# Where linear stability breaks down, and critical slowing down > [!info] Before you start > **Instructions** *(TK)* > > 🎯 **[MR] Canvas quiz for this page** *(link TK)* > > Keep the quiz open in another tab while you work. The questions are in the boxes marked **Quiz** below, in the same order as on Canvas. > [!abstract] What this page is for > On Day 1 we took fish out of a pond and watched two equilibria slide toward each other, and the Workbook 1 companion showed that $f'$ at an equilibrium is a rate. This page puts those two ideas together. You will push the pond to the edge in two different ways, see exactly where linear stability analysis stops working, and see what the pond does instead. Budget **45 minutes to an hour**. > > The applets only run on **scottastrong.org**. If you see an empty box where an applet should be, you have landed on the publish.obsidian.md copy of the page, so switch to the scottastrong.org address. > > Related pages are [[MATH310F26(Day 1) - Instructor information and introduction to the course|Day 1]], [[MATH310F26(Day 2) - Modeling in your own words, and two videos|Day 2]] and the [[MATH310F26(Workbook 1 Companion) - The phase line and what f-prime tells you|Workbook 1 companion]]. --- ## Part 0 · Model A, a constant quota The pond from the first week, with a constant harvest $H$ in fish per unit time, $ \frac{dP}{dt} = rP\left(1-\frac{P}{N}\right) - H, \qquad P^{*}_{\mp} = \frac{N}{2}\left(1 \mp \sqrt{1-\frac{4H}{rN}}\right). $ Throughout this page $r = N = 2$, the same pond as the Workbook 1 sheet, so the equilibria are $P^{*}_{\mp} = 1 \mp \sqrt{1 - H}$. In the applet the left panel is $f(P)$ with the phase line drawn on the $P$ axis (filled circle for a sink, open circle for a source, half-filled when the two collide). The right panel shows solutions from ten starting populations, green if they survive and red if they die out. <div class="applet" data-applet="harvest-models" data-models="A" data-bifurcation="off"></div> **Things to do.** Leave $r$ and $N$ at $2$. Use the preset buttons to step through $H = 0$, $H = H^{*}/2$, $H = 0.95H^{*}$, $H = H^{*}$ and $H = 1.1H^{*}$, and then drag the slider slowly across the whole range. Count the circles on the phase line at each stop, read $P_{-}$ and $P_{+}$ from the readout, and watch which solutions turn red. > [!question] Quiz Q1 · multiple choice > With $r = N = 2$, at what harvest rate $H$ do the two equilibria of Model A merge into one? > - $H = 0.5$ > - $H = 1$ > - $H = 2$ > - $H = 4$ > [!question] Quiz Q2 · multiple choice > Set $H = 0.95H^{*}$. The pond sits at $P_{+} \approx 1.22$ and looks healthy. A bad winter knocks it down to $P = 0.6$. What happens next? > - It grows back to $P_{+}$. > - It settles at $P_{-}$ and stays there. > - It keeps falling and reaches $P = 0$ at a definite, finite time. > - It decays toward $P = 0$ but never actually gets there. > [!question] Quiz Q3 · free response > Describe the solution trajectories of Model A before the bifurcation ($H < H^{*}$), at the bifurcation ($H = H^{*}$), and after it ($H > H^{*}$). For each case say which starting populations survive, which do not, and what the survivors settle on. --- ## Part 1 · Model B, a constant effort Instead of a fixed quota, fish with a fixed effort, so the catch is proportional to how many fish there are, $ \frac{dP}{dt} = rP\left(1-\frac{P}{N}\right) - HP . $ Now $H$ is a rate per fish (one over time) rather than fish per unit time, and every term on the right has a factor of $P$. <div class="applet" data-applet="harvest-models" data-models="B" data-bifurcation="off"></div> **Things to do.** Before touching the slider, factor $P$ out of the right-hand side and find the equilibria on paper. Then step through the presets $H = 0$, $H = r/2$, $H = 0.95r$, $H = r$ and $H = 1.1r$ and compare with what you found. Watch the open and filled circles as $H$ passes $r = 2$, and compare with what the circles did in Model A. > [!question] Quiz Q4 · multiple choice > For Model B with $r = N = 2$ and $0 < H < 2$, what are the equilibria? > - $P = 0$ and $P = 2 - H$ > - $P = 1 \mp \sqrt{1 - H}$ > - $P = 2 - H$ only > - $P = 0$ only > [!question] Quiz Q5 · multiple choice > How does the equilibrium structure of Model B change as $H$ increases through its critical value, compared with Model A? > - In Model B the two equilibria pass through each other and swap stability, while in Model A they collide and disappear. > - In both models the two equilibria collide and disappear. > - In Model B the two equilibria collide and disappear, while in Model A they pass through each other and swap stability. > - Neither model changes; only the location of the equilibria moves. > [!question] Quiz Q6 · free response > As the harvest is changed in both models, the phase line changes. Model B, however, rules out something important that can happen in Model A. What is it, and why can it not happen when Model B is harvested at a rate $0 < H < 2$? > [!note] Two names > What Model A does at $H^{*}$, two equilibria colliding and vanishing, is called a **saddle-node** (or **fold**) bifurcation. What Model B does at $H = r$, two equilibria crossing and trading stability, is called a **transcritical** bifurcation. In both cases a **bifurcation** is a change in how many equilibria there are or which ones are stable, caused by turning a parameter. --- ## Part 2 · What linear stability analysis misses Look at the phase line at the moment of each bifurcation. In Model A at $H = H^{*}$, the root of $f$ sits exactly at the top of the parabola. In Model B at $H = r$, the same thing happens at $P = 0$. In both cases the equilibrium is a root of $f$ **and** a local extremum of $f$, so $ f(y^{*}) = 0 \qquad\text{and}\qquad f'(y^{*}) = 0 . $ That is exactly where linear stability analysis has nothing to say. Here is why, and what replaces it. ### The linearization, and when it works Let $y^{*}$ be an equilibrium of $y' = f(y)$ and measure the distance from it, $\eta = y - y^{*}$. Since $y^{*}$ is a constant, $\eta' = y' = f(y^{*} + \eta)$, and a Taylor series about $y^{*}$ gives $ \eta' = f(y^{*}) + f'(y^{*})\,\eta + \tfrac{1}{2}f''(y^{*})\,\eta^{2} + O(\eta^{3}) . $ The first term is zero because $y^{*}$ is an equilibrium. If $f'(y^{*}) \neq 0$ and $\eta$ is small, the $\eta^{2}$ term is much smaller than the $\eta$ term, so drop it, $ \eta' \approx f'(y^{*})\,\eta \qquad\Longrightarrow\qquad \eta(t) \approx \eta_{0}\, e^{f'(y^{*})\,t}. $ When $f'(y^{*}) < 0$ the disturbance decays exponentially, and $\tau = 1/|f'(y^{*})|$ is the **relaxation time**, the time for the disturbance to shrink by a factor of $e$. For Model A, $f'(P_{+}) = -2\sqrt{1 - H}$, so $ \tau = \frac{1}{2\sqrt{1 - H}} , $ which is $0.71$ at $H = 0.5$, $1.58$ at $H = 0.9$, $5.0$ at $H = 0.99$, and grows without bound as $H \to H^{*} = 1$. ### When it fails At a bifurcation $f'(y^{*}) = 0$, the linear term is gone, and the first term that survives is the quadratic one, $ \eta' \approx \tfrac{1}{2}f''(y^{*})\,\eta^{2} = c\,\eta^{2}, \qquad c = \tfrac{1}{2}f''(y^{*}) . $ This is separable, $ \int_{\eta_{0}}^{\eta} \frac{d\eta}{\eta^{2}} = \int_{0}^{t} c\,dt \quad\Longrightarrow\quad \frac{1}{\eta_{0}} - \frac{1}{\eta} = c\,t \quad\Longrightarrow\quad \boxed{\;\eta(t) = \frac{\eta_{0}}{1 - c\,\eta_{0}\,t}\;} $ Two cases, depending on the sign of $c\,\eta_{0}$. - If $c\,\eta_{0} < 0$ (the disturbance is on the side the parabola pushes back from), the denominator grows and $\eta$ decays, but only like $\eta \approx \dfrac{1}{|c|\,t}$ for large $t$. That is **algebraic** decay, not exponential. - If $c\,\eta_{0} > 0$, the denominator reaches zero at $t = 1/(c\,\eta_{0})$ and the disturbance runs away. The equilibrium is **half-stable**, attracting from one side and repelling from the other, which is what the half-filled circle on the phase line means. For our pond both bifurcations land on the same equation. At $H = H^{*}$, Model A is $f(P) = -(P - 1)^{2}$, and at $H = r$, Model B is $f(P) = -P^{2}$. In both $f'' = -2$, so $c = -1$ and a disturbance above the equilibrium decays like $\eta_{0}/(1 + \eta_{0}t)$. ### Exponential against algebraic How long does it take a disturbance $\eta_{0}$ to shrink to some smaller size $\eta$? | | equation for $\eta$ | solution | time to shrink from $\eta_{0}$ to $\eta$ | |---|---|---|---| | away from the bifurcation | $\eta' = -\lambda\eta$ | $\eta_{0}e^{-\lambda t}$ | $\tau \ln(\eta_{0}/\eta)$ with $\tau = 1/\lambda$ | | at the bifurcation | $\eta' = c\,\eta^{2}$, $c\,\eta_{0} < 0$ | $\dfrac{\eta_{0}}{1 + \lvert c\rvert\,\lvert\eta_{0}\rvert\,t}$ | $\dfrac{1}{\lvert c\rvert}\left(\dfrac{1}{\lvert\eta\rvert} - \dfrac{1}{\lvert\eta_{0}\rvert}\right)$ | With exponential decay, shrinking by another factor of ten always costs the same extra time, $\tau \ln 10$. With algebraic decay, shrinking by another factor of ten costs about ten times as long as the last one did. Near a bifurcation the pond is in between. The linear term is still there but small, so $\tau$ is huge, and the recovery from any disturbance gets slower and slower as $H$ approaches $H^{*}$. That is **critical slowing down**. ### See it in the applets The applet below is our pond written in a cleaner form. With $r = N = 2$, $ f(P) = 2P - P^{2} - H = (1 - H) - (P - 1)^{2}, $ so with $a = 1 - H$ (the distance below the threshold) and $y = P$ it is $y' = a - (y - 1)^{2}$. The sink is at $y^{*} = 1 + \sqrt a$, and $a = 0$ is the bifurcation $H = H^{*}$. The orange dashed curve is the linear prediction $\eta_{0}e^{f'(y^{*})t}$ and the magenta dotted curve is the algebraic one $\eta_{0}/(1 + \eta_{0}t)$. <div class="applet" data-applet="critical-slowing"></div> **Things to do.** 1. Leave $y(0)/y^{*}$ at $1.5$, so the pond starts 50% above its equilibrium. Press **a = 1** (no harvest), then **0.01** (a quota of $0.99$), then **a = 0 (at the threshold)**. Each time, read off how long it takes to get back within 1% of equilibrium, and compare the actual solution with the two predictions. 2. At $a = 1$, set the error axes to **semilog**. The actual error is nearly a straight line, which is what exponential decay looks like on these axes. 3. At $a = 0$, set the error axes to **log–log**. Now the actual error is the straight line (slope $-1$, so $\eta \sim 1/t$) and it sits on the magenta curve, while the linear prediction is flat. 4. Still at $a = 0$, drag $y(0)/y^{*}$ below $1$. This is the other side of the half-stable point, and the solution runs away. 5. Optional cross-check in the two harvesting applets above. Set Model A to $H = H^{*}$ and Model B to $H = r$, and look at how slowly the solutions starting above the merged equilibrium flatten out, compared with $H = H^{*}/2$ and $H = r/2$. <div class="applet" data-applet="slowing-time"></div> **Things to do.** This plot puts recovery time against the distance $a$ from the threshold, on log–log axes. Hover over it to read values. Find the slope of the $\tau$ line and compare with $\tau = \frac{1}{2\sqrt a}$. Then find where the actual recovery time stops following $\tau$ and what it levels off at. > [!question] Quiz Q7 · multiple choice > In the critical-slowing applet, starting 50% above equilibrium, about how long does the pond take to get back within 1% of equilibrium at the threshold, $a = 0$? > - about $2$ > - about $13$ > - about $98$ > - It never gets back, because $f' = 0$ there. > [!question] Quiz Q8 · multiple choice > For our pond, $\tau = \dfrac{1}{2\sqrt{1 - H}}$. If the distance to the threshold, $1 - H$, is cut by a factor of 4, what happens to the relaxation time? > - It doubles. > - It quadruples. > - It is cut in half. > - It stays the same, because the equilibrium is still stable. > [!question] Quiz Q9 · free response > What is critical slowing down, and how is it related to linear stability analysis? Your answer should mention what $f'(y^{*})$ tells you, what happens to it as the bifurcation approaches, and what takes over at the bifurcation itself. --- ## Part 3 · A truck, a flatbed, and a sway Watch this demonstration.  > [!quote] Attribution > **"Tongue Weight Safety Demonstration"**, Weigh Safe, [youtube.com/watch?v=w9Dgxe584Ss](https://www.youtube.com/watch?v=w9Dgxe584Ss). The truck and flatbed have more than one steady way of moving. One of them is a **long-lived oscillation**, the flatbed swinging back and forth behind the truck and dragging the truck along with it. That is not an equilibrium point like the ones on our phase lines, but it plays the same role. It is a state the system settles into and stays in. (In two or more dimensions a steady oscillation like this is called a limit cycle.) > [!question] Quiz Q10 · free response > The swaying oscillation is one steady state of the truck and flatbed. What is the other one? Describe what the truck and flatbed are doing in it, and what happens to a small bump or gust when the system is in that state. **Things to do.** Watch the video again and pay attention to what is changed between one run and the next. Only one thing changes, in the same way that only $H$ changed in the pond. > [!question] Quiz Q11 · multiple choice > Which parameter is changed to move the truck and flatbed from one kind of steady state to the other? > - The speed of the truck. > - Where the load sits on the flatbed, which sets how much of its weight presses down on the hitch (the tongue weight). > - The size of the truck's tires. > - The length of the flatbed. That parameter can be changed a little at a time. Somewhere between a setting where the flatbed tracks straight and a setting where it sways, there has to be a value where one behavior gives way to the other. That value plays the role of $H^{*}$, and it is a bifurcation. > [!question] Quiz Q12 · free response > Suppose the parameter is set on the stable side but close to that in-between value. What would critical slowing down look like for the truck and flatbed? Describe what you would see after a small bump, and how it would change as the parameter is moved closer to the bifurcation. --- ## Part 4 · Many variables at once Every system on this page had one state variable. Most systems worth modeling have many. Think of a lake with fish, plankton and nutrients, a power grid, the climate, or even the truck and flatbed, where the angle of the flatbed, how fast it is swinging and how fast the truck drifts sideways all matter at once. Collect the variables into a vector $\mathbf{x} = (x_1, \dots, x_n)$ and the rule becomes $ \mathbf{x}' = \mathbf{F}(\mathbf{x};\, p). $ Equilibria are still the places where $\mathbf{F} = \mathbf{0}$, and linearization still works. Write $\boldsymbol{\eta} = \mathbf{x} - \mathbf{x}^{*}$ and the Taylor series gives $\boldsymbol{\eta}' \approx J\,\boldsymbol{\eta}$, where the Jacobian matrix $J$ (all the partial derivatives $\partial F_i/\partial x_j$ at $\mathbf{x}^{*}$) takes the place of $f'(y^{*})$. A disturbance now breaks into pieces that behave like $e^{\lambda t}$, one for each eigenvalue $\lambda$ of $J$. The equilibrium is stable when every $\lambda$ has negative real part, and the piece whose $\lambda$ is closest to zero is the last to die out. Turning $p$ can change this structure. Equilibria can appear, vanish or trade stability, as in the pond, or a stable state can give way to a lasting oscillation, as with the truck. In every case the change happens when an eigenvalue, or a pair of them, is pushed to zero real part. That is the many-variable version of $f'(y^{*}) = 0$. > [!question] Quiz Q13 · free response > Imagine a system with many interacting variables sitting at a stable equilibrium, while a parameter is slowly turned toward a bifurcation. For this system, describe what critical slowing down would look like. > > (If it helps, you might think about whether every variable would take longer to recover after a small disturbance, or only some combination of them, and what you would expect to see if the eigenvalues approaching zero real part were a complex pair, as for the truck.) --- ## What to carry forward 1. Linear stability analysis answers two questions at once, whether a disturbance dies out and how fast. At a bifurcation it loses the ability to answer either, and the next term in the Taylor series has to step in. 2. The relaxation time $\tau = 1/|f'(y^{*})|$ grows without bound as a bifurcation approaches. The equilibrium can look perfectly healthy while its recovery gets slower and slower. 3. The same picture shows up far from fish ponds. Any system with a parameter, two kinds of steady behavior, and a value in between where one gives way to the other will slow down before it switches.