Uniqueness and grow-up rate of solutions for pseudo

Transcription

Linear
Algebra and
its Applications
466 (2015) 102–116
Applied
Mathematics
Letters
48 (2015) 8–13
Contents
lists
at ScienceDirect
Contents lists
available
at available
ScienceDirect
LinearMathematics
Algebra andLetters
its Applications
Applied
www.elsevier.com/locate/laa
www.elsevier.com/locate/aml
Inverse
eigenvalue
problem of
matrix
Uniqueness and
grow-up
rate of solutions
forJacobi
pseudo-parabolic
n
mixed
data source
equations in Rwith
with
a sublinear
Sujin Khomrutai Ying Wei 1
Department of Mathematics
and Computer
Science, Nanjing
Faculty of
Science, of
Chulalongkorn
University,
Bangkok 10330, Thailand
Department
of Mathematics,
University
Aeronautics and
Astronautics,
Nanjing 210016, PR China
article
info
a r t i c l e
abstract
i n f o
a b s t r a c t
Article history:
In this paper, we solve an open problem appeared in Cao et al. (2009) concerning
Article history:
Received 9 January 2015
In this
the inverse
eigenvalue problem
reconstructing
the uniqueness of solutions
forpaper,
a sublinear
pseudo-parabolic
Cauchyofproblem.
In the
Received
16 January 2014
Received in revised form
15 March
a Jacobi
matrix
from
its eigenvalues,
its leading
principal
zero initial case, we obtain
the class
of all
non-trivial
global solutions,
whereas,
the
Accepted
20
September
2014
2015
submatrix
and part when
of thetheeigenvalues
of itsissubmatrix
uniqueness
of
global
solutions
is
established
initial
condition
non-zero.
Available online 22 October 2014
Accepted 15 March 2015
is considered. The necessary and suﬃcient conditions for
Submitted
by Y. WeiA lower grow-up rate of solutions is also obtained.
Available online 24 March
2015
the existence and uniqueness
of the
solution
are reserved.
derived.
© 2015 Elsevier
Ltd.
All rights
Furthermore, a numerical algorithm and some numerical
MSC:
Keywords:
examples are given.
15A18
Pseudo-parabolic equation
© 2014 Published by Elsevier Inc.
15A57
Sublinear
Mild solutions
Uniqueness
Grow-up
Keywords:
Jacobi matrix
Eigenvalue
Inverse problem
Submatrix
1. Introduction
In this paper, we consider solutions u(x, t) ≥ 0 of the sublinear pseudo-parabolic Cauchy problem

∂t u − △∂t u = △u + up x ∈ Rn , t > 0,
(1.1)
u(x, 0) = u0 (x) ≥ 0
x ∈ Rn ,
where 0 < p < 1 is a constant and n ≥ 1 is a positive integer. This problem was studied in [1] and
the existence of global solutions was established within C([0, ∞); Cb (Rn )). The question of uniqueness of
solutions, however, has been left open. The purpose of this paper is to settle this question.
In recent years, there is a rich literature addressing the existence, or uniqueness of solutions for pseudoparabolic problems inE-mail
bounded,
or [email protected].
unbounded domains, and for periodic solutions. Among many others,
address:
1
Tel.: +86
13914485239.
we mention [2,1,3–6]. From
practical
point of view, we should also mention [7–10], where pseudo-parabolic
problems appear ashttp://dx.doi.org/10.1016/j.laa.2014.09.031
models for porous media flows with or without dynamic capillarity.
Published
by Elsevier
Inc.
Setting u = e−t U0024-3795/©
in (1.1), 2014
we get
the nonlocal
formulation:
∂t U = BU + e(1−p)t BU p , U |t=0 = u0 and
upon integration we obtain the mild formulation:
E-mail address: [email protected].
http://dx.doi.org/10.1016/j.aml.2015.03.008
0893-9659/© 2015 Elsevier Ltd. All rights reserved.
9
S. Khomrutai / Applied Mathematics Letters 48 (2015) 8–13

U (x, t) = MU (x, t) := u0 (x) +
0
t

BU (x, s)ds +
0
t
e(1−p)s BU p (x, s)ds.
(1.2)
B = (1 − △)−1 is the Bessel potential operator given by

Bϕ =
B(x − y)ϕ(y)dy,
B(x) = |x|(2−n)/2 K(n−2)/2 (|x|),
Rn
and Kν is the modified Bessel function of the second kind.
∞ k
Apart from B, we also need the Green operator G(t) = e−t etB = e−t k=0 tk! B k (t > 0). Both B and G(t)
are positive, bounded, linear operators on Cb (Rn ). Note that B(1) = 1 and G(t)(1) = 1. More details can be
found in [11,12].
Let us state the main definition in this work.
Definition 1. A mild solution (resp., super-solution, or sub-solution) of (1.1) is a function u
C([0, T ); Cb (Rn )) for some 0 < T ≤ ∞ such that
U (x, t) = MU (x, t)
∈
(resp., U ≥ MU , or U ≤ MU )
for all x ∈ Rn , t ∈ [0, T ). If T = ∞, such a function u is called a global mild solution (resp., super-solution,
or sub-solution). We note that U = et u.
In this work, we prove the following main results.
Theorem 1. Let u ≥ 0 be a mild super-solution of (1.1) and 0 ≤ u0 ∈ C(Rn ) with u0 ̸≡ 0. Then
u(x, t) ≥ ((1 − p)t)q for all (x, t) ∈ Rn × [0, T ) where q = 1/(1 − p).
Corollary 1. If u0 ≡ 0, then all the non-trivial global mild solutions of (1.1) have the form
u(x, t) = ((1 − p)(t − τ )+ )q
(τ ≥ 0).
(1.3)
Remark 1. It is straightforward to see that, the nontrivial mild solutions (1.3) are obtained by solving the
ordinary differential equation emerging from (1.1), if assuming that u is only time dependent. Also observe
that these solutions are just translations in time of the maximal solution ((1 − p)t)q .
Theorem 2. Let u, v ∈ C([0, T ); Cb (Rn )), u, v ≥ 0, be mild super-solution and sub-solution, respectively,
of (1.1), and u0 , v0 ∈ C α (Rn ) (0 < α < 1) satisfy u0 (x) ≥ v0 (x) ≥ 0, u0 ̸≡ 0. Then u ≥ v on Rn × [0, T ).
Corollary 2. If u0 ∈ C α (Rn ) (0 < α < 1), u0 ≥ 0, and u0 ̸≡ 0, then there exists a unique global mild solution
u to the Cauchy problem (1.1).
2. Lower bound of grow-up rate
Lemma 1. Let u = e−t U ≥ 0 be a mild super-solution of (1.1) and 0 ≤ u0 ∈ C(Rn ), u0 ̸≡ 0. Then for δ > 1,
t0 ∈ (0, T ), there is a constant cδ = c(δ, u0 , t0 ) such that MU |t0 ≥ cδ e−δ|x| .
t
Proof. Since B is monotone and U ≥ 0, we have U ≥ MU ≥ u0 . Then U ≥ MU ≥ 0 Bu0 (x)ds > 0 on
Rn × (0, T ). By the asymptotic behavior of B(x) as |x| → ∞ and |x| → 0 [13], there is a constant b > 0 such
that

|x|(1−n)/2 if n ̸= 2 or |x| ≥ 1,
B(x) ≥ bθ(x)e−|x| where θ(x) =
(2.1)
1 − ln |x| if n = 2 and |x| < 1.
10
Let 0 < ε < t0 . Since MU is continuous, m0 := inf U ≥ min MU > 0 where inf, min are taken over
{|x| ≤ 2, ε ≤ t ≤ t0 }. If |x| > 2 and |y| < 1, then 1 ≤ |x − y| ≤ |x| + 1 hence
 t0
 t0 
MU (x, t0 ) ≥
BU (x, s)ds ≥
B(x − y)U (y, s)dyds ≥ Ce−δ|x| .
ε
|y|<1
ε
′ −δ|x|
For |x| ≤ 2, clearly MU (x, t0 ) ≥ m0 ≥ C e
for some constant C ′ > 0.
Lemma 2. There is a constant d > 0 such that B(e−P |x| ) ≥ d(1 + P )−
n+1
2
e−P |x| (∀ P > 0).


∞
1−n
Proof. By (2.1), B(e−P |x| ) ≥ bωn e−P |x| 0 θ(r)e−(1+P )r rn−1 dr. If n ̸= 2, then θ(r) = r 2 so B e−P |x| ≥
n+1
n−1
bωn e−P |x| L{r 2 }|s=1+P = d1 (1 + P )− 2 e−P |x| where d1 = bωn Γ ( n+1
2 ) and L{·} is the Laplace
1
− 21
transformation. If n = 2, θ(r) = 1 − ln r (0 < r < 1), θ(r) = r
(r ≥ 1). Since 0 θ(r)e−(1+P )r rdr < ∞, we
−P |x|
− 32 −P |x|
by taking C > 0 sufficiently small. Choosing d = min{d1 , bω2 C},
) ≥ bω2 C(1 + P ) e
have B(e
we obtain the desired estimate. Proof of Theorem 1. Let δ > 1 be sufficiently small so that p˜ := δp ∈ (0, 1). We begin by assuming that
u0 (x) ≥ c0 e−δ|x| for some constant c0 > 0.
Step 1. We show that U (x, t) ≥ (ht)q for some constant h > 0. Here we apply repeatedly (1.2). Since B
is monotone and U ≥ 0, we have U ≥ u0 ≥ c0 e−δ|x| . Using Lemma 2, we get
 t
 t 

n+1
˜
˜
U (x, t) ≥
e(1−p)s BU p (x, s)ds ≥
B cp0 e−p|x|
ds ≥ c1 te−p|x|
,
c1 = cp0 d[(1 + p˜)−1 ] 2 .
0
0
Similarly, using that p˜ > p, we get U (x, t) ≥
t
0


2
2
B cp1 sp e−p˜ |x| ds ≥ c2 t1+p e−p˜ |x|
2

 n+1
c2 = cp0 d1+p (1 + p)−1 (1 + p˜)−p (1 + p˜2 )−1 2 .
By induction, we conclude that U (x, t) ≥ ck+1 t1+p+···+p e−p˜
k
ck+1 = cp0
k+1
d1+p+···+p (1 + p)−p
k
k−1
k+1
d1+p+···+p q −(p
k
k−1
|x|
where

 n+1
k
2
.
· · · (1 + p + · · · + pk )−1 (1 + p˜)−p · · · (1 + p˜k+1 )−1
Since 0 < p < 1, 1 + p + · · · + pj ≤ q =
ck+1 ≥ cp0
k+1
1
1−p
and 1 + p˜i ≤ 2 for all i, j ≥ 1. Thus
+pk−2 +···+1) −(pk +pk−1 +···+1) n+1
2
2
≥ 2−q
n+1
2
cp0
k+1
d1+p+···+p (1 − p)q .
k
Thus U (x, t) ≥ 2−q 2 cp0 (1 − p)q (dt)1+p+···+p e−p˜ |x| . By taking k → ∞ and using that p˜k+1 → 0, we
n+1
n+1
obtain U (x, t) ≥ 2−q 2 (1 − p)q (dt)q = (ht)q where h := 2− 2 (1 − p)d.
n+1
k+1
k
k+1
Step 2. Since B(1) = 1, by iteration we have
 t
 t
U (x, t) ≥
BU p (x, s)ds ≥
B (hpq spq ) ds = hpq q −1 tq ,
0
0
 t
 t 

2
2
U (x, t) ≥
BU p (x, s)ds ≥
B hp q q −p spq ds = hp q q −1−p tq ,
0
0
k
and, by induction we have U (x, t) ≥ hp q q −1−p−···−p
((1 − p)t)q .
k−1
tq . Taking k → ∞, we obtain that U (x, t) ≥ q −q tq =
Step 3. Now we prove the theorem for the case u0 ≥ c0 e−δ|x| . Setting U = et u, we get that
u(x, t) ≥ e−t ((1 − p)t)q and it satisfies (see Lemma 3)
 t
u(x, t) ≥ G(t)u0 +
G(t − s)Bup (x, s)ds.
(2.2)
0
11
Since G(t)(1) = 1 and u ≥ e−t ((1 − p)t)q , we get by (2.2) that
 t
 t

 −ps
pq pq
pq
e−pt sq−1 ds = (1 − p)pq+1 e−pt tq
u(x, t) ≥
G(t − s)B e (1 − p) s ds ≥ (1 − p)
0
0
 t


2
2
2
p2 q+p
G(t − s)B e−p s sq−1 ds ≥ (1 − p)p q+p+1 e−p t tq .
u(x, t) ≥ (1 − p)
0
By induction, we have u(x, t) ≥ (1 − p)p q+p +···+p+1 e−p t tq hence by taking k → ∞ we obtain that
u(x, t) ≥ ((1 − p)t)q . This proves the theorem in the case u0 ≥ c0 e−|x| .
k
k−1
k
Step 4. Now we consider the general case that u0 ≥ 0 with u0 ̸≡ 0. For each 0 < τ < T , we have by
Lemma 1 that u
˜(x, τ ) := Mu(x, τ ) ≥ cδ e−δ|x| where cδ = cδ (δ, u0 , τ ) > 0. Define uτ (x, t) = u(x, t + τ )
(0 ≤ t ≤ T − τ ). Then by the semigroup property of G(t), uτ satisfies
 t
uτ (x, t) ≥ G(t)˜
u(x, τ ) +
G(t − s)B (upτ (x, s)) ds,
0
i.e. uτ is a mild super-solution with u
˜(x, τ ) ≥ cδ e−δ|x| . According to the previous case we find that
q
uτ (x, t) ≥ ((1 − p)t) , hence u(x, t + τ ) ≥ ((1 − p)t)q , (x ∈ Rn , 0 ≤ t ≤ T − τ ).
Now for each 0 < t < T , we have for all τ > 0 sufficiently small that
u(x, t) = u(x, (t − τ ) + τ ) ≥ ((1 − p)(t − τ ))q ,
for all x ∈ Rn . Letting τ → 0, u satisfies the desired estimate.
3. The comparison principle, uniqueness of solutions
Proof of Theorem 2. Let w = (v − u)+ . Since v0 ≤ u0 and |ap − bp | ≤ |a − b|p , we have by Lemma 3 that
 t
 t
(v − u)(x, t) ≤
G(t − s)B(v p − up )+ (x, s)ds ≤
G(t − s)B(v − u)p+ (x, s)ds
0
0
hence w(x, t) ≤ 0 G(t − s)Bwp (x, s)ds. Taking the supremum norm,1 then ∥w(t)∥∞ ≤
∥ · ∥∞ denotes the supremum norm on Rn . It follows by Lemma 4 that
t
∥w(t)∥∞ ≤ ((1 − p)t)q
(t > 0).
t
0
∥w(s)∥p∞ ds where
(3.1)
Claim. (v p − up )+ (x, t) ≤ pqt−1 (v − u)+ (x, t).
1 d
(sv + (1 − s)u)p ds = p(v − u)Θ p−1 , where Θ is between v and u. The claim
Proof. We have v p − up = 0 ds
is trivial if v ≤ u. If v ≥ u then Θ ≥ u ≥ ((1 − p)t)q by Theorem 1, hence the claim is also true in this
case. By the claim, we now have that
 t
 t
p
p
G(t − s)B(v − u )+ (x, s)ds ≤ pq
s−1 G(t − s)Bw(x, s)ds.
(v − u)(x, t) ≤
0
Taking the norm, we get
 t
∥w(t)∥∞ ≤ pq
s−1 ∥w(s)∥∞ ds.
0
0
(3.2)
1 Since B, G(t) are monotone operators on C (Rn ), we have G(t)Bϕ(x) ≤ G(t)B(∥ϕ∥ · 1) = ∥ϕ∥ . Thus ∥G(t)Bϕ∥ ≤ ∥ϕ∥ .
b
∞
∞
∞
∞
12
t
′
(t)
Inspired by [14], we define g(t) = pq 0 s−1 ∥w(s)∥∞ ds. Then gg(t)
≤ pqt−1 . Let ε > 0. Then we have for all
pq
t ≥ ε that g(t) ≤ g(ε)(t/ε) . Using (3.1) and the definition of g, then
 ε
 ε
−1
q
s−1+q ds = p(1 − p)q εq .
g(ε) = pq
s ∥w(s)∥∞ ds ≤ pq(1 − p)
0
0
Plugging into the preceding estimate, we have
g(t) ≤ p(1 − p)q εq (t/ε)pq = p(1 − p)q εq(1−p) tpq → 0
as ε → 0.
It follows that g ≡ 0 for all t > 0 so w ≡ 0. Thus v(x, t) ≤ u(x, t) on Rn × [0, T ).
Proof of Corollary 2. The existence follows from [1]. We apply Theorem 2 to conclude the uniqueness.
Proof of Corollary 1. Let τ = inf{t : u(x, t) > 0 for some x ∈ Rn }. It follows by Lemma 1 that u(x, t0 ) > 0
for all x ∈ Rn , t0 > τ . By the semigroup property of G(t), if t0 > τ , then u(x, t) := u(x, t + t0 ) is a mild
solution of (1.1) with u(x, 0) = u(x, t0 ) > 0. Hence we get by Theorem 1 that u(x, t) ≥ ((1 − p)t)q , so
u(x, t) ≥ ((1 − p)(t − t0 )+ )q . This is true for all t0 > τ , so u(x, t) ≥ ((1 − p)(t − τ )+ )q for all x ∈ Rn , t ≥ 0.
If 0 < t0 < τ , then u(x, t) := u(x, t + t0 ) is a mild solution of (1.1) with zero initial values, i.e. u(x, t) =
t
t
G(t
− s)Bup (x, s)ds. Then 0 ≤ u(x, t) ≤ 0 ∥G(t − s)Bup (x, s)∥∞ ds ≤ 0 ∥u(s)∥p∞ ds. Applying Lemma 4,
0
we get u(x, t) ≤ ((1 − p)t)q , so u(x, t) ≤ ((1 − p)(t − t0 )+ )q for all x ∈ Rn , t > 0. This is true for all t0 < τ ,
hence u(x, t) ≤ ((1 − p)(t − τ )+ )q for all x ∈ Rn , t > 0. We conclude that u(x, t) = ((1 − p)(t − τ )+ )q . t
Acknowledgments
The author would like to express his thanks to the anonymous referees for their valuable suggestions and
comments.
Appendix
Lemma 3. If U = et u ≥ 0 is a mild super-solution of (1.2), then u satisfies (2.2).
Remark 2. If V = et v ≥ 0 is a mild sub-solution, then v satisfies the reversed inequality obtained from (2.2).
t
t
Proof. Setting U = et u, then u(x, t) ≥ e−t u0 + 0 e−(t−s) Bu(x, s)ds + 0 e−(t−s) Bup (x, s)ds. Let A = e−t u0 ,
 t −(t−s)
Bϕ ds, and B = up . Then u(x, t) ≥ U (0) + Lu(x, t) where U (0) = A + LB. By iteration,
Lϕ = 0 e
u(x, t) ≥ U (0) + LU (0) =: U (1) and generally,
u(x, t) ≥ U (k) = (1 + L + · · · + Lk )A + (Lk+1 + · · · + L)B (k ≥ 0).
t
t
2
We calculate LA = 0 e−(t−s) B(e−s u0 )ds = e−t tBu0 , L2 A = e−t 0 sB 2 u0 ds = e−t t2 B 2 u0 , hence by
ts
k
induction we get for all k ∈ N that Lk A = e−t tk! B k u0 . By Fubini’s theorem, we note that 0 0 F (r, s)drds =
tt
F (r, s)dsdr. Then we calculate
0 r
 t t
 t
L2 B =
e−(t−r) B 2 up (r)dsdr =
e−(t−r) (t − r)B 2 up (r)dr
0
r
0
 t t
 t
(t − r)2 3 p
3
−(t−r)
3 p
L B=
e
(s − r)B u (r)dsdr =
e−(t−r)
B u (r)dr.
2!
0
r
0
t
k
k+1 p
By induction, we conclude that Lk+1 B = 0 e−(t−r) (t−r)
u (r)dr.
k! B
13
From the calculations above, we conclude that
u(x, t) ≥
∞

k
L A+
k=0
This proves the desired result.
∞

L
k+1

B = G(t)u0 +
k=0
0
t
G(t − s)Bup (x, s)ds.
Lemma 4. Let p ∈ (0, 1) and λ > 0 be constants. If ξ ∈ C([0, T ]), ξ ≥ 0, and
 t
ξ(s)p ds (0 ≤ t ≤ T ),
ξ(t) ≤ λ
0
then ξ(t) ≤ (λ(1 − p)t)q for all t ∈ [0, T ] where q = 1/(1 − p).
t
Proof. Let f (t) = sup[0,t] ξ(τ ). Then f is non-decreasing, f (t) ≤ λ 0 f (s)p ds ≤ λtf (t)p hence f (t) ≤ (λt)q .
t
t
By iteration, we get f (t) ≤ λ 0 (λs)pq ds = (1 − p)(λt)q since pq = q − 1. Next, f (t) ≤ (1 − p)p λq 0 sq−1 ds =
k
(1 − p)p+1 (λt)q . We inductively obtain f (t) ≤ (1 − p)p +···+p+1 (λt)q for all k ≥ 0. Taking k → ∞ we obtain
f (t) ≤ (λ(1 − p)t)q . References
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Uniqueness and grow-up rate of solutions for pseudo

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