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Advances in Nonlinear Analysis

Editor-in-Chief: Radulescu, Vicentiu / Squassina, Marco


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Comparison results for nonlinear divergence structure elliptic PDE’s

Yichen Liu / Monica Marras / Giovanni Porru
Published Online: 2019-06-01 | DOI: https://doi.org/10.1515/anona-2020-0008

Abstract

First we prove a comparison result for a nonlinear divergence structure elliptic partial differential equation. Next we find an estimate of the solution of a boundary value problem in a domain Ω in terms of the solution of a related symmetric boundary value problem in a ball B having the same measure as Ω. For p-Laplace equations, the corresponding result is due to Giorgio Talenti. In a special (radial) case we also prove a reverse comparison result.

Keywords: Quasi-linear equations; Symmetrization; Comparison results

MSC 2010: 35B51; 35J62; 49K20; 49K30

1 Introduction

In the seminal paper [1], Giorgio Talenti established sharp estimates of the solution to a boundary value problem of a second order elliptic partial differential equation in terms of the solution of a related symmetric problem. We refer to the survey [2] for a detailed treatment of the subject. The interest of these results relies on the obvious fact that a symmetric problem reduces to an ordinary differential equation and is easier to be solved. The papers by Talenti have inspired the use of similar methods in numerous investigations involving both linear and nonlinear elliptic problems.

To be more precise, let Ω ⊂ ℝn be a bounded smooth domain, let f : Ω → ℝ be positive and bounded, and let h : ℝ+ → ℝ+ be non-decreasing. Let g be positive and such that g(s2)s is strictly increasing and differentiable for s > 0. Let u be a solution to (g(|u|2)uxi)xi=f(x)h(u),u>0inΩ,u=0onΩ.(1)

Here and in what follows, the summation convention over repeated indices from 1 to n is in effect. If B ⊂ ℝn is the ball centered at the origin with the same measure as Ω and if f is the Schwarz (decreasing) rearrangement of f, let v be a solution to (g(|v|2)vxi)xi=f(x)h(v),v>0inB,v=0onB.(2)

When u(x)v(x)inB?(3)

Under suitable conditions on the function h, the answer is positive for the p-Laplacian, where g(s2) = sp−2, p > 1. Recently, inequality (3) has been proved for the (p, q)-Laplacian, where g(s2) = sp−2+sq−2, p > q > 1, see [3]. In the last decades, many authors have studied (p, q)-Laplace equations, see [4, 5, 6] and references therein. In this paper we show that (3) holds for a wide class of operators under appropriate conditions on g and h.

Let us find conditions on g and h which ensure existence and uniqueness for problems (1) and (2). With G(s) := g(s2)s we assume:

(G0) There are pq > 1 and M ≥ 1 such that 1Msp1G(s)M(sp1+sq1)s>0.

(G1) G(s) is continuous for s ≥ 0, is strictly increasing and continuously differentiable for s > 0.

(G2) With q as in (G0), the function G(s)sq1is non-decreasing for s>0.

(H1) h(t) is a positive non-decreasing function for t > 0.

(H2) There is 1 < α < q such that h(t) t1−α is bounded and non-increasing for t > 0. Here q is the same as in (G0).

Remark 1.1

If 1 < α < q, (G2) implies G(s)sα1 is strictly increasing for s>0.

We note that conditions (G0), (G1) and (G2) hold for a wide class of equations including the p-Laplacian and the (p, q)-Laplacian.

2 Existence of positive solutions

Assuming condition (G0), the natural space for solutions to problem (1) is the Sobolev space W01,p(Ω). The equation in (1) is the Euler equation of the functional I(w)=Ω(0|w|G(s)dsf(x)0wh(s)ds)dx.

It is well-known that a function u that minimizes I(w) for wW01,p(Ω), w ≥ 0, is a solution to the equation in (1) with u ≥ 0. We claim that, under conditions (G1), (G2), (H1) and (H2), a minimum for I(w) cannot be zero in any ball BΩ. Indeed, arguing by contradiction, let be a minimum vanishing on some ball B. Define z = +ϵϕ, where 0 < ϵ < 1 and ϕC01(B) is a positive function in B and vanishing on ΩB. We have I(z)=I(w~)+B(0ϵ|ϕ|G(s)dsf(x)0ϵϕh(s)ds)dx.

We first observe that conditions (G2) and (H2) imply, for 0 < ϵ < 1 and τ > 0, G(ϵτ)ϵq1G(τ),h(ϵτ)ϵα1h(τ).

Putting s = ϵτ with 0 < ϵ < 1 and using these inequalities we find I(z)=I(w~)+Ω(ϵ0|ϕ|G(ϵτ)dτf(x)ϵ0ϕh(ϵτ)dτ)dxI(w~)+Ω(ϵq0|ϕ|G(τ)dτf(x)ϵα0ϕh(τ)dτ)dx=I(w~)+ϵαΩ(ϵqα0|ϕ|G(τ)dτf(x)0ϕh(τ)dτ)dx.

Since q > α, it is clear that I(z) < I() for ϵ small enough. The claim follows. Therefore, we may assume that there exists a solution to (1) with u > 0 almost everywhere in Ω. At the end of the next Section, we will prove that such a solution u is positive in Ω.

In this paper we consider solutions belonging to C1(Ω). We refer to [7, 8, 9, 10] for regularity results.

3 A comparison result

Lemma 3.1

Let G(s) = g(s2)s satisfy conditions (G1) and (G2). For x, y ∈ ℝn and 0 < t ≤ 1 we have g(|x|2)[|x|2(1+(α1)tα)αtα1xy]+g(|y|2)[|y|2(1+(α1)tα)αtα+1xy]0.(4)

In addition, if |x|+|y| > 0 and 0 < t < 1, inequality (4) holds in a strict sense. Here 1 < α < q.

Proof

Recall the generalized Young’s inequality xy0|x|φ(τ)dτ+0|y|G(τ)dτ,(5)

where φ(τ) is the inverse function of G(τ). Replacing x by g(|x|2)tα−1x we find g(|x|2)tα1xy0tα1G(|x|)φ(τ)dτ+0|y|G(τ)dτ.(6)

Similarly, we have xy0|y|φ(τ)dτ+0|x|G(τ)dτ.

Replacing y by g(|y|2)t1−αy in the latter inequality we find g(|y|2)t1αxy0t1αG(|y|)φ(τ)dτ+0|x|G(τ)dτ.(7)

In view of (6) and (7), inequality (4) holds provided Ψ(t):=G(|x|)|x|(1+(α1)tα)α0tα1G(|x|)φ(τ)dτα0|y|G(τ)dτ+G(|y|)|y|(1+(α1)tα)α0t1αG(|y|)φ(τ)dτα0|x|G(τ)dτ0.(8)

In case of t = 1 we have Ψ(1)=α[G(|x|)|x|0G(|x|)φ(τ)dτ0|y|G(τ)dτ+G(|y|)|y|0G(|y|)φ(τ)dτ0|x|G(τ)dτ].(9)

Putting τ = G(s) we find 0G(|x|)φ(τ)dτ=0|x|sG(s)ds=|x|G(|x|)0|x|G(s)ds.(10)

Similarly, we find 0G(|y|)φ(τ)dτ=|y|G(|y|)0|y|G(s)ds.(11)

Insertion of (10) and (11) into (9) yields Ψ(1) = 0. Hence, to prove that Ψ(t) > 0 for 0 < t < 1 when |x|+|y| > 0, it is enough to prove that Ψ′(t) < 0. Since Ψ(t)=α(α1)t[G(|x|)|x|tαφ(tα1G(|x|))tα1G(|x|)G(|y|)|y|tα+φ(t1αG(|y|))t1αG(|y|)],

we must show that G(|x|)|x|tαφ(tα1G(|x|))tα1G(|x|)<G(|y|)|y|tαφ(t1αG(|y|))t1αG(|y|).(12)

Let us prove that the left hand side of (12) is negative when |x| > 0. Indeed, if |x| > 0, the inequality G(|x|)|x|tαφ(tα1G(|x|))tα1G(|x|)<0

is equivalent to t|x|<φ(tα1G(|x|)),

which in turn is equivalent to G(t|x|)<tα1G(|x|).

The latter inequality can be rewritten as G(t|x|)(t|x|)α1<G(|x|)|x|α1,

which holds by Remark 1.1.

Now, let us prove that the right hand side of (12) is positive when |y| > 0, that is G(|y|)|y|tαφ(t1αG(|y|))t1αG(|y|)>0.

Let us write this inequality as t1|y|>φ(t1αG(|y|)),

which can be rewritten as G(t1|y|)>t1αG(|y|).

This inequality is equivalent to the following G(|y|)|y|α1<G(t1|y|)(t1|y|)α1,

which holds by Remark 1.1. Hence, inequality (12) holds when |x|+|y| > 0. It follows that also (8) and (4) hold in a strict sense for 0 < t < 1. The lemma is proved. □

Theorem 3.2

Let G(s) := g(s2)s satisfy (G0), (G1) and (G2), and let h(t) satisfy (H1) and (H2). Let uC1(Ω), u = 0 on ∂Ω and u > 0 everywhere on Ω such that Ωg(|u|2)uϕdxΩf(x)h(u)ϕdxϕC01(Ω),ϕ0.(13)

Let vC1(Ω), v ≥ 0 on ∂Ω and v > 0 everywhere on Ω such that Ωg(|v|2)vϕdxΩf(x)h(v)ϕdxϕC01(Ω),ϕ0.(14)

Then uv in Ω.

Proof

Define A = {xΩ : u(x) > v(x)}. If we prove that A is empty, the assertion of the theorem follows. We argue by contradiction, assuming A is not empty. For ϵ > 0, define uϵ = u+ϵ and vϵ = v+ϵ. Note that we have uϵ(x) > vϵ(x) in A. Using ϕ1(x)=max[uϵα(x)vϵα(x)uϵα1(x),0]

as test function in (13) we obtain Ag(|u|2)u(uϵαvϵαuϵα1)dxAf(x)h(u)uα1(uϵαvϵα)(uuϵ)α1dx.

Similarly, using ϕ2(x)=max[uϵα(x)vϵα(x)vϵα1(x),0]

as test function in (14) we obtain Ag(|v|2)v(uϵαvϵαvϵα1)dxAf(x)h(v)vα1(uϵαvϵα)(vvϵ)α1dx.

Subtracting the latter inequality from the previous one we get A[g(|u|2)u(uϵαvϵαuϵα1)+g(|v|2)v(vϵαuϵαvϵα1)]dxAf(x)[h(u)uα1(uuϵ)α1h(v)vα1(vvϵ)α1](uϵαvϵα)dx.(15)

Since (uϵαvϵαuϵα1)=u+(α1)(vϵuϵ)αuα(vϵuϵ)α1v

and (vϵαuϵαvϵα1)=v+(α1)(uϵvϵ)αvα(uϵvϵ)α1u,

we find A[g(|u|2)u(uϵαvϵαuϵα1)+g(|v|2)v(vϵαuϵαvϵα1)]dx=A{g(|u|2)[|u|2(1+(α1)(vϵuϵ)α)α(vϵuϵ)α1uv]+g(|v|2)[|v|2(1+(α1)(uϵvϵ)α)α(uϵvϵ)α1vu]}dx.

By Lemma 3.1 with x = ∇ u, y = ∇ v and t = vϵuϵ we have g(|u|2)[|u|2(1+(α1)(vϵuϵ)α)α(vϵuϵ)α1uv]+g(|v|2)[|v|2(1+(α1)(uϵvϵ)α)α(uϵvϵ)α1vu]0.(16)

Therefore, using Fatou’s Lemma we find lim infϵ0A[g(|u|2)u(uϵαvϵαuϵα1)+g(|v|2)v(vϵαuϵαvϵα1)]dxA{g(|u|2)[|u|2(1+(α1)(vu)α)α(vu)α1uv]+g(|v|2)[|v|2(1+(α1)(uv)α)α(uv)α1vu]}dx.(17)

On the other hand, using conditions (H1) and (H2) and Lebesgue dominated theorem we find limϵ0Af(x)[h(u)uα1(uuϵ)α1h(v)vα1(vvϵ)α1](uϵαvϵα)dx=Af(x)[h(u)uα1h(v)vα1](uαvα)dx0

In view of the latter inequality and (17), from (15) as ϵ → 0 we find A{g(|u|2)[|u|2(1+(α1)(vu)α)α(vu)α1uv]+g(|v|2)[|v|2(1+(α1)(uv)α)α(uv)α1vu]}dx0.(18)

By (16), (18) and Lemma 3.1, we must have |∇ u| = |∇ v| = 0 in A. Therefore, ∇ (uv) = 0 in A and uv = 0 on A. Then, u(x) = v(x), contradicting the definition of A. The theorem follows. □

As an application of Theorem 3.2, we can show that problem (1) has a (positive) solution. Indeed, we know that there is a solution u such that u > 0 almost everywhere in Ω. Let xΩ and let B be a ball centered at x and contained in Ω. The function u satisfies div(g(|u|2)u)f_h(u)inB,u0onB,

where f is the inferior of f in B. Now, consider a radially symmetric function z such that div(g(|z|2)z)=f_h(z)inB,z=0onB.

The function z satisfies (see the last section of the present paper) rn1g((z)2)z=f_0rsn1h(z(s))ds.

Here r = |yx| for yB. It follows that z′(r) < 0 and z(x) > 0. Now we apply Theorem 3.2 with Ω = B, f = f, u = z and v = u. We find 0 < z(x) ≤ u(x). Since x is arbitrary, we have u(x) > 0 in Ω.

Corollary 3.1

Let G(s) := g(s2)s satisfy (G0), (G1) and (G2), and let h(t) satisfy (H1) and (H2). Problems (1) and (2) have a unique solution.

4 Extension of a Talenti’s result

In what follows we shall use the Hardy-Littlewood inequalities, namely Ωf(x)g(x)dx0|Ω|f(τ)g(τ)dτ,

and Ωf(x)g(x)dx0|Ω|f(τ)g(τ)dτ,

where f and g are non-negative bounded functions, f* and f* are the decreasing and, respectively, the increasing rearrangement of f, see [11].

We also use the Jensen inequality, that is J(1|Σ|Σf(σ)dσ)1|Σ|ΣJ(f(σ))dσ,(19)

where the function J is positive and convex, and f is non-negative and integrable in Σ.

To prove our next result we need a further condition on G, namely

(G3) There are y ≥ 1 and L > 0 such that J(s):=G(sy)is convex for 0<s<L.

Theorem 4.1

Let G(s) := g(s2)s satisfy (G0), (G1), (G2) and (G3), and let h(t) satisfy (H1) and (H2). If uW01,p(Ω) ∩ C1(Ω) and vW01,p(Ω) ∩ C1(Ω) are solutions to (1) and (2) respectively then we have u(x) ≤ v(x) in Ω.

Proof

For t ≥ 0, let Ω(t) = {xΩ : u(x) > t} and μ(t) = |Ω(t)|. If we set Σ(t) = {xΩ : u(x) = t} and integrate the equation in (1) over Ω(t) we find Σ(t)G(|u|)dσ=Ω(t)f(x)h(u(x))dx0μ(t)f(τ)h(u(τ))dτ.(20)

Take L large enough such that |∇ u| ≤ L in Ω and |∇ v| ≤ L in B, and let y as in condition (G3). Inequality (19) with J(s) = G(sy) and f=|u|1y yields G((1|Σ(t)|Σ(t)|u|1ydσ)y)1|Σ(t)|Σ(t)G((|u|1y)y)dσ=1|Σ(t)|Σ(t)G(|u|)dσ.(21)

With r − 1 = 1/y we find G((1|Σ(t)|Σ(t)|u|r1dσ)1r1)1|Σ(t)|Σ(t)G(|u|)dσ.(22)

Now, by Hölder inequality we have |Σ(t)|=Σ(t)|u|r1r|u|r1rdσ(Σ(t)|u|1dσ)r1r(Σ(t)|u|r1dσ)1r,

from which we find |Σ(t)|r1(Σ(t)|u|1dσ)r11|Σ(t)|Σ(t)|u|r1dσ.(23)

On the other hand, using the Federer co-area formula μ(t)=Σ(t)|u|1dσ

and the isoperimetric inequality Cn(μ(t))n1n|Σ(t)|,Cn=nωn1n,

inequality (23) yields (Cn(μ(t))n1nμ(t))r11|Σ(t)|Σ(t)|u|r1dσ,

from which we find G(Cn(μ(t))n1nμ(t))G((1|Σ(t)|Σ(t)|u|r1dσ)1r1).

As usual, we denote by ωn the measure of the unit ball in ℝn. Inserting the previous estimate into (22) we find G(Cn(μ(t))n1nμ(t))1|Σ(t)|Σ(t)G(|u|)dσ.

Using again the isoperimetric inequality, the latter inequality yields Cn(μ(t))n1nG(Cn(μ(t))n1nμ(t))Σ(t)G(|u|)dσ.(24)

Inequalities (24) and (20) yield Cn(μ(t))n1nG(Cn(μ(t))n1nμ(t))0μ(t)f(τ)h(u(τ))dτ.

Putting μ(t) = s and recalling that μ(t) is essentially the inverse function of u*(s), the latter inequality reads as Cnsn1nG(Cnsn1n(duds))0sf(τ)h(u(τ))dτ.(25)

Now, let (g(|z|2)zxi)xi=f(x)h(u)inB,z=0onB.(26)

Since the domain B and the function f(x)h(u(x)) are radially symmetric, problem (26) as a unique solution z(x) which is radially symmetric. As a consequence, all level sets of z(x) are balls. Arguing as in the previous case and observing that the inequalities are now equalities, instead of (25) we find Cnsn1nG(Cnsn1n(dzds))=0sf(τ)h(u(τ))dτ.(27)

From (25) and (27) it follows that Cnsn1nG(Cnsn1n(duds))Cnsn1nG(Cnsn1n(dzds)).(28)

Since G is strictly increasing, we find Cnsn1n(duds)Cnsn1n(dzds),

whence dudsdzds.

Integrating over (s, |B|) and recalling that u*(|B|) = z*(|B|) = 0 we find u(s)z(s),0<s<|B|,

and u(x)z(x),xB.(29)

Insertion of (29) into (26) yields (g(|z|2)zxi)xif(x)h(z)inB,z=0onB.(30)

Theorem 3.2 applied to (2) and (30) yields z(x)v(x).

The latter inequality and (29) yield the desired result. The theorem is proved. □

5 The radial case

In this section we consider the case Ω is a ball B ⊂ ℝn centered at the origin and radius R, and f(x) is a positive radial function. We write f(x) = f(r) with |x| = r. Of course, Theorem 4.1 continuous to hold. Here we prove a reverse comparison result involving the increasing rearrangement of f.

Let h : ℝ+ → ℝ+ be non-decreasing, and let g be positive and such that g(s2)s is strictly increasing for s > 0. Consider the boundary value problems (g(|v|2)vxi)xi=f(x)h(v)inB,v=0onB,(31) (g(|w|2)wxi)xi=f(x)h(w)inB,w=0onB.(32)

Here f is the Schwarz increasing rearrangement of f, that is, f(x) = f*(ωn|x|n), f*(s) = f*(|B|−s).

Theorem 5.1

Let G(s) := g(s2)s satisfy (G0), (G1) and (G2), and let h(t) satisfy (H1) and (H2). If vW01,p(Ω) ∩ C1(Ω) and wW01,p(Ω) ∩ C1(Ω) are solutions to (31) and (32) respectively then we have v(x) ≥ w(x) in B.

Proof

By Theorem 3.2, problems (31) and (32) have a unique positive solution v(x) and w(x) respectively. Hence, since B is radially symmetric, v(x) and w(x) must be radially symmetric. The equation in (31) can be rewritten as g(|v|2)Δv2gv(v)2=f(r)h(v).

Recall what we are writing f(r) = f(x) with r = |x|. We find rn1gv(n1)rn2gv2rn1gv(v)2=rn1f(r)h(v),

which can be written as (rn1gv)=rn1f(r)h(v).

Integration over (0, r) yields rn1gv=0rρn1f(ρ)h(v(ρ))dρ=1nωnB(r)f(x)h(v(x))dx,(33)

where B(r) is the ball concentric with B and radius r. By (33) we have v′(r) < 0 for 0 < r < R. Since v(R) = 0, v(r) is positive and decreasing. Moreover, (33) and the Hardy-Littlewood inequality B(r)f(x)h(v)dxB(r)f(x)h(v)dx

yield rn1gv1nωnB(r)f(x)h(v)dx.(34)

Now, let (g(|z|2)zxi)xi=f(x)h(v)inB,z=0onB.(35)

The solution z is a radial function. Arguing as in the previous case, one finds rn1gz=1nωnB(r)f(x)h(v)dx.

Inequality (34) and the latter equation yield rn1g((v)2)(v)rn1g((z)2)(z).

Since g(s2)s is strictly increasing, it follows that vz.

Integrating over (r, R) and recalling that z(R) = v(R) = 0 we find v(r) ≥ z(r) for 0 < r < R, and v(x)z(x),xB.(36)

Using the latter inequality and (35) we get (g(|z|2)zxi)xif(x)h(z)inB,z=0onB.(37)

Applying Theorem 3.2 to problems (37) and (32) we find z(x)w(x).

The latter inequality and (36) yield the desired result. The theorem is proved. □

Acknowledgement

This work was initiated when Y. Liu visited the University of Cagliari, Italy. He would like to thank Professors M. Marras, G. Porru and S. Vernier-Piro for their hospitality. Y. Liu is supported by the XJTLU’s research fund RDF-17-02-22.

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About the article

Received: 2018-09-30

Accepted: 2018-12-15

Published Online: 2019-06-01

Published in Print: 2019-03-01


Citation Information: Advances in Nonlinear Analysis, Volume 9, Issue 1, Pages 438–448, ISSN (Online) 2191-950X, ISSN (Print) 2191-9496, DOI: https://doi.org/10.1515/anona-2020-0008.

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© 2020 Y. Liu et al., published by De Gruyter. This work is licensed under the Creative Commons Attribution 4.0 Public License. BY 4.0

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