Linearization in Physics/Mechanics

mechanics includes methods for linearizing the generated equations of motion (EOM) about an operating point (also known as the trim condition). Note that this operating point doesn’t have to be an equilibrium position, it just needs to satisfy the equations of motion.

Linearization is accomplished by taking the first order Taylor expansion of the EOM about the operating point. When there are no dependent coordinates or speeds this is simply the jacobian of the right hand side about \(q\) and \(u\). However, in the presence of constraints more care needs to be taken. The linearization methods provided here handle these constraints correctly.

Background

In mechanics we assume all systems can be represented in the following general form:

\[\begin{split}f_{c}(q, t) &= 0_{l \times 1}\\ f_{v}(q, u, t) &= 0_{m \times 1}\\ f_{a}(q, \dot{q}, u, \dot{u}, t) &= 0_{m \times 1}\\ f_{0}(q, \dot{q}, t) + f_{1}(q, u, t) &= 0_{n \times 1}\\ f_{2}(q, u, \dot{u}, t) + f_{3}(q, \dot{q}, u, r, t) + f_{4}(q, \lambda, t) &= 0_{(o-m+k) \times 1}\end{split}\]

where

\[\begin{split}q, \dot{q} & \in \mathbb{R}^n\\ u, \dot{u} & \in \mathbb{R}^o\\ r & \in \mathbb{R}^s\\ \lambda & \in \mathbb{R}^k\end{split}\]

In this form,

  • \(f_{c}\) represents the configuration constraint equations
  • \(f_{v}\) represents the velocity constraint equations
  • \(f_{a}\) represents the acceleration constraint equations
  • \(f_{0}\) and \(f_{1}\) form the kinematic differential equations
  • \(f_{2}\), \(f_{3}\), and \(f_{4}\) form the dynamic differential equations
  • \(q\) and \(\dot{q}\) are the generalized coordinates and their derivatives
  • \(u\) and \(\dot{u}\) are the generalized speeds and their derivatives
  • \(r\) is the system inputs
  • \(\lambda\) is the Lagrange multipliers

This generalized form is held inside the Linearizer class, which performs the actual linearization. Both KanesMethod and LagrangesMethod objects have methods for forming the linearizer using the to_linearizer class method.

A Note on Dependent Coordinates and Speeds

If the system being linearized contains constraint equations, this results in not all generalized coordinates being independent (i.e. \(q_1\) may depend on \(q_2\)). With \(l\) configuration constraints, and \(m\) velocity constraints, there are \(l\) dependent coordinates and \(m\) dependent speeds.

In general, you may pick any of the coordinates and speeds to be dependent, but in practice some choices may result in undesirable singularites. Methods for deciding which coordinates/speeds to make dependent is behind the scope of this guide. For more information, please see [Blajer1994].

Once the system is coerced into the generalized form, the linearized EOM can be solved for. The methods provided in mechanics allow for two different forms of the linearized EOM:

\(M\), \(A\), and \(B\)

In this form, the forcing matrix is linearized into two separate matrices \(A\) and \(B\). This is the default form of the linearized EOM. The resulting equations are:

\[\begin{split}M \begin{bmatrix} \delta \dot{q} \\ \delta \dot{u} \\ \delta \lambda \end{bmatrix} = A \begin{bmatrix} \delta q_i \\ \delta u_i \end{bmatrix} + B \begin{bmatrix} \delta r \end{bmatrix}\end{split}\]

where

\[\begin{split}M &\in \mathbb{R}^{(n+o+k) \times (n+o+k)}\\ A &\in \mathbb{R}^{(n+o+k) \times (n-l+o-m)}\\ B &\in \mathbb{R}^{(n+o+k) \times s}\end{split}\]

Note that \(q_i\) and \(u_i\) are just the independent coordinates and speeds, while \(q\) and \(u\) contains both the independent and dependent coordinates and speeds.

\(A\) and \(B\)

In this form, the linearized EOM are brought into explicit first order form, in terms of just the independent coordinates and speeds. This form is often used in stability analysis or control theory. The resulting equations are:

\[\begin{split}\begin{bmatrix} \delta \dot{q_i} \\ \delta \dot{u_i} \end{bmatrix} = A \begin{bmatrix} \delta q_i \\ \delta u_i \end{bmatrix} + B \begin{bmatrix} \delta r \end{bmatrix}\end{split}\]

where

\[\begin{split}A &\in \mathbb{R}^{(n-l+o-m) \times (n-l+o-m)}\\ B &\in \mathbb{R}^{(n-l+o-m) \times s}\end{split}\]

To use this form set A_and_B=True in the linearize class method.

Linearizing Kane’s Equations

After initializing the KanesMethod object and forming \(F_r\) and \(F_r^*\) using the kanes_equations class method, linearization can be accomplished in a couple ways. The different methods will be demonstrated with a simple pendulum system:

>>> from sympy import symbols, Matrix
>>> from sympy.physics.mechanics import *
>>> q1 = dynamicsymbols('q1')                     # Angle of pendulum
>>> u1 = dynamicsymbols('u1')                     # Angular velocity
>>> q1d = dynamicsymbols('q1', 1)
>>> L, m, t, g = symbols('L, m, t, g')

>>> # Compose world frame
>>> N = ReferenceFrame('N')
>>> pN = Point('N*')
>>> pN.set_vel(N, 0)

>>> # A.x is along the pendulum
>>> A = N.orientnew('A', 'axis', [q1, N.z])
>>> A.set_ang_vel(N, u1*N.z)

>>> # Locate point P relative to the origin N*
>>> P = pN.locatenew('P', L*A.x)
>>> vel_P = P.v2pt_theory(pN, N, A)
>>> pP = Particle('pP', P, m)

>>> # Create Kinematic Differential Equations
>>> kde = Matrix([q1d - u1])

>>> # Input the force resultant at P
>>> R = m*g*N.x

>>> # Solve for eom with kanes method
>>> KM = KanesMethod(N, q_ind=[q1], u_ind=[u1], kd_eqs=kde)
>>> fr, frstar = KM.kanes_equations([pP], [(P, R)])

1. Using the Linearizer class directly:

A linearizer object can be created using the to_linearizer class method. This coerces the representation found in the KanesMethod object into the generalized form described above. As the independent and dependent coordinates and speeds are specified upon creation of the KanesMethod object, there is no need to specify them here.

>>> linearizer = KM.to_linearizer()

The linearized EOM can then be formed with the linearize method of the Linearizer object:

>>> M, A, B = linearizer.linearize()
>>> M
Matrix([
[1,       0],
[0, -L**2*m]])
>>> A
Matrix([
[                 0, 1],
[L*g*m*cos(q1(t)), 0]])
>>> B
Matrix(0, 0, [])

Alternatively, the \(A\) and \(B\) form can be generated instead by specifying A_and_B=True:

>>> A, B = linearizer.linearize(A_and_B=True)
>>> A
Matrix([
[                0, 1],
[-g*cos(q1(t))/L, 0]])
>>> B
Matrix(0, 0, [])

An operating point can also be specified as a dictionary or an iterable of dictionaries. This will evaluate the linearized form at the specified point before returning the matrices:

>>> op_point = {q1: 0, u1: 0}
>>> A_op, B_op = linearizer.linearize(A_and_B=True, op_point=op_point)
>>> A_op
Matrix([
[     0, 1],
[-g/L, 0]])

Note that the same effect can be had by applying msubs to the matrices generated without the op_point kwarg:

>>> assert msubs(A, op_point) == A_op

Sometimes the returned matrices may not be in the most simplified form. Simplification can be performed after the fact, or the Linearizer object can be made to perform simplification internally by setting the simplify kwarg to True.

2. Using the linearize class method:

The linearize method of the KanesMethod class is provided as a nice wrapper that calls to_linearizer internally, performs the linearization, and returns the result. Note that all the kwargs available in the linearize method described above are also available here:

>>> A, B, inp_vec = KM.linearize(A_and_B=True, op_point=op_point, new_method=True)
>>> A
Matrix([
[     0, 1],
[-g/L, 0]])

The additional output inp_vec is a vector containing all found dynamicsymbols not included in the generalized coordinate or speed vectors. These are assumed to be inputs to the system, forming the \(r\) vector described in the background above. In this example there are no inputs, so the vector is empty:

>>> inp_vec
Matrix(0, 0, [])

What’s with the new_method kwarg?

Previous releases of SymPy contained a linearization method for \(KanesMethod`\) objects. This method is deprecated, and will be removed from future releases. Until then, you must set new_method=True in all calls to KanesMethod.linearize. After the old method is removed, this kwarg will no longer be needed.

Linearizing Lagrange’s Equations

Linearization of Lagrange’s equations proceeds much the same as that of Kane’s equations. As before, the process will be demonstrated with a simple pendulum system:

>>> # Redefine A and P in terms of q1d, not u1
>>> A = N.orientnew('A', 'axis', [q1, N.z])
>>> A.set_ang_vel(N, q1d*N.z)
>>> P = pN.locatenew('P', L*A.x)
>>> vel_P = P.v2pt_theory(pN, N, A)
>>> pP = Particle('pP', P, m)

>>> # Solve for eom with Lagrange's method
>>> Lag = Lagrangian(N, pP)
>>> LM = LagrangesMethod(Lag, [q1], forcelist=[(P, R)], frame=N)
>>> lag_eqs = LM.form_lagranges_equations()

1. Using the Linearizer class directly:

A Linearizer object can be formed from a LagrangesMethod object using the to_linearizer class method. The only difference between this process and that of the KanesMethod class is that the LagrangesMethod object doesn’t already have its independent and dependent coordinates and speeds specified internally. These must be specified in the call to to_linearizer. In this example there are no dependent coordinates and speeds, but if there were they would be included in the q_dep and qd_dep kwargs:

>>> linearizer = LM.to_linearizer(q_ind=[q1], qd_ind=[q1d])

Once in this form, everything is the same as it was before with the KanesMethod example:

>>> A, B = linearizer.linearize(A_and_B=True, op_point=op_point)
>>> A
Matrix([
[     0, 1],
[-g/L, 0]])

2. Using the linearize class method:

Similar to KanesMethod, the LagrangesMethod class also provides a linearize method as a nice wrapper that calls to_linearizer internally, performs the linearization, and returns the result. As before, the only difference is that the independent and dependent coordinates and speeds must be specified in the call as well:

>>> A, B, inp_vec = LM.linearize(q_ind=[q1], qd_ind=[q1d], A_and_B=True, op_point=op_point)
>>> A
Matrix([
[     0, 1],
[-g/L, 0]])

Potential Issues

While the Linearizer class should be able to linearize all systems, there are some potential issues that could occur. These are discussed below, along with some troubleshooting tips for solving them.

1. Symbolic linearization with A_and_B=True is slow

This could be due to a number of things, but the most likely one is that solving a large linear system symbolically is an expensive operation. Specifying an operating point will reduce the expression size and speed this up. If a purely symbolic solution is desired though (for application of many operating points at a later period, for example) a way to get around this is to evaluate with A_and_B=False, and then solve manually after applying the operating point:

>>> M, A, B = linearizer.linearize()
>>> M_op = msubs(M, op_point)
>>> A_op = msubs(A, op_point)
>>> perm_mat = linearizer.perm_mat
>>> A_lin = perm_mat.T * M_op.LUsolve(A_op)
>>> A_lin
Matrix([
[     0, 1],
[-g/L, 0]])

The fewer symbols in A and M before solving, the faster this solution will be. Thus, for large expressions, it may be to your benefit to delay conversion to the \(A\) and \(B\) form until most symbols are subbed in for their numeric values.

2. The linearized form has nan, zoo, or oo as matrix elements

There are two potential causes for this. The first (and the one you should check first) is that some choices of dependent coordinates will result in singularities at certain operating points. Coordinate partitioning in a systemic manner to avoid this is beyond the scope of this guide; see [Blajer1994] for more information.

The other potential cause for this is that the matrices may not have been in the most reduced form before the operating point was substituted in. A simple example of this behavior is:

>>> from sympy import sin, tan
>>> expr = sin(q1)/tan(q1)
>>> op_point = {q1: 0}
>>> expr.subs(op_point)
nan

Note that if this expression was simplified before substitution, the correct value results:

>>> expr.simplify().subs(op_point)
1

A good way of avoiding this hasn’t been found yet. For expressions of reasonable size, using msubs with smart=True will apply an algorithm that tries to avoid these conditions. For large expressions though this is extremely time consuming.

>>> msubs(expr, op_point, smart=True)
1

Further Examples

The pendulum example used above was simple, but didn’t include any dependent coordinates or speeds. For a more thorough example, the same pendulum was linearized with dependent coordinates using both Kane’s and Lagrange’s methods: