Abstract
In the last decades, comparison results of Talenti type for Elliptic Problems with Dirichlet boundary conditions have been widely investigated. In this paper, we generalize the results obtained in Alvino et al. (Commun Pure Appl Math, to appear) to the case of p-Laplace operator with Robin boundary conditions. The point-wise comparison, obtained in Alvino et al. (to appear) only in the planar case, holds true in any dimension if p is sufficiently small.
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1 Introduction
Let \(\beta\) be a positive parameter and let \(\Omega\) be a bounded open set of \({\mathbb {R}}^n\), \(n\ge 2\), with Lipschitz boundary.
Let \(f\in L^{p'}(\Omega )\) be a non-negative function. We consider the following problem
A function \(u \in W^{1,p}(\Omega )\) is a weak solution to (1) if
We want to establish a comparison principle with the solution to the following symmetrized problem
where \(\Omega ^\sharp\) is the ball centered in the origin with the same measure of \(\Omega\) and \(f^\sharp\) is the Schwarz rearrangement of f (see next section for its definition).
This kind of problems has been widely investigated in the last decades. The first step is contained in [10], where Talenti proved a pointwise comparison result between \(u^{\sharp }\) and v in the case of Dirichlet Laplacian. After this, several papers generalized the result of Talenti: for instance, the one by Talenti himself [11], in which the operator is a generic non-linear operator in divergence form, or the one by Alvino, Lions and Trombetti [2] in which the authors deal with both elliptic and parabolic cases: in both papers, Dirichlet boundary conditions are considered.
Different kinds of boundary conditions are considered by Alvino, Nitsch and Trombetti in [3], where they establish a comparison between a suitable norm of u and v, respectively, solution to
They found out that if f is a non-negative function in \(L^2(\Omega )\), then
where \({\left\| \cdot \right\| }_{L^{k,q}}\) is the so-called Lorentz norm, whose definition can be found in next section. Moreover, the authors in [3] were able to establish a comparison á la Talenti,
in the case \(f\equiv 1\) and \(n=2\). This will be the starting point of our work: it will be clear that our results coincide with the one in [3] in the case \(p=2\).
For completeness sake, we cite that this wasn’t the first result in this sense, indeed in [6] the authors study a comparison result for the p-torsion, that is the case \(f \equiv 1\), with a completely different argument, obtaining
Another work that is worth to be mentioned is [1], where the authors obtained similar results to [3] in the case of mixed Dirichlet and Robin boundary conditions.
This paper is organized as follows. In the next section, we give some basic notions about rearrangements of functions and Lorentz spaces. Moreover, we list some properties of the solutions to problems (1) and (3). In Sect. 3, we prove the main results about comparison of the two solutions in terms of the Lorentz norm.
In particular, we prove
Theorem 1.1
Let u and v be the solutions to problem (1) and (3), respectively. Then, we have
We observe that from Theorem 1.1, we have that, if \(p \ge n\)
Theorem 1.2
Assume that \(f\equiv 1\) and let u and v be the solutions to (1) and (3), respectively.
-
(i)
If \(\displaystyle {1 \le p \le \frac{n}{n-1}}\) then
$$\begin{aligned} u^\sharp (x) \le v(x) \qquad x \in \Omega ^\sharp , \end{aligned}$$(6) -
(ii)
if \(\displaystyle { p > \frac{n}{n-1}}\) and \(\displaystyle {0 <k \le \frac{n(p-1)}{n(p-1)-p}}\) , then
$$\begin{aligned} \begin{aligned} {\left\| u\right\| }_{L^{k,1}(\Omega )}&\le {\left\| v\right\| }_{L^{k,1}\left( \Omega ^{\sharp }\right) } \\ {\left\| u\right\| }_{L^{pk,p}(\Omega )}&\le {\left\| v\right\| }_{L^{pk,p}\left( \Omega ^{\sharp }\right) } . \end{aligned} \end{aligned}$$(7)
Then, we explicitly observe that in the case \(f\equiv 1\) from Theorem 1.2, we have that
while we have the point-wise comparison only for \(\displaystyle {p \le \frac{n}{n-1}}\).
In Sect. 4, using tools from Sect. 3, we give a new proof of the Faber–Krahn inequality with Robin boundary conditions in the case \(p \ge n\). This topic was already studied in the papers by Bucur, Giacomini, Daners and Trebeschi, [4,5,6] and [7] where the authors proved the Faber–Krahn inequality for the eigenvalues of the Laplacian, or of the p-Laplacian, with Robin boundary conditions, for every \(p >1\). Actually, the results in [4] are more general, since they hold for every \(p>1\), but they are obtained with completely different tools than the ones contained in our paper.
Finally, in Sect. 5, we provide some examples and open problems, and we discuss the optimality of our results.
2 Notions and preliminaries
Definition 2.1
Let \(u: \Omega \rightarrow {\mathbb {R}}\) be a measurable function, the distribution function of u is the function \(\mu : [0,+\infty [\, \rightarrow [0, +\infty [\) defined by
Here, and in the whole paper, \({\left| A\right| }\) stands for the n-dimensional Lebesgue measure of the set A.
Definition 2.2
Let \(u: \Omega \rightarrow {\mathbb {R}}\) be a measurable function, the decreasing rearrangement of u, denoted by \(u^*\), is the distribution function of \(\mu\).
The Schwarz rearrangement of u is the function \(u^\sharp\) whose level sets are balls with the same measure as the level sets of u. The functions \(u^\sharp\) and \(u^*\) are linked by the relation
It is easily checked that u, \(u^*\) e \(u^\sharp\) are equi-distributed, so it follows that
An important propriety of the decreasing rearrangement is the Hardy–Littlewood inequaliy, that is
So, by choosing \(h=\chi _{\left\{ {\left| u\right| }>t\right\} }\), one has
Definition 2.3
Let \(0<p<+\infty\) and \(0<q\le +\infty\). The Lorentz space \(L^{p,q}(\Omega )\) is the space of those functions such that the quantity:
is finite.
Let us observe that for \(p=q\), the Lorentz space coincides with the \(L^p\) space, as a consequence of the well known Cavalieri’s Principle
See [12] for more details on Lorentz space.
Let us consider the functional
defined on \(W^{1,p}(\Omega )\). This functional is well defined and its Euler–Lagrange equation is exactly (1). If we show that the functional admits a minimum, our problem will always have a solution.
-
(1)
Let us show that the functional is bounded from below, indeed using the parametric Young inequality, we have
$$\begin{aligned} \begin{aligned} {\mathfrak {F}}(u)&\ge \frac{1}{p} \int _{\Omega } {\left| \nabla u\right| }^p \, \mathrm{d}x + \frac{\beta }{p}\int _{\partial \Omega } {\left| u\right| }^p \, \mathrm{d}{\mathcal {H}}^{n-1}(x) - \frac{\varepsilon ^p}{p} \int _{\Omega } {\left| u\right| }^p \, \mathrm{d}x - \frac{1}{p' \varepsilon ^{p'}}\int _{\Omega } {\left| f\right| }^{p'} \, \mathrm{d}x\\&\ge \frac{1}{p} \left( \int _{\Omega } {\left| \nabla u\right| }^p \, \mathrm{d}x+ \beta \int _{\partial \Omega } {\left| u\right| }^p\, \mathrm{d}{\mathcal {H}}^{n-1}(x)\right) - \frac{\varepsilon ^p}{p} \int _{\Omega } {\left| u\right| }^p \, \mathrm{d}x -\frac{1}{p' \varepsilon ^{p'}}\int _{\Omega } {\left| f\right| }^{p'} \, \mathrm{d}x\\&\ge \frac{ \lambda _{1,\beta }(\Omega )- \varepsilon ^p}{p} \int _{\Omega } {\left| u\right| }^p \, \mathrm{d}x - \frac{1}{p' \varepsilon ^{p'}}\int _{\Omega } {\left| f\right| }^{p'} \, \mathrm{d}x \end{aligned} \end{aligned}$$In the last inequality, we used the Sobolev inequality with trace term
$$\begin{aligned} \int _\Omega {\left| \nabla u\right| }^p +\beta \int _{\partial \Omega } {\left| u\right| }^p \ge \lambda _{1,\beta }(\Omega ) \int _\Omega {\left| u\right| }^p. \end{aligned}$$In general, the quantity \(\lambda _{1,\beta }(\Omega )\) denotes the first eigenvalue of the p-Laplacian with Robin boundary conditions, whose definition is given in (28), which can be also seen as a trace constant of the set \(\Omega\).
If \(\varepsilon\) is small enough, then the quantity
$$\begin{aligned} \frac{ \lambda _{1,\beta }(\Omega )- \varepsilon ^p}{p} \end{aligned}$$is non negative, and then
$$\begin{aligned} {\mathfrak {F}}(u)\ge - \frac{1}{p' \varepsilon ^{p'}}\int _{\Omega } {\left| f\right| }^{p'} \end{aligned}$$so
$$\begin{aligned} m=\inf _{W^{1,p}} {\mathfrak {F}}(u)> - \infty . \end{aligned}$$ -
(2)
Compactness and lower semicontinuity.
Let \(\left\{ u_i\right\}\) be a minimizing sequence. We can assume that \({\mathfrak {F}}(u_i)\le m+1\), \(\forall i\). Using again the Young inequality, we have
$$\begin{aligned} \begin{aligned} m+1&\ge \frac{1}{p} \int _{\Omega } {\left| \nabla u_i\right| }^p \, \mathrm{d}x + \frac{\beta }{p}\int _{\partial \Omega } {\left| u_i\right| }^p \, \mathrm{d}{\mathcal {H}}^{n-1}(x)- \int _{\Omega } fu_i \, \mathrm{d}x \\&\ge \frac{1}{p} \int _{\Omega } {\left| \nabla u_i\right| }^p \, \mathrm{d}x + \frac{\beta }{p}\int _{\partial \Omega } {\left| u_i\right| }^p \, \mathrm{d}{\mathcal {H}}^{n-1}(x)- \frac{\varepsilon ^p}{p}\int _{\Omega } {\left| u_i\right| }^p\, \mathrm{d}x- \frac{1}{p' \varepsilon ^{p'}}\int _{\Omega } {\left| f\right| }^{p'} \, \mathrm{d}x \end{aligned} \end{aligned}$$Then
$$\begin{aligned} \begin{aligned} m+1 + \frac{1}{p' \varepsilon ^{p'}}\int _{\Omega } {\left| f\right| }^{p'} \, \mathrm{d}x&\ge \frac{1}{2p} \left( \int _{\Omega } {\left| \nabla u_i\right| }^p \, \mathrm{d}x +\beta \int _{\partial \Omega } {\left| u_i\right| }^p \, \mathrm{d}{\mathcal {H}}^{n-1}(x)\right) -\frac{\varepsilon ^p}{p}\int _{\Omega } {\left| u_i\right| }^p\, \mathrm{d}x \\&+\frac{1}{2p} \left( \int _{\Omega } {\left| \nabla u_i\right| }^p \, \mathrm{d}x +\beta \int _{\partial \Omega } {\left| u_i\right| }^p \, \mathrm{d}{\mathcal {H}}^{n-1}(x)\right) \\&\ge \frac{1}{2p} \int _{\Omega } {\left| \nabla u_i\right| }^p \, \mathrm{d}x + \left( \frac{\lambda _{1,\beta }(\Omega )- 2 \varepsilon ^p}{2p}\right) \int _{\Omega } {\left| u_i\right| }^p\, \mathrm{d}x. \end{aligned} \end{aligned}$$Then, the minimizing sequence \(\left\{ u_i \right\}\) is bounded in \(W^{1,p}(\Omega )\), so there exists a subsequence \(\left\{ u_{i_k}\right\}\) weakly converging in \(W^{1,p}(\Omega )\) and strongly in \(L^p(\Omega )\) to a function u. Let us show that u is the minimum.
The function \(t^p\) is strictly convex for \(p>1\), so
$$\begin{aligned} {\left| u_{i_k}\right| }^p\ge & {} {\left| u\right| }^p + p {\left| u\right| }^{p-2} u\left( u_{i_k}-u\right) \end{aligned}$$(8)$$\begin{aligned} {\left| \nabla u_{i_k}\right| }^p\ge & {} {\left| \nabla u\right| }^p + p {\left| \nabla u\right| }^{p-2} \nabla u\left( \nabla u_{i_k}-\nabla u\right) \end{aligned}$$(9)Putting (8) e (9) in \({\mathfrak {F}}(u_{i_k})\), we obtain
$$\begin{aligned} \begin{aligned} \int _{\Omega } f u_{i_k} \, \mathrm{d}x + {\mathfrak {F}}\left( u_{i_k}\right)&\ge \frac{1}{p}\int _{\Omega }{\left| \nabla u\right| }^p \, \mathrm{d}x + \int _{\Omega } {\left| \nabla u\right| }^{p-2} \nabla u\left( \nabla u_{i_k}-\nabla u\right) \\&+\frac{\beta }{p} \int _{\partial \Omega } {\left| u\right| }^p \, \mathrm{d}{\mathcal {H}}^{n-1}(x) + \beta \int _{\partial \Omega } {\left| u\right| }^{p-2} u\left( u_{i_k}-u\right) \, \mathrm{d}{\mathcal {H}}^{n-1} \end{aligned} \end{aligned}$$Passing to the limit for \(k\rightarrow \infty\), by the weak convergence of \(\left\{ u_{i_k}\right\}\) the integral over \(\Omega\) on the right-hand side goes to 0. The integral over \(\partial \Omega\) goes to 0 as well. Indeed, the space \(W^{1,p}(\Omega )\) is compactly embedded in \(L^p(\partial \Omega )\) (for more details, see [9] 2.5), and \(u_{i_k}-u\rightarrow 0\) in \(L^p(\partial \Omega )\). So, we obtain
$$\begin{aligned} m \ge {\mathfrak {F}}(u). \end{aligned}$$This ensures us that u is the minimum of the functional.
The uniqueness of the minimum follows from the fact that \({\mathfrak {F}}(u)\) is the sum of a strictly convex part and a linear part.
We observe that the solutions u and v to (1) and (3), respectively, are both p-superharmonic and then, by the strong maximum principle in [13], it follows that they achieve their minima on the boundary. Denoting by \(u_m\) and \(v_m\) the minimum of u and v, respectively, thanks to the positiveness of \(\beta\) and the Robin boundary conditions, we have that \(u_m \ge 0\) and \(v_m \ge 0\). Hence, u and v are strictly positive in the interior of \(\Omega\). Moreover, we can observe that
in fact, if we consider
A consequence of (10) that will be used in what follows is that
2.1 Useful lemmas
Let u be the solution to (1). For \(t\ge 0\), we denote by
and by
where \({\left| \cdot \right| }\) is the Lebesgue measure on \({\mathbb {R}}^n\) and \(Per(\cdot )\) is the perimeter.
If v is the solution to (3), using the same notations, we set
Because of the invariance of the p-Laplacian and of the Schwarz rearrangement of f by rotation, there exists a radial solution to (3) and, by uniqueness of solutions, this solution is v.
Since v is radial, positive and decreasing along the radius, then for \(0\le t\le v_m\), \(V_t=\Omega ^\sharp\), while for \(v_m<t<\max _{\Omega ^\sharp }v\), \(V_t\) is a ball, concentric to \(\Omega ^\sharp\) and strictly contained in it.
Lemma 2.1
(Gronwall) Let \(\xi (t): [\tau _0 , + \infty [ \,\rightarrow {\mathbb {R}}\) be a continuous and differentiable function satisfying, for some non-negative constant C, the following differential inequality
Then, we have
-
(i)
\(\displaystyle { \xi (\tau ) \le \left( \xi (\tau _0) + \frac{C}{p-1}\right) \left( \frac{\tau }{\tau _0}\right) ^{p-1} - \frac{C}{p-1} \quad \forall \tau \ge \tau _0}\);
-
(ii)
\(\displaystyle { \xi '(\tau ) \le \left( \frac{(p-1)\xi (\tau _0 )+ C}{\tau _0}\right) \left( \frac{\tau }{\tau _0}\right) ^{p-2} \quad \forall \tau \ge \tau _0}\).
Proof
Dividing both sides of the differential inequality by \(\tau ^p\), we obtain
Now, we integrate from \(\tau _0\) to \(\tau\) and we obtain
which gives (i).
In order to obtain (ii), we just take into account (i) in the differential inequality. \(\square\)
Lemma 2.2
Let u and v be solutions to (1) and (3), respectively. Then for almost every \(t >0\), we have
where \(\displaystyle {\gamma _n= \left( n \omega _n^{1/n}\right) ^{\frac{p}{p-1}}}\).
And
Proof
Let \(t >0\) e \(h >0\), we choose the test function
Then,
Dividing by h, using coarea formula and letting h go to 0, we have that for a. e. \(t>0\)
where
So, using the isoperimetric inequality, for a. e. \(t>0\), we have
Then (12) follows. We notice that if v is the solution to (3), than all the inequalities are verified as equalities, so we have (13). \(\square\)
Lemma 2.3
For all \(\tau \ge v_m\), we have
Moreover,
Proof
If we integrate the quantity
from 0 to \(+\infty\), by Fubini theorem, we obtain
where the last equality follows from the fact that u solves (1).
Analogously
Since u is positive, we obtain, \(\forall t\ge 0\),
on the other hand, since \(\partial V_t \cap \partial \Omega ^\sharp\) is empty for \(t\ge v_m\), we have
and the proof of Lemma 2.3 is complete. \(\square\)
Remark 2.1
It can be observed that, since \(\partial V_t \cap \partial \Omega ^\sharp\) is empty for \(t\ge v_m\) and \(\phi (t ) = {\left| \Omega \right| }\) for \(t \le v_m\), for all \(\delta >0\) and for all t, we have
3 Main results
Now, we prove Theorems 1.1 and 1.2.
Proof of Theorem 1.1
Let \(\displaystyle {0< k \le \frac{n(p-1)}{p(n-1)}}\), so \(\displaystyle {\delta = \frac{1}{k} -\frac{(n-1)p}{n(p-1)}}\) is positive.
Multiplying (12) by \(t^{p-1} \mu (t)^\delta\) and integrating from 0 to \(\tau \ge v_m\), by the previous Lemma, we obtain
Setting \(\displaystyle {F(l)= \int _0^l \omega ^\delta \left( \int _0^\omega f^*(s) \, \mathrm{d}s \right) ^{\frac{1}{p-1}} \, \mathrm{d}\omega }\), we can integrate by parts both sides of the last inequality, getting
Setting \(\displaystyle {\xi (\tau )= \int _0^\tau t^{p-2} \left( \int _0^t \gamma _n \mu (s)^{\frac{1}{k}}\, \mathrm{d}s+ F(\mu (t))\right) \, \mathrm{d}t}\) and \(\displaystyle {C=\frac{{\left| \Omega \right| }^\delta }{p\beta ^{\frac{p}{p-1}}} \left( \int _0^{{\left| \Omega \right| }}f^*(s ) \, \mathrm{d}s\right) ^{\frac{p}{p-1}}}\), we are in the hypothesis of Lemma 2.1 (Gronwall), namely
so, choosing \(\tau _0= v_m\), we have
where
The previous inequality becomes an equality if we replace \(\mu (t)\) with \(\phi (t)\). Since \(\mu (t) \le \phi (t)={\left| \Omega \right| }, \quad \forall t \le v_m\), and F(l) is monotone, we obtain
hence
Passing to the limit as \(\tau \rightarrow \infty\), we get
and hence
To prove the inequality (5), it is enough to show that
Let us consider equation (16), let us integrate by parts the first term on the right-hand side from 0 to \(\tau\) and then let us pass to the limit as \(\tau \rightarrow \infty\), we have
Therefore, if we show that
we obtain the (17). To this aim, we multiply (12) by \(\displaystyle {t^{p-1}F(\mu (t)) \mu (t)^{-\frac{(n-1)p}{n(p-1)} }}\) and integrate. First, we observe that, by the choice \(\displaystyle {k \le \frac{n(p-1)}{(n-2)p+n}}\), it follows that the function \(\displaystyle {h(l)= F(l)l^{-\frac{(n-1)p}{n(p-1)}}}\) is non decreasing. Hence, we obtain
If we integrate by parts both sides of the last expression and set
we obtain
where
Setting
then (19) becomes
So lemma 2.1, with \(\tau _0=v_m\), gives
Of course, the inequality holds as an equality if we replace \(\mu (t)\) with \(\phi (t)\), so we get, keeping in mind that \(\mu (t)\le \phi (t)= {\left| \Omega \right| }\) for \(t \le v_m\),
Letting \(\tau \rightarrow \infty\), one has
as \(H_\mu (\tau ), H_\phi (\tau ) \rightarrow 0\). This proves (18), and hence (5).
The fact that both \(H_\mu\) and \(H_\phi\) go to 0 as \(\tau\) goes to infinity can be easily deduced distinguishing the cases.
-
If \(p \ge 2\)
$$\begin{aligned} t^{p-2} \mu (t)&=\int _{u>t} t^{p-2} \, \mathrm{d}x \le \int _{u>t} u^{p-2} \, \mathrm{d}x \le {\left\| u\right\| }_{L^p}^{p-2} \mu (t)^{\frac{2}{p}} \\ \Rightarrow {\left| H_\mu (\tau )\right| }&=\int _{\tau }^{+\infty } t^{p-2} F(\mu (t)) \mu (t)^{-\frac{p(n-1)}{n(p-1)}} \left( \int _0^{\mu (t)} f^*(s) \, \mathrm{d}s \right) (-\mu '(t)) \, \mathrm{d}t \\&\le \left( \int _0^{{\left| \Omega \right| }} f^*(s) \, \mathrm{d}s\right) {\left\| u\right\| }_{L^p}^{p-2} \int _{\tau }^{+\infty } F(\mu (t)) \mu (t)^{\frac{2}{p}-\frac{p(n-1)}{n(p-1)}-1} (-\mu '(t)) \, \mathrm{d}t \xrightarrow {\tau \rightarrow +\infty } 0. \end{aligned}$$ -
If \(p <2\)
$$\begin{aligned} {\left| H_\mu (\tau )\right| }&=\int _{\tau }^{+\infty } t^{p-2} F(\mu (t)) \mu (t)^{-\frac{p(n-1)}{n(p-1)}} \left( \int _0^{\mu (t)} f^*(s) \, s \right) (-\mu '(t)) \, \mathrm{d}t \\&\le \tau ^{p-2} \int _{\tau }^{+\infty } F(\mu (t)) \mu (t)^{-\frac{p(n-1)}{n(p-1)}} \left( \int _0^{\mu (t)} f^*(s) \, s \right) (-\mu '(t)) \, \mathrm{d}t \xrightarrow {\tau \rightarrow +\infty } 0. \end{aligned}$$and analogously for \(H_\phi\), which concludes the proof.\(\square\)
Proof of Theorem 1.2
- (i):
-
Firstly, we observe that \(\displaystyle {\int _0^{\mu (t)} f^*(s) \, \mathrm{d}s= \mu (t)}\), so (12) becomes
$$\begin{aligned} \gamma _n \mu (t)^{\left( 1-\frac{1}{n}-\frac{1}{p}\right) \frac{p}{p-1}} \le - \mu '(t) + \frac{1}{\beta ^{\frac{1}{p-1}}}\int _{\partial U_t^\text {ext}} \frac{1}{u} \, \mathrm{d}{\mathcal {H}}^{n-1}(x). \end{aligned}$$(20)Let us multiply both sides by \(t^{p-1} \mu (t)^\delta\), where \(\delta = -\left( 1- \frac{1}{n}-\frac{1}{p}\right) \frac{p}{p-1}\). We point out that \(\delta \ge 0\) for \(p \le \frac{n}{n-1}\). Hence, integrating from 0 to \(\tau \ge v_m\), we have
$$\begin{aligned} \begin{aligned} \int _0^\tau \gamma _n t^{p-1}&\le \int _0^\tau t^{p-1} \mu (t)^\delta (- \mu '(t)) \, \mathrm{d}t + \frac{1}{\beta ^{ \frac{1}{p-1}}}\int _0^\tau t^{p-1} \mu (t)^\delta \int _{\partial U_t^\text {ext}} \frac{1}{u} \, \mathrm{d}{\mathcal {H}}^{n-1}(x) \\&\le \int _0^\tau t^{p-1} \mu (t)^\delta (- \mu '(t)) \, \mathrm{d}t + \frac{{\left| \Omega \right| }^{\delta +1}}{p \beta ^{\frac{p}{p-1}}} \end{aligned} \end{aligned}$$(21)Taking into account Remark 2.1, if we replace \(\mu (t)\) with \(\phi (t)\) the previous inequality holds as equality.
Hence, we get
Then, an integration by parts gives
Finally, using Gronwall’s Lemma with the function \(\displaystyle {\xi (\tau )=\int _0^\tau s^{p-2}\left( \frac{\mu (s)^{\delta + 1}-\phi (s)^{\delta + 1}}{\delta +1}\right) \, \mathrm{d}s}\) we obtain
The quantity on the right-hand side is non-positive, thanks to (10), so
and, remembering that,
we get the point-wise inequality of the functions.
-
(ii)
Now, we want to show that
$$\begin{aligned} {\left\| u\right\| }_{L^{k,1}(\Omega )} \le {\left\| v\right\| }_{L^{k,1}\left( \Omega ^{\sharp }\right) } \end{aligned}$$so it is enough to show
$$\begin{aligned} \int _0^{+\infty } \mu (t)^{\frac{1}{k}} \, \mathrm{d}t \le \int _0^{+\infty } \phi (t)^{\frac{1}{k}} \, \mathrm{d}t \end{aligned}$$(22)
We multiply (20) by \(t^{p-1} \mu (t)^{\frac{1}{k}-\left( 1- \frac{1}{n}-\frac{1}{p}\right) \frac{p}{p-1}}\) and integrate from 0 to \(\tau \ge v_m\). Then using Lemma 2.3 and Remark 2.1, we obtain
and equality holds if we replace \(\mu\) with \(\phi\). In order to be shorter, we set
We point out that (23) follows by (21) if \(\eta \ge 0\), namely
With these notations and keeping in mind that \(\mu\) is a non increasing function, we have from (23) taht
Let us set \(\displaystyle {G(\ell )=\int _0^{\ell } w^{\eta } \, dw=\frac{\ell ^{\eta +1}}{\eta +1}}\), let us integrate by parts both sides of (24) in order to obtain
Setting
(25) reads as follows
Hence, using Gronwall’s Lemma 2.1 with \(\tau _0=v_m\), we get
where
Again, if we replace \(\mu\) with \(\phi\), the previous inequality holds as an equality and \(\xi (v_m)\) is less or equal than the same quantity obtained by replacing \(\mu\) with \(\phi\), as (10) holds. Keeping in mind (11), we have
Passing to the limit as \(\tau \rightarrow + \infty\), we get
namely (22).
To conclude the proof, we have to show that
that is to say
We consider (24), pass to the limit as \(\tau \rightarrow +\infty\) and integrate by parts the first term on the right-hand side
Hence, it is enough to show that
To this aim, we multiply (20) by \(t^{p-1} G(\mu (t)) \mu (t)^{-\left( 1-\frac{1}{n}-\frac{1}{p}\right) \frac{p}{p-1}}\) and integrate from 0 to \(\tau \ge v_m\)
Since \(\displaystyle {k \le \frac{n(p-1)}{n(p-1)-p}}\), using Lemma 2.3 and the fact that the function \(G(\ell )\ell ^{-\left( 1-\frac{1}{n}-\frac{1}{p}\right) \frac{p}{p-1}}\) is non-decreasing, we obtain
with
If we replace \(\mu\) with \(\phi\) the previous inequality holds as an equality, thanks to (2.1). Now, let us integrate by parts both sides of (26), obtaining
where
Setting
the (27) reads as follows
Again, using Gronwall’s Lemma 2.1, we get
with
Keeping in mind that for \(\phi\) the previous inequalities hold as equality and the fact that G is not decreasing, \(\xi (v_m)\) is less or equal to the same quantity obtained by replacing \(\mu\) with \(\phi\). Hence, we obtain
and passing to the limit as \(\tau \rightarrow +\infty\), we finally get
indeed, as in the proof of Theorem 1.1, \(H_\mu (\tau )\) and \(H_\phi (\tau )\) go to 0 as \(\tau \rightarrow \infty\).
That concludes the proof. \(\square\)
Corollary 3.1
Let u and v be the solutions to (1) and (3), respectively. Then, if \(p \ge n\), we have
Moreover, in the case \(f \equiv 1\), Theorem 1.2 gives
and the point-wise comparison for \(\displaystyle {p \le \frac{n}{n-1}}\).
Proof
If \(p \ge n\) the upper bounds of k, in both cases (4) e (5), are greater than 1 and so we can choose \(k=1\). The assertion follows from the fact that
Analogously if \(f \equiv 1\). \(\square\)
4 Faber–Krahn inequality
We recall that the first eigenvalue of p-Laplace operator with Robin boundary conditions is obtained as the minimum of the Rayleigh quotients, i.e.,
We can observe that if u achieves the minimum of Rayleigh quotients, so does \({\left| u\right| }\). From this we have that u in non-negative. Furthermore, we have if \(u_{\lambda _{1,\beta}} \ge 0\), as a consequence of Harnack inequality, \(u_{\lambda _{1,\beta}} >0\).
Another important thing is that the eigenvalue is simple. Indeed, as shown in [8], if \(\Omega\) is smooth enough and u and v are to eigenfunctions referred to the first eigenvalue, we can choose as test function \(\displaystyle {\varphi _1 = \frac{u^p - v^p}{ u^{p-1}}}\) in
and \(\displaystyle {\varphi _2 = \frac{v^p - u^p}{ v^{p-1}}}\) in
Summing the two equations, we have
Now, using the well known inequalities, which hold true for each \(w_1\) and \(w_2\in {\mathbb {R}}^n\) ,
if we consider the case \(p \ge 2\), we choose \(w_2= \nabla \log u\) and \(w_1= \nabla \log v\), we obtain
Hence, we obtain that \(v \nabla u = u \nabla v\) a.e. in \(\Omega\), and so there exists a constant K for which \(u=Kv\). This means that \(\lambda _{1,\beta}\) is simple.
For the proof of (29), we refer to [8].
The following corollary of Theorem 1.1 holds true, that is Faber–Krahn inequality.
Corollary 4.1
Let u and v be the solutions to (1) and (3), respectively. Then , if \(p \ge n\), we have
Proof
Let u an eigenfunction referred to the first eigenvalue of (1), then it solves
Now, let z be a solution to the following problem
In that case, Corollary 3.1 gives
and hence, by Hölder inequality
Therefore, observing that we can write the eigenvalue \(\lambda _{1,\beta }(\Omega )\) in the following way, we obtain
\(\square\)
5 Conclusions
We have been able to extend the results obtained for the Laplacian to the p-Laplacian. Many problems remain open, such as
Open Problem In the assumptions of Theorem 1.2, does the point-wise comparison hold also for \(p>\frac{n}{n-1}\)?
We have already observed in the Corollary 3.1 that if \(p\ge n\) we have an estimate on the \(L^p\) norms of u and v. Can we generalize this estimate also for \(q\ne p\)? We know for sure that for \(q=\infty\) this can’t be done, as it can be seen in the following example.
Example 5.1
Let \(\Omega \subseteq {\mathbb {R}}^n\) be the union of two disjoint balls, \(B_1\) and \(B_r\) with radii 1 and r, respectively. We choose \(\displaystyle {\beta < \left( \frac{n-1}{p-1} \right) ^{p-1} }\) with \(p\ne n\), and we fix \(f=1\) on \(B_1\) and \(f=0\) on \(B_r\). Both u and v can be explicitly computed. We have \({\left\| u\right\| }_\infty - {\left\| v\right\| }_\infty = C r^n + o(r^n)\), where C is a positive constant.
Proof
We want an explicit expression of u and v, respectively.
Starting from u, it is a solution to
with \(f|_{B_1}= 1\) and \(f|_{B_r}= 0\).
It’s clear that \(u |_{B_r}=0\) and \(u(x)=u({\left| x\right| })\) it’s radial on \(B_1\).
So, the equation (1) becomes
and then
We set \(c=0\), in order to have a \(C^1\)-solution.
If we integrate, we obtain
The Robin boundary conditions become
now we can compute the value of A
So
As u is decreasing, we have
Now, let us compute v(s). We will do this firstly for \(s\in (0,1)\), then for \(s\in (1, {\overline{r}})\) where \({\overline{r}}= (1 + r^n)^{\frac{1}{n}}\) is determined by the condition \({\left| \Omega \right| }= \vert \Omega ^\sharp \vert\).
Let \(s<1\)
Now, we can’t determine B as before, as v is not identically 0 in the anulus \(B_{{\overline{r}}} \backslash B_1\).
Let \(s >1\) and \(p \ne n\)
by imposing the continuity of the derivative for \(s=1\), we obtain that \(C= -1/n\)
and by Robin conditions
By imposing the continuity of v for \(s=1\), we have
that is to say
For convenience’s sake, we set \(\displaystyle {h=\frac{1}{(n\beta )^\alpha } \left( {\overline{r}}^{-\frac{n-1}{p-1}}-1\right) + \frac{p-1}{n^\alpha (p-n)}\left( {\overline{r}}^{\frac{p-n}{p-1}} -1\right) }\).
So, we have
By using Taylor expansion of the function \((1+r^n)^{\delta }\), we get
so, if we choose \(\beta < \left( \frac{n-1}{p-1}\right) ^{p-1}\), we get
Next example 5.2 is a counterexample to the Corollary 3.1 in the case \(n > p\).
Example 5.2
Let \(\Omega \subseteq {\mathbb {R}}^n\), \(p<n\) be the union of two disjoint balls \(B_1\) and \(B_r\) with radii 1 and r, respectively. We choose \(\displaystyle {\beta \le \left( \frac{n-p}{p(p-1)} \right) ^{p-1} }\) and we fix \(f=1\) on \(B_1\) and \(f=0\) on \(B_r\). Both u and v can be explicitly computed. We have \({\left\| u\right\| }_p^p - {\left\| v\right\| }_p^p = C r^n + o(r^n)\), where C is a positive constant.
Proof
Let us consider the Taylor expansion of \((1+y)^p\), we get
Moreover,
as if \(1<s<{\overline{r}}\)
thus
and by integration we obtain the value of the norm in \(L^p(B_{{\overline{r}}} \backslash B_1)\).
So,
and recalling that
we get
We have to understand whether
If we choose \(\displaystyle {\beta < \left( \frac{n-1}{p-1} \right) ^{p-1} }\), we have \(\displaystyle {- \frac{1}{(n\beta )^\alpha } \frac{n-1}{n(p-1)} + \frac{1}{n^{\alpha +1}}<0}\). In order to have (30), we need
If we show that
then
We just have to verify (31)
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Amato, V., Gentile, A. & Masiello, A.L. Comparison results for solutions to p-Laplace equations with Robin boundary conditions. Annali di Matematica 201, 1189–1212 (2022). https://doi.org/10.1007/s10231-021-01153-y
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DOI: https://doi.org/10.1007/s10231-021-01153-y