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Chapter 7: Systems of Linear DifferentialEquations
Philip Gressman
University of Pennsylvania
Philip Gressman Math 240 002 2014C: Chapter 7 1 / 23
The Beginning
Definition
Every vector function (with values in Rn) which is k-timescontinuously differentiable on an interval I is said to belong toC k(I ,Rn).
Basic Fact
The space C k(I ,Rn) is a vector space over the reals underpointwise addition and scalar multiplication. This vector space isinfinite-dimensional.
Important Transformations
Differentiation maps C k(I ,Rn) to C k−1(I ,Rn) for k > 0 and mapsC∞(I ,Rn) to itself. If A(t) is a k-times differentiablematrix-valued function, then multiplication by A on the left alsomaps C k(I ,Rn) to itself.
Philip Gressman Math 240 002 2014C: Chapter 7 2 / 23
Linear Independence
Definition
Vector functions x1, . . . , x` are, as always, called linearlyindependent when there are no constants c1, . . . , c` for which
c1x1 + · · · c`x` = 0
except c1 = · · · = c` = 0.
IMPORTANT: Remember that when we say that a vectorfunction equals zero, that means it equals the old-fashioned zerovector at every single point.Linear DEPENDENCE is hard: if x1, . . . , x` are linearlyindependent at even a single point, then as vector functions theyare linearly independent. The converse is not true: they mighteven be linearly dependent at every point and still be linearlyindependent as vector functions.
Philip Gressman Math 240 002 2014C: Chapter 7 3 / 23
Linear Independence
Linear Independence
We say that time-dependent vectors ~X1, . . . , ~Xn are linearlyindependent on an interval I when the only constants c1, . . . , cnsuch that
c1~X1(t) + c2~X2(t) + · · ·+ cn~Xn(t) ≡ 0
on the entire interval are all zeros.
Ind:
[1t
],
[01
],
[sin tcos t
]Dep:
[sin2 t
1
],
[1
1 + t
],
[cos2 tt
]
Ind:
[0t
],
[0t2
].
Philip Gressman Math 240 002 2014C: Chapter 7 4 / 23
The Wronskian
Given time-dependent column vectors ~X1, . . . , ~Xn each of length n,we form the Wronskian to be the determinant of the matrix whosecolumns are exactly ~X1, . . . , ~Xn, i.e.,
W (~X1, . . . , ~Xn) = det(~X1, . . . , ~Xn).
FACT: If the Wronskian is nonzero even at a single point, then~X1, . . . , ~Xn must be linearly independent. In fact, they might evenbe linearly independent when the Wronskian is always zero, but forsolutions to first-order systems of ODEs this pathology does nothappen.
Philip Gressman Math 240 002 2014C: Chapter 7 5 / 23
Linear ODE Systems: §7.1–7.2
First-Order Linear Systems
A first-order linear system may be written in the form
d
dt~X = A(t)~X + ~G (t).
Here A is an n × n matrix whose entries may or may not dependon t. ~G (t) is a column vector of length n which is fully describedin the problem itself, and ~X (t) is an unknown column vector oflength n whose entries may depend on t.
General Solutions
The general solution is a complete listing of all solution vectors.
Initial Value Problem
This is the specific solution for which X (0) is prescribed.
Philip Gressman Math 240 002 2014C: Chapter 7 6 / 23
Important General Facts about First-Order Systems
• Higher-order systems of ODEs can always be recast as asystem of first-order ODEs with more unknown functions.
• Systems of ODEs can always be solved by elimination; this is,however, a labor-intensive way to do it since unknownconstants will be related and you’ll have to do a lot of linearequation solving.
Philip Gressman Math 240 002 2014C: Chapter 7 7 / 23
Homogeneous Equations
Definition
The equation ddt~X = A(t)~X is called homogeneous. If your
equation is given as ddt~X = A(t)~X + ~G (t) for some nonzero ~G (t),
then the equation ddt~X = A(t)~X is called the associated
homogeneous first-order system.
Superposition Principle
If ~X1(t) and ~X2(t) are solutions of the homogeneous ODEddt~X = A(t)~X , then c1~X1 + c2~X2 will also be a solution. The same
can be said for any linear combination of any number of solutions(i.e., more than two solutions).
Philip Gressman Math 240 002 2014C: Chapter 7 8 / 23
7.3: Theory of First-order Systems
Theorem: Existence and Uniqueness
The IVP x(t0) = x0, x ′ = A(t)x(t) + b(t) for x , b vector-valuedfunctions of time and A a matrix-valued function of time, has aunique C 1 solution on any interval I containing x0 when A and bare continuous.
Consequences
Theorem: When x(t) is a time-dependent vector in Rn, thegeneral solution of x ′(t) = A(t)x(t) on any interval is ann-dimensional vector space.Theorem: Solutions x1, . . . , xn are linearly independent if and onlyif the Wronskian is never zero.Theorem: If xp is any solution to x ′ = Ax + b, then the generalsolution to this ODE is given by x = xc + xp where xc ranges overall solutions of the associated homogeneous equation.
Philip Gressman Math 240 002 2014C: Chapter 7 9 / 23
Inhomogeneous Systems
Finding the general solution of an inhomogeneous system is onlyslightly more difficult than solving a homogeneous one.
1 You must first find some solution ~Xp. It is called a particularsolution.
2 The general solution of the inhomogeneous system will alwaysbe of the form
~X = c1~X1 + · · ·+ cn~Xn + ~Xp
Where ~X1, . . . , ~Xn are a fundamental set of solutions (i.e., acomplete set) for the associated homogeneous system.
Philip Gressman Math 240 002 2014C: Chapter 7 10 / 23
Homogeneous Linear Systems: §7.4
We consider a system of ODEs with the form
d
dt~X = A~X
where ~X is a column vector of length n and A is an n × n matrixwith constant entries. We begin by looking for very simplesolutions, then use the superposition principle to describe the morecomplicated ones.
The Simplest Case
An example of a very simple solution is one whose direction doesnot change (only the magnitude). It would be expressible in theform
~X (t) = f (t)~E
where ~E is a constant vector and f is some unknown function of t.
Philip Gressman Math 240 002 2014C: Chapter 7 11 / 23
When you assume that a solution has some special form, it isknown as an ansatz. It’s a completely reasonable question to askand mathematically rigorous because you might end up learningthat no such solutions exist. For us, we plug our ansatz
Our Ansatz
~X (t) = f (t)~E
into the equation and get
f ′(t)~E = f (t)A~E ⇒ A~E =f ′(t)
f (t)~E .
If the equation must be true at all times, then f ′(t)f (t) must be
constant. Call the constant λ. We arrive at the eigenvectorequation...
Philip Gressman Math 240 002 2014C: Chapter 7 12 / 23
The Conclusion
If ~E is an eigenvector of A with eigenvalue λ, then
~X (t) = Ceλt~E
solves the first-order system
d
dt~X = A~X .
Moreover, linearly independent eigenvectors give linearlyindependent solutions of the system.
If A is n× n and has n linearly independent eigenvectors ~E1, . . . , ~En
with eigenvalues λ1, . . . , λn, then the general solution of thesystem will be
~X (t) = C1eλ1t~E1 + · · ·+ Cne
λnt~En.
Philip Gressman Math 240 002 2014C: Chapter 7 13 / 23
Complex Eigenvalues
If A is a real matrix with complex eigenvalue λ = α + iβ andeigenvector ~E = ~Ere + i~Eim, Then
~X (t) = eαt+iβt(~Ere + i~Eim)
will be a solution. This can only happen if the real parts andimaginary parts are each solutions by themselves. We conclude
~Xre(t) = eαt(cosβt)~Ere − eαt(sinβt)~Eim
~Xim(t) = eαt(sinβt)~Ere + eαt(cosβt)~Eim
are linearly independent real solutions of the system of ODEs.
Philip Gressman Math 240 002 2014C: Chapter 7 14 / 23
“Missing” Eigenvectors
If A does not have n eigenvectors, the ansatz gives only a partialanswer and we end up missing some solutions. We fix this bymaking a better ansatz (with increasing complexity depending onhow bad the situation is). For example:
New Ansatz
~X (t) = eλt~E2 + teλt~E .
Plug it into ddt~X = A~X , and we get
eλt(λ~E2 + (1 + λt)~E
)= eλt
(A~E2 + tA~E
).
We must have A~E = λ~E and (A− λI ) ~E2 = ~E .
We must take ~E to be an eigenvector, but ~E2 satisfies a differentequality and is called a generalized eigenvector.
Philip Gressman Math 240 002 2014C: Chapter 7 15 / 23
“Missing” Eigenvectors in General
General Ansatz
~X (t) = eλt[tn
n!~En + · · ·+ t~E1 + ~E0
]Generalized Eigenvectors
The general ansatz will solve the system when
(A− λI )~En = 0,
(A− λI )~En−1 = ~En,
...
(A− λI )~E0 = ~E1.
Philip Gressman Math 240 002 2014C: Chapter 7 16 / 23
§7.8: Solution by diagonalization
Given the first-order system
d
dt~X = A~X
one useful technique you should be able to use is solution bydiagonalization. Here the idea is like substitution: you assume~X = P ~Y for some matrix P and then try to solve for Y instead ofX :
d
dtP ~Y = A(P ~Y )⇒ d
dt~Y =
(P−1AP
)~Y .
So if A is diagonalizable, you can do the following:
1 Solve the systemd
dt~Y = D ~Y
where D is the diagonalization of A.
2 To solve the original system, simply set ~X = P ~Y .
Philip Gressman Math 240 002 2014C: Chapter 7 17 / 23
§7.8: Solution by exponentiation
Matrix exponentiation
eAt := I + tA +t2
2A2 +
t3
3!A3 + · · ·
Solution of the IVP
There is exactly one solution to the IVP
d
dt~X (t) = A~X (t), ~X (0) = ~V
and it equals ~X (t) = eAt ~V .
There are several tricks that you might use to carry out the infinitesum and write down a simple formula that equals eAt .
Philip Gressman Math 240 002 2014C: Chapter 7 18 / 23
§7.8: Solution by exponentiation
1 Use diagonalization to find a pattern for the powersA,A2,A3,A4, . . .. The exponential of a diagonal matrix issimply the exponential of each of the diagonal entries.
2 Solve it like a system of equations: You can writeeAt = b0(t)I + b1(t)A + · · ·+ bn−1(t)An−1 for unknownfunctions b0, . . . , bn−1. Often you can solve for thesefunctions using the fact that
eλt = b0(t) + b1(t)λ+ · · ·+ bn−1(t)λn−1
for each of the eigenvalues λ (note that you will be able tosolve when you have n distinct eigenvalues).
3 For 2× 2: if there is only one eigenvalue and only oneeigenvector, then the matrix exponential will take the form
eAt = eλt [I + t(A− λI )] .
Philip Gressman Math 240 002 2014C: Chapter 7 19 / 23
Phase Portraits
A phase portrait is a simultaneous plotting of several solutions ofan ODE. The axes are the coordinates of the vector and the timevariable is suppressed.
Two Eigenvals. < 0 Two Eigenvals. > 0 Mixed Signs
Pictures from Paul’s Online Math Notes
Philip Gressman Math 240 002 2014C: Chapter 7 20 / 23
Phase Portraits for Complex Eigenvalues
When eigenvalue λ = α + iβ:
α < 0 α = 0 α > 0
Pictures from Paul’s Online Math Notes
Philip Gressman Math 240 002 2014C: Chapter 7 21 / 23
Phase Portraits for “Missing” Eigenvectors
λ < 0 λ > 0
Pictures from Paul’s Online Math Notes
Philip Gressman Math 240 002 2014C: Chapter 7 22 / 23
Inhomogeneous Systems/Undetermined Coefficients
Just like for single inhomogeneous ODEs, one can often make aneducated guess about the form of the particular solution:
d
dt~X =
[2 −10 3
]~X +
[1 + e−t
2
]⇒ ~Xp = ~V1 + ~V2e
−t
The structure of the method is still the same:
1 Expand the inhomogeneous terms to look like vectors timesconstants, exponentials, powers of t, and/or sines and cosines.
2 Use the tables from undetermined coeffs and/or your intuitionto guess the form of the particular solution.
3 Instead of multiplying the terms in ~Xp by unknown constants,multiply by unknown vectors.
4 Try to solve for the unknown vectors. If it doesn’t work, tryincluding more terms in your guess with higher powers of tattached.
Philip Gressman Math 240 002 2014C: Chapter 7 23 / 23