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Polygons of differential equations for finding exact solutions
Chaos, Solitons and Fractals 33 (2007) 1480–1496 www.elsevier.com/locate/chaos
Polygons of differential equations for finding exact solutions Nikolai A. Kudryashov *, Maria V. Demina Department of Applied Mathematics, Moscow Engineering and Physics Institute (State University), 31 Kashirskoe Shosse, 115409 Moscow, Russian Federation Accepted 22 February 2006
Abstract A method for finding exact solutions of nonlinear differential equations is presented. Our method is based on the application of polygons corresponding to nonlinear differential equations. It allows one to express exact solutions of the equation studied through solutions of another equation using properties of the basic equation itself. The ideas of power geometry are used and developed. Our approach has a pictorial interpretation, which is illustrative and effective. The method can be also applied for finding transformations between solutions of differential equations. To demonstrate the method application exact solutions of several equations are found. These equations are: the Korteveg– de Vries–Burgers equation, the generalized Kuramoto–Sivashinsky equation, the fourth-order nonlinear evolution equation, the fifth-order Korteveg–de Vries equation, the fifth-order modified Korteveg–de Vries equation and the sixth-order nonlinear evolution equation describing turbulent processes. Some new exact solutions of nonlinear evolution equations are given. Ó 2006 Elsevier Ltd. All rights reserved.
1. Introduction One of the most important problems of nonlinear models analysis is construction of their partial solutions. Nowadays this problem is widely discussed. We know that the inverse scattering transform [1–3] and the Hirota method [3–6] are very useful in looking for solutions of exactly solvable nonlinear equations, while most of nonlinear differential equations describing various processes in physics, biology, economics and other fields of science do not belong to the class of exactly solvable equations. Certain substitutions containing special functions are usually used for determination of partial solutions of nonintegrable equations. The most famous algorithms are the following: the singular manifold method [7–13], the Weierstrass function method [13–15], the tanh–function method [16–21], the Jacobian elliptic function method [22–24], the trigonometric function method [25,26]. Lately it was made an attempt to generalize most of these methods and as a result the simplest equation method appeared [27,28]. Two ideas lay in the basis of this method. The first one was to use an equation of lesser order with known general solution for finding exact solutions. The second one was to take into account possible movable singularities of the original equation. Virtually both ideas existed though not evidently in some methods suggested earlier. *
Corresponding author. Tel./fax: +7 95 3241181. E-mail address: [email protected] (N.A. Kudryashov).
0960-0779/$ - see front matter Ó 2006 Elsevier Ltd. All rights reserved. doi:10.1016/j.chaos.2006.02.012
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However the method introduced in [27,28] has one essential disadvantage concerning with an indeterminacy of the simplest equation choice. This disadvantage considerably decreases the effectiveness of the method. In this paper we present a new method for finding exact solutions of nonlinear differential equations, which greatly expands the method [27,28] and which is free from the disadvantage mentioned above. The method does not postulate the simplest equation but it allows one to find its structure using the properties of the equation studied itself. Our method develops the ideas of power geometry [29–32]. With a help of the power geometry we show that the search of the simplest equation becomes illustrative and effective. Also it is important to mention that the results obtained are sufficiently general and can be applied not only to finding exact solutions but also to constructing transformations for nonlinear differential equations. This paper outline is as follows. The method algorithm is discussed in Section 2. The application of our method to the Korteveg–de Vries–Burgers equation, the fourth-order evolution equation, and the generalized Kuramoto–Sivashinsky equations is presented in Sections 3–5, accordingly. Self-similar solutions of the fifth-order Korteveg–de Vries equations are discussed in Sections 6 and 7. Solitary waves and periodic solutions of the sixth-order nonlinear evolution equation are found in Sections 8 and 9.
2. Method applied To begin with let us introduce several definitions. The expression of the form C 1 zq1 y q2 is called an ordinary monok mial. Here C1 is an arbitrary constant. The product of ordinary monomial and finite amount of derivatives ddzky ; ðk 2 NÞ is called a differential monomial. Under the term differential sum we will mean a sum of ordinary and differential monomials. Let us assume that we look for exact solutions of a nonlinear n-order ODE M n ½yðzÞ;y z ðzÞ;y zz ðzÞ; . . . ;z ¼ 0;
ð2:1Þ
where Mn[y(z), yz(z), yzz(z), . . . , z] is a differential sum. Every monomial can be associated with a point in the plane according to the following rules: C 1 zq1 y q2 ! ðq1 ;q2 Þ;
C2
dky ! ðk;1Þ: dzk
ð2:2Þ
Here again C1, C2 are arbitrary constants. When monomials are multiplied their coordinates are added. The set of points corresponding to all monomials of a differential equation forms its carrier. Having connected the points of the carrier into the convex figure we obtain a convex polygon called the polygon of differential equation. Thus nonlinear ODE (2.1) can be characterized by the polygon L1 in the plane. The periphery of the polygon consists of apexes and ð0Þ ð1Þ edges. By fCj g let us denote the apexes and by fCj g the edges. Most of edges and apexes of the polygon define power or nonpower asymptotics and power expansions for the solutions of the equation [29–32]. Now let us assume that a solution y(z) of the basic equation can be expressed through solutions Y(z) of another equation. The latter equation is called the simplest equation. Consequently we have a relation between y(z) and Y(z) yðzÞ ¼ F ðY ðzÞ;Y z ðzÞ; . . . ;zÞ:
ð2:3Þ
The main problem is to find the simplest equation. Substituting (2.3) into basic equation (2.1) yields a transformed differential equation, which is in its turn characterized by the polygon L2. By L3 denote the polygon corresponding to the ð1Þ simplest equation. Analyzing the polygon L2 we should construct the polygon L3. Let the edge C1 belong to L2 and the ð1Þ ð1Þ ð1Þ e e edge C 1 belong to L3. Assume that the edges C1 and C 1 generate power asymptotics of the transformed differential ð1Þ e ð1Þ have collinear external normal equation and the simplest equation, accordingly. It can be proved that if C1 and C 1 vectors, then corresponding asymptotics coincide accurate to the numerical parameter. Under the term external vector we mean the vector that goes out of the polygon. The edges possessing collinear external normal vectors necessarily are parallel. Consequently, suitable polygon L3 has all or certain part of edges parallel to those of L2. Besides that when the polygon L3 moves along the plane his apexes should cover the carrier of the transformed differential equation. Let us suppose that we have found such polygon L3. Then we can write out the simplest equation Em ðY ðzÞ;Y z ðzÞ; . . . ;zÞ ¼ 0:
ð2:4Þ
It is important to mention that the choice of the simplest equation is not unique. If the following correlation b m ðY ;Y z ; . . . ;zÞ M n ðF ðY ;Y z ; . . . ;zÞÞ ¼ RE
ð2:5Þ
b is a differential operator) is true then it means that for any solution Y(z) of the simplest Eq. (2.4) there exists a (where R solution of (2.1). Generally speaking, any differential equation can be the simplest equation. The only requirement is the
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following: the order of the simplest equation should be lesser than the order of the transformed differential equation. But the most important simplest equations are those that are integrable equations. If the general solution (or a partial solution) of the simplest equation can be found, then we get explicit representation for a solution of the equation studied. Otherwise we have only the relation (2.3) between solutions of (2.1) and (2.4). The most useful examples of the simplest equations are the following: the Riccati equation Y z þ Y 2 aðzÞY bðzÞ ¼ 0
ð2:6Þ
and the equations for elliptic functions R2z ¼ 4R4 þ aR3 þ bR2 þ cR þ d;
ð2:7Þ
R2z
ð2:8Þ
3
2
¼ 4R þ aR þ bR þ c:
Solutions of Eqs. (2.6), (2.7), and (2.8) do not have movable critical points. Now we are able to state our method. It is composed of six steps. The first step. Construction of the polygon L1, which corresponds to the equation studied. The second step. Determination of the movable pole order for solutions of the basic equation and transformation of this equation using the expression (2.3). The third step. Construction of the polygon L2 corresponding to the transformed equation. The fourth step. Construction of the polygon L3, which will characterize the simplest equation. This polygon should posses properties discussed above. The fifth step. Selection of the simplest equation with unknown parameters that generates the polygon L3. The sixth step. Determination of the undefined coefficients, which are present in the transformation F and in the simplest equation.
Remark 1. Very often suitable simplest equations can be found without making transformation (2.3). In this case (2.3) is an identity substitution and we set y(z) Y(z), L2 L1. Remark 2. In some cases the transformation can be included into the simplest equation. Then again (2.3) is an identity substitution. Remark 3. The most powerful transformations, i.e. the transformations that generate new classes of exact solutions, are those that change the pole order of y(z).
3. Exact solutions of the Korteveg–de Vries–Burgers equation To demonstrate our method application let us find exact solutions of the Korteveg–de Vries–Burgers equation ut þ uux þ buxxx muxx ¼ 0:
ð3:1Þ
It has the travelling wave reduction uðx;tÞ ¼ wðzÞ;
z ¼ x C 0 t;
ð3:2Þ
where w(z) satisfies the equation 1 E½w ¼ bwzz mwz þ w2 C 0 w þ C 1 ¼ 0: 2
ð3:3Þ
Here C0, C1, m, and b 5 0 are constants. In the case, C0 ¼
6m2 ; 25b
C 1 ¼ 0:
ð3:4Þ
Eq. (3.3) is integrable and was solved by Painleve´ [5]. Its general solution is expressed via the Jacobi elliptic function. However Eq. (3.3) also describes a solitary wave. The solution in the form of this solitary wave was first obtained in [10] and later it was rediscovered a lot of times.
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Let us formulate the following theorem. Theorem 3.1. Let Y(z) be a solution of the equation Yz þ Y2
m2 ¼ 0: 10b
ð3:5Þ
Then yðzÞ ¼ C 0 þ
3m2 12m Y ðzÞ 12bY ðzÞ2 25b 5
ð3:6Þ
is a solution of Eq. (3.3) in the case, C1 ¼
C 20 18m4 : 2 625b2
ð3:7Þ
Proof. At the first step we should find the polygon L1 corresponding to Eq. (3.3). For monomials of this equation we have points: M1 = (2, 1), M2 = (1, 1), M3 = (0, 2), M4 = (0, 1), M5 = (0, 0). The carrier of Eq. (3.3) is defined by four points: Q1 = M1, Q2 = M3, Q3 = M4, and Q4 = M5. Their convex hull is the triangle L1 (Fig. 1). This triangle contains ð0Þ ð1Þ ð1Þ ð1Þ three apexes Cj ¼ Qj , (j = 1, 2, 4) and three edges C1 ¼ ½Q1 ;Q2 , C2 ¼ ½Q2 ;Q4 , C3 ¼ ½Q1 ;Q4 . Solutions of Eq. (3.3) have the second-order singularity. Let us make the following transformation: wðzÞ ¼ A0 þ A1 Y ðzÞ þ A2 Y ðzÞ2 ; where Y(z) satisfies the first-order differential equation and has the first-order singularity. Substituting (3.8) into Eq. (3.3), we obtain 1 M 2 ½w½Y ¼ ðbA1 þ 2bA2 Y ÞY zz ðmA1 þ 2mA2 Y ÞY z þ 2bA2 Y 2z þ A0 A2 C 0 A2 þ A21 Y 2 þ A1 A2 Y 3 2 1 2 4 1 2 þ A2 Y þ ðA0 A1 C 0 A1 ÞY þ A0 þ C 1 C 0 A0 2 2 ¼ 0:
ð3:8Þ
ð3:9Þ
The following points correspond to the monomials of this equation: M1 = (2, 1), M2 = (2, 2), M3 = (2, 2), M4 = (1, 1), M5 = (1, 2), M6 = (0, 0), M7 = (0, 2), M8 = (0, 4), M9 = (0, 1), M10 = (0, 2), M11 = (0, 3), M12 = (0, 0), M13 = (0, 1), M14 = (0, 2), M15 = (0, 0). The carrier of Eq. (3.9) is determined by seven points: Q1 = M1, Q2 = M2 = M3, Q3 = M8, Q4 = M11, Q5 = M7 = M14, Q6 = M9 = M13, and Q7 = M6 = M12 = M15. Their convex hull is the quadð0Þ ð1Þ ð1Þ rangle (Fig. 2). This quadrangle has four apexes Cj ¼ Qj , (j = 1, 2, 3, 4) and four edges C1 ¼ ½Q1 ;Q2 , C2 ¼ ½Q2 ;Q3 , ð1Þ ð1Þ C3 ¼ ½Q3 ;Q7 , C4 ¼ ½Q1 ;Q7 . So we have constructed the polygon L2. Following our method we should find the polygon L3 with some part of edges parallel to those of L2. Besides that when the polygon L3 moves along the plane, his apexes should cover the carrier of Eq. (3.9). We see that the triangle with apexes M4, Q5, and Q7 satisfies these requirements. Thus the simplest equation is the Riccati equation with constant coefficients E1 ½Y ¼ Y z þ Y 2 b ¼ 0:
ð3:10Þ
Substituting Y z ðzÞ ¼ E1 ½Y Y 2 þ b
ð3:11Þ
Fig. 1. The polygon L1 corresponding to the Eq. (3.3).
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Fig. 2. The polygons L2 and L3 corresponding to the Eqs. (3.9) and (3.10), accordingly.
into Eq. (3.9) and equating coefficients at powers of Y(z) to zero, yields algebraic equations for parameters A2, A1, A0, b and C1. Solving these equations, we get A2 ¼ 12b;
A1 ¼
12m ; 5
A0 ¼ C 0 þ
3m2 ; 25b
b¼
m2 ; 100b2
C1 ¼
C 20 18m4 2 625b2
ð3:12Þ
and the relation b 1 ½Y ; M 2 ½wðY Þ ¼ RE
ð3:13Þ
b is a differential operator. This completes the proof. h where R From equality (3.13) we see that if Y(z) is a solution of Eq. (3.11), then w(z) in formula (3.8) is a solution of Eq. (3.3). Hence we have found the solitary wave in the form of a kink 2 6m2 3m2 mðx C 0 t þ u0 Þ wðzÞ ¼ C 0 þ 1 þ tanh : ð3:14Þ 10b 25b 25b Here u0 is an arbitrary constant.
4. Exact solutions of the nonlinear fourth-order evolution equation Let us look for exact solutions of the following fourth-order evolution equation [33]: ut 2ux uxx u2 uxx 2uu2x þ uxxxx ¼ 0:
ð4:1Þ
Using the travelling wave reduction (3.2) and integrating with respect to z, we get wzzz w2z w2 wz C 0 w þ C 1 ¼ 0:
ð4:2Þ
Later let us prove the following theorem. Theorem 4.1. Let Y(z) be a solution of the equation Y zz YY z C 0 ¼ 0:
ð4:3Þ
Then w(z) = Y(z) is a solution of Eq. (4.2), provided that C1 = 0. Proof. Let us find the polygon that corresponds to Eq. (4.2). The following points are assigned to the monomials of this equation: M1 = (3, 1), M2 = (2, 2), M3 = (1, 3), M4 = (0, 1), M5 = (0, 0). The carrier of the equation contains all these points: Q1 = M1, Q2 = M2, Q3 = M3, Q4 = M4, and Q5 = M5. Their convex hull is the quadrangle L1 with four ð0Þ ð1Þ ð1Þ ð1Þ ð1Þ apexes Cj ¼ Qj , (j = 1, 3, 4, 5) and four edges C1 ¼ ½Q1 ;Q3 , C2 ¼ ½Q3 ;Q4 , C3 ¼ ½Q4 ;Q5 , C4 ¼ ½Q1 ;Q5 (see Fig. 3). Solutions of Eq. (4.2) have the first-order pole. Suitable polygon L3 we find without making the transformation. It is the triangle in Fig. 3. Thus we set wðzÞ Y ðzÞ;
ð4:4Þ
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Fig. 3. The polygons L1 and L3 corresponding to the Eqs. (4.2) and (4.5), accordingly.
where Y(z) is a solution of the second-order equation E2 ½Y ¼ Y zz aYY z b ¼ 0:
ð4:5Þ
Substituting (4.4) and Y zz ¼ E2 ½Y þ aYY z þ b
ð4:6Þ
into Eq. (4.2) and equating coefficients at powers of Y(z) to zero yields algebraic equations for parameters a, b and C1. Hence we get a ¼ 1;
b ¼ C0 ;
C 1 ¼ 0:
ð4:7Þ
We also obtain the relation b 2 ½Y ; M 3 ½wðY Þ ¼ RE
ð4:8Þ
b is a differential operator. This completes the proof. h where R Integrating Eq. (4.3) with respect to z yields the Riccati equation Yz
Y2 C 0 z þ 2C 3 ¼ 0; 2
ð4:9Þ
where C3 is a constant of integration. Setting Y ðzÞ ¼
2Wz W
in (4.9) we get C0 Wzz þ z C 3 W ¼ 0: 2
ð4:10Þ
ð4:11Þ
This equation is equivalent to the Airy equation. Its general solution is 1=3 1=3 WðzÞ ¼ C 1 Aif21=3 C 1=3 C 0 z þ C 4 g; 0 z þ C 4 g þ C 2 Bif2
ð4:12Þ
where Ai(f) and Bi(f) are the Airy functions and C1, C2, and C4 are arbitrary constants. Substituting expression (4.12) into (4.10) we get exact solutions of Eq. (4.2) with two arbitrary constants.
5. Exact solutions of the generalized Kuramoto–Sivashinsky equation Let us look for exact solutions of the generalized Kuramoto–Sivashinsky equation, which can be written as ut þ aum ux þ duxx þ buxxx þ cuxxxx ¼ 0:
ð5:1Þ
Eq. (5.1) at m = 1 is the famous Kuramoto–Sivashinsky equation [34,35], which describes turbulent processes. Its exact solutions at b = 0 were first found in [34]. Its solitary waves at b 5 0 were obtained in [10] and periodical solutions of
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this equation also at b 5 0 were first presented in [13]. Recently it was shown [36,37] that this equation did not have other solutions except found before. Using the variables sffiffiffi pffiffiffiffiffi d d2 b a dc 0 x ¼x ; t0 ¼ t ; u0 ¼ u; r ¼ pffiffiffiffiffi ; a0 ¼ 2 ; ð5:2Þ c c cd d we get the equation in the form (the primes are omitted) ut þ aum ux þ uxx þ ruxxx þ uxxxx ¼ 0:
ð5:3Þ
Using the travelling wave reduction uðx;tÞ ¼ yðzÞ;
z ¼ x C0t
ð5:4Þ
and integrating with respect to z, we get the equation M 3 ½y ¼ y zzz þ ry zz þ y z C 0 y þ
a y mþ1 ¼ 0: mþ1
ð5:5Þ
Here a constant of integration is equated to zero. Let us present our result in the following theorem. Theorem 5.1. Let y(z) be a solution of the equation 13 9a 3 mþ3 y 3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi y: yz ¼ 3 2 2m þ 11m2 þ 18m þ 9 2m þ 18m þ 27
ð5:6Þ
pffiffi 3 , and Then y(z) is also a solution of Eq. (5.5), provided that m 5 0, m 5 1, m 6¼ 32, m 5 3, m 6¼ 93 2
C0 ¼
3ð2m2 þ 9m þ 9Þ ð2m2
þ 18m þ 27Þ
3=2
;
3ð3 þ mÞ r ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi : 2m2 þ 18m þ 27
ð5:7Þ
Proof. Monomials of Eq. (5.5) are determined by the following points: M1 = (3, 1), M2 = (2, 1), M3 = (1, 1), M4 = (0, 1) and M5 = (0, m + 1). The polygon L1 that corresponds to Eq. (5.5) is the major triangle in Fig. 4((a)
Fig. 4. The polygons L1 and L3 corresponding to the Eqs. (5.5) and (5.8), accordingly.
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m > 0 and (b) m < 0). Suitable polygon L3 can be found without making the transformation. It is the smaller triangle in Fig. 4. This triangle contains the points: Q1 = (1, 1), Q2 = M4 = (0, 1), and Q3 ¼ ð0; mþ3 Þ. 3 Consequently the simplest equation corresponding to L3 can be written as E1 ½y ¼ y z Ay
mþ3 3
By ¼ 0;
ð5:8Þ
where A and B are parameters to be found. The general solution of (5.8) takes the form 3 Bmz A m yðzÞ ¼ C 1 exp : 3 B
ð5:9Þ
Substituting y z ¼ E1 ½y þ Ay
mþ3 3
þ By
ð5:10Þ
into Eq. (5.5) and equating coefficients at powers of y(z) to zero yields algebraic equations for parameters A, B, C0 and r in the form A3 ð2m þ 3Þðm þ 3Þðm þ 1Þ þ 9a ¼ 0;
ð5:11Þ
ðm þ 1ÞðB3 þ rB2 þ B C 0 Þ ¼ 0;
ð5:12Þ
2
A ð3 þ mÞðm þ 1ÞðBm þ 3B þ rÞ ¼ 0;
ð5:13Þ
ðm þ 1Þð2r2 m2 þ 18r2 m þ 27r2 9m2 54m 81Þ ¼ 0:
ð5:14Þ
Eqs. (5.11)–(5.14) can be solved and we obtain A1 ¼ ð9aÞ1=3 ðm3 þ 11m2 þ 18m þ 9Þ1=3 ;
ð5:15Þ
A ¼ ð9aÞ1=3 ð2m3 þ 11m2 þ 18m þ 9Þ1=3 : C 0 ¼ BðB2 þ rB þ 1Þ;
B¼
r ; 3þm
ð5:16Þ
3ð3 þ mÞ r ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi : 2m2 þ 18m þ 27
ð5:17Þ
So we have found parameters of Eq. (5.9), conditions (5.7) and the relation b 1 ½y; M 3 ½y ¼ RE
ð5:18Þ
b is a differential operator. This completes the proof. h where R The general solution of Eq. (5.7) can be presented in the form m3 mz yðzÞ ¼ C þ C 1 exp pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; 2m2 þ 18m þ 27
ð5:19Þ
where C1 is an arbitrary constant and C is determined by expression p ffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 9a 2m2 þ 18m þ 27 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi : C¼ p 3 3 ðm þ 3Þð2m2 þ 5m þ 3Þ
ð5:20Þ
pffiffiffiffiffi We would like to note that the value of solitary wave velocity (5.19) tends to C 0 ¼ 1= 27 as m ! 0, but at the same time C0 ! 0 as m ! 1. Assuming m = 1 in (5.19) we obtain the known solitary wave
yðzÞ ¼
pffiffiffiffiffip ffiffiffiffiffiffiffiffiffiffiffiffiffiffi !3 x C0t 47 3 225a þ C 1 exp pffiffiffiffiffi ; 30 47
C0 ¼
60 pffiffiffiffiffi 47 47
ð5:21Þ
pffiffiffiffiffi and the value of the parameter r : r ¼ 12= 47. This solution of the Kuramoto–Sivashinsky equation is the kink [10–14].
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6. Self-similar solutions of the fifth-order Korteveg–de Vries equation Let us find exact solutions of the fifth-order Korteveg–de Vries equation using our approach. This equation can be written as ut þ uxxxxx 10uuxxx 20ux uxx þ 30u2 ux ¼ 0:
ð6:1Þ
It has the self-similar solution uðx;tÞ ¼
1 ð5tÞ
2=5
yðzÞ;
z¼
x ð5tÞ1=5
;
ð6:2Þ
where y(z) satisfies the nonlinear ODE of the form y zzzzz 10yy zzz 20y z y zz þ 30y 2 y z zy z 2y ¼ 0:
ð6:3Þ
Our results are summarized in the following theorem. Theorem 6.1. Let Y(z) be a solution of the equation Y zzzzz 40YY zz Y z 10Y 2 Y zzz 10Y 3z þ 30Y 4 Y z Y zY z ¼ 0:
ð6:4Þ
yðzÞ ¼ Y z Y 2
ð6:5Þ
Then
is a solution of Eq. (6.3). Proof. The following points correspond to the monomials of this equation: M1 = (5, 1), M2 = (3, 2), M3 = (3, 2), M4 = (1, 3), M5 = (0, 1) and M6 = (0, 1). These points generate the carrier of Eq. (6.3): Q1 = M1 = (5, 1), Q2 = M4 = (1, 3), Q3 = M5 = M6 = (0, 1) and Q4 = M2 = M3 = (3, 2). Now we can plot the polygon L1 (see Fig. 5). Solutions of Eq. (6.3) have the second-order pole. Thus let us make the following transformation: y ¼ A1 Y þ A2 Y 2 þ A3 Y z ;
ð6:6Þ
where Y(z) is a function of the first-order pole. Substituting (6.6) into the equation studied, we obtain A1 Y zzzzz þ A2 ðY 2 Þzzzzz þ A3 Y zzzzzz 10ðA1 Y þ A2 Y 2 þ A3 Y z ÞðA1 Y zzz þ A2 ðY 2 Þzzz þ A3 Y zzzz Þ 20ðA1 Y z þ A2 ðY 2 Þz þ A3 Y zz ÞðA1 Y zz þ A2 ðY 2 Þzz þ A3 Y zzz Þ þ 30ðA1 Y þ A2 Y 2 þ A3 Y z Þ2 ðA1 Y z þ A2 ðY 2 Þz þ A3 Y zz Þ zðA1 Y z þ A2 ðY 2 Þz þ A3 Y zz Þ 2A1 Y 2A2 Y 2 2A3 Y z ¼ 0:
ð6:7Þ
The carrier of this equation is defined by ten points: Q1 = (6, 1), Q2 = (5, 2), Q3 = (4, 3), Q4 = (3, 4), Q5 = (2, 5), Q6 = (1, 6), Q7 = (0, 2), Q8 = (0, 1), Q9 = (1, 1), Q10 = (5, 1). In this case, the polygon L2 is the quadrangle presented in Fig. 6. Studying this polygon we can find the polygon L3 for the simplest equation. It is the triangle also presented in Fig. 6. Consequently the simplest equation for (6.3) is E5 ½Y ¼ Y zzzzz þ m1 YY zz Y z þ m2 Y 2 Y zzz þ m3 Y 3z þ m4 Y 4 Y z þ m5 Y þ m6 zY z ¼ 0; where m1, m2, m3 m4, m5 and m6 are unknown parameters to be found.
Fig. 5. The polygon L1 corresponding to the Eq. (6.3).
ð6:8Þ
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Fig. 6. The polygons L2 and L3 corresponding to the Eqs. (6.7) and (6.8), accordingly.
Substituting Y zzzzz ¼ E5 ½Y m1 YY zz Y z m2 Y 2 Y zzz m3 Y 3z m4 Y 4 Y z m5 Y m6 zY z
ð6:9Þ
into Eq. (6.7) and equating coefficients at different powers of Y(z) to zero yields algebraic equations for parameters m1, m2, m3, m4, m5 and m6. As a result we get m1 ¼ 40;
m2 ¼ 10;
m3 ¼ 10;
m4 ¼ 30;
m5 ¼ 1;
m6 ¼ 1
ð6:10Þ
and the relation b 5 ½Y ; M 5 ½yðY Þ ¼ RE
ð6:11Þ
b is a differential operator. This completes the proof. h where R Eq. (6.4) can be integrated in z and we get the fourth-order analogue to the second Painleve´ equation Y zzzz 10Y 2 Y zz 10YY 2z þ 6Y 5 zY b ¼ 0;
ð6:12Þ
where b is an arbitrary constant. Thus we have expressed solutions of Eq. (6.3) (and consequently of Eq. (6.1)) through solutions of (6.12).
7. Self-similar solutions of the fifth-order modified Korteveg–de Vries equation Let us find self-similar solutions of the fifth-order modified Korteveg–de Vries equation ut 10u2 uxxx 40ux uxx 10u3x þ 30u4 ux þ uxxxxx ¼ 0:
ð7:1Þ
It has the scaling reduction [38–40] uðx;tÞ ¼ ð5tÞ1=5 wðzÞ;
z ¼ xð5tÞ1=5 ;
ð7:2Þ
where w(z) satisfies M 4 ½w ¼ wzzzz 10w2 wzz 10ww2z þ 6w5 zw b ¼ 0:
ð7:3Þ
Let us present our result in the theorem. Theorem 7.1. Let Y(z) be a solution of the equation z Y zzz þ 2bYY zz bY 2z 6b2 Y 2 Y z 3b3 Y 4 þ ¼ 0: 2b
ð7:4Þ
Then wðzÞ ¼ bY ðzÞ is a solution of Eq. (7.3), provided that b = 1/2.
ð7:5Þ
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Proof. To begin with the polygon corresponding to Eq. (7.3) should be found. The following points: M1 = (4, 1), M2 = (2, 3), M3 = (2, 3), M4 = (0, 5), M5 = (1, 1), M6 = (0, 0) are assigned to the monomials of the studied equation. The carrier of the equation contains five points Q1 = M1, Q2 = M4, Q3 = M5, Q4 = M6, and Q5 = M2 = M3. Their conð0Þ ð1Þ ð1Þ vex hull is the quadrangle L1 with four apexes Cj ¼ Qj , (j = 1, 2, 3, 4) and four edges C1 ¼ ½Q1 ;Q2 , C2 ¼ ½Q2 ;Q3 , ð1Þ ð1Þ C3 ¼ ½Q3 ;Q4 , C4 ¼ ½Q1 ;Q4 (Fig. 7). ð1Þ The leading members of Eq. (7.1) are located on the edge C1 ¼ ½Q1 ;Q2 . Substituting w(z) = a0zp into the equation wzzzz 10w2 wzz 10ww2z þ 6w5 ¼ 0;
ð7:6Þ
we get fours families of power asymptotics for solutions of (7.3): (p, a0) = (1, 1), (1, 1), (1, 2), and (1, 2). Again taking into account the pole order of solutions and the results of Section 2, we can look for exact solutions of Eq. (7.3) in the form wðzÞ ¼ A1 Y ðzÞ;
A1 6¼ 0;
ð7:7Þ
where Y(z) is a solution of the following simplest equation Y zzz aYY zz bY 2z cY 2 Y z dY 4 mz ¼ 0:
ð7:8Þ
It is important to mention that the transformation (7.7) does not change the quadrangle L1 and besides that the polygon L3 of the simplest Eq. (7.8) possesses all necessary properties. In other words L1 L2 and L3 is the triangle (see Fig. 7). The parameters a, b, c, d, and m can be found. Substituting (7.7) and Y zzz ¼ E3 ½Y þ aYY zz þ bY 2z þ cY 2 Y z þ dY 4 þ mz ¼ 0
ð7:9Þ
into Eq. (7.3) and equating coefficients at different powers of Y(z) to zero yields algebraic equations for parameters a, b, c, d, m, b and A1. Finally, we obtain a ¼ 2b;
c ¼ 6b2 ;
d ¼ 3b3 ;
m¼
1 ; 2b
A1;2 ¼ b;
b¼
1 2
ð7:10Þ
and the relation [39] b 3 ½Y ; M 4 ½wðY Þ ¼ RE
ð7:11Þ
b is a differential operator. This completes the proof. h where R Consequently there exist solutions of the studied equation expressed through the function Y(z) at b = 1/2 or b = 1/2. Let us show that solutions of Eq. (7.4) are associated with solutions of the first Painleve´ equation vzz ¼ Cv2 þ Dz:
ð7:12Þ
Making the transformation vðzÞ ¼ Aw2 þ Bwz
Fig. 7. The polygons L1 and L3 corresponding to the Eqs. (7.3) and (7.8), accordingly.
ð7:13Þ
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in Eq. (7.12), yields the equation Bwzzz þ 2Awwzz þ ð2A CB2 Þw2z 2CABw2 wz CA2 w4 Dz ¼ 0:
ð7:14Þ
We see that the polygon corresponding to this equation coincides with the triangle L3 at Fig. 7. Combining (7.8) and (7.14), we obtain wðzÞ Y ðzÞ;
ð7:15Þ 2
provided that A = b, B = 1, C = 3b, and D = 1/(2b). In other words, if Y(z) is a solution of (7.4), then v(z) = bY + Yz is a solution of vzz ¼ 3bv2
1 z: 2b
ð7:16Þ
We would like to mention that without loss of generality it can be set b = ±1.
8. Solitary waves of the sixth-order nonlinear evolution equation Let us find exact solutions of the sixth-order nonlinear evolution equation, which describes turbulent processes [41,42] ut þ uux þ buxx þ duxxxx þ euxxxxxx ¼ 0:
ð8:1Þ
It is known that Kuramoto–Sivashinskiy equation and the Ginzburg–Landau equation used to describe the turbulence are nonintegrable because they do not pass the Painleve´ test [5]. However, these equations have a list of special solutions [5,10,13]. Eq. (8.1) does not pass the Painleve´ test as well and thus this equation is also nonintegrable. However one can expect that Eq. (8.1) has some special solutions. This equation admits the travelling waves reduction uðx;tÞ ¼ yðzÞ;
ð8:2Þ
z ¼ x C 0 t;
where y(z) satisfies the equation 1 C 1 C 0 y þ y 2 þ by z þ dy zzz þ ey zzzzz ¼ 0: 2
ð8:3Þ
C1 is a constant of integration. As a result we get the following theorem. Theorem 8.1. Let Y(z) be a solution of the equation Y z þ Y 2 ak ¼ 0
ðk ¼ 1;2; . . . ;6Þ:
ð8:4Þ
yðzÞ ¼ 30240eY 5 þ
2520 2520 1260 12600 d2 d 50400eak Y 3 þ dak þ 20160ea2k þ b Y þ C0 11 11 251 30371 e
ð8:5Þ
Then
is a solution of Eq. (8.3), provided that 4112640 d5 w4k d5 w5k 5080320 d3 w3k b 9999360 11 251 e2 e3 e3 5 3 3 2 55460160 d wk 660240 d wk b 25200 d5 w2k þ 30371 2761 e2 30371 e3 e3 2 3 1260 b dwk 12600 bd wk 1 2 þ þ C0; 251 30371 e2 2 e 3 3 213811840e ak 10204656de2 a2k 2045d3 92400d2 eak b¼ ; 121eð9240eak þ 79dÞ
C1 ¼
ð8:6Þ ð8:7Þ
where ak and wk are found from ak ¼
dwk e
w1 ¼
ðk ¼ 1;2; . . . ;6Þ;
1 ; 220
w2 ¼
5 ; 176
ð8:8Þ w3 ¼
1 ; 440
ð8:9Þ
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w4 ¼
1 46031 557 þm ; 52800 m
pffiffiffi ! 1 46031 m i 3 46031 mþ þ 557 ; w5;6 ¼ 2 52800 2m 2 m pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 m ¼ ð113816753 þ 1260 8221079733Þ3 610966:
ð8:10Þ ð8:11Þ ð8:12Þ
Proof. The following points correspond to the monomials of Eq. (8.3): Q1 = (5, 1), Q2 = (0, 2), Q3 = (0, 0), Q4 = (3, 1), and Q5 = (1, 1). The polygon L1 of the equation is the triangle (Fig. 8). Solutions of Eq. (8.3) have the fifth-order pole. Let us express them through a function Y(z) possessing the first-order pole yðzÞ ¼ A0 þ A1 Y þ A2 Y 2 þ A3 Y 3 þ A4 Y 4 þ A5 Y 5 :
ð8:13Þ
Substituting transformation (8.13) into (8.3), we get new equation and the corresponding polygon L2 in the form of quadrangle (Fig. 9). One of the suitable polygons L3 is the triangle in Fig. 9. This triangle is assigned to the Riccati equation with constant coefficients E1 ½Y ¼ Y z þ Y 2 a ¼ 0:
ð8:14Þ
Substituting Y z ¼ E1 ½Y Y 2 þ ak
ð8:15Þ
into Eq. (8.4) and equating coefficients at different powers of Y(z) to zero yields algebraic equations for parameters A5, A4, A3, A2, A1, A0, C1, b, ak and wk. Solving these equations we have 2520d 50400ea; A2 ¼ 0; A1 11 2520 1260 12600 d2 da þ 20160ea2 þ b ; A0 ¼ C 0 : ¼ 11 251 30371 e
A5 ¼ 30240e;
A4 ¼ 0;
A3 ¼
ð8:16Þ
For the parameters C1, b, ak, and wk we obtain expressions (8.6)–(8.10), and (8.12). We also see that the following relation: b 1 ½Y M 5 ½yðY Þ ¼ RE
ð8:17Þ
b is a differential operator. This completes the proof. h holds, where R The general solution of Eq. (8.4) is the following: pffiffiffiffiffi pffiffiffiffiffi Y ðzÞ ¼ ak tanhð ak z þ u0 Þ; ðk ¼ 1;2; . . . ;6Þ:
ð8:18Þ
Substituting (8.13) into (8.5) and taking into account that ak = wkd/e (k = 1, . . . , 6) we get six different solitary waves of Eq. (8.3).
Fig. 8. The polygon L1 corresponding to the Eq. (8.3).
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Fig. 9. The polygons L2 and L3 corresponding to the transformed Eq. (8.3) and to the Eq. (8.14), accordingly.
9. Exact periodic solutions of Eq. (8.1) It was mentioned in the previous section that solutions of Eq. (8.3) have the fifth-order pole. Thus let us make the following transformation yðzÞ ¼ B1 þ B2 RðzÞ þ B3 Rz þ B4 R2 þ B5 RRz ;
ð9:1Þ
where Bk (k = 1, . . . , 5) are constants and R = R(z) is a function of the second-order pole. Now let us prove the following theorem. Theorem 9.1. Let R(z) be a solution of the equation R2z
Then
pffiffiffiffiffi pffiffiffiffiffi 3 1 2 1 Rd2 1 R 21d2 1 3 13 d3 1 1 ad2 21d a ¼ 2R þ aR a R þ þ þ 6 726 e2 2541 108 359370 e3 119790 e3 4356 e2 e2 pffiffiffiffiffi 1 a 21d2 : 15246 e2 3
2
d yðzÞ ¼ C 0 þ 630 ea þ 6eR Rz 11
ð9:2Þ
ð9:3Þ
is a solution of Eq. (8.1), provided that 10 d2 ; 121 e pffiffiffiffiffi 5 10854 d5 1 2 2484 21d ð1;2Þ þ C : C1 ¼ 161051 e3 2 0 161051 e3
b¼
ð9:4Þ ð9:5Þ
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Proof. Substituting (9.1) into Eq. (8.3) we get a new equation of the sixth-order. The polygon L2 of this equation is presented in Fig. 10. Again we should construct suitable polygon L3. Let us take the triangle corresponding to the equation for the elliptic function E1 ½R ¼ R2z þ 2R3 aR2 2bR d ¼ 0:
ð9:6Þ
This triangle satisfies necessary requirements. Substituting R2z ¼ E1 ½R 2R3 þ aR2 þ 2bR þ d; 1 oE1;z ½R Rzz ¼ 3R2 þ aR þ b 2Rz oz
ð9:7Þ ð9:8Þ
into Eq. (8.3) and equating coefficients at different powers of R(z) and Rz to zero yields algebraic equations for the coefficients B5, B4, B3, B2, B1, and B0. As a result we obtain the following expressions: B5 ¼ 3780e; B2 ¼ 0;
B4 ¼ 0;
B3 ¼ 630ea þ
630 d; 11
B1 ¼ C 0 ;
ð9:9Þ
2
10 d ; 121 e pffiffiffiffiffi 2 1 1 d2 1 21d ; b1;2 ¼ a2 þ 12 1452 e2 5082 e2pffiffiffiffiffi pffiffiffiffiffi 1 3 13 d3 1 1 ad2 1 a 21d2 21d3 a þ d 1;2 ¼ ; 108 359370 e3 119790 e3 4356 e2 15246 e2
b¼
ð9:10Þ ð9:11Þ ð9:12Þ
and the relation b 1 ½R; M 5 ½yðRÞ ¼ RE
ð9:13Þ
b is a differential operator. This completes the proof. h where R If R1, R2, and R3 such that R1 P R2 P R3 are real roots of the equations pffiffiffiffiffi ! pffiffiffiffiffi pffiffiffiffiffi 1 2 d2 d2 21 1 3 13 d3 d3 21 ad2 ad2 21 3 2 a a 2R aR þ þ ¼ 0; R 6 108 359379 e3 119790e3 4356e2 15246e2 726e2 2541e2 then the general solution of (9.2) can be written as rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ! R1 R3 R1 R2 RðzÞ ¼ R2 þ ðR1 R2 Þcn2 z;S ; S 2 ¼ : 2 R1 R3
Fig. 10. The polygons L2 and L3 corresponding to the transformed Eq. (8.3) and to the Eq. (9.6), accordingly.
ð9:14Þ
ð9:15Þ
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Thus we have found several solutions of Eq. (8.1) at different values of the parameters. These solution are solitary and periodic waves (see expressions (8.5) and (9.3)).
10. Conclusion In this paper a new method for finding exact solutions of nonlinear differential equations is presented. Our aim was to express solutions of the equation studied through solutions of the simplest equations. The method is more powerful than many other methods because the structure of a solution is not fixed but can be found using the stated algorithm. Thus one can obtain quite general classes of exact solutions. The method is based on the ideas of power geometry. More exactly, in order to find the suitable simplest equation we construct the polygons of nonlinear differential equations. It is important to mention that transformations between nonlinear differential equations can be also found with a help of our method. As an example of our method application exact solutions of some nonlinear differential equations were found. In particular, solutions of the fourth-order evolution equation (4.1) were expressed in terms of the Airy functions; the connection between self-similar solutions of the fifth-order Korteveg–de Vries equation (6.1) and the fifth-order modified Korteveg–de Vries equation (7.1) was constructed; one-parametric family of exact solutions for the generalized Kuramoto–Sivashinsky equation (5.1) was found. Besides that we obtained solitary waves and periodic solutions of the sixth-order evolution equation (8.1).
Acknowledgements The authors are grateful to Professor M.S. El Naschie for his attention to this work and remarks. This work was supported by the International Science and Technology Center under Project B 1213.
References [1] Gardner CS, Greene JM, Kruskal MD, Miura RR. Phys Rev Lett 1967;19:1095–7. [2] Ablowitz MJ, Kaup DJ, Newell AC, Segur H. Phys Rev Lett 1973;31:125. [3] Ablowitz MJ, Clarcson PA. Solitons, nonlinear evolution equations and inverse scattering. Cambridge: Cambridge University Press; 1991. [4] Hirota R. Phys Rev Lett 1971;27:1192–4. [5] Kudryashov NA. Analytical theory of nonlinear differential equations. Moscow-Izhevsk: Institute of Computer Investigations; 2004. p. 360 [in Russian]. [6] Polyanin AD, Zaitsev VF, Zhurov AI. Methods of solving nonlinear equations of mathematical physics and mekhanics. Moscow: Fismatlit; 2005. p. 256 [in Russian]. [7] Weiss J, Tabor M, Carnevalle G. J Math Phys 1983;24:522. [8] Conte R, Musette M. J Phys A: Math Gen 1989;22:169–77. [9] Choudhary SR. Phys Lett A 1991;159:311–7. [10] Kudryashov NA. J Appl Math Mekh 1988;52:361–5. [11] Kudryashov NA. Rep USSR Acad Sci 1989;308:294–8 [in Russian]. [12] Kudryashov NA. Phys Lett A 1991;155:269–75. [13] Kudryashov NA. Phys Lett A 1990;147:287–91. [14] Kudryashov NA. J Appl Math Mekh 1990;54:372–6. [15] Yan ZY. Chaos, Solitons & Fractals 2004;21:1013. [16] Lou SY, Huang G, Ruan H. J Phys A: Math Gen 1991;24:587–90. [17] Parkes EJ, Duffy BR. Comput Phys Commun 1996;98:288–300. [18] Elwakil SA, El-labany SK, Zahran MA, Sabry R. Phys Lett A 2002;299:179–88. [19] Fan EG. Phys Lett A 2000;277:212–8. [20] Fan EG. Phys Lett A 2002;282:18–22. [21] Kudryashov NA, Zargaryan ED. J Phys A: Math Gen 1996;29:8067–77. [22] Liu GT, Fan TY. Phys Lett A 2005;345:161–6. [23] Liu SK, Fu ZT, Liu SD, Zhao Q. Phys Lett A 2001;289:69–74. [24] Fu Z, Zhang L, Liu S, Liu S. Phys Lett A 2004;325:363–9. [25] Fu Z, Liu S, Liu S. Phys Lett A 2004;326:364–74. [26] Yan CT. Phys Lett A 1996;224:77. [27] Kudryashov NA. Phys Lett A 2005;324:99–106.
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Kudryashov NA. Chaos, Solitons & Fractals 2005;24:1217–31. Bruno AD. Power geometry in algebraic and differential equations. Moscow: Nauka, Fizmatlit; 1998. p. 288 [in Russian]. Bruno AD. Russ Math Surveys 2004;59:429–80. Kudryashov NA, Efimova OYu. Chaos, Solitons & Fractals 2006;30(1):110–24. Demina MV, Kudryashov NA. Chaos, Solitons & Fractals, in press. doi:10.1016/j.chaos.2005.10.079. Weiss J. J Math Phys 1983;24:1405. Kuramoto Y, Tsuzuki T. Prog Theor Phys 1976;55:356. Sivashinsky GI. Physica D 1982;4:227–35. Eremenko A. arXiv: nlin.SI/0504053, v.1 25 Apr. 2005, 1–10. Hone ANW. Physica D 2005;205:292–306. Hone ANW. Physica D 1998;118:1–16. Kudryashov NA. Phys Lett A 1997;224(6):353–60. Kudryashov NA, Soukharev MB. Phys Lett A 1998;237:206–16. Kudryashov NA. Math Model 1989;1:1–6 [in Russian]. Beresnev LA, Nikolaevskiy VN. Physica D 1993;66:206–16.
Polygons of differential equations for finding exact solutions Nikolai A. Kudryashov *, Maria V. Demina Department of Applied Mathematics, Moscow Engineering and Physics Institute (State University), 31 Kashirskoe Shosse, 115409 Moscow, Russian Federation Accepted 22 February 2006
Abstract A method for finding exact solutions of nonlinear differential equations is presented. Our method is based on the application of polygons corresponding to nonlinear differential equations. It allows one to express exact solutions of the equation studied through solutions of another equation using properties of the basic equation itself. The ideas of power geometry are used and developed. Our approach has a pictorial interpretation, which is illustrative and effective. The method can be also applied for finding transformations between solutions of differential equations. To demonstrate the method application exact solutions of several equations are found. These equations are: the Korteveg– de Vries–Burgers equation, the generalized Kuramoto–Sivashinsky equation, the fourth-order nonlinear evolution equation, the fifth-order Korteveg–de Vries equation, the fifth-order modified Korteveg–de Vries equation and the sixth-order nonlinear evolution equation describing turbulent processes. Some new exact solutions of nonlinear evolution equations are given. Ó 2006 Elsevier Ltd. All rights reserved.
1. Introduction One of the most important problems of nonlinear models analysis is construction of their partial solutions. Nowadays this problem is widely discussed. We know that the inverse scattering transform [1–3] and the Hirota method [3–6] are very useful in looking for solutions of exactly solvable nonlinear equations, while most of nonlinear differential equations describing various processes in physics, biology, economics and other fields of science do not belong to the class of exactly solvable equations. Certain substitutions containing special functions are usually used for determination of partial solutions of nonintegrable equations. The most famous algorithms are the following: the singular manifold method [7–13], the Weierstrass function method [13–15], the tanh–function method [16–21], the Jacobian elliptic function method [22–24], the trigonometric function method [25,26]. Lately it was made an attempt to generalize most of these methods and as a result the simplest equation method appeared [27,28]. Two ideas lay in the basis of this method. The first one was to use an equation of lesser order with known general solution for finding exact solutions. The second one was to take into account possible movable singularities of the original equation. Virtually both ideas existed though not evidently in some methods suggested earlier. *
Corresponding author. Tel./fax: +7 95 3241181. E-mail address: [email protected] (N.A. Kudryashov).
0960-0779/$ - see front matter Ó 2006 Elsevier Ltd. All rights reserved. doi:10.1016/j.chaos.2006.02.012
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However the method introduced in [27,28] has one essential disadvantage concerning with an indeterminacy of the simplest equation choice. This disadvantage considerably decreases the effectiveness of the method. In this paper we present a new method for finding exact solutions of nonlinear differential equations, which greatly expands the method [27,28] and which is free from the disadvantage mentioned above. The method does not postulate the simplest equation but it allows one to find its structure using the properties of the equation studied itself. Our method develops the ideas of power geometry [29–32]. With a help of the power geometry we show that the search of the simplest equation becomes illustrative and effective. Also it is important to mention that the results obtained are sufficiently general and can be applied not only to finding exact solutions but also to constructing transformations for nonlinear differential equations. This paper outline is as follows. The method algorithm is discussed in Section 2. The application of our method to the Korteveg–de Vries–Burgers equation, the fourth-order evolution equation, and the generalized Kuramoto–Sivashinsky equations is presented in Sections 3–5, accordingly. Self-similar solutions of the fifth-order Korteveg–de Vries equations are discussed in Sections 6 and 7. Solitary waves and periodic solutions of the sixth-order nonlinear evolution equation are found in Sections 8 and 9.
2. Method applied To begin with let us introduce several definitions. The expression of the form C 1 zq1 y q2 is called an ordinary monok mial. Here C1 is an arbitrary constant. The product of ordinary monomial and finite amount of derivatives ddzky ; ðk 2 NÞ is called a differential monomial. Under the term differential sum we will mean a sum of ordinary and differential monomials. Let us assume that we look for exact solutions of a nonlinear n-order ODE M n ½yðzÞ;y z ðzÞ;y zz ðzÞ; . . . ;z ¼ 0;
ð2:1Þ
where Mn[y(z), yz(z), yzz(z), . . . , z] is a differential sum. Every monomial can be associated with a point in the plane according to the following rules: C 1 zq1 y q2 ! ðq1 ;q2 Þ;
C2
dky ! ðk;1Þ: dzk
ð2:2Þ
Here again C1, C2 are arbitrary constants. When monomials are multiplied their coordinates are added. The set of points corresponding to all monomials of a differential equation forms its carrier. Having connected the points of the carrier into the convex figure we obtain a convex polygon called the polygon of differential equation. Thus nonlinear ODE (2.1) can be characterized by the polygon L1 in the plane. The periphery of the polygon consists of apexes and ð0Þ ð1Þ edges. By fCj g let us denote the apexes and by fCj g the edges. Most of edges and apexes of the polygon define power or nonpower asymptotics and power expansions for the solutions of the equation [29–32]. Now let us assume that a solution y(z) of the basic equation can be expressed through solutions Y(z) of another equation. The latter equation is called the simplest equation. Consequently we have a relation between y(z) and Y(z) yðzÞ ¼ F ðY ðzÞ;Y z ðzÞ; . . . ;zÞ:
ð2:3Þ
The main problem is to find the simplest equation. Substituting (2.3) into basic equation (2.1) yields a transformed differential equation, which is in its turn characterized by the polygon L2. By L3 denote the polygon corresponding to the ð1Þ simplest equation. Analyzing the polygon L2 we should construct the polygon L3. Let the edge C1 belong to L2 and the ð1Þ ð1Þ ð1Þ e e edge C 1 belong to L3. Assume that the edges C1 and C 1 generate power asymptotics of the transformed differential ð1Þ e ð1Þ have collinear external normal equation and the simplest equation, accordingly. It can be proved that if C1 and C 1 vectors, then corresponding asymptotics coincide accurate to the numerical parameter. Under the term external vector we mean the vector that goes out of the polygon. The edges possessing collinear external normal vectors necessarily are parallel. Consequently, suitable polygon L3 has all or certain part of edges parallel to those of L2. Besides that when the polygon L3 moves along the plane his apexes should cover the carrier of the transformed differential equation. Let us suppose that we have found such polygon L3. Then we can write out the simplest equation Em ðY ðzÞ;Y z ðzÞ; . . . ;zÞ ¼ 0:
ð2:4Þ
It is important to mention that the choice of the simplest equation is not unique. If the following correlation b m ðY ;Y z ; . . . ;zÞ M n ðF ðY ;Y z ; . . . ;zÞÞ ¼ RE
ð2:5Þ
b is a differential operator) is true then it means that for any solution Y(z) of the simplest Eq. (2.4) there exists a (where R solution of (2.1). Generally speaking, any differential equation can be the simplest equation. The only requirement is the
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following: the order of the simplest equation should be lesser than the order of the transformed differential equation. But the most important simplest equations are those that are integrable equations. If the general solution (or a partial solution) of the simplest equation can be found, then we get explicit representation for a solution of the equation studied. Otherwise we have only the relation (2.3) between solutions of (2.1) and (2.4). The most useful examples of the simplest equations are the following: the Riccati equation Y z þ Y 2 aðzÞY bðzÞ ¼ 0
ð2:6Þ
and the equations for elliptic functions R2z ¼ 4R4 þ aR3 þ bR2 þ cR þ d;
ð2:7Þ
R2z
ð2:8Þ
3
2
¼ 4R þ aR þ bR þ c:
Solutions of Eqs. (2.6), (2.7), and (2.8) do not have movable critical points. Now we are able to state our method. It is composed of six steps. The first step. Construction of the polygon L1, which corresponds to the equation studied. The second step. Determination of the movable pole order for solutions of the basic equation and transformation of this equation using the expression (2.3). The third step. Construction of the polygon L2 corresponding to the transformed equation. The fourth step. Construction of the polygon L3, which will characterize the simplest equation. This polygon should posses properties discussed above. The fifth step. Selection of the simplest equation with unknown parameters that generates the polygon L3. The sixth step. Determination of the undefined coefficients, which are present in the transformation F and in the simplest equation.
Remark 1. Very often suitable simplest equations can be found without making transformation (2.3). In this case (2.3) is an identity substitution and we set y(z) Y(z), L2 L1. Remark 2. In some cases the transformation can be included into the simplest equation. Then again (2.3) is an identity substitution. Remark 3. The most powerful transformations, i.e. the transformations that generate new classes of exact solutions, are those that change the pole order of y(z).
3. Exact solutions of the Korteveg–de Vries–Burgers equation To demonstrate our method application let us find exact solutions of the Korteveg–de Vries–Burgers equation ut þ uux þ buxxx muxx ¼ 0:
ð3:1Þ
It has the travelling wave reduction uðx;tÞ ¼ wðzÞ;
z ¼ x C 0 t;
ð3:2Þ
where w(z) satisfies the equation 1 E½w ¼ bwzz mwz þ w2 C 0 w þ C 1 ¼ 0: 2
ð3:3Þ
Here C0, C1, m, and b 5 0 are constants. In the case, C0 ¼
6m2 ; 25b
C 1 ¼ 0:
ð3:4Þ
Eq. (3.3) is integrable and was solved by Painleve´ [5]. Its general solution is expressed via the Jacobi elliptic function. However Eq. (3.3) also describes a solitary wave. The solution in the form of this solitary wave was first obtained in [10] and later it was rediscovered a lot of times.
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Let us formulate the following theorem. Theorem 3.1. Let Y(z) be a solution of the equation Yz þ Y2
m2 ¼ 0: 10b
ð3:5Þ
Then yðzÞ ¼ C 0 þ
3m2 12m Y ðzÞ 12bY ðzÞ2 25b 5
ð3:6Þ
is a solution of Eq. (3.3) in the case, C1 ¼
C 20 18m4 : 2 625b2
ð3:7Þ
Proof. At the first step we should find the polygon L1 corresponding to Eq. (3.3). For monomials of this equation we have points: M1 = (2, 1), M2 = (1, 1), M3 = (0, 2), M4 = (0, 1), M5 = (0, 0). The carrier of Eq. (3.3) is defined by four points: Q1 = M1, Q2 = M3, Q3 = M4, and Q4 = M5. Their convex hull is the triangle L1 (Fig. 1). This triangle contains ð0Þ ð1Þ ð1Þ ð1Þ three apexes Cj ¼ Qj , (j = 1, 2, 4) and three edges C1 ¼ ½Q1 ;Q2 , C2 ¼ ½Q2 ;Q4 , C3 ¼ ½Q1 ;Q4 . Solutions of Eq. (3.3) have the second-order singularity. Let us make the following transformation: wðzÞ ¼ A0 þ A1 Y ðzÞ þ A2 Y ðzÞ2 ; where Y(z) satisfies the first-order differential equation and has the first-order singularity. Substituting (3.8) into Eq. (3.3), we obtain 1 M 2 ½w½Y ¼ ðbA1 þ 2bA2 Y ÞY zz ðmA1 þ 2mA2 Y ÞY z þ 2bA2 Y 2z þ A0 A2 C 0 A2 þ A21 Y 2 þ A1 A2 Y 3 2 1 2 4 1 2 þ A2 Y þ ðA0 A1 C 0 A1 ÞY þ A0 þ C 1 C 0 A0 2 2 ¼ 0:
ð3:8Þ
ð3:9Þ
The following points correspond to the monomials of this equation: M1 = (2, 1), M2 = (2, 2), M3 = (2, 2), M4 = (1, 1), M5 = (1, 2), M6 = (0, 0), M7 = (0, 2), M8 = (0, 4), M9 = (0, 1), M10 = (0, 2), M11 = (0, 3), M12 = (0, 0), M13 = (0, 1), M14 = (0, 2), M15 = (0, 0). The carrier of Eq. (3.9) is determined by seven points: Q1 = M1, Q2 = M2 = M3, Q3 = M8, Q4 = M11, Q5 = M7 = M14, Q6 = M9 = M13, and Q7 = M6 = M12 = M15. Their convex hull is the quadð0Þ ð1Þ ð1Þ rangle (Fig. 2). This quadrangle has four apexes Cj ¼ Qj , (j = 1, 2, 3, 4) and four edges C1 ¼ ½Q1 ;Q2 , C2 ¼ ½Q2 ;Q3 , ð1Þ ð1Þ C3 ¼ ½Q3 ;Q7 , C4 ¼ ½Q1 ;Q7 . So we have constructed the polygon L2. Following our method we should find the polygon L3 with some part of edges parallel to those of L2. Besides that when the polygon L3 moves along the plane, his apexes should cover the carrier of Eq. (3.9). We see that the triangle with apexes M4, Q5, and Q7 satisfies these requirements. Thus the simplest equation is the Riccati equation with constant coefficients E1 ½Y ¼ Y z þ Y 2 b ¼ 0:
ð3:10Þ
Substituting Y z ðzÞ ¼ E1 ½Y Y 2 þ b
ð3:11Þ
Fig. 1. The polygon L1 corresponding to the Eq. (3.3).
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Fig. 2. The polygons L2 and L3 corresponding to the Eqs. (3.9) and (3.10), accordingly.
into Eq. (3.9) and equating coefficients at powers of Y(z) to zero, yields algebraic equations for parameters A2, A1, A0, b and C1. Solving these equations, we get A2 ¼ 12b;
A1 ¼
12m ; 5
A0 ¼ C 0 þ
3m2 ; 25b
b¼
m2 ; 100b2
C1 ¼
C 20 18m4 2 625b2
ð3:12Þ
and the relation b 1 ½Y ; M 2 ½wðY Þ ¼ RE
ð3:13Þ
b is a differential operator. This completes the proof. h where R From equality (3.13) we see that if Y(z) is a solution of Eq. (3.11), then w(z) in formula (3.8) is a solution of Eq. (3.3). Hence we have found the solitary wave in the form of a kink 2 6m2 3m2 mðx C 0 t þ u0 Þ wðzÞ ¼ C 0 þ 1 þ tanh : ð3:14Þ 10b 25b 25b Here u0 is an arbitrary constant.
4. Exact solutions of the nonlinear fourth-order evolution equation Let us look for exact solutions of the following fourth-order evolution equation [33]: ut 2ux uxx u2 uxx 2uu2x þ uxxxx ¼ 0:
ð4:1Þ
Using the travelling wave reduction (3.2) and integrating with respect to z, we get wzzz w2z w2 wz C 0 w þ C 1 ¼ 0:
ð4:2Þ
Later let us prove the following theorem. Theorem 4.1. Let Y(z) be a solution of the equation Y zz YY z C 0 ¼ 0:
ð4:3Þ
Then w(z) = Y(z) is a solution of Eq. (4.2), provided that C1 = 0. Proof. Let us find the polygon that corresponds to Eq. (4.2). The following points are assigned to the monomials of this equation: M1 = (3, 1), M2 = (2, 2), M3 = (1, 3), M4 = (0, 1), M5 = (0, 0). The carrier of the equation contains all these points: Q1 = M1, Q2 = M2, Q3 = M3, Q4 = M4, and Q5 = M5. Their convex hull is the quadrangle L1 with four ð0Þ ð1Þ ð1Þ ð1Þ ð1Þ apexes Cj ¼ Qj , (j = 1, 3, 4, 5) and four edges C1 ¼ ½Q1 ;Q3 , C2 ¼ ½Q3 ;Q4 , C3 ¼ ½Q4 ;Q5 , C4 ¼ ½Q1 ;Q5 (see Fig. 3). Solutions of Eq. (4.2) have the first-order pole. Suitable polygon L3 we find without making the transformation. It is the triangle in Fig. 3. Thus we set wðzÞ Y ðzÞ;
ð4:4Þ
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Fig. 3. The polygons L1 and L3 corresponding to the Eqs. (4.2) and (4.5), accordingly.
where Y(z) is a solution of the second-order equation E2 ½Y ¼ Y zz aYY z b ¼ 0:
ð4:5Þ
Substituting (4.4) and Y zz ¼ E2 ½Y þ aYY z þ b
ð4:6Þ
into Eq. (4.2) and equating coefficients at powers of Y(z) to zero yields algebraic equations for parameters a, b and C1. Hence we get a ¼ 1;
b ¼ C0 ;
C 1 ¼ 0:
ð4:7Þ
We also obtain the relation b 2 ½Y ; M 3 ½wðY Þ ¼ RE
ð4:8Þ
b is a differential operator. This completes the proof. h where R Integrating Eq. (4.3) with respect to z yields the Riccati equation Yz
Y2 C 0 z þ 2C 3 ¼ 0; 2
ð4:9Þ
where C3 is a constant of integration. Setting Y ðzÞ ¼
2Wz W
in (4.9) we get C0 Wzz þ z C 3 W ¼ 0: 2
ð4:10Þ
ð4:11Þ
This equation is equivalent to the Airy equation. Its general solution is 1=3 1=3 WðzÞ ¼ C 1 Aif21=3 C 1=3 C 0 z þ C 4 g; 0 z þ C 4 g þ C 2 Bif2
ð4:12Þ
where Ai(f) and Bi(f) are the Airy functions and C1, C2, and C4 are arbitrary constants. Substituting expression (4.12) into (4.10) we get exact solutions of Eq. (4.2) with two arbitrary constants.
5. Exact solutions of the generalized Kuramoto–Sivashinsky equation Let us look for exact solutions of the generalized Kuramoto–Sivashinsky equation, which can be written as ut þ aum ux þ duxx þ buxxx þ cuxxxx ¼ 0:
ð5:1Þ
Eq. (5.1) at m = 1 is the famous Kuramoto–Sivashinsky equation [34,35], which describes turbulent processes. Its exact solutions at b = 0 were first found in [34]. Its solitary waves at b 5 0 were obtained in [10] and periodical solutions of
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this equation also at b 5 0 were first presented in [13]. Recently it was shown [36,37] that this equation did not have other solutions except found before. Using the variables sffiffiffi pffiffiffiffiffi d d2 b a dc 0 x ¼x ; t0 ¼ t ; u0 ¼ u; r ¼ pffiffiffiffiffi ; a0 ¼ 2 ; ð5:2Þ c c cd d we get the equation in the form (the primes are omitted) ut þ aum ux þ uxx þ ruxxx þ uxxxx ¼ 0:
ð5:3Þ
Using the travelling wave reduction uðx;tÞ ¼ yðzÞ;
z ¼ x C0t
ð5:4Þ
and integrating with respect to z, we get the equation M 3 ½y ¼ y zzz þ ry zz þ y z C 0 y þ
a y mþ1 ¼ 0: mþ1
ð5:5Þ
Here a constant of integration is equated to zero. Let us present our result in the following theorem. Theorem 5.1. Let y(z) be a solution of the equation 13 9a 3 mþ3 y 3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi y: yz ¼ 3 2 2m þ 11m2 þ 18m þ 9 2m þ 18m þ 27
ð5:6Þ
pffiffi 3 , and Then y(z) is also a solution of Eq. (5.5), provided that m 5 0, m 5 1, m 6¼ 32, m 5 3, m 6¼ 93 2
C0 ¼
3ð2m2 þ 9m þ 9Þ ð2m2
þ 18m þ 27Þ
3=2
;
3ð3 þ mÞ r ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi : 2m2 þ 18m þ 27
ð5:7Þ
Proof. Monomials of Eq. (5.5) are determined by the following points: M1 = (3, 1), M2 = (2, 1), M3 = (1, 1), M4 = (0, 1) and M5 = (0, m + 1). The polygon L1 that corresponds to Eq. (5.5) is the major triangle in Fig. 4((a)
Fig. 4. The polygons L1 and L3 corresponding to the Eqs. (5.5) and (5.8), accordingly.
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m > 0 and (b) m < 0). Suitable polygon L3 can be found without making the transformation. It is the smaller triangle in Fig. 4. This triangle contains the points: Q1 = (1, 1), Q2 = M4 = (0, 1), and Q3 ¼ ð0; mþ3 Þ. 3 Consequently the simplest equation corresponding to L3 can be written as E1 ½y ¼ y z Ay
mþ3 3
By ¼ 0;
ð5:8Þ
where A and B are parameters to be found. The general solution of (5.8) takes the form 3 Bmz A m yðzÞ ¼ C 1 exp : 3 B
ð5:9Þ
Substituting y z ¼ E1 ½y þ Ay
mþ3 3
þ By
ð5:10Þ
into Eq. (5.5) and equating coefficients at powers of y(z) to zero yields algebraic equations for parameters A, B, C0 and r in the form A3 ð2m þ 3Þðm þ 3Þðm þ 1Þ þ 9a ¼ 0;
ð5:11Þ
ðm þ 1ÞðB3 þ rB2 þ B C 0 Þ ¼ 0;
ð5:12Þ
2
A ð3 þ mÞðm þ 1ÞðBm þ 3B þ rÞ ¼ 0;
ð5:13Þ
ðm þ 1Þð2r2 m2 þ 18r2 m þ 27r2 9m2 54m 81Þ ¼ 0:
ð5:14Þ
Eqs. (5.11)–(5.14) can be solved and we obtain A1 ¼ ð9aÞ1=3 ðm3 þ 11m2 þ 18m þ 9Þ1=3 ;
ð5:15Þ
A ¼ ð9aÞ1=3 ð2m3 þ 11m2 þ 18m þ 9Þ1=3 : C 0 ¼ BðB2 þ rB þ 1Þ;
B¼
r ; 3þm
ð5:16Þ
3ð3 þ mÞ r ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi : 2m2 þ 18m þ 27
ð5:17Þ
So we have found parameters of Eq. (5.9), conditions (5.7) and the relation b 1 ½y; M 3 ½y ¼ RE
ð5:18Þ
b is a differential operator. This completes the proof. h where R The general solution of Eq. (5.7) can be presented in the form m3 mz yðzÞ ¼ C þ C 1 exp pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; 2m2 þ 18m þ 27
ð5:19Þ
where C1 is an arbitrary constant and C is determined by expression p ffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 9a 2m2 þ 18m þ 27 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi : C¼ p 3 3 ðm þ 3Þð2m2 þ 5m þ 3Þ
ð5:20Þ
pffiffiffiffiffi We would like to note that the value of solitary wave velocity (5.19) tends to C 0 ¼ 1= 27 as m ! 0, but at the same time C0 ! 0 as m ! 1. Assuming m = 1 in (5.19) we obtain the known solitary wave
yðzÞ ¼
pffiffiffiffiffip ffiffiffiffiffiffiffiffiffiffiffiffiffiffi !3 x C0t 47 3 225a þ C 1 exp pffiffiffiffiffi ; 30 47
C0 ¼
60 pffiffiffiffiffi 47 47
ð5:21Þ
pffiffiffiffiffi and the value of the parameter r : r ¼ 12= 47. This solution of the Kuramoto–Sivashinsky equation is the kink [10–14].
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6. Self-similar solutions of the fifth-order Korteveg–de Vries equation Let us find exact solutions of the fifth-order Korteveg–de Vries equation using our approach. This equation can be written as ut þ uxxxxx 10uuxxx 20ux uxx þ 30u2 ux ¼ 0:
ð6:1Þ
It has the self-similar solution uðx;tÞ ¼
1 ð5tÞ
2=5
yðzÞ;
z¼
x ð5tÞ1=5
;
ð6:2Þ
where y(z) satisfies the nonlinear ODE of the form y zzzzz 10yy zzz 20y z y zz þ 30y 2 y z zy z 2y ¼ 0:
ð6:3Þ
Our results are summarized in the following theorem. Theorem 6.1. Let Y(z) be a solution of the equation Y zzzzz 40YY zz Y z 10Y 2 Y zzz 10Y 3z þ 30Y 4 Y z Y zY z ¼ 0:
ð6:4Þ
yðzÞ ¼ Y z Y 2
ð6:5Þ
Then
is a solution of Eq. (6.3). Proof. The following points correspond to the monomials of this equation: M1 = (5, 1), M2 = (3, 2), M3 = (3, 2), M4 = (1, 3), M5 = (0, 1) and M6 = (0, 1). These points generate the carrier of Eq. (6.3): Q1 = M1 = (5, 1), Q2 = M4 = (1, 3), Q3 = M5 = M6 = (0, 1) and Q4 = M2 = M3 = (3, 2). Now we can plot the polygon L1 (see Fig. 5). Solutions of Eq. (6.3) have the second-order pole. Thus let us make the following transformation: y ¼ A1 Y þ A2 Y 2 þ A3 Y z ;
ð6:6Þ
where Y(z) is a function of the first-order pole. Substituting (6.6) into the equation studied, we obtain A1 Y zzzzz þ A2 ðY 2 Þzzzzz þ A3 Y zzzzzz 10ðA1 Y þ A2 Y 2 þ A3 Y z ÞðA1 Y zzz þ A2 ðY 2 Þzzz þ A3 Y zzzz Þ 20ðA1 Y z þ A2 ðY 2 Þz þ A3 Y zz ÞðA1 Y zz þ A2 ðY 2 Þzz þ A3 Y zzz Þ þ 30ðA1 Y þ A2 Y 2 þ A3 Y z Þ2 ðA1 Y z þ A2 ðY 2 Þz þ A3 Y zz Þ zðA1 Y z þ A2 ðY 2 Þz þ A3 Y zz Þ 2A1 Y 2A2 Y 2 2A3 Y z ¼ 0:
ð6:7Þ
The carrier of this equation is defined by ten points: Q1 = (6, 1), Q2 = (5, 2), Q3 = (4, 3), Q4 = (3, 4), Q5 = (2, 5), Q6 = (1, 6), Q7 = (0, 2), Q8 = (0, 1), Q9 = (1, 1), Q10 = (5, 1). In this case, the polygon L2 is the quadrangle presented in Fig. 6. Studying this polygon we can find the polygon L3 for the simplest equation. It is the triangle also presented in Fig. 6. Consequently the simplest equation for (6.3) is E5 ½Y ¼ Y zzzzz þ m1 YY zz Y z þ m2 Y 2 Y zzz þ m3 Y 3z þ m4 Y 4 Y z þ m5 Y þ m6 zY z ¼ 0; where m1, m2, m3 m4, m5 and m6 are unknown parameters to be found.
Fig. 5. The polygon L1 corresponding to the Eq. (6.3).
ð6:8Þ
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1489
Fig. 6. The polygons L2 and L3 corresponding to the Eqs. (6.7) and (6.8), accordingly.
Substituting Y zzzzz ¼ E5 ½Y m1 YY zz Y z m2 Y 2 Y zzz m3 Y 3z m4 Y 4 Y z m5 Y m6 zY z
ð6:9Þ
into Eq. (6.7) and equating coefficients at different powers of Y(z) to zero yields algebraic equations for parameters m1, m2, m3, m4, m5 and m6. As a result we get m1 ¼ 40;
m2 ¼ 10;
m3 ¼ 10;
m4 ¼ 30;
m5 ¼ 1;
m6 ¼ 1
ð6:10Þ
and the relation b 5 ½Y ; M 5 ½yðY Þ ¼ RE
ð6:11Þ
b is a differential operator. This completes the proof. h where R Eq. (6.4) can be integrated in z and we get the fourth-order analogue to the second Painleve´ equation Y zzzz 10Y 2 Y zz 10YY 2z þ 6Y 5 zY b ¼ 0;
ð6:12Þ
where b is an arbitrary constant. Thus we have expressed solutions of Eq. (6.3) (and consequently of Eq. (6.1)) through solutions of (6.12).
7. Self-similar solutions of the fifth-order modified Korteveg–de Vries equation Let us find self-similar solutions of the fifth-order modified Korteveg–de Vries equation ut 10u2 uxxx 40ux uxx 10u3x þ 30u4 ux þ uxxxxx ¼ 0:
ð7:1Þ
It has the scaling reduction [38–40] uðx;tÞ ¼ ð5tÞ1=5 wðzÞ;
z ¼ xð5tÞ1=5 ;
ð7:2Þ
where w(z) satisfies M 4 ½w ¼ wzzzz 10w2 wzz 10ww2z þ 6w5 zw b ¼ 0:
ð7:3Þ
Let us present our result in the theorem. Theorem 7.1. Let Y(z) be a solution of the equation z Y zzz þ 2bYY zz bY 2z 6b2 Y 2 Y z 3b3 Y 4 þ ¼ 0: 2b
ð7:4Þ
Then wðzÞ ¼ bY ðzÞ is a solution of Eq. (7.3), provided that b = 1/2.
ð7:5Þ
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Proof. To begin with the polygon corresponding to Eq. (7.3) should be found. The following points: M1 = (4, 1), M2 = (2, 3), M3 = (2, 3), M4 = (0, 5), M5 = (1, 1), M6 = (0, 0) are assigned to the monomials of the studied equation. The carrier of the equation contains five points Q1 = M1, Q2 = M4, Q3 = M5, Q4 = M6, and Q5 = M2 = M3. Their conð0Þ ð1Þ ð1Þ vex hull is the quadrangle L1 with four apexes Cj ¼ Qj , (j = 1, 2, 3, 4) and four edges C1 ¼ ½Q1 ;Q2 , C2 ¼ ½Q2 ;Q3 , ð1Þ ð1Þ C3 ¼ ½Q3 ;Q4 , C4 ¼ ½Q1 ;Q4 (Fig. 7). ð1Þ The leading members of Eq. (7.1) are located on the edge C1 ¼ ½Q1 ;Q2 . Substituting w(z) = a0zp into the equation wzzzz 10w2 wzz 10ww2z þ 6w5 ¼ 0;
ð7:6Þ
we get fours families of power asymptotics for solutions of (7.3): (p, a0) = (1, 1), (1, 1), (1, 2), and (1, 2). Again taking into account the pole order of solutions and the results of Section 2, we can look for exact solutions of Eq. (7.3) in the form wðzÞ ¼ A1 Y ðzÞ;
A1 6¼ 0;
ð7:7Þ
where Y(z) is a solution of the following simplest equation Y zzz aYY zz bY 2z cY 2 Y z dY 4 mz ¼ 0:
ð7:8Þ
It is important to mention that the transformation (7.7) does not change the quadrangle L1 and besides that the polygon L3 of the simplest Eq. (7.8) possesses all necessary properties. In other words L1 L2 and L3 is the triangle (see Fig. 7). The parameters a, b, c, d, and m can be found. Substituting (7.7) and Y zzz ¼ E3 ½Y þ aYY zz þ bY 2z þ cY 2 Y z þ dY 4 þ mz ¼ 0
ð7:9Þ
into Eq. (7.3) and equating coefficients at different powers of Y(z) to zero yields algebraic equations for parameters a, b, c, d, m, b and A1. Finally, we obtain a ¼ 2b;
c ¼ 6b2 ;
d ¼ 3b3 ;
m¼
1 ; 2b
A1;2 ¼ b;
b¼
1 2
ð7:10Þ
and the relation [39] b 3 ½Y ; M 4 ½wðY Þ ¼ RE
ð7:11Þ
b is a differential operator. This completes the proof. h where R Consequently there exist solutions of the studied equation expressed through the function Y(z) at b = 1/2 or b = 1/2. Let us show that solutions of Eq. (7.4) are associated with solutions of the first Painleve´ equation vzz ¼ Cv2 þ Dz:
ð7:12Þ
Making the transformation vðzÞ ¼ Aw2 þ Bwz
Fig. 7. The polygons L1 and L3 corresponding to the Eqs. (7.3) and (7.8), accordingly.
ð7:13Þ
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in Eq. (7.12), yields the equation Bwzzz þ 2Awwzz þ ð2A CB2 Þw2z 2CABw2 wz CA2 w4 Dz ¼ 0:
ð7:14Þ
We see that the polygon corresponding to this equation coincides with the triangle L3 at Fig. 7. Combining (7.8) and (7.14), we obtain wðzÞ Y ðzÞ;
ð7:15Þ 2
provided that A = b, B = 1, C = 3b, and D = 1/(2b). In other words, if Y(z) is a solution of (7.4), then v(z) = bY + Yz is a solution of vzz ¼ 3bv2
1 z: 2b
ð7:16Þ
We would like to mention that without loss of generality it can be set b = ±1.
8. Solitary waves of the sixth-order nonlinear evolution equation Let us find exact solutions of the sixth-order nonlinear evolution equation, which describes turbulent processes [41,42] ut þ uux þ buxx þ duxxxx þ euxxxxxx ¼ 0:
ð8:1Þ
It is known that Kuramoto–Sivashinskiy equation and the Ginzburg–Landau equation used to describe the turbulence are nonintegrable because they do not pass the Painleve´ test [5]. However, these equations have a list of special solutions [5,10,13]. Eq. (8.1) does not pass the Painleve´ test as well and thus this equation is also nonintegrable. However one can expect that Eq. (8.1) has some special solutions. This equation admits the travelling waves reduction uðx;tÞ ¼ yðzÞ;
ð8:2Þ
z ¼ x C 0 t;
where y(z) satisfies the equation 1 C 1 C 0 y þ y 2 þ by z þ dy zzz þ ey zzzzz ¼ 0: 2
ð8:3Þ
C1 is a constant of integration. As a result we get the following theorem. Theorem 8.1. Let Y(z) be a solution of the equation Y z þ Y 2 ak ¼ 0
ðk ¼ 1;2; . . . ;6Þ:
ð8:4Þ
yðzÞ ¼ 30240eY 5 þ
2520 2520 1260 12600 d2 d 50400eak Y 3 þ dak þ 20160ea2k þ b Y þ C0 11 11 251 30371 e
ð8:5Þ
Then
is a solution of Eq. (8.3), provided that 4112640 d5 w4k d5 w5k 5080320 d3 w3k b 9999360 11 251 e2 e3 e3 5 3 3 2 55460160 d wk 660240 d wk b 25200 d5 w2k þ 30371 2761 e2 30371 e3 e3 2 3 1260 b dwk 12600 bd wk 1 2 þ þ C0; 251 30371 e2 2 e 3 3 213811840e ak 10204656de2 a2k 2045d3 92400d2 eak b¼ ; 121eð9240eak þ 79dÞ
C1 ¼
ð8:6Þ ð8:7Þ
where ak and wk are found from ak ¼
dwk e
w1 ¼
ðk ¼ 1;2; . . . ;6Þ;
1 ; 220
w2 ¼
5 ; 176
ð8:8Þ w3 ¼
1 ; 440
ð8:9Þ
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w4 ¼
1 46031 557 þm ; 52800 m
pffiffiffi ! 1 46031 m i 3 46031 mþ þ 557 ; w5;6 ¼ 2 52800 2m 2 m pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 m ¼ ð113816753 þ 1260 8221079733Þ3 610966:
ð8:10Þ ð8:11Þ ð8:12Þ
Proof. The following points correspond to the monomials of Eq. (8.3): Q1 = (5, 1), Q2 = (0, 2), Q3 = (0, 0), Q4 = (3, 1), and Q5 = (1, 1). The polygon L1 of the equation is the triangle (Fig. 8). Solutions of Eq. (8.3) have the fifth-order pole. Let us express them through a function Y(z) possessing the first-order pole yðzÞ ¼ A0 þ A1 Y þ A2 Y 2 þ A3 Y 3 þ A4 Y 4 þ A5 Y 5 :
ð8:13Þ
Substituting transformation (8.13) into (8.3), we get new equation and the corresponding polygon L2 in the form of quadrangle (Fig. 9). One of the suitable polygons L3 is the triangle in Fig. 9. This triangle is assigned to the Riccati equation with constant coefficients E1 ½Y ¼ Y z þ Y 2 a ¼ 0:
ð8:14Þ
Substituting Y z ¼ E1 ½Y Y 2 þ ak
ð8:15Þ
into Eq. (8.4) and equating coefficients at different powers of Y(z) to zero yields algebraic equations for parameters A5, A4, A3, A2, A1, A0, C1, b, ak and wk. Solving these equations we have 2520d 50400ea; A2 ¼ 0; A1 11 2520 1260 12600 d2 da þ 20160ea2 þ b ; A0 ¼ C 0 : ¼ 11 251 30371 e
A5 ¼ 30240e;
A4 ¼ 0;
A3 ¼
ð8:16Þ
For the parameters C1, b, ak, and wk we obtain expressions (8.6)–(8.10), and (8.12). We also see that the following relation: b 1 ½Y M 5 ½yðY Þ ¼ RE
ð8:17Þ
b is a differential operator. This completes the proof. h holds, where R The general solution of Eq. (8.4) is the following: pffiffiffiffiffi pffiffiffiffiffi Y ðzÞ ¼ ak tanhð ak z þ u0 Þ; ðk ¼ 1;2; . . . ;6Þ:
ð8:18Þ
Substituting (8.13) into (8.5) and taking into account that ak = wkd/e (k = 1, . . . , 6) we get six different solitary waves of Eq. (8.3).
Fig. 8. The polygon L1 corresponding to the Eq. (8.3).
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Fig. 9. The polygons L2 and L3 corresponding to the transformed Eq. (8.3) and to the Eq. (8.14), accordingly.
9. Exact periodic solutions of Eq. (8.1) It was mentioned in the previous section that solutions of Eq. (8.3) have the fifth-order pole. Thus let us make the following transformation yðzÞ ¼ B1 þ B2 RðzÞ þ B3 Rz þ B4 R2 þ B5 RRz ;
ð9:1Þ
where Bk (k = 1, . . . , 5) are constants and R = R(z) is a function of the second-order pole. Now let us prove the following theorem. Theorem 9.1. Let R(z) be a solution of the equation R2z
Then
pffiffiffiffiffi pffiffiffiffiffi 3 1 2 1 Rd2 1 R 21d2 1 3 13 d3 1 1 ad2 21d a ¼ 2R þ aR a R þ þ þ 6 726 e2 2541 108 359370 e3 119790 e3 4356 e2 e2 pffiffiffiffiffi 1 a 21d2 : 15246 e2 3
2
d yðzÞ ¼ C 0 þ 630 ea þ 6eR Rz 11
ð9:2Þ
ð9:3Þ
is a solution of Eq. (8.1), provided that 10 d2 ; 121 e pffiffiffiffiffi 5 10854 d5 1 2 2484 21d ð1;2Þ þ C : C1 ¼ 161051 e3 2 0 161051 e3
b¼
ð9:4Þ ð9:5Þ
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Proof. Substituting (9.1) into Eq. (8.3) we get a new equation of the sixth-order. The polygon L2 of this equation is presented in Fig. 10. Again we should construct suitable polygon L3. Let us take the triangle corresponding to the equation for the elliptic function E1 ½R ¼ R2z þ 2R3 aR2 2bR d ¼ 0:
ð9:6Þ
This triangle satisfies necessary requirements. Substituting R2z ¼ E1 ½R 2R3 þ aR2 þ 2bR þ d; 1 oE1;z ½R Rzz ¼ 3R2 þ aR þ b 2Rz oz
ð9:7Þ ð9:8Þ
into Eq. (8.3) and equating coefficients at different powers of R(z) and Rz to zero yields algebraic equations for the coefficients B5, B4, B3, B2, B1, and B0. As a result we obtain the following expressions: B5 ¼ 3780e; B2 ¼ 0;
B4 ¼ 0;
B3 ¼ 630ea þ
630 d; 11
B1 ¼ C 0 ;
ð9:9Þ
2
10 d ; 121 e pffiffiffiffiffi 2 1 1 d2 1 21d ; b1;2 ¼ a2 þ 12 1452 e2 5082 e2pffiffiffiffiffi pffiffiffiffiffi 1 3 13 d3 1 1 ad2 1 a 21d2 21d3 a þ d 1;2 ¼ ; 108 359370 e3 119790 e3 4356 e2 15246 e2
b¼
ð9:10Þ ð9:11Þ ð9:12Þ
and the relation b 1 ½R; M 5 ½yðRÞ ¼ RE
ð9:13Þ
b is a differential operator. This completes the proof. h where R If R1, R2, and R3 such that R1 P R2 P R3 are real roots of the equations pffiffiffiffiffi ! pffiffiffiffiffi pffiffiffiffiffi 1 2 d2 d2 21 1 3 13 d3 d3 21 ad2 ad2 21 3 2 a a 2R aR þ þ ¼ 0; R 6 108 359379 e3 119790e3 4356e2 15246e2 726e2 2541e2 then the general solution of (9.2) can be written as rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ! R1 R3 R1 R2 RðzÞ ¼ R2 þ ðR1 R2 Þcn2 z;S ; S 2 ¼ : 2 R1 R3
Fig. 10. The polygons L2 and L3 corresponding to the transformed Eq. (8.3) and to the Eq. (9.6), accordingly.
ð9:14Þ
ð9:15Þ
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Thus we have found several solutions of Eq. (8.1) at different values of the parameters. These solution are solitary and periodic waves (see expressions (8.5) and (9.3)).
10. Conclusion In this paper a new method for finding exact solutions of nonlinear differential equations is presented. Our aim was to express solutions of the equation studied through solutions of the simplest equations. The method is more powerful than many other methods because the structure of a solution is not fixed but can be found using the stated algorithm. Thus one can obtain quite general classes of exact solutions. The method is based on the ideas of power geometry. More exactly, in order to find the suitable simplest equation we construct the polygons of nonlinear differential equations. It is important to mention that transformations between nonlinear differential equations can be also found with a help of our method. As an example of our method application exact solutions of some nonlinear differential equations were found. In particular, solutions of the fourth-order evolution equation (4.1) were expressed in terms of the Airy functions; the connection between self-similar solutions of the fifth-order Korteveg–de Vries equation (6.1) and the fifth-order modified Korteveg–de Vries equation (7.1) was constructed; one-parametric family of exact solutions for the generalized Kuramoto–Sivashinsky equation (5.1) was found. Besides that we obtained solitary waves and periodic solutions of the sixth-order evolution equation (8.1).
Acknowledgements The authors are grateful to Professor M.S. El Naschie for his attention to this work and remarks. This work was supported by the International Science and Technology Center under Project B 1213.
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