# Administrative Information - **Class Meeting**: Wednesday, September 2, 2026, 4:00–4:50pm, **EB120**, Section D - Previous meeting: [[MATH225F26(Day 4) - Initial value problems, intervals of definition, and when the theorem is silent]] - **Slides for today**: `02_initial_value_problems_and_theoretical_concerns` and `03_slope_field_and_phase_line_analysis` - **Live notes**: `tinyurl.com/225F26-Live` --- # Lecture Boards + Transcript + GenAI **Wednesday, September 2, 2026 — the fifth lecture.** A slide-heavy day, and the one where the theorem finally acquired a picture. The morning's animation was walked through in class; then the two-check process got a *reason* rather than a recipe; then a single innocent-looking equation, $y' = y^2$, turned out to have solutions that quit before time runs out. The last twenty minutes left solving behind entirely and asked what a solution **looks like** when you never find it — ending on a vertical line with two arrows on it. ## From the video library Matching lectures from [[Teaching/MATH225/05 Video Library/Introduction to Differential Equations|the course video library]] — slope fields, and the first look at phase lines. > [!abstract]- The relevant videos, in order > - **[[Teaching/MATH225/05 Video Library/Part 1 — First-order equations/1.1 — Definitions, theory, and the geometry of slope fields and phase lines/1.1.07 — Pure time space problems, slope field definition and visualization|1.1.07 — Slope fields, defined and drawn]]** (12:28) — slope fields defined and drawn, starting from pure-time problems. [▶ watch](https://youtu.be/LfcBnSs4LVQ) > - **[[Teaching/MATH225/05 Video Library/Part 1 — First-order equations/1.1 — Definitions, theory, and the geometry of slope fields and phase lines/1.1.07.01 — A Brief on Slope Fields|1.1.07.01 — A brief on slope fields]]** (1:30) — the 90-second version, if you just want the picture. [▶ watch](https://youtu.be/LY3aWTZP2Bs) > - **[[Teaching/MATH225/05 Video Library/Part 1 — First-order equations/1.1 — Definitions, theory, and the geometry of slope fields and phase lines/1.1.08 — Slope fields for an autonomous ODE, equilibrium solutions and trajectories|1.1.08 — Slope fields for an autonomous ODE]]** (6:33) — slope fields for an **autonomous** equation, and where the equilibria show up. [▶ watch](https://youtu.be/pz_UWwethvo) > - **[[Teaching/MATH225/05 Video Library/Part 1 — First-order equations/1.1 — Definitions, theory, and the geometry of slope fields and phase lines/1.1.09 — Slope field MC questions, structures based on independent and dependent variables|1.1.09 — Slope field multiple choice]]** (8:46) — multiple-choice practice matching structures to fields. [▶ watch](https://youtu.be/n4AYqMttDek) > - **[[Teaching/MATH225/05 Video Library/Part 1 — First-order equations/1.1 — Definitions, theory, and the geometry of slope fields and phase lines/1.1.10 — Autonomous slope fields, equilibrium solutions and introduction to phase lines|1.1.10 — Autonomous fields through to the phase line]]** (20:41) — autonomous fields all the way through to the phase line (20:41). [▶ watch](https://youtu.be/ilexNqdYwTM) --- ## 1. The shortest thing anyone can say about a first-order ODE Offered at the top, as the one sentence to carry if everything else falls out: > [!important] Two ways to say the same question > **If I gave you the [velocity](https://en.wikipedia.org/wiki/Velocity) of your car for the whole trip, could you infer the [displacement](https://en.wikipedia.org/wiki/Displacement_(geometry))?** > > Or, geometrically: **if I told you the tangent structure — the slope line at every point in the plane — could you reconstruct the curve?** The field of little tangent lines has a name — a [slope field](https://en.wikipedia.org/wiki/Slope_field). That is the entire subject of first-order differential equations, and it is why the next stretch of the course is about pictures rather than formulas. ## 2. The two-check process, said once, cleanly $\frac{dy}{dt} = \underbrace{\;\cdots\;}_{f(t,\,y)}, \qquad y(t_0) = y_0$ Once the first derivative is **isolated**, everything on the right-hand side is your $f(t,y)$. That $f$ plus the initial condition is an [initial value problem](https://en.wikipedia.org/wiki/Initial_value_problem), and the [existence-and-uniqueness theorem](https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem) applies to exactly this shape. Then two checks, and only two: 1. **Check $f$** near the initial condition. 2. **Check the [partial derivative](https://en.wikipedia.org/wiki/Partial_derivative) $\dfrac{\partial f}{\partial y}$** near the initial condition. > [!tip] How you actually spot a failure > *"We don't get into the weeds associated with continuity."* You will notice a hypothesis failing because the function **lacks definition** somewhere — you are hunting for points where the functions *"start having conniptions."* And the logic from Day 4 stands: **if they are not defined, they certainly could not be continuous.** **Both true simultaneously** → exactly one curve through that initial condition. **Either one fails** → the theorem is silent. ## 3. Where these notes come from Said to the room, unprompted, and worth having on the page it describes: > After class the audio is transcribed locally, the PDFs of the lecture notes are collected, and both go to an LLM that generates a typeset set of notes for the day — the discussion, the mathematics, and the questions that came up. With a caveat delivered in the same breath: because it is transcribed audio and *"I didn't really gatekeep a lot,"* things surface that were never meant to be mathematics. The example given was the **dogs versus cats** exchange from Day 2 — which the notes had not only recorded but *adjudicated*, citing Charlie from Day 1 as the deciding evidence. That is the trade: you get the room, jokes included. It is also how image work reaches you, which is what the next section is. ## 4. Watching a hypothesis fail ![[Media/Day 5/MATH225F26-Day5-Anim-SlopeField-EU.gif]] The slope field of $\boxed{\;\frac{dy}{dt} = -y^{a}\;}$ on $0 \le t \le 2$, $0 \le y \le 1$, with **$a$ sweeping $0 \to 2$**. Every arrow is the slope the equation demands at that point; colour is arrow length, so warmer means steeper. Changing $a$ changes the differential equation, and changing the equation changes the field. Three frames were pulled out of it. ### $a = 0$ ![[Media/Day 5/MATH225F26-Day5-Frame-a0.png]] *"When $a$ is equal to zero, what for realsies is the right-hand side?"* — $-1$. So the slope is $-1$ **regardless of $t$ and regardless of $y$**, and every arrow in the picture is the same arrow. Note what these slope lines do at the bottom of the frame: they **slice straight through the horizontal axis** without noticing it. ### $a = 0.5$ ![[Media/Day 5/MATH225F26-Day5-Frame-a05.png]] Now $\dfrac{dy}{dt} = -\sqrt{y}$. As an $f$, this is perfectly well behaved for $y \ge 0$ — *"this function doesn't really care."* The trouble is the second check: $\frac{\partial f}{\partial y} = -\frac{1}{2}y^{-1/2} = -\frac{1}{2\sqrt{y}}$ and **for $y \approx 0$ this is large.** Large means: *as you move vertically, the slopes are changing a lot.* Zoom in near the axis on this frame and you find radical change in the slope lines, and slope lines **capable of crashing through the horizontal axis**. > [!important] Why the second check is the one about vertical change > The whole purpose of the $\partial f/\partial y$ hypothesis is to ask **how fast the slopes change as you move vertically** — *"because if they get too aggressive, I could have a situation happening like this where things are allowed to crash through each other."* > > That is the intuition Day 4 deferred, and this animation is the answer to it. ### $a = 1$ ![[Media/Day 5/MATH225F26-Day5-Frame-a1.png]] *"And then if I increase $a$ up to 1 proper, we find out that no longer does this issue persist."* > [!warning] What the silence does and does not mean > Where the check fails, **we cannot keep crashes from happening** — so existence and uniqueness says nothing. Asked directly whether that means crashes *must* happen: **it does not.** The theorem is only saying *"stuff is changing so much in that zone there that you could have slope lines crashing through each other."* Permission, not prediction. ## 5. What "crashing" means, concretely Pressed on the word, the picture got sharpened twice. **On the board:** if slope lines point at one another, trajectories can converge and cross. *"When existence and uniqueness is granted, solution trajectories will never cross each other. We want that. We're going to be granted that like 99% of the time."* **And in the room:** > [!quote] Leaving by the door > *"If we all decide to leave the room, we're going to follow paths that go out that door. And if the existence and uniqueness theorem is granted, that means our little velocity lines or arrows won't cross each other. If the existence and uniqueness theorem is silent, it says they might cross each other — which would mean us as styrofoam peanuts smashed through each other on our way out."* > [!example]- Board — the $a=0.5$ computation, and the key point > ![[Media/Day 5/MATH225F26-Day5-Board-1.png]] ## 6. Uniqueness says nothing about *how long* A second, separate limitation, and the pivot of the day: > [!important] Two different questions > Existence and uniqueness asks: **are we all going to find the same dependent variable?** It does **not** ask how long that solution lasts. *"Solutions, however, may not exist for all time."* **The example.** $\;y' = y^2$. What is $f$? Just $y^2$ — and notice the independent variable is not in the game at all, so it cannot cause trouble. Now run the two checks the informal way: - $f = y^2$ is a [parabola](https://en.wikipedia.org/wiki/Parabola). *"Can you draw a parabola without lifting up your pen?"* Yes → continuous. - $\dfrac{\partial f}{\partial y} = 2y$ is a line. Also drawable without lifting the pen → continuous. **Both hypotheses hold at every point of the plane.** Existence and uniqueness is absolutely met — no exceptions, no bad initial conditions. And yet the solution is $y(x) = \frac{-1}{x + C}$ Recognize the shape before doing anything with it: cover the minus sign and set $C=0$ and it is $1/x$, a [hyperbola](https://en.wikipedia.org/wiki/Hyperbola) in quadrants I and III with a **vertical [asymptote](https://en.wikipedia.org/wiki/Asymptote)**. The minus flips it; the $+C$ **shifts the asymptote left or right** — a horizontal shift, which drew a correction from the floor, where "flip" was offered first. ## 7. The interval of definition, with a singularity Impose $y(0) = 3$. Everywhere there is an $x$, put $0$: $3 = \frac{-1}{0 + C} \quad\Longrightarrow\quad C = -\tfrac13 \quad\Longrightarrow\quad \boxed{\;y(x) = \frac{-1}{x - \tfrac13}\;}$ Now the question from Day 4 returns, and this time it bites. **Which $x$ are allowed?** Not $x = \tfrac13$ — and that single missing point **breaks the real line into two pieces**: $\left(-\infty,\ \tfrac13\right) \qquad\text{and}\qquad \left(\tfrac13,\ \infty\right)$ A solution must be [continuous](https://en.wikipedia.org/wiki/Continuous_function) on its domain, and it cannot live through a vertical asymptote, so **one of these two intervals has to be chosen.** > [!warning] The choice is not arbitrary, and the room guessed wrong first > Asked which, the room picked $\left(\tfrac13,\infty\right)$. *"I disagree."* > > The initial condition was imposed at $x = 0$. **Which interval contains $0$?** The first one. **The interval of definition is the piece that contains the initial condition** — here $\left(-\infty,\tfrac13\right)$. And the point worth carrying: > [!idea] Nothing in the equation warned you > *"There's nothing about this problem that's telling me that there's any sort of weirdness going on. It wasn't until I got the solution that I saw that the solution had a singularity."* > > $y' = y^2$ passes both hypotheses everywhere. The theorem promised a unique solution and delivered one — it just never promised it would last. Every initial condition gives a different hyperbola, and of each hyperbola's two branches **you keep exactly one**: the one carrying your initial value. ## 8. The general initial height The worksheet asks the same problem with $y(0) = y_0$. The algebra is identical: $y_0 = \frac{-1}{0+C} \quad\Longrightarrow\quad C = -\frac{1}{y_0} \quad\Longrightarrow\quad y(x) = \frac{-1}{x - \frac{1}{y_0}}$ so the break in the real line now sits at $x = \dfrac{1}{y_0}$, and which side you keep depends on the sign of $y_0$: | | $\dfrac{1}{y_0}$ is | interval containing $x=0$ | |---|---|---| | $y_0 > 0$ | positive | $\left(-\infty,\ \dfrac{1}{y_0}\right)$ | | $y_0 < 0$ | negative | $\left(\dfrac{1}{y_0},\ \infty\right)$ | **Idea.** The constant adapts to the initial condition, and so does the interval — but you always catch exactly one of the hyperbola's two pieces. ## 9. Solution curves without the solution The turn of the day. Where the section is going: **sketch slope fields and use them to sketch solutions**; draw **[phase lines](https://en.wikipedia.org/wiki/Phase_line_(mathematics))** and classify **[equilibrium points](https://en.wikipedia.org/wiki/Equilibrium_point)**; and from phase lines, draw solutions. The emphasis throughout is on *sketching*. Start with something you can already do. Given $\dfrac{dy}{dx} = x^2+1$, take the [antiderivative](https://en.wikipedia.org/wiki/Antiderivative) of both sides and get a cubic. Or: ask a computer to visit a lot of points, evaluate the right-hand side, and draw *a little mini tangent line* at each. Repeat. Then be the [styrofoam peanut](https://en.wikipedia.org/wiki/Slope_field) in the flow the arrows define. *"It's like, yo, go up and to the right. And you're like, I do."* Follow the arrows and the cubic appears. > [!note] But that one was a Calc 1 problem > *"We're not in the business of these differential equations."* When the right-hand side holds only $x$, you antidifferentiate and you are done. **The interesting equations are the ones you cannot solve that way** — which is where the geometry stops being a novelty and starts being the method. ## 10. Autonomous equations, and a line that cannot be crossed $\frac{dy}{dx} = 2y(1-y)$ Independent variable $x$; dependent variable $y$; and **look at what is missing from the right-hand side.** No $x$. So if you stand at a point and move horizontally, the slope does not change — move right again, again, again, and it is *"a cut and paste of another arrow on that horizontal strip."* Set the right-hand side to zero and read off the special heights — the [equilibria](https://en.wikipedia.org/wiki/Equilibrium_point): $y = 0$ and $y = 1$. At those heights the peanut neither rises nor falls, so it keeps trucking along horizontally forever — **the constant functions $y \equiv 0$ and $y \equiv 1$ are solutions.** Everything above $y=1$ descends toward it; everything below $y=0$ runs away; and everything in between **increases**. ![[Media/Day 5/MATH225F26-Day5-Fig-TwoEquilibria.png]] > [!important] Why the rising curves do not blast through $y=1$ > The question came from the floor, and the answer is the whole week folding shut. > > $f = 2y(1-y)$ is a quadratic in $y$ — drawable without lifting the pen. $\dfrac{\partial f}{\partial y} = 2-4y$ is a line — also drawable. **So existence and uniqueness is valid at every point of the plane.** And $y \equiv 1$ is already a solution. > > A rising trajectory therefore **cannot cross $y=1$**, because crossing would mean two solution curves touching, and the theorem forbids it. *"The best they can do is asymptotically approach it."* > > Asked whether that counts as a crash: **no.** *"They will all stay separated… these curves are all just diving in and getting closer and closer, but they don't crash through."* Asked whether they ever actually touch $y=1$: *"They'll never actually touch it."* ## 11. Reading a slope field backwards Four rounds of matching a picture to its equation, and the same diagnostic each time. It is worth writing down as a procedure: > [!tip] How to identify an equation from its slope field > 1. **Move vertically.** Do the slopes change? If **no**, there is no $y$ on the right-hand side. > 2. **Move horizontally.** Do the slopes change? If **no**, there is no $t$ (or $x$). > 3. **Find the special places** — where the slopes are exactly **horizontal**. Those are the values that make the right-hand side zero. > 4. **Check signs on either side** of those places, and match against the candidates. **Round 1.** Slopes constant vertically → no $y$. Changing horizontally → there is a $t$. Horizontal slopes at $t=1$, getting more negative as $t$ increases past it. That is $\dfrac{dy}{dt} = 1-t$. A large group chose the option that vanishes at $y=1$ instead of $t=1$ — *"we just gotta keep the dependent and independent variable situated here."* And since the right-hand side holds only $t$, integrating both sides gives parabolas, which you can see in the arrows. **Round 2.** Unchanged horizontally → no $t$. Changing vertically → a $y$. Horizontal slopes at **both** $y=1$ and $y=-1$. That narrows it to two candidates, and the tiebreak is the **sign between them**: all the slopes between $-1$ and $1$ are negative. **Round 3.** Slopes change **both** horizontally and vertically, so the right-hand side needs both variables — and only one candidate had both. **Round 4.** Both variables again, and the discriminator is a special place: horizontal slopes along $y=1$. The winning right-hand side is the one that vanishes there. Worth noticing what those curves do — they descend, **shallow out, and come back up**, making horseshoe shapes, *"very different than that discussion we had where some solutions were like trying to land on that horizontal."* > [!quote]- On trusting the room > *"It's pretty amazing when you've looked through a lot of stuff. Like for instance, if I'm looking through 500 exams and I look and see most of the answers are pocketing in here, then I check. And I'm like, yeah, that is the right answer. It's like machine learning, but we're the machines."* ## 12. The phase line The last five minutes, and the bridge to Friday. $y' = y$ *"Find me a function where if I take the derivative of that function, I get the exact same thing."* The [exponential](https://en.wikipedia.org/wiki/Exponential_function) — $y(x) = Ce^{x}$, for any $C$. Do not worry yet about where that came from; *"we're gonna work on building exactly those things."* **This equation is [autonomous](https://en.wikipedia.org/wiki/Autonomous_system_(mathematics))**: the right-hand side is $f(y)$, not $f(x,y)$. And autonomy has a precise visual meaning: > [!important] What autonomy buys you > If the right-hand side holds only $y$, then one **vertical strip** of the slope field is identical to the strip to its right, and the one to its right, and so on. **You only need one vertical strip.** So compress the whole field into that one strip and draw it as a vertical line — a **phase line**. Set $f(y) = y = 0$: the constant function $y \equiv 0$ is a solution (check it — both sides are zero), and it goes on the line as a circle. Then read off the arrows: - **Above $y = 0$:** $y$ is positive, so $y' = y$ is positive, so the solution is **increasing** — an up arrow. - **Below $y = 0$:** $y$ is negative, so $y'$ is negative, so it is **decreasing** — a down arrow. ![[Media/Day 5/MATH225F26-Day5-Fig-PhaseLine.png]] Now transfer that to the $xy$-plane. The constant solution is the horizontal line at $y=0$. Start above it: increasing forward in $x$, and — pressing rewind — decreasing backward. Start below it: the mirror image. > [!idea] What just got drawn > *"Do you see that we've basically just drawn the exponential function?"* > > A vertical line with two arrows on it reproduced the shape of $Ce^x$ — **without solving anything.** That is what a phase line is for, and Friday starts there before the course turns to separation of variables. --- # Q+A from the room The live question tool was up but the queue produced no export again, so everything below was asked out loud. **On the theorem and what it forbids** > [!question]- What are the three numbered things on the side? > The **definition of continuity**, in three parts: the function exists at the point, the limit exists there, and the two agree. It was put up as a reminder — *"we don't really use it a whole lot"* — because the two checks are both continuity checks. > [!question]- What is the effect of slope lines overlapping with one another? > Slope lines pointing at each other means trajectories **could converge and cross**. When existence and uniqueness is granted, they never do. *"We're going to be granted that like 99% of the time."* > [!question]- So if the theorem holds, then solutions **will** cross? > The opposite. **When existence and uniqueness is true, solutions never cross.** > [!question]- What does a "crash" actually mean? > At that point the **slope lines touch each other**, and if the slope lines cross then the solution trajectories definitely cross. The picture is §5: everyone leaving the room by the door, and whether the paths are allowed to pass through one another. **On the $y'=y^2$ example** > [!question]- Do we just assume the independent variable is $x$? > Here, yes — but only because the *solution* was handed to you written in $x$. At the stage where you only see $y'$, the tick mark tells you nothing about whether the variable is $x$ or $t$; when someone writes $y(x)$, they have told you. > [!question]- Where did $y(0)=3$ come from? > *"I just made it up. I just chose."* **Initial conditions are things you choose** — like parameters. Absent a domain restriction, pretty much anything you want. > [!question]- Why can't we keep **both** intervals, $\left(-\infty,\tfrac13\right)$ and $\left(\tfrac13,\infty\right)$? > Because keeping both would be *"tantamount to this function going and going and then blowing up where it's no longer defined and can't be continuous there, and then magically picking up down below and continuing that hyperbola."* There is no reason to continue past a vertical asymptote. **Keep the piece that contains the initial condition.** > [!question]- Would a different initial condition give a different answer? > **100%, absolutely.** $C$ changes with the initial condition, and so may the interval — see §8. > [!question]- What is an IC? What is a singularity? > **IC** = initial condition. A **[singularity](https://en.wikipedia.org/wiki/Singularity_(mathematics))** is any point where the function diverges — here, where the hyperbola achieves infinite slope at the vertical asymptote. **On the autonomous example** > [!question]- Is that a crash — those curves piling up near $y=1$? > **No.** They stay separated and just keep decreasing (or increasing) toward the line. *"The best they can do is asymptotically approach it horizontally."* A crash would be two trajectories literally touching, and that is forbidden here. > [!question]- Why can the curves never actually reach $y=1$? > Because $y\equiv1$ is itself a solution and the theorem's hypotheses hold everywhere, so no other solution may touch it. **They approach asymptotically and never arrive.** > [!question]- What does adding $C$ inside the argument do to a graph? > A **horizontal shift** — the answer offered from the floor was "flip," and the correction matters here, because the whole point is that $+C$ slides the vertical asymptote left or right. --- > [!info]- What did not happen today > The **phase line was drawn but not argued** — *"we'll go over this argumentation more next time."* Friday picks up the phase line properly, classifies equilibrium points, and then the course finally turns to **solving**, starting with separation of variables. --- > [!example]- All lecture boards (Day 5) > ![[Media/Day 5/MATH225F26-Day5-Board-1.png]] > > 📄 [[MATH225F26-Day5-Boards-2026-09-02.pdf|Full Notability export, as of this day (7 pages, PDF)]] > > *A slide-heavy day: only one Notability page was written. The closing phase-line sketch was drawn live and did not make it into the export — the figure in §12 stands in for it.*