Some isoperimetric inequalities with respect to monomial weights

TL;DR

This paper establishes isoperimetric inequalities for weighted measures in the upper half-plane, identifying explicit minimizers and deriving implications for weighted Cheeger constants and eigenvalue bounds.

Contribution

It provides explicit solutions to a class of weighted isoperimetric problems with monomial weights, extending classical results to weighted settings.

Findings

Weighted perimeter minimized by symmetric sets

Explicit form of optimal sets derived

Implications for Cheeger constants and eigenvalues

Abstract

We solve a class of isoperimetric problems on $\mathbb{R}^2_+ :=\left\{ (x,y)\in \mathbb{R} ^2 : y>0 \right\}$ with respect to monomial weights. Let $\alpha $ and $\beta $ be real numbers such that $0\le \alpha <\beta+1$, $\beta\le 2 \alpha$. We show that, among all smooth sets $\Omega$ in $\mathbb{R} ^2_+$ with fixed weighted measure $\iint_{\Omega } y^{\beta} dxdy$, the weighted perimeter $\int_{\partial \Omega } y^\alpha \, ds$ achieves its minimum for a smooth set which is symmetric w.r.t. to the $y$--axis, and is explicitly given. Our results also imply an estimate of a weighted Cheeger constant and a lower bound for the first eigenvalue of a class of nonlinear problems.

Figures

Equations

$$(\mathbf{P})\inf \left\{\frac{{\int }_{\partial \Omega }{y}^{\alpha }ds}{{\left[{\iint }_{\Omega }{y}^{\beta }dxdy\right]}^{(\alpha +1)/(\beta +2)}}:0<{\iint }_{\Omega }{y}^{\beta }dxdy<+\infty \right\}=:\mu (\alpha ,\beta ).$$

$$0\le \alpha <\beta +1$$

$$\beta \le 2\alpha .$$

$${\Omega }^{\star }$$

$$f(y)$$

$$\mu (\alpha ,\beta )={\gamma }^{\frac{\alpha +1}{\beta +2}-1}\cdot {\left[\frac{(\beta +1)(\beta +2)}{\alpha +1}\right]}^{\frac{\alpha +1}{\beta +2}}\cdot {\left[B\left(\frac{\alpha +1}{2\gamma },\frac{1}{2}\right)\right]}^{\frac{\gamma }{\beta +2}},$$

$$\mu (\alpha ,2\alpha )=\sqrt{\frac{2\pi (2\alpha +1)}{\alpha +1}}.$$

$$\beta +1>0\text{mbox}and\alpha \ge 0.$$

$$\Omega (y)$$

$${\Omega }^{\prime }$$

$$A_{\beta }(\Omega ):={\iint }_{\Omega }{y}^{\beta }dxdy,$$

$$P_{\alpha }(\Omega ):=\sup \left\{{\iint }_{{\mathbb{R}}_{+}^2}\text{mbox}div\left(y^{\alpha }\mathbf{v}\right)dxdy:\mathbf{v}\in C_0^1({\mathbb{R}}_{+}^2,{\mathbb{R}}^2),|\mathbf{v}|\le 1\text{mbox}in\Omega \right\}.$$

$$P_{\alpha }(\Omega )={\int }_{\partial \Omega \cap {\mathbb{R}}_{+}^2}{y}^{\alpha }d{\mathcal{H}}_1.$$

$$P_{\alpha }(\Omega )=\underset{n\to \infty }{\lim }{P}_{\alpha }({\Omega }_n).$$

$$P_{\alpha }(\Omega )\le \underset{n\to \infty }{\mathrm{liminf}}{P}_{\alpha }({\Omega }_n)$$

$${\mathcal{R}}_{\alpha ,\beta }(\Omega ):=\frac{P_{\alpha }(\Omega )}{{\left[A_{\beta }(\Omega )\right]}^{(\alpha +1)/(\beta +2)}},(0<A_{\beta }(\Omega )<+\infty ).$$

$$(\mathbf{P})\inf \left\{{\mathcal{R}}_{\alpha ,\beta }(\Omega ):0<A_{\beta }(\Omega )<+\infty \right\}=:\mu (\alpha ,\beta ).$$

$$\mu (\alpha ,\beta )=0,$$

$${\int }_{\partial \Omega (t)\cap {\mathbb{R}}_{+}^2}{y}^{\alpha }ds$$

$$\underset{\Omega (t)}{\iint }{y}^{\beta }dxdy$$

$${\mathcal{R}}_{\alpha,\beta}(\Omega(t))=\frac{t+(2/(\alpha+1))}{[t/(\beta+1)]^{(\alpha+1)/(\beta+2)}}\longrightarrow 0,\mbox{ as $t\to+\infty$,}$$

$${\mathcal{R}}_{\alpha ,\beta }(U)<{\mathcal{R}}_{\alpha ,\beta }(\Omega ).$$

$$P_{\alpha }(\Omega )$$

$$\mu (\beta +1,\beta )=\beta +1,$$

$$\underset{t\to \infty }{\lim }{\mathcal{R}}_{\beta +1,\beta }(\Omega (t))=\beta +1.$$

$$\mathcal{C}=\{(\xi (t),\eta (t)):0\le t\le a\},(a\in 0,+\infty ),$$

$$A_{\beta }(\Omega )$$

$$P_{\beta +1}(\Omega )>(\beta +1)A_{\beta }(\Omega ).$$

$$P_{\beta +1}(\Omega )$$

$${\int }_{\partial B_1(z(t))}{y}^{\alpha }ds$$

$$\underset{B_1(z(t))}{\iint }{y}^{\beta }dxdy$$

Full text

Key words: isoperimetric inequality, weighted Cheeger set, eigenvalue problems

2000 Mathematics Subject Classification: 51M16, 46E35, 46E30, 35P15

1. Introduction

The last two decades have seen a growing interest in isoperimetric inequalities with respect to weights.

In most cases, volume and perimeter in those inequalities carried the same weight, because such a setting corresponds to manifolds with density. However, most research dealt with inequalities where both the volume functional and perimeter functional carry the same weight, see for instance [5], [7], [8], [11], [24], [14], [15], [16], [17], [20], [23], [32], [33], [36], [37], [38] and the references therein.

More recently, also problems with different weight functions for perimeter and volume were studied, see for example [2], [3], [4], [6], [22], [25], [26], [29], [34], [35], [40] and the references therein. However, there is only a sparse literature on situations where the isoperimetric sets are not radial, see [26], [19], [1].

In this paper we study the following isoperimetric problem:

Minimize $\displaystyle\int_{\partial\Omega}y^{\alpha}ds$ among all smooth sets $\Omega\subset\mathbb{R}^{2}_{+}$ satisfying $\displaystyle\iint_{\Omega}y^{\beta}\,dxdy=1$

or equivalently

[TABLE]

Our main result, proved in Section 2, is the following:

Theorem 1.1**.**

Assume that

[TABLE]

and

[TABLE]

Then problem (P) has a minimizer which is given by

[TABLE]

Moreover, we have

[TABLE]

where $\gamma:=\beta+1-\alpha$ and $B$ denotes the Beta function. In particular,

[TABLE]

Remark 1.1**.**

(a) First observe that $\Omega^{\star}$ is the half-circle when $\alpha=\beta$. Therefore Theorem 1.1 includes the result obtained by Maderna and Salsa in [33] (see also [14], [10]).

(b) Let $B_{1}^{+}:=\{(x,y)\in\mathbb{R}^{2}:\,x^{2}+y^{2}<1,\,y>0\}$. It is elementary to verify that,

1. if $\beta-\alpha<0$ then $\Omega\supseteq B_{1}^{+}$,

2. if $\beta-\alpha>0$ then $\Omega\subseteq B_{1}^{+}$,

3. if $\beta-\alpha=0$ then $\Omega\equiv B_{1}^{+}$; see Figure 1.

Theorem 1.1 also allows to obtain a Faber-Krahn - type inequality for the so-called weighted Cheeger constant, and in turn a lower bound for the first eigenvalue for a degenerate elliptic operators. For similar results see also [9], [12], [13], [18], [30], [39], [41].

2. Isoperimetric inequality in the upper half plane

Let $\mathbb{R}_{+}^{2}:=\{(x,y)\in\mathbb{R}^{2}:\,y>0\}$. Throughout this paper, we assume that $\alpha,\beta\in\mathbb{R}$ and

[TABLE]

If $\Omega\subset\mathbb{R}_{+}^{2}$ is measurable, we set

[TABLE]

Further, we define the weighted area of $\Omega$ by

[TABLE]

and the weighted relative perimeter of $\Omega$ as

[TABLE]

It is well-known that, if $\Omega$ is an open, rectifiable set, then the following equality holds

[TABLE]

($\mathscr{H}_{1}$ denotes $1$-dimensional Hausdorff-measure.)

Remark 2.1**.**

The following properties of the perimeter are well-known:

Let $\Omega$ be measurable with $0<A_{\beta}(\Omega)<+\infty$ and $P_{\alpha}(\Omega)<+\infty$.

Then there exists a sequence of open, rectifiable sets $\{\Omega_{n}\}$ with $\displaystyle\lim_{n\to\infty}A_{\beta}(\Omega\Delta\Omega_{n})=0$ and

[TABLE]

Further, we have

[TABLE]

for any sequence of open, rectifiable sets $\{\Omega_{n}\}$ satisfying $\displaystyle\lim_{n\to\infty}A_{\beta}(\Omega\Delta\Omega_{n})=0$.

We define the ratio

[TABLE]

Remark 2.2**.**

We have ${\mathcal{R}}_{\alpha,\beta}(t\Omega)={\mathcal{R}}_{\alpha,\beta}(\Omega)$ for every $t>0$.

We study the following isoperimetric problem:

[TABLE]

Our first aim is to reduce the class of admissible sets in the isoperimetric problem (P).

Throughout our proofs let $C$ denote a generic constant which may vary from line but does not depend on the other parameters.

The first two Lemmata give necessary conditions for a minimizer to exist.

Lemma 2.1**.**

If $\alpha>\beta+1$, then

[TABLE]

and (P) has no minimizer.

Proof: Let $\Omega(t):=(0,t)\times(0,1)$, ($t>0$). Then

[TABLE]

Hence

[TABLE]

and the assertion follows. $\hfill\Box$

Lemma 2.2**.**

Assume $\alpha<\beta+1$. Further, let $\Omega\subset\mathbb{R}^{2}_{+}$ be a nonempty, open and rectifiable set, which is not simply connected. Then there exists a nonempty, open and rectifiable set $U\subset\mathbb{R}^{2}_{+}$ which is simply connected, such that

[TABLE]

Proof: (i) First assume that $\Omega$ is connected. Let $G$ be the unbounded component of $\mathbb{R}^{2}\setminus\overline{\Omega}$ and set $U:=\mathbb{R}^{2}\setminus\overline{G}$. Then $U$ is simply connected with $\Omega\subset U\subset\mathbb{R}_{+}^{2}$ and $\partial U\subset\partial\Omega$, so that (2.9) follows.

(ii) Next, let $\Omega=\cup_{k=1}^{m}\Omega_{k}$, with mutually disjoint, nonempty, open, connected and rectifiable sets $\Omega_{k}$, ($k=1,\ldots,m$, $m\geq 2$). We set ${\mathcal{R}}_{\alpha,\beta}(\Omega)=:\lambda$. Let us assume that ${\mathcal{R}}_{\alpha,\beta}(\Omega_{k})\geq\lambda$ for every $k\in\{1,\ldots,m\}$. Then we have, since $(\alpha+1)/(\beta+2)<1$,

[TABLE]

a contradiction. Hence there exists a number $k_{0}\in\{1,\ldots,m\}$ with ${\mathcal{R}}_{\alpha,\beta}(\Omega_{k_{0}})<\lambda$. Then, repeating the argument of part (i), with $\Omega_{k_{0}}$ in place of $\Omega$, we again arrive at (2.9). $\hfill\Box$

Lemma 2.3**.**

There holds

[TABLE]

but (P) has no open rectifiable minimizer.

Proof: With $\Omega(t)$ as in the proof of Lemma 2.1, we calculate ${\mathcal{R}}_{\beta+1,\beta}(\Omega(t))>\beta+1$ and

[TABLE]

Let $\Omega\subset\mathbb{R}_{+}^{2}$ be open, rectifiable and simply connected. Then $\partial\Omega$ is a closed Jordan curve ${\mathcal{C}}$ with counter-clockwise representation

[TABLE]

where $\xi,\eta\in C[0,a]\cap C^{0,1}(0,a)$, $\xi(0)=\xi(a)$, $\eta(0)=\eta(a)$, and $(\xi^{\prime})^{2}+(\eta^{\prime})^{2}>0$ on $[0,a]$. Using Green’s Theorem we evaluate

[TABLE]

Equality in (2.12) can hold only if $\eta^{\prime}\equiv 0$ and $\xi^{\prime}\leq 0$ on $\partial\Omega\cap\mathbb{R}^{2}_{+}$, that is, if $\partial\Omega\cap\mathbb{R}^{2}_{+}$ is a single straight segment which is parallel to the $x$-axis. But this is impossible. Hence we find that

[TABLE]

To show the assertion in the general case, we proceed similarly as in the proof of Lemma 2.2:

Assume first that $\Omega$ is connected and define the sets $G$ and $U$ as in the last proof. Using (2.13), with $U$ in place of $\Omega$, we obtain

[TABLE]

Finally, let $\Omega$ be open and rectifiable. Then $\Omega=\cup_{k=1}^{m}\Omega_{k}$, with mutually disjoint, connected sets $\Omega_{k}$, ($k=1,\ldots,m$). Then (2.14) yields

[TABLE]

Now the assertion follows from (2.15) and (2.11). $\hfill\Box$

Lemma 2.4**.**

Let $2\alpha<\beta$. Then (2.8) holds and (P) has no minimizer.

Proof: Let $z(t):=(0,t)$, ($t\geq 2$). Then we have for all $t\geq 2$,

[TABLE]

This implies

[TABLE]

and the assertion follows. $\hfill\Box$

Next we recall the definition of the Steiner symmetrization w.r.t. the $x$-variable. If $\Omega$ is measurable, we set

[TABLE]

where $\Omega(y)$, $\Omega^{\prime}$ are defined in (2.2) and

[TABLE]

Note that $S(\Omega)(y)$ is a symmetric interval with $\mathscr{L}^{1}(S(\Omega)(y))=\mathscr{L}^{1}(\Omega(y))$.

Since the weight functions in the functionals $P_{\alpha}$ and $A_{\beta}$ do not depend on $x$, we have the following well-known properties, see [28], Proposition 3.

Lemma 2.5**.**

Let $\Omega\subset\mathbb{R}_{+}^{2}$ be measurable. Then

[TABLE]

For nonempty open sets $\Omega$ with $\Omega=S(\Omega)$ we set $\Omega_{t}:=\{(x,y)\in\Omega:\,y>t\}$ and

[TABLE]

Remark 2.3. Assume that $\Omega\subset\mathbb{R}_{+}^{2}$ is a bounded, open and rectifiable set with $0<A_{\beta}(\Omega)<+\infty$ and $\Omega=S(\Omega)$. Then it has the following representation,

[TABLE]

Lemma 2.6**.**

Let $\Omega$ be a nonempty, bounded, open and rectifiable set with $\Omega=S(\Omega)$. Then we have

[TABLE]

where $f$ is given by (2).

Proof: Assume first that $\Omega$ is represented by (2) where

[TABLE]

Then we have for every $y\in(y^{-},y^{+})$,

[TABLE]

Furthermore, there holds

[TABLE]

In the general case the assertions follow from these calculations by approximation with sets $\Omega$ of the type given by (2), (2.24). $\hfill\Box$

Lemma 2.7**.**

Assume that

[TABLE]

and let $\Omega$ be a bounded, open and rectifiable set with $A_{\beta}(\Omega)=1$, $\Omega=S(\Omega)$ and $P_{\alpha}(\Omega)<\mu(\alpha,\beta)+1$. Then there exist positive numbers $C_{1}$ and $C_{2}$ which depend only on $\alpha$ and $\beta$ such that

[TABLE]

Proof: By (2.22) and (2.23) we have

[TABLE]

It follows that

[TABLE]

Setting $z:=\frac{y^{-}}{y^{+}}(\in[0,1))$, we obtain from (2.28) and (2.30),

[TABLE]

which implies that

[TABLE]

with a constant $C$ which depends only on $\alpha$ and $\beta$. By (2.26) we have that

[TABLE]

Hence it follows that

[TABLE]

Using (2.31) and (2.32) this leads to (2.27). $\hfill\Box$

Lemma 2.8**.**

Assume (2.25) and (2.26). Then problem (P) has a minimizer $\Omega^{\star}$ which is symmetric w.r.t. the $y$-axis.

Proof: We proceed in $4$ steps.

Step 1: A minimizing sequence :

Let $\{\Omega_{n}\}$ a minimizing sequence, that is, $\lim_{n\to\infty}\mathcal{R}_{\alpha,\beta}(\Omega_{n})=\mu(\alpha,\beta)$. In view of the Remarks 2.1 and 2.2 and the Lemmata 2.2 and 2.6 we may assume that $\Omega_{n}$ is open, simply connected and rectifiable with $\Omega_{n}=S(\Omega_{n})$, $A_{\beta}(\Omega_{n})=1$ and $P_{\alpha}(\Omega_{n})\leq\mu(\alpha,\beta)+\frac{1}{n}$, ($n\in\mathbb{N}$).

Step 2: Parametrization of $\partial\Omega_{n}$ :

Denote ${\mathcal{C}}_{n}:=\overline{\partial\Omega_{n}\cap\{(x,y):\,x>0,\,y>0\}}$. It is clear that ${\mathcal{C}}_{n}$ a simple smooth curve with

[TABLE]

where $s$ denotes the usual arclength parameter, $L_{n}\in(0,+\infty)$, $x,y\in C[0,L_{n}]\cap C^{0,1}(0,L_{n})$, $x_{n}(s)>0$ and $y_{n}(s)>0$ for every $s\in(0,L_{n})$, ($n\in\mathbb{N}$). We orientate ${\mathcal{C}}_{n}$ in such a way that the mapping $s\mapsto y_{n}(s)$ is nonincreasing and $x_{n}(0)=0$. Setting $y_{n}(0)=:y_{n}^{+}$ and $y_{n}(L_{n})=:y_{n}^{-}$, we have by Lemma 2.6,

[TABLE]

where $C_{1}$ and $C_{2}$ do not depend on $n$. Note that $x_{n}(L_{n})=0$ in case that $y_{n}^{-}>0$. Further, Lemma 2.6, (2.23) shows that

[TABLE]

For our purposes it will be convenient to work with another parametrization of ${\mathcal{C}}_{n}$: We set

[TABLE]

Then $X_{n},Y_{n}\in C[0,1]\cap C^{0,1}(0,1)$, and we evaluate

[TABLE]

Step 3: Limit of the minimizing sequence :

Since $(x_{n}^{\prime}(s))^{2}+(y_{n}^{\prime}(s))^{2}\equiv 1$, we obtain from (2.34) and (2.39) that the family $\{Y_{n}^{\alpha+1}\}$ is equibounded and uniformly Lipschitz continuous on $(0,1)$. Hence there is a function $Y\in C[0,1]$ with $Y^{\alpha+1}\in C^{0,1}(0,1)$ such that, up to a subsequence,

[TABLE]

Moreover, setting $Y(0)=:y^{+}$, $Y(1)=:y^{-}$, the bounds (2.27) are in place and $\sigma\mapsto Y(\sigma)$ is nonincreasing.

Let

[TABLE]

Then from (2.35) and (2.38) we obtain that the families $\{X_{n}\}$ and $\{dX_{n}/d\sigma\}$ are equibounded on every closed subset of $[0,\sigma_{0})$. Hence there exists a function $X\in C[0,\sigma_{0})$ which is locally Lipschitz continuous on $[0,\sigma_{0})$, such that, up to a subsequence,

[TABLE]

Moreover, from (2.35) and (2.38) we find that

[TABLE]

Let $\Omega$ be the set in $\mathbb{R}^{2}_{+}$ with $\Omega=S(\Omega)$ such that $\partial\Omega\cap\{(x,y):\,x>0,y>0\}$ is represented by the pair of functions

[TABLE]

In view of (2.43) we have that

[TABLE]

Step 4: A minimizing set :

We prove that

[TABLE]

In order to prove the first equality, since $A_{\beta}(\Omega_{n})=1$, we prove that

[TABLE]

Fix some $\delta\in(0,\sigma_{0})$. Then (2.38), (2.39), (2.43), (2.40) and (2.41) yield

[TABLE]

Further, (2.38) and (2.43) give

[TABLE]

Now (2.48), (2.49), (2) and (2.44) yield (2.46) and therefore the first of the equalities in (2.45).

Now we prove the second inequality in (2.45). With $\delta$ as above we also have

[TABLE]

In view of (2.40) and (2.41) it follows that $\displaystyle\lim_{n\to\infty}I_{n,\delta}^{1}=0$.

On the other hand, we have

[TABLE]

Define

[TABLE]

Letting $\delta\to 0$ we obtain

[TABLE]

Hence $\Omega$ is a minimizing set.

Note that $\Omega$ must be simply connected in view of Lemma 2.2, which implies that there is a number $\sigma_{1}\in[0,\sigma_{0})$ such that

[TABLE]

This finishes the proof of Lemma 2.8. $\hfill\Box$

          Next we obtain differential equations for the functions $X$ and $Y$ in the proof in Lemma 2.8.

Lemma 2.9**.**

Assume (2.25) and (2.26). Then the minimizer $\Omega$ obtained in Lemma 2.11 is bounded, and its boundary given parametrically by

[TABLE]

where the functions $x,y\in C^{2}[0,L]$ satisfy the following equations:

[TABLE]

together with the boundary conditions

[TABLE]

for some numbers $\lambda>0$ and $0\leq y^{-}<y^{+}$. Finally, the curve (2.9) is strictly convex.

Proof: We proceed in 4 steps.

Step 1: Euler equations :

After a rescaling the parameter $\sigma$, we see that the functions $X(\sigma)$ and $Y(\sigma)$ in the previous proof annihilate the first variation of the functional

[TABLE]

under the constraint

[TABLE]

Hence $X$ and $Y$ satisfy the Euler equations

[TABLE]

where $\lambda$ is a Lagrangian multiplier, and

[TABLE]

In addition, the following boundary conditions are satisfied:

[TABLE]

Step 2: Boundedness:

It will be more convenient to rewrite the above conditions in terms of the arclength parameter $s$: Set

[TABLE]

where $L\in(0,+\infty]$. Then we have $(x^{\prime}(s))^{2}+(y^{\prime}(s))^{2}=1$ so that (2.57), (2.58) yield the system of equations (2.52), (2.53). Integrating (2.53) we obtain

[TABLE]

for some $d\in\mathbb{R}$.

Assume first that $x(s)$ is unbounded. Then $L=+\infty$, and in view of (2.42) we have that $\lim_{s\to\infty}y(s)=0$. If $\alpha=0$, then this would imply $P_{0}(\Omega)=+\infty$, which is impossible. Hence we may restrict ourselves to the case $\alpha>0$.

There is a sequence $s_{n}\to+\infty$ such that $\displaystyle\lim_{n\to\infty}x^{\prime}(s_{n})=1$. Using $s=s_{n}$ in (2.63) and passing to the limit $n\to\infty$ gives $d=0$. Plugging this into (2.52), we find

[TABLE]

Multiplying (2.64) with $y^{\alpha}y^{\prime}$ and integrating, we obtain

[TABLE]

or equivalently,

[TABLE]

for some $D\in\mathbb{R}$. Using $s=s_{n}$ in (2.65) and taking into account that $\lim_{n\to+\infty}y^{\prime}(s_{n})=0$, $\beta+1-\alpha>0$ and $\alpha>0$, we arrive again at a contradiction. Hence $x(s)$ is bounded, and we deduce the boundary conditions (2.55)-(2.9).

Step 3: $\lambda$ is positive :

Multiplying (2.52) with $y$ and integrating from $s=0$ to $s=L$ gives

[TABLE]

Using integration by parts this yields

[TABLE]

The first two boundary terms in this identity vanish due to the boundary conditions (2.55)-(2.9) and the two integrals are positive since $y^{\prime}\leq 0$ and $y^{\prime}\not\equiv 0$. It follows that

[TABLE]

Now, considering equation (2.63) at $s=L$ and taking into account the boundary conditions (2.9), we find

[TABLE]

which implies that

[TABLE]

Step 4: Strict convexity :

From (2.52) and (2.53) we obtain

[TABLE]

Hence, using (2.63), we find for the curvature $\kappa(s)$ of the curve $(x(s),y(s))$, ($s\in(0,L)$),

[TABLE]

The last expression is positive by (2.66) and (2.67), which means that $\Omega$ is strictly convex. The Lemma is proved. $\hfill\Box$

Lemma 2.10**.**

Assume (2.25) and (2.26), and let $\partial\Omega$ be given by (2.9)-(2.9). Then $y^{-}=0$.

Proof: Supposing that $y^{-}>0$, we will argue by contradiction. We proceed in 3 steps.

Step 1: Another parametrization of $\partial\Omega$:

Let

[TABLE]

Since $\Omega$ is strictly convex, there are functions $u_{1},u_{2}\in C^{2}[0,x_{0})\cap C[0,x_{0}]$ such that

[TABLE]

Furthermore, the Euler equations (2.52), (2.53) lead to

[TABLE]

Using (2.71) and the fact that

[TABLE]

lead to

[TABLE]

Finally, the boundary conditions (2.70)–(2.72) lead to the following formulas:

[TABLE]

Step 2: Curvature:

In the following, we will refer to points $(x,u_{1}(x))$ as points of the ’lower curve’ and to points $(x,u_{2}(x))$ as points of the ’upper curve’, ($x\in(0,x_{0})$).

The signed curvature $\kappa$ (see (2)) can be expressed in terms of the functions $u_{1}$ and $u_{2}$. More precisely, we have

[TABLE]

Accordingly, we will write

[TABLE]

Finally, let $s_{0}\in(0,L)$ be taken such that $y(s_{0})=y_{0}$ and $x(s_{0})=x_{0}$. Then formula (2.63) taken at $s=s_{0}$ leads to

[TABLE]

Plugging this into (2) we find

[TABLE]

Differentiating (2.83) we evaluate

[TABLE]

Since $y^{\prime}(s)<0$ for $s\in(0,L)$, this in particular implies

[TABLE]

Step 4:

We claim that

[TABLE]

First observe that (2.86) immediately follows from (2.85) if $\alpha\geq\beta$. Thus it remains to consider the case

[TABLE]

From (2.84) and the fact that $y(s)<y_{0}$ for $s\in[s_{0},L]$ we find that

[TABLE]

This means that

[TABLE]

for some (small) $\varepsilon>0$. Now assume that (2.86) does not hold. By (2.89) there exists a number $x_{1}\in(0,x_{0})$ such that

[TABLE]

We claim that (2.90) implies that

[TABLE]

To prove (2.92), observe first that

[TABLE]

Then, integrating (2.90) over $(x_{1},x_{0})$ leads to

[TABLE]

which implies (2.92). Now, from (2.92) we deduce

[TABLE]

Together with (2.77) and (2.78) we obtain from this

[TABLE]

or, equivalently

[TABLE]

Furthermore, multiplying (2.73) by $(u_{1})^{-\alpha}$, respectively (2.74) by $(u_{2})^{-\alpha}$, adding both equations and taking into account (2.91) leads to

[TABLE]

Using once more (2.77) and (2.78) then gives

[TABLE]

or equivalently,

[TABLE]

From this and (2.93) we then obtain

[TABLE]

Setting

[TABLE]

this leads to

[TABLE]

But this contradicts Lemma A (Appendix). This finishes the proof of (2.86).

Step 5:

Since $\lim_{x\to x_{0}}u_{1}^{\prime}(x)=-\lim_{x\to x_{0}}u_{2}^{\prime}(x)=+\infty$, (2.86) implies that

[TABLE]

which contradicts the boundary conditions (2.72). Hence we must have that $y^{-}=0$. $\hfill\Box$

         Now we are in a position to give a

Proof of Theorem 1.1: We split into two cases.

Case 1: Assume that

[TABLE]

By Lemma 2.10 we have $y^{-}=0$. Now, since $x^{\prime}(L)=0$, equation (2.53) at $s=L$ yields

[TABLE]

so that

[TABLE]

In view of Remark 2.2 we may rescale $\Omega$ in such a way that $y^{+}=1$. Then (2.98) at $s=0$ gives $\lambda=1$. Since $y^{\prime}(s)<0$ for $s\in(0,L)$, we have $s=g(y)$ with a decreasing function $g\in C^{1}(0,1)$. Writing

[TABLE]

we obtain

[TABLE]

and integrating this leads to (1.3) and (1.4).

Case (ii) Now assume that

[TABLE]

Since the case $\alpha=0$ is trivial, we may assume $\alpha>0$. Let us fix such $\alpha$.

First observe that for every smooth domain $U\subset\mathbb{R}^{2}_{+}$, the mapping

[TABLE]

is continuous. Furthermore, from Case (i) we see that the mapping

[TABLE]

is continuous, and the limit

[TABLE]

exists.

Now let $\Omega^{\star}$ be the domain that is given by formulas (1.3), (1.4), with $\beta=2\alpha$. Then we also have

[TABLE]

which implies that $Z\geq\mu(\alpha,2\alpha)$.

Assume that

[TABLE]

Then there is a smooth set $\Omega^{\prime}\subset\mathbb{R}^{2}_{+}$ such that also

[TABLE]

But by (2.101) this implies that

[TABLE]

when $\beta<2\alpha$ and $|\beta-2\alpha|$ is small, which is impossible. Hence we have that

[TABLE]

This finishes the proof of the Theorem.

          Finally we evaluate $\mu(\alpha,\beta)$. Put $\gamma:=\beta+1-\alpha(>0)$. With the Beta function $B$ and the function $f$ given by (1.4) we have

[TABLE]

and

[TABLE]

Using the identity

[TABLE]

we obtain

[TABLE]

which is (1.5). In case of $\beta=2\alpha$ this leads to

[TABLE]

Remark 2.3**.**

It is also well-known that the isoperimetric inequality is equivalent to the following functional inequality, (see [1], Lemma 3.5).

[TABLE]

3. Applications

In this section we firstly show that our isoperimetric inequality implies a sharp estimate of the so-called weighted Cheeger constant.

Then we deduce an estimate of the first eigenvalue to a degenerate elliptic Dirichlet boundary values problem. We begin by introducing some function spaces that will be used in the sequel.

Let $\Omega$ be an open subset of $\mathbb{R}^{2}_{+}$ and $p\in[1,+\infty)$.

By $L^{p}(\Omega;y^{\beta})$ we denote the weighted Hölder space of measurable functions $u:\Omega\to\mathbb{R}$ such that

[TABLE]

Then let $W^{1,p}(\Omega;y^{\alpha},y^{\beta})$ be the weighted Sobolev space of all functions $u\in L^{p}(\Omega;y^{\beta})$ possessing weak first partial derivatives which belong to $L^{p}(\Omega;y^{\beta})$. A norm in $W^{1,p}(\Omega;y^{\alpha},y^{\beta})$ is given by

[TABLE]

For any function $u\in L^{1}(\Omega;y^{\beta})$ we write

[TABLE]

Then let $BV(\Omega;y^{\alpha},y^{\beta})$ be the weighted BV-space of all functions $u\in L^{1}(\Omega;y^{\beta})$ such that $|Du|(\Omega;y^{\alpha})<+\infty$. A norm on $BV(\Omega;y^{\alpha},y^{\beta})$ is given by

[TABLE]

Let us explicitly remark that for an open bounded set $\Omega\subset\mathbb{R}^{2}_{+}$ the following equality holds

[TABLE]

Finally let $X$ be the set of all the functions $w\in C^{1}(\overline{\Omega})$ that vanish in a neighborhood of $\partial\Omega\cap\mathbb{R}_{+}^{2}$ . Then $V^{p}(\Omega;y^{\alpha},y^{\beta})$ will denote the closure of $X$ in the norm of $W^{1,p}(\Omega;y^{\alpha},y^{\beta})$.

Finally we denote by $\Omega^{\bigstar}$ the set $t\Omega^{\star}$, for $t>0$, such that $A_{\beta}(\Omega)=A_{\beta}(\Omega^{\bigstar})$.

3.1. Weighted Cheeger sets

We define the weighted Cheeger constant of an open bounded set $\Omega\subset\mathbb{R}^{2}_{+}$ as

[TABLE]

(see also [30], [41])

We firstly prove that the existence of an admissible set which realizes the minimum in (3.1) (see also [41]).

Lemma 3.1**.**

Assume $0\leq\alpha<\beta+1$ and $\beta\leq 2\alpha$. For any open bounded set $\Omega\subset\subset\mathbb{R}^{2}_{+}$, there exists at least one set $M\subseteq\Omega$, the so-called weighted Cheeger set, such that

[TABLE]

Proof: Since $\Omega$ is open, $h_{\alpha,\beta}(\Omega)$ is finite: Indeed, it is easy to verify that for any ball $B$ with $B\subset\subset\Omega$, the ratio $\frac{P_{\alpha}(B)}{A_{\beta}(B)}$ is finite.

Let $\{E_{k}\}$ be a minimizing sequence for (3.1). Since $\Omega$ is bounded, we have

[TABLE]

Now fix $\varepsilon>0$. There exists an index $\overline{k}$ such that

[TABLE]

Since $\Omega$ is bounded, for all $k>\overline{k}$, we get

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This implies

[TABLE]

Hence $\{E_{k}\}$ is an equibounded family in weighted $BV(\Omega;y^{\alpha},y^{\beta})-$norm. Thus by Lemma B (Appendix), up to subsequences, $\{\chi(E_{k})\}$ converges in the weighted $L^{1}(\Omega;y^{\beta})-$norm and pointwise a.e. to a function $u$. Moreover there exists a subset $M\subseteq\Omega$ such that $u=\chi_{M}$.

Since $\{E_{k}\}$ is a minimizing sequence, by lower semicontinuity of perimeter $P_{\alpha}$ and Lebesgue dominated convergence theorem, we get

[TABLE]

It remains to prove that $M$ is an admissible set, that is we need to prove that

[TABLE]

Assume by contradiction that $A_{\beta}(M)=0$. This implies that $\displaystyle\lim_{k\to+\infty}A_{\beta}(E_{k})=0$.

Now for a fixed $\eta>0$ consider the set

[TABLE]

Now the following inequality holds true

[TABLE]

where $\delta=\eta$, if $\beta\leq 0$ and $\delta=R$ for a suitable $R>0$ such that a ball $B_{R}$ of radius $R$ contains $\Omega$, if $\beta>0$. Denote by $B_{r_{k},\eta}$ a ball of radius $r_{k}$ having the same Lebesgue measure $A_{0}(E_{k,\eta})$ of $E_{k,\eta}$. By the classical isoperimetric inequality, we get

[TABLE]

This yields a contradiction. Therefore (3.4) holds true and the conclusion follows. $\hfill\Box$

Once we have proved the existence of a weighted Cheeger set, we can obtain the following result.

Theorem 3.1**.**

Assume $0\leq\alpha<\beta+1$ and $\beta\leq 2\alpha$ and $\alpha<\beta+1$, and let $\Omega$ be a bounded open subset $\mathbb{R}^{2}_{+}$. Then the following estimate holds true

[TABLE]

Proof: Let $E$ be a nonempty subset of $\Omega$ with $P_{\alpha}(E)<+\infty$. By our isoperimetric inequality Theorem 1.1 and since $E^{\star}\subset\Omega^{\bigstar}$ with $A_{\beta}(E)=A_{\beta}(E^{\bigstar})$, we have that

[TABLE]

It remains to prove the equality in (3.5). Let $F$ be a nonempty subset of $\Omega^{\star}$. Then we have $F^{\bigstar}\subset\Omega^{\bigstar}$ and

[TABLE]

for all $t>0$. Since $F^{\bigstar}\subset\Omega^{\bigstar}$, there exists $t\geq 1$ such that $tF^{\star}=\Omega^{\bigstar}$. Therefore

[TABLE]

which proves the equality in (3.5). $\hfill\Box$

Remark 3.1. Theorem 3.1 could be stated as an estimate of the first eigenvalue of the weighted 1-laplacian.

3.2. A nonlinear eigenvalue problem

Let $\Omega\subset\mathbb{R}_{+}^{2}$ be a bounded domain and let $p\in(1,+\infty).$ We consider the following weighted eigenvalue problem

[TABLE]

where

[TABLE]

By a solution to problem (3.8) we mean a function  $u\in V^{p}\left(\Omega;y^{p\gamma_{1}},y^{\frac{p\gamma_{2}}{p-1}}\right)$ such that

[TABLE]

for all function $\psi\in C^{1}(\overline{\Omega})$ such that $\psi=0$ on $\partial\Omega\cap\mathbb{R}_{+}^{2}.$

Let us denote by $T$ the range of values of $\alpha$ and $\beta$ for which the isoperimetric inequality holds true. We have that

[TABLE]

if and only if

[TABLE]

Furthermore the smallest eigenvalue of problem (3.8), $\lambda_{1,p}^{\gamma_{1},\gamma_{2}}(\Omega),$ has the following variational characterization

[TABLE]

Indeed, see e.g. Theorem 8.9 in [27], for any $(\gamma_{1},\gamma_{2})\in U,$ the following compact embedding holds true

[TABLE]

By adapting the arguments used in [21], [31], we obtain the following result

Theorem 3.2**.**

Let $(\gamma_{1},\gamma_{2})\in U,$ then the following estimate holds true

[TABLE]

Remark 3.2. If $\alpha=\beta>0$ and $p=2$, a Faber-Krahn type inequality for $\lambda_{1,p}^{\gamma_{1},\gamma_{2}}(\Omega)$ holds true (see [33]). Indeed in this case, we have $\alpha=2\gamma_{1}=2\gamma_{2}=\beta.$

Proof of Theorem 3.2 We claim that

[TABLE]

where $h_{\alpha,\beta}\left(\Omega\right)$ is defined in (3.1). Let $u$ be an eigenfunction corresponding to $\lambda_{1,p}^{\gamma_{1},\gamma_{2}}(\Omega)$. Hölder’s inequality gives

[TABLE]

and therefore

[TABLE]

If we set $f:=u^{p}$, then the previous inequality becomes

[TABLE]

On the other hand Coarea formula yields

[TABLE]

[TABLE]

[TABLE]

[TABLE]

The claim is hence proved. It immediately implies Theorem 3.2, thanks to (3.5).

4. Appendix

We prove two technical results.

Lemma A: Let $\alpha,\beta\in\mathbb{R}$ with $0<\alpha\leq\beta\leq 2\alpha$. Then

[TABLE]

Proof: We fix $\alpha>0$ and define

[TABLE]

Then

[TABLE]

Hence $h^{\prime}(1)=0$ which together with (4.3) and (4.4) implies that

[TABLE]

Furthermore we have

[TABLE]

Together with (4.2) and (4.4) this implies

[TABLE]

which is (4.1). $\hfill\Box$

**Lemma B: **Let $\{u_{n}\}\subset BV(\Omega;y^{\alpha},y^{\beta})$ be a bounded sequence. Then there exists a subsequence that converges in $L^{1}(\Omega;y^{\beta})$ and a.e. in $\Omega$ to some function $u$.

Proof: Put $\gamma=\frac{\beta+2}{\alpha+1}(>1)$ and let $\Omega_{\varepsilon}=\Omega\cap\{(x,y)\,:\,y>\varepsilon\}$ for any $\varepsilon>0$.

Let $k\in\mathbb{N}$. By a classical compactness result in the unweighted case, there exists a function $u^{k}\in L^{1}(\Omega_{2^{-k}};y^{\beta})$ and an increasing sequence of integers $\{a(k,m)\}_{m\geq 1}$ such that

[TABLE]

By choosing $\{a(k+1,m)\}$ to be a subsequence of $\{a(k,m\}$, ($k\in\mathbb{N}$), we can achieve that $u^{k}=u^{k+1}$ in $\Omega_{2^{-k}}$, $k\in\mathbb{N}$.

Now put

[TABLE]

In view of our isoperimetric inequality, the sequence $u_{n}$ is equibounded in $L^{\gamma}(\Omega;y^{\beta})$. We have the following estimate:

[TABLE]

with constants that do not depend on $k$. This implies that $u\in L^{1}(\Omega;y^{\beta})$.

Let $\varepsilon>0$. Choose $k$ large enough such that

[TABLE]

and then $m$ large enough such that

[TABLE]

Then we obtain

[TABLE]

where $C$ does not depend on $k$. From this the assertion follows. $\hfill\Box$

**Acknowledgement: ** We are grateful to M. van den Berg for giving us useful suggestions. The first, third, fourth and fifth authors are members of the Gruppo Nazionale per l’Analisi Matematica, la Probabilità e le loro Applicazioni (GNAMPA) of the Istituto Nazionale di Alta Matematica (INdAM) and they thank this institution for the support. The second author was partially supported by Leverhulme Trust ref. VP1-2017-004. We are also grateful to the Departments of Mathematics of Swansea University and of the University of Naples Federico II, and to South China University of Technology (ISAM) at Guangzhou for visiting appointments and their colleagues for their kind hospitality.