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Module 3/Applications

Dynamical Systems: Discrete & Continuous

Use eigenvalues and diagonalization to solve discrete systems x(t+1)=Ax(t)x(t+1)=Ax(t) and continuous systems dxdt=Ax\frac{dx}{dt}=Ax in closed form, and to read off their stability and long-run behavior directly from the eigenvalues.

Before you start — give these a try

Attempting first primes your brain for the lesson — even if you miss. Nothing is graded or saved; it's just a warm-up.

For which of these discrete systems x(t+1)=Ax(t)x(t+1)=Ax(t) do all trajectories converge to 00 as t→∞t\to\infty? Select all that apply. (Each option lists the eigenvalues of AA.)

A discrete system x(t+1)=Ax(t)x(t+1)=Ax(t) has eigenpairs λ1=2, v1=[11]\lambda_1=2,\ v_1=\begin{bmatrix}1\\1\end{bmatrix} and λ2=1, v2=[1−1]\lambda_2=1,\ v_2=\begin{bmatrix}1\\-1\end{bmatrix}. The initial state is x0=3v1+2v2x_0=3v_1+2v_2 (so c1=3c_1=3, c2=2c_2=2). Compute x(2)x(2), giving your answer as [x1x2]\begin{bmatrix}x_1\\x_2\end{bmatrix} (order: first component, then second component).

What you’ll be able to do

  • Model a discrete linear dynamical system x(t+1)=Ax(t)x(t+1)=Ax(t) and write its closed-form solution x(t)=c1λ1tv1+⋯+cnλntvn=Atx0=SDtS−1x0x(t)=c_1\lambda_1^t v_1+\cdots+c_n\lambda_n^t v_n = A^t x_0 = SD^tS^{-1}x_0.
  • Solve a continuous system dxdt=Ax\frac{dx}{dt}=Ax with the eigensolutions x(t)=c1eλ1tv1+⋯+cneλntvnx(t)=c_1 e^{\lambda_1 t}v_1+\cdots+c_n e^{\lambda_n t}v_n, finding the cic_i as coordinates of x0x_0 in the eigenbasis.
  • Determine stability from the eigenvalues: ∣λ∣|\lambda| versus 11 in the discrete case, Re(λ)\mathrm{Re}(\lambda) versus 00 in the continuous case.
  • Classify a 2D equilibrium as a node or a saddle, and identify the dominant eigenvalue/eigenvector that governs the long-run behavior and ratios.
  • Compute the half-life of a continuous exponential decay from its rate constant.

In your course

· MATH2015 · Linear Algebra & Probability
§7.4 Applications
  • Theorem 7.11Discrete Dynamical Systems
    If AA is diagonalizable, x(t+1)=Ax(t)x(t+1)=Ax(t) has solution x(t)=c1λ1tv1+⋯+cnλntvn=Atx0=SDtS−1x0x(t)=c_1\lambda_1^t v_1+\cdots+c_n\lambda_n^t v_n=A^t x_0=SD^tS^{-1}x_0, with the cic_i fixed by x(0)=x0x(0)=x_0.
  • Theorem 7.12Continuous Dynamical Systems
    If AA has a real eigenbasis, dxdt=Ax\frac{dx}{dt}=Ax has solution x(t)=c1eλ1tv1+⋯+cneλntvnx(t)=c_1 e^{\lambda_1 t}v_1+\cdots+c_n e^{\lambda_n t}v_n, where the cic_i are the coordinates of x0x_0 in that eigenbasis.
  • Example 7.10Bird population model (discrete; λ=1,−12\lambda=1,-\tfrac12; long-run juvenile:adult ratio =2=2)
  • Example 7.11Radioactive decay and half-life t∗=−ln⁡2/kt^{\ast}=-\ln 2/k
  • Example 7.12Coupled ODEs with A=[3113]A=\begin{bmatrix}3&1\\1&3\end{bmatrix} (λ=4,2\lambda=4,2; unstable node)
Figures 7.1 and 7.2 give the discrete and continuous phase portraits (unstable node, saddle point, stable node). The scalar case dxdt=kx\frac{dx}{dt}=kx (solution x0ektx_0 e^{kt}) supplies the stability vocabulary, and the same eigenbasis method extends to n×nn\times n systems (e.g. Example 7.13).
1

Discrete systems: iteration is a matrix power

A discrete linear dynamical system updates its state by x(t+1)=A x(t)x(t+1)=A\,x(t), so after tt steps x(t)=Atx0x(t)=A^t x_0. The updating matrix AA encodes the step-to-step rule. When AA is diagonalizable with eigenpairs (λi,vi)(\lambda_i,v_i), expand the initial state in the eigenbasis, x0=c1v1+⋯+cnvnx_0=c_1 v_1+\cdots+c_n v_n. Because Atvi=λitviA^t v_i=\lambda_i^t v_i, each eigen-direction evolves independently, giving x(t)=c1λ1tv1+⋯+cnλntvnx(t)=c_1\lambda_1^t v_1+\cdots+c_n\lambda_n^t v_n. Equivalently x(t)=Atx0=SDtS−1x0x(t)=A^t x_0=SD^tS^{-1}x_0, where S=[ v1  ⋯  vn ]S=[\,v_1\;\cdots\;v_n\,], D=diag(λ1,…,λn)D=\mathrm{diag}(\lambda_1,\dots,\lambda_n), and S−1x0=(c1,…,cn)TS^{-1}x_0=(c_1,\dots,c_n)^T. Diagonalization replaces the awkward power AtA^t with nn scalar powers λit\lambda_i^t.

2

Continuous systems: exponential eigensolutions

A continuous linear system dxdt=Ax\frac{dx}{dt}=Ax is solved by seeking exponential eigensolutions x(t)=eλtvx(t)=e^{\lambda t}v. Substituting gives λeλtv=eλtAv\lambda e^{\lambda t}v=e^{\lambda t}Av, which holds iff Av=λvAv=\lambda v — so (λ,v)(\lambda,v) must be an eigenpair. With a real eigenbasis, superposition combines these into the general solution x(t)=c1eλ1tv1+⋯+cneλntvnx(t)=c_1 e^{\lambda_1 t}v_1+\cdots+c_n e^{\lambda_n t}v_n, where c1,…,cnc_1,\dots,c_n are the coordinates of x0x_0 in that eigenbasis. The one-dimensional case dxdt=kx\frac{dx}{dt}=kx has solution x(t)=x0ektx(t)=x_0 e^{kt}, the building block for every eigen-direction.

3

Stability and the dominant eigenvalue

The eigenvalues decide whether each mode grows or decays. Discrete (λt\lambda^t): ∣λ∣<1|\lambda|<1 decays to 00, ∣λ∣>1|\lambda|>1 grows without bound, ∣λ∣=1|\lambda|=1 is neutral; the system is asymptotically stable iff every ∣λ∣<1|\lambda|<1. Continuous (eλte^{\lambda t}): Re(λ)<0\mathrm{Re}(\lambda)<0 decays, Re(λ)>0\mathrm{Re}(\lambda)>0 grows; asymptotically stable iff every Re(λ)<0\mathrm{Re}(\lambda)<0. In 2D, when both eigenvalues share a sign the origin is a node (stable node if both negative, unstable node if both positive); with opposite signs it is a saddle point. The dominant eigenvalue — largest ∣λ∣|\lambda| (discrete) or largest Re(λ)\mathrm{Re}(\lambda) (continuous) — governs long-run behavior: as t→∞t\to\infty a generic trajectory becomes nearly parallel to the dominant eigenvector, while for t→−∞t\to-\infty it aligns with the weakest one.

Theorem 7.11 — Discrete Dynamical Systems

Consider x(t+1)=Ax(t)x(t+1)=Ax(t). If AA is diagonalizable, the general solution is x(t)=c1λ1tv1+c2λ2tv2+⋯+cnλntvnx(t)=c_1\lambda_1^t v_1+c_2\lambda_2^t v_2+\cdots+c_n\lambda_n^t v_n, where (λi,vi)(\lambda_i,v_i) are the eigenpairs of AA and the constants cic_i are determined by x(0)=x0x(0)=x_0. Equivalently, x(t)=Atx0=SDtS−1x0x(t)=A^t x_0=SD^tS^{-1}x_0.

Intuition. Diagonalizing turns the hard matrix power AtA^t into nn independent scalar powers λit\lambda_i^t. Writing x0x_0 in the eigenbasis decouples the system, so each coordinate is simply multiplied by its own eigenvalue at every step.
Theorem 7.12 — Continuous Dynamical Systems

Consider dxdt=Ax\frac{dx}{dt}=Ax with x(0)=x0x(0)=x_0. If AA has a real eigenbasis v1,…,vnv_1,\dots,v_n with eigenvalues λ1,…,λn\lambda_1,\dots,\lambda_n, then the general solution is x(t)=c1eλ1tv1+⋯+cneλntvnx(t)=c_1 e^{\lambda_1 t}v_1+\cdots+c_n e^{\lambda_n t}v_n. The scalars c1,…,cnc_1,\dots,c_n are the coordinates of x0x_0 with respect to the eigenbasis.

Intuition. Along each eigen-direction the system behaves like the scalar equation dydt=λy\frac{dy}{dt}=\lambda y, whose solution is eλte^{\lambda t}. The eigenbasis decouples the coupled ODEs into nn independent exponentials, and superposition recombines them.

Worked examples

Example 1

(Example 7.10) A species of bird: each female is a juvenile for one year and then becomes an adult, and only adults lay eggs. (i) The juveniles hatched in a year equal 2×2\times the adults alive the year before; (ii) half the adult females survive to the next year; (iii) one quarter of juveniles survive into adulthood. Initially there are 100100 adult and 4040 juvenile females. Find the population after tt years and the long-run ratio of juveniles to adults.

  1. 1

    Let x(t)=[atjt]x(t)=\begin{bmatrix}a_t\\j_t\end{bmatrix} (adults, juveniles). The rules give at+1=12at+14jta_{t+1}=\tfrac12 a_t+\tfrac14 j_t and jt+1=2atj_{t+1}=2a_t, so A=[1/21/420]A=\begin{bmatrix}1/2&1/4\\2&0\end{bmatrix}, x0=[10040]x_0=\begin{bmatrix}100\\40\end{bmatrix}, and x(t)=Atx0x(t)=A^t x_0.

  2. 2

    Characteristic equation: λ2−12λ−12=0\lambda^2-\tfrac12\lambda-\tfrac12=0, i.e. (λ−1)(λ+12)=0(\lambda-1)(\lambda+\tfrac12)=0, so λ1=1\lambda_1=1 and λ2=−12\lambda_2=-\tfrac12.

  3. 3

    Eigenvectors: v1=[12]v_1=\begin{bmatrix}1\\2\end{bmatrix} for λ1=1\lambda_1=1 and v2=[−14]v_2=\begin{bmatrix}-1\\4\end{bmatrix} for λ2=−12\lambda_2=-\tfrac12.

  4. 4

    Write x0=c1v1+c2v2x_0=c_1 v_1+c_2 v_2: solving c1−c2=100c_1-c_2=100 and 2c1+4c2=402c_1+4c_2=40 gives c1=2203c_1=\tfrac{220}{3}, c2=−803c_2=-\tfrac{80}{3}.

  5. 5

    By Theorem 7.11, x(t)=c1(1)tv1+c2(−12)tv2x(t)=c_1(1)^t v_1+c_2\left(-\tfrac12\right)^t v_2, so at=2203+803(−12)ta_t=\tfrac{220}{3}+\tfrac{80}{3}\left(-\tfrac12\right)^t and jt=4403−3203(−12)tj_t=\tfrac{440}{3}-\tfrac{320}{3}\left(-\tfrac12\right)^t; the total is at+jt=220−80(−12)ta_t+j_t=220-80\left(-\tfrac12\right)^t.

  6. 6

    As t→∞t\to\infty, (−12)t→0\left(-\tfrac12\right)^t\to 0 (since ∣−12∣<1|-\tfrac12|<1), so jtat→440/3220/3=2\dfrac{j_t}{a_t}\to\dfrac{440/3}{220/3}=2.

Answer. at=2203+803(−12)ta_t=\tfrac{220}{3}+\tfrac{80}{3}\left(-\tfrac12\right)^t and jt=4403−3203(−12)tj_t=\tfrac{440}{3}-\tfrac{320}{3}\left(-\tfrac12\right)^t (total population →220\to 220). The long-run juvenile-to-adult ratio is 22. The dominant eigenvalue λ=1\lambda=1 is why the population stabilizes rather than growing or dying out.
Example 2

(Example 7.11) The radioactive decay of an isotope obeys dxdt=kx\frac{dx}{dt}=kx with k<0k<0, where x(t)x(t) is the mass remaining at time tt. Find x(t)x(t) and the half-life t∗t^{\ast} (the time for half the sample to decay).

Example 3

(Example 7.12) Solve the coupled system dx1dt=3x1+x2\frac{dx_1}{dt}=3x_1+x_2, dx2dt=x1+3x2\frac{dx_2}{dt}=x_1+3x_2, and classify the equilibrium at the origin.