THE DOUBLE PENDULUM

NUMERICAL SOLUTION IN THE LAGRANGE FORMULATION

Ross Bannister, June 2001

The double pendulum consists of two sections. Suspended from a fixed point is a mass, constrained to be a distance from that point, but is allowed to vary its orientation ( from the vertical). A second mass, is suspended a further distance from the first mass and it too has an angular degree of freedom, being an angle from the vertical at any instant. The physical system is illustrated in Fig. 1 (the system is in a gravitational field of magnitude , directed downwards).

Figure 1: The double pendulum set-up. The lengths and , and the masses and are fixed. The system has the degrees of freedom and .

1. CONSTRUCT THE LAGRANGIAN

The Lagrangian is defined as , where and are expressions for the kinetic and potential energies respectively. In the Lagrange formulation, satisfies Lagrange's equations which determine the time evolution of the system. These are given later. The components of are,

equation

equation

In Eq. (1.2), the and refer to the positions of the respective masses (the dots refer to their rates of change and so the choice of origin is irrelevant). We require Eq. (1.2) to be written in terms of angles and so we exploit the transformations (we now choose the origin to be the fixed point),

equation

equation

equation

equation

Inserting Eqs. (1.3) to (1.6) into Eq. (1.2) (and simplifying), the Lagrangian becomes,

equation

equation

2. THE EQUATIONS OF MOTION

For each degree of freedom, there is an equation of motion. For degree this is,

equation

where in the case of the double pendulum, . The and the represent separate variables. Eqs. (2.1) are coupled second order equations. This is apparent if the total time derivative is expanded by the chain rule,

equation

where . The sets of partial derivatives can be represented as matrices and . Let,

equation

(the components of these matrices, and the right-hand-sides of Eqs. (2.2) are given in the appendix). Equations 2.2 can then be written as,

equation

with the different represented by the different rows of this matrix equation.

3. TRANSFORMING TO FIRST ORDER

For the purposes of numerical solution, the equations of motion are best represented as a set of first order equations. Let . This has the consequence that . Writing the state vector with the four components and , and the equation set becomes,

equation

In Eq. (3.1), each quadrant of the large matrix is itself a matrix. The top two rows are simply a definition of the change of variables, and the bottom two rows are the Lagrangian equations of motion (Eq. (2.4) rearranged). The inverse of is easy to calculate as the matrix is of only small rank (see the appendix).

4. RUNGE-KUTTA SOLUTION

For convenience we will write Eq. (3.1) in the simplified form of,

equation

where the vector represents the four variables pertaining to the angles and their rates of change. The matrix and the vector are functions of the state variable (denoted by the subscript). We will solve the equations using the standard fourth order Runge-Kutta solution [1]. This has five stages,

1. Calculate :

2. Calculate :

3. Calculate :

4. Calculate :

5. Calculate :

The value is set by the initial conditions (as specified).

5. APPENDIX

We note the partial derivatives of (Eq. (1.7)),

equation

equation

equation

equation

equation

equation

equation

equation

equation

equation

Noteworthy also is the inverse of matrix . If,

equation

then the inverse is,

equation

6. REFERENCE

[1] Press W.H., Teukolsky S.A., Vettering W.T., Flannery B.P., Numerical Recipes in Fortran (section 16.1), Cambridge University Press (1994).



Link to Ross' Home Page
Link to Ross' Mechanics Page