Linear systems

The linear system $\dot{\mathbf{x}} = A\mathbf{x}$ is solved completely by one formula, $\mathbf{x}(t) = e^{tA}\mathbf{x}_0$.

1. Linear systems and the matrix exponential (Def. 1.1, Prop. 1.2); $e^{tA}\mathbf{x}_0$ is the unique solution (Thm 1.3).
2. Fundamental matrices and the Wronskian (Def. 1.4, Prop. 1.5).
3. Computing $e^{tA}$: diagonalisable, complex and repeated eigenvalues (Prop. 1.6, Example 1.7).
4. Inhomogeneous systems by variation of parameters (Thm 1.8).
5. Higher-order linear ODEs as systems (Prop. 1.9).

A linear system is $\dot{\mathbf{x}} = A(t)\mathbf{x} + \mathbf{g}(t)$; it is homogeneous if $\mathbf{g} = 0$ and has constant coefficients if $A$ is constant. For a constant matrix $A$, the matrix exponential is

which converges absolutely for all $t$, since $\big\|\frac{t^kA^k}{k!}\big\|\le\frac{(|t|\|A\|)^k}{k!}$ in the operator norm.

1. $\frac{\mathrm{d}}{\mathrm{d}t}e^{tA} = Ae^{tA} = e^{tA}A$.
2. If $AB = BA$ then $e^{A + B} = e^Ae^B$. In particular $e^{(s+t)A} = e^{sA}e^{tA}$ and $(e^{tA})^{-1} = e^{-tA}$.
3. $e^{tPAP^{-1}} = Pe^{tA}P^{-1}$.
4. $\det e^{tA} = e^{t\operatorname{tr}A}$.

For constant $A$, the initial value problem $\dot{\mathbf{x}} = A\mathbf{x}$, $\mathbf{x}(0) = \mathbf{x}_0$ has the unique solution $\mathbf{x}(t) = e^{tA}\mathbf{x}_0$ for all $t\in\mathbb{R}$. The solutions form an $n$-dimensional vector space.

For $\dot{\mathbf{x}} = A(t)\mathbf{x}$, a fundamental matrix $\Phi(t)$ is an $n\times n$ matrix whose columns are $n$ linearly independent solutions; then every solution is $\Phi(t)\mathbf{c}$. Its determinant $W(t) = \det\Phi(t)$ is the Wronskian. For constant $A$, $e^{tA}$ is the fundamental matrix with $\Phi(0) = I$, and any fundamental matrix satisfies $e^{tA} = \Phi(t)\Phi(0)^{-1}$.

If $\Phi$ solves $\dot\Phi = A(t)\Phi$ then $W(t) = W(t_0)\exp\big(\int_{t_0}^t\operatorname{tr}A(s)\,\mathrm{d}s\big)$. In particular $W$ is either never zero or always zero, so solutions independent at one time are independent at all times.

• If $A = PDP^{-1}$ with $D = \operatorname{diag}(\lambda_i)$, then $e^{tA} = P\operatorname{diag}(e^{\lambda_it})P^{-1}$.
• For a real $2\times2$ matrix with eigenvalues $\alpha\pm i\beta$ ($\beta\ne0$), $A$ is similar to $\begin{pmatrix}\alpha & -\beta\\\beta & \alpha\end{pmatrix}$, whose exponential is $e^{\alpha t}\begin{pmatrix}\cos\beta t & -\sin\beta t\\\sin\beta t & \cos\beta t\end{pmatrix}$.
• In general, with Jordan form $A = PJP^{-1}$, $e^{tA} = Pe^{tJ}P^{-1}$ and each block gives $e^{\lambda t}\big(I + tN + \cdots + \frac{t^{m-1}}{(m-1)!}N^{m-1}\big)$; so solutions are combinations of $t^je^{\lambda t}$.

$A = \begin{pmatrix}1 & -1\\1 & 3\end{pmatrix}$ has $\chi_A = (x - 2)^2$ and $N = A - 2I = \begin{pmatrix}-1 & -1\\1 & 1\end{pmatrix}\ne0$, $N^2 = 0$. Since $2I$ and $N$ commute,

The solution with $\mathbf{x}(0) = (1, 0)$ is $e^{2t}(1 - t, t)$. No change of basis was needed: $A = \lambda I + N$ with $N$ nilpotent works whenever $A$ has a single eigenvalue.

For continuous $\mathbf{g}$, the solution of $\dot{\mathbf{x}} = A\mathbf{x} + \mathbf{g}(t)$, $\mathbf{x}(0) = \mathbf{x}_0$ is

More generally, with a fundamental matrix $\Phi$ of $\dot{\mathbf{x}} = A(t)\mathbf{x}$, a particular solution is $\Phi(t)\int\Phi(s)^{-1}\mathbf{g}(s)\,\mathrm{d}s$.

The equation $y^{(n)} + a_{n-1}y^{(n-1)} + \dots + a_0y = 0$ is equivalent to $\dot{\mathbf{x}} = C\mathbf{x}$ with $\mathbf{x} = (y, y', \dots, y^{(n-1)})$ and $C$ the companion matrix (ones on the superdiagonal, last row $-a_0, \dots, -a_{n-1}$). The characteristic polynomial of $C$ is the auxiliary polynomial $\lambda^n + a_{n-1}\lambda^{n-1} + \dots + a_0$, and each eigenvalue of $C$ has a single Jordan block.

• Find eigenvalues; decide diagonalisable, complex pair, or repeated (Jordan).
• Verify $e^{0A} = I$ and $\frac{\mathrm{d}}{\mathrm{d}t}e^{tA}|_{t=0} = A$.
• Check $\det e^{tA} = e^{t\operatorname{tr}A}$.
• For forcing, try undetermined coefficients first; Duhamel always works.