Math 6342 Homework 5 hints
2. Let w := u − v. Note that w satisfies wt + Lw ≤ g(x, t)w in ΩT , where
g(x, t) =
f (u(x, t)) − f (v(x, t))
u(x, t) − v(x, t)
if u(x) 6= v(x).
(g can be anything in {u = v} since gw = 0 in this set anyway.) By assumptions on u, v, f ,
we have |g(x, t)| < M in ΩT ∩ {u 6= v} for some finite M . (why?) Try to show that
h(x, t) = e−2M t w cannot achieve any positive local max in ΩT \ ΓT .
3. First check that I[u(t)] is non-increasing in time (why?), so I[u0 ] < 0 implies I[u(t)] < 0.
R
p+1
d
Then let J[u(t)] := ku(t)k2L2 , and check that dt
J[u(t)] = −c1 I[u(t)] + c2 up+1 dx ≥ c3 J 2
for some c1 , c2 , c3 > 0, which should lead to the desired blow-up result for J[u(t)].
R
4. First note that the mass m := ρ(x, t)dx is conserved during the evolution. Then do a
d
1
computation to show that dt
M2 [u(t)] = 4m − 2π
m2 . If m > 8π, what would go wrong in
finite time?
d
M2 [u(t)], it isR a bit tricky to deal with the term with convolution.
(Note: When computing dt
First apply divergence theorem to get − u(x)∇(N ∗ u)(x) · (2x)dx, then write it as a doubleR
integral − π1 u(x)u(y) (x−y)·x
dxdy. Now, if you exchange dx with dy, the integral shouldn’t
|x−y|2
change since these are dummy variables. Taking the average of the original integral with the
exchanged integral, what do you get?)
1