13.1.1.3 Conversion from implicit to parametric form

There are trade-offs between the implicit and parametric ways to express differential constraints. The implicit representation is more general; however, the parametric form is more useful because it explicitly gives the possible actions. For this reason, it is often desirable to derive a parametric representation from an implicit one. Under very general conditions, it is theoretically possible. As will be explained shortly, this is a result of the implicit function theorem. Unfortunately, the theoretical existence of such a conversion does not help in actually performing the transformations. In many cases, it may not be practical to determine a parametric representation.

To model a mechanical system, it is simplest to express constraints in the implicit form and then derive the parametric representation ${\dot q}=f(q,u)$. So far there has been no appearance of $u$ in the implicit representation. Since $u$ is interpreted as an action, it needs to be specified while deriving the parametric representation. To understand the issues, it is helpful to first assume that all constraints in implicit form are linear equations in ${\dot q}$ of the form

$$g_1(q) {\dot q}_1 + g_2(q) {\dot q}_2 + \cdots + g_n(q) {\dot q}_n = 0,$$ (13.5)

which are called Pfaffian constraints. These constraints are linear only under the assumption that $q$ is known. It is helpful in the current discussion to imagine that $q$ is fixed at some known value, which means that each of the $g_i(q)$ coefficients in (13.5) is a constant.

Suppose that $k$ Pfaffian constraints are given for $k \leq n$ and that they are linearly independent.^13.1 Recall the standard techniques for solving linear equations. If $k=n$, then a unique solution exists. If $k < n$, then a continuum of solutions exists, which forms an $(n-k)$-dimensional hyperplane. It is impossible to have $k > n$ because there can be no more than $n$ linearly independent equations.

If $k=n$, only one velocity vector satisfies the constraints for each $q \in {\cal C}$. A vector field can therefore be derived from the constraints, and the problem is not interesting from a planning perspective because there is no choice of velocities. If $k < n$, then $n-k$ components of ${\dot q}$ can be chosen independently, and then the remaining $k$ are computed to satisfy the Pfaffian constraints (this can be accomplished using linear algebra techniques such as singular value decomposition [399,961]). The components of ${\dot q}$ that can be chosen independently can be considered as $n-k$ scalar actions. Together these form an $(n-k)$-dimensional action vector, $u = (u_1,\ldots,u_{n-k})$. Suppose without loss of generality that the first $n-k$ components of ${\dot q}$ are specified by $u$. The configuration transition equation can then be written as

$${\dot q}_1 = u_1 \qquad {\dot q}_{n-k+1} = f_{n-k+1}(q,u)$$

$${\dot q}_2 = u_2 \qquad {\dot q}_{n-k+2} = f_{n-k+2}(q,u)$$

$$\vdots \qquad \vdots$$ (13.6)

$${\dot q}_{n-k} = u_{n-k} \qquad {\dot q}_n = f_n(q,u) ,$$

in which each $f_i$ is a linear function of $u$ and is derived from the Pfaffian constraints after substituting $u_i$ for ${\dot q}_i$ for each $i$ from $1$ to $n-k$ and then solving for the remaining components of ${\dot q}$. For some values of $q$, the constraints may become linearly dependent. This only weakens the constraints, which means the dimension of $u$ can be increased at any $q$ for which independence is lost. Such points are usually isolated and will not be considered further.

Example 13..2 (Pfaffian Constraints)   Suppose that ${\cal C}= {\mathbb{R}}^3$, and there is one constraint of the form (13.5)

$$2 {\dot q}_1 - {\dot q}_2 - {\dot q}_3 = 0.$$ (13.7)

For this problem, $n=3$ and $k=1$. There are two action variables because $n-k = 2$. The configuration transition equation is

$$\begin{split}{\dot q}_1 & = u_1 {\dot q}_2 & = u_2 {\dot q}_3 &= 2 u_1 - u_2, \end{split}$$ (13.8)

in which the last component was obtained by substituting $u_1$ and $u_2$, respectively, for ${\dot q}_1$ and ${\dot q}_2$ in (13.7) and then solving for ${\dot q}_3$.

The constraint given in (13.7) does not even depend on $q$. The same ideas apply for more general Pfaffian constraints, such as

$$(\cos q_3) {\dot q}_1 - (\sin q_3) {\dot q}_2 - {\dot q}_3 = 0.$$ (13.9)

Following the same procedure, the configuration transition equation becomes

$$\begin{split}{\dot q}_1 & = u_1 {\dot q}_2 & = u_2 {\dot q}_3 &= (\cos q_3) u_1 - (\sin q_3) u_2 . \end{split}$$ (13.10)

$\blacksquare$

The ideas presented so far naturally extend to equality constraints that are not linear in ${\dot x}$. At each $q$, an $(n-k)$-dimensional set of actions, $U(q)$, is guaranteed to exist if the Jacobian $\partial (g_1,\ldots,g_k)/\partial ({\dot q}_1,\ldots,{\dot q}_n)$ (recall (6.28) or see [508]) of the constraint functions has rank $k$ at $q$. This follows from the implicit function theorem [508].

Suppose that there are inequality constraints of the form $g(q,{\dot q}) \leq 0$, in addition to equality constraints. Using the previous concepts, the actions may once again be assigned directly as ${\dot q}_i = u_i$ for all $i$ such that $1 \leq i \leq n-k$. Without inequality constraints, there are no constraints on $u$, which means that $U = {\mathbb{R}}^n$. Since $u$ is interpreted as an input to some physical system, $U$ will often be constrained. In a physical system, for example, the amount of energy consumed may be proportional to $u$. After performing the ${\dot q}_i = u_i$ substitutions, the inequality constraints indicate limits on $u$. These limits are expressed in terms of $q$ and the remaining components of ${\dot q}$, which are the variables ${\dot q}_{n-k+1}$, $\ldots$, ${\dot q}_n$. For many problems, the inequality constraints are simple enough that constraints directly on $U$ can be derived. For example, if $u_1$ represents scalar acceleration applied to a car, then it may have a simple bound such as $\vert u_1\vert \leq 1$.

One final complication that sometimes occurs is that the action variables may already be specified in the equality constraints: $g(q,{\dot q},u) = 0$. In this case, imagine once again that $q$ is fixed. If there are $k$ independent constraints, then by the implicit function theorem, ${\dot q}$ can be solved to yield ${\dot q}=f(q,u)$ (although theoretically possible, it may be difficult in practice). If the Jacobian $\partial (f_1,\ldots,f_n)/\partial (u_1,\ldots,u_k)$ has rank $k$ at $q$, then actions can be applied to yield any velocity on a $k$-dimensional hyperplane in $T_q({\cal C})$. If $k=n$, then there are enough independent action variables to overcome the constraints. Any velocity in $T_q({\cal C})$ can be achieved through a choice of $u$. This is true only if there are no inequality constraints on $U$.