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A Harnack type inequality for an elliptic equation with singular source
Samy Skander Bahoura
To cite this version:
Samy Skander Bahoura. A Harnack type inequality for an elliptic equation with singular source. 2021.
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A Harnack type inequality for an elliptic equation with singular source.
Samy Skander Bahoura
∗Equipe d’Analyse Complexe et Géométrie.
Université Pierre et Marie Curie, 75005 Paris, France.
Abstract
We derive a Harnack type inequality for an equation with logarithmic and singular weight having interior singularity.
Keywords: singular weight, logarithmic singularity, interior singularity, ideal point, a priori estimate, maximum principle,sup,inf,sup =f(inf).
MSC: 35J60, 35B44, 35B45, 35B50
1 Introduction and Main Results
We set∆ =∂11+∂22 on open bounded domainΩofR2. We consider the following equation:
(P)
−∆u=−log|x|
2dV eu inΩ⊂R2,
u≥0 inΩ.
Here:
0< a≤V ≤b <+∞, u∈L∞loc(Ω), and,
d=diam(Ω), 0∈Ω
This equation is defined in the sense of the distributions. Equations of the previous type were studied by many authors, with or without the boundary condition, also for Riemannian surfaces, see [1–20], where one can find some existence and compactness results.
Among other results, we can see in [12] the following important Theorem Theorem A(Brezis-Merle [12]).If (ui)is a sequence of solutions of problem (P)with (Vi) satisfying 0< a≤Vi≤b <+∞and without the term −log|x|
2d, then, for any compact subsetK ofΩ, it holds:
∗e-mails: [email protected], [email protected]
sup
K ui≤c, with c depending ona, b, K,Ω
One can find in [12] an interior estimate if we assumea= 0, but we need an assumption on the integral ofeui, namely, we have:
Theorem B(Brezis-Merle [12]).For (ui)i and (Vi)i two sequences of func- tions relative to the problem(P)without the term −log|x|
2d and with,
0≤Vi≤b <+∞and Z
Ω
euidy≤C,
then for all compact set K ofΩit holds;
sup
K
ui≤c,
with cdepending on b, C, K andΩ.
If we assumeV with more regularity, we can have another type of estimates, asup + inf type inequalities. It was proved by Shafrir see [20], that, if(ui)i is a sequence of functions solutions of the previous equation without assumption on the boundary withVisatisfying0< a≤Vi ≤b <+∞, then we have asup + inf inequality.
The point 0 ∈ Ω is a conical singularity and an interior singularity and called a divisor and an ideal point. See for example the papers of Hulin and Troyanov, [16, 17, 22].
Here, we have:
Theorem 1.1 For sequences(ui)iand(Vi)iof the Problem(P), for all com- pact subsetK of Ωwe have:
||ui||L∞(K)≤c(a, b, K,Ω).
Consider a positive numberM >0, if we assumeui ≥ −M, we can extend the previous result to any functionubounded from below by−M. If we consider the function vi = ui+M, then vi satisfies all the condition of the previous theorem.
Here we have, if we replace the conditionui≥0byui≥ −M:
Corollary 1.2 For sequences (ui)i and (Vi)i of the first equation of (P), for all compact subset K ofΩwe have:
||ui||L∞(K)≤c(a, b, M, K,Ω).
1) We have an example of the previous problem with ui ≥ 0 if we add the Dirichlet condition and use the maximum principle (in this case we assume u∈ W01,1(Ω) and −log|x|
2deu ∈L1(Ω) or u ∈W01,2(Ω) ⇒ eu ∈Lk(Ω),∀k ≥1 as a solution of a variational problem). From the counterexample constructed in the paper of Brezis and Merle, see [12], one can have an example of solutions with Dirichlet condition (but blowing-up solutions on the boundary).
2) We can replace the assumptionu∈L∞loc(Ω) by:
u∈L1loc(Ω) and (−log|x|
2d)eu∈L1loc(Ω), this imply thatu∈L∞loc(Ω).
Indeed, by solving a Dirichlet Problem and use Theorem 1 of the paper of Brezis and Merle, see [12], and Weyl theorem one can havee|u|∈Lkloc(Ω), ∀k≥ 1, the resut follow by the elliptic estimates of Agmon.
A consequence of the previous corollary we have:
Corollary 1.3 For a solution u with V of the first equation of (P), for all compact subsetK of Ωwe have:
sup
K
u≤c(a, b,inf
Ω u, K,Ω).
It is an estimate of the maximum on each compact subset ofΩof the solutions by mean of the infimum on Ω and a, b, K and Ω. (Also we have an a priori estimate).
We ask the following questions about explicit dependance ofsupKuin terms ofinfΩuand inequality of typesup + inf, as in the work of Shafrir [20] and the work of Tarantello, see [21] and Bartolucci-Tarantello, see [9]:
Problems. 1) Consider the Problem(P)without the boundary condition (without Dirichlet condition) and assume that:
0< a≤V ≤b <+∞,
Does exists constantsC1=C1(a, b, K,Ω), C2=C2(a, b, K,Ω)such that:
sup
K
u+C1inf
Ω u≤C2, for all solutionuof(P)?
2) If we add the condition||∇V||∞≤A, can we have a sharp inequality:
sup
K
u+ inf
Ω u≤c(a, b, A, K,Ω)?
2 Proof of the Theorem
We have:
ui∈L∞loc(Ω).
Thus by the local boundedness elliptic esitmates of Agmon and the Sobolev embedding, see [1, 2], we have:
ui∈Wloc2,k(Ω)∩C1,ǫ(Ω), k >2.
Step 1: For all x0 ∈ Ω, there is r > 0 such that: R
B(x0,r)−log|x|
2deuidx is bounded.
Let us considerx0 ∈Ω and set φ1 the first eigenfunction of the Laplacian with Dirichlet condition and with corresponding eigenvalue λ1 >0 for the ball
B(x0,2r) ⊂ Ω. We use integration by parts between ui and φ1. The Stokes formula gives:
Z
B(x0,2r)
−log|x|
2dVieuiφ1dx=λ1
Z
B(x0,2r)
uiφ1dx− Z
∂B(x0,2r)
∂νφ1uidx,
We write:
uiφ1=ui
r
−log|x|
2dφ1× v u u t
1
−log|x|
2d φ1,
We use Cauchy-Schwarz inequality to have:
Z
B(x0,2r)
uiφ1≤C||(−log|x|
2d)u2iφ1||1/2L1(B(x0,2r)), But,ui≥0and∂νφ1≥0, thus:
Z
B(x0,2r)
−log|x|
2deuiφ1dx≤C, sinceφ1>0the result follows.
Step 2: u+i =ui is locally bounded inL1. We have,ui≥0 and we write:
ui=ui(−log|x|
2d)×( 1
−log|x|
2d
)≤Cui(−log|x|
2d)≤Ceui(−log|x|
2d)
By the result of step 1, we obtain:
||ui||L1(B(x0,r0))≤C.
Since,
Z
B(y,r)
−log|x|
2dVieuidx≤C, We have a convergence to a nonegative measureµ:
Z
B(y,r)
−log|x|
2dVieuiφdx→ Z
B(y,r)
φdµ, ∀φ∈Cc(B(y, r)).
We setS the following set:
S={x∈B(y, r),∃(xi)∈Ω, xi →x, ui(xi)→+∞}.
We say that x0 is a regular point of µ if there function ψ ∈ Cc(B(y, r)), 0≤ψ≤1, withψ= 1in a neighborhood ofx0 such that:
Z
ψdµ <4π. (1)
We can deduce that a pointx0 is non-regular if and only ifµ(x0)≥4π.
A consequence of this fact is that ifx0 is a regular point then:
∃ R0>0such that one can bound(ui) = (u+i )in L∞(BR0(x0)). (2) We deduce(2)from corollary 4 of Brezis-Merle paper, and−log|x|2d ∈Lr,∀1≤ r <+∞, we have by step 2 the inequality:
||u+i ||1=||ui||1≤c.
We denote byΣthe set of non-regular points.
Step 3: S =Σ.
We haveS⊂Σ. Let’s considerx0∈Σ. Then we have:
∀ R >0, lim||u+i ||L∞(BR(x0)) = +∞. (3) Suppose contrary that:
||u+i||L∞(BR0(x0))≤C.
Then:
||euik||L∞(BR0(x0))≤C, and
Z
BR(x0)
−log|x|
2dVikeuik =o(1).
ForRsmall enough, which imply(1)for a functionψandx0will be regular, contradiction. Then we have (3). We chooseR0 >0 small such that BR0(x0) contain onlyx0 as non -regular point. Σ. Let’sxi∈BR(x0)scuh that:
u+i (xi) = max
BR(x0)u+i →+∞.
We havexi →x0. Else, there exists xik → x¯ 6=x0 and x¯ 6∈ Σ, i.e. x¯ is a regular point. It is impossible because we would have(2).
Since the measure is finite, if there are blow-up points, or non-regular points, S= Σ is finite.
Step 4: Σ ={∅}.
Now: suppose contrary that there exists a non-regular pointx0. We choose a radius R >0 such that BR(x0) contain only x0 as non-regular point. Thus outsideΣwe have local unfirorm boundedness ofui, also inC1norm. Also, we have weak *-convergence ofVi toV ≥0 withV ≤b.
Let’s consider (by a variational method):
zi∈W01,2(BR(x0)),
−∆zi=fi =−log|x|
2dVieui in BR(x0), et zi= 0 on ∂BR(x0).
By a duality theorem:
zi∈W01,q(BR), ||∇zi||q ≤Cq. By the maximum principle,ui≥ziin BR(x0).
Z
−log|x|
2dezi ≤ Z
−log|x|
2deui≤C. (4)
On the other hand,zi→za.e. (uniformly on compact sets ofBR(x0)−{x0}) withz solution of :
−∆z=µ in BR(x0), et z= 0 on ∂BR(x0).
Also, we have up to a subsequence, zi → z in W01,q(BR(x0)),1 ≤ q < 2 weakly, and thusz∈W01,q(BR(x0)).
Then by Fatou lemma:
Z
−log|x|
2dez≤C. (5)
As x0 ∈ S is not regular point we have µ({x0}) ≥ 4π, which imply that, µ≥4πδx0and by the maximum principle inW01,1(BR(x0))(obtainded by Kato’s inequality)
z(x)≥2 log 1
|x−x0|+O(1) if x→x0. Because,
z1≡2 log 1
|x−x0|+ 2 logR∈W01,s(BR(x0)), 1≤s <2.
Thus,
−log|x|
2dez≥
−Clog|x|
2d
|x−x0|2 , C >0.
Both in the casesx0= 0andx06= 0we have:
Z
BR(x0)
−log|x|
2dez=∞.
However, by(5):
Z
−log|x|
2dez≤C.
which is a contradiction.
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