REGULARITY AND BLOW-UP CONSTANTS OF SOLUTIONS FOR NONLINEAR DIFFERENTIAL EQUATION
Meng-Rong Li and Zing-Hung Lin
Abstract. In this paper we gain some results on the regularity and also the blow-up rates and constants of solutions to the equation u− up= 0 under
some different situations. The blow-up rate and blow-up constant of u(2n)are (p − 2n + 2) and (±) (p − 2n + 2) · Πn−1
i=0 (p − 2i + 2) (p − 2i + 1) E (0) p/2
respectively; blow-up rate and blow-up constant of u(2n+1) are(p − 2n + 1) and (p−2n+2) Πn−1i=0 (p−2i+2)· (p−2i+1) E (0)p−n respectively, where
E(0) = u(0)2− 2 p+1u(0)
p+1.
0. INTRODUCTION
In this paper, we deal with the estimate of blow-up rate and blow-up constant of
u(n)and the regularity of solutions for the nonlinear ordinary differential equation
(0.1) u− up = 0
where p >1.
Our motivation on the problem is based on the studying properties of solutions of the semi-linear wave equation u+ f(u) = 0 [2, 3] with particular cases in zero space dimension and the blow-up phenomena of the solution to equation (0.1)[4].
In this paper, if p = rs, r ∈ N, s ∈ 2N + 1, (r, s) = 1 ( common factor ) we say that p is odd(even respectively) if r is odd ( even, respectively ) .
For p ∈ Q and p ≥ 1, the function up is locally Lipschitz, therefore by stan-dard theory for ordinary differential equation there exists exactly one local classical solution to the equation(0.1) together with initial values u (0) = u0, u(0) = u1.
Received February 11, 2004, accepted April 7, 2005. Communicated by Tai-Ping Liu.
2000 Mathematics Subject Classification: 34, 34A12, 34C99.
Key words and phrases: Estimate, Life-span, Blow-up, Blow-up constant, Regularity.
This work is financed by Grand Hall Company. 777
Notations and Fundamental Lemmata
For a given function u in this work we use the following abbreviations
au(t) = u (t)2, Eu(t) = u(t)2− 2 p+ 1u(t) p+1, J u(t) = au(t)− p−1 4 .
Definition. A function g : R → R has a blow-up rate r means that g exists only in finite time, that is, there is a finite number T∗ such that the following holds
lim
t→T∗g(t)
−1 = 0
and there exists a non-zero β ∈ R with lim
t→T∗(T
∗− t)r
g(t) = β,
in this case β is called the blow-up constant of g .
One can find the detail in[4] for the lemmas given as follows without rigorous argumentations.
Lemma 1. Suppose that u is the solution of(0.1) , then we have
(0.2) E(t)u= Eu(0) , (0.3) (p + 3) u(t)2 = (p + 1) Eu(0) + au(t) , (0.4) Ju(t) = p 2− 1 4 Eu(0) Ju(t) p+3 p−1 and (0.5) Ju (t)2=Ju (0)2−(p−1) 2 4 Eu(0) Ju(0) 2(p+1) p−1 +(p−1) 2 4 Eu(0) Ju(t) 2(p+1) p−1 .
Lemma 2. Suppose that c1 and c2 are real constants and u∈ C2(R) satisfies the inequality
u+ c1u+ c2u≤ 0, u ≥ 0, u(0) = 0, u(0) = 0,
Lemma 3. If g(t) and h (t, r) are continuous with respect to their variables
and the limitlimt→T
g(t) 0 h(t, r) dr exists, then lim t→T g(t) 0 h(t, r) dr = g(T ) 0 h(T, r) dr.
Lemma 4. If T is the life-span of u and u is the solution of the problem
(0.1) with Eu(0) < 0 and p > 1 then T is finite, that is, u is only a local solution of(0.1) . Further, for au(0) ≥ 0, we have the following estimates
(0.6) Ju (t) = −p− 1 2 k1+ Eu(0) Ju(t)k2 ≤ J(0) ∀t ≥ 0, (0.7) Ju(0) Ju(t) dr k1+ Eu(0) rk2 = p− 1 2 t ∀t ≥ 0 and (0.8) T ≤ T1∗(u0, u1, p) = 2 p− 1 Ju(0) 0 dr k1+ Eu(0) rk2 .
For au(0) < 0, there is a constant t0(u0, u1, p) such that
(0.9) Ju (t) = −p− 1 2 k1+ Eu(0) Ju(t)k2 ∀t ≥ t0(u0, u1, p) , Ju (t) = p− 1 2 k1+ Eu(0) Ju(t)k2 ∀t ∈ [0, t0(u0, u1, p)] and (0.10) Ju(0) Ju(t) dr k1+ Eu(0) rk2 = p−1 2 (t−t0(u0, u1, p)) ∀t≥t0(u0, u1, p) , Ju(t0) Ju(0) dr k1+ Eu(0) rk2 = p− 1 2 t0(u0, u1, p) . Also we have (0.11) T ≤ T2∗(u0, u1, p) = 2 p− 1 k 0 dr k1+ Eu(0) rk2 + k J(0) dr k1+ Eu(0) rk2 ,
where k1:= 2 p+ 1, k2 := 2p + 2 p− 1 and k:= 2 p+ 1 −1 Eu(0) p−1 2p+2 . Furthermore, if Eu(0) = 0 and au(0) > 0, then
(0.12) Ju(t) = au(0)−p−14 −p− 1 4 au(0)− p−1 4 −1au(0) t, au(t) = au(0) p+3 p−1 au(0) −p− 1 4 au(0) t − 4 p−1
for each t≥ 0, and
(0.13) T ≤ T3∗(u0, u1, p) := 4 p− 1
au(0) au(0).
Lemma 5. If T is the life-span of u and u is the solution of the problem
(0.1) with Eu(0) > 0, then T is finite; that is, u is only a local solution of (0.1) . If one of the following is valid
(i) au(0)2 >4a u(0) Eu(0) or (ii) a u(0)2= 4au(0) Eu(0) and u1>0 or (iii) a u(0)2= 4au(0) Eu(0), u1<0 and p is odd. Further, in case of (i) , we have the estimate
(0.14) T ≤ T4∗(u0, u1, p) = 2 p− 1 Ju(0) 0 dr k1+ Eu(0) rk2 , and (0.15) a(0) ≥ 0.
In the case of(ii) , we have also
(0.16) T ≤ T5∗(u0, u1, p) = 2 p− 1 ∞ 0 dr k1+ Eu(0) rk2 .
In case of (iii) , we get
(0.17) T ≤ T6∗(u0, u1, p) = 2 p− 1 ∞ 0 dr k1+ Eu(0) rk2 .
Lemma 6. Suppose that u is the solution of the problem(0.1) with one of the following property (i) Eu(0) > 0, au(0)2 <4au(0) Eu(0) or (ii) a u(0)2 = 4au(0) Eu(0), u1 <0 and p is odd. Then T0 given by (0.18) T0(u0, u1, p) = −u(T0 ) −u0 dr Eu(0) − 2rp+1/(p + 1) ,
where −u (T0) = ((p + 1) Eu(0) /2)1/(p+1) is the critical point of u, and u0 must be non-positive.
Remark. Under condition(i)u0 must be negative and p must be even. If u is the solution of the problem(0.1) with Eu(0) = 0 and au(0) = 0, then u must be null.
Lemma 8. Suppose that u is the solution of the problem(0.1) with Eu(0) > 0 and one of the following holds
(i) a
u(0)2 <4au(0) Eu(0) .
(ii) a
u(0)2 = 4au(0) Eu(0) and u1<0, p is even.
Then u possesses a critical point T0(u0, u1, p) given by (0.18) , provided condition
(ii) holds or condition (i) together with au(0) > 0 holds; and under (i) , there exists z <∞ such that
a(z) = 0. For a(0) ≤ 0, we have the null point (zero) z1 of a,
z1(u0, u1, p) = p2− 1 √ 2 4au(0) (p2−1 )Eu(0) 0 dr 2 − (p − 1) k2 3rp+1 , and T ≤ T7∗(u0, u1, p) := z1(u0, u1, p) + T5∗(u0, u1, p) . where k3= p2−1 4 Eu(0) p−1 4 . Furthermore, we also have
(0.19) lim
t→z1au(t) (z1− t)
−2 = E u(0) ,
(0.20) limt→z1(z1− t)
−1a(t) = −2E (0) ,
limt→z1au(t) = 2Eu(0) , and au(t) blows up at T7∗(u0, u1, p) ; that is, limt→T∗
7 1/au(t) = 0. For au(0) > 0, we have the null point z2 of au
z2(u0, u1, p)= p2−1 √ 2 2p+11 (p−1)− 1p+1k3− 2p+1 0 dr 2 − (p−1) k2 3rp+1 + 2p+11 (p−1)− 1p+1k3− 2p+1 2a(0)1/2(p2−1)−1/2Eu(0)−1/2 dr 2−(p−1)k2 3rp+1 and T ≤ T8∗(u0, u1, p) := z2(u0, u1, p) + T6∗(u0, u1, p) . Furthermore, we also have
(0.21) lim t→z2au(t) (z2(u0, u1, p) − t) −2= E u(0) , (0.22) lim t→z2(z2− t) −1a u(t) = −2Eu(0) , lim t→z2a u(t) = 2Eu(0) , and au(t) blows up at T8∗(u0, u1, p) ; that is, limt→T∗
8(u0,u1,p)1/au(t) = 0. Further, under the condition(ii) , we have the null point z3(u0, u1, p) of a,
z3(u0, u1, p) = 2T0(u0, u1, p) ,
T ≤ T9∗(u0, u1, p) = z3(u0, u1, p) + T5∗(u0, u1, p) and au(t) blows up at T9∗(u0, u1, p). Furthermore we have
(0.23) lim t→z3au(t) (z3(u0, u1, p) − t) −2= E u(0) , (0.24) lim t→z3(u0,u1,p)(z3(u0 , u1, p) − t)−1au(t) = −2Eu(0) ,
(0.25) lim
t→z3a
u(t) = 2Eu(0) .
In Section I, we consider the regularity of solution u of equation(1) for p ∈ N and gain the expansion of u(n) in terms of u(k), k < n; in section II, we consider
the regularity of solution u as p∈ Q − N. In the last section, we study the blow-up rates and blow-up constants of u(n) as t approach to life-span T∗ and null point (zero) z under some situations.
1. REGULARITY OFSOLUTION TO THE EQUATION(0.1) WITHp∈ N
In this section we study the regularity of the solution u of the nonlinear equation (0.1) as p ∈ N. First, we see that the well-defined function up is locally Lipschitz,
hence we have the local existence and uniqueness of solution to the equation (1.1)
u= up,
u(0) = u0, u(0) = u1.
Therefore, we rewrite au(t) = a (t) , Ju(t) = J (t) and Eu(t) = E (t) for
convenience. Using(0.2) we have
(1.2) u(t)2= E (0) + 2 p+ 1u(t)
p+1.
1.1 Regularity of Solution to the Equation (1.1) with p∈ N
Now we consider problem(1.1) with p ∈ N, we have the following results: Theorem 1. If u is the solution of the problem(1.1) with the life-span T ∗ and p∈ N, then u ∈ Cq(0, T∗) for any q ∈ N and
(1.3) u(2n)= Cn0 p+1 i=0 En iuCn i, (1.4) u(2n+1) = Cn0 p+1 i=0 En iCn iuCn i−1u = Cn0 p+1 i=0 On iuCn i−1u
for positive integer n, where
Cn0
p+1
denotes the Gaussian integer number of p+1Cn0,
Cn i= (n − i) (p + 1) − 2n + 1, On i = En iCn i, E00= 1 and En0= O(n−1)0 2 p+ 1 C(n−1)0− 1+ 1 = E(n−1)0C(n−1)0 2 p+ 1 C(n−1)0− 1+ 1 , En(n−1) = O(n−1)(n−2)C(n−1)(n−2)− 1E(0) = E(n−1)(n−2)C(n−1)(n−2)C(n−1)(n−2)− 1E(0) , Enk = O(n−1)(k−1)C(n−1)(k−1)− 1E(0) + O(n−1)k 2 p+ 1 C(n−1)k− 1+ 1 = E(n−1)(k−1)C(n−1)(k−1)C(n−1)(k−1)− 1E(0) + E(n−1)kC(n−1)k 2 p+ 1 C(n−1)k− 1+ 1 ,
for positive integer k and 0 < k < n.
Proof. Let vn be the n-th derivative of u ; that is vn:= u(n), then v0n= un, v0 = u, v1 = u, v2 = u, v12 = (u)2. To prove (1.3) we use mathematical induction. When n= 1, we have
v2 = C10 p+1 i=0 E1 iuC1 i = E 10uC10 = v0p, C00= (0 − 0) (p + 1) − 2 × 0 + 1 = 1, C10 = p and E10= E00C00 2 p+ 1(C00− 1) + 1 = 1. Suppose v2n = Cn0 p+1 i=0 En i· v0Cn i, n∈ N. Then
v2n+1 = Cn0 p+1 i=0 En iCn i· v0Cn i−1· v1 and v2n+2 = Cn0 p+1 i=0 En iCn i v0Cn i−1· v2+ (Cn i− 1) v0Cn i−2· v21 . By (1.2) we obtain v2n+2 = Cn0 p+1 i=0 On i· 2 p+ 1(Cn i− 1) + 1 vC0n i+p−1 + Cn0 p+1 i=0 On i· (Cn i− 1) · E (0) v0Cn i−2 = Cn0 p+1 i=0 On i· 2 p+ 1(Cn i− 1) + 1 vC0(n+1) i + Cn0 p+1 i=0 On i· (Cn i− 1) · E (0) v0C(n+1)(i+1) = On0· 2 p+ 1 (Cn0− 1) + 1 v0C(n+1)0 + On0· (Cn0− 1) · E (0) v0C(n+1) 1 + On1· 2 p+ 1(Cn1− 1) + 1 vC0(n+1) 1 + On1· (Cn1− 1) · E (0) v0C(n+1)2 + On2· 2 p+ 1(Cn2− 1) + 1 vC0(n+1)2+ · · · + ... + OnCn0 p+1 · CnCn0 p+1 − 1 · E (0) v C (n+1) Cn0 p+1 +1 0 . Hence
v2n+2 = C(n+1)0 p+1 i=0 E(n+1) i· vC0(n+1) i,
which completes the induction procedures and we obtain(1.3). Using (1.3), we get (1.4).
1.2. The Properties of u(n)
Drawing the graphs of the u(n)is not easy, so in this section we choose a spacial index p= 2.
We consider only on the properties of the solution u to the case that E(0) = 0 for the equation
(1.5) u= u2, u(0) = 1, u(0) = 2 3.
The solution of equation(1.5) can be solved explicitly
u(t) = √ 6
6 − t2
and this affords the graphs of u, u, u, u(3)and u(4) below.
With the help of graphing with maple we find that the n-th derivative u(n) is smooth and that the blow-up rate of u(n) is increasing in n. Here we do not give rigorous proof, we will illustrate this in section III.
2. REGULARITY OFSOLUTION TO THEEQUATION (0.1) WITHp∈ Q − N
According to the preceding section we obtain the solution u∈ Cq(0, T ) of (0.1) with p∈ N for any q ∈ N. In this section we consider the equation of (0.1) with
p∈ Q − N.
Except the null points (zeros) of u, u(q) are also differentiable for any q ∈ N. We have
Theorem 2. If u is the solution of the problem(0.1) with p ∈ Q − N, p ≥ 1 and the followings do not hold
(i) a(0)2<4a (0) E (0) , E (0) > 0,
(ii) a(0)2= 4a (0) E (0) , E (0) > 0 and u
1 <0, p is even, then u ∈ Cq(0, T ) for any q ∈ N. Further, we have
(2.1) u(2n)= n−1 i=0 En iuCn i and (2.2) u(2n+1) = n−1 i=0 En iCn iuCn i−1u =n−1 i=0 On iuCn i−1u.
Proof. Same as the procedures given in the proof of Theorem 1, to prove(2.1)
and(2.2) by mathematical induction. If t0 is the null (zero) point of u, then limt→t0ucn i(t0)−1= 0
for i > n(p−1)+1p+1 = Cn0
p+1 since that Cn i <0, for i > Cp+1n0. By lemma 8 we know
that u possesses the null point (zero) only in the case(i) or (ii) . Hence, we obtain the assertions by Theorem 1.
Similarly, by the same arguments above, we have also a result as following: Theorem 3. If u is the solution of the problem(0.1) with p ∈ Q − N, p ≥ 1 and one of the followings holds
(i) a(0)2<4a (0) E (0) , E (0) > 0
(ii) a(0)2= 4a (0) E (0) , E (0) > 0andu
1 <0, p is even.
Then u ∈ C[p]+2(0, T ), where [p] mean that Gaussian integer number of p. Further, we have (2.3) u(2n)= n−1 i=0 En iuCn i, for n≤ p 2 + 1 and u(2n+1) = n−1 i=0 En iCn iuCn i−1u =n−1 i=0 On iuCn i−1u, for n≤ p 2 + 1. (2.4)
Proof. Same as the proof of Theorem 1, we obtain also the identities(2.3) and
(2.4).
By lemma 8, we know that u possesses the null point (zero) in the case (i) or (ii) . (Figure 2.1) If t0 is the null point of u then limt→t0u−cn i(t) = 0 for
Cn i < 0. Hence, in the case of (i) or (ii) , we should find the range of n with Cn i≥ 0 as i = n − 1, and then u(2n) exists only in such situation.
Here
Cn i= (p + 1) (n − i) − 2n + 1.
Let
Cn(n−1)= (p + 1) (n − (n − 1)) − 2n + 1 ≥ 0,
then we get n≤ p2 + 1. Since n is an integer, we have n ≤ p2 + 1.
Now u(2n) exists for n ≤ p2 + 1 in the case (i) or (ii) ; thus we obtain that
u ∈ C[p]+2(0.T ) .
Example 2.1. To draw the graphs of u(n) for p ∈ Q − N is not easy, so we choose a special index p= 73.
We consider on the properties of the solution u to the case that E(0) > 0 for the equation
(2.5)
u= u73,
Fig. 2.1.
Because the solution of equation (2.5) can not be solved explicitly, we solve this ode numerically and obtain the graphs of u, u, u, u(3) u(4) and u(5) below by Maple.
Fig. 2.3. Graphs of u(3) in solid, u(4) in dash, u(5) in dots.
By Theorem 3, we know that u∈ C4(0, T ) . With the help of graph with maple, we find the t0∼ 1.4 of the null point of u (Figure 2.2) and the u(5) is close in infinite as t approach to 1.4(Figure 2.3) . Hence we know that u(5)(t) does not exist for t = t0 by the graph. The blow-up rate of u(n)is increasing in n. It will be illustrate in the next section .
3. THEBLOW-UPRATE ANDBLOW-UP CONSTANT
Finding out the blow-up rate and blow-up constant of u(n)of the equation(0.1) given as follows is our main result, we have the following results:
Theorem 4. If u is the solution of the problem (0.1) with one of the following properties that (i) E (0) < 0 or (ii) E (0) = 0, a(0) > 0 or (iii) E (0) > 0, a(0)2>4a (0) E (0) or (iv) E (0) > 0, a(0)2= 4a (0) E (0) , u 1>0 or (v) E (0) > 0, a(0)2= 4a (0) E (0) , u 1<0 and p is odd.
Then the blow-up rate of u(2n) is p−12 + 2n, and the blow-up constant of u(2n) is |En0
√ 2(P +1) p−1 2 p−1+2n
; that is, for n∈ N, m ∈ {1, 2, 3, 4, 5, 6}
(3.1) lim t→Tm∗ u(2n)(Tm∗ − t)p−12 +2n = (±1)Cn0E n0 √ 2(P +1) p−1 2 p−1+2n := K2n
The blow-up rate of u(2n+1) is 2
p−1 + 2n + 1, and the blow-up constant of u(2n+1) is
En0Cn0 2 p+1 √ 2(P +1) p−1 2 p−1+2n+1
; that is, for n∈ N, m ∈ {1, 2, 3, 4, 5, 6}
(3.2) lim t→Tm∗ u(2n+1)(Tm∗ − t)p−12 +2n = (±)Cn0E n0Cn0 2 p+1 √ 2(P +1) p−1 2 p−1+2n+1 := K2n+1 where Cn0= (p − 1) n + 1, En0= Πn−1i=0 2 (p − 1)2i2+ (p − 1) i p+ 1 + (p − 1) i + 1 .
Proof. Under condition(i), E (0) < 0, a(0) ≥ 0 by (0.7) and (0.8), we obtain that (3.3) J(t) 0 1 T1∗− t dr k1+ E (0) rk2 = p− 1 2 ∀t ≥ 0. Using lemma 3 and(3.3) we have
limt→T1∗ 1 √ k1 J(t) T1∗− t = p− 1
2 ; (see appendix A.1) in other words, (3.4) limt→T∗ 1 a(t) (T ∗ 1 − t) 4 p−1 = 2 (p − 1)√k1 4 p−1 and then (3.5) limt→T∗ 1u(t) (T1∗− t) 2 p−1 = ± 2 (p − 1)√k1 2 p−1 .
Here Cn i= p + (n − 1 − i) (p + 1) − 2 (n − 1), hence we have Cn i > Cn j as i < j. By(2.1) and (3.5) , we obtain limt→T1∗u(2n)(T1∗− t) 2 p−1×Cn0 = (±1)Cn0E n0 2 (p − 1)√k1 2 p−1×Cn0 .
Since 2
p−1 × Cn0=p−12 + 2n and k1= p+12 , so we get(3.1) for m = 1.
By(0.6) , we find that (3.6) lim t→T1∗ J(t) = −√p− 1 2p + 2 and 2√2 √ p+ 1 = limt→T1∗ a(t) (T1∗− t)p−14 −p−1 4 −1 · lim t→T1∗a (t) (T∗ 1 − t) 4 p−1×p+34 .
Together(3.4) and (2.2) we obtain that
(3.7) lim t→T1∗ u(t) (T1∗− t)p−12 +1= ±k1 2 (p − 1)√k1 2 p−1+1 and lim t→T1∗ u(2n+1)(T1∗− t)p−12 Cn0+1 = lim t→T1∗ n−1 i=0 En iCn iuCn i−1· u· (T∗ 1 − t) 2 p−1Cn0+1 = lim t→T1∗En0Cn0u Cn0−1· u· (T∗ 1 − t) 2 p−1Cn0+1 = lim t→T1∗ En0Cn0uCn0−1· (T∗ 1 − t) 2 p−1Cn0−1· u· (T∗ 1 − t) 2 p−1+1 = lim t→T1∗(±) Cn0E n0Cn0 k1 2 (p − 1)√k1 2 p−1Cn0+1 ; thus(3.2) for m = 1 is proved.
For E(0) < 0, a(0) < 0, by (0.10) we have (3.8) J(t) 0 dr (T∗ 2 − t) k1+ E (0) rk2 = p− 1 2 ∀t ≥ t0.
Using lemma 3,(3.8) and (2.1), therefore we gain the estimate (3.1) for m = 2, and by(0.9) we get the estimate (3.2) for m = 2. (see appendix A.2)
Under(ii) , E (0) = 0, a(0) > 0, inducing (0.12) , we have (3.9) a(t) = a (0)p+3p−1 p− 1 4 a(0) (T3∗− t) − 4 p−1 ∀t ≥ 0.
Using(0.12), we also obtain J(t) = J(0) ∀t ≥ 0 and lim t→T1∗a(t) −p−1 4 −1a(t) = −p− 1 4 a(0)− p−1 4 −1a(0) . By(3.9) and (2.2) , the estimate (3.2) for m = 3 is completely proved.
Under(iii) or (iv) or (v) , the proofs of estimates (3.1) and (3.2) for m = 4, 5, 6 are similar to the above arguments, we omit the argumentations.
Theorem 5. Suppose that u is the solution of the problem(0.1) with E (0) > 0 and one of the following properties holds
(i) a(0)2<4a (0) E (0) and a(0) ≤ 0.
(ii) a(0)2<4a (0) E (0) and a(0) > 0.
(iii) a(0)2= 4a (0) E (0) and u
1<0, p is even. Then we have
(3.10) lim t→z1 u(2n)(t) (zm− t)−Cn(n−1) = (±)Cn(n−1)E n(n−1)E(0) Cn(n−1) 2 and (3.11) lim t→z1u (2n+1)(t) (z m− t)−Cn(n−1)+1 = En(n−1)Cn(n−1)E(0)Cn(n−1)−1 for n∈ N, m ∈ {1, 2, 3}, where zmis the null point (zero) of u and
Cn(n−1)= p − 2n + 2,
En(n−1)= Πn−1i=0 (p − 2i + 2) (p − 2i + 1) E (0)n−1. Proof. Under(i) using (0.19) and (0.20) we get
(3.12) limt→z1u(t) (z1− t)−1= ±E (0)12 and
(3.13) limt→z1u(t) (z1− t)−1= ∓E (0)12. By(2.1) and (3.12) we obtain that
limt→z1u(2n)(z1− t)−Cn(n−1) = limt→z1 n−1 i=0 En iuCn i(z 1− t)−Cn(n−1) = limt→z1En(n−1)uCn(n−1)(z1− t)−Cn(n−1) = (±1)Cn(n−1)E n(n−1)E(0) Cn(n−1) 2 .
Therefore,(3.10) for m = 1 is proved. From(2.2), (3.12) and (3.13), we obtain
limt→z1u(2n+1)(z1− t)−Cn(n−1)+1 = limt→z1 n−1 i=0 EniCniuCni−1u(z 1− t)−Cn(n−1)+1 = limt→z1En(n−1)Cn(n−1)uCn(n−1)−1u(z1− t)−Cn(n−1)+1 = En(n−1)Cn(n−1)E(0)Cn(n−1).
Thus,(3.11) for m = 1 is obtained.
Under the (ii) or (iii), the proofs of estimations (3.10) and (3.11) for m = 2, 3 are similar to the above arguments, we do not bother them again.
Appendix Proof of Theorem 4 A.1 Lemma Lemma A1. If J(t) 0 1 T∗− t dr √ k1+E(0)rk2 = p−1
2 for each t≥ 0, then limt→T∗ √1 k1 J(t) T∗− t= p− 1 2 .
Proof. Let r= (T∗− t) s, then using lemma 3, we conclude
limt→T∗ J(t) (T ∗−t) 0 ds k1+ E (0) (T∗− t)k2sk2 =√1 k1 limt→T ∗ (T ∗ −t)J(t) 0 ds= limt→T∗ √1 k1 J(t) T∗− t. A.2. Lemma
Lemma A2. If u is the solution of the problem(0.1) with E (0) < 0 and a(0) < 0, then(3.1) and (3.2) hold for m = 2.
Proof. By lemma A1.
ACKNOWLEDGMENT
We want to thank Prof. Tsai Long-Yi and Prof. Klaus Schmitt for their continuous encouragement and their discussions of this work, and to Grand Hall for their financial support.
REFERENCES
1. R. Bellman. Stability Theory of Differential Equation, McGraw-Hill Book Company. 1953.
2. M. R. Li, Nichtlineare Wellengleichungen 2. Ordnung auf beschr ¨ankten Gebieten, PhD-Dissertation T ¨ubingen 1994.
3. M. R. Li, Estimation for The Life-span of solutions for Semi-linear Wave Equations.
Proceedings of the Workshop on Differential Equations V., National Tsing-Hua
Uni-versity Hsinchu, Taiwan, Jan. 10-11, 1997,
4. M. R. Li, On the Differential Equation u= up, Taiwanese Math. J., 9(1) (2005), 39-65.
Meng-Rong Li and Zing-Hung Lin Department of Mathematic Sciences, National Chengchi University, Taipei 116, Taiwan, R.O.C. E-mail: Liwei@math.nccu.edu.tw
