Hand in y our solutions un til W ednesda y , Decem b er 13, 2.15 pm (PO b o x of y our T A in V3-128)
total points: 20
Prof. Dr. Moritz Kaßmann Fakultät für Mathematik
Wintersemester 17/18 Universität Bielefeld
Partial Differential Equations II Solutions to exercise sheet IX, December 6
Exercise IX.1 (4 points)
Assumeq > d2,R >0andf ∈Lq(BR). As discussed in the lecture (and below) it can be shown that functionsu∈H1(BR)satisfying−∆u≤f inBRin the weak sense, satisfy
sup
BR/2
u≤cRθ1 Z
BR
u212
+cRθ2kfkLq(BR) (1)
for some constantsc, θ1, θ2 independent ofuandR. Use scaling arguments and computeθ1, θ2.
For the remaining questions assume that Ω ⊂ Rd is open, q > d2, and f ∈ Lq(Ω). Let u 7→ Lu = div(a(·)∇u) denote a strictly elliptic differential operator. The idea of the two exercises is to use techniques that are very similar to the proof in the lecture. Apply the standard definition of weak subsolutions.
Exercise IX.2 (8 points)
Show that there is a constantc≥1such that, if0< ρ < σ,Bρ(x0)bBσ(x0)⊂Ω, for every function u∈H1(Ω)satisfying−Lu≤f inΩand everyk∈R, the following Caccioppoli inequality holds true:
Z
Aρ(x0;k)
|∇u|2 ≤c(σ−ρ)−2 Z
Aσ(x0;k)
(u−k)2+ckfk2Lq(Ω)|(Aσ(x0;k))|1−1/q. (2)
Exercise IX.3 (8 points)
Assume thatu∈H1(Ω)is a function that satisfies (2) for everyBρ(x0)bBσ(x0)⊂Ωand everyk∈R.
Show that then (1) holds.
Hint: You may use without proof the auxiliary iteration lemma given in the lecture.
Solutions Exercise IX.2
Considerv= (u−k)+and a cut-off functionτ ∈CC∞(Bσ(x0))such that0≤τ ≤1, τ ≡1in Bρ(x0)and|∇τ| ≤2(σ−ρ)−1. Setϕ=vτ2as the test function. Notev=u−k,∇v=∇u almost everywhere in{u > k}andv= 0,∇v= 0almost everywhere in{u≤k}. Therefore, if we substitute such ϕin
Z
Ω
aijDiuDjϕ≤ Z
Ω
f ϕ
we integrate over {u > k} ∩Bσ(x0) =Aσ(x0;k). Using the strict ellipticity ofL and the
Young inequality we obtain Z
aijDiuDjϕ= Z
aijDiuDjvτ2+ 2aijDiuDjτ vτ
≥λ Z
|∇v|2τ2−2Λ Z
|∇v||∇τ|vτ
=λ Z
|∇v|2τ2− Z h
|∇v|τ√ λ
i
√2Λ λ|∇τ|v
≥ λ 2
Z
|∇v|2τ2−2Λ2 λ
Z
|∇τ|2v2. Hence we have
Z
|∇v|2τ2 ≤C Z
v2|∇τ|2+ Z
|f|vτ2
from which the estimate Z
|∇(vτ)|2 ≤C Z
v2|∇τ|2+ Z
|f|vτ2
(3) follows by the product rule.
Our aim is now to estimate the integral R
f vτ2. We claim the following:
Z
|f|vτ2 ≤δk∇(vτ)k2L2+Ckfk2Lq|Aσ(x0;k)|1−1q (4) for some smallδ >0.
We distinguish two cases.
(i) Let |Aσ(x0;k)|<1. Recall the Sobolev inequality forvτ ∈H0(Ω):
Z
Ω
(vτ)2∗
2 2∗
≤C Z
Ω
|∇(vτ)|2,
where2∗= d−22d ifd >2and2∗>2is arbitrary ford= 2. Now the Hölder inequality and the Young inequality yield
Z
|f|vτ2 ≤ kfkLqkvτkL2∗|Aσ(x0;k)|1−21∗−1q
≤CkfkLqk∇(vτ)|L2|Aσ(x0;k)|12+1d−1q
≤δk∇(vτ)k2L2 +Ckfk2Lq|Aσ(x0;k)|1+2d−2q,
where δ >0 is small. Note that the assumption q > d2 implies1 + 2d−2q >1−1q.
Thus Z
|f|vτ2 ≤δk∇(vτ)k2L2+Ckfk2Lq|Aσ(x0;k)|1−1q.
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(ii) LetAσ(x0;k)≥1. Then with Young, Hölder and Poincaré Z
|f|vτ2≤ Z
|f|vτ
≤C Z
f2+δ Z
(vτ)2
≤C Z
f2+Cδ Z
|∇(vτ)|2
≤C Z
fq 2q
|Aσ(x0;k)|1−2q +δC Z
|∇(vτ)|2,
whereδ >0 is small. Now we use1−2q <1−1q and obtain Z
|f|vτ2 ≤δk∇(vτ)k2L2 +Ckfk2Lq|Aσ(x0;k)|1−1q
and our claim is proved.
Combining (3) and (4) we have Z
|∇(vτ)|2≤C Z
v2|∇τ|2+kfk2Lq|Aσ(x0;k)|1−1q
(5) Using the properties of our chosen cut-off function τ this yields the inequality (2).
Exercise IX.3
We show the assertion for R= 1. Otherwise the constant in (2) would depend onR and we would not be able to recover the constants θ1 and θ2 from exercise IX.1. Let v and τ be as in exercise IX.2 but with 12 ≤ ρ ≤ σ ≤ 1. Using the Hölder and the Sobolev inequality we obtain
Z
(vτ)2≤ Z
(vτ)2∗ 2
2∗
|Aσ(x0;k)|1−22∗
≤C Z
|∇(vτ)|2|Aσ(x0;k)|2d.
Together with (5) this leads to Z
(vτ)2 ≤C Z
v2|∇τ|2|Aσ(x0;k)|d2 +kfk2Lq|Aσ(x0;k)|1+2d−1q
. (6)
Setε= 2d−1q. Then (6) yields Z
(vτ)2≤C Z
v2|∇τ|2|Aσ(x0;k)|ε+kfk2Lq|Aσ(x0;k)|1+ε
(7)
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if|Aσ(x0;k)|<1. This can be achieved if we choosek≥k0for a constantk0 >ku+kL2(B1), because then
|Aσ(x0;k)|< 1 k
Z
Aσ(x0;k)
u+≤ 1
kku+kL2(B1)p
|Aσ(x0;k)|,
i.e.,
|Aσ(x0;k)|< 1
k2ku+k2L2(B1)<1.
Now the properties ofτ imply the following assertion: There exists a constantC≥1so that∀k≥k0 and all0< ρ < σ≤1:
Z
Aρ(x0;k)
(u−k)2 ≤C|Aσ(x0;k)|ε
(σ−ρ)−2 Z
Aσ(x0;k)
(u−k)2+kfk2Lq|Aσ(x0;k)|
.
Using the properties of the setAσ(x0;k)that are known from the lecture the last inequality implies for any h > k≥k0 and 12 ≤ρ < σ≤1:
Z
Aρ(x0;k)
(u−h)2≤C 1
(σ−ρ)2 + kfk2Lq
(h−k)2
1 (h−k)2ε
Z
Aσ(x0;h)
(u−k)2
1+ε
or
k(u−h)+kL2(Bρ)≤C 1
σ−ρ +kfkLq h−k
1
(h−k)εk(u−k)+k1+εL2(B
σ).
Now we can use the iteration lemma from the lecture withϕ(k, ρ) =k(u−k)+kL2(Bρ), α1 =ε, α2 = 1, β= 1+ε, A1=C, A2=CkfkLq(B1). Note that our choise ofk0guarantees us that the assumption regarding the monotonicity ofϕare satisfied. The lemma gives us
sup
B1 2
u≤d+k0
≤Cϕ(k0,1) +CkfkLq(B1)+kukL2(B1)
≤C kukL2(B1)+kfkLq(B1)
,
since (obviously)ϕ(k0,1)≤ kukL2(B1).
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