Chapter 7 · Section 7.6

7.6 The Matrix Method: Two Equations

In Section 7.4 we solved homogeneous $2 \times 2$ systems by substituting $x = A e^{λ t}$ , $y = B e^{λ t}$ and working out the algebra each time. Now we redo the same thing in the language of Section 7.5. Nothing new is assumed, but the matrix notation does three things for us: it turns the algebra into a standard eigenvalue problem, it makes the three cases (distinct real, complex, repeated eigenvalues) look clean, and it prepares the ground for $n$ equations in Section 7.7.

Throughout this section we study

x^{'} = A x, A = (a_{11} a_{21} a_{12} a_{22}), x (t) = (x_{1} (t) x_{2} (t)),

with real constants $a_{ij}$ . In the notation of Section 7.4, $x_{1} = x$ and $x_{2} = y$ . In the second half of the section we look at the solutions as curves in the plane and classify the possible pictures.

Exponential solutions and the eigenvalue problem

Look for a solution of the form $x = e^{λ t} v$ , where $λ$ is a number and $v \neq = 0$ is a constant vector. Then

x^{'} = λ e^{λ t} v, A x = e^{λ t} A v .

These are equal for all $t$ exactly when $A v = λ v$ . So we have proved:

Theorem 7.6.1 (Eigenvector solutions)

If $λ$ is an eigenvalue of $A$ and $v$ a corresponding eigenvector, then $x (t) = e^{λ t} v$ is a solution of $x^{'} = A x$ on $(- \infty, \infty)$ .

To build the general solution we need two solutions that are linearly independent. Section 7.3 gave the test for $2 \times 2$ systems, and Section 7.8 proves it for any size: two solutions $x_{1}, x_{2}$ of $x^{'} = A x$ are independent if and only if their Wronskian

W (t) = det [x_{1} (t) x_{2} (t)]

is nonzero, and it is enough to check this at one single point, say $t = 0$ . Then every solution can be written as $x = c_{1} x_{1} + c_{2} x_{2}$ . The eigenvalues are the roots of

λ^{2} - (tr A) λ + det A = 0,

and, as for the characteristic equation in Chapter 4, there are three cases.

Case 1: two distinct real eigenvalues

Theorem 7.6.2 (Distinct real eigenvalues)

If $A$ has real eigenvalues $λ_{1} \neq = λ_{2}$ with eigenvectors $v_{1}$ and $v_{2}$ , the general solution of $x^{'} = A x$ is

x (t) = c_{1} e^{λ_{1} t} v_{1} + c_{2} e^{λ_{2} t} v_{2} .

Why. Both terms are solutions by Theorem 7.6.1. At $t = 0$ the Wronskian is $det [v_{1} v_{2}]$ , which is nonzero because eigenvectors of distinct eigenvalues are independent (Theorem 7.5.4).

Example 7.6.1

Solve $x_{1}^{'} = x_{1} + 2 x_{2}$ , $x_{2}^{'} = 3 x_{1}$ , with $x_{1} (0) = 3$ , $x_{2} (0) = - 2$ .

Solution

Matrix form. $x^{'} = A x$ with $A = (1320)$ . (The second equation has no $x_{2}$ term, so $a_{22} = 0$ .)
Eigenvalues. $tr A = 1$ , $det A = - 6$ : $λ^{2} - λ - 6 = (λ - 3) (λ + 2) = 0, λ_{1} = 3, λ_{2} = - 2.$
Eigenvectors. $A - 3 I = (- 2 3 2 - 3)$ gives $v_{1} = v_{2}$ , so $v_{1} = (1, 1)^{T}$ . $A + 2 I = (3322)$ gives $3 v_{1} + 2 v_{2} = 0$ , so $v_{2} = (2, - 3)^{T}$ .
General solution. $x (t) = c_{1} e^{3 t} (11) + c_{2} e^{- 2 t} (2 - 3) .$
Initial condition. At $t = 0$ : $c_{1} (1, 1)^{T} + c_{2} (2, - 3)^{T} = (3, - 2)^{T}$ , i.e. $c_{1} + 2 c_{2} c_{1} - 3 c_{2} = 3, = - 2.$ Subtracting, $5 c_{2} = 5$ , so $c_{2} = 1$ and $c_{1} = 1$ .
Answer. $x_{1} = e^{3 t} + 2 e^{- 2 t}, x_{2} = e^{3 t} - 3 e^{- 2 t} .$ For large $t$ the term $e^{3 t}$ dominates and the solution grows along the direction $(1, 1)^{T}$ (Figure 7.6.1).

Figure 7.6.1. Solutions of Example 7.6.1 drawn in the

x_{1} x_{2}

-plane. Along the line of

v_{2} = (2, - 3)^{T}

(

λ = - 2

) points move toward the origin; along the line of

v_{1} = (1, 1)^{T}

(

λ = 3

) they move away. The highlighted curve is the solution with

x (0) = (3, - 2)^{T}

: it comes in near the first line and leaves along the second.

Our second example has an engineering origin. Two connected compartments (two tanks, two rooms, or two organs exchanging a drug) are often modeled by $x^{'} = A x$ where $x_{j}$ is the amount in compartment $j$ . The off-diagonal entry $a_{ij}$ ( $i \neq = j$ ) is the rate at which material moves from compartment $j$ into compartment $i$ ; the diagonal entry $a_{jj}$ is minus the total rate at which material leaves compartment $j$ .

Example 7.6.2 (Two compartments)

Compartment 1 loses material at the total rate $4 x_{1}$ : the part $2 x_{1}$ flows into compartment 2 and the rest leaves the system. Compartment 2 loses material at the rate $3 x_{2}$ : the part $x_{2}$ flows into compartment 1 and the rest leaves. Initially compartment 1 holds $6$ units and compartment 2 is empty. Find $x_{1} (t)$ and $x_{2} (t)$ .

Solution

Model. Balance = inflow − outflow for each compartment: $x_{1}^{'} x_{2}^{'} = - 4 x_{1} + x_{2}, = 2 x_{1} - 3 x_{2}, A = (- 4 2 1 - 3), x (0) = (60) .$
Eigenvalues. $tr A = - 7$ , $det A = 12 - 2 = 10$ : $λ^{2} + 7 λ + 10 = (λ + 2) (λ + 5) = 0, λ_{1} = - 2, λ_{2} = - 5.$
Eigenvectors. $A + 2 I = (- 2 2 1 - 1)$ gives $v_{2} = 2 v_{1}$ : $v_{1} = (1, 2)^{T}$ . $A + 5 I = (1212)$ gives $v_{2} = - v_{1}$ : $v_{2} = (1, - 1)^{T}$ .
General solution. $x = c_{1} e^{- 2 t} (1, 2)^{T} + c_{2} e^{- 5 t} (1, - 1)^{T}$ .
Initial condition. $c_{1} + c_{2} = 6$ and $2 c_{1} - c_{2} = 0$ . Adding: $3 c_{1} = 6$ , so $c_{1} = 2$ , $c_{2} = 4$ .
Answer. $x_{1} = 2 e^{- 2 t} + 4 e^{- 5 t}, x_{2} = 4 e^{- 2 t} - 4 e^{- 5 t} .$
Interpretation. Both amounts tend to $0$ , since material keeps leaving the system. Compartment 2 first fills up, then empties. Its maximum is where $x_{2}^{'} = - 8 e^{- 2 t} + 20 e^{- 5 t} = 0$ , i.e. $e^{3 t} = \frac{5}{2}$ , $t = \frac{1}{3} ln \frac{5}{2} \approx 0.31$ , and the maximum value is $x_{2} \approx 1.30$ (Figure 7.6.2).

Figure 7.6.2. The amounts in Example 7.6.2. After the fast mode

e^{- 5 t}

has died out, both compartments decay like

e^{- 2 t}

with the fixed ratio

x_{2} : x_{1} = 2 : 1

, the ratio of the components of

v_{1}

Case 2: complex eigenvalues

Let the eigenvalues be $λ = α \pm i β$ with $β \neq = 0$ . Theorem 7.6.1 works just as well with complex numbers, so for an eigenvector $v$ of $α + i β$ the function $e^{(α + i β) t} v$ is a solution. It is complex-valued, while we want real solutions. The way out is the same as in Chapter 4: take real and imaginary parts.

Theorem 7.6.3 (Real and imaginary parts)

Let $A$ be a real matrix. If $w (t) = u (t) + i y (t)$ , with $u$ and $y$ real, is a solution of $x^{'} = A x$ , then $u$ and $y$ are both real solutions.

Proof. $w^{'} = A w$ reads $u^{'} + i y^{'} = A u + i A y$ . Since $A$ is real, $A u$ and $A y$ are real vectors. Two complex vectors are equal only if their real parts and their imaginary parts agree, so $u^{'} = A u$ and $y^{'} = A y$ . $■$

Now split the eigenvector into real and imaginary parts, $v = a + i b$ , and expand with Euler's formula $e^{i βt} = cos βt + i sin βt$ :

e^{(α + i β) t} (a + i b) = e^{α t} (cos βt + i sin βt) (a + i b) = e^{α t} (a cos βt + i a sin βt + i b cos βt - b sin βt) = real part e^{α t} (a cos βt - b sin βt) + i imaginary part e^{α t} (a sin βt + b cos βt) .

Theorem 7.6.4 (Complex eigenvalues)

If $A$ has eigenvalues $α \pm i β$ ( $β \neq = 0$ ) and $a + i b$ is an eigenvector for $α + i β$ , the general solution of $x^{'} = A x$ is

x (t) = c_{1} e^{α t} (a cos βt - b sin βt) + c_{2} e^{α t} (a sin βt + b cos βt) .

Both functions are solutions by Theorem 7.6.3. At $t = 0$ they equal $a$ and $b$ , and these are independent: $v = a + i b$ and $\overset{ˉ}{v} = a - i b$ are eigenvectors of the distinct eigenvalues $α \pm i β$ , hence independent, and $a = \frac{1}{2} (v + \overset{ˉ}{v})$ , $b = \frac{1}{2 i} (v - \overset{ˉ}{v})$ . (If $a$ and $b$ were dependent, then $v$ and $\overset{ˉ}{v}$ , being combinations of them, would be too.) The conjugate eigenvalue $α - i β$ gives nothing new, so we never need to compute its eigenvector.

Example 7.6.3

Solve $x^{'} = (12 - 5 - 5) x$ with $x (0) = (5, 4)^{T}$ .

Solution

Eigenvalues. $tr A = - 4$ , $det A = - 5 + 10 = 5$ : $λ^{2} + 4 λ + 5 = 0, λ = \frac{- 4 \pm 16 - 20}{2} = - 2 \pm i .$ So $α = - 2$ , $β = 1$ .
Eigenvector for $- 2 + i$ . $A - (- 2 + i) I = (3 - i 2 - 5 - 3 - i) .$ The first row says $(3 - i) v_{1} = 5 v_{2}$ . Choose $v_{1} = 5$ , $v_{2} = 3 - i$ : $v = (5, 3 - i)^{T}$ . Check with row 2: $2 \cdot 5 + (- 3 - i) (3 - i) = 10 - (9 + 1) = 0$ .
Split. $v = (53) + i (0 - 1)$ , so $a = (5, 3)^{T}$ , $b = (0, - 1)^{T}$ .
Real solutions. With $β = 1$ , $u_{1} u_{2} = e^{- 2 t} (a cos t - b sin t) = e^{- 2 t} (5 cos t 3 cos t + sin t), = e^{- 2 t} (a sin t + b cos t) = e^{- 2 t} (5 sin t 3 sin t - cos t) .$
Initial condition. $x (0) = c_{1} a + c_{2} b = (5 c_{1}, 3 c_{1} - c_{2})^{T} = (5, 4)^{T}$ , so $c_{1} = 1$ and $c_{2} = - 1$ .
Answer. $x_{1} = 5 e^{- 2 t} (cos t - sin t), x_{2} = e^{- 2 t} (4 cos t - 2 sin t) .$ Both components oscillate with angular frequency $1$ inside the decaying envelope $e^{- 2 t}$ .

A second-order equation becomes a $2 \times 2$ system if we take the unknown and its derivative as the two components. So the matrix method also re-derives the results of Chapters 4 and 5.

Example 7.6.4 (A damped spring as a system)

The free motion of a damped spring satisfies $x^{''} + 2 x^{'} + 5 x = 0$ . Write it as a system for $x_{1} = x$ (position) and $x_{2} = x^{'}$ (velocity) and solve it by the matrix method.

Solution

System. $x_{1}^{'} = x^{'} = x_{2}$ and $x_{2}^{'} = x^{''} = - 5 x - 2 x^{'} = - 5 x_{1} - 2 x_{2}$ : $x^{'} = (0 - 5 1 - 2) x .$
Eigenvalues. $λ^{2} + 2 λ + 5 = 0$ , the characteristic equation of the original ODE; $λ = - 1 \pm 2 i$ .
Eigenvector for $- 1 + 2 i$ . The first row of $A - λ I$ is $(1 - 2 i, 1)$ . By the shortcut of Section 7.5, $v = (1, - 1 + 2 i)^{T}$ , so $a = (1, - 1)^{T}$ and $b = (0, 2)^{T}$ .
General solution. With $α = - 1$ , $β = 2$ : $x = c_{1} e^{- t} (cos 2 t - cos 2 t - 2 sin 2 t) + c_{2} e^{- t} (sin 2 t - sin 2 t + 2 cos 2 t) .$
Compare. The first component is $x = e^{- t} (c_{1} cos 2 t + c_{2} sin 2 t)$ , exactly what Section 4.2 gives, and the second component is its derivative, as it must be.

Case 3: a repeated eigenvalue

Suppose $λ^{2} - (tr A) λ + det A$ has a double root $λ_{1}$ . There are two very different situations.

(a) Two independent eigenvectors. For a $2 \times 2$ matrix this happens only when $A = λ_{1} I$ (if every vector is an eigenvector, then $A e_{1} = λ_{1} e_{1}$ and $A e_{2} = λ_{1} e_{2}$ , which fixes both columns of $A$ ). The system is then uncoupled, $x_{1}^{'} = λ_{1} x_{1}$ , $x_{2}^{'} = λ_{1} x_{2}$ , and the general solution is simply $x = e^{λ_{1} t} (c_{1}, c_{2})^{T}$ .

(b) Only one independent eigenvector $v$ . This is the typical case (Example 7.5.8). We get one solution $x_{1} = e^{λ_{1} t} v$ and need a second. Remembering $x e^{r x}$ from Chapter 4, one might try $t e^{λ_{1} t} v$ . But

(t e^{λ_{1} t} v)^{'} = e^{λ_{1} t} v + λ_{1} t e^{λ_{1} t} v, A (t e^{λ_{1} t} v) = λ_{1} t e^{λ_{1} t} v,

and the term $e^{λ_{1} t} v$ on the left has nothing to cancel against. So we add a correction and try

x_{2} = e^{λ_{1} t} (t v + p)

with an unknown constant vector $p$ . Compute both sides:

x_{2}^{'} A x_{2} = λ_{1} e^{λ_{1} t} (t v + p) + e^{λ_{1} t} v = e^{λ_{1} t} (λ_{1} t v + λ_{1} p + v), = e^{λ_{1} t} (t A v + A p) = e^{λ_{1} t} (λ_{1} t v + A p) .

The terms with $t$ cancel. What remains is $λ_{1} p + v = A p$ , that is

(A - λ_{1} I) p = v .

Definition 7.6.1 (Generalized eigenvector)

A vector $p$ with $(A - λ_{1} I) p = v$ , where $v$ is an eigenvector for $λ_{1}$ , is called a generalized eigenvector of $A$ for $λ_{1}$ .

Theorem 7.6.5 (Repeated eigenvalue, one eigenvector)

Let $λ_{1}$ be a double eigenvalue of the $2 \times 2$ matrix $A$ with only one independent eigenvector $v$ , and let $p$ solve $(A - λ_{1} I) p = v$ . Then the general solution of $x^{'} = A x$ is

x (t) = c_{1} e^{λ_{1} t} v + c_{2} e^{λ_{1} t} (t v + p) .

Two questions remain. Does $p$ exist? The matrix $A - λ_{1} I$ is singular, so the system $(A - λ_{1} I) p = v$ could, in principle, be inconsistent. It never is here. A $2 \times 2$ matrix satisfies its own characteristic equation (this is the Cayley–Hamilton theorem; for $2 \times 2$ you can check $A^{2} - (tr A) A + (det A) I = 0$ by multiplying out). With a double root that equation is $(A - λ_{1} I)^{2} = 0$ . Now take any vector $q$ that is not an eigenvector. Then $w = (A - λ_{1} I) q \neq = 0$ , and $(A - λ_{1} I) w = (A - λ_{1} I)^{2} q = 0$ , so $w$ is an eigenvector, hence $w = k v$ for some $k \neq = 0$ . Then $p = q / k$ works. In practice you just solve the $2 \times 2$ system: its two equations are proportional, and any solution will do.

Are the two solutions independent? At $t = 0$ the Wronskian is $det [v p]$ . If $p$ were a multiple of $v$ , then $(A - λ_{1} I) p$ would be $0$ , not $v$ . So $v, p$ are independent and the Wronskian is nonzero.

Common mistake

The second solution is $e^{λ_{1} t} (t v + p)$ , not $t e^{λ_{1} t} v$ . And $p$ solves $(A - λ_{1} I) p = v$ with the same $v$ that appears in $t v$ . If you rescale $v$ , rescale $p$ too.

Example 7.6.5

Solve $x^{'} = (- 1 1 - 4 - 5) x$ with $x (0) = (3, 2)^{T}$ .

Solution

Eigenvalues. $tr A = - 6$ , $det A = 5 + 4 = 9$ : $λ^{2} + 6 λ + 9 = (λ + 3)^{2}$ . Double eigenvalue $λ_{1} = - 3$ .
Eigenvectors. $A + 3 I = (21 - 4 - 2)$ . Both rows say $v_{1} = 2 v_{2}$ , so there is only one independent eigenvector, $v = (2, 1)^{T}$ .
Generalized eigenvector. Solve $(A + 3 I) p = v$ : $2 p_{1} - 4 p_{2} p_{1} - 2 p_{2} = 2, = 1.$ The first equation is twice the second. Take $p_{2} = 0$ : $p = (1, 0)^{T}$ .
General solution. $x = c_{1} e^{- 3 t} (21) + c_{2} e^{- 3 t} [t (21) + (10)] = e^{- 3 t} (2 c_{1} + c_{2} (2 t + 1) c_{1} + c_{2} t) .$
Initial condition. At $t = 0$ : $2 c_{1} + c_{2} = 3$ and $c_{1} = 2$ , so $c_{2} = - 1$ .
Answer. $x_{1} = (3 - 2 t) e^{- 3 t}, x_{2} = (2 - t) e^{- 3 t} .$ Check: $x_{1}^{'} = (- 2 - 9 + 6 t) e^{- 3 t} = (6 t - 11) e^{- 3 t}$ and $- x_{1} - 4 x_{2} = (- 3 + 2 t - 8 + 4 t) e^{- 3 t} = (6 t - 11) e^{- 3 t}$ .

Method in a nutshell

Write the system as $x^{'} = A x$ . Compute $tr A$ , $det A$ and solve $λ^{2} - (tr A) λ + det A = 0$ .
Distinct real $λ_{1}, λ_{2}$ : find eigenvectors $v_{1}, v_{2}$ ; $x = c_{1} e^{λ_{1} t} v_{1} + c_{2} e^{λ_{2} t} v_{2}$ .
Complex $α \pm i β$ : find one eigenvector $a + i b$ for $α + i β$ ; $x = c_{1} e^{α t} (a cos βt - b sin βt) + c_{2} e^{α t} (a sin βt + b cos βt)$ .
Double $λ_{1}$ : if $A = λ_{1} I$ , $x = e^{λ_{1} t} c$ . Otherwise find $v$ , solve $(A - λ_{1} I) p = v$ ; $x = c_{1} e^{λ_{1} t} v + c_{2} e^{λ_{1} t} (t v + p)$ .
For an initial-value problem, set $t = 0$ and solve the $2 \times 2$ linear system for $c_{1}, c_{2}$ .

Practice with the solver below. Type any real $2 \times 2$ matrix (rows separated by a semicolon) and, if you like, an initial vector. It shows the eigenvalues and eigenvectors, the general solution, the solution of the initial-value problem and the phase portrait with its classification. Compare each step with your hand computation.

The phase plane

A solution $x (t) = (x_{1} (t), x_{2} (t))^{T}$ can be drawn in two ways: as two graphs against $t$ (Figure 7.6.2), or as one curve in the $x_{1} x_{2}$ -plane, traced as $t$ increases (Figure 7.6.1). The plane is called the phase plane, the curves are trajectories (or orbits), and a picture showing a representative set of them, with arrows for the direction of motion, is a phase portrait. The phase portrait shows at a glance how all solutions behave, which is what an engineer usually wants to know: does the system return to rest, oscillate, or run away?

Some general facts hold for every $x^{'} = A x$ :

The constant function $x \equiv 0$ is a solution. A point where the system is at rest is an equilibrium (or critical point). If $det A \neq = 0$ , the origin is the only equilibrium, because $A x = 0$ then forces $x = 0$ . We assume $det A \neq = 0$ from now on.
Trajectories never cross. Through each point passes exactly one trajectory, because of the uniqueness theorem (Section 7.8). In particular, no trajectory reaches the origin in finite time.
For a real eigenvalue $λ$ with eigenvector $v$ , the solutions $c e^{λ t} v$ move along the straight line through the origin in the direction of $v$ : toward the origin if $λ < 0$ , away from it if $λ > 0$ . These straight-line solutions are the skeleton of the portrait (compare Figure 7.5.2).

The possible pictures

Saddle ( $λ_{1} < 0 < λ_{2}$ ). Solutions come in along the eigenline of $λ_{1}$ and leave along the eigenline of $λ_{2}$ . Every other trajectory approaches the origin for a while and then turns away, as in Figure 7.6.1. Only the solutions that start exactly on the "incoming" line tend to $0$ .

Node ( $λ_{1}, λ_{2}$ real, distinct, same sign). Suppose both are negative, $λ_{1} < λ_{2} < 0$ . Every solution tends to $0$ . As $t \to \infty$ , the term with the less negative eigenvalue $λ_{2}$ decays more slowly and dominates, so trajectories enter the origin tangent to $v_{2}$ (the "slow" eigenvector). As $t \to - \infty$ the fast term dominates, so far from the origin trajectories run parallel to $v_{1}$ . In Example 7.6.2 the slow eigenvector is $(1, 2)^{T}$ . If both eigenvalues are positive, the picture is the same with the arrows reversed.

Spiral or focus ( $α \pm i β$ with $α \neq = 0$ ). The factors $cos βt$ , $sin βt$ make the trajectories wind around the origin, and $e^{α t}$ makes them shrink ( $α < 0$ ) or grow ( $α > 0$ ). To find the sense of rotation, compute the velocity $A x$ at one point, for example at $(1, 0)^{T}$ , where $A x = (a_{11}, a_{21})^{T}$ : if $a_{21} > 0$ the motion there is upward, so the rotation is counterclockwise; if $a_{21} < 0$ it is clockwise. In Example 7.6.3, $a_{21} = 2 > 0$ .

Center ( $\pm i β$ , i.e. $α = 0$ ). All solutions are periodic with period $2 π / β$ ; the trajectories are closed curves (in fact ellipses) around the origin. The motion neither decays nor grows: an undamped oscillator.

Degenerate (improper) node (double $λ$ , one eigenvector). There is only one eigenline. From the general solution, $x = t e^{λ t} (c_{2} v + \frac{1}{t} (c_{1} v + c_{2} p))$ , so for $c_{2} \neq = 0$ the direction of $x (t)$ approaches that of $\pm v$ as $t \to \pm \infty$ . All trajectories become tangent to the single eigenline at the origin, and they look like a node "squeezed" from two lines into one.

Star (proper) node ( $A = λ I$ ). Every solution is $e^{λ t} c$ , a straight ray into the origin ( $λ < 0$ ) or out of it ( $λ > 0$ ).

Figure 7.6.3. The main types of phase portraits. Thick straight lines are eigenlines. Matrices used: saddle

(1320)

(Example 7.6.1); stable node

(- 4 2 1 - 3)

(Example 7.6.2); unstable node

(4 - 2 - 1 3)

; stable spiral

(- 1 4 - 8 - 1)

; center

(11 - 2 - 1)

; degenerate node

(- 1 1 - 4 - 5)

(Example 7.6.5). Note how the trajectories of the two nodes enter or leave the origin tangent to the slow eigenline.

Stability

Definition 7.6.2 (Stability of the origin)

The equilibrium $0$ of $x^{'} = A x$ is

stable if every solution that starts close enough to $0$ stays close to $0$ for all $t \geq 0$ (precisely: for every $ε > 0$ there is $δ > 0$ such that $∥ x (0) ∥ < δ$ implies $∥ x (t) ∥ < ε$ for all $t \geq 0$ );
asymptotically stable if it is stable and, in addition, every solution starting close enough to $0$ tends to $0$ as $t \to \infty$ ;
unstable if it is not stable.

For linear systems, "close enough" does not matter: if solutions near $0$ tend to $0$ , then all solutions do, because any solution is a constant multiple of one that starts near $0$ . Reading off the general solutions found above gives:

Theorem 7.6.6 (Stability from the eigenvalues)

Let $A$ be a real $2 \times 2$ matrix with $det A \neq = 0$ .

If both eigenvalues have negative real parts, $0$ is asymptotically stable.
If the eigenvalues are purely imaginary ( $\pm i β$ ), $0$ is stable but not asymptotically stable.
If at least one eigenvalue has a positive real part, $0$ is unstable.

Everything is summarized in Table 7.6.1. Write $τ = tr A$ and $Δ = det A$ (the numbers called $τ$ and $δ$ in Section 7.4); the eigenvalues are $λ = \frac{1}{2} (τ \pm τ^{2} - 4Δ)$ , so the signs of $Δ$ , $τ$ and $τ^{2} - 4Δ$ decide the type without computing any eigenvector.

Table 7.6.1: Classification of the origin for

x^{'} = A x

det A \neq = 0

Eigenvalues	Type of the origin	Stability	Test with $τ$ , $Δ$
real, $λ_{1} < 0 < λ_{2}$	saddle	unstable	$Δ < 0$
real, $λ_{1} < λ_{2} < 0$	node	asymptotically stable	$Δ > 0$ , $τ^{2} > 4Δ$ , $τ < 0$
real, $0 < λ_{1} < λ_{2}$	node	unstable	$Δ > 0$ , $τ^{2} > 4Δ$ , $τ > 0$
$α \pm i β$ , $α < 0$	spiral (focus)	asymptotically stable	$τ^{2} < 4Δ$ , $τ < 0$
$α \pm i β$ , $α > 0$	spiral (focus)	unstable	$τ^{2} < 4Δ$ , $τ > 0$
$\pm i β$	center	stable, not asymptotically	$τ = 0$ , $Δ > 0$
$λ_{1} = λ_{2}$ , one eigenvector	degenerate node	asympt. stable if $λ < 0$ , unstable if $λ > 0$	$τ^{2} = 4Δ$ , $A \neq = \frac{τ}{2} I$
$λ_{1} = λ_{2}$ , $A = λ I$	star node	asympt. stable if $λ < 0$ , unstable if $λ > 0$	$A = \frac{τ}{2} I$

Figure 7.6.4. The trace–determinant plane. Each matrix is a point

(τ, Δ)

. Lower half-plane: saddles (unstable). Upper half-plane: stable to the left of the

Δ

-axis, unstable to the right. Centers sit exactly on the positive

Δ

-axis and repeated eigenvalues exactly on the parabola, so a tiny change in

A

can turn them into spirals or nodes.

When $det A = 0$ one eigenvalue is zero and every point of the corresponding eigenline is an equilibrium. This borderline case is not in the table; you can still solve the system with the methods above (Theorem 7.6.2 applies if the other eigenvalue is nonzero).

Example 7.6.6

Classify the origin for $x^{'} = A x$ and decide its stability, for (a) $(35 - 1 - 1)$ , (b) $(- 2 1 1 - 2)$ , (c) $(2 - 1 5 - 2)$ , (d) $(1232)$ .

Solution

(a) $τ = 2$ , $Δ = - 3 + 5 = 2$ , $τ^{2} - 4Δ = - 4 < 0$ : complex eigenvalues ( $1 \pm i$ ) with positive real part. Unstable spiral; since $a_{21} = 5 > 0$ it turns counterclockwise.
(b) $τ = - 4$ , $Δ = 3$ , $τ^{2} - 4Δ = 4 > 0$ : real eigenvalues $- 1$ and $- 3$ . Stable node (asymptotically stable).
(c) $τ = 0$ , $Δ = - 4 + 5 = 1 > 0$ : eigenvalues $\pm i$ . Center, stable but not asymptotically stable; all solutions are periodic with period $2 π$ .
(d) $Δ = 2 - 6 = - 4 < 0$ : saddle, unstable (the eigenvalues are $4$ and $- 1$ ).

Exercises

In Exercises 1–8 find the general solution of $x^{'} = A x$ .

$A = (42 - 3 - 1)$
Answer
$x = c_{1} e^{t} (1, 1)^{T} + c_{2} e^{2 t} (3, 2)^{T}$
$A = (- 5 2 2 - 2)$
Answer
$x = c_{1} e^{- t} (1, 2)^{T} + c_{2} e^{- 6 t} (2, - 1)^{T}$
$A = (15 - 1 - 3)$
Answer
$λ = - 1 \pm i$ , eigenvector $(1, 2 - i)^{T}$ ; $x = c_{1} e^{- t} (cos t, 2 cos t + sin t)^{T} + c_{2} e^{- t} (sin t, 2 sin t - cos t)^{T}$
$A = (34 - 2 - 1)$
Answer
$λ = 1 \pm 2 i$ , eigenvector $(1, 1 - i)^{T}$ ; $x = c_{1} e^{t} (cos 2 t, cos 2 t + sin 2 t)^{T} + c_{2} e^{t} (sin 2 t, sin 2 t - cos 2 t)^{T}$
$A = (15 - 2 - 1)$
Answer
$λ = \pm 3 i$ , eigenvector $(2, 1 - 3 i)^{T}$ ; $x = c_{1} (2 cos 3 t, cos 3 t + 3 sin 3 t)^{T} + c_{2} (2 sin 3 t, sin 3 t - 3 cos 3 t)^{T}$
$A = (2 - 1 14)$
Answer
$λ = 3$ (double), $v = (1, 1)^{T}$ , $p = (0, 1)^{T}$ ; $x = c_{1} e^{3 t} (1, 1)^{T} + c_{2} e^{3 t} (t, t + 1)^{T}$
$A = (- 7 - 1 4 - 3)$
Answer
$λ = - 5$ (double), $v = (2, 1)^{T}$ , $p = (- 1, 0)^{T}$ ; $x = c_{1} e^{- 5 t} (2, 1)^{T} + c_{2} e^{- 5 t} (2 t - 1, t)^{T}$
$A = (- 2 0 0 - 2)$ . What do the trajectories look like?
Answer
$x = e^{- 2 t} (c_{1}, c_{2})^{T}$ ; every trajectory is a straight ray into the origin (stable star node).
Solve $x^{'} = (23 - 1 - 2) x$ , $x (0) = (1, 5)^{T}$ .
Answer
$x = - e^{t} (1, 1)^{T} + 2 e^{- t} (1, 3)^{T}$ , i.e. $x_{1} = - e^{t} + 2 e^{- t}$ , $x_{2} = - e^{t} + 6 e^{- t}$
Solve $x^{'} = (- 1 2 - 2 - 1) x$ , $x (0) = (1, 3)^{T}$ .
Answer
$x_{1} = e^{- t} (cos 2 t - 3 sin 2 t)$ , $x_{2} = e^{- t} (sin 2 t + 3 cos 2 t)$
Classify the origin and state its stability: (a) $(1 - 2 4 - 3)$ , (b) $(3220)$ , (c) $(51 - 1 3)$ , (d) $(- 1 - 5 21)$ , (e) $(- 3 1 1 - 3)$ .
Answer
(a) $- 1 \pm 2 i$ : stable spiral; (b) $4, - 1$ : saddle, unstable; (c) $4$ double, $A \neq = 4 I$ : unstable degenerate node; (d) $\pm 3 i$ : center, stable; (e) $- 2, - 4$ : stable node.
The overdamped oscillator $x^{''} + 4 x^{'} + 3 x = 0$ is written as a system with $x_{1} = x$ , $x_{2} = x^{'}$ . Find the matrix, the general solution of the system and the type of the origin.
Answer
$A = (0 - 3 1 - 4)$ ; $x = c_{1} e^{- t} (1, - 1)^{T} + c_{2} e^{- 3 t} (1, - 3)^{T}$ ; stable node.