WELL-POSEDNESS FOR A HIGHER-ORDER, NONLINEAR, DISPERSIVE EQUATION ON A QUARTER PLANE

. The focus of the current paper is the higher order nonlinear dispersive equation which models unidirectional propagation of small amplitude long waves in dispersive media. The speciﬁc interest is in the initial-boundary value problem where spatial variable lies in R + , namely, quarter plane problem. With proper requirement on initial and boundary condition, we show local and global well posedness.

1. Introduction. A class of higher-order models for unidirectional water wave of the form η t + η x − γ 1 βη xxt + γ 2 βη xxx + δ 1 β 2 η xxxxt + δ 2 β 2 η xxxxx has recently been derived by Bona, Carvajal, Panthee and Scialom [4]. With appropriate choices of the parameters γ 1 , γ 2 , δ 1 , δ 2 and γ, this equation serves as a model for the propagation of small-amplitude, long-crested surface waves moving to the direction of increasing values of the spatial variable x. Here α is a typical ratio of wave amplitude to depth, β is a representative value of the square of the depth to wavelength and t is proportional to elapsed time. The dependent variable η = η(x, t) is a real-valued function of x ∈ R, t ≥ 0 representing the deviation of the free surface from its undisturbed position at the point corresponding to x at time t. This model subsists on the assumptions that α and β are comparably-sized small quantities while η and its first few partial derivatives are of order one. Moreover, γ 1 and γ 2 are restricted by γ 1 + γ 2 = 1 6 , see Bona et al [4]. If the terms quadratic in α and β moved to the right-hand side, this model takes the form η t + η x − γ 1 βη xxt + γ 2 βη xxx + 3 4 α(η 2 ) x = O(β 2 , αβ, α 2 ) = O(α 2 ) = O(β 2 ).

HONGQIU CHEN
If the higher-order terms are neglected entirely, we obtain the well-known BBM-KdV equation Using the relation after neglecting quadratic terms in α and β, we obtain the well-known KdV-equation and the BBM-equation respectively, (see [1], [13]). At the formal level, the error involved in using (2) or (3) to approximate the full, inviscid water-wave problem is locally of order α 2 ∼ β 2 while the error in the higher-order model (1) is of order α 3 ∼ β 3 .
As [4] pointed out that boundary-value problem may be the most practically interesting, in some circumstances, see also [1] [2], [3], [6], [7], [8] [9] and the references in them. So it is our purpose here to address this issue.
Considered in the current study is the equation posed for x, t ≥ 0, namely, quarter plane problem. This system describes model for long water waves of small but finite amplitude, generated by a wave-maker at the left-hand end and propagating to the right in a uniform, open channel. The channel is taken to be infinitely long to avoid dealing with a boundary condition at the right. Indeed, the model equation (5) is only valid for waves moving to the right. In a real channel, as soon as the wave motion reaches the right-hand end of the channel, comparison between theory and experiment needs to cease because the reflected waves are not described by this one-way equation (see [9] and references in [9]). So it is sometime called wave maker problem. When the BBM-equation (3) is used to model the wave-maker problem, a single boundary data η(0, t), which can be measured by right equipments, in addition to an initial data η(x, 0), is sufficient. However, the equation (5) has fourth order derivative with respect to spatial variable x, an additional boundary data is needed.
Here is the proposed initial-boundary condition with the compatibility condition giving that G and J are order one functions, independent of parameters α and β. As mentioned earlier, Bona et al. [4] showed that the problem (1) with pure initial condition is linearly ill posed when δ 1 , the coefficient of the η xxxxt term, is negative (whilst δ 1 = 0 is not relative to the modeling issue,) hence, it is naturally assumed δ = δ 1 − δ 2 , the coefficient of η xxxxt term in (5), to be positive throughout the current paper. Our attention now turns to the initial-boundary-value problem (5)-(6)- (7).
For the analysis that follows, make variable rescaling:x = β − 1 2 x,t = β − 1 2 t and u(x,t) = αη(x, t). Under these new variable, the problem (5)-(6) reduces to the following non-dimensional 5th order, nonlinear, dispersive equation forx,t ≥ 0. Notice that, α, β only appear in the auxiliary data. Drop the tildes to obtain the BBM-type equation

Remark 1.
As η 0 is order 1 function and independent of the physical parameters α, β which are very small, so that the initial condition αη 0 (β 1 2 x) is a wave form of small amplitude with long wave length.
Since (5)- (6) and (9) are equivalent, the equation in (9) has less number of parameters, we aim to bring the theory for (9) more closely into line with that appearing in [4] for the pure initial-value problem. The main result in this paper is as follows.
Theorem 1.1. Problem (9) is well posed locally in time if the boundary data G, J ∈ C([0, ∞)), the initial data η 0 lies in Sobolev space H s (R + ) for some s ≥ 1 and is twice continuously differentiable locally at x = 0, and G, J and η 0 satisfy the compatibilty condition (7). Moreover, if γ = 7 48 and s ≥ 2, G, J ∈ L 1 (R + ) ∩ L 3 (R + ) and the value η 0 is of order 1, then (9) is well posed globally in time, the solution u lies in space C([0, ∞); H s (R + )), its H s (R + )-norm has the following growth bound in t, The next section is to derive an integral equation equivalent to (9), and provides a detailed statement of the results in view. Analyses leading to the conclusions advertised in the main result are presented in Sections 3 and 4. Section 3 consists of preliminaries and local well-posedness, while Section 4 provides conditions and proof of global well-posedness.
2.1. Preliminary calculations. We commence this section with the following boundary value problem of a second order ordinary differential equation where ρ is a positive number. It is straightforward calculation that the problem has a unique solution where Proposition 1. For any ρ 1 , ρ 2 > 0, and ρ 1 = ρ 2 , the following fourth order ordinary differential equation has a unique solution where . Hence,ṽ Elementary calculus reduces the last integral to the desired result in (14).

Proposition 2.
For any ρ > 0, the following fourth order equation The proof is similar to that of Proposition 1.

2.2.
Derivation of integral equation of problem (9). The problem under consideration is the initial-boundary value problem (9) with the compatibility conditions (7). In this section, the letter f represents the following and denote to reduce messy form involving α and β. Then the differential equation in (9) is briefed as The operator I − γ 1 ∂ xx + δ∂ xxxx is going to be inverted based on values γ 1 and δ taken in following three cases.
Case I. It is assumed that δ ∈ (0, δ, so (25) is written alternatively as where ρ 1 , ρ 2 > 0 are given as Apply Proposition 1 to (26) coupled with the auxiliary dada in (25), it obtains the following differential-integral equation where G 1 is defined in (15). Take (27) into consideration, then Hence, Integrations by parts with respect to z yield the following equation where Integration with respect to t yields the following integral equation, Case II. When δ = By Proposition 2, Integration by parts in (33) with respect to z and then integrating with respect to t, it follows where Case III. The last consideration is for the value of δ strictly greater than

HONGQIU CHEN
Again, application of Proposition 3 to (25) leads to In (38), integrations by parts with respect to z first and then integration with t together with (37) gives the following form where (41) Summarize the three cases just discussed, the quarter plane problem (9) now is converted to the following integral equation, and the operator K is defined as with the integral kernel Remark 2. φ 1 and φ 2 satisfy the following fourth order ordinary differential equations, respectively, and 3. Function classes and local well-posedness. The following notation for function classes is used throughout the discussion. All functions are real-valued and if a spatial domain is not specified for a function class, it is presumed to be R + = [0, ∞). Notation C(R + ) is a standard continuous function space. However, its standard Fréchet-space topologies will not intervene in the analysis. Similarly, consists of all functions in C m (R + ) whose m th derivatives are locally uniformly Hölder continuous with exponent σ.
For any Banach space X, its norm will be denoted · X with the exceptions noted below.
consists of all continuous and bounded functions defined on R + whose derivatives up to order m are also continuos and bounded. The norm on this space is The Sobolev space of L 2 -functions whose distributional derivatives up to order m ≥ 0 also lie in L 2 is denoted simply H m . These spaces carry their standard Hilbertspace structures with its norm · m defined as When m = 0, H 0 = L 2 , its norm is simply denoted · . 3. For s > 0 to be an non-integer, the Sobolev space H s (R + ) is the set of restriction of functions from H s (R) to R + equipped with the norm with the norm When s = m is a positive integer, the norm 4. If X is a Banach space, the space C(0, T ; X) of continuous functions from [0, T ] to X will be found useful. The norm on this space is the usual one, viz.
is said to be admissible if all three functions ψ, g, j are continuous locally at the origin 0, and moreover, ψ is second order differentiable locally at 0, they satisfy (See (7) as how the class from which the auxiliary data is drawn, ψ, g and j are not required to be continuous everywhere.) Our attention now turns to the integral equation (42), in which we begin with properties of the operator K given in (46).
where P k is a k th degree polynomial function.
it implies that for k ≥ 2, and inductively, where P k−1 , P k−2 are polynomial functions of degree k − 1, k − 2 respectively. This together with (50) completes (51). and On the other hand, by property (50), It still lies in C b (R + ) and It is seen that the operator K maps C b (R + ) to C 3 b (R + ) continuously and Kv has regularity exactly three orders higher than that of v.
implies that for some constant c k . By induction, K maps C k b (R + ) to C k+3 b (R + ) continuously for all k = 0, 1, 2, · · · .
If v ∈ C b (R + ) and lim x→∞ v(x) = 0, then for any > 0, there exists M = M > 0 such that |v(x)| ≤ for |x| > M. Whence The proposition is established.
Proof. Introduce a functionK : R → R as follows, in which ρ = δ − 1 4 , and ρ 1 , ρ 2 , θ depend on δ and γ 1 are given in the corresponding context in (27) and (37), respectively. Then when x and z are restrictd to R + , If v is a function from R + to R, then for x ∈ R + , Since the integral kernel and its first derivatives are continuous functions, However, the second order derivative of the integral kernel has a jump discontinuity at x = z as follows, It is seen that for v ∈ L 2 (R + ), for some constant c > 0. The first part of the proposition is established. SinceK Which is equivalent to say that Kv ∈ H 2 (R + ) with for some constant c > 0. It means that K maps L 1 (R + ) to H 2 (R + ) continuously. The proposition is complete.
Lemma 3.1. If boundary data g, j ∈ C(R + ), the initial data u 0 ∈ C 2 b (R + ) and the triple (u 0 , g, j) is admissible, then the integral equation (42) is locally well posed in time. Precisely, there is a small value T that depends on u 0 , g and j such that (42) has a unique solution u which lies in space C(0, T ; C 2 b (R + )). Moreover, the correspondence between initial-boundary data (u 0 , g, j) and the associated solution u of (42) is a Lipschitz continuous mapping from any bounded set of is admissible} into C(0, T ; C 2 b (R + )), and u satisfies u(0, 0) = g(0), u xx (0, 0) = j(0). Proof. Let T > 0 be arbitrary and u ∈ C(0, T ; C 2 b (R + )), then it is clear that (69) and fix T as where . It is straightforward to check that A is a contraction mapping on the ball B R (0) = {u ∈ C(0, T ; C 2 b (R + )) : u C(0,T ;C 2 b (R + )) ≤ R}. Since B R (0) is a complete metric space under the metric · C(0,T ;C 2 b (R + )) , there is a unique u ∈ C(0, T ; C 2 b (R + )) such that Au = u.
Because the solution obtained is the fixed point of a contractive mapping, naturally the mapping from the initial-boundary data to the solution (u 0 , g, j) → u is Lipschitz continuous. To see the point, let u andũ be the solutions of (42) corresponding to the auxiliary data (u 0 , g, j) and (ũ 0 ,g,j) respectively. Then on the time interval [0, T ], we have where A (u0,g,j) is contractive. It provides that

HONGQIU CHEN
where θ < 1 is the contractive constant for the operator K (u0,g,j) . It follows readily that . The Lipschitz continuity from (u 0 , g, j) to u has been established. It remains to check the initial and boundary conditions. By (42), we have It together with lim x→0 ∞ 0 |K(x, z)| dz = 0, φ 1 (0) = 1 and φ 2 (0) = 0 indicates readily that lim x→0 u(x, t) = g(t) uniformly for t ∈ [0, T ]. In a same fashion, The lemma is established.
Corollary 1. If the triple (u 0 , g, j) is admissible, and moreover, the initial value u 0 ∈ C k b (R + ) ∩ C k,s (R + ) and g, j ∈ C m,σ (R + ) for some non-negative integers m ≥ 0, k ≥ 2 and 0 < s, σ ≤ 1, then the solution u of (42) is a member of C m,σ (0, T ; C k b (R + ) ∩ C k,s (R + )), where T only depends on the value of u 0 C 2 b (R + ) , and properties of g and j near t = 0, not their derivatives.
Proof. The assumptions on the auxiliary data mean that u 0 ∈ C 2 b (R + ) and g, j ∈ C(R + ). By the last theorem, equation (42) has a unique solution u which lies in space where T is provided in (70) which only depends on u 0 C 2 b (R + ) , and properties of g and j near t = 0, see (68)-(69)-(70). Repeat (42) in an alternative form and notice that µ defined in (43) lies in C m,σ (0, ∞; H ∞ (R + )). The smoothing property of K implies that the right-hand side is a member of C 1 (0, T ; C 3 b (R + )). That says u ∈ C 0,σ (0, T ; , and therefore, the solution u of (42) is a member of C(0, T ; C k b (R + ) ∩ C k,s (R + )). The corollary is established, Theorem 3.2. Suppose that boundary data g, j ∈ C 1 (R + ), the initial data u 0 ∈ C 4 b (R + ) and the triple (u 0 , g, j) is admissible. Then the solution u of the integral equation (42) which lies in space C(0, T ; C 2 b (R + )) for some T > 0, (see detail in Lemma 3.1), is, in fact, a classical solution of (25). More precisely, The continuous function Proof. Since u ∈ C(0, T ; C 2 b (R + )) is the solution of the integral equation (42), φ 1 , φ 2 ∈ C ∞ b (R + ) and g, j ∈ C 1 (R + ), lies in space C(0, T ; C b (R + )). Hence, for any ∆t = 0, Kf (x, s) ds.
The limit of the right hand-side as ∆t → 0 exists, that is and it lies in C(0, T ; C b (R + )). From Proposition 5, the operator K is an order three smoothening operator, so all lie in C(0, T ; C b (R + )). It transpires that f x ∈ C(0, T ; C b (R + )) as well. By (58), Take derivative with respect to t, it follows that Both ∂ 4 x u and ∂ 4 x ∂ t u belong to C(0, T ; C b (R + )). Because φ 1 and φ 2 satisfy (44) and (45), respectively, It is seen that the continuous function is identically equal to zero for all (x, t) ∈ R + × [0, T ]. Whence, u is the classical solution of the equation The proof of lim is as same as that in proof of Lemma 3.1.
Remark. The solution u cannot acquire more spatial and time regularity than that of the initial data u 0 (x) and boundary data g(t) and j(t), respectively. Under the weak regularity assumptions for g, j and u 0 , it is no longer expected that there is a classical solution to equation (25). Instead, we search for solutions of (42) in the space C(0, T ; H s ) for appropriate s. When s ≤ 5 2 , the second order derivative u xx of solutions u obtained may not exist pointwise, it will turn out that it has trace at x = 0, t = 0, and it satisfies the initial-boundary condition provided that the triple (u 0 , g, j) is admissible. This is not obvious. However, the issue will become clear later.
A function u(x, t) solves the initial-boundary value problem (25) on the halfline R + in the integral sense if u satisfies (42). Elementary considerations reveal if g, j ∈ C(R + ), u 0 ∈ H s for s ≥ 2 and (u 0 , g, j) is admissible, then solutions of the integral equation (42) solve the initial-boundary-value problem (25) in the sense of distributions. Indeed, it is clear that a solution of (42) satisfies the initial condition since g(t) → g(0), j(t) → j(0) and the double integral term vanishes as t → 0. Because lim x→0 K(x, z) = lim x→0 K xx (x, z) = 0 for all z > 0, the dominated convergence theorem implies that the double integral and its second order derivative with respect to x again tend to zero in the limit x → 0. Since it thus follows that the boundary condition is also satisfied. The fact that a solution of (42) is a distributional solution of (25) follows from the fact that u(z, t), u z (z, t), u zz (z, t)) dz, and by the fundamental theorem of calculus see (71) together with identities (48) and (49). Of course, g , j and f x are taken in the sense of distributions. Conversely, a distributional solution u of the initial-boundary-value problem (25) with (u 0 , g, j) being admissible is a solution of the integral equation (42). This latter fact is seen by following the steps outlined earlier for the derivation of the integral equation. Thus the two problems are equivalent as far as admissible solutions are concerned and we will not distinguish between them further.
The following theorem is one of the principal results of the current paper.
Theorem 3.3. If boundary data g, j ∈ C(R + ), the initial data u 0 ∈ H 2 = H 2 (R + ) and (u 0 , g, j) is admissible, then the integral equation (42) is locally well posed in time. Precisely, there is a small value T depending on u 0 2 and local property of g and j at t = 0 such that (42) has unique solution u which lies in space C(0, T ; H 2 ). Moreover, the correspondence between initial and boundary data (u 0 , g, j) and the associated solution u of (42) is a uniformly Lipschitz continuous mapping from any bounded subset of H 2 ×C(0, T )×C(0, T ) into C(0, T ; H 2 ), lim t→0 u(·, t)−u 0 2 = 0, and u is twice differentiable near x = 0 satisfying u(0, t) = g(t) and u xx (0, t) = j(t).
Proof. Like the proof of Lemma 3.1, we show that the operator A defined in (42) has a fixed point in C(0, T ; H 2 ) for some small T > 0. First, let T > 0 be any number and u ∈ C(0, T ; H 2 ), then the nonlinear function f = f (u, u x , u xx ) ∈ C(0, T ; L 2 ), so Kf ∈ C(0, T ; H 3 ), and therefore, A maps C(0, T ; H 2 ) to itself because u 0 ∈ H 2 .
It is straightforward to verify that for any u ∈ B R (0), and for any u, v ∈ B R (0), It is seen that the operator A is a contraction mapping on B R (0), and whence there is a unique point u ∈ B R (0) such that The argument of showing that the mapping from initial and boundary data (u 0 , g, j) to the associated solution u of (42) is a uniformly Lipschitz continuous mapping from any bounded subset of H 2 × C(0, T ) × C(0, T ) into C(0, T ; H 2 ) is as same as that in the proof of Lemma 3.1, so it is not repeated here. What remains is to check the initial and boundary conditions.
Theorem 3.4. In Theorem 3.3, if the initial data u 0 ∈ H s for some s > 2, and the boundary data g, j ∈ C m,σ (0, ∞) for some m ∈ N and 0 < σ ≤ 1, then the solution u lies in space C m,σ (0, T ; H s ).
Proof. Since it is already established that the solution u ∈ C(0, T ; H 2 ) in last theorem, and the fact that f ∈ C(0, T ; L 2 ) and Kf ∈ C(0, T ; H 3 ) indicates that Kf (x, s) ds is a member of C 1 (0, T ; H 3 ). By bootstrapping or mathematical induction, Therefore, u 0 ∈ H s and g, j ∈ C m,σ (0, ∞) imply u ∈ C m,σ (0, T ; H s ).
The theorem is established.
Remark. The time length T depends on u 0 2 (instead of u 0 s ) and local property of g, j at t = 0 as described in Theorem 3.3. The proof is basically applying smoothing property of the operator K. Now we turn our attention to the case where the initial data u 0 ∈ H 1 (R + ), the operator A does not map C(0, T ; H 1 (R + )) to itself due to the terms K(u 2 ) xx and Ku xx . To see this point, using integrations by parts and noticing that K( and similarly, It turns to show that the operator A Σ has a unique fixed point in C(0, T ; H 2 ) for sufficiently small value of T. We now attempt to find a proper space in which the operator A Σ is contractive mapping. By Proposition 7, is a member of C(0, ∞; H 2 ). It is seen that A Σ maps C(0, T ; H 2 ) to itself for any and fix T as It is straightforward to verify that B R (0) is a complete metric space. What remains is to show that A Σ maps B R (0) to itself and is contractive. To do so, for any w ∈ B R (0), we need to check that A Σ w lies in C(0, T ; H 2 ). Let t ≤ T, then, Σw 2 (·, s) ds + 2|γ| t 0 K(·, 0) 2 |(Σw) x (0, s)| + κ 4,1 Σw (·, s) 1 ds.

HONGQIU CHEN
It follows readily that Indeed, A Σ maps B R (0) to itself. For any w 1 , w 2 ∈ B R (0), take the C(0, T ; H 2 )-norm, In consequence, the operator A Σ is a contraction mapping on B R (0), and hence there is a unique point w ∈ B R (0) such that Therefore, u = w + Σ ∈ C(0, T ; H 1 ) is the unique solution of (42). The Lipschitz continuity follows from the contraction mapping principle. The proof is as same as that in the proof of Lemma 3.1. We now check the initial and boundary conditions.
Notice that the fixed point w satisfies for a cubic polynomial function P (R), it follows It implies the initial condition lim t→0 u(x, t) = u 0 (x) uniformly for x ∈ R + . To estimate the boundary condition, reorganize (77) by singling γ 2 , γ-terms out as follows, The Dominated convergence theorem implies that lim x→0 w(x, t) = 0 uniformly for t ∈ [0, T ] due to K(0, 0) = 0 and lim x→0 K(x, z) = lim x→0 K xx (x, z) = 0 for all z > 0. In consequence, uniformly for t ∈ [0, T ]. In (78), taking ∂ xx on both hand sides, applying the property of (74), it follows that Applying the results in (58), It is seen that w xx is in fact continuous pointwise, the dominated convergence theorem together with properties that lim x→0 K xx (x, 0) = − 1 δ and lim x→0 K(x, z) = lim x→0 K xx (x, z) = 0 for any z > 0 yield lim x→0 w xx (x, t) = 0.

Corollary 2.
In the last theorem, if u 0 ∈ H s where 1 ≤ s < 2, then the solution u ∈ C(0, T ; H s ).
4. Global well-posedness. Local well-posedness study in last section is irrelevant to size of α and β. However, it matters in global well-posedness. So we turn our attention back to the initial-boundary value problem (9) where the parameters α and β as of order o(1) constants are in explicit display. Here is our theorem on global well-posedness.
That is to say the problem (9)