In this paper, we consider the inviscid, incompressible planar flows in a bounded domain with a hole and construct stationary classical solutions with single vortex core, which is closed to the hole. This is carried out by constructing solutions to the following semilinear elliptic problem
{−Δψ=λ(ψ−κ4πlnλ)p+,inΩ,ψ=ρλ,on∂O0,ψ=0,on∂Ω0,(1)
where p>1, κ is a positive constant, ρλ is a constant, depending on λ, Ω=Ω0∖ˉO0 and Ω0, O0 are two planar bounded simply-connected domains. We show that under the assumption (lnλ)σ≤ρλ≤(lnλ)1−σ for some σ>0 small, (1) has a solution ψλ, whose vorticity set {y∈Ω:ψ(y)−κ+ρλη(y)>0} shrinks to the boundary of the hole as λ→+∞.
Citation: Shusen Yan, Weilin Yu. Planar vortices in a bounded domain with a hole[J]. Electronic Research Archive, 2021, 29(6): 4229-4241. doi: 10.3934/era.2021081
[1] | Shusen Yan, Weilin Yu . Planar vortices in a bounded domain with a hole. Electronic Research Archive, 2021, 29(6): 4229-4241. doi: 10.3934/era.2021081 |
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In this paper, we consider the inviscid, incompressible planar flows in a bounded domain with a hole and construct stationary classical solutions with single vortex core, which is closed to the hole. This is carried out by constructing solutions to the following semilinear elliptic problem
{−Δψ=λ(ψ−κ4πlnλ)p+,inΩ,ψ=ρλ,on∂O0,ψ=0,on∂Ω0,(1)
where p>1, κ is a positive constant, ρλ is a constant, depending on λ, Ω=Ω0∖ˉO0 and Ω0, O0 are two planar bounded simply-connected domains. We show that under the assumption (lnλ)σ≤ρλ≤(lnλ)1−σ for some σ>0 small, (1) has a solution ψλ, whose vorticity set {y∈Ω:ψ(y)−κ+ρλη(y)>0} shrinks to the boundary of the hole as λ→+∞.
In this paper, we consider a planar incompressible flow in a bounded smooth domain
Ω=Ω0∖ˉO0, |
where
{∇⊥ψ⋅∇ω=0,inΩ,−Δψ=ω,inΩ,ψ=constant,on∂O0,ψ=0,on∂Ω0, | (2) |
where
An existence result obtained by Smets and Schaftingen [15] via a variational method shows that (2) has a solution
∫∂O0∂ψλ∂ν=0, |
and
∫Ωωλ→κ, |
and the vorticity set
Note that in [15], the value
∫∂O0∂ψ∂ν=0. |
The first equation in (2) suggests that
{−Δψ=λ(ψ−κ4πlnλ)p+,inΩ,ψ=ρλ,on∂O0,ψ=0,on∂Ω0, | (3) |
where
In this paper, we mainly focus on the solvability of (3), and the effect from the constant
Let
Δη=0inΩ,η=1on∂O0,η=0on∂Ω0. | (4) |
Making the change of
{−ε2Δu=(u−κ+λεη)p+,inΩ,u=0,on∂Ω. | (5) |
Now we want to find a solution to (3) by constructing a solution for (5), whose vorticity set is close to the boundary of
(lnλ)σ≤ρλ≤(lnλ)1−σ. |
It is easy to see that if
1|lnε|γ2≤λε≤1|lnε|γ1. |
Before we state the main result, we give the following definition.
Definition 1.1. Let
Our main result of this paper can be stated as follows.
Theorem 1.2. Let
{y∈Ω|ψλ(y)>κ4πlnλ}⊂BL√λ(xλ), |
where
Moreover, as
dist(xλ,∂Ω)→0,λ∫Ω(ψλ−κ4πlnλ)p+→κ. |
In particular, if
dist(xλ,x(I))→0. |
As a result of Theorem 1.2, we obtain a flow in
Theorem 1.2 is proved via the following theorem.
Theorem 1.3. Let
{y∈Ω|uε(y)−κ+λεη(y)>0}⊂BLε|lnε|p−12(xε), |
where
dist(xε,∂Ω)→0. |
In particular, if there is a minimum interval
dist(xε,x(I))→0. |
To prove Theorem 1.3, we will use a finite reduction argument as in [6,7,8,9]. Since we consider the vortex nears the boundary, it turns out that more delicate estimates are needed in the proof of Theorem 1.3 than those estimates in [6,7,8,9]. In particular, we need the estimates of the functions
G(y,x)=12πln1|y−x|−H(y,x),x,y∈Ω, | (6) |
and
The stationary incompressible Euler equations have been studied by many authors, see for instance [1,3,4,5,6,7,8,9,10,15,16,17] and references therein. Roughly speaking, there are two commonly used methods to study the existence of the stationary incompressible Euler equations: the vorticity method and the stream-function method. The vorticity method was first established by Arnold [2] and further developed by Burton [4,5] and Tukington [16]. This argument roughly consists in maximizing the kinetic energy under a constrained sublevel set of
This paper is organized as follows. In section 2, we construct approximate solutions for (5). We will carry out a reduction procedure in section 3 and the results of existence will be proved in section 4. In appendix A, we give the estimates for the radius of vortex core.
In this section, following [10], we will construct an approximate solution for (5).
For
−Δϕ=ϕp,ϕ>0,ϕ∈H10(B1(0)). | (7) |
Moreover,
∫B1(0)ϕp+1=π(p+1)2|ϕ′(1)|2,∫B1(0)ϕp=2π|ϕ′(1)|. |
Let
{−ε2Δu=(u−a)p+,y∈BR(0),u=0,y∈∂BR(0), | (8) |
where
Uε,a(y)={a+(εsε)2p−1ϕ(|y|sε),|y|≤sε,aln|y|R/lnsεR,sε≤|y|≤R, | (9) |
where and
(εsε)2p−1ϕ′(1)=alnsεR, | (10) |
which gives an expansion for
sε=(|ϕ′(1)|a)p−12ε|lnε|p−12(1+O(ln|lnε||lnε|)). | (11) |
For any
{−ε2Δu=(Uε,x,a−a)p+,y∈Ω,u=0,y∈∂Ω. |
Then
PUε,x,a(y)=Uε,x,a(y)−alnRsεg(y,x), | (12) |
where
Take
d1=r1|lnε|λε,d2=r2|lnε|λε, | (13) |
where
S=:(∂O0)d2∖¯(∂O0)d1, | (14) |
where
r(x)=|lnε|λεd(x). | (15) |
Then we can rewrite
S={x∈Ω:r1<r(x)<r2}. | (16) |
In the rest of this paper, we assume that
u=PUε,x,aε+ωε, | (17) |
where
(εsε)2p−1ϕ′(1)=aεlnsεR,aε=κ−λεη(x)+aεlnRsεg(x,x). | (18) |
Let us now show that (18) is solvable for
η(x)=η(ˆx)−∂η(ˆx)∂νd(x)+O(d2(x))=1−∂η(ˆx)∂νd(x)+O(d2(x)), |
where
φ(x)=12πln12d(x)+O(d(x)). |
Then by our assumption on
PUε,x,aε(y)−κ+λεη(y)=Uε,x,aε(y)−κ+λεη(y)−aεlnRsεg(y,x)=Uε,x,aε(y)−aε+λε⟨∇η(x),y−x⟩−aεlnRsε⟨∇g(x,x),y−x⟩+O(s1+σε), | (19) |
where
Remark 1. It is not difficult to get the following expansions:
1lnRsε=1Aε+O(λε|lnε|2),aε=(1+lnR|lnε|)κ−λε+λε∂η(ˆx)∂νd(x)+κ|lnε|ln12d(x)+O(1|lnε|1+σ), | (20) |
where
Aε=|lnε|−p−12ln|lnε|+lnR+p−12lnκ|ϕ′(1)|. |
Moreover, in view of
∂aε∂xi=O(λε),∂sε∂xi=O(λεsε),i=1,2. | (21) |
By (9) and (21), we have the following expansion, which will be used in the rest of this paper,
∂Uε,x,aε(y)∂xi={aε|ϕ′(1)|1lnRsεϕ′(|y−x|sε)xi−yi|y−x|1sε+O(λε|lnε|),y∈Bsε(x),aεlnRsεxi−yi|y−x|2+O(λεln|y−x||lnε|),y∈Ω∖Bsε(x). | (22) |
In this section, we reduce the problem of finding a solution for (5) of the form (17) to a finite dimension problem.
Define
Eε,x={u:u∈W2,p(Ω)∩H10(Ω),∫ΩuΔ∂PUε,x,aε∂xidy=0,i=1,2}, |
and
Fε,x={u:u∈Lp(Ω),∫Ωu∂PUε,x,aε∂xidy=0,i=1,2}. |
For any
Qεu=:u+2∑j=1bjε2Δ∂PUε,x,aε∂xj, |
where
2∑j=1bjε2∫Ω∇∂PUε,x,aε∂xj∇∂PUε,x,aε∂xi=∫Ωu∂PUε,x,aε∂xi. | (23) |
The existence of
ε2∫Ω∇∂PUε,x,aε∂xj∇∂PUε,x,aε∂xi=p∫Ω(Uε,x,aε−aε)p−1+(∂Uε,x,aε∂xj−∂aε∂xj)∂PUε,x,aε∂xi=c(δij+o(1))1|lnε|p+1, | (24) |
where
Set
Lεu=−ε2Δu−p(PUε,x,aε−κ+λεη)p−1+u. | (25) |
We have the following result for the operator
Proposition 1. There are constants
s2−2pεε2‖QεLεu‖Lp(BLsε(x))≥σ0‖u‖L∞(Ω). |
As a consequence,
Proof. Suppose to the contrary that there are
s2−2pεnε2n‖QεnLεnun‖Lp(BLsεn(xn))≤1n. | (26) |
First of all, we estimate
QεnLεnun=Lεnun+2∑j=1bj,nε2nΔ∂PUεn,xn,aεn∂xj, | (27) |
where
2∑j=1bj,nε2n∫Ω∇∂PUεn,xn,aεn∂xj∇∂PUεn,xn,aεn∂xi=∫Ω∂PUεn,xn,aεn∂xiLεnun. | (28) |
From (19), (21) and Lemma A.1, we have
∫Ω∂PUεn,xn,aεn∂xiLεnun=∫ΩunLεn∂PUεn,xn,aεn∂xi=p∫Ω[(Uεn,xn,aεn−aεn)p−1+(∂Uεn,xn,aεn∂xi−∂aεn∂xi)−(PUεn,xn,aεn−κ+λεnη)p−1+∂PUεn,xn,aεn∂xi]un=p∫Ω[(Uεn,xn,aεn−aεn)p−1+∂Uεn,xn,aεn∂xi−(Uεn,xn,aεn−aεn+O(λεnsεn))p−1+∂PUεn,xn,aεn∂xi]un+O(λεns2εn|lnεn|p−1)=O(λεns2εn|lnεn|p−1). | (29) |
Then by (24), (28) and (29), we obtain
bj,n=O(λεns2εn|lnεn|2). | (30) |
Write
−ε2nΔun=p(PUεn,xn,aεn−κ+λεnη)p−1+un+fn, | (31) |
where
fn=QεnLεnun−2∑j=1bj,nε2nΔ∂PUεn,xn,aεn∂xj, |
Define
˜un(y)=un(sεny+xn),˜fn(y)=fn(sεny+xn). |
Then we have
−Δ˜un=ps2εnε2n(~PUεn,xn,aεn−κ+λεn˜η)p−1+˜un+s2εnε2n˜f2,n. | (32) |
From (26), (30) and Lemma A.1, we find
s2εnε2n‖˜fn‖Lploc(R2)=on(1)+O(s2−2pεn‖2∑j=1bj,nΔ∂PUεn,xn,aεn∂xj‖Lp(Ω))=on(1). | (33) |
Since the right hand side of (32) is bounded in
−Δ˜u−pwp−1+˜u=0,inR2, | (34) |
where
w(y)={ϕ(|y|),|y|≤1,ϕ′(1)ln|y|,|y|>1. |
So there exist constants
˜u=b1∂w∂y1+b2∂w∂y2. | (35) |
From
∫B1(0)ϕp−1∂ϕ∂yj˜u=0,j=1,2. |
So we get
un=on(1),inC(BLsεn(xn)). | (36) |
By our assumption,
QεnLεnun=0,inΩ∖BLsεn(xn). |
We find that
Δun=0,inΩ∖BLsεn(xn). |
However,
un=on(1),inΩ∖BLsεn(xn). |
So, we have proved that
‖un‖L∞(Ω)=on(1), |
which contradicts
Using the same argument as in Proposition 3.2 in [6], it is not difficult to prove that
We now want to find a solution for (5) of the form
Lεω=lε+Rε(ω), | (37) |
where
lε=(PUε,x,aε−κ+λεη)p+−(Uε,x,aε−aε)p+, | (38) |
and
Rε(ω)=(PUε,x,aε+ω−κ+λεη)p+−(PUε,x,aε−κ+λεη)p+−p(PUε,x,aε−κ+λεη)p−1+ω. | (39) |
From (37), we see
QεLεω=Qεlε+QεRε(ω). | (40) |
Fot
ω=Gεω=:(QεLε)−1Qε(lε+Rε(ω)). | (41) |
We have the following estimates for
Lemma 3.1. It holds
‖lε‖Lp(BLsε(x))=O(λεs1+2pε|lnε|p−1), |
and if
‖Rε(ω)‖Lp(BLsε(x))=O(s2pε|lnε|p−2‖ω‖2L∞(Ω)). |
Proof. For any
|lε|=|(PUε,x,aε−κ+λεη)p+−(Uε,x,aε−aε)p+|=|(Uε,x,aε−aε+O(λεsε))p+−(Uε,x,aε−aε)p+|≤C(λεsε(Uε,x,aε−aε)p−1++(λεsε)p)≤Cλεsε|lnε|p−1. |
So we get
‖lε‖Lp(BLsε(x))≤s2pε‖lε‖L∞(BLsε(x))≤Cλεs1+2pε|lnε|p−1. |
Similarly, using (19) and Lemma A.1, it ie east to see that
‖Rε(ω)‖Lp(BLsε(x))≤C‖ω‖2L∞(Ω)‖(PUε,x,aε−κ+λεη)p−2+‖Lp(BLsε(x))≤Cs2pε|lnε|p−2‖ω‖2L∞(Ω). |
So we complete the proof.
Using Lemma 3.1, we can solve (41) in a standard way and obtain the following proposition.
Proposition 2. There is
‖ωε,x‖L∞(Ω)=O(λεsε). |
Furthermore,
In this section, we will prove our main results.
Define
I(u)=ε22∫Ω|∇u|2−1p+1∫Ω(u−κ+λεη)p+1+, |
and set
F(x)=I(PUε,x,aε+ωε,x), |
where
Lemma 4.1. There holds:
F(x)=I(PUε,x,aε)+O(λ2εs4ε|lnε|p−1). |
Proof. Since
I(PUε,x,aε+ωε,x)=I(PUε,x,aε)+⟨I′(PUε,x,aε),ωε,x⟩+12⟨I′′(PUε,x,aε)ωε,x,ωε,x⟩+˜Rε(ωε,x)=I(PUε,x,aε)−∫Ωlεωε,x+12∫Ωωε,xLεωε,x+˜Rε(ωε,x)=I(PUε,x,aε)−∫Ωlεωε,x+12∫Ω(lε+Rε(ωε,x)−2∑j=1bjε2Δ∂PUε,x,aε∂xj)ωε,x+˜Rε(ωε,x)=I(PUε,x,aε)−12∫Ωlεωε,x+12∫ΩRε(ωε,x)ωε,x+˜Rε(ωε,x), |
where
˜Rε(ωε,x)=−1p+1∫Ω[(PUε,x,aε+ωε,x−κ+λεη)p+1+−(PUε,x,aε−κ+λεη)p+1+−(p+1)(PUε,x,aε−κ+λεη)p+ωε,x−p(p+1)2(PUε,x,aε−κ+λεη)p−1+ω2ε,x]. |
One sees
|˜Rε(ωε,x)|≤C∫Ω(PUε,x,aε−κ+λεη)p−2+ω3ε,x≤Cλ3εs5ε|lnε|p−2. |
Then by Lemma 3.1, we can easily check that
F(x)=I(PUε,x,aε)+O(λ2εs4ε|lnε|p−1). |
Lemma 4.2. We have
I(PUε,x,aε)=Cε+πκε2Aε(2λε∂η(ˆx)∂νd(x)+κ|lnε|ln12d(x))+O(ε2|lnε|2+σ), |
where
Proof. We have
ε2∫Ω|∇PUε,x,aε|2=∫Ω(−ε2ΔPUε,x,aε)PUε,x,aε=∫Ω(Uε,x,aε−aε)p+PUε,x,aε=∫Ω(Uε,x,aε−aε)p+1++aε∫Ω(Uε,x,aε−aε)p+−aεlnRsε∫Ωg(y,x)(Uε,x,aε−aε)p+=π(p+1)2ε2(aεlnRsε)2+2πε2aεlnRsε(κ−λεη(x))+O(λεs3ε|lnε|p). |
On the other hand, by (19)
∫Ω(PUε,x,aε−κ+λεη)p+1+=∫Ω(Uε,x,aε−aε+O(λεsε))p+1+=∫Ω(Uε,x,aε−aε)p+1++O(λεs3ε|lnε|p)=π(p+1)2ε2(aεlnRsε)2+O(λεs3ε|lnε|p). |
Then we get that
I(PUε,x,aε)=π(p−1)4ε2(aεlnRsε)2+πε2aεlnRsε(κ−λεη(x))+O(λεs3ε|lnε|p). |
Therefore, since
I(PUε,x,aε)=Cε+πκε2Aε(2λε∂η(ˆx)∂νd(x)+κ|lnε|ln12d(x))+O(ε2|lnε|2+σ), |
where
Cε=πε2Aε((p−1)κ24Aε+(κ−λε)((1+lnR|lnε|)κ−λε)). |
Proof of Theorem 1.3. By Lemma 4.1 and Lemma 4.2, we have that for
F(x)=Cε+πκε2Aε(2λε∂η(ˆx)∂νd(x)+κ|lnε|ln12d(x))+O(ε2|lnε|2+σ). | (42) |
Consider the following minimization problem
min{F(x):x∈ˉS}. | (43) |
Let
ρ(x)=2λε∂η(ˆx)∂νd(x)+κ|lnε|ln12d(x). | (44) |
The Hopf lemma shows that for any
m1=κ2miny∈∂Ω{(∂η(y)∂ν)−1},m2=κ2maxy∈∂Ω{(∂η(y)∂ν)−1}, |
and choose
r(x0)=κ2(∂η(ˆx0)∂ν)−1, |
where
ρ(x0)=2λε∂η(ˆx0)∂νd(x0)+κ|lnε|ln12d(x0)=κ|lnε|(1+ln12r(x0)+ln(λε|lnε|)).. | (45) |
For
ρ(x)=2λε∂η(ˆx)∂νd(x)+κ|lnε|ln12d(x)=2|lnε|∂η(ˆx)∂νr1+κ|lnε|ln12r1+κ|lnε|ln(λε|lnε|)≥κ|lnε|(r1m2+ln12r1+ln(λε|lnε|))>h(x0), | (46) |
if
Similarly, for
ρ(x)≥κ|lnε|(r2m2+ln12r2+ln(λε|lnε|))>h(x0), | (47) |
if
Therefore, from (42), (46) and (47), for any
F(x0)<F(x). |
Thus there is a minimum point
By our construction, we find
Bsε(1−Lsσε)(xε)⊂{y∈Ω:uε(y)−κ+λεη(y)≥0}⊂Bsε(1+Lsσε)(xε), |
and as
We remark that if there is a minimum interval
In this appendix, we give the estimates for the radius of vortex core. The proofs of such results can be found in [6] Lemma A.1.
In the following, we assume that
Lemma A.1. Let
PUε,x,aε(y)−κ+λεη(y)>0,y∈Bsε(1−Lsσε)(x), |
while
PUε,x,aε(y)−κ+λεη(y)<0,y∈Ω∖Bsε(1+Lsσε)(x). |
Lemma A.2. Suppose that
‖ω‖L∞(Ω)=O(sε). |
Then there is
PUε,x,aε(y)+ω(y)−κ+λεη(y)>0,y∈Bsε(1−Lsσε)(x), |
while
PUε,x,aε(y)+ω(y)−κ+λεη(y)<0,y∈Ω∖Bsε(1+Lsσε)(x). |
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