International Journal of Non-Linear Mechanics 55 (2013) 147–152

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International Journal of Non-Linear Mechanics

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Exact solutions of a third-order ODE from thin film flowusing λ-symmetry method

A.H. Abdel Kader, M.S. Abdel Latif n, H.M. NourEngineering Mathematics and Physics Department, Faculty of Engineering, Mansoura University, Egypt

a r t i c l e i n f o

Article history:Received 28 April 2013Received in revised form21 May 2013Accepted 26 May 2013Available online 6 June 2013

Keywords:Thin filmThird-order ODEλ-symmetryContact angle

62/$ - see front matter & 2013 Elsevier Ltd. Ax.doi.org/10.1016/j.ijnonlinmec.2013.05.013

esponding author. Tel.: +20 1068172760.ail addresses: Leaderabass87@gmail.com (A.H@hotmail.com (M.S. Abdel Latif), hanour@ma

a b s t r a c t

We obtain exact implicit solutions of the third order ordinary differential equation y000 ¼ y−n for the first

time. These exact solutions are obtained using λ-symmetry reduction method, when n¼(5/2) and n¼(5/4).The obtained solutions are valid for the general initial conditions yðαÞ ¼ γ; y′ðαÞ ¼ δ; y″ðαÞ ¼ β. It is shownthat the contact angle |¼901 when n¼(5/2), and it varies with the initial conditions when n¼(5/4).

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1. Introduction

In this paper, λ-symmetry reduction method is used to obtainimplicit exact solutions of the third order ordinary differentialequation:

y000 ¼ y−n; ð1:1Þ

with the general initial conditions

yðαÞ ¼ γ y′ðαÞ ¼ δ y″ðαÞ ¼ β; ð1:2Þwhere, n, α, γ, δ and β are constants. ′¼ d=dx, and y is the height ofa steady thin film on a solid surface.

The ODE Eq. (1.1) can be obtained from investigating steadysolutions of the generalized lubrication equation [15,16]

∂h∂t

¼ −∂∂x

hq∂3h∂x3

� �; h≥0: ð1:3Þ

The non-negative function h(x,t) models the evolution of thefree surface of a thin film spreading on a solid substrate. Thehorizontal coordinate is denoted by x and the time by t. Wheninvestigating steady solutions of Eq. (1.3) the constants n and q areequivalent. Boatto et al. [17] obtain Eq. (1.1) by consideringtraveling wave solutions admitted by Eq. (1.3). In this case theconstants n and q are related by n¼q−1. For q¼1 Eq. (1.3) modelsthe thickness of a thin film in a Hele-Shaw cell [18–22].

The ODE Eqs. (1.1) and (1.2) has not been investigated beforeanalytically [1]. Myers [2] and Tuck [3] presented a review and

ll rights reserved.

. Abdel Kader),ns.edu.eg (H.M. Nour).

discussed some third-order ODEs occurring in thin film flow. Troy[4] and Bernis [5] have studied the existence and uniqueness ofsolutions for Eq. (1.1). Autonomous integrals and Phase planes ofEq. (1.1), obtained by the Lie reduction method, have beeninvestigated by Momoniat [6]. Numerical solutions of Eq. (1.1)are investigated using finite difference schemes in [7,8] by theshooting method in [9]. A power series approximate solution isobtained using Adomian decomposition method by Momoniat,Selway, and Jina [10]. Parametric solutions for Eq. (1.1) are given ingeneral in [11].

Momoniat [6] investigated the symmetries of Eq. (1.1) andobtained the following infinitesimal generators

X1 ¼ ∂x; X2 ¼ ðnþ 1Þx∂x þ 3y∂y:

Eq. (1.1) can be reduced to a second order ODE by using X1; thecanonical coordinates corresponding to X1 are

r¼ y; vðrÞ ¼ x;

in this case we have

y′¼ 1_v; y″¼−

€v_v3

; y000 ¼ ðð3€v2=_vÞ− ⃛vÞ

_v4; ð1:4Þ

where, the prime : denotes the derivative with respect to r:Substituting (1.4) into (1.1) and (1.2), we obtain

rnð3€v2−_v€vÞ_v5

¼ 1; ð1:5Þ

vðγÞ ¼ α; v′ðγÞ ¼ 1δ; v″ðγÞ ¼−

β

δ3; ð1:6Þ

by introducing the transformation _v¼ ð1= ffiffiffiu

p Þ, we obtain

3.0y

A.H. Abdel Kader et al. / International Journal of Non-Linear Mechanics 55 (2013) 147–152148

€u−2ffiffiffiffiu

prn

¼ 0: ð1:7Þ

In the following section, λ-symmetry method will be used tosolve Eq. (1.7). Figs. 1–8

0.5 1.0 1.5 2.0x

1.5

2.0

2.5

Fig. 3. Plot of solution (3.3) with n¼ ð5=4Þ; β¼ 0:

1.0y

2. λ-Symmetry method

The notion of λ-symmetry has been introduced in 2001 byMuriel and Romero [12,13]. In the case of equations with a lack ofpoint symmetries, Muriel and Romero have shown that many ofthe order-reduction processes can be explained by the invarianceof the equation under λ-symmetries. In fact, if an equation isinvariant under a λ-symmetry, one can obtain a complete set offunctionally independent invariants and reduce the order of theequation by one as for Lie symmetries. Soon afterwards, Pucci andSaccomandi have clarified the meaning of λ-prolongation bymeans of classical theory of characteristics of vector fields [23].Gaeta and Morando extend the concept of λ-symmetries to thecase of partial differential equations [24], and also there are manyother studies that are associated with λ-symmetry includingfurther extensions of λ-symmetry to variational problems [25,26]and to difference equations [27]. Several applications of the λ-symmetry approach to relevant equations of the mathematicalphysics appear in [28–30].

Here we investigate λ-symmetries of Eq. (1.7) at n ¼ ð5=2Þ andn ¼ ð5=4Þ.

0.5 1.0 1.5 2.0x

1.5

2.0

2.5

3.0

y

Fig. 1. Plot of solution (3.1) with n¼ ð5=2Þ; β¼ 0:

x

y1.0

0.8

0.6

0.4

0.2

-1.5 -0.5-1.0

Fig. 2. Plot of solution (3.2) with n¼ ð5=2Þ: The contact angle in this case|¼ 901; xn ¼ −1:6986; β¼ 0.

-1.5 -1.0 -0.5x

0.2

0.4

0.6

0.8

Fig. 4. Plot of solution (3.4) with n¼ ð5=4Þ: The contact angle in this case|¼ ArcTanð2Þ; xn ¼ −1:767; β¼ 0:

When n ¼ ð5=2Þ. in this case Eq. (1.7) takes the form

€u−2ffiffiffiffi

up

rð5=2Þ¼ 0: ð2:1Þ

Following Muriel and Romero [12,13], the prolongation formula

X½λ;ð2Þ� ¼ ξ∂r þ η∂u þ η½λ;ð1Þ�∂ _u þ η½λ;ð2Þ�∂ €u;

η½λ;ð1Þ� ¼Drη−Drξ _uþ λðη−ξ _uÞ;

η½λ;ð2Þ� ¼Drη½λ;ð1Þ�−Drξ €uþ λðη½λ;ð1Þ�−ξ €uÞ;

is applied to Eq. (2.1).Where, Dr denotes the total derivativeoperator with respect to r: After solving the obtained determiningequations we obtained a vector field X ¼ ∂r þ ðu=rÞ∂u which is aλ-symmetry of Eq. (2.1) with λ¼ ð2=rÞ.

The first integral of X½λ;ð1Þ� can be obtained by solving thefollowing equation:

wr þurwu þ

ur2

−_ur

� �w _u ¼ 0:

This equation admits a solution in the form

w¼ Gur;−uþ r _u

h i:

-0.5 0.5 1.0 1.5 2.0 2.5x

0.5

1.0

1.5

y

Fig. 6. Plot of solution (3.6) with n¼ ð5=2Þ: The contact angle in this case|¼ 901; xn ¼ −0:6652; β¼ −0:9142:

-1.0 -0.5 0.5 1.0x

1.5

2.0

2.5

3.0

y

Fig. 7. Plot of solution (3.7) with n¼ ð5=4Þ; β¼ 1:6861.

-0.5 0.5 1.0 1.5x

0.5

1.0

1.5

y

Fig. 8. Plot of solution (3.8) with n¼ ð5=4Þ: The contact angle in this case|¼ 66:02611; xn ¼ −0:6653; β¼ −1:1861:

-0.5 0.5 1.0 1.5x

1.5

2.0

2.5

3.0

y

Fig. 5. Plot of solution (3.5) with n¼ ð5=2Þ; β¼ 1:9142

A.H. Abdel Kader et al. / International Journal of Non-Linear Mechanics 55 (2013) 147–152 149

Let

Z ¼ −uþ r _u; R¼ ur; ð2:2Þ

substituting (2.2) into (2.1), we obtain

Z′Z−2=ðffiffiffiR

pÞ¼ 0; ð2:3Þ

where, the prime ′ denotes the derivative with respect to R.Eq. (2.3) has a solution Z2 ¼ 8

ffiffiffiR

p. Substitute into (2.2), to obtain

ð−uþ r _uÞ2 ¼ 8ffiffiffiffiffiffiffiffiu=r

p; ð2:4Þ

this first order ODE has the following two solutions

u1ðrÞ ¼34=3ð−2

ffiffiffi2

pþ rC1Þ4=3

28=3r1=3or u2ðrÞ ¼

34=3ð2ffiffiffi2

pþ rC1Þ4=3

28=3r1=3:

Considering the first solution u1ðrÞ we obtain the followingexact implicit solution for Eq. (1.1)

v¼ x¼Z

1ffiffiffiffiffiu1

p dy

¼ 27=3y1=6ðð−2ffiffiffi2

pþ yC1Þ1=3 þ ð−1Þ2=3

ffiffiffi2

p2F1ðð1=6Þ; ð2=3Þ; ð7=6Þ; ðyC1=2

ffiffiffi2

pÞÞÞ

32=3C1þ c2;

ð2:5Þ

where, 2F1 is the hypergeometric function [14], which can bedefined as

2F1ða; b; c; xÞ ¼ГðcÞ

ГðbÞГðc−bÞZ 1

0tb−1ð1−tÞc−b−1ð1−t xÞ−adt:

The initial conditions (1.6) give

C1 ¼6

ffiffiffi2

pþ 4γ1=4δ3=2

3γ;

c2 ¼

α−γ7=6ð4γ1=12δ1=2 þ ð−1Þ2=3211=631=3

2F1ðð1=6Þ; ð2=3Þ; ð7=6Þ; ð1þ ð1=3ÞÞffiffiffi2

pγ1=4δ3=2ÞÞ

3ffiffiffi2

pþ 2γ1=4δ3=2

;

1.5 1.0 0.5x*

70

75

80

85

90

�

Fig. 4.2. Relationship between ϕ and xn .

-2 -1 1 2 3 4 5

66

68

70

72

�

�

Fig. 4.1. Relationship between ϕ and β.

A.H. Abdel Kader et al. / International Journal of Non-Linear Mechanics 55 (2013) 147–152150

β¼ffiffiffiδ

pð2

ffiffiffi2

pþ γ1=4δ3=2Þ

2γ5=4:

Considering the second solution u2ðrÞ we obtain the followingexact implicit solution for Eq. (1.1)

v¼ x¼Z

1ffiffiffiffiffiu2

p dy

¼ 27=3y1=6ð−ffiffiffi2

p2F1ðð1=6Þ; ð2=3Þ; ð7=6Þ; ð−ðyC1=2

ffiffiffi2

pÞÞ þ ð2

ffiffiffi2

pþ yC1Þ1=3ÞÞ

32=3C1þ c2:

ð2:6Þ

The initial conditions (1.6) give

c1 ¼−6

ffiffiffi2

pþ 4γ1=4δ3=2

3γ;

c2 ¼

α−γ7=6ð−4γ1=12

ffiffiffiδ

pþ 211=631=3

2F1ðð1=6Þ; ð2=3Þ; ð7=6Þ; ð1−ð1=3Þffiffiffi2

pγ1=4δ3=2ÞÞÞ

3ffiffiffi2

p−2γ1=4δ3=2

;

β¼ffiffiffiδ

p ð−2ffiffiffi2

pþ γ1=4δ3=2Þ

2γ5=4:

C2 ¼ αþ4γ

ffiffiffiδ

pð−1þ2F1ðð2=3Þ;1; ð4=3Þ; ð1=16Þð16þ ðδ3=γ3=4Þ þ ððδ3=2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi32þ ðδ3=γ3=4Þ

q=ðγ3=8ÞÞÞÞÞ

δ3=2 þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi32γ3=4 þ δ3

q ;

The contact angle |, which is the angle made by the free surfaceof the thin film with the solid substrate at the contact line

(yðxnÞ ¼ 0), can be computed as follows [6]:

tan |¼ y′ðxnÞ ¼ 1_v¼

ffiffiffiffiffiffiffiffiffiuðyÞ

p;

but, we cannot obtain a contact angle from solution (2.5) since ycannot equal zero. Therefore, from Eq. (2.6), we obtain

y′ðxÞ ¼ 32=3ð2ffiffiffi2

pþ yC1Þ2=3

24=3y1=6;

and in this case when y¼ 0, y′ðxnÞ-∞; which implies that

|¼ 901;

for any initial conditions (1.2).When n ¼ ð5=4Þ. In this case Eq. (1.7) takes the form

€u−2ffiffiffiffi

up

r5=4¼ 0: ð2:7Þ

The vector field X ¼ ∂u is a λ-symmetry of Eq. (2.7) withλ¼ −ð2=ð ffiffi

up

r1=4ÞÞ−ð _u=2Þðu=2Þ−r _u .

The first integral of X½λ;ð1Þ� can be obtained by solving thefollowing equation

wu þ λ w _u ¼ 0;

which have a solution in the form

w¼ G r;−8

ffiffiffiu

p−ur1=4 _uþ r5=4 _u2

r5=4

" #:

Let

Z ¼ −8ffiffiffiu

p−ur1=4 _uþ r5=4 _u2

r5=4; _Z ¼DrZ; ð2:8Þ

substituting (2.8) in (2.7), we obtain

r _Z þ Z ¼ 0: ð2:9ÞEq. (2.9) has the trivial solution Z ¼ 0: Substitute into (2.8), to

obtain

−8ffiffiffiu

p−ur1=4 _uþ r5=4 _u2

r5=4¼ 0: ð2:10Þ

This first order ODE has the following two solutions

u1ðrÞ ¼ ð−2ffiffiffi2

pe−3C1=8ðe3C1=4−r3=4ÞÞ4=3;

or

u2ðrÞ ¼ ð2ffiffiffi2

pe−3C1=8ðe3C1=4−r3=4ÞÞ4=3:

Considering the first solution u1ðrÞ we obtain the followingexact implicit solution for Eq. (1.1)

v¼ x¼Z

1ffiffiffiffiffiu1

p dy

¼−eC1=4ð−e3C1=4 þ y3=4Þ1=3y1=4ð−1þ2F1ðð2=3Þ;1; ð4=3Þ; e−3C1=4y3=4ÞÞ þ C2:

ð2:11ÞThe initial conditions (1.6) give

C1 ¼−83ArcSinh

δ3=2

4ffiffiffi2

pγ3=8

� �þ lnðγÞ;

β¼ffiffiffi2

peArcCoshðð4

ffiffi2

pγ3=8Þ=ðδ3=2ÞÞ ffiffiffi

δp

γ5=8:

A.H. Abdel Kader et al. / International Journal of Non-Linear Mechanics 55 (2013) 147–152 151

Considering the second solution u2ðrÞ we obtain the followingexact implicit solution for Eq. (1.1)

v¼ x¼Z

1ffiffiffiffiffiu2

p dy

¼ eC1=4ðe3C1=4−y3=4Þ1=3y1=4ð−1þ 2F1ðð2=3Þ;1;ð4=3Þ; e−3C1=4y3=4ÞÞ þ C2: ð2:12ÞThe initial conditions (1.6) give

C1 ¼83ArcSinh

δ3=2

4ffiffiffi2

pγ3=8

� �þ lnðγÞ;

C2 ¼ α−18γ1=4δ1=2 δ3=2 þ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi32γ3=4 þ δ3

q� �

� −1þ2F123;1;

43;

32γ3=4

ðδ3=2 þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi32γ3=4 þ δ3

qÞ2

0B@

1CA

0B@

1CA

β¼ −8

ffiffiffiδ

p

γ1=4ðδ3=2 þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi32γ3=4 þ δ3

qÞ:

x¼ 22=3y1=6ð2ð−3ffiffiffi2

pþ ð2þ 3

ffiffiffi2

pÞyÞ1=3 þ ð−1Þ2=327=631=3

2F1ðð1=6Þ; ð2=3Þ; ð7=6Þ; yþ ðffiffiffi2

py=3ÞÞÞ

2þ 3ffiffiffi2

p

−4þ 211=631=3ð−1Þ2=32F1ðð1=6Þ; ð2=3Þ; ð7=6Þ; ð1=3Þð3þ

ffiffiffi2

pÞÞ

2þ 3ffiffiffi2

p ; ð3:5Þ

In this case the contact angle |, is given by

tan |¼ y′ðxnÞ ¼ 1_v¼

ffiffiffiffiffiffiffiffiffiuðyÞ

p;

but, we cannot obtain a contact angle from solution (2.11) sincey cannot equal zero. Therefore, from Eq. (2.12), we obtain

y′ðxÞ ¼ ð2ffiffiffi2

pe−3C1=8ðe3C1=4−y3=4ÞÞ2=3;

and in this case when y¼ 0, we obtain

|¼ ArcTanð2eC1=4Þ ¼ ArcTanð2eð2=3ÞArcCschð4ffiffi2

pγ3=8=δ3=2Þγ1=4Þ; ð2:13Þ

which implies that the value of the contact angle | depends on thevalues of the initial conditions ðδ; βÞ.

By substituting y¼ 0 in Eq. (2.12) and using Eq. (2.13) we obtain

xn ¼ α−14γ1=4 −1þ2F1

23;1;

43;8γ3=4Cot3ðϕÞ

� �� �� tan ðϕÞð−8γ3=4 þ tan 3ðϕÞÞ1=3: ð2:14Þ

3. Numerical examples

In this section we will consider some numerical values for theinitial conditions (1.2).

Example 1. Consider the initial conditions in [6]:yð0Þ ¼ 1; y′ð0Þ ¼ 0; y″ð0Þ ¼ β:

When n¼ ð5=2Þ, we have two solutions:

x¼ 24=3y1=6ðð−1þ yÞ1=3 þ ð−1Þ2=32F1ðð1=6Þ; ð2=3Þ; ð7=6Þ; yÞÞ32=3

−27=3ð−1Þ2=3 ffiffiffi

πp

Γð7=6Þ37=6Γð2=3Þ

; ð3:1Þ

or

x¼ 24=3y1=6ð−ð1−yÞ1=3 þ 2F1ð1=6; ð2=3Þ; ð7=6Þ; yÞÞ32=3 −

27=3 ffiffiffiπ

pΓð7=6Þ

37=6 Γð2=3Þð3:2Þ

When n¼ ð5=4Þ, we have two solutions:

x¼−ð−1þ y3=4Þ1=3y1=4 −1þ2F123;1;

43; y3=4

� �� �

−ð−1Þ2=32

ffiffiffi3

pπ Γ ð1=3Þ

ðΓð−1=3ÞÞ2; ð3:3Þ

or

x¼ ð1−y3=4Þ1=3y1=4 −1þ2F123;1;

43; y3=4

� �� �

þ 2πΓð1=3Þffiffiffi3

pΓð−1=3ÞΓð2=3Þ

: ð3:4Þ

Example 2. Consider the initial conditions in [10]:yð0Þ ¼ 1; y′ð0Þ ¼ 1; y″ð0Þ ¼ β:

When n¼ ð5=2Þ, we have two solutions:or

x¼ 25=3 y1=6ð−ð3ffiffiffi2

pþ ð2−3

ffiffiffi2

pÞyÞ1=3 þ 31=321=6

2F1ðð1=6Þ; ð2=3Þ; ð7=6Þ; ðr−ðffiffiffi2

pr=3ÞÞÞ

−2þ 3ffiffiffi2

p

−4−211=631=3

2F1ðð1=6Þ; ð2=3Þ; ð7=6Þ; ð1−ðffiffiffi2

p=3ÞÞÞ

2−3ffiffiffi2

p : ð3:6Þ

When n¼ ð5=4Þ, we have two solutions:

x¼−e−43ArcCschð4

ffiffi2

pÞ −1þ 1

16ð17þ

ffiffiffiffiffiffi33

pÞy3=4

� �1=3

�y1=4 −1þ2F123;1;

43;116

ð17þffiffiffiffiffiffi33

pÞy3=4

� �� �

þ12

12ð1þ

ffiffiffiffiffiffi33

pÞ

� �1=3

e−ð4=3ÞArcCschð4ffiffi2

pÞ

� −1þ 2F123;1;

43;116

ð17þffiffiffiffiffiffi33

pÞ

� �� �; ð3:7Þ

x¼ 18ð1þ

ffiffiffiffiffiffi33

pÞ2=3ð17þ

ffiffiffiffiffiffi33

p−16r3=4Þ1=3r1=4ð−1þ2F1

� 23;1;

43;−

116

ð−17þffiffiffiffiffiffi33

pÞr3=4Þ

� �

−18ð1þ

ffiffiffiffiffiffi33

pÞ −1þ2F1

23;1;

43;116

ð17−ffiffiffiffiffiffi33

pÞ

� �� �: ð3:8Þ

4. Conclusion and discussion

In this paper, we obtained exact implicit solutions of the thirdorder ordinary differential equation y

000 ¼ y−n. For our bestknowledge these implicit solutions (when n ¼ ð5=2Þ andn¼ ð5=4Þ) are published for the first time.The obtained solutions (2.5), (2.6), (2.11) and (2.12) are valid forthe general initial conditions yðαÞ ¼ γ; y′ðαÞ ¼ δ; y″ðαÞ ¼ β. So, wecan obtain some interesting relationships such asThe contact angle |¼ 901; when n ¼ ð5=2Þ.When γ ¼ 1 we have the following relationship

δ¼−2þ

ffiffiffi2

p ffiffiffiffiffiffiffiffiffiffiffiffiffiffi2þ β3

qβ

;

which implies that β≥ffiffiffiffiffiffi−23

p, to insure that δ is real.

When n ¼ ð5=4Þ, the contact angle | is given by Eq. (2.13) and itis clear that its value varies with the values of the initial

A.H. Abdel Kader et al. / International Journal of Non-Linear Mechanics 55 (2013) 147–152152

conditions. For example when γ ¼ 1, we can obtain the followingrelationship between ϕ and β: See Fig. 4.1 and Fig. 4.2

ϕ¼ ArcTanð2eðð2ArcSinhðððð−2þffiffi2

p ffiffiffiffiffiffiffiffi2þβ3

pÞ=ðβÞÞ3=2=ð4

ffiffi2

pÞÞÞÞ=ð3ÞÞÞ;

From Fig. 4.1, we can notice that ArcTanð2Þ≤ϕ≤90:From Eq. (2.14) we can see that the value of the contact angle ϕ(when n¼ ð5=4Þ) depends on the position of the contact line xn.For example when α¼ 0; γ ¼ 1.

xn ¼−14

−1þ 2F123;1;

43;8 Cot3ðϕÞ

� �� �tan ðϕÞð−8þ tan 3ðϕÞÞ1=3

Acknowledgments

The authors would like to thank the anonymous referees fortheir useful comments and suggestions to improve the paper.

References

[1] E. Momoniat, F.M. Mahomed, Symmetry reduction and numerical solution of athird-order ODE from thin film flow, Mathematical and ComputationalApplications 15 (4) (2010) 709–719.

[2] T.G. Myers, Thin film with high surface tension, SIAM Review 40 (1998)441–462.

[3] E.O. Tuck, L.W. Schwartz, Numerical and asymptotic study of some third-orderordinary differential equations relevant to draining and coating flows, SIAMReview 32 (1990) 453–469.

[4] W.C. Troy, Solutions of third-order differential equations relevant to drainingand coating flows, SIAM Journal on Mathematical Analysis 24 (1993) 155–171.

[5] F. Bernis, L.A. Peletier, Two problems from draining flows involving third orderordinary differential equations, SIAM Journal on Mathematical Analysis 27(1996) 515–527.

[6] E. Momoniat, Symmetries, first integrals and phase planes of a third-orderordinary differential equation from thin film flow, Mathematical and Compu-ter Modeling 49 (2009) 215–225.

[7] E. Momoniat, Numerical investigation of a third-order ODE from thin filmflow, Meccanica 46 (2011) 313–323.

[8] Ebrahim Momoniat, An investigation of an Emden–Fowler equation from thinfilm flow, Acta Mechanica Sinica 28 (2) (2012) 300–307.

[9] E. Momoniat, on the determination of the steady film profile for a non-Newtonian thin droplet, Computers and Mathematics with Applications 62(2011) 383–391.

[10] E. Momoniat, T.A. Selway, K. Jina, Analysis of Adomian decomposition appliedto a third-order ordinary differential equation from thin film flow, NonlinearAnalysis 66 (2007) 2315–2324.

[11] A. D. Polyanin and V. F. Zaitsev, Handbook of Exact Solutions for OrdinaryDifferential Equations, CRC Press, Boca Raton–New York, 2003.

[12] C. Muriel, J.L. Romero, First integrals, integrating factors and λ-symmetries ofsecond order differential equation, Journal of Physics A: Mathematical andTheoretical 42 (2009) 365207–365217.

[13] C. Muriel, J.L. Romero, New methods of reduction for ordinary differentialequations IMA, Journal of Applied Mathematics 66 (2001) 111–125.

[14] Olver, F.W.J., Lozier, D.W., Boisvert, R.F., and Clark, C.W., eds., NIST Handbook ofMathematical Functions, Cambridge University Press, Cambridge, 2010.

[15] H.P. Greenspan, On the motion of a small viscous droplet that wets a surface,Journal of Fluid Mechanics 84 (1978) 125–143.

[16] T.G. Myers, Thin films with high surface tension, SIAM Review 40 (1998)441–462.

[17] S. Boatto, L.P. Kadanoff, P. Olla, Traveling-wave solutions to thin film equations,Physical Review E 48 (1993) 4423–4431.

[18] R. Almgren, Singularity formation in Hele-Shaw bubbles, Physics of Fluids 8(1996) 344–352.

[19] P. Constantin, T.F. Dupont, R.E. Goldstein, L.P. Kadanoff, M.J. Shelley,Su-Min Zhou, Droplet breakup in a model of the Hele-Shaw cell, PhysicalReview E 47 (1993) 4169–4181.

[20] T.F. Dupont, R.E. Goldstein, L.P. Kadanoff, Su-Min Zhou, Finite-time singularityformation in Hele-Shaw systems, Physical Review E 47 (1993) 4182–4196.

[21] R.E. Goldstein, A.I. Pesci, M.J. Shelley, Instabilities and singularities inHele-Shaw flow, Physics of Fluids 10 (1998) 2701–2723.

[22] A.I. Pesci, R.E. Goldstein, M.J. Shelley, Domain of convergence of perturbativesolutions for Hele-Shaw flow near interface collapse, Physics of Fluids 11(1999) 2809–2811.

[23] E. Pucci, G. Saccomandi, on the reduction methods for ordinary differentialequations, Journal of Physics A: Mathematical and General Volume 35 (2002)6145–6155.

[24] G. Gaeta, P. Morando, on the geometry of lambda-symmetries and PDEsreduction, Journal of Physics A 37 (2004) 6955–6975.

[25] G. Cicogna, G. Gaeta, Noether theorem for μ-symmetries, Journal of Physics A:Mathematical and Theoretical 40 (2007) 11899–11921, arXiv: 0708.3144.

[26] C. Muriel, J.L. Romero, P.J. Olver, Variational C∞-symmetries and Euler-Lagrange equations, Journal of Differential Equations 222 (2006) 164–184.

[27] D. Levi, M.A. Rodriguez, λ-symmetries for discrete equations, Journal ofPhysics A: Mathematical and Theoretical 43 (2010) 9 292001,arXiv:1004.4808.

[28] A. Bhuvaneswari, R.A. Kraenkel, M. Senthilvelan, Application of the λ-symmetries approach and time independent integral of the modified Emdenequation, Nonlinear Analysis: Real World Applications 13 (2012) 1102–1114.

[29] A. Bhuvaneswari, R.A. Kraenkel, M. Senthilvelan, Lie point symmetries and thetime-independent integral of the damped harmonic oscillator, Physica Scripta83 (2011) 5 055005.

[30] E. Yasar. Integrating factors and first integrals for Lienard type and frequency-damped oscillators, Mathematical Problems in Engineering 2011 (2011), 10,916437.