# Painlevé Analysis and Abundant Meromorphic Solutions of a Class of Nonlinear Algebraic Differential Equations

Authors: Yongyi Gu; Xiaoxiao Zheng; Fanning Meng
Journal: Mathematical Problems in Engineering
 (2019)

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## Abstract

In this paper, a class of nonlinear algebraic differential equations (NADEs) is studied. The Painlevé analysis of the NADEs is considered. Abundant meromorphic solutions of the NADEs are obtained by means of the complex method. Then, meromorphic exact solutions of the Schamel-Korteweg-de Vries (S-KdV) equation and(2+1)-dimensional sine-Gordon equation are derived via the applications of the NADEs.


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## Body

## 1. Introduction and Main Results

Nonlinear differential equations are universally applied in plasma physics, fluid dynamics, nonlinear optics, solid state physics, biology, chemistry, etc. For example, the impulsive phenomena [1–4] and singular behaviors [5–8] often show some blow-up properties [9–13] which happen in a lot of complex physical processes. To solve various differential equations, some analytical tools as well as symbolic calculation techniques were established, for instance, fixed point theorems [14–17], variational methods [18–21], topological degree method [22–25], iterative techniques [26–29], modified Kudryashov method [30, 31], exp function method [32, 33], sine-Gordon expansion method [34–37], and complex method [38–41]. More works about the differential equations can be read in [42–52].A meromorphic functionf is said to belong to the class W, if f is a rational function of z or an elliptic function or a rational function of eαz,α∈C. In this paper, we use Painlevé analysis [53, 54] and the complex method to study a class of NADEs as follows:(1)Aww′′+w′2+Bnwn+Bn-1wn-1+⋯+B1w+B0=0,where A, Bn(n=1,2,3,⋯) are constants and we give some applications of this equation. Consequently, meromorphic exact solutions of the S-KdV equation and (2+1)-dimensional sine-Gordon equation are obtained.Theorem 1.
Ifn≥5, then (1) does not pass the Painlevé test.Theorem 2.
Ifn=4, AB4≠0, then the meromorphic solutions w of (1) belong to the class W. In addition, (1) has the following classes of solutions.
(I) The rational function solutions(2)w1r,1z=-3AB4B41z-z0-3B310B4,where B3=5B4B2/3, B1=B2B4B2/4B4, B0=B22/48B4,(3)w1r,2z=-3AB4B41z-z0-1z-z1-z0,where B3=10-3AB4/3z1, B2=0, B1=0, B0=0, z1≠0,(4)w1r,3z=-3AB4B41z-z0-1z-z1-z0-3z1,where B3=20-3AB4/3z1, B2=-36A/z12, B1=0, B0=0, z1≠0, z0∈C.
(II) The simply periodic solutions(5)w1s,1z=-5Aα22B3coth⁡α2z-z0-1,where B2=-2Aα2, 25Aα2B4=-3B32, B1=0, B0=0,(6)w1s,2z=-5Aα22B3tanh⁡α2z-z0-1,where B2=-2Aα2, 25Aα2B4=-3B32, B1=0, B0=0,(7)w1s,3z=-3Aα2B3coth⁡α2z-z0,where B3=0, B2=Aα2, B1=0, B0=-9A3α6(9Aα2+4B4)/16B43,(8)w1s,4z=-10Aα2B3coth⁡αz-z0-1,where B4=-3B32/100Aα2, B2=-8Aα2, B1=0, B0=0,(9)w1s,5z=-3Aα2B41sinh⁡αz-z0,where B3=0, B2=-2Aα2, B1=0, B0=0,(10)w1s,6z=-3Aα2B4tanh⁡αz-z0,where B3=0, B2=4Aα2, B1=0, B0=3A2α4/B4,(11)w1s,7z=-3Aα2B4coth⁡αz-z0,where B3=0, B2=4Aα2, B1=0, B0=3A2α4/B4, z0∈C.
(III) The exponential function solutions(12)w1ez=-5Aα2B3eαz-z0eαz1+1eαz-z0-eαz1eαz-z0-1,where B4=-3z1-12B32/25Aα2z1+12, B2=-2Aα2, B1=0, B0=0, z1≠0, z0∈C.
(IV) The elliptic function solutions(13)w1dz=-3AB4B4-℘+E4E℘2+4E2℘+2F℘′-g2℘-Eg212E2-g2℘+4E3-3Eg2℘′+4℘3+12E℘2-3g2℘-Eg2F,where g3=0, F2=4E3-g2E, E and g2 are arbitrary.Theorem 3.
Ifn=3, AB3≠0, then the meromorphic solutions w of (1) belong to the class W. In addition, (1) has the following classes of solutions.
(I) The rational function solutions(14)w2r,1z=-10AB31z-z02-5B212B3,where B1=5B22/16B3, B0=25B23/864B32, z0∈C.
(II) The simply periodic solutions(15)w2s,1z=-5A2B3α2coth2α2z-z0+5A3α2-16B3,w2s,2z=-5A2B3α2tanh2α2z-z0+5A3α2-16B3,and(16)w2s,3z=-5A2B3α2tanh2α2z-z0-5A2B3α2coth2α2z-z0+5A3α2-13B3,where z0∈C. B2=-2Aα2+2A, B1=(-10A2α2+5A2)/4B32, B0=-25A33α2-1/108B32, in the first two cases; B2=(20A2α2+5A2+3B1B3)/5A, B1=-5(4α2-1A2/B3), B0=-50A36α2-1/27B32 in the third case.
(III) The elliptic function solutions(17)w2dz=-10AB3℘z-z0,5B22-16B1B3240A2,432B0B32-120B1B2B3+25B2343200A3-5B212B3,where z0∈C.


## 2. Preliminaries

Setm∈N≔1,2,3,…,, rj∈0,1,2,⋯,j=0,1,…,m, r=(r0,r1,…,rm), and(18)Krwz≔∏j=0mwjzrj;then the degree of Kr[w] is defined by d(r)≔∑j=0mrj. The differential polynomial is denoted by(19)Pw,w′,…,wm≔∑r∈JarKrw,in which J is the finite index set; then deg⁡P(w,w′,…,wm)≔maxr∈J⁡dr is the degree of P(w,w′,…,w(m)).Consider the following ODE:(20)Pw,w′,…,wm=bwn+d,where b≠0,d are constants, and n∈N.Letp,q∈N, and suppose that the meromorphic solutions w of (20) have at least one pole. Substituting the Laurent series(21)wz=∑τ=-q∞βτzτ,q>0,β-q≠0,into (20) to determine p distinct Laurent singular parts,(22)∑τ=-q-1βτzτ,then we say that (20) satisfies the weak p,q condition.Weierstrass elliptic function℘(z)≔℘(z,g2,g3) with double periods satisfies the equation as (23)℘′z2=4℘z3-g2℘z-g3and has the following addition formula:(24)℘z-z0=-℘z+14℘′z+℘′z0℘z-℘z02-℘z0.Them-order Briot-Bouquet equation (BBEq) was studied by Eremenko et al. [55] and is given as follows:(25)Pw,wm=∑i=0nPiwwmi=0,in which Pi(w) are constant coefficient polynomials, and m∈N.Lemma 4 (see [53, 56]).
Letp,s,m,n∈N,deg⁡P(w,w(m))<n. If m-order BBEq(26)Pw,wm=cwn+dsatisfies the weak p,q condition, then the meromorphic solutions w belong to the class W. Suppose, for some values of parameters such solutions w exist, then other meromorphic solutions should form 1 parameter family w(z-z0),z0∈C. Then, each elliptic solution with a pole at z=0 has the following form:(27)wz=∑i=1s-1∑j=2q-1jβ-ijj-1!dj-2dzj-214℘′z+Ci℘z-Di2-℘z+∑i=1s-1β-i12℘′z+Ci℘z-Di+∑j=2q-1jβ-sjj-1!dj-2dzj-2℘z+β0,where β-ij are determined by (21), Ci2=4Di3-g2Di-g3 and ∑i=1sβ-i1=0.Each rational function solution has the following form:(28)Rz=∑i=1s∑j=1qβijz-zij+β0,and it has s(≤p) distinct poles of multiplicity q.Each simply periodic solution hass(≤p) distinct poles of multiplicity q, and it is a rational function R(η) of η=eαz(α∈C), which is(29)Rη=∑i=1s∑j=1qβijη-ηij+β0.For the convenience of readers, we now show the complex method as follows.Step 1.
Substitute the transformT:u(x,t)→w(z), (x,t)→z into the given PDE to obtain ODE.Step 2.
Substitute (21) into (20) to obtain the weak 〈p,q〉 condition.Step 3.
Substitute the indeterminacy forms which are proposed in Lemma4 into (20) and then obtain its meromorphic solutions with the pole at z=0.Step 4.
Applying the addition formula and Lemma4, we obtain the meromorphic solutions w(z-z0).Step 5.
Putting the inverse transformT-1 into w(z-z0), we obtain the exact solutions to the original PDE.


## 3. Painlevé Analysis

The Painlevé test for NADEs is powerful approach to test the integrable differential equations [53, 54].The dominant term determines the multiplicityk of meromorphic solutions w of D(z,w). The dominant part D^(z,w) consists of all dominant terms. Every term of D^(z,w) has the same multiplicity, which is denoted by I(k).The derivative ofD^(z,w) with respect to w is defined by(30)D^′z,w=limλ→0⁡D^z,w+λv-D^z,wλv,which is a linear operator. The root of the equation(31)Pi=limz→0z-i+IkD^′z,wzi-k=0,is called the Fuchs index.The leading members of (1) have the form(32)Aww′′+Aw′2+Bnwn=0.Substituting(33)w=c-kz-kinto (32), we obtain the value of the pole k=2/(n-2) of the general solution. When n≥5, the value of the pole is less than one: k≤2/3. Therefore (1) does not pass the Painlevé test when n≥5. If n=4, then k=1, c-1=-3AB4/B4. If n=3, then k=2, c-2=-10A/B3.We will get the Fuchs indices of these two cases:(34)Aww′′+w′2+B4w4+B3w3+B2w2+B1w+B0=0,(35)Aww′′+w′2+B3w3+B2w2+B1w+B0=0.The dominant part of (34) is D^(z,w)=Aww′′+Aw′2+B4w4, and we have I1(k)=4k, (36)Pi=limz→0z-i+I1kD^′z,c-kz-kzi-k=limz→0z-i+4kAc-kz-kd2dz2+Ac-kz-k′′+2Aw′ddz+4B4c-kz-k3zi-k=Ac-ki-ki-k-1+Ac-kk+1k+2Ac-kki-k+4B4c-k3=0.Substituting k=1, c-1=-3AB4/B4 into (36), we obtain i1=-1, i2=6.The dominant part of (35) is D^(z,w)=Aww′′+Aw′2+B3w3, and we have I2(k)=3k, (37)Pi=limz→0z-i+I2kD^′z,c-kz-kzi-k=limz→0z-i+3kAc-kz-kd2dz2+Ac-kz-k′′+2Aw′ddz+3B3c-kz-k2zi-k=Ac-ki-ki-k-1+Ac-kk+1k+2Ac-kki-k+3B3c-k2=0.Substituting k=2, c-2=-10A/B3 into (37), we obtain i1=-1, i2=10.Considering these Fuchs indices, we know that (34) and (35) can pass the Painlevé test. Therefore, we can have the meromorphic exact solutions of the NADEs. We will show these solutions in the following sections.Remark 5.
From above analysis, we know that Theorem1 holds.


## 4. Proof of Theorem2

Substituting (21) into (34), we obtain p=2, q=1, β-1=-3AB4/B4, β0=-3B3/10B4, β1=3A25BD-9C2/100-3BA3/2, β2=(500B2E-225BCD+54C3)/6000AB2, and β3 is an arbitrary constant.Multiplying (34) by ww′ and integrating it yields(38)A2ww′2+B46w6+B35w5+B24w4+B13w3+B02w2+δ=0,where δ is an integrable constant.Hence, (38) is a first-order BBEq which satisfies the weak 2,1 condition. Therefore, by Lemma 4, the meromorphic solutions of (38) w∈W. So we obtain that the meromorphic solutions of (34) w∈W.By (28), we know that the indeterminate rational solutions of (34) with pole at z=0 are R1(z). We substitute R1(z) into (34) to obtain the following three forms:(39)R11,1z=-3AB4B41z-3B310B4,where B3=5B4B2/3, B1=B2B4B2/4B4, B0=B22/48B4,(40)R11,2z=-3AB4B41z-1z-z1,where B3=10-3AB4/3z1, B2=0, B1=0, B0=0, z1≠0,(41)R11,3z=-3AB4B41z-1z-z1-3z1,where B3=20-3AB4/3z1, B2=-36A/z12, B1=0, B0=0, z1≠0.So the rational solutions of (34) are(42)w1r,1z=-3AB4B41z-z0-3B310B4,where B3=5B4B2/3, B1=B2B4B2/4B4, B0=B22/48B4,(43)w1r,2z=-3AB4B41z-z0-1z-z1-z0,where B3=10-3AB4/3z1, B2=0, B1=0, B0=0, z1≠0,(44)w1r,3z=-3AB4B41z-z0-1z-z1-z0-3z1,where B3=20-3AB4/3z1, B2=-36A/z12, B1=0, B0=0, z1≠0, z0∈C.Letη=eαz. To derive simply periodic solutions, we substitute w=R(η) into (34) to yield(45)Aα2RηR′+η2R′′+Aα2R′η2+B4w4+B3w3+B2w2+B1w+B0=0.We substituteR2(z) into (45) to obtain that(46)R12,1η=-5Aα2B31η-1,where B2=-2Aα2, 25Aα2B4=-3B32, B1=0, B0=0,(47)R12,2η=5Aα2B31η+1,where B2=-2Aα2, 25Aα2B4=-3B32, B1=0, B0=0,(48)R12,3η=-3Aα2B3η+1η-1,where B3=0, B2=Aα2, B1=0, B0=-9A3α6(9Aα2+4B4)/16B43,(49)R12,4η=-20Aα2B31η-1η+1,where B4=-3B32/100Aα2, B2=-8Aα2, B1=0, B0=0,(50)R12,5η=-12Aα2B4ηη-1η+1,where B3=0, B2=-2Aα2, B1=0, B0=0,(51)R12,6η=-3Aα2B4η2-1η2+1,where B3=0, B2=4Aα2, B1=0, B0=3A2α4/B4,(52)R12,7η=-3Aα2B4η2+1η-1η+1,where B3=0, B2=4Aα2, B1=0, B0=3A2α4/B4,(53)R12,8η=-5Aα2B3ηη1+1η-η1η-1,where B4=-3z1-12B32/25Aα2z1+12, B2=-2Aα2, B1=0, B0=0, z1≠0.Substitutingη=eαz into (46)-(53), we achieve simply periodic solutions and exponential function solutions to (34) with pole at z=0(54)w1s0,1z=-5Aα2B31eαz-1=-5Aα22B3coth⁡α2z-1,w1s0,2z=5Aα2B31eαz+1=-5Aα22B3tanh⁡α2z-1,w1s0,3z=-3Aα2B3eαz+1eαz-1=-3Aα2B3coth⁡α2z,w1s0,4z=-20Aα2B31eαz-1eαz+1=-10Aα2B3coth⁡αz-1,w1s0,5z=-12Aα2B4eαzeαz-1eαz+1=-3Aα22B41sinh⁡α/2zcoth⁡α/2z=-3Aα2B41sinh⁡αz,w1s0,6z=-3Aα2B4e2αz-1e2αz+1=-3Aα2B4tanh⁡αz,w1s0,7z=-3Aα2B4e2αz+1eαz-1eαz+1=-3Aα2B4coth⁡αz,and(55)w1e0z=-5Aα2B3eαzeαz1+1eαz-eαz1eαz-1.Thus simply periodic solutions of (34) are(56)w1s,1z=-5Aα22B3coth⁡α2z-z0-1,where B2=-2Aα2, 25Aα2B4=-3B32, B1=0, B0=0,(57)w1s,2z=-5Aα22B3tanh⁡α2z-z0-1,where B2=-2Aα2, 25Aα2B4=-3B32, B1=0, B0=0,(58)w1s,3z=-3Aα2B3coth⁡α2z-z0,where B3=0, B2=Aα2, B1=0, B0=-9A3α6(9Aα2+4B4)/16B43,(59)w1s,4z=-10Aα2B3coth⁡αz-z0-1,where B4=-3B32/100Aα2, B2=-8Aα2, B1=0, B0=0,(60)w1s,5z=-3Aα2B41sinh⁡αz-z0,where B3=0, B2=-2Aα2, B1=0, B0=0,(61)w1s,6z=-3Aα2B4tanh⁡αz-z0,where B3=0, B2=4Aα2, B1=0, B0=3A2α4/B4,(62)w1s,7z=-3Aα2B4coth⁡αz-z0,where B3=0, B2=4Aα2, B1=0, B0=3A2α4/B4, z0∈C.Thus exponential function solutions of (34) are(63)w1ez=-5Aα2B3eαz-z0eαz1+1eαz-z0-eαz1eαz-z0-1,where B4=-3z1-12B32/25Aα2z1+12, B2=-2Aα2, B1=0, B0=0, z1≠0, z0∈C.From (27), we obtain the indeterminate relations to elliptic solutions of (34) with pole at z=0(64)w1d0z=β-12℘′z+C1℘z-D1+β20,where C12=4D13-g2D1-g3. Considering the results obtained above, we infer that β20=0, g3=0, C1=D1=0. So we obtain(65)w1d0z=-3AB4B4℘′z℘z,where g3=0.So, the elliptic function solutions of (34) are(66)w1dz=-3AB4B4℘′z-z0,g2,0℘z-z0,g2,0,where z0∈C, g3=0, g2 is arbitrary. We can apply the addition formula to rewrite it as(67)w1dz=-3AB4B4-℘+E4E℘2+4E2℘+2F℘′-g2℘-Eg212E2-g2℘+4E3-3Eg2℘′+4℘3+12E℘2-3g2℘-Eg2F,where g3=0, F2=4E3-g2E, and E and g2 are arbitrary.The properties of the solutions are shown in Figures1–3.Figure 1
The 3D and 2D surfaces ofw1r,3(z) by considering z=x+t, A=-1, B4=3, z1=1, z0=1, and t=0 for the 2D graphic.Figure 2
The 3D and 2D surfaces ofw1s,1(z) by considering z=x+t, A=-2, B3=5, z0=1, α=1, and t=0 for the 2D graphic.Figure 3
The 3D and 2D surfaces ofw1s,5(z) by considering z=x+t, A=-3, B4=3, z0=1, α=1, and t=0 for the 2D graphic.


## 5. Proof of Theorem3

Substituting (21) into (35) we obtain p=1, q=2, β-2=-10A/B3,β-1=0, β0=-5B2/12B3, β1=0, β2=(16B1B3-5B22)/480AB3, β3=0, and β4 is an arbitrary constant.Multiplying (35) by ww′ and integrating it yields(68)A2ww′2+B35w5+B24w4+B13w3+B02w2+λ=0,where λ is an integrable constant.Hence, (68) is a first-order BBEq which satisfies the weak 1,2 condition. Therefore, by Lemma 4, all meromorphic solutions of (68) w∈W. So we know that all meromorphic solutions of (35) w∈W.By (28), we know that the indeterminate rational solutions of (35) with pole at z=0 is R1(z). We substitute R1(z) into (35) to obtain the following form:(69)R21,1z=-10AB31z2-5B212B3,where B2=45B1B3/5, B0=(25/27)(B1B1B3/B3).Therefore the rational solutions of (35) are(70)w2r,1z=-10AB31z-z02-5B212B3,where B2=45B1B3/5, B0=(25/27)(B1B1B3/B3), z0∈C.Letη=eαz. To derive simply periodic solutions, we substitute w=R(η) into (35) to yield(71)Aα2RηR′+η2R′′+Aα2R′η2+B3w3+B2w2+B1w+B0=0.We substituteR2(z) into (71) to obtain that(72)R22,1η=-10AB3α2η-12-10AB3α2η-1-5A6B3,where B2=-2Aα2+2A, B1=(-10A2α2+5A2)/4B32, B0=-25A33α2-1/108B32,(73)R22,2η=-10AB3α2η+12+10AB3α2η+1-5A6B3,where B2=-2Aα2+2A, B1=(-10A2α2+5A2)/4B32, B0=-25A33α2-1/108B32,(74)R22,3η=-10AB3α2η-12-10AB3α2η-1-10AB3α2η+12+10AB3α2η+1-5A3B3,where B2=(20A2α2+5A2+3B1B3)/5A, B1=-5(4α2-1A2/B3), B0=-50A36α2-1/27B32.Substitutingη=eαz into (72)-(74), we achieve simply periodic solutions to (35) with pole at z=0(75)w2s0,1z=-10AB3α2eαz-12-10AB3α2eαz-1-5A6B3=-10AB3α2eαzeαz-12-5A6B3=-5A2B3α2coth2α2z+5A3α2-16B3,w2s0,2z=-10AB3α2eαz+12+10AB3α2eαz+1-5A6B3=10AB3α2eαzeαz+12-5A6B3=-5A2B3α2tanh2α2z+5A3α2-16B3,and(76)w2s0,3z=-10AB3α2eαz-12-10AB3α2eαz-1-10AB3α2eαz+12+10AB3α2eαz+1-5A3B3=-10AB3α2eαzeαz-12+10AB3α2eαzeαz+12-5A3B3=-5A2B3α2tanh2α2z-5A2B3α2coth2α2z+5A3α2-13B3.Thus simply periodic solutions of (35) are(77)w2s,1z=-5A2B3α2coth2α2z-z0+5A3α2-16B3,w2s,2z=-5A2B3α2tanh2α2z-z0+5A3α2-16B3,and(78)w2s,3z=-5A2B3α2tanh2α2z-z0-5A2B3α2coth2α2z-z0+5A3α2-13B3,where z0∈C. B2=-2Aα2+2A, B1=(-10A2α2+5A2)/4B32, B0=-25A33α2-1/108B32, in the first two cases and B2=(20A2α2+5A2+3B1B3)/5A, B1=-5(4α2-1A2/B3), B0=-50A36α2-1/27B32 in the third case.From (27) of Lemma 4, the elliptic solutions of (35) are expressed as(79)w2d0z=β-2℘z+β30,with pole at z=0.Puttingw2d0(z) into (35), we get(80)w2d0z=-10AB3℘z-5B212B3,where g2=-(16B1B3-5B22)/240A2, and g3=(432B0B32-120B1B2B3+25B23)/43200A3.So, the elliptic solutions of (35) are(81)w2dz=-10AB3℘z-z0,5B22-16B1B3240A2,432B0B32-120B1B2B3+25B2343200A3-5B212B3,where z0∈C.The properties of the solutions are shown in Figures4 and 5.Figure 4
The 3D and 2D surfaces ofw2r,1(z) by considering z=x+t, A=-1, B2=120, B3=10, z0=1, and t=0 for the 2D graphic.Figure 5
The 3D and 2D surfaces ofw2s,3(z) by considering z=x+t, A=-2, B3=5, z0=1, α=1, and t=0 for the 2D graphic.


## 6. Some Applications of the Theorems

### 6.1. S-KdV Equation

S-KdV equation [57] is given by(82)ut+au1/2u+buux+cuxxx=0.where c≠0, a, and b are constants.Substituting traveling wave transform(83)ux,t=ν2x,t,νx,t=wz,z=kx-lt,into (82), and integrating it yields(84)ck2ww′′+w′2+b4w4+a3w3-l2w2+d=0,where d is the integral constant.Equation (84) is converted to (34) where(85)A=ck2,B4=b4,B3=a3,B2=-l2,B1=0,B0=d.Thus, by Theorem 2, meromorphic solutions of (84) w∈W.


### 6.2.(2+1)-Dimensional Sine-Gordon Equation

In 2002, Fan and Hon [58] used generalized tanh method to study the following (2+1)-dimensional sine-Gordon equation:(86)utt-uxx-uyy+μ2sin⁡u=0.where δ is a constant.Substituting the transformations(87)vx,t=wz,v=eiu,z=x+ly+rt,into (86) yields(88)2r2-l2-1ww′′+w′2+μ2w3-μ2w=0.Equation (88) is converted to (35) where(89)A=2r2-l2-1,B3=μ2,B2=0,B1=-μ2,B0=0.Thus, by Theorem 3, meromorphic solutions of (88) w∈W.Remark 6.
By using Theorems2 and 3, we achieve meromorphic exact solutions of the S-KdV equation and (2+1)-dimensional sine-Gordon equation. Here, we do not show in detail for simplicity.


## 6.1. S-KdV Equation

S-KdV equation [57] is given by(82)ut+au1/2u+buux+cuxxx=0.where c≠0, a, and b are constants.Substituting traveling wave transform(83)ux,t=ν2x,t,νx,t=wz,z=kx-lt,into (82), and integrating it yields(84)ck2ww′′+w′2+b4w4+a3w3-l2w2+d=0,where d is the integral constant.Equation (84) is converted to (34) where(85)A=ck2,B4=b4,B3=a3,B2=-l2,B1=0,B0=d.Thus, by Theorem 2, meromorphic solutions of (84) w∈W.


## 6.2.(2+1)-Dimensional Sine-Gordon Equation

In 2002, Fan and Hon [58] used generalized tanh method to study the following (2+1)-dimensional sine-Gordon equation:(86)utt-uxx-uyy+μ2sin⁡u=0.where δ is a constant.Substituting the transformations(87)vx,t=wz,v=eiu,z=x+ly+rt,into (86) yields(88)2r2-l2-1ww′′+w′2+μ2w3-μ2w=0.Equation (88) is converted to (35) where(89)A=2r2-l2-1,B3=μ2,B2=0,B1=-μ2,B0=0.Thus, by Theorem 3, meromorphic solutions of (88) w∈W.Remark 6.
By using Theorems2 and 3, we achieve meromorphic exact solutions of the S-KdV equation and (2+1)-dimensional sine-Gordon equation. Here, we do not show in detail for simplicity.


## 7. Conclusions

In this article, we have studied a class of NADEs which are very important algebraic differential equations since lots of nonlinear differential equations can be converted to it. We showed that the NADEs do not pass the Painlevé test whenn≥5. However, the NADEs can pass the Painlevé test in case of n=4 and n=3. The complex method was utilized to construct meromorphic solutions of the NADEs and then we obtained meromorphic exact solutions of the S-KdV equation and (2+1)-dimensional sine-Gordon equation. To our knowledge, the solutions of this study have not been reported in former literature.



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