# On the Exact Solutions of Two (3+1)-Dimensional Nonlinear Differential Equations

Authors: Fanning Meng; Yongyi Gu
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this article, exact solutions of two (3+1)-dimensional nonlinear differential equations are derived by using the complex method. We change the (3+1)-dimensional B-type Kadomtsev-Petviashvili (BKP) equation and generalized shallow water (gSW) equation into the complex differential equations by applying traveling wave transform and show that meromorphic solutions of these complex differential equations belong to classW, and then, we get exact solutions of these two (3+1)-dimensional equations.


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## Body

## 1. Introduction and Main Results

Nonlinear differential equations (NLDEs) play an important role in the research of nonlinear science, which has attracted a lot of attentions of the researchers [1–8]. The investigation of NLDEs is helpful for well understanding of nonlinear physical phenomena [9–16]. Numerous methods have been developed for seeking traveling wave exact solutions to NLDEs, such as sine-Gordon expansion method [17], Kudryashov method [18], modified simple equation method [19], Jacobi elliptic function expansion [20], exp(−ψ(z))-expansion method [21, 22], modified extended tanh method [23, 24], generalized (G'/G) expansion method [25], and improved F-expansion method [26].In recent years, Yuan et al. [27] introduced an efficient method named complex method to get exact solutions for NLDEs. The complex method is developed by complex analysis and complex differential equations. More details about the complex method can be found in [28–34]. In this work, we will utilize the complex method to achieve exact solutions of the following two (3+1)-dimensional NLDEs.The (3+1)-dimensional BKP equation [35] is given by
(1)uxxxy+θuxuyx+ux+uy+ust−uxx+uss=0,where θ is a constant.The (3+1)-dimensional gSW equation [36] is given by
(2)uxxxy+3uxxuy+3uxuxy−uyt−uxs=0.ClassW consists of elliptic function or their degeneration. Substituting traveling wave transform
(3)ux,y,s,t=Uz,z=n1x+n2y+n3s+n4t,into Eq. (1), and then integrating it we get
(4)n13n2U′′′+θn12n2U′2+n4n1+n2+n3−n12−n32U′+r=0,where r is the integration constant.Theorem 1.
Ifθn1n2≠0, then meromorphic solutions w of Eq. (4) belong to class W and Eq. (4) has the following solutions where cii=1,2,3,4 are the integral constants.
(i)
The rational function solutions(5)Urz=6n1θ1z−z0+n4n1+n2+n3−n12−n322θn12n2z−z0+c1,where r=−n4n1+n2+n3−n12−n322/4θn12n2,z0∈ℂ.
(ii)
The simply periodic solutions(6)U1sz=−3n1μθcothμz−z02−3n1μ2θlncothμ/2z−z0−1cothμ/2z−z0+1+n4n1+n2+n3−n12−n322−2μ2n13n22θn12n2z−z0+c2,U2sz=−3n1μθtanhμz−z02−3n1μ2θlntanhμ/2z−z0−1tanhμ/2z−z0+1+n4n1+n2+n3−n12−n322−2μ2n13n22θn12n2z−z0+c3,where r=μ4n16n22−n4n1+n2+n3−n12−n322/4θn12n2, z0∈ℂ.
(iii)
The elliptic function solutions(7)Udz=−6n1θζz−ζz0−3n1θ℘′z+G℘z−H+n4n1+n2+n3−n12−n322θn12n2z−z0+c4,where G2=4H3−g2H−g3, g2=4n12n2rθ+n4n1+n2+n3−n12−n322/12n16n22, H and g3 are arbitrary.Substituting traveling wave transform(8)ux,y,s,t=Vz,z=m1x+m2y+m3s+m4t,into Eq. (2), and then integrating it we get
(9)m13m2V′′′+3m12m2V′2−m1m3+m2m4V′+λ=0,where λ is the integration constant.Theorem 2.
Ifm1m2≠0, then meromorphic solutions w of Eq. (9) belong to the class W and Eq. (9) has the following solutions where cii=1,2,3,4 are the integral constants.
(i)
The rational function solutions(10)Vrz=2m1z−z0+m1m3+m2m46m12m2z−z0+c1,where λ=m1m3+m2m42/12m12m2,z0∈ℂ.
(ii)
The simply periodic solutions(11)V1sz=−4m1μcothμz−z02−4m1μlncothμ/2z−z0−1cothμ/2z−z0+1+2m13m2μ2+m1m3+m2m46m12m2z−z0+c2,V2sz=−4m1μtanhμz−z02−4m1μlntanhμ/2z−z0−1tanhμ/2z−z0+1+2m13m2μ2+m1m3+m2m46m12m2z−z0+c3,where λ=m1m3+m2m42−m16m22μ4/12m12m2, z0∈ℂ..
(iii)
The elliptic function solutions(12)Vdz=−2m1ζz−ζz0−m1℘′z+E℘z−F+m1m3+m2m46m12m2z−z0+c4,where E2=4F3−g2F−g3, g2=m1m3+m2m42−12λm12m2/12m16m22, F and g3 is arbitrary.


## 2. Preliminaries

Setm∈ℕ≔1,2,3,⋯, ri∈0,1,2,⋯, i=0,1,⋯,m, r=r0,r1,⋯,rm, and
(13)KrUz≔∏i=0mUizri.The degree ofKrU is defined as dr≔∑i=0mri. The differential polynomial is given by
(14)PU,U′,⋯,Um≔∑r∈JarKrU,where J is a finite index set, then degPU,U′,⋯,Um≔maxr∈Jdr is the degree of PU,U′,⋯,Um.Considering the following equation:(15)PU,U′,⋯,Um=aUn+d,where n∈ℕ, and a≠0,d are constants.Assume that meromorphic solutionsw of Eq. (13) have at least one pole and let p, q ∈ N. Substitute the Laurent series
(16)Uz=∑k=−q∞βkzk,β−q≠0,q>0,into Eq. (15) to determine p distinct Laurent principal parts
(17)∑k=−q−1βkzk,then, Eq. (15) is said to satisfy weak <p,q> condition.It is know that Weierstrass elliptic function℘z≔℘z,g2,g3 has double periods and satisfies:
(18)℘′z2=4℘z3−g2℘z−g3.Weierstrass zeta functionζz is a meromorphic function which satisfies
(19)℘z=−ζ′z.These two Weierstrass functions have the following addition formulas:(20)℘z−z0=−℘z+14℘′z+℘′z0℘z−℘z02−℘z0,ζz−z0=ζz−ζz0+12℘′z+℘′z0℘z−℘z0.Eremenko et al. [37] had investigated the following m-order Briot-Bouquet equation (BBEq)
(21)PU,Um=∑j=0nPjUUmj=0,in which m ∈ N, and Pj(U) are constant coefficient polynomials.Lemma 1 [38–40].
Letm, n, p, s ∈ N, degP(U,U(m)) < n. If the m-order BBEq
(22)PU,Um=aUn+csatisfies weak <p,q> condition; then, meromorphic solutions w belong to class W. Assume that some values of parameters such solutions w exist; then, other meromorphic solutions should form 1 parameter family Uz−z0,z0∈C. In addition, each elliptic solution U with a pole at z=0 is.
(23)Uz=∑i=1s−1∑j=2q−1jβ−ijj−1!dj−2dzj−214℘′z+Di℘z−Bi2−℘z+∑i=1s−1β−i12℘′z+Di℘z−Bi+∑j=2q−1jβ−sjj−1!dj−2dzj−2℘z+β0,where β−ij are determined by (16), Di2=4Bi3−g2Bi−g3 and ∑i=1sβ−i1=0.
Each rational function solution is(24)Rz=∑i=1s∑j=1qβijz−zij+β0,which contains s≤p distinct poles of multiplicity q.
Each simply periodic solution is a rational functionRη of η=eαzα∈ℂ, that is
(25)Rη=∑i=1s∑j=1qβijη−ηij+β0,which contains s≤p distinct poles of multiplicity q.


## 3. Proofs of Main Results

Proof of Theorem 1.
Letu=U′, then Eq. (4) becomes
(26)n13n2u′′+θn12n2u2+n4n1+n2+n3−n12−n32u+r=0.
Substituting (16) into Eq. (4), we have p=1,q=2,β−2=6n1/θ,β−1=0,β0=n4n1+n2+n3−n12−n32/2θn12n2,β1=0,β2=4n12n2rθ+n4n1+n2+n3−n12−n322/40θn15n22,β3=0, and β4 is an arbitrary constant. Thus, Eq. (26) is a second-order BBEq as well as satisfies weak h1,2i condition. Therefore, by Lemma 1, we know that the meromorphic solutions of Eq. (26) belong to class W.From (23) of Lemma 1, we have the form of elliptic solutions of Eq. (26)
(27)ud0z=β−2℘z+β10with pole at z=0.Putud0z into Eq. (26) to yield
(28)ud0z=6n1θ℘z+n4n1+n2+n3−n12−n322θn12n2,where g2=4n12n2rθ+n4n1+n2+n3−n12−n322/12n16n22 and g3 is arbitrary.Therefore, the elliptic solutions of Eq. (26) with arbitrary pole are
(29)udz=6n1θ℘z−z0+n4n1+n2+n3−n12−n322θn12n2,where z0∈C.Therefore, the solutions of Eq. (4) are
(30)Udz=∫udzdz=∫6n1θ℘z−z0+n4n1+n2+n3−n12−n322θn12n2dz=6n1θζz−z0+n4n1+n2+n3−n12−n322θn12n2z−z0+c4=−6n1θζz−ζz0−3n1θ℘′z+G℘z−H+n4n1+n2+n3−n12−n322θn12n2z−z0+c4,where G2=4H3−g2H−g3, g2=4n12n2rθ+n4n1+n2+n3−n12−n322/12n16n22, c4 is the integral constant, and H and g3 are arbitrary.By (24), we infer that the indeterminant rational solutions of Eq.(26) are
(31)R1z=β11z2+β12z+β20with pole at z=0.SubstituteR1z into Eq. (26) to yield
(32)R1z=6n1θ1z2+n4n1+n2+n3−n12−n322θn12n2,where r=−n4n1+n2+n3−n12−n322/4θn12n2.So the rational solutions of Eq. (26) with arbitrary pole are
(33)urz=6n1θ1z−z02+n4n1+n2+n3−n12−n322θn12n2.Therefore, the solutions of Eq. (4) are
(34)Urz=∫urzdz=∫6n1θ1z−z02+n4n1+n2+n3−n12−n322θn12n2dz=−6n1θ1z−z0+n4n1+n2+n3−n12−n322θn12n2z−z0+c1,where c1 is the integral constant, r=−n4n1+n2+n3−n12−n322/4θn12n2,z0∈ℂ.Letη=eμz. To obtain simply periodic solutions, we insert u=Rη into Eq. (26) and get
(35)n13n2μ2ηR′+η2R′′+θn12n2R2+n4n1+n2+n3−n12−n32R+r=0.SubstitutingR2z into the Eq.(35), we obtain that
(36)R21z=6n1θμ2η−12+6n1θμ2η−1+μ2n13n2+n4n1+n2+n3−n12−n3222θn12n2,(37)R22z=6n1θμ2η+12−6n1θμ2η+1+μ2n13n2+n4n1+n2+n3−n12−n3222θn12n2,where r=μ4n16n22−n4n1+n2+n3−n12−n322/4θn12n2.Substitutingη=eμz into Eq. (36) and Eq. (37) yields simply periodic solutions to Eq. (26) with pole at z=0(38)u1s0z=6n1θμ2eμz−12+6n1θμ2eμz−1+μ2n13n2+n4n1+n2+n3−n12−n3222θn12n2=6n1θμ2eμzeμz−12+μ2n13n2+n4n1+n2+n3−n12−n3222θn12n2=3n12θμ2coth2μz2+n4n1+n2+n3−n12−n322−2μ2n13n22θn12n2,u2s0z=6n1θμ2eμz+12−6n1θμ2eμz+1+μ2n13n2+n4n1+n2+n3−n12−n3222θn12n2=−6n1θμ2eμzeμz+12+μ2n13n2+n4n1+n2+n3−n12−n3222θn12n2=3n12θμ2tanh2μz2+n4n1+n2+n3−n12−n322−2μ2n13n22θn12n2,where r=μ4n16n22−n4n1+n2+n3−n12−n322/4θn12n2.So simply periodic solutions of Eq. (4) with arbitrary pole are
(39)u1sz=3n12θμ2coth2μz−z02+n4n1+n2+n3−n12−n322−2μ2n13n22θn12n2,and
(40)u2sz=3n12θμ2tanh2μz−z02+n4n1+n2+n3−n12−n322−2μ2n13n22θn12n2,where r=μ4n16n22−n4n1+n2+n3−n12−n322/4θn12n2.Therefore, the solutions of Eq. (4) are
(41)U1sz=∫u1szdz=∫3n12θμ2coth2μz−z02+n4n1+n2+n3−n12−n322−2μ2n13n22θn12n2dz=−3n1μθcothμz−z02−3n1μ2θlncothμ/2z−z0−1cothμ/2z−z0+1+n4n1+n2+n3−n12−n322−2μ2n13n22θn12n2z−z0+c2,and
(42)U2sz=∫u2szdz=∫3n12θμ2tanh2μz−z02+n4n1+n2+n3−n12−n322−2μ2n13n22θn12n2dz=−3n1μθtanhμz−z02−3n1μ2θlntanhμ/2z−z0−1tanhμ/2z−z0+1+n4n1+n2+n3−n12−n322−2μ2n13n22θn12n2z−z0+c3,where c2 and c3 are the integral constants, r=μ4n16n22−n4n1+n2+n3−n12−n322/4θn12n2, z0∈ℂ.Proof of Theorem 2.
Letv=V′, then Eq. (9) becomes
(43)m13m2v′′+3m12m2v2−m1m3+m2m4v+λ=0.
Substituting (16) into Eq.(9), we have p=1,q=2,β−2=−2m1,β−1=0,β0=m1m3+m2m4/6m12m2,β1=0,β2=12λm12m2−m1m3+m2m42/120m15m2,β3=0, and β4 is an arbitrary constant. Thus, Eq. (43) is a second-order BBEq as well as satisfies weak h1,2i condition. Therefore, by Lemma 1, we know that the meromorphic solutions of Eq. (43) belong to class W.From (23) of Lemma 1, we have the form of elliptic solutions of Eq. (43)
(44)vd0z=β−2℘z+β10with pole at z=0.Putvd0z into Eq. (43) to yield
(45)vd0z=−2m1℘z+m1m3+m2m46m12m2,where g2=m1m3+m2m42−12λm12m2/12m16m22 and g3 is arbitrary.Therefore, the elliptic solutions of Eq. (43) with arbitrary pole are
(46)vdz=−2m1℘z−z0+m1m3+m2m46m12m2,where z0∈C.Therefore, the solutions of Eq. (9) are
(47)Vdz=∫vdzdz=∫−2m1℘z−z0+m1m3+m2m46m12m2dz=−2m1ζz−z0+m1m3+m2m46m12m2z−z0+c4=−2m1ζz−ζz0−m1℘′z+E℘z−F+m1m3+m2m46m12m2z−z0+c4,where E2=4F3−g2F−g3, g2=m1m3+m2m42−12λm12m2/12m16m22, c4 is the integral constant, and F and g3 are arbitrary.By (24), we infer that the indeterminant rational solutions of Eq. (43) are
(48)R1z=β11z2+β12z+β20with pole at z=0.SubstituteR1z into Eq. (43) to yield
(49)R1z=−2m1z2+m1m3+m2m46m12m2,where λ=m1m3+m2m42/12m12m2.So the rational solutions of Eq. (43) with arbitrary pole are
(50)vrz=−2m1z−z02+m1m3+m2m46m12m2.Therefore, the solutions of Eq. (9) are
(51)Vrz=∫vrzdz=∫−2m1z−z02+m1m3+m2m46m12m2dz=2m1z−z0+m1m3+m2m46m12m2z−z0+c1,where c1 is the integral constant, λ=m1m3+m2m42/12m12m2,z0∈ℂ.Letη=eμz. To obtain simply periodic solutions, we insert v=Rη into Eq. (43) and get
(52)m13m2μ2ηR′+η2R′′+3m12m2R2−m1m3+m2m4R+λ=0.SubstitutingR2z into the Eq. (35), we obtain that
(53)R21z=−2m1μ2η−12−2m1μ2η−1+m1m3+m2m4−m13m2μ26m12m2,(54)R22z=−2m1μ2η+12+2m1μ2η+1+m1m3+m2m4−m13m2μ26m12m2,where λ=m1m3+m2m42−m16m22μ4/12m12m2.Substitutingη=eμz into Eq. (53) and Eq. (54) yields simply periodic solutions to Eq. (43) with pole at z=0(55)v1s0z=−2m1μ2eμz−12−2m1μ2eμz−1+m1m3+m2m4−m13m2μ26m12m2=−2m1μ2eμzeμz−12+m1m3+m2m4−m13m2μ26m12m2=−2m1μ2coth2μz2+2m13m2μ2+m1m3+m2m46m12m2,v2s0z=−2m1μ2eμz+12+2m1μ2eμz+1+m1m3+m2m4−m13m2μ26m12m2=2m1μ2eμzeμz+12+m1m3+m2m4−m13m2μ26m12m2=−2m1μ2tanh2μz2+2m13m2μ2+m1m3+m2m46m12m2,where λ=m1m3+m2m42−m16m22μ4/12m12m2.So simply periodic solutions of Eq. (9) with arbitrary pole are
(56)v1sz=−2m1μ2coth2μz−z02+2m13m2μ2+m1m3+m2m46m12m2,and
(57)v2sz=−2m1μ2tanh2μz−z02+2m13m2μ2+m1m3+m2m46m12m2,where λ=m1m3+m2m42−m16m22μ4/12m12m2.Therefore, the solutions of Eq. (9) are
(58)V1sz=∫v1szdz=∫−2m1μ2coth2μz−z02+2m13m2μ2+m1m3+m2m46m12m2dz=−4m1μcothμz−z02−4m1μlncothμ/2z−z0−1cothμ/2z−z0+1+2m13m2μ2+m1m3+m2m46m12m2z−z0+c2,and
(59)V2sz=∫v2szdz=∫−2m1μ2tanh2μz−z02+2m13m2μ2+m1m3+m2m46m12m2dz=−4m1μtanhμz−z02−4m1μlntanhμ/2z−z0−1tanhμ/2z−z0+1+2m13m2μ2+m1m3+m2m46m12m2z−z0+c3,where c2 and c3 are the integral constants, λ=m1m3+m2m42−m16m22μ4/12m12m2, z0∈ℂ.



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