The nonlinear differential equation governing the periodic motion of the one-dimensional, undamped, and unforced cubic-quintic Duffing oscillator is solved exactly by obtaining the period and the solution. The period is given in terms of the complete elliptic integral of the first kind and the solution involves Jacobian elliptic functions. We solve the cubic-quintic Duffing equation under arbitrary initial conditions. Physical applications are provided. The solution to the mixed parity Duffing oscillator is also formally derived. We illustrate the obtained results with concrete examples. We give high accurate trigonometric approximations to the Jacobian function cn.

1. Introduction

It is well known that many engineering problems are not linear and their analytical solutions are not easy to obtain. Disturbance methods are among the known methods for solving nonlinear problems, which are based on the existence of small/large parameters, the so-called disturbance parameters. Our approach is different from known solutions to this problem [15]. On the contrary, here, we express the solutions without imaginary quantities: both frequency and modulus are real numbers. The quintic term appearing in the cubic-quintic Duffing equation makes this nonlinear oscillator not only more complex but also more interesting to study.

2. The Analytical Solution to the Cubic-Quintic Duffing Equation

Let , , , , and be arbitrary real numbers. We will solve the initial value problem:

We will assume that . Multiplying (1) by and integrating it with respect to gives

Letwhere the parameter values , , , , , and are to be determined. If the solution in (3) is periodic, it will have the same period as the function cn and this period may be evaluated by means of the formula

In the case, when , we may approximate the value of using the formula

The error for this approximation is given by

On the contrary, we may obtain approximate trigonometric solution making use of the following approximation formula:

See Table 1, for the error = .

A more accurate trigonometric approximation may be obtained using the formulabeing

See Table 2, for the error = .


Introduce the notation:

Definition 1. The discriminant to the i.v.p. (1) is defined as

2.1. First Case:

We define Inserting the ansatz (3) into (10), we obtain

Equating to zero the coefficients of in the numerator of the last expression gives an algebraic system. Solving it, we obtain

For these choices, we obtain the following system:

Eliminating from system (15) gives the sexticwhere

Sextic (16) is solvable by radicals. Indeed, let


The discriminant to cubic (19) is

Cubic (19) has three real roots and at least one of them must be positive. Indeed, let , , and be the roots to this cubic. Then,

Since , at least one of the numbers , , and must be positive. We choose the closest to positive root to cubic (19) so that, in view of (17), the number will be the closest to zero real root to the sextic in (15). Observe also that the condition implies that . Thus, if , then . Moreover, the discriminant to the cubic (17) and have the same sign.

The numbers and are determined from the initial conditions:

The number is a solution to the sextic:

Example 1. LetSextic (15) readsThe roots to this sextic areWe choose the value . The values of and are determined from the initial conditions. They readThe exact solution to the i.v.p.,readsThe solution is periodic with period . The approximate trigonometric solution is given byThe error of this trigonometric approximant compared with the exact solution on equals 0.00352816, see Figure 1.

2.2. Second Case:

Let . Inserting the ansatz (3) into (10), we obtain

Equating to zero the coefficients of gives an algebraic system. Solving it gives

The system reduces to

Eliminating from this system gives the sexticwhere

Sextic (34) has at least one real root. Indeed, let () be its roots. Then,

We will choose the closest to zero real root to sextic (34). The values for and are determined from the initial conditions:

The number is found from the algebraic equation:

Example 2. Let . The i.v.p. problem to be solved isThis problem has a negative discriminant . Sextic (35) readsThe roots areWe choose . The values for and areThe exact solution is given bySee Figure 2, for a comparison with the numerical solution.

2.3. Third Case:

In this case, we have two roots to the cubic:

At least one of these two numbers must be positive. So, we proceed the same way as we did for a positive discriminant.

2.4. Two Particular Cases
2.4.1. First Particular Case:


The discriminant to the i.v.p. (46) equals

If , then the exact solution is given bywhere

Assume now a negative discriminant. The exact solution reads

We claim that . Indeed, let

Then, . However,

From the last identity, it is clear that .

Assume now that . In this case, we have the following solutions: