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Systems of ODEs | Mathematics - Wyatt's Notes

4.1 First-Order Linear Systems

A system of first-order linear ODEs can be written in matrix form:

x=Ax+f(t)\mathbf{x}' = A\mathbf{x} + \mathbf{f}(t)

Where AA is an n×nn \times n matrix and x,fRn\mathbf{x}, \mathbf{f} \in \mathbb{R}^n.

4.2 Homogeneous Systems with Constant Coefficients

For x=Ax\mathbf{x}' = A\mathbf{x}Try x=veλt\mathbf{x} = \mathbf{v}e^{\lambda t}:

λv=Av\lambda \mathbf{v} = A\mathbf{v}

So λ\lambda is an eigenvalue of AA and v\mathbf{v} is the corresponding eigenvector.

Case 1: AA has nn distinct real eigenvalues. The general solution is

x=c1v1eλ1t++cnvneλnt\mathbf{x} = c_1 \mathbf{v}_1 e^{\lambda_1 t} + \cdots + c_n \mathbf{v}_n e^{\lambda_n t}

Case 2: AA has a repeated eigenvalue λ\lambda with algebraic multiplicity mm and geometric Multiplicity k<mk \lt m. Include terms involving tjeλtt^j e^{\lambda t} where generalized Eigenvectors fill out the solution space.

Case 3: Complex eigenvalues λ=α±iβ\lambda = \alpha \pm i\beta with eigenvector v=a±ib\mathbf{v} = \mathbf{a} \pm i\mathbf{b}. The real solutions are eαt(acos(βt)bsin(βt))e^{\alpha t}(\mathbf{a}\cos(\beta t) - \mathbf{b}\sin(\beta t)) and eαt(asin(βt)+bcos(βt))e^{\alpha t}(\mathbf{a}\sin(\beta t) + \mathbf{b}\cos(\beta t)).

4.3 The Matrix Exponential

Definition. eAt=k=0Aktkk!e^{At} = \sum_{k=0}^{\infty} \frac{A^k t^k}{k!}.

Theorem 4.1. The solution to x=Ax\mathbf{x}' = A\mathbf{x} with x(0)=x0\mathbf{x}(0) = \mathbf{x}_0 is x(t)=eAtx0\mathbf{x}(t) = e^{At}\mathbf{x}_0.

Proposition 4.2. If AA is diagonalizable as A=PDP1A = PDP^{-1} Then eAt=PeDtP1e^{At} = Pe^{Dt}P^{-1} Where eDt=diag(eλ1t,,eλnt)e^{Dt} = \mathrm{diag}(e^{\lambda_1 t}, \ldots, e^{\lambda_n t}).

4.4 Worked Example: Distinct Real Eigenvalues

Problem. Solve x=(0123)x\mathbf{x}' = \begin{pmatrix} 0 & 1 \\ -2 & -3 \end{pmatrix}\mathbf{x}.

Solution. Characteristic equation: det(AλI)=λ2+3λ+2=(λ+1)(λ+2)=0\det(A - \lambda I) = \lambda^2 + 3\lambda + 2 = (\lambda + 1)(\lambda + 2) = 0. Eigenvalues: λ1=1\lambda_1 = -1, λ2=2\lambda_2 = -2.

For λ1=1\lambda_1 = -1: (A+I)v=(1122)v=0(A + I)\mathbf{v} = \begin{pmatrix} 1 & 1 \\ -2 & -2 \end{pmatrix}\mathbf{v} = \mathbf{0}. v1=(11)\mathbf{v}_1 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}.

For λ2=2\lambda_2 = -2: (A+2I)v=(2121)v=0(A + 2I)\mathbf{v} = \begin{pmatrix} 2 & 1 \\ -2 & -1 \end{pmatrix}\mathbf{v} = \mathbf{0}. v2=(12)\mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}.

x(t)=c1(11)et+c2(12)e2t\mathbf{x}(t) = c_1 \begin{pmatrix} 1 \\ -1 \end{pmatrix} e^{-t} + c_2 \begin{pmatrix} 1 \\ -2 \end{pmatrix} e^{-2t}. \blacksquare

4.5 Worked Example: Complex Eigenvalues

Problem. Solve x=(0210)x\mathbf{x}' = \begin{pmatrix} 0 & -2 \\ 1 & 0 \end{pmatrix}\mathbf{x}.

Solution

Solution. det(AλI)=λ2+2=0\det(A - \lambda I) = \lambda^2 + 2 = 0. Eigenvalues: λ=±i2\lambda = \pm i\sqrt{2}.

For λ=i2\lambda = i\sqrt{2}: (i221i2)v=0\begin{pmatrix} -i\sqrt{2} & -2 \\ 1 & -i\sqrt{2} \end{pmatrix}\mathbf{v} = \mathbf{0}.

From the first row: i2v12v2=0-i\sqrt{2}\, v_1 - 2v_2 = 0 So v2=i22v1v_2 = -\frac{i\sqrt{2}}{2}v_1.

With v1=2v_1 = 2: v=(20)+i(02)\mathbf{v} = \begin{pmatrix} 2 \\ 0 \end{pmatrix} + i\begin{pmatrix} 0 \\ -\sqrt{2} \end{pmatrix}.

So a=(20)\mathbf{a} = \begin{pmatrix} 2 \\ 0 \end{pmatrix}, b=(02)\mathbf{b} = \begin{pmatrix} 0 \\ -\sqrt{2} \end{pmatrix}.

x(t)=c1[acos(2t)bsin(2t)]+c2[asin(2t)+bcos(2t)]\mathbf{x}(t) = c_1\left[\mathbf{a}\cos(\sqrt{2}\, t) - \mathbf{b}\sin(\sqrt{2}\, t)\right] + c_2\left[\mathbf{a}\sin(\sqrt{2}\, t) + \mathbf{b}\cos(\sqrt{2}\, t)\right]

=c1(2cos(2t)2sin(2t))+c2(2sin(2t)2cos(2t))= c_1 \begin{pmatrix} 2\cos(\sqrt{2}\, t) \\ \sqrt{2}\sin(\sqrt{2}\, t) \end{pmatrix} + c_2 \begin{pmatrix} 2\sin(\sqrt{2}\, t) \\ -\sqrt{2}\cos(\sqrt{2}\, t) \end{pmatrix}. \blacksquare

4.6 Worked Example: Repeated Eigenvalues

Problem. Solve x=(2114)x\mathbf{x}' = \begin{pmatrix} 2 & 1 \\ -1 & 4 \end{pmatrix}\mathbf{x}.

Solution

Solution. det(AλI)=(2λ)(4λ)+1=λ26λ+9=(λ3)2=0\det(A - \lambda I) = (2 - \lambda)(4 - \lambda) + 1 = \lambda^2 - 6\lambda + 9 = (\lambda - 3)^2 = 0.

Repeated eigenvalue λ=3\lambda = 3 with algebraic multiplicity 2.

(A3I)=(1111)(A - 3I) = \begin{pmatrix} -1 & 1 \\ -1 & 1 \end{pmatrix}.

Eigenvector: (11)\begin{pmatrix} 1 \\ 1 \end{pmatrix}. Only one eigenvector (geometric multiplicity 1), so we need a generalized eigenvector.

Find w\mathbf{w} such that (A3I)w=v1=(11)(A - 3I)\mathbf{w} = \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}:

(1111)(w1w2)=(11)\begin{pmatrix} -1 & 1 \\ -1 & 1 \end{pmatrix}\begin{pmatrix} w_1 \\ w_2 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}

w1+w2=1-w_1 + w_2 = 1. Choose w1=0w_1 = 0 Then w2=1w_2 = 1. So w=(01)\mathbf{w} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}.

x(t)=c1(11)e3t+c2[(11)te3t+(01)e3t]\mathbf{x}(t) = c_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} e^{3t} + c_2 \left[\begin{pmatrix} 1 \\ 1 \end{pmatrix} t e^{3t} + \begin{pmatrix} 0 \\ 1 \end{pmatrix} e^{3t}\right]

=e3t[c1(11)+c2(tt+1)]= e^{3t}\left[c_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2 \begin{pmatrix} t \\ t + 1 \end{pmatrix}\right]. \blacksquare

4.7 Fundamental Matrix

Definition. A fundamental matrix Φ(t)\Phi(t) for the system x=Ax\mathbf{x}' = A\mathbf{x} is an n×nn \times n matrix whose columns form a fundamental set of solutions.

Proposition 4.3. Φ(t)\Phi(t) satisfies Φ=AΦ\Phi' = A\Phi And the general solution is x(t)=Φ(t)c\mathbf{x}(t) = \Phi(t)\mathbf{c} for cRn\mathbf{c} \in \mathbb{R}^n.

Proposition 4.4. The matrix exponential eAte^{At} is a fundamental matrix with eA0=Ie^{A \cdot 0} = I. Any fundamental matrix can be written as Φ(t)=eAtΦ(0)\Phi(t) = e^{At}\Phi(0).

4.8 Matrix Exponential Properties

Theorem 4.5. The matrix exponential satisfies:

  1. eA0=Ie^{A \cdot 0} = I
  2. ddteAt=AeAt=eAtA\frac{d}{dt}e^{At} = Ae^{At} = e^{At}A
  3. eAteAs=eA(t+s)e^{At}e^{As} = e^{A(t+s)}
  4. (eAt)1=eAt(e^{At})^{-1} = e^{-At}
  5. If AB=BAAB = BA Then eA+B=eAeBe^{A+B} = e^A e^B

Proof of (1). eA0=k=0Ak0kk!=Ie^{A \cdot 0} = \sum_{k=0}^{\infty} \frac{A^k 0^k}{k!} = I. \blacksquare

Proof of (2). ddteAt=k=1Aktk1(k1)!=Aj=0Ajtjj!=AeAt\frac{d}{dt}e^{At} = \sum_{k=1}^{\infty} \frac{A^k t^{k-1}}{(k-1)!} = A\sum_{j=0}^{\infty} \frac{A^j t^j}{j!} = Ae^{At}. Since AA commutes with itself, AeAt=eAtAAe^{At} = e^{At}A. \blacksquare

Proof of (4). From (3) with s=ts = -t: eAteAt=eA(tt)=e0=Ie^{At}e^{-At} = e^{A(t-t)} = e^0 = I. \blacksquare

4.9 Phase Portrait Analysis for 2D Systems

For the linear system x=Ax\mathbf{x}' = A\mathbf{x} with AA a 2×22 \times 2 matrix, the qualitative Behaviour near the origin is determined by the eigenvalues:

EigenvaluesPhase PortraitStability
λ1,λ2<0\lambda_1, \lambda_2 \lt 0Real, distinctStable nodeAsymptotically stable
λ1,λ2>0\lambda_1, \lambda_2 > 0Real, distinctUnstable nodeUnstable
λ1<0<λ2\lambda_1 \lt 0 \lt \lambda_2Saddle pointUnstable
λ=α±iβ\lambda = \alpha \pm i\beta, α<0\alpha \lt 0Stable spiralAsymptotically stable
λ=α±iβ\lambda = \alpha \pm i\beta, α>0\alpha > 0Unstable spiralUnstable
λ=±iβ\lambda = \pm i\betaCenter(Marginally) stable

Remark. The trace-determinant plane provides a convenient classification. Let τ=tr(A)\tau = \mathrm{tr}(A) and Δ=det(A)\Delta = \det(A). The eigenvalues satisfy λ2τλ+Δ=0\lambda^2 - \tau\lambda + \Delta = 0 So:

λ=τ±τ24Δ2\lambda = \frac{\tau \pm \sqrt{\tau^2 - 4\Delta}}{2}

  • τ24Δ>0\tau^2 - 4\Delta > 0: real eigenvalues (node or saddle)
  • τ24Δ<0\tau^2 - 4\Delta \lt 0: complex eigenvalues (spiral or center)
  • τ24Δ=0\tau^2 - 4\Delta = 0: repeated eigenvalues (proper or improper node)

Stability is determined by the sign of τ\tau: stable if τ<0\tau \lt 0Unstable if τ>0\tau > 0.

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The trace-determinant plane classifies 2D linear systems. The parabola τ2=4Δ\tau^2 = 4\Delta separates real from complex eigenvalues; the τ=0\tau = 0 line separates stable from unstable. The x-axis represents the trace τ\tau and the y-axis represents Δ\Delta.

4.10 Nonhomogeneous Systems

For x=Ax+f(t)\mathbf{x}' = A\mathbf{x} + \mathbf{f}(t)If Φ(t)\Phi(t) is a fundamental matrix for the Homogeneous system, the general solution is

x(t)=Φ(t)c+Φ(t)Φ1(s)f(s)ds\mathbf{x}(t) = \Phi(t)\mathbf{c} + \Phi(t)\int \Phi^{-1}(s)\mathbf{f}(s)\, ds

Worked Example. Solve x=(1002)x+(et0)\mathbf{x}' = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}\mathbf{x} + \begin{pmatrix} e^t \\ 0 \end{pmatrix}.

Solution

Solution. Eigenvalues: 11 and 22. Φ(t)=(et00e2t)\Phi(t) = \begin{pmatrix} e^t & 0 \\ 0 & e^{2t} \end{pmatrix}.

Φ1(s)=(es00e2s)\Phi^{-1}(s) = \begin{pmatrix} e^{-s} & 0 \\ 0 & e^{-2s} \end{pmatrix}.

Φ1(s)f(s)=(es00e2s)(es0)=(10)\Phi^{-1}(s)\mathbf{f}(s) = \begin{pmatrix} e^{-s} & 0 \\ 0 & e^{-2s} \end{pmatrix}\begin{pmatrix} e^s \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}.

Φ1(s)f(s)ds=(t0)\int \Phi^{-1}(s)\mathbf{f}(s)\, ds = \begin{pmatrix} t \\ 0 \end{pmatrix}.

xp=Φ(t)(t0)=(tet0)\mathbf{x}_p = \Phi(t)\begin{pmatrix} t \\ 0 \end{pmatrix} = \begin{pmatrix} te^t \\ 0 \end{pmatrix}.

x(t)=c1(et0)+c2(0e2t)+(tet0)\mathbf{x}(t) = c_1 \begin{pmatrix} e^t \\ 0 \end{pmatrix} + c_2 \begin{pmatrix} 0 \\ e^{2t} \end{pmatrix} + \begin{pmatrix} te^t \\ 0 \end{pmatrix}. \blacksquare

4.11 Common Mistakes

flowchart TD
    A[4_Systems Of Odesx] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Intuition

Systems of ODEs describe how multiple quantities evolve together, each potentially influencing the others. A linear system x=Ax\mathbf{x}' = A\mathbf{x} decomposes along the eigenvectors of AA: each eigenvector defines a natural direction of motion, and its eigenvalue determines whether the solution grows, decays, or oscillates along that direction. The matrix exponential eAte^{At} is the fundamental object — it maps initial conditions to solutions at time tt. When eigenvalues are complex, the solution spirals; when real and negative, it decays toward the origin; when real and positive, it explodes. The trace-determinant plane classifies all 2D linear systems into nodes, saddles, spirals, and centres.

4.11 Common Mistakes

Mistake 1: Forgetting to account for all initial conditions When solving a system of nn ODEs, the general solution must contain nn arbitrary constants. Students often omit constants or assume the initial conditions are automatically satisfied. Always apply all initial conditions to determine every constant, and verify the solution satisfies each equation in the system.

Mistake 2: Using the wrong eigenvalue-eigenvector pairs When a matrix has repeated eigenvalues, the eigenvectors may not span the full space, requiring generalised eigenvectors. A common error is writing the same eigenvector for both occurrences of a repeated eigenvalue instead of constructing the Jordan chain. Check the geometric multiplicity against the algebraic multiplicity.

Mistake 3: Confusing the matrix exponential eAte^{At} with eAete^A e^{t} The identity eAt=eAte^{At} = e^{At} holds trivially, but eA+B=eAeBe^{A+B} = e^A e^B only when AB=BAAB = BA. Students sometimes write eAt=(eA)te^{At} = (e^A)^t or attempt to exponentiate component-wise. For diagonalisable matrices, use eAt=PeDtP1e^{At} = Pe^{Dt}P^{-1} where DD is the diagonal matrix of eigenvalues.

Cross-References

  • Stability and Phase Plane Analysis: The phase portrait classification of 2D systems uses the eigenvalue analysis developed for matrix systems.
  • Second-Order Linear ODEs: Systems of first-order ODEs can be derived from a single second-order ODE by introducing new variables.
  • Laplace Transforms: The Laplace transform converts systems of ODEs into algebraic equations in the frequency domain.
  • Partial Derivatives: The Jacobian matrix and matrix exponential are fundamental tools from linear algebra used throughout systems theory.