Comparison results for nonlinear divergence structure elliptic PDE ’ s

Abstract: First we prove a comparison result for a nonlinear divergence structure elliptic partial di erential equation. Next we nd 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 samemeasure as Ω. For pLaplace equations, the corresponding result is due to Giorgio Talenti. In a special (radial) case we also prove a reverse comparison result.


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 di erential 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 di erential 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 ⊂ R n 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(s )s is strictly increasing and di erentiable for s > . Let u be a solution to − g(|∇u| )ux i x i = f (x)h(u), u > in Ω, u = on ∂Ω.
Here and in what follows, the summation convention over repeated indices from to n is in e ect. If B ⊂ R n 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 When Under suitable conditions on the function h, the answer is positive for the p-Laplacian, where g(s ) = s p− , p > . Recently, inequality (3) has been proved for the (p, q)-Laplacian, where g(s ) = s p− + s q− , p > q > , 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 nd conditions on g and h which ensure existence and uniqueness for problems (1) and (2). With G(s) := g(s )s we assume: (G1) G(s) is continuous for s ≥ , is strictly increasing and continuously di erentiable for s > .
(G2) With q as in (G0), the function G(s) s q− is non-decreasing for s > .
(H1) h(t) is a positive non-decreasing function for t > .
(H2) There is < α < q such that h(t)t −α is bounded and non-increasing for t > . Here q is the same as in (G0).
We note that conditions (G0), (G1) and (G2) hold for a wide class of equations including the p-Laplacian and the (p,q)-Laplacian.

Existence of positive solutions
Assuming condition (G0), the natural space for solutions to problem (1) is the Sobolev space W ,p (Ω). The equation in (1) is the Euler equation of the functional It is well-known that a function u that minimizes I(w) for w ∈ W ,p (Ω), w ≥ , is a solution to the equation in (1) with u ≥ . We claim that, under conditions (G1), (G2), (H1) and (H2), a minimum for I(w) cannot be zero in any ball B ⊂ Ω. Indeed, arguing by contradiction, letw be a minimum vanishing on some ball B. De ne z =w + ϵϕ, where < ϵ < and ϕ ∈ C (B) is a positive function in B and vanishing on Ω \ B. We have We rst observe that conditions (G2) and (H2) imply, for < ϵ < and τ > , Putting s = ϵτ with < ϵ < and using these inequalities we nd Since q > α, it is clear that I(z) < I(w) for ϵ small enough. The claim follows. Therefore, we may assume that there exists a solution to (1) with u > almost everywhere in Ω. At the end of the next Section, we will prove that such a solution u is positive in Ω.
Proof. Recall the generalized Young's inequality where φ(τ) is the inverse function of G(τ). Replacing x by g(|x| )t α− x we nd Similarly, we have Replacing y by g(|y| )t −α y in the latter inequality we nd In view of (6) and (7), inequality (4) holds provided In case of t = we have Putting τ = G(s) we nd Similarly, we nd Insertion of (10) and (11) into (9) yields Ψ( ) = . Hence, to prove that Ψ(t) > for < t < when |x| + |y| > , it is enough to prove that Ψ (t) < . Since we must show that Let us prove that the left hand side of (12) is negative when |x| > . Indeed, if |x| > , the inequality The latter inequality can be rewritten as which holds by Remark 1.1. Now, let us prove that the right hand side of (12) is positive when |y| > , that is Let us write this inequality as which can be rewritten as This inequality is equivalent to the following which holds by Remark 1.1. Hence, inequality (12) holds when |x| + |y| > . It follows that also (8) and (4) hold in a strict sense for < t < . The lemma is proved.
Let v ∈ C (Ω), v ≥ on ∂Ω and v > everywhere on Ω such that Then u ≤ v in Ω.
If we prove that A is empty, the assertion of the theorem follows. We argue by contradiction, assuming A is not empty. For ϵ > , de ne uϵ = u + ϵ and vϵ = v + ϵ. Note that we , as test function in (13) we obtain Similarly, using , as test function in (14) we obtain Subtracting the latter inequality from the previous one we get Since By Lemma 3.1 with x = ∇u, y = ∇v and t = vϵ uϵ we have Therefore, using Fatou's Lemma we nd On the other hand, using conditions (H1) and (H2) and Lebesgue dominated theorem we nd In view of the latter inequality and (17), from (15) as ϵ → we nd  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 > almost everywhere in Ω. Let x ∈ Ω and let B be a ball centered at x and contained in Ω. The function u satis es where f is the inferior of f in B. Now, consider a radially symmetric function z such that The function z satis es (see the last section of the present paper) −r n− g((z ) )z = f r s n− h(z(s)) ds.
Here r = |y − x| for y ∈ B. It follows that z (r) < and z(x) > . Now we apply Theorem 3.2 with Ω = B, f = f , u = z and v = u. We nd < z(x) ≤ u(x). Since x is arbitrary, we have u(x) > in Ω.

Extension of a Talenti's result
In what follows we shall use the Hardy-Littlewood inequalities, namely and 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 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 γ ≥ and L > such that  (1) over Ω(t) we nd Take L large enough such that |∇u| ≤ L in Ω and |∇v| ≤ L in B, and let γ as in condition (G3). Inequality (19) with J(s) = G(s γ ) and f = |∇u| γ yields With r − = /γ we nd Now, by Hölder inequality we have from which we nd Integrating over (s, |B|) and recalling that u * (|B|) = z * (|B|) = we nd u * (s) ≤ z * (s), < s < |B|, Insertion of (29) into (26) yields Theorem 3.2 applied to (2) and (30) yields The latter inequality and (29) yield the desired result. The theorem is proved.

The radial case
In this section we consider the case Ω is a ball B ⊂ R n 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(s )s is strictly increasing for s > . Consider the boundary value problems Here f is the Schwarz increasing rearrangement of f , that is, f (x) = f * (ωn|x| n ), f * (s) = f * (|B| − s). 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 Recall what we are writing f (r) = f (x) with r = |x|. We nd −r n− gv − (n − )r n− gv − r n− g v (v ) = r n− f (r)h(v), which can be written as − r n− gv = r n− f (r)h(v).
Integration over ( , r) yields −r n− gv = r ρ n− f (ρ)h(v(ρ))dρ = nωn B(r) f (x)h(v(x))dx, where B(r) is the ball concentric with B and radius r. By (33) we have v (r) < for < r < R. Since v(R) = , v(r) is positive and decreasing. Moreover, (33) and the Hardy-Littlewood inequality The solution z is a radial function. Arguing as in the previous case, one nds −r n− gz = nωn B(r) f (x)h(v)dx.
Since g(s )s is strictly increasing, it follows that −v ≥ −z .
Integrating over (r, R) and recalling that z(R) = v(R) = we nd v(r) ≥ z(r) for < r < R, and Using the latter inequality and (35) we get − g(|∇z| )zx i x i ≥ f (x)h(z) in B, z = on ∂B.
The latter inequality and (36) yield the desired result. The theorem is proved.