A Few Thoughts on the Derivative

Robert, a friend and colleague of mine from the University of Tennessee, has written multiple posts on automobiles. In a most elegant prose, he describes how the car is personified and how it eventually becomes a part of who you are. It's your best mate, your friend, something that you respect and adore. Your car can even shape your personality, thoughts and ideas. It speaks to you.

This is where I draw the inspiration for this post. While cars are certainly a passion of mine, they don't generate even close to the same emotions as it does for my friend. What does, however, is mathematics. When you live and breathe the subject day in and day out, you come to appreciate not only the most complicated elements of mathematics but also the simplistic notions as well. And why not? This is the medium by which we communicate with nature, mother Earth and the cosmos. It connects us, as humans, to everything that surrounds us. And just as we develop the bond with nature by utilizing the awesome power within this universal language, we also establish a significant connection with the language itself. Mathematics begins speaking to us and before we know it, we have personified the entire field. Each mathematical element becomes the friendly neighbor we interact with on a daily basis. One that's always there to provide interesting conversation that guides and assists you in all your endeavors. And each character is respected, admired and even feared all at the same time. For the most part, however, their countless complexities tend to bring a smile to your face. So which character shall we extol upon today? To the derivative: "Sing in me, Muse, and through me tell the story."

We start with how the derivative is defined. Take a function $f(x)$ and define it's derivative as \[ \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \frac{df}{dx} = f'(x). \] If the limit exists at a point $(a,f(a))$, then we may construct the tangent line at that point using the derivative evaluated at $x=a$ as the slope. Take for example the function $f(x) = \sin x$. We find that the derivative is $f'(x) = \cos x$. We'd like to construct the tangent line at $(\pi, 0)$. Since $f'(\pi) = \cos(\pi) = -1$ this now acts as my slope for the tangent line at the aforementioned point. So we find that the equation for the line is $g(x) = -x + \pi$. Graphing $f(x)$ and $g(x)$, we see the following.

Okay, so what? We can draw the tangent line at a point on the graph of $f(x)$ just by determining it's derivative, but does this provide us a bigger picture for functions in general? Take another look at the previous graph and you'll notice something. See how the slope of the tangent line was negative ($f'(\pi)=-1$)? Notice how the graph of the function $f(x)=\sin x$ is decreasing at that point? In fact, just by examining the derivative between particular intervals, we can determine if the original function is increasing or decreasing. If the derivative is strictly positive on some interval $I=(a,b)$, then the function is increasing within that interval. Similarly, if the derivative is strictly negative on $I$, then the function is decreasing.

What about the critical values of a function (i.e. where the function exhibits a local maximum or minimum)? Remember, the way the derivative is defined tells us the slope of the tangent line at any point. Therefore, it makes sense that the evaluation of the derivative at critical values would yield zero, since this is where the function changes direction. Thus, if I set the derivative equal to zero, I should be able to locate the critical points of the function.

Let us return to our function $f(x)=\sin x$. Since $f(x)$ is $2\pi$-periodic, we will just stay on the interval $(0,2\pi)$. Now take the derivative and set equal to zero. \[ f'(x)=\cos x=0, \ \ \ \text{on} \ \ (0,2\pi) \] This gives us $x = \pi/2$ and $x=3\pi/2$. Evaluating our function $f(x)=\sin x$ at these values gives us $f(\pi/2)=1$ and $f(3\pi/2)=-1$. So, we should have our critical points at $(\pi/2,1)$ and $(3\pi/2,-1)$ on the interval $(0,2\pi)$. You may see these coordinates plotted below along our function $f(x)=\sin x$.

Now that we know how to find the critical points of a function, how can we determine whether they are local maximums or minimums? Notice that when we take a second derivative, we obtain \[ f''(x)=\lim_{h \to 0}\frac{f'(x+h)-f'(x)}{h} \] just by using the definition of the derivative. Now suppose $x$ in the above is a critical value and $f''(x) > 0$. Then we have that \[ 0 < f''(x)=\lim_{h \to 0}\frac{f'(x+h)-f'(x)}{h}=\lim_{h \to 0}\frac{f'(x+h)}{h}. \] This means that for sufficiently small $h$, we have \[ \frac{f'(x+h)}{h} > 0. \] So, if $h \to 0^-$, then $f'(x+h)<0$ and we are decreasing. If $h \to 0^+$, then $f'(x+h)>0$ and we are increasing. This tells us that for $f''(x) > 0$ at a critical $x$, that point must be a local minimum. Similar results hold for when $f''(x) < 0$ in that it tells us that the critical value $x$ is a local maximum. This is what's known as the Second Derivative Test. Gathering all the information the derivative has given us, let's go one step further and see if we can't break in on the ground floor of ordinary differential equations (ODE). Simply put, an ODE is a function $F$ of $y=y(x)$ and all of its derivatives set equal to some other function $f(x,y)$ or \[ F(y,y',y'',y''',...,y^{(n)})=f(x,y). \] With the skill level that we've developed so far, let's focus on an ODE of order one \[ \frac{dy}{dx} = f(x,y) \] on some region $I=(a,b) \times (c,d)$. The goal now, is to determine a solution of the ODE. Not knowing any particular methods of solving the above equation, let's see if we can't just determine graphical qualities of $y=y(x)$. This shouldn't be too much of a stretch for us since we've been given that the derivative is set equal to some function $f(x,y)$. By evaluating $f(x,y)$ at a coordinate within $I$, we know the slope of the tangent line for $y=y(x)$ at that coordinate. Furthermore, we can construct the tangent line at that point. Now imagine if we evaluated $f(x,y)$ at multiple coordinates within $I$ and constructed multiple tangent lines. We start to get a better picture of the behavior of our solution $y=y(x)$ without finding the explicit solution. For example, take \[ \frac{dy}{dx}=\cos (x+y) \] and let's draw multiple tangent lines in the region $I=(0,2\pi) \times (0,2\pi)$ by using the evaluation of $\cos (x+y)$ at multiple coordinates within $I$ as the slope of our tangent lines. For simplicity, we shorten the length of each tangent line and construct a unit vector in the direction of the tangent line at each coordinate. What we see is the following.

By following the vectors, we now have an idea as to the behavior of our solution. We now display a stream plot below to provide an even better understanding of our solution $y=y(x)$.

So, without even solving for the solution $y=y(x)$, we have painted a picture of the function and examined its behavior. To conclude our story, let's start by recalling all the information the derivative has given us.

1.) We can now draw the graph of the tangent line at points where $f(x)$ is defined and the derivative exists.

2.) We can determine intervals of increase and decrease.

3.) We can find critical points as well as determine whether they are local maximums or minimums.

4.) Does this allow us to paint a picture of the behavior of the function as well as graph it? Yes it does.

5.) And we have broken in on the ground floor of a formidable subject just by using the above properties of the derivative.

In other words, the derivative has given us a wealth of information about our function $f(x)$. The power that this mathematical concept unleashes to the mathematician is incredible and yet we tend to forget just how lucky we are to possess it. So next time you're analyzing different functions or simply trying to graph them, don't forget: like a good neighbor, the derivative is there.

Kepler & Newton

I must confess that I haven't taken a single physics course while at university. In fact, the last time I took physics was in high school and this introduced me to the subject at the elementary level, but no further. Since PDEs are ubiquitous across all fields of science, I can't help but think that this must have been huge mistake on my part. Alas, the freedoms of the undergraduate. Still, the connections between mathematics and physics continue to fascinate me to the point that I persist in reading and writing about it as you have seen through my two previous posts. Even though I keep telling myself that maybe I should embark on writing a post more finely tuned towards pure mathematics, I can't help but to indulge once more on these two fields and elaborate on yet another fruitful relationship.

In 1609, Johannes Kepler published his Astronomia nova which contained his first two laws of planetary motion.

All planets move in ellipses with the Sun at one focus.

Planets sweep out equal areas in equal times.

He established these laws with the use of Tycho Brahe's observational data of the orbit of Mars. In fact, Kepler had actually set out to show that Mars traveled in an ovoid (egg-shaped) path around the Sun since he thought an elliptical trajectory was just too simple for other astronomers to overlook. As you would expect, his measurements came close to Brahe's data but was still off by a small margin. It was in 1605 that Kepler returned to the idea of the elliptical orbit and...eureka!

As for his second law, he also used the observational data on the orbit of Mars and constructed a formula in which a planet's rate of motion is inversely proportional to its distance from the Sun. To verify this formula, however, would have required extensive and tedious work and so Kepler reformulated the above via his second law.

Now enter a most beloved mathematician and physicist, Sir Isaac Newton. In 1687, Newton presented to the world his Philosophiae Naturalis Principia Mathematica which contained not only the very beginnings of calculus, but also two crucial laws of motion. One was his second fundamental law that

The net force on an object is equal to the rate of change of its linear momentum.

The second being his universal law of gravitation which states that

Any two bodies in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Mathematically speaking, we have that his second law of motion gives \[ \vec{F}=\frac{d\vec{p}}{dt}, \] where $\vec{p}=m\vec{v}$ with $m$ being the mass and $\vec{v}$ the velocity. His universal law of gravitation is written as \[ \vec{F}=-\frac{GMm}{r^2}\vec{u}, \ \ r=\|\vec{r}\|, \ \ \vec{u}=\frac{\vec{r}}{\|\vec{r}\|}, \] where $G=6.6720 \times 10^{-11} Nm^2/kg^2$ is Newton's gravitational constant, $M$ and $m$ are the two point masses and $r$ is the distance between them. The theme for the today's classroom is to show that we can derive Kepler's first two observational laws on planetary motion from the previous two laws of motion given to us by Newton. We begin by showing that planets travel in elliptical paths around the Sun. First, we note that \begin{align*} &\frac{d}{dt} (\vec{x} \cdot \vec{y})=\frac{d\vec{x}}{dt} \cdot \vec{y}+\vec{x}\cdot \frac{d\vec{y}}{dt} \\ &\frac{d}{dt} (\vec{x} \times \vec{y})=\frac{d\vec{x}}{dt} \times \vec{y}+\vec{x}\times \frac{d\vec{y}}{dt}. \end{align*} Now let's assume that our planet is revolving around the Sun and that $\vec{r}(t)$ is tracing out the planet's trajectory. We will now show that the planet's motion is planar and the Sun lies within this plane of motion. To do this, we will show that $\vec{r} \times \vec{v}$ is constant ($\vec{v}=\vec{r}'(t)$). So we have that \[ \frac{d}{dt} (\vec{r} \times \vec{v})=\frac{d\vec{r}}{dt} \times \vec{v}+\vec{r}\times \frac{d\vec{v}}{dt}=\vec{v} \times \vec{v}+\vec{r}\times\vec{a}=\vec{r} \times \vec{a}, \] where $\vec{a}=\vec{r}''(t)$. Taking into consideration Newton's two above laws, letting $M$ be the mass of the Sun and $m$ the mass of our planet we obtain \[ m\vec{a}=-\frac{GMm}{r^2}\vec{u}. \] Dividing both sides out by $m$, we have \[ \vec{a}=-\frac{GM}{r^3}\vec{r}. \] Thus, we have that \[ \frac{d}{dt} (\vec{r} \times \vec{v})=\vec{r} \times \vec{a}=\vec{r} \times \left(-\frac{GM}{r^3}\vec{r}\right)=0. \] This gives us that $\vec{r} \times \vec{v}=\vec{c}$, where $\vec{c}$ is a constant vector. Since this is true for all time $t$, we have that the motion of our planet is planar. Now let this planar motion be in the $xy$-plane. We will also find another expression for $\vec{c}$. Noting $\vec{r}=r\vec{u}$, we have that \[ \vec{v}=\frac{d}{dt}(r\vec{u})=r\frac{d\vec{u}}{dt}+\frac{dr}{dt}\vec{u}. \] From the above, we have \[ \vec{c}=\vec{r} \times \vec{v}=(r\vec{u}) \times \left(r\frac{d\vec{u}}{dt}+\frac{dr}{dt}\vec{u}\right)=r^2\left(\vec{u}\times\frac{d\vec{u}}{dt}\right)+r\frac{dr}{dt}(\vec{u}\times\vec{u})=r^2\left(\vec{u}\times\frac{d\vec{u}}{dt}\right). \] Taking note that the magnitude of $\vec{r}(t)$ is the polar coordinate $r$ of the planet's position, we now set out to find a polar coordinate description of the planet's trajectory. We have the following expression for $\vec{a} \times \vec{c}$. \begin{align*} \vec{a} \times \vec{c} &= \left(-\frac{GM}{r^2}\vec{u}\right) \times r^2\left(\vec{u}\times\frac{d\vec{u}}{dt}\right) \\ &=-GM\left[ \vec{u} \times \left(\vec{u}\times\frac{d\vec{u}}{dt}\right)\right] \\ &=GM\left[\left(\vec{u}\times\frac{d\vec{u}}{dt}\right) \times \vec{u}\right] \\ &=GM\left[(\vec{u} \cdot \vec{u})\frac{d\vec{u}}{dt}-\left(\vec{u}\cdot \frac{d\vec{u}}{dt}\right)\vec{u}\right] \\ &=GM\frac{d\vec{u}}{dt} \\ &=\frac{d}{dt}(GM\vec{u}). \end{align*} On the other hand, we have \begin{align*} \vec{a} \times \vec{c} &= \frac{d\vec{v}}{dt} \times \vec{c} \\ &=\frac{d\vec{v}}{dt} \times \vec{c} + \vec{v} \times \frac{d\vec{c}}{dt} \\ &=\frac{d}{dt}(\vec{v} \times \vec{c}). \end{align*} So we have that \[ \vec{a}\times\vec{c}=\frac{d}{dt}(GM\vec{u})=\frac{d}{dt}(\vec{v} \times \vec{c}). \] Integrating with respect to $t$, we find \[ \vec{v} \times \vec{c} = GM\vec{u} + \vec{d} \] for some arbitrary constant vector $\vec{d}$. Since both $\vec{v} \times \vec{c}$ and $\vec{u}$ lie in the same plane, then so must $\vec{d}$. Furthermore, we have that \[ \vec{u} \cdot \vec{d}=\|\vec{u}\|\|\vec{d}\|\cos \theta = d\cos \theta, \] where $d$ is the magnitude of $\vec{d}$ and $\theta$ is the angle between vectors $\vec{u}$ and $\vec{d}$. Letting $c=\|\vec{c}\|$, we have that \begin{align*} c^2 &= \vec{c} \cdot \vec{c} \\ &=(\vec{r} \times \vec{v}) \cdot \vec{c} \\ &=\vec{r} \cdot (\vec{v} \times \vec{c}) \\ &=r\vec{u} \cdot (GM\vec{u}+\vec{d}). \end{align*} Hence, \[ c^2=GMr+rd\cos \theta. \] Solving for $r$, we obtain \[ r=\frac{c^2}{GM+d\cos \theta}. \] Dividing out by $GM$ on both numerator and denominator and letting $p=c^2/GM$ and $e=d/GM$, we have \[ r=\frac{p}{1+e\cos \theta}. \] Multiplying by $1+e\cos\theta$ and solving for $r$ again yields \[ r=p-er\cos\theta. \] Since $x=r\cos\theta$ and $r^2=x^2+y^2$, we have \[ x^2+y^2=p^2-2pex+e^2x^2. \] This gives us \[ (1-e^2)x^2+2pex+y^2=p^2, \] which describes an ellipse for $0<|e|<1$, a parabola for $e=\pm 1$ or a hyperbola for $|e|>1$. Now the parabolic and hyperbolic trajectories are possible for comets or other deep space objects of the same type. Planets, however, exhibit a closed orbit around the Sun which forces us to conclude that our orbit must be elliptical. Cleaning up the above by completing the square and assuming $0<|e|<1$, we obtain \[ \frac{(x+pe/(1-e^2))^2}{p^2/(1-e^2)^2}+\frac{y^2}{p^2/(1-e^2)}=1. \] Thus, we have an ellipse centered at $(-pe/(1-e^2)^2,0)$ with semimajor axis of length $a=p/(1-e^2)$ and semiminor axis of length $b=p/\sqrt{1-e^2}$. The foci of the ellipse are at distance \[ \sqrt{a^2-b^2} = \frac{p|e|}{1-e^2} \] from the center and we see that one of them must be the location of the Sun.

Proving the Kepler's second law, will be much easier than the first. Fix a point $P_0$ on the planet's orbit. Then we have that the area swept out between $P_0$ and a moving point $P$ is given by the integral \[ A(\theta)=\int_{\theta_0}^\theta\frac12 r^2 d\varphi. \] To show that planets sweep out equal area in equal time, all we need to show is that the time derivative of $A$ is constant. By chain rule, we have that \[ \frac{dA}{dt}=\frac{dA}{d\theta}\frac{d\theta}{dt}. \] First, we calculate \[ \frac{dA}{d\theta} = \frac12 \left[r(\theta)\right]^2 \] via the Fundamental Theorem of Calculus. So we have that \[ \frac{dA}{dt}=\frac12 r^2\frac{d\theta}{dt}. \] Now let's recall that $\vec{u}=\vec{r}/r$ and $\vec{r}=(r\cos\theta,r\sin \theta)$. Thus, \begin{align*} &\vec{u}=(\cos\theta,\sin\theta) \\ &\frac{d\vec{u}}{dt}=\left(-\sin\theta\frac{d\theta}{dt},\cos\theta\frac{d\theta}{dt}\right). \end{align*} By a calculation of a cross product, we have that \[ \vec{c}=r^2\left(\vec{u}\times\frac{d\vec{u}}{dt}\right)=\left(0,0,r^2\frac{d\theta}{dt}\right). \] So we have that the magnitude of $\vec{c}$ is \[ \|\vec{c}\|=c=r^2\frac{d\theta}{dt}, \] which implies that \[ \frac{dA}{dt}=\frac12 c. \] This concludes our proof of both Kepler's first and second laws of planetary motion. $\Box$

So there you have it, a mathematical derivation of Kepler's laws from Newton's. Until next time, I leave you with a quote from Newton that should be the watchword of all mathematicians.

If I have seen further it is by standing on the shoulders of giants.

Lord Kelvin

We return now to the concept of an ideal fluid. Let's recall that this type of fluid exhibits only a pressure force that acts orthogonal to the boundary, which neglects any tangential stresses (or viscous forces). Furthermore, we assume the fluid is incompressible. This gives us the following equations governing its motion \[ \begin{cases} \partial_t \vec{u}+\vec{u}\cdot\nabla \vec{u}=-\vec{\nabla p} \\ \nabla \cdot \vec{u}=0, \end{cases} \] which are known to us as the Euler equations.

In fluid mechanics, Kelvin's circulation theorem states that in a barotropic ideal fluid with conservative body forces, the circulation around a closed curve (which encloses the same fluid elements) moving with the fluid remains constant with time. So sayeth our Lord Kelvin in 1869. Today's discussion will center around his theorem and hopefully provide us with more insight as to the formation of particular physical phenomena. So sit back, relax, pour yourself a glass of sherry and enjoy.

Before we start with the proof of Kelvin's theorem, let's recall an important fundamental concept in fluid mechanics: our particle trajectories. They are defined in the following manner (keeping in mind that we are in three dimensions)

\[ \begin{cases} \partial_t\vec{\varphi}(\vec{x},t)=\vec{u}(\vec{\varphi}(\vec{x},t),t) \\ \vec{\varphi}(\vec{x},0)=\vec{x}. \end{cases} \] Let $C$ be a simple closed contour in the fluid at $t=0$ and let $C_t=\vec{\varphi}_t(C)$ be the contour carried along by the flow. Then we have the circulation around $C_t$ to be defined as the line integral \[ \Gamma_{C_t}=\oint_{C_t}\vec{u}\cdot d\vec{s}. \] Now we are ready to prove the circulation theorem.

It suffices to show that \[ \frac{d}{dt}\Gamma_{C_t}=0. \] Let $\vec{x}(s)$ be a parametrization of $C$, $s \in [0,1]$. Then a parametrization of $C_t$ is $\vec{\varphi}(\vec{x}(s),t)$, $s\in [0,1]$. Thus, we have that % \begin{align*} \frac{d}{dt}\oint_{C_t}\vec{u}\cdot d\vec{s}&=\frac{d}{dt}\int_0^1\vec{u}(\vec{\varphi}(\vec{x}(s),t),t)\cdot\frac{\partial}{\partial s}\vec{\varphi}(\vec{x}(s),t)ds \\ &=\int_0^1\frac{D\vec{u}}{Dt}(\vec{\varphi}(\vec{x}(s),t))\cdot \frac{\partial}{\partial s}\vec{\varphi}(\vec{x}(s),t)ds + \int_0^1\vec{u}(\vec{\varphi}(\vec{x}(s),t),t)\cdot \frac{\partial}{\partial t}\frac{\partial}{\partial s}\vec{\varphi}(\vec{x}(s),t)ds \\ &=\int_0^1\frac{D\vec{u}}{Dt}(\vec{\varphi}(\vec{x}(s),t))\cdot \frac{\partial}{\partial s}\vec{\varphi}(\vec{x}(s),t)ds + \int_0^1\vec{u}(\vec{\varphi}(\vec{x}(s),t),t)\cdot \frac{\partial}{\partial s}\vec{u}(\vec{\varphi}(\vec{x}(s),t),t)ds \\ &=\int_0^1\frac{D\vec{u}}{Dt}(\vec{\varphi}(\vec{x}(s),t))\cdot \frac{\partial}{\partial s}\vec{\varphi}(\vec{x}(s),t)ds \\ &=\oint_{C_t}\frac{D\vec{u}}{Dt}\cdot d\vec{s} \\ &=-\oint_{C_t}\nabla p\cdot d\vec{s}=0. \ \ \ \Box \end{align*} Here we had $\frac{D}{Dt}=\partial_t + \vec{u} \cdot \nabla$ as our material derivative.

If we now take note of Stokes' Theorem, then we see that \[ \frac{d}{dt}\oint_{C_t}\vec{u}\cdot d\vec{s}=\frac{d}{dt}\int\int_{\Sigma_t}\vec{\omega}\cdot d\vec{A}=0, \] where $\vec{\omega}=\nabla \times \vec{u}$ is our vorticity. This tells us that not only is circulation conserved, but also the flux of vorticity. Now before proceeding, we define vortex lines and vortex tubes.

"A vortex line or vorticity line is a line which is everywhere tangent to the local vorticity vector. A vortex tube is the surface in the continuum formed by all vortex-lines passing through a given closed curve in the continuum."

Furthermore, the strength of a vortex tube is flux of vorticity across a cross-section of the tube.

It was in 1858 that Hermann von Helmholtz published his result on vortex tubes stating that the strength of a vortex tube for inviscid fluid motion remains constant with time, which can certainly be validated now by the work of Lord Kelvin (the flux of vorticity is conserved). From here, we observe that if a vortex tube is stretched and its cross-sectional area decreases, then the magnitude of our vorticity vector $\vec{\omega}$ must increase. To see this, consider the time evolution of a vortex tube. According to Helmholtz' theorem, tubes move with the fluid. Moreover, their strength does not change with time. Therefore, if the area of a cross-section of a tube should become very small, vorticity would have to amplify proportionally. Since the fluid is incompressible, however, the volume between two sections of the tube remains constant. Therefore, any shrinking of the cross-sectional area must be accompanied by a longitudinal stretching. We conclude that the local stretching of a vortex tube gives rise to a proportional amplification of the magnitude of vorticity. This is the so-called "vortex stretching mechanism."

We are now able to understand the basics of tornadogenesis. Here one envisions a circular shear flow parallel to the earth's surface. Any circular cylinder centered at the center of flow will have the flow field tangent to its boundary, and the vortex lines are straight up. Such a cylinder is therefore a vortex tube.

Now suppose there is a vertical density gradient caused by heating of the earth's surface and a resulting updraft caused by the buoyancy effect. In this, the air moves rapidly upward in localized regions, and to satisfy conservation of mass, other air has to move rapidly in those regions. The density variation is actually minor although the convective winds resulting from it are not minor. So the usual assumption of incompressibility is okay.

This movement is superimposed as a perturbation of the basic circular shear flow mentioned above. The vortices are now the prefered regions of updraft. So we have a vortex together with an updraft in the same direction. The updraft has the effect of stretching the vortex tube. In stretching the vortex tube, cross-sectional areas contract, which results in an intensification and localization of the vorticity, hence a strengthening of the associated swirling winds. A tornado is now born.

For further reading, I recommend

A. Chorin, J. Marsden, A Mathematical Introduction to Fluid Mechanics, Springer-Verlag, (1979).

P. Fife, A Gentle Introduction to the Physics and Mathematics of Incompressible Flow, Course Notes, 2000.

Capturing the Hearts and Minds

I write to you now from my office at the University of Notre Dame. It has been a long five years and as I sit here at my desk, I reflect back on teaching mathematics to the students of this incredible university. One of these reflections centers around maintaining the mathematical scrutiny of my class and their drive to discover.

Why are they here in the first place, you might wonder. Well, when your class consists primarily of engineers, about the only thing you can see in their eyes are dollar signs. Their thought process: Memorize the formula, excel in your technological prowess and you will be well on your way to a very comfortable lifestyle. And while this may to some extent be true, they are not sitting in the classroom for the right reason. They are not asking the right question: why? Why do these equations and these mathematical techniques exist? Where do they come from?

I say these things because I once thought this way and that's certainly not the right route to take when learning about science and mathematics. Let me take you back in time about two years ago when I was down for the holidays visiting with my family in the low country of Georgia. I was spending time with a good friend of mine in the area who works at the Federal Law Enforcement Training Center in Brunswick. We were sitting around sipping on some of his fine scotch when he started asking me questions about my research.

That's when I gave him the following set of equations \[ \begin{cases} \partial_tu+u\partial_xu+(1-\partial_x^2)^{-1}\partial_x\left[u^2+\frac{1}2(\partial_xu)^2+\frac{\sigma}2\rho^2\right]=0 \\ \partial_t\rho+\partial_x(u\rho)=0. \end{cases} \] I told him that here $u(x,t)$ represents fluid velocity and that $\rho(x,t)$ represents the fluid surfaces' deviation from equilibrium. From there, I began to explain everything that I've proven about this system of partial differential equations, but before I could finish he asked the right question: why? Why am I looking at this particular set of equations and where does it originate? To be honest, I didn't have an answer for him.

So, I hopped on a plane back to South Bend and my journey in search of the origins of these equations began. It didn't take long to discover that they stem directly from Newton's second law: Force equals the change of momentum or \[ \vec{F}=\frac{d\vec{p}}{dt}=\frac{d(m\vec{v})}{dt}=m\vec{a}, \] where $\vec{p}=m\vec{v}$ is momentum, $\vec{a}$ is acceleration and mass is understood to be constant. It was in 1687, that Sir Isaac Newton wrote down his three fundamental laws of motion and has been the basis for all classical mechanics ever since.

Then in 1757, Leonard Euler took it a step further and wrote down the governing equations for the motion of an ideal, inviscous fluid \[ \begin{cases} \partial_t\vec{u}+\vec{u}\cdot\nabla\vec{u}=-\vec{\nabla p} \\ \nabla \cdot \vec{u}=0, \end{cases} \] where $\vec{u}(\vec{x},t)$ is the velocity vector field and $p(\vec{x},t)$ is the scalar pressure acting orthogonal to the fluid region.

Unfortunately for Euler, very few, if any, fluids are inviscous and it took until 1821 for a gentleman by the name of Claude Navier to jump in and add viscosity to the set of equations and another 24 years till a rigorous derivation of the following set of equations that Navier produced \[ \begin{cases} \partial_t\vec{u}+\vec{u}\cdot\nabla\vec{u}=-\vec{\nabla p}+\nu\Delta\vec{u} \\ \nabla \cdot \vec{u}=0. \end{cases} \] was given by Sir George Gabriel Stokes. Here, we have $\nu$ is our viscosity coefficient.

It was at this moment that I saw where my set of equations came from. By taking the divergence ($\text{div}=\nabla \cdot$) of the Euler equations and solving a Poisson problem, one can see the similarity between the set of equations I posed to my friend and those that I just mentioned. It's all about the physical context that you are viewing the equations. A particular scientific lense that one looks through to view the birth of the set of equations that I work with from Euler, its father.

Returning from this short digression to the matter at hand, we must always ask our students the same questions my friend was asking me. Have them travel further on down the rabbit hole in search of the truth. To mathematically scrutinize all the models, methods and scientific madness that fills their courses. For it's not just a memory game of who can store all the equations in their head. It's the one who thinks about the situation, gets enveloped in the physical context and learns something.

We all have different opinions as to what fills our life with great purpose. For me, it's about learning as much as I can of the world that surrounds me and beyond. What better way to do so than through the looking glass of science. My only hope is that through the time I devote lecturing my students about this particular philosophy, some will take it to heart, break free from the temptations to just memorize formulas and actually participate in the field of mathematics.