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A highly accurate computational approach to solving the diffusion equation of a fractional order.

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Bin Jebreen, H. A Highly Accurate Computational Approach to Solving the Diffusion Equation of a Fractional Order. Mathematics 2024 , 12 , 1965. https://doi.org/10.3390/math12131965

Bin Jebreen H. A Highly Accurate Computational Approach to Solving the Diffusion Equation of a Fractional Order. Mathematics . 2024; 12(13):1965. https://doi.org/10.3390/math12131965

Bin Jebreen, Haifa. 2024. "A Highly Accurate Computational Approach to Solving the Diffusion Equation of a Fractional Order" Mathematics 12, no. 13: 1965. https://doi.org/10.3390/math12131965

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Epiperimetric inequalities in the obstacle problem for the fractional Laplacian

• Open access
• Published: 25 June 2024
• Volume 63 , article number  150 , ( 2024 )

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• Matteo Carducci   ORCID: orcid.org/0009-0001-8045-2263 1  

Using epiperimetric inequalities approach, we study the obstacle problem \(\min \{(-\Delta )^su,u-\varphi \}=0,\) for the fractional Laplacian \((-\Delta )^s\) with obstacle \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , \(k\ge 2\) and \(\gamma \in (0,1)\) . We prove an epiperimetric inequality for the Weiss’ energy \(W_{1+s}\) and a logarithmic epiperimetric inequality for the Weiss’ energy \(W_{2m}\) . Moreover, we also prove two epiperimetric inequalities for negative energies \(W_{1+s}\) and \(W_{2m}\) . By these epiperimetric inequalities, we deduce a frequency gap and a characterization of the blow-ups for the frequencies \(\lambda =1+s\) and \(\lambda =2m\) . Finally, we give an alternative proof of the regularity of the points on the free boundary with frequency \(1+s\) and we describe the structure of the points on the free boundary with frequency 2 m , with \(m\in \mathbb {N}\) and \(2\,m\le k.\)

Avoid common mistakes on your manuscript.

1 Introduction

1.1 obstacle problem for the fractional laplacian.

Let \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , \(k\ge 2\) and \(\gamma \in (0,1)\) , that decays rapidly at infinity, we consider a solution of the obstacle problem for the fractional Laplacian with obstacle \(\varphi \) , that is a function \(u:\mathbb {R}^n\rightarrow \mathbb {R}\) such that

with \(s\in (0,1)\) . The fractional Laplacian \((-\Delta )^s\) defined as

where \(c_{n,s}=2^{2\,s}s\frac{\Gamma (\frac{n+2\,s}{2})}{\Gamma (1-s)}\pi ^{-\frac{n}{2}}\) is a normalization constant.

The aim of the paper is to established the optimal regularity of the solution and to describe the structure and the regularity of the free boundary

is the contact set.

1.2 The extension operator \(L_a\)

To study this problem, we will use the Caffarelli-Silvestre extension of u . As in [ 6 ], we consider the function \({\widetilde{u}}:\mathbb {R}^n\times \mathbb {R}\rightarrow \mathbb {R}\) satisfying

where \(X=(x,y)\in \mathbb {R}^{n+1}_+:=\mathbb {R}^n\times (0,+\infty )\) and

According to [ 6 ] (see also [ 24 ]), if we choose \(a:=1-2s\in (-1,1)\) , we get

with \(d_s>0\) , i.e. \((-\Delta )^s\) is a Dirichlet-to-Neumann map for \(L_a\) . Now, since

in distributional sense (see Proposition 9.1 ), we get that the problem ( 1 ) is equivalent to

where \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , \(k\ge 2\) and \(\gamma \in (0,1)\) .

In particular, when \(s=\frac{1}{2}\) , i.e. \(a=0\) and \(L_a=\Delta \) , the problem ( 2 ) is the thin obstacle problem (also know as Signorini problem).

Localizing the problem in \(B_1\subset \mathbb {R}^{n+1}\) , the solution of ( 2 ) can be obtained by minimizing the functional

among the admissible functions

where \(B_1':=B_1\cap \{y=0\}\) and c is the trace on \(\partial B_1\) of u .

Here we denote by \(H^1(\Omega ,a):=H^1(\Omega ,|y|^a)\) the weighted Sobolev space. Similarly, \(L^2(\Omega ,a):=L^2(\Omega ,|y|^a)\) is the weighted Lebesgue space.

In the following, with a slight abuse of notation, we denote by u the \(L_a\) -extension in \(\mathbb {R}^{n+1}\) of u , and we suppose that \(u\in H^1_{loc}(\mathbb {R}^{n+1},a)\) .

Moreover, with a slight abuse of notation, we also denote by \(x_0\in \mathbb {R}^n\) the point \((x_0,0)\in \mathbb {R}^{n}\times \{0\}.\)

1.3 Obstacle \(\varphi \equiv 0\)

We will say that u is a solution with 0 obstacle, if solves ( 2 ) with \(\varphi \equiv 0\) , i.e. if u satisfies

Moreover, we denote by

the set of admissible function with \(\varphi \equiv 0\) , then u is a minimum of the functional ( 3 ) in the class \(\mathcal {K}_c\) , with \(c=u|_{\partial B_1}\) .

1.4 Reduction to 0 obstacle

Let u be a solution of ( 2 ) with obstacle \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , \( k\ge 2\) and \(\gamma \in (0,1)\) . Let \(q_k^{x_0}(x)\) be the k -th Taylor polynomial of \(\varphi \) at \(x_0\in \Gamma (u)\) and \({\widetilde{q}}_k^{x_0}(x,y)\) be a polynomial of degree k and the \(L_a\) -harmonic extension of \(q_k^{x_0}(x)\) (see Lemma 9.3 ). Then \({\widetilde{q}}_k^{x_0}(x,y)\) solves the following problem

We can define

then \(u^{x_0}\) solves the following problem

where \(h(x,y)=h^{x_0}(x,y):= \Delta _x(\varphi (x)-q_k^{x_0}(x))\) .

Notice that starting from an obstacle problem with obstacle \(\varphi \not \equiv 0\) , we have reduced the problem to the case \(\varphi \equiv 0\) , where the right-hand side in the third and fourth line of ( 5 ) is not 0. However, the function \(h=h^{x_0}\) is very small near \(x_0\) . Precisely, by the \(C^{k,\gamma }\) -regularity of \(\varphi \) ,

Notice that the function \(u^{x_0}\) inherits the local behavior of u . In what follows we will study the properties of the functions \(u^{x_0}\) at all the points \(x_0\) on the boundary \(\Gamma (u)\) of the contact set \(\Lambda (u)\) .

1.5 State of art

In this section we give a brief overview on the state of the art of the obstacle for the fractional Laplacian; for more details we refer to [ 9 ] for \( s\in (0,1)\) , and to [ 11 , 23 ] for the case \(s=\frac{1}{2}\) .

The obstacle problem for the fractional Laplacian was studied by Silvestre [ 25 ], where it was established the almost-optimal regularity \(C^{1,\alpha }\) of the solution for all \(\alpha \in (0,s)\) , in the case \(\varphi \in C^{2,1}(\mathbb {R}^n)\) . Moreover, in the same paper, it is proved that if \(\varphi \in C^{1,\beta }(\mathbb {R}^n)\) , then \(u\in C^{1,\alpha }(\mathbb {R}^n)\) for all \(\alpha \in (0,\min \{s,\beta \})\) .

Later, in [ 7 ], Caffarelli, Salsa and Silvestre proved the optimal regularity \(C^{1,s}\) of the solution for \(\varphi \in C^{2,1}(\mathbb {R}^n)\) , thus generalizing the result previously obtained by Athanasopoulos and Caffarelli [ 1 ] in the case \(s=\frac{1}{2}\) and \(\varphi \equiv 0\) . They use a modification of the following “pure” Almgren’s frequency function

which is monotone in r provided that \(\varphi \equiv 0\) . Precisely, setting

they introduce the following generalized Almgren’s frequency function:

with \(k=2\) , \(\gamma =1\) and \(p=1\) . This function is monotone in r also when \(\varphi \not \equiv 0\) , with \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) . Here we denoted by \(v=u^{x_0}\) , with \(x_0\in \Gamma (u)\) , a solution to a similar problem to ( 5 ).

As a consequence of the monotonicity of \(\Phi ^{x_0}\) , they obtain that if \(x_0\in \Gamma (u)\) is a free boundary point such that

for some \(\lambda \in (0,2)\) , then the rescalings

converge, up to a subsequence, to a function \(v_{0,x_0}\) which is a \(\lambda \) -homogeneous solution of ( 4 ) (with 0 obstacle) (see Lemma 6.1 and Lemma 6.2 in [ 7 ]).

Thus, the following set of free boundary points is well-defined:

for every \(\lambda \in (0,2)\) . Moreover, it was shown in [ 7 ] that \(\Gamma _\lambda (u)=\emptyset \) for all \(\lambda \in (0,2){\setminus }\{1+s\}\) .

We will call \(\Gamma _{1+s}(u)\) the set of regular points and we will denote it by \(\text{ Reg }(u)\) since it is known to be locally a \(C^{1,\alpha }\) submanifold of dimension \(n-1\) (see [ 7 ]). This is a generalization of the result of Athanasopoulos, Caffarelli and Salsa obtained in [ 2 ] in the case \(s=\frac{1}{2}\) and \(\varphi \equiv 0\) .

In the case \(\varphi \equiv 0\) , we can consider the “pure” Almgren’s frequency function as in ( 7 ), and the free boundary can be decomposed as

where \(\text{ Reg }(u)=\Gamma _{1+s}(u)\) are the regular points, \(\text{ Sing }(u)=\bigcup _{m\in \mathbb {N}}\Gamma _{2m}\) are the so-called singular points, and \(\text{ Other(u) }\) are all the remaining points in \(\Gamma (u)\) .

For the singular points, in the case \(s=\frac{1}{2}\) , in [ 18 ], Garofalo and Petrosyan proved that \(\text{ Sing }(u)\) is contained in the union of at most countably many submanifolds of class \(C^{1}\) . In [ 21 ], using the monotonicity of the generalized Almgren’s frequency for \(k\ge 2\) , \(\gamma \in (0,1)\) and p small enough, Garofalo and Ros-Oton extended the structure of singular set to any \(s\in (0,1)\) . Indeed in the case \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , \(k\ge 2\) and \(\gamma \in (0,1)\) , they proved that

is contained in the union of countably many submanifolds \(C^{1}\) , where the bound \(2m\le k\) is needed in order to assure the existence of blow-ups. Thus, they improved the result previously obtained in [ 18 ] (in the case \(s=\frac{1}{2}\) ), where more regularity of the obstacle \(\varphi \) was required.

Moreover, Focardi and Spadaro described the entire free boundary, up to sets of null \(\mathcal {H}^{n-1}\) measure, in the case \(\varphi \equiv 0\) in [ 14 ] and in the case \(\varphi \not \equiv 0\) , with \(\varphi \) suitably smooth and decaying fast at infinity, in [ 15 ].

In general case, even \(\varphi \in C^\infty \) , the free boundary might contain points of infinite contact order, as shown explicitly by Fernández-Real and Ros-Oton in [ 12 ]. However, Fernández-Real and co-authors in [ 12 , 16 ] proved that, “generically”, the set where the free boundary is not smooth is at most \((n-2)\) -dimensional and the non-regular set has zero \(\mathcal {H}^{n-3-\alpha _0}\) measure for a “generic” solution. This solves the analogue of a conjecture of Schaeffer in \(\mathbb {R}^3\) and \(\mathbb {R}^4\) for the thin obstacle problem.

An alternative proof of the regularity and structure of the free boundary uses epiperimetric inequalities approach for the Weiss’ energy

In the case \(s=\frac{1}{2}\) and \(\varphi \equiv 0\) , Garofalo, Petrosyan and Smit Vega Garcia [ 20 ] and Focardi and Spadaro [ 13 ] proved an epiperimetric inequality for \(W_{1+s}\) to deduce the regularity \(C^{1,\alpha }\) of the regular points \(\text{ Reg }(u)\) . In the case \(s\in (\frac{1}{2},1)\) and \(\varphi \not \equiv 0\) , using epiperimetric inequalities approach, the same regularity for \(\Gamma _{1+s}(u)\) was established by Garofalo, Petrosyan, Pop and Smit Vega Garcia in [ 19 ] (see also [ 17 ] for \(\varphi \equiv 0\) and \(s\in (0,1)\) ). The regularity of the free boundary \(\Gamma _{1+s}(u)\) in the case of more general degenerate elliptic operators for variable coefficients was established recently by Banerjee et al. [ 3 ], using again an epiperimetric inequality.

Following epiperimetric inequalities approach, Colombo et al. [ 8 ] give an alternative proof of the structure of singular set \(\text{ Sing }(u)\) in the case \(s=\frac{1}{2}\) and \(\varphi \equiv 0\) . They improved the regularity of the manifolds that contains the singular set up to \(C^{1,\log }\) , due to the logarithmic epiperimetric inequality for \(W_{2m}\) .

1.6 Main results

The goal of this paper is to generalize the epiperimetric inequalities that we already know for the thin obstacle problem \(s=\frac{1}{2}\) , to any \(s\in (0,1)\) . With this generalization, we can deduce the previous results of regularity and structure of the free boundary, even for non-zero obstacles \(\varphi \not \equiv 0\) , with \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , \(k\ge 2\) and \(\gamma \in (0,1)\) . In particular, we prove an epiperimetric inequality for \(W_{1+s}\) and a logarithmic epiperimetric inequality for \(W_{2m}\) , for each \(s\in (0,1)\) .

Before we state our main results (Theorems 1.1 and 1.2 below), we recall that

is the set of admissible function, for each \(c\in H^1(\partial B_1,a)\) . We will also denote by \(W_\lambda (u)\) the Weiss’ energy \(W_\lambda ^{x_0}(r,u)\) , when \(x_0=0\) and \(r=1\) .

Theorem 1.1

(Epiperimetric inequality for \(W_{1+s}\) ) Let \(\mathcal {K}_c\) defined in ( 10 ) and \(z=r^{1+s}c(\theta )\in \mathcal {K}_c\) be the \((1+s)\) -homogeneous extension in \(\mathbb {R}^{n+1}\) of a function \(c\in H^1(\partial B_1,a)\) . Therefore there is \(\zeta \in \mathcal {K}_c\) such that

with \(\kappa =\frac{1+a}{2n+a+5}.\)

Theorem 1.2

(Logarithmic epiperimetric inequality for \(W_{2m}\) ) Let \(\mathcal {K}_c\) defined in ( 10 ) and \(z=r^{2m}c(\theta )\in \mathcal {K}_c\) be the 2 m -homogeneous extension in \(\mathbb {R}^{n+1}\) of a function \(c\in H^1(\partial B_1,a)\) . We also suppose that exists a constant \(\Theta >0\) such that

then there is \(\zeta \in \mathcal {K}_c\) such that

with \(\beta =\frac{n-1}{n+1}\) and \({\varepsilon }={\varepsilon }(n,m,a)>0\) small enough.

The first inequality was originally proved in [ 20 ] and in [ 13 ] for \(s=\frac{1}{2}\) and generalized to any \(s\in (0,1)\) in [ 19 ] and in [ 17 ]. For the proof, we use an alternative method, that follows the idea in [ 8 ], decomposing the datum \(c\in H^1(\partial B_1,a)\) in terms of eigenfunctions of \(L_a\) restricted to \(\partial B_1\) .

The second inequality was originally proved in [ 8 ] for \(s=\frac{1}{2}\) , but this is a new result for each \(s\in (0,1)\) .

Moreover, we will give a proof of two epiperimetric inequalities for negative energies Footnote 1 for \(W_{1+s}\) and for \(W_{2m}\) (see Theorems 5.1 and 5.2 ), originally proved for \(s=\frac{1}{2}\) in [ 5 ] and in [ 8 ] respectively. Using this two epiperimetric inequalities for negative energies, together with the first two epiperimetric inequalities in Theorems 1.1 and 1.2 , we deduce a frequency gap.

Proposition 1.3

(Frequency gap) Let u be a solution of ( 2 ) and \(v=u^{x_0}\) be the solution of ( 5 ) with \(x_0\in \Gamma (u)\) , \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , \(k\ge 2\) and \(\gamma \in (0,1)\) .

If v is \(\lambda \) -homogeneous with \(\lambda <k+\gamma \) , then

for some constants \(c^\pm _{m,a}>0\) , that depend only on n ,  m and a .

In particular, if u is a \(\lambda \) -homogeneous solution of ( 4 ) (with 0 obstacle) with \(\lambda >0\) , then the same conclusion hold for u .

Notice that \(\lambda \not \in \bigcup _{m\in \mathbb {N}}\left( (2m-c^-_{m,a})\cup (2m+c^+_{m,a})\right) \) is a new result for any \(s\in (0,1)\) and it is originally proved in [ 8 ] for \(s=\frac{1}{2}\) and \(\varphi \equiv 0\) . Observe that the function \(-|y|^{2s}\) is a solution of ( 4 ) (with 0 obstacle), then we are able to prove the best frequency gap around \(1+s\) .

Furthermore, we use the epiperimetric inequalities to deduce a characterization of the \(\lambda \) -homogeneous solutions of ( 4 ) (with 0 obstacle), in the case \(\lambda =1+s\) and \(\lambda =2m\) , as described in the following proposition.

Proposition 1.4

Let u be a \(\lambda \) -homogeneous solution of ( 4 ) (with 0 obstacle).

If \(\lambda =1+s\) , then \(u=Ch_e^s\) , for some \(C\ge 0\) and \(e\in \partial B'_1\) , where

If \(\lambda =2m\) , then \(u=p_{2m}\) , for some \(p_{2m}\) that is a polynomial 2 m -homogeneous and \(L_a\) -harmonic, with \(p_{2m}\ge 0\) on \(B'_1\) .

In particular we characterized the blow-ups of a solution u at \(x_0\in \Gamma (u)\) with frequency \(\lambda =1+s\) or \(\lambda =2m.\)

Finally we use the epiperimetric inequality for \(W_{1+s}\) in Theorem 1.1 and the logarithmic epiperimetric inequality for \(W_{2m}\) in Theorem 1.2 to get the regularity and the structure of the free boundary. In particular we prove the regularity of the points on the free boundary with frequency \(1+s\) , denoted by \(\Gamma _{1+s}(u)\) , and we describe the structure of the points with frequency 2 m , denoted by \(\Gamma _{2m}(u)\) , with \(2m\le k.\) See ( 19 ) below for the definition of \(\Gamma _{\lambda }(u)\) .

Theorem 1.5

Let u be a solution of ( 2 ) with obstacle \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , \(k\ge 2\) and \(\gamma \in (0,1)\) .

\(\Gamma _{1+s}(u)\) is locally a \(C^{1,\alpha }\) submanifold of dimension \(n-1\) , for some \(\alpha >0\) , i.e. for all \(x_0\in \Gamma _{1+s}(u)\) , there is \(\rho >0\) and \(g:U\subset \mathbb {R}^{n-1}\rightarrow \mathbb {R}\) of class \(C^{1,\alpha }\) such that

\(\Gamma _{2m}(u)\) is contained in the union of at most countably many submanifolds of class \(C^{1,\log }\) , for all \(2\,m\le k\) . In particular, for such m , we have that for every \(j\in \{0,\ldots n-1\},\)

is locally contained in a \(C^{1,\log }\) submanifold of dimension j , where \(d_{2m}^{x_0}\) is defined in ( 13 ) below. In particular, when \(\varphi \in C^{\infty }(\mathbb {R}^n)\) , the singular sets \(\cup _{m\in \mathbb {N}}\Gamma _{2m}^j(u)\) is contained in the union of countably many j -dimensional submanifolds of class \(C^{1,\log }\) . Moreover, the singular set \(\text{ Sing }(u)=\cup _{m\in \mathbb {N}}\Gamma _{2\,m}(u)\) is contained in the union of countably many submanifolds of class \(C^{1,\log }\) .

Here we have defined

to be the dimension of the “tangent plane”, where \(p_{2m}^{x_0}\) is the unique homogeneous blow-up (see Proposition 1.4 ) of u at \(x_0\in \Gamma _{2m}(u)\) .

In particular, we improve the regularity of submanifolds that contain \(\Gamma _{2m}^j(u)\) from \(C^{1}\) (proved in [ 21 ]) to \(C^{1,\log }\) .

1.7 Structure of the paper

The paper is organized as follows.

In Sect.  2 , we recall the generalized Almgren’s frequency function, the Weiss’ energy W for 0 obstacle and the Weiss’ energy \({\widetilde{W}}\) for non-zero obstacle. We also introduce the operator \(L_a^S\) , i.e. \(L_a\) restricted to \(\partial B_1\) , its eigenfunctions and its relation to the Weiss’ energy W . Finally, we recall the properties of the function \(h_e^s\) defined in ( 12 ).

In Sect.  3 , we prove Theorem 1.1 , i.e. the epiperimetric inequality for \(W_{1+s}\) .

In Sect.  4 , we prove Theorem 1.2 , i.e. the logarithmic epiperimetric inequality for \(W_{2m}\) .

In Sect.  5 , we state and prove two epiperimetric inequalities for negative energies \(W_{1+s}\) and \(W_{2m}\) (Theorems 5.1 and 5.2 ).

In Sect.  6 , using the four epiperimetric inequalities proved in the previous sections, we establish a frequency gap in Proposition 1.3 .

In Sect.  7 , using Theorems 1.1 and 1.2 , we prove Proposition 1.4 , i.e. the characterization of the \(\lambda \) -homogeneous solutions of ( 4 ) (with 0 obstacle), in the case \(\lambda =1+s\) and \(\lambda =2m\) .

In Sect.  8 , we use Theorems 1.1 and 1.2 to prove our main result on the regularity and the structure of the free boundary (Theorem 1.5 ).

2 Preliminaries

2.1 generalized almgren’s frequency function.

The original generalized Almgren’s frequency function in [ 7 ] must be modified in the case \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , as in the following proposition.

We follow [ 21 ], but a similar generalized Almgren’s frequency can be found in [ 7 ] or in [ 4 ].

First we recall the function \(H^{x_0}(r):=H^{x_0}(r,v)\) as in ( 8 ), with \(v=u^{x_0}\) . We drop the dependence on \(x_0\) if it is not ambiguous.

Proposition 2.1

(Monotonicity of generalized Almgren’s frequency) Let u be a solution of ( 2 ) and \(v=u^{x_0}\) be the solution of ( 5 ) with \(x_0\in \Gamma (u)\) , \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , \(k\ge 2\) and \(\gamma \in (0,1)\) . We define

for \(p\in (0,\gamma )\) and C large enough. Therefore, there is \(r_0>0\) such that the function \(r\mapsto \Phi ^{x_0}(r,v)\) is monotone increasing for \(r\in (0,r_0)\) .

The proof is technical and we skip it, since it is standard and proved in many variations. We refer to Proposition 6.1 in [ 21 ] for the complete proof. \(\square \)

By the previous proposition, if \(v=u^{x_0}\) , it is well-defined

for \(x_0\in \Gamma (u)\) .

In the case of obstacle \(\varphi \equiv 0\) , if the “pure” Almgren’s frequency \(N^{x_0}(0^+,u)=\lambda \) , with \(N^{x_0}\) as in ( 7 ), then we can consider a blow-up of u at \(x_0\) , that is a \(\lambda \) -homogeneous solution (see [ 2 ]).

We want a similar result for \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) . Precisely, we will show that if \(\Phi ^{x_0}(0^+,v)=n+a+2\lambda \) , then the blow-up of v at \(x_0\) is a \(\lambda \) -homogeneous solution. But this is only true if \(\lambda <k+\gamma \) , and this is the motivation for defining a new generalized Almgren’s frequency, which was originally introduced in [ 7 ].

To prove this result, we need the following two lemmas from [ 21 ].

Let u be a solution of ( 2 ) and \(v=u^{x_0}\) be the solution of ( 5 ) with \(x_0\in \Gamma (u)\) , \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , \(k\ge 2\) and \(\gamma \in (0,1)\) . If \(\Phi ^{x_0}(0^+,v)=n+a+2\lambda \) with \(\lambda <k+\gamma \) , then

for all \(r\in (0,r_0).\) Moreover, for every \({\varepsilon }>0\) there is \(r_{\varepsilon }\) such that

for all \(r\in (0,r_{\varepsilon })\) . In particular, \(\Phi ^{x_0}(0^+,v)\) does not depend on \(p\in (0,\gamma )\) .

The proof of the first part follows by the monotonicity of generalized Almgren’s frequency and an integration from r to \(r_0\) . The second part is similar. See Lemma 6.4 in [ 21 ] for more details. \(\square \)

Now we proceed with the second lemma.

By ( 6 ), we obtain

where in the last inequality we used ( 14 ). \(\square \)

Now we are ready to prove the existence of a \(\lambda \) -homogeneous blow-up, when \(\lambda <k+\gamma \) . We denote by

Proposition 2.4

Let u be a solution of ( 2 ) and \(v=u^{x_0}\) be the solution of ( 5 ) with \(x_0\in \Gamma (u)\) , \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , \(k\ge 2\) and \(\gamma \in (0,1)\) . If \(\Phi ^{x_0}(0^+,v)=n+a+2\lambda \) with \(\lambda <k+\gamma \) , then, up to subsequences, the rescalings

converge in \(C^{1,\alpha }_a(B_R^+)\) for all \(R>0\) to a blow-up \(v_{0,x_0}\) , as \(r\rightarrow 0^+\) , which is a solution of ( 4 ) (with 0 obstacle) and it is \(\lambda \) -homogeneous.

We can proceed as in the proof of Proposition 6.6 in [ 21 ].

We want to point out that, since the values of \(\Phi ^{x_0}(0^+,v)\) do not depend on \(p\in (0,\gamma )\) , by Lemma 2.2 , we can choose \(p\in (0,\gamma )\) small enough such that

for some \({\varepsilon }>0\) . Hence there is \(r_1>0\) such that \(H(r)\ge r^{n+a+2(k+\gamma -p)}\) for all \(r\in (0,r_1)\) , by ( 15 ).

In particular, the monotonicity of the function \(r\mapsto r(1+Cr^p)\frac{H'(r)}{H(r)}\) is equivalent to the monotonicity of the function \(r\mapsto \Phi ^{x_0}(r,v)\) for r small enough. \(\square \)

Notice that the blow-up \(v_{0,x_0}\) is not a priori unique. Still, all the blow-ups at \(x_0\) have the same homogeneity. Therefore, for \(\lambda <k+\gamma \) , the following set is well-defined:

Equivalently, \(\Gamma _\lambda (u)\) is the set of all points \(x_0\in \Gamma (u)\) such that \(\Phi ^{x_0}(0^+,v)=n+a+2\lambda \) , with \(\lambda <k+\gamma \) .

As in the proof of the last proposition, we get

for r small enough. Then, by ( 15 ) and ( 16 ), we obtain

2.2 Properties of the Weiss’ energy \({\widetilde{W}}\)

In the case \(\varphi \not \equiv 0\) , we can consider the following Weiss’ energy, which is a small modification of the Weiss’ energy W for the obstacle \(\varphi \equiv 0\) , defined in ( 9 ). Precisely, we define

and we drop the dependence on \(x_0\) if \(x_0=0\) .

Let u be a solution of ( 2 ) and \(v=u^{x_0}\) be the solution of ( 5 ) with \(x_0\in \Gamma (u)\) , \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , \(k\ge 2\) and \(\gamma \in (0,1)\) . If

We give only the idea of the proof; for more details we refer to [ 19 ]. First we can change the variables in the expression of H ( r ) and with a standard computation we get ( 21 ).

Now, if \(X=(x,y)\) , then using

and an integration by parts, we get ( 22 ). \(\square \)

We next show that also the Weiss’ energy \({\widetilde{W}}\) satisfies a monotonicity formula.

Proposition 2.7

(Monotonicity of the Weiss’ energy \({\widetilde{W}}\) ) Let u be a solution of ( 2 ) and \(v=u^{x_0}\) be the solution of ( 5 ) with \(x_0\in \Gamma (u)\) , \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , \(k\ge 2\) and \(\gamma \in (0,1)\) . If \(\lambda <k+\gamma \) , then there exists contants \(C,r_0>0\) such that for all \(x_0\in \Gamma (u)\) and for all \(r\in (0,r_0)\) , it holds

i.e. the function \(r\mapsto {\widetilde{W}}^{x_0}_\lambda (r,v)+Cr^{k+\gamma -\lambda }\) is increasing for all \(r\in (0,r_0)\) .

The original proof is done in [ 19 ], Proposition 3.5, in the case \(\lambda =1+s\) and with a different exponent for r in left-hand side. We generalize this result to any \(\lambda \) and to any \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) . \(\square \)

Since we have

by ( 21 ) and ( 22 ), we obtain

Now, we can use ( 16 ) to get

which proves the claim.

Proposition 2.8

Let u be a solution of ( 2 ) and \(v=u^{x_0}\) be the solution of ( 5 ) with \(x_0\in \Gamma (u)\) , \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , \(k\ge 2\) and \(\gamma \in (0,1)\) . If \(\lambda <k+\gamma \) and \(x_0\in \Gamma _{\lambda }(u)\) , then

In particular, using ( 23 ), we get

for all \(r\in (0,r_0)\) .

We can write

Hence, using ( 15 ) and Remark 2.5 , we get

which concludes the proof. \(\square \)

2.3 Homogeneous rescalings and homogeneous blow-up

The sequence ( 18 ) has good rescaling properties with respect to the Almgren’s frequency function. Now we consider another sequence of rescalings, the homogeneous rescalings, which have good rescaling properties with respect to the Weiss’ energy.

Proposition 2.9

Let u be a solution of ( 2 ) and \(v=u^{x_0}\) be the solution of ( 5 ) with \(x_0\in \Gamma (u)\) , \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , \(k\ge 2\) and \(\gamma \in (0,1)\) . Suppose that \(\Phi ^{x_0}(0^+,v)=n+a+2\lambda \) with \(\lambda <k+\gamma \) . Let \(v^{(\lambda )}_{r,x_0}\) be the homogeneous rescalings of v at \(x_0\in \Gamma (u)\) , defined as

Then, up to a subsequence, the homogeneous rescalings converge in \(C^{1,\alpha }_a(B_R^+)\) for all \(R>0\) (defined in ( 17 )), as \(r\rightarrow 0^+\) , to a blow-up \(v^{(\lambda )}_{0,x_0}\) , which is \(\lambda \) -homogeneous and is a solution to ( 4 ) (with 0 obstacle).

By the Poincaré inequality ( 70 ), in order to show the boundness of \(v_r\) in \(H^1(B_R,a)\) , it is sufficient to prove the boundness of \(v_r\) in \(L^2(\partial B_R,a)\) and the boundness of \(|\nabla v_r|\) in \(L^2(B_R,a)\) . The first bound follows by ( 14 ). In fact

Also the second bound follows from ( 14 ):

where in the last inequality we used ( 15 ), ( 16 ) and the monotonicity of the function \(r\mapsto \Phi ^{x_0}(r,v)\) . Thus, for r small enough,

as in the proof of Proposition 2.4 .

Hence, up to subsequences, \(v_r\) converges to some \(v_0\) weakly in \(H^1(B_R,a)\) and

by ( 16 ), with an equality in \(\mathbb {R}^{n+1}{\setminus }(\{v_r=0\}\cap \{y=0\})\) . Therefore, we can use the estimate in [ 7 ] (Proposition 4.3 and Lemma 4.4) to get the convergence in \(C^{1,\alpha }_a\) , since \(v_r\) are solutions of ( 5 ), where the right-hand sides in the third and the fourth lines are as in ( 27 ). Moreover, \(v_0\) is a solution of ( 4 ) (with 0 obstacle), since we can send \(r\rightarrow 0^+\) in ( 27 ).

Finally, we show that \(v_0\) is homogeneous. Indeed, by ( 23 ), we have that for all \(0<R_1<R_2<r_0\)

where sending \(r_k\rightarrow 0^+\) the left-hand side vanishes, by ( 24 ). Thus, by arbitrariness of \(R_1\) and \(R_2\) , we obtain that \(v_0\) is homogeneous, since \(\nabla v_0\cdot x=\lambda v_0\) on \(\partial B_r\) for all \(r\in (0,r_0)\) . \(\square \)

2.4 The operator \(L_a^S\)

The strategy to prove the epiperimetric inequalities is to decompose a trace \(c\in H^1(\partial B_1,a)\) in terms of the eigenfunctions of the operator \(L_a\) restricted to \(\partial B_1\) . The restriction \(L_a^S\) is defined for any function \(\phi \in H^1(\partial B_1,a)\) as

where \(\Phi (x)=\phi \left( \frac{x}{|x |}\right) \) .

Remark 2.10

Since in spherical coordinates we have

if and only if

where \(\lambda ^a(\alpha )=\alpha (\alpha +n+a-1)\) .

By Liouville Theorem 9.5 , if we suppose that \(\phi \) is even in the y direction, then \(r^\alpha \phi (\theta )\) is \(L_a\) -harmonic if and only if \(r^\alpha \phi (\theta )\) is a polynomial \(L_a\) -harmonic with \(\alpha \in \mathbb {N}\) .

Using the theory of compact operators, we can prove that there exist an increasing sequence of eigenvalues \(\{\lambda ^a_k\}_{k\in \mathbb {N}}\subset \mathbb {R}_{\ge 0}\) and a sequence of eigenfunctions \(\{\phi _k\}_{k\in \mathbb {N}}\subset H^1(\partial B_1,a)\) normalized in \(L^2(\partial B_1,a)\) , such that

with \(\{\phi _k\}_{k\in \mathbb {N}}\) orthonormal basis of \(H^1(\partial B_1,a)\) .

The (normalized) eigenspace corresponding to eigenvalue \(\lambda ^a\) is

for all \(\lambda ^a\subset \{\lambda _k^a\}_{k\in \mathbb {N}}\) .

We denote by \(\alpha _k\in \mathbb {N}\) the grade of the polynomial that corresponds to the eigenvalue \(\lambda ^a_k\) , i.e. the only natural number such that \(\lambda ^a(\alpha _k)=\lambda ^a_k\) .

In particular \( \lambda ^a_1=\lambda ^a(0)=0\) and \(E(\lambda ^a_1)\) is the space of constant functions, while \( \lambda ^a_2=\ldots =\lambda ^a_{n+2}=\lambda ^a(1)=n+a\) and \(E(\lambda ^a_2)\) is the space of linear functions. Finally \(\lambda ^a_k \ge \lambda ^a(2)\) for \(k\ge n+3\) .

2.5 The Weiss’ energy W and eigenfunctions of \(-L_a^s\)

The following lemma is a generalization of Lemma 2.3 and Lemma 2.4 in [ 8 ], for the Weiss’ energy W with weight. It will be used in several proof later.

Let \(\phi \in H^1(\partial B_1,a)\) with

where \(\phi _k\) normalized eigenfunctions of \(-L_a^S\) as above, and let \(r^{\mu }\phi (\theta )\) be the \(\mu \) -homogeneous extension, then

Moreover, if

Finally, if \(c\in H^1(\partial B_1)\) such that \(r^{\mu +t}c\) is a solution of ( 4 ) (with 0 obstacle), then

The proof is very similar to the one in [ 8 ], but we briefly recall it for the sake of completeness.

By ( 65 ) we get

from where ( 30 ) and ( 32 ) follow.

Now if \(r^{\mu +t}c\) is a solution, then \(W_{\mu +t}(r^{\mu +t}c)=0\) , by ( 68 ). Therefore ( 33 ) holds.

Finally, the proof of ( 34 ) follows by ( 66 ), ( 67 ) and ( 33 ). \(\square \)

2.6 Properties of \(h_e^s\)

Finally we recall the properties of the function \(h_e^s\) defined in ( 12 ), which is the only \((1+s)\) -homogeneous solution of ( 4 ) (with 0 obstacle). The latter follows from Proposition 1.4 , but it was originally proved in [ 7 ].

Proposition 2.12

Let \(e\in \partial B_1'\) and \(h_e^s\) as in ( 12 ), then \(h_e^s\) is a \((1+s)\) -homogeneous solution of ( 4 ) (with 0 obstacle), \(h_e^s=0\) on \(B'_1\cap \{x\cdot e\le 0\}\) and it holds

with \(c_s=2^{1-s}(1+s)\) . Finally, the \(L^2(\partial B_1,a)\) projection on linear functions of \(h_e^s\) has the form \(C(x\cdot e)\) for some \(C>0\) .

The proof is a simple computation, hence it will be omitted. \(\square \)

3 Epiperimetric inequality for \(W_{1+s}\)

The proof of the epiperimetric inequality for \(W_{1+s}\) follows the ideas of the proof from [ 8 ] in the case \(s=\frac{1}{2}\) , i.e. we decompose the trace \(c\in H^1(\partial B_1,a)\) in terms of eigenfunction of \(L_a^S\) .

3.1 Decomposition of c

Let \(c\in H^1(\partial B_1,a)\) even in the y direction and such that \(c\ge 0\) in \(\partial B'_1\) . We decompose c using eigenfunctions of the operator \(-L_a^S\) defined in ( 28 ).

The projection on linear functions \(E(\lambda _2^a)\) of c has the form \(c_1(x\cdot e)\) for some \(e\in \partial B'_1\) , then the projection of \(h_e^s\) on \(E(\lambda _2^a)\) has the same form \(C(x\cdot e)\) for \(C>0\) . Thus we can choose \(C>0\) such that c and \(Ch_e^s\) has the same projection on \(E(\lambda _2^a)\) .

Notice that the function \(u_0(x,y)=|y |^{1+s}\) restricted on \(\partial B_1\) has 0 projection on \(E(\lambda _2^a)\) . Therefore we can choose \(c_0\in \mathbb {R}\) such that the projections of \(c-Ch_e^s\) and \(u_0\) on the constant functions \(E(\lambda ^a_1) \) are the same. Then

Hence we can decompose z as

and we can define the competitor \(\zeta \) as

which is an admissible function ( \(\zeta \in \mathcal {K}_c\) ), since \(\zeta (r,\theta )\ge r^2c(\theta )\) on \(B'_1\) .

3.2 Proof of Theorem 1.1

Let’s start with a lemma, that will allow us to compute the Weiss’ energy \(W_{1+s}\) of z and \(\zeta \) .

If \(\psi =r^\alpha \phi (\theta )\) , with \(\phi \in H^1(\partial B_1,a)\) , then

By Proposition 2.12 and ( 68 ), we obtain

Additionally, since \(u_0\) is \((1+s)\) -homogeneous and

we have that

Finally, we notice that

which, together with the previous identities, gives the claim. \(\square \)

Proof of Theorem 1.1

By this lemma, we can easily conclude the proof of the epiperimetric inequality for \(W_{1+s}\) . Indeed, if \(\kappa =\kappa ^a_{2,1+s}=\frac{1+a}{2n+a+5}\) , we can use

with ( 32 ), ( 35 ) and ( 36 ) to conclude. \(\square \)

If the equality in the epiperimetric inequality holds, then \(c_0=0\) and, by ( 32 ), \(\phi \) is an eigenfunction corresponding to the eigenvalue \(\lambda ^a(2)\) . Therefore

Furthermore, since \(z\ge 0\) on \(B'_1\) and \(h_e^s=0\) on \(B'_1\cap \{x\cdot e<0\}\) , we have that \(r^{1+s}\phi \ge 0\) on \(B'_1\cap \{x\cdot e<0\}\) , but \(r^{1+s}\phi \) is even in the y direction, so \(r^{1+s}\phi \ge 0\) on \(B'_1\) .

4 Logarithmic epiperimetric inequality for \(W_{2m}\)

The proof of the logarithmic epiperimetric inequality for \(W_{2m}\) follows the ideas of the proof from [ 8 ] in the case \(s=\frac{1}{2}\) . The strategy is the same as the one of the proof of Theorem 1.1 .

4.1 Construction of \(h_{2m}\)

For the proof of the logarithmic epiperimetric inequality, we need to build an eigenfunction of \(-L_a^S\) as follows.

There is a 2 m -homogeneous \(L_a\) -harmonic polynomial \(h_{2m}\) such that \(h_{2m}\equiv 1\) on \(\partial B'_1\) .

The polynomial is given by

where the constants \(C_k\) are yet to be chosen.

Notice that

Thus, \(h_{2m}\) is \(L_a\) -harmonic if and only if

Therefore, we can choose

for \(k\in \{1,\ldots m\}\) and \(C_0=1\) , which concludes the construction of \(h_{2m}\) .

4.2 Decomposition of c

Let \(c\in H^1(\partial B_1,a)\) , we can decompose

where \(\phi _k(\theta )\) are the normalized eigenfunctions of \(-L_a^S\) with eigenvalues \(\lambda _k^a\) and corresponding homogeneity \(\alpha _k\) , then

Let \(z=r^{2m}c\) be the 2 m -homogeneous extension of c and let \(h_{2m}\) as in Remark 4.1 , therefore

where \(M=\max \{P_-(\theta '):\theta '\in \partial B'_1\}\) and \(P_-(\theta )\) is the negative part of \(P(\theta )\) .

We choose a competitor \(\zeta \) extending with homogeneity \(\alpha >2m\) the high modes on the sphere and leaving the rest unchanged, i.e.

for some \(2m<\alpha <2m+\frac{1}{2}\) , then

on \(B'_1\) , since we have chosen \(M>0\) such that \(P(\theta )+M\ge 0\) \(\forall \theta \in \partial B'_1\) . This means that \(\zeta \in \mathcal {K}_c\) .

Defining \(\kappa _{\alpha ,2m}^a\) as in ( 31 ), we will choose \(\varepsilon =\varepsilon (n,m)>0\) small enough and \(2m<\alpha <2m+\frac{1}{2}\) such that

for \(\Theta >0\) and \(\beta =\frac{n-1}{n+1}\) . Notice that to be able to choose such \(\alpha \) , we must have an estimate of the type

for some \(C_{n,m}>0\) , that should depend only on n , m and a , since we want that \(\alpha \) depends only on n , m and a . For this reason we ask for the bounds \(\Vert c\Vert _{L^2(\partial B_1),a}^2\le \Theta \) and \(|W_{2\,m}(z)|\le \Theta \) .

4.3 Proof of Theorem 1.2

First we want to compute the term \(W_{2m}(\zeta )-(1-\kappa _{\alpha ,2m}^a)W_{2m}(z)\) , and we see that for \(\alpha \) near \(2\,m\) and \(\alpha >2\,m\) , it is negative. This is contained in the following lemmas.

In the hypotheses of Theorem 1.2 , we have

for some \( C_1,C_2>0\) depending only on n , m and a .

We introduce the following functions:

Then, we can write \(z=\psi +H_{2m}+\varphi _{2m}\) and \(\zeta =\psi +H_{2m}+\varphi _\alpha \) .

With this decomposition, \(\psi \) is orthogonal in \(L^2(B_1,a)\) and in \(H^1(B_1,a)\) to \(\varphi _\mu \) for \(\mu =\alpha ,2m\) . Additionally, \(H_{2m}\) is \(L_a\) -harmonic and 2 m -homogeneous, therefore using ( 68 ), we get

Notice now that, since \(\psi \) has only frequencies lower than 2 m , we have \(W_{2m}(\psi )<0\) . Thus, using ( 30 ), we get

where in the second equality we used ( 32 ) with \(C_1\) and \({\overline{C}}\) depending only on n , m and a .

Observe that

where we used

and we have chosen

Hence, we conclude by choosing \(C_2={\overline{C}} {\overline{C}}_2\) and \(C_1\) as above. \(\square \)

In the same hypotheses of the Theorem 1.2 , we have

for some \(C_3>0\) that depends only on n , m and a .

Since \(\phi _k\) are \(C^1\) on \(\partial {B_1}\) and only a finite number of k is such that \(\alpha _k\le 2m\) , we have

for all k such that \(\alpha _k\le 2m\) and for some \(L_m>0\) .

Moreover, the coefficients \(c_k\) corresponding to P are bounded by \(\Theta ^\frac{1}{2}\) , since \(\Vert P\Vert _{L^2(\partial B_1,a)}^2\le \Vert c\Vert _{L^2(\partial B_1,a)}^2\le \Theta \) , we get P is L -Lipschitz continuous on \(\partial {B_1}\) , with \(L=L_{n,m,a}\Theta ^\frac{1}{2}\) .

Now, since \(\phi ^2(\theta )\ge P_-^2(\theta )\) for all \(\theta \in \partial B'_1\) , it follows that

by Proposition 9.6 , and since \(\phi \) contains only eigenfunctions corresponding to eigenvalue \(\lambda ^a_k>\lambda ^a(2m)\) .

Finally we claim that

in fact the norm in \(L^2\) of \(P_-^2(\theta )\) is controlled by the volume of an \((n-1)\) -dimensional cone with height \(M^2\) and radius of the base \(\frac{M}{L}\) , since the graph of \(P_-\) must be above this cone, by the Lipschitz continuity of \(P_-\) .

Thus, since \(1-\beta =\frac{2}{n+1}\) and \(L=L_{n,m}\Theta ^\frac{1}{2}\) , we obtain

which is precisely ( 41 ). \(\square \)

Proof of Theorem 1.2

Notice that we can suppose \(W_{2m}(z)>0\) , otherwise we have done.

First, as already observed, we must show the estimate ( 38 ).

In fact, as in ( 40 ) with \(\alpha =2m\) , we have

where in the last equality we used ( 30 ).

Using the orthogonality of P and \(\phi \) and again ( 30 ), we obtain

where we used that \(\Vert c\Vert _{L^2(\partial B_1)}^2\le \Theta \) and \(|W_{2m}(z)|\le \Theta \) , that concludes the estimate.

If we choose \(\varepsilon <\frac{C_2}{C_1C_3}\) , where \(C_1,C_2,C_3\) are the constant in ( 39 ) and ( 41 ) and if we choose \(\kappa _{\alpha ,2m}\) as in ( 37 ), we deduce that

that conclude the proof. \(\square \)

We have proved a stronger version of the logarithmic epiperimetric inequality, that is

for \(\varepsilon _1=\frac{C_2\varepsilon }{2}\) , choosing \(\varepsilon <\frac{C_2}{2C_1C_3}\) in ( 42 ).

5 Epiperimetric inequalities for negative energies

In this section we prove two epiperimetric inequalities for negative energies Footnote 2 \(W_{1+s}\) and \(W_{2m}\) .

These epiperimetric inequalities are a generalization for the case \(s=\frac{1}{2}\) and they allow us to prove the backward frequency gap in Proposition 1.3 .

5.1 Epiperimetric inequality for negative energies \(W_{1+s}\)

In the case \(s=\frac{1}{2}\) , the epiperimetric inequality for negative energies \(W_{1+s}\) was proved in [ 5 ]. We follow the same idea.

Theorem 5.1

(Epiperimetric inequality for negative energies \(W_{1+s}\) ) Let \(\mathcal {K}_c\) be defined as in ( 10 ) and \(z=r^{1+s}c(\theta )\in \mathcal {K}_c\) be the \((1+s)\) -homogeneous extension in \(\mathbb {R}^{n+1}\) of a function \(c\in H^1(\partial B_1,a)\) . Then, there is \(\zeta \in \mathcal {K}_c\) such that

with \(\varepsilon =\frac{1+a}{2n-a+3}.\)

Let z be the \((1+s)\) -homogeneous extension of its trace \(c\in H^1(\partial B_1,a)\) . Then, we can decompose z as

with \(u_0(x,y)=|y|^{2s}\) , as in the proof of Theorem 1.1 . Therefore the explicit competitor is

which is an admissible function, since \(\zeta =z\ge 0\) on \(B'_1\) , i.e. \(\zeta \in \mathcal {K}_c\) .

Now we want to compute the Weiss’ energy of \(Cr^{1+s}h^s_e(\theta )+c_0r^\alpha u_0(\theta )+r^{1+s}\phi (\theta )\) , for \(\alpha =2s,{1+s}\) . By ( 68 ), we have \(W_{1+s}(h^s_e)=0\) . Then

where in the last equality we used that \(u_0\equiv \) 0 on \(B'_1\) , combined with Proposition ( 2.12 ).

Integrating by parts, we get that

For I , we notice that the function \(-r^{2\,s}u_0(\theta )=-|y |^{2\,s} \) is a solution of ( 4 ) (with 0 obstacle), then using ( 33 ), we obtain

and using ( 34 ), we get

since \(\varepsilon =\frac{1+a}{2n-a+3}\) , with a simple calculation.

For J , using ( 30 ), we deduce that

since \(\lambda ^a(2)\ge \lambda ^a({1+s})\) .

For K , we integrate by parts

Now, by ( 29 ), we get

with \(\lambda ^a(2\,s)=2ns\) and \(\lambda ^a({1+s})=(1+s)n+(1+s)(1-s)\) , then

where in the last equality we used ( 64 ). Hence, we obtain

For L , by Proposition ( 2.12 ), we have

where we used that \(h_e^s(\theta )=u_0(\theta )=0\) for \(\theta \in B'_1\cap \{x\cdot e<0\}\) and that \(\phi =c\ge 0\) in \(B'_1\cap \{x\cdot e<0\}\) .

Finally, since \(I,J,K,L\le 0\) , we conclude by using ( 44 ). \(\square \)

5.2 Epiperimetric inequality for negative energies \(W_{2m}\)

In the case \(s=\frac{1}{2}\) , the epiperimetric inequality for negative energies \(W_{2m}\) was proved in [ 8 ]. We follow the same idea.

Theorem 5.2

(Epiperimetric inequality for negative energies \(W_{2m}\) ) Let \(\mathcal {K}_c\) be defined as in ( 10 ) and let \(z=r^{2m}c(\theta )\in \mathcal {K}_c\) be the 2 m -homogeneous extension in \(\mathbb {R}^{n+1}\) of a function \(c\in H^1(\partial B_1,a)\) such that \(\Vert c\Vert _{L^2(\partial B_1,a)}^2\le 1\) . Then, there is \(\zeta \in \mathcal {K}_c\) such that

with \(\varepsilon ={\varepsilon }(n,m,a)>0\) small enough.

Notice that we can suppose \(W_{2m}(z)<0\) , otherwise we have done.

We can decompose \(c\in H^1(\partial B_1,a)\) as

where \(\phi _k(\theta )\) are as above, with

We decompose \(z=r^{2m}c\) , the 2 m -homogeneous extension of c , as

where \(h_{2m}\) as in Remark 4.1 and \(M=\max \{P_-(\theta ): \theta \in \partial B_1'\}\) as above.

We define \(\varepsilon =\kappa ^a_{2m,\alpha }\) with \(\kappa ^a_{2m,\alpha }\) as in ( 31 ) and \(\alpha \in (2m-\frac{1}{2},2m)\) and we will choose \(\varepsilon =\varepsilon (n,m,a)>0\) small enough (which corresponds to choosing \(\alpha \) close to \(2\,m\) and \(\alpha <2\,m\) ).

The explicit competitor is

where we notice that \(\zeta \in \mathcal {K}_c\) since \(\zeta (r,\theta )\ge r^{2m} c(\theta )\ge 0\) on \(B'_1\) .

As above, we can define

and we write \(z=\psi _{2m}+H_{2m}+\varphi \) and \(\zeta =\psi _\alpha +H_{2m}+\varphi \) .

Moreover, since the functions \(\psi _\mu \) are orthogonal to \(\varphi \) in \(L^2( B_1,a)\) and in \(H^1( B_1,a)\) , since \(H_{2m}\) is \(L_a\) -harmonic and 2 m -homogeneous, and since \(W_{2m}(\varphi )>0\) by ( 30 ), we have that

where in the second equality we used ( 32 ).

Furthermore, if \(C_1=\lambda ^a(2m-\frac{1}{2})-\lambda ^a(2m-1)\) , then

where the last inequality follows by

where we used the Cauchy–Schwartz inequality and that all norms in a finite dimensional space are equivalent.

We deduce that

where we have chosen \(\varepsilon =\varepsilon (n,m,a)>0\) small enough. \(\square \)

6 Frequency gap

Once we have proved the epiperimetric inequalities in Theorems 1.1 , 1.2 , 5.1 and 5.2 , the proof of the frequency gap is standard, as done in [ 8 ].

First notice that without loss of generality we can consider a \(\lambda \) -homogeneous solution of ( 4 ) (with 0 obstacle). In fact, let \(v=u^{x_0}\) be a \(\lambda \) -homogeneous solution of ( 5 ) with obstacle \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , \(k\ge 2\) and \(\gamma \in (0,1)\) . If \(\lambda < k+\gamma \) , then \(\Phi ^{x_0}(0^+,v)=n+a+2\lambda \) by Remark 2.5 . Thus we can consider z , the blow-up of v at \(x_0\in \Gamma (u)\) , that is a solution of ( 4 ) (with 0 obstacle).

Proof of Proposition 1.3

Step 1. To prove the frequency gap around \(1+s\) , it is sufficient to check that if \(\lambda =1+s+t\) with \(t>0\) , then \(t\ge 1-s\) , while if \(t<0\) , then \(t\le -(1-s)\) .

Let \(c\in H^1(\partial B_1,a)\) be a trace of a \((1+s+t)\) -homogeneous solution, say \(r^{{1+s}+t}c(\theta )\) .

If \(t>0\) , then \(W_{1+s}(r^{{1+s}+t}c)=t\Vert c\Vert ^2_{L^2(\partial B_1,a)}>0,\) by ( 33 ). Thus, using the epiperimetric inequality for \(W_{1+s}\) (Theorem 1.1 ), we obtain

where in the last equality we used ( 34 ). Therefore we deduce that

which implies that \(t\ge 1-s\) .

If \(t<0\) , then \(W_{{1+s}}(r^{{1+s}+t}c)=t\Vert c\Vert _{L^2(\partial B_1,a)}^2<0 \) by ( 33 ). Thus, using the epiperimetric inequality for negative energies \(W_{1+s}\) , i.e. Theorem 5.1 , we obtain

where we used ( 34 ). Hence, since we have a negative energy, we get

which gives \(t\le - ({1+s})\) .

Step 2. Let \(c\in H^1(\partial B_1,a)\) be a trace of a \((2m+t)\) -homogeneous solution, say \(r^{2m+t}c(\theta )\) , with \(\Vert c\Vert _{L^2(\partial B_1,a)}^2=1\) . It is sufficient to check that if \(t>0\) , then \(t\ge c_m^+\) , for some \(c_m^+>0\) , while if \(t<0\) , then \(t\le - c_m^-\) , for some \(c_m^->0\) .

Let \(t>0\) , then

by ( 33 ). Using the logarithmic epiperimetric inequality for \(W_{2m}\) , i.e. Theorem 1.2 , and ( 45 ), we obtain

where in the last equality we used ( 34 ). Therefore, since we have a negative energy, we have

which gives that \(t\ge c_m^+\) , for some explicit constant \(c_m^+>0\) .

If \(t<0\) then \( W_{2m}(r^{2m+t}c)=t<0 \) by ( 33 ), therefore using the epiperimetric inequality for negative energies \(W_{2m}\) , i.e. Theorem 5.2 , we obtain

where in the last equality we used ( 34 ). Thus we get

which is \(t\le -c_m^-\) for some explicit constant \(c_m^->0.\)

\(\square \)

7 Characterization of blow-ups

The epiperimetric inequality approach allows us to give an alternative proof of the characterization of blow-ups, in the spirit of [ 8 ]. For the original proof we refer to [ 7 ] and [ 21 ].

Proof of Proposition 1.4

Step 1. We want to prove that if z is a solution of ( 4 ) (with 0 obstacle) and it is \((1+s)\) -homogeneous, then \(z=Ch_e^s\) , \(e\in \partial B'_1\) and \(C\ge 0\) .

In fact, suppose that \(c\in H^1(\partial B_1,a)\) is the trace of z and let \(\zeta \) be the competitor in the epiperimetric inequality for \(W_{1+s}\) . Then \(W_{1+s}(z)=0\) , by Proposition 2.12 . Therefore

i.e. the epiperimetric inequality is an equality. Thus, by Remark 3.2 , \(c(\theta )=Ch_e^s(\theta )+\phi (\theta )\) , for some \(e\in \partial B'_1\) , \(C\ge 0\) and \(\phi \) eigenfunction corresponding to eigenvalue \(\lambda ^a(2)\) with \(\phi \ge 0\) on \(B'_1\) .

Hence, using Proposition 2.12 with an integration by parts, we get

where we used ( 30 ) and ( 68 ). In particular the last inequality is actually an equality. Then, by ( 30 ), we get that \(\phi \equiv 0\) , i.e. \(z=Ch_e^s\) for some \(C\ge 0\) .

Step 2. Suppose that z is a solution of ( 4 ) (with 0 obstacle) and that z is 2 m -homogeneous. Then we claim that \(z=p_{2m}\) for some polynomial \(p_{2m}\) which is \(L_a\) -harmonic.

Let \(c\in H^1(\partial B_1,a)\) be the trace of z and let \(\zeta \) be the competitor in the logarithmic epiperimetric inequality for \(W_{2m}\) . Without loss of generality, we can suppose \(\Vert c\Vert _{L^2(\partial B_1,a)}=1\) . Hence, by the strong version of the log-epiperimetric inequality 43 , we have

i.e. \(\Vert \nabla _\theta \phi \Vert _{L^2(\partial B_1,a)}=0\) .

Thus \(\phi \equiv 0\) and c contains only eigenfunctions corresponding to eigenvalues \(\lambda ^a_k\le \lambda ^a(2m)\) , i.e.

Using ( 30 ), we obtain

i.e. the frequencies \(\alpha _k<2m\) must vanish. Therefore c is an eigenfunction corresponding to eigenvalue \(\lambda ^a(2m)\) and it follows that the homogeneous extension z is a 2 m -homogeneous \(L_a\) -harmonic polynomial. \(\square \)

8 Regularity of \(\Gamma _{1+s}(u)\) and structure of \(\Gamma _{2m}(u)\)

We conclude this paper with the most important application of the epiperimetric inequalities in Theorems 1.1 and 1.2 , i.e. the proof of Theorem 1.5 . The proof following a standard argument, for instance see [ 19 ] for the case \(s\in (0,1)\) or [ 8 , 13 , 20 ], for the case \(s=\frac{1}{2}.\)

In this last section, we recall the results from [ 19 ], where the claim (1) was proved in the case \(s\in (\frac{1}{2},1)\) . The regularity assumption for \(\varphi \) , that is \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , allows us to generalize the result to any \(s\in (0,1)\) by using the same argument.

Indeed, in [ 19 ] it was proved that the function \(r\mapsto {\widetilde{W}}^{x_0}_{\lambda }(r,v)+Cr^{2s-1}\) is increasing (with a slight difference in the definition of v ), where \(2s-1>0\) . We notice that in the case \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , we have that the function \(r\mapsto {\widetilde{W}}^{x_0}_\lambda (r,v)+Cr^{k+\gamma -\lambda }\) is increasing, with \(k+\gamma -\lambda >0\) , by ( 23 ), with \(v=u^{x_0}\) the solution of ( 5 ).

For the case (2), we give the complete proof, which is based on similar arguments and uses ideas from [ 8 , 18 ]. Notice that in the case \(\lambda =2m\) , the condition \(\lambda < k+\gamma \) become \(\lambda =2\,m\le k\) since \(2\,m\) and k are integers.

8.1 Decay of the Weiss’ energy \({\widetilde{W}}\)

The first result, that follows by the epiperimetric inequalities, is the decay of the Weiss’ energy \({\widetilde{W}}\) for obstacle \(\varphi \not \equiv 0.\)

Proposition 8.1

(Decay of the Weiss’ energy \({\widetilde{W}}\) ) Let u be a solution of ( 2 ) and \(v=u^{x_0}\) be the solution of ( 5 ) with \(x_0\in \Gamma (u)\) , \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , \(k\ge 2\) and \(\gamma \in (0,1)\) . Let \(K\subset \Gamma _\lambda (u)\cap \mathbb {R}^{n+1}\) be a compact set.

If \(\lambda =1+s\) , then there is \(\alpha \in (0,1)\) such that

for all \(x_0\in \Gamma _{1+s}(u)\cap K\) and for all \(r\in (0,r_0)\) , for some \(r_0>0\) .

If \(\lambda =2\,m\le k\) and \(\beta \in (0,1)\) is the constant in Theorem 1.2 , then

for all \(x_0\in \Gamma _{2m}(u)\cap K\) and for all \(r\in (0,r_0)\) , for some \(r_0>0\) .

Let \(c_r\) be the \(\lambda \) -homogeneous extension of \(v_r|_{\partial B_1}\) . As in Lemma 5.1 in [ 19 ], we obtain

Now we separate the case \(\lambda =1+s\) and \(\lambda =2\,m\) .

Step 1. In the case \(\lambda =1+s\) , we have that

by ( 48 ), ( 49 ) and by epiperimetric inequality for \(W_{1+s}\) (Theorem 1.1 ).

Now for all \(\alpha \in (0,1)\) , we have

where we have chosen \(\alpha <c\) and we used ( 25 ). Integrating from r to \(r_0\) , we obtain ( 46 ).

Step 2. In the case \(\lambda =2m\) , since \(W_{2\,m}(1,c_r)\ge W_{2\,m}(1,\zeta _r)\ge 0\) , we have that

by epiperimetric inequality (Theorem 1.2 ) for \(W_{2m}\) (see Remark 8.2 below). Then

where we used ( 48 ), ( 49 ) and where we suppose \({\widetilde{W}}^{x_0}_{2m}(r,v)\ge 0\) , since otherwise we were done.

Let \(F(r)=\max \{{\widetilde{W}}^{x_0}_{2\,m}(r,v),\left( \frac{2}{c}\right) ^ {\frac{1}{1+\beta }}Cr^{\frac{k+\gamma -\lambda }{1+\beta }}\}\) , then we claim that

for all \(r\in (0,r_0)\) , up to decreasing \(r_0\) . In fact, if \(F'(r)=\frac{d}{dr}{\widetilde{W}}^{x_0}_{2m}(r,v)\) , then

by ( 50 ) and since \({\widetilde{W}}^{x_0}_{2\,m}(r,v)\ge \left( \frac{2}{c}\right) ^{\frac{1}{1+\beta }} Cr^{\frac{k+\gamma -\lambda }{1+\beta }}.\) While, if

since \({\widetilde{W}}^{x_0}_{2\,m}(r,v)\le \left( \frac{2}{c}\right) ^{\frac{1}{1+\beta }} Cr^{\frac{k+\gamma -\lambda }{1+\beta }},\) that is the claim

Now notice that the function \(r\mapsto -F(r)^{-\beta }-\frac{c}{2}\beta \log (r)\) is increasing for \(r\in (0,r_0)\) , by the previous inequality.

Therefore, if we choose \(r_0>0\) such that \({\widetilde{W}}_{2m}^{x_0}(r_0,v)\ge 0\) , we have that ( 47 ) holds, if \(r\le r_0^2\) .

In the Step 2 of the proof of Proposition 8.1 , we used the logarithmic epiperimetric inequality for the rescaled \(c_r\) , but to use Theorem 1.2 , we have to check that the conditions ( 11 ) hold with \(\Theta >0\) that does not depend on r .

In fact \(\Vert c_r\Vert ^2_{L^2(\partial B_1,a)}=\Vert v_r\Vert ^2_{L^2(\partial B_1,a)}\le \Theta \) as in the proof of Proposition 2.9 .

Moreover, by ( 66 )

where in the equality we used ( 65 ) and in the last inequality we used the boundness of \(v_r\) in \(H^1(B_1,a)\) , as in the proof of Proposition ( 2.9 ).

8.2 Decay of homogeneous rescalings

The decay of the Weiss’ energy allows us to prove a decay of the norm in \(L^1(\partial B_1,a)\) of the homogeneous rescalings. As a consequence, we get the uniqueness of the homogeneous blow-up.

Proposition 8.3

(Decay of homogeneous rescalings) Let u be a solution of ( 2 ) and \(v=u^{x_0}\) be the solution of ( 5 ) with \(x_0\in \Gamma (u)\) , \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , \(k\ge 2\) and \(\gamma \in (0,1)\) . Let \(v^{(\lambda )}_{r,x_0} \) be the homogeneous rescalings from ( 26 ), and \(K\subset \Gamma _\lambda (u)\cap \mathbb {R}^{n+1}\) be a compact set.

If \(\lambda =1+s\) , then, up to decreasing \(\alpha \) in Proposition 8.1 , we have

for all \(x_0\in \Gamma _{1+s}(u)\cap K\) and for all \(0<r'<r<r_0\) , for some \(r_0>0\) .

If \(\lambda =2m\le k\) , and \(\beta \in (0,1)\) is the constant in Theorem 1.2 , then

for all \(x_0\in \Gamma _{2m}(u)\cap K\) and for all \(0<r'<r<r_0\) , for some \(r_0>0\) .

In particular, the homogeneous blow-up \(v^{(\lambda )}_{0,x_0} \) is unique and the whole sequence \(v^{(\lambda )}_{r,x_0} \) converges to \(v^{(\lambda )}_{0,x_0} \) as \(r\rightarrow 0^+\) .

Dropping the dependence on x and \(\lambda \) , for \(0<r'<r<r_0\) , where \(r_0\) is as in the previous Proposition, using ( 23 ), we get

as in [ 19 ].

The conclusion follows by a dyadic decomposition as in [ 13 , 20 ] or [ 8 ], and by using ( 46 ) for \(\lambda =1+s\) , and ( 47 ) for \(\lambda =2m\) . \(\square \)

8.3 Non-degeneracy of homogeneous blow-up

Another consequence of the decay of the Weiss’ energy is the non-degeneracy of the homogeneous blow-ups, i.e. the homogeneous blow-ups cannot vanish identically.

Proposition 8.4

(Non-degeneracy of homogeneous blow-up) Let u be a solution of ( 2 ) and \(v=u^{x_0}\) be the solution of ( 5 ) with \(x_0\in \Gamma (u)\) , \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , \(k\ge 2\) and \(\gamma \in (0,1)\) . If \(\lambda =1+s\) or \(\lambda =2m\le k\) , then

for all \(x_0\in \Gamma _\lambda (u)\) and for all \(r\in (0,r_0)\) , for some \(r_0>0\) , where

In particular the homogeneous blow-up \(v^{(\lambda )}_{0,x} \) is non-trivial, since

Note that the inequality follows by the proof of ( 14 ), in fact we can obtain that the function \(r\mapsto \frac{H^{x_0}(r)}{r^{n+a+2\lambda }}\) is increasing for r small enough. Hence it is sufficient to prove that \(H_0^{x_0}>0\) .

If by contradiction \(H_0^{x_0}=0\) , we can arguing as in Lemma 7.2 in [ 8 ]. \(\square \)

8.4 Regularity of blow-ups

Roughly speaking, we prove the regularity of the map that to any point \(x\in \Gamma _\lambda (u)\cap K\) associates the homogeneous blow-up of \(v=u^{x}\) at x , where \(\lambda =1+s\) or \(\lambda =2m\) . We are able to prove the regularity with an explicit modulus of continuity that depend on the right-hand side of ( 51 ) and ( 52 ).

Proposition 8.5

Let u be a solution of ( 2 ) and \(v=u^{x}\) be the solution of ( 5 ) with \(x\in \Gamma _\lambda (u)\) , \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) , \(k\ge 2\) and \(\gamma \in (0,1)\) . Let \(v_{0,x}^{(\lambda )}\) be the \(\lambda \) -homogeneous blow-up of \(v=u^{x}\) at x and \(K\subset \Gamma _\lambda (u)\cap \mathbb {R}^{n+1}\) be a compact set.

If \(\lambda =1+s\) and

with \(\lambda _x>0\) , \(e(x)\in \partial B'_1\) and \(h_e^s\) as in ( 12 ), then up to decreasing \(\alpha \) in Proposition 8.3 , it holds

for all \(x_1,x_2\in \Gamma _{1+s}(u)\cap K\) .

If \(\lambda =2m\le k\) and

with \(\lambda _x>0\) and \(p_x\) a 2 m -homogeneous polynomial such that \(\Vert p_x\Vert _{L^2(\partial B_1,a)}=1\) , then

for all \(x_1,x_2\in \Gamma _{2m}(u)\cap K\) .

We sketched the proof, similar to the one done in [ 19 ].

First note that the characterization of blow-up and \(\lambda _x>0\) follows by Proposition 1.4 and Proposition 8.4 .

Secondly, for some dimensional constant \(c_0>0\) , we have

where we denote by

the modulus of continuity. We indicate explicitly the dependence on \(\alpha \) since it must be reduced to obtain the final claim.

Let now \(x_1,x_2\in \Gamma _\lambda (u)\cap K\) and \(r=|x_1-x_2|^\sigma \) with \(\sigma \in (0,1)\) to choose. Let \(w_r\) be the homogeneous rescalings in ( 26 ) of \(w=u-\varphi \) . Using the regularity of the solution and \(\nabla w(x_2,0)=0\) , we deduce that

where we can choose, for example, \(\sigma =\frac{1}{2m}\) for \(\lambda =2m\) .

Let \(Q ^{x_i}\) as in Lemma ( 9.7 ), we have

where in the last inequality we have used Lemma 9.7 for the first term, and the same computation in ( 58 ) for the second term. Therefore, recalling the definition of \(u^{x_i}\) that is a solution of ( 4 ), we deduce that

if \(\sigma \in (0,1)\) such that \((k+\gamma -\lambda )\sigma =1-\sigma \) in the case \(\lambda =1+s\) .

By the estimates ( 57 ) and ( 59 ) (with the regularity of the solution), we obtain

if we choose \(\delta =\alpha \sigma \wedge (1-\sigma )\) .

Furthermore, we denote by \(u_{0,x_i}\) the homogeneous blow-up of the rescalings in ( 26 ) of the function \(u^{x_i}\) . By ( 51 ), ( 52 ) and ( 59 ), we get

since \(\delta =\alpha \sigma \wedge (1-\sigma )\) .

Hence, by the homogeneity of \(u_{0,x_1}-u_{0,x_2}\) and by Proposition 9.6 , it follows that

where in the last inequality we used ( 60 ) (with the regularity of the solution).

Using ( 53 ), ( 55 ) and ( 61 ), we get

where we used the \(\omega \) -continuity of the function \(x\mapsto \lambda _x\) for \(x\in K\) to estimate from below \(\lambda _{x_1}\) .

Now, by ( 62 ) and the definition of \(h_e^s\) in ( 12 ), for \(\lambda =1+s\) we have

which implies ( 54 ).

Finally, for \(\lambda =2m\) , since all norms in a finite dimensional space are equivalent, we get

By a similar computation as in ( 62 ), we obtain

where we used ( 60 ), then we conclude ( 56 ) \(\square \)

8.5 Proof of Theorem 1.5

For the regularity of \(\Gamma _{1+s}(u)\) and the structure of \(\Gamma _{2m}(u)\) for \(2m\le k\) we can proceed with a standard argument, as in [ 8 , 13 , 18 , 19 , 20 ].

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Acknowledgements

I would like to thank Bozhidar Velichkov for all the useful discussion and encouragement. The author was partially supported by the European Research Council (ERC), EU Horizon 2020 programme, through the project ERC VAREG - Variational approach to the regularity of the free boundaries (No. 853404).

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In the following proposition, we see that a solution u of ( 4 ) (with 0 obstacle) is a solutions of \(L_a u=f\) in the whole ball \(B_1\) with right-hand side f which is a measure depending on u .

Proposition 9.1

Let \(u\in H^1(B_1,a)\) be a solution of \(L_au=0\) in \(B_1^+\) and even in the y direction. Then

This is a simple consequence of an integration by parts, but for the sake of completeness we give the complete proof.

We compute \(\text{ div }(|y|^a\nabla u)\) in distributional sense, as following

where we used the integration by parts

and \(-\text{ div }(|y|^a\nabla u)=0\) in \(B_1^+\cap \{y\ge \varepsilon \}\) . \(\square \)

In the following lemma we prove some useful identities.

Let \(w\in H^1(B_1,a)\) be \(\lambda \) -homogeneous. Then

where \(\nabla _\theta w(x)\) is the gradient of w on \(\partial B_1\) .

Moreover, if w is also a solution of ( 4 ) (with 0 obstacle), then

in particular

The proof is a straightforward computation. \(\square \)

The following lemma allows us to reduce the problem ( 2 ) for obstacle \(\varphi \not \equiv 0\) , to the problem ( 5 ) with 0 obstacle and right-hand side h .

Let \(q_k(x)\) be an homogeneous polynomial of degree k . Then there is a unique \({\widetilde{q}}_k(x,y)\) homogeneous polynomial of degree k such that

In particular, an homogeneous polynomial of degree k , that satisfies ( 69 ) and vanishes identically on \(\{y=0\}\) , must vanish identically on \(\mathbb {R}^{n+1}\) .

See Lemma 5.2 in [ 21 ]. \(\square \)

The following theorem is a generalization of Poincaré Theorem in weighted Sobolev space \(H^1(B_R,a).\)

Theorem 9.4

Let \(w\in H^1(B_R,a)\) . Then

with \(C=C(n,a)>0.\)

See Lemma 2.10 in [ 7 ]. \(\square \)

We recall the following generalization of the Liouville’s Theorem for entire \(L_a\) -harmonic functions.

Theorem 9.5

(Liouville Theorem) Let w be a global solution of \(L_aw(x,y)=0\) for \((x,y)\in \mathbb {R}^n\times \mathbb {R}\) such that w is even in the y direction and

Then, w is a polynomial.

See Lemma 2.7. in [ 7 ]. \(\square \)

In the following Proposition, we show an embedding from \(H^1(B_1,a)\) , the weighted Sobolev space in \(B_1\) , to \(L^2(B'_1)\) , the Lebesgue space on \(B'_1\) .

Proposition 9.6

If \(a\in (-1,1)\) , then there is a bounded operator \(T:H^1(B_1,a)\rightarrow L^2(B'_1) \) , i.e.

for all \(w\in H^1(B_1,a)\) . Moreover

for all \(\phi \in H^1(\partial B_1,a)\) .

See Theorem 2.8. in [ 22 ]. \(\square \)

Let \(\varphi \in C^{k,\gamma }(\mathbb {R}^n)\) and \(x_1,x_2\in \mathbb {R}^n\) . Let \(Q ^{x_i}(x,y)={\widetilde{q}}_k^{x_i}(x,y)-q_k^{x_i}(x)\) , where \(q_k^{x_i}\) is the k -th Taylor polynomial of \(\varphi \) at \(x_i\) and \({\widetilde{q}}_k^{x_i}\) is the extension according to Lemma 9.3 . If \(r=|x_1-x_2|^\sigma \) for some \(\sigma \in (0,1)\) , then

where \(Q^{x_i}_{r,x_1}\) are the rescalings of \(Q^{x_i}\) as in ( 26 ).

We denote by q the function \(q_k\) and we denote by \({\widetilde{q}}\) the function \({\widetilde{q}}_k\) . Moreover we consider \(q^{x_i}_{r,x_i}\) the rescalings of \(q^{x_i}\) as in ( 26 ).

Notice that, by \(C^{k,\gamma }\) -regularity of \(\varphi ,\) we get

where we used ( 72 ) in the last inequality.

Notice that the space of \(L_a\) -harmonic polynomials in \(\mathbb {R}^{n+1}\) of degree k which are even in the y direction is a finite dimensional space. Then

for each p which is a polynomial of degree k , \(L_a\) -harmonic and even in the y direction. Notice that the right-hand side is a norm by Lemma 9.3 .

Consider the operation of extension as in Lemma 9.3 and the operation of rescaling as in ( 26 ). These operations commute, by the explicit formulation of the extension, defined in Lemma 5.2 in [ 21 ].

Therefore, we have

by ( 72 ), which concludes the proof. \(\square \)

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Carducci, M. Epiperimetric inequalities in the obstacle problem for the fractional Laplacian. Calc. Var. 63 , 150 (2024). https://doi.org/10.1007/s00526-024-02767-9

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Published : 25 June 2024

DOI : https://doi.org/10.1007/s00526-024-02767-9

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Solve equations with fractions

Here you will learn about how to solve equations with fractions, including solving equations with one or more operations. You will also learn about solving equations with fractions where the unknown is the denominator of a fraction.

Students will first learn how to solve equations with fractions in 7th grade as part of their work with expressions and equations and expand that knowledge in 8th grade.

What are equations with fractions?

Equations with fractions involve solving equations where the unknown variable is part of the numerator and/or denominator of a fraction.

The numerator (top number) in a fraction is divided by the denominator (bottom number).

To solve equations with fractions, you will use the “balancing method” to apply the inverse operation to both sides of the equation in order to work out the value of the unknown variable.

The inverse operation of addition is subtraction.

The inverse operation of subtraction is addition.

The inverse operation of multiplication is division.

The inverse operation of division is multiplication.

For example,

\begin{aligned} \cfrac{2x+3}{5} \, &= 7\\ \colorbox{#cec8ef}{$\times \, 5$} \; & \;\; \colorbox{#cec8ef}{$\times \, 5$} \\\\ 2x+3&=35 \\ \colorbox{#cec8ef}{$-\,3$} \; & \;\; \colorbox{#cec8ef}{$- \, 3$} \\\\ 2x & = 32 \\ \colorbox{#cec8ef}{$\div \, 2$} & \; \; \; \colorbox{#cec8ef}{$\div \, 2$}\\\\ x & = 16 \end{aligned}

Common Core State Standards

How does this relate to 7th grade and 8th grade math?

• Grade 7: Expressions and Equations (7.EE.A.1) Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients.
• Grade 8: Expressions and Equations (8.EE.C.7) Solve linear equations in one variable.
• Grade 8: Expressions and Equations (8.EE.C.7b) Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.

How to solve equations with fractions

In order to solve equations with fractions:

Identify the operations that are being applied to the unknown variable.

• Apply the inverse operations, one at a time, to both sides of the equation .
• When you have the variable on one side, you have the final answer.
• Check the answer by substituting the answer back into the original equation .

[FREE] Solve Equations with Fractions Worksheet (Grade 6 to 8)

Use this worksheet to check your grade 6 to 8 students’ understanding of solving equations with fractions. 15 questions with answers to identify areas of strength and support!

Solve equations with fractions examples

Example 1: equations with one operation.

Solve for x \text{: } \cfrac{x}{5}=4 .

The unknown is x.

Looking at the left hand side of the equation, the x is divided by 5.

\cfrac{x}{5}

2 Apply the inverse operations, one at a time, to both sides of the equation.

The inverse of “dividing by 5 ” is “multiplying by 5 ”.

You will multiply both sides of the equation by 5.

3 When you have the variable on one side, you have the final answer.

The final answer is x=20.

4 Check the answer by substituting the answer back into the original equation.

You can check the answer by substituting the answer back into the original equation.

\cfrac{20}{5}=20\div5=4

Example 2: equations with one operation

Solve for x \text{: } \cfrac{x}{3}=8 .

Looking at the left hand side of the equation, the x is divided by 3.

\cfrac{x}{3}

Apply the inverse operations, one at a time, to both sides of the equation.

The inverse of “dividing by 3 ” is “multiplying by 3 ”.

You will multiply both sides of the equation by 3.

Write the final answer, checking that it is correct.

The final answer is x=24.

\cfrac{24}{3}=24\div3=8

Example 3: equations with two operations

Solve for x \text{: } \cfrac{x \, + \, 1}{2}=7 .

Looking at the left hand side of the equation, 1 is added to x and then divided by 2 (the denominator of the fraction).

\cfrac{x \, + \, 1}{2}

First, clear the fraction by multiplying both sides of the equation by 2.

Then, subtract 1 from both sides.

The final answer is x=13.

\cfrac{13 \, +1 \, }{2}=\cfrac{14}{2}=14\div2=7

Example 4: equations with two operations

Solve for x \text{: } \cfrac{x}{4}-2=3 .

Looking at the left hand side of the equation, x is divided by 4 and then 2 is subtracted.

\cfrac{x}{4}-2

First, add 2 to both sides of the equation.

Then, multiply both sides of the equation by 4.

\cfrac{20}{4}-2=20\div4-2=5-2=3

Example 5: equations with three operations

Solve for x \text{: } \cfrac{3x}{5}+1=7 .

Looking at the left hand side of the equation, x is multiplied by 3, then divided by 5 , and then 1 is added.

\cfrac{3x}{5}+1

First, subtract 1 from both sides of the equation.

Then, multiply both sides of the equation by 5.

Finally, divide both sides by 3.

The final answer is x=10.

\cfrac{3 \, \times \, 10}{5}+1=\cfrac{30}{5}+1=6+1=7

Example 6: equations with three operations

Solve for x \text{: } \cfrac{2x-1}{7}=3 .

Looking at the left hand side of the equation, x is multiplied by 2, then 1 is subtracted, and the last operation is divided by 7 (the denominator).

\cfrac{2x-1}{7}

First, multiply both sides of the equation by 7.

Next, add 1 to both sides.

The final answer is x=11.

\cfrac{2 \, \times \, 11-1}{7}=\cfrac{22-1}{7}=\cfrac{21}{7}=3

Example 7: equations with the unknown as the denominator

Solve for x \text{: } \cfrac{24}{x}=6 .

Looking at the left hand side of the equation, x is the denominator. 24 is divided by x.

\cfrac{24}{x}

You need to multiply both sides of the equation by x.

Then, you can divide both sides by 6.

The final answer is x=4.

\cfrac{24}{4}=24\div4=6

Example 8: equations with the unknown as the denominator

Solve for x \text{: } \cfrac{18}{x}-6=3 .

Looking at the left hand side of the equation, x is the denominator. 18 is divided by x , and then 6 is subtracted.

\cfrac{18}{x}-6

First, add 6 to both sides of the equation.

Then, multiply both sides of the equation by x.

Finally, divide both sides by 9.

The final answer is x=2.

\cfrac{18}{2}-6=9-6=3

Teaching tips for solving equations with fractions

• When students first start working through practice problems and word problems, provide step-by-step instructions to assist them with solving linear equations.
• Introduce solving equations with fractions with one-step problems, then two-step problems, before introducing multi-step problems.
• Students will need lots of practice with solving linear equations. These standards provide the foundation for work with future linear equations in Algebra I and II.
• Provide opportunities for students to explain their thinking through writing. Ensure that they are using key vocabulary, such as, absolute value, coefficient, equation, common factors, inequalities, simplify, etc.

Easy mistakes to make

• The solution to an equation can be any type of number The unknowns do not have to be integers (whole numbers and their negative opposites). The solutions can be fractions or decimals. They can also be positive or negative numbers.
• The unknown of an equation can be on either side of the equation The unknown, represented by a letter, is often on the left hand side of the equations; however, it doesn’t have to be. It could also be on the right hand side of an equation.

• Lowest common denominator (LCD) It is common to get confused between solving equations involving fractions and adding and subtracting fractions. When adding and subtracting, you need to work out the lowest/least common denominator (sometimes called the least common multiple or LCM). When you solve equations involving fractions, multiply both sides of the equation by the denominator of the fraction.

Related math equations lessons

• Math equations
• Rearranging equations
• How to find the equation of a line
• Substitution
• Linear equations
• Writing linear equations
• Solving equations
• Identity math
• One step equations

Practice solve equations with fractions questions

1. Solve: \cfrac{x}{6}=3

You will multiply both sides of the equation by 6, because the inverse of “dividing by 6 ” is “multiplying by 6 ”.

The final answer is x = 18.

\cfrac{18}{6}=18 \div 6=3

2. Solve: \cfrac{x \, + \, 4}{2}=7

Then subtract 4 from both sides.

The final answer is x = 10.

\cfrac{10 \, + \, 4}{2}=\cfrac{14}{2}=14 \div 2=7

3. Solve: \cfrac{x}{8}-5=1

First, add 5 to both sides of the equation.

Then multiply both sides of the equation by 8.

The final answer is x = 48.

\cfrac{48}{8}-5=48 \div 8-5=1

4. Solve: \cfrac{3x \, + \, 2}{4}=2

First, multiply both sides of the equation by 4.

Next, subtract 2 from both sides.

The final answer is x = 2.

\cfrac{3 \, \times \, 2+2}{4}=\cfrac{6 \, + \, 2}{4}=\cfrac{8}{4}=8 \div 4=2

5. Solve: \cfrac{4x}{7}-2=6

Then multiply both sides of the equation by 7.

Finally, divide both sides by 4.

The final answer is x = 14.

\cfrac{4 \, \times \, 14}{7}-2=\cfrac{56}{7}-2=56 \div 7-2=6

6. Solve: \cfrac{42}{x}=7

Then you divide both sides by 7.

The final answer is x = 6.

\cfrac{42}{6}=42 \div 6=7

Solve equations with fractions FAQs

Yes, you still follow the order of operations when solving equations with fractions. You will start with any operations in the numerator and follow PEMDAS (parenthesis, exponents, multiply/divide, add/subtract), followed by any operations in the denominator. Then you will solve the rest of the equation as usual.

The next lessons are

• Inequalities
• Types of graphs
• Math formulas
• Coordinate plane
• Number patterns
• Algebraic expressions

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Course: Algebra 1   >   Unit 2

• Why we do the same thing to both sides: Variable on both sides
• Intro to equations with variables on both sides
• Equations with variables on both sides: 20-7x=6x-6
• Equations with variables on both sides

Equation with variables on both sides: fractions

• Equations with variables on both sides: decimals & fractions
• Equation with the variable in the denominator

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Study Guides > Prealgebra

Solving equations by clearing fractions, learning outcomes.

• Use the least common denominator to eliminate fractions from a linear equation before solving it
• Solve equations with fractions that require several steps

Solve equations by clearing the Denominators

• Find the least common denominator of all the fractions in the equation.
• Multiply both sides of the equation by that LCD. This clears the fractions.
• Isolate the variable terms on one side, and the constant terms on the other side.
• Simplify both sides.
• Use the multiplication or division property to make the coefficient on the variable equal to [latex]1[/latex].

Licenses & Attributions

Cc licensed content, original.

• Question ID 142514, 142542. Authored by: Lumen Learning. License: CC BY: Attribution . License terms: IMathAS Community License, CC-BY + GPL.
• Solve a Linear Equation with Parentheses and a Fraction 2/3(9x-12)=8+2x. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution .

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• Ex 1: Solve an Equation with Fractions with Variable Terms on Both Sides. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution .
• Question ID 71948. Authored by: Alyson Day. License: CC BY: Attribution . License terms: IMathAS Community License, CC-BY + GPL.

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8.4 Solve Equations with Fraction or Decimal Coefficients

Learning objectives.

By the end of this section, you will be able to:

Solve equations with fraction coefficients

• Solve equations with decimal coefficients

Be Prepared 8.10

Before you get started, take this readiness quiz.

Multiply: 8 · 3 8 . 8 · 3 8 . If you missed this problem, review Example 4.28

Be Prepared 8.11

Find the LCD of 5 6 and 1 4 . 5 6 and 1 4 . If you missed this problem, review Example 4.63

Be Prepared 8.12

Multiply: 4.78 4.78 by 100 . 100 . If you missed this problem, review Example 5.18

Solve Equations with Fraction Coefficients

Let’s use the General Strategy for Solving Linear Equations introduced earlier to solve the equation 1 8 x + 1 2 = 1 4 . 1 8 x + 1 2 = 1 4 .

This method worked fine, but many students don’t feel very confident when they see all those fractions. So we are going to show an alternate method to solve equations with fractions. This alternate method eliminates the fractions.

We will apply the Multiplication Property of Equality and multiply both sides of an equation by the least common denominator of all the fractions in the equation. The result of this operation will be a new equation, equivalent to the first, but with no fractions. This process is called clearing the equation of fractions . Let’s solve the same equation again, but this time use the method that clears the fractions.

Example 8.37

Solve: 1 8 x + 1 2 = 1 4 . 1 8 x + 1 2 = 1 4 .

Try It 8.73

Solve: 1 4 x + 1 2 = 5 8 . 1 4 x + 1 2 = 5 8 .

Try It 8.74

Solve: 1 6 y − 1 3 = 1 6 . 1 6 y − 1 3 = 1 6 .

Notice in Example 8.37 that once we cleared the equation of fractions, the equation was like those we solved earlier in this chapter. We changed the problem to one we already knew how to solve! We then used the General Strategy for Solving Linear Equations.

Solve equations with fraction coefficients by clearing the fractions.

• Step 1. Find the least common denominator of all the fractions in the equation.
• Step 2. Multiply both sides of the equation by that LCD. This clears the fractions.
• Step 3. Solve using the General Strategy for Solving Linear Equations.

Example 8.38

Solve: 7 = 1 2 x + 3 4 x − 2 3 x . 7 = 1 2 x + 3 4 x − 2 3 x .

We want to clear the fractions by multiplying both sides of the equation by the LCD of all the fractions in the equation.

Try It 8.75

Solve: 6 = 1 2 v + 2 5 v − 3 4 v . 6 = 1 2 v + 2 5 v − 3 4 v .

Try It 8.76

Solve: −1 = 1 2 u + 1 4 u − 2 3 u . −1 = 1 2 u + 1 4 u − 2 3 u .

In the next example, we’ll have variables and fractions on both sides of the equation.

Example 8.39

Solve: x + 1 3 = 1 6 x − 1 2 . x + 1 3 = 1 6 x − 1 2 .

Try It 8.77

Solve: a + 3 4 = 3 8 a − 1 2 . a + 3 4 = 3 8 a − 1 2 .

Try It 8.78

Solve: c + 3 4 = 1 2 c − 1 4 . c + 3 4 = 1 2 c − 1 4 .

In Example 8.40 , we’ll start by using the Distributive Property. This step will clear the fractions right away!

Example 8.40

Solve: 1 = 1 2 ( 4 x + 2 ) . 1 = 1 2 ( 4 x + 2 ) .

Try It 8.79

Solve: −11 = 1 2 ( 6 p + 2 ) . −11 = 1 2 ( 6 p + 2 ) .

Try It 8.80

Solve: 8 = 1 3 ( 9 q + 6 ) . 8 = 1 3 ( 9 q + 6 ) .

Many times, there will still be fractions, even after distributing.

Example 8.41

Solve: 1 2 ( y − 5 ) = 1 4 ( y − 1 ) . 1 2 ( y − 5 ) = 1 4 ( y − 1 ) .

Try It 8.81

Solve: 1 5 ( n + 3 ) = 1 4 ( n + 2 ) . 1 5 ( n + 3 ) = 1 4 ( n + 2 ) .

Try It 8.82

Solve: 1 2 ( m − 3 ) = 1 4 ( m − 7 ) . 1 2 ( m − 3 ) = 1 4 ( m − 7 ) .

Solve Equations with Decimal Coefficients

Some equations have decimals in them. This kind of equation will occur when we solve problems dealing with money and percent. But decimals are really another way to represent fractions. For example, 0.3 = 3 10 0.3 = 3 10 and 0.17 = 17 100 . 0.17 = 17 100 . So, when we have an equation with decimals, we can use the same process we used to clear fractions—multiply both sides of the equation by the least common denominator .

Example 8.42

Solve: 0.8 x − 5 = 7 . 0.8 x − 5 = 7 .

The only decimal in the equation is 0.8 . 0.8 . Since 0.8 = 8 10 , 0.8 = 8 10 , the LCD is 10 . 10 . We can multiply both sides by 10 10 to clear the decimal.

Try It 8.83

Solve: 0.6 x − 1 = 11 . 0.6 x − 1 = 11 .

Try It 8.84

Solve: 1.2 x − 3 = 9 . 1.2 x − 3 = 9 .

Example 8.43

Solve: 0.06 x + 0.02 = 0.25 x − 1.5 . 0.06 x + 0.02 = 0.25 x − 1.5 .

Look at the decimals and think of the equivalent fractions.

0.06 = 6 100 , 0.02 = 2 100 , 0.25 = 25 100 , 1.5 = 1 5 10 0.06 = 6 100 , 0.02 = 2 100 , 0.25 = 25 100 , 1.5 = 1 5 10

Notice, the LCD is 100 . 100 .

By multiplying by the LCD we will clear the decimals.

Try It 8.85

Solve: 0.14 h + 0.12 = 0.35 h − 2.4 . 0.14 h + 0.12 = 0.35 h − 2.4 .

Try It 8.86

Solve: 0.65 k − 0.1 = 0.4 k − 0.35 . 0.65 k − 0.1 = 0.4 k − 0.35 .

The next example uses an equation that is typical of the ones we will see in the money applications in the next chapter. Notice that we will distribute the decimal first before we clear all decimals in the equation.

Example 8.44

Solve: 0.25 x + 0.05 ( x + 3 ) = 2.85 . 0.25 x + 0.05 ( x + 3 ) = 2.85 .

Try It 8.87

Solve: 0.25 n + 0.05 ( n + 5 ) = 2.95 . 0.25 n + 0.05 ( n + 5 ) = 2.95 .

Try It 8.88

Solve: 0.10 d + 0.05 ( d − 5 ) = 2.15 . 0.10 d + 0.05 ( d − 5 ) = 2.15 .

ACCESS ADDITIONAL ONLINE RESOURCES

• Solve an Equation with Fractions with Variable Terms on Both Sides
• Ex 1: Solve an Equation with Fractions with Variable Terms on Both Sides
• Ex 2: Solve an Equation with Fractions with Variable Terms on Both Sides
• Solving Multiple Step Equations Involving Decimals
• Ex: Solve a Linear Equation With Decimals and Variables on Both Sides
• Ex: Solve an Equation with Decimals and Parentheses

Section 8.4 Exercises

Practice makes perfect.

In the following exercises, solve the equation by clearing the fractions.

1 4 x − 1 2 = − 3 4 1 4 x − 1 2 = − 3 4

3 4 x − 1 2 = 1 4 3 4 x − 1 2 = 1 4

5 6 y − 2 3 = − 3 2 5 6 y − 2 3 = − 3 2

5 6 y − 1 3 = − 7 6 5 6 y − 1 3 = − 7 6

1 2 a + 3 8 = 3 4 1 2 a + 3 8 = 3 4

5 8 b + 1 2 = − 3 4 5 8 b + 1 2 = − 3 4

2 = 1 3 x − 1 2 x + 2 3 x 2 = 1 3 x − 1 2 x + 2 3 x

2 = 3 5 x − 1 3 x + 2 5 x 2 = 3 5 x − 1 3 x + 2 5 x

1 4 m − 4 5 m + 1 2 m = −1 1 4 m − 4 5 m + 1 2 m = −1

5 6 n − 1 4 n − 1 2 n = −2 5 6 n − 1 4 n − 1 2 n = −2

x + 1 2 = 2 3 x − 1 2 x + 1 2 = 2 3 x − 1 2

x + 3 4 = 1 2 x − 5 4 x + 3 4 = 1 2 x − 5 4

1 3 w + 5 4 = w − 1 4 1 3 w + 5 4 = w − 1 4

3 2 z + 1 3 = z − 2 3 3 2 z + 1 3 = z − 2 3

1 2 x − 1 4 = 1 12 x + 1 6 1 2 x − 1 4 = 1 12 x + 1 6

1 2 a − 1 4 = 1 6 a + 1 12 1 2 a − 1 4 = 1 6 a + 1 12

1 3 b + 1 5 = 2 5 b − 3 5 1 3 b + 1 5 = 2 5 b − 3 5

1 3 x + 2 5 = 1 5 x − 2 5 1 3 x + 2 5 = 1 5 x − 2 5

1 = 1 6 ( 12 x − 6 ) 1 = 1 6 ( 12 x − 6 )

1 = 1 5 ( 15 x − 10 ) 1 = 1 5 ( 15 x − 10 )

1 4 ( p − 7 ) = 1 3 ( p + 5 ) 1 4 ( p − 7 ) = 1 3 ( p + 5 )

1 5 ( q + 3 ) = 1 2 ( q − 3 ) 1 5 ( q + 3 ) = 1 2 ( q − 3 )

1 2 ( x + 4 ) = 3 4 1 2 ( x + 4 ) = 3 4

1 3 ( x + 5 ) = 5 6 1 3 ( x + 5 ) = 5 6

In the following exercises, solve the equation by clearing the decimals.

0.6 y + 3 = 9 0.6 y + 3 = 9

0.4 y − 4 = 2 0.4 y − 4 = 2

3.6 j − 2 = 5.2 3.6 j − 2 = 5.2

2.1 k + 3 = 7.2 2.1 k + 3 = 7.2

0.4 x + 0.6 = 0.5 x − 1.2 0.4 x + 0.6 = 0.5 x − 1.2

0.7 x + 0.4 = 0.6 x + 2.4 0.7 x + 0.4 = 0.6 x + 2.4

0.23 x + 1.47 = 0.37 x − 1.05 0.23 x + 1.47 = 0.37 x − 1.05

0.48 x + 1.56 = 0.58 x − 0.64 0.48 x + 1.56 = 0.58 x − 0.64

0.9 x − 1.25 = 0.75 x + 1.75 0.9 x − 1.25 = 0.75 x + 1.75

1.2 x − 0.91 = 0.8 x + 2.29 1.2 x − 0.91 = 0.8 x + 2.29

0.05 n + 0.10 ( n + 8 ) = 2.15 0.05 n + 0.10 ( n + 8 ) = 2.15

0.05 n + 0.10 ( n + 7 ) = 3.55 0.05 n + 0.10 ( n + 7 ) = 3.55

0.10 d + 0.25 ( d + 5 ) = 4.05 0.10 d + 0.25 ( d + 5 ) = 4.05

0.10 d + 0.25 ( d + 7 ) = 5.25 0.10 d + 0.25 ( d + 7 ) = 5.25

0.05 ( q − 5 ) + 0.25 q = 3.05 0.05 ( q − 5 ) + 0.25 q = 3.05

0.05 ( q − 8 ) + 0.25 q = 4.10 0.05 ( q − 8 ) + 0.25 q = 4.10

Everyday Math

Coins Taylor has $2.00 $2.00 in dimes and pennies. The number of pennies is 2 2 more than the number of dimes. Solve the equation 0.10 d + 0.01 ( d + 2 ) = 2 0.10 d + 0.01 ( d + 2 ) = 2 for d , d , the number of dimes.

Stamps Travis bought $9.45 $9.45 worth of 49-cent 49-cent stamps and 21-cent 21-cent stamps. The number of 21-cent 21-cent stamps was 5 5 less than the number of 49-cent 49-cent stamps. Solve the equation 0.49 s + 0.21 ( s − 5 ) = 9.45 0.49 s + 0.21 ( s − 5 ) = 9.45 for s , s , to find the number of 49-cent 49-cent stamps Travis bought.

Writing Exercises

Explain how to find the least common denominator of 3 8 , 1 6 , and 2 3 . 3 8 , 1 6 , and 2 3 .

If an equation has several fractions, how does multiplying both sides by the LCD make it easier to solve?

If an equation has fractions only on one side, why do you have to multiply both sides of the equation by the LCD?

In the equation 0.35 x + 2.1 = 3.85 , 0.35 x + 2.1 = 3.85 , what is the LCD? How do you know?

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?

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Solving Equations with Fractions

I know fractions are difficult, but with these easy step-by step instructions you'll be solving equations with fractions in no time.

Do you start to get nervous when you see fractions? Do you have to stop and review all the rules for adding, subtracting, multiplying and dividing fractions?

If so, you are just like almost every other math student out there! But... I am going to make your life so much easier when it comes to solving equations with fractions!

Our first step when solving these equations is to get rid of the fractions because they are not easy to work with!

Let see what happens with a typical two-step equation with the distributive property.

In this problem, we would typically distribute the 3/4 throughout the parenthesis and then solve. Let's see what happens:

Yuck! That just made this problem worse! Now we have two fractions to contend with and that means subtracting fractions and multiplying fractions.

So... let's stop here and say,

We DO NOT want to do this! DO NOT distribute fractions.

We are going to learn how to get rid of the fractions and make this much more simple!

So... what do we do? We are going to get rid of just the denominator in the fraction, so we will be left with the numerator, or just an integer!

I know, easier said than done! It's really not hard, but before I get into it, I want to go over one algebra definition.

We need to discuss the word term.

In Algebra, each term within an equation is separated by a plus (+) sign, minus (-) sign or an equals sign (=). Variable or quantities that are multiplied or divided are considered the same term.

That last example is the most important to remember. If a quantity is in parentheses, it it considered one term!

Let's look at a few examples of how to solve these crazy looking problems!

Example 1 - Equations with Fractions

Take a look at this example on video if you are feeling overwhelmed.

Hopefully you were able to follow that example. I know it's tough, but if you can get rid of the fraction, it will make these problems so much easier. Keep going, you'll get the hang of it!

In the next example, you will see two fractions.  Since they have the same denominator, we will multiply by the denominator and get rid of both fractions.

Example 2 - Equations with Fractions with the Same Denominator

Did you notice how multiplying by 2 (the denominator of both fractions) allowed us to get rid of the fractions?  This is the best way to deal with equations that contain fractions.

In the next example, you will see what happens when you have 2 fractions that have different denominators. 

We still want to get rid of the fractions all in one step. Therefore, we need to multiply all terms by the least common multiple.  Remember how to find the LCM?  If not, check out the LCM lesson here .

Example 3 - Equations with Two Fractions with Different Denominators

Yes, the equations are getting harder, but if you take it step-by-step, you will arrive at the correct solution. Keep at it - I know you'll get it!

• Solving Equations
• Equations with Fractions

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Algebra: Fraction Problems

Related Topics: More Algebra Word Problems

In these lessons, we will learn how to solve fraction word problems that deal with fractions and algebra. Remember to read the question carefully to determine the numerator and denominator of the fraction.

We will also learn how to solve word problems that involve comparing fractions, adding mixed numbers, subtracting mixed numbers, multiplying fractions and dividing fractions.

Fraction Word Problems using Algebra

Example: 2/3 of a number is 14. What is the number?

Answer: The number is 21.

Example: The numerator of a fraction is 3 less than the denominator. When both the numerator and denominator are increased by 4, the fraction is increased by fraction.

Solution: Let the numerator be x, then the denominator is x + 3, and the fraction is \(\frac{x}{{x + 3}}\) When the numerator and denominator are increased by 4, the fraction is \(\frac{{x + 4}}{{x + 7}}\) \(\frac{{x + 4}}{{x + 7}} - \frac{x}{{x + 3}} = \frac{{12}}{{77}}\) 77(x + 4)(x + 3) – 77x(x+7) = 12(x + 7)(x + 3) 77x 2 + 539x + 924 – 77x 2 – 539x = 12x 2 + 120x + 252 12x 2 + 120x – 672 = 0 x 2 + 10x – 56 = 0 (x – 4)(x + 14) = 0 x = 4 (negative answer not applicable in this case)

How to solve Fraction Word Problems using Algebra? Examples: (1) The denominator of a fraction is 5 more than the numerator. If 1 is subtracted from the numerator, the resulting fraction is 1/3. Find the original fraction. (2) If 3 is subtracted from the numerator of a fraction, the value of the resulting fraction is 1/2. If 13 is added to the denominator of the original fraction, the value of the new fraction is 1/3. Find the original fraction. (3) A fraction has a value of 3/4. When 14 is added to the numerator, the resulting fraction has a value equal to the reciprocal of the original fraction, Find the original fraction.

Algebra Word Problems with Fractional Equations Solving a fraction equation that appears in a word problem Example: One third of a number is 6 more than one fourth of the number. Find the number.

Fraction and Decimal Word Problems How to solve algebra word problems with fractions and decimals? Examples: (1) If 1/2 of the cards had been sold and there were 172 cards left, how many cards were printed? (2) Only 1/3 of the university students wanted to become teachers. If 3,360 did not wan to become teachers, how many university were there? (3) Rodney guessed the total was 34.71, but this was 8.9 times the total. What was the total?

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