A system of first-order linear ODEs can be written in matrix form:
x ′ = A x + f ( t ) \mathbf{x}' = A\mathbf{x} + \mathbf{f}(t) x ′ = A x + f ( t )
Where A A A n × n n \times n n × n x , f ∈ R n \mathbf{x}, \mathbf{f} \in \mathbb{R}^n x , f ∈ R n
For x ′ = A x \mathbf{x}' = A\mathbf{x} x ′ = A x x = v e λ t \mathbf{x} = \mathbf{v}e^{\lambda t} x = v e λ t
λ v = A v \lambda \mathbf{v} = A\mathbf{v} λ v = A v
So λ \lambda λ A A A v \mathbf{v} v
Case 1: A A A n n n The general solution is
x = c 1 v 1 e λ 1 t + ⋯ + c n v n e λ n t \mathbf{x} = c_1 \mathbf{v}_1 e^{\lambda_1 t} + \cdots + c_n \mathbf{v}_n e^{\lambda_n t} x = c 1 v 1 e λ 1 t + ⋯ + c n v n e λ n t
Case 2: A A A λ \lambda λ m m m k < m k \lt m k < m Include terms involving t j e λ t t^j e^{\lambda t} t j e λ t
Case 3: Complex eigenvalues λ = α ± i β \lambda = \alpha \pm i\beta λ = α ± i β v = a ± i b \mathbf{v} = \mathbf{a} \pm i\mathbf{b} v = a ± i b The real solutions are e α t ( a cos ( β t ) − b sin ( β t ) ) e^{\alpha t}(\mathbf{a}\cos(\beta t) - \mathbf{b}\sin(\beta t)) e α t ( a cos ( β t ) − b sin ( β t )) e α t ( a sin ( β t ) + b cos ( β t ) ) e^{\alpha t}(\mathbf{a}\sin(\beta t) + \mathbf{b}\cos(\beta t)) e α t ( a sin ( β t ) + b cos ( β t ))
Definition. e A t = ∑ k = 0 ∞ A k t k k ! e^{At} = \sum_{k=0}^{\infty} \frac{A^k t^k}{k!} e A t = ∑ k = 0 ∞ k ! A k t k
Theorem 4.1. The solution to x ′ = A x \mathbf{x}' = A\mathbf{x} x ′ = A x x ( 0 ) = x 0 \mathbf{x}(0) = \mathbf{x}_0 x ( 0 ) = x 0 x ( t ) = e A t x 0 \mathbf{x}(t) = e^{At}\mathbf{x}_0 x ( t ) = e A t x 0
Proposition 4.2. If A A A A = P D P − 1 A = PDP^{-1} A = P D P − 1 e A t = P e D t P − 1 e^{At} = Pe^{Dt}P^{-1} e A t = P e D t P − 1 e D t = d i a g ( e λ 1 t , … , e λ n t ) e^{Dt} = \mathrm{diag}(e^{\lambda_1 t}, \ldots, e^{\lambda_n t}) e D t = diag ( e λ 1 t , … , e λ n t )
Problem. Solve x ′ = ( 0 1 − 2 − 3 ) x \mathbf{x}' = \begin{pmatrix} 0 & 1 \\ -2 & -3 \end{pmatrix}\mathbf{x} x ′ = ( 0 − 2 1 − 3 ) x
Solution. Characteristic equation: det ( A − λ I ) = λ 2 + 3 λ + 2 = ( λ + 1 ) ( λ + 2 ) = 0 \det(A - \lambda I) = \lambda^2 + 3\lambda + 2 = (\lambda + 1)(\lambda + 2) = 0 det ( A − λ I ) = λ 2 + 3 λ + 2 = ( λ + 1 ) ( λ + 2 ) = 0 λ 1 = − 1 \lambda_1 = -1 λ 1 = − 1 λ 2 = − 2 \lambda_2 = -2 λ 2 = − 2
For λ 1 = − 1 \lambda_1 = -1 λ 1 = − 1 ( A + I ) v = ( 1 1 − 2 − 2 ) v = 0 (A + I)\mathbf{v} = \begin{pmatrix} 1 & 1 \\ -2 & -2 \end{pmatrix}\mathbf{v} = \mathbf{0} ( A + I ) v = ( 1 − 2 1 − 2 ) v = 0 v 1 = ( 1 − 1 ) \mathbf{v}_1 = \begin{pmatrix} 1 \\ -1 \end{pmatrix} v 1 = ( 1 − 1 )
For λ 2 = − 2 \lambda_2 = -2 λ 2 = − 2 ( A + 2 I ) v = ( 2 1 − 2 − 1 ) v = 0 (A + 2I)\mathbf{v} = \begin{pmatrix} 2 & 1 \\ -2 & -1 \end{pmatrix}\mathbf{v} = \mathbf{0} ( A + 2 I ) v = ( 2 − 2 1 − 1 ) v = 0 v 2 = ( 1 − 2 ) \mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix} v 2 = ( 1 − 2 )
x ( t ) = c 1 ( 1 − 1 ) e − t + c 2 ( 1 − 2 ) e − 2 t \mathbf{x}(t) = c_1 \begin{pmatrix} 1 \\ -1 \end{pmatrix} e^{-t} + c_2 \begin{pmatrix} 1 \\ -2 \end{pmatrix} e^{-2t} x ( t ) = c 1 ( 1 − 1 ) e − t + c 2 ( 1 − 2 ) e − 2 t ■ \blacksquare ■
Problem. Solve x ′ = ( 0 − 2 1 0 ) x \mathbf{x}' = \begin{pmatrix} 0 & -2 \\ 1 & 0 \end{pmatrix}\mathbf{x} x ′ = ( 0 1 − 2 0 ) x
Solution Solution. det ( A − λ I ) = λ 2 + 2 = 0 \det(A - \lambda I) = \lambda^2 + 2 = 0 det ( A − λ I ) = λ 2 + 2 = 0 λ = ± i 2 \lambda = \pm i\sqrt{2} λ = ± i 2
For λ = i 2 \lambda = i\sqrt{2} λ = i 2 ( − i 2 − 2 1 − i 2 ) v = 0 \begin{pmatrix} -i\sqrt{2} & -2 \\ 1 & -i\sqrt{2} \end{pmatrix}\mathbf{v} = \mathbf{0} ( − i 2 1 − 2 − i 2 ) v = 0
From the first row: − i 2 v 1 − 2 v 2 = 0 -i\sqrt{2}\, v_1 - 2v_2 = 0 − i 2 v 1 − 2 v 2 = 0 v 2 = − i 2 2 v 1 v_2 = -\frac{i\sqrt{2}}{2}v_1 v 2 = − 2 i 2 v 1
With v 1 = 2 v_1 = 2 v 1 = 2 v = ( 2 0 ) + i ( 0 − 2 ) \mathbf{v} = \begin{pmatrix} 2 \\ 0 \end{pmatrix} + i\begin{pmatrix} 0 \\ -\sqrt{2} \end{pmatrix} v = ( 2 0 ) + i ( 0 − 2 )
So a = ( 2 0 ) \mathbf{a} = \begin{pmatrix} 2 \\ 0 \end{pmatrix} a = ( 2 0 ) b = ( 0 − 2 ) \mathbf{b} = \begin{pmatrix} 0 \\ -\sqrt{2} \end{pmatrix} b = ( 0 − 2 )
x ( t ) = c 1 [ a cos ( 2 t ) − b sin ( 2 t ) ] + c 2 [ a sin ( 2 t ) + b cos ( 2 t ) ] \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] x ( t ) = c 1 [ a cos ( 2 t ) − b sin ( 2 t ) ] + c 2 [ a sin ( 2 t ) + b cos ( 2 t ) ]
= c 1 ( 2 cos ( 2 t ) 2 sin ( 2 t ) ) + c 2 ( 2 sin ( 2 t ) − 2 cos ( 2 t ) ) = 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} = c 1 ( 2 cos ( 2 t ) 2 sin ( 2 t ) ) + c 2 ( 2 sin ( 2 t ) − 2 cos ( 2 t ) ) ■ \blacksquare ■
Problem. Solve x ′ = ( 2 1 − 1 4 ) x \mathbf{x}' = \begin{pmatrix} 2 & 1 \\ -1 & 4 \end{pmatrix}\mathbf{x} x ′ = ( 2 − 1 1 4 ) x
Solution Solution. det ( A − λ I ) = ( 2 − λ ) ( 4 − λ ) + 1 = λ 2 − 6 λ + 9 = ( λ − 3 ) 2 = 0 \det(A - \lambda I) = (2 - \lambda)(4 - \lambda) + 1 = \lambda^2 - 6\lambda + 9 = (\lambda - 3)^2 = 0 det ( A − λ I ) = ( 2 − λ ) ( 4 − λ ) + 1 = λ 2 − 6 λ + 9 = ( λ − 3 ) 2 = 0
Repeated eigenvalue λ = 3 \lambda = 3 λ = 3
( A − 3 I ) = ( − 1 1 − 1 1 ) (A - 3I) = \begin{pmatrix} -1 & 1 \\ -1 & 1 \end{pmatrix} ( A − 3 I ) = ( − 1 − 1 1 1 )
Eigenvector: ( 1 1 ) \begin{pmatrix} 1 \\ 1 \end{pmatrix} ( 1 1 )
Find w \mathbf{w} w ( A − 3 I ) w = v 1 = ( 1 1 ) (A - 3I)\mathbf{w} = \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} ( A − 3 I ) w = v 1 = ( 1 1 )
( − 1 1 − 1 1 ) ( w 1 w 2 ) = ( 1 1 ) \begin{pmatrix} -1 & 1 \\ -1 & 1 \end{pmatrix}\begin{pmatrix} w_1 \\ w_2 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} ( − 1 − 1 1 1 ) ( w 1 w 2 ) = ( 1 1 )
− w 1 + w 2 = 1 -w_1 + w_2 = 1 − w 1 + w 2 = 1 w 1 = 0 w_1 = 0 w 1 = 0 w 2 = 1 w_2 = 1 w 2 = 1 w = ( 0 1 ) \mathbf{w} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} w = ( 0 1 )
x ( t ) = c 1 ( 1 1 ) e 3 t + c 2 [ ( 1 1 ) t e 3 t + ( 0 1 ) e 3 t ] \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] x ( t ) = c 1 ( 1 1 ) e 3 t + c 2 [ ( 1 1 ) t e 3 t + ( 0 1 ) e 3 t ]
= e 3 t [ c 1 ( 1 1 ) + c 2 ( t t + 1 ) ] = e^{3t}\left[c_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} + c_2 \begin{pmatrix} t \\ t + 1 \end{pmatrix}\right] = e 3 t [ c 1 ( 1 1 ) + c 2 ( t t + 1 ) ] ■ \blacksquare ■
Definition. A fundamental matrix Φ ( t ) \Phi(t) Φ ( t ) x ′ = A x \mathbf{x}' = A\mathbf{x} x ′ = A x n × n n \times n n × n
Proposition 4.3. Φ ( t ) \Phi(t) Φ ( t ) Φ ′ = A Φ \Phi' = A\Phi Φ ′ = A Φ x ( t ) = Φ ( t ) c \mathbf{x}(t) = \Phi(t)\mathbf{c} x ( t ) = Φ ( t ) c c ∈ R n \mathbf{c} \in \mathbb{R}^n c ∈ R n
Proposition 4.4. The matrix exponential e A t e^{At} e A t e A ⋅ 0 = I e^{A \cdot 0} = I e A ⋅ 0 = I Φ ( t ) = e A t Φ ( 0 ) \Phi(t) = e^{At}\Phi(0) Φ ( t ) = e A t Φ ( 0 )
Theorem 4.5. The matrix exponential satisfies:
e A ⋅ 0 = I e^{A \cdot 0} = I e A ⋅ 0 = I d d t e A t = A e A t = e A t A \frac{d}{dt}e^{At} = Ae^{At} = e^{At}A d t d e A t = A e A t = e A t A e A t e A s = e A ( t + s ) e^{At}e^{As} = e^{A(t+s)} e A t e A s = e A ( t + s ) ( e A t ) − 1 = e − A t (e^{At})^{-1} = e^{-At} ( e A t ) − 1 = e − A t If A B = B A AB = BA A B = B A e A + B = e A e B e^{A+B} = e^A e^B e A + B = e A e B Proof of (1). e A ⋅ 0 = ∑ k = 0 ∞ A k 0 k k ! = I e^{A \cdot 0} = \sum_{k=0}^{\infty} \frac{A^k 0^k}{k!} = I e A ⋅ 0 = ∑ k = 0 ∞ k ! A k 0 k = I ■ \blacksquare ■
Proof of (2). d d t e A t = ∑ k = 1 ∞ A k t k − 1 ( k − 1 ) ! = A ∑ j = 0 ∞ A j t j j ! = A e A t \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} d t d e A t = ∑ k = 1 ∞ ( k − 1 )! A k t k − 1 = A ∑ j = 0 ∞ j ! A j t j = A e A t A A A A e A t = e A t A Ae^{At} = e^{At}A A e A t = e A t A ■ \blacksquare ■
Proof of (4). From (3) with s = − t s = -t s = − t e A t e − A t = e A ( t − t ) = e 0 = I e^{At}e^{-At} = e^{A(t-t)} = e^0 = I e A t e − A t = e A ( t − t ) = e 0 = I ■ \blacksquare ■
For the linear system x ′ = A x \mathbf{x}' = A\mathbf{x} x ′ = A x A A A 2 × 2 2 \times 2 2 × 2
Eigenvalues Phase Portrait Stability λ 1 , λ 2 < 0 \lambda_1, \lambda_2 \lt 0 λ 1 , λ 2 < 0 Stable node Asymptotically stable λ 1 , λ 2 > 0 \lambda_1, \lambda_2 > 0 λ 1 , λ 2 > 0 Unstable node Unstable λ 1 < 0 < λ 2 \lambda_1 \lt 0 \lt \lambda_2 λ 1 < 0 < λ 2 Saddle point Unstable λ = α ± i β \lambda = \alpha \pm i\beta λ = α ± i β α < 0 \alpha \lt 0 α < 0 Stable spiral Asymptotically stable λ = α ± i β \lambda = \alpha \pm i\beta λ = α ± i β α > 0 \alpha > 0 α > 0 Unstable spiral Unstable λ = ± i β \lambda = \pm i\beta λ = ± i β Center (Marginally) stable
Remark. The trace-determinant plane provides a convenient classification. Let τ = t r ( A ) \tau = \mathrm{tr}(A) τ = tr ( A ) Δ = det ( A ) \Delta = \det(A) Δ = det ( A ) λ 2 − τ λ + Δ = 0 \lambda^2 - \tau\lambda + \Delta = 0 λ 2 − τ λ + Δ = 0
λ = τ ± τ 2 − 4 Δ 2 \lambda = \frac{\tau \pm \sqrt{\tau^2 - 4\Delta}}{2} λ = 2 τ ± τ 2 − 4Δ
τ 2 − 4 Δ > 0 \tau^2 - 4\Delta > 0 τ 2 − 4Δ > 0 τ 2 − 4 Δ < 0 \tau^2 - 4\Delta \lt 0 τ 2 − 4Δ < 0 τ 2 − 4 Δ = 0 \tau^2 - 4\Delta = 0 τ 2 − 4Δ = 0 Stability is determined by the sign of τ \tau τ τ < 0 \tau \lt 0 τ < 0 τ > 0 \tau > 0 τ > 0
Trace-Determinant Plane: Stability Classification
The trace-determinant plane classifies 2D linear systems. The parabola τ 2 = 4 Δ \tau^2 = 4\Delta τ 2 = 4Δ τ = 0 \tau = 0 τ = 0 x-axis represents the trace τ \tau τ y-axis represents Δ \Delta Δ
For x ′ = A x + f ( t ) \mathbf{x}' = A\mathbf{x} + \mathbf{f}(t) x ′ = A x + f ( t ) Φ ( t ) \Phi(t) Φ ( t )
x ( t ) = Φ ( t ) c + Φ ( t ) ∫ Φ − 1 ( s ) f ( s ) d s \mathbf{x}(t) = \Phi(t)\mathbf{c} + \Phi(t)\int \Phi^{-1}(s)\mathbf{f}(s)\, ds x ( t ) = Φ ( t ) c + Φ ( t ) ∫ Φ − 1 ( s ) f ( s ) d s
Worked Example. Solve x ′ = ( 1 0 0 2 ) x + ( e t 0 ) \mathbf{x}' = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}\mathbf{x} + \begin{pmatrix} e^t \\ 0 \end{pmatrix} x ′ = ( 1 0 0 2 ) x + ( e t 0 )
Solution Solution. Eigenvalues: 1 1 1 2 2 2 Φ ( t ) = ( e t 0 0 e 2 t ) \Phi(t) = \begin{pmatrix} e^t & 0 \\ 0 & e^{2t} \end{pmatrix} Φ ( t ) = ( e t 0 0 e 2 t )
Φ − 1 ( s ) = ( e − s 0 0 e − 2 s ) \Phi^{-1}(s) = \begin{pmatrix} e^{-s} & 0 \\ 0 & e^{-2s} \end{pmatrix} Φ − 1 ( s ) = ( e − s 0 0 e − 2 s )
Φ − 1 ( s ) f ( s ) = ( e − s 0 0 e − 2 s ) ( e s 0 ) = ( 1 0 ) \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} Φ − 1 ( s ) f ( s ) = ( e − s 0 0 e − 2 s ) ( e s 0 ) = ( 1 0 )
∫ Φ − 1 ( s ) f ( s ) d s = ( t 0 ) \int \Phi^{-1}(s)\mathbf{f}(s)\, ds = \begin{pmatrix} t \\ 0 \end{pmatrix} ∫ Φ − 1 ( s ) f ( s ) d s = ( t 0 )
x p = Φ ( t ) ( t 0 ) = ( t e t 0 ) \mathbf{x}_p = \Phi(t)\begin{pmatrix} t \\ 0 \end{pmatrix} = \begin{pmatrix} te^t \\ 0 \end{pmatrix} x p = Φ ( t ) ( t 0 ) = ( t e t 0 )
x ( t ) = c 1 ( e t 0 ) + c 2 ( 0 e 2 t ) + ( t e t 0 ) \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} x ( t ) = c 1 ( e t 0 ) + c 2 ( 0 e 2 t ) + ( t e t 0 ) ■ \blacksquare ■