Uniqueness and grow-up rate of solutions for pseudo
Linear
Algebra and
its Applications
466 (2015) 102–116
Applied
Mathematics
Letters
48 (2015) 8–13
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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 weiyingb@gmail.com.
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: sujin.k@chula.ac.th.
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
S. Khomrutai / Applied Mathematics Letters 48 (2015) 8–13
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
S. Khomrutai / Applied Mathematics Letters 48 (2015) 8–13
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
S. Khomrutai / Applied Mathematics Letters 48 (2015) 8–13
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
ts
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 =
tt
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
S. Khomrutai / Applied Mathematics Letters 48 (2015) 8–13
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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