Electronic Journal of Differential Equations, Vol. 2013 (2013), No. 250, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
ASYMPTOTIC BEHAVIOR OF POSITIVE SOLUTIONS OF THE
NONLINEAR DIFFERENTIAL EQUATION t2 u00 = un
MENG-RONG LI, HSIN-YU YAO, YU-TSO LI
Abstract. In this article we study properties of positive solutions of the ordinary differential equation t2 u00 = un for 1 < n ∈ N, we obtain conditions for
their blow-up in finite time, and some properties for global solutions. Equations containing more general nonlinear terms are also considered.
1. Introduction
Some interesting results on the blow-up, blow-up rates, and estimates for the lifespan of solutions of the Emden-Fowler equation and the semi-linear wave equation
u + f (u) = 0 have been obtained, as shown in the references.
Here we wish to study the Emden-Fowler type wave equation, i.e. solutions,
independent of the space variable x, of the equation t2 utt − ∆u = un for n > 1.
The existence and uniqueness of local solutions of the initial-value problem
t2 u00 = un ,
1 < n ∈ N,
u(1) = u0 ,
u0 (1) = u1 ,
(1.1)
follow by standard arguments. Considering the transformation t = es , u(t) = v(s),
we have t2 u00 (t) = −vs (s) + vss (s), v(s)n = −vs (s) + vss (s) and v(0) = u(1) = u0 ;
vs (0) = u0 (1) = u1 , the problem (1.1) can be transformed into
vss (s) − vs (s) = v n (s),
v(0) = u0 ,
(1.2)
vs (0) = u1 .
Thus, the existence of local solutions u for (1.1) in (1, T ) is equivalent to the
existence of local solutions v for (1.2) in (0, ln T ). In this article, we give estimates
for the life-span T ∗ of positive solutions u of (1.1) in three different cases. The
main results are as follows:
1−n
2
2 , ε ∈ (0, 1);
(a) u1 = 0, u0 > 0: T ∗ ≤ ek1 , k1 := s0 + 2(n+3)
8− n−1 v(s0 )
(b) u1 > 0, u0 > 0:
q
(i) E(0) ≥ 0, T ∗ ≤ ek2 , k2 :=
1−n
2
2
n+1
n−1
2 u0
2 u0
;
n−1 u1
;
(ii) E(0) < 0, T ∗ ≤ ek3 , k3 :=
1
(c) u1 < 0, u0 ∈ (0, (−u1 ) n ): u(t) ≤ (u0 − u1 − un0 ) + (u1 + un0 )t − un0 ln t.
2000 Mathematics Subject Classification. 34A34, 34C11, 34C60.
Key words and phrases. Nonlinear differential equation; Emden-Fowler equation; blow-up rate.
c
2013
Texas State University - San Marcos.
Submitted October 5, 2013. Published November 20, 2013.
1
2
M.-R. LI, H.-Y. YAO, Y.-T. LI
EJDE-2013/250
where E(0) is defined in the next section and s0 is given by (3.3).
In Section 6, we replace the nonlinear term v n by a more general increasing
function f (v).
2. Notation and some lemmas
For a given function v, we use the following functions
n−1
2
a(s) := v(s)2 , E(0) := u21 −
un+1
, J(s) := a(s)− 4 ,
0
n+1
where u0 and u1 are the given initial conditions.
By an easy calculation we can obtain the following two Lemmas; we shall omit
the proof of the first lemma.
Lemma 2.1. Suppose that v ∈ C 2 [0, T ] is the solution of (1.2), then
Z s
2
v(s)n+1 = E(0),
E(s) = vs (s)2 − 2
vs (r)2 dr −
n+1
0
Z s
2
00
0
(n + 3)vs (s) = (n + 1)E(0) + a (s) − a (s) + 2(n + 1)
vs (r)2 dr,
0
Z s
n+3
n2 − 1
a0 (s)
00
n−1
J (s) =
(E(0) −
J(s)
+2
vs (r)2 dr),
4
n+1
0
2(n+1)
2(n+1)
(n − 1)2
E(0)(J(s) n−1 − J(0) n−1 )
J 0 (s)2 = J 0 (0)2 +
4
Z s
2(n+1)
(n − 1)2
J(s) n−1
vs (r)2 dr.
+
2
0
(2.1)
(2.2)
(2.3)
(2.4)
Lemma 2.2. For u0 > 0, the positive solution v of (1.2) satisfies:
(i) If u1 ≥ 0, then vs (s) > 0 for all s > 0
1
n
(ii) If u1 < 0, u0 ∈ (0, (−u1 ) ), then vs (s) < 0 for all s > 0.
(2.5)
(2.6)
un0
Proof. (i) Since vss (0) = u1 +
> 0, we know that vss (s) > 0 in [0, s1 ) and vs (s)
is increasing in [0, s1 ) for some s1 > 0. Moreover, since v and vs are increasing
in [0, s1 ), vss (s1 ) = vs (s1 ) + v(s1 )n > vs (0) + v(0)n > 0 for all s ∈ [0, s1 ) and
vs (s1 ) > vs (s) > 0 for all s ∈ [0, s1 ), we know that there exists a positive number
s2 > 0, such that vs (s) > 0 for all s ∈ [0, s1 + s2 ). Continuing this process, we
obtain vs (s) > 0 for all s > 0, for which the solution exists.
(ii) Since vss (0) = vs (0) + v(0)n = u1 + un0 < 0, there exists a positive number
s1 > 0 such that vss (s) < 0 in [0, s1 ), vs (s) is decreasing in [0, s1 ); therefore,
vs (s) < vs (0) = u1 < 0 for all s ∈ [0, s1 ) and v(s) is decreasing in [0, s1 ). Moreover,
since v and vs are decreasing in [0, s1 ), vss (s) = vs (s) + v(s)n < vs (0) + v(0)n < 0
for all s ∈ [0, s1 ) and vs (s1 ) < vs (s) < 0 for all s ∈ [0, s1 ), we know that there exists
a positive number s2 > 0, such that vs (s) < 0 for all s ∈ [0, s1 + s2 ). Continuing
this process, we obtain vs (s) < 0 for all s > 0 in the interval of existence.
3. Life-span of positive solutions of (1.1) when u1 = 0, u0 > 0
In this section we want to estimate the life-span of a positive solution u of (1.1)
if u1 = 0, u0 > 0. Here the life-span T ∗ of u means that u is the solution of
equation (1.1) and u exists only in [0, T ∗ ) so that the problem (1.1) possesses a
positive solution u ∈ C 2 [0, T ∗ ).
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ASYMPTOTIC BEHAVIOR
3
Theorem 3.1. For u1 = 0, u0 > 0, the positive solution u of (1.1) blows up in
finite time; that is, there exists T ∗ < ∞ so that
u(t)−1 → 0
as t → T ∗ .
Proof. By (2.5), we know that vs (s) > 0, a0 (s) > 0 for all s > 0 provided that
u1 = 0, u0 > 0. By Lemma 2.1,
a00 (s) − a0 (s) = 2(vs (s)2 + v(s)n+1 ),
(a0 (s)e−s )0 = e−s (a00 (s) − a0 (s)) = 2e−s (vs (s)2 + v(s)n+1 ),
Z s
Z s
n+1
0
−s
−r
2
n+1
a (s)e = 2
e (vs (r) + v(r)
)dr ≥ 4
e−r vs (r)v(r) 2 dr,
0
0
and a0 (0) = 0, hence we have
Z s
8
(v(r)(n+3)/2 e−r |sr=0 +
v(r)(n+3)/2 e−r dr)
n+3
0
Z s
8
8
(n+3)/2 −s
(n+3)/2
(v(s)
e − v(0)
)+
v(r)(n+3)/2 e−r dr.
=
n+3
n+3 0
a0 (s)e−s ≥
Since a0 (s) > 0 for all s > 0, v is increasing on (0, ∞) and
Z s
8
8
(v(s)(n+3)/2 e−s − v(0)(n+3)/2 ) +
v(0)(n+3)/2 e−r dr
n+3
n+3 0
8
8
=
(v(s)(n+3)/2 e−s − v(0)(n+3)/2 ) +
v(0)(n+3)/2 (1 − e−s ),
n+3
n+3
8
8
(n+3)/2
a0 (s) ≥
(v(s)(n+3)/2 − v(0)(n+3)/2 ) =
(v(s)(n+3)/2 − u0
).
n+3
n+3
(3.1)
Using u1 = 0 and integrating (1.2), we obtain
Z s
vs (s) = v(s) − u0 +
v(r)n dr,
0
Z s
vs (s) ≥ v(s) − u0 +
v(0)n dr = v(s) − u0 + un0 s,
(3.2)
0
−s
−s
−s n
(e v(s))s = e (vs (s) − v(s)) ≥ e (u0 s − u0 ),
8
8
(n+3)/2
).
a0 (s) ≥
(v(s)(n+3)/2 − v(0)(n+3)/2 ) =
(v(s)(n+3)/2 − u0
n+3
n+3
a0 (s)e−s ≥
According to (3.2), and since v 0 (s) > 0, v(s)(n+3)/2 ≥ (u0 + un0 (es − 1 − s))(n+3)/2 ,
for all ∈ (0, 1), we obtain
v(s)(n+3)/2 ≥ (u0 + un0 (es − 1 − s))(n+3)/2 ,
(n+3)/2
v(s)(n+3)/2 − 8u0
(n+3)/2
≥ (u0 + un0 (es − 1 − s))(n+3)/2 − 8u0
(n+3)/2
≥ (u0
n(n+3)
2
+ u0
(n+3)/2
= ( − 8)u0
n+3
(es − 1 − s)(n+3)/2 ) − 8u0 2
n(n+3)
2
+ u0
(es − 1 − s)(n+3)/2 .
Now, we want to find a number s0 > 0 such that
es0 − s0 = 1 +
8 − n+3
(1−n) 2/(n+3)
u0 2
.
(3.3)
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M.-R. LI, H.-Y. YAO, Y.-T. LI
EJDE-2013/250
This means that there exists a number s0 > 0 satisfying (3.3) with ∈ (0, 1)
such that
(n+3)/2
v(s)(n+3)/2 − 8u0
≥ 0 for all s ≥ s0 .
From (3.1), it follows that
a0 (s) ≥
8
8
(n+3)/2
v(s)(n+3)/2 −
u
n+3
n+3 0
(n+3)/2
v(s)(n+3)/2 − 8u0
8−
v(s)(n+3)/2 +
n+3
n+3
8−
≥
v(s)(n+3)/2 , for all s ≥ s0 .
n+3
For all s ≥ s0 , ∈ (0, 1), we obtain that
=
8−
v(s)(n+3)/2 ,
n+3
n+1
8−
v(s)− 2 vs (s) ≥
,
2(n + 3)
1−n
2
8−
(v(s) 2 )s ≥
1−n
2(n + 3)
2v(s)vs (s) ≥
and hence
8− 1−n
.
2(n + 3) 2
Integrating the above inequality, we conclude that
1−n
1−n
8− n−1
v(s) 2 ≤ v(s0 ) 2 −
(s − s0 ).
2(n + 3) 2
(v(s)
1−n
2
)s ≤
Thus, there exists a number
s∗1 ≤ s0 +
1−n
2(n + 3) 2
v(s0 ) 2 =: k1
8− n−1
such that v(s)−1 → 0 for s → s∗1 , that is, u(t)−1 → 0 as t → ek1 , which implies
that the life-span T ∗ of a positive solution u is finite and T ∗ ≤ ek1 .
4. Life-span of positive solutions of (1.1) when u1 > 0, u0 > 0
In this section we estimate the life-span of a positive solution u of (1.1) whenever
u1 > 0, u0 > 0.
Theorem 4.1. For u1 > 0, u0 > 0, the positive solution u of (1.1) blows up in
finite time; that is, there exists a number T ∗ < ∞ so that
u(t)−1 → 0
as t → T ∗ .
Proof. We separate the proof into two parts depending on whether E(0) ≥ 0 or
E(0) < 0.
(i) Assume that E(0) ≥ 0. By (2.1) and (2.5) we have
2
v(s)n+1 ≥ E(0),
n+1
r
2
2
2
n+1
vs (s) ≥
v(s)
+ E(0), vs (s) ≥
v(s)n+1 + E(0).
n+1
n+1
vs (s)2 −
EJDE-2013/250
ASYMPTOTIC BEHAVIOR
5
Since E(0) ≥ 0, we obtain
r
n+1
2
v(s) 2 ,
n+1
r
2
− n+1
2
v(s)
· vs (s) ≥
,
n+1
r
1−n
1−n
2
(v(s) 2 )s ≤
.
2
n+1
vs (s) ≥
Integrating the above inequality, we obtain
v(s)
1−n
2
1−n
2
≤ u0
1−n
+
2
r
2
s.
n+1
Thus, there exists
r
n + 1 1−n
2
u0 2 =: k2
≤
n−1
2
such that v(s)−1 → 0 for s → s∗2 ; that is, u(t)−1 → 0 as t → ek2 , which means that
the life-span T ∗ of a positive solution u is finite and T ∗ ≤ ek2 .
s∗2
(ii ) Assume that E(0) < 0. From (2.1) and (2.5) we obtain that J 0 (s) =
a0 (s) > 0, vs (s) > 0 for all s > 0 and
s
Z s
n−1
2
n+1
n+1
0
J (s) = −
+ E(0)a(s)− 2 + 2a(s)− 2
vs (r)2 dr
2
n+1
0
r
2
n−1
n+1
+ E(0)a(s)− 2 ,
≤−
2
n+1
Z r
n−1 s
2
n+1
J(s) ≤ J(0) −
+ E(0)a(r)− 2 dr.
2
n
+
1
0
− n+3
4 a0 (s),
− n−1
4 a(s)
Since E(0) < 0 and a0 (s) > 0 for all s > 0,
Z r
n−1 s
2
n+1
J(s) ≤ J(0) −
+ E(0)a(0)− 2 dr
2
n+1
0
r
n−1
n
−
1
2
n+1
= a(0)− 4 −
+ E(0)a(0)− 2 s.
2
n+1
Thus, there exists a number
s∗3 ≤
n−1
n+1
1
2
2
a(0)− 4 (
+ E(0)a(0)− 2 )− 2 =: k3
n−1
n+1
such that J(s∗3 ) = 0 and a(s)−1 → 0 for s → s∗3 ; that is, u(t)−1 → 0 as t → ek3 .
This means that the life-span T ∗ of u is finite and T ∗ ≤ ek3 .
5. Life-span of positive solutions of (1.1) when u1 < 0
Finally, we estimate the life-span of a positive solution u of (1.1) when u1 < 0.
1
Theorem 5.1. For u1 < 0, u0 ∈ (0, (−u1 ) n ) we have
u(t) ≤ (u0 − u1 − un0 ) + (u1 + un0 )t − un0 ln t,
6
M.-R. LI, H.-Y. YAO, Y.-T. LI
EJDE-2013/250
and in particular, if E(0) ≥ 0, we have
1−n
u(t) ≤ (u0 2 +
n−1
2
r
2
2
ln t) 1−n .
n+1
Proof. (i) By (1.2) Rand integrating this equation with respect to s, we get vs (s) =
s
(u1 − u0 ) + v(s) + 0 v(r)n dr. By (2.6), we have that v is decreasing and
Z s
vs (s) ≤ (u1 − u0 ) + v(s) +
v(0)n dr = (u1 − u0 ) + v(s) + un0 s,
0
e
−s
s
Z
v(s) − u0 ≤ (u1 − u0 )
e
−r
dr +
un0
s
Z
re−r dr
0
0
= (u1 − u0 )(1 − e−s ) + un0 (−se−s − e−s + 1);
that is,
u(t) ≤ (u0 − u1 ) + u1 t + un0 (t − 1 − ln t)
= (u0 − u1 − un0 ) + (u1 + un0 )t − un0 ln t.
(ii) If E(0) ≥ 0, by (2.1), we have
Z s
2
n+1
v(s)
= E(0) + 2
vs (r)2 dr ≥ E(0),
vs (s) −
n+1
0
2
2
n+1
2
v(s)
≥
v(s)n+1 .
vs (s) ≥ E(0) +
n+1
n+1
q
q
n+1
1−n
2
2
2
By (2.6), we obtain that −vs (s) ≥ n+1
v(s) 2 , n−1
(v(s) 2 )s ≥ n+1
and
r
1−n
1−n
2
2
s≤
(v(s) 2 − v(0) 2 ),
n+1
n−1
r
1−n
1−n
2
n−1
2
v(s) 2 ≥ (u0 +
s).
2
n+1
2
Then, we know that
1−n
2
v(s) ≤ (u0
n−1
+
2
r
2
2
s) 1−n ,
n+1
for all s ≥ 0;
that is,
1−n
2
u(t) ≤ (u0
n−1
+
2
r
2
2
ln t) 1−n
n+1
for all t ≥ 1.
6. A generalization of Theorem 4.1
In this section we want to extent the blow-up result for the following generalization of (1.2),
vss (s) − vs (s) = f (v),
(6.1)
v(0) = v0 , vs (0) = v1 ,
where f is an increasing continuous function with f (0) = 0. We have the following
result.
EJDE-2013/250
ASYMPTOTIC BEHAVIOR
7
Theorem 6.1. Suppose that f is an increasing
R v function with f (0) = 0 and suppose
v is a positive solution of (6.1). If F (v) := 0 f (r)dr, then
Z s
vs (r)2 dr − 2F (v(s))
(6.2)
Ē(s) := vs (s)2 − 2
0
is constant. Furthermore, if there exists a positive constant k such that F (s) ≥
ksp+1 , p > 1 for all s ≥ 0, and v1 > 0, then the life span of v is finite.
Proof. By an argument similar to that used in proving (2.1), we easily obtain that
Ē(s) is a constant. Since f is increasing, we have
vf (v) = (v − 0) · (f (v) − f (0)) ≥ 0
for v ≥ 0,
thus
2
Z
2
(v )s − 2v (s) ≥ 2v0 (v1 − v0 ) + 2
s
vs2 (r)dr.
(6.3)
0
By (6.2) and (6.3), we have Ē(s) = v12 − 2F (v0 ) := Ē, and
v 2 (s) ≥ v0 v1 e2s − v0 (v1 − v0 ),
Z s
2
2
(v )s − 2v (s) ≥ 2v0 (v1 − v0 ) + 2
vs2 (r)dr
0
Z s
Z r
= 2v0 (v1 − v0 ) + 2
(Ē + 2F (v(r)) + 2
vs (η)2 dη)dr
0
0
Z s
≥ 2v0 (v1 − v0 ) + 2Ēs + 4k
v p+1 (r)dr
0
Z s
p+1
1−p
≥ 2v0 (v1 − v0 ) + 2Ēs + 4ks (
v 2 (r)dr) 2 .
(6.4)
(6.5)
0
Let
Rs
0
2
v (r)dr := b(s), e
−s
b(s) = B(s). Then
b(s)00 − 2b(s)0 ≥ 2v0 (v1 − v0 ) + 2Ēs + 4ks1−p b(s)
p+1
2
for s > 0
and by (6.5), we have
(e−s (b(s)0 − b(s)))0
= e−s (b(s)00 − 2b(s)0 + b(s))
≥ 2v0 (v1 − v0 ) + 2Ēs + 4ks1−p e−s b
p+1
2
(6.6)
+ e−s b(s)
= 2v0 (v1 − v0 ) + 2Ēs + 4ks1−p (e−s b(s))
p+1
2
e
p−1
2 s
+ e−s b(s) ≥ 0,
(e−s b(s))00 = (e−s (b(s)0 − b(s)))0
p+1
es p+1 −s
) 2 (e b(s)) 2 + e−s b(s)
s2
p+1
p+1
≥ 2v0 (v1 − v0 ) + 2Ēs + 22− 2 k(e−s b(s)) 2 + e−s b(s),
≥ 2v0 (v1 − v0 ) + 2Ēs + 4k(
p+1
p+1
B(s)00 ≥ 2v0 (v1 − v0 ) + 2Ēs + 22− 2 kB(s) 2 + B(s).
From (6.4) it follows that
v0 v1 2s
b(s) ≥
(e − 1) − v0 (v1 − v0 )s,
2
v 0 v1 s
B(s) ≥
(e − e−s ) − v0 (v1 − v0 )se−s ,
2
(6.7)
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M.-R. LI, H.-Y. YAO, Y.-T. LI
2v0 (v1 − v0 ) + 2Ēs +
EJDE-2013/250
B(s)
≥ 0,
2
s ≥ s0
for some s0 > 0. Therefore,
B(s)
v0 v1 s
≥
e ,
2
5
B(s)00 ≥
B 0 (s) ≥
s ≥ s0 ,
v0 v1 s
(e − es0 ) + B 0 (s0 ) > 0, s ≥ s1
5
for some s1 > s0 .
By (6.7), for all s ≥ s1 ,
((B(s)0 )2 )0 = 2B(s)0 B(s)00
p−1
2
p+1
kB(s) 2 B(s)0
p−1
p+3
2
= 22− 2 k
(B(s) 2 )0 ,
p+3
≥ 22−
p−1
p+3
23− 2 k p+3
(B ) − B (s1 ) ≥
(B 2 − B(s1 ) 2 ),
p+3
0 2
0
2
p−1
p+3
23− 2 k p+3
(B 2 − B(s1 ) 2 ) + B 0 (s1 )2
p+3
23− p−1
3− p−1
2 k
2 k
p+3
p+3
p+3
2
B 2 +
B 2 − 2B(s1 ) 2 + B 0 (s1 )2 ,
=
2(p + 3)
2(p + 3)
(B 0 )2 ≥
7−p
p+3
2 4
B 4
B ≥√
p+3
0
for s ≥ s2 ,
for some s2 > s1 ; hence, for s ≥ s2 ,
7−p
p+3
1−p
2 4
4
(B 4 )0 = B −4 B 0 (s) ≥ √
,
1−p
p+3
7−p
B(s)
1−p
4
≤ B(s2 )
1−p
4
−
p−1 2 4
√
(s − s2 )
4
p+3
for all s ≥ s2 > 0.
Thus B(s) blows up at a finite s∗ . Since b(s) = es B(s), b(s) also blows up at s∗ .
Further, since v 2 (s) = b0 (s) ≥ 2b(s), v(s) blows up at s∗ , as well.
Acknowledgements. Thanks are due to Professors Tai-Ping Liu, Ton Yang and
Shih-Shien Yu for their continuous encouragement; to the anonymous referee for
his/her helpful comments; and to Professor K. Schmitt for his comments and suggestions on Theorem 6.1. The authors want to thank Metta Education, Grand Hall
and Auria Solar for their financial assistance.
References
[1] M. R. Li; Nichtlineare Wellengleichungen 2. Ordnung auf beschränkten Gebieten. PhDDissertation, Tübingen 1994.
[2] M. R. Li; Estimates for the life-span of solutions of semilinear wave equations. Comm. Pure
Appl. Anal., 2008, 7(2): 417-432.
[3] M. R. Li; On the blow-up time and blow-up rate of positive solutions of semilinear wave
equations u − up = 0 in 1-dimensional space. Comm.Pure Appl. Anal., to appear.
[4] M. R. Li; On the semilinear wave equations. Taiwanese J. Math., 1998, 2(3): 329-345.
[5] M. R. Li, L. Y. Tsai; On a system of nonlinear wave equations. Taiwanese J. Math., 2003,
7(4): 555-573.
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[6] M. R. Li, L. Y. Tsai; Existence and nonexistence of global solutions of some systems of
semilinear wave equations. Nonlinear Analysis, 2003, 54: 1397-1415.
[7] Meng-rong Li, Jenet Pai; Quenching problem in some semilinear wave equations. Acta Math.
Sci., 2008, 28B(3): 523-529.
[8] M. R. Li; On the generalized Emden-Fowler Equation u00 (t)u(t) = c1 + c2 u0 (t)2 with c1 ≥ 0,
c2 ≥ 0. Acta Math. Sci., 2010 30B(4): 1227-1234.
[9] M. R. Li; Blow-up results and asymptotic behavior of the Emden-Fowler Equation. Acta
Math. Sci., 2007, 4: 703-734.
[10] Meng-Rong Li, Yue-Loong Chang, Yu-Tso Li; A Mathematical Model of Enterprise Competitive Ability and Performance through Emden-Fowler Equation (II), Acta Mathematica
Scientia, 2013, 33(4): 1127-1140.
Meng-Rong Li
Department of Mathematical Sciences, National Chengchi University, Taipei, Taiwan
E-mail address: liwei@math.nccu.edu.tw
Hsin-Yu Yao
Department of Mathematical Sciences, National Chengchi University, Taipei, Taiwan
E-mail address: diadia0914@gmail.com
Yu-Tso Li
Department of Aerospace and Systems Engineering, Feng Chia University, Taichung,
Taiwan
E-mail address: joycelion74@gmail.com