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On the Generalized Simplest Equations: Toward the Solution of Nonlinear Differential Equations with Variable Coefficients

By Gunawan Nugroho, Purwadi Agus Darwito, Ruri Agung Wahyuono and Murry Raditya

Submitted: November 13th 2020Reviewed: December 22nd 2020Published: January 14th 2021

DOI: 10.5772/intechopen.95620

Downloaded: 104

Abstract

The simplest equations with variable coefficients are considered in this research. The purpose of this study is to extend the procedure for solving the nonlinear differential equation with variable coefficients. In this case, the generalized Riccati equation is solved and becomes a basis to tackle the nonlinear differential equations with variable coefficients. The method shows that Jacobi and Weierstrass equations can be rearranged to become Riccati equation. It is also important to highlight that the solving procedure also involves the reduction of higher order polynomials with examples of Korteweg de Vries and elliptic-like equations. The generalization of the method is also explained for the case of first order polynomial differential equation.

Keywords

  • the simplest equation
  • Riccati equation
  • nonlinear differential equations
  • reduction of polynomial

1. Introduction

Despite the advent of supercomputers in numerical methods, increasing activities are devoted to solving nonlinear differential equations by analytical method in recent years [1, 2, 3]. Analytical solutions have their own importance concerning the physical phenomena as they are often pave the way to the construction of right theory [4]. Many methods have been proposed concerning this important problem and generally, for the problems with constant coefficients. One of the useful methods is the method of simplest Equation [5] or for some authors, the auxiliary Equations [6]. The method is built by the utilization of the first integral of simplest nonlinear differential equations, such as Bernoulli and Riccati Equations [7]. The method had produced many new solutions of the considered nonlinear differential equations, generally with constant coefficients [8, 9].

For the more general cases, we have found that the method can be extended such as involving the solution of the nonlinear differential equations with variable coefficients. The nature of variable coefficients often arises in the equation describing the heterogenous media and composites [10] or in other cases are produced by the coordinate transformation of the partial differential Equations [11]. Those two categories are developed rapidly in recent years with the capacity of high-speed super computers which sufficient for computing nonlinear problem with complex geometries [12, 13], as sometimes desired by engineering design activities. The role of analytical solutions is as a benchmark to validate the computer algorithm with simpler geometries as it is usually performed [14].

In this chapter, the solution method of the simplest equations is different from the cases of constant coefficients except, on Bernoulli equation. Hence, we will start from Riccati equation instead of Bernoulli equation as the simplest equation to highlight the novelty of the procedure. The method is then followed by examples and conclusion.

2. The first integral of the simplest equations with variable coefficients

2.1 Riccati equation

Consider the Riccati equation with variable coefficients as follows,

Aξ=a1ξA2+a2ξA+a3ξE1

Let A=β1β2and the above equation can be rearranged as,

β2β1ξ+β1β2ξ=a1β12β22+a2β1β2+a3orβ2β1ξa1β12β22a2β1β2=β1β2ξ+a3=γβ1β2.

and is separated as

β1ξa1β12β2a2+γβ1=0andβ2ξ+γβ2a3β1=0E2

The solutions for β1and β2are

β2=eξa2+γξeξa2+γa1β2+C11

and

β3=eξγdξξeξγdξa3β1+C2E3

The relation for A=β1β2is thus,

A=β1β2=eξa2ξeξa2+γa1β2+C11ξeξγdξa3β1+C2E4

Without loss of generality, suppose that β1=eξγdξand thus the above relation is performed as,

β1β2=eξa2ξeξa2a1β1β2+C11ξa3+C2E5

Rearrange Eq. (5) and integrate once,

a1eξa2β1β2ξeξa2a1β1β2+C1=a1e2ξa2ξa3+C2

or

ξeξa2a1β1β2+C12=2ξa1e2ξa2ξa3+C2+C3

The solution for Ais then,

A=β1β2=22eξa2ξa3+C2ξa1e2ξa2ξa3+C2+C312E6

where the coefficients aiwill be determined later from the substitution into the considered nonlinear differential equations.

2.2 The Jacobi and Weierstrass equations

It is interesting to know that other simplest equations can also be rearranged into the Riccati-type equations. The famous examples are Jacobi [15] and Weierstrass Equations [16], which can solve a large class of nonlinear differential equations. Let us consider Jacobi type equation with variable coefficients,

ϕξ2=b1ξϕ4+b2ξϕ3+b3ξϕ2+b4ξϕ+b5ξE7

and the Weierstrass equation as follows,

ϕξ2=b1ξϕ3+b2ξϕ2+b3ξϕ+b4ξE8

Here, the reader should not be confused by the coefficients which represent different functions with the same index. Take ϕ=1ν+aξand the Weierstrass equation becomes Jacobi equation which admits the similar method of solution.

Concerning the search for obtaining solution of (7) and (8), the balancing principle suggests the substitution of the first order series ϕ=b6+b7Aas in the following,

b6ξ+b7ξA+b7Aξ2=b1b74A4+4b1b6b73+b2b73A3+6b1b62b72+3b2b6b72+b3b72A2+4b1b63b7+3b2b62b7+2b3b6b7+b4b7A+b1b64+b2b63+b3b62+b4b6+b5E9

Performing the Riccati equation Aξ=a1b7A2+a2A+a3into (9) and we generate the following expression,

b6ξ+b7ξA+b7Aξ2=a1b72A2+a2b7+b7ξA+a3b7+b6ξ2=a12b74A4+2a1b7a2b7+b7ξA3+2a1b7a3b7+b6ξ+a2b7+b7ξ2A2+2a2b7+b7ξa3b7+b6ξA+a3b7+b6ξ2

The coefficients of polynomial are then related with the coefficients in (9) in order to determine a1,a2,a3,b6,b7as functions of the known b1,b2,b3,b4and b5as follows,

b1b74=a12b744b1b6b73+b2b73=2a1b7a2b7+b7ξ6b1b62b72+3b2b6b72+b3b72=2a1b7a3b7+b6ξ+a2b7+b7ξ24b1b63b7+3b2b62b7+2b3b6b7+b4b7=2a2b7+b7ξa3b7+b6ξb1b64+b2b63+b3b62+b4b6+b5=a3b7+b6ξ2E10

Hence, the first equation gives,

a1=f0b1E11

and the second equation is then,

a2b7+b7ξ=4b1b6b72+b2b722f0E12

The next relation produces,

a3b7+b6ξ=6b1b62b7+3b2b6b7+b3b72f018f034b1b6+b22b73E13

Eqs. (12) and (13) are thus substituted into the fourth relation of (12) to form the third order polynomial equation in term of b6as follows,

32f04b1+64b13b74192f02b12b72b63+24f04b2+32b12b2b7416b12b2b7496f02b1b2b7248f02b1b2b72b62+16f04b3+4b1b22b74+8b1b22b7232f02b1b3b7224f02b22b72b6+8f04b4+b23b748f02b2b3b72=0E14

which the roots will determine the solution for b6as functions of b1,b2,b3,b4,b7, or b6=f1b7in simple unknown variable. The step now is to find the polynomial expression for b7from the last relation of (10) as,

b1f14b7+b2f13b7+b3f12b7+b4f1b7+b5=6b1f12b7b7+3b2f1b7b7+b3b72f018f034b1f1b7+b22b732E15

Therefore, the last equation gives the expression for b7as polynomial equation of higher order, and the generated polynomial is,

anb7n+an1b7n1+an2b7n2+an3b7n3+..+a2b72+a1b7+a0=0E16

In this case the higher order polynomial will be solved by reducing the order.

3. Reduction of higher order polynomial

Consider the sixth order polynomial equation as in the following,

b76+a4b75+a5b74+a6b73+a7b72+a8b7+a9=0

First, multiply the above equation with the function αand rearranged as,

B6+a4αB5+a5α2B4+a6α3B3+a7α4B2+a8α5B+a9α6+φ=φE17

where B=αb7. The polynomial equation is cut as in the following,

B4+b1B3+b2B2+b3B+b4B2+b5B2B4+b1B3+b2B2+b3B+b4b6=φ

Note that, the coefficients biin this section is different from the previous section. Expanding for the new coefficients,

B6+b1B5+b2b5B4+b3b1b5B3+b4+b5b2b5B2b3b5Bb4b6=φE18

Hence, the relation for coefficients is,

b1=a1αb2b6=a2α2b3b1b6=a3α3b4+b5b2b6=a4α4orb3b6=a5α5b4b6=a6α6+φb6=φb5b1=a1αb2b6=a2α2b3a1αb2a2α2=a3α3b4+b5b2a2α22=a4α4+a2α2b2a2α2a3α3+a1αb2a2α2b2a2α2=a5α5a4α4+a2α2b2a2α2+b2a2α22b5b2a2α2=a6α6+φb2a2α2=a4α4+a2α2b2a2α2+b2a2α22b5b2a2α2+a6α6b5.

The fifth coefficient relation is rearranged as,

b2a2α22+a3a1α2b2a2α2+a5a1α4=0E19

and the roots are,

b2a2α2=12α2a3a1±a32a124a5a112=f0α2E20

Also, the last relation is rewritten as,

b2a2α23+a2α2b2a2α22+a4α4b2a2α2+a6α6=0E21

Note that performing (19) into (21) will remove b5and α. Thus, it is necessary to take other relation, i.e. a6α6=α12+b5, which will produce the cubic equation as follows,

α12+f03+a2f02+a4f0α6+b5=0E22

which has the roots as,

α6=12f03+a2f02+a4f0±12f03+a2f02+a4f024b512E23

Substituting back into a6α6=α12+b5to get,

12f03+a2f02+a4f0±12f03+a2f02+a4f024b5122+b5=12a6f03+a2f02+a4f0±12a6f03+a2f02+a4f024b512or
14f03+a2f02+a4f024b512f03+a2f02+a4f0f03+a2f02+a4f024b512+14f03+a2f02+a4f02+b5=12a6f03+a2f02+a4f0±12a6f03+a2f02+a4f024b512or
b5=14f03+a2f02+a4f0214f03+a2f02+a4f02+a6f03+a2f02+a4f0f03+a2f02+a4f0+a62E24

Therefore, αis also determined by (24) and so all the coefficients, b1,b2,b3,b4,b6,φ. The polynomial equation of sixth order is then re-expressed as,

B4+b1B3+b2B2+b3B+b4B2b6=b5B2φb5E25

as reduced into the quartic equation the roots can be obtained by radical solution.

The procedure described by (17–25) can be applied and iterated into (16) until the polynomial equation of b7is reduced into quartic equation. Hence, all the coefficients for Riccati equation the first order series, i.e. a1,a2,a3,b6,b7are determined and produce the solution as,

ϕ=b6+b7Aorϕ=b6+22b7eξa2ξa3+C2ξa1b7e2ξa2ξa3+C2+C312E26

Thus, following the method explained by (2–6) and (17–25), we have arrived at the solution of Jacobi and Weierstrass equations with variable coefficients.

4. Solution examples

4.1 The elliptic-like equation

As an application, consider the elliptic-like equation with forcing function,

ϕξξ+b1ξϕ3+b2ξϕ=b3ξE27

The balance principle suggests that the solution should be in the form,

ϕ=b4+b5AE28

Substituting into (27) will reproduce the following expression,

b5Aξξ+2b5ξAξ+b1b53A3+3b1b4b52A2+b5ξξ+3b1b42b5+b2b5A+b4ξξ+b1b43+b2b4=b3E29

The next step is to differentiate the Riccati equation once,

Aξξ=2a12A3+3a1a2+a1ξA2+2a1a3+a22+a2ξA+a2a3+a3ξ

Substituting into Eq. (29) and it will produce the polynomial equation as in the following,

2a12b5+b1b53A3+3a1a2b5+a1ξb5+2a1b5ξ+3b1b4b52A2+2a1a3b5+a22b5+a2ξb5+2a2b5ξ+b5ξξ+3b1b42b5+b2b5A+a2a3b5+a3ξb5+2a3b5ξ+b4ξξ+b1b43+b2b4=b3

or the next step is to relate the coefficients as,

2a12b5+b1b53=03a1a2b5+a1ξb5+2a1b5ξ+3b1b4b52=02a1a3b5+a22b5+a2ξb5+2a2b5ξ+b5ξξ+3b1b42b5+b2b5=0a2a3b5+a3ξb5+2a3b5ξ+b4ξξ+b1b43+b2b4=b3

In this case, the first equation gives,

a1=fb1b5E30

For the second equation,

3fa2b5+fξb5+3fb5ξ+3b1b4b5=0

Thus, provide the expression for b5as,

b5=C4f13eξa2+b1b4fE31

The third and fourth equations produce,

a3b5=b51eξa2ξb5eξa2b3b4ξξb1b43b2b4+C4E32

Substituting (32) into (31),

a22a2ξ2a2b5ξb5b5ξξb53b1b42b2=2fb51eξa2ξb5eξa2b3b4ξξb1b43b2b4+C4

Replace b5with (10),

23fξfa2+13fξξf49fξf2b1b4f2+b1b4fξ3b1b42b2=2feξb1b4fξeξb1b4fb3b4ξξb1b43b2b4+C4E33

which then solves a2regardless of b4. In this case, we take b4as the chosen fundamental variable and the resulted coefficients, b5,a1,a2,a3depend on b4and with the known coefficients b1,b2,b3. Therefore, the solution of (27) is generated as,

ϕ(ξ)=b5(b4){22eξa2(b4)dξ(ξa3(b4)dξ+C2)[ξa1(b4)e2ξa2(b4)dξ(ξa3(b4)dξ+C2)dξ+C3]12}+b4E34

4.2 Korteweg de Vries equation

The next example is for the Korteweg de Vries type equation,

ϕξξξ+b1ξϕϕξ+b2ξϕξ+b3ξϕ+b4ξ=0E35

The balancing principle with application of Riccati equation will determined the ansatz,

ϕ=b5+b6A+b7A2E36

Performing into (35) will produce,

2b7AAξξξ+b6Aξξξ+6b7AξAξξ+6b7ξAξ2+6b7ξAAξξ+2b1b72A3Aξ+3b1b6b7A2Aξ+6b7ξξ+b1b62+2b1b5b7+2b2b7AAξ+3b6ξAξξ+3b6ξξ+b1b5b6+b2b6Aξ+b1b7b7ξA4+b1b6ξb7+b1b6b7ξA3+b7ξξξ+b1b5b7ξ+b1b6b6ξ+b1b5ξb7+b2b7ξ+b3b7A2+b6ξξξ+b1b5b6ξ+b1b5ξb6+b2b6ξ+b3b6A+b5ξξξ+b1b5b5ξ+b2b5ξ+b3b5+b4=0E37

Performing the Riccati equation into (37) will produce the following polynomial,

12b7a13+2b1b72a1+12a13b7A5+18a1a1ξb7+54a12a2b7+6a13b6+b1b7b7ξ+2a2b1b72+18a12b7ξ+3a1b1b6b7A4+40a12a3b7+32a1a22b7+12a1ξa2b7+18a1a2ξb7+2a1ξξb7+6a1a1ξb6+12a12a2b6+b1b6ξb7+b1b6b7ξ+2a3b1b72+6a23b7+30a1a2b7ξ+6a1ξb7ξ+6a12b6ξ+3a2b1b6b7+6a1b7ξξ+a1b1b62+2a1b1b5b7+2a1b2b7A3+52a1a2a3b7+14a1ξa3b7+10a1a3ξb7+12a2a2ξb7+2a2ξξb7+8a23b7+8a12a3b6+7a1a22b6+3a1ξa2b6+6a1a2ξb6+a1ξξb6+24a1a3b7ξ+12a22b7ξ+6a2ξb7ξ+9a1a2b6ξ+3a1ξb6ξ+b7ξξξ+b1b5b7ξ+b1b6b6ξ+b1b5ξb7+b2b7ξ+b3b7+3a3b1b6b7+6a2b7ξξ+a2b1b62+2a2b1b5b7+2a2b2b7+3a1b6ξξ+a1b1b5b6+a1b2b6A2+16a1a32b7+14a22a3b7+10a2ξa3b7+8a2a3ξb7+2a3ξξb7+8a1a2a3b6+4a1ξa3b6+2a1a3ξb6+3a2a2ξb6+a2ξξb6+a23b6+18a2a3b7ξ+6a3ξb7ξ+6a1a3b6ξ+3a22b6ξ+3a2ξb6ξ+b6ξξξ+b1b5b6ξ+b1b5ξb6+b2b6ξ+b3b6+6a3b7ξξ+a3b1b62+2a3b1b5b7+2a3b2b7+3a2b6ξξ+a2b1b5b6+a2b2b6A+2a1a32b6+a22a3b6+2a2ξa3b6+a2a3ξb6+a3ξξb6+6a2a32b7+6a3a3ξb7+3a2a3b6ξ+3a3ξb6ξ+b5ξξξ+b1b5b5ξ+b2b5ξ+6a32b7ξ+3a3b6ξξ+a3b1b5b6+a3b2b6+b3b5+b4=0E38

From this step on, there is a little hope to solve all the coefficients as they are equal to zero. As it has also to be reduced, it is important to note that the problem of reduction here is different from the case of Jacobi equation since all the coefficients are in principle solvable in algebraic form. In this case, it is not practical to reduce the fifth order polynomial as the even highest power, i.e., as a tenth order polynomial equation. The calculation will become too tedious as the detail expression is needed in the reduced polynomial equation. The next sub section will illustrate the reduction of an odd highest power polynomial equation.

4.3 Reduction of fifth order polynomial

Consider Eq. (38) as follows,

d1A5+d2A4+d3A3+d4A2+d5A+d6=0

Multiply by the function, βand rearrange,

d1B5+d2βB4+d3β2B3+d4β3B2+d5β4B+d6β5+φ=φE39

where, B=βA. Rearranged Eq. (39) as given by,

b1B3+b2B2+b3B+b4B2+b5B2b1B3+b2B2+b3B+b4b6=φE40

Expanding the all the coefficients as,

b1B5+b2B4+b3b1b6B3+b4+b5b2b6B2b3b6Bb4b6=φ

Relate the coefficients as in the following,

b1=d1b2=d2βb3b1b6=d3β2b4+b5d2β1d1b3d3β2=d4β31d12b3d3β22=d5β4+d3β21d1b3d3β2d4β3+d2β1d1b3d3β2b51d1b3d3β2=d6β5+φ1d1b3d3β2=d4β3+d2β1d1b3d3β2b51d1b3d3β2+d5β5b5E41

The fifth equation of (41) gives the roots as,

1d1b3d3β2=12β2d3±d324d512=β2f0E42

Moving to the last equation, the functions b5and βdisappear from the operation. In this case we will consider the test function, b5+β10=d6β5, and will perform as,

β10+d4β31d1b3d3β2+d2β1d12b3d3β22+b5=0E43

Substituting for b3, the expression for βis,

β10+d4f0+d2f02d4β5+b5=0β5=12d4f0+d2f02±12d4f0+d2f0224b512E44

Substitute back to, b5+β10=d6β5as follows

12d4f0+d2f02±12d4f0+d2f0224b5122+b5=12d6d4f0+d2f02±12d6d4f0+d2f0224b512or
14d4f0+d2f0224b512d4f0+d2f02d4f0+d2f0224b512+14d4f0+d2f022+b5=12d6d4f0+d2f02±12d6d4f0+d2f0224b512or
b5=14d4f0+d2f02214d4f0+d2f022+d6d4f0+d2f02d4f0+d2f02+d62E45

which then solves b5, β, φand thus generates all the coefficients of bi. The polynomial is then rewritten as,

b1B3+b2B2+b3B+b4B2b6=b5B2φb5

which is reduced as,

d1A3+d2A2+12d1d3±d324d512+d3A+d4+12d2d3±d324d512=0E46

Eq. (46) dictates that the relations, d1=d2=d3=d4=0will satisfy for the solution. Hence, the coefficients are then,

12b7a13+2b1b72a1+12a13b7=018a1a1ξb7+54a12a2b7+6a13b6+b1b7b7ξ+2a2b1b72+18a12b7ξ+3a1b1b6b7=040a12a3b7+32a1a22b7+12a1ξa2b7+18a1a2ξb7+2a1ξξb7+6a1a1ξb6+12a12a2b6+b1b6ξb7+b1b6b7ξ+2a3b1b72+6a23b7+30a1a2b7ξ+6a1ξb7ξ+6a12b6ξ+3a2b1b6b7+6a1b7ξξ+a1b1b62+2a1b1b5b7+2a1b2b7=052a1a2a3b7+14a1ξa3b7+10a1a3ξb7+12a2a2ξb7+2a2ξξb7+8a23b7+8a12a3b6+7a1a22b6+3a1ξa2b6+6a1a2ξb6+a1ξξb6+24a1a3b7ξ+12a22b7ξ+6a2ξb7ξ+9a1a2b6ξ+3a1ξb6ξ+b7ξξξ+b1b5b7ξ+b1b6b6ξ+b1b5ξb7+b2b7ξ+b3b7+3a3b1b6b7+6a2b7ξξ+a2b1b62+2a2b1b5b7+2a2b2b7+3a1b6ξξ+a1b1b5b6+a1b2b6=0E47

The first equation gives,

b7=fb1a12E48

The second equation is rewritten as,

18a1ξfa13+54fa13a2+6a13b6+b1ffξa14+2b1f2a13a1ξ+2a2b1f2a14+18fξa14+18fa13a1ξ+3b1b6fa12=0or
18a1ξf+54fa2+6b6+b1ffξa1+2b1f2a1ξ+2a2b1f2a1+18fξa1+18fa1ξ+3b1b6f=0or
36f+2b1f2a1ξ+54fa2+6b6+b1ffξ+2a2b1f2+18fξa1+3b1b6f=0

The solution for b6is then,

36f+2b1f2a1ξ=b1ffξ+2a2b1f2+18fξa154fa26+3fb1b6b6=16+3fb1b1ffξ+2a2b1f2+18fξa1+36f+2b1f2a1ξ+54fa2=h1a1a2E49

The third equation will produce,

40a12a3f+32a12a22f+12a1ξa2fa1+18a2ξfa12+2a1ξξfa1+6a1ξb6+12a1a2b6+b1b6ξfa1+b1b6fξa1+2fa1ξb1b6+2a3b1f2a13+6a23fa1+30fξa12a2+30ffξa12a2+60f2a1ξa1a2+6a1ξfξa1+12fa1ξ2+6a1b6ξ+3a2b1b6fa1+6fξξa12+24fξa1ξa1+12fa1ξξa1+12fa1ξ2+b1b62+2b1b5fa12+2b2fa12=0

Take the expression for b5as,

b5=12b1fa12h2a1a2+40fa12+2b1f2a13a3

with,

h2a1a2=32a12a22f+12a1ξa2fa1+18a2ξfa12+2a1ξξfa1+6a1ξh1+12a1a2h1+b1h1ξfa1+b1h1fξa1+24fξa1ξa1+6a23fa1+30fξa12a2+30ffξa12a2+60f2a1ξa1a2+6a1ξfξa1+12fa1ξ2+6a1h1ξ+3a2b1h1fa1+6fξξa12+2fa1ξb1h1+12fa1ξξa1+12fa1ξ2+b1h12+2b2fa12E50

The fourth relation of (47) will generate,

h3a1a2=24fξa13+48fa12a1ξ+3b1b6fa12a3+6fa1ξa1ξξ12a2a2ξfa12+2a2ξξfa12+8a23fa12+7a1a22h1+3a1ξa2h1+6a1a2ξh1+a1ξξh1+12fξa12a22+24fa1a1ξa22+6fξa12a2ξ+12fa1a1ξa2ξ+9a1a2h1ξ+3a1ξh1ξ+fξξξa12+2fξξa1a1ξ+4fξξa1a1ξ+6fξa1ξ2+6fξa1a1ξξ+2fa1a1ξξξ+b1h1h1ξ+fξa12b2+2fa1a1ξb2+b3fa12+2a2b2fa12+3a1h1ξξ+a1b2h1+6fξξa12a2+24fξa1a1ξa2+12fa1ξ2a2+12fa1a1ξξa2+a2b1h12+h22b1fa12+b1fa12h22b1fa12ξE51

which then produce the solution of a3.

Note that a1and a2are the chosen fundamental variables and according to (48–51) and with the known coefficients b1,b2,b3,b4, they will define, b5,b6,b7,a3,β. Therefore, the solution of Korteweg de Vries equation is generated as,

ϕξ=b5a1a2+b6a1a222eξa2ξa3a1a2+C2ξa1e2ξa2ξa3a1a2+C2+C312+b7a1a212e2ξa2ξa3a1a2+C22ξa1e2ξa2ξa3a1a2+C2+C3E52

Since only a few of the considered equation has a special polynomial to be solved by equating all the variable coefficients to zero, it is important to note that the reduction of polynomial order would be an important step. Solving all coefficients to zero often be an obstacle because the difficulty would be the same or even more than the original nonlinear ODEs. In this case, the reduction of polynomial manipulates and reduces the need for solving all coefficients.

However, it is possible not to search for the expression of variable coefficients, i.e., b5,b6,b7,a1,a2and a3. First the roots of Eq. (37) are determined first as ϕ, and then Eq. (1) is decomposed as,

Dξ=a1BD2+a2BξBD+a3BE53

with A=BD. The solution of (53) is then,

D=22Beξa2ξa3B+C2ξa1Be2ξa2ξa3B+C2+C312or
A=BD=22eξa2ξa3B+C2ξa1Be2ξa2ξa3B+C2+C312E54

The definition for Bis determined by substituting the polynomial solution, ϕinto (54) as in the following,

A=ϕ=22eξa2ξa3B+C2ξa1Be2ξa2ξa3B+C2+C312

Rearranging the above equation as,

ϕ2ξa1Be2ξa2ξa3B+C2+C3=12e2ξa2ξa3B+C22

Differentiating once,

a1Be2ξa2ξa3B+C2=a2ϕ2e2ξa2ϕξϕ3e2ξa2ξa3B+C22+a3B1ϕ2e2ξa2ξa3B+C2or
a1B=a2ϕ2ϕξϕ3ξa3B+C2+a3B1ϕ2E55

Eq. (55) is a first order ODE in Band can be easily solved, which then prove that A=ϕwithout establishing the explicit expression for variable coefficients.

5. Generalized method

In this section, the method of solution to the Riccati equation is extended for the class of the first order polynomial differential equation as,

Aξ=anAn+an1An1+an2An2++a3A3+a2A2+a1A+a0E56

The above equation can be always re-expressed as,

Aξ=(bnAn2+bn1An3+bn2An4++b3A+b2)A2+b1A+(bnAn2+bn1An3+bn2An4++b3A+b2)b0

or

Aξ=bnAn+bn1An1+(bn2+bnb0)An2++(b3+bn1b0)A3+(b2+bn2b0)A2+(b1+b3b0)A+b2b0E57

which the coefficients will be reformulated as,

bn=an,bn1=an1,b2b0=a0bn2+bnb0=an2b3+bn1b0=a3b2+bn2b0=a2b1+b3b0=a1orbn=an,bn1=an1,b2b0=a0a2b22b23+ana0=an2a0b2b3+an1a0b2=a3b22+bn2a0=a2b2b1+b3a0b2=a1E58

In this case, we will always obtain the new coefficients bi. Proceeding into the other equations andmultiply the equation by the function α, to get,

αAξ=bnAn2+bn1An3+bn2An4++b3A+b2αA2+b1αA+bnAn2+bn1An3+bn2An4++b3A+b2αb0or
Bξ=(bnα2nBn2+bn1α3nBn3+bn2α4nBn4++b3α1B+b2)α1B2  +(b1+αtα)B+(bnα2nBn2+bn1α3nBn3+bn2α4nBn4++b3α1B+b2)αb0E59

where B=αA. Then, all the new coefficients in biwill be determined. The step is now to solve the Riccati equation. Let B=β2β3, the equation can be rearranged as,

β3β2ξ+β2β3ξ=bnα2nBn2+bn1α3nBn3+bn2α4nBn4++b3α1B+b2α1β22β32+b1+αξαβ2β3+bnα2nBn2+bn1α3nBn3+bn2α4nBn4++b3α1B+b2αb0or
β3β2ξbnα2nBn2+bn1α3nBn3+bn2α4nBn4++b3α1B+b2α1β22β32b1+αξαβ2β3=β2β3ξ+bnα2nBn2+bn1α3nBn3+bn2α4nBn4++b3α1B+b2αb0=γβ2β3E60

and is separated as,

β2ξbnα2nBn2+bn1α3nBn3+bn2α4nBn4++b3α1B+b2α2β22β3b1+αξα+γβ2=0and
β3ξ+γβ31β2bnα2nBn2+bn1α3nBn3+bn2α4nBn4++b3α1B+b2αb0=0E61

The solutions for β2and β3are,

β2=eξb1+αξα+γxξetb1+αξα+γbnα2nBn2+bn1α3nBn3+bn2α4nBn4++b3α1B+b2α1β3+C11and
β3=eξγdξξeξγdξbnα2nBn2+bn1α3nBn3+bn2α4nBn4++b3α1B+b2αb0β2+C2E62

The relation for B=β2β3is thus,

B=β2β3=eξ(b1+αξα)dξ[ξeξ(b1+αξα+γ)dξ+C1(bnα2nBn2+bn1α3nBn3+bn2α4nBn4+(+b3α1B+b2)α1β3dξ]1x[ξeξγdξ(bnα2nBn2+bn1α3nBn3+bn2α4nBn4++b3α1B+b2)αb0dξ+C2]

Without loss of generality, suppose that β2=φeξγdξand the above relation is performed as,

β2β3=eξb1+αξαξeξb1+αξαbnα2nBn2+bn1α3nBn3+bn2α4nBn4++b3α1B+b2α1φβ2β3+C11ξbnα2nBn2+bn1α3nBn3+bn2α4nBn4++b3α1B+b2αφ1b0+C2or
β2β3ξeξb1+αξαbnα2nBn2+bn1α3nBn3+bn2α4nBn4++b3α1B+b2α1φβ2β3+C1=eξb1+αξαtbnα2nBn2+bn1α3nBn3+bn2α4nBn4++b3α1B+b2αφ1b0+C2

Rearrange the above equation as,

eξb1bnα2nBn2+bn1α3nBn3+bn2α4nBn4++b3α1B+b2φβ2β3ξeξb1bnyn2+bn1yn3+bn2yn4++b3y+b2φβ2β3+C1=e2ξb1αφbnα2nBn2+bn1α3nBn3+bn2α4nBn4++b3α1B+b2ξbnα2nBn2+bn1α3nBn3+bn2α4nBn4++b3α1B+b2αφ1b0+C2

Let e2ξb1αφ=αφ1b0and integrate the above equation to get,

ξeξb1bnα2nBn2+bn1α3nBn3+bn2α4nBn4++b3α1B+b2φβ2β3dt+C12=ξbnα2nBn2+bn1α3nBn3+bn2α4nBn4++b3α1B+b2αφ1b0+C22or
eξb1bnα2nBn1+bn1α3nBn2+bn2α4nBn3++b3α1B2+b2Bφ2=bnα2nBn2+bn1α3nBn3+bn2α4nBn4++b3α1B+b2αb0

The solution for Bis then reduced into the solution of the polynomial equation. Thus, let A=α1B=ϕ, where ϕis the expression from the solution of the resulting polynomial equation which is similar to (38). The expression for αcan be determined by the inverse method as in (53–55) for the first order polynomial differential Eq. (56).

6. Conclusion

In this chapter, we propose the method of the simplest or the auxiliary equation to solve the nonlinear differential equation with variable coefficients. The method is based on the solution of the generalized Riccati equation as the simplest equation. It is found that the other known simplest equations, i.e., Jacobi and Weierstrass equation, are also solved by the Riccati equation. The applications with the variable coefficients elliptic-like and Korteweg de Vries equations show that the problem of solving nonlinear differential equations with variable coefficients are simplified, especially by the reduction of the resulting polynomial equation in solving the Korteweg de Vries equation. The generalization of the method is also derived in detail.

Conflict of interest

Authors declare that there is no conflict of interest.

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Gunawan Nugroho, Purwadi Agus Darwito, Ruri Agung Wahyuono and Murry Raditya (January 14th 2021). On the Generalized Simplest Equations: Toward the Solution of Nonlinear Differential Equations with Variable Coefficients [Online First], IntechOpen, DOI: 10.5772/intechopen.95620. Available from:

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