Pengfei's Notes — GeistHaus

Solve IVPs using Laplace Transform – a problematic example

Pengfei Jan 12, 2024

Laplace transform has been used to solve Initial Value Problems. This is a topic covered in a standard differential equation course at most universities. In this post I will describe a strange situation: the solution of 2nd order ODE obtained by using Laplace transform appears to violate one of the two initial conditions. Then a […]

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Laplace transform has been used to solve Initial Value Problems. This is a topic covered in a standard differential equation course at most universities. In this post I will describe a strange situation: the solution of 2nd order ODE obtained by using Laplace transform appears to violate one of the two initial conditions. Then a short justification will be given. It turns out to be a very simple problem, just one may get caught off guard.

Background. Consider a 2nd order linear constant coefficient differential equation:
$y'' + p y' + q y = f(t)$ , $y(0)=a$ , $y'(0)=b$ .
Applying Laplace transform, we get
$(s^2 Y(s) - sa -b) + p(sY(s) -a) +q Y(s) = F(s)$ , where the capital-case functions are the Laplace transforms of the corresponding lowercase functions: $F(s)=L[f(t)]$ and $Y(s)=L[y(t)]$ . It follows that
$(s^2 + p s + q)Y(s) = F(s) + a s + pa$ , or equally, $Y(s) =\frac{F(s) + a s + pa}{s^2 + p s + q}$ .
Applying the inverse transform, we obtain the solution of the IVP.

Laplace Transform of the delta functions. Let $\delta_c(t)$ be the delta function at $c\ge 0$ . We know that $L[\delta_c(t)] = e^{-c s}$ . In particular, $L[\delta_0(t)] = 1$ .

A problematic example. Consider the IVP $y''+y=\delta_0(t)$ , $y(0)=0$ , $y'(0)=0$ .
Laplace Transform: $(s^2+1)Y(s) = 1$ , or equally, $Y(s) = \frac{1}{s^2+1}$ .
Inverse Transform: $y(t)= L^{-1}[Y(s)] = \sin t, t\ge 0$ .

The above process is fairly straightforward. Most of us would stop here and move onto the next problem. Some of you may double check:
(1) $y'' + y =0$ for $t \neq 0$ , where the delta function $\delta_0(t)$ also vanishes;
(2) $y(0) = \sin 0 =0$ , the first initial condition checked;
(3) $y'=\cos t$ , $y'(0)=\cos 0 =1$ , wait, what? — it violates the second initial condition.

You can check the above computation. There is no error. How is it a solution since it does not satisfy the initial conditions that we started with? Should we call it a solution? Is Laplace transform not applicable here? These were my initial reactions.

To resolve this inconsistency, we need to properly interpret the initial conditions $y(0)=0$ and $y'(0)=0$ , since the external force $f(t) = \delta_0(t)$ also happens at this moment. A better way to properly state the initial conditions is $y(0-)=0$ and $y'(0-)=0$ . Then the external force is applied and the solution $y(t)$ reacts accordingly. Since the force is a delta function, it actually results in a jump of the velocity at $t=0+$ : $y'(0+)=1$ .

A few paragraphs can be added to fully justify the correctness of the solution. For example, one can extend the solution $y(t)=0$ for $t < 0$ , and then do a better job of explaining how the solution satisfies the differential equation. But I believe the above description is enough to convince you that a discontinuity is created after the delta force and the solution is indeed a correct one.

http://pfzhang.wordpress.com/?p=1770

Extensions

Fixed points

Pengfei Jun 4, 2021

Let be a neighborhood of , be a local diffeomorphism fixing . Then is said to be stable if every open neighborhood of contains an invariant neighborhood of . It is said to be unstable if there is an open neighborhood of such that . It is said to be mixed if it is neither […]

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Let $U$ be a neighborhood of $O=(0,0)$ , $f:U\to \mathbb{R}^2$ be a local diffeomorphism fixing $O$ . Then $O$ is said to be stable if every open neighborhood $V\subset U$ of $O$ contains an invariant neighborhood of $O$ . It is said to be unstable if there is an open neighborhood $V\subset U$ of $O$ such that $\bigcap_{n\in\mathbb{Z}}f^nV=\{O\}$ . It is said to be mixed if it is neither stable nor unstable.

Example. (1) When $f$ is a rotation, then $O$ is stable. (2) When $O$ is a hyperbolic fixed point, then it is unstable. (3) Using complex coordinate $z$ , let $f(z)=\frac{z}{1-z}$ . Then $f^n(z)=\frac{z}{1-n z}$ . Picking $z=\frac{1}{n}$ and letting $n\to 0$ , we see that $O$ is not stable; picking $z=i/n$ and letting $n\to 0$ , we see that $O$ is not unstable. Therefore, $O$ is mixed.

Example. Let $\phi(x,y)=(x^2+y^2)^n$ . Consider the rotation $(\dot x, \dot y)= (y, -x)$ and its nonlinear perturbation $(\dot x, \dot y)= (y+\phi(x,y)x, -x+\phi(x,y)y)$ . Then $\frac{d}{dt} (x^2+y^2) =2x\dot x+ 2y\dot x = 2\phi(x,y)(x^2+y^2)>0$ whenever $(x,y)\neq O$ . It follows that $O$ is unstable.

http://pfzhang.wordpress.com/?p=1755

Extensions

When the suspension flow is Reeb

Pengfei Aug 30, 2020

Let be a homeomorphism on a compact space . Given a ceiling function , we consider the mapping torus and the suspension flow on , which is just the flow on , respecting the equivalence relation . Let be an exact symplectic manifold, be a Hamiltonian diffeomorphism. Then for any primitive 1-form , that is, […]

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Let $f$ be a homeomorphism on a compact space $X$ . Given a ceiling function $\tau: X \to (0, \infty)$ , we consider the mapping torus $X_{\tau}$ and the suspension flow $f_t$ on $X_{\tau}$ , which is just the flow on $X\times \mathbb{R}$ , $(x,s) \mapsto (x, s+t)$ respecting the equivalence relation $(x, s) \sim (f(x), s- \tau(x))$ .

Let $(M, \omega)$ be an exact symplectic manifold, $f$ be a Hamiltonian diffeomorphism. Then for any primitive 1-form $\lambda$ , that is, $d\lambda =\omega$ , consider the 1-form $\lambda + ds$ on $M\times \mathbb{R}$ . When does it descend to a 1-form on $M_{\tau}$ ? If so, then $\lambda + ds$ is a contact form on $M_{\tau}$ and $f_t$ is the corresponding Reeb flow on $(M_{\tau}, \lambda + ds)$ .

Recall that $f^{\ast}\lambda -\lambda = d\phi$ for some function $\phi$ on $M$ . Therefore, $\lambda + ds$ descends to a 1-form on $M_{\tau}$ if and only if $\tau - \phi$ is constant. Note that different choices of $\lambda$ result in an additional coboundary of $\phi$ and hence an additional coboundary of the ceiling function $\tau$ on $M$ .

A serious issue about the assumption of exactness of $M$ is that closed symplectic manifolds are never exact by Stokes’s Theorem. So in order to construct a contact flow using a symplectic map $f$ on a closed symplectic manifold $(M, \omega)$ , one can blow up the manifold $M$ at an elliptic fixed point or along an elliptic periodic orbit, and denote the new manifold by $\bar M$ . Then we extend $f$ to $\bar f$ on $\bar M$ , construct the suspension $(\bar M)_{f}$ . It is a contact manifold with boundary. One can glue a solid torus twisted along the boundary component of $(\bar M)_{f}$ and obtain a closed contact manifold.

http://pfzhang.wordpress.com/?p=1748

Extensions

Radon measures

Pengfei Aug 3, 2020

Let be a set, and be a -algebra on . A function is called a (signed) measure if it satisfies and -additive. Note that is automatic whenever there exists with finite measure. Now suppose is a topological space. There is a natural -algebra, the Borel -algebra on . Then a measure is called Borel if […]

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Let $X$ be a set, and $\Sigma$ be a $\sigma$ -algebra on $X$ . A function $\mu: \Sigma \to \mathbb{R}$ is called a (signed) measure if it satisfies $\mu(\varnothing)=0$ and $\sigma$ -additive. Note that $\mu(\varnothing)=0$ is automatic whenever there exists $E\in \Sigma$ with finite measure.

Now suppose $X$ is a topological space. There is a natural $\sigma$ -algebra, the Borel $\sigma$ -algebra $\mathcal{B}$ on $X$ . Then a measure $\mu$ is called Borel if $\mathcal{B} \subset \Sigma(\mu)$ .

Regularity. A Borel measure $\mu$ is inner regular if for any open subset $U$ , $\mu(U)=\sup \mu(K)$ over compact subsets $K\subset U$ . It is outer regular if for any Borel subset $E$ , $\mu(E) =\inf \mu(U)$ over open subsets $U \supset E$ . It is locally finite if for any $x\in X$ , $\mu(U)$ is finite for some neighborhood $U\ni x$ . Note that if $X$ is locally compact, then being locally finite is equivalent to that $\mu(K)$ is finite for any compact subset $K$ .

There are different ways to define Radon measures.

Definition 1. A Borel measure $\mu$ is called a Radon measure if it is inner regular, outer regular and locally finite.

Definition 2. A Radon measure is a (positive) continuous linear functional $L$ on $C_c(X, \mathbb{R})$ , where $C_c(X, \mathbb{R})$ is the vector space of real-valued continuous functions with compact support.

By Riesz-Markov representation theorem, the two definitions are equivalent.

http://pfzhang.wordpress.com/?p=1745

Extensions

Generating functions

Pengfei Jul 21, 2020

Let be the universal cover of the annulus , be a diffeomorphism on preserving the 2-form , be a lift of . Suppose is monotone twist. That is, given , the function is strictly increasing. Let be the set of points such that . It follows that we can reinterpret the function as . In […]

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Let $\hat A=\mathbb{R} \times [0,1]$ be the universal cover of the annulus $A= \mathbb{T} \times [0,1]$ , $f$ be a diffeomorphism on $A$ preserving the 2-form $\omega = dx \wedge dy$ , $F$ be a lift of $f$ . Suppose $f$ is monotone twist. That is, given $F(x,y)=(x_1, y_1)$ , the function $y\mapsto x_1$ is strictly increasing.

Let $B$ be the set of points $(x, x') \in \mathbb{R}^2$ such that $F(I_x) \cap I_{x'} \neq \emptyset$ . It follows that we can reinterpret the function $F$ as $(x, x') \in B \mapsto (y,y')$ . In fact, there exists a function $h: B\to \mathbb{R}$ such that $y=-\partial_1 h(x,x')$ and $y'=\partial_2 h(x,x')$ . Such a function is called a generating function of $f$ .

For the existence of a generating function, a necessary and sufficient condition is $\partial_2 y(x,x') + \partial_1 y'(x,x')=0$ . The last equation holds since $f$ preserves the 2-form $\omega = dx \wedge dy$ .

Then we compare the two expressions $h(x,x')=\int y'(x,t)dt + \phi(x)$ and $h(x,x')=\int y(t,x')dt + \psi(x')$ and get a formula for the generating function.

In the other direction, if such a generating function exists, then $f$ preserves the 2-form $\omega =dx \wedge dy$ .

http://pfzhang.wordpress.com/?p=1739

Extensions

Mechanics

Pengfei Jul 16, 2020

Newton’s laws of motion are three physical laws that, together, laid the foundation for classical mechanics. The second law provides a differential equation for the motion: given an initial position and an initial velocity, one can find the position at any given time . After Newton, one might wonder why the nature behaves this way. […]

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After Newton, one might wonder why the nature behaves this way. When we throw a stone in the air, the stone doesn’t know any math and still ‘knows’ where to land. This leads to the least action principle in Lagrangian mechanics: among all the possible choices of paths from one point to another (in the configuration space), the one with least action is the physical one. This leads to the Euler-Lagrange equation.

The Euler-Lagrange equation is a second order different equation about the motion $x(t)$ . One can introduce the velocity variable $v=\dot x$ and reduces E-L to a system of first order differential equations on the phase space $(x,v)$ . However, the different equation about $v$ is implicit.

The Hamiltonian mechanics is kind of dual to the Lagrangian mechanics. It involves the momentum $p$ instead of the velocity $v$ . This is not an artificial change: the new space admits a natural (symplectic) structure, and the differential equations of motion become explicit: it is given by the Hamiltonian vector field of the Hamiltonian function.

http://pfzhang.wordpress.com/?p=1734

Extensions

Lipschitz invariant curves

Pengfei Jul 16, 2020

Let be an ellipse with , be the billiard map on the phase space . Note that is a monotone twist map. The rotation interval of is . Moreover, for any , , there is a unique invariant curve of rotation number . The case is special: it consists of a periodic orbit of period […]

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Let $\Gamma=E(a,b)$ be an ellipse with $a > b > 0$ , $F$ be the billiard map on the phase space $M= \Gamma\times [0,\pi]$ . Note that $F$ is a monotone twist map. The rotation interval of $F$ is $\rho(F)=[0,1]$ . Moreover, for any $\rho \in [0,1]$ , $\rho\neq 1/2$ , there is a unique invariant curve of rotation number $\rho$ . The case $\rho=1/2$ is special: it consists of a periodic orbit $\{p_0, p_1\}$ of period $2$ and two pairs of heterolinic connections between them, say the upper pair $(U_0, U_1)$ and the lower pair $(L_0, L_1)$ . These pairs are invariant but not smooth: both are singular at the periodic orbits. However, they are the only two ways to form invariant curves of rotation number $1/2$ : $(U_j, L_{1-j})$ is smooth, but not invariant.

Now let $m,n\ge 1$ , $M_{j,k}=j\Gamma \times [0,k\pi]$ be a covering space of $M$ that is $j$ copies in the horizontal direction and then $k$ copies in the vertical direction. It can be viewed as a standard annulus. Let $F_{j,k}$ be the lift of $F$ to the new annulus $M_{j,k}$ . Then the rotation interval is $\rho(F_{j,k})=[0, k/j]$ . For each $\rho=\frac{k'+\frac{1}{2}}{j}$ , $k'=0,\dots, k-1$ , there are exactly two invariant curves of rotation number $\rho$ , both having $2j$ singular points.

http://pfzhang.wordpress.com/?p=1727

Extensions

The unit tangent bundle of S^2

Pengfei Jul 8, 2020

Let be the stand sphere with the induced Riemannian metric, be the unit tangent bundle of . There is a natural map from to . That is, let , set . It is easy to see that . Moreover, is bijective and hence a homeomorphism. Let , be the equator, which is a simple closed […]

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Let $S^2 \subset \mathbb{R}^3$ be the stand sphere with the induced Riemannian metric, $M =UTS^2$ be the unit tangent bundle of $S^2$ . There is a natural map $f$ from $M$ to $M_3(\mathbb{R})$ . That is, let $(p, v)\in M$ , set $f(p,v)=(p, v, p\times v)$ . It is easy to see that $f(M) \subset SO(3)$ . Moreover, $f: M\to SO(3)$ is bijective and hence a homeomorphism.

Let $\gamma: S^1 \to S^2$ , $t\mapsto (\cos t, \sin t, 0)$ be the equator, which is a simple closed geodesic. Let $D\gamma \subset M$ be the lifted closed orbit of the geodesic flow on $S^2$ , and $\hat \gamma = f(D\gamma) \subset SO(3)$ be the corresponding curve of matrices. Note that $\hat\gamma(t) = [R_t, 1]$ .

Let $\gamma_s$ , $s\in[0, 1]$ , be a family of closed curves that deform $\gamma_0 = \gamma$ to the trivial curve at the north pole $N=(0,0,1)$ . That is, $\gamma_s(t)=(r\cos t, r\sin t, s)$ , where $r^2 + s^2 =1$ . For each $s\in (0, 1)$ , let $D'\gamma_s \subset M$ be the curve induced by $D\gamma_s$ . Let $\hat\gamma_s = f(D'\gamma_s)$ , $0\le s <1$ . This process is not defined directly for $s=1$ . But the limit $\lim_{s\to 1}\hat\gamma_s$ does exist. Therefore, we denote the limit as $\hat\gamma_1$ . It is easy to see that $\hat\gamma_1(t)=[(0,0,1), (\cos t, \sin t, 0), (-\sin t, \cos t, 0)]$ , which coincides with $f(UT_NS^2)$ .

Similarly, one can deform $\hat\gamma$ to the unit tangent bundle at the south pole $S(0,0,-1),$ say $\hat\gamma_{-1}$ . Note that both cycles wrap around the $z$ -axis counterclockwise. However, the two normal directions (aka the orientation) at $N$ and $S$ are opposite. So we have $[\hat\gamma_1 + \hat\gamma_{-1}] = 0 \in \pi_1(M)$ . It follows that $[2\hat\gamma] = 0 \in \pi_1(M)$ , while $[\hat\gamma] \neq 0$ . More generally, one can show that a smooth cycle $\hat\gamma \subset M$ with transverse self-intersections is contractible in $M$ if and only if it has an odd number of self-intersections. In particular, $\pi_1(M)=\mathbb{Z}_2$ . This is a topological invariant. Hence it holds for all metrics on $S^2$ .

http://pfzhang.wordpress.com/?p=1719

Extensions

Non-autonomous flow

Pengfei Jun 11, 2020

Consider a smooth one-parameter family of diffeomorphisms on a manifold . It is a flow if and for every , , . Set . This generates a vector field . More generally, one consider a time-dependent vector field. That is, , where may change in time. This in turn genertes a one-parameter family (non-autonomous) of […]

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Consider a smooth one-parameter family ${f_t}$ of diffeomorphisms on a manifold $M$ . It is a flow if $f_0(x)=x$ and $f_{t}\circ f_{s}(x) = f_{s+t}(x)$ for every $x\in M$ , , $t ,s \in \mathbb{R}$ . Set $X(x)=\lim\limits_{h\to 0} \frac{1}{h}(f_h(x) - x)$ . This generates a vector field $X: M \to TM$ .

More generally, one consider a time-dependent vector field. That is, $X: M\times \mathbb{R} \to TM$ , where $X(x,t) \in T_xM$ may change in time. This in turn genertes a one-parameter family (non-autonomous) of maps $f_t: M\times \mathbb{R} \to M$ , $t \in \mathbb{R}$ , such that $f_{t'}(f_{t}(x,s), t+s) = f_{t' +t}(x,s)$ .

http://pfzhang.wordpress.com/?p=1714

Extensions

Perron numbers

Pengfei Jun 3, 2020

Consider a monic polynomial with integer coefficients: , .The complex roots of such polynomials are called algebraic integers. For example, integers and the roots of integers are algebraic integers. Note that the Galois conjugates of an algebraic integer are also algebraic integers. Consider a square matrix with positive integer entries. By Perron-Frobenius Theorem, there is […]

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Consider a monic polynomial with integer coefficients: $p(x)=x^d + a_1 x^{d-1} + \cdots + a_{d-1}x + a_d$ , $a_j \in \mathbb{Z}$ .
The complex roots of such polynomials are called algebraic integers. For example, integers and the roots of integers are algebraic integers. Note that the Galois conjugates of an algebraic integer are also algebraic integers.

Consider a square matrix $A$ with positive integer entries. By Perron-Frobenius Theorem, there is a unique positive eigenvalue $\lambda >0$ . Moreover, it admits an eigenvector of all positive entries, and satisfies $\lambda > |\lambda'|$ for any other eigenvalue $\lambda'$ of $A$ . In particular, $\lambda > |\lambda_{\sigma}|$ for any of its Galois conjugates. Such a number is called a Perron number. More generally, a weak Perron number is a real algebraic integer whose modulus is greater than or equal to that of all of its Galois conjugates.

Let $f$ be a continuous map on the interval $I=[0,1]$ , and $h_{\text{top}}(f)$ the topological entropy. Assume $f$ is postcritically finite: $\text{PC}(f)=\{f^n c: c\in \text{Crit}(f)\}$ is a finite set. Then the partition $\alpha$ of $I$ along the postcritical set $\text{PC}(f)$ is a Markov partition for $f$ , since (1) the endpoints are sent to endpoints, (2) every folding corresponds to a critical point of $f$ . Therefore, $\lambda(f)=e^{h_{\text{top}}(f)}$ is the leading eigenvalue of the incidence matrix $A_{f,\alpha}$ associated to the Markov partition $\alpha$ . It follows that $\lambda(f)$ is a weak Perron number.

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Extensions