Existence, Uniqueness & Bifurcations

Both hypotheses earn their keep: $y' = y^{2/3}$, $y(0) = 0$ has the two solutions $y \equiv 0$ and $y = (t/3)^3$ — the derivative $\tfrac23y^{-1/3}$ blows up at $0$ and uniqueness genuinely fails. And existence can be merely local: $y' = y^2$, $y(0) = 1$ solves to $y = \tfrac{1}{1-t}$, alive only until $t = 1$ — solutions can end in finite time without warning. In fact there are infinitely many solutions through $(0, 0)$: stay at $0$ until any time $c \ge 0$, then follow $y = \left(\tfrac{t - c}{3}\right)^3$.

Plotting equilibria against $r$ gives the bifurcation diagram — the parabola $r = -y^2$ — and the whole family's fate in one picture.

Step 1 — equilibria: $y^2 - y + h = 0$, so $y = \tfrac{1 \pm \sqrt{1 - 4h}}{2}$.

Step 2 — read the cases: $h < \tfrac14$: two equilibria (upper stable — sustainable); $h = \tfrac14$: they merge at $y = \tfrac12$; $h > \tfrac14$: none, $y' < 0$ everywhere, extinction in finite time.

Step 3 — the moral: the maximum sustainable harvest is the peak of the growth curve, $\tfrac14$ — and passing it is not a gradual decline but a saddle-node cliff. This one phase-line computation is the backbone of real fisheries mathematics.

Why: write $y = y^* + u$ with $u$ small. Then $u' = f(y^* + u) \approx f(y^*) + f'(y^*)u = f'(y^*)u$, so to first order $u = u_0e^{f'(y^*)t}$, which shrinks when $f'(y^*) < 0$ and grows when $f'(y^*) > 0$. The sign of $f'(y^*)$ also says how $f$ crosses zero: decreasing through $y^*$ means $f > 0$ just below (flow up, toward $y^*$) and $f < 0$ just above (flow down, toward $y^*$) — the phase-line picture of stability.

Step 1 — derivative test: $f'(1) = -2 < 0$, so $y = 1$ is stable; $f'(3) = 2 > 0$, so $y = 3$ is unstable.

Step 2 — the phase line agrees: $f > 0$ below $1$ (up), $f < 0$ between $1$ and $3$ (down), $f > 0$ above $3$ (up).

Step 3 — the starts $0$, $2$ and $2.9$ all end at $1$; near $1$ the gap decays like $e^{-2t}$, the rate $|f'(1)| = 2$.

Step 4 — the start $3.1$ runs away. Separating variables gives $\dfrac{y - 3}{y - 1} = Ce^{2t}$ with $C = \tfrac{0.1}{2.1} = \tfrac1{21}$, and $y$ is infinite when $Ce^{2t} = 1$: at $t = \tfrac12\ln 21 \approx 1.52$. A start of $2.9$ and a start of $3.1$ differ by $0.2$, yet one settles and the other escapes in finite time — the unstable equilibrium is a watershed.

Step 1 — $f'(k\pi) = \cos k\pi = (-1)^k$: the even multiples $0, \pm2\pi, \ldots$ are unstable, the odd multiples $\pm\pi, \pm3\pi, \ldots$ stable. They alternate, as they must: between two consecutive zeros $f$ keeps one sign, so each interval flows from one equilibrium into the next.

Step 2 — $y(0) = 1$ lies in $(0, \pi)$, where $\sin y > 0$, so $y$ rises toward $\pi$.

Step 3 — exactly: separating, $\int\dfrac{dy}{\sin y} = \ln\left|\tan\tfrac y2\right|$, so $\tan\tfrac y2 = \tan\tfrac12\cdot e^t$ and $y = 2\arctan\left(\tan\tfrac12\cdot e^t\right) \to 2\cdot\tfrac\pi2 = \pi$.

Step 4 — the approach is exponential with rate $|f'(\pi)| = 1$: the gap $\pi - y$ shrinks by the factor $e^{-1}$ per unit time, eventually.

Step 1 — go back to signs: $f(y) = (y-2)^2 > 0$ on both sides of $2$. Below $2$ solutions rise toward it; above $2$ they rise away from it. The equilibrium is semistable.

Step 2 — exactly: $\dfrac{dy}{(y-2)^2} = dt$ gives $y = 2 - \dfrac{1}{t + C}$. From $y(0) = 1.9$: $C = 10$, and $y = 2 - \tfrac{1}{t + 10} \to 2$, but only like $\tfrac1t$ — far slower than any exponential. From $y(0) = 2.1$: $C = -10$, so $y = 2 + \tfrac{1}{10 - t}$, which blows up at $t = 10$.

Step 3 — the moral: $f'(y^*) = 0$ means the linear approximation is flat, and the behaviour is decided by the next term of the Taylor expansion, here $(y - 2)^2$.

Why only those points: where $f_y \ne 0$, the implicit function theorem says the equation $f(y; r) = 0$ defines $y$ as a smooth function of $r$ nearby, so the branch just continues and keeps its stability. Something can happen only where $f_y = 0$. Which shape appears depends on structure: with no special structure the fold (saddle-node) is typical; an equilibrium present for every $r$ (like $y = 0$) forces a crossing; a symmetry $y \to -y$ forces a pitchfork.

Step 1 — on $y = 0$: $f_y = r$, so this branch is stable for $r < 0$ and unstable for $r > 0$.

Step 2 — on $y = r$: $f_y = r - 2r = -r$, unstable for $r < 0$ and stable for $r > 0$.

Step 3 — at $r = 0$ the branches meet ($f = 0$ and $f_y = 0$ together) and exchange stability: the solid parts of the diagram form an X with its left half on $y = 0$ and its right half on $y = r$.

Step 4 — no equilibrium is created or destroyed. This shape appears whenever an equilibrium exists for every parameter value — here $y = 0$, as in a population that cannot appear from nothing.

Step 1 — $f_y = 1 - E - 2y$. At $y = 1 - E$: $f_y = -(1 - E)$, stable while $E < 1$. At $y = 0$: $f_y = 1 - E$, unstable while $E < 1$. At $E = 1$ the branches cross: a transcritical bifurcation, and for $E > 1$ the stock dies out.

Step 2 — the sustained catch is the harvest at the stable equilibrium: $Y(E) = E(1 - E)$.

Step 3 — maximize: $Y'(E) = 1 - 2E = 0$ at $E = \tfrac12$, giving $Y = \tfrac14$ with the stock at $\tfrac12$.

Step 4 — compare the constant-harvest model of the first section: the same maximum catch $\tfrac14$, but there it sits exactly at the saddle-node cliff, while here the stock at $E = \tfrac12$ has recovery rate $|f_y| = \tfrac12$. Setting effort instead of quota is safer for the same catch.

Step 1 — the function $g(y) = y + e^{-y}$ has $g'(y) = 1 - e^{-y}$, negative for $y < 0$ and positive for $y > 0$, so its minimum value is $g(0) = 1$.

Step 2 — hence: for $r < 1$ no equilibria (every solution decreases forever, since $f < 0$); at $r = 1$ one, $y = 0$; for $r > 1$ two, one on each side of $0$.

Step 3 — stability: $f_y = e^{-y} - 1$ is negative for $y > 0$ (stable branch) and positive for $y < 0$ (unstable).

Step 4 — for example at $r = 2$: the equilibria solve $y + e^{-y} = 2$, numerically $y \approx 1.84$ (stable) and $y \approx -1.15$ (unstable). The diagram is the curve $r = y + e^{-y}$ turned on its side, with its fold at $(1, 0)$ — the saddle-node shape without any square root in sight.

Continuity of $\partial f/\partial y$ makes $f$ Lipschitz in $y$, so successive iterates differ by a factor $\le L|t - t_0|$ each round — a geometric squeeze forcing convergence to a solution, and forcing any two solutions together (uniqueness) by the same estimate. For $y' = y$, the iterates are exactly the Taylor partial sums of $e^t$. $\blacksquare$

For $f = r + y^2$: $y^2 = -r$ and $2y = 0$ give $y = 0$, $r = 0$ — the collision point — and near it the equilibria $\pm\sqrt{-r}$ exist only on one side. Two conditions, one parameter: bifurcation points are isolated, which is why they are events. $\blacksquare$

Step 1 — equilibria: $y = 0$, $y = r$, $y = 1$.

Step 2 — sign of $f$ on each interval (test points): for $y < 0$: $(-)(-)(-) < 0$, moving down; on $(0, r)$: $(+)(-)(-) > 0$, moving up; on $(r, 1)$: $(+)(-)(+) < 0$, down; for $y > 1$: $(+)(+)(+) > 0$, up.

Step 3 — read stability: arrows converge on $r$ (stable) and diverge from $0$ and $1$ (unstable).

Step 4 — as $r \to 1$ the stable equilibrium $r$ moves up to meet the unstable one at $1$. At $r = 1$ they merge: $f = y(y - 1)^2$ is positive on both sides of $1$, so $y = 1$ is semistable. For $r > 1$ they have passed through each other and swapped roles: now $f'(1) = 1 - r < 0$, so $1$ is stable, and $f'(r) = r(r - 1) > 0$, so $r$ is unstable. Nothing is destroyed — this is a transcritical bifurcation, an exchange of stability, not a saddle-node.

Step 5 — verify with $f'$: $f'(r) = r(r - 1) < 0$ ✓ stable; $f'(0) = r > 0$ and $f'(1) = 1 - r > 0$ ✓ unstable.

Step 1 — separate: $\dfrac{dy}{y^2} = dt$, so $-\dfrac1y = t + C$.

Step 2 — data: $-\tfrac12 = C$, hence $y = \dfrac{1}{\tfrac12 - t} = \dfrac{2}{1 - 2t}$.

Step 3 — the solution blows up as $t \to \tfrac12^-$: it exists only on $[0, \tfrac12)$, even though $f(y) = y^2$ is perfectly smooth everywhere.

Step 4 — the moral: Picard guarantees existence only locally; a smooth right-hand side that grows faster than linearly can drive solutions to infinity in finite time. Compare $y' = y$ (exists forever) — the difference between linear and quadratic growth is the difference between eternity and half a second.