Lump solutions of the fractional Kadomtsev–Petviashvili equation

Abstract

Of concern is the fractional Kadomtsev–Petviashvili (fKP) equation and its lump solution. As in the classical Kadomtsev–Petviashvili equation, the fKP equation comes in two versions: fKP-I (strong surface tension case) and fKP-II (weak surface tension case). We prove the existence of nontrivial lump solutions for the fKP-I equation in the energy subcritical case $\alpha >\frac{4}{5}$ by means of variational methods. It is already known that there exist neither nontrivial lump solutions belonging to the energy space for the fKP-II equation [9] nor for the fKP-I when $\alpha \le \frac{4}{5}$ [26]. Furthermore, we show that for any $\alpha >\frac{4}{5}$ lump solutions for the fKP-I equation are smooth and decay quadratically at infinity. Numerical experiments are performed for the existence of lump solutions and their decay. Moreover, numerically, we observe cross-sectional symmetry of lump solutions for the fKP-I equation.

1 Introduction

The present paper is devoted to the study of fully localized solitary solutions (also known as lump solutions) of the fractional Kadomtsev–Petviashvili (fKP) equation

$$\begin{aligned} u_t+uu_x-\textrm{D}^\alpha _x u_x+\sigma \partial _x^{-1}u_{yy}=0. \end{aligned}$$

(1.1)

Here the real function $u=u(t,x,y)$ depends on the spatial variable $(x,y)\in \mathbb {R}^2$ and the temporal variable $t\in \mathbb {R}_+$. The linear operator $\textrm{D}^\alpha _x$ denotes the Riesz potential of order $\alpha \in \mathbb {R}$ in x-direction, which is defined by multiplication with $|\cdot |^\alpha$ on the frequency space, that is

$$\begin{aligned} {\mathcal {F}}(\textrm{D}^\alpha _x f)(t,\xi _1, \xi _2)=|\xi _1|^\alpha {\hat{f}}(t,\xi _1, \xi _2), \end{aligned}$$

where the operator ${\mathcal {F}}$ denotes the extension to the space of tempered distributions ${\mathcal {S}}'({\mathbb {R}^n})$ of the Fourier transform ${\mathcal {F}}(f)(\xi ):=\int _{\mathbb {R}^n} f(x)e^{-\textrm{i}\xi x}\, dx$ on the Schwartz space ${\mathcal {S}}({\mathbb {R}^n})$ with inverse ${\mathcal {F}}^{-1}(f):=\frac{1}{2\pi }{\mathcal {F}}(f)(-\cdot )$. We also write ${{\hat{f}}}:={\mathcal {F}}(f)$. The operator $\partial _x^{-1}$ is defined as a Fourier multiplier operator on the x-variable as ${\mathcal {F}}(\partial _x^{-1}f)(t,\xi _1,\xi _2)=\frac{1}{\textrm{i}\xi _1}{{\hat{f}}}(t,\xi _1,\xi _2).$ In the case $\alpha =2$ equation (1.1) becomes the classical Kadomtsev–Petviashvili (KP) equation which was introduced by Kadomtsev & Petviashvili [19] as a weakly two-dimensional extension of the celebrated Korteweg–de Vries (KdV) equation,

$$\begin{aligned} u_t+uu_x+u_{xxx}=0, \end{aligned}$$

which is a spatially one-dimensional equation appearing in the context of small-amplitude shallow water-wave model equations. The KP equation comes in two versions: For $\sigma =-1$ it is called KP-I and for $\sigma =1$ it is called KP-II. Roughly speaking, the KP-I equation represents the case of strong surface tension, while the KP-II equation appears as a model equation for weak surface tension. Analogously to the classical case, the fKP equation is a two-dimensional extension of the fractional Korteweg–de Vries (fKdV) equation

$$\begin{aligned} u_t+uu_x-\textrm{D}^\alpha _x u_x=0 \end{aligned}$$

and (1.1) is referred to as the fKP-I equation when $\sigma =-1$ and as the fKP-II equation when $\sigma =1$. Notice that for $\alpha =1$ in (1.1) we recover the KP-version of the Benjamin–Ono equation. During the last decade there has been a growing interest in fractional regimes such as the fKdV or the fKP equation (see for example [1, 4, 14,15,16, 20, 21, 23, 26, 27, 29, 31, 32, 34, 38] and the references therein). Even though most of these equations are not derived by asymptotic expansions from governing equations in fluid dynamics they can be thought of as dispersive corrections.

Formally, the fKP equation does not only conserve the $L^2$–norm

$$\begin{aligned} M(u)=\int _{\mathbb {R}^2} u^2\,\textrm{d}(x,y), \end{aligned}$$

but also the energy

$$\begin{aligned} E_\alpha (u):=\int _{\mathbb {R}^2}\left( \frac{1}{2} (\textrm{D}^\frac{\alpha }{2}_x u)^2-\tfrac{1}{6}u^3-\tfrac{1}{2}\sigma (\partial _x^{-1}u_y)^2 \right) \, \textrm{d}(x,y). \end{aligned}$$

Notice that the corresponding energy space

$$\begin{aligned} X_\frac{\alpha }{2}(\mathbb {R}^2):=\{u\in L^2(\mathbb {R}^2)\mid \textrm{D}^\frac{\alpha }{2}_xu , \partial _x^{-1}u_y\in L^2(\mathbb {R}^2)\} \end{aligned}$$

equipped with the norm

$$\begin{aligned} ||\phi ||_\frac{\alpha }{2}^2:=\left\Vert \phi \right\Vert _{L^2(\mathbb {R}^2)}^2 +\left\Vert \textrm{D}_x^\frac{\alpha }{2}\phi \right\Vert _{L^2(\mathbb {R}^2)}^2 +\left\Vert \partial _x^{-1}\partial _y\phi \right\Vert _{L^2(\mathbb {R}^2)}^2, \end{aligned}$$

includes a zero-mass constraint with respect to x. We refer to [26] for derivation issues and well-posedness results for the Cauchy problem associated with (1.1). The fKP equation is invariant under the scaling

$$\begin{aligned} u_\lambda (t,x,y)=\lambda ^\alpha u(\lambda ^{\alpha +1}t,\lambda x, \lambda ^{\frac{\alpha +2}{2}}y), \end{aligned}$$

and $\Vert u_\lambda \Vert _{L^2}=\lambda ^{\frac{3\alpha -4}{4}}\Vert u\Vert _{L^2}$. Thus, $\alpha =\frac{4}{3}$ is the $L^2$-critical exponent for the fKP equation. The ranges $\alpha >\frac{4}{3}$ and $\alpha <\frac{4}{3}$ are called sub- and supercritical, respectively. Due to the embedding $X_{\frac{\alpha }{2}}\subset L^3(\mathbb {R}^2)$ for $\alpha \ge \frac{4}{5}$ (cf. [26, Lemma 1.1]), we call $\alpha =\frac{4}{5}$ the energy critical exponent for the fKP equation.

A traveling-wave solution $u(t,x,y)=\phi (x-ct,y)$ of the fKP equation propagating in x-direction with wave speed $c>0$, satisfies the steady equation

$$\begin{aligned} -c\phi +\frac{1}{2}\phi ^2-\textrm{D}^\alpha _x \phi +\sigma \partial _x^{-2}\phi _{yy}=0. \end{aligned}$$

(1.2)

Lump solutions are traveling-wave solutions decaying to 0 as $|(x,y)|\rightarrow \infty$.

1.1 Main results

Our aim is to study the existence and spatial decay of lump solutions for the fKP equation. Since it is known [9, 26] that the fKP-II equation for any $\alpha$ as well as the fKP-I equation for $\alpha \le \frac{4}{5}$ do not admit any lump solutions in $X_\frac{\alpha }{2}\cap L^3(\mathbb {R}^2)$, the study of this paper is concerned with traveling waves for the fKP-I equation for $\alpha >\frac{4}{5}$. We prove the following two main theorems. Moreover, we study lump solutions and some of their properties numerically.

Theorem 1

(Existence of lump solutions) For any $\tfrac{4}{5}< \alpha$ there exists a lump solution $\phi \in X_\frac{\alpha }{2}$ of (1.2) with $\sigma =-1$.

Theorem 2

(Decay of lump solutions) Any lump solution $\phi \in X_{\frac{\alpha }{2}}$ of (1.2) with $\sigma =-1$ is smooth and satisfies

$$\begin{aligned} \varrho ^2\phi \in L^\infty (\mathbb {R}^2),\qquad \text{ where }\quad \varrho ^2(x,y)=x^2+y^2. \end{aligned}$$

The classical KP-I equation possesses an explicit lump solution of the form

$$\begin{aligned} \phi _e(x-ct,y)= 8c\frac{1-\tfrac{c}{3}(x-ct)^2+\tfrac{c^2}{3}y^2}{\left( 1+\tfrac{c}{3}(x-ct)^2+\tfrac{c^2}{3}y^2\right) ^2}. \end{aligned}$$

(1.3)

We would like to point out that De Bouard & Saut [9] studied the existence of of lump solutions for the generalized KP-I equation

$$\begin{aligned} (u_t+u^pu_x-u_{xxx})_x-u_{yy}=0 \end{aligned}$$

(1.4)

where $p=m/n\ge 1$, m, n relatively prime and n odd. Furthermore, in their continuation paper [10], de Bouard & Saut investigated the symmetry and decay of lump solutions for (1.4) and showed that for all $p\ge 1$ the decay is quadratic. Our studies follow a similar approach as in [9, 10]. However, special attention needs to be given to the nonlocal operator $\textrm{D}_x^\alpha$. While many proofs can be adapted with a bit more technical effort due to the nonlocal operator, the result on decay of lump solutions in the supercritical case $\frac{4}{5}<\alpha <\frac{4}{3}$ (which includes the Benjamin–Ono KP version for $\alpha =1$) needs a modified approach, since in the supercritical case the symbol of an operator related to the linear dispersion is no longer $L^2$-integrable.

On the existence result: We give a brief outline of the existence proof for lump solutions of (1.4) in [9] by variational methods, since we will be using the same strategy to prove existence of lump solutions of the fKP-I equation (1.1). First consider the constrained minimization problem

$$\begin{aligned} I_\mu =\inf \left\{ \left\Vert \phi \right\Vert _Y^2:\phi \in Y , \int _{\mathbb {R}^2}\phi ^{p+2}=\mu \right\} \end{aligned}$$

for $\mu >0$ fixed, where Y is the closure of $\partial _x(C_0^\infty (\mathbb {R}^2))$ (the space of functions of the form $\partial _x\varphi$ with $\varphi \in C_0^\infty (\mathbb {R}^2)$) with respect to the norm

$$\begin{aligned} \left\Vert \partial _x\varphi \right\Vert _Y^2=\left\Vert \nabla \varphi \right\Vert _{L^2(\mathbb {R}^2)}^2 +\left\Vert \partial _x^2\varphi \right\Vert _{L^2(\mathbb {R}^2)}^2. \end{aligned}$$

Via the Lagrange multiplier principle one finds (after rescaling) that solutions of the constrained minimization problem $I_\mu$ are lump solutions of (1.4). The task is then to prove existence of solutions of $I_\mu$ and this is achieved using the concentration-compactness theorem (cf. Theorem 3). The variational formulation associated with $I_\mu$ has several good properties. The functional being minimized is just the norm of the space Y. It is therefore immediate that it is coercive, bounded from below and weakly lower semi-continuous; properties which are all advantageous in the context of minimization problems, see [39, Theorem 1.2]. Furthermore, since the norm is homogeneous, it is easily shown that $I_\mu$ is subadditive as a function of $\mu$ and this property is essential in proving that the dichotomy scenario in the concentration-compactness theorem does not occur.

We prove Theorem 1 by extending the strategy of [9], outlined above, to the fractional case. Generally speaking, the fractional derivative and the fact that we are allowing for weak dispersion makes the proof of Theorem 1 more technical than its classical local counterpart $(\alpha =2)$. A key ingredient in the proof is the anisotropic Sobolev inequality [26, Lemma 1.1] (see also Proposition 1 (ii)), which in particular says that for $\tfrac{4}{5}\le \alpha$, the space $X_\frac{\alpha }{2}$ is continuously embedded in $L^3(\mathbb {R}^2)$. This result is what determines the values of $\alpha$ for which we can prove existence of solitary waves. In fact, for $\alpha \le \tfrac{4}{5}$ there exist no nontrivial lump solutions of for the fKP-I equation in $X_\frac{\alpha }{2}\cap L^3(\mathbb {R}^2)$ [26, Proposition 1.2].

We would like to mention that there are several existence results on lump solutions using variational approaches for other two-dimensional equations. The full water-wave problem admits lump solutions both for strong [5, 18] and weak [6] surface tension. In the strong surface tension case the lump solutions can be approximated by rescalings of KP-I lumps, while in the weak surface tension case the lump solutions can be approximated by rescalings of Davey-Stewartson type solitary waves. The full dispersion KP (FDKP) equation was introduced in [25, chapter 8] as a model for weakly transversal three dimensional water-waves which preserves the dispersion relation of the full water-wave problem. A comparison of the fDKP equation with the KP equation for the propagation of water waves is given in [28]. Just as for the classical and fractional KP equation, the FDKP equation can be considered for both strong (FDKP-I) and weak (FDKP-II) surface tension. In [11] it was shown that the FDKP-I equation admits lump solutions and later on in [12] it was shown that also the FDKP-II equation possesses lump solutions. This is in contrast to the fKP-II equation, which does not admit any lump solutions [26]. Just like for the full water-wave problem, in the strong surface tension case the lump solutions can be approximated by rescalings of KP-I lumps, while in the weak surface tension case the lump solutions can be approximated by rescalings of Davey–Stewartson type solitary-waves.

On the decay result: The proof of Theorem 2 on the decay properties of lump solutions is closely related to that of [3, Theorem 3.1.2] and [10, Theorem 4.1]. The steady equation (1.2) can be rewritten as a convolution equation of the form

$$\begin{aligned} \phi =\frac{1}{2}K_\alpha *\phi ^2,\qquad {{\hat{K}}}_\alpha (\xi _1,\xi _2)= m_\alpha (\xi _1,\xi _2), \end{aligned}$$

where the symbol $m_\alpha$ is given by $m_\alpha (\xi _1,\xi _2)=\frac{\xi _1^2}{|\xi |^2+\xi _1^{\alpha +2}}.$

Remark 1

An immediate consequence of the discontinuity of the symbol $m_\alpha$ at the origin is that any nontrivial, continuous lump solution of (1.2) decays at most quadratically. Let us assume for a contradiction that $\phi$ is a nontrivial, continuous lump solution, which decays at infinity as $|\cdot |^{-\delta }$ for some $\delta >2$. Then $\phi \in L^1(\mathbb {R}^2)$, which implies that the Fourier transformation of $\phi$ is continuous. But ${\hat{\phi }} = \frac{1}{2}m_{\alpha }\hat{\phi ^2}$ cannot be continuous at the origin, since $\hat{\phi ^2}(0,0)>0$ and $m_\alpha$ is discontinuous at the origin. We conclude that the singularity of the symbol $m_\alpha$ induced by the transverse direction forces the decay of any nontrivial, continuous lump solution to be at most quadratic.

Remark 2

In view of Remark 1 the decay rate in Theorem 2 is optimal.

The idea is to study the kernel function $K_\alpha$ and to show that it has exactly quadratic decay at infinity (independent of $\alpha$). Then the decay properties of $K_\alpha$ are used to show that also $\phi$ decays quadratically at infinity.

On the numerics: We conduct numerical experiments to observe the lump solutions and some of their properties. For this purpose, we generate the solutions numerically by using Petviashvili iteration method. The method was proposed first by Petviashvili [37] to compute the lump solutions of the KP-I equation. The convergence of the method for the KP-equation was later discussed in [35] and now it is widely used to numerically evaluate traveling wave solutions of evolution equations (see for example [2, 33, 36] and the references therein).

Applying the Fourier transform to (1.2) with respect to the space variables (x, y) we obtain

$$\begin{aligned} c{\widehat{\phi }}-\frac{1}{2}\widehat{\phi ^2}+|\xi _1|^{\alpha } {\widehat{\phi }}+\frac{\xi _2^2}{\xi _1^2} {\widehat{\phi }}=0. \end{aligned}$$

(1.5)

An iterative algorithm for the equation (1.5) can be proposed as

$$\begin{aligned} {\widehat{\phi }}_{n+1}(\xi _1,\xi _2)=\frac{\widehat{\phi ^2_n}(\xi _1,\xi _2)}{2(c+\frac{\xi _2^2}{\xi _1^2}+|\xi _1|^{\alpha })} ,~~~~ n=1,2,\dots , \end{aligned}$$

(1.6)

where $\phi _n$ is the $n^{th}$ iteration of the numerical solution. Since (1.6) is generally divergent, the Petviashvili iteration is given as

$$\begin{aligned} {\widehat{\phi }}_{n+1}(\xi _1,\xi _2)=\frac{(M_n)^{\nu }}{2(c+\frac{\xi _2^2}{\xi _1^2}+|\xi _1|^{\alpha })} \widehat{\phi ^2_n}(\xi _1,\xi _2),~~~~ n=1,2,\dots , \end{aligned}$$

(1.7)

by introducing the stabilizing factor

$$\begin{aligned} M_n=\frac{ \int _{\mathbb {R}^2} 2 (c+\frac{\xi _2^2}{\xi _1^2}+|\xi _1|^{\alpha })~(\widehat{\phi _n})^2 \textrm{d}(\xi _1,\xi _2) }{ \int _{\mathbb {R}^2} \widehat{\phi ^2_n}~\widehat{\phi _n}~ \textrm{d}(\xi _1,\xi _2) }. \end{aligned}$$

Here the free parameter $\nu$ is chosen as 2 for the fastest convergence. To evaluate the term $1/ \xi _1^2$ for $\xi _1=0$, we regularize it as $1/( \xi _1+i\lambda )^2$, where $\lambda = 2.2 \times 10^{-16}$ as in [22, 24]. We control the iterative process by the error between two consecutive iterations

$$\begin{aligned} \text{ error }(n)=\Vert \phi _n-\phi _{n-1}\Vert _\infty ,~~~~ n=1,2,\dots , \end{aligned}$$

by the stabilization factor error $|1-M_n|$, and the residual error

$$\begin{aligned} Res(n)=\Vert {\mathcal {S}}\phi _n\Vert _\infty ,~~~~ n=1,2,\dots \end{aligned}$$

where

$$\begin{aligned} {\mathcal {S}}=\left( -c\phi +\frac{1}{2}\phi ^2-\textrm{D}^\alpha _x \phi \right) _{xx}-\phi _{yy}. \end{aligned}$$

We make sure that the errors are of order less than $10^{-5}$. In addition, we control the decay of Fourier coefficients $\hat{\phi }(\xi _1, \xi _2)$ in the numerical experiments.

1.2 Notation and organization of the paper

We first introduce a notation, which is frequently used in the sequel. Let f and g be two positive functions. We write $f\lesssim g$ ($f\gtrsim g$) if there exists a constant $c>0$ such that $f\le c g$ ($f\ge cg$). Moreover, we use the notation $f\eqsim g$ whenever $f\lesssim g$ and $f\gtrsim g$.

We conclude the introduction by the organization of the paper: In Section 2 we prove existence of lump solutions for the fKP-I equation (Theorem 1) via a variational approach. We also present numerically generated lump solutions and observe the cross-sectional symmetry of the solutions numerically. Section 3 is devoted to the proof of Theorem 2, which relies upon a careful study of the decay and regularity of the kernel function $K_\alpha$. The appendix contains some technical results which are needed for the analysis in Section 3.

2 Existence of solitary wave solutions

We consider the (rescaled) traveling wave fKP-I equation:

$$\begin{aligned} \phi +\textrm{D}_x^\alpha \phi +\partial _x^{-2}\partial _y^2\phi -\frac{\phi ^2}{2}=0. \end{aligned}$$

(2.1)

Equation (2.1) can be realized as a constrained minimization problem. Indeed, let

$$\begin{aligned} {\mathcal {L}}(\phi )=\frac{1}{2}\int _{\mathbb {R}^2} \left( \phi ^2+(\textrm{D}_x^\frac{\alpha }{2}\phi )^2 +(\partial _x^{-1}\partial _y\phi )^2 \ \right) \textrm{d}(x,y),\ {\mathcal {N}}(\phi )=\frac{1}{6}\int _{\mathbb {R}^2}\phi ^3\ \textrm{d}(x,y), \end{aligned}$$

which we study in the space $X_\frac{\alpha }{2}$ and consider the constrained minimization problem

$$\begin{aligned} I_\mu =\inf \{{\mathcal {L}}(\phi ):\phi \in X_\frac{\alpha }{2},\ {\mathcal {N}}(\phi )=\mu \}. \end{aligned}$$

(2.2)

In order to find nontrivial solutions we assume that $\mu \ne 0$ and without loss of generality we may further assume that $\mu > 0$.

Let $\phi$ be a solution of (2.2). Then there exists a Lagrange multiplier $\lambda \in \mathbb {R}$ such that

$$\begin{aligned} \textrm{d}{\mathcal {L}}(\phi )-\lambda \textrm{d}{\mathcal {N}}(\phi )=0. \end{aligned}$$

(2.3)

Since

$$\begin{aligned} \textrm{d}{\mathcal {L}}(\phi )=\phi +\textrm{D}_x^\alpha \phi +\partial _x^{-2}\partial _y^2\phi ,\qquad \textrm{d}{\mathcal {N}}(\phi )=\frac{1}{2}\phi ^2, \end{aligned}$$

equation (2.3) becomes

$$\begin{aligned} \phi +\textrm{D}_x^\alpha \phi +\partial _x^{-2} \partial _y^2\phi -\lambda \frac{\phi ^2}{2}=0. \end{aligned}$$

By rescaling $\phi (x,y)=\lambda ^{-1}\tilde{\phi }(x,y),$ we find that $\tilde{\phi }$ satisfies the equation

$$\begin{aligned} \tilde{\phi }+\textrm{D}_x^\alpha \tilde{\phi }+\partial _x^{-2} \partial _y^2\tilde{\phi }-\frac{\tilde{\phi }^2}{2}=0, \end{aligned}$$

which is (2.1). Therefore, in order to prove the existence of the solutions of equation (2.1), we will prove existence of solutions of the constrained minimization problem (2.2).

In the sequel, let us fix $\mu >0$ (this will ensure that $I_\mu >0$, see Corollary 1) and let $\{\phi _n\}_{n\in \mathbb {N}}\subset X_{\frac{\alpha }{2}}$ be a minimizing sequence such that ${\mathcal {N}}(\phi _n)=\mu$ and $\lim _{n\rightarrow \infty }{\mathcal {L}}(\phi _n)=I_\mu$. We aim to show that there exists a subsequence (not relabeled) of $\{\phi _n\}_{n\in \mathbb {N}}$, which converges to a function $\phi \in X_{\frac{\alpha }{2}}$ satisfying ${\mathcal {L}}(\phi )=I_\mu$ and ${\mathcal {N}}(\phi )=\mu$.

Let us set

$$\begin{aligned} e_n=\frac{1}{2}\left( \phi _n^2+(\textrm{D}_x^\frac{\alpha }{2}\phi _n) ^2+(\partial _x^{-1}\partial _y\phi _n)^2\right) \end{aligned}$$

and note that

$$\begin{aligned} {\mathcal {L}}(\phi _n)=\int _{\mathbb {R}^2}e_n\ \textrm{d}(x,y). \end{aligned}$$

We will use the following version of the concentration–compactness theorem for the sequence $\{e_n\}_{n\in \mathbb {N}}$ and show that the concentration scenario occurs. This is then used to construct a convergent subsequence of $\{\phi _n\}_{n\in \mathbb {N}}$, converging to a solution of (2.2)

Theorem 3

Let $d\in \mathbb {N}$. Any sequence $\{e_n\}_{n\in \mathbb {N}}\subset L^1(\mathbb {R}^d)$ of non-negative functions such that

$$\begin{aligned} \lim _{n\rightarrow \infty }\int _{\mathbb {R}^d}e_n\ \textrm{d}x=I>0, \end{aligned}$$

admits a subsequence, denoted again by $\{e_n\}_{n\in \mathbb {N}}$, for which one of the following phenomena occurs:

• Vanishing: For each $r>0$, one has

  $$\begin{aligned} \lim _{n\rightarrow \infty }\left( \sup _{x\in \mathbb {R}^d}\int _{B_r(x)}e_n\ \textrm{d}x\right) =0. \end{aligned}$$

• Dichotomy: There are sequences $\{x_n\}_{n\in \mathbb {N}}\subset \mathbb {R}^d$, $\{M_n\}_{n\in \mathbb {N}}, \{N_n\}_{n\in \mathbb {N}}\subset \mathbb {R}$ and $I^*\in (0,I)$ such that $M_n, N_n \rightarrow \infty ,\ \frac{M_n}{N_n}\rightarrow 0$ and

  $$\begin{aligned} \lim _{n\rightarrow \infty }\int _{B_{M_n}(x_n)}e_n\ \textrm{d}x=I^*,\quad \lim _{n\rightarrow \infty }\int _{B_{N_n}(x_n)}e_n\ \textrm{d}x=I^*. \end{aligned}$$

• Concentration: There exists a sequence $\{x_n\}_{n\in \mathbb {N}}\subset \mathbb {R}^d$ with the property that for each $\varepsilon >0$, there exists $r>0$ with

  $$\begin{aligned} \int _{B_r(x_n)}e_n\ \textrm{d}x\ge I-\varepsilon ,~ \text { for all }~n\in \mathbb {N}. \end{aligned}$$

Interpreting I as a mass, Theorem 3 says that $\{e_n\}_{n\in \mathbb {N}}$ admits a subsequence for which one of the following occur: The mass spreads out in $\mathbb {R}^n$ (vanishing), it splits into two parts (dichotomy) or the mass is uniformly concentrated in $\mathbb {R}^n$ (concentration).

2.1 Preliminary results

In this subsection we will gather some of the results we need in order to apply Theorem 3.

Proposition 1

Let $\phi \in X_\frac{\alpha }{2}$. Then,

• (i) ${\mathcal {L}}(\phi )=\frac{1}{2}\left\Vert \phi \right\Vert _\frac{\alpha }{2}^2$,

• (ii) for $\frac{4}{5}\le \alpha \le 2$, one has the anisotropic Sobolev inequality

  $$\begin{aligned} \left\Vert \phi \right\Vert _{L^3(\mathbb {R}^2)}^3\lesssim \left\Vert \phi \right\Vert _{L^2(\mathbb {R}^2)}^ \frac{5\alpha -4}{\alpha +2}\left\Vert \textrm{D}_x^\frac{\alpha }{2}\phi \right\Vert _{L^2(\mathbb {R}^2)}^\frac{18-5\alpha }{2(\alpha +2)} \left\Vert \partial _x^{-1}\partial _y\phi \right\Vert _{L^2(\mathbb {R}^2)}^\frac{1}{2}. \end{aligned}$$

  In particular $X_{\frac{\alpha }{2}}\subset L^3(\mathbb {R}^2)$ and $\left\Vert \phi \right\Vert _{L^3(\mathbb {R}^2)}\lesssim \left\Vert \phi \right\Vert _\frac{\alpha }{2}$ for all $\alpha \ge \tfrac{4}{5}$.

Proof

Part (i) is immediate while part (ii) can be found in [26, Lemma 1.1]. $\square$

Corollary 1

The minimum $I_\mu$ is positive.

Proof

By Proposition 1 we have that

$$\begin{aligned} \mu ={\mathcal {N}}(\phi )\lesssim \left\Vert \phi \right\Vert _{L^3}^3\lesssim \left\Vert \phi \right\Vert _\frac{\alpha }{2}^3\eqsim {\mathcal {L}}(\phi )^\frac{3}{2}. \end{aligned}$$

$\square$

Corollary 1 ensures that the minimizer is not given by the trivial solution.

Lemma 1

For any $\alpha >0$, the space $X_\frac{\alpha }{2}$ is compactly embedded in $L_{\text {loc}}^2(\mathbb {R}^2)$.

Proof

The proof follows essentially the lines in [10, Lemma 3.3]. We include it here for the sake of completeness.

For $\phi \in X_\frac{\alpha }{2}$, let $\varphi =\partial _x^{-1}\phi$. From the definition of $X_\frac{\alpha }{2}$ we find that $\partial _x\varphi ,\ \partial _y\varphi \in L^2(\mathbb {R}^2)$, that is, $\varphi \in {\dot{H}}^1(\mathbb {R}^2)$. From Poincare’s inequality we have that ${\dot{H}}^1(\mathbb {R}^2)$ is continuously embedded in $\text {BMO}(\mathbb {R}^2)$. It follows from this that $\varphi \in \text {BMO}(\mathbb {R}^2)\subset L_{\text {loc}}^q(\mathbb {R}^2)$ for all $0<q<\infty$. Let $\{\phi _n\}_{n=1}^\infty$ be a bounded sequence in $X_\frac{\alpha }{2}$. We will show that for any $R>0$ there exists a subsequence $\{\phi _{n_k}\}_{n=1}^\infty$, which converges in $L^2(B_R)$, where $B_R$ is the ball of radius R centered at the origin in $\mathbb {R}^2$. Let $\varphi _n=\partial _x^{-1}\phi _n$. Since we are only interested in convergence in $L^2(B_R)$, we may assume that $\varphi _n$ is supported on $B_{2R}$ by multiplying $\varphi _n$ with a smooth cutoff function $\psi$ such that $\psi \equiv 1$ in $B_R$ and $\text {supp}(\psi )\subset B_{2R}$. It follows then that $\phi _n$ is supported on $B_{2R}$ as well.

Since $\{\phi _n\}_{n=1}^\infty$ is bounded in $X_\frac{\alpha }{2}$ we can extract a subsequence, which we still denote by $\{\phi _n\}_{n=1}^\infty$, such that $\phi _n\rightharpoonup \phi$, for some $\phi \in X_\frac{\alpha }{2}$. Moreover, by replacing $\phi _n$ with $\phi _n-\phi$, we may assume that $\phi =0$. Our aim is then to show that

$$\begin{aligned} \int _{\mathbb {R}^2}|\phi _n|^2\ \textrm{d}(x,y)\rightarrow 0,\ ~\text {as }~n\rightarrow \infty . \end{aligned}$$

Let $R_1>0$. We have

$$\begin{aligned} \int _{\mathbb {R}^2}|\phi _n|^2\ \textrm{d}(x,y)&=\int _{\mathbb {R}^2}|\hat{\phi }_n|^2\ \textrm{d}(\xi _1,\xi _2)\nonumber \\&=\int _{\{|\xi _1|\le R_1,\ |\xi _2|\le R_1^2\}}| \hat{\phi }_n|^2\ \textrm{d}(\xi _1,\xi _2)+\int _{\{|\xi _1|\ge R_1\}}| \hat{\phi }_n|^2\ \textrm{d}(\xi _1,\xi _2)\nonumber \\&\quad +\int _{\{|\xi _1|\le R_1,\ |\xi _2|\ge R_1^2\}}|\hat{\phi }_n|^2\ \textrm{d}(\xi _1,\xi _2). \end{aligned}$$

(2.4)

We proceed to estimate each integral on the right-hand side of (2.4) separately. For the third integral we can write

$$\begin{aligned} \int _{\{|\xi _1|\le R_1,\ |\xi _2|\ge R_1^2\}}|\hat{\phi }_n|^2\ \textrm{d}(\xi _1,\xi _2)&=\int _{\{|\xi _1|\le R_1,\ |\xi _2|\ge R_1^2\}} \tfrac{\xi _1^2}{\xi _2^2}|{\mathcal {F}}(\partial _x^{-1}\partial _y\phi _n)|^2 \ \textrm{d}(\xi _1,\xi _2)\\&\le \frac{R_1^2}{R_1^4}\left\Vert \partial _x^{-1}\partial _y\phi _n\right\Vert _{L^2(\mathbb {R}^2)}^2\\&=\frac{1}{R_1^2}\left\Vert \partial _x^{-1}\partial _y\phi _n\right\Vert _{L^2(\mathbb {R}^2)}^2 \end{aligned}$$

and for the second one

$$\begin{aligned} \int _{\{|\xi _1|\ge R_1\}}|\hat{\phi }_n|^2\ \textrm{d}(\xi _1,\xi _2)&=\int _{\{|\xi _1|\ge R_1\}} \frac{1}{|\xi _1|^\alpha }|{\mathcal {F}}(\textrm{D}_x^ \frac{\alpha }{2}\phi _n)|^2\ \textrm{d}(\xi _1,\xi _2)\\&\le \frac{1}{R_1^\alpha }\left\Vert \textrm{D}_x^\frac{\alpha }{2}\phi _n\right\Vert _{L^2(\mathbb {R}^2)}^2. \end{aligned}$$

From these estimates we conclude that, given $\varepsilon >0$ we can choose $R_1$ sufficiently large such that

$$\begin{aligned} \int _{\{|\xi _1|\ge R_1\}}|\hat{\phi }_n|^2\ \textrm{d}(\xi _1,\xi _2)+\int _{\{|\xi _1|\le R_1,\ |\xi _2|\ge R_1^2\}}|\hat{\phi }_n|^2\ \textrm{d}(\xi _1,\xi _2)<\varepsilon . \end{aligned}$$

In order to deal with the first integral, we first note that since $\phi _n\rightharpoonup 0$ in $X_\frac{\alpha }{2}$, we have

$$\begin{aligned} \hat{\phi }_n(\xi _1,\xi _2)=\int _{B_{2R}}\textrm{e}^{ -\textrm{i}(x\xi _1+y\xi _2)}\phi _n(x,y)\ \textrm{d}(x,y)\rightarrow 0\ ~\text {as }~n\rightarrow \infty . \end{aligned}$$

Moreover,

$$\begin{aligned} |\hat{\phi }_n(\xi _1,\xi _2)|\le \left\Vert \phi _n\right\Vert _{L^1(B_{2R})}\lesssim \left\Vert \phi _n\right\Vert _{L^2(B_{2R})}. \end{aligned}$$

Since $\{\phi _n\}_{n=1}^\infty$ is bounded in $X_\frac{\alpha }{2}$ we can conclude that $\{\hat{\phi }_n\}_{n=1}^\infty$ is bounded in $L^\infty (\mathbb {R}^2)$, so by the dominated convergence theorem

$$\begin{aligned} \int _{\{|\xi _1|\le R_1,\ |\xi _2|\le R_1^2\}}|\hat{\phi }_n|^2\ \textrm{d}(\xi _1,\xi _2)\rightarrow 0,\ \text {as }n\rightarrow \infty . \end{aligned}$$

$\square$

Next we prove that $I_\mu$ is subadditive as a function of $\mu$, a property which will be crucial when proving that the dichotomy scenario in Theorem 3 does not occur.

Proposition 2

The infimum $I_\mu$ is strictly increasing and subadditive as a function of $\mu$, that is

$$\begin{aligned} I_{\mu _1+\mu _2}<I_{\mu _1}+I_{\mu _2},~\text { for all }~\mu _1,\mu _2>0. \end{aligned}$$

Proof

Let $h\in X_\frac{\alpha }{2}$ be such that ${\mathcal {N}}(h)=1$ and let $\phi =\mu ^\frac{1}{3}h$. Then $N(\phi )=\mu$ and ${\mathcal {L}}(\phi )=\mu ^\frac{2}{3}{\mathcal {L}}(h)$, which implies

$$\begin{aligned} I_\mu =\mu ^\frac{2}{3}I_1, \end{aligned}$$

from which the statement in the proposition directly follows. $\square$

When applying Theorem 3 we will be taking integrals over bounded domains. It is therefore useful to consider the norm $\left\Vert \cdot \right\Vert _\frac{\alpha }{2}$ restricted to a bounded domain $\Omega \subset \mathbb {R}^2$:

$$\begin{aligned} ||\phi ||_{\frac{\alpha }{2},\Omega }^2=\left\Vert \phi \right\Vert _{ L^2(\Omega )}^2+\left\Vert \textrm{D}_x^\frac{\alpha }{2}\phi \right\Vert _ {L^2(\Omega )}^2+\left\Vert \partial _x^{-1}\partial _y\phi \right\Vert _{L^2(\Omega )}^2. \end{aligned}$$

We also make the following definition.

Definition 1

Let $\Omega$ be a bounded domain in $\mathbb {R}^2$. For $f\in L_{\text {loc}}^1(\mathbb {R}^2)$, let

$$\begin{aligned} f_\Omega :=f-M_\Omega (f),\end{aligned}$$

where $M_\Omega (f):=\frac{1}{|\Omega |}\int _\Omega f \ \textrm{d}(x,y)$ is the mean of f over $\Omega$.

When proving that the vanishing scenario does not occur we will make use of the following result.

Proposition 3

Let $\phi \in X_\frac{\alpha }{2}$, $\varphi =\partial _x^{-1}\phi$ and let $\psi$ be a smooth cutoff function supported on a bounded domain¹$\Omega =\{(x,y)\in \mathbb {R}^2 \mid y\in (a,b), x\in (h_1(y),h_2(y))\}$, for some $a,b\in \mathbb {R}$ and $h_i\in C([a,b])$, $i=1,2$. Define

$$\begin{aligned} F_\Omega (\phi )=\partial _x(\psi (\varphi _\Omega )). \end{aligned}$$

Then,

$$\begin{aligned} \left\Vert F_\Omega (\phi )\right\Vert _{\frac{\alpha }{2}}\lesssim \left\Vert \phi \right\Vert _{\frac{\alpha }{2},\Omega }. \end{aligned}$$

Proof

We have that

$$\begin{aligned} \left\Vert F_\Omega (\phi )\right\Vert _{\frac{\alpha }{2}}^2= \left\Vert \partial _x(\psi \varphi _\Omega )\right\Vert _{L^2(\mathbb {R}^2)}^2 +\left\Vert \partial _y(\psi \varphi _\Omega )\right\Vert _{L^2(\mathbb {R}^2)}^2 +\left\Vert \textrm{D}_x^\frac{\alpha }{2} \partial _x(\psi \varphi _\Omega )\right\Vert _{L^2(\mathbb {R}^2)}^2. \end{aligned}$$

We consider each of these terms separately. The first term can be estimated as

$$\begin{aligned} \left\Vert \partial _x(\psi \varphi _\Omega )\right\Vert _{L^2(\mathbb {R}^2)} =\left\Vert \psi _x\varphi _\Omega +\psi \phi \right\Vert _{L^2(\mathbb {R}^2)}\le \left\Vert \psi _x\varphi _\Omega \right\Vert _{L^2(\mathbb {R}^2)}+\left\Vert \psi \phi \right\Vert _{L^2(\mathbb {R}^2)}, \end{aligned}$$

and $\left\Vert \psi \phi \right\Vert _{L^2(\mathbb {R}^2)}\lesssim \left\Vert \phi \right\Vert _{L^2(\Omega )}$, while

$$\begin{aligned} \left\Vert \psi _x\varphi _\Omega \right\Vert _{L^2(\mathbb {R}^2)}&\lesssim \left\Vert \varphi _\Omega \right\Vert _{L^2(\Omega )}\\&\lesssim \left\Vert \varphi _x\right\Vert _{L^2(\Omega )} +\left\Vert \varphi _y\right\Vert _{L^2(\Omega )}\\&=\left\Vert \phi \right\Vert _{L^2(\Omega )}+\left\Vert \partial _ x^{-1}\partial _y\phi \right\Vert _{L^2(\Omega )}, \end{aligned}$$

where we used Poincar’s inequality and the definition $\varphi = \partial _x^{-1}\phi$. Hence,

$$\begin{aligned} \left\Vert \partial _x(\psi \varphi _\Omega )\right\Vert _{L^2}^2\lesssim \left\Vert \phi \right\Vert _{L^2(\Omega )}+\left\Vert \partial _x^{-1}\partial _y\phi \right\Vert _{L^2(\Omega )} \end{aligned}$$

(2.5)

and in the same way we find

$$\begin{aligned} \left\Vert \partial _y(\psi \varphi _\Omega )\right\Vert _{L^2}^2\lesssim \left\Vert \phi \right\Vert _{L^2(\Omega )}+\left\Vert \partial _x^{-1}\partial _y\phi \right\Vert _{L^2(\Omega )}. \end{aligned}$$

(2.6)

Recall that $\Omega =\{(x,y)\in \mathbb {R}^2 \mid y\in (a,b), x\in (h_1(y),h_2(y))\}$ and set $\Omega _y:=(h_1(y),h_2(y))$. By the Leibniz rule for fractional derivatives (see e.g. [17, Theorem 7.6.1]), we can estimate

$$\begin{aligned} \left\Vert \textrm{D}_x^\frac{\alpha }{2}\partial _x(\psi \varphi _\Omega )\right\Vert _{L^2(\mathbb {R}^2)}^2&= \int _{\mathbb {R}} \left\Vert \textrm{D}_x^\frac{\alpha }{2}\partial _x(\psi \varphi _\Omega )(\cdot , y)\right\Vert _{L^2(\Omega _y)}^2\, \textrm{d}y\\&\lesssim \int _{a}^b \left\Vert \textrm{D}_x^\frac{\alpha }{2}\partial _x \psi (\cdot , y)\right\Vert _{L^\infty (\Omega _y)}^2\left\Vert \varphi _\Omega (\cdot , y)\right\Vert _{L^2(\Omega _y)}^2\\&\quad + \left\Vert \psi (\cdot , y)\right\Vert _{L^\infty (\Omega _y)}^2 \left\Vert \textrm{D}_x^\frac{\alpha }{2}\partial _x \varphi _\Omega (\cdot , y)\right\Vert _{L^2(\Omega _y)}^2 \textrm{d}y. \end{aligned}$$

Using that $\psi$ is a smooth function, we conclude by Poincaré’s inequality that

$$\begin{aligned} \begin{aligned} \left\Vert \textrm{D}_x^\frac{\alpha }{2}\partial _x(\psi \varphi _\Omega )\right\Vert _{L^2(\mathbb {R}^2)}^2&\lesssim \left\Vert \varphi _{\Omega }\right\Vert _{L^2(\Omega )}^2 + \left\Vert \textrm{D}_x^\frac{\alpha }{2}\partial _x \varphi _\Omega \right\Vert _{L^2(\Omega )}^2\\&\lesssim \left\Vert \varphi _x\right\Vert _{L^2(\Omega )}+\left\Vert \varphi _y\right\Vert _{L^2(\Omega )}+ \left\Vert \textrm{D}_x^\frac{\alpha }{2}\partial _x \varphi \right\Vert _{L^2(\Omega )}^2 \\&=\left\Vert \phi \right\Vert _{L^2(\Omega )}+\left\Vert \partial _x^{-1}\partial _y\phi \right\Vert _{L^2(\Omega )}+ \left\Vert \textrm{D}_x^\frac{\alpha }{2}\phi \right\Vert _{L^2(\Omega )}^2. \end{aligned} \end{aligned}$$

(2.7)

Gathering (2.5), (2.6), and (2.7), we have shown that

$$\begin{aligned} \left\Vert F_\Omega (\phi )\right\Vert _{\frac{\alpha }{2}}\lesssim \left\Vert \phi \right\Vert _{\frac{\alpha }{2},\Omega }. \end{aligned}$$

$\square$

Eventually, when excluding the dichotomy scenario we will make use of the following lemma, which provides a Poincaré-like inequality.

Lemma 2

( [9], Lemma 3.1) Let $2\le p<\infty$ and $R>0$. Then there exists a positive constant C such that for all $f\in L_\text {loc}^1(\mathbb {R}^2)$ one has that

$$\begin{aligned} \Vert f_{A_{2R,R}}\Vert _{L^p(A_{2R,R})} \le C R^\frac{2}{p}\Vert \nabla f\Vert _{L^2(A_{2R,R})}, \end{aligned}$$

where $A_{2R,R}\subset \mathbb {R}^2$ denotes the annulus centered at the origin of radii $2R>R$.

2.2 Existence of minimizers

Let $\{\phi _n\}_{n\in \mathbb {N}}\subset X_\frac{\alpha }{2}$ be a minimizing sequence for the constrained minimization problem (2.2), that is, ${\mathcal {N}}(\phi _n)=\mu$ and $\lim _{n\rightarrow \infty } {\mathcal {L}}(\phi _n)=I_\mu$. We will apply Theorem 3 to the sequence

$$\begin{aligned} e_n=\frac{1}{2}\left( \phi _n^2+(\textrm{D}_x^ \frac{\alpha }{2}\phi _n)^2+(\partial _x^{-1}\partial _y\phi _n)^2\right) . \end{aligned}$$

Recall that

$$\begin{aligned} \int _{\mathbb {R}^2}e_n\ \textrm{d}(x,y)={\mathcal {L}}(\phi _n). \end{aligned}$$

We will show in Proposition 4 and Proposition 5 that the vanishing and dichotomy scenarios in Theorem 3 does not occur and then use the concentration scenario to construct a convergent subsequence of $\{\phi _n\}_{n\in \mathbb {N}}$, converging to a solution $\phi$ of (2.2).

Proposition 4

(Excluding “vanishing”) No subsequence of $\{e_n\}_{n\in \mathbb {N}}$ has the vanishing property in Theorem 3.

Proof

Assume for a contradiction that vanishing does occur, that is

$$\begin{aligned} \lim _{n\rightarrow \infty }\left( \sup _{(x,y)\in \mathbb {R}^2}\int _{B_r(x,y)}e_n\ \textrm{d}(x,y)\right) =0 \end{aligned}$$

for each $r>0$. Let us cover $\mathbb {R}^2$ with balls $B_{1,j}$, $j\in \mathbb {N}$, of radius 1 such that each point in $\mathbb {R}^2$ is contained in at most three balls. Let $\{\psi _j\}_{n\in \mathbb {N}}$ be a smooth partition of unity such that $\text {supp}(\psi _j)\subset B_{1,j}$. Using Proposition 1 (ii) and Proposition 3 we find

$$\begin{aligned} |{\mathcal {N}}(\phi _n)|&\lesssim \left\Vert \phi _n\right\Vert _{L^3(\mathbb {R}^2)}^3\\&\lesssim \sum _{j\in \mathbb {N}}\left\Vert F_{B_{1,j}}(\phi _n)\right\Vert _{L^3(\mathbb {R}^2)}^3\\&\le \sup _{j\in \mathbb {N}}\left\Vert F_{B_{1,j}}(\phi _n)\right\Vert _{L^3(\mathbb {R}^2)} \sum _{j\in \mathbb {N}}\left\Vert F_{B_{1,j}}(\phi _n)\right\Vert _{L^3(\mathbb {R}^2)}^2\\&\lesssim \sup _{j\in \mathbb {N}}\left\Vert F_{B_{1,j}}(\phi _n)\right\Vert _{\frac{\alpha }{2}} \sum _{j\in \mathbb {N}}\left\Vert F_{B_{1,j}}(\phi _n)\right\Vert _{\frac{\alpha }{2}}^2\\&\lesssim \sup _{j\in \mathbb {N}}\left\Vert \phi _n\right\Vert _{\frac{\alpha }{2},B_{1,j}} \sum _{j\in \mathbb {N}}\left\Vert \phi _n\right\Vert _{\frac{\alpha }{2}, B_{1,j}}^2\\&\lesssim \sup _{j\in \mathbb {N}}\left( \int _{B_{1,j}}e_n\ \textrm{d}(x,y)\right) ^\frac{1}{2}\left\Vert \phi _n\right\Vert _\frac{\alpha }{2}^2. \end{aligned}$$

By letting $n\rightarrow \infty$ we get ${\mathcal {N}}(\phi _n)\rightarrow 0$, which contradicts the fact that ${\mathcal {N}}(\phi _n)=\mu >0$. $\square$

Proposition 5

(Excluding “dichotomy”) No subsequence of $\{e_n\}_{n\in \mathbb {N}}$ has the dichotomy property in Theorem 3.

Proof

Throughout the proof we will use $B_R$ to denote the ball in $\mathbb {R}^2$ centered at the origin of radius $R>0$ and $A_{R_1,R_2}$ to denote the annulus centered at the origin of radii $R_1>R_2>0$.

Assume for a contradiction that the dichotomy scenario in Theorem 3 occurs, that is there exist sequences $\{(x_n,y_n)\}_{n\in \mathbb {N}}\subset \mathbb {R}^2, \{M_n\}_{n\in \mathbb {N}}, \{N_n\}_{n\in \mathbb {N}}\subset \mathbb {R}$ and $I^* \in (0,I_\mu )$ with $M_n, N_n, \frac{N_n}{M_n} \rightarrow \infty$ for $n\rightarrow \infty$ and

$$\begin{aligned} \lim _{n\rightarrow \infty }\int _{B_{M_n}(x_n,y_n)}e_n\ \textrm{d}x=I^*,\quad \lim _{n\rightarrow \infty }\int _{B_{N_n}(x_n,y_n)}e_n\ \textrm{d}x=I^*. \end{aligned}$$

(2.8)

We will show that this leads to a contradiction, by proving that provided (2.8) holds, there exists two sequences $\{\omega _n^{(1)}\}_{n\in \mathbb {N}}, \{\omega _n^{(2)}\}_{n\in \mathbb {N}}$, which have in the limit $n\rightarrow \infty$ disjoint support and

• (i) ${\mathcal {N}}(\omega _n^{(1)}) + {\mathcal {N}}(\omega _n^{(2)}) -{\mathcal {N}}(\omega _n)\rightarrow 0$,

• (ii) ${\mathcal {L}}(\omega _n^{(1)})\rightarrow I^*$ and ${\mathcal {L}}(\omega _n^{(2)})\rightarrow (I_\mu -I^*)$,

where $\omega _n=\phi _n(\cdot + (x_n,y_n))$ is the shift of $\phi _n$ by $(x_n,y_n)$. We shift the function $\phi _n$ for reasons of convenience in order to work with balls and annuli centered at the origin instead of at $(x_n,y_n)$. Notice that if (i) and (ii) hold we obtain a contradiction due to the subadditivity of the $I_\mu$ stated in Proposition 2: Set

$$\begin{aligned} \mu _{1,n}:= {\mathcal {N}}(\omega _n^{(1)})\qquad \text{ and }\qquad \mu _{2,n}:= {\mathcal {N}}(\omega _n^{(2)}), \end{aligned}$$

and $\mu _i:=\lim _{n\rightarrow \infty }\mu _{i,n}$ for $i=1,2$. Then (i) implies that $\mu _1+\mu _2=\mu$, since ${\mathcal {N}}(\omega _n)=\mu$ for all $n\in \mathbb {N}$. First we show that $\mu _1\ne 0$. If $\mu _1= 0$, then $\mu _2= \mu$. By setting

$$\begin{aligned} {\tilde{\omega }}_n^{(2)}:= \left( \frac{\mu }{\mu _{2,n}}\right) ^\frac{1}{3}\omega _n^{(2)} \end{aligned}$$

we find ${\mathcal {N}}({\tilde{\omega }}_n^{(2)})=\mu$ for all $n\in \mathbb {N}$ and

$$\begin{aligned} \left| {\mathcal {L}}({\tilde{\omega }}_n^{(2)})-{\mathcal {L}}( \omega _n^{(2)})\right| \rightarrow 0 \qquad \text{ for }\quad n\rightarrow \infty , \end{aligned}$$

since $\lim _{n\rightarrow \infty } \frac{\mu }{\mu _{2,n}}=1$. But then by using (ii) we obtain

$$\begin{aligned} I_\mu \le {\mathcal {L}}({\tilde{\omega }}_n^{(2)}) \rightarrow I_\mu -I^* < I_\mu \qquad \text{ for }\quad n\rightarrow \infty , \end{aligned}$$

which is a contradiction. Hence, $\mu _1\ne 0$ and similarly we find $\mu _2\ne 0$. Thus, $|\mu _i|>0$ for $i=1,2$ and we can define the rescaled functions

$$\begin{aligned} {\bar{\omega }}_n^{(i)}:= \left( \frac{|\mu _i|}{\mu _{i,n}}\right) ^\frac{1}{3}\omega _n^{(i)}\qquad \text{ for }\quad i=1,2, \end{aligned}$$

which satisfy ${\mathcal {N}}({\bar{\omega }}_n^{(i)})=|\mu _i|$ for all $n\in \mathbb {N}$ and

$$\begin{aligned} \lim _{n\rightarrow \infty } {\mathcal {L}}({\bar{\omega }}_n^{(1)})=I^*,\qquad \lim _{n\rightarrow \infty } {\mathcal {L}}({\bar{\omega }}_n^{(1)})=I_\mu -I^*, \end{aligned}$$

by (ii) together with $\lim _{n\rightarrow \infty } \left| \frac{|\mu _i|}{\mu _{i,n}}\right| =1$. Combining this with the subadditivity of $I_\mu$ for $\mu >0$, which is stated in Proposition 2, we find the contradiction

$$\begin{aligned} I_\mu \le I_{|\mu _1|+|\mu _2|}<I_{|\mu _1|}+I_{|\mu _2|}\le \lim _{n\rightarrow \infty } \left( {\mathcal {L}}({\bar{\omega }}_n^{(1)})+{\mathcal {L}}(\bar{\omega }_n^{(2)})\right) = I_\mu . \end{aligned}$$

We are left to show that there exists two sequences $\{\omega _n^{(1)}\}_{n\in \mathbb {N}}, \{\omega _n^{(2)}\}_{n\in \mathbb {N}}$, which have in the limit $n\rightarrow \infty$ disjoint support and satisfy (i), (ii). To this end, let $\varphi _n=\partial _x^{-1}\phi _n$ and let $\chi :\mathbb {R}^2\rightarrow [0,1]$ be a smooth cutoff function such that $\chi (x,y)=1$ for $|(x,y)|\le 1$ and $\chi (x,y)=0$ for $|(x,y)|\ge 2$. Next let $\sigma _n:=\varphi _n(\cdot +(x_n,y_n))$ and

$$\begin{aligned} \sigma _n^{(1)}:=\chi _{1n}\sigma _{n, A_{2M_n, M_n}},\quad \sigma _n^{(2)}:=\chi _{2n}\sigma _{n,A_{N_n,N_n/2}}, \end{aligned}$$

where

$$\begin{aligned} \chi _{1n}(x,y):=\chi \left( \frac{1}{M_n}(x,y)\right) ,\quad \chi _{2n}:=1-\chi \left( \frac{2}{N_n}(x,y)\right) . \end{aligned}$$

Eventually, we define

$$\begin{aligned} \omega _n:=\partial _x\sigma _n,\quad \omega _n^{(i)}:=\partial _x\sigma _n^{(i)},\ i=1,2. \end{aligned}$$

We remark that by definition $\omega _n=\phi _n(\cdot + (x_n,y_n))$. Furthermore,

$$\begin{aligned} \text {supp}(\omega _n^{(1)})\subset B_{2M_n}\qquad \text{ and } \qquad \text {supp}(\omega _n^{(2)})\subset \mathbb {R}^2\backslash B_{\frac{N_n}{2}}. \end{aligned}$$

See Figure 1 for an illustration of the supports for $\omega _n^{(i)}, i=1,2$.

Fig. 1

For n large enough the supports of $\omega _n^{(1)}$ and $\omega _n^{(2)}$, given by $B_{2M_n}$ and $\mathbb {R}^2\setminus B_{\frac{N_n}{2}}$, are disjoint. On $B_{M_n}$ and $\mathbb {R}^2\setminus B_{N_n}$ the functions $\omega _n^{(1)}$ and $\omega _n^{(2)}$ coincide with $\omega _n$, respectively

Roughly speaking the dichotomy assumption implies that the mass of $e_n$, which is given by ${\mathcal {L}}(\phi _n)=\frac{1}{2}\Vert \phi _n\Vert _{X_\frac{\alpha }{2}}^2$ splits into two disjoint regions. To be more precise, (2.8) yields

$$\begin{aligned} \begin{aligned} \left\Vert \omega _n\right\Vert _{\frac{\alpha }{2}, A_{N_n,M_n}}^2&=\left\Vert \omega _n\right\Vert _{\frac{\alpha }{2}, B_{N_n}}^2 -\left\Vert \omega _n\right\Vert _{\frac{\alpha }{2}, B_{M_n}}^2\\&=2\left( \int _{(x_n,y_n)+B_{N_n}}e_n\ \textrm{d}(x,y)-\int _{(x_n,y_n)+B_{M_n}}e_n\ \textrm{d}(x,y)\right) \rightarrow 0, \end{aligned} \end{aligned}$$

(2.9)

as $n\rightarrow \infty$. Using this result together with Proposition 1 (ii) and Proposition 3 we also find that

$$\begin{aligned} \left\Vert \omega _n\right\Vert _{L^3(A_{N_n,M_n})}\rightarrow 0 \qquad \text{ for }\quad n\rightarrow \infty . \end{aligned}$$

(2.10)

In what follows we will prove that the statements (i) and (ii) hold true.

• (i) Consider

  $$\begin{aligned} \begin{aligned}&\left| {\mathcal {N}}(\omega _n^{(1)}) + {\mathcal {N}}(\omega _n^{(2)}) -{\mathcal {N}}(\omega _n)\right| \\&= \left| \int _{\mathbb {R}^2}(\omega _n^{(1)})^3\ \textrm{d}(x,y)+\int _{\mathbb {R}^2}(\omega _n^{(2)})^3\ \textrm{d}(x,y)-\int _{\mathbb {R}^2}\omega _n^3\ \textrm{d}(x,y)\right| \\&=\bigg |\int _{A_{2M_n,M_n}}(\omega _n^{(1)})^3\textrm{d}(x,y)\quad +\int _{A_{N_n,\frac{N_n}{2}}}(\omega _n^{(2)})^3\ \textrm{d}(x,y)\\&\quad -\int _{A_{N_n,M_n}}\omega _n^3\ \textrm{d}(x,y)\bigg |, \end{aligned} \end{aligned}$$

  (2.11)

  where we used that $\omega _n^{(1)}=\omega _n$ on $B_{M_n}$ and $\omega _n^{(2)}=\omega _n$ on $\mathbb {R}^2 \setminus B_{N_n}$. The term $\int _{A_{N_n,M_n}}\omega _n^3\ \textrm{d}(x,y)$ tends to zero in view of (2.10) and

  $$\begin{aligned} \left\Vert w_n^{(1)}\right\Vert _{L^3(A_{2M_n,M_n})}&= \left\Vert \partial _x \sigma _n^{(1)}\right\Vert _{L^3(A_{2M_n,M_n})}\\&\le \frac{1}{M_n}\left\Vert \partial _x\chi _{1n}\sigma _{n,A_{2M_n, M_n}}\right\Vert _{L^3(A_{2M_n,M_n})}\\&\quad +\left\Vert \chi _{1n}\partial _x\sigma _{n,A_{2M_n, M_n}}\right\Vert _{L^3(A_{2M_n,M_n})}\\&=\frac{1}{M_n}\left\Vert \partial _x\chi _{1n}\sigma _{n,A_{2M_n, M_n}}\right\Vert _{L^3(A_{2M_n,M_n})}\\&\quad +\left\Vert \chi _{1n}\omega _n\right\Vert _{L^3(A_{2M_n,M_n})}, \end{aligned}$$

  where we used that $\partial _x \sigma _{n,A_{2M_n, M_n}}= \partial _x \sigma _n = \omega _n$. Using Lemma 2, the smoothness of $\chi _{1,n}$, and (2.9), the first term on the right-hand side above can be estimated by

  $$\begin{aligned} \begin{aligned} \frac{1}{M_n}&\left\Vert \partial _x\chi _{1n}\sigma _{n,A_{2M_n, M_n}}\right\Vert _{L^3(A_{2M_n,M_n})}\lesssim \frac{1}{M_n}\left\Vert \sigma _{n,A_{2M_n, M_n}}\right\Vert _{L^3(A_{2M_n,M_n})}\\&\lesssim M_n^{-\frac{2}{3}}\left\Vert \nabla \sigma _{n,A_{2M_n, M_n}}\right\Vert _{L^2(A_{2M_n,M_n})}\\&\le M_n^{-\frac{2}{3}}\left( \left\Vert \partial _x\sigma _{n,A_{2M_n, M_n}}\right\Vert _{L^2(A_{2M_n,M_n})} +\left\Vert \partial _y\sigma _{n,A_{2M_n, M_n}}\right\Vert _{L^2(A_{2M_n,M_n})}\right) \\&=M_n^{-\frac{2}{3}}\left( \left\Vert \omega _n\right\Vert _{L^2(A_{2M_n,M_n})} +\left\Vert \partial _y\partial _x^{-1}\omega _n\right\Vert _{L^2(A_{2M_n,M_n})}\right) \\&\le M_n^{-\frac{2}{3}}\Vert \omega _n\Vert _{\frac{\alpha }{2}, A_{N_n,M_n}} \rightarrow 0 \end{aligned} \end{aligned}$$

  (2.12)

  as $n\rightarrow \infty$. The second term tends to zero as $n\rightarrow \infty$ due to (2.10) and the boundedness of $\chi _{1,n}$. We conclude

  $$\begin{aligned} \int _{\mathbb {R}^2}(\omega _n^{(1)})^3\ \textrm{d}(x,y)\rightarrow 0\qquad \text{ and }\qquad \int _{\mathbb {R}^2}(\omega _n^{(2)})^3\ \textrm{d}(x,y)\rightarrow 0 \end{aligned}$$

  as $n\rightarrow \infty$, where the second assertion can be shown in the same way. Together with (2.10), equation (2.11) finishes the proof of statement (i).

• (ii) We proceed to investigate the limit

  $$\begin{aligned}&\lim _{n\rightarrow \infty }{\mathcal {L}}(\omega _n^{(1)})\\&=\frac{1}{2}\lim _{n\rightarrow \infty }\left( \left\Vert \omega _n^{(1)}\right\Vert _{L^2(\mathbb {R}^2)}^2 +\left\Vert \textrm{D}_x^\frac{\alpha }{2}\omega _n^{(1)}\right\Vert _{L^2(\mathbb {R}^2)}^2 +\left\Vert \partial _x^{-1}\partial _y\omega _n^{(1)}\right\Vert _{L^2(\mathbb {R}^2)}^2\right) \end{aligned}$$

  and show that $\lim _{n\rightarrow \infty }{\mathcal {L}}(\omega _n^{(1)}) = I^*$. First consider

  $$\begin{aligned} \begin{aligned}&\left\Vert \omega _n^{(1)}\right\Vert _{L^2(\mathbb {R}^2)}^2=\left\Vert \frac{1}{M_n}\partial _x \chi _{1n}\sigma _{n,B_{M_n}}+\chi _{1n}\omega _n\right\Vert _{L^2(\mathbb {R}^2)}^2\\&\quad =\frac{1}{M_n^2}\left\Vert \partial _x\chi _{1n}\sigma _{n,A_{2M_n, M_n}}\right\Vert _{L^2(\mathbb {R}^2)}^2+\frac{2}{M_n}\langle \partial _x\chi _{1n} \sigma _{n,A_{2M_n, M_n}},\chi _{1n}\omega _n\rangle _{L^2(\mathbb {R}^2)}\\&\quad +\left\Vert \chi _{1n}\omega _n\right\Vert _{L^2(\mathbb {R}^2)}^2. \end{aligned} \end{aligned}$$

  (2.13)

  Since $\partial _x \chi _{1,n}$ has support in $A_{2M_n,M_n}$ a similar argument as in (2.12) shows

  $$\begin{aligned} \frac{1}{M_n}\left\Vert \partial _x\chi _{1n}\sigma _{n,A_{2M_n, M_n}}\right\Vert _{L^2(\mathbb {R}^2)}\lesssim \left\Vert \omega _n\right\Vert _{\frac{\alpha }{2}, A_{2M_n,M_n}}\rightarrow 0,\text { as }n\rightarrow \infty , \end{aligned}$$

  (2.14)

  by using Lemma 2 and (2.9). Hence, we find that both the first and second term on the right-hand side of (2.13) tend to zero as $n\rightarrow \infty$. For the third term on the right-hand side of (2.13) we have

  $$\begin{aligned} \left\Vert \chi _{1n}\omega _n\right\Vert _{L^2(\mathbb {R}^2)}^2=\left\Vert \omega _n\right\Vert _{L^2(B_{M_n})} ^2+\left\Vert \chi _{1n}\omega _n\right\Vert _{L^2(A_{2M_n,M_n})}^2, \end{aligned}$$

  where we used that $\text{ supp } (\chi _{1,n})\subset B_{2M_n}$ and $\chi _{1,n}=1$ on $B_{M_n}$. Due to (2.9) we find

  $$\begin{aligned} \left\Vert \chi _{1n}\omega _n\right\Vert _{L^2(A_{2M_n,M_n)}}\lesssim \left\Vert \omega _n\right\Vert _{\frac{\alpha }{2},A_{2M_n,M_n}}\rightarrow 0. \end{aligned}$$

  We conclude

  $$\begin{aligned} \lim _{n\rightarrow \infty }\left| \left\Vert \omega _n^{(1)}\right\Vert _ {L^2(\mathbb {R}^2)}^2-\left\Vert \omega _n\right\Vert _{L^2(B_{M_n})}^2\right| =0. \end{aligned}$$

  (2.15)

  In the same way we can show

  $$\begin{aligned} \lim _{n\rightarrow \infty }\left| \left\Vert \partial _x^{-1}\partial _y\omega _n^{(1)}\right\Vert _{L^2(\mathbb {R}^2)} -\left\Vert \partial _x^{-1}\partial _y\omega _n\right\Vert _{L^2(B_{M_n})}^2\right| =0, \end{aligned}$$

  (2.16)

  so that we are only left to study

  $$\begin{aligned} \begin{aligned} \left\Vert \textrm{D}_x^\frac{\alpha }{2}\omega _{n}^{(1)}\right\Vert _{L^2(\mathbb {R}^2)}^2&=\left\Vert \textrm{D}_x^\frac{\alpha }{2}\left( \frac{1}{M_n}\partial _x\chi _{1n}\sigma _{n,A_{2M_n, M_n}}+\chi _{1n}\omega _n\right) \right\Vert _{L^2(\mathbb {R}^2)}^2\\&=\frac{1}{M_n^2}\left\Vert \textrm{D}_x^\frac{\alpha }{2}(\partial _x\chi _{1n}\sigma _{n,A_{2M_n, M_n}})\right\Vert _{L^2(\mathbb {R}^2)}^2\\&\quad +\frac{2}{M_n}\langle \textrm{D}_x^\frac{\alpha }{2}(\partial _x\chi _{1n} \sigma _{n,A_{2M_n, M_n}}),\textrm{D}_x^\frac{\alpha }{2}(\chi _{1n}\omega _n)\rangle _{L^2(\mathbb {R}^2)}\\&\quad +\left\Vert \textrm{D}_x^\frac{\alpha }{2}(\chi _{1n}\omega _n)\right\Vert _{L^2(\mathbb {R}^2)}^2. \end{aligned} \end{aligned}$$

  (2.17)

  We show first

  $$\begin{aligned} \Vert \textrm{D}_x^\frac{\alpha }{2}\left( \partial _x \chi _{1,n} \sigma _{n,A_{2M_n,M_n}}\right) \Vert _{L^2(\mathbb {R}^2)}\lesssim M_n \Vert \omega _n\Vert _{X_\frac{\alpha }{2},A_{2M_n,M_n}}, \end{aligned}$$

  (2.18)

  which implies by (2.9), the smoothness of $\chi _{1,n}$ and the boundedness of $\omega _n$ in $X_\frac{\alpha }{2}$ that the first two terms on the right-hand side of (2.17) tend to zero as $n\rightarrow \infty$. As in the proof of Proposition 3, an application of Leibniz’ rule for fractional derivatives yields

  $$\begin{aligned}&\Vert \textrm{D}_x^\frac{\alpha }{2}\left( \partial _x \chi _{1,n} \sigma _{n,A_{2M_n,M_n}}\right) \Vert _{L^2(\mathbb {R}^2)} \\&\lesssim \Vert \sigma _{n,A_{2M_n, M_n}}\Vert _{L^2(A_{2M_n,M_n})} + \Vert \textrm{D}_x^\frac{\alpha }{2} \sigma _{n,A_{2M_n,M_n}}\Vert _{L^2(A_{2M_n,M_n})}\\&\le 2\Vert \sigma _{n,A_{2M_n, M_n}}\Vert _{L^2(A_{2M_n,M_n})} + \Vert \textrm{D}_x^\frac{\alpha }{2} \omega _n\Vert _{L^2(A_{2M_n,M_n})}, \end{aligned}$$

  where we used interpolation and $\partial _x \sigma _{n,A_{2M_n,M_n}}= \omega _n$ in the last inequality. Using Lemma 2, the first term on the right-hand side above can by estimated by $M_n\Vert \omega _n\Vert _{X_\frac{\alpha }{2},A_{2M_n,M_n}}$ in the same spirit as in (2.12), while the second term is bounded by $\Vert \omega _n\Vert _{X_\frac{\alpha }{2},A_{2M_n,M_n}}$. Hence, (2.18) holds true and the first two terms in (2.17) tend to zero as $n\rightarrow \infty$. It remains to investigate the third term in (2.14), given by

  $$\begin{aligned} \Vert \textrm{D}_x^\frac{\alpha }{2} \left( \chi _{1,n}\omega _n\right) \Vert _{L^2(\mathbb {R}^2)}^2 = \Vert \textrm{D}_x^\frac{\alpha }{2} \omega _n\Vert _{L^2(B_{M_n})}^2 + \Vert \textrm{D}_x^\frac{\alpha }{2} \left( \chi _{1,n}\omega _n\right) \Vert _{L^2(A_{2M_n,M_n})}^2. \end{aligned}$$

  Again, by applying Leibniz’ rule for fractional derivatives and using that $\chi _{1,n}$ is smooth, we find that

  $$\begin{aligned} \Vert \textrm{D}_x^\frac{\alpha }{2} \left( \chi _{1,n}\omega _n\right) \Vert _{L^2(A_{2M_n,M_n})}^2&\lesssim \Vert \omega _n\Vert _{L^2(A_{2M_n,M_n})}^2 + \Vert \textrm{D}_x^\frac{\alpha }{2}\omega _n\Vert _{L^2(A_{2M_n,M_n})}^2\\&\le \Vert \omega \Vert _{X_\frac{\alpha }{2}, A_{2M_n,M_n}} \rightarrow 0, \end{aligned}$$

  by (2.9). This allows us to conclude

  $$\begin{aligned} \lim _{n\rightarrow \infty }\bigg |\left\Vert \textrm{D}_x^\frac{\alpha }{2} \omega _n^{(1)}\right\Vert -\left\Vert \textrm{D}_x^\frac{\alpha }{2}\omega _n\right\Vert _{L^2(B_{M_n})}\bigg |=0. \end{aligned}$$

  (2.19)

  Gathering (2.15), (2.16), and (2.19) we have shown

  $$\begin{aligned} {\mathcal {L}}(\omega _n^{(1)})\rightarrow I^* \qquad \text{ for }\quad n\rightarrow \infty . \end{aligned}$$

  In the same way we can obtain

  $$\begin{aligned} {\mathcal {L}}(\omega _n^{(2)})\rightarrow I_\mu -I^*\qquad \text{ for }\quad n\rightarrow \infty , \end{aligned}$$

  which proves statement (ii).

$\square$

By Proposition 4 and Proposition 5 the scenarios of "vanishing" and "dichotomy" in Theorem 3 are ruled out and the only possibility left is the concentration scenario. Hence, there exists $\{(x_n,y_n)\}_{n\in \mathbb {N}}\subset \mathbb {R}^2$ such that for each $\varepsilon >0$ there exists $r>0$ with

$$\begin{aligned} \int _{B_r(x_n,y_n)}e_n\ \textrm{d}(x,y)\ge I_\mu -\varepsilon ,\text { for all }n\in \mathbb {N}. \end{aligned}$$

This implies that, for r sufficiently large,

$$\begin{aligned} \int _{\mathbb {R}^2\backslash B_r(x_n,y_n)}e_n<\varepsilon . \end{aligned}$$

(2.20)

By taking a subsequence we may assume that (2.20) holds for all $n\in \mathbb {N}$. This implies in particular that

$$\begin{aligned} \left\Vert \phi _n(\cdot -(x_n,y_n))\right\Vert _{L^2(|(x,y)|>r)}<\varepsilon . \end{aligned}$$

(2.21)

Since $\{\phi _n\}_{n\in \mathbb {N}}$ is a bounded sequence in $X_\frac{\alpha }{2}$ we may assume $\phi _n\rightharpoonup \phi \in X_\frac{\alpha }{2}$. From Proposition 1 we know that $X_\frac{\alpha }{2}$ is compactly embedded in $L_{\text {loc}}^2(\mathbb {R}^2)$, and therefore $\phi _n\rightarrow \phi$ strongly in $L_{\text {loc}}^2(\mathbb {R}^2)$. By combining this with (2.21) we can use Cantors diagonal extraction process to extract a subsequence, still denoted by $\phi _n$, converging strongly in $L^2(\mathbb {R}^2)$. Proposition 1 (ii) implies that for $\alpha > \tfrac{4}{5}$, $\phi _n\rightarrow \phi$ in $L^3(\mathbb {R}^2)$ as well, which yields that ${\mathcal {N}}(\phi )=\mu$. Finally, since ${\mathcal {L}}(\phi )=\frac{1}{2}\left\Vert \phi \right\Vert _\frac{\alpha }{2}$ and the norm is weakly lower semicontinuous, we find

$$\begin{aligned} {\mathcal {L}}(\phi )\le \liminf {\mathcal {L}}(\phi _n)=I_\mu . \end{aligned}$$

Hence, $\phi$ is a solution of the minimization problem (2.2).

Next, we investigate the lump solutions numerically. For the implementation of the iteration scheme (1.7), we consider the space interval as $[-1024,1024]\times [-1024,1024]$ and we set $c=1$ in all the experiments. We use $\phi _0(x,y)=\exp (-x^2-y^2 )$ as the initial guess for the iteration. To test the efficiency of the numerical scheme we first consider the KP-I case (i.e., $\alpha =2$) where the exact analytical lump solution is given in (1.3).

Fig. 2

The variation of the iteration, stabilization factor, and the residual errors with the number of iterations in the semi-log scale (top left), numerically generated lump solution of KP-I equation (top right), x-cross section $\phi (x,0)$ (bottom left) and y-cross section $\phi (0,y)$ (bottom right) of both numerical and analytical solutions

In Figure 2, we represent the numerically generated lump solution of the KP-I equation and the cross sections $\phi (x,0)$ and $\phi (0,y)$ of both numerical and analytical solutions. Choosing the number of grid points as $N_x=N_y=2^{13}$ for both x and y coordinates we see that the $L^\infty$-norm of the difference of numerical and exact solutions is approximately of order $10^{-5}$ after 50 iterations. In Figure 2 we also present the variation of three different errors with the number of iterations in a semi-log scale. The two-dimensional geometry of the solutions and the periodic setting in both directions for implementing the numerical scheme cause a slow convergence rate. In Table 1, we present errors for increasing values of $N_x$ and $N_y$. We observe that the errors get smaller for larger values of $N_x$ and $N_y$.

In the next experiment, we consider some examples for the fractional case. Figure 3 depicts the profiles of the numerical solutions for $\alpha =1.7$ and $\alpha =1.35$, respectively. It can be seen from the numerical results that the lump solutions become more peaked for smaller values of $\alpha$. Therefore, to ensure the required numerical accuracy we need to increase the number of grid points to $2^{14}$ for both x and y directions when $\alpha =1.35$. In this case the Fourier coefficients go down to $10^{-5}$. To obtain the same numerical accuracy for smaller values of $\alpha$, we need to increase the number of Fourier modes even more, which is not accessible due to the limits of computation.

We also observe the cross-sectional symmetry of the lump solutions of the fKP-I equation numerically. We present several x and y-cross sections of the solutions for various $\alpha$. We consider the cases $\alpha =2$, $\alpha =1.7$ and $\alpha =1.35$ in Figure 4, Figure 5, and Figure 6, respectively. The numerical results indicate symmetry in both x and y directions.

Table 1 Numerical errors for several values of $N_x$ and $N_y$ when $\alpha =2$ after 50 iterations

Fig. 3

Lump solutions for $\alpha =1.7$ (left panel) and $\alpha =1.35$ (right panel)

Fig. 4

Several x and $y-$cross sections of the numerically generated lump solution for $\alpha =2$

Fig. 5

Several x and $y-$cross sections of the numerically generated lump solution for $\alpha =1.7$

Fig. 6

Several x and $y-$cross sections of the numerically generated lump solution for $\alpha =1.35$

3 Decay of lump solutions

Throughout this section, unless specifically stated otherwise, we assume that $\alpha >\frac{4}{5}$. The existence of lump solutions $u(t,x,y)=\phi (x-ct,y)$ for the fKP-I equation, where $\phi \in X_{\frac{\alpha }{2}}$, was proved in the previous section. The function $\phi$ satisfies the (rescaled) traveling wave fKP-I equation

$$\begin{aligned} -\phi _{xx}-\phi _{yy}- \textrm{D}_x^\alpha \phi _{xx}+\frac{1}{2} (\phi ^2)_{xx}=0, \end{aligned}$$

(3.1)

which can be written in convolution form as

$$\begin{aligned} \phi =\frac{1}{2}K_\alpha *\phi ^2,\qquad {{\hat{K}}}_\alpha (\xi _1,\xi _2)= m_\alpha (\xi _1,\xi _2), \end{aligned}$$

(3.2)

where the symbol $m_\alpha$ is given by

$$\begin{aligned} m_\alpha (\xi _1,\xi _2)=\frac{\xi _1^2}{|\xi |^2+|\xi _1|^{\alpha +2}}. \end{aligned}$$

Let us recall from Remark 1 that any nontrivial, continuous solution $\phi$ of (3.2) decays at most quadratically. In this section, we show that any nontrivial solution $\phi \in X_{\frac{\alpha }{2}}$ of (3.2) decays indeed exactly quadratically, that is we prove Theorem 2.

The idea is to study the kernel function $K_\alpha$ and to show that it has quadratic decay at infinity (independent of $\alpha$). Then the decay properties of $K_\alpha$ are used to show that also $\phi$ decays quadratically at infinity.

In the sequel we denote by $\varrho :\mathbb {R}^2 \rightarrow \mathbb {R}$ the function

$$\begin{aligned}\varrho (x,y)=|(x,y)|=(x^2+y^2)^{\frac{1}{2}}.\end{aligned}$$

It will be useful to note that for all $a\ge 1$ we have that $\varrho ^a$ is convex, so that

$$\begin{aligned} \varrho ^a(x,y)\lesssim \varrho ^a(x-{\bar{x}}, y-{\bar{y}})+\varrho ^a(\bar{x}, {\bar{y}})\qquad \text{ for } \text{ all }\quad (x,y),({\bar{x}}, {\bar{y}}) \in \mathbb {R}^2. \end{aligned}$$

(3.3)

Notice that by (3.3) and Young’s inequality

$$\begin{aligned} \Vert \varrho ^2 \phi \Vert _\infty&\lesssim \Vert \varrho ^2K_\alpha *\phi ^2\Vert _\infty + \Vert K_\alpha * \varrho ^2\phi ^2\Vert _\infty \\ {}&\lesssim \Vert \varrho ^2K_\alpha \Vert _\infty \Vert \phi \Vert _{L^2(\mathbb {R}^2)}^2 + \Vert K_\alpha \Vert _{L^q(\mathbb {R}^2)}\Vert \varrho ^2\phi ^2\Vert _{L^{q^\prime }(\mathbb {R}^2)} \end{aligned}$$

for some $1\le q,q^\prime \le \infty$ with $1=\frac{1}{q}+\frac{1}{q^\prime }$, so that the statement of Theorem 2 is proved provided that

• (A) $\varrho ^2K_\alpha \in L^\infty (\mathbb {R}^2)$

• (B) there exists $1\le q\le \infty$ such that $K_\alpha \in L^q(\mathbb {R}^2)$ and $\varrho ^2\phi ^2 \in L^{q^\prime }(\mathbb {R}^2)$, where $q^\prime$ is the dual conjugate of q.

Before studying the properties of the kernel function $K_\alpha$, we state the following two lemmata, which yield some a priori regularity of lump solutions in the energy space.

Lemma 3

Any solution $\phi$ of (3.2) in the energy space $X_{\frac{\alpha }{2}}$ satisfies $\phi \in L^r(\mathbb {R}^2)$ for all $2\le r<\infty$ and $\phi \in H^\infty (\mathbb {R}^2).$ In particular, $\phi$ is uniformly continuous and decays to zero at infinity.

Proof

Let us start by repeating the Hörmander–Mikhilin multiplier theorem [30], which states that if $f:\mathbb {R}^2 \rightarrow \mathbb {R}$ is a function, which is smooth outside the origin and

$$\begin{aligned} \xi \mapsto \xi _1^{k_1}\xi _2^{k_2} \frac{\textrm{d}^k}{\textrm{d} \xi _1^{k_1}\textrm{d} \xi _2^{k_2}}f(\xi ) \end{aligned}$$

is bounded on $\mathbb {R}^2$ for all $k_1,k_2 \in \{0,1\}$ with $k=k_1+k_2 \in \{0,1,2\}$, then f is a Fourier multiplier on $L^p(\mathbb {R}^2)$ for all $1<p<\infty$, i.e. the operator $T_f$ defined by $T_f g= {\mathcal {F}}^{-1}\left( f {{\hat{g}}}\right) = {\mathcal {F}}^{-1}(f)*g$ is bounded on $L^p(\mathbb {R}^2)$. By the Hörmander–Mikhilin multiplier theorem, it is easy to check that the functions

$$\begin{aligned} \xi \mapsto m_\alpha (\xi ),\qquad \xi \mapsto |\xi _1|^\alpha m_\alpha (\xi ),\qquad \xi \mapsto \xi _2m_\alpha (\xi ) \end{aligned}$$

are Fourier multipliers on $L^p(\mathbb {R}^2)$ for $1<p<\infty$. Let $\phi \in X_{\frac{\alpha }{2}}$. Due to Proposition 1 (ii), we have that $\phi \in L^3(\mathbb {R}^2)$, which implies $\phi ^2 \in L^{\frac{3}{2}}(\mathbb {R}^2)$. Since

$$\begin{aligned} \phi = \frac{1}{2}{\mathcal {F}}^{-1}(m_\alpha )*\phi ^2, \ \textrm{D}^\alpha _x \phi = \frac{1}{2}{\mathcal {F}}^{-1}(|\xi _1|^\alpha m_\alpha )*\phi ^2,\ \phi _y = -\frac{{{\,\mathrm{\textrm{i}}\,}}}{2}{\mathcal {F}}^{-1}(\xi _2m_\alpha )*\phi ^2, \end{aligned}$$

we find

$$\begin{aligned} \phi , \textrm{D}^\alpha _x \phi , \phi _y \in L^{\frac{3}{2}}(\mathbb {R}^2). \end{aligned}$$

In particular, $\phi$ belongs to the anisotropic Sobolev space $W^{\mathbf {\alpha }, \frac{3}{2}}(\mathbb {R}^2)$ for $\mathbf {\alpha }= (\alpha ,1)$, where

$$\begin{aligned} W^{\mathbf {\alpha }, q}(\mathbb {R}^2):=\{\phi \in L^q(\mathbb {R}^2)\mid \textrm{D}_x^{\alpha _1} \phi , \textrm{D}_y^{\alpha _2}\phi \in L^q(\mathbb {R}^2)\}. \end{aligned}$$

We use the following anisotropic Gagliardo–Nirenberg inequality for fractional derivatives [13, Theorem 1.1]: If $\phi \in W^{\mathbf {\alpha }, q}(\mathbb {R}^2)$ with $A:=\frac{1}{q}\left( \frac{1}{\alpha _1}+\frac{1}{\alpha _2}\right) -1>0$ and $M:=1+(\frac{1}{p}-\frac{1}{q})\left( \frac{1}{\alpha _1}+\frac{1}{\alpha _2}\right) >0$, then

$$\begin{aligned} \Vert \phi \Vert _{L^r(\mathbb {R}^2)}\lesssim \Vert \phi \Vert _{L^p(\mathbb {R}^2)}^{1-\theta }\Vert \textrm{D}_x^{\alpha _1} \phi \Vert _{L^q(\mathbb {R}^2)}^{\theta _1}\Vert \textrm{D}_y^{\alpha _2} \phi \Vert _{L^q(\mathbb {R}^2)}^{\theta _2}, \end{aligned}$$

(3.4)

for all

$$\begin{aligned} p\le r<r_*:=\frac{1}{A }\left( \frac{1}{\alpha _1}+\frac{1}{\alpha _2}\right) , \end{aligned}$$

where $\theta = \theta _1 + \theta _2$ and $\theta _i= (\frac{1}{p}-\frac{1}{r})(\alpha _i M)^{-1}$. Applied to the situation at hand, we can choose $p=2$, $q=\frac{3}{2}$ and $\mathbf {\alpha }=(\alpha ,1)$ for $\alpha = \frac{4}{5+4\varepsilon }$ with $\varepsilon >0$ arbitrarily small, which yields $A= \frac{1}{2}+\frac{2}{3}\varepsilon$ and $M=\frac{5}{8}-\frac{1}{6}\varepsilon$. Due to (3.4) we find that

$$\begin{aligned} \phi \in L^r(\mathbb {R}^2)\qquad \text{ for } \text{ all } \quad 2\le r < \frac{9}{2}. \end{aligned}$$

Repeating the same argument for $\phi \in L^{\frac{9}{2}-2\varepsilon }(\mathbb {R}^2)$ with $\phi ^2 \in L^{\frac{9-4\varepsilon }{4}}(\mathbb {R}^2)$, we find that $\phi \in W^{(\alpha ,1), \frac{9-4\varepsilon }{4}}(\mathbb {R}^2)$ and again by the fractional Gagliardo–Nirenberg inequality for $p=2$, $q=\frac{9-4\varepsilon }{4}$, $\mathbf {\alpha }=(\alpha ,1)$ for $\alpha = \frac{4}{5+4\varepsilon }$, we obtain that $A=\frac{8\varepsilon }{9-4\varepsilon }$ and $M=\frac{9}{8}+\frac{\varepsilon }{2}\frac{4\varepsilon +7}{4\varepsilon -9}$, so that

$$\begin{aligned} \phi \in L^r(\mathbb {R}^2)\qquad \text{ for } \text{ all } \quad 2\le r < \infty , \end{aligned}$$

(3.5)

by letting $\varepsilon \rightarrow 0$. This proves the first assertion. The relation (3.5) implies by the Fourier multiplier theorem that

$$\begin{aligned} \phi , \textrm{D}_x^\alpha \phi , \phi _y \in L^r(\mathbb {R}^2)\qquad \text{ for } \text{ all } \quad 2\le r < \infty .\end{aligned}$$

Next, we aim to bootstrap the smoothness. By Hölder’s inequality it is clear also that $(\phi ^2)_y \in L^r(\mathbb {R}^2)$ for all $2\le r<\infty$. Due to the Leibniz rule for fractional derivatives (see e.g. [17, Theorem 7.6.1]) and Hölder’s inequality we can estimate

$$\begin{aligned} \Vert \textrm{D}_x^\alpha \phi ^2\Vert _{L^r(\mathbb {R}^2)}^r&= \int _{\mathbb {R}} \Vert \textrm{D}_x^\alpha \phi (\cdot , y)\Vert _{L^r(\mathbb {R})}^r\, \textrm{d}y\\&\lesssim \int _{\mathbb {R}}\Vert \textrm{D}_x^\alpha \phi (\cdot , y)\Vert _{L^{2r}(\mathbb {R})}^r\Vert \phi (\cdot , y)\Vert _{L^{2r}(\mathbb {R})}^r\, \textrm{d}y\\&\le \left( \int _{\mathbb {R}}\Vert \textrm{D}_x^\alpha \phi (\cdot , y)\Vert _{L^{2r}(\mathbb {R})}\,\textrm{d}y\right) ^\frac{1}{2}\left( \int _{\mathbb {R}}\Vert \phi (\cdot , y)\Vert _{L^{2r}(\mathbb {R})}^{2r}\,\textrm{d}y\right) ^\frac{1}{2}\\&= \Vert \textrm{D}_x^\alpha \phi \Vert _{L^{2r}(\mathbb {R}^2)}^\frac{r}{2}\Vert \phi \Vert _{L^{2r}(\mathbb {R}^2)}^\frac{r}{2}, \end{aligned}$$

which yields that also $\textrm{D}_x^\alpha \phi ^2 \in L^r(\mathbb {R}^2)$ for all $2\le r <\infty$. This can be used to bootstrap the smoothness of $\phi$, by using

$$\begin{aligned} \textrm{D}^\alpha _x \phi = \frac{1}{2}K_\alpha * \textrm{D}^\alpha _x \phi ^2\qquad \text{ and } \qquad \phi _y = \frac{1}{2}K_\alpha *(\phi ^2)_y. \end{aligned}$$

Reiterating the argument yields

$$\begin{aligned} \phi , \textrm{D}_x^\alpha \phi , \phi _y, \textrm{D}_x^{2\alpha } \phi ,\textrm{D}_x^{\alpha }\phi _y, \phi _{yy} \in L^r(\mathbb {R}^2)\qquad \text{ for } \text{ all } \quad 2 \le r < \infty . \end{aligned}$$

and eventually $\textrm{D}_x^k \phi \in L^r(\mathbb {R}^2)$ for all $2\le r <\infty$ and $k\in \mathbb {N}$, which implies that $\phi \in H^\infty (\mathbb {R}^2)$. Eventually, since $H^\infty (\mathbb {R}^2)$ is embedded into the space of uniformly continuous functions on $\mathbb {R}^2$ and $\phi$ is $L^2(\mathbb {R}^2)$-integrable, we deduce that $\phi$ decays to zero at infinity. $\square$

Lemma 4

Any solution $\phi$ of (3.1) in the energy space $X_{\frac{\alpha }{2}}$ satisfies

$$\begin{aligned} \int _{\mathbb {R}^2} (x^2+y^2)(|\phi _x|^2+|\phi _y|^2+|\textrm{D}^ \frac{\alpha }{2}_x\phi _x|^2)\,\textrm{d}(x,y)<\infty . \end{aligned}$$

Proof

The proof follows essentially the lines in [10, Lemma 3.1]. Here, we proceed formally by omitting the truncation function at infinity. First, let us multiply (3.1) by $x^2\phi$ and integrate over $\mathbb {R}^2$. Then

$$\begin{aligned} 0= \int _{\mathbb {R}^2} x^2 \phi \left( -\phi _{xx}-\phi _{yy} -\textrm{D}^\alpha _x \phi _{xx}+\frac{1}{2}\left( \phi ^2\right) _{xx}\right) \,\textrm{d}(x,y). \end{aligned}$$

Using integration by parts we find that

$$\begin{aligned} \int _{\mathbb {R}^2} x^2 \phi \left( -\phi _{xx}-\phi _{yy}+\frac{1}{2}\left( \phi ^2\right) _{xx}\right) \,\textrm{d}(x,y)&=\int _{\mathbb {R}^2} x^2 \left( \phi _x^2+ \phi _y^2-\phi \phi _x^2\right) \\&-\phi ^2 +\frac{2}{3}\phi ^3 \,\textrm{d}(x,y). \end{aligned}$$

In view of Lemma 8, the nonlocal part can be written as

$$\begin{aligned} -\int _{\mathbb {R}^2}x^2\phi \textrm{D}^\alpha _x \phi _{xx}\, d(x,y)&=-\int _{\mathbb {R}^2} x^2\phi _{xx}\textrm{D}^\alpha _x \phi \,\textrm{d}(x,y)-\int _{\mathbb {R}^2} 4x\phi _x \textrm{D}^\alpha _x \phi \,d(x,y)\\&\quad -\int _{\mathbb {R}^2} 2\phi \textrm{D}^\alpha _x \phi \,\textrm{d}(x,y)\\&=\int _{\mathbb {R}^2}x^2\left( \textrm{D}^{\frac{\alpha }{2}}\phi _x\right) ^2\,\textrm{d}(x,y) - \frac{1}{4}(\alpha +2)^2\\&\quad \times \int _{\mathbb {R}^2}\left( \textrm{D}^{\frac{\alpha }{2}}\phi \right) ^2\,\textrm{d}(x,y). \end{aligned}$$

Adding the above equalities we obtain that

$$\begin{aligned} \int _{\mathbb {R}^2} x^2 \left( \phi _x^2+ \phi _y^2+\left( \textrm{D}^{\frac{\alpha }{2}}\phi _x\right) ^2\right) \,\textrm{d}(x,y)&=\int _{\mathbb {R}^2}x^2\phi \phi _x^2 + \phi ^2 -\frac{2}{3}\phi ^3\nonumber \\&\quad +\frac{1}{4}(\alpha +2)^2\left( \textrm{D}^ {\frac{\alpha }{2}}\phi \right) ^2\,\textrm{d}(x,y) \end{aligned}$$

(3.6)

Multiplying (3.1) by $y^2\phi$ instead yields

$$\begin{aligned} \int _{\mathbb {R}^2} y^2 \phi \left( -\phi _{xx}-\phi _{yy} -\textrm{D}^\alpha _x \phi _{xx}+\frac{1}{2}\left( \phi ^2\right) _{xx}\right) \,\textrm{d}(x,y)=0. \end{aligned}$$

Again, using integration by parts, we find that

$$\begin{aligned} \int _{\mathbb {R}^2}y^2 \left( \phi _x^2+\phi _y^2+\left( \textrm{D}^\frac{\alpha }{2}_x\phi _x\right) ^2\right) \, \textrm{d}(x,y) = \int _{\mathbb {R}^2} \phi ^2+y^2\phi \phi _x^2\,\textrm{d}(x,y). \end{aligned}$$

(3.7)

Adding (3.6) and (3.7), while keeping in mind that $\phi \in X_{\frac{\alpha }{2}}$, we can estimate

$$\begin{aligned} \int _{\mathbb {R}^2} (x^2+y^2)(|\phi _x|^2+|\phi _y|^2+|\textrm{D}^\frac{\alpha }{2}_x\phi _x|^2)\, \textrm{d}(x,y)\lesssim 1 + \int _{\mathbb {R}^2} (x^2+y^2)\phi \phi _x^2\,\textrm{d}(x,y). \end{aligned}$$

Using that $\phi$ is continuous and tends to zero at infinity (see Lemma 3), there exists $R>0$ such that $\phi (x,y)\le \frac{1}{2}$ for $|(x,y)|\ge R$ and we conclude

$$\begin{aligned} \int _{\mathbb {R}^2} (x^2+y^2)(|\phi _x|^2+|\phi _y|^2+|\textrm{D}^\frac{\alpha }{2}_x\phi _x|^2)\lesssim _R 1. \end{aligned}$$

$\square$

Properties of the kernel function $K_\alpha$. We will first concentrate on the regularity properties of $K_\alpha$.

Lemma 5

$m_\alpha \in L^p(\mathbb {R}^2)$ if and only if $p>\frac{2}{\alpha }+\frac{1}{2}$.

Proof

It is clear that $m_\alpha \in L^\infty (\mathbb {R}^2)$. Let us compute

$$\begin{aligned} \Vert m_\alpha \Vert _{L^p(\mathbb {R}^2)}^p&= \int _{\mathbb {R}^2} \frac{1}{\left( 1+|\xi _1|^\alpha + \frac{\xi _2^2}{\xi _1^2}\right) ^p}\, \textrm{d}(x,y)\\&= \int _{\mathbb {R}} \frac{1}{\left( 1+|\xi _1|^\alpha \right) ^p} \int _{\mathbb {R}} \frac{1}{\left( 1+\frac{\xi _2^2}{(1+|\xi _1|^\alpha )\xi _1^2}\right) ^p} \, \textrm{d}\xi _2\,\textrm{d}\xi _1 \\&=\int _{\mathbb {R}} \frac{|\xi _1|}{\left( 1+|\xi _1|^\alpha \right) ^{p-\frac{1}{2}}} \,\textrm{d}\xi _1 \int _{\mathbb {R}} \frac{1}{\left( 1+z^2\right) ^p}\,\textrm{d}z, \end{aligned}$$

where we used the change of variables $z=\tfrac{\xi _2}{|\xi _1|(1+|\xi _1|^\alpha )^\frac{1}{2}}$. Since the second integral above is clearly convergent for any $p>\tfrac{1}{2}$, we find that $m_\alpha \in L^p(\mathbb {R}^2)$ if and only if

$$\begin{aligned} p>\frac{2}{\alpha }+\frac{1}{2}. \end{aligned}$$

$\square$

Remark 3

The above lemma implies that $m_\alpha \in L^2(\mathbb {R}^2)$ if and only if $\alpha >\frac{4}{3}$, which is the $L^2$-critical exponent. In this case it follows immediately, that also $K_\alpha \in L^2(\mathbb {R}^2)$ and the proof of Theorem 2 can be done essentially by following the lines in [10]. In the supercritical case $\tfrac{4}{5}<\alpha \le \frac{4}{3}$, which in particular includes the Benjamin–Ono KP equation for $\alpha =1$, the symbol $m_\alpha$ belongs to an $L^p$-space with $p>2$ so that the integrability properties of the kernel $K_\alpha$ are a priori not clear.

Lemma 6

The kernel function $K_\alpha$ is smooth outside the origin.

Proof

Let $\chi : \mathbb {R}^2 \rightarrow \mathbb {R}$ be a compactly supported, radial, smooth function with $\chi (0,0)=1$. Set ${\bar{m}}_\alpha :=\chi m_\alpha$. Then ${\bar{m}}_\alpha$ has compact support and ${\mathcal {F}}^{-1}( \bar{m}_\alpha )$ is real analytic. Now, set $\tilde{m}_\alpha :=(1-\chi )m_\alpha$. Then ${\tilde{m}}_\alpha$ is smooth. Let us fix $(x_0,y_0) \ne (0,0)$ and let $\psi :\mathbb {R}^2\rightarrow \mathbb {R}$ be a compactly supported, smooth function with $\psi (x,y)=1$ in an arbitrarily small neighborhood of $(x_0,y_0)$ and $\psi (0,0)=0$. Then also

$$\begin{aligned} \Psi _k(x,y)=|(x,y)|^{-2k}\psi \end{aligned}$$

is smooth and compactly supported. Notice that ${\hat{\Psi }}_k = -\Delta ^{-k} {\hat{\psi }}$ and

$$\begin{aligned} {\tilde{m}}_\alpha * {\hat{\psi }} = -{\tilde{m}}_\alpha * \Delta ^k{\hat{\Psi }}_k = -(\Delta ^k {\tilde{m}}_\alpha * {\hat{\Psi }}_k). \end{aligned}$$

Since $\Psi _k$ is smooth with compact support, we know that $\hat{\Psi }_k \in {\mathcal {S}}(\mathbb {R}^2)$. Furthermore $\Delta ^k \tilde{m}_\alpha$ is smooth with $\Delta ^k {\tilde{m}}_\alpha (\xi ) \lesssim \tfrac{1}{|\xi |^{\alpha + 2k}}$ for $|\xi | \rightarrow \infty$. Since the convolution of two integrable, smooth functions is smooth and decays at least as fast as the function with the lower decay, we deduce that ${\tilde{m}}_\alpha *{\hat{\psi }}$ is smooth and decays at least as $\frac{1}{|\cdot |^{\alpha +2k}}$ at infinity for an arbitrary choice of $k\in \mathbb {N}$. In particular, ${\mathcal {F}}^{-1}({\tilde{m}}_\alpha *\hat{\psi })= {\mathcal {F}}^{-1}({\tilde{m}}_\alpha ) \psi$ is smooth, which yields that ${\mathcal {F}}^{-1}({\tilde{m}}_\alpha )$ is smooth outside the origin. We conclude that

$$\begin{aligned} K_\alpha = {\mathcal {F}}^{-1}( {\bar{m}}_\alpha ) + {\mathcal {F}}^{-1}({\tilde{m}}_\alpha ) \end{aligned}$$

is smooth outside the origin. $\square$

Let us now investigate the behavior of $K_\alpha$ at infinity. We show that the decay is quadratic, independently of the value of $\alpha >0$.

Proposition 6

For any $\alpha >0$, we have that $\varrho ^2K_\alpha$ belongs to $L^\infty (\mathbb {R}^2)$.

Proof

We have $K_\alpha ={\mathcal {F}}^{-1}(m_\alpha )$, so that

$$\begin{aligned} K_\alpha (x,y)&=\int _{\mathbb {R}^2}\frac{\xi _1^2}{\xi _1^2 + \xi _2^2 +| \xi _1|^{\alpha +2}}e^{\textrm{i}x\xi _1 + \textrm{i}y\xi _2}\,\textrm{d}\xi _1\, \textrm{d}\xi _2\\&=\int _\mathbb {R}\frac{|\xi |}{(1+|\xi |^\alpha )^{\frac{1}{2}}} e^{-|y||\xi |(1+|\xi |^\alpha )^{\frac{1}{2}}}e^{\textrm{i}x\xi }\, \textrm{d}\xi , \end{aligned}$$

where we used that

$$\begin{aligned} {\mathcal {F}}\left( \frac{1}{a^2+(\cdot )^2}\right) (y)=\int _{\mathbb {R}} \frac{1}{a^2+\xi _2^2}e^{-\textrm{i}\xi _2 y}\, \textrm{d}\xi _2 = \frac{1}{a}e^{-a|y|} \end{aligned}$$

and $a^2=\xi _1^2+|\xi _1|^{\alpha +2}$. Let us consider the case where $\xi \ge 0$ (the proof works similarly for $\xi <0$). Assume for the moment that $y\ne 0$. Setting

$$\begin{aligned} K_\alpha ^+(x,y):=\int _0^\infty \frac{\xi }{(1+\xi ^\alpha )^{\frac{1}{2}}}e^{-|y| \xi (1+\xi ^\alpha )^{\frac{1}{2}}}e^{\textrm{i}x\xi }\, \textrm{d}\xi , \end{aligned}$$

we can write

$$\begin{aligned} K_\alpha ^+(x,y)=\int _0^\infty \frac{\xi }{(1+\xi ^\alpha )^{\frac{1}{2}}}\frac{1}{G^\prime (\xi )}\frac{d}{d\xi }\left( e^{G(\xi )}\right) \, \textrm{d}\xi , \end{aligned}$$

where $G(\xi ):=\textrm{i}x\xi - |y|\xi (1+\xi ^\alpha )^\frac{1}{2}$. Using integration by parts, we obtain

$$\begin{aligned} K_\alpha ^+(x,y)= -\int _0^\infty \frac{\textrm{d}}{\textrm{d}\xi } \left( \frac{\xi }{(1+\xi ^\alpha )^{\frac{1}{2}}}\frac{1}{G^\prime (\xi )} \right) e^{G(\xi )}\,\textrm{d}\xi . \end{aligned}$$

Applying again integration by parts, we find

$$\begin{aligned} K_\alpha ^+(x,y)&=-\left[ \frac{\textrm{d}}{\textrm{d}\xi } \left( \frac{\xi }{(1+\xi ^\alpha )^{\frac{1}{2}}}\frac{1}{G^\prime (\xi )} \right) \frac{1}{G^\prime (\xi )} e^{G(\xi )} \right] _0^\infty \\&\quad +\int _0^\infty \frac{\textrm{d}}{\textrm{d}\xi }\left( \frac{\textrm{d}}{\textrm{d}\xi } \left( \frac{\xi }{(1+\xi ^\alpha )^{\frac{1}{2}}}\frac{1}{G^\prime (\xi )} \right) \frac{1}{G^\prime (\xi )}\right) e^{G(\xi )}\,\textrm{d}\xi \\&=\frac{\textrm{d}}{\textrm{d}\xi } \left( \frac{\xi }{(1+\xi ^\alpha )^{\frac{1}{2}}}\frac{1}{G^\prime (\xi )} \right) \frac{1}{G^\prime (\xi )} \Big |_{\xi =0}\\&\quad +\int _0^\infty \frac{\textrm{d}}{\textrm{d}\xi }\left( \frac{\textrm{d}}{\textrm{d}\xi } \left( \frac{\xi }{(1+\xi ^\alpha )^{\frac{1}{2}}}\frac{1}{G^\prime (\xi )} \right) \frac{1}{G^\prime (\xi )}\right) e^{G(\xi )}\,\textrm{d}\xi . \end{aligned}$$

In order to lighten the notation, we set

$$\begin{aligned} F(\xi ):=\frac{\textrm{d}}{\textrm{d}\xi } \left( \frac{\xi }{(1+\xi ^\alpha )^{\frac{1}{2}}}\frac{1}{G^\prime (\xi )} \right) \frac{1}{G^\prime (\xi )} , \end{aligned}$$

so that

$$\begin{aligned} K_\alpha ^+(x,y)= F(0)+\int _0^\infty F^\prime (\xi )e^{G(\xi )}\,\textrm{d}\xi . \end{aligned}$$

Using Lemma 9, we find

$$\begin{aligned} |K_\alpha ^+(x,y)|\le \frac{1}{x^2+y^2}. \end{aligned}$$

We are left to consider the case when $y=0$, that is

$$\begin{aligned} K_\alpha (x,0)=\int _\mathbb {R}\frac{|\xi |}{(1+|\xi |^\alpha )^\frac{1}{2}}e^{\textrm{i}x\xi }\,\textrm{d}\xi . \end{aligned}$$

Notice that $x^2K_\alpha (x,0)={-}{\mathcal {F}}^{-1} \left( \frac{\textrm{d}^2}{\textrm{d}\xi ^2}\frac{|\xi |}{(1+|\xi |^\alpha )^\frac{1}{2}}\right)$ and

$$\begin{aligned} \frac{\textrm{d}^2}{\textrm{d}\xi ^2}\frac{|\xi |}{(1+|\xi |^\alpha )^\frac{1}{2}}=2\delta _0(\xi )+g(\xi ), \end{aligned}$$

where $\delta _0$ denotes the delta distribution centered at zero and $g\in L^1(\mathbb {R})$. Thus $x\mapsto x^2K_\alpha (x,0)$ belongs to $L^\infty (\mathbb {R})$. $\square$

In order to determine the $L^p$-regularity of $K_\alpha$, it is left to investigate the behaviour of the kernel function close to the origin. To do so, we will use that $|\nabla m_\alpha |\lesssim h_\alpha$, where

$$\begin{aligned} h_\alpha (\xi _1,\xi _2):=\frac{\xi _1}{|\xi |^2+\xi _1^{\alpha +2}}. \end{aligned}$$

(3.8)

Notice also that $\widehat{\partial _x^{-1} K_\alpha }(\xi _1,\xi _2) =-{{\,\mathrm{\textrm{i}}\,}}h(\xi _1,\xi _2)$ and (3.2) can be written as

$$\begin{aligned} \phi =-\frac{{{\,\mathrm{\textrm{i}}\,}}}{2}H_\alpha *(\phi ^2)_x,\qquad {{\hat{H}}}_\alpha (\xi _1,\xi _2)= h_\alpha (\xi _1,\xi _2). \end{aligned}$$

(3.9)

Lemma 7

(The symbol $h_\alpha$) We have that

• a) $h_\alpha \in L^p(\mathbb {R}^2)$ if and only if $\frac{1}{2}+\frac{3}{2(1+\alpha )}<p<2$ and

  $$\begin{aligned} H_\alpha \in L^{p^\prime }(\mathbb {R}^2)\qquad \text{ for } \quad 2<p^\prime < \frac{4+\alpha }{2-\alpha }. \end{aligned}$$

• b) $\varrho H_\alpha \in L^\infty (\mathbb {R}^2)$.

Proof

Similar as in the proof of Lemma 5 we compute

$$\begin{aligned} \Vert h_\alpha \Vert _{L^p(\mathbb {R}^2)}^p&= \int _{\mathbb {R}^2} \frac{1}{\left( |\xi _1|+|\xi _1|^{\alpha +1}+\frac{\xi _2^2}{|\xi _1|}\right) ^p}\,\textrm{d}(\xi _1,\xi _2)\\&=\int _{\mathbb {R}} \frac{1}{|\xi _1|^p\left( 1+|\xi _1|^{\alpha }\right) ^p} \int _{\mathbb {R}} \frac{1}{ \left( 1+\frac{\xi _2^2}{\xi _1^2(1+|\xi _1|^{\alpha })}\right) ^p}\, \textrm{d}\xi _2 \,\textrm{d}\xi _1\\&=\int _{\mathbb {R}} \frac{1}{|\xi _1|^{p-1}\left( 1+|\xi _1|^{\alpha }\right) ^{p-\frac{1}{2}}} \,\textrm{d}\xi _1 \int _{\mathbb {R}} \frac{1}{ \left( 1+z^2\right) ^p}\, \textrm{d}z, \end{aligned}$$

where we used the change of variables $z=\frac{\xi _2}{|\xi _1| (1+|\xi _1|^\alpha )^\frac{1}{2}}$. Since the last integral above is bounded for all $p>1$, we find that $h_\alpha \in L^p(\mathbb {R}^2)$ if and only if

$$\begin{aligned} \frac{1}{2}+\frac{3}{2(1+\alpha )}<p<2. \end{aligned}$$

Since the Fourier transform is a bounded function from $L^p(\mathbb {R}^2)$ to $L^{p^\prime }(\mathbb {R}^2)$ for $p\in [1,2]$ and $p^\prime$ being the dual conjugate to p, we obtain immediately that

$$\begin{aligned} H_\alpha \in L^{p^\prime }(\mathbb {R}^2)\qquad \text{ for } \quad 2<p^\prime < \frac{4+\alpha }{2-\alpha }. \end{aligned}$$

Thereby, part (a) is proved. In order to prove part (b) we proceed as in the proof of Proposition 6. We have that

$$\begin{aligned} H_\alpha (x,y)&=\int _{\mathbb {R}^2}\frac{\xi _1}{\xi _1^2+\xi _2^2 +|\xi |^{\alpha +2}}e^{\textrm{i}x\xi _1+\textrm{i}y\xi _2}\, \textrm{d}(\xi _1,\xi _2)\\&= \int _{\mathbb {R}} \xi _1 e^{\textrm{i}x\xi _1}\int _{\mathbb {R}} \frac{1}{\xi _1^2+\xi _2^2+|\xi _1|^{\alpha +2}}e^{\textrm{i}\xi _2y}\,\textrm{d}\xi _2\, \textrm{d}\xi _1\\&= \int _{\mathbb {R}} \frac{\xi }{|\xi |(1+|\xi |^\alpha )^\frac{1}{2} } e^{\textrm{i}x\xi -|\xi |(1+|\xi |^\alpha )^\frac{1}{2}|y|}\,\textrm{d}\xi \\&=\int _{\mathbb {R}} \text{ sgn }(\xi )\frac{1}{(1+|\xi |^\alpha )^\frac{1}{2} }e^{\textrm{i}x\xi -|\xi |(1+|\xi |^\alpha )^\frac{1}{2}|y|}\,\textrm{d}\xi . \end{aligned}$$

Let us consider the positive part of the integral, the negative part can be estimated analogously. Assume for the moment that $y\ne 0$ and set

$$\begin{aligned} H^+_\alpha (x,y):=\int _0^\infty \frac{1}{(1+|\xi |^\alpha )^\frac{1}{2} }e^{\textrm{i}x\xi -|\xi |(1+|\xi |^\alpha )^\frac{1}{2}|y|}\,\textrm{d}\xi . \end{aligned}$$

With $E(\xi ):= \frac{1}{(1+\xi ^\alpha )^\frac{1}{2}}\frac{1}{G^\prime (\xi )}$, we obtain after integration by parts

$$\begin{aligned} H^+_\alpha (x,y)= -E(0)- \int _0^\infty E^\prime (\xi )e^{G(\xi )}\, \textrm{d}\xi , \end{aligned}$$

where $G(\xi )=\textrm{i}x\xi -|\xi |(1+\xi ^\alpha )^\frac{1}{2}|y|$. In view of Lemma 10 we find that

$$\begin{aligned} |H^+_\alpha (x,y)|\lesssim \frac{1}{\sqrt{x^2+y^2}}. \end{aligned}$$

(3.10)

If $y=0$, we have

$$\begin{aligned} H_\alpha (x,0)=\int _\mathbb {R}\frac{\xi }{|\xi |(1+|\xi |^\alpha )^\frac{1}{2}}e^{\textrm{i}x\xi }\,\textrm{d}\xi . \end{aligned}$$

Notice that $\textrm{i}xH_\alpha (x,y)=-{\mathcal {F}}^{-1} \left( \frac{\textrm{d}}{\textrm{d}\xi } \frac{\xi }{|\xi |(1+|\xi |^\alpha )^\frac{1}{2}}\right)$ and

$$\begin{aligned} \frac{\textrm{d}}{\textrm{d}\xi }\frac{\xi }{|\xi |(1+|\xi |^\alpha )^\frac{1}{2}} =\frac{1}{(1+|\xi |^\alpha )^\frac{1}{2}} \delta _0(\xi ) + g(\xi ), \end{aligned}$$

where $\delta _0$ denotes the delta distribution centered at zero and $g\in L^1(\mathbb {R})$. We deduce that $x\mapsto |x|H_\alpha (x,0)$ is a bounded function. Together with (3.10) this proves the claim that $\varrho H_\alpha \in L^\infty (\mathbb {R}^2)$. $\square$

Proposition 7

The kernel function $K_\alpha$ satisfies the regularity

$$\begin{aligned} K_\alpha \in L^{r}(\mathbb {R}^2)\qquad \text{ for }\quad 1<r< \frac{8+2\alpha }{8-\alpha }. \end{aligned}$$

Proof

We know already from Lemma 6 and Proposition 6 that $K_\alpha$ is smooth outside the origin and $\varrho ^2K_\alpha \in L^\infty (\mathbb {R}^2)$. Introducing a smooth truncation function $\vartheta :\mathbb {R}_+\rightarrow \mathbb {R}_+$, which is compactly supported in a neighborhood of zero, denoted by $B\subset \mathbb {R}^2$, with $\vartheta (0)=1$, we find that

$$\begin{aligned} (1-\vartheta (\varrho ))K_\alpha \in L^s(\mathbb {R}^2)\qquad \text{ for } \text{ all }\quad s>1. \end{aligned}$$

(3.11)

In order to determine the regularity of $K_\alpha$ close to zero, recall that $|\nabla m_\alpha |\lesssim |h_\alpha |$, where $h_\alpha$ defined in (3.8) so that

$$\begin{aligned} |\nabla m_\alpha |\in L^q(\mathbb {R}^2)\qquad \text{ for }\qquad \frac{1}{2}+\frac{3}{2(1+\alpha )}<q<2, \end{aligned}$$

due to Lemma 7 (a). Now, we use that the Fourier transformation is a bounded operator from $L^p(\mathbb {R}^2)$ to $L^{p^\prime }(\mathbb {R}^2)$ when $p\in [1,2]$ and $p^\prime$ is the dual conjugate of p and obtain that

$$\begin{aligned} \Vert \varrho K_\alpha \Vert _{L^{q^\prime }(\mathbb {R}^2)}&\lesssim \Vert xK\Vert _{L^{q^\prime }(\mathbb {R}^2)}+\Vert yK\Vert _{L^{q^\prime }(\mathbb {R}^2)}\\&\le \Vert \partial _{\xi _1}m_\alpha \Vert _{L^{q}(\mathbb {R}^2)}+\Vert \partial _{\xi _2}m_\alpha \Vert _{L^{q}(\mathbb {R}^2)}\\&\lesssim 2\Vert |\nabla m_\alpha |\Vert _{L^{q}(\mathbb {R}^2)}, \end{aligned}$$

so that

$$\begin{aligned} \varrho K_\alpha \in L^{q^\prime }(\mathbb {R}^2)\qquad \text{ for }\qquad 2<q^\prime < \frac{4+\alpha }{2-\alpha }, \end{aligned}$$

and in fact $\varrho K_\alpha \in L^{s}(B)$ for $1\le s<\frac{4+\alpha }{2-\alpha }$, by Hölder’s inequality and the boundedness of B. Then, we estimate

$$\begin{aligned} \Vert \vartheta (\varrho )K_\alpha \Vert _{L^r(\mathbb {R}^2)}\lesssim \Vert \varrho ^{-1}\Vert _{L^t(B)} \Vert \varrho K_\alpha \Vert _{L^{s}(B)}, \end{aligned}$$

where $\frac{1}{r}=\frac{1}{t}+\frac{1}{s}$. Since $\varrho ^{-1}\in L^t(B)$ if and only if $1\le t<2$, we conclude that

$$\begin{aligned} \vartheta (\varrho )K_\alpha \in L^r(\mathbb {R}^2)\qquad \text{ for } \quad 1\le r<\frac{8+2\alpha }{8-\alpha }. \end{aligned}$$

In view of (3.11) we deduce that

$$\begin{aligned} K_\alpha \in L^{r}(\mathbb {R}^2)\qquad \text{ for }\quad 1<r< \frac{8+2\alpha }{8-\alpha }. \end{aligned}$$

$\square$

A non-optimal decay rate. First notice that $\phi$ inherits the integrability properties of $K_\alpha$, since

$$\begin{aligned} \Vert \phi \Vert _{L^r(\mathbb {R}^2)}= \frac{1}{2}\Vert K_\alpha *\phi ^2\Vert _{L^r(\mathbb {R}^2)} \lesssim \Vert K_\alpha \Vert _{L^r(\mathbb {R}^2)}\Vert \phi ^2\Vert _{L^1(\mathbb {R}^2)}\lesssim \Vert K_\alpha \Vert _{L^r(\mathbb {R}^2)}\Vert \phi \Vert _{L^2(\mathbb {R}^2)}^2, \end{aligned}$$

by the $L^2$-integrability of $\phi$. Interpolating between the boundedness of $\phi$, which is due to Lemma 3, and the $L^r$-integrability of $\phi$ for $1<r<\frac{8+2\alpha }{8-\alpha }$, we actually find that

$$\begin{aligned} \phi \in L^p(\mathbb {R}^2)\qquad \text{ for } \text{ all }\quad p>1. \end{aligned}$$

(3.12)

Proposition 8

(A priori decay estimate) If $\phi$ is a solution of (3.1) in the energy space $X_{\frac{\alpha }{2}}$, then

$$\begin{aligned} \varrho \phi \in L^\infty (\mathbb {R}^2). \end{aligned}$$

Proof

Recall from (3.9) that

$$\begin{aligned} \phi =-\frac{\textrm{i}}{2}H_\alpha *(\phi ^2)_x \end{aligned}$$

so that by Young’s inequality

$$\begin{aligned} \Vert \varrho \phi \Vert _\infty&\le \Vert \varrho H*\phi \phi _x\Vert _\infty + \Vert H* \varrho \phi \phi _x\Vert _\infty \\&\lesssim \Vert \varrho H\Vert _\infty \Vert \phi \Vert _{L^2(\mathbb {R})}\Vert \phi _x\Vert _{L^2(\mathbb {R})} + \Vert H\Vert _{L^{q^\prime }(\mathbb {R}^2)}\Vert \varrho \phi _x\Vert _{L^2(\mathbb {R}^2)} \Vert \phi \Vert _{L^s(\mathbb {R}^2)}, \end{aligned}$$

where $\frac{1}{s}+\frac{1}{2}=\frac{1}{q}$ for $\frac{1}{2}+\frac{3}{2(1+\alpha )}<q<2$ and $q^\prime$ being the dual of q. Now, the statement follows from Lemma 7 and Lemma 4. $\square$

Proposition 9

(A non-optimal decay rate) If $\phi$ is a solitary solution of (3.1), then

$$\begin{aligned} \varrho ^{1+\delta } \phi \in L^\infty (\mathbb {R}^2) \end{aligned}$$

for any $0\le \delta <1$.

Proof

We use the regularity in Proposition 8 to improve the decay rate by estimating

$$\begin{aligned} \Vert \varrho ^{1+\delta }\phi \Vert _\infty \lesssim \Vert \varrho ^{1+\delta }K_\alpha *\phi ^2\Vert _\infty + \Vert K_\alpha * \varrho ^{1+\delta }\phi ^2\Vert _\infty , \end{aligned}$$

where we also used the convexity of $\varrho ^{1+\delta }$. The first norm on the right-hand side above is clearly bounded by Young’s inequality, Proposition 6 and the $L^2$-integrability of $\phi$. For the second norm, let $\varepsilon >0$ be a small constant so that $0<\delta<\frac{1}{1+\varepsilon }<1$. Using that $K_\alpha \in L^{1+\varepsilon }(\mathbb {R}^2)$ for $\varepsilon >0$ small enough, we estimate

$$\begin{aligned} \Vert K_\alpha * \varrho ^{1+\delta }\phi ^2\Vert _\infty \lesssim \Vert K_\alpha \Vert _{L^{1+\varepsilon }(\mathbb {R}^2)}\Vert \varrho ^{1+\delta }\phi ^2\Vert _{L^{\frac{1+\varepsilon }{\varepsilon }}(\mathbb {R}^2)}. \end{aligned}$$

Notice that

$$\begin{aligned} \Vert \varrho ^{1+\delta }\phi ^2\Vert _{L^{\frac{1+\varepsilon }{\varepsilon }}(\mathbb {R}^2)}^{\frac{\varepsilon }{1+\varepsilon }}&= \int _{\mathbb {R}^2} |\varrho \phi |^{\frac{(1+\delta )(1+\varepsilon )}{\varepsilon }} |\phi |^{(1-\delta )\frac{1+\varepsilon }{\varepsilon }}\, \textrm{d}(x,y)\\&\le \Vert \varrho \phi \Vert _\infty ^{\frac{(1+\delta )(1+\varepsilon )}{\varepsilon }} \int _{\mathbb {R}^2} |\phi |^{(1-\delta )\frac{1+\varepsilon }{\varepsilon }}\, \textrm{d}(x,y). \end{aligned}$$

By our choice of $\varepsilon >0$, we have $(1-\delta )\frac{1+\varepsilon }{\varepsilon }>1$, so the above norm is bounded by (3.12), which concludes the proof of the statement. $\square$

Proof of Theorem 2

In view of the discussion at the beginning of this section and Lemma 3, we obtain our main result

$$\begin{aligned} \varrho ^2\phi \in L^\infty (\mathbb {R}^2) \end{aligned}$$

provided that (A) and (B) at the beginning of the section are satisfied. The statement in (A) is proved in Proposition 6, while the first part of statement (B) follows from Proposition 7, where it is shown that

$$\begin{aligned} K_\alpha \in L^{r}(\mathbb {R}^2)\qquad \text{ for }\quad 1<r< \frac{8+2\alpha }{8-\alpha }. \end{aligned}$$

Now, we make use of the non-optimal decay estimate in Proposition 9 to show that indeed $\varrho ^2\phi ^2 \in L^{r^\prime }(\mathbb {R}^2)$, where $r^\prime$ is the dual conjugate to r. For any $0\le \delta <1$, we have that

$$\begin{aligned} \int _{\mathbb {R}^2} |\varrho ^2\phi ^2|^{r^\prime }\,\textrm{d}(x,y) \le \Vert \varrho ^{1+\delta }\phi \Vert _\infty ^{\frac{2r^\prime }{1+\delta }}\int _{\mathbb {R}^2} \phi ^{2r^\prime \frac{\delta }{1+\delta }}\,\textrm{d}(x,y). \end{aligned}$$

Choosing $\delta =r-1 \in (0,1)$ we find that $\frac{\delta }{1+\delta }r^\prime = \frac{r-1}{r} r^\prime =1$ and the boundedness of $\varrho ^2\phi ^2$ in $L^{r^\prime }(\mathbb {R}^2)$ follows from the $L^2$-integrability of $\phi$. Hence, statement (B) is shown, which concludes the proof of Theorem 2.

Remark 4

(Benjamin–Bona–Mahony KP equation) The decay result in Theorem 2 is equally valid for lump solutions of the fractional BBM-KP equation, which is when the term $\textrm{D}_x^\alpha u_x$ in (1.1) is replaced by $\textrm{D}^\alpha _xu_t$.

Remark 5

(Rotation modified KP equation) Lump solutions $u(t,x,y)=\phi (x-ct,y)$ of the rotation modified KP equation

$$\begin{aligned} (u_t + uu_x - \beta u_{xxx})_x+u_{yy}=\gamma u, \end{aligned}$$

where $\beta \in \mathbb {R}$ determines the type of dispersion and $\gamma >0$ is the Coriolis parameter due to the Earth’s rotation, exist for $\beta >0$ and $c<2\sqrt{\gamma \beta }$, cf. [7, Theorem 2.2, Remark 2.4]. They satisfy the convolution equation

$$\begin{aligned} \phi = -\frac{1}{2}K*\phi ^2, \qquad {\hat{K}}(\xi _1,\xi _2)=m(\xi _1,\xi _2), \end{aligned}$$

where

$$\begin{aligned}m(\xi _1,\xi _2)=\frac{\xi _1^2}{-c\xi _1^2+\beta \xi _1^4+\xi _2^2+\gamma }. \end{aligned}$$

Due to the Coriolis parameter $\gamma >0$, the symbol m is smooth at the origin, which allows lump solutions to decay exponentially at infinity (cf. [8, Theorem 1.6]). If the dispersive term $\beta u_{xxx}$ were replaced by the fractional term $-\beta |\textrm{D}_x|^\alpha u_x$, which would lead to an fractional rotation modified KP equation, we’d expect a decay of lump solutions, which depends on $\alpha$ (in a similar way as we see it for the fractional KdV equation [14, 20]).

Remark 6

(Full dispersion KP equation) The full dispersion KP equation is given by

$$\begin{aligned} u_t - l(\textrm{D})u_x + uu_x=0,\qquad l(\textrm{D})=(1+\beta |\textrm{D}|^2)^\frac{1}{2}\left( \frac{\tanh (|\textrm{D}|)}{|\textrm{D}|}\right) ^\frac{1}{2} \left( 1+\frac{\textrm{D}_y^2}{\textrm{D}_x^2}\right) ^\frac{1}{2}. \end{aligned}$$

Here $\beta >0$ is the surface tension coefficient. Existence of lump solutions is shown in [11, 12]. In the same way as for the fKP-I equation, the transverse direction induces a discontinuity at the origin of the symbol $m(\xi _1,\xi _2)= \frac{1}{c+l(\xi _1,\xi _2)}$. Therefore, the decay of lump solutions is also at most quadratic.

To visualize the decay rate of lumps for the fKP-I equation we consider the product of numerically generated lumps with $\varrho ^2(x,y)=x^2+y^2$. Figure 7 shows x and $y-$cross sections of this product for $\alpha =2$, $\alpha =1.7$, and $\alpha =1.35$. As the decay rate is quadratic, the result approaches a constant value for increasing |x| and |y|. We observe that the behavior is similar for all $\alpha$ values but the aforementioned constant becomes smaller for smaller values of $\alpha$.

Fig. 7

The $x-$cross section (solid line) and the $y-$cross section (dashed line) of the product of numerically generated lump solutions with $\varrho ^2(x,y)=(x^2+y^2)$ for $\alpha =2$ (top left panel), $\alpha =1.7$ (top right panel) and $\alpha =1.35$ (bottom panel)

4 Auxiliary results

Lemma 8

(Fractional integration by parts) Let $\alpha \ge 0$. Then,

$$\begin{aligned} \int _{\mathbb {R}} \phi \textrm{D}^\alpha _x \phi \,\textrm{d}x = \int _{\mathbb {R}^2}\left( \textrm{D}^{\frac{\alpha }{2}}\phi \right) ^2\,\textrm{d}x, \qquad \int _{\mathbb {R}}x \phi _x \textrm{D}^\alpha _x \phi \, dx=\frac{\alpha -1}{2}\int _{\mathbb {R}} \left( \textrm{D}^\frac{\alpha }{2}_x\phi \right) ^2\, \textrm{d}x \end{aligned}$$

and

$$\begin{aligned} \int _{\mathbb {R}} x^2\phi _{xx} \textrm{D}^\alpha _x \phi \,\textrm{d}x = -\int _{\mathbb {R}} x^2 \left( \textrm{D}^\frac{\alpha }{2}_x\phi _x\right) ^2\,\textrm{d}x + \tfrac{1}{4}(\alpha -2)^2 \int _{\mathbb {R}}\left( \textrm{D}^\frac{\alpha }{2}_x\phi \right) ^2\,\textrm{d}x. \end{aligned}$$

Proof

The first assertion follows immediately by

$$\begin{aligned} \int _{\mathbb {R}} \phi \textrm{D}^\alpha _x \phi \,\textrm{d}x = \langle \phi , \overline{\textrm{D}^\alpha _x \phi }\rangle = \langle \hat{\phi }, |\xi |^\alpha \overline{{\hat{\phi }} }\rangle =\langle |\xi |^\frac{\alpha }{2}{\hat{\phi }}, \overline{|\xi |^\frac{\alpha }{2} {\hat{\phi }} }\rangle = \int _{\mathbb {R}}\left( \textrm{D}^\frac{\alpha }{2}_x\phi \right) ^2\,\textrm{d}x. \end{aligned}$$

For the second statement, notice first that

$$\begin{aligned} \langle {\hat{\phi }} _\xi , \overline{|\xi |^\alpha \xi {\hat{\phi }}} \rangle&=- \langle {\hat{\phi }}, (\alpha +1) |\xi |^\alpha \overline{{\hat{\phi }} }+ |\xi |^\alpha \xi \overline{{\hat{\phi }}_\xi }\rangle \\&= -(\alpha +1)\int _{\mathbb {R}^2}\left( \textrm{D}^\frac{\alpha }{2}_x\phi \right) ^2\, \textrm{d}x - \langle {\hat{\phi }}_\xi , \overline{|\xi |^\alpha \xi \hat{\phi }}\rangle , \end{aligned}$$

therefore

$$\begin{aligned} \langle {\hat{\phi }} _\xi ,\overline{ |\xi |^\alpha \xi {\hat{\phi }}} \rangle =-\tfrac{\alpha +1}{2} \int _{\mathbb {R}}\left( \textrm{D}^\frac{\alpha }{2}_x\phi \right) ^2\, \textrm{d}x, \end{aligned}$$

(4.1)

which implies

$$\begin{aligned} \int _{\mathbb {R}} x \phi _x \textrm{D}^\alpha _x \phi \,\textrm{d}x&= -\langle (\xi {\hat{\phi }})_\xi , |\xi |^\alpha \overline{{\hat{\phi }}} \rangle \\&= - \langle {\hat{\phi }}, |\xi |^\alpha \overline{{\hat{\phi }}} \rangle - \langle {\hat{\phi }}_\xi , |\xi |^\alpha \xi \overline{{\hat{\phi }}} \rangle \\&=-\int _{\mathbb {R}} \left( \textrm{D}^\frac{\alpha }{2}_x\phi \right) ^2\, \textrm{d}x + \frac{\alpha +1}{2}\int _{\mathbb {R}}\left( \textrm{D}^\frac{\alpha }{2}_x\phi \right) ^2\, \textrm{d}x. \end{aligned}$$

Turning to the third statement, notice first that

$$\begin{aligned} \int _{\mathbb {R}}x^2 \left( \textrm{D}^{\frac{\alpha }{2}}_x\phi \right) ^2\,\textrm{d}x&=-\langle (|\xi |^\frac{\alpha }{2}\xi {\hat{\phi }})_{\xi \xi }, |\xi |^\frac{\alpha }{2}\xi \overline{{\hat{\phi }}}\rangle \nonumber \\&\quad = -\left( \tfrac{\alpha }{2}+1\right) \tfrac{\alpha }{2} \int _{\mathbb {R}}\left( \textrm{D}^\frac{\alpha }{2}_x\phi \right) ^2\, \textrm{d}x+ \langle {\hat{\phi }}_\xi , |\xi |^\alpha \xi ^2 \overline{{\hat{\phi }}_\xi } \rangle , \end{aligned}$$

(4.2)

where we used (4.1). Now,

$$\begin{aligned} \int _{\mathbb {R}} x^2\phi _{xx} \textrm{D}^\alpha _x \phi \,\textrm{d}x&= \langle (\xi ^2 {\hat{\phi }} )_{\xi \xi }, |\xi |^\alpha {\hat{\phi }} \rangle \\&= 2\langle {\hat{\phi }} , |\xi |^\alpha \overline{{\hat{\phi }}} \rangle + 4\langle {\hat{\phi }}_\xi , |\xi |^\alpha \xi \overline{{\hat{\phi }}} \rangle + \langle {\hat{\phi }}_{\xi \xi }, |\xi |^\alpha \xi ^2\overline{{\hat{\phi }}} \rangle \\&= 2\langle {\hat{\phi }} , |\xi |^\alpha \overline{{\hat{\phi }}} \rangle + (2-\alpha )\langle {\hat{\phi }}_\xi , |\xi |^\alpha \xi \overline{\hat{\phi }} \rangle - \langle {\hat{\phi }}_{\xi }, |\xi |^\alpha \xi ^2\overline{{\hat{\phi }}_\xi } \rangle \\&= -\int _{\mathbb {R}} x^2 \left( \textrm{D}^\frac{\alpha }{2}_x\phi _x\right) ^2\,\textrm{d}x + \tfrac{1}{4}(\alpha -2)^2 \int _{\mathbb {R}}\left( \textrm{D}^\frac{\alpha }{2}_x\phi \right) ^2\,\textrm{d}x, \end{aligned}$$

by (4.1) and (4.2). $\square$

Lemma 9

(Properties of F) Let $\alpha >0$, $G(\xi )=\textrm{i}x\xi - |y|\xi (1+\xi ^\alpha )^\frac{1}{2}$ for $\xi \ge 0$ and $y\ne 0$. The function

$$\begin{aligned} F(\xi )=\frac{\textrm{d}}{\textrm{d}\xi } \left( \frac{\xi }{(1+\xi ^\alpha )^{\frac{1}{2}}}\frac{1}{G^\prime (\xi )} \right) \frac{1}{G^\prime (\xi )}\qquad \text{ for }\quad \xi \ge 0 \end{aligned}$$

satisfies

• (a) $F(0)=\frac{1}{[G^\prime (0)]^2}=(\textrm{i}x-|y|)^{-2}$

• (b) $|F^\prime (\xi )|\lesssim T(\xi ) \frac{1}{x^2+y^2}$, for some function T such that $Te^{G}\in L^1(\mathbb {R}_+).$

Proof

Let us first summarize all needed derivatives for the function G:

$$\begin{aligned} G(\xi )&=\textrm{i}x\xi - |y|\xi (1+\xi ^\alpha )^\frac{1}{2}\\ G^\prime (\xi )&= \textrm{i}x -|y|\left( (1+\xi ^\alpha )^\frac{1}{2} + \frac{\alpha }{2}\xi ^\alpha (1+\xi ^\alpha )^{-\frac{1}{2}}\right) \\&= \frac{1}{2}(1+\xi ^\alpha )^{-\frac{1}{2}} \left( 2\textrm{i}x (1+\xi ^\alpha )^{\frac{1}{2}}- |y|\left( 2+ (2+\alpha )\xi ^\alpha \right) \right) \\ G^{\prime \prime }(\xi )&=-\frac{\alpha }{4}|y|(1+\xi ^\alpha ) ^{-\frac{3}{2}}\xi ^{\alpha -1}\left( 2(1+\alpha )+(2+\alpha )\xi ^\alpha \right) \\ G^{\prime \prime \prime }(\xi )&= \frac{\alpha }{8}|y|(1+\xi ^\alpha )^{-\frac{5}{2}}\xi ^{\alpha -2} \left( 4(1+\xi ^\alpha )^2-\alpha ^2(\xi ^{2\alpha }+2\xi ^\alpha +4)\right) \end{aligned}$$

Then, we compute F as

$$\begin{aligned} F(\xi )= \frac{1}{(1+\xi ^\alpha )^\frac{1}{2}[G^\prime (\xi )]^2} - \frac{\xi \left( \frac{\alpha }{2}(1+\xi ^\alpha )^{-\frac{1}{2}}\xi ^{\alpha -1}G^\prime (\xi ) +(1+\xi ^\alpha )^{\frac{1}{2}}G^{\prime \prime }(\xi ) \right) }{(1+\xi ^\alpha )\left[ G^\prime (\xi )\right] ^3}, \end{aligned}$$

which yields $F(0)=\frac{1}{[G^\prime (0)]^2}=(\textrm{i}x-|y|)^{-2}$ and proves part (a). A tedious, but straightforward computation yields that the derivative F is given by

$$\begin{aligned} F'(\xi )&=-\frac{\alpha }{2}\frac{(1+\alpha )\xi ^{\alpha -1} +(1-\frac{\alpha }{2})\xi ^{2\alpha -1}}{(1+\xi ^\alpha )^\frac{5}{2} G'(\xi )^2}\\&\quad +\frac{(\frac{3}{2}\alpha \xi ^\alpha -3(1+\xi ^\alpha )) G''(\xi )-(1+\xi ^\alpha )\xi G'''(\xi )}{(1+\xi ^\alpha )^ \frac{3}{2}G'(\xi )^3}+\frac{3\xi G''(\xi )^2}{(1+\xi ^\alpha ) ^\frac{1}{2}G'(\xi )^4} \\&=: T_1(\xi )+T_2(\xi )+T_3(\xi ). \end{aligned}$$

Now, we insert the expressions for $G^\prime , G^{\prime \prime },$ and $G^{\prime \prime \prime }$. We have

$$\begin{aligned} |G^\prime (\xi )|^2&=\frac{1}{4}(1+\xi ^\alpha )^{-1} \left( 4x^2(1+\xi ^\alpha )+y^2(2+(2+\alpha )\xi ^\alpha )^2\right) \gtrsim x^2+y^2,\\ |G^{\prime \prime }(\xi )|&\eqsim |y|(1+\xi ^\alpha )^{-\frac{1}{2}}\xi ^{\alpha -1},\\ |G^{\prime \prime \prime }(\xi )|&\eqsim |y|(1+\xi ^\alpha )^{-\frac{5}{2}}\xi ^{\alpha -2} \left| (4-\alpha ^2)(1+\xi ^\alpha )^2-3\alpha ^2\right| . \end{aligned}$$

Starting with $T_1$ we estimate

$$\begin{aligned} |T_1(\xi )| \lesssim \frac{\xi ^{\alpha -1}}{(1+\xi ^\alpha )^\frac{3}{2}(x^2+y^2)}. \end{aligned}$$

(4.3)

For $T_2$ we find

$$\begin{aligned} |T_2(\xi )|\lesssim |y|\frac{ \xi ^{\alpha -1}}{(1+\xi ^\alpha )(x^2+y^2)^\frac{3}{2}}. \end{aligned}$$

(4.4)

Eventually, we estimate $T_3$ as

$$\begin{aligned} |T_3(\xi )|\lesssim y^2 \frac{\xi ^{2\alpha -1}}{(1+\xi ^\alpha )^\frac{3}{2}(x^2+y^2)^2}. \end{aligned}$$

(4.5)

Summarizing (4.3)-(4.5), we find that $|F^\prime (\xi )|\lesssim T(\xi ) \frac{1}{x^2+y^2}$, where $Te^{G}\in L^1(\mathbb {R}_+)$ which proves part (b).

$\square$

Lemma 10

(Properties of E) Let $\alpha >0$, $G(\xi )=\textrm{i}x\xi - |y|\xi (1+\xi ^\alpha )^\frac{1}{2}$ for $\xi \ge 0$ and $y\ne 0$. The function

$$\begin{aligned} E(\xi )=\frac{1}{(1+\xi ^\alpha )}\frac{1}{G^\prime (\xi )}\qquad \text{ for }\quad \xi \ge 0 \end{aligned}$$

satisfies

• (a) $E(0)=\frac{1}{G^\prime (0)}=\textrm{i}x-|y|$

• (b) $|E^\prime (\xi )|\lesssim S(\xi ) \frac{1}{x^2+y^2}$, for some function S such that $Se^{G}\in L^1(\mathbb {R}_+)$.

Proof

The proof follows by direct computation. Recall that

$$\begin{aligned} |G^\prime (\xi )|^2 =\frac{1}{4}(1+\xi ^\alpha )^{-1}\left( 4x^2(1+\xi ^\alpha ) +y^2(2+(2+\alpha )\xi ^\alpha )^2\right) \gtrsim x^2+y^2 \qquad \end{aligned}$$

(4.6)

Part (a) follows immediately from $G^\prime (0)=\textrm{i}x-|y|$. For part (b), we compute the derivative of $E^\prime$ and use (4.6) to estimate

$$\begin{aligned} E^\prime (\xi )=\frac{\alpha }{4} \frac{\xi ^{\alpha -1}}{(1+\xi ^\alpha )^{2}[G^\prime (\xi )]^2}\lesssim \frac{\xi ^{\alpha -1}}{(1+\xi ^\alpha )^2}\frac{1}{x^2+y^2}, \end{aligned}$$

which yields the statement. $\square$