# Higher-Order Stationary Dispersive Equations on Bounded Intervals

Authors: N. A. Larkin; J. Luchesi
Journal: Advances in Mathematical Physics
 (2018)

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## Abstract

A boundary value problem for a stationary nonlinear dispersive equation of2l+1 order on an interval (0,L) was considered. The existence, uniqueness, and continuous dependence of a regular solution have been established.


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## Body

## 1. Introduction

This work concerns the existence, uniqueness, and continuous dependence of regular solutions to a boundary value problem for one class of nonlinear stationary dispersive equations posed on bounded intervals,(1)au+∑j=1l-1j+1Dx2j+1u+uux=fx,l∈N, where a is a positive constant. This class of stationary equations appears naturally while one wants to solve a corresponding evolution equation(2)ut+∑j=1l-1j+1Dx2j+1u+uux=0,l∈N, making use of an implicit semidiscretization scheme:(3)un-un-1h+∑j=1l-1j+1Dx2j+1un+unuxn=0,l∈N,where h>0, [1]. Comparing (3) with (1), it is clear that a=1/h>0 and f(x)=un-1/h.Forl=1, we have the well-known generalized KdV equation and for l=2 the Kawahara equation. Initial value problems for the Kawahara equation, which had been derived in [2] as a perturbation of the Korteweg-de Vries (KdV) equation, have been considered in [3–12] and attracted attention due to various applications of those results in mechanics and physics such as dynamics of long small-amplitude waves in various media [13–15]. On the other hand, last years appeared publications on solvability of initial-boundary value problems for dispersive equations (which included the KdV and Kawahara equations) in bounded and unbounded domains [16–23]. In spite of the fact that there is not some clear physical interpretation for the problems on bounded intervals, their study is motivated by numerics [24]. The KdV and Kawahara equations have been developed for unbounded regions of wave propagations; however, if one is interested in implementing numerical schemes to calculate solutions in these regions, there arises the issue of cutting off a spatial domain approximating unbounded domains by bounded ones. In this case, some boundary conditions are needed to specify a solution. Therefore, precise mathematical analysis of mixed problems in bounded domains for dispersive equations is welcome and attracts attention of specialists in this area [16–19, 21, 25].As a rule, simple boundary conditions atx=0 and x=1 such as u=ux=0x=0,u=ux=uxx=0x=1 for the Kawahara equation were imposed. Different kind of boundary conditions was considered in [19, 26]. Obviously, boundary conditions for (1) are the same as for (2). Because of that, study of boundary value problems for (1) helps to understand solvability of initial-boundary value problems for (2).Last years, publications on dispersive equations of higher orders appeared [7, 9, 10, 21, 27]. Here, we propose (1) as a stationary analog of (2) because the last equation includes classical models such as the KdV and Kawahara equations.The goal of our work is to formulate a correct boundary value problem for (1) and to prove the existence, uniqueness, and continuous dependence on perturbations of f(x) for regular solutions.The paper has the following structure. Section1 is Introduction. Section 2 contains formulation of the problem and main results of the article. In Section 3 we give some useful facts. Section 4 is devoted to the boundary value problem for a complete linear equation, necessary to prove in Section 5 the existence of regular solutions for the original problem. Finally, in Section 6 uniqueness is proved which provided certain restriction on f as well as continuous dependence of solutions.


## 2. Formulation of the Problem and Main Results

For reala>0, consider the following one-dimensional stationary higher-order equation:(4)au+∑j=1l-1j+1D2j+1u+uDu=fin0,Lsubject to boundary conditions(5)Dku0=DkuL=DluL=0,k=0,…,l-1,where 0<L<∞, l∈N, Di=di/dxi, D1≡D are the derivatives of order i∈N, and f∈L2(0,L) is the given function.Throughout this paper we adopt the usual notation(·,·), ·, and ·Hi, i∈N, for the inner product and the norm in L2(0,L) and the norm in Hi(0,L), respectively [28]. Symbols C0,Ci,i∈N, mean positive constants appearing during the text.The main results of the article are the following theorem.Theorem 1.
Letf∈L2(0,L). Then for fixed a>0, problem (4)-(5) admits at least one regular solution u=u(x)∈H2l+1(0,L) such that (6)uH2l+1≤C61+x,f21/2 with C6 depending only on L, l, a, and ((1+x),f2). Moreover, if l≥2 and 1+x,f21/2<2a/3aβ/2 with β=min⁡{a/2,1}, then the solution is uniquely defined and depends continuously on f. For l=1 the uniqueness and continuous dependence are satisfied if 1+x,f21/2 is sufficiently small.


## 3. Preliminary Results

Lemma 2.
For allu∈H1(0,L), such that u(x0)=0 for some x0∈[0,L], one has (7)supx∈0,L⁡ux≤2u1/2Du1/2.Proof.
Letx0∈[0,L], such that u(x0)=0. Then for any x∈(0,L)(8)u2x=∫x0xDu2ξdξ≤2∫x0xuξDξdξ≤2∫0LuxDuxdx≤2uDu.From this, the result follows immediately.We will use the following version of the Gagliardo-Nirenberg’s inequality [29–31].Theorem 3.
For1≤q,r≤∞ suppose u belongs to Lq(0,L) and its derivatives of order m belong to Lr(0,L). Then for the derivatives Diu,0≤i<m the following inequalities hold: (9)DiuLp≤K1DmuLrθuLq1-θ+K2uLq, where (10)1p=i+θ1r-m+1-θ1q, for all θ∈[i/m,1] (the constants K1, K2 depending only on L, m, i, q, and r).We will use the following fixed point theorem [32].Theorem 4 (Schaefer’s fixed point theorem).
LetX be a real Banach Space. Suppose B:X→X is a compact and continuous mapping. Assume further that the set (11)u∈X∣u=λBufor  some0≤λ≤1 is bounded. Then B has a fixed point.We start with the linearized version of (4), (5).


## 4. Linear Problem

Consider the linear equation(12)Au≡au+∑j=1l-1j+1D2j+1u=fin0,Lsubject to boundary conditions (5).Theorem 5.
Letf∈L2(0,L). Then problem (12)-(5) admits a unique regular solution u=u(x)∈H2l+1(0,L) such that (13)uH2l+1≤C3f with C3 depending only on L and a.Proof.
Denote(14)Uu≡Id2l+1u0⋮Dl-1u0uL⋮DluL, where Id2l+1 is the identity matrix of order 2l+1. Suppose f∈C([0,L]) and consider the following problem:(15)Au=f,Uu=0as well as the associated homogeneous problem(16)Au=0,(17)Uu=0.It is known, [33, 34], that (15) has a unique classical solution if and only if (16)-(17) has only the trivial solution.
Letu∈C2l+1([0,L]) be a nontrivial solution of (16)-(17). Multiplying (16) by u and integrating over (0,L), we have (18)au2+∑j=1l-1j+1D2j+1u,u=0. By integration by parts and the principle of finite induction, we calculate (19)D2j+1u,u=∑k=1j-1k+1Dk-1uxD2j+1-kuxx=0x=L+-1j12Djux2x=0x=L for all j∈N. Fixing l∈N and making use of (5), we find that (20)D2j+1u,u=0forj∈1,…,l-1,D2l+1u,u=-1l+112Dlu02.Thus (21)∑j=1l-1j+1D2j+1u,u=-1l+1-1l+112Dlu02=12Dlu02; therefore(22)au2+12Dlu02=0which implies au2≤0. Since a>0, it follows that u≡0.
Therefore, (15) has a unique classical solution u∈C2l+1([0,L]) given by (23)ux=∫0LGx,ξfξdξ, where G:[0,L]×[0,L]→R is Green’s function associated with problem (16)-(17), [33, 34]. That is, (24)Gx,ξ=vx-ξ+∑k=02lukxdkξ,0≤ξ≤x≤L∑k=02lukxdkξ,0≤x<ξ≤L, with (25)ukx=v2l-kx+∑s=k+12lb2l+1-svs-k-1x,k=0,…,2l, where b2l+1-s are the coefficients of (12). The function v is a unique solution to the following initial value problem: (26)au+∑j=1l-1j+1D2j+1u=0inRD2lu0=1,Diu0=0,i=0,…,2l-1, and the continuous real functions dk are determined by uk, v, and (5).
We prove the following estimates.
Estimate I. Multiplying (12) by u, we obtain(27)au2+12Dlu02=f,u.By Cauchy-Schwarz’s inequality, we get(28)u≤1af.
Estimate II. Multiplying (12) by (1+x)u and integrating over (0,L) we have(29)au,1+xu+∑j=1l-1j+1D2j+1u,1+xu=f,1+xu.Integration by parts and the principle of finite induction give (30)D2j+1u,1+xu=∑k=1j-1k+11+xDk-1uxD2j+1-kuxx=0x=L+∑k=1j-1kkDk-1uxD2j-kuxx=0x=L+-1j1+x2Djux2x=0x=L+-1j+12j+12Dju2 for all j∈N. Fixing l∈N and making use of (5), we get (31)D2j+1u,1+xu=-1j+12j+12Dju2forj∈1,…,l-1,D2l+1u,1+xu=-1l+112Dlu02+-1l+12l+12Dlu2.Therefore (32)∑j=1l-1j+1D2j+1u,1+xu=∑j=1l2j+12Dju2+12Dlu02. Applying Schwarz’s inequality on the right-hand side of (29), we conclude(33)uH0l≤C0fwith C0 depending only on L and a.
Estimate III. Rewriting (12) in the form (34)-1l+1D2l+1u=f-au-∑j=1l-1-1j+1D2j+1u, we estimate(35)D2l+1u≤f+au+∑j=1l-1D2j+1u.For l=1 we have ∑j=1l-1-1j+1D2j+1u=0 and for l≥2 denote J={1,…,l-1} and (36)I1=j∈J∣2j+1≤l,I2=j∈J∣l<2j+1<2l+1. Hence we can write (37)∑j=1l-1D2j+1u=∑j∈I1D2j+1u+∑j∈I2D2j+1u. Then (35) becomes(38)D2l+1u≤f+au+∑j∈I1D2j+1u+∑j∈I2D2j+1u.Making use of (33), we get(39)au+∑j∈I1D2j+1u≤a+lC0f.On the other hand, l<2j+1<2l+1 for all j∈I2. Hence, by Theorem 3, there are K1j, K2j, depending only on L and l, such that (40)D2j+1u≤K1jD2l+1uθju1-θj+K2ju,withθj=2j+12l+1. Making use of Young’s inequality with pj=1/θj, qj=1/1-θj, and arbitrary ϵ>0, we get (41)D2j+1u≤ϵD2l+1u+C1jϵu+K2ju, where C1j(ϵ)=qjpjϵ/(K1j)pjqj/pj-1. Summing over j∈I2 and making use of (28), we find(42)∑j∈I2D2j+1u≤lϵD2l+1u+1a∑j∈I2C1jϵ+K2jf.Substituting (39), (42) into (38), we obtain (43)D2l+1u≤lϵD2l+1u+1+a+lC0+1a∑j∈I2C1jϵ+K2jf. Taking ϵ=1/2l, we conclude(44)D2l+1u≤C2f,where C2 depends only on L, l, and a.
Again by Theorem3, for all i=l+1,…,2l, there are K1i, K2i depending only on L and l such that (45)Diu≤K1iD2l+1uθiu1-θi+K2iu,withθi=i2l+1. Making use of (28), (44), we obtain(46)Diu≤K1iC2θia1-θi+K2iaf,fori=l+1,…,2l.Taking into account (33), (44), and (46), we conclude that u∈H2l+1(0,L) and(47)uH2l+1≤C3fwith C3 depending only on L, l, and a. Uniqueness of u follows from (28). In fact, such calculations must be performed for smooth solutions and the general case can be obtained via density arguments. Therefore, the proof of the Theorem 5 is complete.


## 5. Nonlinear Case

Givenu∈H01(0,L), set F≔f-uDu. Clearly, F∈L2(0,L) and by Lemma 2, (48)F≤f+uDu≤f+supx∈0,L⁡ux21/2Du≤f+2u1/2Du3/2. By the Young inequality with p=4, q=4/3, and ϵ=1, we obtain(49)F≤f+uH012.Let w∈H2l+1(0,L) be a unique solution of the linear equation(50)aw+∑j=1l-1j+1D2j+1w=Fin0,Lsubject to boundary conditions (5). By (47), we know additionally that(51)wH2l+1≤C3F≤C3f+uH012. Let us henceforth write Bu=w whenever w is derived from u via (50), (5). We assert that B:H01(0,L)→H01(0,L) is compact and continuous.Indeed, if{uk} is a bounded sequence in H01(0,L), then, in view of estimate (51), we have that sequence {wk} is bounded in H2l+1(0,L). Since H2l+1(0,L) is compactly embedded in H01(0,L), then there exists a convergent in H01(0,L) subsequence {Buks}s=1∞; therefore B is compact.Similarly, letuk→u in H01(0,L), then there are a subsequence {wks}s=1∞ and a function w∈H01(0,L) such that wks→w in H01(0,L). Write (50) in the form (52)awks+∑j=1l-1j+1D2j+1wks=f-uksDuks for all s∈N. Consequently by (51), passing to the limit as s→∞, we find (53)aw+∑j=1l-1j+1D2j+1w=f-uDu. Thus w=Bu. Hence, uk→u in H01(0,L) implies Buk→Bu in H01(0,L). This proves that B is continuous.Finally, we must show that the set(54)u∈H010,L∣u=λBufor  some0≤λ≤1 is bounded in H01(0,L). Assume u∈H01(0,L) such that (55)u=λBufor  some0<λ≤1, then (56)auλ+∑j=1l-1j+1D2j+1uλ=f-uDuin0,L;Dkuλ0=DkuλL=DluλL=0,k=0,…,l-1,that is,(57)au+∑j=1l-1j+1D2j+1u+λuDu=λfin0,Land u satisfies boundary conditions (5).


### 5.1. A Priori Estimates

Estimate IV. Multiplying (57) by u and integrating over (0,L), we have (58)au2+∑j=1l-1j+1D2j+1u,u+λuDu,u=λf,u. Integrating by parts and using boundary conditions (5), we get (59)λuDu,u=0. Hence, similar to (28), we obtain(60)u≤1af.Estimate V. Multiplying (57) by (1+x)u and integrating over (0,L), we have (61)au,1+xu+∑j=1l-1j+1D2j+1u,1+xu+λuDu,1+xu=λf,1+xu. It is easy to verify that (62)∑j=1l-1j+1D2j+1u,1+xu=∑j=1l2j+12Dju2+12Dlu02. Integrating by parts, using boundary conditions (5) and Lemma 2, we get (63)-λuDu,1+xu=λ3∫0Lu3xdx≤13∫0Luxux2dx≤13supx∈0,L⁡ux∫0Lux2dx≤23u5/2Du1/2. By the Young inequality, with p=4, q=4/3, and ϵ=21/4, we obtain (64)λuDu,1+xu≥-12Du2-bu10/3, where b=2-5/33-1/2.Moreover, by the Young inequality with arbitraryϵ>0, we get (65)f,1+xu≤ϵ21+x,u2+12ϵ1+x,f2; therefore (66)a-ϵ21+x,u2+Du2+∑j=2l2j+12Dju2+12Dlu02≤bu10/3+12ϵ1+x,f2. Since (67)∫0L1+xf2dx=f2+∫0Lxf2dx≥f2, it follows from (28) that (68)a-ϵ21+x,u2+Du2+∑j=2l2j+12Dju2+12Dlu02≤12ϵ+ba10/31+x,f22/31+x,f2. Taking ϵ=a>0, we conclude(69)uH0l≤C41+x,f21/2,where (70)C4=1β12a+ba10/31+x,f22/31/2,β=min⁡a2,1.Remark 6.
Note that estimate (69) does not depend on L∈(0,∞). This estimate may be used to prove the existence of a weak solution, u∈H0l(0,L).Estimate VI. Rewriting (57) in the form (71)-1l+1D2l+1u=λf-au-∑j=1l-1-1j+1D2j+1u-λuDu, we estimate(72)D2l+1u≤f+au+∑j∈I1D2j+1u+∑j∈I2D2j+1u+uDu.By (69),(73)au+∑j∈I1D2j+1u≤a+lC41+x,f21/2,uDu≤2u1/2Du3/2≤uH012≤C421+x,f2.Acting in the same way as we have proved (42),(74)∑j∈I2D2j+1u≤lϵD2l+1u+1a∑j∈I2C1jϵ+K2j1+x,f21/2.Substituting (73), (74) into (72), we obtain (75)D2l+1u≤lϵD2l+1u+1a∑j∈I2C1jϵ+K2j1+x,f21/2+1+a+lC41+x,f21/2.Setting ϵ=1/2l, we conclude(76)D2l+1u≤C51+x,f21/2,where C5 depends only on L, l, a, and ((1+x),f2).Making use of (60), (76), and Theorem 3, we obtain(77)Diu≤K1iC5θia1-θi+K2ia1+x,f21/2,fori=l+1,…,2l.Taking into account (69), (76), and (77), we finally conclude(78)uH2l+1≤C61+x,f21/2with C6 depending only on L, l, a, and ((1+x),f2).Applying Theorem4, we complete the proof of the existence part of Theorem 1.


## 5.1. A Priori Estimates

Estimate IV. Multiplying (57) by u and integrating over (0,L), we have (58)au2+∑j=1l-1j+1D2j+1u,u+λuDu,u=λf,u. Integrating by parts and using boundary conditions (5), we get (59)λuDu,u=0. Hence, similar to (28), we obtain(60)u≤1af.Estimate V. Multiplying (57) by (1+x)u and integrating over (0,L), we have (61)au,1+xu+∑j=1l-1j+1D2j+1u,1+xu+λuDu,1+xu=λf,1+xu. It is easy to verify that (62)∑j=1l-1j+1D2j+1u,1+xu=∑j=1l2j+12Dju2+12Dlu02. Integrating by parts, using boundary conditions (5) and Lemma 2, we get (63)-λuDu,1+xu=λ3∫0Lu3xdx≤13∫0Luxux2dx≤13supx∈0,L⁡ux∫0Lux2dx≤23u5/2Du1/2. By the Young inequality, with p=4, q=4/3, and ϵ=21/4, we obtain (64)λuDu,1+xu≥-12Du2-bu10/3, where b=2-5/33-1/2.Moreover, by the Young inequality with arbitraryϵ>0, we get (65)f,1+xu≤ϵ21+x,u2+12ϵ1+x,f2; therefore (66)a-ϵ21+x,u2+Du2+∑j=2l2j+12Dju2+12Dlu02≤bu10/3+12ϵ1+x,f2. Since (67)∫0L1+xf2dx=f2+∫0Lxf2dx≥f2, it follows from (28) that (68)a-ϵ21+x,u2+Du2+∑j=2l2j+12Dju2+12Dlu02≤12ϵ+ba10/31+x,f22/31+x,f2. Taking ϵ=a>0, we conclude(69)uH0l≤C41+x,f21/2,where (70)C4=1β12a+ba10/31+x,f22/31/2,β=min⁡a2,1.Remark 6.
Note that estimate (69) does not depend on L∈(0,∞). This estimate may be used to prove the existence of a weak solution, u∈H0l(0,L).Estimate VI. Rewriting (57) in the form (71)-1l+1D2l+1u=λf-au-∑j=1l-1-1j+1D2j+1u-λuDu, we estimate(72)D2l+1u≤f+au+∑j∈I1D2j+1u+∑j∈I2D2j+1u+uDu.By (69),(73)au+∑j∈I1D2j+1u≤a+lC41+x,f21/2,uDu≤2u1/2Du3/2≤uH012≤C421+x,f2.Acting in the same way as we have proved (42),(74)∑j∈I2D2j+1u≤lϵD2l+1u+1a∑j∈I2C1jϵ+K2j1+x,f21/2.Substituting (73), (74) into (72), we obtain (75)D2l+1u≤lϵD2l+1u+1a∑j∈I2C1jϵ+K2j1+x,f21/2+1+a+lC41+x,f21/2.Setting ϵ=1/2l, we conclude(76)D2l+1u≤C51+x,f21/2,where C5 depends only on L, l, a, and ((1+x),f2).Making use of (60), (76), and Theorem 3, we obtain(77)Diu≤K1iC5θia1-θi+K2ia1+x,f21/2,fori=l+1,…,2l.Taking into account (69), (76), and (77), we finally conclude(78)uH2l+1≤C61+x,f21/2with C6 depending only on L, l, a, and ((1+x),f2).Applying Theorem4, we complete the proof of the existence part of Theorem 1.


## 6. Uniqueness and Continuous Dependence

We separated two cases.(1) Case
l
≥
2. Let u1 and u2 be two distinct solutions of (4)-(5). Then the difference u=u1-u2 satisfies the equation(79)au+∑j=1l-1j+1D2j+1u+u1Du+uDu2=0and boundary conditions (5).Multiplying (79) by u and integrating over (0,L), we have(80)au2+∑j=1l-1j+1D2j+1u,u+u1Du,u+uDu2,u=0.Integrating by parts and using boundary conditions (5), we get (81)u1Du,u≥-12supx∈0,L⁡Du1xu2. Similarly, (82)uDu2,u≥-supx∈0,L⁡Du2xu2. We reduce (80) to the inequality(83)a-12supx∈0,L⁡Du1x-supx∈0,L⁡Du2xu2≤0.For i=1,2, we have ui∈H2l+1(0,L) and Dui(L)=Dui(0)=0. By Lemma 2 and estimate (69), we obtain (84)supx∈0,L⁡Duix≤2Dui1/2D2ui1/2≤22Dui+D2ui≤2uiH0l≤2C41+x,f21/2 with (85)C4=1β12a+2-5/33-1/3a10/31+x,f22/31/2. Therefore (83) can be rewritten as(86)a-322C41+x,f21/2u2≤0.Since (87)2a3aβ2<234a7/4∀a>0, it follows that if 1+x,f21/2<2a/3aβ/2, for fixed a>0, then (88)C4<1aβ1/2 and consequently (89)a-322C41+x,f21/2>0. Hence, (86) implies u=0 and uniqueness is proved for l≥2.To show continuous dependence of solutions of perturbations off(x), let fi∈L2(0,L) such that(90)1+x,fi21/2<2a3aβ2,i=1,2.Consider u1 and u2 solutions of (4)-(5) with the right-hand sides f1 and f2, respectively. Then, similar to (83), u1-u2 satisfies the following inequality: (91)a-12supx∈0,L⁡Du1x-supx∈0,L⁡Du2xu1-u2≤f1-f2 which can be rewritten as (92)a-322C4MMu1-u2≤f1-f2, where (93)M=max⁡1+x,f121/2,1+x,f221/2,C4M=1β12a+ba10/3M4/31/2.Making use of (90), we obtain (94)u1-u2≤C7f1-f2 with C7=a-32/2C4MM-1>0. Hence f1-f2→0 implies u1-u2→0. This proves the continuous dependence for l≥2.(2) Case
l
=
1. For l=1, problem (4)-(5) becomes(95)au+D3u+uDu=fin0,L(96)u0=uL=DuL=0.Let u1 and u2 be two distinct solutions of (95)-(96). Then the difference u=u1-u2 satisfies the equation(97)au+D3u+12Du12-u22=0and boundary conditions (96).Multiplying (97) by u and integrating over (0,L), we have(98)au2+12Du02+12Du12-u22,u=0.Integrating by parts and using the boundary conditions (96), we get (99)12Du12-u22,u=12Du1+u2u,u=-12u1+u2u,Du=-14u1+u2,Du2=14Du1+u2,u2≤14supx∈0,L⁡Du1+u2xu2 and (98) becomes(100)a-14supx∈0,L⁡Du1x+supx∈0,L⁡Du2xu2≤0. By (60), (69), it follows that(101)D3ui≤2f+C421+x,f2,i=1,2.According to Theorem 3 and (60), (101), we estimate for i=1,2(102)supx∈0,L⁡Duix≤K1D3ui1/2ui1/2+K2ui≤K12D3ui+K12+K2ui≤K12C421+x,f2+K1+K12a+K2af. Suppose 1+x,f21/2<1, then ((1+x),f2)<1+x,f21/2; therefore (103)supx∈0,L⁡Duix≤K12C42+K1+K12a+K2a1+x,f21/2,i=1,2. Hence we can rewrite (100) as follows:(104)a-K14C42+K12+K14a+K22a1+x,f21/2u2≤0.For a fixed a>0 assume that (105)1+x,f21/2<min⁡234a7/4,4a2β1+2aβ+βK1+2βK2. Then C42<1/aβ and (106)a-K14C42+K12+K14a+K22a1+x,f21/2>0; hence (104) implies u=0. Assuming that, for l=1, (107)1+x,f21/2<min⁡1,234a7/4,4a2β1+2aβ+βK1+2βK2, we complete the proof of uniqueness. The continuous dependence for this case follows in the same manner as it has been done for the case l≥2 provided 1+x,f21/2 is sufficiently small.This completes the proof of the uniqueness and continuous dependence part of Theorem1.



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*Source: 7874305-2018-01-23.xml*