Classical solutions for the generalized Kadomtsev–Petviashvili I equations

Abstract

Purpose

This paper is devoted to the generalized Kadomtsev–Petviashvili I equation. This study aims to propose a new approach for investigation for the existence of at least one global classical solution and the existence of at least two nonnegative global classical solutions. The main arguments in this paper are based on some recent theoretical results.

Design/methodology/approach

This paper is devoted to the generalized Kadomtsev–Petviashvili I equation. This study aims to propose a new approach for investigation for the existence of at least one global classical solution and the existence of at least two nonnegative global classical solutions. The main arguments in this paper are based on some recent theoretical results.

Findings

This paper is devoted to the generalized Kadomtsev–Petviashvili I equation. This study aims to propose a new approach for investigation for the existence of at least one global classical solution and the existence of at least two nonnegative global classical solutions. The main arguments in this paper are based on some recent theoretical results.

Originality/value

This article is devoted to the generalized Kadomtsev–Petviashvili I equation. This study aims to propose a new approach for investigation for the existence of at least one global classical solution and the existence of at least two nonnegative global classical solutions. The main arguments in this paper are based on some recent theoretical results.

Citation

Georgiev, S., Boukarou, A., Bouhali, K. and Zennir, K. (2023), "Classical solutions for the generalized Kadomtsev–Petviashvili I equations", Arab Journal of Mathematical Sciences, Vol. ahead-of-print No. ahead-of-print. https://doi.org/10.1108/AJMS-08-2022-0195

1. Introduction

The Soviet physicists Boris Kadomtsev and Vladimir Petviashvili derived the equation that now bears their name, the Kadomtsev–Petviashvili equation (shortly the KP equation), as a model that describes the evolution of long ion-acoustic waves of small amplitude propagating in plasmas under the effect of long transverse perturbations. A particular case of this model is the Korteweg-de Vries (KdV) equation in the case of the absence of transverse dynamics. The KP equation is an extension of the classical KdV equation to two spatial dimensions, and it was used by Ablowitz and Segur for modeling of surface and internal water waves and for modeling in nonlinear optics, as well as in other physical settings.

The KP equation I can consider as a nonlinear partial differential equation in two spatial and one temporal coordinate. There are two distinct versions of the KP equation, which can be written in a normalized form in the following way:

ut+6uux+uxxxx+3σ2uyy=0,

where u = u (x, y, t) is a scalar function, x and y are the longitudinal and transverse spatial coordinates, subscripts x, y, t denote partial derivatives and σ² = ±1. When σ = 1, this equation is known as the KPII equation, and in this case, it models water waves with small surface tension. In the case when σ = −1, this equation is known as the KPI equation, and in this case, it models waves in thin films with high surface tension. In the references, the equation is often written with different coefficients in front of the various terms. Note that the particular values are inessential and they can be modified by appropriate rescaling of the dependent variables and of the independent variables.

This paper is devoted to the IVP for the generalized Kadomtsev–Petviashvilli I (gKP I) equation

(1.1)∂x∂tu+u∂xu+μ2∂xlu+ν∂yyu=0,t>0,(x,y)∈R2,u(0,x,y)=u0(x,y),(x,y)∈R2,

where

H1.

l∈N, l ≥ 5, u0∈Cl+1(R2), |u₀(x, y)| ≤ B, (x,y)∈R2, B > 0 is a given constant.

H2.

ν = ±1, μ > 0.

In the particular case, when l = 5, equation (1.1) is reduced to the fifth-order KP I equation and in the case l = 3, equation (1.1) is reduced to the KP equation.

In Ref. [1], when l = 3, the authors the local well-posedness for the Cauchy problem for the KP equation in certain Sobolev spaces. Generically, the solution of the KP equation develops a singularity in finite time t. It is discussed in Refs [2, 3] that this singularity develops at a point where the derivatives become divergent in all directions except one.

In Ref. [4], the authors established the local well-posedness of the Cauchy problem for the gKP I equation in anisotropic Sobolev spaces Hs1,s2(R) when s1>−α−14,s2≥0 and α ≥ 4, and global well-posedness in Hs1,0(R) when s1>−(α−1)(3α−4)4(5α+3) and 4 ≤ α ≤ 5, as well as when s1>−α(3α−4)4(5α+4) and α > 5.

Mechanical systems with impact, heartbeats, blood flows, population dynamics, industrial robotics, biotechnology, economics, etc. are real-world and applied science phenomena which are abruptly changed in their states at some time instants due to short time perturbations whose duration is negligible in comparison with the duration of these phenomena. A natural framework for mathematical simulation of such phenomena is differential equations when more factors are taken into account, please see Refs [5–8].

This paper aims to investigate the IVP (1.1) for the existence of at least one and at least two global classical solutions. In addition of (H1) and (H2), suppose

H3.

g∈C(0,∞×R2) is a positive function on 0,∞×R2 such that

g(0,x,y)=g(t,0,y)=g(t,x,0)=0,(t,x,y)∈0,∞×R2,

and

22l+4(l+1)!(1+t)2(∑r=0l+1|x|r)(∑r=0l+1|y|r)∫0t∫0x∫0yg(t1,x1,y1)dy1dx1dt1≤A,(t,x,y)∈0,∞×R2,

for some constant A > 0, and

H4.

ϵ ∈ (0, 1), A and B satisfy the inequalities ϵB₁(1 + A) < B and AB₁ < 1.

H5.

Let m > 0 be large enough and A, B, r, L, R₁ be positive constants that satisfy the following conditions

r<L<R1,ϵ>0,R1>25m+1L,AB1<L5.

In the last section, we will give an example of a function g and constants ϵ, A, B, B₁, r, L, R₁ and m that satisfy (H3)–(H5).

With X=C1(0,∞×Cl+1(R2)), we denote the space of all continuous functions on 0,∞×R2 so that u_t, ∂xru, ∂_txu, ∂_yu, ∂y2u, r = 1, …, l + 1, exist and are continuous on 0,∞×R2.

Our main result for existence of at least one global classical solution is as follows.

Theorem 1.1.

Suppose that (H1)–(H4) hold. Then the IVP (1.1) has at least one solution u ∈ X.

Next theorem is our result for the existence of at least two global nonnegative classical solutions.

Theorem 1.2.

Suppose that (H1)–(H3) and (H5) hold. Then the IVP (1.1) has at least two nonnegative solutions u₁, u₂ ∈ X.

The main idea for the proof of our main results is as follows. First, we find an integral representation of the solutions of the IVP (1.1). Then we construct a pair of operators so that any fixed point of their sum is a solution of the IVP (1.1). We find some a-priori estimates of the defined operators and using some fixed point theorems we conclude the existence of at least one global classical solution and the existence of at least two nonnegative classical solutions of the IVP (1.1).

The paper is organized as follows. In the next section, we give some auxiliary results. In Section 3, we give some preliminary results. In Section 4, we will prove our main results. In Section 5, we give an example to illustrate our main results.

2. Auxiliary results

In this section, as in Ref. [9], we will give some basic definitions and facts which will be used in this paper. Moreover, we will formulate the basic fixed-point theorems which we explore to prove our main results. For more details, we refer the reader to the papers [10–14] and references therein. To prove the existence of at least one global classical solution for the IVP (1.1), we will use the following fixed-point theorem.

Theorem 2.1.

([9, 12, 13]) Suppose that the constants ϵ and B are positive constants. Let E be a Banach space and define the set X = {x ∈ E: ‖x‖ ≤ B} and the operator Tx = −ϵx, x ∈ X. Assume that the operator S : X → E is a continuous operator and the set (I − S)(X) resides in a compact subset of E. Let also,

(2.1) {x∈E:x=λ(I−S)x,‖x‖=B}=∅

for any λ∈0,1ϵ. Then there exists x* ∈ X for which one has

Tx*+Sx*=0.

Below, assume that X is a real Banach space. Now, we will recall the definition of a completely continuous operator in a Banach space.

Definition 2.2.

[9] A map K : X → X is called a completely continuous map if it is continuous and it maps any bounded set into a relatively compact set.

For completeness, we will recall the definition of the Kuratowski measure of noncompactness, which will be used to be define l-set contraction mappings when l∈N0.

Definition 2.3.

[9] With Ω_X we will denote the class of all bounded sets of X. Then the Kuratowski measure of noncompactness α:ΩX→0,∞ is defined by

α(Y)=infδ>0:Y=⋃j=1mYjanddiam(Yj)≤δ,j∈{1,…,m}.

here, with diam (Y_j) = sup{‖x − y‖_X : x, y ∈ Y_j} we will denote the diameter of Y_j, j ∈ {1, …, m}.

For the main properties of the measure of noncompactness, we refer the reader to Ref. [10]. Now, we are ready to define an l-set contraction in a Banach space for any l∈N0.

Definition 2.4.

[9] A map K : X → X is called an l-set contraction if it is continuous, bounded and there exists a constant l ≥ 0 for which one has the following inequality

α(K(Y))≤lα(Y)

for any bounded set Y ⊂ X. The map K will be called a strict set contraction map if l < 1.

Note that any completely continuous mapping K : X → X is a 0-set contraction (see Ref. [11], p. 264). Next, for our main results, we have a need for a definition of an expansive operator.

Definition 2.5.

[9] Let X and Y be real Banach spaces. A map K : X → Y is called expansive if there exists a constant h > 1 for which one has the following inequality

‖Kx−Ky‖Y≥h‖x−y‖X

for any x, y ∈ X.

Now, we will recall the definition of a cone in a Banach space.

Definition 2.6.

[9] A closed, convex set P in X is said to be cone if

1. αx∈P for any α ≥ 0 and for any x∈P,

2. x,−x∈P implies x = 0.

Denote P*=P\{0}. The next result is a fixed-point theorem which we will use to prove the existence of at least two nonnegative global classical solutions of the IVP (1.1). For its proof, we refer the reader to the paper [14].

Theorem 2.7.

Let P be a cone of a Banach space E; Ω a subset of P and U₁, U₂ and U₃ three open bounded subsets of P such that U¯1⊂U¯2⊂U3 and 0 ∈ U₁. Assume that T:Ω→P is an expansive mapping with constant h > 1, S:U¯3→E is a k-set contraction with 0 ≤ k < h − 1 and S(U¯3)⊂(I−T)(Ω). Suppose that (U2\U¯1)∩Ω≠∅,(U3\U¯2)∩Ω≠∅, and there exists u0∈P* such that the following conditions hold:

1. Sx ≠ (I − T)(x − λu₀), for all λ > 0 and x ∈ ∂U₁ ∩ (Ω + λu₀),

2. There exists ϵ ≥ 0 such that Sx ≠ (I − T)(λx), for all λ ≥ 1 + ϵ,      x ∈ ∂U₂ and λx ∈ Ω,

3. Sx ≠ (I − T)(x − λu₀), for all λ > 0 and x ∈ ∂U₃ ∩ (Ω + λu₀).

Then T + S has at least two non-zero fixed points x1,x2∈P such that

x1∈∂U2∩Ω and x2∈(U¯3\U¯2)∩Ω

or

x1∈(U2\U1)∩Ω and x2∈(U¯3\U¯2)∩Ω.

3. Preliminary results

In this section, we will define suitable operators and we will deduct some a-priori estimates which we will use to prove our main results. Let X be endowed with the norm

‖u‖=maxsup(t,x,y)∈0,∞×R2|u(t,x,y)|,sup(t,x,y)∈0,∞×R2|∂tu(t,x,y)|,sup(t,x,y)∈0,∞×R2|∂xru(t,x,y)|,sup(t,x,y)∈0,∞×R2|∂yku(t,x,y)|,r=1,…,l+1,k=1,2,sup(t,x,y)∈0,∞×R2|∂txu(t,x,y)|,

provided it exists. For u ∈ X, define the operator

S1u(t,x,y)=u(t,x,y)−u0(x,y)+∫0t−∂tu(t1,x,y)+∂t∂xu(t1,x,y)+∂xu(t1,x,y)2+u(t1,x,y)∂xxu(t1,x,y)+μ2∂xl+1u(t1,x,y)+ν∂yyu(t1,x,y)dt1,(t,x,y)∈0,∞×R2.

In the next lemma, we will establish that any solution of an integral equation is a solution to the IVP (1.1).

Lemma 3.1.

Suppose that (H1) and (H2) hold. Let u ∈ X be a solution of the equation

(3.1) S1u(t,x,y)=0,(t,x,y)∈0,∞×R2.

Then it is a solution of the IVP (1.1).

Proof. Let u ∈ X be a solution of equation (3.1). Using the definition of S₁, we get the following integral equation

0=u(t,x,y)−u0(x,y)+∫0t−∂tu(t1,x,y)+∂t∂xu(t1,x,y)+∂xu(t1,x,y)2+u(t1,x,y)∂xxu(t1,x,y)+μ2∂xl+1u(t1,x,y)+ν∂yyu(t1,x,y)dt1,(t,x,y)∈0,∞×R2.

For the last integral equation, we differentiate one time with respect to t and we get the following integral equation

0=∂tu(t,x,y)−∂tu(t,x,y)+∂t∂xu(t,x,y)+∂xu(t,x,y)2+u(t,x,y)∂xxu(t,x,y)+μ2∂xl+1u(t,x,y)+ν∂yyu(t,x,y),(t,x,y)∈0,∞×R2.

Therefore

0=∂t∂xu(t,x,y)+∂xu(t,x,y)2+u(t,x,y)∂xxu(t,x,y)+μ2∂xl+1u(t,x,y)+ν∂yyu(t,x,y),(t,x,y)∈0,∞×R2.

Thus, u satisfies the first equation of (1.1). Now, we put t = 0 and we arrive at the equality

u(0,x,y)=u0(x,y),(x,y)∈R2.

From here, we conclude that u satisfies the second equation of (1.1). Consequently, u is a solution to the IVP (1.1). This completes the proof.□

Now, we will give an a-priori estimate of the operator S₁. For this aim, we define the constant

B1=2+|ν|+μ2B+2B2.

Lemma 3.2.

Suppose that (H1) and (H2) hold. Let u ∈ X be such that ‖u‖ ≤ b, for some constant b > 1. Then one has

|S1u(t,x,y)|≤B1(1+t),(t,x,y)∈0,∞×R2.

Proof. By the definition of the operator S₁, one gets

|S1u(t,x)|=u(t,x,y)−u0(x,y)+∫0t−∂tu(t1,x,y)+∂t∂xu(t1,x,y)+∂xu(t1,x,y)2+u(t1,x,y)∂xxu(t1,x,y)+μ2∂xl+1u(t1,x,y)+ν∂yyu(t1,x,y)dt1≤|u(t,x,y)|+|u0(x,y)|+∫0t|∂tu(t1,x,y)|+|∂t∂xu(t1,x,y)|+∂xu(t1,x,y)2+|u(t1,x,y)‖∂xxu(t1,x,y)|+μ2|∂xl+1u(t1,x,y)|+|ν||∂yyu(t1,x,y)|dt1≤2B+∫0tB+B+B2+B2+μ2B+|ν|Bdt1=2B+2+|ν|+μ2B+2B2t≤B1(1+t),(t,x,y)∈0,∞×R2.

This completes the proof.□

For u ∈ X, we define the operator

S2u(t,x,y)=∫0t∫0x∫0y(t−t1)(x−x1)l+1(y−y1)l+1g(t1,x1)S1u(t1,x1,y1)dy1dx1dt1,

(t,x,y)∈0,∞×R2. In the next lemma, we will give an estimate of the norm of the operator S₂.

Lemma 3.3.

Suppose that (H1)–(H3) hold. For u ∈ X, ‖u‖ ≤ B, one has the following estimate

‖S2u‖≤AB1.

Proof. We will use the inequality (v + w)^q ≤ 2^q(v^q + w^q), q > 0, v, w > 0, to find estimates for S₂u and its derivatives. Then, we will deduct the desired estimate for the norm of S₂u. We have

|S2u(t,x,y)|=∫0t∫0x∫0y(t−t1)(x−x1)l+1(y−y1)l+1g(t1,x1)S1u(t1,x1,y1)dy1dx1dt1≤∫0t∫0x∫0y(t−t1)|x−x1|l+1|y−y1|l+1g(t1,x1)|S1u(t1,x1,y1)|dy1dx1dt1≤B1(1+t)∫0t∫0x∫0y(t−t1)|x−x1|l+1|y−y1|l+1g(t1,x1)dy1dx1dt1≤22l+4B1(1+t)2|x|l+1|y|l+1∫0t∫0x∫0yg(t1,x1,y1)dy1dx1dt1≤(l+1)!22l+4B1(1+t)2∑r=0l+1|x|r∑r=0l+1|y|r×∫0t∫0x∫0yg(t1,x1,y1)dy1dx1dt1≤AB1,(t,x,y)∈0,∞×R2.

Now, we will estimate the first derivative with respect to t of S₂u. For it, one has

|∂tS2u(t,x,y)|=∫0t∫0x∫0y(x−x1)l+1(y−y1)l+1g(t1,x1)S1u(t1,x1,y1)dy1dx1dt1≤∫0t∫0x∫0y|x−x1|l+1|y−y1|l+1g(t1,x1)|S1u(t1,x1,y1)|dy1dx1dt1≤B1∫0t∫0x∫0y|x−x1|l+1|y−y1|l+1g(t1,x1)dy1dx1dt1≤22l+4B1(1+t)2|x|2l+1|y|2l+1∫0t∫0x∫0yg(t1,x1,y1)dy1dx1dt1≤(l+1)!22l+4B1(1+t)2∑r=0l+1|x|r∑r=0l+1|y|r×∫0t∫0x∫0yg(t1,x1,y1)dy1dx1dt1≤AB1,(t,x,y)∈0,∞×R2.

For the derivatives of S₂u with respect to x, one deduct

|∂xrS2u(t,x,y)|=(l+1)…(l−r+2)×∫0t∫0x∫0y(t−t1)(x−x1)l−r+1(y−y1)l+1g(t1,x1)S1u(t1,x1,y1)dy1dx1dt1≤(l+1)…(l−r+2)×∫0t∫0x∫0y(t−t1)|x−x1|l−r+1|y−y1|l+1g(t1,x1)|S1u(t1,x1,y1)|dy1dx1dt1≤(l+1)!B1(1+t)∫0t∫0x∫0y(t−t1)|x−x1|l−r+1|y−y1|l+1g(t1,x1)dy1dx1dt1≤(l+1)!22l−r+4B1(1+t)2|x|l−r+1|y|l+1∫0t∫0x∫0yg(t1,x1,y1)dy1dx1dt1≤(l+1)!22l+4B1(1+t)2∑r=0l+1|x|r∑r=0l+1|y|r×∫0t∫0x∫0yg(t1,x1,y1)dy1dx1dt1≤AB1,(t,x,y)∈0,∞×R2,r=1,…,l+1.

As above, one can get the following estimates

|∂ykS2u(t,x,y)|≤AB1,(t,x,y)∈0,∞×R2,k=1,2.

Note that for the mixed derivative ∂_txS₂u, one has

|∂txS2u(t,x,y)|=(l+1)∫0t∫0x∫0y(x−x1)l(y−y1)l+1g(t1,x1)S1u(t1,x1,y1)dy1dx1dt1≤(l+1)∫0t∫0x∫0y|x−x1|l|y−y1|l+1g(t1,x1)|S1u(t1,x1,y1)|dy1dx1dt1≤(l+1)B1(1+t)∫0t∫0x∫0y|x−x1|l|y−y1|l+1g(t1,x1)dy1dx1dt1≤(l+1)!B122l+3B1(1+t)2|x|lyl+1∫0t∫0x∫0yg(t1,x1,y1)dy1dx1dt1≤(l+1)!22l+4B1(1+t)2∑r=0l+1|x|r∑r=0l+1|y|r×∫0t∫0x∫0yg(t1,x1,y1)dy1dx1dt1≤AB1,(t,x,y)∈0,∞×R2.

Thus,

‖S2u‖≤AB1.

This completes the proof.□

In the next result, we will give other integral equations whose solutions are solutions to the IVP (1.1).

Lemma 3.4.

Suppose (H1), (H2) and let g∈C(0,∞×R) be a positive function almost everywhere on 0,∞×R2. If u ∈ X satisfies the equation

(3.2) S2u(t,x,y)=0,(t,x,y)∈0,∞×R2,

then u is a solution to the IVP (1.1).

Proof. We differentiate two times with respect to t, five times with respect to x and five times with respect to y equation (3.2) and we find

g(t,x,y)S1u(t,x,y)=0,(t,x,y)∈0,∞×(R2\{{x=0}∪{y=0}}),

whereupon

S1u(t,x,y)=0,(t,x,y)∈0,∞×(R2\{{x=0}∪{y=0}}).

since S₁u(⋅, ⋅, ⋅) is a continuous function on 0,∞×R2, we get

0=limt→0S1u(t,0,0)=limx→0S1u(0,x,0)=limy→0S1u(0,0,y)=limt,x→0S1u(t,x,0)=limt,y→0S1u(t,0,y)=limx,y→0S1u(0,x,y)=limt,x,y→0S1u(t,x,y).

thus,

S1u(t,x,y)=0,(t,x,y)∈0,∞×R2.

hence and Lemma 3.1, we conclude that u is a solution to the IVP (1.1). This completes the proof.□

4. Proof of the main results

4.1 Proof of Theorem 1.1

Let Ỹ̃̃ denote the set of all equi-continuous families in X with respect to the norm ‖ ⋅‖. Let also, Ỹ̃=Ỹ̃̃¯ be the closure of Ỹ̃̃, Ỹ=Ỹ̃∪{u0},

Y={u∈Ỹ:‖u‖≤B}.

Note that Y is a compact set in X. For u ∈ X, define the operators

Tu(t,x,y)=−ϵu(t,x,y),Su(t,x,y)=u(t,x,y)+ϵu(t,x,y)+ϵS2u(t,x,y),(t,x,y)∈0,∞×R2.

For u ∈ Y, using Lemma 3.3, we have

‖(I−S)u‖=‖ϵu−ϵS2u‖≤ϵ‖u‖+ϵ‖S2u‖≤ϵB1+ϵAB1=ϵB1(1+A)<B

Thus, S : Y → E is continuous and (I − S)(Y) resides in a compact subset of E. Now, suppose that there is a u ∈ E so that ‖u‖ = B and

u=λ(I−S)u,

or

1λu=(I−S)u=−ϵu−ϵS2u,

or

1λ+ϵu=−ϵS2u,

for some λ∈0,1ϵ. Hence, ‖S₂u‖ ≤ AB₁ < B,

ϵB<1λ+ϵB=1λ+ϵ‖u‖=ϵ‖S2u‖<ϵB,

which is a contradiction. Hence Theorem 2.1 follows that the operator T + S has a fixed point u* ∈ Y. Therefore,

u*(t,x,y)=Tu*(t,x,y)+Su*(t,x,y)=−ϵu*(t,x,y)+u*(t,x,y)+ϵu*(t,x,y)+ϵS2u*(t,x,y),(t,x,y)∈0,∞×R2,

where

0=S2u*(t,x,y),(t,x,y)∈0,∞×R2.

from here and from Lemma 3.4, it follows that u is a solution to the IVP (1.1). This completes the proof.

4.2 Proof of Theorem 1.2

Let X be the space used in the previous section. Let also,

P̃={u∈X:u≥0on0,∞×R2}.

with P we will denote the set of all equi-continuous families in P̃. For v ∈ X, define the operators

T1v(t,x,y)=(1+mϵ)v(t,x,y)−ϵL10,S3v(t,x,y)=−ϵS2v(t,x,y)−mϵv(t,x,y)−ϵL10,

t∈0,∞, (x,y)∈R2. Note that any fixed point v ∈ X of the operator T₁ + S₃ is a solution to the IVP (1.1). Define

U1=Pr={v∈P:‖v‖<r},U2=PL={v∈P:‖v‖<L},U3=PR1={v∈P:‖v‖<R1},R2=R1+AmB1+L5m,Ω=PR2¯={v∈P:‖v‖≤R2}.

1. Let v₁, v₂ ∈ Ω. Then, we get

‖T1v1−T1v2‖=(1+mε)‖v1−v2‖.

from the last equality, we conclude that the operator T₁ : Ω → X is an expansive operator with a constant h = 1 + mɛ > 1.

1. Take v∈P¯R1 arbitrarily. Then

‖S3v‖≤ε‖S2v‖+mε‖v‖+εL10≤εAB1+mR1+L10.

from the last inequality, we conclude that the set S3(P¯R1) is uniformly bounded. Because the operator S3:P¯R1→X is a continuous operator, we get that S3(P¯R1) is equi-continuous. Therefore, the operator S3:P¯R1→X is a 0-set contraction.

1. Take v1∈P¯R1 arbitrarily. Set

v2=v1+1mS2v1+L5m.

note that S2v1+L5≥0 on 0,∞×R2. Therefore, v₂ ≥ 0 on 0,∞×R2 and we have the following estimate

‖v2‖≤‖v1‖+1m‖S2v1‖+L5m≤R1+AmB1+L5m=R2.

consequently v₂ ∈ Ω. Moreover,

−εmv2=−εmv1−εS2v1−εL10−εL10

or

(I−T1)v2=−εmv2+εL10=S3v1.

Therefore, S3(P¯R1)⊂(I−T1)(Ω).

1. Suppose that for any v0∈P*, there exist λ > 0 and z∈∂Pr∩(Ω+λv0) or z∈∂PR1∩(Ω+λv0) such that

S3z=(I−T1)(z−λv0).

hence,

−ϵS2z−mϵz−ϵL10=−mϵ(z−λv0)+ϵL10

or

−S2z=λmv0+L5.

from the last equation, we arrive at

‖S2z‖=λmv0+L5>L5.

this is a contradiction.

1. Assume that for any ϵ₁ ≥ 0 small enough there exist a x1∈∂PL and λ₁ ≥ 1 + ϵ₁ such that λ1x1∈P¯R1 and

(4.1)S3x1=(I−T1)(λ1x1).

When ϵ1>25m, one has x1∈∂PL, λ1x1∈P¯R1, λ₁ ≥ 1 + ϵ₁ and (4.1) holds. Since x1∈∂PL and λ1x1∈P¯R1, it follows that

25m+1L<λ1L=λ1‖x1‖≤R1.

in addition,

−ϵS2x1−mϵx1−ϵL10=−λ1mϵx1+ϵL10,

or

S2x1+L5=(λ1−1)mx1.

hence,

2L5≥S2x1+L5=(λ1−1)m‖x1‖=(λ1−1)mL,

and

25m+1≥λ1,

which is a contradiction.

Therefore, all conditions of Theorem 2.7 hold and the IVP (1.1) has at least two solutions u₁ and u₂ so that

‖u1‖=L<‖u2‖<R1,

or

r<‖u1‖<L<‖u2‖<R1.

5. An example

Below, we will illustrate our main results. Let B = μ = ν = 1 and

R1=10,L=5,r=4,m=1050,A=15B1,ϵ=15B1(1+A).

let also,

u0(x,y)=11+x2+y2,(x,y)∈R2.

then

B1=2+2+2=6,

and

AB1=15<B,ϵB1(1+A)<1,

i.e. (H4) holds. Next,

r<L<R1,ϵ>0,R1>25m+1L,AB1<L5.

i.e. (H5) holds. Take

h(s)=log1+sl+12+s2l+21−sl+12+s2l+2,l(s)=arctansl+121−s2l+2,s∈R,s≠±1.

then

h′(s)=22(l+1)sl(1−s2l+2)(1−sl+12+s2l+2)(1−sl+12+s2l+2),l′(s)=(l+1)2sl(1+s2l+2)1+s4l+4,s∈R,s≠±1.

therefore

lims→±∞∑r=0l+1srh(s)=lims→±∞h(s)1∑r=0l+1sr=lims→±∞h′(s)−∑r=0l(r+1)sr∑r=0l+1sr2=−lims→±∞22(l+1)sl(1−s2l+2)∑r=0l+1sr2∑r=0l(r+1)sr(1−sl+12+s2l+2)(1−sl+12+s2l+2)≠±∞

and

lims→±∞∑r=0l+1srl(s)=lims→±∞l(s)1∑r=0l+1sr=lims→±∞l′(s)−∑r=0l(r+1)sr∑r=0l+1sr2=−lims→±∞(l+1)2sl(1+s2l+2)∑r=0l+1sr21+s4l+4∑r=0l(r+1)sr≠±∞.

consequently

−∞<lims→±∞∑r=0l+1srh(s)<∞,−∞<lims→±∞∑r=0l+1srl(s)<∞.

hence, there exists a positive constant C₁ so that

∑r=0l+1|s|r1(4l+4)2log1+sl+12+s2l+21−sl+12+s2l+2+1(2l+2)2arctansl+121−s2l+2≤C1,

s∈R. Note that lims→±1l(s)=π2 and by Ref. [15] (p. 707, Integral 79), we have

∫dz1+z4=142log1+z2+z21−z2+z2+122arctanz21−z2.

let

Q(s)=sl(1+s4l+4),s∈R,

and

g1(t,x,y)=Q(t)Q(x)Q(y),t∈0,∞,(x,y)∈R2.

then there exists a constant C > 0 such that

22l+4(l+1)!(1+t)2∑r=0l+1|x|r∑r=0l+1|y|r×∫0t∫0x∫0yg1(t1,x1,y1)dy1dx1dt1≤C,(t,x,y)∈[0,∞)×R2.

let

g(t,x,y)=ACg1(t,x,y),(t,x,y)∈0,∞×R3.

then

22l+4(l+1)!(1+t)2∑r=0l+1|x|r∑r=0l+1|y|r×∫0t∫0x∫0yg(t1,x1,y1)dy1dx1dt1≤A,(t,x,y)∈[0,∞)×R2.

i.e. (H3), holds. Therefore, for the IVP

∂x∂tu+u∂xu+μ2∂xlu+ν∂yyu=0,t>0,(x,y)∈R2,u(0,x,y)=11+x2+y2,(x,y)∈R2,

are fulfilled all conditions of Theorem 1.1 and Theorem 1.2.