remark that (F, G) ∈ E l, p (d, µ, δ, α) for almost
all q ∈ C, Req = µ, (F, G) ∈ H (l, p, δ) and F(x', x n) = 0 for
x n≥α
It is evident that there exists a unique function v(ξ', x n),
such that L(x n, ξ', D n , q, d) v(ξ', x n) =n – 1 F(ξ', x n , q),
v(ξ', x n) = 0, for sufficiently large x n Denote the map: (F, G) v Then is a
linear bounded map from L p(0, α) to H 2m, p(+) contin-uous in (ξ', q) Set
4 Consider the problem
e c x n
D n
jΦk(ξ' xn, ,q) L p
p
k= 1
m
∑
j= 0
2m
detBj(ξ' Dn, , ,q d)Φk(ξ' xn, ,q) x
n= 0≠0
for ξ' + q ≠0, 0≤ ≤j m–1, 1≤ ≤k m.
+
n
+
n
E l p,
α∪> 0 +
n
+
n
E l p,
l> 0
∩
+
n
R0(F G, ) n 1
1 [θGξ0' q, (n 1F(ξ' xn, ,q))
= + 1( –θ)ξ' q1, (n 1F(ξ' x, ,n q)n 1G(ξ' q, ))],
θ θ ξ(( ' q, )δ,d) 1 if (ξ' q, )δ,d<1
0 if (ξ' q, )δ,d≥2
⎩
⎨
⎧
∂t
- d
, , ,
n
, t>0,
∈
=
∂t
- d
, ,
x n= 0 = g0(x' t, ), t>0,
j = 1 2, , ,… m.
Trang 4Theorem 2 Suppose that the operator (x n , D, q, d)
is elliptic Then for any ψ ∈ (n ) there exists an ε > 0,
such that for 0 < ε≤ε0, there exists a linear operator R
with the following properties:
(i) for any a ≥ 0: R is bounded from the space
(d, µ, δ, a, × (0, +∞), n – 1× (0, +∞)) to
P l, p (d, µ, δ, × (0, +∞));
(ii) for any a > 0, ( f, g) ∈ P l, p (d, µ, δ, a, × (0, +∞),
n – 1× (0, +∞))
the operator T: P l, p (d, µ, δ, × (0, +∞), n – 1×
(0, +∞)) → (d, µ, δ, n – 1× (0, +∞)); is bounded;
(iii) if u ∈ P l, p (d, µ, δ, × (0, +∞)) and u(x', x n , t) = 0,
for x n≥ a , then
5 Let Ω be a bounded set in n with smooth
bound-ary ∂Ω Consider the problem
(3)
(4)
where {U i, ϕi = 1 is a partition of unity in Ω and
the functions ψi∈ (Ω) are such that suppψi⊂ U i and
ψi (x) = 1 in a neighborhood of suppϕi
After the Laplace transform problem (3), (4) goes to
the problem:
C0∞
P l p,
a> 0
n
+
n
+
n
0 x n D ∂
∂t
- d
, , ,
∂t
- d
, , ,
⎝
⎛ +
⎝
⎛
–0 x n D ∂
∂t
- d
, , ,
⎠
⎞
⎠
⎞R f g( , ) = (f g, )+T f g( , ),
where ψε( )x ψ ε–γ1
x1 … ε–γn
x n
, ,
=
+
n
P l p,
l= 0
∞
∩
+
n
∂t
- d
, , ,
⎛ ⎞ ψε( )x x D ∂
∂t
- d
, , ,
⎝
⎛ +
⎝
⎛
–0 x n D ∂
∂t
- d
, , ,
⎠
⎞
⎠
⎞u = u.
∂t
- d
, , ,
⎛ ⎞u x t( , ) = f x t( , ), x∈Ω, t>0,
∂t
- d
, , ,
j(x t, ), x∈∂Ω,
=
t>0, j = 1 2, , ,… m,
}iN=1
C0∞
L x D q d( , , , )u x q( , ) = f x q( , ), x∈Ω,
B j(x D q d, , , )u x q( , ) = g j(x q, ),
x∈∂Ω, j = 1 2, , ,… m.
The operator (x, D, q, d) = (L(x, D, q, d), B j (x, D, q, d)|∂Ω is elliptic at a point y ∈ Ω, if for y ∉∂Ω there is a
neighborhood U ∩ ∂Ω = , y∈ U, in which ϕi Lψi has the principal part
for y ∈∂Ω there is a neighborhood U i∩ ∂Ω ≠ , y ∈ U i,
in which ϕi Lψi has the principal part
and ϕi B jψi has the principal part
Then the trivial solution of the Cauchy problem
is the only its bounded solution
Theorem 3 Let the operator (x, D, q, d) be elliptic.
Then for all sufficiently large µ = Req and for any ( f, g) ∈ P l + 1, p (d, µ, δ, Ω× (0, +∞), ∂Ω × (0, +∞)), prob-lem (3), (4) has a unique solution u ∈ P l, p (d, µ, δ,
Ω× (0, +∞))
6 Consider the problem
(5)
(6)
(7)
where f, q j satisfy the following conditions:
(i) the maps (x, t, u) f (x, t, u) and (x, t, u j) gj (x,
t, u j) satisfy the Caratheodory conditions, i.e they are
L i
0
x, , ,ζ q d
( ) ϕ( )x a iα(x,ξ)qβ
α +dβ =2m
=
for x∈U i, ξ + q ≠0;
L i0(x, ,ξ q)
= ϕi( )x a iα(x,ξ')x n lαξαqβ≠0
α +dβ =2m
γ α ,
〈 〉+dβ–lα=2m
∑
for x∈U i\∂Ω, ξ + q ≠0,
B ij0(x', , ,ξ q d) b jkl(x',ξ')ξn
k
q l
k+∑dl=m j
=
L i0(0 x, , ,n ξ' D n, ,q d)v( )x n = 0,
B ij
0
0, ,ξ' Dn, ,q d
( )v( )0 = 0,
j = 1 2, , ,… m, for ξ' + q ≠0
∂t
- d
, , ,
= f x t u x t( , ) … D x
2m 1
u u t( , )
∂t
- d
, , ,
∂Ω
= g j x' t u x t( , ) … D m x j 1
u x t( , )
j = 1 2, , ,… m;
∂k
u x t( , )
∂t k
-t= 0
d
- , , , ,
Trang 5continuous in u = (u1, u2, …, u N), uj = (u1, u2, …, )
for almost all x, and measurable in x, t for all u, u j,
(ii) the maps u(x, t) ( f (x, t, u(x, t)), g j (x, t,
uj (x, t))), where
from P l, p (d, µ, δ, Ω × (0, +∞)) to P l, p (d, µ, δ, Ω × (0, +∞), ∂Ω × (0, +∞)) are compact;
(iii) there exists a r > 0, such that
,
where
Theorem 4 If (x, D, q, d) is elliptic, f, g j satisfy
conditions (i)–(iii), then for all sufficiently large µ =
Req, problem (5)–(7) has a solution u ∈ (P l, p (d, µ, δ,
Ω × (0, +∞)).
Proof For any w ∈ P l, p (d, µ, δ, Ω × (0, +∞)) the
problem
where µ = Req , has a unique solution u ∈ P l, p (d, µ, δ,
Ω × (0, +∞)), such that
By condition (iii) there are r > 0 and µ0 > 0, such that
for w ∈ P l, p (d, µ, δ, Ω × (0, +∞)) with
= r, ∀µ≥µ0, we have –1 x, D, , d × ||( f, g j)||λ, µ≤ 1 Thus ≤ r.
By (ii) the map w u is compact from the closed
sphere
to the ball
Therefore, by the Rothe theorem, this map posses a
fixed point in the ball B r of the space P l, p (d, µ, δ, Ω ×
(0, +∞)), which is a solution to the problem (5)–(7)
REFERENCES
1 M S Agranovich and M I Vishik, Usp Mat Nauk 19
(3), 53–161 (1964)
2 E Rothe, Compositio Math 5, 177–197 (1937).
3 Yu V Egorov, Nguyen Minh Chuong, and Dang Anh
Tuan, C R Acad Sci Paris, Ser 1 337, 451–456 (2003).
4 Yu V Egorov, Nguyen Minh Chuong, and Dang Anh
Tuan, Dokl Math 74, 874–877 [Dokl Akad Nauk 411,
732–735 (2006)]
u N
j
u x t( , ) = (u x t( , ) …, ,D x 2m 1u x t( , );
uj(x t, ) u x t( , ) … D x m j 1
u x t( , ) , ,
=
∂t
- d
, , ,
µ → + ∞
sup
f g, j
( ) λ µ,
= sup 1
r
- (f x t u x t( , , ( , )),g j(x t u, , j(x t, ))) P l p, (d, , , µ δ Ω × ( 0 + , ∞ ) , ∂Ω × 0 + , ∞ ) ) u P
l p, (d, , , µ δ Ω × ( 0 + , ∞ ) ) = r
∂t
- d
, , ,
⎛ ⎞u x t( , ) = f x t w x t( , , ( , )),
x∈Ω, t>0,
∂t
- d
, , ,
∂Ω = g j(x t w x t, , ( , )),
x∈∂Ω, t>0,
j = 1 2, , ,… m,
u P
l p, (d, , , µ δ Ω × ( 0 + , ∞ ) ) 1
∂t
- d
, , ,
≤
× (f x t w x t( , , ( , )),g j(x t w, , j(x t, ))) P l p, (d, , , µ δ Ω × ( 0 + , ∞ ) , ∂Ω × ( 0 + , ∞ ) )
l p, (d, , , µ δ Ω × ( 0 + , ∞ ) ) - ⎝⎛
-∂
∂t
- - u P l p, (d, , , µ δ Ω × ( 0 + , ∞ ) )
S r = {w∈P l p, (d, , ,µ δ Ω×(0 +, ∞))
w P l p, (d, , , µ δ Ω × ( 0 + , ∞ ) ) = r}
B r = {w∈P l p, (d, , ,µ δ Ω×(0 +, ∞))
l p, (d, , , µ δ Ω × 0 + , ∞ ) )≤r}