[ALGEBRA] Linear ODE systems

This page solves coupled first-order equations $\mathbf{x}'=A\mathbf{x}$ by the continuous analogue of matrix powers. An eigenvector gives a direction whose time dependence is a single exponential and its eigenvalue is the rate, read for growth or decay by stability of systems.

The page links the two strands. The Calculus ODE pages develop scalar methods, but the exponential modes needed here are derived below, so the page can be read first and compared with the scalar methods later.

First-order systems

With a single equation, the system is a first-order linear equation, which that page solves with an integrating factor.

[DEFINITION]

First-order linear equation

A first-order linear ODE in normal form is

y'+p(x)y=q(x).

[1]:Integrating factor

[DEFINITION]

First-order linear system

A first-order linear system with constant coefficient matrix $A$ has the form

\mathbf{x}'(t)=A\mathbf{x}(t),

where $\mathbf{x}(t)=(x_1(t),\ldots,x_n(t))^{\mathsf T}$ and $A$ is an $n\times n$ matrix. Specifying $\mathbf{x}(0)=\mathbf{x}_0$ makes this an initial-value problem.

Writing the equations as one matrix equation records all couplings in $A$ . For example,

\begin{aligned} x_1'&=x_1+3x_2,\\ x_2'&=3x_1+x_2 \end{aligned} \qquad\Longleftrightarrow\qquad \mathbf{x}'= \begin{pmatrix} 1&3\\ 3&1 \end{pmatrix}\mathbf{x}.

Eigenvectors as exponential modes

An eigenvector is a direction that $A$ does not mix with other directions, and its eigenvalue is the rate attached to that direction. An eigenbasis writes every initial vector as a combination of such modes.

[DEFINITION]

Diagonalisation

An $n\times n$ matrix $A$ over a field $F$ is diagonalisable over $F$ if there are an invertible matrix $P\in M_n(F)$ and a diagonal matrix $D\in M_n(F)$ such that

P^{-1}AP=D, \qquad\text{equivalently}\qquad A=PDP^{-1}.

The columns of $P$ are the eigenvectors in the chosen eigenbasis and the corresponding eigenvalues appear on the diagonal of $D$ .

[DEFINITION]

Eigenvalue

Let $T:V\to V$ be a linear map over a field $F$ . A scalar $\lambda\in F$ is an eigenvalue of $T$ if there is a non-zero vector $v\in V$ such that

T(v)=\lambda v.

[1]:Linear map
[2]:Eigenvectors for distinct eigenvalues
[3]:A real matrix with complex eigenvalues
[4]:Invertibility and eigenvalues
[5]:Diagonalisation
15 more

[THEOREM]

Exponential modes

For a constant matrix $A$ and a non-zero vector $\mathbf u$ , the function

\mathbf{x}(t)=\mathbf{u}e^{\lambda t}

is a solution of $\mathbf{x}'=A\mathbf{x}$ if and only if $\lambda$ is an eigenvalue of $A$ and $\mathbf{u}$ is a corresponding eigenvector. The zero solution is handled separately by taking $\mathbf u=\mathbf0$ .

When $A$ is real, the calculation may take place in $\mathbb C^n$ to use complex eigenvalues and eigenvectors. Combining a complex solution with its conjugate then gives real solutions, as in the oscillatory example below.

[PROOF]

Differentiating the proposed function gives

\mathbf{x}'(t)=\lambda\mathbf{u}e^{\lambda t}.

The equation $\mathbf{x}'=A\mathbf{x}$ therefore becomes

\lambda\mathbf{u}e^{\lambda t} =A\mathbf{u}e^{\lambda t}.

Since $e^{\lambda t}\neq0$ , this is equivalent to $A\mathbf{u}=\lambda\mathbf{u}$ . The reverse implication is the same calculation read backwards.

[1]:Exponential modes

Superposing independent modes

[PROPOSITION]

Linear combinations of solutions

If $\mathbf{x}_1(t)$ and $\mathbf{x}_2(t)$ solve $\mathbf{x}'=A\mathbf{x}$ , then every linear combination

\alpha\mathbf{x}_1(t)+\beta\mathbf{x}_2(t)

also solves the system.

Differentiation and multiplication by $A$ are both linear:

\frac{d}{dt}(\alpha\mathbf{x}_1+\beta\mathbf{x}_2) =\alpha A\mathbf{x}_1+\beta A\mathbf{x}_2 =A(\alpha\mathbf{x}_1+\beta\mathbf{x}_2).

[1]:First-order linear system

[METHOD]

Solve a system by eigenmodes

When $A$ has an eigenbasis $v_1,\ldots,v_n$ with eigenvalues $\lambda_1,\ldots,\lambda_n$ :

find the eigenvalues and corresponding eigenvectors;
write the initial vector as $\mathbf{x}_0=c_1v_1+\cdots+c_nv_n$ ;
replace each eigenmode $v_j$ by $e^{\lambda_jt}v_j$ ;
combine the modes: $\mathbf{x}(t) =c_1e^{\lambda_1t}v_1+\cdots+c_ne^{\lambda_nt}v_n.$

This is the matrix-power calculation with the discrete factor $\lambda_j^n$ replaced by the continuous factor $e^{\lambda_jt}$ . A matrix without an eigenbasis instead produces terms such as $te^{\lambda t}$ , as in the repeated-root case for scalar equations.

[1]:First-order linear system
[2]:Exponential modes
[3]:Diagonalisation
[4]:A system with a limiting state
[5]:Complex eigenvalues and a real solution
[6]:A two-variable economic model

[EXAMPLE]

A system with a limiting state

Consider

\mathbf{x}'= \begin{pmatrix} -3&3\\ 3&-3 \end{pmatrix}\mathbf{x}, \qquad \mathbf{x}(0)= \begin{pmatrix}4\\0\end{pmatrix}.

The eigenvalues are $0$ and $-6$ , with eigenvectors

v_1=\begin{pmatrix}1\\1\end{pmatrix}, \qquad v_2=\begin{pmatrix}1\\-1\end{pmatrix}.

Since

\begin{pmatrix}4\\0\end{pmatrix} =2\begin{pmatrix}1\\1\end{pmatrix} +2\begin{pmatrix}1\\-1\end{pmatrix},

the solution is

\mathbf{x}(t) =2\begin{pmatrix}1\\1\end{pmatrix} +2e^{-6t}\begin{pmatrix}1\\-1\end{pmatrix} = \begin{pmatrix} 2+2e^{-6t}\\ 2-2e^{-6t} \end{pmatrix}.

The $e^{-6t}$ mode decays to zero, so $\mathbf{x}(t)$ tends to the equilibrium $(2,2)^{\mathsf T}$ . The solution does not tend to zero, because the $\lambda=0$ mode stays constant.

[routine]

[1]:Solve a system by eigenmodes

[EXAMPLE]

Complex eigenvalues and a real solution

For

\mathbf{x}'= \begin{pmatrix} 1&-2\\ 2&1 \end{pmatrix}\mathbf{x}, \qquad \mathbf{x}(0)= \begin{pmatrix}1\\0\end{pmatrix},

the eigenvalues are $1+2i$ and $1-2i$ . Corresponding eigenvectors may be chosen as

v_+=\begin{pmatrix}i\\1\end{pmatrix}, \qquad v_-=\begin{pmatrix}-i\\1\end{pmatrix}.

At $t=0$ , the coefficients must satisfy

c_+\begin{pmatrix}i\\1\end{pmatrix} +c_-\begin{pmatrix}-i\\1\end{pmatrix} =\begin{pmatrix}1\\0\end{pmatrix}.

Solving these two scalar equations gives $c_+=-i/2$ and $c_-=i/2$ . Therefore

\mathbf{x}(t) = -\frac{i}{2}e^{(1+2i)t}\begin{pmatrix}i\\1\end{pmatrix} +\frac{i}{2}e^{(1-2i)t}\begin{pmatrix}-i\\1\end{pmatrix}.

Using Euler's formula $e^{i\theta}=\cos\theta+i\sin\theta$ with $\theta=2t$ and simplifying yields the real solution

\mathbf{x}(t) =e^t \begin{pmatrix} \cos(2t)\\ \sin(2t) \end{pmatrix}.

Direct differentiation checks that $\mathbf{x}'=A\mathbf{x}$ and $\mathbf{x}(0)=(1,0)^{\mathsf T}$ .

The factor $e^t$ makes the amplitude grow, while the sine and cosine rotate the state with angular frequency $2$ . The trajectory therefore spirals outward from the origin: the imaginary parts of the eigenvalues produce the oscillation and their positive real part produces the instability.

[exam]

[1]:Solve a system by eigenmodes

Higher-order equations as systems

Constant-coefficient scalar equations are solved with a characteristic equation and its roots. Each such equation can also be rewritten as a first-order system.

[DEFINITION]

Constant-coefficient equation

A second-order constant-coefficient linear ODE has the form

ay''+by'+cy=f(x), \qquad a\neq0,

where $a$ , $b$ and $c$ are constants. It is homogeneous when $f(x)\equiv0$ .

[DEFINITION]

Characteristic equation

For the homogeneous equation

ay''+by'+cy=0,

the exponential trial function $y=e^{rx}$ gives the characteristic equation

ar^2+br+c=0.

[METHOD]

Read the characteristic roots

Solve the characteristic equation and choose the real homogeneous modes according to the roots:

If $r_1\neq r_2$ are distinct real roots, use $y_h=Ae^{r_1x}+Be^{r_2x}.$
If $r$ is a repeated real root, use $y_h=(A+Bx)e^{rx}.$
If the roots are $r=\alpha\pm i\beta$ with $\beta\neq0$ , use $y_h=e^{\alpha x} \bigl(A\cos(\beta x)+B\sin(\beta x)\bigr).$

[1]:Characteristic equation
[2]:Two distinct real roots
[3]:A repeated real root
[4]:Initial data for distinct modes
[5]:First- and second-order price models

[EXAMPLE]

Converting a higher-order equation

The third-order equation

y'''+ay''+by'+cy=0

becomes a first-order system after setting

x_1=y,\qquad x_2=y',\qquad x_3=y''.

Then

\frac{d}{dt} \begin{pmatrix} x_1\\x_2\\x_3 \end{pmatrix} = \begin{pmatrix} 0&1&0\\ 0&0&1\\ -c&-b&-a \end{pmatrix} \begin{pmatrix} x_1\\x_2\\x_3 \end{pmatrix}.

The first two rows record $x_1'=x_2$ and $x_2'=x_3$ , while the third row is the original equation solved for $y'''$ .

[routine]

[1]:First-order linear system

A coupled economic model

[EXAMPLE]

A two-variable economic model

Let $y_1$ be the inflation level and $y_2$ the interest-rate level. Consider the model

\frac{dy_1}{dt}=\frac12y_1-\frac15y_2, \qquad \frac{dy_2}{dt}=\frac35y_1-\frac25y_2,

with

\mathbf{y}(0)= \begin{pmatrix}4/100\\6/100\end{pmatrix}, \qquad A= \begin{pmatrix} 1/2&-1/5\\ 3/5&-2/5 \end{pmatrix}.

The characteristic polynomial is

\det(A-\lambda I) =\lambda^2-\frac1{10}\lambda-\frac2{25},

\lambda_1=\frac{1+\sqrt{33}}{20}, \qquad \lambda_2=\frac{1-\sqrt{33}}{20}.

Eigenvectors with second component $1$ are

v_1= \begin{pmatrix} (9+\sqrt{33})/12\\1 \end{pmatrix}, \qquad v_2= \begin{pmatrix} (9-\sqrt{33})/12\\1 \end{pmatrix}.

Writing the initial vector as $\mathbf{y}(0)=c_1v_1+c_2v_2$ gives

c_1=\frac{33-\sqrt{33}}{1100}, \qquad c_2=\frac{33+\sqrt{33}}{1100}.

Hence

\mathbf{y}(t) =\frac{33-\sqrt{33}}{1100} e^{(1+\sqrt{33})t/20}v_1 +\frac{33+\sqrt{33}}{1100} e^{(1-\sqrt{33})t/20}v_2.

The first exponent is positive and the second is negative, so one eigenmode grows and the other decays.

The growing mode is the model's unstable direction: any non-zero component of the initial state along $v_1$ eventually dominates. The decaying mode along $v_2$ is a short-run adjustment.

[exam]

[THEOREM]

Continuous stability from real parts

For a characteristic root $r=\alpha+i\beta$ ,

|e^{rt}|=e^{\alpha t}.

Therefore $\alpha<0$ gives decay, $\alpha>0$ gives an unbounded mode and, for a simple root, $\alpha=0$ gives a non-decaying oscillatory or constant mode. A repeated root on the imaginary axis introduces a term $t e^{rt}$ , whose magnitude is unbounded.

For a mode $e^{\lambda t}$ , $\operatorname{Re}(\lambda)$ controls growth or decay, as in the continuous stability criterion for scalar characteristic roots. The time factor is the same exponential that appears in second-order constant-coefficient equations.

Linear ODE systems

First-order linear equation

Diagonalisation

Eigenvalue

Constant-coefficient equation

Characteristic equation

Read the characteristic roots

Continuous stability from real parts

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