Monday, December 10, 2012

Differential Geometry Part 1

To review for my upcoming final in differential geometry, I will go over my last two midterms.

1. Parametrize the curve $y^2+y=x^2$ using as the parameter $t$ slope of the line from the origin to a point on it, i.e. $y=tx$.

The first thing we should do is substitute $tx$ for $y$ in the first equation to obtain $$\left(t^2-1\right)x^2+tx=0.$$ By the quadratic equation, we find that $$x=-\frac{t}{t^2-1}\quad\text{or}\quad x=0.$$ To find $y$ in terms of $t$, we substitute this $x$ into the second equation to obtain $$y=-\frac{t^2}{t^2-1}\quad\text{or}\quad y=0.$$ Therefore, the parametric equation of this curve is $$r\left(t\right)=\begin{cases} -\frac{t}{t^2-1}\partial_x+-\frac{t^2}{t^2-1}\partial_y & \text{ if } t\in\mathbb{R}\setminus\{-1,1\}, \\ 0\partial_x+0\partial_y & \text{ if } t\in\{-1,1\}. \end{cases}$$ 2. The evolute of the parabola $y=x^2/2$ is $y=1+3/2x^{2/3}$. How many normals can be drawn to this parabola from $\left(1,-1\right)$? Explain.

Substituting this point into the evolute's equation yields $$y\left(1\right)=\frac52>-1.$$ In other words, the point is below the evolute.

Some time ago, we concluded that under the evolute or on its cusps, only one normal can be drawn; above the evolute, only three normals can be drawn; and on the evolute, only two normals can be drawn.

Therefore, only one normal can be drawn.

3. Consider the curve $r\left(t\right)=e^t\left(\sin t-\cos t\right)\partial_x+e^t\left(\sin t+\cos t\right)\partial_y$ with $t\in\mathbb R$.

a. Find the speed.

The speed is defined as follows: $$v=\left|r'\right|.$$ Therefore, we have that \begin{align} r'\left(t\right)&=2e^t\sin t\partial_x+2e^t\cos t\partial_y,\\ \left|r'\right|&=2e^t=v. \end{align} b. Find a natural parameter.

The natural parameter is defined as follows: $$s=\int v\,dt.$$ Therefore, we have that $$s=2e^t+c.$$ c. Find involutes.

Involutes are defined as follows: $$h=r-T\int v\,dt.$$ Therefore, we have that \begin{align} T&=\frac{r'}{v}=\sin t\partial_x+\cos t\partial_y,\\ h&=-\left[\left(e^t+c\right)\sin t+e^t\cos t\right]\partial_x+\left[e^t\sin t-\left(e^t+c\right)\cos t\right]\partial_y \end{align} 4. Consider the curve $r\left(t\right)=t\partial_x+\ln\cos t\partial_y$ with $t\in\left(-\pi/2,\pi/2\right).$

a. Find the speed.
Hint: $1+\tan^2t=\sec^2t$.

As before, we have that \begin{align} r'\left(t\right)&=\partial_x-\frac{1}{\cos t}\sin t=\partial_x-\tan t\partial_y,\\ v&=\sqrt{1+\tan^2t}=\sqrt{\sec^2t}=\sec t. \end{align} b. Find the unit tangent.

As before, we have that $$T=\cos t\partial_x-\sin t\partial_y.$$ c. Find the right unit normal.

The right unit normal is defined as follows: $$\tilde N=-T_2\partial_x+T_1\partial_y.$$ Therefore, we have that $$\tilde N=\sin t\partial_x+\cos t\partial_y.$$ d. Find the signed curvature.

The signed curvature is defined as follows: $$\tilde\varkappa=\frac{r'\times r''}{v^3}$$ Therefore, we have that \begin{align} r''\left(t\right)&=-\sec^2t\partial_y,\\ r'\times r''&=\begin{vmatrix} 1 & -\tan t\\ 0 & -\sec^2t \end{vmatrix}=-\sec^2t,\\ \tilde\varkappa&=-\cos t. \end{align} e. Find the principal unit normal.

The principal unit normal is defined as follows $$N=\text{sign}\left(\tilde\varkappa\right)\tilde N.$$ Therefore, we have that $$N=-\sin t\partial_x-\cos t\partial_y.$$ f. Find the evolute.

The evolute is defined as follows: $$c=r+\frac{1}{\tilde\varkappa}\tilde N.$$ Therefore, we have that $$c=\left(t-\tan t\right)\partial_x+\left(\ln\cos t-1\right)\partial_y.$$ 5. Fill in the blanks to form a true statement.

a. If the speed is $3$ and $T=\partial_x$, then the velocity is "$3\partial_x$."

b. Natural parameter is unique up to "shift and sign change."

c. A plane has "two" unit normals.

d. Inflection point is where a curve "has $\varkappa=0$."

e. If the curvature is $\varkappa=e^{2t}+1$ and the natural parameter is $s=e^t$, then the natural equation of the curve is "$\varkappa\left(s\right)=s^2+1$."

f. Lines and circles are the only plane curves that "have constant curvature."

g. The unit tangent to a curve is drawn below, draw and label $N$ and $\tilde N$.

h. Tangents to a plane curve are normal to "its involutes."

i. Wavefronts of a plane curve have cusps on "its evolute."

j. Involute of the evolute is "a wavefront of the plane curve."

6. Prove that $\ddot T\cdot T=-\varkappa^2$, where $T$ is the unit tangent and $\varkappa$ is the curvature of a curve.
Hint: Differentiate $\dot T\cdot T$.

Proof. $\dot T=\varkappa N\Longrightarrow\dot T\cdot T=\varkappa N\cdot T=0$. \begin{align} \frac{d}{dt}\left(\dot T\cdot T\right)&=\ddot T\cdot T+\dot T\cdot\dot T=\ddot T\cdot T+\left|\dot T\right|^2\\ &=\ddot T\cdot T+\left|\varkappa N\right|^2=\ddot T\cdot T+\varkappa^2=0\\ &\Longrightarrow\ddot T\cdot T=-\varkappa^2.\qquad\square \end{align} 7. Let $r\left(t\right)$, $w\left(t\right)$ be two curves that are normal to the segment connecting them at each point, i.e. $r'\perp\left(w-r\right)$ and $w'\perp\left(w-r\right)$. Prove that they are equidistant, i.e. $\left|w-r\right|=\text{const}$.
Hint: Differentiate $\left|w-r\right|^2=\left(w-r\right)\cdot\left(w-r\right)$.

Proof. \begin{align} \frac d{dt}\left[\left|w-r\right|^2\right]&=\frac d{dt}\left[\left(w-r\right)\cdot\left(w-r\right)\right]\\ &=2\left(w'-r'\right)\cdot\left(w-r\right)\\ &=2\left[w'\cdot(w-r)-r'\cdot(w-r)\right]=0.\qquad\square \end{align}

Plane Curves

Consider the equation of a circle $$x^2+y^2=R^2,$$ where $R$ is its radius. How do we parametrize it?

Recall that a function $f(t)$ can be parametrized an infinite number of ways. Moreover, the standard parametrization is $$r(t)=t\partial_x+f(t)\partial_y.$$ However, the most common way to parametrize our circle is as follows: $$r(t)=R(\sin t\partial_x+\cos t\partial_y),$$ where $t\in[0,2\pi).$

Note: this is not its standard parametrization.

If we want to find its perimeter, then we compute the following integral: \begin{align} P&=\int_{0}^{2\pi}|r'(t)|\,dt\\ &=\int_{0}^{2\pi}\sqrt{R^2\cos^2t+R^2\sin^2t}\,dt\\ &=2\pi R, \end{align} which should be familiar to you if you have some basic knowledge in geometry.

Finally, we can use the indefinite form of the above integral to express our parametric equation in terms of its arc length $s$, as follows: $$r(s)=R\left(\sin\frac{s}{R}\partial_x+\cos\frac{s}{R}\partial_y\right).$$ If we wish to compute the curvature $\varkappa$ of the circle, then we can use the above natural parametrization as follows: \begin{align} T&=\dot r\\ &=\cos\frac{s}{R}\partial_x-\sin\frac{s}{R}\partial_y,\\ \varkappa&=|\dot T|\\ &=\left|\frac{1}{R}\left(-\sin\frac{s}{R}\partial_x-\cos\frac{s}{R}\partial_y\right)\right|\\ &=\frac{1}{R}. \end{align} This makes sense since the circle is the only plane curve with constant curvature.

Saturday, December 8, 2012

Fourier Transforms

Let $f\left(x\right)$ be piecewise smooth, and $$\int_{-\infty}^\infty|f\left(x\right)|\,dx<\infty.$$ The Fourier transform and inverse Fourier transform of $f$ are respectively defined as follows: \begin{align} F\left(\omega\right)&\equiv\frac1{2\pi}\int_{-\infty}^\infty f\left(x\right)e^{i\omega x}\,dx,\\ f\left(x\right)&\equiv\int_{-\infty}^\infty F\left(\omega\right)e^{-i\omega x}\,d\omega. \end{align} Together, they are called a Fourier transform pair.

Fourier transforms are frequently used to solve partial differential equations defined over infinite spatial domains. So, let us use this Fourier transform pair to solve the following heat equation \begin{align} u_t&=ku_{xx},\qquad -\infty<x<\infty,\\ u\left(x,0\right)&=f\left(x\right). \end{align} To introduce some notation, let $\mathcal F$ and $\mathcal{F}^{-1}$ stand for the Fourier transform and inverse Fourier transform operators, respectively. It follows that \begin{align} \mathcal F\left(u_t\right)&=\mathcal F\left(ku_{xx}\right),\\ \frac1{2\pi}\int_{-\infty}^\infty \frac\partial{\partial t}u\left(x,t\right)e^{i\omega x}\,dx&=\frac k{2\pi}\int_{-\infty}^\infty \frac{\partial^2}{\partial x^2}u\left(x,t\right)e^{i\omega x}\,dx,\\ \frac\partial{\partial t}U\left(\omega,t\right)&=-k\omega^2U\left(\omega,t\right). \end{align} This is an ordinary differential equation whose solution is $$U\left(\omega,t\right)=c\left(\omega\right)\exp\left(-k\omega^2t\right).$$ Applying the Fourier transform to the initial condition yields \begin{align} \mathcal F\left[u\left(x,0\right)\right]&=\mathcal F\left[f\left(x\right)\right],\\ U(\omega,0)&=\frac1{2\pi}\int_{-\infty}^\infty f\left(x\right)e^{i\omega x}\,dx\\ &=c\left(\omega\right). \end{align} To be able to apply the inverse Fourier transform to find $u$, we must know what the inverse Fourier transform of a Gaussian is, namely, $$G(\omega)=\exp\left(-k\omega^2t\right).$$ Applying the definition yields \begin{align} \mathcal{F}^{-1}[G(\omega)]&=\mathcal{F}^{-1}\left[\exp\left(-k\omega^2t\right)\right],\\ g(x)&=\int_{-\infty}^\infty\exp\left(-k\omega^2t\right)\exp\left(-i\omega x\right)\,d\omega\\ &=\sqrt{\frac{\pi}{kt}}\exp\left(-\frac{x^2}{4kt}\right). \end{align} We must also know that convolution is defined as follows: $$F(\omega)G(\omega)=\frac1{2\pi}\int_{-\infty}^\infty f(\bar x)g(x-\bar x)\,d\bar x.$$ Finally, the inverse Fourier transform of $$U\left(\omega,t\right)=\exp\left(-k\omega^2t\right)c\left(\omega\right)$$ turns out to be the following: $$u(x,t)=\frac1{2\sqrt{\pi kt}}\int_{-\infty}^\infty\exp\left(-\frac{\bar{x}^2}{4kt}\right)f(x-\bar x)\,d\bar x.$$

Friday, December 7, 2012

Complex Form of Fourier Series

I will talk about Fourier transforms in the next entry. To do so, I will first introduce a way to convert our Fourier series currently defined in terms of sines and cosines into Fourier series defined in terms of complex exponentials.

First of all, recall that $$f(x)\sim a_0+\sum_{n=1}^{\infty}\left(a_n\cos\frac{n\pi x}L+b_n\sin\frac{n\pi x}L\right),\tag{1}$$ where, by orthogonality principles, \begin{align} a_0&=\frac1{2L}\int_{-L}^Lf(x)\,dx,\\ a_n&=\frac1L\int_{-L}^Lf(x)\cos\frac{n\pi x}L\,dx,\\ b_n&=\frac1L\int_{-L}^Lf(x)\sin\frac{n\pi x}L\,dx. \end{align} Now, recall Euler's formulas: $$\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}2,\qquad\text{and}\qquad\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}.$$ It follows that we can rewrite equation $(1)$ as follows: $$f(x)\sim a_0+\frac12\sum_{n=1}^{\infty}(a_n-ib_n)e^{\frac{n\pi ix}L}+\frac12\sum_{n=1}^{\infty}(a_n+ib_n)e^{-\frac{n\pi ix}L}.$$ Now, let's change the index of summation in the first term from $n$ to $-n$: $$f(x)\sim a_0+\frac12\sum_{n=-1}^{-\infty}(a_{-n}-ib_{-n})e^{-\frac{n\pi ix}L}+\frac12\sum_{n=1}^{\infty}(a_n+ib_n)e^{-\frac{n\pi ix}L}.$$ It follows from the definition of $a_n$ and $b_n$ that $a_{-n}=a_n$ and $b_{-n}=-b_n$. Therefore, if we let \begin{align} c_0&=a_0,\\ c_n&=\frac{a_n+ib_n}2, \end{align} we will have the following Fourier series: $$f(x)\sim\sum_{n=-\infty}^{\infty}c_ne^{-\frac{in\pi x}{L}},$$ where the coefficients are $$c_n=\frac1{2L}\int_{-L}^Lf(x)e^{\frac{in\pi x}L}\,dx.$$

Vibrating String with Fixed Ends

Until now, we have been working only with the heat equation. Today, we are going to solve the one-dimensional wave equation with homogeneous boundary conditions and no sources, namely, \begin{align} u_{tt}&=c^2u_{xx},\\ u(0,t)&=0,\quad u(L,t)=0,\\ u(x,0)&=f(x),\quad u_t(x,0)=g(x). \end{align} As before, we want to use the method of separation of variables by setting $u(x,t)=\phi(x)h(t)$ and substituting above: $$\phi h''=c^2\phi''h.$$ We want to ignore trivial solutions. This implies our boundary conditions are $\phi(0)=0$, and $\phi(L)=0$.

Separating variables yields $$\frac{\phi''}{\phi}=\frac{h''}{c^2h}\color{blue}{=-\lambda},$$ Because functions of distinct independent variables can only be equal if they equate to the same constant, we have introduced the equality in blue (the negative sign is purely out of convenience later on).

We now have a system of two ordinary differential equations: \begin{align} \phi''&=-\lambda\phi,\tag{1}\\ h''&=-\lambda c^2h.\tag{2} \end{align} Equation $(1)$ has three cases:

1. $\lambda<0$:
2. $$\phi(x)=c_1\sinh\sqrt{-\lambda}x+c_2\cosh\sqrt{-\lambda}x.$$ Applying the boundary conditions to this yields \begin{align} \phi(0)&=c_2=0,\\ \phi(L)&=c_1\sinh\sqrt{-\lambda}L=0\Longrightarrow c_1=0. \end{align} This is the trivial solution, so we drop it.
3. $\lambda=0$:
4. $$\phi(x)=c_1x+c_2.$$ Applying the boundary conditions to this yields \begin{align} \phi(0)&=c_2=0,\\ \phi(L)&=c_1L=0\Longrightarrow c_1=0. \end{align} This is the trivial solution, so we drop it.
5. $\lambda>0$:
6. $$\phi(x)=c_1\sin\sqrt{\lambda}x+c_2\cos\sqrt{\lambda}x.$$ Applying the boundary conditions to this yields \begin{align} \phi(0)&=c_2=0,\\ \phi(L)&=c_1\sin\sqrt{\lambda}L=0\\ &\Longrightarrow\lambda_n=\left(\frac{n\pi}{L}\right)^2,\\ \phi_n(x)&=C_n\sin\frac{n\pi x}{L}. \end{align}
Now that we know the value of $\lambda_n$, we can use it to solve equation $(2)$: $$h_n(t)=A_n\sin \frac{n\pi ct}{L}+B_n\cos \frac{n\pi ct}{L}.$$ Therefore, by the principle of superposition, we find that the solution to this partial differential equation is $$\color{blue}{u(x,t)=\sum_{n=1}^{\infty}\left(G_n\sin \frac{n\pi ct}{L}+F_n\cos \frac{n\pi ct}{L}\right)\sin\frac{n\pi x}{L}.}$$ The coefficients $F_n$ and $G_n$ can be found by applying the initial conditions: $$u(x,0)=\sum_{n=1}^{\infty}F_n\sin\frac{n\pi x}{L}=f(x).$$ \begin{align} u_t(x,t)&=\sum_{n=1}^{\infty}\frac{n\pi c}{L}\left(G_n\cos \frac{n\pi ct}{L}-F_n\sin \frac{n\pi ct}{L}\right)\sin\frac{n\pi x}{L},\\ u_t(x,0)&=\sum_{n=1}^{\infty}G_n\frac{n\pi c}{L}\sin\frac{n\pi x}{L}=g(x). \end{align} Finally, applying orthogonality principles to the above equations to find $F_n$ and $G_n$ yields \begin{align} \color{blue}{F_n}&\color{blue}{=\frac{2}{L}\int_{0}^{L}f(x)\sin\frac{n\pi x}{L}\,dx,}\\ \color{blue}{G_n}&\color{blue}{=\frac{2}{n\pi c}\int_{0}^{L}g(x)\sin\frac{n\pi x}{L}\,dx.} \end{align} Letting $c=1$, $L=1$, $f(x)=\sin x$ and $g(x)=0$ gives us the following wave (the approximation is very low):

Regular Sturm-Liouville Eigenvalue Problem

Definition

A regular Sturm-Liouville eigenvalue problem consists of the Sturm-Liouville differential equation $$\frac{d}{dx}\left[p(x)\frac{d\phi}{dx}\right]+q(x)\phi+\lambda\sigma(x)\phi=0,\qquad a<x<b,$$ subject to the boundary conditions \begin{align} \beta_1\phi(a)+\beta_2\frac{d\phi}{dx}(a)&=0,\\ \beta_3\phi(b)+\beta_4\frac{d\phi}{dx}(b)&=0,\qquad\beta_i\in\mathbb{R}, \end{align} where $p$, $q$, and $\sigma$ are real and continuous, and both $p>0$ and $\sigma>0$. Moreover, only Dirichlet, Neumann and Robin boundary conditions are considered.

Theorems

The following theorems about it have been derived:

1. Each eigenvalue $\lambda_n\in\mathbb{R}$
2. $\lambda_1<\lambda_2<\cdots$
3. $\lambda_n$ has eigenfunction $\phi_n(x)$, and for $a<x<b$, $\phi_n$ has $n-1$ zeros.
4. The $\phi_n$ form a complete set, i.e., $$f(x)\sim\sum_{n=1}^{\infty}a_n\phi_n,$$ where $f$ is piecewise smooth. Also, with properly chosen $a_n$, this converges to $[f(x^+)+f(x^-)]/2$ on $a<x<b$.
5. If $\lambda_n\neq\lambda_m$,
6. $$\int_a^b\phi_n\phi_m\sigma\,dx=0.$$
The following is the Rayleigh quotient: $$\lambda=\frac{-p\phi\,d\phi/dx|_a^b+\int_a^b[p(d\phi/dx)^2-q\phi^2]\,dx}{\int_a^b\phi^2\sigma\,dx}.$$ Most of these theorems can be proved using Green's formula $$\int_a^b[uL(v)-vL(u)]\,dx=p\left(u\frac{dv}{dx}-v\frac{du}{dx}\right)\Bigg|_a^b,$$ where $$L\equiv\frac d{dx}\left(p\frac d{dx}\right)+q.$$ The Rayleigh quotient proves that $\lambda\geq0$ if $-p\phi\,d\phi/dx|_a^b\geq0$ and $q\leq0$.

Enlace Cube Puzzle

My brother gave me this other puzzle (see the previous one here):

Which is the longest line?

I assume this is a square box.

To make things simpler, without loss of generality, let this be a unit box, that is, let its side have length one.

Observe how the lines take circular segment shapes on the faces of the box. So, the lines can be described in terms of a sum of quarter circles. Mathematically, $$L=\frac\pi2\sum_{i=1}^{n}1=\frac{n\pi}2.$$ In other words, we don't even have to do any math at all; all we have to do is count the number of quarter circles and whichever line has the most is the longest one.

To me, it looks like the blue line has $8$ quarter circles while the red one has $7$. So, the blue line is the longest line.

By the way, bro:

Spoiler:
Mi hermano, han pasado tantas cosas este año que tendría que sentarme contigo por un día entero para poder contarte más o menos cómo me ha estado yendo. Ahorita la universidad me está medio matando con exámenes finales, pero tengo planeado escribirte cuando los termine la próxima semana (¡esta vez sí!). :)