Existence, Uniqueness & Bifurcations

Study Sheet

Existence, Uniqueness & Bifurcations

Picard's theorem, phase lines with parameters, harvesting collapse

When Solutions Exist, Split, and Vanish

Tip
Picard's theorem: the rules of the game

If ff and ∂f/∂y\partial f/\partial y are continuous near (t0,y0)(t_0, y_0), the IVP y′=f(t,y)y' = f(t, y), y(t0)=y0y(t_0) = y_0 has exactly ONE solution near t0t_0.

Both hypotheses earn their keep: y′=y2/3y' = y^{2/3}, y(0)=0y(0) = 0 has the two solutions y≡0y \equiv 0 and y=(x/3)3y = (x/3)^3 — the derivative 23y−1/3\tfrac23y^{-1/3} blows up at 00 and uniqueness genuinely fails. And existence can be merely LOCAL: y′=y2y' = y^2, y(0)=1y(0) = 1 solves to y=11−ty = \tfrac{1}{1-t}, alive only until t=1t = 1 — solutions can end in finite time without warning.

Tip
Bifurcations: how equilibria are born and die

Add a parameter: y′=r+y2y' = r + y^2. For r<0r < 0: two equilibria (±−r\pm\sqrt{-r}, lower stable, upper unstable). At r=0r = 0 they COLLIDE into one half-stable point; for r>0r > 0 none remain — a SADDLE-NODE bifurcation, the generic way equilibria appear and annihilate. The signature: f=0f = 0 and f′=0f' = 0 simultaneously.

Plotting equilibria against rr gives the bifurcation diagram — the parabola r=−y2r = -y^2 — and the whole family's fate in one picture.

Example
Worked: harvesting a logistic population to collapse

Model y′=y(1−y)−hy' = y(1-y) - h: logistic growth minus constant harvest hh.

Step 1 — equilibria: y2−y+h=0y^2 - y + h = 0, so y=1±1−4h2y = \tfrac{1 \pm \sqrt{1 - 4h}}{2}.

Step 2 — read the cases: h<14h < \tfrac14: two equilibria (upper stable — sustainable); h=14h = \tfrac14: they merge at y=12y = \tfrac12; h>14h > \tfrac14: NONE, y′<0y' < 0 everywhere, extinction in finite time.

Step 3 — the moral: the maximum sustainable harvest is the peak of the growth curve, 14\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.

Side note
Significance: tipping points are bifurcations

Fishery collapses, sudden lake eutrophication, and climate tipping points are saddle-node bifurcations in the wild: a slowly moving parameter annihilates the stable state, and the system jumps — with no gentle warning and no easy way back (hysteresis). The mathematics of this topic is how ecologists quantify "how close to the edge are we?"

Try it
Try it: a bifurcation diagram in your head

For y′=ry−y3y' = ry - y^3: equilibria as rr varies? Work: y(r−y2)=0y(r - y^2) = 0: always y=0y = 0; for r>0r > 0 also y=±ry = \pm\sqrt r. At r=0r = 0 the origin loses stability and hands it to the new pair — a PITCHFORK bifurcation, the symmetric sibling of the saddle-node, and the standard model of spontaneous symmetry breaking.

Proofs & Why It Matters

Tip
Proof idea: Picard iteration

Rewrite the IVP as the integral equation y(t)=y0+∫t0tf(s,y(s)) dsy(t) = y_0 + \int_{t_0}^t f(s, y(s))\,ds and iterate: yn+1(t)=y0+∫t0tf(s,yn(s)) dsy_{n+1}(t) = y_0 + \int_{t_0}^t f(s, y_n(s))\,ds starting from y0(t)≡y0y_0(t) \equiv y_0.

Continuity of ∂f/∂y\partial f/\partial y makes ff Lipschitz in yy, so successive iterates differ by a factor ≤L∣t−t0∣\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′=yy' = y, the iterates are exactly the Taylor partial sums of ete^t. ■\blacksquare

Tip
Proof: the saddle-node signature

At a bifurcation the equilibrium equation f(y;r)=0f(y; r) = 0 must have a DOUBLE root — a simple root moves smoothly as rr varies (implicit function theorem) and cannot vanish. A double root means f=0f = 0 and fy=0f_y = 0 together.

For f=r+y2f = r + y^2: y2=−ry^2 = -r and 2y=02y = 0 give y=0y = 0, r=0r = 0 — the collision point — and near it the equilibria ±−r\pm\sqrt{-r} exist only on one side. Two conditions, one parameter: bifurcation points are isolated, which is why they are EVENTS. ■\blacksquare

Going Deeper: Worked Problems

Example
Worked: a full phase-line analysis with a parameter

For y′=y(y−1)(y−r)y' = y(y - 1)(y - r) with 0<r<10 < r < 1, find and classify all equilibria and describe what happens as r→1r \to 1.

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

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

Step 3 — read stability: arrows converge on rr (stable) and diverge from 00 and 11 (unstable).

Step 4 — as r→1r \to 1 the stable equilibrium collides with the unstable one at 11 and they annihilate (a saddle-node): for r>1r > 1 the region just above 00 flows all the way up unboundedly.

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

Example
Worked: finite-time blowup, exactly

Solve y′=y2y' = y^2, y(0)=2y(0) = 2, and find the time at which the solution ceases to exist.

Step 1 — separate: dyy2=dt\dfrac{dy}{y^2} = dt, so −1y=t+C-\dfrac1y = t + C.

Step 2 — data: −12=C-\tfrac12 = C, hence y=112−t=21−2ty = \dfrac{1}{\tfrac12 - t} = \dfrac{2}{1 - 2t}.

Step 3 — the solution blows up as t→12−t \to \tfrac12^-: it exists only on [0,12)[0, \tfrac12), even though f(y)=y2f(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′=yy' = y (exists forever) — the difference between linear and quadratic growth is the difference between eternity and half a second.