Systems of Differential Equations

Study Sheet

Systems of Differential Equations

Eigenvalues run the phase portrait

Linear Systems

Tip
Solve along eigenvectors

For u′=Au\mathbf u' = A\mathbf u, each eigenpair gives a straight-line solution eλtve^{\lambda t}\mathbf v, and the general solution mixes them. The eigenvalue SIGNS shape the picture: both negative — everything decays to the origin (sink); both positive — source; opposite signs (det⁡A<0\det A < 0) — a SADDLE.

Concept
Complex eigenvalues spiral

Eigenvalues α±βi\alpha \pm \beta i make trajectories rotate with rate β\beta while eαte^{\alpha t} scales them: spirals in or out, or pure circles when α=0\alpha = 0. The system x′=yx' = y, y′=−xy' = -x is the cleanest case: x(t)=cos⁡tx(t) = \cos t from (1,0)(1,0) — a rotation, forever.

Example
Decoupled warm-up

x′=2xx' = 2x, y′=3yy' = 3y solves coordinate-wise: (e2t,e3t)(e^{2t}, e^{3t}). Diagonalization is the art of choosing coordinates that make EVERY system look like this one.

Side note
Significance: coupled everything

Predator-prey cycles, coupled pendulums, chemical reaction networks, epidemic compartments (S-I-R), and multi-loop circuits are systems of ODEs. The eigenvalue portrait — sink, source, saddle, spiral — is the universal vocabulary for "what happens in the long run" across all of them.

Try it
Try it: classify at a glance

Classify u′=(−311−3)u\mathbf u' = \begin{pmatrix}-3&1\\1&-3\end{pmatrix}\mathbf u. Work: trace −6<0-6 < 0, det =9−1=8>0= 9 - 1 = 8 > 0, discriminant 36−32=4>036 - 32 = 4 > 0: two real negative eigenvalues (−2-2 and −4-4) — a stable NODE; everything decays to the origin, fastest along the λ=−4\lambda = -4 eigendirection (1,−1)(1,-1).

Proofs & Why It Matters

Tip
Proof: e^λtv solves u′ = Au

Differentiate u(t)=eλtv\mathbf u(t) = e^{\lambda t}\mathbf v: u′=λeλtv\mathbf u' = \lambda e^{\lambda t}\mathbf v. Apply the matrix instead: Au=eλtAv=eλtλvA\mathbf u = e^{\lambda t}A\mathbf v = e^{\lambda t}\lambda\mathbf v using Av=λvA\mathbf v = \lambda\mathbf v.

The two agree, so every eigenpair yields a straight-line solution; independence of eigenvectors makes their combinations the general solution. ■\blacksquare

Tip
Proof: x′ = y, y′ = −x moves in circles

Let E(t)=x(t)2+y(t)2E(t) = x(t)^2 + y(t)^2. Then E′=2xx′+2yy′=2xy−2yx=0E' = 2xx' + 2yy' = 2xy - 2yx = 0: EE is CONSTANT, so trajectories stay on circles about the origin. Differentiating x′=yx' = y gives x′′=y′=−xx'' = y' = -x, simple harmonic motion, so from (1,0)(1,0): x=cos⁡tx = \cos t, y=−sin⁡ty = -\sin t — uniform rotation, confirming the circle. ■\blacksquare

Going Deeper: Explanations & Worked Problems

Concept
From eigenpairs to the phase portrait, step by step

To solve u′=Au\mathbf u' = A\mathbf u: find the eigenpairs of AA; each gives a straight-line solution eλtve^{\lambda t}\mathbf v; the general solution is c1eλ1tv1+c2eλ2tv2c_1e^{\lambda_1t}\mathbf v_1 + c_2e^{\lambda_2t}\mathbf v_2; initial data fixes c1,c2c_1, c_2 by writing u(0)\mathbf u(0) in the eigenbasis.

The PICTURE follows from the signs alone: both λ<0\lambda < 0 — every trajectory slides to the origin (stable node); both >0> 0 — everything escapes (source); opposite signs — a SADDLE: trajectories sweep in along the negative eigenline and out along the positive one, and det⁡A=λ1λ2<0\det A = \lambda_1\lambda_2 < 0 detects this without computing anything else. Complex α±βi\alpha \pm \beta i: spirals, inward if α<0\alpha < 0, outward if α>0\alpha > 0, closed ellipses if α=0\alpha = 0. Trace and determinant of a 2×22\times2 thus classify the whole portrait before any solving happens.

Example
Worked: a full system solve with initial data

Solve u′=(2112)u\mathbf u' = \begin{pmatrix}2&1\\1&2\end{pmatrix}\mathbf u, u(0)=(3,1)\mathbf u(0) = (3, 1).

Step 1 — eigenpairs (computed in the eigenvalue topic): λ=3\lambda = 3 with (1,1)(1,1); λ=1\lambda = 1 with (1,−1)(1,-1).

Step 2 — general solution: u=c1e3t(1,1)+c2et(1,−1)\mathbf u = c_1e^{3t}(1,1) + c_2e^{t}(1,-1).

Step 3 — initial data: (3,1)=c1(1,1)+c2(1,−1)(3, 1) = c_1(1,1) + c_2(1,-1) means c1+c2=3c_1 + c_2 = 3, c1−c2=1c_1 - c_2 = 1: c1=2c_1 = 2, c2=1c_2 = 1.

Step 4 — the answer: u(t)=2e3t(1,1)+et(1,−1)\mathbf u(t) = 2e^{3t}(1,1) + e^{t}(1,-1), i.e. x=2e3t+etx = 2e^{3t} + e^t, y=2e3t−ety = 2e^{3t} - e^t.

Step 5 — behavior: both eigenvalues are positive, so the origin is a source; as tt grows the e3te^{3t} term dominates and the trajectory hugs the line y=xy = x — the fast eigendirection — while for t→−∞t \to -\infty it aligned with (1,−1)(1,-1). Eigenvectors are the skeleton every trajectory drapes itself on.

Example
Worked: classify without solving

Classify the equilibrium of u′=Au\mathbf u' = A\mathbf u for A=(1221)A = \begin{pmatrix}1&2\\2&1\end{pmatrix}, then for B=(01−10)B = \begin{pmatrix}0&1\\-1&0\end{pmatrix}, using only trace and determinant.

For AA: det⁡=1−4=−3<0\det = 1 - 4 = -3 < 0 — eigenvalues have opposite signs (their product is negative): a SADDLE, unstable, no further computation required; indeed λ=3,−1\lambda = 3, -1 with eigenlines y=xy = x (outflow) and y=−xy = -x (inflow). For BB: tr=0\mathrm{tr} = 0, det⁡=1\det = 1: eigenvalues satisfy λ2+1=0\lambda^2 + 1 = 0, so λ=±i\lambda = \pm i — pure imaginary: closed orbits, a CENTER; and indeed x′=y,y′=−xx' = y, y' = -x conserves x2+y2x^2 + y^2, so trajectories are literal circles. Two matrices, four numbers total examined, both portraits fully determined.