Integrable Abel Equations and Vein's Abel Equation

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4-2016
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S.C. Mancas
Munich University of Applied Sciences, [email protected]
Haret C. Rosu
Instituto Potosino de Investigacion Cientifica y Tecnologica
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Mancas, S., & Rosu, H. C. (2016). Integrable Abel Equations and Vein's Abel Equation. Mathematical
Methods in the Applied Sciences, 39(6). https://doi.org/10.1002/mma.3575
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Math. Meth. Appl. Sci. 39 (2016) 1376-1387
Integrable Abel equations and Vein’s Abel equation
Stefan C. Mancas
∗
Fakult¨at fu¨r Informatik und Mathematik
Hochschule Mu¨nchen - Munich University of Applied Sciences, Germany
Haret C. Rosu
†
IPICyT, Instituto Potosino de Investigacion Cientifica y Tecnologica,
Camino a la presa San Jos´e 2055, Col. Lomas 4a Secci´on, 78216 San Luis Potos´ı, S.L.P., Mexico
(Dated: online at Math. Meth. Appl. Sci. – 7/28/2015)
We first reformulate and expand with several novel findings some of the basic results in the
integrability of Abel equations. Next, these results are applied to Vein’s Abel equation whose
solutions are expressed in terms of the third order hyperbolic functions and a phase space analysis
of thecorresponding nonlinear oscillator is also provided.
Keywords: Abel equation; Appell invariant; normal form; canonical form; third order hyperbolic
function.
I. INTRODUCTION
The first order Abel nonlinear equation of the second kind has the form
dw
(w+s) +p+q w+q w2+rw3 =0 , (1)
1 2
dx
where p, q , q , r, and s are all some functions of x. The case r = 0 has been introduced more than two hundred
1 1
years ago by Abel [1]. Many solvable equations of this type are collected in [2] and other ones can be found in more
recent works [3–9]. The transformation 1/y =w+s converts this equation in the form
dy ds
=r+(q 3rs)y+(q 2q s+3rs2)y2+(p q s+q s2 rs3)y3 , (2)
2 1 2 1 2
dx − − dx − − −
which is Abel’s equation of the first kind. We see that Abel’s original equation of the second kind is actually a
homogeneous case of that of the first kind.
One can also write (2) in the canonical form
dy
=f (x)+f (x)y+f (x)y2+f (x)y3 =F(x,y) , (3)
0 1 2 3
dx
where only the case s=0 leads to the simple identifications: f =r, f =q , f =q , and f =p.
0 1 2 2 1 3
TheintegrabilityfeaturesofAbel’sequationinthecanonicalformareextremelyimportantbecauseofitsconnection
withnonlinearsecondorderdifferentialequationsthatphenomenologicallydescribeawideclassofnonlinearoscillators
which in this way can be treated analytically. In particular, Vein [10] introduced an Abel equation with f = 0 and
0
rational forms of f , f , and f whose solutions are expressed in terms of third order hyperbolic functions. However,
1 2 3
to the best of our knowledge, the paper of Vein went unnoticed for many years and only recently Yamaleev [11, 12]
provided a generalization in the framework of third order multicomplex algebra. Needless to say, the corresponding
Vein’snonlinearoscillatorhasnotbeenstudiedintheliterature. Thiswasthemainmotivationforwritingthispaper,
which is organized as follows. In section II, we present a lemma that encodes the connection of Abel equation with
nonlinear oscillator equations and some related simple results. In section III, we reformulate several integrability
results for Abel’s equation with the purpose to apply them to Vein’s Abel equation, which is the subject of section
IV.ThedynamicalsystemsanalysisofVein’soscillatorisdevelopedinsectionV,andweendupwiththeconclusions.
∗Electronicaddress: [email protected]
†Electronicaddress: [email protected]
2
II. CONNECTION WITH SECOND ORDER NONLINEAR ODES
The known importance of Abel’s equation in its canonical form (3) stems from the fact that its integrability leads
to closed form solutions to a general nonlinear ODE of the form
x +f (x)x +f (x)+f (x)x2+f (x)x3 =0 , (4)
ζζ 2 ζ 3 1 ζ 0 ζ
where the variables x(ζ) and y(x(ζ)) are some parametric solutions that depend on a generalized coordinate ζ. This
can be expressed by the following lemma [9]
Lemma 1. Solutions to a general second-order ODE of type (4) may be obtained via the solutions to
Abel’s equation (3) and vice versa using the following relationship
dx
=v(x(ζ)) . (5)
dζ
Proof. To show the equivalence, one just needs the chain rule
d2x dvdx dv
= =v (6)
dζ2 dxdζ dx
which turns (4) into the Abel equation of the second kind in canonical form
dv
v +f (x)+f (x)v+f (x)v2+f (x)v3 =0. (7)
3 2 1 0
dx
Via the inverse transformation
1
v(x(ζ))= . (8)
y(x(ζ))
of the dependent variable, equation (7) becomes (3). Moreover, the linear term in (3) can always be eliminated via
the transformation
y(x)=z(x)e f1(x)dx (9)
R
which gives
dz
=h (x)+h (x)z2+h (x)z3 =H(x,z) , (10)
0 2 3
dx
where
h0(x)=f0(x)e− f1(x)dx
R
h (x)=f (x)e f1(x)dx (11)
2 2
R
h (x)=f (x)e2 f1(x)dx .
3 3
R
The case where f = 0 r = 0 is seen from (2) to be the one actually considered by Abel, for in this case the
0
→
reduced Abel equation
dw
(w+s) +p+q w+q w2 =0 , (12)
1 2
dx
can be always put in the form
dz
=h (x)z2+h (x)z3 (13)
2 3
dx
where h (x) and h (x) are reduced functions and are given by the expressions
2 3
ds
h (x)=(q 2q s)e q2(x)dx
2 1 2
− dx − R (14)
h (x)=(p q s+q s2)e2 q2(x)dx .
3 1 2
− R
3
Using the lemma, the reduced Abel equation corresponds to a linear ODE without higher-order dissipative terms
x +h (x)x +h (x)=0 . (15)
ζζ 2 ζ 3
On the other hand, in the case of f =0, the Abel equation of the first-kind (3) becomes a Riccati equation, while
3
the second-kind homogeneous Abel equation (7) is reduced to the Riccati equation
dv
= f (x) f (x)v f (x)v2 , (16)
2 1 0
dx − − −
which is equivalent to
dy
=f (x)+f (x)y+f (x)y2 . (17)
0 1 2
dx
This correspondsto thesimple factthataninversepowerofaRiccatisolutionalsosatisfiesaRiccatiequationwith
redistributed coefficients. Now, we eliminate the linear part to obtain the reduced Riccati equation
dz
=h (x)+h (x)z2 (18)
0 2
dx
which corresponds to a nonlinear ODE with higher-order dissipative terms
x +h (x)x +h (x)x3 =0 , (19)
ζζ 2 ζ 0 ζ
where now the coefficients are
h0(x)=re− (q2−3rs)dx
R (20)
h2(x)=(q1 s′ 2q2s+3rs2)e (q2−3rs)dx .
− − R
III. SOME ABEL INTEGRABILITY CASES
A. Abel’s equation of constant coefficients
Denote the coefficients f (x) = A of (3), where A R are constants and A = 0 so that F(x,y) = F(y). It is
i i i 3
∈ 6
obvious that the roots of the equation F(y)=0 are themselves solutions of (3). More generally, the general solution
of (3) is obtained via factorization of the denominator in the right-hand side of
dy
=A dx , (21)
y3+ A2y2+ A1y+ A0 3
A3 A3 A3
which leads to the following cases:
(i) If y =y =y R , so integration of (21) has the form
1 2 3
6 6 ∈
dy
=(y y1)y2−y3(y y2)y3−y1(y y3)y1−y2 =c1eA3x , (22)
(y y )(y y )(y y ) − − −
Z − 1 − 2 − 3
(ii) If y =y =y R, then the integration of (21) is of the form
1 2 3
6 ∈
dy 1 1 y y
1
= + ln − =A x+c , (23)
(y y )(y y )2 (y y )(y y ) (y y )2 |y y | 3 2
Z − 1 − 2 1− 2 − 2 1− 2 − 2
(iii) If y =y =y R , so (21) becomes
1 2 3
∈
dy 1
= =A x+c , (24)
(y y )3 2(y y )2 3 3
Z − 1 − − 1
(iv) If y =y =α+iβ C and y R, the integration of (21) leads to
1 2 3
∈ ∈
dy α y y α
=ln (y α)2+β2 + − 1 arctan − =A x+c . (25)
3 4
(y α)2+β2 (y y ) − β β
Z − − 3 (cid:12)p (cid:12)
(cid:12) (cid:12)
(cid:16) (cid:17) (cid:12) (cid:12)
4
B. Integrability based on the normal form of Abel’s equation
If the following transformations as given in Kamke’s book [13]
y(x)=ω(x)η(ξ(x)) f2(x) , ω(x)=e (cid:18)f1−3ff223(cid:19)dx ,
R
− 3f (x)
3 (26)
ξ(x)= f ω2dx
3
Z
are applied to equation (3), then one obtains Abel’s equation in normal form
dη
=η3+I(x) , (27)
dξ
where the invariant I(x) is given by
f + 1 d f2 f1f2 + 2f23
I(x)= 0 3dx f3 − 3f3 27f32 . (28)
(cid:16) f(cid:17)ω3
3
Thus,weconcludethatifI(x) constant,then(27)isintegrablesinceitisseparable. Ifonechoosesrelationsbetween
≡
the functions f such that the invariant is null and letting f (x)/f (x)=n(x), then
i 2 3
dn 2
+ f n2 f n+3f =0 (29)
2 1 0
dx 9 −
is a Riccati equation, which is alwaysintegrablebecause it is obtained fromAbel’s equationin normalform, which is
integrable. Thus, the solution to (27) is
1
η = , (30)
√c 2ξ
−
and explicitly, the solution to (3) with null invariant I(x) is
e (cid:18)f1−3ff223(cid:19)dx f
R 2
y = . (31)
sc−2 f3 eR(f1−3ff223)dx 2 − 3f3
Z (cid:18) (cid:19)
C. Integrability of Abel equation with non-constant coefficients and f0 =0
In this subsection, we will consider the original Abel equation (2) of the first-kind
dy
=f (x)y+f (x)y2+f (x)y3 , (32)
1 2 3
dx
which corresponds to (4) without the cubic nonlinearity.
Let us use the transformation
e f1(x)dx
y(x)= R , (33)
ν(x)
which is (9) with z = 1 that allows us to put (32) into a differential form
ν
νdν+(P +Qν)dx=0 , (34)
where
P(x)=f (x)e2 f1(x)dx
3
R (35)
Q(x)=f (x)e f1(x)dx .
2
R
5
Thus, the original equation (1) is considerably reduced only by having f =0.
0
For equation (34), if there exists an integrating factor [14]
µ(x,ν)=e−k(ν+ Q(x)dx) , (36)
R
where k is a constant, then
(P +Qν)µdx+νµdν =0 , (37)
is integrable, provided that
∂µ ∂µ
Qµ+(P +Qν) =ν , (38)
∂ν ∂x
which leads to Q=kP. Hence, the necessary condition for the integration of (32) is
d f f dn
kf (x)e f1(x)dx =f (x) 2 =f 2 =f n (39)
3 2 1 1
R → dx f f → dx
3 3
(cid:0) (cid:1)
Using µ(x,ν)=e kνe k2 P(x)dx, we find that there exists the potential Ψ(x,ν)=c , which satisfies
− − Ψ
R
∂Ψ
=(1+kν)Pµ ,
∂x
(40)
∂Ψ
=νµ.
∂ν
Therefore, the solution to (34) is
Ψ(x,ν)=(1+kν)e−k(ν+k Pdx) =cΨ, (41)
R
or in terms of the integrating factor
Ψ(x,ν)=(1+kν)µ(x,ν)=c . (42)
Ψ
By substituting the condition (39), the invariant (28) has the form
I(x)= 2k3e ff232dx . (43)
27 R
D. Canonical form of Abel’s equation and the integrating factor
AccordingtothebookofKamke[13],forequationsofthe type(32)forwhichthereisnocondition(39),oneshould
change the variables according to
y(x)=ω˜(x)η˜(ζ(x)) , ω˜ =e f1(x)dx ,
R
(44)
ζ(x)= f (x)ω˜(x)dx ,
2
Z
which lead to the canonical form
dη˜
=η˜2+g(ζ)η˜3, (45)
dζ
where
f (x)
g(ζ(x))= 3 e f1(x)dx (46)
f (x) R
2
is the Appell invariant. Then, the integrating factor µ can be used to formulate the following interesting result:
6
Lemma 2. Any Abel equation in the canonical form (45) is integrable as long as the invariant is constant, with
g(ζ)= 1, and solution given by (41).
k
Proof. In (34), let ν(x)= 1 , where dζ(x) =Q(x), which leads to
η˜(ζ(x)) dx
dη˜ =η˜2+ P(x)η˜3 =η˜2+ f3(x)eRf1(x)dxη˜3 =η˜2+ 1η˜3, (47)
dζ Q(x) f (x) k
2
that is separable, with solution
1 1 1 1
ln + =ζ+c+ , (48)
k η˜ k η˜
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
which after simplification leads back to (41). (cid:12) (cid:12)
IV. VEIN’S ABEL EQUATION
The following Abel equation:
dy 2b 3(ax+b2) x3 3abx a3 b3
= − y+ y2+ − − − y3 , (49)
dx bx+a2 bx+a2 bx+a2
where a,b R, is known from a paper of Vein to be integrable [10]. We first notice that if we define the following
∈
determinants,
x a b
− − x b a b
D3 =(cid:12) −b x −a (cid:12) , D2 = b −a , D1 =|−b| , D = −x −a ,
(cid:12) a b x (cid:12) (cid:12)− − (cid:12) (cid:12) − (cid:12)
(cid:12) − − (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
then they allow us to(cid:12)write the coeffi(cid:12) cients of Vein’s Equation (49) as f1 = 2D1/D, f2(x) = 3D2/D, and f3(x) =
−
D /D. Thus, (49) can be written in the compact form
3
dy 2D 3D D
= 1y 2y2+ 3y3 . (50)
dx D − D D
Next, using the transformation given by (26) with the coefficients from (49), we have
a2+bx
ω = (51)
x3 3abx a3 b3
− − −
and
dξ
=f (x)ω2 =ω. (52)
3
dx
Using these expressions, the invariant I(x) becomes
I(x)= ddx3ff23 − f31ff32 + 227ff2332 = −ω2 1 . (53)
f ω3 1ω3 ≡−
3 ω
Therefore, the normal form of Abel’s Equation (45) is
dη
=η3 1. (54)
dξ −
Using (26) and integrating (54), we obtain
1 (η 1)2 √3 √3
ξ(η) c= ln − arctan (1+2η) . (55)
− 6 1+η+η2 − 3 3
7
But because ξ(x)= ωdx, we find
R 2
1 x (a+b) √3 √32x+a+b
ξ(x) c= ln − arctan . (56)
− 6 x2+(a+b)x+a2 ab+b2 − 3 3 a b
(cid:0) −(cid:1) −
From the last two equations, one might infer that
x+b
η(x)= (57)
a b
−
and consequently using (26) would get
b
y(x)= . (58)
x (a+b) (a b)
− −
(cid:0) (cid:1)
Unfortunately, because for (57), the argument of arctan is satisfied, but the argument of ln is not satisfied, then we
conclude that (58) is not the solution of (49).
To see how the implicit solutions (55), (57) can be untangled and written explicitly, we will proceed as Vein did.
Firstly, we show that there are actually three solutions of the type (58), which are the solutions of (49) and can be
put in the form [10]:
b t b t b t
=(ax+b2) (bx+a2) 3 , =(ax+b2) (bx+a2) 1 , =(ax+b2) (bx+a2) 2 , (59)
y − t y − t y − t
1 (cid:20) 1(cid:21) 2 (cid:20) 2(cid:21) 3 (cid:20) 3(cid:21)
with the t-functions defined cyclically as
t =cφ (s)+φ (s) , t =cφ (s)+φ (s) , t =cφ (s)+φ (s) ,
1 1 2 2 2 3 3 3 1
where c is an arbitrary constant as it does not appear in the differential equation. The φ functions are the following
triad:
1 x√3
φ (x)= ex+2e x/2cos ,
1 −
3 2
1(cid:2) x√3(cid:3) π
φ (x)= ex 2e x/2cos + , (60)
2 −
3 − 2 3
1(cid:2) (cid:0)x√3 π(cid:1)(cid:3)
φ3(x)= ex 2e−x/2cos ,
3 − 2 − 3
(cid:2) (cid:0) (cid:1)(cid:3)
also known as the third-order hyperbolic functions, and are plotted in Fig. 1.
FIG. 1: The third-orderhyperbolicfunctions φi,i=1,2,3 of Equation (60).
8
They are independent because their Wronskian W(φ ,φ ,φ )=1, and they satisfy the following relationships
1 2 3
φ +φ +φ =ex ,
1 2 3
φ3+φ3+φ3 3φ φ φ =1 .
(cid:26) 1 2 3− 1 2 3
They also fulfill the following relationships
dφ /dx=φ , d2φ /dx2 =φ , d3φ /dx3 =φ ,
1 3 1 2 1 1
dφ /dx=φ , d2φ /dx2 =φ , d3φ /dx3 =φ , (61)
2 1 2 3 2 2
dφ /dx=φ , d2φ /dx2 =φ , d3φ /dx3 =φ ,
3 2 3 1 3 3
which are cyclic. That is, the fourth-order derivatives have the same property as the first ones, the fifth-order
derivatives as the second ones, and so forth. In particular,they are independent solutions of the differential equation
d3φ/dx3 =φ, but also of d6φ/dx6 =φ and of any d3+3mφ/dx3+3m =φ, with m=0,1,2,....
Let
t=cφ (s)+φ (s) , (62)
1 2
where c is a constant. Then, t fulfills
t′ =cφ3(s)+φ1(s) ,
t =cφ (s)+φ (s) , (63)
′′ 2 3
t =t
′′′
with =d/ds. Suppose s is defined implicitly as a multi-valued function of x as follows
′
t t aφ (s)+bcφ (s)+(ac+b)φ (s)
′ ′′ 1 2 3
x=a +b = =Φ(s) (64)
t t cφ (s)+φ (s)
(cid:18) (cid:19) (cid:18) (cid:19) 1 2
and denote the inverse relation by the explicit formula
s=Φ 1(x) . (65)
−
Then y =ds/dx=dΦ 1(x)/dx satisfies Abel’s equation (49).
−
Indeed, from (64), differentiating with respect to s and by the usage of (63), one obtains
2
dx t t t t
′′ ′ ′ ′′
=a +b 1 .
ds t − t − t t
(cid:20) (cid:18) (cid:19) (cid:21) (cid:20) (cid:18) (cid:19)(cid:18) (cid:19)(cid:21)
The quotient t′′ can be eliminated by means of (64), which leads to
t
dx t
b =(ax+b2) (bx+a2) ′ . (66)
ds − t
Differentiating again, this time with respect to x we have,
bd2s t t t 2 ds
dx2 =a b ′ (bx+a2) ′′ ′ .
− ds 2 − t − t − t dx
dx (cid:18) (cid:19) (cid:20) (cid:18) (cid:19) (cid:21)
Eliminating the t-quotients by mea(cid:0)ns(cid:1)of (64) and (66), one can find after some reduction the following equation
d2s ds ds 2 ds 3
(bx+a2) = 2b +3(ax+b2) +(x3 3abx a3 b3) .
dx2 − dx dx − − − dx
(cid:18) (cid:19) (cid:18) (cid:19)
Substitutingy = ds anddividingby(bx+a2)leadtoAbel’sequation(49). Thus,thesolutioncanbewrittenexplicitly
dx
as
b cφ (Φ 1(x))+φ (Φ 1(x))
=(ax+b2) (bx+a2) 3 − 1 − , (67)
y − cφ (Φ 1(x))+φ (Φ 1(x))
(cid:20) 1 − 2 − (cid:21)
with c an arbitrary constant.
The other two solutions can be obtained by cyclically replacing the φ’s in the t function
t =cφ (s)+φ (s) , t =cφ (s)+φ (s) . (68)
2 2 3 3 3 1
9
V. VEIN’S NONLINEAR OSCILLATOR
To see what kind of oscillator corresponds to Vein’s Abel equation, we will proceed as follows. First, we will
eliminate the linear term from (49) using
2b dx z
y =ze− bx+a2 = , (69)
R (bx+a2)2
to obtain
dz
=h (x)z2+h (x)z3, (70)
2 3
dx
where
3(ax+b2)
h (x)= ,
2 (bx+a2)3
(71)
x3 3abx a3 b3
h (x)= − − − .
3 (bx+a2)5
Then using lemma 1 with x =v(x(ζ))= 1 , we obtain
ζ z(x(ζ))
x +h (x)x +h (x)=0 . (72)
ζζ 2 ζ 3
which represents a nonlinear oscillator with rational friction and nonlinearity that are plotted in Figures 2 and 3,
respectively, for the following values (i) a=1, b= 2; (ii) a=1, b= 1; (iii) a=2, b= 1; (iv) a= 1, b=1.
− − − −
FIG. 2: The friction function h2(x) of Vein’soscillator at thefixed points ( 1,0),(0,0),(1,0), and (0,0).
−
Writing equation (72) as a dynamical system
x =v =M(x,v)
ζ (73)
v = h (x)v h (x)=N(x,v)
ζ 2 3
− −
and because (73) cannot be put in the form of (40), because h3(x) = const., we will be using instead the standard
h2(x) 6
methods of phase-plane analysis and use the linear approximation at the equilibrium points of (73) to classify them
instead of solving the equation by finding the potential Ψ.
Also, notice that the system is non-Hamiltonian because there is no potential Ψ(x,v)=const. such that
∂Ψ =v
∂v (74)
∂Ψ =h (x)v+h (x).
∂x 2 3