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# Snowfall problem.
Let us look at an example of where a differential equation emerges to **model a problem** (by describing rates of change). We will see to do this we would need to make copious amount of **assumptions**. And at the end, we will still need to **solve the differential equation**, **use the conditions given**, and **answer the question**.
Alright, consider the following poem, Snowfall:
>*Snow falls before noon,*
>*falling at a constant rate.*
>*Ted starts cleaning at noon,*
>*cleaning at a constant rate.*
>*The first hour Ted cleans one mile,*
>*but the next, only half a mile.*
>*When did snow fall begin?*
## Modeling the problem.
Looking at this, one might feel lost. But let us analyze it. Since this describes Ted cleaning snow, and moving some amount of distance, perhaps that is some unknown function we like to consider. So say $x(t)$ describes the distance travelled by Ted as a function of time $t$. Now these things all need units, so say $x$ is in miles and $t$ is in hours, and that $t=0$ is noon, and $x=0$ is the initial position of Bob.
Now, we might as well assume Ted is travelling on a perfectly flat and straight road of some constant width $w$. The fact that Ted travels less distance as times goes on, is because snow kept piling up in front of him! Since this height faced by Ted is also changing in time, let us denote $h(t)$ as the height of snow in front of Ted at time $t$.
Woah -- this is a lot of unknowns, but let us think. the poem says snow falls at a constant rate, so the snow piling up should be increasing at a constant rate. We know what kind of function increases at a constant rate -- a linear function. So we know $$
h(t) = a t + b
$$
for some constants $a,b$. Now, since snow is piling up, we should have the rate $a > 0$. Also we know something about $b$ -- this is $h(0)$, This is the height of snow faced by Ted at time $t = 0$. Since Ted started his cleaning at noon, and snow had begun falling before noon, the height of snow faced by Ted at noon is positive, so $b > 0$.
Ok, what else have we not used? Ted is cleaning snow at a constant rate -- and it is this cleaning that is allowing him to move forward. How do we reckon this? Let us denote $V(t)$ to be the total amount of snow cleaned by Ted at time $t$. What we do know is $$
\frac{dV}{dt} = c,
$$for some constant $c$. We now will try to relate all these together, using an **infinitesimal** method -- analyze what happens in a small amount of time, say $\Delta t$, and then taking the limit to zero. Let me show you.
Say at time $t$ Ted is at position $x(t)$ and the height in front of Ted is $h(t)$. Now say a small amount of time $\Delta t$ has elapsed since. Ted is now at the position $x(t+\Delta t)$ and the height in front of him is now $h(t + \Delta t) = at +a\Delta t + b$ . If you draw a picture (do it!) we now have a trapezoid that is the cross section of snow that would be cleaned in this $\Delta t$ time interval. Now, actually not quite, since the motion forward is not constant, so this is only an approximation, but should be good enough. So we can say that the amount of snow cleaned between $t$ and $t+\Delta t$ is approximately $$
\Delta V \approx \frac{1}{2}w \cdot [(at+b) +(at + a\Delta t +b)] \Delta x
$$Now, let us divide by $\Delta t$, we get $$
\frac{\Delta V}{\Delta t} \approx \frac{1}{2} w \cdot [2at + 2 b + a\Delta t] \frac{\Delta x}{\Delta t}
$$Now you can probably smell what's next -- follow your nose! Take the limit as $\Delta t \to 0$, and we get $$
\frac{dV}{dt} = w \cdot (at + b) \frac{dx}{dt} = c,
$$which we know it is a constant $c$. What we get here is a **differential equation**, an equation about some unknown function and its derivative: $$
\frac{dx}{dt} = \frac{c / w}{at + b}
$$ Huzzah! Before we move on, just briefly reflect that in order to obtain this differential equation, some assumptions are made, and they may or may not be good assumptions, but it is a start. But after that we have a purely mathematical problem to tackle.
## Solving the differential equation.
Can we solve for this unknown function $x(t)$? Yes, note that we actually are given the derivative of $x(t)$, so we can directly employ **fundamental theorem of calculus**, and get that $$
x(t) - x(0) = \int_{s=0}^{t} \frac{c}{w(as+b)} ds = \frac{c}{aw} [\ln(at+b)-\ln(b)].
$$
And since the initial position $x(0) = 0$, with some simplification we get $$
x(t) = \frac{c}{aw} \ln\left( \frac{a}{b}t+1 \right).
$$Great! We've solved the differential equation for the position of Ted $x(t)$, but really just its form -- what it looks like. Did we answer the question? We want the time when snow starts falling, call it $t^{\ast}$ Since snow starts falling before noon, and that noon time is $t = 0$, we see that $t^{\ast} < 0$.
## Using conditions given and answering the question.
Where might we be able to infer this time? The height of the snow piling up! At this time $t^{\ast}$, the height of the snow should be zero! So $h(t^{\ast}) = at^{\ast}+ b = 0$, in other words $$
t^{\ast} = - \frac{b}{a}.
$$
Ok, so far so good. But we still don't know about $a$ nor $b$. Are there any information we have not yet used? Indeed, we know the positions of Ted at time $t=1$ and $t=2$, which are $x=1$ and $x= 3 / 2$ respectively. So we know $$
\frac{c}{aw} \ln\left( \frac{a}{b} + 1 \right) = 1\quad,\quad \frac{c}{aw}\ln\left( \frac{2a}{b} +1\right) = \frac{3}{2}.
$$
Note we are interested in $t^{\ast}$, which is some ratio of $a$ and $b$, so let us treat $\frac{a}{b} = r$ as a single variable. Now, dividing the two equations we get $$
\frac{\ln(r +1)}{\ln(2r+1)} = \frac{2}{3}
$$which simplifies to the algebraic equation $$
(r+1)^{3} = (2r+1)^{2}.
$$
Expanding this with **binomial theorem**, and simplifying, we get $$
r^{3} - r^{2} -r = 0
$$giving three solutions, $$
r = 0, \frac{1+\sqrt{5}}{2} , \frac{1-\sqrt{5}}{2}
$$Since $t^{\ast} = - \frac{b}{a} =-\frac{1}{r}$, only one of them makes sense (think about why!). We deduce that $$
t^{\ast} = \frac{-2}{1+\sqrt{5}} \approx - 0.6180 \text{(hr)},
$$
which translates to $37$ minutes before noon, or $11:23$ AM!
## Some remarks.
Let us take stock of what we saw.
- When dealing with a real world problem, we may be able to obtain a differential equation if we can some how describe the rates of change. In doing so many assumptions are made. Whether they are "good" assumptions will ultimately depend on verification against actual data. Nevertheless, once we have a differential equation it is now a math problem, which is what we will focus on more. We will visit more of these kinds of "derivation for a differential equation" together, just to get the hang of it. One day, you may need to come up with a model yourself!
- The differential equation we obtained has the form $$
y'(t) = f(t)
$$for some unknown function $y=y(t)$, and $f$ is a function of $t$ alone. In this case we can use FTC directly, obtaining $$
y(t)-y(t_{0}) = \int_{t_{0}}^{t}f(s)ds
$$
- However as we will soon see, differential equations can take other forms where FTC is not directly applicable. For example, what if the unknow function $y(t)$ satisfies $y' = y$? or $y'=y + t$? It is not clear what integrating both sides directly gives. For instance if we have $y'=y$, naively integrating both sides give $y(t)-y(0) = \int_{0}^{t}y(s)ds$. But we don't even know what $y$ is, how do we know what is the result of the RHS integral? (What is interesting, however, is that we turned our differential equation into an equivalent **integral equation**.) We will develop some strategies for these.
- A differential equation problem often comes with conditions, and we should use them to solve the problem. Pay attention to them. A special type of condition is called initial values, which we will look at them later. They are special in the sense that under the right condition we can ensure unique solution.
- Which brings up another point, how do we know the solution we get is unique? Or if one even exist at all? We will look at this as well in the form of existence and uniqueness theorem. Here is an example of a differential equation problem with multiple valid solutions: $y'(x)=y^{2/3}$ with $y(0) = 0$. One can show that each of the following is in fact a solution on the entire real interval $(-\infty,\infty)$:
- $y(x)=0$ for all $x \in (-\infty,\infty)$ is a solution.
- $y(x)=\frac{1}{27}x^{3}$ for all $x \in (-\infty,\infty)$ is a solution.
- $y(x) = \begin{cases}0 & \text{if }x< 0\\ \frac{1}{27}x^{3} & \text{if } x\ge 0 \end{cases}$ for all $x \in(-\infty,\infty)$ is a solution.
- $y(x)= \begin{cases} \frac{1}{27}x^{3} & \text{if } x \le 0\\0 & \text{if } x\in(0,T) \\ \frac{1}{27}(x-T)^{3} & \text{if } x \ge T\end{cases}$ is a solution on $(-\infty,\infty)$, for any fixed positive $T$!
- Plot those out! Can you come up with others?
- Or, what does it even mean to be a solution to a differential equation? What qualifies as a solution? What if we can't solve it exactly, can we approximate it?
Okay, a lot ahead of us. Let us start by establish some basic terminology and classification for differential equations.
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