ODE-Project Linear Systems in Higher Dimensions

Section 3.8 Linear Systems in Higher Dimensions

Objectives

To understand we can solve the system $x^{'} = A x$ by finding eigenvalues and eigenvectors for $A,$ where $A$ is an $n \times n$ matrix.
To understand that the Principle of Superposition holds for higher-order systems.
To understand that the geometry of a system in $R^{3}$ is characterized by stable lines and stable planes.

Suppose that we have two masses on a table,

m_{1}

and

m_{2},

connected by three springs with the outside springs connected to two walls (Figure 3.8.1), and the masses are free to move horizontally. We will assume that the springs are uniform and all have the same spring constant

k .

The horizontal displacements of the springs are denoted by

x_{1} (t)

and

x_{2} (t)

for the masses

m_{1}

and

m_{2},

respectively. Assuming that there is no damping, the only forces acting on mass

m_{1}

at time

t

are those of left and middle springs. The force from the left spring will be

- k x_{1}

while the force from middle spring will be

k (x_{2} - x_{1}) .

By Newton’s Second Law of motion, we have

🔗

m_{1} x_{1}^{″} = - k x_{1} + k (x_{2} - x_{1}) .

Similarly, the only forces acting on the second mass,

m_{2},

will come from middle and right springs. Again using Newton’s Second Law,

🔗

m_{2} x_{2}^{″} = - k (x_{2} - x_{1}) - k x_{2} .

🔗

Figure 3.8.1. A double spring-mass system
🔗

We now have a system of two second-order linear equations

🔗

\begin{aligned} m_{1} x_{1}^{″} & = - 2 k x_{1} + k x_{2} \\ m_{2} x_{2}^{″} & = k x_{1} - 2 k x_{2} . \end{aligned}

If we define

x_{3}

and

x_{4}

x_{3} = x_{1}^{'}

and

x_{4} = x_{2}^{'},

we now have first-order linear system of four equations,

🔗

\begin{aligned} x_{1}^{'} & = x_{3} \\ x_{2}^{'} & = x_{4} \\ x_{3}^{'} & = - 2 \frac{k}{m_{1}} x_{1} + \frac{k}{m_{1}} x_{2} \\ x_{4}^{'} & = \frac{k}{m_{2}} x_{1} - 2 \frac{k}{m_{2}} x_{2} \end{aligned}

We can represent this system in the matrix form

x^{'} = A x,

where

🔗

A = (\begin{matrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ - 2 k / m_{1} & k / m_{1} & 0 & 0 \\ k / m_{2} & - 2 k / m_{2} & 0 & 0 \end{matrix}) .

We will learn how to analyze and solve such systems in the next two sections.

🔗

Subsection 3.8.1 Higher-Order Linear Systems

We can write the system

🔗

\begin{aligned} x_{1}^{'} & = a_{11} x_{1} + \dots a_{1 n} x_{n} \\ x_{2}^{'} & = a_{21} x_{1} + \dots a_{2 n} x_{n} \\ ⋮ \\ x_{n}^{'} & = a_{n 1} x_{1} + \dots a_{n n} x_{n} \end{aligned}

in matrix form

x^{'} = A x,

where

🔗

A = (\begin{matrix} a_{11} & a_{12} & \dots & a_{1 n} \\ a_{21} & a_{22} & \dots & a_{2 n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{n 1} & a_{n 2} & \dots & a_{n n} \end{matrix}) and x = (\begin{matrix} x_{1} \\ x_{2} \\ ⋮ \\ x_{n} \end{matrix}) .

🔗

The strategy for finding solutions to the system

x^{'} = A x

is the same as for systems of two equations. If

λ

is an eigenvalue of

A

with eigenvector

v,

🔗

x (t) = e^{λ t} v

is a solution for our system. Indeed,

🔗

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

🔗

Example 3.8.2.

The system

🔗

\begin{aligned} x^{'} & = - 5 x - 8 y - 2 z \\ y^{'} & = 5 x + 12 y + 4 z \\ z^{'} & = - 11 x - 19 y - 5 z \end{aligned}

can be rewritten as

🔗

x^{'} = (\begin{matrix} x^{'} \\ y^{'} \\ z^{'} \end{matrix}) = (\begin{matrix} - 5 & - 8 & - 2 \\ 5 & 12 & 4 \\ - 11 & - 19 & - 5 \end{matrix}) (\begin{matrix} x \\ y \\ z \end{matrix}) = A x .

We can compute the eigenvalues of

A

by finding the roots of its characteristic polynomial

🔗

det (A - λ I) = - λ^{3} - 2 λ^{2} - λ + 2 = (λ - 1) (λ + 1) (λ - 2) .

Thus, the eigenvalues of

A

are

λ = 2,

λ = 1,

and

λ = - 1 .

To find an eigenvector for

λ = 2,

we must find a nontrivial solution for system of equations

(A - 2 I) x = 0,

🔗

\begin{aligned} - 7 x - 8 y - 2 z & = 0 \\ - 5 x + 10 y + 4 z & = 0 \\ - 11 x - 19 y - 7 z & = 0. \end{aligned}

It is easy to check that

(2, - 3, 5)

is a solution. Similarly, we can determine that

(5, - 7, 13)

is an eigenvector for

λ = 1

and

(1, - 1, 2)

is an eigenvector for

λ = - 1 .

Thus, we have found three solutions for the system

x^{'} = A x,

🔗

x_{1} (t) = e^{2 t} (\begin{matrix} 2 \\ - 3 \\ 5 \end{matrix}), x_{2} (t) = e^{t} (\begin{matrix} 5 \\ - 7 \\ 13 \end{matrix}), and x_{3} (t) = e^{- t} (\begin{matrix} 1 \\ - 1 \\ 2 \end{matrix}) .

🔗

The Principle of Superposition also holds for higher-order systems. If

x_{1} (t)

and

x_{2} (t)

are solutions for

x^{'} = A x,

then

🔗

c_{1} x_{1} (t) + c_{2} x_{2} (t)

is a solution for the system, since

🔗

\begin{aligned} \frac{d}{d t} [c_{1} x_{1} (t) + c_{2} x_{2} (t)] & = c_{1} \frac{d}{d t} x_{1} (t) + c_{2} \frac{d}{d t} x_{2} (t) \\ = c_{1} A x_{1} (t) + c_{2} A x_{2} (t) \\ = A [c_{1} x_{1} (t) + c_{2} x_{2} (t)] . \end{aligned}

Consequently,

🔗

x (t) = c_{1} e^{2 t} (\begin{matrix} 2 \\ - 3 \\ 5 \end{matrix}) + e^{t} (\begin{matrix} 5 \\ - 7 \\ 13 \end{matrix}) + c_{3} e^{- t} (\begin{matrix} 1 \\ - 1 \\ 2 \end{matrix})

is a solution for our system. This is, in fact, the general solution for the system.

🔗

Although we shall not cover the notions of linear independence, canonical matrices, and change of coordinates for

R^{n},

the same ideas that we used for systems of first-order linear differential equations in

R^{2}

carry over to

R^{n} .

The necessary linear algebra is covered in any good linear algebra course. Also, see Chapters 5 and 6 in [12]. In addition, finding eigenvalues for matrices greater than

2 \times 2,

we will need to find the roots of a characteristic polynomial of degree greater than two, which can be difficult. A good course in linear algebra will cover techniques of finding eigenvalues for larger matrices.

🔗

Example 3.8.3.

Let us see how the linear algebra works in the previous example. If we form the matrix

🔗

T = (\begin{matrix} 2 & 5 & 1 \\ - 3 & - 7 & - 1 \\ 5 & 13 & 2 \end{matrix})

from the eigenvectors of

A,

we can convert the system

x^{'} = A x

to the system

🔗

\begin{aligned} y^{'} & = (T^{- 1} A T) y \\ (\begin{array}{c} 1 & - 3 & - 2 \\ - 1 & 1 & 1 \\ 4 & 1 & - 1 \end{array}) (\begin{array}{c} - 5 & - 8 & - 2 \\ 5 & 12 & 4 \\ - 11 & - 19 & - 5 \end{array}) (\begin{array}{c} 2 & 5 & 1 \\ - 3 & - 7 & - 1 \\ 5 & 13 & 2 \end{array}) y, \\ = (\begin{array}{c} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{array}) y, \end{aligned}

which we can immediately solve:

🔗

y (t) = c_{1} e^{2 t} (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) + c_{2} e^{t} (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}) + c_{3} e^{- t} (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) .

Multiplying our solution by

T

yields the general solution

🔗

x (t) = T y = c_{1} e^{2 t} (\begin{matrix} 2 \\ - 3 \\ 5 \end{matrix}) + e^{t} (\begin{matrix} 5 \\ - 7 \\ 13 \end{matrix}) + c_{3} e^{- t} (\begin{matrix} 1 \\ - 1 \\ 2 \end{matrix}),

the solution to the system

x^{'} A x .

🔗

Subsection 3.8.2 The Geometry of Solutions

In Section 3.6, we classified all of the geometry of the solutions for planar systems using the trace-determinant plane. The geometry for linear systems in three variables is a bit more complicated. For a system

🔗

\begin{aligned} x^{'} & = a_{1} x + a_{1} y + a_{3} z \\ y^{'} & = b_{1} x + b_{2} y + b_{3} z \\ z^{'} & = c_{1} x + c_{2} y + c_{3} z, \end{aligned}

our solution curves live in

R^{3},

and there is simply a lot more room to move around in three dimensions than in two dimensions. The origin is still an equilibrium solution for a system of linear differential equations in three variables. The origin is a stable equilibrium solution if any solution

x (t)

approaches

0 = (0, 0, 0)

t \to \infty;

otherwise,

0

is an unstable equilibrium solution. In the case of planar systems, an unstable solution is a nodal saddle, a nodal source, a spiral source, or a source with a single unstable line. In the case of

R^{3},

we could have a stable line of solutions and an unstable plane of solutions. In this case, all solutions of the system with initial condition lying on the stable line would approach the origin as

t \to \infty,

but all solutions with initial conditions that are a nonzero point on the unstable plane would move away from the origin.

🔗

Example 3.8.4.

In Example 3.8.2, we had solutions

🔗

x_{1} (t) = e^{2 t} (\begin{matrix} 2 \\ - 3 \\ 5 \end{matrix}), x_{2} (t) = e^{t} (\begin{matrix} 5 \\ - 7 \\ 13 \end{matrix}), and x_{3} (t) = e^{- t} (\begin{matrix} 1 \\ - 1 \\ 2 \end{matrix}) .

The straight line through the origin and the point

(1, - 1, 2)

is a stable line. That is, for any initial condition

x (0) = (x_{0}, y_{0}, z_{0})

lying on this line, our solution will tend toward the origin as

t \to \infty .

On the other hand, the plane spanned by

(2, - 3, 5)

and

(5, - 7, 13)

is unstable plane. Solutions on this plane move away from the origin as

t \to \infty .

Of course,

(0, 0, 0)

is an equlibrium solution for our system. We say that the origin is a saddle in this example (Figure 3.8.5).

🔗

a stable line and an unstable plane in three dimensions with a transformation to another unstable plane and stable line — Figure 3.8.5. A saddle in $R^{3}$

🔗

Example 3.8.6.

For the system

🔗

x^{'} = (\begin{matrix} 0 & 1 & 0 \\ - 1 & 0 & 0 \\ 0 & 0 & - 1 \end{matrix}) x

we have a very different type of unstable equilibrium solution. The eigenvalues of this matrix are

λ = \pm i

and

λ = - 1 .

Thus, a solution satisfying the initial condition

x (0) = (x_{0}, y_{0}, z_{0})

is given by

🔗

x (t) = x_{0} (\begin{matrix} \cos t \\ - \sin t \\ 0 \end{matrix}) + y_{0} (\begin{matrix} \sin t \\ \cos t \\ 0 \end{matrix}) + z_{0} e^{- t} (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) .

z_{0} = 0,

then our initial condition is in the

x y

-plane and all of the solutions lie on circles centered at the origin. On the other hand, if

x_{0} = 0

and

y_{0} = 0,

we have a stable line of solutions lying along the

z

-axis. In fact, each solution that does not lie on the stable line lies on a cylinder in

R^{3}

given by

x^{2} + y^{2} = r^{2}

for some constant

r > 0 .

These solutions spiral towards the circular solution of radius

r

in the

x y

-plane if

z_{0} \neq 0

(Figure 3.8.7).

🔗

Figure 3.8.7. A spiral center in

R^{3}

🔗

Example 3.8.8.

For an example of a stable plane and an unstable line, let us consider the system

🔗

x^{'} = (\begin{matrix} - 1 & 1 & 0 \\ - 1 & - 1 & 0 \\ 0 & 0 & 1 \end{matrix}) x = A x .

The characteristic equation of the matrix

A

🔗

λ^{3} + λ^{2} - 2 = (λ - 1) (λ^{2} + 2 λ + 2) = 0.

Thus, the eigenvalues of

A

are

λ = 1

and

λ = - 1 \pm i .

Solving

🔗

[A - (- 1 + i)] x = 0

gives us an eigenvector

(1, i, 0)

for

λ = - 1 + i .

Since

🔗

(\begin{matrix} 1 \\ i \\ 0 \end{matrix}) = (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}) + i (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}),

we will let

v_{1} = (1, 0, 0)

and

v_{2} = (0, 1, 0) .

Since

v_{3} = (0, 0, 1)

is an eigenvector for

λ = 1,

our system has solution

🔗

x (t) = x_{0} e^{- t} (\begin{matrix} \cos t \\ - \sin t \\ 0 \end{matrix}) + y_{0} e^{- t} (\begin{matrix} \sin t \\ \cos t \\ 0 \end{matrix}) + z_{0} e^{t} (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}),

where

x (0) = (x_{0}, y_{0}, z_{0}) .

z_{0} = 0,

our initial condition lies in the

x y

-plane and solution curves spiral in towards the origin. Thus, we have a stable plane. On the other hand, if

x_{0} = 0

and

y_{0} = 0

but

z_{0} \neq 0,

then our solution approaches

\pm \infty

t \to \infty .

In this case, the

z

-axis is an unstable line (Figure 3.8.9).

🔗

Figure 3.8.9. A spiral saddle in

R^{3}

🔗

For an example of a stable equilibrium solution at the origin, consider the system

🔗

x^{'} = (\begin{matrix} λ_{1} & 0 & 0 \\ 0 & λ_{2} & 0 \\ 0 & 0 & λ_{3} \end{matrix}) x,

where

λ_{3} < λ_{2} < λ_{1} < 0 .

For an initial condition

(x_{0}, y_{0}, z_{0})

with at least one coordinate nonzero, the corresponding solution tends towards the origin tangentially to the

x

-axis as

t \to \infty

(Figure 3.8.10).

🔗

Figure 3.8.10. A sink in

R^{3}

🔗

Changing the system that in ExampleExample 3.8.8 to be

🔗

x^{'} = (\begin{matrix} - 1 & 1 & 0 \\ - 1 & - 1 & 0 \\ 0 & 0 & - 1 \end{matrix}) x,

we obtain the solution satisfying the initial condition

x (0) = (x_{0}, y_{0}, z_{0})

to be

🔗

x (t) = x_{0} e^{- t} (\begin{matrix} \cos t \\ - \sin t \\ 0 \end{matrix}) + y_{0} e^{- t} (\begin{matrix} \sin t \\ \cos t \\ 0 \end{matrix}) + z_{0} e^{- t} (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}),

where

x (0) = (x_{0}, y_{0}, z_{0}) .

In this case, all solutions will approach the origin as

t \to \infty .

🔗

Activity 3.8.1. Solving Higher Order Systems.

Find the eigenvalues of $A .$ You may find Sage useful.
Find the eigenvectors for each eigenvalue of $A .$
What is the general solution of $d x / d t = A x ?$
Describe the nature of the nature of the solution space in $R^{3} .$ Are there stable lines or planes?

🔗

(a)

A = (\begin{matrix} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{matrix})

🔗

(b)

A = (\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix})

🔗

(c)

A = (\begin{matrix} - 13 & - 8 & - 4 \\ 12 & 7 & 4 \\ 24 & 16 & 7 \end{matrix})

🔗

Subsection 3.8.3 The Double Spring-Mass Systems Revisited

Let us return to our spring-mass system

x^{'} = A x,

where

🔗

A = (\begin{matrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ - 2 k / m_{1} & k / m_{1} & 0 & 0 \\ k / m_{2} & - 2 k / m_{2} & 0 & 0 \end{matrix}) .

To keep matters simple, we will assume that

m_{1} = m_{2} = k = 1 .

Thus, our matrix now becomes

🔗

A = (\begin{matrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ - 2 & 1 & 0 & 0 \\ 1 & - 2 & 0 & 0 \end{matrix}) .

The characteristic polynomial of

A

🔗

det (A - λ I) = λ^{4} + 4 λ^{2} + 3 = (λ^{2} + 1) (λ^{2} + 3) .

Thus, the eigenvalues of

A

are

λ = \pm i

and

λ = \pm i \sqrt{3} .

We can find eigenvectors

🔗

v_{1} = (\begin{matrix} i / \sqrt{3} \\ - i / \sqrt{3} \\ - 1 \\ 1 \end{matrix}), v_{2} = (\begin{matrix} - i / \sqrt{3} \\ i / \sqrt{3} \\ - 1 \\ 1 \end{matrix}), v_{3} = (\begin{matrix} - i \\ - i \\ 1 \\ 1 \end{matrix}), v_{4} = (\begin{matrix} i \\ i \\ 1 \\ 1 \end{matrix}),

corresponding to the eigenvalues

λ_{1} = i \sqrt{3},

λ_{2} = - i \sqrt{3},

λ_{3} = i,

λ_{4} = - i,

respectively. Consequently, the general solution to our system is

🔗

x (t) = c_{1} e^{i \sqrt{3} t} v_{1} + c_{1} e^{- i \sqrt{3} t} v_{2} + c_{1} e^{i t} v_{3} + c_{1} e^{- i t} v_{4};

however, this form of the solution is not very useful. By examining real and imaginary parts of

e^{i \sqrt{3} t} v_{1}

and

c_{1} e^{i t} v_{3},

we can rewrite the solution as

🔗

x (t) = c_{1} (\begin{matrix} \cos \sqrt{3} t \\ - \cos \sqrt{3} t \\ - \sqrt{3} \sin \sqrt{3} t \\ \sqrt{3} \sin \sqrt{3} t \end{matrix}) + c_{2} (\begin{matrix} - \sin \sqrt{3} t \\ \sin \sqrt{3} t \\ - \sqrt{3} \cos \sqrt{3} t \\ \sqrt{3} \cos \sqrt{3} t \end{matrix}) + c_{3} (\begin{matrix} \cos t \\ \cos t \\ - \sin t \\ - \sin t \end{matrix}) + c_{4} (\begin{matrix} \sin t \\ \sin t \\ \cos t \\ \cos t \end{matrix}) .

If we have the following initial conditions,

🔗

\begin{aligned} x_{1} (0) & = 0 \\ x_{2} (0) & = 0 \\ x_{1}^{'} (0) = x_{3} (0) & = 2 \\ x_{2}^{'} (0) = x_{4} (0) & = 2, \end{aligned}

we can determine

c_{1} = c_{2} = c_{3} = 0

and

c_{4} = 2 .

Thus, the solution to our initial value problem is

🔗

x (t) = (\begin{matrix} 2 \sin t \\ 2 \sin t \\ 2 \cos t \\ 2 \cos t \end{matrix}),

and the two masses will oscillate with a frequency of one and an amplitude of two. We leave the details as an exercise.

🔗

Subsection 3.8.4 Important Lessons

$\begin{aligned} x_{1}^{'} & = a_{11} x_{1} + \dots a_{1 n} x_{n} \\ x_{2}^{'} & = a_{21} x_{1} + \dots a_{2 n} x_{n} \\ ⋮ \\ x_{n}^{'} & = a_{n 1} x_{1} + \dots a_{n n} x_{n} \end{aligned}$

can be written in matrix form $x^{'} = A x,$ where

$A = (\begin{matrix} a_{11} & a_{12} & \dots & a_{1 n} \\ a_{21} & a_{22} & \dots & a_{2 n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{n 1} & a_{n 2} & \dots & a_{n n} \end{matrix}) and x = (\begin{matrix} x_{1} \\ x_{2} \\ ⋮ \\ x_{n} \end{matrix}) .$
As in the case of $R^{2},$ we can solve the system $x^{'} = A x$ by finding eigenvalues and eigenvectors for $A .$
The Principle of Superposition holds for higher-order systems. If $x_{1} (t)$ and $x_{2} (t)$ are solutions for $x^{'} = A x,$ then

$c_{1} x_{1} (t) + c_{2} x_{2} (t)$

is a solution for the system.
The geometry for a system in $R^{3}$ is more complicated than the planar case. However, the solutions are usually characterized by stable lines or stable planes.

🔗

Reading Questions 3.8.5 Reading Questions

1.

Is it possible for a

3 \times 3

to have a line of stable solutions and a plane of unstable solutions? Explain.

🔗

Exercises 3.8.6 Exercises

Finding Solutions of $3 \times 3$ Systems.

For each of the linear systems

d x / d t = A x

in Exercise Group 3.8.6.1–8

🔗

Find the eigenvalues of $A .$ You may find Sage useful.
Find the eigenvectors for each eigenvalue of $A .$
What is the general solution of $d x / d t = A x ?$
Describe the nature of the nature of the solution space in $R^{3} .$ Are there stable lines or planes?