BLOW-UP BEHAVIOR FOR A SEMILINEAR HEAT EQUATION
WITH A NONLINEAR BOUNDARY CONDITION
SHENG-CHENFU, JONG-SHENQGUO ANDJE-CHIANGTSAI
(Received December 28, 2001, revised September 9, 2002)
Abstract. We study the blow-up behaviors of solutions of a semilinear heat equation with a nonlinear boundary condition. Under certain conditions, we prove that the blow-up point occurs only at the boundary. Then, by applying the well-known method of Giga-Kohn, we derive the time asymptotic of solutions near the blow-up time. Finally, we prove that the blow-up is complete.
1. Introduction. In this paper, we study the following initial boundary value problem ut = uxx+ up, x ∈ (0, 1) , t > 0 , (1.1) ux(0, t) = 0 , ux(1, t) = uq(1, t) , t > 0 , (1.2) u(x, 0) = u0(x) , x ∈ [0, 1] , (1.3)
wherep, q are positive constants, and u0(x) is a positive smooth function. For convenience, we always assume that
u0(0) = 0 , u0(1) = uq0(1) .
We say that the solutionu of the problem (1.1) – (1.3) blows up if there is a finite time T such that max0≤x≤1u(x, t) → ∞ as t ↑ T . It has been shown in [11] that the solution u of the problem (1.1) – (1.3) blows up if and only if max{p, q} > 1. In [11], they also studied the blow-up set and derived the upper and lower bounds for blow-up rate under certain conditions. A pointx0is said to be a blow-up point foru if there is a sequence {(xn, tn)} such that xn → x0,tn → T , and u(xn, tn) → ∞ as n → ∞. Under certain conditions, it is shown in [11] that the boundary pointx = 1 is the only blow-up point. This phenomenon of blow-up on the boundary has been observed and studied by many authors. We refer the readers to two nice survey papers [4] and [2] and the references cited therein. See also the references cited in [11].
We are concerned with the blow-up behaviors of solutions of the problem (1.1) – (1.3). Hence throughout this work we always assume that max{p, q} > 1. In the sequel, we shall assume that the solutionu of the problem (1.1) – (1.3) blows up at T < ∞. First, we study the blow-up set. We prove that blow-up point occurs only at the boundary pointx = 1, if u0 ≥ 0 in [0, 1]. This improves the results of [11], where the monotonicity of u in time is
2000 Mathematics Subject Classification. Primary 35K60; Secondary 34A12, 35B40, 35K20.
This work was partly supported by the National Science Council of the Republic of China under the contract NSC 89-2115-M-003-014.
assumed. Furthermore, by deriving some a priori estimates, with the help of the lower and upper bounds for blow-up rate, we can apply the well-known Giga-Kohn transformation (cf. [8]) to derive the time asymptotic of solutions. This gives a more precise information of the blow-up behaviors. We remark here that a similar problem for the case when the heat operator is replaced by the porous medium operator in the half real line has been studied by de Pablo, Quirós and Rossi [3].
The next question is the possibility of continuation of solutions beyond the blow-up time. For more references on this subject, we refer the readers to the paper [5] and some references listed there. We show that the blow-up for the problem (1.1) – (1.3) is complete, i.e., solutions blowing up in finite time will be infinite identically after the blow-up time.
This paper is organized as follows. We study the blow-up set in Section 2. Some a priori estimates are given in Section 3. In Section 4, we study the self-similar solution for the critical case by an ordinary differential equation approach. Then we study the time asymptotic of the solution in Section 5. Finally, in Section 6 we prove that the blow-up is complete.
We thank the referee for helpful comments which improve some results in Section 2.
2. Blow-up set. Letu be the solution of the problem ut = uxx+ up, x ∈ (0, 1) , 0< t < T , (2.1) ux(0, t) = 0 , ux(1, t) = uq(1, t) , 0< t < T , (2.2) u(x, 0) = u0(x) , x ∈ [0, 1] , (2.3)
whereT is the blow-up time of u. Here as usual we always assume that u0(0) = 0 and u0(1) = uq0(1). We shall assume that u0≥ 0 in [0, 1], so that by the maximum principle we haveux > 0 in (0, 1] × (0, T ). We shall modify the method of Friedman and McLeod ([7]) to study the blow-up set.
THEOREM 2.1. Suppose thatp > 1. If u0≥ 0 in [0, 1], then the blow-up occurs only atx = 1.
PROOF. Suppose that there is a blow-up pointa ∈ [0, 1). Then there is a sequence
{(xn, tn)} such that xn → a, tn → T , and u(xn, tn) → ∞ as n → ∞. Fix a constant d ∈ (a, 1). By comparing the solution u with the function
u(xn, tn) sin[π(x − d)/(1 − d)] exp{−[π/(1 − d)]2(t − tn)} for eachn sufficiently large, it is easy to show that
u(x, t) ≥ u(xn, tn) sin[π(x − d)/(1 − d)] exp{−[π/(1 − d)]2(t − tn)} forx ∈ [d, 1] and t ≥ tn. Hence
lim
t→Tu(x, t) = ∞ (2.4)
Now, we fix a compact subset[b, c] of (d, 1). We take any r ∈ (1, p) and consider the function
J (x, t) = ux(x, t) − g (x)ur(x, t) ,
where g(x) = ε sin[π(x − b)/(c − b)] for some ε > 0 to be determined. Using (2.4), there is
at0∈ (0, T ) such that
Jt− Jxx− (pup−1+ 2rgur−1)J ≥ 0
(2.5)
in[b, c] × [t0, T ) for any ε > 0. By choosing ε > 0 sufficiently small such that J (x, t0) ≥ 0 for anyx ∈ [b, c], it follows from the maximum principle that J ≥ 0 in [b, c]×[t0, T ). Hence we have
u−r(x, t)ux(x, t) ≥ g (x) in [b, c] × [t0, T ) . (2.6)
An integration of (2.6) leads to a contradiction. Hence the theorem follows. 2 Ifp ≤ 1, then q > 1, since max{p, q} > 1.
LEMMA 2.2. Let 0< p ≤ 1. If u0≥ 0 in [0, 1], then x = 1 is the only blow-up point. PROOF. Sinceu0(1) > 0, there is a constant δ ∈ (0, 1) such that u0> 0 in [1 − δ, 1]. Set η = inf 1−δ≤x≤1{u 0(x)/uq0(x)}, M = (q − p) sup [1−δ,1]×[0,T )u p−1.
Then 0< η ≤ 1 and 0 < M < ∞. Choose a positive integer n ≥ 3 such that n ≥ M and a positive numberε < min{η, δn}. Define g (x) = (x − 1 + ε1/n)n if 1− ε1/n ≤ x ≤ 1; g (x) = 0, otherwise. It is easy to see that g ∈ C2([0, 1]) and satisfies
0≤ g ≤ ε, g≥ 0, g≥ 0, g≥ Mg . (2.7)
Then, by using the factq > 1 and the maximum principle, it is easy to show that g (x)uq(x, t) ≤ ux(x, t)
for 0≤ x ≤ 1 and 0 ≤ t < T . Hence
u−q(x, t)ux(x, t) ≥ g (x) (2.8)
for 0 ≤ x ≤ 1 and 0 ≤ t < T . An integration of (2.8) shows that u cannot blow up at any
pointx < 1. This proves the lemma. 2
Indeed, the conditionu0≥ 0 in [0, 1] can be removed if 0 < p ≤ 1. THEOREM 2.3. If 0< p ≤ 1, then x = 1 is the only blow-up point.
PROOF. We first extend the functionu(x, t) to w(x, t) defined on [−1, 1] × [0, T ) so thatw(x, t) = u(x, t) and w(−x, t) = w(x, t) for x ∈ [0, 1] and t ∈ [0, T ). Then w satisfies
wt = wxx+ wp, x ∈ (−1, 1) , 0< t < T ,
Arguing as Lemma 1.2 of [9], there existst∗∈ (0, T ) such that n(t) := #{a ∈ [−1, 1] | wx(a, t) = 0}
is a constant for allt ≥ t∗. Moreover, there areC1functionss0(t), . . . , s±l(t), l ≥ 0, from [t∗, T ) to [−1, 1] such that
s−l(t) < · · · < s0(t) < · · · < sl(t) , s0(t) ≡ 0 ,
{a ∈ [−1, 1] | wx(a, t) = 0} = {s−l(t), . . . , s0(t), . . . , sl(t)} for t ≥ t∗,
and the limitsi := limt↑T si(t) exists for all i. Since n(t) is constant in [t∗, T ), it follows
from Theorem C of [1] thatwxx(si(t), t) = 0 for all t ∈ [t∗, T ). Note that for each i there
is a fixed sign forwxx(si(t), t) for all t ∈ [t∗, T ). Also, it suffices to consider the so-called
maximum curve, i.e., the curve for whichwxx(si(t), t) < 0 on [t∗, T ).
Ifl = 0, then ux(x, t) > 0 on (0, 1] × [t∗, T ). Hence the conclusion follows from
Lemma 2.2. Suppose thatl > 0. Set mi(t) := w(si(t), t). Notice that mi(t) < mi(t)p on
[t∗, T ), if w
xx(si(t), t) < 0 on [t∗, T ). Since 0 < p ≤ 1, mi(t) remains bounded near T .
This implies thatw cannot blow up at any point in (−1, 1). The theorem is proved. 2 3. Some a priori estimates. In this section, we will derive some a priori estimates which will be used in Section 5 to prove the time asymptotic results. Letu be the solution
of (2.1) – (2.3) with blow-up timeT . From now on we shall always assume that u0 ≥ 0 in
[0, 1], so that by the maximum principle we have ux > 0 in (0, 1] × (0, T ). Notice that u(1, t) = max0≤x≤1u(x, t).
The following lemma is given in [11] under the assumptionu0+ u
p
0 ≥ a > 0 in [0, 1]. Indeed, we have the following lemma.
LEMMA 3.1. Ifu0+ up0 ≥ 0 in [0, 1], then ut ≥ 0 in [0, 1] × [0, T ). PROOF. Setv = ut. Thenv satisfies
vt = vxx+ pup−1v , 0< x < 1 , 0< t < T ,
vx(0, t) = 0 , vx(1, t) = quq−1(1, t)v(1, t) , 0< t < T , v(x, 0) = u0+ up0 ≥ 0 , 0 ≤ x ≤ 1 .
For any fixedτ ∈ (0, T ), let
L = max 0≤x≤1,0≤t≤τ 1 2qu q−1(x, t) , M = 2L + 4L2+ max 0≤x≤1,0≤t≤τ{pu p−1(x, t)} .
Setw(x, t) = e−Mt−Lx2v(x, t). Then w satisfies
wt = wxx+ 4Lxwx+ cw , 0 < x < 1 , 0 < t ≤ τ , wx(0, t) = 0 , wx(1, t) = dw(1, t) , 0< t ≤ τ , w(x, 0) ≥ 0, 0≤ x ≤ 1 ,
wherec = c(x, t) ≤ 0 and d = d(t) ≤ 0. By the maximum principle, we obtain that w ≥ 0
Recall from [11] that ifut ≥ 0, then there are positive constants c and A such that c(T − t)−α≤ u(1, t) ≤ A(T − t)−α,
(3.1)
where the exponent is given by
α =
1/(p − 1) if p ≥ 2q − 1 ,
1/[2(q − 1)] if p < 2q − 1 .
Hereafter we shall assume thatu0 ≥ 0 and u0+ up0 ≥ 0 in [0, 1]. Therefore, we have ux≥ 0 and ut ≥ 0. We now make the following Giga-Kohn transformation
y = √1− x
T − t , s = − ln(T − t) , (3.2)
w(y, s) = (T − t)αu(x, t) , (3.3)
whereα is defined as in (3.1). Let
W = {(y, s) | 0 < y < es/2, s > − ln T } . Then forp > 2q − 1 we have
ws = wyy−y2wy− αw + wp in W , (3.4) wy(0, s) = −eγ sw(0, s)q, wy(es/2, s) = 0 , s > − ln T , (3.5) w(y, − ln T ) = Tαu0(1 − y√T ) , 0≤ y ≤ 1/√T , (3.6)
whereγ = [(2q − 1) − p]/[2(p − 1)] < 0; for p = 2q − 1 we have ws = wyy−y2wy− αw + wp in W , (3.7) wy(0, s) = −w(0, s)q, wy(es/2, s) = 0, s > − ln T , (3.8) w(y, − ln T ) = Tαu0(1 − y√T ) , 0≤ y ≤ 1/√T , (3.9)
while forp < 2q − 1 we have
ws = wyy−y2wy− αw + eσ swp in W , (3.10) wy(0, s) = −w(0, s)q, wy(es/2, s) = 0, s > − ln T , (3.11) w(y, − ln T ) = Tαu0(1 − y√T ) , 0≤ y ≤ 1/√T , (3.12) whereσ = [p − (2q − 1)]/[2(q − 1)] < 0. We have the following a priori estimates forw. LEMMA 3.2. w and wyare bounded in ¯W .
PROOF. The fact thatw is bounded follows from (3.1).
It follows from Lemma 3.1 thatuxx ≥ −up in[0, 1] × [0, T ). Multiplying the above inequality byux ≥ 0 and integrating it from x to 1 , we obtain
u2x(x, t) ≤ u2q(1, t) + 2 p + 1u
p+1(1, t) . (3.13)
Note thatwy(y, s) = −(T − t)α+1/2ux(x, t).
Recall (3.1). Forp ≥ 2q − 1, it follows from (3.13) and Lemma 3.1 that w2y(y, s) ≤ (T − t)2α+1u2q(1, t) + 2 p + 1(T − t) 2α+1up+1(1, t) = [(T − t)αu(1, t)]p+1u2q−1−p(1, t) + 2 p + 1[(T − t) αu(1, t)]p+1 ≤ Ap+1u2q−1−p 0 (1) + Ap+1.
Forp < 2q − 1, it also follows from (3.13) and Lemma 3.1 that w2y(y, s) ≤ (T − t)2α+1u2q(1, t) + 2 p + 1(T − t) 2α+1up+1(1, t) = [(T − t)αu(1, t)]2q+ 2 p + 1[(T − t) αu(1, t)]2qup−2q+1(1, t) ≤ A2q+ 2 p + 1A 2qup−2q+1 0 (1) .
Hence the lemma is proved. 2
LEMMA 3.3. There is a positive constantC such that |ws(y, s)| ≤ C(1 + y) and
|wyy(y, s)| ≤ C(1 + y) in ¯W .
PROOF. It follows from Lemma 3.2 that|ws(y, s) − wyy(y, s)| ≤ C(1 + y) in ¯W for some positive constantC. The lemma follows by applying the standard theory of parabolic equations, e.g., Theorem 6.44, Theorem 4.30 and Theorem 4.31 in [10]. 2
4. Self-similar solution. In this section, we shall study the self-similar solution of (1.1) – (1.2) for the casep = 2q − 1, i.e., q = (p + 1)/2. We are concerned with the existence and uniqueness of global positive monotone decreasing solution of the initial value problem (P): w−1 2yw − αw + wp= 0 , y > 0 , (4.1) w(0) = −wq(0) , (4.2)
wherew = w(y) and α = 1/(p − 1). We always assume that p > 1. The existence result has been obtained before by Wang and Wang in [12]. Here we present a different proof for the existence. Some of the lemmas will be useful for the proof of uniqueness.
Given anyη > 0, there is a unique local solution w(y; η) of (4.1) – (4.2) with w(0) = η. Letρ(y) = exp{−y2/4} and f (w) = −αw + wp. Thenw satisfies
(ρw)(y) = −ηq−
y
0 ρ(s)f (w(s))ds . (4.3)
Letκ be the unique positive solution of f (w) = 0. Note that w< 0 as long as w ≥ κ. Set κ0= p + 1 p + 3 α κ . (4.4) Note that 0< κ0< κ.
LEMMA 4.1. Letη ≥ κ0. Thenw < 0 as long as w > 0. PROOF. Let
G(y) = 1 2[w
(y)]2+ F (w(y)) , whereF (w) =κwf (s)ds. Note that
G(0) ≥ F (0) if and only if η ≥ κ0. (4.5) Since G(y) = 1 2y[w (y)]2,
and the problem (P) has no non-trivial constant solution,G is strictly increasing.
Ifw is not monotone decreasing, then there is the first critical point y0 > 0 of w such thatw(y0) = 0 and w > 0 in [0, y0]. Notice that w(y0) < κ. Hence
G(y0) = F (w(y0)) < F (0) .
On the other hand, by (4.5) we haveG(0) ≥ F (0), since η ≥ κ0. This implies thatG(0) >
G(y0), a contradiction. Therefore, the lemma follows. 2
Suppose thatw > 0 and w < 0 in [0, ∞). Let l = limy→∞w(y). Then l ∈ [0, κ) and there is a sequence{yn} such that w(yn) → 0 as n → ∞. Dividing Equation (4.1) by y and integrating it from 1 to ynfor anyn large, as n → ∞, this leads to a contradiction, if l ∈ (0, κ). Hence l = 0.
LEMMA 4.2. For η ≥ κ0, the solutionw is monotone decreasing to zero at some finiteR.
PROOF. Otherwise, by Lemma 4.1 and the above observation,w(y) → 0 as y → ∞ and there is a sequence{yn} such that w(yn) → 0 as n → ∞. Then G(yn) → F (0) as n →
∞. Since G is monotone increasing, its limit must be greater than G(0), i.e., F (0) > G(0), a
contradiction to (4.5). This proves the lemma. 2
We now turn to the case whenη is small. First, let η0be a positive constant such that
−f (w) ≥ αw/2 for all w ∈ [0, η0]. Notice that η0 < κ. Choose η1 ∈ (0, η0) such that η1−q > e1/4 for allη ∈ (0, η1). Now, given any fixed η ∈ (0, η1), suppose that w < 0 in
[0, R] and w(R) = 0 for some R = R(η) > 0. Since, by (4.3), ρ(y)w(y) ≥ −ηq for all y ∈ [0, R), we have
η = −
R
0
Let g(y) = yey2/4. Since g is strictly monotone increasing, we conclude thatR = R(η) > 1,
ifη < η1.
LEMMA 4.3. There is a small positive constantη∗ such that w vanishes before w vanishes, if η < η∗.
PROOF. Letη∗be a positive constant such thatη∗< η1and η1−q >1+ α/4
α/2 e
1/4 for all η ∈ (0, η
∗) . (4.6)
Suppose that there is anη ∈ (0, η∗) such that the lemma does not hold. Then the correspond-ing solutionw must have the property that w < 0 in [0, y0] for some y0> 1. From (4.3) it follows thatρ(y)w(y) ≥ −ηqfor ally ∈ [0, y0] and so
w(y) = η +
y
0
w(s)ds ≥ η − ηqyey2/4 for all y ∈ [0, y0] . (4.7)
Then from (4.3), (4.7), and noting thatη < η0, we obtain that ρ(y)w(y) ≥ −ηq+α 2 y 0 ρ(s)w(s)ds ≥ −ηq+α 2 y 0 e−s2/4[η − ηqses2/4]ds ≥ − 1+α 4y 2 ηq+α 2ye −y2/4 η
for ally ∈ (0, y0). In particular, for y = 1 we have e−1/4w(1) ≥ − 1+α 4 ηq+α 2e −1/4η > 0 ,
sinceη < η∗. This is a contradiction and the lemma is proved. 2 Now, we define
I1= {η > 0 | w(y; η) is decreasing to zero at some finite R} I2= {η > 0 | w(y; η) vanishes before w(y; η) vanishes}
Notice that w and w cannot vanish at the same time. HenceI1 andI2 are disjoint. Lemmas 4.2 and 4.3 imply that[κ0, ∞) ⊂ I1and(0, η∗) ⊂ I2.
LEMMA 4.4. The setI2is open.
PROOF. Letη0 ∈ I2. Thenη0 < κ0 < κ and there is the first point y0 > 0 such that w0 > 0 in [0, y0], w0 < 0 in [0, y0) and w0(y0) = 0, where w0(y) = w(y; η0). Since w0(y0) > 0, there is a positive constant δ such that w0(y) > 0 for y ∈ (y0, y0+ δ]. Let ε > 0, ε < w0(y0)/2, and ε < w0(y0+ δ)/2. By the continuous dependence of initial value, there is a positive constantγ such that |w(y; η) − w0(y)| < ε and |w(y; η) − w0(y)| < ε for ally ∈ [0, y0+ δ], if η ∈ (η0− γ, η0+ γ ). This implies that (η0− γ, η0+ γ ) ⊂ I2and so
To prove thatI1is open, we consider the quantity
H (y) = αw(y) +1
2yw (y) . (4.8)
ThenH satisfies the equation H(y) = 1 2yH (y) + 1 2 + α w(y) −1 2yw p(y) . (4.9)
Suppose thatH (y0) < 0 for some y0 ≥ 0. Then w(y0) < 0 by (4.8) and H(y0) < 0 by (4.9). Hence, by (4.9) again,H(y) < 0 and H (y) < 0 for all y ≥ y0as long asw > 0.
LEMMA 4.5. The setI1is open.
PROOF. First, we claim that if there is a pointy0 ≥ 0 such that H (y0) < 0, then w is decreasing aftery0and vanishes at some finiteR > y0. Otherwise, ifw > 0 in [0, ∞), thenH (y) < 0 and H(y) < 0 for all y ≥ y0. Hence there is a positive constantδ such that H (y) ≤ −δ for all y ≥ y0. By an integration, we obtain that
w(y) ≤ (y0/y)2αw(y0) − δ α+ δ α(y0/y) 2α → −δ α as y → ∞ , a contradiction.
Now, letη0∈ I1andw0(y) = w(y; η0). Then there is a finite R0> 0 such that w0 < 0 andw0> 0 in [0, R0). Since w0(R0) = 0 and w0(R0) < 0, there is a positive constant δ such thatH0(R0− δ) < 0, where H0(y) = αw0(y) + yw0(y)/2. It follows from the theory of continuous dependence on initial value that there is a positive constantγ such that w(y; η) > 0,w(y; η) < 0 for y ∈ [0, R0− δ], and H (R0− δ) < 0, if η ∈ (η0− γ, η0+ γ ). Then w is decreasing afterR0− δ and vanishes at some finite R > R0− δ, if η ∈ (η0− γ, η0+ γ ).
Hence the lemma is proved. 2
We now state and prove an existence theorem as follows.
THEOREM 4.6. There is a global positive monotone decreasing solution of (P). PROOF. Set ¯η = inf I1. Then the corresponding solution ¯w(y) = w(y; ¯η) must be a
global positive monotone decreasing solution of (P). 2
Indeed, for anyη ∈ I1∪ I2, the corresponding solutionw(y; η) is a global positive monotone decreasing solution of (P) satisfyingw(y; η) → 0 as y → ∞.
We have from Lemma 4.2 the estimate ¯η < κ0. Also, the initial valueη < κ0for any global positive monotone decreasing solution of (P). To derive a better estimate, we need the following generalized version of Pohozaev Identity, which is inspired by Lemma 2.1 of [13] (see also [14]).
LEMMA 4.7. Supposew(y) is a solution of (P) and define J (y) := ρ(y)(w(y))2−y
2ρ(y)w (y)w(y) +1 4 − α ρ(y)w2(y) + 2 p + 1ρ(y)w p+1(y) ,
whereρ(y) = exp[−y2/4]. Then the following identity holds: J (y) = p + 3 p + 1w p+1(0) + p − 5 4(p − 1)w 2 (0) + y 0 sρ(s) p − 1 2(p + 1)w p−1(s) −1 8 w2(s)ds . PROOF. DifferentiatingJ (y) and using (4.1), we obtain
J(y) = yρ(y) p − 1 2(p + 1)w p−1(y) − 1 8 w2(y) . IntegratingJ(y) from 0 to y and noting that
J (0) = p + 3 p + 1w p+1(0) + p − 5 4(p − 1)w 2 (0) ,
we get the desired identity. 2
COROLLARY 4.8. Suppose thatw(y) is a global positive solution of (P) satisfying w(y) → 0 as y → ∞. Then ∞ 0 sρ(s) 1 8− p − 1 2(p + 1)w p−1(s) w2(s)ds = p + 3 p + 1w p+1(0) + p − 5 4(p − 1)w 2(0) . (4.10)
PROOF. Sincew(y) is a global positive solution of (P) and limy→∞w(y) = 0, there is a sequenceyn→ ∞ such that limn→∞w(yn) = 0. Using Lemma 4.7, (4.10) follows. 2 Definef1(w) = {1/8 − [(p − 1)/(2(p + 1))]wp−1}w2and let ¯κ = (α/2)α. Then it is easy to check that
max
w∈[0,∞)f1(w) = f1( ¯κ) .
The following lemma gives an upper bound for any global positive solution of (P) which tends to zero asy → ∞.
LEMMA 4.9. Suppose thatw(y) is a global positive solution of (P) with w(0) = η such thatw(y) → 0 as y → ∞. Then η < ¯κ. In particular, we have ¯η < ¯κ.
PROOF. For contradiction, we assume thatη ≥ ¯κ. It follows from the definition of ¯κ
that ∞ 0 sρ(s)f1(w(s))ds < ∞ 0 sρ(s)f1( ¯κ)ds = p − 1 4(p + 1)¯κ 2 . On the other hand, sincew(0) = η ≥ ¯κ, we have
p + 3 p + 1w p+1(0) + p − 5 4(p − 1)w 2(0) ≥ p + 3 2(p + 1)(p − 1)w 2(0) + p − 5 4(p − 1)w 2(0) = p − 1 4(p + 1)w 2 (0) ≥ p − 1 4(p + 1)¯κ 2 ,
a contradiction to(4.10). This completes the proof. 2
THEOREM 4.10. If 1 < p ≤ 2, then there is a unique global positive monotone de-creasing solution of (P).
PROOF. For contradiction, we suppose that there are two distinct global positive mono-tone decreasing solutionsw1andw2of (P). Note thatwi(y) → 0 as y → ∞ for i = 1, 2.
First, we claim thatw1andw2must intersect each other at least once. Multiplying the equation
(ρwi)(y) = −ρ(y)f (wi(y)), i = 1, 2 ,
byw2fori = 1; by w1fori = 2, respectively, and integrating by parts, we end up with
∞
0
ρ(s)w1(s)w2(s)(w1p−1(s) − wp−12 (s))ds = w1(0)w2(0)(wq−12 (0) − wq−11 (0)) , usingwi(0) = −wqi(0), i = 1, 2. Hence they must intersect each other at least once.
Without loss of generality, we may assume thatw1(y) > w2(y) in [0, y0) and w1(y0) = w2(y0) for some y0 > 0. Then we have w1(y0) < w2(y0). Hence there exists y1 > y0such thatw1(y1) < w2(y1). Define v(y) := w1(y) − w2(y). Then it follows from (4.1) that v(y) satisfies
v−y 2v
+ [pξp−1(y) − α]v = 0 , (4.11)
for someξ(y) ∈ [min{w1(y), w2(y)}, max{w1(y), w2(y)}]. Since limy→∞wi(y) = 0, i = 1, 2, there exists y2> y1such that| v(y2) |<| v(y1) | /2.
Now, let ¯y ∈ [0, y2] be a minimal point of v in [0, y2]. Then ¯y ∈ (0, y2) and v( ¯y) < 0. Since ¯y is an interior extreme point of v, we have
v( ¯y) = 0 , v( ¯y) ≥ 0 . (4.12)
From Lemma 4.9 and ξ( ¯y) ≤ max{w1(0), w2(0)}, it follows that pξp−1( ¯y) − α < 0 , (4.13)
if 1< p ≤ 2. Then, by (4.11), (4.12) and (4.13), we obtain that 0=v( ¯y) − ¯y
2v
( ¯y) + [pξp−1( ¯y) − α]v( ¯y)
≥ [pξp−1( ¯y) − α]v( ¯y) > 0 ,
a contradiction. This completes the proof. 2
We conjecture that Theorem 4.10 should hold for anyp > 1. Unfortunately we are unable to prove it now, so we left it as an open problem.
5. Time asymptotic analysis. In this section, we shall study the time asymptotic of the solutions of the problem (2.1) – (2.3) for various cases. The method is the same as the one used in [8] with some modifications. Hence we shall only give the outline of the proofs.
THEOREM 5.1. Forp > 2q − 1, we have
(T − t)αu(1 − y√T − t, t) → κ
PROOF. As in [8], we take any increasing sequence{sj} in (0, ∞) such that sj +1−sj →
∞ as j → ∞. For each j ∈ N, we define
wj(y, s) = w(y, s+sj) for all (y, s) ∈ Wj ≡ {(y, s) | 0 ≤ y ≤ e(s+sj)/2, s ≥ −sj−ln T } . Note that∞j =1Wj = [0, ∞) × R and W1 ⊂ W2 ⊂ · · · . Recall Lemmas 3.2 and 3.3. By the Ascoli-Arzela Theorem and a diagonal process, we can get a subsequence (still denoted bywj) such that wj(y, s) → w∞(y, s) as j → ∞ uniformly on any compact subset of
[0, ∞) × R and that for any integer m we have wj,y(y, m) → w∞,y(y, m) as j → ∞ pointwise fory ∈ [0, ∞) for some function w∞defined on[0, ∞) × R. It is easy to see that w∞is a classical solution of the equation
ws = wyy− 1
2ywy− αw + w
p in [0, ∞) × R .
Now, we claim thatw∞,s(y, s) ≡ 0 in [0, ∞) × R. Introduce the energy function E[w](s) =1 2 s 0 ρw2ydy + α 2 s 0 ρw2dy − 1 p + 1 s 0 ρwp+1dy ,
whereρ(y) = e−y2/4. By a simple computation, we get
−d dsE[w](s) = s 0 ρws2dy − G(s) , (5.1) where G(s) =ρ(s) 1 2w 2 y(s, s) + α 2w 2 (s, s) − 1 p + 1w p+1(s, s) + w y(s, s)ws(s, s) + exp (2q − 1) − p 2(p − 1) s w(0, s)qws(0, s) .
Lets0= max{2 ln 2, − ln T }. Note that
{(y, s) | 0 ≤ y ≤ s, s ≥ s0} ⊆ ¯W .
Integrating both sides of (5.1) fromm + sj tom + sj +1for anym, j ∈ Z with m + sj ≥ s0, we obtain
m+sj+1
m+sj
s
0
ρ(y)w2s(y, s)dyds
= Em+sj[w](m + sj) − Em+sj+1[w](m + sj +1) +
m+sj+1
m+sj
G(s)ds .
By a change of variable, we get
m+sj+1−sj
m
s+sj
0
ρ(y)w2j,s(y, s)dyds
= Em+sj[wj](m) − Em+sj+1[wj +1](m) +
m+sj+1
m+sj
Since G(s) ≤C exp(2q − 1) − p 2(p − 1) s (1 + s) , it follows that ∞ s0 G(s)ds < ∞ . Proceeding as in [8], we get M m ∞ 0
ρw2∞,sdyds = 0 for all m, M ∈ Z, m < M .
Hencew∞,s≡ 0 and so w∞(y, s) = w∞(y) for all y ∈ [0, ∞) and s.
Note thatw∞(0) > 0. Also, from wj,y(0, s) = −eγ (s+sj)wj(0, s)q, where γ = [(2q − 1) − p]/[2(p − 1)] < 0 ,
it follows thatw∞ (0) = 0. Therefore, w∞is a bounded positive global solution of
w−1
2yw
− αw + wp= 0
and sow∞≡ κ (cf. [8]). Since the sequence {sj} is arbitrary, the theorem follows. 2
Recall from [6] that there is a unique bounded positive global solution (denoted byV (y))
of
w−1
2yw
− αw = 0 , w(0) = −wq(0) .
(5.2)
THEOREM 5.2. Forp < 2q − 1, we have
(T − t)αu(1 − y√T − t, t) → V (y)
ast → T uniformly for y ∈ [0, C] for any C > 0. Here α = 1/[2(q − 1)].
PROOF. Letsj, wj, w∞be defined as in Theorem 5.1. Then it is easy to see thatw∞is a classical solution of
ws = wyy− 1
2ywy− αw , y ∈ [0, ∞) , s ∈ R . Next, we introduce the energy function
E[w](s) = 1 2 s 0 ρwy2dy +α 2 s 0 ρw2dy − 1 q + 1w q+1(0, s) .
Proceeding as in the proof of Theorem 5.1, we obtain thatw∞,s ≡ 0 and so w∞(y, s) = w∞(y). Since wy(0, s) = −wq(0, s), we get w∞(0) = −wq∞(0). Recall w∞(0) > 0. Hence
w∞(y) = V (y) and the theorem follows. 2
Finally, we shall consider the critical case, i.e., the casep = 2q − 1. Suppose that ¯w(y) (as defined in Section 4) is the unique global positive monotone decreasing solution of (4.1). Then the same argument as above leads to the following conclusion. Note thatwy < 0 for y ≥ 0. Hence the limit function satisfies w∞≤ 0.
THEOREM 5.3. Letp = 2q − 1. If 1 < p ≤ 2, then we have (T − t)αu(1 − y√T − t, t) → ¯w(y) ast → T uniformly for y ∈ [0, C] for any C > 0. Here α = 1/(p − 1).
6. Complete blow-up. Suppose that the solutionu of the problem (1.1) – (1.3) blows up at the finite timeT and x = 1 is the only blow-up point. This is the case if max{p, q} > 1 andu0≥ 0 in [0, 1]. Let
f(n)(s) = min{sq, nq} , g(n)(s) = min{sp, np} , s ≥ 0 , n ∈ N , and letu(n)be the solution of the problem (B(n)):
u(n)t = u(n)xx + g(n)(u(n)) , x ∈ (0, 1) , t > 0 , (6.1) u(n)x (0, t) = 0, u(n)x (1, t) = f(n)(u(n)(1, t)) , t > 0 , (6.2) u(n)(x, 0) = u0(x) , x ∈ [0, 1] . (6.3)
We shall follow the method used in [5] to prove that the blow-up is complete, i.e., asn → ∞, u(n)(x, t) → ∞ for all (x, t) ∈ [0, 1] × (T , ∞).
Let K = max0≤x≤1u0(x). Since f(n) and g(n) are locally Lipschitz in (0, K] and u0(1) = f(n)(u0(1)) for n > K, the solution u(n)of (B(n)) isC1up to the boundary.
Suppose thatv(n)is a positive smooth supersolution andw(n)is a smooth subsolution of (B(n)). Then it is easy to show by the maximum principle thatv(n) ≥ w(n)for 0 ≤ x ≤ 1, t > 0, if n > K. Note that the function K + (np+ nq)t + nqx2/2 is a supersolution of (B(n)). Therefore, for any positive integern > K, the problem (B(n)) has a unique positive global (in time) solutionu(n)such thatu(n) ≤ u(n+1)for(x, t) ∈ [0, 1] × [0, ∞) and u(n)≤ u for (x, t) ∈ [0, 1] × [0, T ).
Now, we define
v(x, t) = lim n→∞u
(n)(x, t) , 0≤ x ≤ 1 , t > 0 .
Then we can show thatv(x, t) = u(x, t) for 0 ≤ t < T . Note that v(1, T ) = ∞. Further-more, we have
LEMMA 6.1. Ifq ≥ 1, then v(1, t) = ∞ for all t ≥ T .
PROOF. For anyM > 1, there is a smooth function U such that U (1) = M, U(1) = Mq,U (ξ ) = 0, and U+ Up= 0 in (ξ, 1] for some ξ ∈ (0, 1), since q ≥ 1. We extend the functionU to be linear on [0, ξ ] so that U ∈ C2([0, 1]). Let M > max{1, u0(1), u0
1/q
∞ }.
Thenu0intersectsU exactly once.
Sincev(1, T ) = ∞, there is a positive integer k > K such that u(k)(1, t0) > M for some t0 ∈ (0, T ). Then there is t1 ∈ (0, t0) such that u(k)(1, t1) = M and u(k)(1, t) < M for all t ∈ [0, t1). Since u(k)(0, t) > U (0) and u(k)(1, t) < U (1) for all t ∈ [0, t1), it implies that u(k)(·, t) intersects U at least once. Note that
(u(k)− U)t = (u(k)− U)xx+ c(x, t)(u(k)− U) , c(x, t) =
g(k)(u(k)) + Uxx u(k)− U .
Sinceu(k)(x, t) is bounded away from zero in [0, 1] × [0, t1), the function c(x, t) is bounded. Applying Theorem D of [1],u(k)(·, t) intersects U exactly once for t < t1. Lets(k)(t) be the function such that
(u(k)− U)(s(k)(t), t) = 0 for all t ∈ [0, t1) . Applying Theorem D of [1] again, we have
(u(k)− U)x(s(k)(t), t) = 0 for all t ∈ [0, t1) .
Therefore, by the Implicit Function Theorem, the functions(k)(t) is continuous in [0, t1). Now, we will show that
u(k)(x, t1) ≥ U (x) for all x ∈ [0, 1] . (6.4)
To prove (6.4), we consider two cases. First, we suppose that limt→t−
1 s
(k)(t) exists. we claim
that limt→t−
1 s
(k)(t) = 1. For contradiction, we assume that 0 ≤ lim t→t1−s
(k)(t) < 1. Recall
that(u(k)− U)(1, t1) = 0 and (u(k)− U)(x, t) ≤ 0 for all x ∈ [s(k)(t), 1] and t ∈ [0, t1]. By
the Hopf Boundary Point Lemma,(u(k)− U)x(1, t1) > 0, a contradiction. Since
(u(k)− U)(x, t) ≥ 0 , x ∈ [0, s(k)(t)] , t ∈ [0, t1) , by lettingt → t1, the inequality (6.4) follows.
Next, we suppose that limt→t−
1 s
(k)(t) does not exist. We assume that
a ≡ lim inf t→t1−
s(k)(t) < lim sup t→t1−
s(k)(t) ≡ b.
Thenξ < a < b ≤ 1. It is easy to see that (u(k)− U)(x, t1) = 0 for all x ∈ [a, b] and (u(k)− U)(x, t1) > 0 for all x ∈ [0, a). If b = 1, then (6.4) follows immediately. Suppose
thatb < 1. For contradiction, we assume that (u(k)− U)(x0, t1) < 0 for some x0∈ (b, 1). Since lim supt→t−
1 s
(k)(t) = b < x
0, there existst2 ∈ [0, t1) such that s(k)(t) < x0for all
t ∈ (t2, t1). Hence
(u(k)− U)(x, t) ≤ 0 , for all (x, t) ∈ {[x0, 1] × [t2, t1]} ∪ {[s(k)(t), 1] × [0, t2]} . It follows from the Hopf Boundary Point Lemma that(u(k)− U)x(1, t1) > 0, a contradiction. Hence (6.4) follows.
Since u(k)(ξ, t) > U (ξ ) for all t ≥ t1, it follows from the maximum principle that
u(k)(x, t) ≥ U (x) for all (x, t) ∈ [ξ, 1] × [t1, ∞). In particular, u(n)(1, t) ≥ M for all t ≥ T andn ≥ k. Hence v(1, t) = ∞ for all t ≥ T and the lemma follows. 2
Now, by the representation formula of solution of (B(n)), u(n)(x, t) = 1 0 u(n)(y, t1)Γ (x, t; y, t1)dy + t t1 f(n)(u(n)(1, τ ))Γ (x, t; 1, τ )dτ − t t1 u(n)(1, τ )Γy(x, t; 1, τ )dτ + t t1 u(n)(0, τ )Γy(x, t; 0, τ )dτ + t t1 1 0 g(n)(u(n)(y, τ ))Γ (x, t; y, τ )dydτ
forx ∈ (0, 1) and t > t1≥ 0, where
Γ (x, t; y, τ ) = √ 1 4π(t − τ )exp −(x − y)2 4(t − τ ) ,
and the jump relation ofu(n)(0, t)
1 2u (n)(0, t) = 1 0 u(n)(y, t1)Γ (0, t; y, t1)dy + t t1 f(n)(u(n)(1, τ ))Γ (0, t; 1, τ )dτ − t t1 u(n)(1, τ )Γy(0, t; 1, τ )dτ + t t1 1 0 g(n)(u(n)(y, τ ))Γ (0, t; y, τ )dydτ
fort > t1 ≥ 0, we conclude that v(x, t) ≡ ∞ for all t > T . This proves that the blow-up is complete whenq ≥ 1.
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SHENG-CHENFU: JONG-SHENQGUO ANDJE-CHIANGTSAI:
DEPARTMENT OFMATHEMATICALSCIENCES DEPARTMENT OFMATHEMATICS NATIONALCHENGCHIUNIVERSITY NATIONALTAIWANNORMALUNIVERSITY 64 SEC. 2, ZHI-NANROAD 88, SEC. 4, TINGCHOUROAD
TAIPEI116 TAIPEI117
