Classical Solutions for the Generalized Kawahara-KdV System

1. Introduction

In this paper, we investigate the Cauchy problem for the generalized Kawahara-KdV system

$$\begin{array}{ccc}{\displaystyle {\partial }_{t}u+\sum _{k=0}^{N_1}\sum _{l=0}^{N_1-k}{\partial }_x\left\{\sum _{m=0}^{N_1-k}{\partial }_x^m{u}^p{P}_{k,l,m}\left({\partial }_x^{l}v\right)\right\}+\sum _{k=1}^{N_2}{a}_k(t,x){\partial }_x^{2k+1}u}\hfill & =\hfill & 0\hfill \\ \\ {\displaystyle {\partial }_{t}v+\sum _{k=0}^{N_3}\sum _{l=0}^{N_3-k}{\partial }_x^k\left\{\sum _{m=0}^{N_3-k}{\partial }_x^m{v}^p{Q}_{k,l,m}\left({\partial }_x^{l}u\right)\right\}+\sum _{k=1}^{N_4}{b}_k(t,x){\partial }_x^{2k+1}v}\hfill & =\hfill & 0,\hfill \\ \\ (t,x)\in [0,\infty )\times \mathbb{R},u(0,x)=u_0\left(x\right),v(0,x)=v_0\left(x\right),x\in \mathbb{R},\hfill & \hfill \end{array}$$

(1)

where

 (Hyp1) 

$u_0,v_0\in {\mathcal{C}}^q\left(\mathbb{R}\right)$, $0\le u_0,v_0\le \mathcal{B}$ on $\mathbb{R}$ for some positive constant $\mathcal{B}>1$, $a_j$, $b_k\in \mathcal{C}([0,\infty )\times \mathbb{R})$, $0\le |a_j|,|b_k|\le \mathcal{B}$ on $[0,\infty )\times \mathbb{R}$, $j=1,\dots ,N_2$, $k=1,\dots ,N_4$,

$$\begin{array}{ccc}\hfill P_{k,l,m}\left(z\right)& =& \sum _{r=0}^{N_5}{c}_{k,l,m,r}(t,x)z^r,\hfill \\ \\ \hfill Q_{k,l,m}\left(z\right)& =& \sum _{r=0}^{N_6}{d}_{k,l,m,r}(t,x)z^r,(t,x)\in [0,\infty )\times \mathbb{R},z\in \mathbb{R},\hfill \end{array}$$

$c_{k,l,m,j},d_{k,l,m,r}\in \mathcal{C}([0,\infty ),{\mathcal{C}}^q\left(\mathbb{R}\right))$,

$$0\le \left|{\partial }_x^{p_1}{c}_{k,l,m,j}\right|,\left|{\partial }_x^{p_1}{d}_{k,l,m,r}\right|\le \mathcal{B},$$

on $[0,\infty )\times \mathbb{R}$, $j=1,\dots ,N_5$, $r=1,\dots ,N_6$, $p_1=1,\dots ,N_1$, p, $N_1,N_2,N_3,N_4,N_5,N_6\in \mathbb{N}$,

$$q=max\{N_1,2N_2+1,N_3,2N_4+1\}.$$

Kondo and Pes [1] proved this system is local well-posedness in analytic Gevrey spaces $G^{\sigma ,s}\left(\mathbb{R}\right)$ with $s\ge 2N+1/2$ where $N=max\{N_2,N_4\}$.

The range of equations that this model encompasses is obviously broad and can represent many physical situations. As examples, we consider the nonlinear term

$$\sum _{k=0}^{N_1}\sum _{\ell =0}^{N_1-k}{\partial }_x^k\left\{\sum _{m=0}^{N_1-k}{\partial }_x^m{u}^p{P}_{k,\ell ,m}\left({\partial }_x^{\ell }v\right)\right\}.$$

If $N_1=1$, then we have that this term is equal to

$$\begin{array}{cc}\hfill & u^p{P}_{0,0,0}\left(v\right)+{\partial }_x{u}^p{P}_{0,0,1}\left(v\right)+u^p{P}_{0,1,0}\left({\partial }_{x}v\right)+{\partial }_x{u}^p{P}_{0,1,1}\left({\partial }_{x}v\right)\hfill \\ \hfill & +{\partial }_x\left[u^p{P}_{1,0,0}\left(v\right)+{\partial }_x{u}^p{P}_{1,0,1}\left(v\right)\right].\hfill \end{array}$$

(2)

When $N_1=N_2=1$ and $p=1$, taking $P_{0,0,0}=P_{0,0,1}=P_{0,1,0}=P_{0,1,1}=$ $P_{1,0,1}\equiv 0,P_{1,0,0}\left(x\right)=x^2,a_1=1$ and doing the same choices to the polynomial $Q_{k,\ell ,m}$ with $N_3=N_4=1$ and $b_1=1$, we have a couple system of modified Korteweg-de Vries equations (see [2,3])

$$\left\{\begin{array}{cc}{\partial }_{t}u+{\partial }_x^{3}u+{\partial }_x\left(u{v}^2\right)=0,\hfill & \\ {\partial }_{t}v+{\partial }_x^{3}v+{\partial }_x\left(v{u}^2\right)=0,\hfill & \\ u(x,0)=u_0\left(x\right),v(x,0)=v_0\left(x\right).\hfill \end{array}\right.$$

Considering in (2) the case when $p=q\in \mathbb{N}$ and $P_{1,0,0}\left(x\right)=x^{q+1}$, also the remaining null polynomials, since that $N_1=N_2=N_3=N_4=1$, we obtain a more general system, studied in [4], as follows

$$\left\{\begin{array}{cc}{\partial }_{t}u+{\partial }_x^{3}u+{\partial }_x\left(u^q{v}^{q+1}\right)=0,\hfill & \\ {\partial }_{t}v+{\partial }_x^{3}v+{\partial }_x\left(v^q{u}^{q+1}\right)=0,\hfill & \\ u(x,0)=u_0\left(x\right),v(x,0)=v_0\left(x\right).\hfill \end{array}\right.$$

On the other hand, to find systems more complicated, we can consider $N_1=2$ and $p=1$, then in this case the term nonlinear is more general, as we can see bellow

$$\begin{array}{cc}\hfill u& P_{0,0,0}\left(v\right)+{\partial }_{x}u{P}_{0,0,1}\left(v\right)+{\partial }_x^{2}u{P}_{0,0,2}\left(v\right)\hfill \\ & +u{P}_{0,1,0}\left({\partial }_{x}v\right)+{\partial }_{x}u{P}_{0,1,1}\left({\partial }_{x}v\right)+{\partial }_x^{2}u{P}_{0,1,2}\left({\partial }_{x}v\right)\hfill \\ & +u{P}_{0,2,0}\left({\partial }_x^{2}v\right)+{\partial }_{x}u{P}_{0,2,1}\left({\partial }_x^{2}v\right)+{\partial }_x^{2}u{P}_{0,2,2}\left({\partial }_x^{2}v\right)\hfill \\ & +{\partial }_x\left[u{P}_{1,0,0}\left(v\right)+{\partial }_{x}u{P}_{1,0,1}\left(v\right)+{\partial }_x^{2}u{P}_{1,0,2}\left(v\right)\right.\hfill \\ & \left.+u{P}_{1,1,0}\left({\partial }_{x}v\right)+{\partial }_{x}u{P}_{1,1,1}\left({\partial }_{x}v\right)+{\partial }_x^{2}u{P}_{1,1,2}\left({\partial }_{x}v\right)\right]\hfill \\ & +{\partial }_x^2\left[u{P}_{2,0,0}\left(v\right)+{\partial }_{x}u{P}_{2,0,1}\left(v\right)+{\partial }_x^{2}u{P}_{2,0,2}\left(v\right)\right].\hfill \end{array}$$

Observe that, changing v by u, and considering again all identical null polynomials, except $P_{0,0,1}\left(x\right)=x^k$, we obtain the Kawahara equation [5]

$${\partial }_{t}u+{\partial }_x^{3}u+{\partial }_x^{5}u+u^k{\partial }_{x}u=0.$$

The aim of this paper is to investigate the IVP (1) for existence of global classical solutions.

Theorem 1. 

Suppose $\left(\right(Hyp1)$. Then the IVP (1) has at least one solution

$$(u,v)\in {\left({\mathcal{C}}^1([0,\infty ),{\mathcal{C}}^q\left(\mathbb{R}\right))\right)}^2.$$

Theorem 2. 

Suppose $\left(\right(Hyp1)$. Then the IVP (1) has at least two nonnegative solutions

$$(u_1,v_1),(u_2,v_2)\in {\left({\mathcal{C}}^1([0,\infty ),{\mathcal{C}}^q\left(\mathbb{R}\right))\right)}^2.$$

The paper is organized as follows. In the next section, we give some auxiliary results. In Section 3, we prove Theorem 1. In Section 4, we prove Theorem 2. In Section 5, we give an example to illustrate our main results.

2. Preliminary Results

To prove our existence result we will use the following fixed point theorem.

Theorem 3. 

Let $ϵ>0$, $\mathcal{B}>0$, $\mathcal{E}$ be a Banach space and $\mathcal{X}=\{x\in \mathcal{E}:\parallel x\parallel \le \mathcal{B}\}$. Let also, $\mathcal{T}x=-ϵx$, $x\in \mathcal{X}$, $S:\mathcal{X}\to \mathcal{E}$ is a continuous, $(I-S)\left(\mathcal{X}\right)$ resides in a compact subset of $\mathcal{E}$ and

$$\{x\in \mathcal{E}:x=\lambda (I-S)x,\parallel x\parallel =\mathcal{B}\}=\varnothing ,\forall \lambda \in \left(0,\frac{1}{ϵ}\right).$$

(3)

Then there exists $x^{*}\in \mathcal{X}$ so that

$$\mathcal{T}{x}^{*}+S{x}^{*}=0.$$

Proof. 

Define

$$r\left(-\frac{1}{ϵ}x\right)=\left\{\begin{array}{c}-\frac{1}{ϵ}x\mathrm{if}\parallel x\parallel \le \mathcal{B}ϵ\hfill \\ \\ \frac{\mathcal{B}x}{\parallel x\parallel }\mathrm{if}\parallel x\parallel >\mathcal{B}ϵ.\hfill \end{array}\right.$$

Then $r\left(-\frac{1}{ϵ}(I-S)\right):\mathcal{X}\to \mathcal{X}$ is continuous and compact. Hence and the Schauder fixed point theorem, it follows that there exists $x^{*}\in \mathcal{X}$ so that

$$r\left(-\frac{1}{ϵ}(I-S)x^{*}\right)=x^{*}.$$

Assume that $-\frac{1}{ϵ}(I-S)x^{*}\notin \mathcal{X}$. Then

$$\parallel (I-S)x^{*}\parallel >\mathcal{B}ϵ,\frac{\mathcal{B}}{\parallel (I-S)x^{*}\parallel }<\frac{1}{ϵ},$$

and

$$x^{*}=\frac{\mathcal{B}}{\parallel (I-S)x^{*}\parallel }(I-S)x^{*}=r\left(-\frac{1}{ϵ}(I-S)x^{*}\right),$$

and hence, $\parallel x^{*}\parallel =\mathcal{B}$. This contradicts with (3). Therefore $-\frac{1}{ϵ}(I-S)x^{*}\in \mathcal{X}$ and

$$x^{*}=r\left(-\frac{1}{ϵ}(I-S)x^{*}\right)=-\frac{1}{ϵ}(I-S)x^{*},$$

or

$$-ϵx^{*}+S{x}^{*}=x^{*},$$

or

$$\mathcal{T}{x}^{*}+S{x}^{*}=x^{*}.$$

This completes the proof. □

Let $\mathcal{X}$ be a real Banach space.

Definition 1. 

A mapping $\mathcal{K}:\mathcal{X}\to \mathcal{X}$ is said to be completely continuous if it is continuous and maps bounded sets into relatively compact sets.

The concept for l-set contraction is related to that of the Kuratowski measure of noncompactness which we recall for completeness.

Definition 2. 

Let ${\Omega }_{\mathcal{X}}$ be the class of all bounded sets of $\mathcal{X}$. The Kuratowski measure of noncompactness $\alpha :{\Omega }_{\mathcal{X}}\to [0,\infty )$ is defined by

$$\alpha \left(\mathcal{Y}\right)=inf\left\{\delta >0:\mathcal{Y}=\bigcup _{j=1}^m{\mathcal{Y}}_j\mathrm{and}\mathrm{diam}\left({\mathcal{Y}}_j\right)\le \delta ,j=1,\dots ,m\right\},$$

where $\mathrm{diam}\left({\mathcal{Y}}_j\right)={sup\{\parallel x-y\parallel }_{\mathcal{X}}:x,y\in {\mathcal{Y}}_j\}$ is the diameter of ${\mathcal{Y}}_j$, $j=1,\dots ,m$.

For the main properties of measure of noncompactness we refer the reader to [6].

Definition 3. 

A mapping $\mathcal{K}:\mathcal{X}\to \mathcal{X}$ is said to be l-set contraction if it is continuous, bounded and there exists a constant $l\ge 0$ such that

$$\alpha \left(\mathcal{K}\right(\mathcal{Y}\left)\right)\le l\alpha \left(\mathcal{Y}\right),$$

for any bounded set $\mathcal{Y}\subset \mathcal{X}$. The mapping $\mathcal{K}$ is said to be a strict set contraction if $l<1$.

Obviously, if $\mathcal{K}:\mathcal{X}\to \mathcal{X}$ is a completely continuous mapping, then $\mathcal{K}$ is 0-set contraction (see [7], pp. 264).

Definition 4. 

Let $\mathcal{X}$ and $\mathcal{Y}$ be real Banach spaces. A mapping $\mathcal{K}:\mathcal{X}\to \mathcal{Y}$ is said to be expansive if there exists a constant $h>1$ such that

$${\parallel \mathcal{K}x-\mathcal{K}y\parallel }_{\mathcal{Y}}\ge h{\parallel x-y\parallel }_{\mathcal{X}},$$

for any $x,y\in \mathcal{X}$.

Definition 5. 

A closed, convex set $\mathcal{P}$ in $\mathcal{X}$ is said to be cone if

1.

$\alpha x\in \mathcal{P}$ for any $\alpha \ge 0$ and for any $x\in \mathcal{P}$,

2.

$x,-x\in \mathcal{P}$ implies $x=0$.

Denote ${\mathcal{P}}^{*}=\mathcal{P}\setminus \left\{0\right\}$.

Lemma 4. 

Let $\mathcal{X}$ be a closed convex subset of a Banach space $\mathcal{E}$ and $\mathcal{U}\subset \mathcal{X}$ a bounded open subset with $0\in \mathcal{U}.$ Assume there exists $\epsilon >0$ small enough and that $\mathcal{K}:\overline{\mathcal{U}}\to \mathcal{X}$ is a strict k-set contraction that satisfies the boundary condition:

$$\mathcal{K}x\notin \{x,\lambda x\}forallx\in \partial \mathcal{U}and\lambda \ge 1+\epsilon .$$

Then the fixed point index $i(\mathcal{K},\mathcal{U},\mathcal{X})=1.$

Proof. 

Consider the homotopic deformation $\mathcal{H}:[0,1]\times \overline{\mathcal{U}}\to \mathcal{X}$ defined by

$$\mathcal{H}(t,x)=\frac{1}{\epsilon +1}t\mathcal{K}x.$$

The operator $\mathcal{H}$ is continuous and uniformly continuous in t for each $x,$ and the mapping $\mathcal{H}(t,.)$ is a strict set contraction for each $t\in [0,1]$. In addition, $\mathcal{H}(t,.)$ has no fixed point on $\partial \mathcal{U}$. On the contrary,

• If $t=0$, there exists some $x_0\in \partial \mathcal{U}$ such that $x_0=0$, contradicting $x_0\in \mathcal{U}.$

• If $t\in (0,1]$, there exists some $x_0\in \mathcal{P}\cap \partial \mathcal{U}$ such that $\frac{1}{\epsilon +1}t\mathcal{K}{x}_0=x_0$; then $\mathcal{K}{x}_0=\frac{1+\epsilon }{t}{x}_0$ with $\frac{1+\epsilon }{t}\ge 1+\epsilon ,$ contradicting the assumption. From the invariance under homotopy and the normalization properties of the index, we deduce

$$i(\frac{1}{\epsilon +1}\mathcal{K},\mathcal{U},\mathcal{X})=i(0,\mathcal{U},\mathcal{X})=1.$$

New, we show that

$$i(\mathcal{K},\mathcal{U},\mathcal{X})=i(\frac{1}{\epsilon +1}\mathcal{K},\mathcal{U},\mathcal{X}).$$

We have

$$\frac{1}{\epsilon +1}\mathcal{K}x\ne x,\forall x\in \partial \mathcal{U}.$$

(4)

Then there exists $\gamma >0$ such that

$$\parallel x-\frac{1}{\epsilon +1}\mathcal{K}x\parallel \ge \gamma ,\forall x\in \partial \mathcal{U}.$$

In other hand, we have $\frac{1}{ϵ+1}\mathcal{K}x\to \mathcal{K}x$ as $ϵ\to 0,$ for $x\in \overline{\mathcal{U}}.$ So for $\epsilon$ small enough

$$\parallel \mathcal{K}x-\frac{1}{\epsilon +1}\mathcal{K}x\parallel <\frac{\gamma }{2},\forall x\in \partial \mathcal{U}.$$

Define the convex deformation $G:[0,1]\times \overline{\mathcal{U}}\to \mathcal{X}$ by

$$G(t,x)=t\mathcal{K}x+(1-t)\frac{1}{\epsilon +1}\mathcal{K}x.$$

The operator G is continuous and uniformly continuous in t for each $x,$ and the mapping $G(t,.)$ is a strict set contraction for each $t\in [0,1]$ (since $t+\frac{1}{\epsilon +1}(1-t)<t+1-t=1$).In addition, $G(t,.)$ has no fixed point on $\partial \mathcal{U}$. In fact, for all $x\in \partial \mathcal{U}$, we have

$$\begin{array}{ccc}\parallel x-G(t,x)\parallel \hfill & =\hfill & \parallel x-t\mathcal{K}x-(1-t)\frac{1}{\epsilon +1}\mathcal{K}x\parallel \hfill \\ & \ge \hfill & \parallel x-\frac{1}{\epsilon +1}\mathcal{K}x\parallel -t\parallel \mathcal{K}x-\frac{1}{\epsilon +1}\mathcal{K}x\parallel \hfill \\ & >\hfill & \gamma -\frac{\gamma }{2}>\frac{\gamma }{2}.\hfill \end{array}$$

Then our claim follows from the invariance property by homotopy of the index.

□

Proposition 5. 

Let $\mathcal{P}$ be a cone in a Banach space $\mathcal{E}$. Let also, $\mathcal{U}$ be a bounded open subset of $\mathcal{P}$ with $0\in \mathcal{U}.$ Assume that $\mathcal{T}:\Omega \subset \mathcal{P}\to \mathcal{E}$ is an expansive mapping with constant $h>1,$$S:\overline{\mathcal{U}}\to \mathcal{E}$ is a l-set contraction with $0\le l<h-1$, and $S\left(\overline{\mathcal{U}}\right)\subset (I-\mathcal{T})(\Omega ).$ If there exists $\epsilon \ge 0$ such that

$$Sx\notin \left\{\right(I-\mathcal{T}\left)\right(x),(I-\mathcal{T}\left)\right(\lambda x\left)\right\}forallx\in \partial \mathcal{U}\cap \Omega and\lambda \ge 1+\epsilon ,$$

then the fixed point index $i_{*}(\mathcal{T}+S,\mathcal{U}\cap \Omega ,\mathcal{P})=1.$

Proof. 

The mapping ${(I-\mathcal{T})}^{-1}S:\overline{\mathcal{U}}\to \mathcal{P}$ is a strict set contraction and it is readily seen that the following condition is satisfied

$${(I-\mathcal{T})}^{-1}Sx\notin \{x,\lambda x\}\mathrm{for}\mathrm{all}x\in \partial \mathcal{U}\mathrm{and}\lambda \ge 1+ϵ.$$

Our claim then follows from the definition of $i_{*}$ and the following Lemma 4. □

The following result will be used to prove our main result.

Theorem 6. 

Let $\mathcal{P}$ be a cone of a Banach space $\mathcal{E}$; Ω a subset of $\mathcal{P}$ and ${\mathcal{U}}_1,{\mathcal{U}}_{2}and{\mathcal{U}}_3$ three open bounded subsets of $\mathcal{P}$ such that ${\overline{\mathcal{U}}}_1\subset {\overline{\mathcal{U}}}_2\subset {\mathcal{U}}_3$ and $0\in {\mathcal{U}}_1.$ Assume that $\mathcal{T}:\Omega \to \mathcal{P}$ is an expansive mapping with constant $h>1,$ $S:{\overline{\mathcal{U}}}_3\to \mathcal{E}$ is a k-set contraction with $0\le k<h-1$ and $S\left({\overline{\mathcal{U}}}_3\right)\subset (I-\mathcal{T})(\Omega ).$ Suppose that $({\mathcal{U}}_2\setminus {\overline{\mathcal{U}}}_1)\cap \Omega \ne \varnothing ,({\mathcal{U}}_3\setminus {\overline{\mathcal{U}}}_2)\cap \Omega \ne \varnothing ,$ and there exists $u_0\in {\mathcal{P}}^{*}$ such that the following conditions hold:

 (i) 

$Sx\ne (I-\mathcal{T})(x-\lambda u_0),$ for all $\lambda >0$ and $x\in \partial {\mathcal{U}}_1\cap (\Omega +\lambda u_0),$

 (ii) 

there exists $ϵ\ge 0$ such that $Sx\ne (I-\mathcal{T})\left(\lambda x\right),$ for all $\lambda \ge 1+ϵ,x\in \partial {\mathcal{U}}_2$ and $\lambda x\in \Omega$,

 (iii) 

$Sx\ne (I-\mathcal{T})(x-\lambda u_0),$ for all $\lambda >0$ and $x\in \partial {\mathcal{U}}_3\cap (\Omega +\lambda u_0).$

Then $\mathcal{T}+S$ has at least two non-zero fixed points $x_1,x_2\in \mathcal{P}$ such that

$$x_1\in \partial {\mathcal{U}}_2\cap \Omega and{x}_2\in ({\overline{\mathcal{U}}}_3\setminus {\overline{\mathcal{U}}}_2)\cap \Omega ,$$

or

$$x_1\in ({\mathcal{U}}_2\setminus {\mathcal{U}}_1)\cap \Omega and{x}_2\in ({\overline{\mathcal{U}}}_3\setminus {\overline{\mathcal{U}}}_2)\cap \Omega .$$

Proof. 

If $Sx=(I-\mathcal{T})x$ for $x\in \partial {\mathcal{U}}_2\cap \Omega$, then we get a fixed point $x_1\in \partial {\mathcal{U}}_2\cap \Omega$ of the operator $\mathcal{T}+S$. Suppose that $Sx\ne (I-\mathcal{T})x$ for any $x\in \partial {\mathcal{U}}_2\cap \Omega$. Without loss of generality, assume that $\mathcal{T}x+Sx\ne xon\partial {\mathcal{U}}_1\cap \Omega and\mathcal{T}x+Sx\ne xon\partial {\mathcal{U}}_3\cap \Omega$, otherwise the conclusion has been proved. By [8] and Proposition 5, we have

$$i_{*}(\mathcal{T}+S,{\mathcal{U}}_1\cap \Omega ,\mathcal{P})=i_{*}(\mathcal{T}+S,{\mathcal{U}}_3\cap \Omega ,\mathcal{P})=0and{i}_{*}(\mathcal{T}+S,{\mathcal{U}}_2\cap \Omega ,\mathcal{P})=1.$$

The additivity property of the index yields

$$i_{*}(\mathcal{T}+S,({\mathcal{U}}_2\setminus {\overline{\mathcal{U}}}_1)\cap \Omega ,\mathcal{P})=1and{i}_{*}(\mathcal{T}+S,({\mathcal{U}}_3\setminus {\overline{\mathcal{U}}}_2)\cap \Omega ,\mathcal{P})=-1.$$

Consequently, by the existence property of the index, $\mathcal{T}+S$ has at least two fixed points $x_1\in ({\mathcal{U}}_2\setminus {\mathcal{U}}_1)\cap \Omega and{x}_2\in ({\overline{\mathcal{U}}}_3\setminus {\overline{\mathcal{U}}}_2)\cap \Omega .$ □

3. Proof of Theorem 1

Let ${\mathcal{X}}_1={\mathcal{C}}^1([0,\infty ),{\mathcal{C}}^q\left(\mathbb{R}\right))$ be endowed with the norm

$$\begin{array}{ccc}\hfill {\parallel u\parallel }_1& =& max\left\{\underset{(t,x)\in [0,\infty )\times \mathbb{R}}{sup}\right|u(t,x)|,\underset{(t,x)\in [0,\infty )\times \mathbb{R}}{sup}|{\partial }_{t}u(t,x)|,\hfill \\ \\ & & \underset{(t,x)\in [0,\infty )\times \mathbb{R}}{sup}|{\partial }_x^{j}u(t,x)|,j\in \{1,\dots ,q\}\},\hfill \end{array}$$

provided it exists. Define $\mathcal{X}={\mathcal{X}}_1\times {\mathcal{X}}_1$ with the norm

$$\parallel (u,v)\parallel =max\{\parallel u{\parallel }_1,\parallel v{\parallel }_1\}.$$

For $(u,v)\in \mathcal{X}$, define

$$\begin{array}{ccc}\hfill {\displaystyle Q_1(u,v)(t,x)}& =& \sum _{k=0}^{N_1}\sum _{l=0}^{N_1-k}{\partial }_x\left\{\sum _{m=0}^{N_1-k}{\partial }_x^m{u}^p{P}_{k,l,m}\left({\partial }_x^{l}v\right)\right\},\hfill \\ \\ \hfill {\displaystyle Q_2(u,v)(t,x)}& =& \sum _{k=0}^{N_3}\sum _{l=0}^{N_3-k}{\partial }_x^k\left\{\sum _{m=0}^{N_3-k}{\partial }_x^m{v}^p{Q}_{k,l,m}\left({\partial }_x^{l}u\right)\right\},(t,x)\in [0,\infty )\times \mathbb{R}.\hfill \end{array}$$

Then the IVP (1) can be rewritten in the form

$$\begin{array}{ccc}{\displaystyle {\partial }_{t}u+Q_1(u,v)+\sum _{k=1}^{N_2}{a}_k(t,x)+{\partial }_x^{2k+1}u}\hfill & =\hfill & 0,\hfill \\ \\ {\displaystyle {\partial }_{t}v+Q_2(u,v)+\sum _{k=1}^{N_4}{b}_k(t,x){\partial }_x^{2k+1}v}\hfill & =\hfill & 0,(t,x)\in [0,\infty )\times \mathbb{R},\hfill \\ \\ u(0,x)=u_0\left(x\right),v(0,x)\hfill & =\hfill & v_0\left(x\right),x\in \mathbb{R}.\hfill \end{array}$$

(5)

Let

$$\begin{array}{ccc}\hfill C_1& =& \{\sum _{k=0}^{N_1}\sum _{l=0}^{N_1-k}\sum _{m=0}^{N_1-k}\sum _{r=0}^k\left(\begin{array}{c}k\\ r\end{array}\right){\left((k-r+m)!\right)}^{2}p!{\mathcal{B}}^p\sum _{j=0}^{N_5}\sum _{i=0}^r\left(\begin{array}{c}r\\ i\end{array}\right){(i!)}^{2}j!{\mathcal{B}}^{j+1},\hfill \\ \\ & & \sum _{k=0}^{N_3}\sum _{l=0}^{N_3-k}\sum _{m=0}^{N_3-k}\sum _{r=0}^k\left(\begin{array}{c}k\\ r\end{array}\right){\left((k-r+m)!\right)}^{2}p!{\mathcal{B}}^p\sum _{j=0}^{N_6}\sum _{i=0}^r\left(\begin{array}{c}r\\ i\end{array}\right){(i!)}^{2}j!{\mathcal{B}}^{j+1}\}.\hfill \end{array}$$

Lemma 7. 

Suppose $\left(\right(Hyp1)$. If $(u,v)\in \mathcal{X}$ and $\parallel (u,v)\parallel \le \mathcal{B}$, then

$$|Q_1(u,v)(t,x)|,|Q_2(u,v)(t,x)|\le C_1,(t,x)\in [0,\infty )\times \mathbb{R}.$$

Proof. 

We have

$$\begin{array}{c}\hfill {\displaystyle {\partial }_x\left\{\sum _{m=0}^{N_1-k}{\partial }_x^m{u}^p{P}_{k,l,m}\left({\partial }_x^{l}v\right)\right\}}\\ \\ & =& \sum _{m=0}^{N_1-k}{\partial }_x^k\left({\partial }_x^m{u}^p{P}_{k,l,m}\left({\partial }_x^{l}v\right)\right)\hfill \\ \\ & =& \sum _{m=0}^{N_1-k}\sum _{r=0}^k\left(\begin{array}{c}k\\ r\end{array}\right){\partial }_x^{k-r+m}{u}^p{\partial }_x^r{P}_{k,l,m}\left({\partial }_x^{l}v\right)\hfill \\ \\ & =& \sum _{m=0}^{N_1-k}\sum _{r=0}^k\left(\begin{array}{c}k\\ r\end{array}\right){\partial }_x^{k-r+m}{u}^p{\partial }_x^r\sum _{j=0}^{N_5}{c}_{k,l,m,j}{\left({\partial }_x^{l}v\right)}^j\hfill \\ \\ & =& \sum _{m=0}^{N_1-k}\sum _{r=0}^k\left(\begin{array}{c}k\\ r\end{array}\right){\partial }_x^{k-r+m}{u}^p\sum _{j=0}^{N_5}\sum _{i=0}^r\left(\begin{array}{c}r\\ i\end{array}\right){\partial }_x^{r-i}{c}_{k,l,m,j}{\partial }_xî{\left({\partial }_x^{l}v\right)}^j,\hfill \end{array}$$

$k\in \mathbb{N}$, $0\le k\le N_1$. Since $\mathcal{B}>1$, we have

$$\left|{\partial }_x^{r_1}{u}^{r_2}\right|\le {(r_1!)}^2{r}_2!{\mathcal{B}}^{r_2},$$

for any $r_1,r_2\in \mathbb{N}$, $r_1\le q$. Then

$$\begin{array}{c}\hfill {\displaystyle \left|{\partial }_x\left\{\sum _{m=0}^{N_1-k}{\partial }_x^m{u}^p{P}_{k,l,m}\left({\partial }_x^{l}v\right)\right\}\right|}\\ \\ & \le & \sum _{m=0}^{N_1-k}\sum _{r=0}^k\left(\begin{array}{c}k\\ r\end{array}\right)\left|{\partial }_x^{k-r+m}{u}^p\right|\sum _{j=0}^{N_5}\sum _{i=0}^r\left(\begin{array}{c}r\\ i\end{array}\right)\left|{\partial }_x^{r-i}{c}_{k,l,m,j}\right|\left|{\partial }_x^i{\left({\partial }_x^{l}v\right)}^j\right|\hfill \\ \\ & \le & \sum _{m=0}^{N_1-k}\sum _{r=0}^k\left(\begin{array}{c}k\\ r\end{array}\right){\left((k-r+m)!\right)}^{2}p!{\mathcal{B}}^p\sum _{j=0}^{N_5}\sum _{i=0}^r\left(\begin{array}{c}r\\ i\end{array}\right){(i!)}^{2}j!{\mathcal{B}}^{j+1},\hfill \end{array}$$

on $[0,\infty )\times \mathbb{R}$, $0\le k\le N_1$, and then

$$\begin{array}{ccc}\hfill |Q_1(u,v)|& \le & \sum _{k=0}^{N_1}\sum _{l=0}^{N_1-k}\sum _{m=0}^{N_1-k}\sum _{r=0}^k\left(\begin{array}{c}k\\ r\end{array}\right){\left((k-r+m)!\right)}^{2}p!B^p\sum _{j=0}^{N_5}\sum _{i=0}^r\left(\begin{array}{c}r\\ i\end{array}\right){(i!)}^{2}j!{\mathcal{B}}^{j+1}\hfill \\ \\ & \le & C_1,\hfill \end{array}$$

on $[0,\infty )\times \mathbb{R}$. As above,

$$|Q_2(u,v)|\le C_1,$$

on $[0,\infty )\times \mathbb{R}$. This completes the proof. □

For $(u,v)\in \mathcal{X}$, define the operators

$$\begin{array}{ccc}\hfill S_1^1(u,v)(t,x)& =& u(t,x)-u_0\left(x\right)\hfill \\ \\ & & +{\int }_0^t\left(Q_1(u,v)(s,x)+\sum _{k=1}^{N_2}{a}_k(s,x){\partial }_x^{2k+1}u(s,x)\right)ds,\hfill \\ \\ \hfill S_1^2(u,v)(t,x)& =& v(t,x)-v_0\left(x\right)\hfill \\ \\ & & +{\int }_0^t\left(Q_2(u,v)(s,x)+\sum _{k=1}^{N_4}{b}_k(s,x){\partial }_x^{2k+1}v(s,x)\right)ds,\hfill \\ \\ \hfill S(u,v)(t,x)& =& \left(S_1^1(u,v)(t,x),S_1^2(u,v)(t,x)\right),\hfill \end{array}$$

$(t,x)\in [0,\infty )\times \mathbb{R}$.

Lemma 8. 

Suppose $\left(\right(Hyp1)$. If $(u,v)\in \mathcal{X}$ satisfies the equation

$$S_1(u,v)(t,x)=0,(t,x)\in [0,\infty )\times \mathbb{R},$$

then $(u,v)$ is a solution to the IVP (1).

Proof. 

We have

$$\begin{array}{ccc}0\hfill & =\hfill & u(t,x)-u_0\left(x\right)\hfill \\ \\ \hfill & \hfill & {\displaystyle +{\int }_0^t\left(Q_1(u,v)(s,x)+\sum _{k=1}^{N_2}{a}_k(s,x){\partial }_x^{2k+1}u(s,x)\right)ds,}\hfill \\ \\ 0\hfill & =\hfill & v(t,x)-v_0\left(x\right)\hfill \\ \\ \hfill & \hfill & {\displaystyle +{\int }_0^t\left(Q_2(u,v)(s,x)+\sum _{k=1}^{N_4}{b}_k(s,x){\partial }_x^{2k+1}v(s,x)\right)ds,}\hfill \end{array}$$

(6)

$(t,x)\in [0,\infty )\times \mathbb{R}$, which we differentiate with respect to t and we get (5). We put $t=0$ in (6) and we obtain

$$\begin{array}{ccc}\hfill 0& =& u(0,x)-u_0\left(x\right)\hfill \\ \\ \hfill 0& =& v(0,x)-v_0\left(x\right),x\in \mathbb{R}.\hfill \end{array}$$

Thus, $(u,v)$ is a solution to the IVP (1). This completes the proof. □

Let

$${\mathcal{B}}_1=max\{2\mathcal{B},C_1+N_2{\mathcal{B}}^2,C_1+N_4{\mathcal{B}}^2\}.$$

Lemma 9. 

Suppose $\left(\right(Hyp1)$. If $(u,v)\in \mathcal{X}$ and $\parallel (u,v)\parallel \le \mathcal{B}$, then

$$\begin{array}{ccc}\hfill |S_1^1(u,v)(t,x)|& \le & {\mathcal{B}}_1(1+t),\hfill \\ \\ \hfill |S_1^2(u,v)(t,x)|& \le & {\mathcal{B}}_1(1+t),(t,x)\in [0,\infty )\times \mathbb{R}.\hfill \end{array}$$

Proof. 

We have

$$\begin{array}{ccc}\hfill |S_1^1(u,v)(t,x)|& =& |u(t,x)-u_0\left(x\right)\hfill \\ \\ & & +{\int }_0^t\left(Q_1(u,v)(s,x)+\sum _{k=1}^{N_2}{a}_k(s,x){\partial }_x^{2k+1}u(s,x)\right)ds|\hfill \\ \\ & \le & \left|u(t,x)\right|+|u_0\left(x\right)|\hfill \\ \\ & & +{\int }_0^t\left(|Q_1(u,v)(s,x)|+\sum _{k=1}^{N_2}|a_k(s,x)\left|\right|{\partial }_x^{2k+1}u(s,x)|\right)ds\hfill \\ \\ & \le & 2\mathcal{B}+{\int }_0^t(C_1+N_2{\mathcal{B}}^2)ds\hfill \\ \\ & \le & {\mathcal{B}}_1(1+t),(t,x)\in [0,\infty )\times \mathbb{R}.\hfill \end{array}$$

As above,

$$|S_1^2(u,v)(t,x)|\le {\mathcal{B}}_1(1+t),(t,x)\in [0,\infty )\times \mathbb{R}.$$

This completes the proof. □

Below, suppose that

 (Hyp2) 

there exist a function $g\in \mathcal{C}\left(\right[0,\infty )\times \mathbb{R})$, $g>0$ on $(0,\infty )\times (\mathbb{R}\setminus \{0\left\}\right)$, $g(0,x)=g(t,0)=0$, $(t,x)\in [0,\infty )\times \mathbb{R}$, and a positive constant $\mathcal{A}$ such that

$$q!·2^{q+1}(1+t+t^2){(1+|x|+\cdots +|x|}^q){\int }_0^t\left|{\int }_0^{x}g(t_1,x_1)d{x}_1\right|d{t}_1\le \mathcal{A},$$

$(t,x)\in [0,\infty )\times \mathbb{R}$.

In the last section we will give examples for g and $\mathcal{A}$ that satisfy $\left(Hyp2\right)$. For $(u,v)\in \mathcal{X}$, define the operators

$$\begin{array}{ccc}\hfill S_2^1(u,v)(t,x)& =& {\int }_0^t{\int }_0^x(t-t_1){(x-x_1)}^{q}g(t_1,x_1)S_1^1(u,v)(t_1,x_1)d{x}_{1}d{t}_1,\hfill \\ \\ \hfill S_2^2(u,v)(t,x)& =& {\int }_0^t{\int }_0^x(t-t_1){(x-x_1)}^{q}g(t_1,x_1)S_1^2(u,v)(t_1,x_1)d{x}_{1}d{t}_1,\hfill \\ \\ \hfill S_2(u,v)(t,x)& =& (S_2^1(u,v)(t,x),S_2^2(u,v)(t,x)),(t,x)\in [0,\infty )\times \mathbb{R}.\hfill \end{array}$$

Lemma 10. 

Suppose $\left(\right(Hyp1)$ and $\left(Hyp2\right)$. If $(u,v)\in \mathcal{X}$ satisfies the equation

$$S_2(u,v)(t,x)=0,(t,x)\in [0,\infty )\times \mathbb{R},$$

then $(u,v)$ is a solution to the IVP (1).

Proof. 

We differentiate two times in t and $q+1$ times in x the equation (10) and we get

$$g(t,x)S_1^1(u,v)(t,x)=g(t,x)S_1^2(u,v)(t,x)=0,(t,x)\in [0,\infty )\times \mathbb{R}.$$

Hence,

$$S_1^1(u,v)(t,x)=S_1^2(u,v)(t,x)=0,(t,x)\in (0,\infty )\times (\mathbb{R}\setminus \left\{0\right\}).$$

Since $S_1^1(u,v)(·,·)$ and $S_1^2(u,v)(·,·)$ are continuous functions on $[0,\infty )\times \mathbb{R}$, we have

$$\begin{array}{ccc}\hfill 0& =& S_1^1(u,v)(0,x)=S_1^2(u,v)(0,x)\hfill \\ \\ & =& \underset{t\to 0}{lim}{S}_1^1(u,v)(t,x)=\underset{t\to 0}{lim}{S}_1^2(u,v)(t,x)\hfill \\ \\ & =& \underset{x\to 0}{lim}{S}_1^1(u,v)(t,x)=\underset{x\to 0}{lim}{S}_1^2(u,v)(t,x)\hfill \\ \\ & =& S_1^1(u,v)(t,0)=S_1^2(u,v)(t,0),(t,x)\in [0,\infty )\times \mathbb{R}.\hfill \end{array}$$

Therefore

$$S_1^1(u,v)(t,x)=S_1^2(u,v)(t,x)=0,(t,x)\in [0,\infty )\times \mathbb{R}.$$

Now, applying Lemma 8, we get the desired result. □

Lemma 11. 

Suppose $\left(\right(Hyp1)$ and $\left(Hyp2\right)$. If $(u,v)\in \mathcal{X}$, $\parallel (u,v)\parallel \le \mathcal{B}$, then

$$\parallel S_2(u,v)\parallel \le \mathcal{A}{\mathcal{B}}_1.$$

Proof. 

We will use the inequality ${(z+w)}^r\le 2^r(z^r+w^r)$, $w,z,q\ge 0$. We have

$$\begin{array}{ccc}\hfill |S_2^1(u,v)(t,x)|& =& \left|{\int }_0^t{\int }_0^x(t-t_1){(x-x_1)}^{q}g(t_1,x_1)S_1^1(u,v)(t_1,x_1)d{x}_{1}d{t}_1\right|\hfill \\ \\ & \le & {\int }_0^t\left|{\int }_0^x(t-t_1)|x-x_1{|}^{q}g(t_1,x_1)\left|S_1^1(u,v)(t_1,x_1)\right|d{x}_1\right|d{t}_1\hfill \\ \\ & \le & {\mathcal{B}}_1{\int }_0^t\left|{\int }_0^x(t-t_1)(1+t_1){|x-x_1|}^{q}g(t_1,x_1)d{x}_1\right|d{t}_1\hfill \\ \\ & \le & {\mathcal{B}}_{1}t(1+t)2^{q+1}{\left|x\right|}^q{\int }_0^t\left|{\int }_0^{x}g(t_1,x_1)d{x}_1\right|d{t}_1\hfill \\ \\ & \le & \mathcal{A}{\mathcal{B}}_1,(t,x)\in [0,\infty )\times \mathbb{R},\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill |{\partial }_t{S}_2^1(u,v)(t,x)|& =& \left|{\int }_0^t{\int }_0^x{(x-x_1)}^{q}g(t_1,x_1)S_1^1(u,v)(t_1,x_1)d{x}_{1}d{t}_1\right|\hfill \\ \\ & \le & {\int }_0^t\left|{\int }_0^x|x-x_1{|}^{q}g(t_1,x_1)\left|S_1^1(u,v)(t_1,x_1)\right|d{x}_1\right|d{t}_1\hfill \\ \\ & \le & {\mathcal{B}}_1{\int }_0^t\left|{\int }_0^x(1+t_1){|x-x_1|}^{q}g(t_1,x_1)d{x}_1\right|d{t}_1\hfill \\ \\ & \le & {\mathcal{B}}_1(1+t)2^{q+1}{\left|x\right|}^q{\int }_0^t\left|{\int }_0^{x}g(t_1,x_1)d{x}_1\right|d{t}_1\hfill \\ \\ & \le & \mathcal{A}{\mathcal{B}}_1,(t,x)\in [0,\infty )\times \mathbb{R},\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill |{\partial }_x{S}_2^1(u,v)(t,x)|& =& q\left|{\int }_0^t{\int }_0^x(t-t_1){(x-x_1)}^{q-1}g(t_1,x_1)S_1^1(u,v)(t_1,x_1)d{x}_{1}d{t}_1\right|\hfill \\ \\ & \le & q{\int }_0^t\left|{\int }_0^x(t-t_1)|x-x_1{|}^{q-1}g(t_1,x_1)\left|S_1^1(u,v)(t_1,x_1)\right|d{x}_1\right|d{t}_1\hfill \\ \\ & \le & q{\mathcal{B}}_1{\int }_0^t\left|{\int }_0^x(t-t_1)(1+t_1){|x-x_1|}^{q-1}g(t_1,x_1)d{x}_1\right|d{t}_1\hfill \\ \\ & \le & q{\mathcal{B}}_{1}t(1+t)2^q{\left|x\right|}^{q-1}{\int }_0^t\left|{\int }_0^{x}g(t_1,x_1)d{x}_1\right|d{t}_1\hfill \\ \\ & \le & \mathcal{A}{\mathcal{B}}_1,(t,x)\in [0,\infty )\times \mathbb{R},\hfill \end{array}$$

and so on. As above,

$$\begin{array}{ccc}\hfill |S_2^2(u,v)(t,x)|\le \mathcal{A}{\mathcal{B}}_1,\left|{\partial }_t{S}_2^2(u,v)(t,x)\right|& \le & \mathcal{A}{\mathcal{B}}_1,\hfill \\ \\ \hfill |{\partial }_x^j{S}_2^2(u,v)(t,x)|\le \mathcal{A}{\mathcal{B}}_1,j\in \{1,\dots ,q\}.\end{array}$$

$(t,x)\in [0,\infty )\times \mathbb{R}$. Thus,

$$\parallel S_2(u,v)\parallel \le \mathcal{A}{\mathcal{B}}_1.$$

This completes the proof. □

Below, suppose

 (Hyp3) 

$ϵ\in (0,1)$, $\mathcal{A}$, $\mathcal{B}$ and ${\mathcal{B}}_1$ satisfy the inequalities $ϵ{\mathcal{B}}_1(1+\mathcal{A})<1$ and $\mathcal{A}{\mathcal{B}}_1<\mathcal{B}$.

Let $\tilde{\widetilde{\tilde{\mathcal{Y}}}}$ denote the set of all equi-continuous families in $\mathcal{X}$ with respect to the norm $\parallel ·\parallel$. Let also, $\widetilde{\tilde{\mathcal{Y}}}=\overline{\tilde{\widetilde{\tilde{\mathcal{Y}}}}}$ be the closure of $\tilde{\widetilde{\tilde{\mathcal{Y}}}}$, $\tilde{\mathcal{Y}}=\widetilde{\tilde{\mathcal{Y}}}\cup \left\{(u_0,v_0)\right\}$,

$$\mathcal{Y}=\{(u,v)\in \tilde{\mathcal{Y}}:(u,v)\ge 0,\parallel (u,v)\parallel \le \mathcal{B}\}.$$

Note that $\mathcal{Y}$ is a compact set in $\mathcal{X}$. For $(u,v)\in \mathcal{X}$, define the operators

$$\begin{array}{ccc}\hfill \mathcal{T}(u,v)(t,x)& =& -ϵ(u,v)(t,x),\hfill \\ \\ \hfill S(u,v)(t,x)& =& (u,v)(t,x)+ϵ(u,v)(t,x)+ϵS_2(u,v)(t,x),(t,x)\in [0,\infty )\times \mathbb{R}.\hfill \end{array}$$

For $(u,v)\in \mathcal{Y}$, using Lemma 10, we have

$$\begin{array}{ccc}\hfill \parallel (I-S)(u,v)\parallel & =& \parallel ϵ(u,v)-ϵS_2(u,v)\parallel \hfill \\ \\ & \le & ϵ\parallel (u,v)\parallel +ϵ\parallel S_2(u,v)\parallel \hfill \\ \\ & \le & ϵ{\mathcal{B}}_1+ϵ\mathcal{A}{\mathcal{B}}_1\hfill \\ \\ & =& ϵ{\mathcal{B}}_1(1+\mathcal{A})\hfill \\ \\ & <& \mathcal{B}.\hfill \end{array}$$

Thus, $S:\mathcal{Y}\to \mathcal{X}$ is continuous and $(I-S)\left(\mathcal{Y}\right)$ resides in a compact subset of $\mathcal{X}$. Now, suppose that there is a $(u,v)\in \mathcal{X}$ so that $\parallel (u,v)\parallel =\mathcal{B}$ and

$$(u,v)=\lambda (I-S)(u,v),$$

or

$$\frac{1}{\lambda }(u,v)=(I-S)(u,v)=-ϵ(u,v)-ϵS_2(u,v),$$

or

$$\left(\frac{1}{\lambda }+ϵ\right)(u,v)=-ϵS_2(u,v),$$

for some $\lambda \in \left(0,\frac{1}{ϵ}\right)$. Hence, $\parallel S_2(u,v)\parallel \le \mathcal{A}{\mathcal{B}}_1<\mathcal{B}$,

$$ϵ\mathcal{B}<\left(\frac{1}{\lambda }+ϵ\right)\mathcal{B}=\left(\frac{1}{\lambda }+ϵ\right)\parallel (u,v)\parallel =ϵ\parallel S_2(u,v)\parallel <ϵ\mathcal{B},$$

which is a contradiction. Hence and Theorem 3, it follows that the operator $\mathcal{T}+S$ has a fixed point $(u^{*},v^{*})\in \mathcal{Y}$. Therefore

$$\begin{array}{ccc}\hfill (u^{*},v^{*})(t,x)& =& \mathcal{T}(u^{*},v^{*})(t,x)+S(u^{*},v^{*})(t,x)\hfill \\ \\ & =& -ϵ(u^{*},v^{*})(t,x)+(u^{*},v^{*})(t,x)+ϵ(u^{*},v^{*})(t,x)+ϵS_2(u^{*},v^{*})(t,x),\hfill \end{array}$$

$(t,x)\in [0,\infty )\times \mathbb{R}$, whereupon

$$0=S_2(u^{*},v^{*})(t,x),(t,x)\in [0,\infty )\times \mathbb{R}.$$

From here and from Lemma 10, it follows that $(u^{*},v^{*})$ is a solution to the IVP (1). This completes the proof.

4. Proof of Theorem 2

Let $\mathcal{X}$ be the space used in the previous section. Suppose

 (Hyp4) 

Let $m>0$ be large enough and $\mathcal{A}$, $\mathcal{B}$, r, L, $R_1$ be positive constants that satisfy the following conditions

$$r<L<R_1\le \mathcal{B},ϵ>0,R_1>\left(\frac{2}{5m}+1\right)L,$$

$$\mathcal{A}{\mathcal{B}}_1<\frac{L}{5}.$$

Let

$$\tilde{P}=\{(u,v)\in \mathcal{X}:(u,v)\ge 0\mathrm{on}[0,\infty )\times \mathbb{R}\}.$$

With $\mathcal{P}$ we will denote the set of all equi-continuous families in $\tilde{P}$. For $(u,v)\in \mathcal{X}$, define the operators

$$\begin{array}{ccc}\hfill {\mathcal{T}}_1(u,v)(t,x)& =& (1+mϵ)(u,v)(t,x)-\left(ϵ\frac{L}{10},ϵ\frac{L}{10}\right),\hfill \\ \\ \hfill S_3(u,v)(t,x)& =& -ϵS_2(u,v)(t,x)-mϵ(u,v)(t,x)-\left(ϵ\frac{L}{10},ϵ\frac{L}{10}\right),\hfill \end{array}$$

$t\in [0,\infty )$. Note that any fixed point $(u,v)\in \mathcal{X}$ of the operator ${\mathcal{T}}_1+S_3$ is a solution to the IVP (1). Define

$$\begin{array}{ccc}\hfill {\mathcal{U}}_1& =& {\mathcal{P}}_r=\{(u,v)\in \mathcal{P}:\parallel (u,v)\parallel <r\},\hfill \\ \\ \hfill {\mathcal{U}}_2& =& {\mathcal{P}}_L=\{(u,v)\in \mathcal{P}:\parallel (u,v)\parallel <L\},\hfill \\ \\ \hfill {\mathcal{U}}_3& =& {\mathcal{P}}_{R_1}=\{(u,v)\in \mathcal{P}:\parallel (u,v)\parallel <R_1\},\hfill \\ \\ \hfill R_2& =& R_1+\frac{\mathcal{A}}{m}{\mathcal{B}}_1+\frac{L}{5m},\hfill \\ \\ \hfill \Omega & =& \overline{{\mathcal{P}}_{R_2}}=\{(u,v)\in \mathcal{P}:\parallel (u,v)\parallel \le R_2\}.\hfill \end{array}$$

1.

For $(u_1,v_1),(u_2,v_2)\in \Omega$, we have

$$\begin{array}{c}\hfill \parallel {\mathcal{T}}_1(u_1,v_1)-{\mathcal{T}}_1(u_2,v_2)\parallel =(1+m\epsilon )\parallel (u_1,v_1)-(u_2,v_2)\parallel ,\end{array}$$

whereupon ${\mathcal{T}}_1:\Omega \to \mathcal{X}$ is an expansive operator with a constant $h=1+m\epsilon >1$.

2.

For $(u,v)\in {\overline{\mathcal{P}}}_{R_1}$, we get

$$\begin{array}{ccc}\hfill \parallel S_3(u,v)\parallel & \le & \epsilon \parallel S_2(u,v)\parallel +m\epsilon \parallel (u,v)\parallel +\epsilon \frac{L}{10}\hfill \\ \\ & \le & \epsilon \left(\mathcal{A}{\mathcal{B}}_1+m{R}_1+\frac{L}{10}\right).\hfill \end{array}$$

Therefore $S_3\left({\overline{\mathcal{P}}}_{R_1}\right)$ is uniformly bounded. Since $S_3:{\overline{\mathcal{P}}}_{R_1}\to \mathcal{X}$ is continuous, we have that $S_3\left({\overline{\mathcal{P}}}_{R_1}\right)$ is equi-continuous. Consequently $S_3:{\overline{\mathcal{P}}}_{R_1}\to \mathcal{X}$ is a 0-set contraction.

3.

Let $(u_1,v_1)\in {\overline{\mathcal{P}}}_{R_1}$. Set

$$(u_2,v_2)=(u_1,v_1)+\frac{1}{m}{S}_2(u_1,v_1)+\left(\frac{L}{5m},\frac{L}{5m}\right).$$

Note that $S_2{u}_1+\frac{L}{5}\ge 0$, $S_2{v}_1+\frac{L}{5}\ge 0$ on $[0,\infty )\times \mathbb{R}$. We have $u_2,v_2\ge 0$ on $[0,\infty )\times \mathbb{R}$ and

$$\begin{array}{ccc}\hfill \parallel (u_2,v_2)\parallel & \le & \parallel (u_1,v_1)\parallel +\frac{1}{m}\parallel S_2(u_1,v_1)\parallel +\frac{L}{5m}\hfill \\ \\ & \le & R_1+\frac{\mathcal{A}}{m}{\mathcal{B}}_1+\frac{L}{5m}\hfill \\ \\ & =& R_2.\hfill \end{array}$$

Therefore $(u_2,v_2)\in \Omega$ and

$$-\epsilon m(u_2,v_2)=-\epsilon m(u_1,v_1)-\epsilon S_2(u_1,v_1)-\epsilon \left(\frac{L}{10},\frac{L}{10}\right)-\epsilon \left(\frac{L}{10},\frac{L}{10}\right)$$

or

$$\begin{array}{ccc}\hfill (I-{\mathcal{T}}_1)(u_2,v_2)& =& -\epsilon m(u_2,v_2)+\epsilon \left(\frac{L}{10},\frac{L}{10}\right)\hfill \\ \\ & =& S_3(u_1,v_1).\hfill \end{array}$$

Consequently $S_3\left({\overline{\mathcal{P}}}_{R_1}\right)\subset (I-{\mathcal{T}}_1)(\Omega )$.

4.

Assume that for any $(u_0,v_0)\in {\mathcal{P}}^{*}$ there exist $\lambda \ge 0$ and $(u,v)\in \partial {\mathcal{P}}_r\cap (\Omega +\lambda (u_0,v_0))$ or $v\in \partial {\mathcal{P}}_{R_1}\cap (\Omega +\lambda (u_0,v_0))$ such that

$$S_3(u,v)=(I-{\mathcal{T}}_1)((u,v)-\lambda (u_0,v_0)).$$

Then

$$-ϵS_2(u,v)-mϵ(u,v)-ϵ\left(\frac{L}{10},\frac{L}{10}\right)=-mϵ((u,v)-\lambda (u_0,v_0))+ϵ\left(\frac{L}{10},\frac{L}{10}\right),$$

or

$$-S_2(u,v)=\lambda m(u_0,v_0)+\left(\frac{L}{5},\frac{L}{5}\right).$$

Hence,

$$\parallel S_{2}v\parallel =∥\lambda m(u_0,v_0)+\left(\frac{L}{5},\frac{L}{5}\right)∥>\frac{L}{5}.$$

This is a contradiction.

5.

Suppose that for any ${ϵ}_1\ge 0$ small enough there exist a $(u_1,v_1)\in \partial {\mathcal{P}}_L$ and ${\lambda }_1\ge 1+{ϵ}_1$ such that ${\lambda }_1(u_1,v_1)\in {\overline{\mathcal{P}}}_{R_1}$ and

$$S_3(u_1,v_1)=(I-{\mathcal{T}}_1)\left({\lambda }_1(u_1,v_1)\right).$$

(7)

In particular, for ${ϵ}_1>\frac{2}{5m}$, we have $(u_1,v_1)\in \partial {\mathcal{P}}_L$, ${\lambda }_1(u_1,v_1)\in {\overline{\mathcal{P}}}_{R_1}$, ${\lambda }_1\ge 1+{ϵ}_1$ and (7) holds. Since $(u_1,v_1)\in \partial {\mathcal{P}}_L$ and ${\lambda }_1(u_1,v_1)\in {\overline{\mathcal{P}}}_{R_1}$, it follows that

$$\left(\frac{2}{5m}+1\right)L<{\lambda }_{1}L={\lambda }_1\parallel (u_1,v_1)\parallel \le R_1.$$

Moreover,

$$-ϵS_2(u_1,v_1)-mϵ(u_1,v_1)-ϵ\left(\frac{L}{10},\frac{L}{10}\right)=-{\lambda }_{1}mϵ(u_1,v_1)+ϵ\left(\frac{L}{10},\frac{L}{10}\right),$$

or

$$S_2(u_1,v_1)+\left(\frac{L}{5},\frac{L}{5}\right)=({\lambda }_1-1)m(u_1,v_1).$$

From here,

$$2\frac{L}{5}\ge ∥S_2(u_1,v_1)+\left(\frac{L}{5},\frac{L}{5}\right)∥=({\lambda }_1-1)m\parallel (u_1,v_1)\parallel =({\lambda }_1-1)mL,$$

and

$$\frac{2}{5m}+1\ge {\lambda }_1,$$

which is a contradiction.

Therefore all conditions of Theorem 2 hold. Hence, the IVP (1) has at least two solutions $(u_1,v_1)$ and $(u_2,v_2)$ so that

$$\parallel (u_1,v_1)\parallel =L<\parallel (u_2,v_2)\parallel <R_1,$$

or

$$r<\parallel (u_1,v_1)\parallel <L<\parallel (u_2,v_2)\parallel <R_1.$$

5. Example

Below, we will illustrate our main results. Let $\mathcal{B}=1$ and

$$R_1=10,L=5,r=4,m={10}^{50},\mathcal{A}=\frac{1}{10{\mathcal{B}}_1},ϵ=\frac{1}{5{\mathcal{B}}_1(1+\mathcal{A})},$$

$N_j=5$, $j\in \{1,\dots ,4\}$, $p=10$. Then

$$\mathcal{A}{\mathcal{B}}_1=\frac{1}{10}<\mathcal{B},ϵ{\mathcal{B}}_1(1+\mathcal{A})<1,$$

i.e., $\left(Hyp3\right)$ holds. Next,

$$r<L<R_1,ϵ>0,R_1>\left(\frac{2}{5m}+1\right)L,\mathcal{A}{\mathcal{B}}_1<\frac{L}{5}.$$

i.e., $\left(Hyp4\right)$ holds. Take

$$h\left(s\right)=log\frac{1+s^{q+1}\sqrt{2}+s^{2q+2}}{1-s^{q+1}\sqrt{2}+s^{2q+2}},l\left(s\right)=arctan\frac{s^{q+1}\sqrt{2}}{1-s^{2q+2}},s\in \mathbb{R},s\ne \pm 1.$$

Then

$$\begin{array}{ccc}\hfill h^{\prime }\left(s\right)& =& \frac{2\sqrt{2}(q+1)s^q(1-s^{2q+2})}{(1-s^{q+1}\sqrt{2}+s^{2q+2})(1-s^{q+1}\sqrt{2}+s^{2q+2})},\hfill \\ \\ \hfill l^{\prime }\left(s\right)& =& \frac{(q+1)\sqrt{2}{s}^q(1+s^{2q+2})}{1+s^{4q+4}},s\in \mathbb{R},s\ne \pm 1.\hfill \end{array}$$

Therefore

$$\begin{array}{c}\hfill {\displaystyle \underset{s\to \pm \infty }{lim}\sum _{r=0}^{q+1}{s}^{r}h\left(s\right)=\underset{s\to \pm \infty }{lim}\frac{h\left(s\right)}{\frac{1}{{\sum }_{r=0}^{l+1}{s}^r}}}\\ \\ & =& \underset{s\to \pm \infty }{lim}\frac{h^{\prime }\left(s\right)}{-\frac{{\sum }_{r=0}^q(r+1)s^r}{{\left({\sum }_{r=0}^{q+1}{s}^r\right)}^2}}\hfill \\ \\ & =& -\underset{s\to \pm \infty }{lim}\frac{2\sqrt{2}(q+1)s^q(1-s^{2q+2}){\left({\sum }_{r=0}^{q+1}{s}^r\right)}^2}{\left({\sum }_{r=0}^q(r+1)s^r\right)(1-s^{q+1}\sqrt{2}+s^{2q+2})(1-s^{q+1}\sqrt{2}+s^{2q+2})}\hfill \\ \\ & \ne & \pm \infty ,\hfill \end{array}$$

and

$$\begin{array}{c}\hfill {\displaystyle \underset{s\to \pm \infty }{lim}\sum _{r=0}^{q+1}{s}^{r}l\left(s\right)=\underset{s\to \pm \infty }{lim}\frac{l\left(s\right)}{\frac{1}{{\sum }_{r=0}^{q+1}{s}^r}}}\\ \\ & =& \underset{s\to \pm \infty }{lim}\frac{l^{\prime }\left(s\right)}{-\frac{{\sum }_{r=0}^q(r+1)s^r}{{\left({\sum }_{r=0}^{q+1}{s}^r\right)}^2}}\hfill \\ \\ & =& -\underset{s\to \pm \infty }{lim}\frac{(q+1)\sqrt{2}{s}^q(1+s^{2q+2}){\left({\sum }_{r=0}^{q+1}{s}^r\right)}^2}{\left(1+s^{4q+4}\right)\left({\sum }_{r=0}^q(r+1)s^r\right)}\hfill \\ \\ & \ne & \pm \infty .\hfill \end{array}$$

Consequently

$$\begin{array}{ccc}\hfill -\infty & <& \underset{s\to \pm \infty }{lim}\left(\sum _{r=0}^{q+1}{s}^r\right)h\left(s\right)<\infty ,\hfill \\ \\ \hfill -\infty & <& \underset{s\to \pm \infty }{lim}\left(\sum _{r=0}^{q+1}{s}^r\right)l\left(s\right)<\infty .\hfill \end{array}$$

Hence, there exists a positive constant $C_2$ so that

$$\begin{array}{ccc}\hfill \sum _{r=0}^{q+1}{\left|s\right|}^r\left(\frac{1}{(4q+4)\sqrt{2}}log\frac{1+s^{q+1}\sqrt{2}+s^{2q+2}}{1-s^{q+1}\sqrt{2}+s^{2q+2}}+\frac{1}{(2q+2)\sqrt{2}}arctan\frac{s^{q+1}\sqrt{2}}{1-s^{2q+2}}\right)& \le & C_2,\hfill \end{array}$$

$s\in \mathbb{R}$. Note that $\underset{s\to \pm 1}{lim}l\left(s\right)=\frac{\pi }{2}$ and by [9] (pp. 707, Integral 79), we have

$$\int \frac{dz}{1+z^4}=\frac{1}{4\sqrt{2}}log\frac{1+z\sqrt{2}+z^2}{1-z\sqrt{2}+z^2}+\frac{1}{2\sqrt{2}}arctan\frac{z\sqrt{2}}{1-z^2}.$$

Let

$$Q\left(s\right)=\frac{s^q}{(1+s^{4q+4})},s\in \mathbb{R},$$

and

$$g_1(t,x,y)=Q\left(t\right)Q\left(x\right),t\in [0,\infty ),x\in \mathbb{R}.$$

Then there exists a constant $C>0$ such that

$$\begin{array}{c}\hfill 2^{q+1}(q+1)!(1+t+t^2)\left(\sum _{r=0}^q{\left|x\right|}^r\right){\int }_0^t\left|{\int }_0^x{g}_1(t_1,x_1)d{x}_1\right|d{t}_1\le C,(t,x)\in [0,\infty )\times \mathbb{R}.\end{array}$$

Let

$$g(t,x)=\frac{\mathcal{A}}{C}{g}_1(t,x),(t,x)\in [0,\infty )\times \mathbb{R}.$$

Then

$$\begin{array}{c}\hfill 2^{q+1}q!(1+t+t^2)\left(\sum _{r=0}^q{\left|x\right|}^r\right){\int }_0^t\left|{\int }_0^{x}g(t_1,x_1)d{x}_1\right|d{t}_1\le \mathcal{A},(t,x)\in [0,\infty )\times \mathbb{R}.\end{array}$$

i.e., $\left(Hyp3\right)$ holds. Therefore for the IVP

$$\begin{array}{ccc}\hfill {\partial }_{t}u+\sum _{k=0}^5\sum _{l=0}^{5-k}{\partial }_x\left\{\sum _{m=0}^{5-k}{\partial }_x^m{u}^{10}{\partial }_x^{l}v\right\}+\sum _{k=1}^5\frac{1}{(1+t^{2k})(1+x^{2k})}{\partial }_x^{2k+1}u& =& 0\hfill \\ \\ \hfill {\partial }_{t}v+\sum _{k=0}^5\sum _{l=0}^{5-k}{\partial }_x^k\left\{\sum _{m=0}^{5-k}{\partial }_x^m{v}^{10}{\partial }_x^{l}u\right\}+\sum _{k=1}^5\frac{1}{(2+t^{4k})(3+x^{6k})}{\partial }_x^{2k+1}v& =& 0,\hfill \\ \\ \hfill (t,x)\in [0,\infty )\times \mathbb{R},u(0,x)=\frac{1}{1+x^4},v(0,x)=\frac{1}{3+4x^8},x\in \mathbb{R},& \end{array}$$

are fulfilled all conditions of Theorem 1 and Theorem 2.