Systems of Equations

Let $I$ be an interval, $A:I\rightarrow\mathbb{R}^{\,n\times n}$ be a matrix-valued function with continuous entries and $\underline{g}:I\rightarrow\mathbb{R}^n$ be continuous. A linear system of first-order equations and its associated homogeneous system are

$$\underline{y}'=A(t)\,\underline{y}+\underline{g}(t),\qquad\underline{y}'=A(t)\,\underline{y},$$

where the independent variable is written $t$ and derivatives of vector- and matrix-valued functions are taken entrywise. Every initial value problem $\underline{y}\left(t_0\right)=\underline{y}_0$ has exactly one solution on all of $I.$ The superposition principle holds: the solutions of the homogeneous system form a vector space, which has dimension $n$ since $\underline{y}\mapsto\underline{y}\left(t_0\right)$ is an isomorphism onto $\mathbb{R}^n,$ and the general solution of the inhomogeneous system is the sum of a particular solution and the general solution of the homogeneous system. A linear equation of order $n$ is the special case in which $A$ is the companion matrix with ones on its superdiagonal, $-a_0,\ldots,-a_{n-1}$ in its last row and zeros elsewhere.

A fundamental matrix of the homogeneous system is a matrix-valued function $\Phi:I\rightarrow\mathbb{R}^{\,n\times n}$ whose columns are a basis of the solution space, so that $\Phi'=A\,\Phi$ and $\Phi(t)$ is invertible for every $t.$ The Wronskian $W=\det\Phi$ satisfies Liouville’s formula,

$$W(t)=W\left(t_0\right)\exp\left(\int_{t_0}^{t}\mathrm{tr}\,A(s)\,ds\right),$$

so $n$ solutions form a basis if and only if their Wronskian is non-zero at one, and hence every, point. The general solution of the homogeneous system is $\Phi(t)\,\underline{c}$ for $\underline{c}\in\mathbb{R}^n,$ the solution of the initial value problem is $\Phi(t)\,\Phi\left(t_0\right)^{-1}\underline{y}_0$ and variation of parameters gives the solution of the inhomogeneous initial value problem.

$$\underline{y}(t)=\Phi(t)\,\Phi\left(t_0\right)^{-1}\underline{y}_0+\Phi(t)\int_{t_0}^{t}\Phi(s)^{-1}\underline{g}(s)\,ds$$

Let $A\in\mathbb{R}^{\,n\times n}$ be a constant matrix. The matrix exponential of $A$ is the sum of the exponential series with $A$ in place of the scalar.

$$e^{A}:=\sum_{k=0}^{\infty}\frac{A^k}{k!}$$

The series converges absolutely in the Frobenius norm, since $\lVert A^k\rVert_F\leq\lVert A\rVert_F^k$ and $\sum_{k=0}^{\infty}\frac{\lVert A\rVert_F^k}{k!}=e^{\lVert A\rVert_F},$ and $\mathbb{R}^{\,n\times n}$ with this norm is complete. For $A,B\in\mathbb{R}^{\,n\times n},$ invertible $P$ and $s,t\in\mathbb{R}$ the matrix exponential has the following properties:

$$e^{0}=I_n,\qquad e^{\left(s+t\right)A}=e^{sA}e^{tA},\qquad\left(e^{tA}\right)^{-1}=e^{-tA},\qquad\frac{d}{dt}e^{tA}=A\,e^{tA}=e^{tA}A$$ $$e^{A+B}=e^{A}e^{B}\,\,\mathrm{if}\,\,AB=BA,\qquad e^{PAP^{-1}}=P\,e^{A}P^{-1},\qquad\det e^{A}=e^{\mathrm{tr}\,A}.$$

The exponential law can fail for non-commuting matrices. Since $\frac{d}{dt}e^{tA}=A\,e^{tA},$ the matrix $e^{tA}$ is a fundamental matrix of the system with constant coefficients and the solution of its initial value problem is

$$\underline{y}'=A\,\underline{y},\qquad\underline{y}(t)=e^{\left(t-t_0\right)A}\,\underline{y}_0.$$

If $A$ is diagonalisable as $A=P\Lambda P^{-1}$ then $e^{tA}=P\,e^{t\Lambda}P^{-1}$ where $e^{t\Lambda}$ is diagonal with entries $e^{\lambda_it}.$ Equivalently, if $\underline{v}$ is an eigenvector of $A$ with eigenvalue $\lambda$ then $\underline{y}=e^{\lambda t}\underline{v}$ is a solution, and an eigen-basis gives a fundamental set of solutions. A pair of complex conjugate eigenvalues $\alpha\pm i\beta$ with eigenvectors $\underline{u}\pm i\underline{w}$ gives, by Euler’s formula, the real solutions

$$e^{\alpha t}\left(\underline{u}\cos\beta t-\underline{w}\sin\beta t\right),\qquad e^{\alpha t}\left(\underline{u}\sin\beta t+\underline{w}\cos\beta t\right).$$

If $A$ is defective then the eigenvectors do not span $\mathbb{R}^n.$ A generalised eigenvector of rank $k$ for the eigenvalue $\lambda$ is a vector $\underline{w}$ with $\left(A-\lambda I_n\right)^k\underline{w}=\underline{0}$ and $\left(A-\lambda I_n\right)^{k-1}\underline{w}\neq\underline{0}.$ Since $\lambda I_n$ and $A-\lambda I_n$ commute, $e^{tA}\underline{w}=e^{\lambda t}e^{t\left(A-\lambda I_n\right)}\underline{w}$ and the second series terminates, which gives the solution

$$\underline{y}(t)=e^{\lambda t}\sum_{j=0}^{k-1}\frac{t^j}{j!}\left(A-\lambda I_n\right)^j\underline{w}.$$

Every matrix has a basis of generalised eigenvectors, so these solutions, with polynomial factors of degree less than the algebraic multiplicity, always give a fundamental set. The companion matrix of $y''-2y'+y=0$ has the repeated eigenvalue $1$ with a single eigenvector, and the generalised eigenvector recovers the solution $t\,e^t.$

Let $U\subseteq\mathbb{R}^n$ be open and $\underline{F}:U\rightarrow\mathbb{R}^n$ be $C^1.$ An autonomous system is $\underline{y}'=\underline{F}\left(\underline{y}\right),$ whose right-hand side is a vector field on the phase space $U.$ The image of a solution in $U$ is a trajectory or orbit, which is tangent to the vector field at every point, and the phase portrait of the system is the collection of its trajectories. By uniqueness, distinct trajectories do not intersect and a trajectory that returns to a point is a closed curve corresponding to a periodic solution. A zero $\underline{y}^*$ of $\underline{F}$ is an equilibrium point, whose trajectory is the single point.

Let $A\in\mathbb{R}^{\,2\times 2}$ be invertible with eigenvalues $\lambda_1,\lambda_2,$ so that $\underline{0}$ is the only equilibrium point of $\underline{y}'=A\,\underline{y}.$ Its phase portrait is classified by the eigenvalues, which are determined by the trace $\tau=\lambda_1+\lambda_2$ and determinant $\Delta=\lambda_1\lambda_2$ through $\lambda^2-\tau\lambda+\Delta=0:$

• if $\Delta<0$ the eigenvalues are real with opposite signs and the equilibrium point is a saddle: trajectories approach along the eigenvector of the negative eigenvalue and leave along the eigenvector of the positive one;
• if $\Delta>0$ and $\tau^2\geq 4\Delta$ the eigenvalues are real with the same sign and the equilibrium point is a node, towards which every trajectory moves if $\tau<0$ and away from which every trajectory moves if $\tau>0;$
• if $\Delta>0$ and $\tau^2<4\Delta$ the eigenvalues are complex with real part $\frac{\tau}{2}$ and the equilibrium point is a spiral, inwards if $\tau<0$ and outwards if $\tau>0,$ or a centre surrounded by closed trajectories if $\tau=0.$

Stability of an equilibrium point $\underline{y}^*$ of an autonomous system is defined as for a single equation with the norm in place of the absolute value: $\underline{y}^*$ is stable if for every $\varepsilon>0$ there exists $\delta>0$ such that every solution with $\left\lVert\underline{y}\left(t_0\right)-\underline{y}^*\right\rVert<\delta$ exists for all $t\geq t_0$ and satisfies $\left\lVert\underline{y}(t)-\underline{y}^*\right\rVert<\varepsilon,$ asymptotically stable if it is stable and every solution that starts sufficiently close to it converges to it as $t\rightarrow\infty,$ and unstable if it is not stable. For the linear system $\underline{y}'=A\,\underline{y}$ the origin is asymptotically stable if and only if every eigenvalue of $A$ has negative real part, in which case $\left\lVert e^{tA}\right\rVert_F\rightarrow 0$ exponentially, and unstable if some eigenvalue has positive real part; a centre is stable but not asymptotically stable.

The linearisation of the autonomous system at an equilibrium point $\underline{y}^*$ is the linear system $\underline{u}'=J_{\underline{F}}\left(\underline{y}^*\right)\underline{u}$ for the deviation $\underline{u}=\underline{y}-\underline{y}^*,$ obtained by discarding the remainder of the first-order Taylor expansion of $\underline{F}.$ The equilibrium point is a hyperbolic equilibrium point if no eigenvalue of $J_{\underline{F}}\left(\underline{y}^*\right)$ has zero real part. The linearisation theorem states that a hyperbolic equilibrium point is asymptotically stable if every eigenvalue of the Jacobian matrix has negative real part and unstable if some eigenvalue has positive real part, and the Hartman–Grobman theorem states that near a hyperbolic equilibrium point the phase portrait is the image of the phase portrait of the linearisation under a homeomorphism. Linearisation is inconclusive at non-hyperbolic equilibrium points: the pendulum equation $\theta''+\sin\theta=0,$ written as the system $\theta'=v,$ $v'=-\sin\theta,$ has a centre of the linearisation at $\left(0,0\right),$ which is stable since the energy $\frac{1}{2}v^2-\cos\theta$ is constant along trajectories, and a saddle at $\left(\pi,0\right).$

Let $V:U\rightarrow\mathbb{R}$ be $C^1$ with $V\left(\underline{y}^*\right)=0$ and $V>0$ elsewhere near $\underline{y}^*.$ Along a solution the chain rule gives $\frac{d}{dt}V\left(\underline{y}(t)\right)=\nabla V\left(\underline{y}\right)\cdot\underline{F}\left(\underline{y}\right).$ $V$ is a Lyapunov function for $\underline{y}^*$ if $\nabla V\cdot\underline{F}\leq 0$ near $\underline{y}^*,$ and Lyapunov’s stability theorem states that an equilibrium point with a Lyapunov function is stable, and asymptotically stable if $\nabla V\cdot\underline{F}<0$ away from $\underline{y}^*.$ The energy of the pendulum, shifted by its minimum, is a Lyapunov function at $\left(0,0\right).$