Classical Solutions for the Generalized Kawahara-KdV System

In this article we investigate the generalized Kawahara-KdV system. A new topological approach is applied to prove the existence of at least one classical solution and at least two nonnegative classical solutions. The arguments are based upon recent theoretical results.

Keywords:

Subject: Physical Sciences - Mathematical Physics

MSC: 37C25, 47H10

1. Introduction

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

\begin{matrix} \partial_{t} u + \sum_{k = 0}^{N_{1}} \sum_{l = 0}^{N_{1} - k} \partial_{x} \{\sum_{m = 0}^{N_{1} - k} \partial_{x}^{m} u^{p} P_{k, l, m} (\partial_{x}^{l} v)\} + \sum_{k = 1}^{N_{2}} a_{k} (t, x) \partial_{x}^{2 k + 1} u & = & 0 \\ \partial_{t} v + \sum_{k = 0}^{N_{3}} \sum_{l = 0}^{N_{3} - k} \partial_{x}^{k} \{\sum_{m = 0}^{N_{3} - k} \partial_{x}^{m} v^{p} Q_{k, l, m} (\partial_{x}^{l} u)\} + \sum_{k = 1}^{N_{4}} b_{k} (t, x) \partial_{x}^{2 k + 1} v & = & 0, \\ (t, x) \in [0, \infty) \times R, u (0, x) = u_{0} (x), v (0, x) = v_{0} (x), x \in R, \end{matrix}

(1)

where

(Hyp1): $u_{0}, v_{0} \in C^{q} (R)$ , $0 \leq u_{0}, v_{0} \leq B$ on $R$ for some positive constant $B > 1$ , $a_{j}$ , $b_{k} \in C ([0, \infty) \times R)$ , $0 \leq | a_{j} |, | b_{k} | \leq B$ on $[0, \infty) \times R$ , $j = 1, \dots, N_{2}$ , $k = 1, \dots, N_{4}$ ,

$\begin{matrix} P_{k, l, m} (z) & = & \sum_{r = 0}^{N_{5}} c_{k, l, m, r} (t, x) z^{r}, \\ Q_{k, l, m} (z) & = & \sum_{r = 0}^{N_{6}} d_{k, l, m, r} (t, x) z^{r}, (t, x) \in [0, \infty) \times R, z \in R, \end{matrix}$

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

$0 \leq |\partial_{x}^{p_{1}} c_{k, l, m, j}|, |\partial_{x}^{p_{1}} d_{k, l, m, r}| \leq B,$

on $[0, \infty) \times R$ , $j = 1, \dots, N_{5}$ , $r = 1, \dots, N_{6}$ , $p_{1} = 1, \dots, N_{1}$ , p, $N_{1}, N_{2}, N_{3}, N_{4}, N_{5}, N_{6} \in N$ ,

$q = max {N_{1}, 2 N_{2} + 1, N_{3}, 2 N_{4} + 1} .$

Kondo and Pes [1] proved this system is local well-posedness in analytic Gevrey spaces

G^{σ, s} (R)

with

s \geq 2 N + 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_{ℓ = 0}^{N_{1} - k} \partial_{x}^{k} \{\sum_{m = 0}^{N_{1} - k} \partial_{x}^{m} u^{p} P_{k, ℓ, m} (\partial_{x}^{ℓ} v)\} .

N_{1} = 1

, then we have that this term is equal to

\begin{matrix} u^{p} P_{0, 0, 0} (v) + \partial_{x} u^{p} P_{0, 0, 1} (v) + u^{p} P_{0, 1, 0} (\partial_{x} v) + \partial_{x} u^{p} P_{0, 1, 1} (\partial_{x} v) \\ + \partial_{x} [u^{p} P_{1, 0, 0} (v) + \partial_{x} u^{p} P_{1, 0, 1} (v)] . \end{matrix}

(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} (x) = x^{2}, a_{1} = 1

and doing the same choices to the polynomial

Q_{k, ℓ, 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])

\{\begin{matrix} \partial_{t} u + \partial_{x}^{3} u + \partial_{x} (u v^{2}) = 0, \\ \partial_{t} v + \partial_{x}^{3} v + \partial_{x} (v u^{2}) = 0, \\ u (x, 0) = u_{0} (x), v (x, 0) = v_{0} (x) . \end{matrix}

Considering in (2) the case when

p = q \in N

and

P_{1, 0, 0} (x) = 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

\{\begin{matrix} \partial_{t} u + \partial_{x}^{3} u + \partial_{x} (u^{q} v^{q + 1}) = 0, \\ \partial_{t} v + \partial_{x}^{3} v + \partial_{x} (v^{q} u^{q + 1}) = 0, \\ u (x, 0) = u_{0} (x), v (x, 0) = v_{0} (x) . \end{matrix}

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{matrix} u & P_{0, 0, 0} (v) + \partial_{x} u P_{0, 0, 1} (v) + \partial_{x}^{2} u P_{0, 0, 2} (v) \\ + u P_{0, 1, 0} (\partial_{x} v) + \partial_{x} u P_{0, 1, 1} (\partial_{x} v) + \partial_{x}^{2} u P_{0, 1, 2} (\partial_{x} v) \\ + u P_{0, 2, 0} (\partial_{x}^{2} v) + \partial_{x} u P_{0, 2, 1} (\partial_{x}^{2} v) + \partial_{x}^{2} u P_{0, 2, 2} (\partial_{x}^{2} v) \\ + \partial_{x} [u P_{1, 0, 0} (v) + \partial_{x} u P_{1, 0, 1} (v) + \partial_{x}^{2} u P_{1, 0, 2} (v) \\ + u P_{1, 1, 0} (\partial_{x} v) + \partial_{x} u P_{1, 1, 1} (\partial_{x} v) + \partial_{x}^{2} u P_{1, 1, 2} (\partial_{x} v)] \\ + \partial_{x}^{2} [u P_{2, 0, 0} (v) + \partial_{x} u P_{2, 0, 1} (v) + \partial_{x}^{2} u P_{2, 0, 2} (v)] . \end{matrix}

Observe that, changing v by u, and considering again all identical null polynomials, except

P_{0, 0, 1} (x) = 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

((H y p 1)

. Then the IVP (1) has at least one solution

(u, v) \in {(C^{1} ([0, \infty), C^{q} (R)))}^{2} .

Theorem 2.

Suppose

((H y p 1)

. Then the IVP (1) has at least two nonnegative solutions

(u_{1}, v_{1}), (u_{2}, v_{2}) \in {(C^{1} ([0, \infty), C^{q} (R)))}^{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

B > 0

E

be a Banach space and

X = {x \in E : ∥ x ∥ \leq B}

. Let also,

T x = - ϵ x

x \in X

S : X \to E

is a continuous,

(I - S) (X)

resides in a compact subset of

E

and

{x \in E : x = λ (I - S) x, ∥ x ∥ = B} = \emptyset, \forall λ \in (0, \frac{1}{ϵ}) .

(3)

Then there exists

x^{*} \in X

so that

T x^{*} + S x^{*} = 0 .

Proof.

Define

r (- \frac{1}{ϵ} x) = \{\begin{matrix} - \frac{1}{ϵ} x if ∥ x ∥ \leq B ϵ \\ \frac{B x}{∥ x ∥} if ∥ x ∥ > B ϵ . \end{matrix}

Then

r (- \frac{1}{ϵ} (I - S)) : X \to X

is continuous and compact. Hence and the Schauder fixed point theorem, it follows that there exists

x^{*} \in X

so that

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

Assume that

- \frac{1}{ϵ} (I - S) x^{*} \notin X

. Then

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

and

x^{*} = \frac{B}{∥ (I - S) x^{*} ∥} (I - S) x^{*} = r (- \frac{1}{ϵ} (I - S) x^{*}),

and hence,

∥ x^{*} ∥ = B

. This contradicts with (3). Therefore

- \frac{1}{ϵ} (I - S) x^{*} \in X

and

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

- ϵ x^{*} + S x^{*} = x^{*},

T x^{*} + S x^{*} = x^{*} .

This completes the proof. □

Let

X

be a real Banach space.

Definition 1.

A mapping

K : X \to 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

Ω_{X}

be the class of all bounded sets of

X

. The Kuratowski measure of noncompactness

α : Ω_{X} \to [0, \infty)

is defined by

α (Y) = inf \{δ > 0 : Y = ⋃_{j = 1}^{m} Y_{j} and diam (Y_{j}) \leq δ, j = 1, \dots, m\},

where

diam (Y_{j}) = {sup {∥ x - y ∥}_{X} : x, y \in Y_{j}}

is the diameter of

Y_{j}

j = 1, \dots, m

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

Definition 3.

A mapping

K : X \to X

is said to be l-set contraction if it is continuous, bounded and there exists a constant

l \geq 0

such that

α (K (Y)) \leq l α (Y),

for any bounded set

Y \subset X

. The mapping

K

is said to be a strict set contraction if

l < 1

Obviously, if

K : X \to X

is a completely continuous mapping, then

K

is 0-set contraction (see [7], pp. 264).

Definition 4.

Let

X

and

Y

be real Banach spaces. A mapping

K : X \to Y

is said to be expansive if there exists a constant

h > 1

such that

{∥ K x - K y ∥}_{Y} \geq h {∥ x - y ∥}_{X},

for any

x, y \in X

Definition 5.

A closed, convex set

P

X

is said to be cone if

1.: $α x \in P$ for any $α \geq 0$ and for any $x \in P$ ,
2.: $x, - x \in P$ implies $x = 0$ .

Denote

P^{*} = P ∖ {0}

Lemma 4.

Let

X

be a closed convex subset of a Banach space

E

and

U \subset X

a bounded open subset with

0 \in U .

Assume there exists

ε > 0

small enough and that

K : \bar{U} \to X

is a strict k-set contraction that satisfies the boundary condition:

K x \notin {x, λ x} f o r a l l x \in \partial U a n d λ \geq 1 + ε .

Then the fixed point index

i (K, U, X) = 1 .

Proof.

Consider the homotopic deformation

H : [0, 1] \times \bar{U} \to X

defined by

H (t, x) = \frac{1}{ε + 1} t K x .

The operator

H

is continuous and uniformly continuous in t for each

x,

and the mapping

H (t, .)

is a strict set contraction for each

t \in [0, 1]

. In addition,

H (t, .)

has no fixed point on

\partial U

. On the contrary,

• If

t = 0

, there exists some

x_{0} \in \partial U

such that

x_{0} = 0

, contradicting

x_{0} \in U .

• If

t \in (0, 1]

, there exists some

x_{0} \in P \cap \partial U

such that

\frac{1}{ε + 1} t K x_{0} = x_{0}

; then

K x_{0} = \frac{1 + ε}{t} x_{0}

with

\frac{1 + ε}{t} \geq 1 + ε,

contradicting the assumption. From the invariance under homotopy and the normalization properties of the index, we deduce

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

New, we show that

i (K, U, X) = i (\frac{1}{ε + 1} K, U, X) .

We have

\frac{1}{ε + 1} K x \neq x, \forall x \in \partial U .

(4)

Then there exists

γ > 0

such that

∥ x - \frac{1}{ε + 1} K x ∥ \geq γ, \forall x \in \partial U .

In other hand, we have

\frac{1}{ϵ + 1} K x \to K x

ϵ \to 0,

for

x \in \bar{U} .

So for

ε

small enough

∥ K x - \frac{1}{ε + 1} K x ∥ < \frac{γ}{2}, \forall x \in \partial U .

Define the convex deformation

G : [0, 1] \times \bar{U} \to X

G (t, x) = t K x + (1 - t) \frac{1}{ε + 1} 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}{ε + 1} (1 - t) < t + 1 - t = 1

).In addition,

G (t, .)

has no fixed point on

\partial U

. In fact, for all

x \in \partial U

, we have

\begin{matrix} ∥ x - G (t, x) ∥ & = & ∥ x - t K x - (1 - t) \frac{1}{ε + 1} K x ∥ \\ \geq & ∥ x - \frac{1}{ε + 1} K x ∥ - t ∥ K x - \frac{1}{ε + 1} K x ∥ \\ > & γ - \frac{γ}{2} > \frac{γ}{2} . \end{matrix}

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

□

Proposition 5.

Let

P

be a cone in a Banach space

E

. Let also,

U

be a bounded open subset of

P

with

0 \in U .

Assume that

T : Ω \subset P \to E

is an expansive mapping with constant

h > 1,

S : \bar{U} \to E

is a l-set contraction with

0 \leq l < h - 1

, and

S (\bar{U}) \subset (I - T) (Ω) .

If there exists

ε \geq 0

such that

S x \notin {(I - T) (x), (I - T) (λ x)} f o r a l l x \in \partial U \cap Ω a n d λ \geq 1 + ε,

then the fixed point index

i_{*} (T + S, U \cap Ω, P) = 1 .

Proof.

The mapping

{(I - T)}^{- 1} S : \bar{U} \to P

is a strict set contraction and it is readily seen that the following condition is satisfied

{(I - T)}^{- 1} S x \notin {x, λ x} for all x \in \partial U and λ \geq 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

P

be a cone of a Banach space

E

; Ω a subset of

P

and

U_{1}, U_{2} a n d U_{3}

three open bounded subsets of

P

such that

{\bar{U}}_{1} \subset {\bar{U}}_{2} \subset U_{3}

and

0 \in U_{1} .

Assume that

T : Ω \to P

is an expansive mapping with constant

h > 1,

S : {\bar{U}}_{3} \to E

is a k-set contraction with

0 \leq k < h - 1

and

S ({\bar{U}}_{3}) \subset (I - T) (Ω) .

Suppose that

(U_{2} ∖ {\bar{U}}_{1}) \cap Ω \neq \emptyset, (U_{3} ∖ {\bar{U}}_{2}) \cap Ω \neq \emptyset,

and there exists

u_{0} \in P^{*}

such that the following conditions hold:

(i): $S x \neq (I - T) (x - λ u_{0}),$ for all $λ > 0$ and $x \in \partial U_{1} \cap (Ω + λ u_{0}),$
(ii): there exists $ϵ \geq 0$ such that $S x \neq (I - T) (λ x),$ for all $λ \geq 1 + ϵ, x \in \partial U_{2}$ and $λ x \in Ω$ ,
(iii): $S x \neq (I - T) (x - λ u_{0}),$ for all $λ > 0$ and $x \in \partial U_{3} \cap (Ω + λ u_{0}) .$

Then

T + S

has at least two non-zero fixed points

x_{1}, x_{2} \in P

such that

x_{1} \in \partial U_{2} \cap Ω a n d x_{2} \in ({\bar{U}}_{3} ∖ {\bar{U}}_{2}) \cap Ω,

x_{1} \in (U_{2} ∖ U_{1}) \cap Ω a n d x_{2} \in ({\bar{U}}_{3} ∖ {\bar{U}}_{2}) \cap Ω .

Proof.

S x = (I - T) x

for

x \in \partial U_{2} \cap Ω

, then we get a fixed point

x_{1} \in \partial U_{2} \cap Ω

of the operator

T + S

. Suppose that

S x \neq (I - T) x

for any

x \in \partial U_{2} \cap Ω

. Without loss of generality, assume that

T x + S x \neq x o n \partial U_{1} \cap Ω a n d T x + S x \neq x o n \partial U_{3} \cap Ω

, otherwise the conclusion has been proved. By [8] and Proposition 5, we have

i_{*} (T + S, U_{1} \cap Ω, P) = i_{*} (T + S, U_{3} \cap Ω, P) = 0 a n d i_{*} (T + S, U_{2} \cap Ω, P) = 1 .

The additivity property of the index yields

i_{*} (T + S, (U_{2} ∖ {\bar{U}}_{1}) \cap Ω, P) = 1 a n d i_{*} (T + S, (U_{3} ∖ {\bar{U}}_{2}) \cap Ω, P) = - 1 .

Consequently, by the existence property of the index,

T + S

has at least two fixed points

x_{1} \in (U_{2} ∖ U_{1}) \cap Ω a n d x_{2} \in ({\bar{U}}_{3} ∖ {\bar{U}}_{2}) \cap Ω .

□

3. Proof of Theorem 1

Let

X_{1} = C^{1} ([0, \infty), C^{q} (R))

be endowed with the norm

\begin{matrix} {∥ u ∥}_{1} & = & max {sup_{(t, x) \in [0, \infty) \times R} | u (t, x) |, sup_{(t, x) \in [0, \infty) \times R} | \partial_{t} u (t, x) |, \\ sup_{(t, x) \in [0, \infty) \times R} | \partial_{x}^{j} u (t, x) |, j \in {1, \dots, q}}, \end{matrix}

provided it exists. Define

X = X_{1} \times X_{1}

with the norm

∥ (u, v) ∥ = max {∥ u ∥_{1}, ∥ v ∥_{1}} .

For

(u, v) \in X

, define

\begin{matrix} Q_{1} (u, v) (t, x) & = & \sum_{k = 0}^{N_{1}} \sum_{l = 0}^{N_{1} - k} \partial_{x} \{\sum_{m = 0}^{N_{1} - k} \partial_{x}^{m} u^{p} P_{k, l, m} (\partial_{x}^{l} v)\}, \\ Q_{2} (u, v) (t, x) & = & \sum_{k = 0}^{N_{3}} \sum_{l = 0}^{N_{3} - k} \partial_{x}^{k} \{\sum_{m = 0}^{N_{3} - k} \partial_{x}^{m} v^{p} Q_{k, l, m} (\partial_{x}^{l} u)\}, (t, x) \in [0, \infty) \times R . \end{matrix}

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

\begin{matrix} \partial_{t} u + Q_{1} (u, v) + \sum_{k = 1}^{N_{2}} a_{k} (t, x) + \partial_{x}^{2 k + 1} u & = & 0, \\ \partial_{t} v + Q_{2} (u, v) + \sum_{k = 1}^{N_{4}} b_{k} (t, x) \partial_{x}^{2 k + 1} v & = & 0, (t, x) \in [0, \infty) \times R, \\ u (0, x) = u_{0} (x), v (0, x) & = & v_{0} (x), x \in R . \end{matrix}

(5)

Let

\begin{matrix} C_{1} & = & {\sum_{k = 0}^{N_{1}} \sum_{l = 0}^{N_{1} - k} \sum_{m = 0}^{N_{1} - k} \sum_{r = 0}^{k} (\begin{matrix} k \\ r \end{matrix}) {((k - r + m)!)}^{2} p! B^{p} \sum_{j = 0}^{N_{5}} \sum_{i = 0}^{r} (\begin{matrix} r \\ i \end{matrix}) {(i!)}^{2} j! B^{j + 1}, \\ \sum_{k = 0}^{N_{3}} \sum_{l = 0}^{N_{3} - k} \sum_{m = 0}^{N_{3} - k} \sum_{r = 0}^{k} (\begin{matrix} k \\ r \end{matrix}) {((k - r + m)!)}^{2} p! B^{p} \sum_{j = 0}^{N_{6}} \sum_{i = 0}^{r} (\begin{matrix} r \\ i \end{matrix}) {(i!)}^{2} j! B^{j + 1}} . \end{matrix}

Lemma 7.

Suppose

((H y p 1)

. If

(u, v) \in X

and

∥ (u, v) ∥ \leq B

, then

| Q_{1} (u, v) (t, x) |, | Q_{2} (u, v) (t, x) | \leq C_{1}, (t, x) \in [0, \infty) \times R .

Proof.

We have

\begin{matrix} \partial_{x} \{\sum_{m = 0}^{N_{1} - k} \partial_{x}^{m} u^{p} P_{k, l, m} (\partial_{x}^{l} v)\} \\ = & \sum_{m = 0}^{N_{1} - k} \partial_{x}^{k} (\partial_{x}^{m} u^{p} P_{k, l, m} (\partial_{x}^{l} v)) \\ = & \sum_{m = 0}^{N_{1} - k} \sum_{r = 0}^{k} (\begin{matrix} k \\ r \end{matrix}) \partial_{x}^{k - r + m} u^{p} \partial_{x}^{r} P_{k, l, m} (\partial_{x}^{l} v) \\ = & \sum_{m = 0}^{N_{1} - k} \sum_{r = 0}^{k} (\begin{matrix} k \\ r \end{matrix}) \partial_{x}^{k - r + m} u^{p} \partial_{x}^{r} \sum_{j = 0}^{N_{5}} c_{k, l, m, j} {(\partial_{x}^{l} v)}^{j} \\ = & \sum_{m = 0}^{N_{1} - k} \sum_{r = 0}^{k} (\begin{matrix} k \\ r \end{matrix}) \partial_{x}^{k - r + m} u^{p} \sum_{j = 0}^{N_{5}} \sum_{i = 0}^{r} (\begin{matrix} r \\ i \end{matrix}) \partial_{x}^{r - i} c_{k, l, m, j} \partial_{x} î {(\partial_{x}^{l} v)}^{j}, \end{matrix}

k \in N

0 \leq k \leq N_{1}

. Since

B > 1

, we have

|\partial_{x}^{r_{1}} u^{r_{2}}| \leq {(r_{1}!)}^{2} r_{2}! B^{r_{2}},

for any

r_{1}, r_{2} \in N

r_{1} \leq q

. Then

\begin{matrix} |\partial_{x} \{\sum_{m = 0}^{N_{1} - k} \partial_{x}^{m} u^{p} P_{k, l, m} (\partial_{x}^{l} v)\}| \\ \leq & \sum_{m = 0}^{N_{1} - k} \sum_{r = 0}^{k} (\begin{matrix} k \\ r \end{matrix}) |\partial_{x}^{k - r + m} u^{p}| \sum_{j = 0}^{N_{5}} \sum_{i = 0}^{r} (\begin{matrix} r \\ i \end{matrix}) |\partial_{x}^{r - i} c_{k, l, m, j}| |\partial_{x}^{i} {(\partial_{x}^{l} v)}^{j}| \\ \leq & \sum_{m = 0}^{N_{1} - k} \sum_{r = 0}^{k} (\begin{matrix} k \\ r \end{matrix}) {((k - r + m)!)}^{2} p! B^{p} \sum_{j = 0}^{N_{5}} \sum_{i = 0}^{r} (\begin{matrix} r \\ i \end{matrix}) {(i!)}^{2} j! B^{j + 1}, \end{matrix}

[0, \infty) \times R

0 \leq k \leq N_{1}

, and then

\begin{matrix} | Q_{1} (u, v) | & \leq & \sum_{k = 0}^{N_{1}} \sum_{l = 0}^{N_{1} - k} \sum_{m = 0}^{N_{1} - k} \sum_{r = 0}^{k} (\begin{matrix} k \\ r \end{matrix}) {((k - r + m)!)}^{2} p! B^{p} \sum_{j = 0}^{N_{5}} \sum_{i = 0}^{r} (\begin{matrix} r \\ i \end{matrix}) {(i!)}^{2} j! B^{j + 1} \\ \leq & C_{1}, \end{matrix}

[0, \infty) \times R

. As above,

| Q_{2} (u, v) | \leq C_{1},

[0, \infty) \times R

. This completes the proof. □

For

(u, v) \in X

, define the operators

\begin{matrix} S_{1}^{1} (u, v) (t, x) & = & u (t, x) - u_{0} (x) \\ + \int_{0}^{t} (Q_{1} (u, v) (s, x) + \sum_{k = 1}^{N_{2}} a_{k} (s, x) \partial_{x}^{2 k + 1} u (s, x)) d s, \\ S_{1}^{2} (u, v) (t, x) & = & v (t, x) - v_{0} (x) \\ + \int_{0}^{t} (Q_{2} (u, v) (s, x) + \sum_{k = 1}^{N_{4}} b_{k} (s, x) \partial_{x}^{2 k + 1} v (s, x)) d s, \\ S (u, v) (t, x) & = & (S_{1}^{1} (u, v) (t, x), S_{1}^{2} (u, v) (t, x)), \end{matrix}

(t, x) \in [0, \infty) \times R

Lemma 8.

Suppose

((H y p 1)

. If

(u, v) \in X

satisfies the equation

S_{1} (u, v) (t, x) = 0, (t, x) \in [0, \infty) \times R,

then

(u, v)

is a solution to the IVP (1).

Proof.

We have

\begin{matrix} 0 & = & u (t, x) - u_{0} (x) \\ + \int_{0}^{t} (Q_{1} (u, v) (s, x) + \sum_{k = 1}^{N_{2}} a_{k} (s, x) \partial_{x}^{2 k + 1} u (s, x)) d s, \\ 0 & = & v (t, x) - v_{0} (x) \\ + \int_{0}^{t} (Q_{2} (u, v) (s, x) + \sum_{k = 1}^{N_{4}} b_{k} (s, x) \partial_{x}^{2 k + 1} v (s, x)) d s, \end{matrix}

(6)

(t, x) \in [0, \infty) \times R

, which we differentiate with respect to t and we get (5). We put

t = 0

in (6) and we obtain

\begin{matrix} 0 & = & u (0, x) - u_{0} (x) \\ 0 & = & v (0, x) - v_{0} (x), x \in R . \end{matrix}

Thus,

(u, v)

is a solution to the IVP (1). This completes the proof. □

Let

B_{1} = max {2 B, C_{1} + N_{2} B^{2}, C_{1} + N_{4} B^{2}} .

Lemma 9.

Suppose

((H y p 1)

. If

(u, v) \in X

and

∥ (u, v) ∥ \leq B

, then

\begin{matrix} | S_{1}^{1} (u, v) (t, x) | & \leq & B_{1} (1 + t), \\ | S_{1}^{2} (u, v) (t, x) | & \leq & B_{1} (1 + t), (t, x) \in [0, \infty) \times R . \end{matrix}

Proof.

We have

\begin{matrix} | S_{1}^{1} (u, v) (t, x) | & = & | u (t, x) - u_{0} (x) \\ + \int_{0}^{t} (Q_{1} (u, v) (s, x) + \sum_{k = 1}^{N_{2}} a_{k} (s, x) \partial_{x}^{2 k + 1} u (s, x)) d s | \\ \leq & | u (t, x) | + | u_{0} (x) | \\ + \int_{0}^{t} (| Q_{1} (u, v) (s, x) | + \sum_{k = 1}^{N_{2}} | a_{k} (s, x) | | \partial_{x}^{2 k + 1} u (s, x) |) d s \\ \leq & 2 B + \int_{0}^{t} (C_{1} + N_{2} B^{2}) d s \\ \leq & B_{1} (1 + t), (t, x) \in [0, \infty) \times R . \end{matrix}

As above,

| S_{1}^{2} (u, v) (t, x) | \leq B_{1} (1 + t), (t, x) \in [0, \infty) \times R .

This completes the proof. □

Below, suppose that

(Hyp2): there exist a function $g \in C ([0, \infty) \times R)$ , $g > 0$ on $(0, \infty) \times (R ∖ {0})$ , $g (0, x) = g (t, 0) = 0$ , $(t, x) \in [0, \infty) \times R$ , and a positive constant $A$ such that

$q! \cdot 2^{q + 1} (1 + t + t^{2}) {(1 + | x | + \dots + | x |}^{q}) \int_{0}^{t} | \int_{0}^{x} g (t_{1}, x_{1}) d x_{1} | d t_{1} \leq A,$

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

In the last section we will give examples for g and

A

that satisfy

(H y p 2)

. For

(u, v) \in X

, define the operators

\begin{matrix} 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}, \\ 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}, \\ 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 R . \end{matrix}

Lemma 10.

Suppose

((H y p 1)

and

(H y p 2)

. If

(u, v) \in X

satisfies the equation

S_{2} (u, v) (t, x) = 0, (t, x) \in [0, \infty) \times 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 R .

Hence,

S_{1}^{1} (u, v) (t, x) = S_{1}^{2} (u, v) (t, x) = 0, (t, x) \in (0, \infty) \times (R ∖ {0}) .

Since

S_{1}^{1} (u, v) (\cdot, \cdot)

and

S_{1}^{2} (u, v) (\cdot, \cdot)

are continuous functions on

[0, \infty) \times R

, we have

\begin{matrix} 0 & = & S_{1}^{1} (u, v) (0, x) = S_{1}^{2} (u, v) (0, x) \\ = & lim_{t \to 0} S_{1}^{1} (u, v) (t, x) = lim_{t \to 0} S_{1}^{2} (u, v) (t, x) \\ = & lim_{x \to 0} S_{1}^{1} (u, v) (t, x) = lim_{x \to 0} S_{1}^{2} (u, v) (t, x) \\ = & S_{1}^{1} (u, v) (t, 0) = S_{1}^{2} (u, v) (t, 0), (t, x) \in [0, \infty) \times R . \end{matrix}

Therefore

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

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

Lemma 11.

Suppose

((H y p 1)

and

(H y p 2)

. If

(u, v) \in X

∥ (u, v) ∥ \leq B

, then

∥ S_{2} (u, v) ∥ \leq A B_{1} .

Proof.

We will use the inequality

{(z + w)}^{r} \leq 2^{r} (z^{r} + w^{r})

w, z, q \geq 0

. We have

\begin{matrix} | 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} | \\ \leq & \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} \\ \leq & B_{1} \int_{0}^{t} | \int_{0}^{x} (t - t_{1}) (1 + t_{1}) {| x - x_{1} |}^{q} g (t_{1}, x_{1}) d x_{1} | d t_{1} \\ \leq & B_{1} t (1 + t) 2^{q + 1} {| x |}^{q} \int_{0}^{t} | \int_{0}^{x} g (t_{1}, x_{1}) d x_{1} | d t_{1} \\ \leq & A B_{1}, (t, x) \in [0, \infty) \times R, \end{matrix}

and

\begin{matrix} | \partial_{t} S_{2}^{1} (u, v) (t, x) | & = & | \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} | \\ \leq & \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} \\ \leq & B_{1} \int_{0}^{t} | \int_{0}^{x} (1 + t_{1}) {| x - x_{1} |}^{q} g (t_{1}, x_{1}) d x_{1} | d t_{1} \\ \leq & B_{1} (1 + t) 2^{q + 1} {| x |}^{q} \int_{0}^{t} | \int_{0}^{x} g (t_{1}, x_{1}) d x_{1} | d t_{1} \\ \leq & A B_{1}, (t, x) \in [0, \infty) \times R, \end{matrix}

and

\begin{matrix} | \partial_{x} S_{2}^{1} (u, v) (t, x) | & = & q | \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} | \\ \leq & q \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} \\ \leq & q B_{1} \int_{0}^{t} | \int_{0}^{x} (t - t_{1}) (1 + t_{1}) {| x - x_{1} |}^{q - 1} g (t_{1}, x_{1}) d x_{1} | d t_{1} \\ \leq & q B_{1} t (1 + t) 2^{q} {| x |}^{q - 1} \int_{0}^{t} | \int_{0}^{x} g (t_{1}, x_{1}) d x_{1} | d t_{1} \\ \leq & A B_{1}, (t, x) \in [0, \infty) \times R, \end{matrix}

and so on. As above,

\begin{matrix} | S_{2}^{2} (u, v) (t, x) | \leq A B_{1}, | \partial_{t} S_{2}^{2} (u, v) (t, x) | & \leq & A B_{1}, \\ | \partial_{x}^{j} S_{2}^{2} (u, v) (t, x) | \leq A B_{1}, j \in {1, \dots, q} . \end{matrix}

(t, x) \in [0, \infty) \times R

. Thus,

∥ S_{2} (u, v) ∥ \leq A B_{1} .

This completes the proof. □

Below, suppose

(Hyp3): $ϵ \in (0, 1)$ , $A$ , $B$ and $B_{1}$ satisfy the inequalities $ϵ B_{1} (1 + A) < 1$ and $A B_{1} < B$ .

Let

\tilde{\tilde{\tilde{Y}}}

denote the set of all equi-continuous families in

X

with respect to the norm

∥ \cdot ∥

. Let also,

\tilde{\tilde{Y}} = \bar{\tilde{\tilde{\tilde{Y}}}}

be the closure of

\tilde{\tilde{\tilde{Y}}}

\tilde{Y} = \tilde{\tilde{Y}} \cup {(u_{0}, v_{0})}

Y = {(u, v) \in \tilde{Y} : (u, v) \geq 0, ∥ (u, v) ∥ \leq B} .

Note that

Y

is a compact set in

X

. For

(u, v) \in X

, define the operators

\begin{matrix} T (u, v) (t, x) & = & - ϵ (u, v) (t, x), \\ S (u, v) (t, x) & = & (u, v) (t, x) + ϵ (u, v) (t, x) + ϵ S_{2} (u, v) (t, x), (t, x) \in [0, \infty) \times R . \end{matrix}

For

(u, v) \in Y

, using Lemma 10, we have

\begin{matrix} ∥ (I - S) (u, v) ∥ & = & ∥ ϵ (u, v) - ϵ S_{2} (u, v) ∥ \\ \leq & ϵ ∥ (u, v) ∥ + ϵ ∥ S_{2} (u, v) ∥ \\ \leq & ϵ B_{1} + ϵ A B_{1} \\ = & ϵ B_{1} (1 + A) \\ < & B . \end{matrix}

Thus,

S : Y \to X

is continuous and

(I - S) (Y)

resides in a compact subset of

X

. Now, suppose that there is a

(u, v) \in X

so that

∥ (u, v) ∥ = B

and

(u, v) = λ (I - S) (u, v),

\frac{1}{λ} (u, v) = (I - S) (u, v) = - ϵ (u, v) - ϵ S_{2} (u, v),

(\frac{1}{λ} + ϵ) (u, v) = - ϵ S_{2} (u, v),

for some

λ \in (0, \frac{1}{ϵ})

. Hence,

∥ S_{2} (u, v) ∥ \leq A B_{1} < B

ϵ B < (\frac{1}{λ} + ϵ) B = (\frac{1}{λ} + ϵ) ∥ (u, v) ∥ = ϵ ∥ S_{2} (u, v) ∥ < ϵ B,

which is a contradiction. Hence and Theorem 3, it follows that the operator

T + S

has a fixed point

(u^{*}, v^{*}) \in Y

. Therefore

\begin{matrix} (u^{*}, v^{*}) (t, x) & = & T (u^{*}, v^{*}) (t, x) + S (u^{*}, v^{*}) (t, x) \\ = & - ϵ (u^{*}, v^{*}) (t, x) + (u^{*}, v^{*}) (t, x) + ϵ (u^{*}, v^{*}) (t, x) + ϵ S_{2} (u^{*}, v^{*}) (t, x), \end{matrix}

(t, x) \in [0, \infty) \times R

, whereupon

0 = S_{2} (u^{*}, v^{*}) (t, x), (t, x) \in [0, \infty) \times 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

X

be the space used in the previous section. Suppose

(Hyp4): Let $m > 0$ be large enough and $A$ , $B$ , r, L, $R_{1}$ be positive constants that satisfy the following conditions

$r < L < R_{1} \leq B, ϵ > 0, R_{1} > (\frac{2}{5 m} + 1) L,$

$A B_{1} < \frac{L}{5} .$

Let

\tilde{P} = {(u, v) \in X : (u, v) \geq 0 on [0, \infty) \times R} .

With

P

we will denote the set of all equi-continuous families in

\tilde{P}

. For

(u, v) \in X

, define the operators

\begin{matrix} T_{1} (u, v) (t, x) & = & (1 + m ϵ) (u, v) (t, x) - (ϵ \frac{L}{10}, ϵ \frac{L}{10}), \\ S_{3} (u, v) (t, x) & = & - ϵ S_{2} (u, v) (t, x) - m ϵ (u, v) (t, x) - (ϵ \frac{L}{10}, ϵ \frac{L}{10}), \end{matrix}

t \in [0, \infty)

. Note that any fixed point

(u, v) \in X

of the operator

T_{1} + S_{3}

is a solution to the IVP (1). Define

\begin{matrix} U_{1} & = & P_{r} = {(u, v) \in P : ∥ (u, v) ∥ < r}, \\ U_{2} & = & P_{L} = {(u, v) \in P : ∥ (u, v) ∥ < L}, \\ U_{3} & = & P_{R_{1}} = {(u, v) \in P : ∥ (u, v) ∥ < R_{1}}, \\ R_{2} & = & R_{1} + \frac{A}{m} B_{1} + \frac{L}{5 m}, \\ Ω & = & \bar{P_{R_{2}}} = {(u, v) \in P : ∥ (u, v) ∥ \leq R_{2}} . \end{matrix}

1.: For $(u_{1}, v_{1}), (u_{2}, v_{2}) \in Ω$ , we have

$\begin{matrix} ∥ T_{1} (u_{1}, v_{1}) - T_{1} (u_{2}, v_{2}) ∥ = (1 + m ε) ∥ (u_{1}, v_{1}) - (u_{2}, v_{2}) ∥, \end{matrix}$

whereupon $T_{1} : Ω \to X$ is an expansive operator with a constant $h = 1 + m ε > 1$ .
2.: For $(u, v) \in {\bar{P}}_{R_{1}}$ , we get

$\begin{matrix} ∥ S_{3} (u, v) ∥ & \leq & ε ∥ S_{2} (u, v) ∥ + m ε ∥ (u, v) ∥ + ε \frac{L}{10} \\ \leq & ε (A B_{1} + m R_{1} + \frac{L}{10}) . \end{matrix}$

Therefore $S_{3} ({\bar{P}}_{R_{1}})$ is uniformly bounded. Since $S_{3} : {\bar{P}}_{R_{1}} \to X$ is continuous, we have that $S_{3} ({\bar{P}}_{R_{1}})$ is equi-continuous. Consequently $S_{3} : {\bar{P}}_{R_{1}} \to X$ is a 0-set contraction.
3.: Let $(u_{1}, v_{1}) \in {\bar{P}}_{R_{1}}$ . Set

$(u_{2}, v_{2}) = (u_{1}, v_{1}) + \frac{1}{m} S_{2} (u_{1}, v_{1}) + (\frac{L}{5 m}, \frac{L}{5 m}) .$

Note that $S_{2} u_{1} + \frac{L}{5} \geq 0$ , $S_{2} v_{1} + \frac{L}{5} \geq 0$ on $[0, \infty) \times R$ . We have $u_{2}, v_{2} \geq 0$ on $[0, \infty) \times R$ and

$\begin{matrix} ∥ (u_{2}, v_{2}) ∥ & \leq & ∥ (u_{1}, v_{1}) ∥ + \frac{1}{m} ∥ S_{2} (u_{1}, v_{1}) ∥ + \frac{L}{5 m} \\ \leq & R_{1} + \frac{A}{m} B_{1} + \frac{L}{5 m} \\ = & R_{2} . \end{matrix}$

Therefore $(u_{2}, v_{2}) \in Ω$ and

$- ε m (u_{2}, v_{2}) = - ε m (u_{1}, v_{1}) - ε S_{2} (u_{1}, v_{1}) - ε (\frac{L}{10}, \frac{L}{10}) - ε (\frac{L}{10}, \frac{L}{10})$

or

$\begin{matrix} (I - T_{1}) (u_{2}, v_{2}) & = & - ε m (u_{2}, v_{2}) + ε (\frac{L}{10}, \frac{L}{10}) \\ = & S_{3} (u_{1}, v_{1}) . \end{matrix}$

Consequently $S_{3} ({\bar{P}}_{R_{1}}) \subset (I - T_{1}) (Ω)$ .
4.: Assume that for any $(u_{0}, v_{0}) \in P^{*}$ there exist $λ \geq 0$ and $(u, v) \in \partial P_{r} \cap (Ω + λ (u_{0}, v_{0}))$ or $v \in \partial P_{R_{1}} \cap (Ω + λ (u_{0}, v_{0}))$ such that

$S_{3} (u, v) = (I - T_{1}) ((u, v) - λ (u_{0}, v_{0})) .$

Then

$- ϵ S_{2} (u, v) - m ϵ (u, v) - ϵ (\frac{L}{10}, \frac{L}{10}) = - m ϵ ((u, v) - λ (u_{0}, v_{0})) + ϵ (\frac{L}{10}, \frac{L}{10}),$

or

$- S_{2} (u, v) = λ m (u_{0}, v_{0}) + (\frac{L}{5}, \frac{L}{5}) .$

Hence,

$∥ S_{2} v ∥ = ∥λ m (u_{0}, v_{0}) + (\frac{L}{5}, \frac{L}{5})∥ > \frac{L}{5} .$

This is a contradiction.
5.: Suppose that for any $ϵ_{1} \geq 0$ small enough there exist a $(u_{1}, v_{1}) \in \partial P_{L}$ and $λ_{1} \geq 1 + ϵ_{1}$ such that $λ_{1} (u_{1}, v_{1}) \in {\bar{P}}_{R_{1}}$ and

$S_{3} (u_{1}, v_{1}) = (I - T_{1}) (λ_{1} (u_{1}, v_{1})) .$

(7)

In particular, for $ϵ_{1} > \frac{2}{5 m}$ , we have $(u_{1}, v_{1}) \in \partial P_{L}$ , $λ_{1} (u_{1}, v_{1}) \in {\bar{P}}_{R_{1}}$ , $λ_{1} \geq 1 + ϵ_{1}$ and (7) holds. Since $(u_{1}, v_{1}) \in \partial P_{L}$ and $λ_{1} (u_{1}, v_{1}) \in {\bar{P}}_{R_{1}}$ , it follows that

$(\frac{2}{5 m} + 1) L < λ_{1} L = λ_{1} ∥ (u_{1}, v_{1}) ∥ \leq R_{1} .$

Moreover,

$- ϵ S_{2} (u_{1}, v_{1}) - m ϵ (u_{1}, v_{1}) - ϵ (\frac{L}{10}, \frac{L}{10}) = - λ_{1} m ϵ (u_{1}, v_{1}) + ϵ (\frac{L}{10}, \frac{L}{10}),$

or

$S_{2} (u_{1}, v_{1}) + (\frac{L}{5}, \frac{L}{5}) = (λ_{1} - 1) m (u_{1}, v_{1}) .$

From here,

$2 \frac{L}{5} \geq ∥S_{2} (u_{1}, v_{1}) + (\frac{L}{5}, \frac{L}{5})∥ = (λ_{1} - 1) m ∥ (u_{1}, v_{1}) ∥ = (λ_{1} - 1) m L,$

and

$\frac{2}{5 m} + 1 \geq λ_{1},$

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

∥ (u_{1}, v_{1}) ∥ = L < ∥ (u_{2}, v_{2}) ∥ < R_{1},

r < ∥ (u_{1}, v_{1}) ∥ < L < ∥ (u_{2}, v_{2}) ∥ < R_{1} .

5. Example

Below, we will illustrate our main results. Let

B = 1

and

R_{1} = 10, L = 5, r = 4, m = 10^{50}, A = \frac{1}{10 B_{1}}, ϵ = \frac{1}{5 B_{1} (1 + A)},

N_{j} = 5

j \in {1, \dots, 4}

p = 10

. Then

A B_{1} = \frac{1}{10} < B, ϵ B_{1} (1 + A) < 1,

i.e.,

(H y p 3)

holds. Next,

r < L < R_{1}, ϵ > 0, R_{1} > (\frac{2}{5 m} + 1) L, A B_{1} < \frac{L}{5} .

i.e.,

(H y p 4)

holds. Take

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

Then

\begin{matrix} h^{'} (s) & = & \frac{2 \sqrt{2} (q + 1) s^{q} (1 - s^{2 q + 2})}{(1 - s^{q + 1} \sqrt{2} + s^{2 q + 2}) (1 - s^{q + 1} \sqrt{2} + s^{2 q + 2})}, \\ l^{'} (s) & = & \frac{(q + 1) \sqrt{2} s^{q} (1 + s^{2 q + 2})}{1 + s^{4 q + 4}}, s \in R, s \neq \pm 1 . \end{matrix}

Therefore

\begin{matrix} lim_{s \to \pm \infty} \sum_{r = 0}^{q + 1} s^{r} h (s) = lim_{s \to \pm \infty} \frac{h (s)}{\frac{1}{\sum_{r = 0}^{l + 1} s^{r}}} \\ = & lim_{s \to \pm \infty} \frac{h^{'} (s)}{- \frac{\sum_{r = 0}^{q} (r + 1) s^{r}}{{(\sum_{r = 0}^{q + 1} s^{r})}^{2}}} \\ = & - lim_{s \to \pm \infty} \frac{2 \sqrt{2} (q + 1) s^{q} (1 - s^{2 q + 2}) {(\sum_{r = 0}^{q + 1} s^{r})}^{2}}{(\sum_{r = 0}^{q} (r + 1) s^{r}) (1 - s^{q + 1} \sqrt{2} + s^{2 q + 2}) (1 - s^{q + 1} \sqrt{2} + s^{2 q + 2})} \\ \neq & \pm \infty, \end{matrix}

and

\begin{matrix} lim_{s \to \pm \infty} \sum_{r = 0}^{q + 1} s^{r} l (s) = lim_{s \to \pm \infty} \frac{l (s)}{\frac{1}{\sum_{r = 0}^{q + 1} s^{r}}} \\ = & lim_{s \to \pm \infty} \frac{l^{'} (s)}{- \frac{\sum_{r = 0}^{q} (r + 1) s^{r}}{{(\sum_{r = 0}^{q + 1} s^{r})}^{2}}} \\ = & - lim_{s \to \pm \infty} \frac{(q + 1) \sqrt{2} s^{q} (1 + s^{2 q + 2}) {(\sum_{r = 0}^{q + 1} s^{r})}^{2}}{(1 + s^{4 q + 4}) (\sum_{r = 0}^{q} (r + 1) s^{r})} \\ \neq & \pm \infty . \end{matrix}

Consequently

\begin{matrix} - \infty & < & lim_{s \to \pm \infty} (\sum_{r = 0}^{q + 1} s^{r}) h (s) < \infty, \\ - \infty & < & lim_{s \to \pm \infty} (\sum_{r = 0}^{q + 1} s^{r}) l (s) < \infty . \end{matrix}

Hence, there exists a positive constant

C_{2}

so that

\begin{matrix} \sum_{r = 0}^{q + 1} {| s |}^{r} (\frac{1}{(4 q + 4) \sqrt{2}} log \frac{1 + s^{q + 1} \sqrt{2} + s^{2 q + 2}}{1 - s^{q + 1} \sqrt{2} + s^{2 q + 2}} + \frac{1}{(2 q + 2) \sqrt{2}} arctan \frac{s^{q + 1} \sqrt{2}}{1 - s^{2 q + 2}}) & \leq & C_{2}, \end{matrix}

s \in R

. Note that

lim_{s \to \pm 1} l (s) = \frac{π}{2}

and by [9] (pp. 707, Integral 79), we have

\int \frac{d z}{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 (s) = \frac{s^{q}}{(1 + s^{4 q + 4})}, s \in R,

and

g_{1} (t, x, y) = Q (t) Q (x), t \in [0, \infty), x \in R .

Then there exists a constant

C > 0

such that

\begin{matrix} 2^{q + 1} (q + 1)! (1 + t + t^{2}) (\sum_{r = 0}^{q} {| x |}^{r}) \int_{0}^{t} | \int_{0}^{x} g_{1} (t_{1}, x_{1}) d x_{1} | d t_{1} \leq C, (t, x) \in [0, \infty) \times R . \end{matrix}

Let

g (t, x) = \frac{A}{C} g_{1} (t, x), (t, x) \in [0, \infty) \times R .

Then

\begin{matrix} 2^{q + 1} q! (1 + t + t^{2}) (\sum_{r = 0}^{q} {| x |}^{r}) \int_{0}^{t} | \int_{0}^{x} g (t_{1}, x_{1}) d x_{1} | d t_{1} \leq A, (t, x) \in [0, \infty) \times R . \end{matrix}

i.e.,

(H y p 3)

holds. Therefore for the IVP

\begin{matrix} \partial_{t} u + \sum_{k = 0}^{5} \sum_{l = 0}^{5 - k} \partial_{x} \{\sum_{m = 0}^{5 - k} \partial_{x}^{m} u^{10} \partial_{x}^{l} v\} + \sum_{k = 1}^{5} \frac{1}{(1 + t^{2 k}) (1 + x^{2 k})} \partial_{x}^{2 k + 1} u & = & 0 \\ \partial_{t} v + \sum_{k = 0}^{5} \sum_{l = 0}^{5 - k} \partial_{x}^{k} \{\sum_{m = 0}^{5 - k} \partial_{x}^{m} v^{10} \partial_{x}^{l} u\} + \sum_{k = 1}^{5} \frac{1}{(2 + t^{4 k}) (3 + x^{6 k})} \partial_{x}^{2 k + 1} v & = & 0, \\ (t, x) \in [0, \infty) \times R, u (0, x) = \frac{1}{1 + x^{4}}, v (0, x) = \frac{1}{3 + 4 x^{8}}, x \in R, \end{matrix}

are fulfilled all conditions of Theorem 1 and Theorem 2.

Funding

This research received no external funding.

Acknowledgments

The authors extend their appreciation to the Deanship of Scientific Research at Imam Mohammad Ibn Saud Islamic University for funding this work through Research Group no. RG-21-09-51

Conflicts of Interest

Authors declare that they have no conflict of interest.

References

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Classical Solutions for the Generalized Kawahara-KdV System

Abstract

1. Introduction

2. Preliminary Results

3. Proof of Theorem 1

4. Proof of Theorem 2

5. Example

Funding

Acknowledgments

Conflicts of Interest

References

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