section 11.1: linear systems (theory)justincc/fall2019/math... · section 11.1: linear systems...

MATH148 c©Justin Cantu

Section 11.1: Linear Systems (Theory)

In this section, we will study homogeneous, linear first-order systems of differential equations withconstant coefficients, that is, systems of the form

dx1dt

= a11x1 + a12x2

dx2dt

= a21x1 + a22x2

where the variables x1 and x2 are functions of t and the coefficients aij are constants (i.e., theydon’t depend on t). This system can be written in matrix form as

dx1dt

dx2dt

=

[a11 a12a21 a22

] [x1x2

]or

dx

dt= Ax

A solution to this system is a pair of functions x1(t), x2(t) (or x(t) = [x1(t), x2(t)]T in vector

function form) that satisfies both differential equations above. Such a pair of functions defines aparametric curve in the x1x2-plane, which has a natural direction indicated by how the curve istraced as t increases.

Example: The curve defined by x(t) = [t2 + 2t, t3 − 4t]T is given below (for 0 ≤ t ≤ 2). Indicatewith an arrow the direction in which the curve is traced as t increases.

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Tangent Vectors:

The vector function x(t) = [x1(t), x2(t)]T defines a parametric curve in the x1x2-plane. The vector

dx

dt=

dx1dt

dx2dt

is the tangent vector to the curve at the point (x(t), y(t)). If we think of the curve as describingthe motion of an object, then a tangent vector describes what direction the motion is trending in,and how fast the object is moving (given by the length).

Example: Let x(t) = 〈t2 + 2t, t3 − 4t〉.

(a) Finddx

dt.

(b) Find a tangent vector to the curve traced by x(t) at the point where t = 1.

(c) The curve traced by x(t) is given below (0 ≤ t ≤ 2). Draw the tangent vector found in (b).

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Vector Fields and Direction Fields:

Consider the system

dx1dt

= x1 − 2x2

dx2dt

= x2

We now use the ideas from the previous subsection to try to visualize solution curves to our systemof differential equations. Namely, suppose that you are standing at a specific point in the x1x2-plane,say (x1, x2) = (2,−1). Our system then tells us what direction we move in:

dx

dt=

dx1dt

dx2dt

=

[2− 2(−1)−1

]=

[4−1

]

We draw this vector at the point (2,-1) in the x1x2-plane. We can do this for more and more points,until we get a vector field that helps us see the shapes of solution curves:

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To visualize these solutions, we only need the direction of these vectors, so we normalize them allto be the same size, giving a direction field:

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Solving Linear Systems:

Consider the linear system of differential equations

dx

dt= Ax

where A is 2× 2 with real entries.

Claim: The vector function x(t) =

[u1e

λt

u2eλt

]= eλt

[u1u2

], where λ, u1, and u2 are constants, is a

solution of the above system for appropriate choices of λ, u1, u2.

Conclusion: The function x(t) = eλt[u1u2

]is a solution of

dx

dt= Ax if λ is an eigenvalue of A with

corresponding eigenvector

[u1u2

].

General Solution of a Homogeneous Linear System: Letdx

dt= Ax where A is a 2 × 2

matrix with two real and distinct eigenvalues λ1 and λ2 with corresponding eigenvectors u1 andu2. Then

x(t) = c1eλ1tu1 + c2e

λ2tu2

is the general solution of this system. The constants c1 and c2 depend on the initial condition.

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Example: Solve the initial-value problem

dx

dt=

[4 71 −2

]x

with x1(0) = −1 and x2(0) = −2.

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Example: Solve the initial-value problem

dx1dt

= −3x1 + 4x2

dx2dt

= −x1 + 2x2

with x1(0) = 1 and x2(0) = 2.

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Equilibria and Stability:

An equilibrium of the linear systemdx

dt= Ax

is a solution x̂ =

[x̂1x̂2

]for which there is no change, that is,

dx̂

dt= Ax̂ = 0

Note: The tangent vector at an equilibrium is [0, 0]T , so that there is “no movement” once thispoint is reached.

To find equilibria, we must solveAx = 0

The zero vector [0, 0]T is clearly a solution, and we call [0, 0]T the trivial equilibrium. If detA 6= 0(i.e., A is nonsingular), then this is the only solution. We will study the stability of the trivial equi-librium in such cases.

Fact: It can be shown (see page 530), that detA = λ1λ2, where λ1, λ2 are the eigenvalues of A. Inparticular, detA 6= 0 if and only if λ1 and λ2 are both nonzero. This leads to the following cases.

Stability of x̂ = [0, 0]T when A is Nonsingular: Consider the linear system

dx

dt= Ax

where A is a 2× 2 matrix with nonzero eigenvalues λ1 and λ2.

Case 1: Distinct Real Eigenvalues (λ1 6= λ2)

(a) If both eigenvalues are negative, then [0, 0]T is called a sink or stable node.

(b) If both eigenvalues are positive, then [0, 0]T is called a source or unstable node.

(c) If the eigenvalues have opposite sign, then [0, 0]T is called a saddle point (unstable).

Case 2: Complex Conjugate Eigenvalues (λ1,2 = a± bi)

(a) If both eigenvalues have negative real parts (a < 0), then [0, 0]T is called a stable spiral.

(b) If both eigenvalues have positive real parts (a > 0), then [0, 0]T is called an unstablespiral.

(c) If both eigenvalues are purely imaginary (a = 0), then [0, 0]T is called a neutral spiral orcenter (neither stable nor unstable).

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The direction fields below illustrate the various cases for stability of [0, 0]T . The lines defined byeigenvectors are plotted in red, and sample solution curves are plotted in blue, green, and black.

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Example: Consider the system

dx1dt

= −2x1 + 4x2

dx2dt

= 2x1 − 5x2

Determine the stability of [0, 0]T and classify the equilibrium.


dx1dt

= 6x1 − 4x2

dx2dt

= −3x1 + 5x2


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dx1dt

= −5x1 − 2x2

dx2dt

= 6x1 + 3x2



dx1dt

= −x1 − 5x2

dx2dt

= 4x1 − 3x2


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dx1dt

= x1 + 3x2

dx2dt

= −2x1 + x2



dx1dt

= −2x2

dx2dt

= 2x1


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section 11.1: linear systems (theory)justincc/fall2019/math... · section 11.1: linear systems...

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