The initial-boundary value problem for the biharmonic Schr\"odinger equation on the half-line

We study the local and global wellposedness of the initial-boundary value problem for the biharmonic Schr\"odinger equation on the half-line with inhomogeneous Dirichlet-Neumann boundary data. First, we obtain a representation formula for the solution of the linear nonhomogenenous problem by using the Fokas method (also known as the \emph{unified transform method}). We use this representation formula to prove space and time estimates on the solutions of the linear model in fractional Sobolev spaces by using Fourier analysis. Secondly, we consider the nonlinear model with a power type nonlinearity and prove the local wellposedness by means of a classical contraction argument. We obtain Strichartz estimates to treat the low regularity case by using the oscillatory integral theory directly on the representation formula provided by the Fokas method. Global wellposedness of the defocusing model is established up to cubic nonlinearities by using the multiplier technique and proving hidden trace regularities.


Introduction
This article studies the local and global wellposedness of the initial -(inhomogeneous) boundary value problem for the biharmonic nonlinear Schrödinger equation (NLS) which is posed on the right half-line: q(x, 0) = q 0 (x), x ∈ R + , ( q(0, t) = g 0 (t), t ∈ (0, T), (1.3) where f (q) = κ|q| p q, κ ∈ C, T, p > 0, and q is a complex valued function.The analysis here is carried out in the L 2 −based fractional Sobolev space H s (R + ) at the spatial level, where throughout the paper (without any restatement) we will assume the following in order to work with a sufficiently nice nonlinearity: (a1) if s is integer, then p ≥ s if p is an odd integer and p ≥ s − 1 if p is non-integer, (a2) if s is non-integer, then p > s if p is an odd integer and p ≥ s if p is noninteger.The fourth-order NLS, in the form was introduced by [26]- [27] to study the stabilizing role of the higher-order dispersive effects.It was shown that the solutions are stable if γ < 0, p ≤ 4 n or if γ −1, . Moreover, solutions were found to be unstable for γ < 0, p ≥ 8 n in which case solutions may cease to exist globally.
In the absence of the Laplacian, the fourth order NLS takes the form and it is called the biharmonic NLS.It was shown by [13] (see also [3] and the references therein) that all solutions of the biharmonic NLS are global if γ > 0.Moreover, it was found that p = 8 n is the critical exponent for singularity formation if γ < 0, and smallness in the mean-square sense is sufficient for global existence if p = 8 n .The biharmonic NLS iu t + γ∆ 2 u + κ|u| p u = 0, with γ, κ ∈ R, is said to be focusing (resp.defocusing) if γκ < 0 (resp.γκ > 0).The rigourous analysis of the solutions of the fourth order Schrödinger equation started with the proof of sharp space-time decay properties for the linear group associated to the operator i∂ t + λ∆ + ∆ 2 , where λ ∈ {−1, 0, 1} [5].One can actually use these properties to obtain Strichartz estimates, which gives the local wellposedness at H 2 -level.
The references given above studied the fourth order Schrödinger equation in the whole Euclidean space, namely the spatial domain was assumed to be equal to R n .The absence of the boundary in these studies simplified the mathematical and physical analysis of the problem to some extent.However, in order to boost the physical reality, it is common to assume that the evolution takes place in a region with boundary, and what happens at the boundary influences the nature of the solutions.This is especially important for a control scientist, since boundary can be used as a control point, particularly when it is difficult or impossible to access the medium of the evolution.This idea motivated some of the recent studies related with the controllability of the the solutions of the linear fourth order Schrödinger equation.For instance, [40], [41], and [42] studied the wellposedness and exact controllability of the linear biharmonic Schrödinger equation on a bounded domain Ω ⊂ R n .Most recently, [1] studied the stabilization of the linear biharmonic Schrödinger equation on a bounded domain with a locally supported internal damping.
From the physical point of view, the model under consideration in this paper corresponds to a situation in which the wave is generated from a fixed source such that the wave train moves into the medium in one specific direction.Wellposedness of similar inhomogenenous initial boundary value problems on the half-line were recently considered for the classical Schrödinger equation; see for example [9], [8], [6], [25], and [16].We prove the corresponding wellposedness theorems for the biharmonic Schrödinger equations, and as far as we know this is the first treatment of the fourth order Schrödinger equations subject to inhomogeneous boundary conditions.
1.1.Main results.In this paper, attention is given only to the biharmonic NLS.More general fourth order Schrödinger equations with mixed dispersion as in (1.5) will be taken into consideration in a further study.Our first main result is the local wellposedness of solutions in fractional Sobolev spaces.More precisely, we prove the following theorem.
2 ).Then, (1.1)-(1.4)has the following local wellposedness properties: (i) Local existence and uniqueness: there exists a unique local solution q ∈ C([0, T 0 ]; H s (R + )) for some T 0 ∈ (0, T], then there is T 0 > 0 such that the flow (q 0 , g 0 , g 1 ) → q is Lipschitz continuous from B into C([0, T 0 ]; H s (R + )), (iii) Blow-up alternative: Let S be the set of all T 0 ∈ (0, T] such that there exists a unique local solution in C([0, T 0 ]; H s (R + )).If T max := sup Remark 1.8.The proof of the above theorem is based on the Fokas method ( [14], [15]) combined with classical contraction arguments.The Fokas method is a unified approach for solving initial-(inhomogeneous) boundary value problems for a general class of linear evolution equations.It has significant advantages over traditional methods.One of these advantages is that the solution formula obtained via the Fokas method is uniformly convergent at the boundary points.This is an important property for numerical studies.
Another remarkable feature of the Fokas method is that one can obtain the necessary space and time estimates for the corresponding linear evolution operator directly from the representation formula by using Fourier analysis.This allows one to easily study the wellposedness of the initial-boundary value problem for corresponding nonlinear models.
Remark 1.10.The global wellposedness problem for the nonlinear fourth order Schrödinger equation is a nontrivial problem in the presence of inhomogeneous boundary conditions as opposed to the case of homogeneous boundary conditions.The main difficulty lies in the fact that one looses all energy conservation and control properties once g 0 and g 1 are non-zero.For instance, the most basic energy identiy, namely the L 2 -energy, satisfies an equality given by 1 2 This identity involves the unknown traces q xxx (0, t) and q xx (0, t).Higher order energy estimates are even more complicated.
The extra regularity result q xx (0, •), q xxx (0, •) ∈ L 2 (0, T) proved in the above theorem is called a hidden trace regularity since in general for an arbitrary H 2 function, these traces do not need to be well-defined in the sense of Sobolev trace theory.Hence, this shows that the biharmonic Schrödinger operator has a hidden regularizing property in the sense of traces.
1.2.Orientation.We prove the main results in several steps: Step 1 -Representation formula.We use the Fokas method (also known as the unified transform method) to obtain a representation formula for the solution of the linear biharmonic Schrödinger equation with interior force and inhomogenenous boundary inputs.The derivation of the representation formula is more complicated than the classical Schrödinger equation due to the higher order nature of the evolution operator.
Step 2 -Cauchy problem.Secondly, we study the Cauchy problem on the spatial domain R. We obtain the necessary space and times estimates on the solutions of the Cauchy problem.These estimates are later used to study the half-line problem with zero boundary inputs by extending the given initial datum from half-line to the whole line.
Step 3 -Half line problem.We use the representation formula obtained in Step 1 to study the half-line problem with zero initial data and inhomogeneous Dirichlet-Neumann boundary inputs.We obtain space and time estimates on the solutions of the half-line problem.The analysis poses more challenges than the Schrödinger equation with only Dirichlet input, because the inhomogeneous Neumann input here requires us to deal with integrals that involve singular integrands.Singularities are treated with cut-off functions.
Step 4 -Operator theoretic formula.In this step, we express the representation formula of the linear problem with internal force in operator theoretic form.Then, we replace the internal source with the given nonlinearity and applying the contraction argument to this form of the representation formula to obtain the local wellposedness.
Step 5 -Global wellposedness.Finally we use the multiplier method to obtain useful energy estimates that involve information on the unknown boundary traces.The classical multipliers for the fourth order Schrödinger equations are not sufficient.Therefore, we use also the control theoretic multipliers to prove hidden trace regularities for the second and third order boundary traces of the solution.We combine these with H 2 energy identities to deduce the global wellposedness of solutions up to cubic powers.The results obtained here are quite interesting compared to solutions of the homogeneous boundary value problem, whose solutions can be shown to be global for all powers in the defocusing case.

Linear model
In this section, we consider the linear fourth-order Schrödinger equation on the right half-line with inhomogeneous Dirichlet-Neumann data in L 2 −based fractional Sobolev spaces: q(0, t) = g 0 (t), t ∈ (0, T), (2.3) 2.1.Representation formula.We will obtain a representation formula for the solution of (2.1)- (2.4).In order to do this, we will use the Fokas (unified transform) method.To this end, let q(k, t) denote the spatial Fourier transform of the solution of (2.1)-(2.4) on the right half-line: Note that the condition Im k ≤ 0 is enforced so that the right hand side of the formula (2.5) will in general converge.Applying this transform to the problem (2.1)-(2.4),after some computations we get Integrating the above equation in the temporal variable, we obtain (2.7) we can rewrite (2.7) as Multiplying both sides of (2.9) by e ik 4 t and then taking the inverse Fourier transform, we get The above formula consists of two unknown boundary traces, namely q xxx (0, t) and q xx (0, t).The idea of the uniform transform method is based on eliminating these unknown quantities from the solution formula.This is achieved in two steps: (1) We deform the contour of integration involving unknown quantities from (−∞, ∞) to another appropriate contour.(2) We take advantage of the invariant properties of equation (2.9) satisfied by q(k, t).
To this end, we first consider the region D described by It is easy to show that the above region can also be written by where the principle argument of a complex number is assumed to be defined in the interval [0, 2π).Now, we split D in two disjoint parts depending on whether k ∈ D is in the upper or lower half-plane: The following lemma follows from the the unified theory given in [15, Proposition 1.1, Chapter 1]: Lemma 2.13 (Deformation).Let q be a solution of (2.1) on Ω ≡ R + × (0, T) such that q is sufficiently smooth up to the boundary of Ω and decays sufficiently fast as x → ∞, uniformly in [0, T].Then, q(x, t) can be represented by where 4 , T) and ∂D + is oriented in such a way that D + is to the left of ∂D + .Now, we will use the invariant properties of the equation (2.9).Replacing k by −k in this equation keeps gj (−ik 4 , T) invariant for j = 0, 3, and one obtains Similarly, replacing k by ik and −ik, one can keep gj (−ik 4 , T) invariant for j = 0, 3.Moreover, we have the identities   The identities (2.15) and (2.17) are valid in D + 1 .Solving these identities for g2 and g3 and using the fact that G j = gj for j = 0, 1, we obtain Similarly, by using the identities (2.15) and (2.16), which are valid on D + 2 , we have 14) together with the fact that G j = gj for j = 0, 1, we deduce the following identity: The second and the seventh integrals at the right hand side of (2.23) vanish by Cauchy's theorem and one obtains the following representation formula where the right hand side includes information coming only from the prescribed initial-boundary-interior data: 2.2.Cauchy problem.In this section, we consider the fourth order linear Schrödinger equation on the whole real line R: for some constant c s ≥ 0.
Proof.We prove this lemma by using arguments similar to the ones given in the proof of [16,Theorem 4], which are based on the Fourier representation of the solution.There are a few crucial differences, particularly in the temporal regularity, compared to the classical Schrödinger equation due to the biharmonic evolution operator considered here.
Upon taking the Fourier transform of (2.25)-(2.26) in the spatial variable, we find ŷ(ξ, t) = e iξ 4 t ŷ0 (ξ) (2.31) where ŷ0 is the Fourier transform of y 0 .Note that and In order to see this, let t, t n ∈ [0, T] be such that t n → t.Then, where lim Note that the right hand side of the above inequality is a nonnegative integrable function since Hence, by the dominated convergence theorem lim n→∞ y(t n ) − y(t) H s (R) = 0.In other words, y(t n ) → y(t) in H s (R).Hence, we have just proved that y ∈ C([0, T]; H s (R)).
In order to prove y ∈ C(R; H 2s+3 8 (0, T)), we first write y = y 1 + y 2 , where and θ is a smooth cut-off function satisfying θ ≡ 1 for |ξ| ≤ 1, 0 ≤ θ ≤ 1 for 1 < |ξ| < 2, and θ ≡ 0 for |ξ| ≥ 2. Taking the j th order time derivative of y 1 with 0 ≤ j ≤ m, using the definition of θ, Cauchy-Schwarz inequality, and the definition of the Sobolev norm, one deduces that there exists a non-negative constant c(s, m) ≥ 0 such that at first for all m ∈ N, and then by interpolation for all m ≥ 0.
In order to obtain a similar estimate for y 2 , we split it into two terms and write y 2 = y 2,1 + y 2,2 , where Let us first consider the integral given by y 2,1 .We change the variables in this integral by setting ξ 4 = −τ and we define z 1 4 for z ∈ R + to be the negative real number which is obtained by taking the argument of z as 4π.Then, y 2,1 can be rewritten as .
The above formula can be thought of as the inverse (temporal) Fourier transform of the function , τ ∈ (−∞, −1).
Note then, The same estimate is also true for y 2,2 by similar arguments.Now, using these two estimates together with (2.33), one obtains y(x, •) for s ≥ − 3 2 .Continuity of the map x → y(x, •) can be shown by using the dominated convergence theorem again.Differentiating the problem (2.25)-(2.26) in x and repeating the above analysis with initial data y 0 ∈ H s−1 (R), we deduce that
Proof.Using the representation formula (2.24), the solution of (2.35)-(2.38) is given by where H 0 and H 1 are defined in terms of the formula (2.18) with data h 0 and h 1 , respectively.
Space estimate.Note that ∂D + 1 = γ 1 ∪ γ 2 and ∂D + 2 = γ 3 ∪ γ 4 (see Figure 3), where γ 1 (k We will start by considering the first component of (2.42), which is given by Let us first consider the case s = 0.In this case, by using the change of variables τ = −4k 4 , we get where the second inequality above is a property of the Laplace transform (see for example [16, Lemma 3.2]).Taking the square root in (2.43), we obtain . (2.44) Similarly, Taking the square root in (2.45), we obtain .
Continuity in t can easily be justified by means of the dominated convergence theorem.Therefore we just proved that z ∈ C([0, T ]; H s (R)) under the given assumptions on s, h 0 and h 1 .
Time estimate.We can rewrite (2.41) as We will start with considering the first component of (2.56), which is given by By using the change of variables τ = k 4 , we get where z 1 4 for z ∈ R − is defined by using the argument equal to π so that τ 1 4 ∈ γ 1 for τ ∈ (−∞, 0).Note that we can regard z 1 (x, •) as the inverse Fourier transform of the function from which it follows that (2.57) taking into account that Im τ ≥ 0 so that |e iτ Now, we will estimate the term In order to do this, we write z 2 = z 2,1 + z 2,2 where and θ is a smooth cut-off function satisfying By using the properties of θ, we can rewrite z 2,1 as Using the same change of variables τ = k 4 as before, we get e iτ .
The j th order time derivative of z 2,1 (x, •) is then written .
But since in particular s < s, we have h 1 . Therefore, we deduce that .
From the definition of the Sobolev norm, one obtains at first for all m ∈ N, and then by interpolation for all m ≥ 0. In particular, by choosing m = 2s+3 8 one has z 2,1 (x) .
(2.60) for s ≥ − 3 2 .Now, let us estimate z 2,2 (x, •).By using the definition of θ, we can rewrite this function as Using the same change of variables τ = k 4 with argument of τ equal to π, we have .
(2.63) Continuity in x can easily be justified by means of the dominated convergence theorem.Therefore we just proved that z ∈ C(R + ; H 2s+3 8 (0, T )) under the given assumptions on h 0 and h 1 .The time estimate for z x can be proven using arguments as in (2.57) after differentiating in x.Note that this will bring an extra factor of |τ| 1/4 into the integrals in the estimates.Therefore one will obtain z x ∈ C(R + ; H 2s+1 8 (0, T )).

2.4.
Representation formula -revisited.Let s ∈ 0, 9  2 \ 1 2 , 3 2 .Let (•) * denote a bounded extension operator from H s (R + ) into H s (R).Existence of such an extension operator is guaranteed by the definition of the Sobolev space and the associated Sobolev norm on the half-line: Now, given q 0 ∈ H s (R + ), we let q * 0 ∈ H s (R) be the extension of q with respect to the fixed extension operator (•) * just defined.Note that by the boundedness of this operator we have Therefore, y(t) = S R (t)q * 0 solves the problem iy t + ∆ 2 y = 0, y(0, t) = q * 0 (x), x ∈ R, t ∈ (0, T) where S R (t) is the evolution operator for the free linear biharmonic Schrödinger equation given in Lemma 2.27.Similarly, given f ∈ L 1 (0, T; H s (R + )), let f * ∈ L 1 (0, T; H s (R)) be the extension of f in the spatial variable with respect to the extension operator (•) * .Then the solution of the non-homogeneous Cauchy problem can be written as For j = 0, 1, we set Note that these traces exist by Lemma 2.27 as elements of H Compatibility conditions.In order for solutions to be continuous at the space-time corner point (x, t) = (0, 0), we can find the necessary (compability) conditions based on the analysis of traces again.Suppose that s ∈ 0, 9  2 9  2 , then 2s+3 8 > 1 2 and therefore g 0 (0) is well-defined, but q(0, 0) = q 0 (0), and hence we must have g 0 (0) = q 0 (0)(= a 0 (0)).By a similar argument, we deduce that if s ∈ 3 2 , 9 2 , then g 1 (0) = q 0 (0)(= a 1 (0)) is a necessary condition.On the other hand, if [p 0 , p 1 ] ∈ H .
For the existence of such extension, see for instance [4,Lemma 2.1].
Therefore, if we define then q = q e | [0,T) solves (2.1)-(2.4).Note that both g 0 − a 0 − b 0 and g 1 − a 1 − b 1 satisfy the necessary comptability conditions given in Section 2.3 since g 0 (0) = a 0 (0), b 0 (0) = b 1 (0) = 0, and g 1 (0) = a 1 (0).Now, we are ready to state the following lemma which follows by combining the space-time estimates proved for the solution generators S R and S b in the previous sections.
Lemma 2.65.Let s ∈ 0, 9  2 \ 1 2 , 3 2 , q 0 ∈ H s (R + ), g 0 ∈ H 2s+3 8 (0, T), g 1 ∈ H 2s+1 8 (0, T), and f ∈ L 1 (0, T; H s (R + )).Suppose also that the initial-boundary data satisfies the necessary compatibility conditions.Then the following estimate holds true for the solution of (2.1)- (2.4).Local existence.Note that a solution of (1.1)-(1.4) is a fixed point of the operator Ψ which is given by where for j = 0, 1, we set and b j (q * )(t) := −i∂ We consider the operator Ψ on the Banach space X T 0 := C([0, T 0 ]; H s (R + )).In order to prove the local existence of solutions we will use the Banach fixed point theorem on a closed ball B R (0) of the function space X T 0 for appropriately chosen R > 0 and T 0 ∈ (0, T].Let us first show that Ψ maps B R (0) onto itself for appropriate R and sufficiently small T 0 .First of all, we have via Lemma 2.27 and the boundedness of the extension operator the estimate which gives S R (•)q * 0 X T 0 q 0 H s (R + ) .
Let us recall the following lemma, which holds true under the given assumptions (a1) and (a2).The proof can be found for instance in [4, Lemma 3.1]: for y, z ∈ H s (R).
Using Lemma 3.3, we have Similarly, The last term in (3.1) is estimated as follows q 0 H s (R + ) .(3.9) In (3.9), the second inequality follows from the fact that ∂ x S R (t)q * 0 is a solution of the linear biharmonic Schrödinger equation on R with initial condition Now, unleashing the Gronwall's inequality we get q 1 (t) − q 2 (t) H s (R + ) = 0, which implies q 1 ≡ q 2 on [0, T 0 ].(0, T 0 ).Let (y 0 , g 0 , g 1 ) ∈ B and (z 0 , h 0 , h 1 ) ∈ B. Let y, z be two solutions on a common time interval (0, T 0 ) corresponding to (y 0 , g 0 , g 1 ) and(z 0 , h 0 , h 1 ), respectively.Then Now, using the linear theory together with the nonlinear H s estimates on the differences, we have Choosing R, as in the proof of the local existence, and T 0 accordingly small enough, we obtain Blow-up alternative.In this section, we want to obtain a condition which guarantees that a given local solution on [0, T 0 ] can be extended globally.Let's consider the set S of all T 0 ∈ (0, T] such that there exists a unique local solution in X s T 0 .We claim that if T max := sup T 0 ∈S T 0 < T, then lim t↑T max q(t) H s (R + ) = ∞.In order to prove the claim, assume to the contrary that lim t↑T max q(t) H s (R + ) = ∞.Then ∃M and t n ∈ S such that t n → T max and q(t n ) H s (R + ) ≤ M. For a fixed n, we know that there is a unique local solution q 1 on [0, t n ].Now, we consider the following model.
We know from the local existence theory that the above model has a unique local solution q 2 on some interval [t n , t n + δ] for some then q is a solution on [0, t n + δ] where t n + δ > T max , which is a contradiction.Hence, the proof of the local wellposedness is complete.

Global wellposedness.
In this section, we assume κ ∈ R − .First we multiply the main equation by q, use integration by parts on R + , and take the imaginary parts.Then, we have 1 2 As we see from the above identity, the conservation of L 2 -energy is lost in the presence of inhomogenenous boundary inputs.Therefore, the global wellposedness is quite a nontrivial problem.In order to prove the global wellposedness, one needs to gather some information on the second and third order traces.This enforces us to use other multipliers of higher order.To this end, we multiply the main equation by q x and use integration by parts on R + .Therefore, we first write We can rewrite the first term as Now, we multiply the main equation by q t and use integration by parts on R + .Therefore, we have = Re q xxx (0, t) ḡ 0 (t) − Re q xx (0, t) ḡ 1 (t) .(3.25) Next, we multiply the main equation by q xxx and use integration by parts on R + .Therefore, we first write ∞ 0 q t qxxx dx − i ∞ 0 ∂ 4 x q qxxx dx = −iκ ∞ 0 |q| p q qxxx dx. (3.26) Taking the imarginary part of the first term at the left hand side of (3.26) and using integration by parts, we obtain Im ∞ 0 q t qxxx dx = −Im g 0 (t) qxx (0,  Combining the three identities above,  Note that by integration by parts ∞ 0 |q x | 2 dx = −q(0, t) qx (0, t) − ∞ 0 q qxx dx, (3.33)

17
) respectively.Let us define two subregions of D + (See Figure 1) by D + 1 := D + ∩ {Re k ≥ 0} and D + 2 := D + ∩ {Re k ≤ 0} and let us also set the following transformation for the given boundary data: