Blow-up of Solutions of Semilinear Parabolic Equations

Remark 2. The assumption ∫ ∞ du
< ∞ is not sufficient for blow-up, if f
1 f(u)
is not convex. (See Fila, Ninomiya and Vázquez [8]).
Remark 3. The proof does not necessarily imply that y(t) → ∞ as t → T .
When f(u) = u p , there appears a critical exponent for the blowup and
global existence of positive solutions.
Theorem 3 (Fujita, 1966 [12]). Consider the equation
u t = ∆u + u p , x ∈ R N , p > 1.
(i) If p < 1 + 2 , all positive solutions blow up.
N
(ii) If p > 1 + 2 , there exist both blowing up and global positive solutions.
N
Hayakawa, 1973 [21], Kobayashi, Sirao, Tanaka, 1977 [25]: The critical
case p = 1 + 2 is as (i).
N
Why is p = 1 + 2 critical? The following is an intuitive explanation.
N
The fundamental solution of u t = ∆u decays like t − N 2 , while the solution of
u t = u p is given by C(T − t) − 1
p−1 . They are balanced if
N
2 = 1
p − 1 ⇔ p = 1 + 2 N .
Better explanation is as follows. Define v(y, s) = e s
p−1 u(e
s
2 y, e s − 1). Then v
solves
v s = ∆v + y 2 · ∇v + 1
p − 1 v + vp , y ∈ R N , s > 0.
Here the principal eigenvalue of Lv := ∆v + y · ∇v in 2 L2 ρ with ρ(y) = e |y|2
4
is given by λ 1 = − N . This implies that the trivial solution v ≡ 0 is stable
2
if N > 1 , and is unstable if N < 1 . As a consequence of instability, the
2 p−1 2 p−1
solution blows up.
Next, consider
u t = ∆u + f(u), x ∈ Ω,
u = 0,
x ∈ ∂Ω,
u(x, 0) = u 0 (x),
where Ω is a bounded domain. For v ∈ C 1 (Ω), define
E(v) = 1 ∫
∫
|∇v| 2 dx − F (v)dx, F (v) =
2
Ω
Ω
4
∫ v
0
f(u)du.