Solutions.
1. Let Ω be the half plane {x2 > 0} in R2 . Let u ∈ C 2 (Ω) ∩ C(Ω) be harmonic:
∆u = 0
in Ω.
Under the additional assumption that u(·) is bounded from above in Ω, prove that
sup u = sup u.
Ω
∂Ω
Note: The additional assumption is needed to exclude examples like u(x1 , x2 ) = x2 .
Hint: Take for ε > 0 the harmonic in Ω function
q
v(x1 , x2 ) = u(x1 , x2 ) − ε log x21 + (x2 + 1)2 .
Apply the maximum principle in an appropriate bounded region. Let ε → 0.
Solution
p
Denote R2+ := {x2 > 0}, Br+ := B(0, r) ∩ R2+ . Denote w(x, t) = log x21 + (x2 + 1)2 ,
2
where x = (x1 , x2 ). Then w ∈ C ∞ (R+ ), and, by explicit calculation, ∆w = 0.
Moreover, w > 0 in R2+ , and w(x) → ∞ as |x| → ∞ in R2+ . Also w(0) = 0.
Note that v = v = u − w. Thus we get ∆v = 0 in R2+ . Then, for any r > 0, we
apply the maximum principle in Br+ and get
max v = max
v .
+
Br+
∂Br
Note that ∂Br+ = Γ0r ∪ Γ1r where Γ0r = B(0, r) ∩ {x2 = 0}, and Γ1r = ∂B(0, r) ∩ R2+ ,
Since u is bounded from above in R2+ , say supR2+ u = M , it follows that for any > 0,
there exists R() such that
v < −M
on Γ1r
for all r > R().
On the other hand, w(0) = 0, thus v (0) = u(0) ≥ −M , and 0 ∈ Γ0r . Thus,
max
v = max
v
+
0
for all r > R(),
max v = max
v
0
for all r > R().
∂Br
Γr
which implies
Br+
Γr
Thus for each > 0, we get
sup v = sup v .
R2+
(1)
∂R2+
Since v(x1 , x2 ) = u(x1 , x2 ) − εw(x1 , x2 ) ≤ u(x1 , x2 ) for each (x1 , x2 ) ∈ R2+ , we get
from (1)
sup v ≤ sup u.
R2+
∂R2+
Thus for each (x1 , x2 ) ∈ R2+
u(x1 , x2 ) − εw(x1 , x2 ) ≤ sup u,
∂R2+
and, sending → 0+. we get
u(x1 , x2 ) ≤ sup u.
∂R2+
Thus
sup u ≤ sup u
R2+
∂R2+
and since the opposite inequality is obviously true (since u is continuous in R2+ ), we
get
sup u = sup u.
R2+
∂R2+
2. Let U ⊂ Rn be open and bounded, let UT = U × (0, T ] be the parabolic cylinder,
and and ΓT = UT \ UT be the parabolic boundary. Let nonnegative u ∈ C12 (UT ) ∩
C(UT ) satisfies
ut = ∆u + cu in UT ,
where c(x, t) is continuous in UT .
(a) Prove that if c(x, t) < 0 in UT , then
max u = max u.
UT
ΓT
Hint. Show that max u cannot be assumed in UT unless max u ≤ 0. In order to do
UT
UT
that, consider the possible signs of ut and ∆u at a point of maximum (x0 , t0 ) ∈ UT .
Note that it is possible that t0 = T .
(b) Show that more generally
max u ≤ eCT max u,
UT
ΓT
where
C = max(0, max c)
UT
γt
Hint. Substitute u(x, t) = e v(x, t), where γ > C.
Solution
(a) We show (as hint suggests) that maxUT u cannot be achieved in UT unless maxUT u ≤
0. This proves assertion (a) since u ≥ 0 in UT , i.e. maxΓT u ≥ 0.
Suppose maxUT u is achieved at (x0 , t0 ) ∈ UT = U × (0, T ]. Then taking into
account that t0 = T is possible, we get ut (x0 , t0 ) ≥ 0. Also, since u(·, t0 ) achieves
its maximum over U at x0 ∈ U , it follows that D2 u(x0 , t0 ) is a nonpositive-definite
matrix, thus ∆u(x0 , t0 ) ≤ 0. Thus we get at (x0 , t0 )
cu = ∂t u − ∆u ≥ 0.
From this and since c < 0 in UT , get u(x0 , t0 ) ≤ 0, as claimed.
(b) Let u(x, t) = eγt v(x, t). Then ut = eγt (vt + γv), so
0 = ut − ∆u − cu = eγt [vt − ∆v − (c − γ)v].
Choose γ > max(0, maxUT c), where we use continuity of c in UT . Then d(x, t) :=
c(x, t) − γ < 0 in UT , and
vt = ∆v + dv
in UT .
Thus, using part (a) and the fact that 0 ≤ v ≤ u ≤ eγT v in UT from definition of v,
get
max u ≤ eγT max v = eγT max v ≤ eγT max u.
UT
ΓT
UT
ΓT
Thus
max u ≤ eγT max u
ΓT
UT
for any γ > max(0, maxUT c). Then the inequality also holds for γ = max(0, maxUT c).
3.(2.5 #17 from Evans book). Let u ∈ C 2 (R × [0, ∞)) solve the initial-value problem
for the wave equation in one dimension:
utt − uxx = 0
in R × (0, ∞),
u = g, ut = h
on R × {t = 0}.
Z
1 ∞ 2
ut (x, t)dx,
Suppose g, h have compact support. The kinetic energy is k(t) =
2
−∞
Z
1 ∞ 2
and the potential energy is p(t) =
u (x, t)dx. Prove:
2 −∞ x
(a) k(t) + p(t) is constant in t;
(b) k(t) = p(t) at all large enough times t.
Solution
We have g(x), h(x) ≡ 0 for all |x| ≥ M . Let
Z x
H(x) =
h(s)ds,
−∞
where the integral is finite for any x since h has compact support. Moreover, H(x) = 0
RM
for all x < −M , and H(x) = const = −M h(s)ds for all x > M .
By uniqueness theorem u(x, t) is defined by d’Alembert’s formula, which can be
written as
1
1
u(x, t) = [g(x + t) + g(x − t)] + [H(x + t) − H(x − t)]for (x, t) ∈ R × [0, ∞).
2
2
From the properties of g, H, get
g(x + t) = g(x − t) = 0 and H(x + t) = H(x − t) if |x| > M + t, where t > 0.
Thus, u(x, t) = 0 if |x| > M + t. That is, for any t > 0 the function u(·, t) has
compact support. Now we compute, integrating by parts with respect to x
Z ∞
d
[k(t) + p(t)] =
ut (x, t)utt (x, t) + ux (x, t)uxt (x, t) dx
dt
−∞
Z ∞
ut (x, t)utt (x, t) − ut (x, t)uxx (x, t) dx = 0
=
−∞
since utt = uxx . This implies assertion (a).
(b) Differentiating d’Alembert’s formula, get
1
1
ux (x, t) = [g 0 (x + t) + g 0 (x − t)] + [h(x + t) − h(x − t)]
2
2
1 0
1
ut (x, t) = [g (x + t) − g 0 (x − t)] + [h(x + t) + h(x − t)].
2
2
Thus,
Z
1 ∞ 2
p(t) − k(t) =
[u (x, t) − u2t (x, t)]dx
2 −∞ x
Z
1 ∞ 0
[g (x + t) + h(x + t)][g 0 (x − t) − h(x − t)]dx.
=
2 −∞
Now, if t > M , we have (x + t) − (x − t) = 2t > 2M , so either |x + t| > M or
|x − t| > M . Thus the last integral is zero for t > M , i.e. p(t) = k(t) for t > M .
4.(2.5 #18 from Evans book). Let u solve
utt − ∆u = 0
in R3 × (0, ∞),
u = g, ut = h on R3 × {t = 0},
where g, h are smooth and have have compact support. Show there exists constant
C such that
|u(x, t)| ≤ C/t
(x ∈ R3 , t > 0).
Solution
First we prove that u(x, t) is given by Kirchhoff’s formula:
Z
u(x, t) = −
th(y) + g(y) + Dg(y) · (y − x) dS(y).
∂B(x,t)
This will follow from the uniqueness of solution of the homogeneous wave equation
in R3 × (0, ∞) when initial data have compact support:
Let u(x, t), v(x, t) be two smooth solutions of the homogeneous wave equation in
R × (0, ∞) with the same initial data f, g with compact support. Suppose R > 0 is
such that supp g(·), h(·) ⊂ B(0, R). Then, by Theorem on finite propagation speed,
supp u(·, t) ⊂ B(0, R + t) for any t > 0, and same for v(·, t). Now let T > 0, and
U = B(0, R + T ), and UT = U × (0, T ], ΓT = UT − UT ≡ (∂U × [0, T )) ∪ (U × {t = 0}).
Since u = v = 0 on ∂U × [0, T ) as we discussed above, we get u(x, t) = v(x, t) on
ΓT . Then uniqueness Theorem in UT implies that u = v in UT . Since u = v = 0 on
(R3 × [0, T ]) \ UT , it follows that u = v in R3 × [0, T ]. Since T > 0 is arbitrary, then
u = v in R3 × (0, ∞), thus solution in R3 × (0, ∞) is unique, thus given by Kirchhoff’s
formula.
For any x ∈ R3 , t > 0, the area of the part of the sphere ∂B(x, t) which is inside
the ball B(0, R) is no greater than the area of the sphere ∂B(0, R). Thus,
3
|∂B(x, t) ∩ B(0, R)| ≤ min(|∂B(0, R)|, |∂B(x, t)|) = 4π min(R2 , t2 ).
Then using Kirchhoff’s formula, and using the fact that |f, g, Dg| ≤ C on R3 for some
C since f, g are smooth with compact support, and also noting that |y − x| = t in the
last term of Kirchhoff’s formula, we get:
Z
C
t|h(y)| + |g(y)| + t|Dg(y)| dS(y)
|u(x, t)| ≤ 2
t ∂B(x,t)∩B(0,R)
Ct + C
Ct + C
|∂B(x, t) ∩ B(0, R)| ≤ 4π min(R2 , t2 )
≤ C1 /t.
≤
2
t
t2