Families of Solutions of Multitemporal Nonlinear Schrödinger PDE

: The multitemporal nonlinear Schrödinger PDE (with oblique derivative) was stated for the ﬁrst time in our research group as a universal amplitude equation which can be derived via a multiple scaling analysis in order to describe slow modulations of the envelope of a spatially and temporarily oscillating wave packet in space and multitime (an equation which governs the dynamics of solitons through meta-materials). Now we exploit some hypotheses in order to ﬁnd important explicit families of exact solutions in all dimensions for the multitime nonlinear Schrödinger PDE with a multitemporal directional derivative term. Using quite effective methods, we discovered families of ODEs and PDEs whose solutions generate solutions of multitime nonlinear Schrödinger PDE. Each new construction involves a relatively small amount of intermediate calculations.

The multitime NLSE has been indicated to manage the evolution of a multitime wave packet in a weakly nonlinear and dispersive medium and has possible applications in diverse fields such as nonlinear optics, water waves and plasma physics. In this paper, it is shown that the multitime NLSE solutions can be expressed analytically by specific methods. Section 2 analyzes solutions of a special multitime NLSE obtained by fixing a multitemporal direction h. Section 3 introduces and studies solutions of the multitime NLSE with a specified f . Section 4 gives solutions of multitime NLSE whose argument is independent of x. Section 4.1 refers to first integral theory in the context of multitime NLSE. Section 4. 2 gives a theorem about the solutions of multitime NLSE with the argument independent of x. Section 5 introduces the multitime NLSE in a Riemannian setting. Section 6 formulates some conclusions that underline the authors' thoughts.
The recent mathematical literature dedicated to the context insists on the following topics: The papers [2][3][4][5] give properties of nonlinear Schrödinger equation; [6,7] underline an exact solution of the single-time Schrödinger equation; [8] introduces and studies the multitime Schrödinger equation; [1,[9][10][11][12] introduce and describe the multitime solitons as solutions of multitime PDEs. The paper [13] refers to solving the time-dependent Schrödinger equation via Laplace transform on t. A single-time Schrödinger equation in Riemannian setting is studied in the paper [14]. The paper [15] comes closest to our style by offering "methods for constructing complex solutions of nonlinear PDEs using simpler solutions".
For possible numerical solutions we can use the techniques from our papers [16][17][18][19], using either discretization with respect to space variable x, or in relation to multitime variable t, or both.

Solutions of a Special Multitime NLSE
We will consider a particular form of the Equation (1), assuming that the vector field h = (h α ) does not depend on x, and the function f and the vector field h are of class C 1 . Consequently, in this case the multitime NLSE has the form (ii) there exists its partial derivative of the second order with respect to x; (iii) we have |u(x, t)| 2 ∈ I 2 for any (x, t) = (x, t 1 , t 2 , . . . , t m ) ∈ I 0 × D 0 ; (iv) and u verifies the relation (2) on I 0 × D 0 (generally, the solutions are not globally defined).
Note that from the relationship (2) it follows that the function ∂ 2 u ∂x 2 is continuous. Suppose m ≥ 2. Let us give some methods to find solutions of multitime NLSE (2), using the orbits of the vector field h. The symmetric ODEs system describes the orbits of the vector field h = (h α ). The first integrals of this ODEs system are C 1 solutions of the PDE h α (t) ∂w ∂t α (t) = 0. (4) Proposition 1. Let k ∈ R be a fixed constant and the subset D 0 ⊆ D be open.
(b) Suppose that the C 1 function ω 0 : D 0 → R verifies the PDE h α (t) ∂ω 0 ∂t α (t) = 1, for any t ∈ D 0 (the local description of functions of type ω 0 is found in Proposition 10, selecting q = 1). Then the function v : If there is an index β ∈ {1, 2, . . . , m} for which the component h β depends only on the variable t β and h β (t β ) = 0, for any t β , and if H is a primitive of the function 1 h β , then we can select ω 0 (t) := H(t β ).
Proof. (a) Replacing the function u in (2), we find Since the function v(·) is a solution of PDE (5), it turns out that the above equality is equivalent to and this last equality is true since the function ϕ is a solution of the PDE (6).
(b) The statements are verified immediately by direct calculation.
Then, the function u : is a solution of multitime NLSE (2).

Corollary 1.
Suppose that the function f does not depend on x, i.e., there exists the C 1 function f : Proof. The constant function ϕ(x, t) := √ q 0 is a solution of PDE (2). The conclusion results immediately by applying Proposition 2.
Proposition 3. Let k ∈ R be a constant and let v : D 0 → C be a C 1 solution of PDE (5), such that |v(t)| = 1, for any t ∈ D 0 (D 0 ⊆ D, D 0 open). Let us consider ϕ(x, t, c), with c = (c 1 , . . . , c r ) ∈ G, a solution of the PDE (6), which depends on r parameters, where G is an open subset of R r .
Hypotheses: the function ϕ : I 0 × D 0 × G → C is of class C 1 with respect to all arguments; there exists ∂ 2 ϕ ∂x 2 and it is continuous with respect to all arguments; furthermore, for each c ∈ G, the function ϕ( · , · , c) : I 0 × D 0 → C is a solution of the multitime NLSE (6) (I 0 subinterval of I 1 ).
Then, the function u : is a solution of the multitime NLSE (2).
In this way we showed that ϕ is a solution of PDE (6). According to Proposition 1, we find that ϕ(x, t) · v(t) is a solution of PDE (2). Due to Proposition 2, it follows that ϕ(x, t) · v(t) · e iΨ r+1 (t) is a solution of the PDE (2).

Proposition 4.
Suppose that G is an open subset of R r . Let us consider ϕ(x, t, c), with c = (c 1 , . . . , c r ) ∈ G, as a solution of the PDE (2) which depends on r parameters.
Hypotheses: the function ϕ : I 0 × D 0 × G → C is of class C 1 with respect to whole arguments; there exists ∂ 2 ϕ ∂x 2 , continuous in all arguments; more, for each c ∈ G, the function ϕ( · , · , c) : Then, the function u : is a solution of multitime NLSE (2).
Now let us use an original ODE that incorporates the De Boer-Ludford ODE in plasma physics and the Stuart-Landau ODE in hydrodynamic stability.

Proposition 5.
Let us consider the second order differential equation where k is a real fixed number (y real solution). Let ξ( · , c 1 , . . . , c r ) : I 0 → R be a solution of the ODE (8), which depends on r parameters, with (c 1 , . . . , c r ) ∈ G, where G is an open subset of R r and I 0 is a subinterval of I 1 . Suppose that the function ξ : I 0 × G → R is of class C 1 with respect to all arguments.
One observes immediately that, for all c = (c 1 , . . . , c r ) ∈ G, the function ϕ( · , · , c) is a solution of the PDE (6). One applies Proposition 3 and we obtain the conclusion.
Applying Proposition 5, for k = 0, v = 1, we find the next result. Proposition 6. Let us consider the second order ODE y + f x, y 2 y = 0.
Impose ξ( · , c 1 , . . . , c r ) as a solution of the Equation (9), which depends on r parameters, ξ( · , c 1 , . . . , c r ) : I 0 → R; with (c 1 , . . . , c r ) ∈ G; where G is an open subset of R r and I 0 is a subinterval of I 1 . Suppose that the function ξ : I 0 × G → R is of class C 1 with respect to all arguments.
We add the C 1 functions E 1 : an open neighborhood of t 0 , such that, for any t ∈ W 0 to have F(t) ∈ U 0 .
We use the auxiliary function ξ : For k ∈ L 3 fixed, the function ξ( · , c 1 , c 2 ; k) is a solution of PDE (8) which depends on two parameters (c 1 , c 2 ) ∈ L 1 × L 2 and has the properties of hypothesis of Proposition 5. From Proposition 5 it follows that the function is a solution of PDE (2); but this solution also depends on one parameter, namely by k. From Proposition 4 (with r = 1, choosing Ψ r+1 = 0) it follows that the function u : is a solution of multitime NLSE (2).

Lemma 1.
Suppose that the function f does not depend on x, i.e., there exists a C 1 function f : I 2 → R, such that f (x, p) = f (p), for all x ∈ R, for all p ∈ I 2 (we have I 1 = R). Let c 1 , c 2 ∈ R, such that c 2 1 ∈ I 2 ; let k := c 2 2 − f (c 2 1 ). (a) The function y : R → C, y(x) = c 1 e ic 2 x , for all x ∈ R, is a solution of the ODE for all x ∈ R, for all t ∈ D, is a solution of multitime NLSE (1).

Proof. (a)
Replacing y in the equality (10) A similar calculation is made with the one from the point a), with k = 0.

Proposition 7.
Let us consider the multitime NLSE (2); suppose there exists the C 1 function f : Then, the function u : is a solution of the multitime NLSE (2).

Lemma 2.
Let k ∈ R be a constant and let g 1 : Let g 2 : I 0 → R be a primitive of the function Then, the function Proposition 8. Let k ∈ R be a constant; let g 1 : I 0 → R \ {0} be a solution of the Equation (11); let g 2 : I 0 → R be a primitive of the function Then, the function u : is a solution of multitime NLSE (2).
Proof. According to Lemma 2, the function y(x) = g 1 (x)e ikg 2 (x) is a solution of equation (12) which is equivalent to the fact that the function ϕ(x, t) := g 1 (x)e ikg 2 (x) is a solution of the Equation (6). From Proposition 1 it follows that u is a solution of multitime NLSE (2).
The function g 1 depends on two constants and the parameter k; the same for the function g 2 . The function v depends also on the parameter k. Using Propositions 4 and 8, a solution can be obtained for the multitime NLSE (2) analogous to Remark 4.
If the function h α 0 depends only on the variable t α 0 and h α 0 (t α 0 ) = 0, for all t α 0 , then In this case we choose ω 0 (t) = t m b m . It is easy to see that the m − 1 functionally independent first integrals for the ODEs system (3) can be chosen in this way The orbits of the parallel vector field b = (b 1 , ..., b m ) are straight lines. Lemma 3. (a) Let a ∈ R be a fixed constant. We consider the ODE y = 2(y 2 − a 2 )y.
The functions (16) are two solutions of the Equation (15) which depend on the parameter c 1 .
(b) Let a ∈ R be a fixed constant. We consider the ODE y = 2(y 2 + a 2 )y.
The functions are two solutions of the Equation (17) which depend on one parameter.
Let a ∈ R be an arbitrary constant. We choose k = 2a 2 − c. The Equation (8) becomes the Equation (15). For this equation we found two solutions of it that depend on one parameter, namely the functions defined by the formulas (16).
Applying Proposition 5, for v(t) := e −ikω 0 (t) , we deduce that the functions are solutions of the Equation (13). (8) becomes (17). Two solutions of it that depend on one parameter are given by relationships (18).
So, two more families of solutions of Equation (13) are obtained: We choose k = −c. The Equation (8) becomes y = 2y 3 . It is easy to see that a solution that depends on one parameter is: Analogously, the functions are solutions of the Equation (13). The functions u a , u a , w a , w a , are solutions of the Equation (13) which depend on the parameter a. According to Proposition 4, it follows that if we replace a by E 3 (F(·)), we obtain again solutions of the Equation (13).
For the functions u a , w a , is more convenient to take a √ 2 as parameter; hence we will replace a √ 2 with E 3 (F(·)). Consequently, we obtain the following solutions of the Equation (13) u( to which the solution θ(x, t) defined by the Formula (19) is added. It is observed that we can give up writing the sign "±" in the above formulas, because replacing function E 2 (F(t)) with function π + E 2 (F(t)) we get the sign change (e iπ = −1).
Using Proposition 7, with v(t; k) := e −ikω 0 (t) (and f (p) = c − 2p), the following solutions of the Equation (13) η( are also obtained. Lemma 4. Let a, B be real constants, such that a + B > 0 and a = 0. We consider the ODE (i) If B < 0, then the functions and are solutions of the ODE (20). The functions g 2,1 , respectively g 2,2 are primitives of functions 1 (ii) If B > 0, then the function is a solution of the Equation (20). We obtain the primitive (iii) If B = 0, then the function is a solution of the Equation (20). We have The functions g 1,1 , g 2,1 are well defined for any x ∈ R. The functions g 1,2 , g 2,2 are well defined for any x = −c 1 √ −B , x ∈ R. The functions g 1,3 , g 2,3 are well defined for any x = −c 1 + π 2 + pπ √ B , with p ∈ Z (x ∈ R). The functions g 1,4 , g 2,4 are well defined for any x = −c 1 , x ∈ R. The functions g 1,1 , g 1,2 , g 1,3 , g 1,4 do not vanish anywhere.
Proof. The functions g 1,1 and g 1,2 are well defined since a + B > 0 and B < 0, hence the expressions that appear under radical in the Formulas (21) and (22) are strictly positive; one remarks immediately that also g 2,1 , g 2,2 are well defined (a = 0, a + B > 0, B < 0).
The function g 1,3 is well defined since in this case a + B > 0 and B > 0, hence the expression that appears below the radical in the Formula (25) is strictly positive; it is immediately noticeable that the function g 2,3 is well defined. In case (iii), we have B = 0 and a + B > 0, i.e., a > 0; we deduce that the functions g 1,4 and g 2,4 are well defined.
From the above it is immediately noticeable that the functions g 1,1 , g 1,2 , g 1,3 , g 1,4 do not vanish anywhere.
We shall show that g 1,1 , g 1,2 , g 1,3 , g 1,4 are solutions of the equation which is equivalent to (since so it is enough to show that g 1,1 , g 1,2 , g 1,3 , g 1,4 are solutions of the Equation (30). Case (i). Denote by T(x) one of the functions: In both situations we have T = √ −B(1 − T 2 ). The relation g 2 1 = a + B − B · T 2 is also true; taking the derivative we obtain The relationship (30) was obtained.
Case (ii). Denote T(x) := tan x √ B + c 1 . We have T = √ B(1 + T 2 ). The relation g 2 1 = a + B + B · T 2 is also true; taking the derivative we obtain From g 2 1 = a + B + B · T 2 , we find , so even in this case the relationship (30) is true. We showed that g 1,1 , g 1,2 , g 1,3 , g 1,4 are solutions of the Equation (30); as we have seen, the relationship (30) is equivalently to the relation (29).
Using the derivative, from the relation (29) we obtain In case (i), from the equality it follows that g 1 vanishes at a finite number of points.
In case (ii), from equality B + c 1 , it follows that zeros of g 1 are isolated points.
In case (iii), from the equality g 1 (x)g 1 (x) = − 1 (x + c 1 ) 3 , it follows that g 1 (x) = 0. Divide the relation (31) by 2g 1 (x), there where g 1 (x) = 0; one obtains It follows from the above that the relationship (32) is fulfilled everywhere, except for some isolated points (possibly). From continuity, it follows that the relation (32) is true throughout (on the definition domain of g 1 ).
For each a ∈ R, we consider the second order algebraic equation (with the unknown B) There exists an interval I such that 0 / ∈ I, c 2 / ∈ I and, for all a ∈ I, to have that, for all a ∈ I, to have (3a − c) 2 − a 3 < 0. The interval can be taken I = (a 0 , ∞). Further we consider a ∈ I. The product of the roots of the Equation (34) is (3a − c) 2 − a 3 < 0. It follows that the roots B 1 (a), B 2 (a) of the Equation (34), are real, distinct and have different signs. These are where D(a) = (a 2 − 6a + 2c) 2 − 4(3a − c) 2 + 4a 3 > 0, a ∈ I. Let us observe that if B is one of the roots of the Equation (34), then a + B = 0. If a + B = 0, then from the equality (33) it follows 3a + B − c = 0, i.e., a = c 2 , what you cannot for a ∈ I. Hence a + B = 0. From the relation (33), we obtain a + B > 0. We proved that for any a ∈ I, the relations a + B 1 (a) > 0 and a + B 2 (a) > 0 are true. Let a ∈ I. We select k = 3a + B 1 (a) − c. From equality (33) it follows k 2 = (a + B 1 (a))a 2 . We notice that the Equation (11) becomes in this case the Equation (20), with B = B 1 (a) < 0. Assumptions of Lemma 4, (i), are satisfied.
For the last example we do not ask anymore (necessarily) a ∈ I. Let a be a real root of the equation a 3 = (3a − c) 2 . There is at least one such solution for that the equation considered has degree three. From a 3 = (3a − c) 2 it follows a ≥ 0 and if a = 0, then c = 0. Hence for c = 0, it follows a > 0. For c = 0, we shall take the root a = 9 > 0.
We select k = 3a − c. From the equality a 3 = (3a − c) 2 we find k 2 = a 3 . One remark that the Equation (11) becomes in this case the Equation (20), with B = 0. Assumptions of Lemma 4, (iii), are satisfied.
Remark 5. From the above solutions of the Equation (13) one obtains solutions corresponding to the PDE
In this Section, we shall determine exact solutions of multitime NLSE (2) of exponential form u(x, t) = ϕ(x, t)e iω(t) , with ϕ(x, t) ∈ R, ω(t) ∈ R, i.e., the argument ω of u does not depend on x. The solutions of this form are suggested by the solutions in the previous sections and by the question: Are there other solutions of this type? These solutions are in fact extensions of spatial solitons in the paper [16].
is a solution of multitime NLSE (2) if and only if ϕ is of class C 1 , there exists ∂ 2 ϕ ∂x 2 , and for any (x, t) ∈ I 0 × D 0 , the following three conditions are satisfied: Proof. We remark that |u| 2 ∈ I 2 if and only if ϕ 2 ∈ I 2 .
(a) If the C 1 function w : D 0 → R verifies, for all t ∈ D 0 , the equation (b) Suppose that the function q is of class C 1 . Let W be an open neighborhood of t 0 , W ⊆ V (as example W = W 0 ). Let us consider a C 1 function E : U → R, U ⊆ R m−1 , U open. Suppose that for any t ∈ W we have F(t) ∈ U. Then, the function w : W ∩ W 0 → R, defined by is of class C 1 , and verifies the relation (42), for all t ∈ W ∩ W 0 .
Proof. (a) Since w is of class C 1 , from the relation (42) we deduce that the function q is continuous. We fix s ∈ J 1 × J 2 × . . . × J m−1 . Hence for any s m ∈ J m , we have ( s, s m ) ∈ W 1 ⊆ H(D 0 ∩ V 0 ). We can define the function From Lemma 5, (b), and the equality (42), it follows We find that the above relationship coincides with the relationship (43). Let t ∈ W 0 . According to Remark 6, we have H(t) = (F(t), t m ) ∈ W 1 . In the Formula (43), we take s = H(t) = (F(t), t m ), i.e., s = F(t), s m = t m , and we obtain the Formula (44) (since G(H(t)) = t).
(b) Since q is of class C 1 , it follows that w is of class C 1 . The variables of the function E are denoted by (s 1 , s 2 , . . . , s m−1 ).
, for all s ∈ W 1 . Then Then, the function Ψ : W → R, Ψ(t) = E F(t) , for all t ∈ W, is a first integral for the ODEs system (3).
Proof. (a) Renumbering the indices, we can assume that α 0 = m.
We choose W 1 and W 0 as in the Remark 6. Let U 0 : Applying Proposition 10, (a), for w = Ψ, q = 0, it follows that, for all t ∈ W 0 , we have (b) A calculation similar to the proof of Proposition 10, (b) is made.

The Theorem about the Solutions with the Argument Independent of x
Let us insist again on solutions of exponential form of the multitime NLSE (2).
We need the functions ϕ : L × D 1 → R, ω : D 1 → R, ω of class C 1 (L subinterval of I 1 , with x 0 ∈ L; D 1 ⊆ D, D 1 open, with t 0 ∈ D 1 ). Suppose that the function is a solution for the multitime NLSE (2). We add the condition ϕ(x 0 , t 0 ) > 0. Let us consider the C 1 functions F 1 , F 2 , . . ., F m−1 : V → R as first integrals of the ODEs system (3), with V ⊆ D, t 0 ∈ V, V open and connected, such that for all t ∈ V, we have h α 0 (t) = 0, and the condition (40) is satisfied.
Let E 1 , E 2 : U 0 → R be defined by we proved the statements iv) and v).
So we also proved the statements from iii).
, for all t ∈ W 0 , for all x ∈ I 0 .
If in the relation (39) it counts x = x 0 , then for any t ∈ W 0 one obtains We define the function E 3 : U 0 → R, , for all s ∈ U 0 .
From the previous formula it follows that E 3 is a continuous function. However, it does not turn out that E 3 is of class C 1 . This remains to be proved.
Since for any t ∈ V we have h m (t) = 0, we deduce that h m has constant sign on V; it follows that, for all s ∈ U 0 , we have E 5 ( s) = 0.
Let ω : , for all t ∈ W 0 . (a) Suppose that the function u : I 0 × W 0 → C is a solution of the multitime NLSE (2), of the form u(x, t) = ϕ(x, t)e iω(t) , for all (x, t) and, for any (x, t) ∈ I 0 × W 0 , we have (b) Converse. Suppose that the relations (51) are true. Then: the function u : I 0 × W 0 → C defined, for any (x, t) ∈ I 0 × W 0 , by the Formula (52) is a solution of the multitime NLSE (2), and satisfies the conditions (50).
The relation (38) is obviously fulfilled because for every x fixed, ϕ(x, · ) is a function that depends on F( · ).
This is a solution similar to those in Remark 4. Proposition 12 and Theorem 1 say that, locally, the solutions of the multitime NLSE (2) whose argument does not depend on x, are only those described in Remark 4 (if we assume in addition that u(x 0 , t 0 ) = 0).

Remark 8.
For m = 1, we easily find that the results obtained in this article remain true under the following conditions: the functions Ψ, respectively Ψ j , which appear in Propositions 2-7, are taken constants; -the functions E j , which appear in Remark 4, in Section 3, in Proposition 12, and in Theorem 1 are taken constant; the functions F 1 , F 2 , . . ., F m−1 are no longer considered in the case m = 1 (instead of E j (F(t)) is put the constant E j ).

Multitime NLSE in Riemannian Setting
From the physical point of view, it would be more important to further study a multitime NLSE in Riemannian setting, which is still an open problem. For that, let (M, g) be a smooth compact Riemannian manifold of dimension n (particularly, n = 2, 3) without boundary. In Riemannian setting, the multitime NLSE is Let dx g denotes the volume element of the compact Riemannian manifold (M, g) and |u| 2 = uū, |∇ g u| 2 g = ∇ g u, ∇ gū g . Open problem Find the minimum of the multitemporal energy functional E(t) = 1 2 M |u| 2 + |∇ g u| 2 g dx g , constrained by the multitime NLSE (54). Hint To solve this problem we need techniques from variational calculus. This multitime NLSE is mainly concerned with the interface between Riemannian geometry and quantum mechanics, but it leads, in a natural way, to questions of functional analysis related to the theory of operators on Hilbert spaces. In some respects these problems are similar to those studied in the standard Euclidean case, but depending on the Riemannian metric g these might go beyond and provide new aspects to the problem.

Conclusions
The exact solutions of multitime NLSE (with oblique derivative) are closely related to the orbits of the direction vector field h. Our techniques for finding these solutions started from this important idea applied to PDEs that contain directional derivative. We follow a different route than those in the papers [1][2][3][4][5][6][7][8][9][10][11][12][13][14][15][16][17][18][19] and carry out the calculations to obtain significative exact solutions of multitime NLSE. This computational paper and the obtained results show that our methods are simple, efficient, straightforward and powerful. Moreover, the presented methods can be employed in many other types of nonlinear PDEs arising in mathematics, mathematical physics, engineering and economics.
The linearization of the multitime NLSE around a solution returns to the linearization of the function f . Linearization offers patterns of solutions that approximate the solutions of the original equation (this corresponds to studying the tangent space at a point of all solutions moduli space).
We hope that this paper will open a door for many readers to a multitime intriguing topic that is an active field of current research, especially in the Riemannian formulation of multitime nonlinear Schrödinger's PDE.