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Math 5AI: Project 4
Linear systems of differential equations
January 22-27, 2010
We next study dynamical systems of the form
dx/dt =
dy/dt =
a11 x + a12 y + f (t),
a21 x + a22 y + g(t),
where the aij ’s are constants, and f (t) and g(t) are well-behaved functions of
t. Dynamical systems of this type are said to be linear , and there is a rather
complete theory for finding their solutions.
A. Homogeneous first order linear systems of differential equations:
We first consider the associated homogeneous system:
dx/dt
dy/dt
= a11 x + a12 y,
= a21 x + a22 y.
(1)
There is a systematic method for solving a system of this form, which has the
virtue of extending to systems with n unknowns. This method consists of two
steps:
Step I. Assume the solution has the special form,
x = Aeλt ,
y = Beλt ,
(2)
where A, B, and λ are constants, and solve for A, B, and λ. Usually, there
will be two values λ1 and λ2 for λ which yield nonzero solutions to the linear
system:
x = A1 e λ 1 t
x = A2 eλ2 t
.
λ1 t and
y = B1 e
y = B2 eλ2 t
Step II. Apply the superposition principle to derive the general solution:
x = c1 A1 eλ1 t + c2 A2 eλ2 t
.
y = c1 B1 eλ1 t + c2 B2 eλ2 t
1
Here c1 and c2 are constants of integration, which can be determined by initial
conditions.
Dangerous curve! If only one value of λ can be found, one is said to be in the
notorious “repeated root case,” and this simple method will not work. If there
is time, we will introduce such a method later.
It is perhaps easiest to understand this method when it is presented in the
context of a simple example.
1. a. Suppose that our system is
dx/dt
dy/dt
= y,
= 2x + y.
According to (2),
dy
dx
= λAeλt ,
= λBeλt .
dt
dt
Substitution into the system of differential equations yields
λAeλt
= Beλt ,
λBeλt
=
(3)
2Aeλt + Beλt ,
and dividing by eλt yields
λA
= B,
λB
=
2A + B.
Determine which values of λ allow for nonzero solutions to this homogenous
linear system for (A, B).
b. For each choice of λ find a corresponding solution to the homogeneous linear
system, thereby obtaining two linearly independent solutions
x = A1 eλ1 t
x = A2 e λ 2 t
.
λ1 t and
y = B1 e
y = B2 eλ2 t
to (2).
To find the general solution, we now make use of the superposition principle: If
x = x1 (t),
y = y1 (t)
and
x = x2 (t),
y = y2 (t)
are solutions to the the homogeneous linear system
dx/dt
dy/dt
= a11 x + a12 y,
= a21 x + a22 y,
then so is the linear superposition
x =
y =
c1 x1 (t) + c2 x2 (t),
c1 y1 (t) + c2 y2 (t),
2
where c1 and c2 are arbitrary constants.
c. Use the superposition principle to find the general solution, involving two
arbitrary constants of integration c1 and c2 , to (2).
d. Suppose that x0 and y0 are given. To find a solution x = x(t), y = y(t) which
satisfies the initial conditions x(0) = x0 , y(0) = y0 , we substitute t = 0 into the
general solution and solve for the constants c1 and c2 . Carry this procedure
out. Note that given any initial point (x0 , y0 ), there is a unique solution curve
to our dynamical system which starts at the point (x0 , y0 ) at the time t = 0.
Step I is often somewhat more complicated to carry out than it was in this
example. Let us consider the general case. Substituting (2) into (1), we obtain
λAeλt
λBeλt
= a11 Aeλt + a12 Beλt ,
= a21 Aeλt + a22 Beλt .
We divide by eλt ,
λA
λB
= a11 A + a12 B,
= a21 A + a22 B,
and simplify to obtain
(a11 − λ)A + a12 B
a21 A + (a22 − λ)B
= 0,
= 0.
(4)
Given λ, this is a linear system in two unknowns A and B.
We ask: for which λ’s does (4) have nontrivial solutions? To answer this
question, we multiply the first of equations (4) by a22 − λ and the second by
a12 :
(a11 − λ)(a22 − λ)A +a12 (a22 − λ)B = 0,
a12 a21 A
+a12 (a22 − λ)B = 0.
Subtract the second of these from the first to obtain
(a11 − λ)(a22 − λ)A − a12 a21 A = 0.
Thus if A 6= 0,
(a11 − λ)(a22 − λ) − a12 a21 = 0.
(5)
One readily checks that the same equation must hold if B 6= 0. Thus λ must
satisfy equation (5) if equation (4) is to have nontrivial solutions.
Equation (5) can be written more simply as
a11 − λ
a12 = 0.
a21
a22 − λ This is known as the characteristic equation, and its roots are called the eigenvalues of the coefficient matrix
a11 a12
A=
.
a21 a22
3
We can now expand somewhat on our description of Step I: We find the
roots of the characteristic equation
a11 − λ
a12 = 0,
a21
a22 − λ and for each choice of characteristic root, we find a nonzero solution to the
corresponding linear system
(a11 − λ)A + a12 B
a21 A + (a22 − λ)B
= 0,
= 0.
We will usually have two characteristic roots, and two corresponding solutions
x = A1 e λ 1 t
x = A2 eλ2 t
.
λ1 t and
y = B1 e
y = B2 eλ2 t
According to the principle of superposition, the general solution is then
x = c1 A1 eλ1 t + c2 A2 eλ2 t ,
y = c1 B1 eλ1 t + c2 B2 eλ2 t .
2. Use the procedure outlined above to find the general solution to
dx/dt =
dy/dt =
3x + 3y,
5x + y.
B. Complex roots:
The characteristic equation of a linear system
dx/dt = a11 x + a12 y,
dy/dt = a21 x + a22 y.
sometimes has complex roots,
ceding section works just fine,
simple tricks make the case of
the case of real roots.
For example, to carry out
homogeneous linear system,
but the method we have described in the preso long as one uses complex arithmetic. A few
complex roots no more difficult to handle than
the procedure described above for solving the
dx/dt = x + y,
dy/dt = −2x + 3y,
we need to solve the characteristic equation
1−λ
1
−2
3−λ
4
= 0.
(6)
or
λ2 − 4λ + 5 = 0.
(1 − λ)(3 − λ) + 2 = 0,
We can solve this equation by using
√ the quadratic formula, or by completing
the square. The result is λ = 2 ± −1 = 2 ± i. For each root λ, we need to
determine the A’s and B’s that satisfy the linear system
(1 − λ)A + 1B
−2A + (3 − λ)B
= 0,
= 0.
In the case where λ = 2 + i, this system is
(−1 − i)A + 1B
−2A + (1 − i)B
=
=
0,
0.
Note that if we multiply the first equation by 1 − i we get the second equation.
If we set A = 1, then it follows from the first equation that B = 1 + i. We thus
obtain a particular solution to (6):
x
y
= e(2+i)t ,
= (1 + i)e(2+i)t .
In order to put this solution into a somewhat more manageable form, we
make use of Euler’s formula eiθ = cos θ + i sin θ. The solution becomes
x =
y =
e2t eit
(1 + i)e2t eit
e2t (cos t + i sin t),
(1 + i)e2t (cos t + i sin t),
=
=
or equivalently,
x =
y =
e2t cos t
+ie2t sin t,
2t
e (cos t − sin t) +ie2t (cos t + sin t).
(7)
This is a complex-valued solution to our system.
3. a. Find a nonzero solution corresponding to the characteristic root λ = 2 − i.
(This should be easy—just change i to −i.)
b. Use the superposition principle to show that the real part and the imaginary
part of the solution (7) is also a solution.
c. Use part b to determine the general solution to (6) in real form.
4. a. Find the general solution to the linear system
dx/dt
= x + 5y,
dy/dt = −x − 3y.
b. Find the particular solution which satisfies the initial conditions
x(0) = 1,
y(0) = 0.
5
C. Phase portraits of linear systems:
We have seen how to find the explicit solutions to the homogeneous linear system
dx/dt
dy/dt
= a11 x + a12 y,
= a21 x + a22 y.
Let us review the procedure. The first step is to calculate the eigenvalues of the
coefficient matrix
a11 a12
A=
.
a21 a22
Usually there will be two distinct eigenvalues λ1 and λ2 . For each eigenvalue
λi , we find a nonzero solution (Ai , Bi ) to the homogeneous linear system
(a11 − λi )A + a12 B
a21 A + (a22 − λi )B
= 0,
= 0.
The general solution is then
x = c1 A1 eλ1 t + c2 A2 eλ2 t ,
y = c1 B1 eλ1 t + c2 B2 eλ2 t .
(8)
We next describe how to sketch the phase portrait for the linear system in the
(x, y)-plane, that is a sketch of the paths traversed by the solution curves in the
(x, y)-plane, in the case where the eigenvalues are real. The line which passes
through the origin x = 0, y = 0 and the point x = Ai , y = Bi is called the
eigenspace corresponding to the eigenvalue λi —we will denote the eigenspace
by Vλi .
5. a. Consider the linear system
dx/dt =
dy/dt =
2x + 4y,
3x − 2y.
(9)
which can be written in the form
dx
= Ax,
dt
where
x=
x
y
and A =
2
3
4
.
−2
Show that the eigenvalues of the matrix A are 4 and −4.
b. Find the eigenspaces
V4 = {x ∈ R2 : (A − 4I)x = 0},
c. Choose nonzero eigenvectors
A1
∈ V4
B1
V−4 = {x ∈ R2 : (A − (−4)I)x = 0}
and
6
A2
B2
∈ V−4 ,
(10)
and write out the general solution
x
A1 4t
A2 −4t
= c1
e + c2
e .
y
B1
B2
d. Sketch the eigenspaces V4 and V−4 in R2 and put arrows along the eigenspaces
to indicate the direction of fluid flow along the eigenspaces. Eliminate the
variable t from x(t) an y(t) and sketch several of the resulting curves relating x
and y, thereby obtaining the phase portrait of the linear system. Is the constant
solution at (0, 0) stable? Why or why not?
6. a. Consider the linear system
dx/dt =
dy/dt =
−x,
2x − 3y.
(11)
which can be written in the form
dx
= Ax,
dt
where
x=
x
y
and A =
−1
2
0
.
−3
(12)
Show that the eigenvalues of the matrix A are −1 and −3.
b. Find the eigenspaces
V−1 = {x ∈ R2 : (A − (−1)I)x = 0},
c. Choose nonzero eigenvectors
A1
∈ V−1
B1
V−3 = {x ∈ R2 : (A − (−3)I)x = 0}
and
A2
B2
∈ V−3 ,
and write out the general solution
x
A1 −t
A2 −3t
= c1
e + c2
e .
y
B1
B2
d. Sketch the eigenspaces V−1 and V−3 in R2 and put arrows along the eigenspaces
to indicate the direction of fluid flow along the eigenspaces. Eliminate the variable t from x(t) an y(t) and sketch several of the resulting curves relating x and
y, thereby obtaining the phase portrait of the linear system. Is the constant
solution at (0, 0) stable? Why or why not?
7. a. Consider the linear system
dx/dt
dy/dt
= −(x − 2),
= 2(x − 2) − 3(y − 3).
7
(13)
Show that this system has a constant solution at the point (2, 3). Can you
sketch the phase portrait of this linear system?
b. Consider the linear system
dx/dt =
dy/dt =
−x + 7,
2x − 3y − 5.
(14)
Find the constant solution to this linear system. Can you sketch its phase
portrait?
8. Mathematical software available on the web can be used to study the qualitative behavior of solutions to homogeneous linear systems for various choices
of coefficient matrix. To do this use the software package (PPLANE 2005.10)
available at:
http://math.rice.edu/∼dfield/dfpp.html
By clicking at a point in the “phase plane” window, you can have the software
sketch a solution curve which starts at that point. Print out some representative
orbits to include in your notebook for each of the following homogeneous linear
systems:
a. Sketch representative orbits for the linear system
dx/dt =
(x − 5) − (y − 2),
dy/dt =
4(x − 5) − (y − 2).
Is the constant solution at (5, 2) stable?
b. Sketch representative orbits for the linear system
dx/dt =
(x − 1) − (y − 1),
dy/dt =
3(x − 1) + 5(y − 1)
Is the constant solution at (1, 1) stable?
c. Sketch representative orbits for the linear system
dx/dt =
(x + 1) − (y + 2),
dy/dt =
5(x + 1) + 5(y + 2)
s the constant solution at (−1, −2) stable?
Remark. A rough idea of the phase portrait to
dx/dt
dy/dt
= a11 (x − x0 ) + a12 (y − y0 ),
= a21 (x − x0 ) + a22 (y − y0 ).
can be inferred from the eigenvalues of the coefficient matrix
a11 a12
A=
.
a21 a22
8
Assuming that the two eigenvalues are distinct and neither is zero, there are
names for the six possible cases:
If the eigenvalues are . . .
positive real numbers
negative real numbers
real numbers with opposite signs
complex with positive real part
complex with negative real part
purely imaginary
the point (x0 , y0 ) is a(n) . . .
expanding node.
contracting node.
saddle point.
expanding spiral.
contracting spiral.
center.
Using the mathematical software, you should be able to come up with pictures
for all six cases.
D. Variation of parameters:
Variations of parameters is a general method which enables us to solve the
nonhomogeneous linear system
dx/dt =
dy/dt =
a11 x + a12 y + f (t),
a21 x + a22 y + g(t),
(15)
once we have found the general solution to the associated homogeneous linear
system
dx/dt = a11 x + a12 y,
dy/dt = a21 x + a22 y.
The idea is simple: Replace the constants of integration c1 and c2 in the general
solution to the associated homogeneous system by variables v1 and v2 , substitute
into (15) and solve for v1 and v2 .
9. a. Find the general solution to the linear system
dx/dt =
−2x − y,
dy/dt =
x − 2y.
Your solution should involve two arbitrary constants of integration.
b. Use variation of parameters to find the general solution to the nonhomogeneous system
dx/dt = −2x − y + e−2t sec t,
dy/dt = x − 2y.
Homework 4. Due Friday, January 29, 2009.
H.4.1.a. Find the general solution to the linear system
dx/dt =
5x − 2y,
dy/dt =
5x − y.
9
b. Find the particular solution which satisfies the initial conditions
x(0) = 0,
y(0) = 1.
10