Gantry crane

As shown on the right, a gantry crane is an overhead crane that is able to move its hook in 3 axes as indicated by the arrows. Intuitively, we can deduce that the crane should be a holonomic system as, for a given movement of its components, it doesn't matter what order or velocity the components move: as long as the total displacement of each component from a given starting condition is the same, all parts and the system as a whole will end up in the same state. Mathematically we can prove this as such:

We can define the configuration space of the system as:

${\displaystyle {\overrightarrow {u}}={\begin{bmatrix}x\\y\\z\end{bmatrix}}}$

We can say that the deflection of each component of the crane from its "zero" position are ${\displaystyle B}$, ${\displaystyle G}$, and ${\displaystyle O}$, for the blue, green, and orange components, respectively. The orientation and placement of the coordinate system does not matter in whether a system is holonomic, but in this example the components happen to move parallel to its axes. If the origin of the coordinate system is at the back-bottom-left of the crane, then we can write the position constraint equation as:

${\displaystyle (x-B)+(y-G)+(z-(h-O))=0}$

Where ${\displaystyle h}$ is the height of the crane. Optionally, we may simplify to the standard form where all constants are placed after the variables:

${\displaystyle x+y+z-(B+G+h-O)=0}$

Because we have derived a constraint equation in holonomic form (specifically, our constraint equation has the form ${\displaystyle f(x,y,z)=0}$ where ${\displaystyle \{x,y,z\}\in {\overrightarrow {u}}}$), we can see that this system must be holonomic.

Pendulum

As shown on the right, a simple pendulum is a system composed of a weight and a string. The string is attached at the top end to a pivot and at the bottom end to a weight. Being inextensible, the string’s length is a constant. This system is holonomic because it obeys the holonomic constraint

${\displaystyle {x^{2}+y^{2}}-L^{2}=0,}$

where ${\displaystyle (x,\ y)}$ is the position of the weight and ${\displaystyle L}$ is length of the string.

Rigid body

The particles of a rigid body obey the holonomic constraint

${\displaystyle (\mathbf {r} _{i}-\mathbf {r} _{j})^{2}-L_{ij}^{2}=0,\,}$

where ${\displaystyle \mathbf {r} _{i}}$, ${\displaystyle \mathbf {r} _{j}}$ are respectively the positions of particles ${\displaystyle P_{i}}$ and ${\displaystyle P_{j}}$, and ${\displaystyle L_{ij}}$ is the distance between them. If a given system is holonomic, rigidly attaching additional parts to components of the system in question cannot make it non-holonomic, assuming that the degrees of freedom are not reduced (in other words, assuming the configuration space is unchanged).

Example

Consider this dynamical system described by a constraint equation in Pfaffian form.

${\displaystyle (\cos \theta )dx+(\sin \theta )dy+(y\cos \theta -x\sin \theta )d\theta =0}$

The configuration space, by inspection, is ${\displaystyle {\overrightarrow {u}}={\begin{bmatrix}x&y&\theta \end{bmatrix}}^{\mathrm {T} }}$. Because there are only three terms in the configuration space, there will be only one test equation needed. We can organize the terms of the constraint equation as such, in preparation for substitution:

${\displaystyle A_{\alpha }=\cos \theta }$

${\displaystyle A_{\beta }=\sin \theta }$

${\displaystyle A_{\gamma }=y\cos \theta -x\sin \theta }$

${\displaystyle u_{\alpha }=dx}$

${\displaystyle u_{\beta }=dy}$

${\displaystyle u_{\gamma }=d\theta }$

Substituting the terms, our test equation becomes:

${\displaystyle (y\cos \theta -x\sin \theta ){\bigg (}{\frac {\partial }{\partial x}}(\sin \theta )-{\frac {\partial }{\partial y}}(\cos \theta ){\bigg )}+(\sin \theta ){\bigg (}{\frac {\partial }{\partial \theta }}(\cos \theta )-{\frac {\partial }{\partial x}}(y\cos \theta -x\sin \theta ){\bigg )}+(\cos \theta ){\bigg (}{\frac {\partial }{\partial y}}(y\cos \theta -x\sin \theta )-{\frac {\partial }{\partial \theta }}(\sin \theta ){\bigg )}=0}$

After calculating all partial derivatives, we get:

${\displaystyle (y\cos \theta -x\sin \theta ){\bigg (}0-0{\bigg )}+(\sin \theta ){\bigg (}-\sin \theta -(-\sin \theta ){\bigg )}+(\cos \theta ){\bigg (}\cos \theta -\cos \theta {\bigg )}=0}$

Simplifying, we find that:

${\displaystyle 0=0}$

We see that our test equation is true, and thus, the system must be holonomic.

We have finished our test, but now knowing that the system is holonomic, we may wish to find the holonomic constraint equation. We can attempt to find it by integrating each term of the Pfaffian form and attempting to unify them into one equation, as such:

${\displaystyle \int (\cos \theta )dx=x\cos \theta +f(y,\theta )}$

${\displaystyle \int (\sin \theta )dy=y\sin \theta +f(x,\theta )}$

${\displaystyle \int (y\cos \theta -x\sin \theta )d\theta =y\sin \theta +x\cos \theta +f(x,y)}$

It's easy to see that we can combine the results of our integrations to find the holonomic constraint equation:

${\displaystyle y\sin \theta +x\cos \theta +C=0}$

where C is the constant of integration.

Constraints of constant coefficients

For a given Pfaffian constraint where every coefficient of every differential is a constant, in other words, a constraint in the form:

${\displaystyle \sum _{j}\ A_{ij}\,du_{j}+A_{i}\,dt=0;\;\{A_{ij},A_{i};\,j=1,2,\ldots$ ;\,i=1,2,\ldots \}\in \mathbb {R} }

the constraint must be holonomic.

We may prove this as follows: consider a system of constraints in Pfaffian form where every coefficient of every differential is a constant, as described directly above. To test whether this system of constraints is holonomic, we use the universal test. We can see that in the test equation, there are three terms that must sum to zero. Therefore, if each of those three terms in every possible test equation are each zero, then all test equations are true and this the system is holonomic. Each term of each test equation is in the form:

${\displaystyle A_{3}\left({\frac {\partial A_{2}}{\partial u_{1}}}-{\frac {\partial A_{1}}{\partial u_{2}}}\right)}$

where:

• ${\displaystyle A_{1}}$, ${\displaystyle A_{2}}$, and ${\displaystyle A_{3}}$ are some combination (with ${\textstyle {\binom {n}{3}}=n(n-1)(n-2)/6}$ total combinations) of ${\displaystyle A_{ij};\,j=1,2,\ldots }$ and ${\displaystyle A_{i}}$ for a given constraint ${\displaystyle i}$.
• ${\displaystyle u_{1}}$, ${\displaystyle u_{2}}$, and ${\displaystyle u_{3}}$ are the corresponding combination of ${\displaystyle u_{j};\,j=1,2,\ldots }$ and ${\displaystyle dt}$.

Additionally, there are ${\displaystyle i}$ sets of test equations.

We can see that, by definition, all ${\displaystyle A_{n}}$ are constants. It is well-known in calculus that any derivative (full or partial) of any constant is ${\displaystyle 0}$. Hence, we can reduce each partial derivative to:

${\displaystyle A_{3}{\big (}0-0{\big )}}$

and hence each term is zero, the left side each test equation is zero, each test equation is true, and the system is holonomic.

Configuration spaces of two or one variable

Any system that can be described by a Pfaffian constraint and has a configuration space or state space of only two variables or one variable is holonomic.

We may prove this as such: consider a dynamical system with a configuration space or state space described as:

${\displaystyle {\overrightarrow {u}}={\begin{bmatrix}u_{1}&u_{2}\end{bmatrix}}^{\mathrm {T} }}$

if the system is described by a state space, we simply say that ${\displaystyle u_{2}}$ equals our time variable ${\displaystyle t}$. This system will be described in Pfaffian form:

${\displaystyle A_{i1}du_{1}+A_{i2}du_{2}=0}$

with ${\displaystyle i}$ sets of constraints. The system will be tested by using the universal test. However, the universal test requires three variables in the configuration or state space. To accommodate this, we simply add a dummy variable ${\displaystyle \lambda }$ to the configuration or state space to form:

${\displaystyle {\overrightarrow {u}}={\begin{bmatrix}u_{1}&u_{2}&\lambda \end{bmatrix}}^{\mathrm {T} }}$

Because the dummy variable ${\displaystyle \lambda }$ is by definition not a measure of anything in the system, its coefficient in the Pfaffian form must be ${\displaystyle 0}$. Thus we revise our Pfaffian form:

${\displaystyle A_{i1}du_{1}+A_{i2}du_{2}+0d\lambda =0}$

Now we may use the test as such, for a given constraint ${\displaystyle i}$ if there are a set of constraints:

${\displaystyle (0){\bigg (}{\frac {\partial }{\partial u_{1}}}(A_{i2})-{\frac {\partial }{\partial u_{2}}}(A_{i1}){\bigg )}+(A_{i2}){\bigg (}{\frac {\partial }{\partial \lambda }}(A_{i1})-{\frac {\partial }{\partial u_{1}}}(0){\bigg )}+(A_{i1}){\bigg (}{\frac {\partial }{\partial u_{2}}}(0)-{\frac {\partial }{\partial \lambda }}(A_{i2}){\bigg )}=0}$

Upon realizing that :${\displaystyle {\frac {\partial }{\partial \lambda }}(f(u_{1},u_{2}))=0}$ because the dummy variable ${\displaystyle \lambda }$ cannot appear in the coefficients used to describe the system, we see that the test equation must be true for all sets of constraint equations and thus the system must be holonomic. A similar proof can be conducted with one actual variable in the configuration or state space and two dummy variables to confirm that one-degree-of-freedom systems describable in Pfaffian form are also always holonomic.

In conclusion, we realize that even though it is possible to model nonholonomic systems in Pfaffian form, any system modellable in Pfaffian form with two or fewer degrees of freedom (the number of degrees of freedom is equal to the number of terms in the configuration space) must be holonomic.

Important note: realize that the test equation failed because the dummy variable, and hence the dummy differential included in the test, will differentiate anything that is a function of the actual configuration or state space variables to ${\displaystyle 0}$. Having a system with a configuration or state space of:

${\displaystyle {\overrightarrow {u}}={\begin{bmatrix}u_{1}&u_{2}&u_{3}\end{bmatrix}}^{\mathrm {T} }}$

and a set of constraints where one or more constraints are in the Pfaffian form:

${\displaystyle A_{i1}du_{1}+A_{i2}du_{2}+0du_{3}=0}$

does not guarantee the system is holonomic, as even though one differential has a coefficient of ${\displaystyle 0}$, there are still three degrees of freedom described in the configuration or state space.