Math 820.
Problems set 1.
Solutions
Problem 1. Let Ω ⊂ Rn be an open bounded set with C 1 boundary, and u ∈ H 1 (Ω) be a
weak solution of the equation
n
X
(aij (x)uxi )xj =
i,j=1
n
X
fxi i + g,
i=1
where f, g ∈ L2 (Ω), f = (f 1 , . . . , f n ). Assume aij = aji and uniform ellipticity, i.e., for
some Λ > λ > 0
n
X
Λ|ξ|2 ≥
aij (x)ξi ξj ≥ λ|ξ|2 for all x ∈ Ω, ξ ∈ Rn .
i,j=1
(i) Show that for any subdomain Ω0 ⊂⊂ Ω the estimate holds
kukH 1 (Ω0 ) ≤ C(kukL2 (Ω) + kf kL2 (Ω) + kgkL2 (Ω) ),
where C = C(n, λ, Λ, d0 ), d0 = dist(Ω0 , ∂Ω).
(ii) Assume in addition that n > 2, and that u satisfies the boundary condition
u − h ∈ H01 (Ω), where h ∈ H 1 (Ω). Show that
kuk
2n
L n−2 (Ω)
≤ C(khkH 1 (Ω) + kf kL2 (Ω) + kgk
2n
L n+2 (Ω)
),
Solution. We first note
Pn that kaij kL∞ ≤ Λ. Indeed, using ξ = ek , i.e. k-th basis vector
for k = 1, . . . , n, we get i,j=1 aij (x)ξi ξj = akk , thus λ ≤ akk ≤ Λ. Next, using ξ = ek + em
and taking into account the symmetry aij = aji , we get
λ ≤ akk + amm + 2akm ≤ Λ,
λ
2
Λ
2
thus −Λ + ≤ akm ≤ − λ. Thus |akm | ≤ Λ.
Now we solve Problem 1. In the argument, the constant C may change from expression
to expression, and depends only on n, λ, Λ, d0 . Also, summation over the repeated indices
is always assumed.
(i) There exists η ∈ C 1 (Rn ) satisfying η ≡ 1 on Ω0 , η ≡ 0 outside Ω, 0 ≤ η ≤ 1 in Rn ,
and kDηkL∞ (Rn ) ≤ d20 . Then ϕ := η 2 u ∈ H01 (Ω), and we use it in the weak form of the
equation:
Z
Z
2
aij uxi uxj η + 2aij (x)uxi uηηxj dx =
f i uxi η 2 + 2f i uηηxi − guη 2 dx
Ω
Ω
η2
λ|Du|2 η 2 .
We use ellipticity to get aij uxi uxj ≥
We estimate other terms as: |aij (x)uxi uηηxj | ≤
1 2
1
λ
Λ(ε|Du|2 η 2 + 4ε
u |Dη|2 ), |f i uxi η 2 | ≤ ε|Du|2 η 2 + 4ε
|f |2 η 2 , etc., and choose ε = 4(Λ+1)
to
get
Z
Z
Z
2 2
2
2
2
2
2 2
|Du| η dx ≤ C
(u + f )(η + |Dη| ) + g η dx ≤ C (u2 + f 2 + g 2 )dx.
Ω
Since η ≡ 1 on
Ω
Ω0 ,
Ω
we get
kDuk2L2 (Ω0 ) ≤ C(kuk2L2 (Ω) + kf k2L2 (Ω) + kgk2L2 (Ω) ).
Adding to this estimate the obvious inequality kuk2L2 (Ω0 ) ≤ kuk2L2 (Ω) , we obtain assertion of
part (i) of the problem.
1
(ii) Since u − h ∈ H01 (Ω), we can use it as the test function in the weak form of the
equation, to get:
Z
Z
aij (u − h)xi (u − h)xj =
− aij hxi (u − h)xj + f i (u − h)xi − g(u − h) dx
Ω
Ω
Then, using also ellipticity in the left-hand side, and and using the coefficients bound in
the calculation similar as in part (i) above to estimate:
Z
Z
Z
C
2
i
(f 2 + |Dh|2 )dx,
− aij hxi (u − h)xj + f (u − h)xi ≤ ε |D(u − h)| dx +
ε
Ω
Ω
Ω
and choosing ε small depending on λ, we get
Z
Z
Z
2
2
2
g(u − h) dx ,
|D(u − h)| dx ≤ C(
(f + |Dh| )dx +
(1)
Ω
Ω
Ω
with C = C(n, λ, Λ). To estimate the last term in the right-hand side, we denote
p=
Then
1
p
+
1
q
2n
,
n−2
q=
2n
.
n+2
= 1, and using Holder inequality we get
Z
Z
1 Z
1
Z
2
Z
2
p
q
p
q
1
p
q
p
q
g(u − h) dx ≤
|u
−
h|
dx
|g|
dx
≤
ε
|u
−
h|
dx
+
|g|
dx
.
4ε
Ω
Ω
Noting that p =
2n
n−2
Ω
Ω
Ω
= 2∗ , and u − h ∈ H01 (Ω), we get by Sobolev inequalities
ku − hk2L2∗ (Ω) ≤ CkD(u − h)k2L2 (Ω) ,
(2)
where C depends on Ω. Using this and the previous estimate in (1), we obtain
ku − hk2L2∗ (Ω) ≤ Cεku − hk2L2∗ (Ω) + Ckf k2L2 (Ω) + Ckhk2H 1 (Ω) +
C
kgk2 2n
.
4ε
L n+2 (Ω)
Choosing ε small so that Cε = 21 , thus ε depends on n, λ, Λ, Ω, we get
ku − hk2 2n
≤ C kuk2L2 (Ω) + kf k2L2 (Ω) + khk2H 1 (Ω) + kgk2
L n−2 (Ω)
2n
L n+2 (Ω)
,
which implies
kuk
2
2n
L n−2 (Ω)
≤ C kuk2L2 (Ω) + kf k2L2 (Ω) + khk2H 1 (Ω) + khk2
2n
L n−2 (Ω)
+ kgk
2
.
2n
L n+2 (Ω)
Finally, we apply Sobolev inequality (2) (with h instead of u − h) to the term khk2
2n
L n−2 (Ω)
in the right-hand side, and obtain the estimate of part (ii) of Problem.
Problem 2. Let Ω ⊂ Rn be an open bounded set. For a function u ∈ H 1 (Ω), we say u ≤ 0
on ∂Ω if its positive part u+ = max(u, 0) satisfies u+ ∈ H01 (Ω). For u, v ∈ H 1 (Ω) we say
u ≤ v on ∂Ω if u − v ≤ 0 on ∂Ω. For u ∈ H 1 (Ω) define
sup u := inf{M | M ∈ R, u ≤ M on ∂Ω}.
∂Ω
Assume that aij (x), Λ, λ are as in Problem 1. Assume that c ∈ L∞ (Ω) and c ≤ 0 in Ω.
(i) Let u ∈ H 1 (Ω) be a weak subsolution of
n
X
(aij (x)uxi )xj + c(x)u ≥ 0
in Ω.
i,j=1
Prove
sup u ≤ sup u+ ,
Ω
∂Ω
assuming that the right-hand side is finite.
Hint: in the weak form of the equation, use test function ϕ = max(u − k, 0), where
k = sup∂Ω u+ .
(ii) Let u1 , u2 ∈ H 1 (Ω) be weak solutions of
n
X
(aij (x)uxi )xj + c(x)u =
i,j=1
n
X
fxi i + g,
i=1
where f, g ∈ L2 (Ω), f = (f 1 , . . . , f n ). Let u1 ≥ u2 on ∂Ω. Prove u1 ≥ u2 in Ω.
(iii) Let h ∈ H 1 (Ω). Prove that there exists at most one weak solution u ∈ H 1 (Ω) of the
equation from part (ii), satisfying u − h ∈ H01 (Ω).
Solution. (i) For k ∈ R, denote
Ωk := Ω ∩ {u > k}.
Next, denoting k = sup∂Ω u+ , we obtain that ϕ := max(u − k, 0) ∈ H01 (Ω). Thus we can
use ϕ as a test function in the weak form of the equation. Then, noting that Dϕ = Du on
Ωk and Dϕ = 0 on Ω \ Ωk , we obtain:
Z X
Z
n
aij uxi uxj dx −
cuϕ dx ≤ 0.
Ωk i,j=1
Ω
Note that k ≥ 0 and ϕ ≥ 0 from their definition, and that uϕ = 0 on {u ≤ k}. Thus uϕ ≥ 0
in Ω. Then, using ellipticity and c ≤ 0 in the inequality above, we get
Z
Z
cuϕ dx ≤ 0.
|Du|2 dx ≤
λ
Ω
Ωk
Thus Du = 0 almost everywhere on Ω ∩ {u > k}. Combined with the fact that u ≤ k on
∂Ω this implies u ≤ k in Ω. Thus part (i) is proved.
(ii) Let v = u2 − u1 . Then v satisfies (weakly) the homogeneous equation
n
X
(aij (x)vxi )xj + c(x)v = 0
in Ω,
i,j=1
and v ≤ 0 on ∂Ω, i.e. v + = 0 on ∂Ω. Then by part (i),
sup v ≤ sup v + = 0.
Ω
∂Ω
Thus v ≤ 0 in Ω, i.e. u2 ≤ u1 in Ω.
(iii) Suppose u1 and u2 are two solutions of the boundary-value problem in part (iii).
Then u1 − u2 ∈ H01 (Ω). Thus we have both u1 ≤ u2 and u2 ≤ u1 on ∂Ω. Then part (ii)
implies u1 ≤ u2 and u1 ≥ u2 in Ω, thus u1 = u2 in Ω.
Problem 3. Let Ω ⊂ Rn be an open bounded set, and u ∈ H 1 (Ω) be a weak solution of
the equation
n
n
X
X
i
(A (x, Du))xi =
fxi i + g,
i=1
(f 1 , . . . , f n ),
L2 (Ω),
i=1
where f, g ∈
f =
and A = (A1 , ..., An ) is smooth. Assume that
A(x, 0) = 0 for all x ∈ Ω. Assume uniform ellipticity, i.e., for some Λ > λ > 0
n
X
2
Λ|ξ| ≥
Aipj (x, p)ξi ξj ≥ λ|ξ|2
for all x ∈ Ω, p, ξ ∈ Rn .
i,j=1
Assume |Dp A(x, p)| ≤ Λ for all x ∈ Ω, p ∈ Rn .
(i) Show that for any subdomain Ω0 ⊂⊂ Ω the estimate holds
kukH 1 (Ω0 ) ≤ C(kukL2 (Ω) + kf kL2 (Ω) + kgkL2 (Ω) ),
with C = C(n, Λ, λ, d0 ), where d0 = dist(Ω0 , ∂Ω).
(ii) In addition to previous conditions, assume that f ≡ 0,
|Aixj (x, p)| ≤ Λ|p|
for all x ∈ Ω, p ∈ Rn , i, j = 1, . . . , n,
and u ∈ H 2 (Ω). Show that for any subdomain Ω0 ⊂⊂ Ω the estimate holds
kukH 2 (Ω0 ) ≤ C(kukL2 (Ω) + kgkL2 (Ω) ),
with C =
C(n, Λ, λ, d0 ),
where d0 = dist(Ω0 , ∂Ω).
Solution. (i) Weak form of the equation is
Z X
Z
n
i
A (x, Du)ϕxi dx =
Ω i=1
f i ϕxi − gϕ dx
Ω
for any ϕ ∈ C01 (Ω). Since A(x, 0) = 0, we have
i
A (x, Du) =
n
X
Z
aij (x)uxj ,
1
Dpj Ai (x, tDu(x)) dt.
where aij (x) =
0
j=1
Thus u is a weak solution of the linear equation
n
n
X
X
(aij (x)uxi )xj =
fxi i + g,
i,j=1
i=1
with aij (x) defined above. Also, ellipticity of the nonlinear equation and expression of aij
imply that coefficients aij are elliptic:
Z 1 X
Z 1
n
n
X
i
aij (x)ξi ξj =
Dpj A (x, tDu(x))ξi ξj dt ≥
λ|ξ|2 dt = λ|ξ|2 ,
i,j=1
0
0
i,j=1
and the upper bound is shown similarly. Also,
Z 1
Z
|aij (x)| ≤
|Dpj Ai (x, tDu(x))| dt ≤
0
1
Λ dt = Λ.
0
Then we use the estimate of Problem 1(i) to conclude.
d0
2 -neighborhood
d0
0
00
(ii) Let subdomain Ω00 be the
(3)
dist(Ω , ∂Ω ) =
2
,
of Ω0 . Then Ω0 ⊂⊂ Ω00 ⊂⊂ Ω, and
dist(Ω00 , ∂Ω) ≥
d0
.
2
Fix k ∈ {1, . . . , n} and ϕ ∈ Cc2 (Ω). In the weak for of the equation use the test function
∂k ϕ, which is possible since ∂k ϕ ∈ C01 (Ω). Then we get:
Z X
Z
n
i
− gϕxk dx.
A (x, Du)ϕxi xk dx =
Ω i=1
Ω
Since u ∈ H 2 (Ω) and A is smooth with A(x, 0) = 0 and |Dp A(x, p)| ≤ Λ, |Dx A(x, p)| ≤ Λ|p|
for all x, p), then we get A(x, Du) ∈ H 1 (Ω). Thus we can integrate by parts with respect
to ∂xk on the left, to get:
Z X
Z X
Z
n
n
gϕxk dx.
Aipj (x, Du)uxk xj ϕxi dx +
Aixk (x, Du)ϕxi dx =
Ω i,j=1
Ω i=1
Ω
Denote
v = uxk and ãij (x) = Aipj (x, Du(x)), f˜i (x) = −Aixk (x, Du(x)) + g(x)δki ,
0
if i 6= k
where δki =
. Then v ∈ H 1 (Ω), fˆ = (fˆ1 , . . . , fˆn ∈ L2 (Ω), and rewriting the
1
if i = k
last equation using these notations, we find that v satisfies
Z X
Z X
n
n
ãij vxj ϕxi dx =
f˜i ϕxi dx in Ω,
Ω i,j=1
Ω i=1
for any ϕ ∈ Cc2 (Ω). By approximation in Sobolev spaces, the last equation holds for any
ϕ ∈ H01 (Ω). Thus v is a weak solution of
n
n
X
X
(ãij (x)vxj )xi =
f˜xi i in Ω.
i,j=1
i=1
Also, from the explicit expressions of ãij (x) and the properties of the nonlinear equation,
we find that coefficients ãij (x) satisfy ellipticity with constants λ, Λ, and kãij kL∞ (Ω) ≤ Λ.
Then, by the resulting estimate of Problem 1(i) applied in the domains Ω0 ⊂⊂ Ω00 and using
(3), we get
(4)
kvkH 1 (Ω0 ) ≤ C(kvkL2 (Ω00 ) + kf˜kL2 (Ω00 ) ),
with C = C(n, λ, Λ, d0 ). We note that kvkL2 (Ω00 ) = kuxk kL2 (Ω00 ) ≤ kDukL2 (Ω00 ) . Also,
kf˜kL2 (Ω00 ) ≤ ΛkDukL2 (Ω00 ) + kgkL2 (Ω00 ) .
Thus (4) implies:
kuxk kH 1 (Ω0 ) ≤ C(kDukL2 (Ω00 ) + kgkL2 (Ω00 ) ).
Since this is true for all k ∈ {1, . . . , n}, we get
kDukH 1 (Ω0 ) ≤ C(kDukL2 (Ω00 ) + kgkL2 (Ω00 ) ).
Adding to this the inequality kukL2 (Ω0 ) ≤ kukL2 (Ω00 ) , we get
kukH 2 (Ω0 ) ≤ C(kukH 1 (Ω00 ) + kgkL2 (Ω) ).
Finally we estimate the term kukH 1 (Ω00 ) in the right-hand side, using the result of part (i)
of this problem in the domains Ω00 ⊂⊂ Ω with use of (3). This concludes the proof.