Dynamical systems notation


by Rasalhague
Tags: dynamical, notation, systems
Rasalhague
Rasalhague is offline
#1
Jul26-11, 02:24 PM
P: 1,400
I have some questions about what I think is a fairly standard and common short-hand notation used in physics.

Today I watched lecture 2 in the nptelhrd series Classical Physics by Prof. V. Balakrishnan. In it, he models a kind of system called a simple harmonic oscillator, I think using [itex]TC = C \times \mathbb{R} = \mathbb{R}^2[/itex] for a state space (He calls it phase space, but I'll use the more general name, as phase space is said elsewhere to have a coordinate called "momentum" whereas he calls the corresponding coordinate "velocity".), where [itex]C[/itex] is the configuration space of the system, and [itex]TC[/itex] the tangent bundle thereon. He labels points in state space with [itex]q[/itex] and [itex]\dot{q}[/itex], thus [itex](q,\dot{q}) \in \mathbb{R}^2[/itex]. So far so good. Then he writes some equations:

[tex]\ddot{q}=-\omega q, \enspace\enspace\enspace V(q)=\frac{1}{2}m\omega^2q^2, \enspace\enspace\enspace m\ddot{q}=-\frac{\mathrm{d} V}{\mathrm{d} q}(q);[/tex]

[tex]\dot{q}=v, \enspace\enspace\enspace \dot{v}=-\frac{V'(q)}{m}, \enspace\enspace\enspace\frac{\mathrm{d} v}{\mathrm{d} q}=-\frac{\omega^2}{v}q.[/tex]

I'm not satisfied that I understand all of these symbols.

I think [itex]\omega = \sqrt{k/m}[/itex] and [itex]m[/itex] are constants (angular velocity and mass). I think [itex]\ddot{q}[/itex] should mean the value at [itex]t[/itex] of the second derivative of some function whose value at [itex]t[/itex] is labelled [itex]q[/itex]. I'm guessing this implicit function is the first component function, [itex]\gamma_1[/itex], of a curve function, [itex]\gamma : \mathbb{R} \rightarrow \mathbb{R}^2 \; |\; t \mapsto (\gamma_1(t),\gamma_2(t))[/itex], whose image is a trajectory in state space, and that this is an arbitrary element of the set of trajectories defined by the differential equation(s). I think [itex]V : \mathbb{R} \rightarrow \mathbb{R}[/itex] is a scalar field on the configuration space [itex]C = \mathbb{R}[/itex].

Does [itex]\dot{q}=v[/itex] mean [itex]\gamma_2(t)=f\circ\gamma_1(t)[/itex] for some unknown function [itex]f:\mathbb{R}\rightarrow \mathbb{R}[/itex]?

If so, does does [itex]\dot{v}[/itex] mean [itex](f\circ\gamma_1)'(t)[/itex] or [itex]f'\circ\gamma_1(t)[/itex]? I'm guessing the latter.

Is [itex]-\frac{V'(q)}{m}[/itex] to be read as [itex]-\frac{(V\circ\gamma_1)'(t)}{m}[/itex] or [itex]-\frac{V'\circ\gamma_1(t)}{m}[/itex]? Again, I'd guess the latter.

How about the final equation?

[tex]\frac{\mathrm{d} v}{\mathrm{d} q}=-\frac{\omega^2}{v}q[/tex]

Is it

[itex]f'\circ\gamma_1(t)=-\frac{\omega^2}{f\circ\gamma_1(t)}\gamma_1(t) \enspace ?[/itex]
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Rasalhague
Rasalhague is offline
#2
Jul26-11, 11:33 PM
P: 1,400
I think I've got it now. He writes points in the image of the curve as (q(t),v(t)), meaning [itex]\gamma[t]=(\gamma_1[t],\gamma_2[t])[/itex]. (I'll use square brackets here around the arguments of a function, to disambiguate them from the rounded brackets used to show order of operations.) His equations [itex]\dot{q}=v[/itex] and [itex]\dot{v}=-V'(q)/m[/itex] mean

[tex]\gamma'[t]:=(\gamma_1'[t],\gamma_2'[t])=\left ( \gamma_2[t],-\frac{V'\circ\gamma_1[t]}{m} \right )[/tex]

So, although "position"coordinate [itex]q[/itex] and "velocity"coordinate [itex]\dot{q}[/itex] are independent variables for functions whose domain is the state space, "position"particle [itex]\gamma_1[/itex] and "velocity"particle [itex]\gamma_2 = \gamma_1'[/itex] are functions related in the familiar way, the latter being always the derivative of the former, in any dynamical system. It just happens the same names, and often symbols, are used for both concepts.

Finally,

[tex]\frac{\mathrm{d} v}{\mathrm{d} q}[/tex]

can be analysed, with the Leibniz notation for the single-variable chain rule in mind, as

[tex]\frac{\mathrm{d} v}{\mathrm{d} t}\frac{\mathrm{d} t}{\mathrm{d} q}[/tex]

which, all being well, means

[tex](\gamma_2\circ(\gamma_1^{-1}))'\circ\gamma_1[t][/tex]

[tex]=(\gamma_2'\circ(\gamma_1)^{-1}\circ\gamma_1)[t]\cdot((\gamma^{-1})'\circ\gamma_1)[t][/tex]

[tex]=\gamma_2'[t]\cdot((\gamma^{-1})'\circ\gamma_1)[t][/tex]

[tex]=\frac{\gamma_2'[t]}{\gamma_1'[t]},[/tex]

so that, in this context, an expression like [itex]\mathrm{d}f[/itex] can be read as another notation for [itex]f'[/itex].
Rasalhague
Rasalhague is offline
#3
Jul26-11, 11:48 PM
P: 1,400
Proof of the identity used in the final step. Let [itex]y=f[x][/itex] such that [itex]x=g[y][/itex], where [itex]x[/itex] and [itex]y[/itex] are arbitrary real numbers. Then

[tex](g\circ f)'[x]=((g'\circ f)[x])\cdot(f'[x]).[/tex]

But [itex]g\circ f[x]=g[y]=x[/itex], so [itex]g\circ f[/itex] is the identity function on [itex]\mathbb{R}[/itex]. So

[tex](Id)'[x]=1=((g'\circ f)[x])\cdot(f'[x]),[/tex]

so, for [itex]f'[x]\neq 0[/itex],

[tex]g'[y]=\frac{1}{f'[x]}.[/tex]

That is:

[tex](f^{-1})'\circ f[x]=\frac{1}{f'[x]}.[/tex]

Hence the notation

[tex]\frac{\mathrm{d} x}{\mathrm{d} y}=\frac{1}{(\frac{\mathrm{d} y}{\mathrm{d} x})}.[/tex]

Rasalhague
Rasalhague is offline
#4
Jul27-11, 01:27 PM
P: 1,400

Dynamical systems notation


Quote Quote by Rasalhague View Post
angular velocity
Oopsh, no not angular velocity, just a parameter, a constant of the system.
Rasalhague
Rasalhague is offline
#5
Jul31-11, 12:22 PM
P: 1,400
I think my interpretation of the ideas is right, but from this thread (see especially Fredrik's post #19), it seems I may have misunderstood what role the notation [itex]q[/itex] and [itex]\dot q[/itex] play, and that they're really synonymous with [itex]\gamma_1[/itex] and [itex]\gamma_2=\gamma_1'[/itex], but are traditionally used sloppily also to denote the value of these functions. However, I'm still troubled by the widespread insistence that they stand for "independent variables" (which obviously isn't the case if they're defined as [itex]\gamma_1[/itex] and [itex]\gamma_2=\gamma_1'[/itex], or even the values of these functions). Balakrishnan talks about them as independent variables, and Roger Penrose calls them independent variables, in the quote in #23 of the thread I linked to. Penrose also seems to be treating them as (natural?) coordinate functions on the state space. But perhaps he's simultaneously letting them denote the coordinate representations of curves through the state space...


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