# On a Class of Two-Dimensional Douglas and Projectively Flat Finsler Metrics

Authors: Guojun Yang
Journal: The Scientific World Journal
 (2013)

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## Abstract

We study a class of two-dimensional Finsler metrics defined by a Riemannian metricα and a 1-form β. We characterize those metrics which are Douglasian or locally projectively flat by some equations. In particular, it shows that the known fact that β is always closed for those metrics in higher dimensions is no longer true in two-dimensional case. Further, we determine the local structures of two-dimensional (α, β)-metrics which are Douglasian, and some families of examples are given for projectively flat classes with β being not closed.


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## Body

## 1. Introduction

Projective Finsler geometry studies equivalent Finsler metrics on the same manifold with the same geodesics as points [1]. Douglas curvature (D) is an important projective invariants in projective Finsler geometry. A Finsler metric is called Douglasian if  D=0 and locally projectively flat if at every point, there are local coordinate systems in which geodesics are straight. It is known that a locally projectively flat Finsler metric can be characterized by D=0 and vanishing Weyl curvature. As we know, the locally projectively flat class of Riemannian metrics is very limited, nothing but the class of constant sectional curvature (Beltrami theorem). However, the class of locally projectively flat Finsler metrics is very rich. It is known that locally projectively flat Finsler metrics must be Douglasian, but Douglas metrics are not necessarily locally projectively flat. Therefore, it is a natural problem to study and classify Finsler metrics which are Douglasian or locally projectively flat. For this problem, we can only investigate some special classes of Finsler metrics.In this paper, we shall consider a special class of Finsler metrics defined by a Riemannian metricα=aij(x)yiyj and a 1-form β=bi(x)yi on a manifold M. Such metrics are called (α,β)-metrics. An (α,β)-metric can be expressed in the following form:
(1)F=αϕ(s),s=βα,
where ϕ(s)>0 is a C∞ function on (-bo,bo). It is known that F is a regular Finsler metric (defined on the whole TM-{0} and positive definite) for any (α,β) with ∥β∥α<bo if and only if
(2)ϕ(s)-sϕ′(s)+(ρ2-s2)ϕ′′(s)>0,(|s|≤ρ<bo),
where bo is a constant. If ϕ does not satisfy (2), then F=αϕ(β/α) is singular.Randers metrics are a special class of(α,β)-metrics. It is known that a Randers metric F=α+β is a Douglas metric if and only if β is closed [2], and it is locally projectively flat if and only if α is locally projectively flat and β is closed [2, 3]. Usually we call F=αϕ(s) with ϕ(s)=ϵs+1+ks2, where k,ϵ are constants, a Finsler metric of Randers type, which is essentially a Randers metric.(
α
,
β
)-metrics are computable, and it has been shown that (α,β)-metrics have a lot of special geometric properties [4–13]. In [5, 10], the authors study and characterize (α,β)-metrics which are, respectively, Douglasian and locally projectively flat in dimension n≥3. However, the two-dimensional case remains open. In this paper, we will solve this problem in two-dimensional case and meanwhile give  their local structures in part.Theorem 1.
LetF=αϕ(s), s=β/α, be a regular (α,β)-metric on an open subset U⊂R2, where ϕ(0)=1. Suppose that β is not parallel with respect to α and F is not of Randers type. Let F be a Douglas metric. Then one has one of the following two cases.   (i)
ϕ(s) satisfies
(3){1+(k1+k3)s2+k2s4}ϕ′′(s)=(k1+k2s2){ϕ(s)-sϕ′(s)},
where k1,k2, and k3 are constants satisfying
(4)k2≠(2k1+3k3)(3k1+2k3)25,k2≠k1k3.
Further, β must be closed.(ii)
F can be written as
(5)F=α~±β~2α~,(α~:=α2-kβ2,β~:=cβ),
where k,c are constants with c≠0. In this case, β is generally not closed.In dimensionn>2 in Theorem 1, it is proved in [5, 10] that the metric in Theorem 1 must be given by (3) with k2≠k1k3 and β must be closed. In Theorems 9 and 11, we give general characterizations for two-dimensional (α,β)-metrics (might be singular) which are Douglasian and locally projectively flat, respectively.Next we consider the local structure of the Douglas metrics in Theorem1. By using some deformations on α and β, we can determine the local structure of two-dimensional regular Douglas (α,β)-metrics, which is shown in the following two theorems. For the local structure of the singular Douglas classes in Theorem 9(iii) and Theorem 9(iv), we will have a discussion in Section 6.Theorem 2.
LetF=α±β2/α be a two-dimensional regular Douglas (α,β)-metric with β≠0. Then α and β can be locally written as
(6)α2=1(1∓B)3×{Bu2+v2[(y1)2+(y2)2]∓9(1±B+B2)β2},(7)β=B(1±2B)3/2uy1+vy2u2+v2,
where u=u(x),v=v(x) are scalar functions such that
(8)f(z)=u+iv,z=x1+ix2,
is a complex analytic function and B=B(x) is a scalar function satisfying 0<B<1 if F=α+β2/α and 0<B<1/2 if F=α-β2/α.We see that in Theorem2 the metric is determined by the triple parametric scalar functions (B,u,v), where u and v are a pair of complex conjugate functions. We will prove Theorem 2 by using Corollary 10 and the result in [14] (also see [15]).Theorem 3.
LetF=αϕ(s), s=β/α, be a two-dimensional regular Douglas (α,β)-metric with β≠0, where ϕ(s) satisfies (3) and (4) and ϕ(0)=1. Then α and β can be locally written as
(9)α=Bc(y1)2+(y2)2u2+v2,β=B(uy1+vy2)c(u2+v2),(c:=e∫0B‍(1/2)((k3+k2t)/(1+(k1+k3)t+k2t2))dt),
where B=B(x)>0,u=u(x),v=v(x) are some scalar functions which satisfy the following PDEs:
(10)u1=v2,u2=-v1,v1+σ1v=uσ2,
where ui:=uxi,vi:=vxi, and σi:=σxi, and σ is defined by
(11)e2σ:=Bc2(u2+v2).We will prove Theorem3 by (32) and the result in [14] (also see [15]). The metric in Theorem 3 is determined by the triple parametric scalar functions (B,u,v) which satisfy (10). It seems hard to obtain the complete solutions of the PDEs (10). However, we can give some special solutions of (10). For example, the following triple is a solution:
(12)σ=x1,u=(c2sinx2-c1cosx2)e-x1,v=(c1sinx2+c2cosx2)e-x1,
where  c1,c2 are constants, and then B is determined by (11).Now we consider the local structure of the locally projectively flat metrics in Theorem1. The local structure of F determined by (3) with k2≠k1k3 (for the dimension n≥2) has been solved in [16] (also see another way in [13]). However, it seems difficult to determine the local structure of F determined by (5). By using (40) and (98) with τ=0, we can construct the following example, which can be directly verified. We omit the details.Example 4.
LetF=α±β2/α be a two-dimensional (α,β)-metric. Suppose α and β take the form
(13)α=eσ(x)(y1)2+(y2)2,β=eσ(x)(ξ(x)y1+η(x)y2),
where ξ=ξ(x),η=η(x),σ=σ(x) are some scalar functions. Define
(14)ξ(x)=±12x2+c2c3∓(x1+c1)2∓(x2+c2)2,η(x)=∓12x1+c1c3∓(x1+c1)2∓(x2+c2)2,σ(x)=ln[c3∓(x1+c1)2∓(x2+c2)2]+c4,
where c1,c2, and c3 are constants with c3>0. Then F is projectively flat with β being not closed.For the singular projectively flat classes in Theorem11(iii) and Theorem 11(iv), we also construct some examples with β being not closed (see Examples 15 and 16 below). As we have shown, it seems an obstacle to determine the local structure of the projectively flat classes when β is not closed.Open Problem.
Determine the local structure of a two-dimensional(α,β)-metric F=α±β2/α which is locally projectively flat.


## 2. Preliminaries

LetF=F(x,y) be a Finsler metric on an n-dimensional manifold M. The geodesic coefficients are defined by
(15)Gi:=14gil{[F2]xkylyk-[F2]xl}.A Finsler metricF=F(x,y) is called a  Douglas metric if the spray coefficients Gi  are in the following form:
(16)Gi=12Γjki(x)yjyk+P(x,y)yi,
where Γjki(x) are local functions on M and P(x,y) is a local positively homogeneous function of degree one in y. It is easy to see that F is a Douglas metric if and only if Giyj-Gjyi is a homogeneous polynomial in (yi) of degree three, which by (16) can be written as [2]
(17)Giyj-Gjyi=12(Γkliyj-Γkljyi)ykyl.According to G. Hamel’s result, a Finsler metricF is projectively flat in U if and only if
(18)Fxmylym-Fxl=0.
The above formula implies that Gi=Pyi with P given by
(19)P=Fxmym2F.For a Riemannian metricα=aijyiyj and a 1-form β=biyi on a manifold M, let ∇β=bi∣jyidxj denote the covariant derivatives of β with respect to α. Put
(20)rij:=12(bi∣j+bj∣i),sij:=12(bi∣j-bj∣i),rj:=birij,sj:=bisij,si:=aiksk,
where bi:=aijbj and (aij) is the inverse of (aij).Consider an(α,β)-metric F=αϕ(β/α). By (15), the spray coefficients Gi of F are given by [3, 4, 8, 10, 11]
(21)Gi=Gαi+αQs0i+α-1Θ(-2αQs0+r00)yi+Ψ(-2αQs0+r00)bi,
where sji=aikskj,s0i=skiyk,si=bkski, and s0=siyi, and
(22)Q:=ϕ′ϕ-sϕ′,Θ:=Q-sQ′2Δ,Ψ:=Q′2Δ,Δ:=1+sQ+(b2-s2)Q′.By (21) one can see that F=αϕ(β/α) is a Douglas metric if and only if
(23)αQ(s0iyj-s0jyi)+Ψ(-2αQs0+r00)(biyj-bjyi)=12(Gkliyj-Gkljyi)ykyl,
where Gkli:=Γkli-γkli, Γkli are given in (16) and γkli:=∂2Gαi/∂yk∂yl.Further,F=αϕ(β/α) is projectively flat on U⊂Rn if and only if
(24)(amlα2-ymyl)Gαm+α3Qsl0+Ψα(-2αQs0+r00)(αbl-syl)=0,
where yl=amlym.


## 3. Equations in a Special Coordinate System

In order to prove Theorems9 and 11, one has to simplify (23) and (24). The main technique is to fix a point and choose a special coordinate system (s,ya) as in [10, 11].Fix an arbitrary pointx∈M, and take an orthonormal basis {ei} at x such that
(25)α=∑i=1n‍(yi)2,β=by1.
Then we change coordinates (yi) to (s,ya) such that
(26)α=bb2-s2α-,β=bsb2-s2α-,
where α-=∑a=2n‍(ya)2. Let
(27)r-10:=r1aya,r-00:=rabyayb,s-0:=saya.
We have s-0=bs-10,s1=bs11=0. The following lemmas are trivial.Lemma 5.
In the special local coordinate system atx as mentioned previously, if b=constant, then r11=0,r1a+s1a=0 at x.Lemma 6 (see [11]).
Forn≥2, suppose p+qα-=0, where p=p(y-) and q=q(y-) are homogeneous polynomials in y-=(ya);  then p=0,q=0.The following two propositions are simple corollaries from [5, 10] (also see (23) and (24)).Proposition 7 (n=2).
An(α,β)-metric F=αϕ(β/α) is a Douglas metric if and only if, at each point x, there is a suitable coordinate system such that at x there exist numbers Gjki  (i,j,k=1,2) which are independent of s such that
(28)s22(b2-s2)(G111-G122-G212)+12G221=bΨ(s2b2-s2r11+r22),(29)1b2-s2[2Ψ(b2-s2)-1]b3Qs12-2bΨr12s=G1122(b2-s2)s3+12(G222-G121-G211)s.Proposition 8 (n=2).
An(α,β)-metric F=αϕ(β/α) is projectively flat if and only if
(30)s22(b2-s2)(-G~111+2G~122)-12G~221=bΨ(s2b2-s2r11+r22),(31)1b2-s2[2Ψ(b2-s2)-1]b3Qs12-2bΨr12s=-G~1122(b2-s2)s3+12(-G~222+2G~121)s,
where G~jki:=∂2Gαi/∂yj∂yk are the connection coefficients of α.Comparing (28) and (30), (29) and (31), it is easy to see that if Gjki=Gkji, then G~jki=-Gjki. So if we can solve Gjki from (28) and (29), then we can solve G~jki from (30) and (31). In the following we only consider (28) and (29), from which we will solve Gjki.


## 4. Douglas(α,β)-Metrics

In this section, we characterize two-dimensional(α,β)-metrics (might be singular) which are Douglas metrics. We have the following theorem.Theorem 9.
LetF=αϕ(s), s=β/α, be an (α,β)-metric on an open subset U⊂R2, where ϕ(0)=1. Suppose that β is not parallel with respect to α and F is not of Randers type. Then F is a Douglas metric on U if and only if F lies in one of the following four classes:   (i)
ϕ(s) satisfy (3) with k2≠k1k3 and β satisfies
(32)bi∣j=2τ{(1+k1b2)aij+(k2b2+k3)bibj},(ii)
ϕ(s) and β satisfy
(33)ϕ(s)=1-ks2+cs21-ks2,(34)rij=2τ{[1+(2c-k)b2]aij-[k+3c-(k+c)kb2]bibj}+d(bisj+bjsi),
where τ=τ(x) is a scalar function, k,c are constants with c≠0 and 1-kb2≥0, and d=d(x) is given by
(35)d=3c-k-(2c-k)kb21-(k+c)b2.(iii)
(b=constant) ϕ(s) and β satisfy
(36)ϕ(s)=b2-s2b+b2-s2∫0s‍c(b2-t2)3/2(t21-kt2)mdt,(37)rij=-1b2(bisj+bjsi),
where c,k are constants and m≥1 is an integer.(iv)
(b=constant) ϕ(s) and β satisfy
(38)ϕ(s)=b2-s2b+b2-s2∫0s‍c(b2-t2)3/2(t21-kt2)m-1/2dt,(39)rij=-1b2(bisj+bjsi),
where c,k are constants and m≥1 is an integer.By Theorem9(ii), we can easily get the following corollary.Corollary 10.
LetF=α±β2/α be a two-dimensional (α,β)-metric. Then F is a Douglas metric if and only if β satisfies
(40)rij=2τ{(1±2b2)aij∓3bibj}+3±1-b2(bisj+bjsi),
where τ=τ(x) is a scalar function. Note that F=α+β2/α is regular if and only if b<1; F=α-β2/α is regular if and only if b<1/2.We prove Theorem9 using Proposition 7. The proof can be divided into two cases (r11,r22)≠(0,0) and (r11,r22)=(0,0).


### 4.1.(r11,r22)≠(0,0)

In this case, we will obtain two classes: Theorem9(i) and Theorem 9(ii).First, (28) can be written in the following form:
(41)2Ψ=λs2+μ(b2-s2)δs2+η(b2-s2),
where λ,μ,δ, and η are numbers independent of s. By (28) and (41), it is easy to prove that if λη-μδ≠0, then for some scalar τ=τ(x), we have (see also [5])
(42)r11=2b2δτ,r22=2b2ητ.
One can see that if an (α,β)-metric F=αϕ(β/α) is not of Randers type, then λη-μδ≠0.Now we put(43)a0:=1,ai:=ϕ(i)(0)i!,i=1,2,….
By (41), it has been proved in [10] that if 2a4+a22=0, then F is of Randers type. Thus we may assume that 2a4+a22≠0. Then there is a scalar ϵ=ϵ(x)≠0 such that
(44)μ=k1ϵ,η=(1+k1b2)ϵ,λ=(k1+k2b2)ϵ,δ=(1+(k1+k3)b2+k2b4)ϵ,(45)2Ψ=k1+k2s21+k1b2+(k3+k2b2)s2,
where k1,k2, and k3 are some constants determined by
(46)k1=2a2,k2=2(a4a22-5a2a6+12a42)2a4+a22,k3=-11a2a4+5a6+3a232a4+a22.
Note that k2-k1k3≠0 is equivalent to 2a4+a22≠0. Since F is not of Randers type, we get k2-k1k3≠0.Plugging (45) into (29), we get
(47)-2bs(b2-s2)(k1+k2s2)r12-2b3(1+k1s2+k3s2+k2s4)Qs12+s(1+k1b2+k3s2+k2b2s2)×{(b2-s2)ξ-G112s2}=0,
where ξ:=G121+G211-G222. Now plug the Taylor expansion of ϕ(s) into (47), and let pi be the coefficients of si in (47). By p1=0,p3=0, and p5=0, we have the following cases.(
i
) If
(48)1+(k1+k3)b2+k2b4≠0,
then
(49)r12=b215(k1(3k12+2k1k3+10k2)b4++(21k12+14k1k3-5k2)b2++18k1-3k3b4)×(1+(k1+k3)b2+k2b4)-1s12,(50)G112=2b15((3k12k3+2k1k32-2k2k3-18k1k2)b4--5(3k12+k2+2k1k3)b2-15k1b4)×(1+(k1+k3)b2+k2b4)-1s12,(51)ξ=2bk115((3k12+2k1k3+10k2)b4+6+6(3k1+2k3)b2+15b4)×(1+(k1+k3)b2+k2b4)-1s12.(ii)
If(52)1+(k1+k3)b2+k2b4=0,then b=constant. By Lemma 5 we get r12+s12=0. If s12≠0, then we get
(53)k2=3+k1b22b4,k3=-5+3k1b22b2.Case 1.
s
12
=
0.  This implies Theorem 9(i).If (48) holds, then r12=0 by (49). If (52) holds, we also get r12=0. In both cases, we have G112=ξ=0 by plugging r12=s12=0 into (29). Thus (29) becomes trivial. By r12=0, (42) and (44), we get the expression of bi∣j in (32). Further, (45) can be written in the form (3) with k2≠k1k3. This class belongs to Theorem 9(i).Case 2.
s
12
≠
0. This implies Theorem 9(ii).Case  2A. Assume that (48) holds. We plug (49), (50), and (51) into (47), and then we obtain
(54)Q=-115{(3k12k3+2k1k32-2k2k3-18k1k2)s4-5-5(3k12+k2+2k1k3)s2-15k1}s×(1+(k1+k3)s2+k2s4)-1.
By (54) and (45), we have
(55)k2=(2k1+3k3)(3k1+2k3)25.
Plug (55) into (54) and we get
(56)ϕ(s)=155+(4k1+k3)s25+(3k1+2k3)s2,
where k1≠k3 since (55) and k2≠k1k3. Letting k1=2c-k and k3=-3c-k in (56), we get (33). Substituting (55) into (49) gives
(57)r12=b2[k1(3k1+2k3)b2+6k1-k3]5+(2k1+3k3)b2s12.
Letting k1=2c-k and k3=-3c-k and using (42), we obtain (34).Case  2B. Assume that (52) holds. Then  r12=-s12 and  (53) holds. It is easy for us to get
(58)ϕ(s)=b+c~s2b2-s2,(59)rij=2τ~(b2aij-bibj)-1b2(bisj+bjsi).
This class is a special case of Theorem 1(ii).


### 4.2.(r11,r22)=(0,0)

Sinceβ is not parallel and (r11,r22)=(0,0), we will see that s12≠0 from the following proof to different cases. It follows from (28) that
(60)G221=0,G111=G122+G212.
Plugging the expressions of Q and Ψ into (29) yields
(61)s(b2-s2)[2br12+G112s2-(b2-s2)ξ]ϕ′′+s[G112s2-(b2-s2)ξ](ϕ-sϕ′)+2b3s12ϕ′=0,
where  ξ:=G121+G211-G222.Let(62)ϕ=a0+∑i=1haisi+o(sh),a0=1,
where h is a sufficiently large integer. Plugging the above Taylor series into (61), we obtain a power series ∑kpksk=0. It is easily seen that the coefficient pk of sk is given by
(63)pk=A1r12+A2G112+A3ξ+A4s12,
where
(64)A1:=b[(k-1)(k-2)ak-1-k(k+1)ak+1b2],A2:=12(k-2)[(k-4)ak-3-(k-1)ak-1b2],A3:=12[k(k+1)ak+1b4-(2k-1)(k-2)ak-1b2+(k-2)(k-4)ak-3b4],A4:=-(k+1)ak+1b3,
and  ai=0  if i<0. In particular, we have
(65)p0=-a1b3s12.
So if s12≠0, then by p0=0, we have a1=0.Case  I. Suppose db≠0. We will prove that one case belongs to Theorem 9(ii) with the scalar τ=τ(x)=0, and other cases are excluded.Solving the systemp1=0,p3=0,p5=0 yields the following three cases.(i)
If2a4≠-a22, then we get (49), (50), and (51) by using (46).(ii)
If2a4=-a22 and 2a6≠a23, then
(66)r12=15(8a2b2-1)s12,G112=-4a2bs12,ξ=165a2bs12.(iii)
If2a4=-a22 and 2a6=a23, then
(67)G112=-4a2bs12,ξ=4a2b1+2a2b2(r12+s12).It follows from (49) or (66) that r12=0 if s12=0. If (67) holds and s12=0, then we have G112=0, and thus we have r12=0 by (29) since F is not of Randers type (also see the proof in [5]). Therefore in this case we have s12≠0.Case  IA. Suppose 2a4=-a22 and 2a6≠a23. Then plug (66) into (61), and by using db≠0 we get ϕ(s)=1+2a2s2. This case is excluded.Case  IB. Suppose a4≠-(1/2)a22. Plugging (49), (50), and (51) into (61) and by using db≠0 and  s12≠0, we obtain three ODEs on ϕ(s) whose discussion of solutions can be divided into the following cases.Ifk3≠k1, then in a similar way as in Section 4.1, we can easily show that this class belongs to Theorem 1(ii) with τ=τ(x)=0.Ifk3=k1≠0,  then we obtain k2=k12,  which is impossible since  k2≠k1k3.Ifk1=k3=0,  then it is easy to see that  F is of Randers type, which is excluded.Case  IC. Suppose a4=-(1/2)a22 and a6=(1/2)a23. Then we have (67). Note that we have a1=0 since s12≠0. We will show that this case is excluded.For the functionf(s)=1+2a2s2, its Taylor coefficients ci of si (i≥0) are given by
(68)c2i+1=0,c2i=C1/2i(2a2)i,
where Cμi are the generalized combination coefficients. So in all a2i+1’s or a2i’s there exists some minimal m such that
(69)a2m+1≠0,(m≥1),ora2m≠C1/2m(2a2)m,(m≥4).CaseIC(1). Assume a2m+1≠0 in (69). Then plugging (67), a2m-3=0 and a2m-1=0 into p2m=0 (see (63)) yields
(70)a1+2m[-2mr12+(4ma2b2-2a2b2-1)s12]=0.
Therefore, it follows from (70) that we have
(71)r12=4ma2b2-2a2b2-12ms12.CaseIC(2). Assume a2m≠C1/2m(2a2)m  in (69). Plugging (67) and
(72)a2m-4=C1/2m-2(2a2)m-2,a2m-2=C1/2m-1(2a2)m-1
into p2m-1=0 (see (63)) yields
(73)[a2m-C1/2m(2a2)m]×[(1-2m)r12+(4ma2b2-4a2b2-1)s12]=0.
Therefore it follows from (73) that we have
(74)r12=4ma2b2-4a2b2-12m-1s12.Finally, plugging (67) and (71) or (74) into (61) and using db≠0 we get ϕ(s)=1+2a2s2. Thus both cases are excluded.Case  II. Suppose db=0. We will obtain Theorem 9(iii) and Theorem 1(iv).By Lemma5 we have
(75)r12=-s12.
For the function f(s)=(1/b)b2-s2, its Taylor coefficients ci of si (i≥0) are given by
(76)c2i+1=0,c2i=C1/2i(-1b2)i.
Since a1=0 and F is not of Randers type, in all a2i+1’s or a2i’s, there exists some minimal m such that
(77)a2m+1≠0,(m≥1),ora2m≠C1/2m(-1b2)m,(m≥1).CaseII(1). Assume a2m+1≠0 in (77). Plugging (75), a2m-3=0 and a2m-1=0 into p2m=0 (see (63)) yields
(78)ξ=-2m-1mbs12.
Plugging (78) and a2m-1=0 into p2m+2=0 yields
(79)G112=1m(2m+1)[(2m+3)ba2m+3ma2m+1+4m2-3b]s12.
Now plug (75), (78), and (79) into (61), and then we obtain
(80)ϕ-sϕ′+(b2-s2)ϕ′′sϕ+(b2-s2)ϕ′=2ms(1-ks2),
where k is a constant determined by a2m+1 and a2m+3. Let
(81)Φ:=sϕ(s)+(b2-s2)ϕ′(s).
Then (80) becomes
(82)Φ′Φ=2ms(1-ks2).
We get
(83)Φ=c(s21-ks2)m,
where c is a constant. Then we can easily get
(84)ϕ=b2-s2∫c(b2-s2)3/2(s21-ks2)mds.
By assumption, ϕ(0)=1, we get (36). Further, since r11=0,r22=0, and r12=-s12, we get (37). This class belongs to Theorem 9(iii).Case  II(2). Assume a2m≠C1/2m(-1/b2)m in (77). Plugging (75) and the expressions of a2m-4 and a2m-2 into p2m-1=0 yields
(85)ξ=-4(m-1)(2m-1)bs12.
Plugging (85) and the expressions of a2m-2 into p2m+1=0 (see (63)) yields
(86)G112=T1T2s12,
where T1 and T2 are defined by
(87)T1:=4m(2m-1)(m-1)C1/2m(-1b2)m-1+2(2m-1)(2m2-2m-1)b2a2m+4(m+1)b4a2m+2,T2:=m(2m-1)2b3[a2m-C1/2m(-1b2)m].
Now plug (75), (85), and (86) into (61), and then we obtain
(88)ϕ-sϕ′+(b2-s2)ϕ′′sϕ+(b2-s2)ϕ′=2m-1s(1-ks2),
where k is a constant. By the same argument, we obtain (38). This gives Theorem 9(iv).


## 4.1.(r11,r22)≠(0,0)

In this case, we will obtain two classes: Theorem9(i) and Theorem 9(ii).First, (28) can be written in the following form:
(41)2Ψ=λs2+μ(b2-s2)δs2+η(b2-s2),
where λ,μ,δ, and η are numbers independent of s. By (28) and (41), it is easy to prove that if λη-μδ≠0, then for some scalar τ=τ(x), we have (see also [5])
(42)r11=2b2δτ,r22=2b2ητ.
One can see that if an (α,β)-metric F=αϕ(β/α) is not of Randers type, then λη-μδ≠0.Now we put(43)a0:=1,ai:=ϕ(i)(0)i!,i=1,2,….
By (41), it has been proved in [10] that if 2a4+a22=0, then F is of Randers type. Thus we may assume that 2a4+a22≠0. Then there is a scalar ϵ=ϵ(x)≠0 such that
(44)μ=k1ϵ,η=(1+k1b2)ϵ,λ=(k1+k2b2)ϵ,δ=(1+(k1+k3)b2+k2b4)ϵ,(45)2Ψ=k1+k2s21+k1b2+(k3+k2b2)s2,
where k1,k2, and k3 are some constants determined by
(46)k1=2a2,k2=2(a4a22-5a2a6+12a42)2a4+a22,k3=-11a2a4+5a6+3a232a4+a22.
Note that k2-k1k3≠0 is equivalent to 2a4+a22≠0. Since F is not of Randers type, we get k2-k1k3≠0.Plugging (45) into (29), we get
(47)-2bs(b2-s2)(k1+k2s2)r12-2b3(1+k1s2+k3s2+k2s4)Qs12+s(1+k1b2+k3s2+k2b2s2)×{(b2-s2)ξ-G112s2}=0,
where ξ:=G121+G211-G222. Now plug the Taylor expansion of ϕ(s) into (47), and let pi be the coefficients of si in (47). By p1=0,p3=0, and p5=0, we have the following cases.(
i
) If
(48)1+(k1+k3)b2+k2b4≠0,
then
(49)r12=b215(k1(3k12+2k1k3+10k2)b4++(21k12+14k1k3-5k2)b2++18k1-3k3b4)×(1+(k1+k3)b2+k2b4)-1s12,(50)G112=2b15((3k12k3+2k1k32-2k2k3-18k1k2)b4--5(3k12+k2+2k1k3)b2-15k1b4)×(1+(k1+k3)b2+k2b4)-1s12,(51)ξ=2bk115((3k12+2k1k3+10k2)b4+6+6(3k1+2k3)b2+15b4)×(1+(k1+k3)b2+k2b4)-1s12.(ii)
If(52)1+(k1+k3)b2+k2b4=0,then b=constant. By Lemma 5 we get r12+s12=0. If s12≠0, then we get
(53)k2=3+k1b22b4,k3=-5+3k1b22b2.Case 1.
s
12
=
0.  This implies Theorem 9(i).If (48) holds, then r12=0 by (49). If (52) holds, we also get r12=0. In both cases, we have G112=ξ=0 by plugging r12=s12=0 into (29). Thus (29) becomes trivial. By r12=0, (42) and (44), we get the expression of bi∣j in (32). Further, (45) can be written in the form (3) with k2≠k1k3. This class belongs to Theorem 9(i).Case 2.
s
12
≠
0. This implies Theorem 9(ii).Case  2A. Assume that (48) holds. We plug (49), (50), and (51) into (47), and then we obtain
(54)Q=-115{(3k12k3+2k1k32-2k2k3-18k1k2)s4-5-5(3k12+k2+2k1k3)s2-15k1}s×(1+(k1+k3)s2+k2s4)-1.
By (54) and (45), we have
(55)k2=(2k1+3k3)(3k1+2k3)25.
Plug (55) into (54) and we get
(56)ϕ(s)=155+(4k1+k3)s25+(3k1+2k3)s2,
where k1≠k3 since (55) and k2≠k1k3. Letting k1=2c-k and k3=-3c-k in (56), we get (33). Substituting (55) into (49) gives
(57)r12=b2[k1(3k1+2k3)b2+6k1-k3]5+(2k1+3k3)b2s12.
Letting k1=2c-k and k3=-3c-k and using (42), we obtain (34).Case  2B. Assume that (52) holds. Then  r12=-s12 and  (53) holds. It is easy for us to get
(58)ϕ(s)=b+c~s2b2-s2,(59)rij=2τ~(b2aij-bibj)-1b2(bisj+bjsi).
This class is a special case of Theorem 1(ii).


## 4.2.(r11,r22)=(0,0)

Sinceβ is not parallel and (r11,r22)=(0,0), we will see that s12≠0 from the following proof to different cases. It follows from (28) that
(60)G221=0,G111=G122+G212.
Plugging the expressions of Q and Ψ into (29) yields
(61)s(b2-s2)[2br12+G112s2-(b2-s2)ξ]ϕ′′+s[G112s2-(b2-s2)ξ](ϕ-sϕ′)+2b3s12ϕ′=0,
where  ξ:=G121+G211-G222.Let(62)ϕ=a0+∑i=1haisi+o(sh),a0=1,
where h is a sufficiently large integer. Plugging the above Taylor series into (61), we obtain a power series ∑kpksk=0. It is easily seen that the coefficient pk of sk is given by
(63)pk=A1r12+A2G112+A3ξ+A4s12,
where
(64)A1:=b[(k-1)(k-2)ak-1-k(k+1)ak+1b2],A2:=12(k-2)[(k-4)ak-3-(k-1)ak-1b2],A3:=12[k(k+1)ak+1b4-(2k-1)(k-2)ak-1b2+(k-2)(k-4)ak-3b4],A4:=-(k+1)ak+1b3,
and  ai=0  if i<0. In particular, we have
(65)p0=-a1b3s12.
So if s12≠0, then by p0=0, we have a1=0.Case  I. Suppose db≠0. We will prove that one case belongs to Theorem 9(ii) with the scalar τ=τ(x)=0, and other cases are excluded.Solving the systemp1=0,p3=0,p5=0 yields the following three cases.(i)
If2a4≠-a22, then we get (49), (50), and (51) by using (46).(ii)
If2a4=-a22 and 2a6≠a23, then
(66)r12=15(8a2b2-1)s12,G112=-4a2bs12,ξ=165a2bs12.(iii)
If2a4=-a22 and 2a6=a23, then
(67)G112=-4a2bs12,ξ=4a2b1+2a2b2(r12+s12).It follows from (49) or (66) that r12=0 if s12=0. If (67) holds and s12=0, then we have G112=0, and thus we have r12=0 by (29) since F is not of Randers type (also see the proof in [5]). Therefore in this case we have s12≠0.Case  IA. Suppose 2a4=-a22 and 2a6≠a23. Then plug (66) into (61), and by using db≠0 we get ϕ(s)=1+2a2s2. This case is excluded.Case  IB. Suppose a4≠-(1/2)a22. Plugging (49), (50), and (51) into (61) and by using db≠0 and  s12≠0, we obtain three ODEs on ϕ(s) whose discussion of solutions can be divided into the following cases.Ifk3≠k1, then in a similar way as in Section 4.1, we can easily show that this class belongs to Theorem 1(ii) with τ=τ(x)=0.Ifk3=k1≠0,  then we obtain k2=k12,  which is impossible since  k2≠k1k3.Ifk1=k3=0,  then it is easy to see that  F is of Randers type, which is excluded.Case  IC. Suppose a4=-(1/2)a22 and a6=(1/2)a23. Then we have (67). Note that we have a1=0 since s12≠0. We will show that this case is excluded.For the functionf(s)=1+2a2s2, its Taylor coefficients ci of si (i≥0) are given by
(68)c2i+1=0,c2i=C1/2i(2a2)i,
where Cμi are the generalized combination coefficients. So in all a2i+1’s or a2i’s there exists some minimal m such that
(69)a2m+1≠0,(m≥1),ora2m≠C1/2m(2a2)m,(m≥4).CaseIC(1). Assume a2m+1≠0 in (69). Then plugging (67), a2m-3=0 and a2m-1=0 into p2m=0 (see (63)) yields
(70)a1+2m[-2mr12+(4ma2b2-2a2b2-1)s12]=0.
Therefore, it follows from (70) that we have
(71)r12=4ma2b2-2a2b2-12ms12.CaseIC(2). Assume a2m≠C1/2m(2a2)m  in (69). Plugging (67) and
(72)a2m-4=C1/2m-2(2a2)m-2,a2m-2=C1/2m-1(2a2)m-1
into p2m-1=0 (see (63)) yields
(73)[a2m-C1/2m(2a2)m]×[(1-2m)r12+(4ma2b2-4a2b2-1)s12]=0.
Therefore it follows from (73) that we have
(74)r12=4ma2b2-4a2b2-12m-1s12.Finally, plugging (67) and (71) or (74) into (61) and using db≠0 we get ϕ(s)=1+2a2s2. Thus both cases are excluded.Case  II. Suppose db=0. We will obtain Theorem 9(iii) and Theorem 1(iv).By Lemma5 we have
(75)r12=-s12.
For the function f(s)=(1/b)b2-s2, its Taylor coefficients ci of si (i≥0) are given by
(76)c2i+1=0,c2i=C1/2i(-1b2)i.
Since a1=0 and F is not of Randers type, in all a2i+1’s or a2i’s, there exists some minimal m such that
(77)a2m+1≠0,(m≥1),ora2m≠C1/2m(-1b2)m,(m≥1).CaseII(1). Assume a2m+1≠0 in (77). Plugging (75), a2m-3=0 and a2m-1=0 into p2m=0 (see (63)) yields
(78)ξ=-2m-1mbs12.
Plugging (78) and a2m-1=0 into p2m+2=0 yields
(79)G112=1m(2m+1)[(2m+3)ba2m+3ma2m+1+4m2-3b]s12.
Now plug (75), (78), and (79) into (61), and then we obtain
(80)ϕ-sϕ′+(b2-s2)ϕ′′sϕ+(b2-s2)ϕ′=2ms(1-ks2),
where k is a constant determined by a2m+1 and a2m+3. Let
(81)Φ:=sϕ(s)+(b2-s2)ϕ′(s).
Then (80) becomes
(82)Φ′Φ=2ms(1-ks2).
We get
(83)Φ=c(s21-ks2)m,
where c is a constant. Then we can easily get
(84)ϕ=b2-s2∫c(b2-s2)3/2(s21-ks2)mds.
By assumption, ϕ(0)=1, we get (36). Further, since r11=0,r22=0, and r12=-s12, we get (37). This class belongs to Theorem 9(iii).Case  II(2). Assume a2m≠C1/2m(-1/b2)m in (77). Plugging (75) and the expressions of a2m-4 and a2m-2 into p2m-1=0 yields
(85)ξ=-4(m-1)(2m-1)bs12.
Plugging (85) and the expressions of a2m-2 into p2m+1=0 (see (63)) yields
(86)G112=T1T2s12,
where T1 and T2 are defined by
(87)T1:=4m(2m-1)(m-1)C1/2m(-1b2)m-1+2(2m-1)(2m2-2m-1)b2a2m+4(m+1)b4a2m+2,T2:=m(2m-1)2b3[a2m-C1/2m(-1b2)m].
Now plug (75), (85), and (86) into (61), and then we obtain
(88)ϕ-sϕ′+(b2-s2)ϕ′′sϕ+(b2-s2)ϕ′=2m-1s(1-ks2),
where k is a constant. By the same argument, we obtain (38). This gives Theorem 9(iv).


## 5. Projectively Flat(α,β)-Metrics

In this section, we characterize two-dimensional(α,β)-metrics (might be singular) which are projectively flat. We have the following theorem.Theorem 11.
LetF=αϕ(s), s=β/α, be an (α,β)-metric on an open subset U⊂R2 with ϕ(0)=1. Suppose that β is not parallel with respect to α and F is not of Randers type. Then F is projectively flat in U with Gi=P(x,y)yi if and only if F lies in one of the following four classes.(i)
ϕ(s) and β satisfy (3) and (32), and the spray coefficients Gαi of α satisfy
(89)Gαi=ρyi-τ(k1α2+k2β2)bi.
In this case, the projective factor P is given by
(90)P=ρ+τα{[1+(k1+k3)s2+k2s4]ϕ′ϕ-(k1+k2s2)sϕ′ϕ}.(ii)
ϕ(s) and β satisfy (33) and (34), and
(91)Gαi=ρyi-τ{(2c-k)α2+(c+k)kβ2}bi+(2c-k)(1-kb2)α2+ckβ21-(c+k)b2si.
In this case, the projective factor P is given by
(92)P=ρ-4c2s31+(c-k)s2τα+σ1σ2s0,
where
(93)σ1:=[(k-2c)(c-k)kb2+4c2-3kc+k2]s2+(2c-k)(1-kb2),σ2:=[1-(c+k)b2][1+(c-k)s2].(iii)
ϕ(s) and β satisfy (36) and (37), and
(94)Gαi=ρyi-(2m-1)α2+kβ22mb2si.
In this case, the projective factor P is given by
(95)P=ρ-s(1-ks2)ϕ′+[(2m-1)+ks2]ϕ2mb2ϕs0.(iv)
ϕ(s) and β satisfy (38) and (39), and
(96)Gαi=ρyi-2(m-1)α2+kβ2(2m-1)b2si.
In this case, the projective factor P is given by
(97)P=ρ-s(1-kb2)ϕ′+[2(m-1)+ks2]ϕ(2m-1)b2ϕs0.In the above, ρ=c1(x)y1+c2(x)y2 is a 1-form.By Theorem11(ii), we can easily get the following corollary.Corollary 12.
LetF=α±β2/α be two-dimensional (α,β)-metric. Then F is locally projectively flat if and only if β satisfies (40) and Gαi satisfy
(98)Gαi=ρyi∓2τα2bi-2α2b2∓1si.
In this case, the projective factor P is given by
(99)P=ρ-4s31±s2τα-2(2s2±1)(b2∓1)(s2±1)s0.To prove Theorem11, it follows from comparing Propositions 7 and 8 that we only need to give the expressions (89)–(97) for each class in Theorem 11.


### 5.1. The Spray Coefficients ofα

In this subsection we will show the expressions of the spray coefficientsGαi for each class in Theorem 11. Note that by G~jki=∂2Gαi/∂yj∂yk, the spray Gαi of α can be expressed as
(100)Gαi=12G~jkiyjyk.Case  I. Suppose that (r11,r22)≠(0,0). It has been proved in [10] that
(101)G~111=2t1-2λb3τ,G~221=-2μb3τ,G~121=t2,G~122=t1,
where t1,t2 are numbers independent of s and τ is given by  (32) or (34). By (50) and (51), we can get G~112 and G~222.Ifβ is closed (s12=0), then it follows from (44) and (101) and the expressions of G~112 and G~222 that (89) holds, where we put ρ as ρ=tiyi. This case has been given by [10] in case of n≥3.Ifβ is not closed, then by putting k1=2c-k and k3=-3c-k we get (91) from (55) and (101) and the expressions of G~112 and G~222.Case  II. Suppose that (r11,r22)=(0,0). Then by (30) we get
(102)G~221=0,G~111=2G~122=2t1,G~122=t1,G~121=t2.Ifdb≠0, then we have shown in Section 4.2 that a4≠-(1/2)a22. In this case, we obtain (91) with τ=0.Ifdb=0, then we get (77). If a2m+1≠0 in (77), then we get from (78), (79) that
(103)G~112=-2m-1+kb2mbs12,G~222=2G~121-2m-1mbs12.
Then it follows from (102) and (103) that (94) holds. If a2m≠C1/2m(-1/b2)m in (77), then we get from (85), (86) that
(104)G~112=-2(2m2-2m-k)m(2m-1)bs12,G~222=2G~121-4(m-1)(2m-1)bs12.
Then it follows from (102) and (104) that (96) holds.


### 5.2. The Projective Factors

In this subsection, we are going to find the expression for the projective factor for each class in Theorem11.Actually, (90) has been proved in [10], since β is closed in Theorem 11(i). So we only show the expressions of P in (92), (95), and (97). In the left three classes, since β may not be closed, it is not easy to show the projective factors P in the initial local projective coordinate system (in such a coordinate system, geodesics are straight lines). However, it is easy to be solved by choosing another local projective coordinate system and then returning to the initial local projective coordinate system, just as that in [10].Fix an arbitrary pointxo∈U⊂R2. By the above idea and a suitable affine transformation, we may assume (U,xi) is a local projective coordinate system satisfying that αxo=(y1)2+(y2)2 and βxo=by1. Then at xo we have
(105)s1=s1=0,s2=s2=bs12,s0=bs12y2,b1=b1=b,b2=b2=0.Suppose (33), (34), and (91) hold in U. Then it is easy to get r00,s0i,Q,Θ, and Ψ. Plug them into (21), and then at xo we see that Gi=Pyi, where P is given by
(106)P=ρ+A1τ+A2bs12y2,
where
(107)A1:=-4c2(by1)3[1+(c-k)b2](y1)2+(y2)2,A2:=((y2)2b2[1+(2c-k)b2][(2c-k)-k(c-k)b2]××(y1)2+(2c-k)(1-kb2)(y2)2b2)×((y2)2b2[1-(c+k)b2]×{[1+(c-k)b2](y1)2+(y2)2})-1.
By using
(108)bs12y2=s0,(y1)2+(y2)2=α2,by1=β,βα=s,
we can transform (106) as (92). It is a direct computation, so the details are omitted. Since xo is arbitrarily chosen, (92) holds in U.The left proofs are similar. So the details are omitted.We have found the projective factor for each class in Theorem11. This also gives a proof to the inverse of Theorem 11.


## 5.1. The Spray Coefficients ofα

In this subsection we will show the expressions of the spray coefficientsGαi for each class in Theorem 11. Note that by G~jki=∂2Gαi/∂yj∂yk, the spray Gαi of α can be expressed as
(100)Gαi=12G~jkiyjyk.Case  I. Suppose that (r11,r22)≠(0,0). It has been proved in [10] that
(101)G~111=2t1-2λb3τ,G~221=-2μb3τ,G~121=t2,G~122=t1,
where t1,t2 are numbers independent of s and τ is given by  (32) or (34). By (50) and (51), we can get G~112 and G~222.Ifβ is closed (s12=0), then it follows from (44) and (101) and the expressions of G~112 and G~222 that (89) holds, where we put ρ as ρ=tiyi. This case has been given by [10] in case of n≥3.Ifβ is not closed, then by putting k1=2c-k and k3=-3c-k we get (91) from (55) and (101) and the expressions of G~112 and G~222.Case  II. Suppose that (r11,r22)=(0,0). Then by (30) we get
(102)G~221=0,G~111=2G~122=2t1,G~122=t1,G~121=t2.Ifdb≠0, then we have shown in Section 4.2 that a4≠-(1/2)a22. In this case, we obtain (91) with τ=0.Ifdb=0, then we get (77). If a2m+1≠0 in (77), then we get from (78), (79) that
(103)G~112=-2m-1+kb2mbs12,G~222=2G~121-2m-1mbs12.
Then it follows from (102) and (103) that (94) holds. If a2m≠C1/2m(-1/b2)m in (77), then we get from (85), (86) that
(104)G~112=-2(2m2-2m-k)m(2m-1)bs12,G~222=2G~121-4(m-1)(2m-1)bs12.
Then it follows from (102) and (104) that (96) holds.


## 5.2. The Projective Factors

In this subsection, we are going to find the expression for the projective factor for each class in Theorem11.Actually, (90) has been proved in [10], since β is closed in Theorem 11(i). So we only show the expressions of P in (92), (95), and (97). In the left three classes, since β may not be closed, it is not easy to show the projective factors P in the initial local projective coordinate system (in such a coordinate system, geodesics are straight lines). However, it is easy to be solved by choosing another local projective coordinate system and then returning to the initial local projective coordinate system, just as that in [10].Fix an arbitrary pointxo∈U⊂R2. By the above idea and a suitable affine transformation, we may assume (U,xi) is a local projective coordinate system satisfying that αxo=(y1)2+(y2)2 and βxo=by1. Then at xo we have
(105)s1=s1=0,s2=s2=bs12,s0=bs12y2,b1=b1=b,b2=b2=0.Suppose (33), (34), and (91) hold in U. Then it is easy to get r00,s0i,Q,Θ, and Ψ. Plug them into (21), and then at xo we see that Gi=Pyi, where P is given by
(106)P=ρ+A1τ+A2bs12y2,
where
(107)A1:=-4c2(by1)3[1+(c-k)b2](y1)2+(y2)2,A2:=((y2)2b2[1+(2c-k)b2][(2c-k)-k(c-k)b2]××(y1)2+(2c-k)(1-kb2)(y2)2b2)×((y2)2b2[1-(c+k)b2]×{[1+(c-k)b2](y1)2+(y2)2})-1.
By using
(108)bs12y2=s0,(y1)2+(y2)2=α2,by1=β,βα=s,
we can transform (106) as (92). It is a direct computation, so the details are omitted. Since xo is arbitrarily chosen, (92) holds in U.The left proofs are similar. So the details are omitted.We have found the projective factor for each class in Theorem11. This also gives a proof to the inverse of Theorem 11.


## 6. Singular Classes in Theorems9 and 11

In this section, we will firstly discuss the local structures of the singular classes in Theorem9(iii) and Theorem 9(iv), and then construct some examples for Theorem 11(iii) and Theorem 11(iv).Since every two-dimensional Riemann metric is locally conformally flat, we may put(109)α=eσ(x)(y1)2+(y2)2,
where x=(x1,x2). Since (59) is equivalent to b2=∥β∥α2=constant in two-dimensional case (see a simple proof in [7]), (59) holds if and only if β is in the form
(110)β=beσ(x)[ξ(x)y1+η(x)y2]ξ(x)2+η(x)2,
where b=∥β∥α=constant. Thus (109) and (110) give all the local solutions of (59). If we put α and β in the forms (109) and (110), then we have (59); that is,
(111)rij=2τ~(b2aij-bibj)-1b2(bisj+bjsi),
where τ~ is given by
(112)2τ~=(ξ2+η2)(ξσ1+ησ2)-ξηη1+ξ2η2+η2ξ1-ξηξ2beσ(ξ2+η2)3/2,
where σ1:=∂σ/∂x1, σ2:=∂σ/∂x2,  and so forth. Further, β is not closed if and only if
(113)(ξ2+η2)(ξσ2-ησ1)-ξ2η1-ξηη2+ξηξ1+η2ξ2≠0.
In particular, if we put ξ=x2,η=-x1, and σ=c[(x1)2+(x2)2], where c≠0 is a constant, then we have τ~=0 by (112) and (113) holds (β is not closed).Proposition 13.
Define a two-dimensional(α,β)-metric F on R2 by
(114)F=bα2+kβ2b2α2-β2,
where k,b are constants with k≠-1/b. Then F is a Douglas metric if and only if α and β can be locally defined by (109) and (110), where ξ,η, and σ are some scalar functions on R2. There are many choices for ξ,η, and σ such that β is not closed.Proposition 14.
LetF=αϕ(s), s=β/α, be a two-dimensional (α,β)-metric, where ϕ(0)=1. Let ϕ(s) be given by (36) or (38) (not given by (59)). Then F is a Douglas metric if and only if α and β can be locally defined by (109) and (110), where ξ,η, and σ are some scalar functions satisfying τ~=0 in (112). There are many choices for ξ,η, and σ such that β is not closed.Next we construct some singular examples for Theorem11(iii) and Theorem 11(iv) which are projectively flat. One can directly verify the following two examples.Example 15.
LetF=αϕ(s), s=β/α, be two-dimensional (α,β)-metric, where ϕ(0)=1. Let ϕ(s) be given by (36) with k=0, and define α and β by (109) and (110), where
(115)ξ=x2,η=-x1,σ=(m-12)ln[(x1)2+(x2)2].
Then F is projectively flat with β being not closed.Example 16.
LetF=αϕ(s), s=β/α, be two-dimensional (α,β)-metric, where ϕ(0)=1. Let ϕ(s) be given by (38) with k=0, and define α and β by (109) and (110), where
(116)ξ=x2,η=-x1,σ=(m-1)ln[(x1)2+(x2)2].
Then F is projectively flat with β being not closed.It might be also an interesting problem to show the local structures of the two classes of Theorem11(iii) and Theorem 11(iv). This problem is still open.


## 7. Proof of Theorem2

In the following proof, our idea is to choose a suitable deformation onα and β such that we can obtain a conformal form on a Riemannian space. Then using the result in [14], we can complete our proof.Define a new Riemann metricα~ and a 1-form β~ by
(117)α~:=ξα2+ηβ2,β~:=β,
where
(118)ξ:=(1∓b2)3(1±2b2)3/2,η:=98b2{(1±2b2)3/2-1∓2b2+4b4(1±2b2)3/2}.
Since F=α±β2/α is a Douglas metric, we have (40). By (117) and (40), a direct computation gives
(119)r~ij=2τ(1∓b2)2(1±2b2)5/2a~ij.
So β~=β is a conformal 1-form with respect to α~.Sinceα~ is a two-dimensional Riemann metric, we can express α~ locally as
(120)α~:=eσ(y1)2+(y2)2,
where σ=σ(x) is a scalar function. We can obtain the local expression of β~=β by (119) and (120) (see [14]). By the result in [14], we have
(121)β~=b~1y1+b~2y2=e2σ(uy1+vy2),
where u=u(x),v=v(x) are a pair of scalar functions such that
(122)f(z)=u+iv,z=x1+ix2,
is a complex analytic function.We can expressσ using u,v,b2 by computing the quantity ∥β∥α~2. Firstly, by (120) and (121) we get
(123)∥β∥α~2=e2σ(u2+v2).
On the other hand, by the definition of α~ in (117), the inverse a~ij of a~ij is given by
(124)a~ij=1ξ(aij-ηbibjξ+ηb2).
Now plug ξ and η into the above, and we obtain
(125)∥β∥α~2=a~ijbibj=b2(1±2b2)3/2.
Thus by (123) and (125) we get
(126)e2σ=1u2+v2b2(1±2b2)3/2.Finally, by plugging (126), (120), and (121) into (117), we easily get α and β given by (6) and (7), respectively, where we define B:=b2.The Riemann metricα expressed in (6) is generally not in the conformally flat form. We show another example which is expressed in a different form. Put
(127)α=eσ(y1)2+(y2)2,β=eσ(ξy1+ηy2),σ=-32ln[1±2(x1)2±2(x2)2]+c,ξ=x2,η=-x1,
where c is a constant. Then β is not closed and satisfies (40) with τ=0. Therefore, F=α±β2/α is a Douglas metric.


## 8. Proof of Theorem3

This proof is similar to that in Theorem2. In the proof of Theorem 2, we make a deformation on α and keep β unchanged. For the proof of Theorem 3 in the following, we will give a deformation on β but keep α unchanged.Define a Riemannian metricα~ and 1-form β~ by
(128)α~:=α,β~:=βc,
where c=c(b2) is defined in (9), where we define B:=b2. By (32) and a direct computation, we can obtain
(129)b~i∣j=2τ(1+k1b2)ca~ij=2τ(1+k1b2)caij.
So β~ is a closed 1-form conformal with respect to α.Now we expressα locally as
(130)α:=eσ(y1)2+(y2)2,
where σ=σ(x) is a scalar function. Then by the result in [14], we have
(131)β~=b~1y1+b~2y2=e2σ(uy1+vy2),
where u=u(x),v=v(x) are a pair of scalar functions such that
(132)f(z)=u+iv,z=x1+ix2,
is a complex analytic function.By the property ofu,v and the fact that β~ given in (131) is closed, it is easily seen that u,v, and σ satisfy the PDEs (10).Now we determineσ in terms of the triple (B,u,v), where B:=b2. Firstly by (128) and then by (130) and (131) we get
(133)∥β~∥α2=b2c-2,∥β~∥α2=e2σ(u2+v2).
Therefore, by (133), we get (11).Finally, we can easily getα and β given by (9) from (11), (128), (130), and (131).



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*Source: 291491-2013-10-22.xml*