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BY 4.0 license Open Access Published by De Gruyter Open Access July 1, 2022

Initial-boundary value problem of fifth-order Korteweg-de Vries equation posed on half line with nonlinear boundary values

Xiangqing Zhao, Chengqiang Wang and Jifeng Bao
From the journal Open Mathematics

Abstract

In this paper, we study the initial boundary problem of fifth-order Korteweg-de Vries equation with nonlinear boundary values. First, we establish a so-called sharp boundary trace regularity associated with the linearized fifth-order Korteweg-de Vries equation. Then, aided by the sharp boundary trace regularity, we verify that initial-boundary value problem of fifth-order Korteweg-de Vries equation with nonlinear boundary conditions is locally well-posed when initial and boundary values are properly chosen.

MSC 2010: 35G16; 35Q53

1 Introduction

Korteweg-de Vries equation

t u ± x 3 u + u x u = 0

origins from shallow water waves that are weakly and nonlinearly interacting, and nowadays is extensively applied in ion acoustic waves in plasma, long internal waves in a density-stratified ocean, and acoustic waves on a crystal lattice, etc. However, under certain conditions, the third-order dispersive term is too weak to describe the physical facts, then fifth-order dispersive term is introduced to strength it (if the angle between the propagation direction and the magnetic-acoustic wave in a cold collision-free plasma and the external magnetic field become critical value, then the third-order dispersive term vanishes and is replaced by the fifth-order dispersive term [1]; a fifth-order term was necessary to model capillary-gravity waves for Bond number near 1 3  [2])

t u ± x 5 u + u x u = 0 .

The Cauchy problem for the fifth-order KdV equation has been extensively studied after Kato smoothing effect discovered in the early 1980s, see for example [3,4,5, 6,7,8]. Compared with pure initial value problems, initial boundary value problems posed on part of entire line with boundaries are more applicable to the reality and can provide more accurate data to physical experiments or practical problems. Although there is less research on initial boundary value problems than that on pure initial value problems, nowadays, more and more attention has been paid for initial boundary value problems [9,10,11, 12,13].

Different from the former works (say, for example, [9,10,11, 12,13]) which deal with the linear boundary values, we investigate the well-posedness in the sense of Hadamard of the initial-boundary value problem

(1.1) t u + x 5 u = u x u , x > 0 , t > 0 , u ( x , 0 ) = ϕ ( x ) , x > 0 , x u ( 0 , t ) = h 1 ( t ) , x 2 u ( 0 , t ) = u ( 0 , t ) 2 + h 2 ( t ) , x 4 u ( 0 , t ) = u ( 0 , t ) 2 + h 3 ( t ) , t > 0 ,

with nonlinear boundary feedback (This often happens in application [14,15], or theory, i.e., feedback stabilization [16]). Due to the presence of the nonlinear boundary condition, the Kato smoothing effect is not strong enough to deal with (1.1). Instead, a so-called sharp boundary trace regularity is needed. More precisely, for any h 1 H 0 s + 1 5 , h 2 H 0 s 5 , h 3 H 0 s 2 5 the corresponding solution u of

(1.2) t u + x 5 u = 0 , x > 0 , t > 0 , u ( x , 0 ) = 0 , x > 0 , x u ( 0 , t ) = h 1 ( t ) , x 2 u ( 0 , t ) = h 2 ( t ) , x 4 u ( 0 , t ) = h 3 ( t ) , t > 0 ,

satisfies:

x k u L ( R + ; H 2 k 5 ( 0 , T ) ) , k = 0 , 1 , 2 , 3 , 4 .

Taking account of sharp boundary trace regularity, we define the solution space as

Y T s u C ( 0 , T ; H s ( R + ) ) k = 0 4 u L R + ; H s + 2 k 5 ( 0 , T ) < .

For the vector consists of initial and boundary values ( φ , h ) ( φ , ( h 1 , h 2 , h 3 ) ) , we defined the corresponding function space as

X T s H s ( R + ) × s ( 0 , T ) ,

with

s ( 0 , T ) H 0 s + 1 5 ( 0 , T ) × H 0 s 5 ( 0 , T ) × H 0 s 2 5 ( 0 , T ) .

The main results can be stated as:

Theorem 1.1

Let s [ 0 , 5 ] , T > 0 be given. For any s-compatible ( φ , h ) X T s , f L 1 ( 0 , T ; H s ( R + ) ) , there exists T ( 0 , T ] such that the initial-boundary value problem

(1.3) t u + x 5 u = f , x > 0 , t > 0 , u ( x , 0 ) = ϕ ( x ) , x > 0 , x u ( 0 , t ) = h 1 ( t ) , x 2 u ( 0 , t ) = u ( 0 , t ) 2 + h 2 ( t ) , x 4 u ( 0 , t ) = u ( 0 , t ) 2 + h 3 ( t ) , t > 0 ,

admits a unique solution u Y T s . Moreover, the solution depends Lipschitz continuously on ( φ , h ) and f in the corresponding space.

Remark 1.2

However, initial boundary value problem (1.1) includes both nonlinear damping and nonlinear feedback, Kato smoothing effect, and sharp boundary trace regularity are not enough to deal with both nonlinearities, and Bourgain’s regularity is needed. Thus, the solution space should be taken as:

Y T s u C ( 0 , T ; H s ( R + ) ) X s , b k = 0 4 u L R + ; H s + 2 k 5 ( 0 , T ) < ,

where X s , b is the Bourgain space.

The paper is organized as follows. In Section 2, we will first derive an explicit integral representation of (1.2), then established sharp boundary trace regularity. The proof of our main result in this paper (Theorem 1.1) will be presented in Section 3. Section 4 includes remarks which show the readers all the possible cases can be proved similarly.

2 Linear estimates

In this part, we will establish linear estimate as follows:

Proposition 2.1

Let T > 0 and 0 s 5 be given. For any ϕ H s ( R + ) , h s ( R + ) , f L 1 ( 0 , T ; H s ( R + ) ) , the initial boundary value problem

(2.1) t u + x 5 u = f ( x , t ) , x > 0 , t > 0 , u ( x , 0 ) = ϕ ( x ) , x > 0 , x u ( 0 , t ) = h 1 ( t ) , x 2 u ( 0 , t ) = h 2 ( t ) , x 4 u ( 0 , t ) = h 3 ( t ) , t > 0

admits a solution u C ( 0 , T ; H s ( R + ) ) L 2 ( 0 , T ; H 2 + s ( R + ) ) satisfying

u Y T s C ( ( ϕ , h ) X T s + f L 1 ( 0 , T ; H s ( R + ) ) ) .

2.1 Representation of the solution

Applying Laplace transformation to (2.1) we obtain

(2.2) s u ˆ ( x , s ) + d 5 u ˆ d x 5 ( x , s ) = 0 , d d x u ˆ ( 0 , s ) = h ˆ 1 ( s ) , d 2 d x 2 u ˆ ( 0 , s ) = h ˆ 2 ( s ) , d 4 d x 4 u ˆ ( 0 , s ) = h ˆ 3 ( s ) .

Then, the solution u ˆ ( x , s ) of (2.2) can be written in the form

u ˆ ( x , s ) = j = 1 3 c j ( s ) e λ j ( s ) x ,

where λ j ( s ) , j = 1 , 2 , 3 , are the solutions of the characteristic equation:

(2.3) s = λ 5 , with Re s > 0

and c j ( s ) , j = 1 , 2 , 3 , solve the linear system

λ 1 c 1 + λ 2 c 2 + λ 3 c 3 = h ˆ 1 , λ 1 2 c 1 + λ 2 2 c 2 + λ 3 2 c 3 = h ˆ 2 , λ 1 4 c 1 + λ 2 4 c 2 + λ 3 4 c 3 = h ˆ 3 .

Then, Cramer’s rule implies that

c j ( s ) = Δ j ( s ) Δ ( s ) , j = 1 , 2 , 3 ,

where Δ ( s ) is the determinant of the coefficient matrix

Δ = λ 1 λ 2 λ 3 λ 1 2 λ 2 2 λ 3 2 λ 1 4 λ 2 4 λ 3 4

and Δ j ( s ) ( j = 1 , 2 , 3 ) are determinants of the matrices that are obtained by replacing the i th-column ( j = 1 , 2 , 3 ) of Δ ( s ) by the column vector ( h ˆ 1 ( s ) , h ˆ 2 ( s ) , h ˆ 3 ( s ) ) .

Inverse Laplace transform of u ˆ ( x , s ) yields

u ( x , t ) = 1 2 π i r i r + i e s t u ˆ ( x , s ) d s = j = 1 3 1 2 π i r i r + i e s t Δ j ( s ) Δ ( s ) e λ j ( s ) x d s .

In order to find the smoothing effect associated with each boundary value function, we divide the solution u of (2.1) into

u ( x , t ) = u 1 ( x , t ) + u 2 ( x , t ) + u 3 ( x , t ) m = 1 3 u m ( x , t ) ,

where u m ( x , t ) solves (2.1) with h j = 0 when j m ( m , j = 1 , 2 , 3 ). Thus, each u m ( x , t ) has the representation:

u m ( x , t ) = j = 1 3 1 2 π i r i r + i e s t u ˆ ( x , s ) d s = j = 1 3 1 2 π i r i r + i e s t Δ j m ( s ) Δ ( s ) e λ j ( s ) x h ˆ m ( s ) , d s ,

where Δ j m ( s ) is obtained from Δ j ( s ) by letting h ˆ m ( s ) = 1 and h ˆ k ( s ) = 0 for k m ( k , m = 1 , 2 , 3 ).

In the last formulas, the right-hand sides are continuous with respect to r for r 0 . As the left-hand sides do not depend on r , we may take r = 0 in these formulas. Hence,

u m ( x , t ) = j = 1 3 1 2 π i 0 i e s t Δ j m ( s ) Δ ( s ) e λ j ( s ) x h ˆ m ( s ) d s + j = 1 3 1 2 π i i 0 e s t Δ j m ( s ) Δ ( s ) e λ j ( s ) x h ˆ m ( s ) d s = j = 1 3 1 2 π 0 e i ρ 5 t Δ j m + ( ρ ) Δ + ( ρ ) e λ j + ( ρ ) x h ˆ m + ( ρ ) 5 ρ 4 d ρ + j = 1 3 1 2 π 0 e i ρ 5 t Δ j m ( ρ ) Δ ( ρ ) e λ j ( ρ ) x h ˆ m ( ρ ) 5 ρ 4 d ρ ,

where h ˆ ± ( ρ ) = h ˆ ( ± i ρ 5 ) , and Δ j m ± ( ρ ) , Δ ± ( ρ ) , λ j ± ( ρ ) are defined the same way. Thus, h ˆ + ( ρ ) ¯ = h ˆ ( ρ ) , Δ + ( ρ ) ¯ = Δ ( ρ ) , λ j + ( ρ ) ¯ = λ j ( ρ ) .

When taking s = i ρ 5 , the roots (characteristic roots) of (2.3) are as follows:

λ 1 ( ρ ) = i ρ , λ 2 ( ρ ) = ρ cos 9 π 10 + i sin 9 π 10 , λ 3 ( ρ ) = ρ cos 13 π 10 + i sin 13 π 10 .

2.2 Boundary smoothing effect

We introduce the following technical lemma

Lemma 2.2

For any f L 2 ( a , ) , let Kf be the function defined by

K f ( x ) = a e γ ( μ ) x f ( μ ) d μ ,

where a R , γ ( μ ) is a continuous complex-valued function defined on ( a , ) satisfying the following two conditions

  1. Re γ ( μ ) < 0 , for μ > a ;

  2. There exist δ > 0 and b > 0 such that

    sup a < μ < a + δ Re γ ( μ ) μ a b ;

  3. There exists a complex number α + i β such that

    lim μ γ ( μ ) μ = α + i β .

Then there exists a constant C such that for all f L 2 ( 0 , ) ,

K f L 2 ( R + ) C f L 2 ( a , ) .

Proof

See Lemma 2.5 in [17].□

Proposition 2.3

Let s > 0 . There exists a constant C such that h s ( R + ) , the solution

u ( x , t ) W b d r ( h )

of initial boundary value problem

(2.4) t u + x 5 u = 0 , x > 0 , t > 0 , u ( x , 0 ) = 0 , x > 0 , x u ( 0 , t ) = h 1 ( t ) , x 2 u ( 0 , t ) = h 2 ( t ) , x 4 u ( 0 , t ) = h 3 ( t ) , t > 0 ,

satisfying

(2.5) sup 0 < t < T u ( , t ) H s ( R + ) + k = 0 4 x k u L R + ; H s + 2 k 5 ( 0 , T ) C h s ( R + ) .

Proof

We have for ρ ,

(2.6) Δ 11 + ( ρ ) Δ + ( ρ ) ρ 1 , Δ 21 + ( ρ ) Δ + ( ρ ) ρ 1 , Δ 31 + ( ρ ) Δ + ( ρ ) ρ 1 ,

(2.7) Δ 12 + ( ρ ) Δ + ( ρ ) ρ 2 , Δ 22 + ( ρ ) Δ + ( ρ ) ρ 2 , Δ 32 + ( ρ ) Δ + ( ρ ) ρ 2 ,

and

(2.8) Δ 13 + ( ρ ) Δ + ( ρ ) ρ 4 , Δ 23 + ( ρ ) Δ + ( ρ ) ρ 4 , Δ 33 + ( ρ ) Δ + ( ρ ) ρ 4

(the tedious proof of the aforementioned estimates (2.6)–(2.8) will be postponed to the appendix).

The solution of (2.4) can be divided into

u = u 1 + u 2 + u 3 ,

where u i is the solution with respect to h i ( i = 1 , 2 , 3 ). For example, u 1 is the solution of

(2.9) t u x 5 u = 0 , x > 0 , t > 0 , u ( x , 0 ) = 0 , x > 0 , x u ( 0 , t ) = h 1 ( t ) , x 2 u ( 0 , t ) = x 4 u ( 0 , t ) = 0 , t > 0 .

It is enough to verify the estimate (2.5) for u 1

(2.10) sup 0 < t < T u 1 ( , t ) H s ( R + ) + k = 0 4 x k u 1 L ( [ 0 , ) ; H s + 2 k 5 ( 0 , T ) ) C h 1 H s + 1 5 ( R + ) .

The other cases are similar, so we omit the details here.

When s = 0 . Observing that

u 1 ( x , t ) = j = 1 3 1 2 π 0 e i ρ 5 t Δ j 1 + ( ρ ) Δ + ( ρ ) e λ j + ( ρ ) x h ˆ 1 + ( ρ ) 5 ρ 4 d ρ + j = 1 3 1 2 π 0 e i ρ 5 t Δ j m ( ρ ) Δ ( ρ ) e λ j ( ρ ) x h ˆ 1 ( ρ ) 5 ρ 4 d ρ u 11 ( x , t ) + u 12 ( x , t ) .

By (2.6), it follows from Lemma 2.2 that for any t 0 ,

u 11 ( , t ) L 2 ( R + ) 2 C j = 1 3 0 Δ j 1 + ( ρ ) Δ + ( ρ ) 2 ( e Re λ j ( ρ ) + 1 ) 2 h ˆ 1 ( ρ ) ( 5 ρ 4 ) 2 d ρ C 0 h ˆ 1 ( ρ ) 2 ρ 6 d ρ = C 0 h ˆ 1 ( i ρ 5 ) 2 ρ 2 d ρ 5 C h 1 H 1 5 ( R + ) 2 .

The same argument applied to u 12 to obtain the same estimate. Thus,

(2.11) u 1 C ( 0 , T ; L 2 ( R + ) ) C h 1 H 1 5 ( R + ) .

Noting that for 0 k 4 , we have

x k u 11 ( x , t ) = j = 1 3 1 2 π i 0 i e i ρ 3 t Δ j 1 + ( ρ ) Δ + ( ρ ) [ λ j ( ρ ) ] k e λ j ( ρ ) x h ˆ 1 ( ρ ) 5 ρ 4 d ρ = j = 1 3 1 2 π i 0 i e i μ t Δ j 1 + ( θ ( μ ) ) Δ + ( θ ( μ ) ) [ λ j ( θ ( μ ) ) ] k e λ j ( θ ( μ ) ) x h ˆ 1 ( θ ( μ ) ) d μ ,

where θ ( μ ) is the real solution of μ = ρ 5 for ρ 0 . By using the Plancherel theorem (with respect to t ), it yields that for any x > 0 ,

x k u 11 H 2 k 5 ( 0 , T ) 2 C j = 1 3 0 ( 1 + μ 2 ) 2 k 5 [ λ j ( θ ( μ ) ) ] k e λ j ( θ ( μ ) ) x Δ j 1 + ( ρ ) Δ + ( ρ ) 2 h ˆ 1 ( i μ ) 2 d μ .

Thus, one finds there is a constant C such that

(2.12) sup x ( 0 , ) x k u 11 H 2 k 5 ( R + ) 2 C j = 1 3 0 ( 1 + μ 2 ) 2 k 5 [ λ j ( θ ( μ ) ) ] k 2 sup x ( 0 , ) e λ j ( θ ( μ ) ) x 2 Δ j 1 + ( ρ ) Δ + ( ρ ) 2 h ˆ 1 ( i μ ) 2 d μ C h 1 H 1 5 ( R + ) .

The following estimates were used in obtaining the last inequality:

sup x ( 0 , ) e λ 1 ( ρ ) x 2 Δ 11 + ( ρ ) Δ + ( ρ ) 2 ρ 1 , sup x ( 0 , ) e λ 2 ( ρ ) x 2 Δ 21 + ( ρ ) Δ + ( ρ ) 2 ρ 1 , sup x ( 0 , ) e λ 3 ( ρ ) x 2 Δ 31 + ( ρ ) Δ + ( ρ ) 2 ρ 1 .

(2.12) leads to

(2.13) k = 0 4 x k u 1 L R + ; H 2 k 5 ( 0 , T ) C h 1 H 1 5 ( R + ) .

Collecting (2.11) and (2.13) to obtain (2.10) for u 1 .

When s = 5 . For the solution u of (2.9), let v = t u . Then v is a solution of

(2.14) t v x 5 v = 0 , x > 0 , t > 0 , v ( x , 0 ) = 0 , x > 0 , x v ( 0 , t ) = h 1 ( t ) , x 2 v ( 0 , t ) = x 4 v ( 0 , t ) = 0 , t > 0 .

Applying (2.13) to (2.14) yields that

(2.15) v C ( 0 , T ; L 2 ( R + ) ) + k = 0 4 v L ( R + ; H 2 k 5 ( R + ) ) C h 1 H 1 5 ( R + ) .

Define

u 1 ( x , t ) = 0 t v ( x , τ ) d τ ,

then we have

x u 1 ( 0 , t ) = 0 t x v ( 0 , τ ) d τ = 0 t h 1 ( τ ) d τ = h 1 ( t ) .

Furthermore, it is easy to verify that

t u 1 + x 5 u 1 = 0 t t v ( x , τ ) d τ 0 t x 5 v ( x , τ ) d τ = 0 t ( t v x 5 v ) d τ = 0 .

Thus, u 1 solves the initial-boundary value problem (2.9). Since

t u 1 = x 5 u 1 ,

it follows that

u 1 C ( 0 , T ; H 5 ( R + ) ) + k = 0 4 u 1 L ( ( R + ) ; H 7 k 5 ( R + ) ) C h 1 H 1 + 5 5 ( R + ) ,

which is the result of Proposition 2.5 for s = 5 .

When 0 < s < 5 . Interpolation leads to the desired result.

For s [ 5 n , 5 ( n + 1 ) ] ( n = 1 , 2 , ), it is enough to repeat the procedure of the case s [ 0 , 5 ] .□

2.3 Extension strategy

For the solution

u ( x , t ) = W R ( t ) ϕ ( x ) + 0 t W R ( t s ) f ( s ) d s

of the initial value problem

(2.16) t u + x 5 u = f ( x , t ) , x R , t > 0 , u ( x , 0 ) = ϕ ( x ) , x R ,

we have the Kato smoothing effects and hidden regularity as follows:

Lemma 2.4

Let T > 0 be given, s [ 0 , 5 ] . Then ϕ H s ( R ) , f L 1 ( 0 , T ; H s ( R ) ) , the solution u of Cauchy problem (2.16) satisfying

u L 2 ( 0 , T ; H loc 2 + s ( R ) ) + sup 0 < t < u ( , t ) H s ( R ) C ( ϕ H s ( R ) + f L 1 ( R + ; H s ( R ) ) ) .

Proof

See [11].□

Lemma 2.5

Let T > 0 be given, s [ 0 , 5 ] . Then ϕ H s ( R ) , f L 1 ( 0 , T ; H s ( R ) ) , the solution u of Cauchy problem (2.16) satisfying

k = 0 4 sup x R x k u H s + 2 k 5 ( 0 , T ) C ( ϕ H s ( R ) + f L 1 ( ( 0 , T ) , H s ( R ) ) ) .

Proof

See [11].□

The solution of the initial-boundary value problem

(2.17) t u + x 5 u = f ( x , t ) , x > 0 , t > 0 , u ( x , 0 ) = ϕ ( x ) , x > 0 , x u ( 0 , t ) = x 2 u ( 0 , t ) = x 4 u ( 0 , t ) = 0 , t > 0

can be represented by

u ( x , t ) = W R ( t ) ϕ + 0 t W R ( t τ ) f ( , τ ) d τ W b d r p W b d r q ,

where ϕ = E ϕ and f = E f are bounded extension of ϕ and f from R + to R respectively, and q = ( q 1 , q 2 , q 3 ) , p = ( p 1 , p 2 , p 3 ) with

q 1 ( t ) = W R ( t ) ϕ _ x = 0 , q 2 ( t ) = x W R ( t ) ϕ x = 0 , q 3 ( t ) = x 2 W R ( t ) ϕ x = 0

and

p 1 ( t ) = 0 t W R ( t τ ) f ( , τ ) d τ x = 0 , p 2 ( t ) = x 0 t W R ( t τ ) f ( , τ ) d τ x = 0 , p 3 ( t ) = x 2 0 t W R ( t τ ) f ( , τ ) d τ x = 0 .

Then, establish the estimates on W b d r q , W b d r p :

Lemma 2.6

Let T > 0 be given, s [ 0 , 5 ] . Then ϕ H s ( R + ) , we have

sup 0 < t < T W b d r q ( , t ) H s ( R + ) + k = 0 4 sup x R W b d r q H s + 2 k 5 ( 0 , T ) C ϕ H s ( R + ) .

Proof

By Proposition 2.3 and Lemmas 2.4–2.5, we have

sup 0 < t < T W b d r q ( , t ) H s ( R ) + k = 0 4 sup x R + W b d r q H s + 2 k 5 ( 0 , T ) W R ϕ H s + 2 5 ( 0 , T ) + x W R ϕ H s + 1 5 ( 0 , T ) + x 2 W R ϕ H s 5 ( 0 , T ) ϕ H s ( R ) C ϕ H s ( R + ) .

Lemma 2.7

Let T > 0 be given, s [ 0 , 5 ] . Then f L 1 ( ( 0 , T ) , H s ( R + ) ) , we have

sup 0 < t < T W b d r p ( , t ) H s ( R + ) + k = 0 4 sup x R W b d r p H s + 2 k 5 ( 0 , T ) f L 1 ( 0 , T ; H s ( R + ) ) .

Proof

By Proposition 2.3 and Lemmas 2.4–2.5, we have

sup 0 < t < W b d r p ( , t ) H s ( R ) + k = 0 4 sup x R W b d r p H s + 2 k 5 ( 0 , T ) 0 t W R ( t τ ) f ( , τ ) d τ H s + 2 5 ( 0 , T ) + x 0 t W R ( t τ ) f ( , τ ) d τ H s + 1 5 ( 0 , T ) + x 2 0 t W R ( t τ ) f ( , τ ) d τ H s 5 ( 0 , T ) f L 1 ( 0 , T ; H s ( R ) ) C f L 1 ( 0 , T ; H s ( R + ) ) .

2.4 Proof of Proposition 2.1.

Proof

Collecting all of the results of Proposition 2.3, Lemma 2.6, Lemma 2.7 together, we obtain

u C ( 0 , T ; H s ( R + ) ) + k = 0 4 sup x R + u H s + 2 k 5 ( 0 , T ) C ( ϕ H s ( R + ) + h s ( R + ) + f L 1 ( 0 , T ; H s ( R + ) ) ) ,

which is

u Y T s C ( ( ϕ , h ) X T s + f L 1 ( 0 , T ; H s ( R + ) ) ) .

3 Well-posedness

For the nonlinear boundary feedback u ( 0 , t ) 2 , we need the following nonlinear estimates:

Lemma 3.1

Let T > 0 be given. Then s [ 0 , 5 ] , we have

(3.1) u v H s 5 ( 0 , T ) C T 3 10 1 s 5 u H s + 2 5 ( 0 , T ) v H s + 2 5 ( 0 , T ) ,

(3.2) u v H s 1 5 ( 0 , T ) C T 5 10 1 s 5 u H s + 2 5 ( 0 , T ) v H s + 2 5 ( 0 , T ) ,

and

(3.3) u v H s 2 5 ( 0 , T ) C T 7 10 1 s 5 u H s + 2 5 ( 0 , T ) v H s + 2 5 ( 0 , T ) .

Proof

We apply the interpolation skill to prove the inequalities (3.1)–(3.3).

Proof of (3.1). When s = 0 . Since H 2 5 L 10 , then Hölder’s inequality leads to

u v L 2 ( 0 , T ) T 3 10 u L 10 ( 0 , T ) v L 10 ( 0 , T ) T 7 10 u H 2 5 ( 0 , T ) v H 2 5 ( 0 , T ) .

When s = 5 . Since H 1 ( 0 , T ) is Banach algebra, we have

u v H 1 ( 0 , T ) u H 1 ( 0 , T ) v H 1 ( 0 , T ) u H 7 5 ( 0 , T ) v H 7 5 ( 0 , T ) .

Taking θ = 1 s 5 , interpolation leads to

[ L 2 ( 0 , T ) , H 1 ( 0 , T ) ] θ = H s 5 , H 2 5 ( 0 , T ) , H 7 5 ( 0 , T ) θ = H s + 2 5

and

u v H s 5 ( 0 , T ) C T 3 10 1 s 5 u H s + 2 5 ( 0 , T ) v H s + 2 5 ( 0 , T ) .

Proof of (3.2), (3.3) is similar, we omit the details.□

Proof of Theorem 1.1.

Proof

Let r > 0 and 0 < τ max { 1 , T } be constant to be determined. Take a ball from Y τ s :

τ , r = { v Y τ s r } ,

which is bounded closed convex subset of Y τ s . Define a map Γ on τ , r by

u = Γ ( v )

being the unique solution of

t u + x 5 u = f , x > 0 , t > 0 , u ( x , 0 ) = ϕ , x > 0 , x u ( 0 , t ) = h 1 ( t ) , x 2 u ( 0 , t ) = v ( 0 , t ) 2 + h 2 ( t ) , x 4 u ( 0 , t ) = v ( 0 , t ) 2 + h 3 ( t ) t > 0 ,

for v τ , r .

Applying Proposition 2.1, Lemma 3.1, we have

Γ ( v ) Y τ s C ϕ H s ( R + ) + h s + v ( 0 , t ) 2 H s 2 5 ( 0 , τ ) + f L 1 ( [ 0 , T ] ; H s ( R + ) ) C ϕ H s ( R + ) + h s + τ 7 10 1 s 5 v ( 0 , t ) H s + 2 5 2 ( 0 , τ ) + f L 1 ( [ 0 , T ] ; H s ( R + ) ) < C ( ϕ H s ( R + ) + h s + f L 1 ( [ 0 , T ] ; H s ( R + ) ) ) + C τ 7 10 1 s 5 v L ( R + ; H s + 2 5 ( 0 , τ ) ) 2 < C ( ϕ H s ( R + ) + h s + f L 1 ( [ 0 , τ ] ; H s ( R + ) ) ) + C τ 7 10 1 s 5 v Y τ s 2 .

Take r = 2 C ( ϕ H s ( R + ) + h s + f L 1 ( [ 0 , τ ] ; H s ( R + ) ) ) , choose 0 < τ 1 so small that C τ 7 10 1 s 5 r 1 3 . Then for any v τ , r , we have

Γ ( v ) Y τ s r 2 + C 2 τ 1 4 r 2 < r 2 + r 3 = 5 6 r ,

which implies that Γ : τ , r τ , r .

For any u , v τ , r , Γ ( u ) Γ ( v ) satisfies

t U x 5 U = 0 , x > 0 , t > 0 , U ( x , 0 ) = 0 , x > 0 , x U ( 0 , t ) = 0 , x 2 U ( 0 , t ) = u ( 0 , t ) 2 v ( 0 , t ) 2 , x 4 U ( 0 , t ) = u ( 0 , t ) 2 v ( 0 , t ) 2 , t > 0 .

Thus, by Lemma 3.2, then Lemma 3.1, we obtain

Γ ( u ) Γ ( v ) Y τ s C u ( 0 , t ) 2 v ( 0 , t ) 2 H s 2 5 ( 0 , τ ) C τ 7 10 1 s 5 u ( 0 , t ) + v ( 0 , t ) H s + 2 5 ( 0 , τ ) u ( 0 , t ) v ( 0 , t ) H s + 2 5 ( 0 , τ ) C τ 7 10 1 s 5 u L ( R + ; H s + 2 5