4

Nonlinear Systems

Partials ยท Nullclines ยท Jacobian ยท Coexistence ยท Hopf

โˆ‚ Partial Derivatives
The idea โ€” change only one variable at a time
A partial derivative $\frac{\partial f}{\partial x}$ asks: "If I change $x$ by a tiny bit while holding every other variable frozen, how does $f$ change?"

How to compute: Treat all variables except $x$ as plain numbers (constants). Then differentiate normally with respect to $x$ only.
Three worked examples
1. $f(x,y)=3x^2y+5y^3$
$\frac{\partial f}{\partial x}=6xy+0=6xy$   (treat $y$ as a constant; the $5y^3$ term has no $x$ โ†’ becomes 0)
$\frac{\partial f}{\partial y}=3x^2+15y^2$   (treat $x$ as a constant; $3x^2$ acts like a coefficient)

2. $g(M,S)=3M-M^2-2MS$
$\frac{\partial g}{\partial M}=3-2M-2S$   (product rule on $2MS$: treat $S$ as a constant, differentiate $M$)
$\frac{\partial g}{\partial S}=-2M$   (only the $-2MS$ term contains $S$; $3M$ and $-M^2$ have no $S$)

3. $h(N,P)=\frac{wN}{d+N}P$
$\frac{\partial h}{\partial N}=\frac{wd}{(d+N)^2}P$   (quotient rule on $\frac{wN}{d+N}$, treating $P$ as constant)
$\frac{\partial h}{\partial P}=\frac{wN}{d+N}$   (the whole fraction is just a constant with respect to $P$; differentiate $P$)
๐Ÿ“ Tangent Plane (Linear Approximation)
The formula
For small changes $\Delta x$ and $\Delta y$ near a point $(x_0,y_0)$: $$\Delta f \approx \frac{\partial f}{\partial x}\bigg|_{(x_0,y_0)}\Delta x + \frac{\partial f}{\partial y}\bigg|_{(x_0,y_0)}\Delta y$$ This is the best linear approximation to $f$ near $(x_0,y_0)$. It's the same idea as the Jacobian, just applied to a single function.
Full exam-style worked example (BMI)
$B(W,H)=703W/H^2$ where $W$=weight (lbs), $H$=height (inches).
Starting point: $W_0=160$, $H_0=60$. Changes: $\Delta W=-9$ (lost 9 lbs), $\Delta H=+1$ (grew 1 inch).

Step 1 โ€” compute partial derivatives at $(160,60)$:
$\frac{\partial B}{\partial W}=\frac{703}{H^2}\bigg|_{H=60}=\frac{703}{3600}\approx 0.195$
$\frac{\partial B}{\partial H}=703W\cdot(-2)H^{-3}\bigg|_{(160,60)}=\frac{-1406\times 160}{216000}\approx -1.041$

Step 2 โ€” apply linear approximation:
$\Delta B\approx(0.195)(-9)+(-1.041)(1)\approx-1.755-1.041\approx\mathbf{-2.8}$

The patient's BMI decreases by approximately 2.8 points.
๐ŸŽฏ Nullclines & Finding All Equilibria
What is a nullcline?
A nullcline for variable $X$ is the set of points where $X'=0$ โ€” i.e., where $X$ is momentarily not changing.
An equilibrium point (EP) is where BOTH $X'=0$ AND $Y'=0$ simultaneously โ€” i.e., the intersection of an $X$-nullcline and a $Y$-nullcline.
๐Ÿ”ฅ Complete procedure for finding all EPs in a 2D system
Set $X'=0$. Factor the expression. This gives you conditions like "$X=0$ OR [some equation relating $X$ and $Y$]."
Set $Y'=0$. Factor. This gives you more conditions.
Consider all combinations. If $X'=0$ gives "$X=0$ or $f(X,Y)=0$" and $Y'=0$ gives "$Y=0$ or $g(X,Y)=0$" โ€” you have $2\times 2=4$ cases to check.
Solve each case. Substitute one condition into the other. Some cases give impossible solutions (discard) or give actual $(X^*,Y^*)$ values.
State all EPs found. A biologically valid EP must have $X^*\geq 0$ and $Y^*\geq 0$.
Full worked example โ€” Frogs & Salamanders (from review session)
$$X'=5X-X^2-XY=X(5-X-Y)$$ $$Y'=10Y-Y^2-6XY=Y(10-Y-6X)$$ Step 1 โ€” Set $X'=0$:
$X(5-X-Y)=0$ โ†’ Either $X=0$ or $5-X-Y=0$ (i.e., $Y=5-X$)

Step 2 โ€” Set $Y'=0$:
$Y(10-Y-6X)=0$ โ†’ Either $Y=0$ or $10-Y-6X=0$ (i.e., $Y=10-6X$)

Step 3 โ€” Four cases:
CaseFrom $X'=0$From $Y'=0$SolveEP found
1$X=0$$Y=0$โ€”$(0,0)$
2$X=0$$Y=10-6X=10$$Y=10$$(0,10)$
3$Y=5-X$$Y=0$$5-X=0$โ†’$X=5$$(5,0)$
4$Y=5-X$$Y=10-6X$$5-X=10-6X$โ†’$5X=5$โ†’$X=1$, $Y=4$$(1,4)$
All four EPs: $(0,0)$, $(0,10)$, $(5,0)$, $(1,4)$.
Second worked example โ€” Mouth Microbiome (M. mutans & S. sanguinis)
$$M'=3M-M^2-2MS=M(3-M-2S)$$ $$S'=4S-2MS-2S^2=S(4-2M-2S)=2S(2-M-S)$$ $M'=0$: $M=0$ or $M=3-2S$
$S'=0$: $S=0$ or $S=2-M$

Four cases:
• $M=0,S=0$: EP = $(0,0)$
• $M=0$: $S=2-0=2$. EP = $(0,2)$
• $S=0$: $M=3-2(0)=3$. EP = $(3,0)$
• $M=3-2S$ and $S=2-M$: sub โ†’ $M=3-2(2-M)=3-4+2M$โ†’$-M=-1$โ†’$M=1$, $S=1$. EP = $(1,1)$

Four EPs: $(0,0)$, $(0,2)$, $(3,0)$, $(1,1)$.
Jacobian stability analysis โ€” procedure
โ‘  Find EPs set X'=0, Y'=0 solve for (X*,Y*) โ‘ก Jacobian J compute โˆ‚f/โˆ‚X etc evaluate at EP โ‘ข Eigenvalues (Jโ‚โ‚+Jโ‚‚โ‚‚), det(J) ฮปยฒโˆ’(a+d)ฮป+det=0 โ‘ฃ Classify det<0โ†’saddle | (Jโ‚โ‚+Jโ‚‚โ‚‚)<0โ†’stable (Jโ‚โ‚+Jโ‚‚โ‚‚)=0, det>0 โ†’ Hopf! โš  fails if any ฮป=0 or Re(ฮป)=0
โš  Schematic illustration only โ€” not drawn to scale or from simulation.
๐Ÿงฎ The Jacobian Matrix
What the Jacobian does
Near an equilibrium point, a nonlinear system behaves approximately like a linear one. The Jacobian $J$ is the matrix of that linear approximation. It tells you the stability of the EP.

For a system $X'=f(X,Y)$, $Y'=g(X,Y)$: $$J=\begin{pmatrix}\frac{\partial f}{\partial X}&\frac{\partial f}{\partial Y}\\\frac{\partial g}{\partial X}&\frac{\partial g}{\partial Y}\end{pmatrix}$$ Each entry is a partial derivative. Then evaluate at the EP $(X^*,Y^*)$.
๐Ÿ”ฅ Step-by-step Jacobian procedure
Write down both equations: $f(X,Y)=X'$ and $g(X,Y)=Y'$
Compute all four partial derivatives: $\partial f/\partial X$, $\partial f/\partial Y$, $\partial g/\partial X$, $\partial g/\partial Y$
Substitute the EP $(X^*,Y^*)$ into each partial derivative to get numbers
Compute $(J_{11}+J_{22})$ (sum of diagonal entries) and $(J_{11}J_{22}-J_{12}J_{21})$
Classify the EP using the diagonal sum/det rules below
Full worked example โ€” Frogs & Salamanders at EP $(1,4)$
$f=X(5-X-Y)=5X-X^2-XY$, $g=Y(10-Y-6X)=10Y-Y^2-6XY$

Partial derivatives:
$\frac{\partial f}{\partial X}=5-2X-Y$,   $\frac{\partial f}{\partial Y}=-X$
$\frac{\partial g}{\partial X}=-6Y$,   $\frac{\partial g}{\partial Y}=10-2Y-6X$

Evaluate at $(X^*,Y^*)=(1,4)$:
$J_{11}=5-2(1)-4=\mathbf{-1}$,   $J_{12}=-1=\mathbf{-1}$
$J_{21}=-6(4)=\mathbf{-24}$,   $J_{22}=10-2(4)-6(1)=10-8-6=\mathbf{-4}$

$$J=\begin{pmatrix}-1&-1\\-24&-4\end{pmatrix}$$ sum of diag $= -1+(-4)=-5$, $(J_{11}J_{22}-J_{12}J_{21})=(-1)(-4)-(-1)(-24)=4-24=-20$

Classify: $(J_{11}J_{22}-J_{12}J_{21})=-20<0$ โ†’ Saddle point (unstable).
๐Ÿ“Š Classifying Equilibria via diagonal sum and det
๐Ÿ— The complete classification rules
ConditionEP TypeStable?
$(J_{11}J_{22}-J_{12}J_{21})<0$Saddle pointNo (always unstable)
$(J_{11}J_{22}-J_{12}J_{21})>0$, $(J_{11}+J_{22})<0$, $(J_{11}+J_{22})^2>4(J_{11}J_{22}-J_{12}J_{21})$Stable nodeYes
$(J_{11}J_{22}-J_{12}J_{21})>0$, $(J_{11}+J_{22})>0$, $(J_{11}+J_{22})^2>4(J_{11}J_{22}-J_{12}J_{21})$Unstable nodeNo
$(J_{11}J_{22}-J_{12}J_{21})>0$, $(J_{11}+J_{22})<0$, $(J_{11}+J_{22})^2<4(J_{11}J_{22}-J_{12}J_{21})$Stable spiralYes
$(J_{11}J_{22}-J_{12}J_{21})>0$, $(J_{11}+J_{22})>0$, $(J_{11}+J_{22})^2<4(J_{11}J_{22}-J_{12}J_{21})$Unstable spiralNo
$(J_{11}J_{22}-J_{12}J_{21})>0$, $(J_{11}+J_{22})=0$Center / Hopf bifurcationBorderline
Shortcut: If $(ad-bc)<0$ โ†’ saddle (done). If $(ad-bc)>0$ โ†’ check (Jโ‚โ‚+Jโ‚‚โ‚‚): negative = stable, positive = unstable, zero = Hopf.
๐Ÿค Coexistence of Populations
Definition and procedure
Two species coexist if both survive long-term โ†’ the stable EP has both coordinates positive.

Find all EPs of the system
For each EP, compute $J$ and find eigenvalues โ†’ identify which EP is stable (all Re(ฮป) < 0)
Check the stable EP: if both coordinates > 0 โ†’ coexistence. If one = 0 โ†’ that species goes extinct. If one < 0 โ†’ biologically impossible, not a real EP.
๐Ÿ”ฅ Common mistake: forgetting to check signs
A "coexistence EP" like $(-2, 5)$ is biologically impossible โ€” populations can't be negative. Always check $X^*\geq 0$ and $Y^*\geq 0$ before calling it coexistence.
Phase portraits โ€” continuous-time stability types
stable node ฮปโ‚,ฮปโ‚‚ < 0 Re(ฮป)<0 (cts) or |ฮป|<1 (disc) unstable node ฮปโ‚,ฮปโ‚‚ > 0 Re(ฮป)>0 (cts) or |ฮป|>1 (disc) saddle point ฮปโ‚<0, ฮปโ‚‚>0 โ†‘ stable manifold โ†’ unstable manifold stable spiral aยฑbi, a<0 a<0 โ†’ spiral inward limit cycle after Hopf (a>0) outside: spiral in inside: spiral out
โš  Schematic illustration only โ€” not drawn to scale or from simulation.
๐Ÿ”€ Finding the Hopf Bifurcation Parameter
Glycolysis example (Class 18) โ€” fully worked
$$S'=1-cSP^2 \qquad P'=cSP^2-P$$ Find EP: $S'=0$: $cSP^2=1$. $P'=0$: $cSP^2=P$ โ†’ $P=1$. Then $cS(1)=1$โ†’$S=1/c$. EP: $(1/c,1)$

Compute Jacobian at EP:
$\frac{\partial S'}{\partial S}=-cP^2\big|_1=-c$    $\frac{\partial S'}{\partial P}=-2cSP\big|_{(1/c,1)}=-2c(1/c)(1)=-2$
$\frac{\partial P'}{\partial S}=cP^2\big|_1=c$    $\frac{\partial P'}{\partial P}=(2cSP-1)\big|_{(1/c,1)}=2-1=1$
$$J=\begin{pmatrix}-c&-2\\c&1\end{pmatrix}$$ Find Hopf: Sum of diagonal entries $= -c+1=1-c$. Set to zero: $1-c=0$ โ†’ $c^*=1$
Verify: $(J_{11}J_{22}-J_{12}J_{21})\big|_{c=1}=(-1)(1)-(-2)(1)=-1+2=1>0$ โœ“

Interpret: For $c<1$: diagonal sum $(1-c)>0$ โ†’ unstable spiral + limit cycle (oscillations!). For $c>1$: diagonal sum $(1-c)<0$ โ†’ stable spiral (no oscillations).
โœฆ Unit 4 Practice Quiz

โœฆ Quiz โ€” Nullclines, Jacobian & Stability

Q1. For $f(M,S)=3M-M^2-2MS$, what is $\partial f/\partial S$?

A) $3-2M$
B) $-2M$
C) $-2S$
D) $3M-2MS$

Q2. For the system $X'=X(5-X-Y)$, $Y'=Y(10-Y-6X)$, how many equilibrium points are there?

A) 1 โ€” only the origin
B) 2 โ€” origin and coexistence EP
C) 4 โ€” $(0,0)$, $(5,0)$, $(0,10)$, $(1,4)$
D) 3 โ€” the three non-trivial EPs only

Q3. A Jacobian at an EP has diagonal sum $(J_{11}+J_{22})=-2$ and $(J_{11}J_{22}-J_{12}J_{21})=-5$. The EP is:

A) Stable spiral (diagonal sum < 0, det > 0)
B) Stable node
C) Saddle point (det < 0 โ†’ always saddle)
D) Unstable spiral

Q4. The diagonal sum $(J_{11}+J_{22})=w-0.3$ at an EP with $(J_{11}J_{22}-J_{12}J_{21})=2>0$. The Hopf bifurcation occurs at:

A) $w=0$
B) $w=0.3$ (set diagonal sum = 0, solve)
C) $w=-0.3$
D) $w=2$ (the det value)

Q5. The only stable EP is $(2,5)$. Do the species coexist?

A) No โ€” there must be an EP where one species is zero
B) Yes โ€” stable EP has both coordinates positive (2>0 and 5>0)
C) Cannot tell without more information
D) No โ€” both species must die out since the EP is an attractor

Q6. The triangular Jacobian at a boundary EP is $J=\begin{pmatrix}-3&7\\0&-2\end{pmatrix}$. The eigenvalues are:

A) $\lambda=\pm\sqrt{6}$ (from det)
B) $\lambda=-3,-2$ (triangular: read the diagonal)
C) $\lambda=3,2$ (absolute values of diagonal)
D) Must compute characteristic polynomial