Date modified: Tue 2025-07-01 . 07:18 PM
Date created: Sun 2025-06-29 . 09:07 PM
# Week 2 Tuesday. ## What the curve?! Parabolic reflectors. - Real world **problem** -> (**formulate** into) -> A mathematical **model** -> (mathematical analysis) -> mathematical **result** -> (**interpret** the answer) -> real world **problem**. - Let us look for a curve (shape) $C$ with the following property: Fix a direction $\vec v$, this curve $C$ would reflect all rays of light parallel to $\vec v$ to the same point $F$. Call this the **focal point**. - We need to model how reflection works: We say a ray reflect off of a point on curve $C$ if it bounces off the tangent line and goes across the normal with the same angle. The normal is perpendicular to the tangent. See picture. - Let us simplify our problem. Without loss, we can say the curve has functional form $y=y(x)$, and the rays are all parallel to the $x$-axis, coming from $+\infty$ from the right to left, and that they are all to reflect to the point $F=(0,0)$. We expect a symmetry above and below the $x$-axis, so consider the rays are above the $x$-axis, and so we take our curve $y(x)$ with $y > 0$. - We now use geometry to help us set up a differential equation model for this problem. - Suppose some light ray at some height hits the curve at $P(x,y)=(x,y(x))$. This ray is then reflected to point $F(0,0)$. - Draw out the tangent line at $P$ and the normal line at $P$. - The tangent line hits the $x$-axis, say at point $A$. Now we have triangle $\triangle AFP$. Using the fact that the light reflects across the normal at equal angles, we can deduce that $\triangle AFP$ is isosceles. And that the angle $\alpha = \angle PAF$ is half of angle $\beta = \angle PF(+\infty)$. - But $\tan(\alpha)$ is the slope of the tangent line to $y$ at $P$, so $\tan \alpha = y'(x)$. - And $\tan(\beta)$ is the slope of the line segment $FP$, so $\tan \beta = \frac{y(x)}{x}$. - As $\beta = 2\alpha$, we can use **double-angle formula for tangent**, $\tan(\beta)=\tan(2\alpha) = \frac{2\tan(\alpha)}{1-\tan^{2}(\alpha)}$. - This gives $\frac{y}{x}=\frac{2y'}{1-(y')^{2}}$. Wow looks scary! But this is quadratic in $y'$. Geometrically $\frac{2y'}{1-(y')^{2}}$ represents the slope of a line that has twice the angle of the tangent line. - Cross multiplying gives $y-y(y')^{2}=2xy'$, so we have $y(y')^{2}+2xy'-y=0$. Applying quadratic formula gives $$ y' = \frac{-2x \pm \sqrt{4x^{2}+2y^{2}}}{2y} $$Still this look scary, but we can write this differential form, and get $$ 2y dy = -2x dx \pm2\sqrt{x^{2}+y^{2}}dx $$or in other words $$ \frac{xdx+ydy}{\sqrt{x^{2}+y^{2}}}=\pm dx. $$ - Now the left-hand side we can recognize its potential (from homework)! We have $$ d(\sqrt{x^{2}+y^{2}}) = \pm dx $$Hence undoing the $d$, we get $$ \sqrt{x^{2}+y^{2}}=\pm x + C $$for some constant $C$. - With some algebra, we square this out and get $$ x^{2} + y^{2} = x^{2} \pm 2xC + C^{2} \implies y^{2}=\pm2xC + C^{2} $$Note the $C$'s are the same $C$, so we cannot change one to another. But we see that these are side ways parabolas! $x = \pm\frac{y^{2}-C^{2}}{2C}$. And since the rays are coming from the right, to the left. We see that we take the positive of these. So $x = \frac{y^{2}-C^{2}}{2C}$ are these parabolas that would reflect all rays parallel to the $x$-axis coming from the right to the left, to the point $F(0,0)$! - It's good to plot these on Demos, and see that they geometrically make sense! - Note: Parabolas reflects a set of parallel ray of light to a common point. Ellipses reflect rays emanating from one fixed point to another fixed point. And hyperbolas reflect rays aiming at one fixed point to another fixed point. ## Autonomous system and phase line analysis of equilibrium solutions. - If a differential equation $y=y(x)$ has a slope field that has no explicit $x$ in it, where $y'(x)=f(y)$, then it is said to be **autonomous**. - In this situation, solutions are **translational invariant**. That is, if $y(x)$ is a solution, then so is $y(x+C)$. Proof. Indeed, since $y' = f(y)$, write $g(x)=y(x+C)$. Then $g'(x)=y'(x+C)=f(y(x+C))=f(g(x))$. That is, $g'=f(g)$. :) $\blacksquare$ - This is also clear from the **slope field** plot of $y'=f(y)$. If one translates horizontally, it has the same slopes. - If the independent variable is **time**, then this means the system is independent of **when** it starts. Starting the system at $t_{0}$ will evolve the same had you start the system at $t_{0}+C$. The curve just shifts left or right. This is usually desirable for a physical system, where the time of experiment is not important. (Not every physical system is like this!) - Examples. $y'=ky$ growth of rabbits. $y'=ky-a$, rabbits with a predator. $y'=ky(A-y)$, rabbits with a carrying capacity of $A$, spread of rumors, diseases. - When the DE is autonomous $y'=f(y)$, we can find **equilibrium** (constant) solutions $y(x)=K$ by setting $y'=0$. This amounts to solving for $y$ where $f(y)=0$. - Example. Find all equilibrium solutions to $y'=(y+2)^{2}(y-1)\ln(y+5)$. - Note this is autonomous, and $y'=0$ when $y = -2$, $y=1$, and $y=-4$. - Equilibrium does not mean stable! We can analyze stability by drawing out something called the **phase line** of the system. - Draw a vertical line denoting the $y$-axis. Label in all the equilibrium points. And draw arrow towards positive if $y' > 0$; and draw arrow towards negative if $y' < 0$. This tells us the **flow** of the system, given a starting initial condition. This is sign analysis! - If an equilibrium point is between two arrow points to it, then it is a **stable equilibrium**. In any other cases we will call it **unstable** in this class (sometimes you can define notions as semi-stable, but it is really just unstable in practice, due to random perturbations. Think of a ball on a hill or in a bowl.) - Example. Classify the equilibrium solutions to $y'=(y+2)^{2}(y-1)\ln(y+5)$ as stable or unstable. - $y=-4$: Stable; $y=-2$: Unstable; $y=1$: Unstable. ## Logistic equation. - An autonomous DE of the form $y'=k(A-y)y$ is often called a **logistic equation**, for the S-shape solutions that it has. No one really knows _why_ it's the word "logistic". - It is often to model population growth with a carrying capacity. - Suppose there is a forest with 100 acres. And the forest initially caught on fire. Let us denote $y(t)$ as the amount of area that is on fire. Then we can crudely model it as $y'=ky(100-y)$. This is saying the rate of change of the fire spread is jointly proportional to what's currently on fire, and what has yet to be burned. - By drawing the slope field out, and the phase line next to it, we can see the predicted behavior of solutions. So if $y(0)=1$, it will flow to the equilibrium solution $y=100$ (if $k > 0$). - Now suppose we introduce a fire fighting effort $r$, some constant rate to decrease the fire. Then our model is now $y'=ky(A-y) - r$. This is often constant or at best constant because there is only so many firefighters in the world. (It does not scale as there is more fire!) - By plotting $y'=f(y)$ the graph of $f(y)=ky(A-y)$ vs $y$, we can see the effect of $-r$. - For example suppose the initial condition is $y(0)=1$, what is $r$ so that $y'=2y(100-y)-r$ will give a solution $y$ such that $y(t^{\ast})=0$, that $y$ has a chance to evolve to $0$. - By graphing it out we see that we need $r$ where $f(1)=0$, so $r=2\cdot98=196$. ## Mixing example. Let's mix solutions. - Keeping track of the relations: mass (M), volume (V), and concentration (C). - A large tank initially holds 10 L of salt water of concentration 2 g/L. - A stream of salt solution of 5 g/L is added to the tank at 4 L / min. - Assume the solution mix instantly - While at the same time, the tank drains out 4 L / min of mixed solution out. - If $C(t)$ denotes the concentration of the solution in the tank, what is $\lim_{t\to\infty} C(t)$? Find an expression of $C(t)$. - Analysis of the problem. - Concentration is mass over volume. The volume of the tank is always 10 L. Denote $M(t)$ to be the mass of salt in the tank. So initially $M(0) = 20$ g. - If we can solve $M(t)$, then $C(t) = \frac{M(t)}{10}$, or $M(t)=10C(t)$ - Let time $t$ be in minutes. Then (5 g/L)(4 L/min) = 20 g /min is the rate of addition of salt to the tank. - But at the same time, a well-mixed solution is leaving, this means we are losing at the rate of $C(t)(4)$ g/min of salt. - So the net rate of change of salt is $\frac{dM}{dt}=20-4C$, so $10 \frac{dC}{dt}=20-4C$, in other words $\frac{dC}{dt} = 2 - \frac{4}{10}C$. - This is an autonomous system with equilibrium solution $2-\frac{4}{10}C=0$, or $C=5$ as expected. And this is a stable solution. We expect $C(t) \to 5$ as $t \to \infty$. - Solving it explicitly, $\frac{dC}{2-\frac{4}{10}C}=dt$ gives $\frac{-10}{4}\ln|2-\frac{4}{10}C|=t+K$, or $2-\frac{4}{10}C(t)=Ae^{-4/10 t}$, or $C(t) = 5-Ae^{-4t/10}$. And if $C(t) = 2$, we get $A=3$. So the expression is $C(t)=5-3e^{-4t/10}$. - **Variations. What if volume is not constant throughout? Then $V=V(t)$ and we need to keep track of it!** - A large tank initially holds 10 L of salt water of concentration 2 g/L. - A stream of salt solution of 5 g/L is added to the tank at 4 L / min. - Assume the solution mix instantly - While at the same time, the tank drains out 3 L / min of mixed solution out. - If $C(t)$ denotes the concentration of the solution in the tank, what is $\lim_{t\to\infty} C(t)$? Find an expression of $C(t)$. - We have $C(t) = \frac{M(t)}{V(t)}$, where $M(t)$ denotes mass of salt in tank, $V(t)$ denotes volume of solution in tank. Let time $t$ be in minutes. - What is the rate of change of the mass of salt added? It is 20 g/ min. - What is the rage of change of mass of salt leaving? It is $C(t)\cdot3$ g/min. - So $\frac{dM}{dt}=20-3C$. But $M(t)=C(t)V(t)$, and $M'(t)=C'(t)V(t) +C(t)V'(t)$. - Do we know $V(t)$? Yes $V(t) =10+t$, from the information given. - So $C'(t)(10+t)+C(t)=20-3C(t)$, $C'(t)=\frac{dC}{dt}=\frac{20-4C}{10+t}$. This is separable so we solve. - $\frac{dC}{20-4C}=\frac{dt}{10+t}$, we have $\frac{-1}{4}\ln(20-4C)=\ln(10+t)+K$. So $20-4C=\frac{A}{(10+t)^{4}}$, in other words $C=5-\frac{A}{(10+t)^{4}}$. Since $C(0)=2$, we have $C(t)=5-\frac{30000}{(10+t)^{4}}$. - **Variation. Double mixing problem.** - A large tank initially holds 10 L of salt water of concentration 2 g/L. - A stream of salt solution of 5 g/L is added to the tank at 4 L / min. - Assume the solution mix instantly - While at the same time, the tank drains out 3 L/min of mixed solution out into second tank. - The second tank initially holds 5 L of salt water of concentration of 3 g/L. - The second tank mixes instantly. - The second tank drains at a rate of 3 L/min. - What is the set up? We have $M_{1}$, $M_{2}$, $V_{1}$, $V_{2}$, $C_{1}$, $C_{2}$. Let $t$ be time in minutes. - $\frac{dM_{1}}{dt}=20-3C_{1}$, and $M_{1}=C_{1}V_{1}$ with $M_{1}'=C_{1}'V_{1}+C_{1}V_{1}'$. And $V_{1}=10+t$. - $\frac{dM_{2}}{dt}=3C_{1}-3C_{2}$, and $M_{2}=5C_{2}$. So $5C_{2}'=3C_{1}-3C_{2}$. - We have $C_{1}=5-\frac{30000}{(10+t)^{4}}$, so $C_{2}'=\frac{3}{5}\left( 5- \frac{30000}{(10+t)^{4}} \right)-\frac{3}{5}C_{2}$. - $C_{2}'=3-\frac{18000}{(10+t)^{4}}-\frac{3}{5}C_{2}$, or $C_{2}'+\frac{3}{5}C_{2}=3-\frac{18000}{(10+t)^{4}}$ - This is linear, however we can only express the answer as an integral - Integrating factor $\mu=e^{3t/5}$. And $$ C_{2}(t) = e^{-3t/5} \left[\int_{0}^{t} e^{-3s/5} (3- \frac{18000}{(10+s)^{t}}ds+K\right] $$Since $C_{2}(0) = 3$, $K=3$. So $$ C_{2}(t) = e^{-3t/5} \left[\int_{0}^{t} e^{-3s/5} (3- \frac{18000}{(10+s)^{t}}ds+3\right]. $$