Blow-up of Solutions of Semilinear Parabolic Equations

where ∫ 1
√
dy
0
then it follows that
is a constant dependent on p. Define C
1−y p+1 p := ∫ 1
0
ϕ(M) = C p M − p−1
2 .
√
Since ϕ is monotone decreasing and the range of ϕ is (0, ∞), ϕ(M) =
has a unique positive solution for every L > 0.
Theorem 6. Consider the problem
⎧
⎨ u t = u xx + u p , |x| < 1, p > 1,
(P) u = 0, |x| = 1,
⎩
u(x, 0) = u 0 (x), |x| < 1.
dy
√1−y p+1 ,
2
L p+1
q.e.d.
If u 0 (x) ≥ v(x), u 0 ≢ v(x), v is the solution of (S), then u blows up in finite
time.
Proof. By the maximum principle, if τ > 0, then u(x, τ) > v(x) for |x| < 1.
Moreover u x (1, τ) < v x (1) and u x (−1, τ) > v x (−1). Hence there is k > 1
such that u(x, τ) > kv(x).
Thus, it is suffcient to consider the case where u 0 (x) = kv(x), k > 1. In
this case, we have
(u t ) t = (u t ) xx + pu p−2 u t , |x| < 1,
u t = 0, |x| = 1,
u t (x, 0) = kv xx + k p v p > k(v xx + v p ) = 0.
By the maximum principle, we obtain u t ≥ 0 as long as u exists.
Suppose u is global. If u is bounded, there exists a solution w(x) of (S)
such that
u(x, t) → w(x) as t → ∞, w(x) > v(x), |x| < 1,
this is a contradiction with Theorem 5.
If u is unbounded, then
u(x, t) ≡ u(−x, t), u x (x, t) < 0 for x ∈ [0, 1),
and either
or
(i) lim
t→∞
u(x, t) < ∞, x ∈ (a, 1], a ∈ (0, 1],
(ii) lim
t→∞
u(x, t) = ∞, |x| < 1,
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