A Note on Symmetry of Solutions for a Class of Singular Semilinear Elliptic Problems

Let u be a solution of (1.1). Assume that the domain Ω is strictly convex with respect to the ν-direction and symmetric with respect to T0ν. Moreover, assume that

g ⁢ ( x ) ≥ g ⁢ ( x 0 ν )   𝑎𝑛𝑑   h ⁢ ( x ) ≤ h ⁢ ( x 0 ν )   for all ⁢ x ∈ Ω 0 ν .

Then, u is symmetric with respect to T0ν and nondecreasing with respect to the ν-direction in Ω0ν. If Ω is a ball, then u is radially symmetric with ∂⁡u∂⁡r⁢(r)<0 for r≠0.

Let u0 be the solution of (1.3). Then, for any a⁢(ν)<λ<λ1⁢(ν), we have

and

Proof.

Let un∈C⁢(Ω¯)∩H01⁢(Ω) be the unique solution of

{ - Δ ⁢ u n = g n ⁢ ( x ) ( u n + 1 n ) γ for x ∈ Ω , u n > 0 for x ∈ Ω , u n = 0 for x ∈ ∂ ⁡ Ω ,

where gn⁢(x)=min⁡(g⁢(x),n). The existence of un was proved in [5] and the uniqueness follows by [7]. Since the problem is no more singular, by standard elliptic estimates it follows that un∈C2⁢(Ω¯). Therefore, we can use the moving plane technique as in [2, 18, 25] to deduce that

u n ⁢ ( x ) < u n λ ν ⁢ ( x )   for all ⁢ x ∈ Ω λ ν .

By [5] we have that un converges to u0 a.e. as n tends to infinity and, therefore, (2.1) follows by passing to the limit. In the same way, we obtain

∂ ⁡ u 0 ∂ ⁡ ν ⁢ ( x ) ≥ 0   for all ⁢ x ∈ Ω λ 1 ⁢ ( ν ) ν

and, therefore, (2.2) follows via the strong maximum principle. ∎

Let u0 be the solution of (1.3). Assume that the domain Ω is strictly convex with respect to the ν-direction and symmetric with respect to T0ν. Then, u0 is symmetric with respect to T0ν and nondecreasing with respect to the ν-direction in Ω0ν. Moreover, if Ω is a ball, then u0 is radially symmetric with ∂⁡u0∂⁡r⁢(r)<0 for r≠0.

Lemma 3.1 (see [8, Lemma 4])

Let γ>0. Consider the function

h γ ⁢ ( x , y , z , t ) := x γ ⁢ ( x + y ) γ ⁢ ( z + t ) γ + x γ ⁢ z γ ⁢ ( z + t ) γ - z γ ⁢ ( x + y ) γ ⁢ ( z + t ) γ - x γ ⁢ z γ ⁢ ( x + y ) γ

and the domain D⊂ℝ4 defined by

D := { ( x , y , z , t ) : 0 ≤ x ≤ z , 0 ≤ t ≤ y } .

Then, it follows that hγ≤0 in D.

Let u be a solution of (1.1) with γ>0 and let w be given by (1.2). Then, it follows that

w > 0   in ⁢ Ω .

Proof.

Since u∈C⁢(Ω¯) and u0∈C⁢(Ω¯), then w∈C⁢(Ω¯)∩H01⁢(Ω). By the hypotheses on f and h it is easy to check that u is a supersolution, in the sense of [7, Definition 2.5], of the equation

- Δ ⁢ v = g ⁢ ( x ) v γ .

Arguing as in [7, Lemma 2.8], we get

u ≥ u 0   in ⁢ Ω

and

w ≥ 0   in ⁢ Ω .

We show that w>0 in the interior of Ω making use of the maximum principle in regions where the problem is not singular. Assume by contradiction that there exists a point x0∈Ω such that w⁢(x0)=0 and let r=r⁢(x0)>0 such that Br(x0)⊂⊂Ω. We have

- Δ ⁢ w = - Δ ⁢ u + Δ ⁢ u 0 = g ⁢ ( x ) ( u 0 + w ) γ + h ⁢ ( x ) ⁢ f ⁢ ( u ) - g ⁢ ( x ) u 0 γ ≥ g ⁢ ( x ) ⁢ ( 1 ( u 0 + w ) γ - 1 u 0 γ )

in Br⁢(x0). Since u0⁢(x0)>0, we can assume that u0 is positive in Br⁢(x0) so that u0+w is also positive in Br⁢(x0). By (Hg) we have

g ⁢ ( x ) ⁢ ( 1 ( u 0 + w ) γ - 1 u 0 γ ) = k ⁢ ( x ) ⁢ w

for some bounded coefficient k⁢(x). Therefore, we can find Λ>0 such that

g ⁢ ( x ) ⁢ ( 1 ( u 0 + w ) γ - 1 u 0 γ ) + Λ ⁢ w ≥ 0

in Br⁢(x0), so that

- Δ ⁢ w + Λ ⁢ w ≥ 0   in ⁢ B r ⁢ ( x 0 ) .

By the strong maximum principle (see [19]) we have w≡0 in Br⁢(x0) and by a covering argument we have w≡0 in Ω. But w≡0 in Ω implies that f≡0, and we get a contradiction. ∎

Let a⁢(ν)<λ<λ1⁢(ν) and let Ω′ be a connected subdomain of Ωλν. Assume that

g ⁢ ( x ) ≥ g ⁢ ( x λ ν )   𝑎𝑛𝑑   h ⁢ ( x ) ≤ h ⁢ ( x λ ν )   for all ⁢ x ∈ Ω ′ .

Let u be a solution of (1.1) and let w be given by (1.2). If

∂ ⁡ w ∂ ⁡ ν ≥ 0   in ⁢ Ω ′ ,

then the alternative

∂ ⁡ w ∂ ⁡ ν > 0   in ⁢ Ω ′

or

∂ ⁡ w ∂ ⁡ ν = 0   in ⁢ Ω ′

holds.

Proof.

We begin by observing that by the monotonicity assumptions on the functions g and h it follows that

∂ ⁡ g ∂ ⁡ ν ≤ 0   and   ∂ ⁡ h ∂ ⁡ ν ≥ 0   a.e. in ⁢ Ω ′ .

Set

g ν := ∂ ⁡ g ∂ ⁡ ν , h ν := ∂ ⁡ h ∂ ⁡ ν , w ν := ∂ ⁡ w ∂ ⁡ ν , u 0 ν := ∂ ⁡ u 0 ∂ ⁡ ν .

Since f′≥0 a.e. by (Hf), we have u0ν≥0 in Ω′ by Proposition 2.1, u≥u0 by Lemma 3.2, and, by differentiating the equation in (1.1), we get that wν solves

We recall now that g∈Lloc∞⁢(Ω) by (Hg) and that u is bounded away from zero in Ω′. Therefore, we can find Λ>0 such that

- Δ ⁢ w ν ≥ - γ ⁢ g ⁢ ( x ) u γ + 1 ⁢ w ν ≥ - Λ ⁢ w ν ,

so that the conclusion follows by the standard strong maximum principle (see [19]). ∎

Let a⁢(ν)<λ<λ1⁢(ν) and let Ω′⊆Ωλν. Assume that

g ⁢ ( x ) ≥ g ⁢ ( x λ ν )   𝑎𝑛𝑑   h ⁢ ( x ) ≤ h ⁢ ( x λ ν )   for all ⁢ x ∈ Ω ′ .

Let u be a solution of (1.1) and let w be given by (1.2). Assume that

w ≤ w λ ν   on ⁢ ∂ ⁡ Ω ′ .

Then, there exists a positive constant δ=δ⁢(u,f) such that, if ℒ⁢(Ω′)≤δ, then

w ≤ w λ ν   in ⁢ Ω ′ .

Proof.

We have

- Δ ⁢ ( u 0 + w ) = 1 ( u 0 + w ) γ + h ⁢ ( x ) ⁢ f ⁢ ( u 0 + w )   in ⁢ Ω

and

- Δ ⁢ ( u 0 λ ν + w λ ν ) = 1 ( u 0 λ ν + w λ ν ) γ + h ⁢ ( x λ v ) ⁢ f ⁢ ( u 0 λ ν + w λ ν )   in ⁢ Ω .

Since (w-wλν)+∈H01⁢(Ω′), we can consider a sequence of positive functions ψn such that

ψ n ∈ C c ∞ ⁢ ( Ω ′ )   and   ψ n → H 0 1 ⁢ ( Ω ′ ) ( w - w λ ν ) + .

We can also assume that suppψn⊆supp(w-wλν)+. Choosing ψn as test function in the weak formulation of the above equations and subtracting, we get

∫ Ω ′ ( ∇ ⁡ ( u 0 + w ) - ∇ ⁡ ( u 0 λ ν + w λ ν ) ) ⁢ ∇ ⁡ ψ n = ∫ Ω ′ ( g ⁢ ( x ) ( u 0 + w ) γ + h ⁢ ( x ) ⁢ f ⁢ ( u 0 + w ) - g ⁢ ( x λ v ) ( u 0 λ ν + w λ ν ) γ - h ⁢ ( x λ v ) ⁢ f ⁢ ( u 0 λ ν + w λ ν ) ) ⁢ ψ n ⁢ 𝑑 x .

Recall that u0 solves (1.3). It follows easily that u0λν solves

{ - Δ ⁢ u 0 λ ν = g ⁢ ( x λ v ) ( u 0 λ ν ) γ in Ω , u 0 λ ν > 0 in Ω , u 0 λ ν = 0 on ∂ ⁡ Ω .

Moreover, since g⁢(x)≥g⁢(xλν) and h⁢(x)≤h⁢(xλν) in Ω′ by assumption, we get

Since u0≤u0λν in Ωλν and w≥wλν on supp⁡ψn, by applying Lemma 3.1 with u0=x, w=y, u0λν=z, and wλν=t, we obtain

( u 0 ) γ ⁢ ( u 0 + w ) γ ⁢ ( u 0 λ ν + w λ ν ) γ + ( u 0 ) γ ⁢ ( u 0 λ ν ) γ ⁢ ( u 0 λ ν + w λ ν ) γ - ( u 0 λ ν ) γ ⁢ ( u 0 + w ) γ ⁢ ( u 0 λ ν + w λ ν ) γ - ( u 0 ) γ ⁢ ( u 0 λ ν ) γ ⁢ ( u 0 + w ) γ ≤ 0 ,

so that

g ⁢ ( x ) ⁢ ( 1 ( u 0 λ ν ) γ - 1 ( u 0 ) γ + 1 ( u 0 + w ) γ - 1 ( u 0 λ ν + w λ ν ) γ ) ≤ 0

since g⁢(x)>0 in Ω. Therefore, by (Hf) and (Hh) there exists C>0 such that

Passing to the limit as n→∞, we get

∫ Ω ′ | ∇ ( w - w λ ν ) + | 2 ≤ C ∫ Ω ′ | ( w - w λ ν ) + | 2

and, by the Poincaré inequality, we can find C′⁢(Ω′)>0 such that

∫ Ω ′ | ∇ ( w - w λ ν ) + | 2 ≤ C ′ ( Ω ′ ) ∫ Ω ′ | ∇ ( w - w λ ν ) + | 2 .

For sufficiently small δ, it follows that C′⁢(Ω′)<1. This shows that (w-wλν)+=0 in Ω′, which gives that w≤wλν  in Ω′. ∎

Let u be a solution of (1.1). Assume that

g ⁢ ( x ) ≥ g ⁢ ( x λ ν )   𝑎𝑛𝑑   h ⁢ ( x ) ≤ h ⁢ ( x λ ν )   for all ⁢ x ∈ Ω λ ν

and for any a⁢(ν)<λ≤λ1⁢(Ω). Let w be given by (1.2) and assume that

w ≤ w λ ν   in ⁢ Ω λ ν

for some a⁢(ν)<λ≤λ1⁢(Ω). Then, w<wλν in Ωλν unless w≡wλν in Ωλν.

Proof.

Let us assume that there exists a point x0∈Ωλν such that w⁢(x0)=wλν⁢(x0) and let r=r⁢(x0)>0 such that Br(x0)⊂⊂Ωλν. Since g≥c>0 in Ω by (Hg), we have g⁢(x)≥g⁢(xλν) and h⁢(x)≤h⁢(xλν) by assumption, w>0 by Lemma 3.2, w≤wλν in Ωλν by assumption, and u0≤u0λν in Ωλν by Proposition 2.1. Using the above, in Br⁢(x0) we have

Since f is nondecreasing by (Hf) and h is nonnegative by (Hh), we get

h ⁢ ( x ) ⁢ ( f ⁢ ( u 0 λ ν + w λ ν ) - f ⁢ ( u 0 + w ) ) ≥ 0 .

Moreover, since the function

h ⁢ ( t ) := a - γ - b - γ + ( b + t ) - γ - ( a + t ) - γ

is increasing in [0,∞) for 0<a≤b, we also have

1 u 0 γ - 1 ( u 0 λ ν ) γ + 1 ( u 0 λ ν + w ) γ - 1 ( u 0 + w ) γ ≥ 0 .

It follows that

- Δ ⁢ ( w λ ν - w ) ≥ g ⁢ ( x ) ⁢ ( 1 ( u 0 λ ν + w λ ν ) γ - 1 ( u 0 λ ν + w ) γ ) .

Since u0λν⁢(x0)>0, arguing as in Lemma 3.2, we find Λ>0 such that, eventually reducing r, we have

g ⁢ ( x ) ⁢ ( 1 ( u 0 λ ν + w λ ν ) γ - 1 ( u 0 λ ν + w ) γ ) + Λ ⁢ ( w λ ν - w ) ≥ 0

in Br⁢(x0), so that

- Δ ⁢ ( w λ ν - w ) + Λ ⁢ ( w λ ν - w ) ≥ 0   in ⁢ B r ⁢ ( x 0 ) .

By the strong maximum principle (see [19]), we get wλν-w≡0 in Br⁢(x0). Using a covering argument, it follows that wλν-w≡0 in Ωλν and the result follows. ∎

Let u be a solution of (1.1) and let w be given by (1.2). Assume that

g ⁢ ( x ) ≥ g ⁢ ( x λ 1 ⁢ ( ν ) ν )   𝑎𝑛𝑑   h ⁢ ( x ) ≤ h ⁢ ( x λ 1 ⁢ ( ν ) ν )   for all ⁢ x ∈ Ω λ 1 ⁢ ( ν ) ν .

Then, for any

a ⁢ ( ν ) < λ < λ 1 ⁢ ( ν ) ,

we have

Moreover,

Finally, (4.1) and (4.2) hold true replacing w by u.

Proof.

Let λ>a⁢(ν). Since w>0 in Ω, by Lemma 3.2 we have

w ≤ w λ ν   on ⁢ ∂ ⁡ Ω λ ν .

Therefore, using Proposition 3.4, for ℒ⁢(Ωλν) sufficiently small, we obtain

and w<wλν in Ωλν by Lemma 3.5. Set

Λ 0 = { λ > a ⁢ ( ν ) : w ≤ w t ν ⁢ in ⁢ Ω t ν ⁢ for all t ∈ ( a ⁢ ( ν ) , λ ] } ,

which is not empty thanks to (4.3), and

λ 0 = sup ⁡ Λ 0 .

By the definition of λ1⁢(ν), to prove our result we have to show that λ0=λ1⁢(ν). Assume that λ0<λ1⁢(ν) and observe that, by continuity, we obtain w≤wλ0ν in Ωλ0ν. By Lemma 3.5 it follows that w<wλ0ν in Ωλ0ν unless w=wλ0ν in Ωλ0ν. Because of the zero Dirichlet boundary conditions, since w>0 in the interior of the domain, the case w≡wλ0ν in Ωλ0ν is not possible. Thus, w<wλ0ν in Ωλ0ν.

We can now consider δ, given by Proposition 3.4, so that the weak comparison principle holds true in any subdomain Ω′ if ℒ⁢(Ω′)≤δ. Fix a compact set 𝒦⊂Ωλ0ν so that ℒ⁢(Ωλ0ν∖𝒦)≤δ2. By compactness we find σ>0 such that

w λ 0 ν - w ≥ 2 ⁢ σ > 0   in ⁢ 𝒦 .

Take now ε¯>0 sufficiently small so that λ0+ε¯<λ1⁢(ν) and, for any 0<ε≤ε¯,

1. wλ0+εν-w≥σ>0 in 𝒦,

2. ℒ⁢(Ωλ0+εν∖𝒦)≤δ.

Taking (a) into account, it is now easy to check that, for any 0<ε≤ε¯, we have w≤wλ0+εν on the boundary of Ωλ0+εν∖𝒦. Consequently, by (b), we can apply the weak comparison principle (Proposition 3.4) to deduce that

w ≤ w λ 0 + ε ν   in ⁢ Ω λ 0 + ε ν ∖ 𝒦 .

Thus, w≤wλ0+εν in Ωλ0+εν and, by applying Lemma 3.5, we have w<wλ0+εν in Ωλ0+εν. We get a contradiction with the definition of λ0 and this shows that λ0=λ1⁢(ν). Thus, (4.1) is proved.

By simple geometric considerations and by (4.1) it follows that w is nondecreasing in Ωλ1⁢(ν)ν in the ν-direction. This gives

∂ ⁡ w ∂ ⁡ ν ⁢ ( x ) ≥ 0   in ⁢ Ω λ 1 ⁢ ( ν ) ν ,

so it is easy to deduce (4.2) from Proposition 3.3.

To prove that (4.1) and (4.2) hold true replacing w with u, just recall that

u = u 0 + w

and exploit Proposition 2.1. ∎

Proof of Theorem 1.1.

We can prove Theorem 1.1 as a consequence of Proposition 4.1. By assumption we have λ1⁢(ν)=0. By Proposition 4.1 we obtain

u ⁢ ( x ) ≤ u 0 ν ⁢ ( x )   for all ⁢ x ∈ Ω 0 ν

and, replacing ν with -ν, we get

u ⁢ ( x ) ≥ u 0 ν ⁢ ( x )   for all ⁢ x ∈ Ω 0 ν .

Then, u⁢(x)=u0ν⁢(x) in Ω. The monotonicity of u follows by (4.2). ∎