Generalization of Exp-function and other standard function methods

Applied Mathematics and Computation 216 (2010) 2380–2393

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Applied Mathematics and Computation

journal homepage: www.elsevier .com/ locate /amc

Generalization of Exp-function and other standard function methods

Zenonas Navickas a, Liepa Bikulciene a, Minvydas Ragulskis b,*

a Department of Applied Mathematics, Kaunas University of Technology, Studentu 50-325, Kaunas LT-51368, Lithuaniab Research Group for Mathematical and Numerical Analysis of Dynamical Systems, Kaunas University of Technology, Studentu 50-222, Kaunas LT-51368, Lithuania

a r t i c l e i n f o

Keywords:The Exp-function methodMultiplicative operatorH-rankStandard functions

0096-3003/$ - see front matter � 2010 Elsevier Incdoi:10.1016/j.amc.2010.03.083

* Corresponding author.E-mail addresses: [email protected] (Z. NaURL: http://www.personalas.ktu.lt/~mragul (M.

a b s t r a c t

Exp-function and other standard function methods for solving nonlinear differential equa-tions are generalized in this paper. An analytical criterion determining if a solution can beexpressed in a form comprising standard functions is derived. New computational algo-rithm for automatic identification of the structure of the solution is constructed. The algo-rithm provides information if the solution can be expressed as a sum of standard functions,a ratio of sums of standard functions, or even a more complex algebraic form involvingstandard functions. Several examples are used to illustrate the proposed concept.

� 2010 Elsevier Inc. All rights reserved.

1. Introduction

The Exp-function method was proposed in 2006 by He and Wu [1] to seek solitary solutions, periodic solutions and com-pacton-like solutions of nonlinear differential equations. It has been demonstrated that the Exp-function method, with thehelp of symbolic computation, provides a powerful mathematical tool for solving high-dimensional nonlinear evolutions inmathematical physics. The Exp-function method has been exploited for the determination of exact solutions of many non-linear differential equations [2–5]; we give only a few of many references available.

Several alternative modifications of the Exp-function method have been developed. The tanh, extended tanh, improvedtanh and generalized tanh function methods [6–9], sech and rational Exp-function method [10], the modified simplest equa-tion method [11] and many similar standard function methods are also successfully used for the construction of solutions ofnonlinear differential equations.

An analytical criterion determining if a solution of a differential equation can be expressed in an analytical form compris-ing exponential functions is developed in [12]. The employment of this criterion does not only give an answer to the above-stated question but gives the structure of the solution so that one does not have to guess what the form of the solution is. Theload of symbolic calculations is brought before the structure of the solution is identified. This is in contrary to the Exp-func-tion type methods where the structure of the solution is first guessed, and then symbolic calculations are exploited for theidentification of parameters.

The object of this paper is to develop and generalize Exp-function type methods. We seek three objectives in this process:

(i) To investigate possibilities to express solutions of differential equations in forms comprising not only exponential,hyperbolic tangent functions, but other standard functions (for example radical functions). The objective is to forma class of standard functions which could be used to express solutions of a widest possible class of differentialequations.

. All rights reserved.

vickas), [email protected] (L. Bikulciene), [email protected] (M. Ragulskis).Ragulskis).

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Z. Navickas et al. / Applied Mathematics and Computation 216 (2010) 2380–2393 2381

(ii) To develop a convenient computational algorithm for the identification of the structure of the solution. These algo-rithms must eliminate the need for guessing how the structure of the solution looks like. In this sense our approachis an alternative (and much more general) method compared to Exp-function type methods where the structure of thesolution is first guessed, and only then symbolic computations are used to calculate appropriate coefficients. Our algo-rithm identifies the structure of the solution automatically and provides information on how the solution is expressedin a form comprising a standard function (if it is a sum of standard functions, a ratio of sums of standard functions, oreven a more complex form involving standard functions).

(iii) It is well known that eigenvalues and eigenvectors play a central role in analysis of linear differential equations. Itwould be of high interest to find analogies of eigenvalues and eigenvectors for nonlinear differential equations anddynamical systems. We propose an insight into nonlinear dynamical systems through the definition of the H-rank,the Hankel characteristic equation and its solutions.

2. Preliminary definitions

2.1. Functions and their extensions

Definition 1. A function f ðx; s0; . . . ; sn�1Þ, where

f :¼ f ðx; s0; . . . ; sn�1Þ ¼Xþ1j¼0

pjðs0; . . . ; sn�1Þðx� cÞj

j!; jx� cj < Rf ð1Þ

is called an extended function if its domain is extended from its region of convergence jx� cj < Rf into a wider region(the real axis or the complex plane) with the exception of a limited number of special singular points x1; . . . ; xm

where limx!xijf ðxÞj ¼ þ1; i ¼ 1;2; . . . ;m; where x; c 2 R; fs0; s1; . . . ; sn�1g is a finite set of parameters; n is the number of

parameters; s0; s1; . . . ; sn�1 2 C; complex functions pjðs0; . . . ; sn�1Þ are polynomials, ratios of polynomials or radicals ofpolynomials.

Fx;s0 ;...;sn�1 is denoted as a set of extended functions.

Definition 2. A function

yðxÞ ¼Xþ1j¼0

qj

j!xj; ð2Þ

where qj; q0 ¼ 1; qj ¼Qj�1

j¼0ðaþ bkÞ; qj – 0; j ¼ 1;2; . . . are coefficients of the expansion, is called a standard function.We will use several standard functions for the construction of generalized solutions of differential equations (we analyze

these solutions only on the real x-axis):

y1ðxÞ ¼Xþ1j¼0

xj

j!; y1ðxÞ ¼ expðxÞ; x 2 R; qj ¼ 1;

y2ðxÞ ¼Xþ1j¼0

xj; jxj < 1; y2ðxÞ ¼1

1� x; x – 1; qj ¼ j!;

y3ðxÞ ¼Xþ1j¼0

ð2j� 1Þ!!j!

xj; ð�1Þ!! ¼ 1; 1!! ¼ 1; 3!! ¼ 1 � � �3; . . . ; jxj < 12

; y3ðxÞ ¼1ffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1� 2xp ; x <

12

; qj ¼ ð2j� 1Þ!!:

ð3Þ

Necessary order derivatives of the described functions do exist and this fact will be not stressed further on.

2.2. Differential and multiplicative operators

Definition 3. The generalized differential operator is defined as [13]:

D :¼ Q 0Ds0 þ � � � þ Q n�1Dsn�1 ; ð4Þ

where Q0; . . . ;Qn�1 2 Fs0 ;...;sn�1 ; Fs0 ;...;sn�1 is a set of extended functions which do not depend on x; Fs0 ;...;sn�1 � Fx;s0 ;...;sn�1 ; Dsjare

conventional linear differential operators (partial differentiation is performed in respect of a parameter sjÞ;D0

sj:¼ 1; j ¼ 0;1; . . . ;n� 1; 1 is an identical linear operator.

2382 Z. Navickas et al. / Applied Mathematics and Computation 216 (2010) 2380–2393

Usual properties of differentiation hold for the generalized differential operator [13]:

DXm

j¼1

cjfj

!¼Xm

j¼1

cj � Dfj;

Dnðfk � flÞ ¼Xn

j¼0

n

j

� �Djfk

� �Dn�jfl

� �;

Df m ¼ mf m�1Df ;

Dfk

fl¼ ðDfkÞfl � fkðDflÞ

f 2l

;

� � �m; n ¼ 0;1; . . . ; cj 2 C; j ¼ 1;2; . . . ;m; k; l ¼ 1;2; . . . ;n; f ; fk; fl 2 Fx;s0 ;...;sn�1 :

ð5Þ

It can be noted that the generalized differential operator sometimes is referred to as a vector field [15].

Definition 4. The multiplicative operator [13] is a linear operator:

G :¼ GD :¼Xþ1j¼0

xj

j!Dj: ð6Þ

It can be noted that the multiplicative operator is denoted as the exponential operator GD :¼ exD in [15,16].We will enumerate several properties of the multiplicative operator:

GXm

j¼1

cjfj ¼Xm

j¼1

cjGfj;

Gf ðc; s0; . . . ; sn�1Þ ¼ f ðGc; Gs0; . . . ;Gsn�1Þ;

Gfkðc; s0; . . . ; sn�1Þflðc; s0; . . . ; sn�1Þ

¼ fkðGc; Gs0; . . . ;Gsn�1ÞflðGc; Gs0; . . . ;Gsn�1Þ

;

ð7Þ

where cj 2 C; j ¼ 1;2; . . . ;m; f ðx; s0; . . . ; sn�1Þ; f kðx; s0; . . . ; sn�1Þ; f lðx; s0; . . . ; sn�1Þ 2 Fx;s0 ;...;sn�1 .For example,

exDc cn ¼ ðxþ cÞn; exDc f ðc; s0; . . . ; sn�1Þ ¼ f ðc þ x; s0; . . . ; sn�1Þ: ð8Þ

2.3. Operator method for solving ordinary differential equations

2.3.1. First order ordinary differential equations, the initial value problemThe exact solution of a differential equation

y0x ¼ P1ðx; yÞ; yðx0; sÞ ¼ s; x0 2 R; y 2 Fx;s0 ð9Þ

reads [13]:

y ¼Xþ1j¼0

ðx� x0Þj

j!ððDc þ P1ðc; sÞDsÞjsÞ

��c¼x0

; ð10Þ

where the superscript denotes a full derivative and the subscript denotes the variable of differentiation. It can be noted thatthe solution is expressed in a form of an infinite order polynomial.

2.3.2. nth order ordinary differential equations, the initial value problemThe exact solution of an initial value problem

yðnÞx ¼ Pn x; y; y0x; . . . ; yðn�1Þx

� ;

yðjÞx ðx; s0; s1; . . . ; sn�1Þ��x¼x0¼ sj; x0 2 R; Pnðx; s0; s1; . . . ; sn�1Þ 2 Fx;s0 ;...;sn�1

ð11Þ

reads:

y ¼Xþ1j¼0

ðx� x0Þj

j!Djs0

� ��c¼x0

; ð12Þ

where

D :¼ Dc þ s1Ds0 þ s2Ds1 þ � � � þ sn�1Dsn�2 þ Pnðc; s0; s1; . . . ; sn�1ÞDsn�1 : ð13Þ

Example 1. Let Dnþ1s0 ¼ 0, but Dns0 – 0. Then, Dnþks0 ¼ 0 for k ¼ 1;2; . . ., and yðx; s0; s1; . . . ; sn�1Þ ¼Pn

j¼0ðx�x0Þj

j! Djs0

� �jc¼x0

;

meaning that the function yðx; s0; s1; . . . ; sn�1Þ is an nth order polynomial of x. It can be noted that fcDnsjc 2 Cg# KerD.


2.3.3. Systems of first order ordinary differential equations, the initial value problemThe exact solution of a system of differential equations

z0x ¼ Rðx; z;u; vÞ; zðx0; s0; s1; s2Þ ¼ s0;

u0x ¼ Pðx; z;u; vÞ; uðx0; s0; s1; s2Þ ¼ s1;

v 0x ¼ Qðx; z;u; vÞ; vðx0; s0; s1; s2Þ ¼ s2;

8><>: ð14Þ

reads:

z ¼Xþ1j¼0

ðx� x0Þj

j!Djs0

� ��c¼x0

;

u ¼Xþ1j¼0

ðx� x0Þj

j!Djs1

� ��c¼x0

;

v ¼Xþ1j¼0

ðx� x0Þj

j!Djs2

� ��c¼x0

;

ð15Þ

where D ¼ Dc þ Rðc; s0; s1; s2ÞDs0 þ Pðc; s0; s1; s2ÞDs1 þ Qðc; s0; s1; s2ÞDs2 .

Example 2. The generalized differential operator for a system(
z0x ¼ 1; zðx0; c; sÞ ¼ c;
u0x ¼ uz ; uðx0; c; sÞ ¼ s;

ð16Þ

reads: D ¼ Dc þ sc Ds. Thus, D0c ¼ c; Dc ¼ 1; D0s ¼ s; Ds ¼ s

c; but D2c ¼ D2s ¼ 0. Thus, the expression on the solution ðz;uÞ reads:

z ¼ c þ ðx� x0Þ;

u ¼ sþ scðx� x0Þ:

ð17Þ

It can be noted that the system (16) is equivalent to the differential equation xy0 ¼ y which general solution is

y ¼ ax; a ¼ sc2 C

� �: ð18Þ

2.4. The H-rank of a sequence

A Hankel matrix [14] for the sequence ðpj; j ¼ 0;1;2; . . .Þ where pj 2 Fx;s0 ;...;sn�1 or pj 2 C is defined as:

Hn :¼

p0 p1 � � � pn�1

p1 p2 � � � pn

� � �pn�1 pn � � � p2n�2

26664

37775; n ¼ 1;2; . . . : ð19Þ

Definition 5. A sequence ðpj; j 2 Z0Þ has an H-rank Hrðpj; j 2 Z0Þ ¼ m; m 2 Z0;m < þ1 if the sequence of determinants ofHankel matrixes has the following form: ðd1; d2; . . . ; dm; 0; 0; . . .Þ; where dm – 0; dmþ1 ¼ dmþ2 ¼ � � � ¼ 0.

Definition 6. The length of a sequence ðpj; j 2 Z0Þ is Lðpj; j 2 Z0Þ ¼ m; m 2 Z0, if the sequence has the formðp0; p1; . . . ; pm; 0; 0; . . .Þ; where pm – 0; pmþ1 ¼ pmþ2 ¼ � � � ¼ 0.

Corollary 1. If Lðpj; j 2 Z0Þ ¼ m; m < þ1, then Hrðpj; j 2 Z0Þ ¼ m.

Definition 7. The characteristic Hankel equation for a sequence ðpj; j 2 Z0Þ which H-rank is equal to m is defined as:

det

p0 p1 � � � pm

p1 p2 � � � pmþ1

� � �pm�1 pm � � � p2m�1

1 q � � � qm

26666664

37777775¼ 0: ð20Þ


Expansion of the determinant in Eq. (20) yields an mth order algebraic equation for determination of roots of the character-istic equation:

qm þ Am�1qm�1 þ � � � þ A1qþ A0 ¼ 0: ð21Þ

Corollary 2. The following statement holds true [14]:Let Hrðpj; j 2 Z0Þ ¼ m and the multiplicity of roots q1;q2; . . . ;ql of the char-

acteristic equation (Eq. (21)) is accordingly m1;m2; . . . ;ml;Pl

r¼1mr ¼ m. Then,

pj ¼Xl

r¼1

Xmr�1

k¼0

lrk

j

k

� �qj�k

r ; ð22Þ

where lrk;qr 2 C.The opposite statement also holds true. If Eq. (22) holds, then

Hrðpj; j 2 Z0Þ ¼ m1 þm2 þ � � � þml: ð23Þ

Definition 8. A sequence is entitled as an algebraic sequence if elements of that sequence ðpj; j 2 Z0Þ are defined by Eq. (22).

We assume that lrkjk

� �qj�k

r ¼ 0 if jk

� �¼ 0 what is true when 0 6 j < k, where j

k

� �¼ j!

k!ðj�kÞ!.

Corollary 3. In case when all roots of the characteristic equation are different, Eq. (22) obtains a more simple form:

pj ¼Xm

r¼1

lrqjr: ð24Þ

It can be noted that coefficients lrk (or just lrÞ can be found solving the linear algebraic system of equations (q1;q2; . . . ;ql

are already determined):

Xl

r¼1

Xmr�1

k¼0

j

k

� �qj�k

r lrk ¼ pj; j ¼ 0;1; . . . ;m� 1: ð25Þ

This system of equations always has the only solution [14].It can be also noted that one can use Eq. (22) or Eq. (24) to calculate all elements of the series ðpj; j 2 Z0Þ starting from p2m

if Hrðpj; j 2 Z0Þ ¼ m and the first 2m elements of that series are known.

Example 3. Lets consider a sequence ðp0; p1; p2; 0; 0; . . .Þ where p2 – 0 and pj ¼ 0; j ¼ 3;4; . . .. Then, det H3 ¼ �p32 – 0; but

det Hj ¼ 0 for j ¼ 4;5; . . .. Thus, Hrðp0; p1; p2; 0; 0; . . .Þ ¼ 3. It is clear that
p0 p1 p2 0p1 p2 0 0p2 0 0 01 q q2 q3
��

��¼ 0; ð26Þ

what yields q3p3 ¼ 0; q1 ¼ q2 ¼ q3 ¼ 0. Therefore, finally:

pj ¼ p00j þ p1j1

� �0j�1 þ p2

j2

� �0j�2; j ¼ 0;1;2; . . . ; where 00 :¼ 1; 01 ¼ 02 ¼ � � � ¼ 0: ð27Þ

2.5. Changing the independent variable of a differential equation

The independent variable of a differential equation can be changed in order to produce a more convenient form of theexact solution of a differential equation.

Let an invertible function is given by following equalities:
x :¼ uðzÞ; z ¼ wðxÞ: ð28Þ
Then following expressions hold:

z0x ¼ w0x ¼1u0z

;

z00xx ¼ �1

u0z� 2 u00zz � z0x ¼ �

u00zz

u0z� 3 ;

� � � :

ð29Þ


Also, for every function

y ¼ yðxÞ ¼ yðuðzÞÞ :¼ xðzÞ ¼ xðwðxÞÞ ¼ x ð30Þ

following equalities hold:

y0x ¼ x0z��z¼wðxÞ � w

0x ¼

1u0z�x0z

��z¼wðxÞ;

y00xx ¼ x00zz

��z¼wðxÞ � w0x

� 2 þx0z��z¼wðxÞ � w

00xx ¼

1

u0z� 2 �x

00zz

��z¼wðxÞ �

u00zz

u0z� 3 �x

0z

��z¼wðxÞ;

� � � :

ð31Þ

Thus,

y0x��

x¼uðzÞ ¼1u0z�x0z;

y00xx

��x¼uðzÞ ¼

1

u0z� 2 �x

00zz �

u00zz

u0z� 3 �x

0z;

� � � :

ð32Þ

Merging Eqs. (29)–(32) into one equality and using symbol s to identify the change of variable in the differential equationyields:

s x; y; y0x; y00xx; . . .�

:¼ xjx¼uðzÞ; yjx¼uðzÞ; y0xjx¼uðzÞ; y00xxjx¼uðzÞ; � � ��

; ð33Þ

or

s x; y; y0x; y00xx; . . .�

:¼ uðzÞ;x;1u0z�x0z;

1

u0z� 2 �x

00zz �

u00zz

u0z� 3 �x

0z; � � �

!: ð34Þ

For example, the image of a second order differential equation (Eq. (11) at n ¼ 2) takes the following form after the variablechange x ¼ uðzÞ:

1

u0z� 2 x00zz �

u00zz

u0z� 3 x0z ¼ P2 uðzÞ; x; 1

u0zx0z

� �: ð35Þ

Example 4. Several typical variable changes are listed below:

(i) Exponential variable change expðaxÞ ¼ z:

s x; y; y0x; y00xx; . . .�

:¼ 1a

ln z; x; azx0z;a2 z2x00zz þ zx0z�

; . . .

� �: ð36Þ

(ii) Logarithmic variable change a ln x ¼ z:

s x; y; y0x; y00xx; . . .�

:¼ expza

� �;x; exp � z

a

� �x0z; exp �2z

a

� �x00zz �x0z�

; . . .

� �: ð37Þ

(iii) Symmetric variable change ax ¼ z:

s x; y; y0x; y00xx; . . .�

:¼ az

; x;� z2

ax0z;

2z3

a2 x0z þz4

a2 x00zz; . . .

� �: ð38Þ

where a is a parameter of the variable change; a 2 R; a – 0.

3. Main theorems

We will prove one of the fundamental properties of the generalized differential operator. This property can be used toexpress a solution of a differential equation in a form comprising standard functions.


3.1. Theorem 1

The formulation of the Theorem.

(a) A generalized differential operator D :¼ Q1Ds1 þ � � � þ Q nDsn ; Q 1; . . . ;Qn 2 Fs1 ;...;sn is used to calculate functionspj :¼ Djs; pj 2 Fs1 ;...;sn ; j ¼ 0;1; . . ..

(b) A rule q0 :¼ 1; qjþ1 ¼Qj

k¼0ðaþ bkÞ; j ¼ 0;1;2; . . . is used to construct a sequence ðqj; j 2 Z0Þ; a; b 2C are such con-stants that qj – 0.

(A) Then, three following statements can have a sense (but can be individually true or not true):(i) pj ¼ qj

Pmr¼1lrk

jr , where kk;lk 2 Fs1 ;...;sn ; kk – kl and lk – 0 when k – l and k; l ¼ 1;2; . . . ;m; m 2 N and is a

fixed constant; j ¼ 0;1;2; . . ..(ii) Hr 1

qjpj; j 2 Z0

� �¼ n; n 2 N and is a fixed constant. Also, roots of Hankel characteristic equation

q1;q2; . . . ;qn 2 Fs1 ;...;sn are all different.(iii) There exists a set of functions c1; c2; . . . ; c�n;r1;r2; . . . ;r�n 2 Fs0 ;...;sn�1 (�n 2 N and is a fixed constant) which sat-

isfy following conditions:

ðaÞck – cl; rk – 0;

ðbÞr1 þ r2 þ � � � þ r�n ¼ s;

ðcÞDrk ¼ arkck;

ðdÞDck ¼ bc2k ;

ð39Þ

where k – l; k ¼ 1;2; . . . ; �n; a; b 2 C.
(B) Statements (i), (ii) and (iii) are equivalent.
Proof

1. The equivalency of statements (i) and (ii) follows from results proven in [14]. Also, m ¼ n and kk ¼ qk whenk ¼ 1;2; . . . ;m.

2. Lets assume that the statement (iii) holds true. Then, p0 ¼ D0s ¼ s ¼ q0s;

p1 ¼ DX�n

r¼1

rr ¼X�n

r¼1

Drr ¼ aX�n

r¼1

rrcr ¼ q1

X�n

r¼1

rrcr:

We make a proposition that Djs ¼ pj ¼Qj�1

l¼0ðaþ blÞP�n

r¼1rrcjr ; j 2 N and is a fixed constant. Then,

Djþ1s ¼ Dpj ¼Yj�1

l¼0

ðaþ blÞX�n

r¼1

ðDrrÞcjr þ ri Dcj

r

� � ¼Yj�1

l¼0

ðaþ blÞX�n

r¼1

arrcjþ1r þ jrrbcjþ1

r

� ¼Yj

l¼0

ðaþ blÞX�n

r¼1

rrcjþ1r ¼ pjþ1:

ð40ÞThus, the statement (ii) (and the statement (i)) holds true.

3. Lets assume that statements (i) and (ii) hold true. Then,

Djþ1s ¼ qjDXm

r¼1

lrkjr ¼ qj

Xm

r¼1

kjrðDlrÞ þ jlrk

j�1r ðDkrÞ

� �: ð41Þ

Thus, the following system of equations is produced:Xm

r¼1

kjrðDlrÞ þ jlrk

j�1r ðDkrÞ

� �¼ 1

qjqjþ1

Xm

r¼1

lrkjþ1r ; ð42Þ

which is satisfied if following identities hold: Dlr ¼ alrkr and Dkr ¼ blrk2r . Thus, the system of equations has at least one

solution. On the other hand, the extended Van-Der-Mond determinant of coefficients at unknowns Dlr and Dkr (when jsweeps over any of natural numbers 0 6 j1 < j2 < � � � < j2mÞ

D ¼

kj11 kj1

2 � � � kj1m j1l1k

j1�11 j1l2k

j1�12 � � � j1lmkj1�1

m

kj21 kj2

2 � � � kj2m j2l1k

j2�11 j2l2k

j2�12 � � � j2lmkj2�1

m

� � � � � � � � � � � � � � � � � � � � � � � �kj2m

1 kj2m2 � � � kj2m

m j2ml1kj2m�11 j2ml2k

j2m�12 � � � j2mlmkj2m�1

m

��

��

¼ l1l2 � . . . � lm

kj11 kj1

2 � � � kj1m j1k

j1�11 j1k

j1�12 � � � j1k

j1�1m

kj21 kj2

2 � � � kj2m j2k

j2�11 j2k

j2�12 � � � j2k

j2�1m

� � � � � � � � � � � � � � � � � � � � � � � �kj2m

1 kj2m2 � � � kj2m

m j2mkj2m�11 j2mkj2m�1

2 � � � j2mkj2m�1m

��

��ð43Þ


cannot be equivocally equal to zero. Thus, the solution of the system of equations is the only one. Finally, the validity of thestatement (iii) follows from the validity of statements (i) and (ii) at m ¼ �n; kk ¼ ck; lk ¼ rk; k ¼ 1;2; . . . ; �n; a ¼ a andb ¼ b. h

It can be noted that Theorem 1 holds only when all roots of the Hankel characteristic equation are different.Theorem 1 forms the theoretical background for the construction of general solutions of differential equations. Let a se-

quence ðDjs; j 2 Z0Þ satisfy equivalent statements (i), (ii) and (iii) of the Theorem 1. Then, a general solutiony ¼ yðx; s0; s1; . . . ; sn�1Þ of a differential equation can be expressed in the following form:

yðx; s0; s1; . . . ; sn�1Þ ¼Xþ1j¼0

ðqj

Xm

r¼1

lrkjrÞðx� x0Þj

j!¼Xm

r¼1

lr

Xþ1j¼0

qj

j!ðkrðx� x0ÞÞj ¼

Xm

r¼1

lrf ðkrðx� x0ÞÞ; ð44Þ

where f ðxÞ is a standard function defined in Definition 2.

3.2. Theorem 2

Roots of the Hankel characteristic equation coincide with roots of the standard characteristic equationkm � am�1k

m�1 � � � � � a1k� a0 ¼ 0 for linear differential equations with constant coefficients.

Proof. The initial value problem reads:

yðmÞx ¼ am�1yðm�1Þx þ � � � þ a1y0x þ a0y; yðrÞx ðx; s0; . . . ; sm�1Þjx¼x0

¼ sr ;

r ¼ 0;1; . . . ;m� 1; a0; a1; . . . ; am�1 2 R:ð45Þ

We select qj ¼ 1; j ¼ 0;1;2; . . ..Then, the generalized differential operator takes the following form (Eq. (13)):

D ¼ s1Ds0 þ s2Ds1 þ � � � þ sm�1Dsm�2 þ ðam�1sm�1 þ � � � þ a1s1 þ a0s0ÞDsm�1 : ð46Þ

It can be noted that �a0s0 � a1Ds0 � � � � � am�1Dm�1s0 þ Dms0 ¼ 0 because�a0s0 � a1s1 � � � � � am�1sm�1 þ am�1sm�1 þ � � � þ a1s1 þ a0s0 ¼ 0. Thus,�a0Djs0 � a1Djþ1s0 � � � � � am�1Djþm�1s0 þ Djþms0 ¼ 0; j ¼ 0;1;2; . . .. The last equalities yield:

dmþ1 ¼ det

D0s0 Ds0 � � � Dms0

Ds0 D2s0 � � � Dmþ1s0

� � � � � � � � � � � �Dms0 Dmþ1s0 � � � D2ms0

26664

37775 ¼ 0: ð47Þ

But

dm ¼ detD0s0 � � � Dm�1s0

� � � � � � � � �Dm�1s0 � � � D2m�2s0

264

375 – 0: ð48Þ

Thus, the sequence of determinants is ðd1; d2; . . . ; dm; 0; 0; . . .Þ. Therefore, HrðDjs; j ¼ 0;1;2; . . .Þ ¼ m. Moreover, b ¼ 0 becauseqj ¼ 1. Next, one has to find roots of the Hankel characteristic equation q1;q2; . . . ;qm.

If all roots are different, then Djs0 ¼Pm

r¼1lrðs0; s1; . . . ; sm�1Þqjr . Thus, Dlr ¼ lrqr ; Dqr ¼ 0; r ¼ 1;2; . . . ;m� 1. Finally,

y ¼Xm

r¼1

lrðs0; s1; . . . ; sm�1Þ expðqrðx� x0ÞÞ: � ð49Þ

It can be noted that in general case not only roots but also their multiplicities of the Hankel characteristic equation coin-cide with roots of the standard characteristic equation if only the equation is a linear differential equation with constantcoefficients.

3.3. The case when some roots of the Hankel characteristic equation are multiple

It is clear that Theorem 1 does not hold when some roots of the Hankel characteristic equation are multiple. In that casethe structure of the solution becomes more complex and cannot be expressed a form comprising only standard functions(the independent variable also figures in the expression then). We will give a short discussion and an example though thisis an object of the future research.

Let Djs ¼ qj

Plr¼1

Pmr�1k¼0 lrk

jk

� �qj�k

r . In other words, multiplicities of roots of the Hankel characteristic equation

q1;q2; . . . ;ql are m1; m2; . . . ; ml accordingly. Then, Eq. (12) yields:


Xþ1j¼0

qj

jk

� �qj�k

rðx� x0Þj

j!

��c¼x0

¼ ðx� x0Þk

k!

Xþ1j¼k

qj

ðj� kÞ! ðqrðx� x0ÞÞj�k��

c¼x0

¼ ðx� x0Þk

k!qkr

Xþ1j¼0

qj

j!ðqrðx� x0ÞÞj

!ðkÞx

��c¼x0

¼ ðx� x0Þk

k!qkrðf ðqrðx� x0ÞÞÞðkÞx

��c¼x0

: ð50Þ

Therefore,

y ¼Pþ1j¼0

qj

Pl

r¼1

Pmr�1

k¼0lrk

j

k

� �qj�k

rðx�x0Þj

j!

��c¼x0

¼Pl

r¼1

Pmr�1

k¼0lrk

Pþ1j¼0

qj

j

k

� �qj�k

rðx�x0Þj

j!

��c¼x0

¼Pl

r¼1

Pmr�1

k¼0lrk

ðx�x0Þk

k!qkrðf ðqrðx� x0ÞÞÞðkÞx

��c¼x0

:

ð51Þ

Example 5. y00 ¼ 4y0 � 4y; y ¼ yðx; s0; s1Þ; initial conditions yð0; s0; s1Þ ¼ s0; ðyðx; s0; s1ÞÞ0xjx¼0 ¼ s1.The generalized differential operator for this differential equations is D ¼ s1Ds0 þ 4ðs1 � s0ÞDs1 .Trivial transformations yield HrðDjs; j ¼ 0;1;2; . . .Þ ¼ 2. Roots of the Hankel characteristic equation are q1 ¼ q2 ¼ 2. Then,

l10 ¼ s0 and l11 ¼ s1 � s0. Thus, pj ¼ s02j þ ðs1 � 2s0Þj1

� �2j�1; and finally,

yðx; s0; s1Þ ¼ s0 expð2xÞ þ ðs1 � 2s0Þx2ðexpð2xÞÞ0x ¼ s0 expð2xÞ þ ðs1 � 2s0Þx expð2xÞ:

This trivial example illustrates the above-mentioned fact that a solution cannot be expressed in a form comprising standardfunctions if roots of the Hankel characteristic equation are all not different.

4. The generalization of the Exp-function method

4.1. The algorithm for the construction of an exact solution of an ordinary differential equation

We will illustrate the algorithm for the initial problem of an nth order ordinary explicit differential equation:

dnydxn ¼ Pn x; y;

dydx; . . . ;

dn�1y

dxn�1

!;

djy

dxjðx; s0; s1; . . . ; sn�1Þ

��x¼x0

¼ sj; x0 2 R; Pnðc; s0; s1; . . . ; sn�1Þ 2 Fx;s1 ;...;sn : ð52Þ

(A) The change of the independent variable x ¼ uðz; aÞ; z ¼ wðx; aÞ; where a is an independent parameter (in general, thevariable change can comprise many independent parameters).It can be noted that the selection of the invertible function uðz; aÞ is in many particular cases dependent on a subjec-tive experience and knowledge of the area of application where the differential equation is used. The famous Exp-function method is based on the variable change uðz; aÞ :¼ expðzÞ. Then the differential equation takes the formdefined by Eq. (34) and the image differential equation takes the form analogous to Eq. (35).

(B) Selection of the parameter b.The sequence ðqj; j 2 Z0Þ is constructed as follows: q0 :¼ 1; qjþ1 ¼

Qjk¼0ðaþ bkÞ; j ¼ 0;1;2; . . .. It can be noted that the

selection of the parameters a and b determines the standard function (Eq. (2)) which will figure in the expression ofthe exact solution (in most cases a ¼ 1 is assumed).

(C) Construction of the generalized differential operator:

D :¼ Dc þ s1Ds0 þ s2Ds1 þ � � � þ sn�1Dsn�2 þ Pnðc; s0; s1; . . . ; sn�1ÞDsn�1 : ð53Þ

(D) Calculation of the first elements of the sequence ðp̂j; j 2 Z0Þ (Eq. (1)):

p̂j :¼ 1qj

Djs; j ¼ 0;1;2; . . . ;2m: ð54Þ

(E) Construction of first elements of the Hankel matrixes sequence:

H1 ¼ ½p̂0�; H2 ¼p̂0 p̂1

p̂1 p̂2

�; . . . ; Hmþ1 ¼

p̂0 p̂1 � � � p̂m

p̂1 p̂2 � � � p̂mþ1

� � � � � � � � � � � �p̂m p̂mþ1 � � � p̂2m

26664

37775: ð55Þ


(F) Calculation of determinants of Hankel matrixes dk ¼ det Hk up to the first zero:

ðd1; d2; . . . ; dm; 0Þ; d1; d2; . . . ; dm – 0; dmþ1 ¼ 0: ð56Þ

(G) Construction of the mth Hankel’s characteristic equation and determination of its roots q1;q2; . . . ;qm:

det

p̂0 p̂1 � � � p̂m

p̂1 p̂2 � � � p̂mþ1

� � � � � � � � � � � �p̂m�1 p̂m � � � p̂2m�1

1 q � � � qm

26666664

37777775¼ 0: ð57Þ

(H) If all roots of the characteristic equation are different, the following linear algebraic system of equations is solved fordetermination of l1; l2; . . . ;lm:

l1qj1 þ l2q

j2 þ � � � þ lmqj

m ¼ p̂j; j ¼ 0;1; . . . ;m� 1: ð58Þ

(I) Following equalities must be tested:

Dlr ¼ lrqr ; Dqr ¼ bq2r ; r ¼ 1;2; . . . ;m: ð59Þ

If these equalities hold, then statements (i), (ii) and (iii) of the Theorem 1 hold true too for the sequenceðDjs; j ¼ 0;1;2; . . .Þ. Moreover, dmþ1 ¼ dmþ2 ¼ � � � ¼ 0 and the exact solution of the differential equation can be ex-pressed in a form comprising the selected standard function.

(J) Construction of the exact solution of the image differential equation in a form of a series:

xðzÞ ¼Xþ1j¼0

qj

j!

Xm

r¼1

lrqjr

!ðz� z0Þj ¼

Xm

r¼1

lr

Xþ1j¼0

qj

j!qj

rðz� z0Þj: ð60Þ

(K) Expression of the exact solution in a form comprising the selected standard function f ðzÞ ¼Pþ1

j¼0qj

j! zj:

xðzÞ ¼Xm

r¼1

lrf ðqrðz� z0ÞÞ: ð61Þ

(L) Construction of the exact solution of the original differential equation:

yðx; s1; . . . ; snÞ ¼Xm

r¼1

lrf ðqrðwðx; aÞ � wðx0; aÞÞÞ: ð62Þ

(M) Checking if the produced solution satisfies the original differential equation.

5. Examples

5.1. The exact solution is expressed in a ratio of sums of exponential functions

The initial problem reads:

y0x ¼ y2 þ y� 6; y ¼ yðx; sÞ; yð0; sÞ ¼ s: ð63Þ

A variable change x ¼ 1a ln z; z ¼ expðaxÞ yields:

x0z ¼1azðx2 þx� 6Þ; x ¼ xðz; sÞ;xð1; sÞ ¼ s: ð64Þ

We select b ¼ 1. Thus, qj ¼ j!; j 2 Z0. The generalized differential operator reads:

D ¼ Dc þ1acðs2 þ s� 6ÞDs ¼ Dc þ

ðsþ 3Þðs� 2Þac

Ds: ð65Þ

Computational symbolic transformations yield ðpj :¼ DjsÞ:

p0 ¼ s; p1 ¼ðsþ 3Þðs� 2Þ

ac; p2 ¼

ðsþ 3Þ2ðs� 2Þ2ð2s� 1Þa2c2 � ðsþ 3Þðs� 2Þ

ac2 ;

p3 ¼ðsþ 3Þðs� 2Þð6s2 þ 6s� 11� 6sa� 3aþ 2a2Þ

a3c3 ; . . . :

ð66Þ


Next, determinants of Hankel matrixes dn ¼ detðHnÞ; n ¼ 1;2; . . . are calculated:

d1 ¼ s; d2 ¼ðsþ 3Þðs� 2Þð�sþ sc � 12Þ

2a2c2 ;

d3 ¼ðsþ 3Þ2ðs� 2Þ2ða� 5Þðaþ 5Þð2a2s� 36aþ 3saþ 25sÞ

144a6c6 :

ð67Þ

It can be noted, that equality a ¼ 5 produces d3 ¼ 0 at any s. Thus, Hr 1j! D

js��a¼5

; j 2 Z0

� �¼ 2. Therefore, the variable change

x ¼ 15 ln z; z ¼ expð5xÞ appears to be appropriate.

Hankel characteristic equation reads:

s Ds 12 D2s

Ds 12 D2s 1

6 D3s

1 q q2

�� ¼ 0: ð68Þ

Elementary transformations yield q2 � s�25c q ¼ 0. Roots are q1 ¼ 0; q2 ¼ s�2

5c . Thus, l1 ¼ �3; l2 ¼ sþ 3. We check the validityof the Theorem 1:

Dl1 ¼ l1q1 ¼ 0; Dl2 ¼ Dðsþ 3Þ ¼ ðsþ 3Þðs� 2Þ5c

¼ l2q2;

Dq1 ¼ 0 ¼ q21; Dq2 ¼ D

s� 25c¼ s� 2

5c

� �2

¼ q22:

ð69Þ

Thus, the sequence ðqj; j 2 Z0Þ and the variable change were selected appropriately. Now,

y ¼Pþ1

j¼0 j! �3 � 0j þ ðsþ 3Þ s�25c

� j� ��

c¼1

ðz�1Þjj! ¼ �3þ ðsþ 3Þ

Pþ1j¼0

s�2ð Þðz�1Þ5

� �j, when jz� 1j < 5

js�2j. Now we expand the pro-duced series into the whole z-axis, except the point z ¼ sþ3

s�2. Then,

y ¼ �3þ 5sþ 155� szþ zþ 2z� 2

¼ 2ð3þ sÞ � 3ð2� sÞz3þ sþ ð2� sÞz : ð70Þ

Finally, the exact solution of the differential equation reads:

yðx; sÞ ¼ 2ð3þ sÞ � 3ð2� sÞ expð5xÞ3þ sþ ð2� sÞ expð5xÞ ¼ 2ð3þ sÞ expð�3xÞ � 3ð2� sÞ expð2xÞ

ð3þ sÞ expð�3xÞ þ ð2� sÞ expð2xÞ : ð71Þ

5.2. The exact solution is expressed in a form comprising a square root

The initial problem reads:

y0x ¼ y3; y ¼ yðx; sÞ; yðc; sÞ ¼ s: ð72Þ

Now we do not make a variable change, but select b :¼ 2 and q0 ¼ 1; qj ¼ ð2j� 1Þ!!; j ¼ 1;2; . . . instead.The generalized differential operator reads D ¼ s3Ds. Then, p0 ¼ s; pj ¼ 1

qjDjs ¼ s2jþ1; j ¼ 1;2; . . ..

It is easy to see that the sequence of determinants of Hankel matrices is ðs; 0; 0; . . .Þ; thus Hrðpj; j 2 Z0Þ ¼ 1. The Hankel

characteristic equation reads s s3

1 q

�� ¼ 0; or q ¼ s2. Then, l ¼ s.

Now we need to check the validity of Theorem 1: Dl ¼ s3Dss ¼ s3 ¼ lq; Dq ¼ s3Dss2 ¼ 2s4 ¼ 2q2. Thus, the selection ofthe sequence q0 ¼ 1; qj ¼ ð2j� 1Þ!!; j ¼ 1;2; . . . is appropriate for this differential equation.

Now, the structure of the standard function yields:

yðxÞ ¼ sþXþ1j¼1

ð2j� 1Þ!!j!

s � s2jðx� cÞj ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� 2s2ðx� cÞ

p ; jx� cj < 12s2 : ð73Þ

Finally, the exact solution of the differential equation reads:

yðx; cÞ ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� 2s2ðx� cÞ

p ; x <1

2s2 þ c if it is required that yðx; cÞ 2 R: ð74Þ

5.3. Partial solutions of the Liouville’s equation

A natural question is if our developed technique could be applied for the construction of an exact analytical solution of anequation which is already solved by the classical Exp-function method. We select the Liouville’s equation which is analyzedin [17]:


y � y00xx � y0x� 2 þ 2y3 ¼ 0; ð75Þ

where the partial solution y ¼ yðxÞ satisfies initial conditions yð0Þ ¼ s and y0xðxÞ��

x¼0 ¼ t. We will seek only such solutionswhich can be expressed in a ratio of finite sums of exponential functions. We perform the variable change defined by Eq.(36) at a :¼ 1 because it helps to simplify technical computations. The image differential equation of Eq. (75) then takesthe following form:

x z2x00zz þ zx0z�

� z2 x0z� 2 þ 2x3 ¼ 0; ð76Þ

where x ¼ xðzÞ; xð1Þ ¼ s and x0zðzÞ��z¼1 ¼ t. The generalized differential operation of the image differential equation reads:

D ¼ Dc þ tDs þc2t2 � cst � 2s3

c2s� Dt: ð77Þ

Then, xðzÞ ¼Pþ1

j¼0ðz�1Þj

j! � Djs��

c¼1and yðxÞ ¼ xðexpðxÞÞ. We select qj :¼ j! (Eq. (3)) and construct the sequence ðp̂j; j ¼ 0;1;2; . . .Þ

where p̂j ¼ 1j! D

js.

The sequence of determinants of Hankel matrixes is:

det H1 ¼ jp̂0j ¼ s;

det H2 ¼p̂0 p̂1

p̂1 p̂2

�� ¼ � c2t2 þ sct þ 2s3

2c2 ;

det H3 ¼ �c2t2 � s2 þ 4s3

144s3c6 ðc4t4 þ 3sc3t3 þ 4c2t2s3 þ 2t2s2c2 þ 6ts4c � 12s6Þ;

det H4 ¼ðc2t2 � s2 þ 4s3Þ2

1036800s8c12

�88t2s7c2 � 72c3t3s6 þ 42c4t4s5 þ 48c5t5s4 þ 10c6t6s3 þ 1112t2s9c2þþ212s6c4t4 þ 360s11 þ 236c2t2s8 þ 360ts9c þ 1440s12 þ c8t8 þ 6sc7t7��18s3c5t5 þ 7s2c6t6 � 44s4c4t4 � 24s5c3t3 þ 1440cts10 þ 456t3s7c3

0B@

1CA:

Thus, det H3jc¼1 ¼ det H4jc¼1 ¼ 0 when

t2 � s2 þ 4s3 ¼ 0; ð78Þ

or t ¼ �sffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� 4sp

; s 6 1=4. The graph of the curve (78) is illustrated in Fig. 1.Thus, the solution of the differential Eq. (75) can be expressed in a ratio of finite sums of exponential functions when vari-

ables of initial conditions s and t satisfy equality (78). Now the algorithm described in Section 4.1 can be used to constructthe function xðzÞ. It is sufficient to check if this function satisfies the original differential equation to conclude that it is apartial solution.

First of all, we construct the Hankel characteristic Eq. (57):

dets Ds 1

2 D2s

Ds 12 D2s 1

6 D3s

1 q q2

264

375�� c ¼ 1t ¼ �s

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� 4sp

¼ 0; ð79Þ

Fig. 1. The graph of the relationship between initial conditions s and t.


which yields q1 ¼ q2 ¼ 12 �1�

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� 4sp� �

. We assume q ¼ q1 ¼ q2 ¼ 12


� 1� �

and construct equalities:

Djsj!

�� c ¼ 1t ¼ s


¼ l1qj þ l2

j

1

� �qj�1; j ¼ 0;1;2; . . . : ð80Þ

Eq. (80) yields a system of linear algebraic equations for determination of coefficients l1 and l2:

12


� 1� �� j

l1 þ12


� 1� �� j�1

� j � l2 ¼Djsj!

�� c ¼ 1t ¼ s


; j ¼ 0;1: ð81Þ

It can be noted that the assumption q ¼ q1 ¼ q2 ¼ 12 �


� 1� �

would yield the same function x ¼ x zð Þ.

The system of Eq. (81) yields l1 ¼ s and l2 ¼ s2 1þ


. Finally,

xðzÞj c ¼ 1t ¼ s


¼Xþ1j¼0

ðz� 1Þj s12


� 1� �� j

þ s2

1þffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� 4sp� �

j12


� 1� �� j�1

!

¼2s 1� 2sþ


z

2szþ 1� 2sþffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� 4sp� �2 ¼

1�2sþffiffiffiffiffiffiffiffi1�4sp

2s � z

zþ 1�2sþffiffiffiffiffiffiffiffi1�4sp

2s

� �2 ð82Þ

Further computations can be simplified introducing an auxiliary parameter b :¼ 1�2sþffiffiffiffiffiffiffiffi1�4sp

2s ; b 2 R. Now it is easy to check that

the function �xðzÞ :¼ xðzÞj c ¼ 1t ¼ s


¼ b�zðbþzÞ2

satisfies differential Eq. (76). Therefore,

yðxÞ ¼ �xðexpðxÞÞ ¼ b expðxÞðexpðxÞ þ bÞ2

ð83Þ

is a partial solutions of the differential Eq. (75) satisfying initial conditions yð0Þ ¼ b

ðbþ1Þ2and y0xðxÞjx¼0 ¼

bðb�1Þðbþ1Þ3

for all b (exceptb ¼ �1).

Thus, Hr 1j! D

js�� c ¼ 1t ¼ �s


; j ¼ 0;1;2; . . .

0B@

1CA ¼ 2 and equalities (80) hold true.

It can be noted, that Eq. (83) coincides with the solution presented in [17]. But we would like to stress that we did notguess the structure of the solution; it has been automatically identified by the direct application of the algorithm describedin Section 4.1. This is the main difference between classical Exp-function type methods and our proposed technique.

But even more important fact is that our approach enabled to show that the solution of the differential Eq. (75) takes theform (83) only when initial conditions satisfy Eq. (78). This is an important fact, and we argue that the conclusion done in[17] is in general incorrect. The partial solution cannot be expressed in the form represented by Eq. (83) for any initial con-ditions. Our technique enabled the identification of the constraint linking two initial conditions. Classical Exp-function typemethods are incapable of finding such constrains and may produce wrong results in general, what is clearly illustrated bythis example.

6. Conclusions

We have constructed an analytical criterion determining if a solution of a nonlinear ordinary differential equation can beexpressed in a form comprising standard functions. This criterion is much more general compared to the criterion presentedin [12]. First of all, not only exponential functions are considered in the exact solution of a nonlinear differential equation. Infact, the new criterion works with any standard function. But the most important result is that this new criterion can be usedto identify the structure of the solution which can be much more complex than a sum of standard functions or a ratio of sumsof standard functions. That opens new possibilities for finding exact solutions of nonlinear differential equations.

New computational algorithm for automatic identification of the structure of the solution is constructed. Several exam-ples are used to illustrate the proposed concept.

Recent developments of the Exp-function method were summarized in [18,19]; the Exp-function method has been usedto solve differential-difference equations and stochastic equations [20–22]. Applicability of our technique for these problemsis a definite object of future research.


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Generalization of Exp-function and other standard function methods

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