• #Lecture 13: Dynamical Symmetry -- 1
  • #Noether's theorem :--
• #Lecture 14: (Finding dyn. symmetry group SU(n))
• #Lecture 15: (Randomness in phase-space)
• #Lecture 16
  • #Non linear Map~:--
  • #Moral of the story
  • #Integral equation :-

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Understanding the relation of symmetry \(\Leftrightarrow\) invariance \(\Rightarrow\) conservation laws. First step in this regard was taken by Emmy Noether.

(conserved quantities \(\vec P,\ \vec L,\ E(H)\ldots\))

Let us take a case of a particle moving in 2-D :--

\[ H(q_1,q_2,p_1,p_2) \;=\; \frac{\vec p^{\,2}}{2m} + V(q_1 q_2) \] \[ p^2 = p_1^2 + p_2^2 \]

If the potential has spherical symmetry (in 2-D its rotational) then we know that \(H\) is C.O.M.

\[ \text{if}\quad V(q_1,q_2) = f(r), \qquad r = \sqrt{q_1^2+q_2^2} \]

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In rotated frame of reference what remains constant :--

1. \(H\)
2. EOM don't change
3. Solution space/set do not change

Q) Does solution remains same/const?

Lets take simplest case

\[ \begin{aligned} x^2 &= 4\\ x^2-4 &= 0\\ (x-2)(x+2) &= 0\\ x &= 2,-2 \qquad \{-2,2\} \end{aligned} \]

after rota transformation

\[ \begin{aligned} x'^2 &= 4 \qquad x' = -x\\ (x'-2)(x'+2) &= 0\\ x' &= -2,+2 \qquad \{-2,2\} \end{aligned} \]

since *\(x' = -x\)*

we deduce that the \(+2\) root changed to \(-2\) and \(-2 \to +2\),

but set of solution remains constant/same

even though the elements of set of solution can convert into one another.

So the def\(^{\text{n}}\) of dynamical symmetry becomes,

\(\ast\) Dynamical symmetry :--

The dynamical symmetry of a given system is a set of transformations of phase space variables, such that EOM don't change and as a consequence, the set of solutions do not change.

(1) Understanding Dynamical symmetry in Lagrangian formalism :--

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EOM do not change if \(L' = L\)

or, Invariance \(\Rightarrow\) \(L' = L\)

\[ \frac{d L(Q(\alpha),\dot Q(\alpha))}{d\alpha} = 0 \] \[ \Rightarrow \quad \frac{\partial L}{\partial Q}\frac{dQ}{d\alpha} + \frac{\partial L}{\partial \dot Q}\frac{d\dot Q}{d\alpha} = 0 \] \[ \Rightarrow \quad \frac{d}{dt}\left(\frac{\partial L}{\partial \dot Q}\right)\frac{dQ}{d\alpha} + \frac{\partial L}{\partial \dot Q}\frac{d}{d\alpha}\left(\frac{dQ}{dt}\right) = 0 \]

\(\alpha, t\) are independent so

\[ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot Q}\right)\frac{dQ}{d\alpha} + \frac{\partial L}{\partial \dot Q}\frac{d}{dt}\left(\frac{dQ}{d\alpha}\right) = 0 \] \[ \Rightarrow \quad \frac{d}{dt}\left(\left(\frac{\partial L}{\partial \dot Q}\right)\cdot\left(\frac{dQ}{d\alpha}\right)\right) = 0 \] \[ \Rightarrow \quad \frac{\partial L}{\partial \dot Q}\left(\frac{dQ}{d\alpha}\right) = \text{const of motion} \]

in our frame

\[ \frac{\partial L}{\partial \dot Q}\left(\frac{dQ}{d\alpha}\right) \equiv \frac{\partial L}{\partial \dot q}\left(\frac{dQ}{d\alpha}\right)\bigg|_{\alpha=0} = p\,\frac{dQ}{d\alpha}\bigg|_{\alpha=0} = \underline{\text{COM}} \]

Remember, that Lagrangian \((L)\) is not unique, we can add total time derivative of some fun. of \(q,t\), therefore it is not necessary that

\[ \text{Invariance} \Rightarrow \underline{L' = L} \]

we can have \(L' \neq L\) still EOM don't change.

Note :-- The crucial point of Noether's theorem is that set of transformations should be continuous, that's why we differentiated with `\(\alpha\)', and it must be connected with Identity (original \((q,\dot q)\).) In other words if we have original coordinates and then we make a transformation continuously from no transformation at all.

Parity is not continuous transformation

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Symmetry :-- If Lagrangian is invariant under continuous set of transformations then Noether's theorem tells us \(\exists\) a conserved quantity for each transformation parameter.

eg: In shift of origin we \((x,y,z) \longrightarrow (x+\alpha,\ y+\beta,\ z+\gamma)\)

we have 3 parameters of transformation i.e. \(\alpha,\beta,\gamma\)

So according to Noethers theorem we will have 3 conserved quantity.

(2) Understanding in Hamiltonian Framework :--

\[ H(q,p): \qquad \dot q_i = \frac{\partial H}{\partial p_i}, \qquad \dot p_i = -\frac{\partial H}{\partial q_i} \qquad \boxed{\dot{\underset{\sim}{X}} = J\nabla H} \] \[ \uparrow \text{(set of EOM)} \] \[ \exists \quad \left(\begin{array}{l} \{q_i,q_j\} = 0 = \{p_i,p_j\}\\ \{q_i,p_j\} = \delta_{ij} \end{array}\right) \]

Q.) What set of transformations leaves Hamiltons EOM unchanged

Ans) canonical transformations.

\[ \dot Q_i = \frac{\partial K}{\partial P_i}, \qquad \dot P_i = -\frac{\partial K}{\partial Q_i} \qquad \boxed{\dot{\underset{\sim}{\xi}} = J\nabla_\xi K} \] \[ \uparrow \text{(EOM)} \]

also

\[ \begin{aligned} \{Q_i,P_j\} &= \delta_{ij}\\ \{Q_i,Q_j\} &= 0 = \{P_i,P_j\} \end{aligned} \]

to see set of solutions remain unchanged, we need only those canonical transformations for which *\(K=H\)*, else it won't be a symmetry. (How?)

Suppose \(\underset{\sim}{X} = (q,p) \;\xrightarrow{\ \text{C.T.}\ }\; \underset{\sim}{\xi} = (Q,P)\)

where \(\underset{\sim}{X},\ \underset{\sim}{\xi}\) are set of variables.

\[ \left. \begin{aligned} \{Q_i, P_j\} &= \delta_{ij} = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial q_k}\frac{\partial P_j}{\partial p_k} - \frac{\partial Q_i}{\partial p_k}\frac{\partial P_j}{\partial q_k}\right)\\ \{Q_i, Q_j\} &= 0 = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial q_k}\frac{\partial Q_j}{\partial p_k} - \frac{\partial Q_i}{\partial p_k}\frac{\partial Q_j}{\partial q_k}\right)\\ \{P_i, P_j\} &= 0 = \sum_{k=1}^{n}\left(\frac{\partial P_i}{\partial q_k}\frac{\partial P_j}{\partial p_k} - \frac{\partial P_i}{\partial p_k}\frac{\partial P_j}{\partial q_k}\right) \end{aligned} \right\} \Rightarrow \left(\frac{\partial \underline{\xi}}{\partial \underline{x}}\right)^{T} J \left(\frac{\partial \underline{\xi}}{\partial \underline{x}}\right) = J \]

all these relations must be valid to qualify for transformation to be canonical.

\[ J = \begin{pmatrix} O_n & I_n \\ -I_n & O_n \end{pmatrix} \qquad I_n,\, O_n \text{ are } n\times n \text{ matrices.} \]

any \(\{A,B\}\) can be written as, \(\{A,B\} = (\nabla A)^{T} J (\nabla B)\)

\(\left(M_{p\times q}\right)^{T} J_{m\times m} M_{p\times q}\), \(q = 2n\), \(p = 2n\)

If a matrix \((M)\) follows

\[ \text{if,}\quad M^{T} J M = J, \]

then \(M\) is called symplectic matrix

Recall,

\[ J^2 = -I, \qquad J^{T} = J^{-1} = -J \]

so symplectic matrix have inverse so canonical transformations Matrix is invertible. \(M = \left(\dfrac{\partial \underline{\xi}}{\partial \underline{x}}\right) = \) non-singular *\(|M| \neq 0\)*

\[ \Updownarrow \]

we know, canonical transformation must be inverse transformations,

also, Product of two symplectic matrix is also a symplectic matrix. (as set of symplectic matrices form a group).

\[ M_1 . M_2 = M_3 \qquad\qquad M_1, M_2, M_3 \to \text{symplectic matrices.} \]

physically it implies that two canonical transformations applied serielly are equivalant to a single canonical transformation.

Note:-- The inverse, composition law suggests set of canonical transformation for n-d.o.f Hamiltonian system form a group. It's the group of \(2n\times 2n\) symplectic matrices. Called

\[ Sp(2n, \mathbb{R}). \]

\(\hookrightarrow\) Group of C.T.

Note: J is also a symplectic matrix

\[ M^{T} J M = J \] \[ \text{LHS,} \Rightarrow J^{T} J J \Rightarrow (-J) J^{2} = (-J)(-I) = J = \underline{\text{RHS}} \]

1. what is C.T. coorresponding Symplectic matrix `J' ? \[ \left.\begin{aligned} Q &= -p \\ P &= q \end{aligned}\right\} \quad \Leftrightarrow \quad (M = J) \]

Note:-- We know that group of orthogonal matrices in 3-diamencions have three parameters, (for rotation)

\[ SO(n) \]

\(\hookrightarrow\) Group of orthogonal matrices in n-diamencion with deterrminent (1)

(Digression:)

1. How many parameters for n-D ?

Ans: (a) rotation in 2-D about point (origin).

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\[ SO(2); \quad \begin{pmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{pmatrix} \]

\(\downarrow\)

orthogonal.

set of transformation is homogeneous (origin unchanged).

2-D rotation is (linear, Homogeneous)

\[ \left(|SO(2)| = \underline{+1}\ ,\ \text{length unchanged.}\right) \]

Defining rotation :--

(for n-diamensions)

• linear
• Homogeneous
• distances unchanged \(\Rightarrow\) Matrice should be orthogonal
• \(|SO(n)| = \underline{1}\) (unimodular)

Note:-- In 3-D if we rotate by a single axis, it is not termed as rotation. as from def\(^{n}\) we have not defined rotation as rotation about a single axis. For Rotation \(T\) should be linear, Homogeneous, orthogonal, and unimodular.

Note:-- If the transformation is homogeneous (origin -- fixed) to achieve distances unchanged the transformation matrix should be orthogonal. and if we want orientation to be unchanged we need it to be unimodular. *\(|M| = 1\)*. So as we know \(\det(\text{orthogonal }m) = \underline{\pm 1}\). we choose *\(+1\)* in rotation. for reflection *\(|M| = -1\)*.

In 3-D we can do rotation in a plane, where two variables are affected at a time. For two variables affected at a time; no. of rotations are \(^{n}C_2\). for n-diamension.

no. of rotation \(\Rightarrow{}^{n}C_2 = \dfrac{n(n-1)}{2}\)

for \(n=3\), no. of rotations \(= \dfrac{3.2}{2} = 3\)

So we think (mistakenly) that rotation is occuring about a axis, but really we mean is rotation in \(xz\) plane, \(yz\) plane, \(xy\) plane.

Hence, only for 3-D \(\left(\dfrac{n(n-1)}{2} = n\right)\)

So \(SO(n)\) has \(^{n}C_2\) generators.

In exactly the same way,

Now we can ask how many generators does this symplectic group \(Sp(2n, R)\) have? (or) How many possible parameters we need to specify all possible canonical transformations.

let us make Infinitensimal transformation.

\[ M = I + \epsilon\, G_{(2n\times 2n)} \longrightarrow \text{Generator of Transformation matrix} \]

\(\hookleftarrow\) parameter

\[ \begin{aligned} & M^{T} J M = J\\ \Rightarrow\ & (I + \epsilon G)^{T} J (I + \epsilon G) = J\\ \Rightarrow\ & (I + \epsilon G^{T}) J (I + \epsilon G) = J\\ \Rightarrow\ & (J + \epsilon G^{T} J)(I + \epsilon G) = J\\ \Rightarrow\ & J + \epsilon G^{T} J + \epsilon J G + \epsilon^{2} G^{T} J G = J \qquad \text{dropping } \underline{\underline{\epsilon^{2}}}\\ \Rightarrow\ & J + \epsilon G^{T} J + \epsilon J G = J\\ \Rightarrow\ & \boxed{G^{T} J = -J G}\\ \Rightarrow\ & G^{T} = -J G J^{-1} \qquad (J^{-1} = -J)\\ \Rightarrow\ & \boxed{G^{T} = J G J} \end{aligned} \]

So Generator of symplectic transformation must satisfy above equation.

so symplectic matrices are generated by these matrices which satisfies \(G^{T} = JGJ\)

exercise \(\to\) Show that \(Sp(2n)\) has \(2(2n+1)\) generators.

what are generators? The way we look at it can be interpreted by looking at infinitesimal transformation.

\[ \begin{aligned} SO(2) = \begin{pmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha\end{pmatrix} \ \xrightarrow{\ (\alpha \to \delta\alpha)\ }\ &\begin{pmatrix} 1 & \delta\alpha \\ -\delta\alpha & 1\end{pmatrix}\\ = &\ I + \begin{pmatrix} 0 & \delta\alpha \\ -\delta\alpha & 0\end{pmatrix}\\ = &\ I + \delta\alpha \begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix} \end{aligned} \]

\(\hookrightarrow\) Generator.

\(\hookrightarrow\) parameter

for \(SO(3)\) we will have such 3 generators.

and for \(Sp(2n)\) we will have \(2(2n+1)\) generators.

Note:-- we actually like to write generators as Hermitian matrix. because we need most of time to exponentiate hermitian matrix. as, \(e^{iH} = \) unitary matrix.

Eg 2-D oscillator :--

\[ H = \tfrac{1}{2}\left(q_1^2 + p_1^2 + w^2 q_2^2 + p_2^2\right) \]

quasi periodic (\(w = \) irrational[?])

periodic but complicated (\(w = \) rational[?])

\(\downarrow\)

(\(w = \) rational no)

if \((w=1)\)

\[ H = \tfrac{1}{2}\left(q_1^2 + p_1^2 + q_2^2 + p_2^2\right) \]

isotropic (same \(k\) in all directio[?])

\(\hookrightarrow\) spring constant[?]

H is invariant under \(SO(4)\).

\(\hookrightarrow\) set of rotations in 4-D phase space.

Note:-- we could have dynamical symmetry arising as a consequence of transformations not just from of potentials (not just of physical coordinate) but it could also involve the momenta (phase--space).

\(\to\) no. of generators \(SO(4) = {}^{4}C_2 = 6\) --(1)

\(\to\) set of canonical transformations \((n=2) \to Sp(2\times 2)\)

\(Sp(4)\), has \(2(4+1) = 10\) generators. --(2)

From (1) and (2) we deduce that all transformations which leaves hamiltonian unchanged need not be canonical transformations, similarly all C.T need not leave `H' unchanged.

For symmetry we need transformation to be canonical and which leaves `H' unchanged. \((K = H)\)

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Dynamical symmetry group is

\[ Sp(4) \cap SO(4). \ \sim\ SU(2) \]

\(\hookrightarrow\) Isomorphic

\(SU(2)\) set of \(2\times 2\) matrices which are unitary and unimodular \((|M| = 1)\)

\[ U \text{ is unitary} \Rightarrow \boxed{U^{\dagger} U = I} \]

Q. Is it true \(U^{\dagger} U = I = U U^{\dagger}\)

Ans, Only true for finite diamentional matrices (as left inverse becomes equal to right inverse). But for infinitesimal matrices \(U^{\dagger}U = I \not\Rightarrow U U^{\dagger}\), But we are concerened with finite--diamentional matrices, so it does not matter.

\[ \begin{aligned} U^{\dagger} U &= I \ \Rightarrow\ \text{partial isometry}\\ U U^{\dagger} &= I \ \Rightarrow\ \text{\qquad `` \qquad ''} \end{aligned} \]

verify that following \(J_1, J_2\) are c.o.m are true.

\[ J_2 = \tfrac{1}{2}\left(q_1 q_2 + p_1 p_2\right) \qquad\qquad J_1 = \tfrac{1}{4}\left(q_1^2 + p_1^2 - q_2^2 - p_2^2\right) \] \[ J_3 = \tfrac{1}{2}\left(q_1 p_2 - q_2 p_1\right) \]

verify, \(\left(\{J_i, H\} = 0, \quad \{J_i, J_j\} = \epsilon_{ijk} J_k\right)\).

Recall 2-D isotropic oscillator :-- \(\left(\begin{aligned} m &= 1\\ w &= 1\end{aligned}\right)\)

\[ H = \tfrac{1}{2}\left(q_1^2 + p_1^2 + q_2^2 + p_2^2\right) \] \[ \begin{aligned} F_1 &= H_1 \qquad (q_1^2 + p_1^2)\\ F_2 &= H_2 \qquad (q_2^2 + p_2^2) \end{aligned} \]

its 2-dof and we have 2 c.o.M which are in involution with each other \((\{F_1, F_2\} = 0\). (Integrable)

Note:--

The dynamical symmetry will be that set of transformations of 4 phase space variables which leaves the H, EOM unchanged and hence preserves the solution set.

\(Sp(2n) = Sp(4) = \) Group of canonical Transformations for this H.

\(SO(4) = SO(4) = \) Group of rotations which leaves' H' unchanged

\(SU(n) = SU(2) = \) Group of C.T. which leaves `H' unchanged.

(also called as the dyn. symmetry of this H).

*Let's focus on \(SU(2)\). :--*

\(SU(2)\) consiste of all \(2\times 2\) matrices, \(M = \begin{pmatrix} a & b\\ c & d\end{pmatrix}\), such that

(\(\Rightarrow\) implies)

\[ \begin{aligned} M^{\dagger}M &= I\\ \det(M) &= +1 \end{aligned} \]

There are \(4\times 2 = 8\) parameters as each entry can be complex no.

\[ \begin{aligned} a &= a_1 + i a_2\\ b &= b_1 + i b_2\\ c &= c_1 + i c_2\\ d &= d_1 + i d_2 \end{aligned} \qquad (a_1, a_2 \text{--}\text{--} d_2) \]

\(\hookrightarrow\) 8 parameters.

using, \(M^{\dagger}M = I\) or \(M^{-1} = M^{\dagger}\)

\(\to\) we will have 4 conditions (one for each element).

Therefore no. of parameters are reduced to 4.

after imposing \(|M| = +1\), one more parameter is lost-- so total of three parameters are left.

(Digression) :--

\[ M^{\dagger} = M^{-1} \qquad\qquad M = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \] \[ M^{\dagger} = \begin{pmatrix} \alpha^{*} & \gamma^{*} \\ \beta^{*} & \delta^{*}\end{pmatrix} \qquad --(1) \] \[ M^{-1} = \frac{1}{|M|}\begin{pmatrix} \delta & -\gamma \\ -\beta & \alpha\end{pmatrix}^{T} = \frac{1}{(|M|)}\begin{pmatrix} \delta & -\beta \\ -\gamma & \alpha\end{pmatrix} \qquad --(2) \]

equating (1) and (2) and \(|M| = \alpha\delta - \beta\gamma = \underline{1}\)

\[ \begin{aligned} \alpha^{*} &= \delta && --(3)\\ \beta^{*} &= -\gamma && --(4)\\ \gamma^{*} &= -\beta && --(5)\\ \delta^{*} &= \alpha && --(6) \end{aligned} \] \[ \alpha^{*} = \delta,\ \delta^{*} = \alpha \ \Rightarrow\ \begin{aligned}(\alpha_1 + i\alpha_2)^{*} &= (\delta_1 + i\delta_2)\\ (\delta_1 + i\delta_2)^{*} &= \alpha_1 + i\alpha_2\end{aligned} \ \Rightarrow\ \begin{aligned}(\alpha_1 &= \delta_1)\\ \alpha_2 &= -\delta_2\end{aligned} \]

so M becomes

\[ M = \begin{pmatrix} \alpha & \beta \\ -\beta^{*} & \alpha^{*}\end{pmatrix}, \qquad \begin{aligned} |M| &= 1\\ \alpha\delta - \beta\gamma &= 1\\ \alpha\alpha^{*} + \beta\beta^{*} &= 1\\ |\alpha|^{2} + |\beta|^{2} &= 1 \end{aligned} \]

Hence, any \((2\times 2)\) unimodulor, unitary matrix can be written is

\[ \begin{pmatrix} \alpha & \beta \\ -\beta^{*} & \alpha^{*}\end{pmatrix} \quad\text{with,}\quad \left(|\alpha|^{2} + |\beta|^{2} = \underline{1}\right) \quad \begin{cases} \alpha = \alpha_1 + i\alpha_2\\ \beta = \beta_1 + i\beta_2\end{cases} \]

The Generators of \(SU(2)\) are c.o.M which are not in involution with each other. and turns out these are \(J_1, J_2, J_3\)

\[ \begin{aligned} J_1 &= \tfrac{1}{4}\left(q_1^2 + p_1^2 - q_2^2 - p_2^2\right)\\ J_2 &= \tfrac{1}{2}\left(q_1 q_2 + p_1 p_2\right)\\ J_3 &= \tfrac{1}{2}\left(q_1 p_2 - p_1 q_2\right) \end{aligned} \qquad \left.\begin{aligned} \{J_1, J_2\} &= J_3\\ \{J_2\, J_3\} &= J_1\\ \{J_3\, J_1\} &= J_2 \end{aligned}\right\} \begin{aligned}(\text{or})\ &\{J_i, J_j\}\\ &= \epsilon_{ijk} J_k\end{aligned} \]

(so 3 parameters require 3 generators )

1. How many independent parameters in \((n\times n)\) unitary matrix

Ans, \(2.n^2 - n^2 = \underline{n^2}\)

So, for \(SU(n)\) we have \((n^2-1)\) independent parameters where 1 parameter got reduced from \(|M| = \pm 1\) condition.

Therefore, i.e. \(\left(SU(3) \text{ has } 3^2 - 1 = 8 \text{ generators}\right)\)

For 3-D Isotropic oscillator :--

\(Sp(2n)\). \(Sp(6)\) is group of C.T.

\(SO(N)\) \(SO(6)\) is group of symmetry transformations

{ dimension of phasespace}

\(SU(8)\) is one required dynamical symmetry group.

The Kepler problem :--

\[ H(\vec{r}, \vec{p}) = \frac{\vec{p}^{\,2}}{2m} + \left(-\frac{k}{r}\right) \qquad \ldots\ \left(\begin{aligned}&\text{assuming interaction}\\ &\text{is attractive.}\end{aligned}\right) \]

Note:--

\(V(r) = \begin{cases} -k/r \\ r^2\end{cases}\) are the only potentials for which bounded orbits are closed curves.

\(\downarrow\)

\(\left(\begin{aligned}&\text{periodic}\\ &\text{orbits}\end{aligned}\right)\).

\(V(r) = r^2\) 3-D oscillator, with \(SU(3)\)

1. what is symmetry group of Kepler's hemiltonian?

C.O.M :-- \(H,\ \vec{L}^{2},\ \vec{L}\cdot\hat{n}\ \longrightarrow\) 3 independent c.o.m which are in involution with each other.

\(\vec{L}\) is c.o.m but \(\{L_i, L_j\} = \epsilon_{ijk} L_k\) \(\left(\begin{aligned}&\text{not in involution}\\ &\text{with each other}\end{aligned}\right)\).

Note

Any central potential problem will have these c.o.m but there is something special with kepler problem as there exist further C.O.M \((\vec{A})\).

\[ \vec{A} = \vec{p}\times\vec{L} - \frac{mk\vec{r}}{r} \qquad \left(A = f(L) \ \begin{aligned}&\text{so its not}\\ &\text{independent}\end{aligned}\right) \]

(laplace-- Runge --lenz vector)

(verifying \(\vec{A}\) is C.O.M) :-- \((\vec{L}\) is c.o.m\()\)

\[ \frac{d\vec{A}}{dt} = \frac{d\vec{p}}{dt}\times\vec{L} + \vec{p}\times\overset{0}{\frac{d\vec{L}}{dt}} + \frac{mk}{r^2}\frac{dr}{dt}\vec{r} - \frac{mk}{r}\frac{d\vec{r}}{dt} \] \[ \begin{aligned} \text{for } V &= -k/r\\ F &= -\nabla V\\ &= -\nabla(-k/r)\\ &= \nabla(k/r)\\ &= -k/r^{2}\,\hat{e}_r\\ \frac{d\vec{p}}{dt} &= F = \frac{-k}{r^{3}}\vec{r} \end{aligned} \] \[ \frac{d\vec{A}}{dt} = \frac{-k}{r^{3}}\vec{r}\times(\vec{r}\times\vec{p}) + \frac{mk}{r^{2}}\frac{dr}{dt}\vec{r} - \frac{mk}{r}\frac{d\vec{r}}{dt} \] \[ \frac{d\vec{A}}{dt} = \frac{-k}{r^{3}}\left(\vec{r}\times(\vec{r}\times\vec{p})\right) + \frac{mk}{r^{2}}\frac{dr}{dt}\vec{r} - \frac{k\vec{p}}{r} \qquad \left(\vec{p} = m\frac{d\vec{r}}{dt}\right) \] \[ \begin{aligned} \bar{a}\times(\bar{b}\times\bar{c}) &= \text{vector in plane of } \bar{b}, \bar{c}.\\ &= \alpha\vec{b} + \beta\vec{c} \end{aligned} \]

LHS is linear, \(\alpha, \beta\) are scalers formed out of \(a, b, c\) as LHS is linear RHS must be linear,

\[ \Rightarrow\ \begin{aligned}\alpha &= \lambda(\vec{a}\cdot\vec{c})\\ \beta &= \mu(\vec{a}\cdot\vec{b})\end{aligned} \]

$\begin{aligned}&\text{`}\lambda \ \& \ \mu\text{' should be}\\ &\text{universal, it can't}\\ &\text{depend on } \bar{A}, \bar{B}, \bar{C}.\end{aligned}$

\[ \bar{a}\times(\bar{b}\times\bar{c}) = \lambda(\bar{a}\cdot\bar{c})\vec{b} + \mu(\vec{a}\cdot\vec{b})\vec{c} \]

interchang \(b, c\).

\[ \vec{a}\times(\vec{c}\times\vec{b}) = -\,\bar{a}\times(\bar{b}\times\bar{c}) \]

only way it can happen is, \(\mu = -\lambda\)

\[ \bar{a}\times(\bar{b}\times\bar{c}) = \lambda\left((\bar{a}\cdot\bar{c})\vec{b} - (\vec{a}\cdot\vec{b})\vec{c}\right) \]

turns out \(\lambda = 1\) using special case \(\hat{\imath}\times(\hat{\imath}\times\hat{\jmath}) = \hat{\imath}\times\hat{k} = -\hat{\jmath}\)

\[ \lambda\left\{(\hat{\imath}\cdot\hat{\jmath})\hat{\imath} - \hat{\jmath}(\hat{\imath}\cdot\hat{\imath})\right\} = -\hat{\jmath} \ \Rightarrow\ \boxed{\lambda = 1} \] \[ \begin{aligned} \frac{d\vec{A}}{dt} &= \frac{-k}{r^{3}}\left[(\vec{r}\cdot\vec{p})\vec{r} - r^{2}\vec{p}\right] + \frac{mk}{r^{2}}\frac{dr}{dt}\vec{r} - \frac{k\vec{p}}{r}\\ &= \frac{-k}{r^{3}}(\vec{r}\cdot\vec{p})\vec{r} + \frac{k\vec{p}}{r} + \frac{mk}{r^{2}}\frac{dr}{dt}\vec{r} - \frac{k\vec{p}}{r}\\ &= \frac{-k(m)}{r^{3}}\left(\vec{r}\cdot\frac{d\vec{r}}{dt}\right)\vec{r} + \frac{mk}{r^{2}}\frac{dr}{dt}\vec{r} \qquad \left(\vec{u}\cdot\frac{d\vec{u}}{dt} = u\frac{du}{dt}\right)\\ &= \frac{-mk}{r^{3}}\,r\frac{dr}{dt}\vec{r} + \frac{mk}{r^{2}}\frac{dr}{dt}\vec{r} \end{aligned} \] \[ \frac{d\vec{A}}{dt} = 0 \qquad \text{therefore } \vec{A} \text{ is C.O.M.} \]

\(\vec{A}\) is c.o.m therefore it remains eunchanged in direction and megnitude.

\[ \vec{A} = \vec{p}\times(\vec{r}\times\vec{p}) - \frac{mk\,\vec{r}}{r} \]

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\begin{tikzpicture}[scale=0.75,>=Stealth]
  % --- left small figure ---
  \begin{scope}
    \draw (0,0) ellipse [x radius=1.7, y radius=0.75];
    \draw (-1.7,0) -- (2.9,0);
    \fill (-1.1,-0.1) circle (1.4pt);
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    \draw[->] (P) -- ++(0.55,0.22) node[right] {$\vec{r}$};
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  % --- right larger figure ---
  \begin{scope}[xshift=6.5cm]
    \draw (0,0) ellipse [x radius=2.9, y radius=1.15];
    \fill (-1.6,-0.15) circle (1.4pt);
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    \draw[->,very thick] (Q) -- ++(-0.1,-1.4);
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    \draw[->] (1.6,1.5) .. controls (2.9,1.5) and (3.3,0.6) .. (3.5,0.25);
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(dir. of \(\vec{A}\) lies along semi major axis.)

The \(\dfrac{d\vec{A}}{dt} = 0\), implies orbit does not show precession.

\(\left[\begin{aligned}&\text{Binary stars precess} \simeq 4^{\circ} \text{ per orbit}\\ &\text{(Precession of mercuury} \simeq 500^{\prime\prime} \text{ per century)}\end{aligned}\right]\)

So in total we got \(H, \vec{L}, \vec{A}\) total of \((1+3+3) = 7\) C.O.M but phase space is only 6-diamentional. It turns out that they are not independent of each other. \(\left(\begin{aligned}&\vec{A}^{2} \text{ is expressed in terms of } H \ \&\ \vec{L}^{2}\\ &\vec{A}\cdot\vec{L} = 0.\end{aligned}\right.\)

Note:-- \(\dfrac{d\vec{A}}{dt} = 0\) independent of sign of `k' so \(\exists\ \vec{A}\) for repulsive forces also.

\(*\) The dynamical symmetry group is \(SU(4)\). instead of \(SU(3)\) This extra symmetry is seen in terms of degeneracy in quantum mechanics.

This extra symmetry is because of \(\vec{A}\).

Problem 1)

Beed on a rotating hoop of wire.

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  \draw[->] (0,-0.6) -- (0,3.1) node[left] {$z$};
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  \draw[->] (1.25,0.8) -- (1.25,0.2);
  \node at (1.45,0.45) [right] {$mg$};
\end{tikzpicture}

\[ V = \begin{cases} 0 & \text{at } z = 0\\ mgz & z = z\end{cases} \] \[ L = T - V \] \[ L = \tfrac{1}{2}m\left(\dot{x}^2 + \dot{y}^2 + \dot{z}^2\right) - mgz \]

constraint \(\longrightarrow\)

\[ y^2 + (z-R)^2 = R^2 \]

The moment we provided \(\vec{w}\) to hoop we need cylindrical symmetry. \((s, \phi, z)\)

\[ \begin{aligned} x &= s\cos\phi, & \dot{x} &= \dot{s}\cos\phi - s\sin\phi\,\dot{\phi}\\ y &= s\sin\phi, & \dot{y} &= \dot{s}\sin\phi + s\cos\phi\,\dot{\phi}\\ z &= z, & \dot{z} &= \dot{z} \end{aligned} \]

so L becomes;

\[ L = \tfrac{1}{2}m\left[\dot{s}^2 + s^2\dot{\phi}^2 + \dot{z}^2\right] - mgz \qquad --(1) \]

constraint becomes

\[ (x^2 + y^2) + (z-R)^2 = R^2 \] \[ \Rightarrow\ s^2 + (z-R)^2 = R^2 \qquad --(2) \] \[ \text{also,}\quad \dot{\phi} = \omega \qquad --(3) \]

eq\(^{n}\) (2) implies, \(z = R \pm \sqrt{R^2 - s^2}\)

we should choose \(z = R - \sqrt{R^2 - s^2}\)

as for \(s = 0\), \(z \to 0\)

\[ \dot{z} = \frac{+(-2s)\,\dot{s}}{2\sqrt{R^2 - s^2}} = \frac{s\dot{s}}{\sqrt{R^2 - s^2}} \qquad --(4) \]

using eqn ③ ④ in ①. `L' becomes

\[ L = \frac{m}{2}\left[\dot{\rho}^2 + \rho^2\omega^2 + \frac{\rho^2\dot{\rho}^2}{R^2-\rho^2}\right] - mg\left(R - \sqrt{R^2-\rho^2}\right) \] \[ = \frac{m}{2}\left[\frac{\dot{\rho}^2R^2 - \dot{\rho}^2\rho^2 + \rho^2\omega^2R^2 - \rho^2\omega^2\rho^2 + \rho^2\dot{\rho}^2}{R^2-\rho^2}\right] - mgR + mg\sqrt{R^2-\rho^2} \] \[ L(\rho,\dot{\rho}) \;=\; \frac{m}{2}\left(\frac{\dot{\rho}^2R^2}{R^2-\rho^2}\right) + \tfrac{1}{2}m\rho^2\omega^2 - mgR + mg\sqrt{R^2-\rho^2} \]

since \(L\) only depends on one dynamical variable of space \((\rho)\) it has only one degree of freedom \((\rho)\).

\[ \boxed{\;\frac{\partial L}{\partial \rho} = \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\rho}}\right)\;} \] \[ p_\rho = \frac{\partial L}{\partial \dot{\rho}} = \frac{mR^2\dot{\rho}}{(R^2-\rho^2)} \qquad \Rightarrow \qquad \dot{\rho} = \frac{p(R^2-\rho^2)}{mR^2} \] \[ \begin{aligned} H(\rho,p) &= p\dot{\rho} - L\\ &= \;p\dot{\rho}\;\frac{mR^2}{(R^2-\rho^2)}\;\frac{p^2(R^2-\rho^2)^2}{m^2R^4} \; - L \end{aligned} \] \[ H(\rho,p) = \frac{p^2(R^2-\rho^2)}{mR^2} - \tfrac{1}{2}m\omega^2\rho^2 + mgR - mg\sqrt{R^2-\rho^2} \] \[ \begin{aligned} \dot{\rho} &= \frac{\partial H}{\partial p} = \frac{2p(R^2-\rho^2)}{mR^2}\\ \dot{p} &= \frac{-\partial H}{\partial \rho} = \frac{+2p^2\rho}{mR^2} + m\omega^2\rho - \frac{mg\rho}{\sqrt{R^2-\rho^2}} \end{aligned} \]

How do we find eqn of phase-trajectory using Hamiltons. eqn of motion.?

\[ H(q,p) \] \[ \begin{aligned} \dot{q}_i &= \frac{\partial H}{\partial p_i}\\ \dot{p}_i &= -\frac{\partial H}{\partial q_i} \end{aligned} \]

Ans, we have to solve these equations; if these eqn are integrable.

for 1-D.O.f, C.O.M \(=H\)

so trajectory is just given by \(H=\) const.

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\draw[->] (0.4,1.05) .. controls (1.0,1.6) and (1.7,0.95) .. (2.4,1.25);
\end{tikzpicture}

for \(n-\)D.O.f, we need \(n-\)C.O.M and find the C.T. which takes to \((I,\theta)\) variables and task is not easy to do.

\(\Big(\) If system is integrable, we can predict what will happen at any time t' in future provided initial phase-space point'.\(\Big)\)

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\begin{tikzpicture}[scale=0.95,>=Stealth]
\node (int) at (0,0) {Integrable};
\draw[->] (1.1,0) -- (2.2,0);
\node[anchor=west] at (2.3,0) {$-$ periodic motion};
\node[anchor=west] at (2.3,-0.7) {$-$ quasi-periodic};
\node[anchor=west] at (2.3,-1.5) {$-$ ergodic motion.};
\node[anchor=west] at (2.3,-2.1) {$-$ mixing};
\node[anchor=west] at (2.3,-2.7) {$-$ exponential};
\draw[->] (7.0,0.4) -- (7.0,-1.6);
\node[anchor=west] at (7.2,-0.5) {increasing complexity.};
\draw[->] (5.6,-2.35) -- (5.9,-2.75);
\node[anchor=west] at (6.0,-2.65) {instability on the average};
\node[anchor=west] at (5.1,-3.15) {arise from lack of enough C.O.M.};
\end{tikzpicture}

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\begin{tikzpicture}[scale=1.0,>=Stealth]
\draw[->] (-1.0,0) -- (3.6,0);
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% closed loop
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\draw (0.55,-0.1) -- (0.95,0.3);
\draw (0.73,2.0) -- (1.13,2.4);
\end{tikzpicture}

final pos phase space volume is closer to initial one

If initial volume element in phase space visits the neighbourhood of every point of part of phase-space, in given sufficient time, then we say motion is ergodic.

Note~:--

Worse can happen to initial volume element, even when volume is preserved.

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\begin{tikzpicture}[scale=1.0,>=Stealth]
\draw[pattern=north east lines] (0,1.05) rectangle ++(0.55,0.55);
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\draw[->] (5.75,0.42) -- (5.9,0.4);
% long thin filament
\draw (6.0,0.35) .. controls (6.6,0.25) and (7.0,0.15) .. (7.6,0.0);
\draw (6.0,0.35) .. controls (6.6,0.15) and (7.0,0.08) .. (7.6,0.0);
% vertical squiggle
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\draw (6.35,0.4) .. controls (6.75,1.2) and (6.9,1.9) .. (6.95,2.5);
\end{tikzpicture}

as time goes on such pieces of volume element can go as far as size of system (arbitrary far) such kind of motion is ergodic too but there is a spacial name to the motion called as mixing.

Mixing~:--

Suppose the volume of full phase space is \(\Omega\).

\(A_0-\) initial phase space volume

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% blob boundary
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\(\mu \to\) measure

If \(A_0\) is completly mixed up, how much of \(A_0\) is present in \(B\).

\[ \frac{\mu(A_n\cap B)}{\mu(A_0)} \;=\; \frac{\mu(B)}{\mu(\Omega)} \]

if, \(\displaystyle \mathop{\mathrm{Lt}}_{n\to\infty}\; \mu(A_n\cap B) = \frac{\mu(A_0)\,\mu(B)}{\mu(\Omega)}\)

then the dynamics is said to be strongly mixing.

Note~:-- mixing implies ergodicity whearas ergodicity does not imply mixing as we can have ergodic motion without mixing (distortion of phase-space element).

\(\bullet\) Rate of mixing~:--

Typically, mixing occurs exponentially fast.

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\node[anchor=west] at (8.1,0.6) {$(\lambda > 0)$};
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system is said to be in, exponential sensitivity to initial conditions.

example \((1-D\) phase-space\() \to\) hypothetical

\[ \dot{x} = x \qquad\qquad x(t) = x(0)\,e^{t} \]

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\draw[->] (-1.5,0.1) -- (-0.85,0.13);
\node at (-1.5,-0.35) {$0$};
% ticks
\draw (1.05,0.05) -- (1.05,0.32);
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\node at (1.0,-0.4) {$x_0$};
\node at (1.4,-0.25) {$\epsilon$};
\node at (2.05,-0.4) {$x_0+\epsilon$};
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\node at (3.7,-0.6) {$x_0e^{t}$};
\node at (5.3,-0.55) {$(x_0+\epsilon)e^{t}$};
% small bracket x0 -> x0+eps
\draw (1.05,0.55) -- (1.05,0.85) -- (3.55,0.85) -- (3.55,0.55);
% big bracket
\draw (1.75,0.95) -- (1.75,1.3) -- (5.2,1.3) -- (5.2,0.55);
\draw[->] (5.2,0.55) -- (5.2,0.4);
\draw[<->] (3.6,0.6) -- (5.15,0.6);
\node at (4.4,0.85) {$\epsilon e^{t}$};
\draw[->] (4.0,2.15) .. controls (4.6,1.9) and (4.7,1.7) .. (4.55,1.35);
\node[anchor=west] at (3.0,2.75) {the initial \emph{sep} \emph{error} is};
\node[anchor=west] at (3.0,2.35) {exponentially increasing};
\end{tikzpicture}

This is just a analogy to understand but keep in mind \(\dot{x}=x\) is integrable and this is not chaos.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
% stable manifold (near horizontal)
\draw (-3.2,0.15) .. controls (-1.5,0.0) and (0.5,-0.1) .. (2.3,-0.15);
\draw[->] (-2.2,0.06) -- (-2.0,0.05);
\draw[->] (-1.4,0.0) -- (-1.2,0.0);
\draw[->] (1.1,-0.12) -- (0.9,-0.12);
% unstable manifold (diagonal)
\draw (-2.3,-1.9) -- (1.6,1.5);
\draw[->] (-1.5,-1.2) -- (-1.3,-1.03);
\draw[->] (-0.6,-0.42) -- (-0.4,-0.25);
\draw[->] (0.5,0.53) -- (0.7,0.7);
% label arrow for unstable manifold
\draw[->] (0.9,1.05) .. controls (1.1,1.75) and (1.9,1.85) .. (2.5,1.75);
\node[anchor=west] at (2.6,1.75) {unstable manifold};
% incoming curve from right bending up
\draw (2.6,0.55) -- (1.1,0.55);
\draw[->] (2.0,0.55) -- (1.8,0.55);
\draw (1.1,0.55) .. controls (0.75,0.6) and (0.8,1.1) .. (1.15,1.35);
% trajectory from upper x0
\filldraw (-3.3,0.45) circle (1.4pt);
\node at (-3.55,0.75) {$x_0$};
\draw (-3.3,0.45) .. controls (-1.6,0.45) and (0.0,0.75) .. (1.0,1.5);
\draw[->] (-2.4,0.45) -- (-2.2,0.46);
% trajectory from lower x0
\filldraw (-3.15,-0.5) circle (1.4pt);
\node at (-3.45,-0.25) {$x_0$};
\draw (-3.15,-0.5) .. controls (-2.3,-0.5) and (-1.9,-0.55) .. (-1.85,-0.95);
\draw[->] (-2.6,-0.5) -- (-2.4,-0.5);
\draw (-1.85,-0.95) .. controls (-1.9,-1.35) and (-2.4,-1.7) .. (-2.9,-2.0);
\draw[->] (-2.15,-1.4) -- (-2.3,-1.55);
\draw (-1.5,-2.0) -- (-0.6,-1.0);
\draw[->] (-1.15,-1.6) -- (-1.3,-1.78);
% stable manifold label
\draw[->] (1.85,-0.35) .. controls (2.1,-0.75) and (2.3,-1.0) .. (2.6,-1.15);
\node[anchor=west] at (2.7,-1.15) {stable manifold \quad (sepretrix)};
\end{tikzpicture}

we can see that _\(x_0\) decides_ the future of two sm phase space points is very different. which are in neighbourhood.

so it is like if we hv sepretrix at every point that would imply chaos.

conditions of having chaos.

1. Bounded \(-\)phase space.
2. Exponential sensitivity to initial conditions.
3. A dense set of unstable periodic orbits.

*Exponential sensitivity in bounded \(-\)phase space*~:--

If we have two different initial cond\(^n\) \(x_0, y_0\) with different phase trejectories. Find seperation between \(x(t), y(t)\).

\[ |x_0-y_0| = \epsilon \]

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\filldraw (0,0.35) circle (1.3pt);
\node at (-0.4,0.5) {$x_0$};
\filldraw (0.05,0.0) circle (1.3pt);
\node at (-0.4,0.0) {$y_0$};
\draw (0,0.35) .. controls (1.4,0.75) and (2.4,1.3) .. (3.4,2.2);
\draw[->] (1.85,1.05) -- (2.0,1.18);
\draw (0.05,0.0) .. controls (1.2,-0.1) and (2.4,-0.7) .. (3.4,-1.7);
\draw[->] (1.65,-0.4) -- (1.8,-0.5);
\draw[->] (0.15,-0.15) .. controls (0.25,-0.6) and (0.1,-0.85) .. (-0.05,-1.0);
\node at (0.15,-1.35) {$x_0+\epsilon$};
\end{tikzpicture}

\[ \lim_{t\to\infty}\frac{1}{t}\ln\frac{|x(t)-y(t)|}{|x(0)-y(0)|} = 0 \]

as as \(t\to\infty\)

\[ |x(t)-y(t)| \leq L \]

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\draw[->] (0,0.5) -- (0,0.05);
\node[anchor=north] at (0,0.0) {system};
\node[anchor=north] at (0,-0.45) {size};
\node[anchor=north] at (0,-0.9) {(bounded)};
\end{tikzpicture}

to avoid such situations let us take \(\epsilon \to 0\)

\[ \mathop{\mathrm{Lt}}_{\epsilon\to 0}\;\mathop{\mathrm{Lt}}_{t\to\infty}\;\frac{1}{t}\,\ln\frac{|x(t)-y(t)|}{|x(0)-y(0)|} \;=\; \lambda(x_0)\;\; 0 \] \[ |x(0)-y(0)| = \epsilon \]

but lets invert the limits,

\[ \mathop{\mathrm{Lt}}_{t\to\infty}\;\lim_{\epsilon\to 0}\;\frac{1}{t}\;\frac{|x(t)-y(t)|}{|x(0)-y(0)|} \;=\; \lambda(x_0) \qquad -\;\mbox{\textcircled{1}} \]

\(\llcorner\) `Liapunov exponent'

For \(n-\)diamentional phase space we need \(n-\)liapunov exponent, so there is a full spectrum of liapunov exponents.

eqn ① provides maximum Liapunov exponent. as if a circle shrinks to line

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\draw (0,0) circle (0.75);
\draw[<->] (0.0,0.72) -- (0.0,-0.72);
\node[anchor=east] at (-0.15,0.0) {$d_0$};
\draw[<->] (0.18,-0.05) -- (0.55,-0.05);
\node at (0.42,0.35) {$D_0$};
% sliver
\draw (0.6,0.5) .. controls (2.5,0.35) and (4.5,0.2) .. (6.1,2.4);
\draw (0.6,-0.5) .. controls (2.5,-0.9) and (4.2,-1.5) .. (5.95,-3.1);
\draw (6.1,2.4) -- (6.05,-3.1);
\draw (5.95,-3.1) .. controls (6.1,-1.0) and (6.2,0.7) .. (6.15,2.4);
\node[anchor=west] at (6.25,2.0) {$D(t)$};
\draw[<->] (6.9,2.45) -- (6.9,-3.05);
\node[anchor=west] at (7.0,-0.3) {$d(t)$};
\end{tikzpicture}

\(\Big(\) \(d_0\) has exponential increment while \(D_0\) has decreased with time.\(\Big)\).

In continous time dynamics, \(N=3\) is minimum required for chaos.

For descrete time dynamics \(N=1\) will do.

*\(N = \) d.o.f.* Particle

Instead of differential eqn we call them difference eqn difference eqn are called maps.

*\(1-D\) map*~:-- \((\Delta x_n = x_{n+1} - x_n = a.)\)

\[ x_{n+1} = f(x_n) \qquad\qquad n = 0,1,2 \cdots \mbox{ discrete time} \]

(i) \((a<1)\) suppose *\(a<1\)*

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.15,>=Stealth]
\draw (-2.9,0) -- (3.1,0);
\draw[->] (-0.12,-1.9) -- (0.06,2.3);
% 45 degree line
\draw (-2.2,-2.2) -- (2.2,2.2);
\node[anchor=south] at (1.5,2.15) {45$^\circ$ line};
\node[anchor=south east] at (0.1,2.35) {$x_{n+1} = a\,x_n$};
% ax line
\draw (-2.9,-1.31) -- (3.1,1.40);
\node[anchor=west] at (3.1,1.45) {$ax$};
% cobweb positive side (a = 0.45)
\draw[dashed] (2.4,0) -- (2.4,1.08);
\draw[dashed,->] (2.4,1.08) -- (1.5,1.08);
\draw[dashed] (1.5,1.08) -- (1.08,1.08);
\draw[dashed] (1.08,1.08) -- (1.08,0.486);
\draw[dashed,->] (1.08,0.486) -- (0.75,0.486);
\draw[dashed] (0.75,0.486) -- (0.486,0.486);
\draw[dashed] (0.486,0.486) -- (0.486,0.22);
\draw[dashed,->] (0.486,0.22) -- (0.35,0.22);
\draw[dashed] (0.35,0.22) -- (0.22,0.22);
\node[anchor=north] at (2.4,-0.05) {$x_1$};
\node[anchor=north] at (1.12,-0.05) {$x_2$};
\node[anchor=north] at (0.5,-0.05) {$x_3$};
\node[anchor=north] at (0.15,-0.05) {$x_4$};
% cobweb negative side
\draw[dashed] (-2.6,0) -- (-2.6,-1.17);
\draw[dashed,->] (-2.6,-1.17) -- (-2.0,-1.17);
\draw[dashed] (-2.0,-1.17) -- (-1.17,-1.17);
\draw[dashed] (-1.17,-1.17) -- (-1.17,-0.53);
\draw[dashed,->] (-1.17,-0.53) -- (-0.85,-0.53);
\draw[dashed] (-0.85,-0.53) -- (-0.53,-0.53);
\draw[dashed] (-0.53,-0.53) -- (-0.53,-0.24);
\end{tikzpicture}

So, \(x=0\) is a stable fixed point

as we went to \(x=0\) from both side

(ii) \((a>1)\)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.15,>=Stealth]
\draw[->] (-2.9,0) -- (3.3,0);
\draw[<->] (0.0,-2.4) -- (0.0,2.4);
% 45 degree line
\draw (-1.9,-1.9) -- (2.3,2.3);
% ax line (a = 1.9)
\draw (-1.35,-2.55) -- (1.3,2.45);
% cobweb outward, x0 = 0.25
\draw[dashed] (0.25,0) -- (0.25,0.475);
\draw[dashed,->] (0.25,0.475) -- (0.38,0.475);
\draw[dashed] (0.38,0.475) -- (0.475,0.475);
\draw[dashed] (0.475,0.475) -- (0.475,0.90);
\draw[dashed,->] (0.475,0.90) -- (0.7,0.90);
\draw[dashed] (0.7,0.90) -- (0.90,0.90);
\draw[dashed] (0.90,0.90) -- (0.90,1.71);
\draw[dashed,->] (0.90,1.71) -- (1.3,1.71);
\draw[dashed] (1.3,1.71) -- (1.71,1.71);
\draw[dashed] (1.71,1.71) -- (1.71,0);
\node[anchor=north] at (0.2,-0.05) {$x_0$};
\node[anchor=north] at (0.5,-0.05) {$x_1$};
\node[anchor=north] at (0.95,-0.05) {$x_2$};
\node[anchor=north] at (1.75,-0.05) {$x_3$};
% negative side
\draw[dashed] (-0.25,0) -- (-0.25,-0.475);
\draw[dashed,->] (-0.25,-0.475) -- (-0.38,-0.475);
\draw[dashed] (-0.38,-0.475) -- (-0.475,-0.475);
\draw[dashed] (-0.475,-0.475) -- (-0.475,-0.90);
\draw[dashed,->] (-0.475,-0.90) -- (-0.7,-0.90);
\draw[dashed] (-0.7,-0.90) -- (-0.90,-0.90);
\draw[dashed] (-0.90,-0.90) -- (-0.90,-1.71);
\draw[dashed,->] (-0.90,-1.71) -- (-1.3,-1.71);
\draw[dashed] (-1.3,-1.71) -- (-1.71,-1.71);
\draw[dashed] (-1.71,-1.71) -- (-1.71,0);
\end{tikzpicture}

So, \(x=0\) in this case is unstable fixed point.

(iii) \(a=1\) map is \(x=x\) (identity map)

(trivial degenerate case)

Where ever the particle is, it does not move at all.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
\draw[->] (-1.2,0) -- (1.6,0);
\draw[->] (0,-0.9) -- (0,1.4);
\draw[->] (-1.0,-1.0) -- (1.2,1.2);
\end{tikzpicture}

(iv) for any arbitrary map.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.5,>=Stealth]
\draw (-1.6,0) -- (2.6,0);
\draw (0,-1.3) -- (0,2.5);
% 45 degree line
\draw (-1.3,-1.3) -- (2.35,2.35);
% map curve
\draw[domain=-1.35:2.25,samples=80] plot (\x,{0.4*\x + 0.4*\x*\x - 0.216});
\draw[->] (2.25,2.81) -- (2.45,3.06);
% fixed points
\node at (-0.3,-0.3) {$\times$};
\node at (1.8,1.8) {$\times$};
% cobweb: left of stable fp
\draw (-1.2,0) -- (-1.2,-0.12);
\draw[->] (-1.2,-0.12) -- (-0.75,-0.12);
\draw (-0.75,-0.12) -- (-0.12,-0.12);
\draw[dashed] (-0.12,-0.12) -- (-0.12,-0.258);
% cobweb: between fps, moving left
\draw[dashed] (1.0,0) -- (1.0,0.584);
\draw[->] (1.0,0.584) -- (0.8,0.584);
\draw (0.8,0.584) -- (0.584,0.584);
\draw[dashed] (0.584,0.584) -- (0.584,0.154);
\draw[->] (0.584,0.154) -- (0.4,0.154);
\draw (0.4,0.154) -- (0.154,0.154);
% cobweb: right of unstable fp, diverging
\draw[dashed] (2.0,0) -- (2.0,2.184);
\draw[->] (2.0,2.184) -- (2.1,2.184);
\draw (2.1,2.184) -- (2.184,2.184);
\draw[dashed] (2.184,2.184) -- (2.184,2.566);
\draw[->] (2.184,2.566) -- (2.35,2.566);
% labels
\draw[->] (1.95,1.85) .. controls (2.3,1.75) and (2.6,1.6) .. (2.85,1.5);
\node[anchor=west] at (2.9,1.5) {unstable};
\draw[->] (-0.28,-0.38) .. controls (-0.2,-0.75) and (-0.15,-0.95) .. (-0.05,-1.15);
\node[anchor=west] at (-0.05,-1.3) {stable};
\end{tikzpicture}

we can also see that slope at stable point is \(<\)1 and slope at unstable fixed point is \(>1\).

so, if `\(x^*\)' is fixed point then,

\[ \begin{aligned} |f'(x^*)| &> 1 \quad \Rightarrow\; x^* \mbox{ is unstable}\\ |f'(x^*)| &< 1 \quad \Rightarrow\; x^* \mbox{ is, stable} \end{aligned} \]

for \(f'(x^*) = -1\) cyclic (loop).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.3,>=Stealth]
\draw (-0.3,0) -- (2.8,0);
\draw (0,-0.2) -- (0,2.6);
\node[anchor=east] at (-0.1,1.4) {$x_{n+1}$};
% 45 line
\draw (0.1,0.1) -- (2.4,2.4);
% map line slope -1 through (1.2,1.2)
\draw (0.25,2.15) -- (2.3,0.1);
% period-2 square cobweb
\draw[dashed] (0.5,0) -- (0.5,1.9);
\draw[->] (0.5,1.9) -- (1.2,1.9);
\draw (1.2,1.9) -- (1.9,1.9);
\draw[dashed] (1.9,1.9) -- (1.9,0.5);
\draw[->] (1.9,0.9) -- (1.9,0.7);
\draw[->] (1.9,0.5) -- (1.2,0.5);
\draw (1.2,0.5) -- (0.5,0.5);
\draw[->] (0.5,0.5) -- (0.5,1.2);
\node[anchor=north] at (0.5,-0.05) {$x_0$};
\end{tikzpicture}

\(x^* = \) marginal fixed point (indifferent fixed point).

(v) Tangent case:--

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.4,>=Stealth]
\draw[->] (-0.4,0) -- (2.9,0);
\draw[->] (0,-0.35) -- (0,2.5);
\node[anchor=east] at (-0.1,1.2) {$x_{n+1}$};
\node[anchor=north] at (2.2,-0.15) {$x_n$};
% 45 line
\draw (-0.3,-0.3) -- (2.4,2.4);
% tangent curve
\draw[domain=0.05:2.1,samples=70] plot (\x,{\x + 0.35*(\x-1)*(\x-1)});
\draw[->] (2.1,2.5) -- (2.35,2.8);
% cobweb approaching from left
\draw (0.35,0.55) -- (0.62,0.55);
\draw[->] (0.62,0.55) -- (0.8,0.55);
\draw (0.8,0.55) -- (0.98,0.55);
\draw (0.98,0.55) -- (0.98,0.98);
\draw[->] (0.98,0.75) -- (0.98,0.9);
% cobweb diverging right
\draw (1.35,1.39) -- (1.35,1.75);
\draw[->] (1.35,1.6) -- (1.35,1.72);
\draw[->] (1.35,1.75) -- (1.6,1.75);
\draw (1.6,1.75) -- (1.78,1.75);
\draw (1.78,1.75) -- (1.78,2.2);
\draw[->] (1.78,2.2) -- (2.05,2.2);
\node[anchor=west] at (2.0,1.05) {(unstable) fixed point.};
\end{tikzpicture}

(vi) (fixed point is missed)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.5,>=Stealth]
\draw[->] (-0.9,0) -- (3.0,0);
\draw[->] (0,-0.55) -- (0,2.6);
\node[anchor=east] at (-0.1,1.5) {$x_{n+1}$};
\node[anchor=north] at (1.3,-0.15) {$x_n$};
\draw (-0.85,-0.85) -- (2.45,2.45);
\draw[domain=0.1:2.2,samples=70] plot (\x,{\x + 0.08 + 0.3*(\x-1)*(\x-1)});
\draw[->] (2.2,2.71) -- (2.45,2.96);
\foreach \a/\b in {0.3/0.527, 0.527/0.674, 0.674/0.786, 0.786/0.935,
                   0.935/1.016, 1.016/1.096, 1.096/1.179, 1.179/1.269,
                   1.269/1.371, 1.371/1.492, 1.492/1.645, 1.645/1.850}{
  \draw[->] (\a,\a) -- (\a,\b);
  \draw[->] (\a,\b) -- (\b,\b);
}
\draw[dashed] (0.3,0) -- (0.3,0.3);
\end{tikzpicture}

(Intermittency)

(Irregularity followed by long burst of regularity)

Exponential divergence of phase trejectory can be characterised by "liapunov exponent".

Note~:--

we can find finite time "liapunov exponent" but it is not relevent for the purpose we have in mind.

So the reason why we need very long time liapunov exponent is following

If we start with variable \(x_0\) \(x_n = x_0\) after `n' time steps.

\[ x_0 \longrightarrow x_1 \longrightarrow x_2 \longrightarrow \cdots x_n \]

*\(x_n\) is not computable* as error is exponentially increasing

\(\downarrow\) \((x_0,x_n)\)

ex. If we write no. in binary and lets say error doubles each steps then error from \(100^{th}\) decimal ste place comes to first decimal place after 100 steps.

If x' represents any physical quantity, there may be some function of x' \(\to \phi(x)\) whose avg. can be written as \(\langle \phi(x)\rangle\)

\[ \langle \phi(x)\rangle = \frac{1}{n}\sum_{j=0}^{n-1}\phi(x_j) \]

\(n \to\) time \(\langle \phi(x)\rangle = \) time average

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\draw (-0.4,0) -- (6.6,0);
\foreach \x in {0,1.7,3.4,5.1}{\draw (\x,-0.45) -- (\x,0.55);}
\node[anchor=south] at (0.55,0.1) {$\phi(x_0)$};
\node[anchor=south] at (2.25,0.1) {$\phi(x_1)$};
\node[anchor=south] at (3.95,0.1) {$\phi(x_2)$};
\node[anchor=south] at (5.75,0.1) {$\phi(x_n)$};
\node[anchor=north] at (0.55,-0.05) {$t_0$};
\node[anchor=north] at (2.25,-0.05) {$t_1$};
\node[anchor=north] at (3.95,-0.05) {$t_2$};
\node[anchor=north] at (5.75,-0.05) {$-\,t_n$};
\end{tikzpicture}

since \(x_j\) is not computable, then \(\langle \phi(x)\rangle\) is meaningless. but we would like to know what is long time avg

\[ \langle \phi(x)\rangle = \lim_{n\to\infty}\frac{1}{n}\sum_{j=0}^{n-1}\phi(x_j) \qquad -\mbox{\textcircled{1}} \]

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\draw[->] (0,1.4) -- (0,-0.6);
\node[anchor=west] at (0.05,0.4) {(Replacing time avg by ensamble avg.)};
\end{tikzpicture}

\[ \langle\langle \phi(x)\rangle\rangle = \int dx\; \rho(x)\,\phi(x) \qquad -\mbox{\textcircled{2}} \]

If the system is ergodic; ① \(=\) ②.

time avg. \(=\) ensamble avg.

Note~:-- \(\rho(x)\) should not vary with evolution and to attain this the dynamics has to seen very very long time, thats why we need long time avg also for this reason we need liapunov exponent for very long time

simplest is \(\to\) Bernoulli map/(shift).

*def\(^n\)*~:--

(Bernoulli map is the map of unit interval to itself.)

\(x_0\) lies in \([0,1]\) \(x_0 \longrightarrow x_1 \longrightarrow x_2 -- \; x_n\)

If \(x_1\) is \(f(x_0)\); \(x_1\) lies in \([0,1]\) \(x_0 \in [0,1]\)

\(x_1 \in [0,1] \Rightarrow x_n \in [0,1]\)

ex~:-- \(x_{n+1} = 2x_n \bmod 1\)

In map of \(f(x)\) fixed point \(x^*\) satisfies~:--

\[ x^* = f(x^*) \]

fixed points are given by intersection of \(x_{n+1}\) with 45\(^\circ\) line. which are \(\{0,1\}\)

\[ f'(0) = f'(1) = 2 > 1 \]

so, \(x^* = 0,1\) are unstable fixed points.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=3.4,>=Stealth]
\draw (-0.18,0) -- (1.25,0);
\draw (0,-0.3) -- (0,1.25);
\node[anchor=east] at (-0.06,0.6) {$x_{n+1}$};
\draw (-0.12,-0.12) -- (1.12,1.12);
\draw (0,0) -- (0.5,1);
\draw (0.5,0) -- (1,1);
\draw[dashed] (0,1) -- (1,1);
\draw[dashed] (0.5,0) -- (0.5,1);
\draw[dashed] (1,0) -- (1,1);
\filldraw (0,0) circle (0.02);
\filldraw (0.5,1) circle (0.02);
\filldraw (1,1) circle (0.02);
\node[anchor=west] at (0.06,0.09) {45$^\circ$};
\node[anchor=north] at (0,-0.03) {$0$};
\node[anchor=north] at (0.5,-0.03) {$(1/2)$};
\node[anchor=north] at (0.72,-0.03) {$x_n$};
\node[anchor=north] at (1.0,-0.03) {$1$};
\end{tikzpicture}

This can also happen

\[ \begin{aligned} f(a) &= b\\ f(b) &= a \end{aligned} \]

first iterator of above map.

\[ f^{(2)}(x) = 4x \bmod 1 \]

now we have four stable points \(\{0,\; 1/3,\; 2/3,\; 1\}\)

slope at each point is 4. which is greater than 1 so fixed points are unstable

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=3.4,>=Stealth]
\draw (-0.18,0) -- (1.25,0);
\draw (0,-0.3) -- (0,1.25);
\node[anchor=east] at (-0.06,0.6) {$f^{(2)}$};
\draw (-0.12,-0.12) -- (1.12,1.12);
\draw (0,0) -- (0.25,1);
\draw (0.25,0) -- (0.5,1);
\draw (0.5,0) -- (0.75,1);
\draw (0.75,0) -- (1,1);
\draw[dashed] (0,1) -- (1,1);
\foreach \x in {0.25,0.5,0.75,1}{\draw[dashed] (\x,0) -- (\x,1);}
\filldraw (0,0) circle (0.02);
\filldraw (0.3333,0.3333) circle (0.02);
\filldraw (0.6667,0.6667) circle (0.02);
\filldraw (1,1) circle (0.02);
\node[anchor=west] at (0.06,0.09) {45$^\circ$};
\node[anchor=north] at (0.25,-0.03) {\scriptsize $1/4$};
\node[anchor=north] at (0.3333,-0.13) {\scriptsize $(1/3)$};
\node[anchor=north] at (0.5,-0.03) {\scriptsize $2/4$};
\node[anchor=north] at (0.6667,-0.13) {\scriptsize $(2/3)$};
\node[anchor=north] at (0.75,-0.03) {\scriptsize $3/4$};
\node[anchor=north] at (1.0,-0.03) {\scriptsize $1$};
\end{tikzpicture}

Hence it seems like slope is increasing for iterated map. and there are not stable fixed points.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\node[anchor=east] at (1.6,0.45) {for \quad $x_0 = $};
\node[draw,circle,inner sep=1pt] (a) at (2.0,0.45) {$1/3$};
\node[anchor=west] at (2.5,0.45) {, \qquad $f(x_0) = 2/3$};
\node[anchor=west] at (0.9,-0.35) {$f^2(x_0) = 4/3 \bmod 1 = $};
\node[draw,circle,inner sep=1pt] at (5.2,-0.35) {$1/3$};
\draw[decorate,decoration={brace,amplitude=6pt}] (6.0,0.7) -- (6.0,-0.6);
\node[anchor=west] at (6.35,0.05) {period `2' cycle};
\end{tikzpicture}

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\node[anchor=east] at (0.8,0) {for \quad $x_0 = $};
\node[draw,circle,inner sep=1pt] (x0) at (1.2,0) {$1/5$};
\node[anchor=west] (x1) at (2.3,0) {$x_1 = 2/5$,};
\node[anchor=west] (x2) at (4.3,0) {$x_2 = 4/5$,};
\node[anchor=west] (x3) at (6.3,0) {$x_3 = \; 8/5 \bmod 1 = 3/5$};
\node[anchor=west] (x4) at (2.8,-1.2) {$x_4 = 6/5 \bmod 1 = $};
\node[draw,circle,inner sep=1pt] (x4c) at (6.4,-1.2) {$1/5$};
\draw[->] (1.65,0.25) to[bend left=45] (2.45,0.3);
\draw[->] (3.7,0.3) to[bend left=40] (4.45,0.3);
\draw[->] (5.7,0.3) to[bend left=40] (6.45,0.3);
\draw[->] (9.45,-0.3) .. controls (10.1,-0.95) and (8.4,-1.7) .. (6.85,-1.25);
\draw[->] (2.7,-1.55) to[bend right=45] (1.0,-0.45);
\node[anchor=west] at (6.8,-2.0) {period `5' cycle.};
\end{tikzpicture}

Any Rational no. is going to be part of periodic orbit. So it is clear we have infinite no. of unstable periodic orbits as well.

Note~:--

Let us write \(x_0, x_1 \; -- \; x_n\) in binary to see why it is called as Bernoulli shift.

\(x_0 \in [0,1]\) \(a_i = \{0\) or \(1\}\)

\[ x_0 = 0.\;a_0\,a_1\,a_2\,a_3\; -- \] \[ x_0 = \frac{a_0}{2^1} + \frac{a_1}{2^2} + \frac{a_3}{2^3} + \cdots \]

for \(x_{n+1} = 2x_n \bmod 1\)

\[ x_1 = a_0 .\, a_1\,a_2\,a_3 \; -- \]

if \(a_0 = 0\) no problem \(\Rightarrow\) \(x_0 < 1/2\)

\(a_0 = 1\) \(x_0 \in (1/2,1]\)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\draw[->] (0,0.5) .. controls (0.4,0.1) and (0.7,0.0) .. (1.2,0.0);
\node[anchor=west] at (1.25,0.0) {$(1.\,a_1a_2a_3\,)\bmod 1 = 0.\,a_1a_2a_3\;-$};
\end{tikzpicture}

So, \(x_1 = 0.\,a_1\,a_2\,a_3 \; \cdots\)

\(\Rightarrow\) we are loosing information in forward direction so we can't invert the map. as map function is non-linear. (linear maps are invertible)

so its the non-linearity which leads to chaos.

"Any rational no is part of periodic orbit".

ex \(x_0 = 1/3 = 0.01010101 \; ----\)

\[ \begin{aligned} x_1 &= \tfrac{2}{3} = 0.1010101\\ x_2 &= \tfrac{1}{3} = 0.01010101 \; -- \; = x_0 \quad \Rightarrow \mbox{ period `2' cycle} \end{aligned} \]

\(\to\) (repetitive in pattern or terminating.)

Because every point is unstable, every rational no is unstable, it should act like a seperatrix which throws out no. on each side.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\draw (0,0) -- (4.5,0);
\draw (0,-0.18) -- (0,0.2);
\draw (4.5,-0.18) -- (4.5,0.2);
\node[anchor=north] at (0,-0.15) {$0$};
\node[anchor=north] at (4.5,-0.15) {$1$};
\end{tikzpicture}

on line from \(0\) to \(1\) there are so many irrational no. than rational no.

and if \(x_0 = \) irrational no. it wanders every where and by filling this interval \([0,1]\) densly.

\(\to\) The system is ergodic.

\(\to\) The system is dissipative (How ?) \(\to\) will prove later

\(\star\) *Liapunov Exponent for (1\(-\)D map)~:--*

\[ \lambda(x_0) = \lim_{n\to\infty}\;\lim_{\epsilon\to 0}\;\frac{1}{n}\,\log\frac{|x_n-y_n|}{|x_0-y_0|} \qquad |x_0-y_0| = \epsilon \] \[ = \lim_{n\to\infty}\left(\lim_{\epsilon\to 0}\frac{1}{n}\ln\left|\frac{f^n(x_n) - f^n(x_0+\epsilon)}{x_0 - (x_0+\epsilon)}\right|\right) \] \[ \begin{aligned} f(x) &= ax\\ f'(x) &= a &&\qquad a \in [0,1]\\ f'(x) &< \underline{1} \quad \Rightarrow \mbox{ stable fixed points.} &&\qquad (0<a<1) \end{aligned} \]

also for, liapunav exponent \(\lambda(x_0) = \ln a < 0\) so for stable fixed point liapunav exponent becomes negative (everything falls to one place).

Q.) what would be liapunav exponent for period 2 cycle?

\[ \begin{aligned} f(a) &= b\\ f(b) &= a \end{aligned} \]

for stability \(|\,f'(a)\,f'(b)\,| < 1\)

(slope \(<1\))

\(\Rightarrow\) liapunov exponent will be negative for period 2 cycle.

Note~:-- The map is random, in the following sense, we agree that the rule is completely deterministic. for each \(x_0\) we have well defined \(x_1\) (with no randomness) and yet this is as random as coin toss.

suppose below is our phase space. In reality we need bins/cells to keep track of phase space points. so we do partition of phase space into cells/bins and numerically calculate how many times the perticular point lies in perticular cell/bin.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
% blob
\draw (0.9,4.3) .. controls (2.1,4.6) and (3.1,3.9) .. (3.0,2.9)
      .. controls (2.9,2.0) and (3.5,1.3) .. (3.3,0.6)
      .. controls (3.1,-0.1) and (1.9,-0.35) .. (1.2,0.2)
      .. controls (0.5,0.75) and (0.75,1.6) .. (0.6,2.4)
      .. controls (0.45,3.2) and (0.2,4.05) .. (0.9,4.3);
% grid lines
\foreach \x in {0.35,1.05,1.75,2.45,3.15}{\draw (\x,-0.5) -- (\x,4.75);}
\foreach \y in {0.5,1.2,1.9,2.6,3.3,4.0}{\draw (-0.15,\y) -- (3.9,\y);}
% dots
\filldraw (1.4,1.55) circle (0.045);
\filldraw (2.1,1.7) circle (0.045);
\filldraw (1.6,0.65) circle (0.045);
% cell label
\draw[->] (3.7,3.55) .. controls (2.9,3.3) and (2.2,2.6) .. (1.6,2.2);
\node[anchor=west] at (3.75,3.6) {`cell'};
\node[anchor=north] at (1.9,-0.6) {(phase$-$space)};
\end{tikzpicture}

For \(x_{n+1} = 2x_n \bmod 1\)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=3.2,>=Stealth]
  % axes
  \draw[->] (-0.05,0) -- (1.15,0);
  \draw (0,-0.02) -- (0,1.1);
  % branches of the doubling map
  \draw[thick] (0,0) -- (0.5,1);
  \draw[thick] (0.5,0) -- (1,1);
  % 45 degree line
  \draw (0,0) -- (1,1);
  \filldraw (1,1) circle (0.012);
  % guide lines
  \draw[dashed] (0,1) -- (0.5,1);
  \draw[dashed] (0.5,1) -- (1,1);
  \draw[dashed] (1,1) -- (1,0);
  \draw[dashed] (0.5,1) -- (0.5,0);
  % labels
  \node[left] at (0,1) {$1$};
  \node[left] at (-0.02,0.62) {$x_{n+1}$};
  \node[below left] at (0,0) {$0$};
  \node[below] at (0.5,-0.02) {$\frac{1}{2}$};
  \node[below] at (1,-0.02) {$1$};
  \node[below right] at (0.62,-0.02) {$x_n$};
  \node[right] at (0.09,0.09) {$45^\circ$};
  \draw (0.07,0) arc (0:45:0.07);
  % braces
  \draw[decorate,decoration={brace,amplitude=4pt,mirror}] (0.02,-0.16) -- (0.48,-0.16);
  \draw[decorate,decoration={brace,amplitude=4pt,mirror}] (0.52,-0.16) -- (0.98,-0.16);
  \node[below] at (0.25,-0.21) {$L$};
  \node[below] at (0.75,-0.21) {$R$};
  \node[below] at (0.25,-0.30) {$\left(\begin{array}{c}\text{left}\\ \text{bin}\end{array}\right)$};
  \node[below] at (0.75,-0.30) {$\left(\begin{array}{c}\text{Right}\\ \text{bin}\end{array}\right)$};
\end{tikzpicture}

we just want to know a particular point lies in left bin or right bin.

Any irrational number from 0 to 1 can be written as non-terminating in binary

\[ x_i = 0.011001111100101011100000100\,1 \cdots \] \[ \begin{aligned} x_0 &= 0.\,a_0\,a_1\,a_2 \cdots a_i \cdots a_n\\ x_1 &= a_0.\,a_1\,a_2\,a_3 \cdots a_{i+1} \cdots a_{n+1} \cdots a_n \end{aligned} \]

Above \(x_i\) is equivalent in asking wheather \(a_i\) is 0' or 1'

\[ \begin{aligned} a_i &= 0 && \text{for} \quad \longleftrightarrow L\\ a_i &= 1 && \text{for} \quad \longleftrightarrow R \end{aligned} \] \[ \begin{aligned} &\text{for } x_i < \tfrac{1}{2} && a_0 = 0 && \longleftrightarrow L\\ &\quad\ 1 > x_i > \tfrac{1}{2} && a_0 = 1 && \longleftrightarrow R \end{aligned} \]

So any arbitrary string of irrational no. can be written as

\[ L\,L\,R\,L\,R\,R\,R\,L\,L \cdots L\,R\,L \cdots L\,L\,R \cdots \]

which can then also be said to be obtained from coin toss experiment.

\[ \rho_2(x_2) = \int \delta(x_2 - f(x_1))\,.\,\delta(x_1 - f(x_0))\,dx_1 \]

any given point; \(y \xrightarrow{\ \text{iteration}\ } f(y)\)

after `n' time steps whats the probability density in \(x\)

\[ \rho_n(x) = \int dy\ \delta(x - f(y))\,\rho_{n-1}(y) \] \[ \Big\downarrow \ \lim (n \to \infty) \qquad\qquad (n = \text{time steps}) \] \[ \left(\begin{array}{l}\text{Frobenious --}\\ \text{Perron equation.}\end{array}\right) \quad \boxed{\rho(x)} = \int dy\ \delta(x - f(y))\ \rho(y) \] \[ \hookrightarrow \text{invariant density.} \]

\(\rho_n(x) = \) density of \(n^{\text{th}}\) iterate does not depends on initial condition.

\(\longrightarrow\) singular integral equation with kernel \(K(x,y) = \delta(x - f(y))\)

\[ \phi(x) = \int dy\ K(x,y)\,\phi(y) \]

\(K(x,y) = \) kernel of the integral operator.

where integral operator is \(\int dy\ K(x,y)\) acting on \(\phi(y)\)

\(\longrightarrow\) *Eigen value \(=1\)*

Normalization condition on \(\rho(x)\)

\[ \int dx\ \rho(x) = 1 \]

[?]

\[ \rho(x) \geq 0 \ , \quad x \in I \] \[ \rho(x) = \int_0^1 dy\ \delta(x - f(y))\,\rho(y) \qquad \begin{aligned} f(y) &= 2y \bmod 1\\ &= \begin{cases} 2y & 0 \leq y < \tfrac{1}{2}\\ 2y-1 & 1 \geq y > \tfrac{1}{2}\end{cases} \end{aligned} \] \[ \rho(x) = \int_0^{1/2} dy\ \delta(x - 2y)\,\rho(y) \;+\; \int_{1/2}^{1} dy\ \delta(x - 2y + 1)\,\rho(y) \] \[ \left\{ \delta(ax) = \frac{1}{|a|}\,\delta(x) \right. \] \[ = \left(\tfrac{1}{2}\right)\int_0^{1/2} dy\ \delta\!\left(y - \tfrac{x}{2}\right)\rho(y) \;+\; \tfrac{1}{2}\int_{1/2}^{1} dy\ \delta\!\left(y - \tfrac{x+1}{2}\right)\rho(y) \] \[ \rho(x) = \tfrac{1}{2}\left[\rho\!\left(\tfrac{x}{2}\right) + \rho\!\left(\tfrac{x+1}{2}\right)\right] \quad \cdots \quad \text{functional equation} \]

\(\rho(x) = \) const from normalizing \(\int_0^1 c\,dx = 1\) \(\left(c = 1\right)\)

\[ \boxed{\rho(x) = 1} \]

Therefore iteratese of irrational no. uniformaly and densly fill up the complete interval.