It is convenient to split things up into homogeneous problems and then look for particular solutions. First
of all, is a definition of the general solution to a homogeneous problem.
Definition 24.5.1Let A be an n×n matrix. The general solution to the homogeneous problem is definedto be all solutions to the equation
′
x = Ax
Note how there is no initial condition. We just look for all solutions to the above differential equation.
The following theorem describes all of these solutions.
Theorem 24.5.2The general solutionto the homogeneous problem x^{′} = Ax consists of all vectorsof the form Φ
(t)
c where c is a vector in F^{n}and Φ
(t)
is the fundamental matrix of A.
Proof: Let x be a solution to the equation. Then x
(0)
= c for some c. Consider Φ
(t)
c and x
(t)
both
solve x^{′} = Ax the first doing so because
Φ′(t)c = AΦ (t) c
Thus Φ
(t)
c and x
(t)
both solve the same differential equation and have the same initial condition.
Therefore, these are the same and this shows that the set of solutions to x^{′} = Ax consists of Φ
(t)
c for
c ∈ F^{n} as claimed. ■
Example 24.5.3Find the general solution to x^{′} = Ax where
This was done by a computer algebra system. Now take invers Laplace transforms of this to get the
fundamental matrix Φ
(t)
=
( et(3t+ 1) tet 2tet tet )
| t t t t |
|| − 3te − e (t− 1) − 2te − te ||
( et − e2t − 6tet et − e2t − 2tet 2et − e2t − 4tet et − e2t − 2tet )
2e2t − 2et + 6tet 2e2t − 2et + 2tet 2e2t − 2et + 4tet 2e2t − et + 2tet
therefore, the general solution is of the form Φ
(t)
c where c ∈ F^{n}. In other words, it is the set of linear
combinations of the columns of Φ
(t)
. Since Φ
(t)
^{−1} = Φ
(− t)
, the columns are linearly independent and
this shows that the dimension of the solution space is n if A is n×n. In the above example, the dimension
of the general solution is 4 because A is 4 × 4.
Now consider the general solution to
x′ = Ax + f
There is a very easy way to describe this. It is just the general solution to x^{′} = Ax added to x_{p} where x_{p} is
any particular solution to the above nonhomogeneous equation.
Theorem 24.5.4The general solution to x^{′} = Ax + fconsists of all solutions to this equation. Itis of the form Φ
(t)
c + x_{p}where x_{p}is a particular solution meaning x_{p}^{′} = Ax_{p} + f.
Proof:Anything of the form Φ
(t)
c + x_{p} is a solution to x^{′} = Ax + f. It remains to verify that this is
the only way it can happen. Let z^{′} = Az + f and consider
(z − xp)
. Then
′ ′ ′
(z − xp )= z − xp = Az +f− (Axp + f) = A (z − xp)
and so z − x_{p} is a solution to x^{′} = Ax. Therefore, from Theorem 24.5.2, there exists c ∈ F^{n} such that
z
(t)
= Φ
(t)
c + x_{p}
(t)
. ■
Example 24.5.5Find the general solution to
′
x = Ax + f
where
( ) ( t )
2 − 4 − 2 e sint
A = |( 3 − 4 − 2 |) , f (t) = |( e−tcost|)
− 3 10 6 t
First I will find the fundamental matrix using the following syntax.
The first line starting with “simplify” will give the fundamental matrix and the second will give a
particular solution. The claim about the first was already considered. As to the second, if x is a particular
solution with zero initial condition,
sX (s) = AX (s)+ F (s)
In the above syntax, the F
(s)
comes from laplace(f). Then
X (s) = (sI − A)−1F (s)
and this involves inv(s*b-c)*laplace(f) in the above syntax. Then you do ilaplace to this thing to get a
particular solution. Try it. You will get a horrendous mess but Matlab has no problem in doing
it.
You see, you can really solve things with these methods quite easily. However, what if you
can’t find the inverse Laplace transform of
(sI − A)
^{−1}F
(s)
? Even in this case, there is no
real problem because you can express the inverse Laplace transform of this expression as an
integral. This involves Formula 19 in the table. This integral is called a convolution integral.
Theorem 24.5.6Suppose F
(s)
is the Laplace transform of f
(t)
and G
(s)
is the Laplace transform ofg
(t)
. Then F
(s)
G
(s)
is the Laplace transform of
∫ ∫
t t
0 f (u)g(t− u)du = 0 f (t− u)g(u)du ≡ f ∗ g(t)
Proof: To be rigorous, you really need to replace improper integrals with integrals over a finite interval
and then take a limit, but the idea is essentially as follows:
∫ ∞ ∫ t ∫ ∞ ∫ ∞
e− st f (t − u )g(u)dudt = e− stf (t− u)g (u) dtdu
0 0 ∫0∞ ∫u∞
= e− s(t−u)f (t − u)e−sug (u )dtdu
∫0 ∫u
∞ ∞ − sr −su
= 0 0 e f (r)e g(u)drdu
∫ ∞ (∫ ∞ )
= e−sug(u) e−srf (r)dr du
∫0∞ 0∫ ∞
= e−srf (r) dr e− sug (u) du
0 0
= F (s)G (s)
The other formula follows from changing the variable. ■
Note that F
(s)
could be a matrix and G
(s)
could be a vector. You simply need the multiplication to
make sense.
Example 24.5.7Find a particular solution to
x ′(t) = Ax (t)+f (t)
where
( )
t
f (t) = |( t |)
(2 )
ln t + 1
and
( )
− 1 0 − 6
A = |( − 2 1 − 5 |)
1 0 4
There is no way you will find a decent closed form solution to this in terms of elementary functions
because of the horrible ln
(t2 + 1)
but this is not really a problem because you can find a particular
solution in terms of a convolution. You just need to find the fundamental matrix which is not hard. I will
use the following to find the fundamental matrix.
syms s t; b=eye(3); c=[-1 0 -6;-2 1 -5;1 0 4];
simplify(ilaplace(inv(s*b-c)))
This yields for Φ
(t)
( )
− et(2et − 3) 0 − 6et(et − 1)
|( − et(t+ et − 1) et − et(2t+ 3et − 3)|)
t t t t
e (e − 1) 0 e (3e − 2)
Then using the above theorem,x_{p}
(t)
=
( )
∫ ( − eu (2eu − 3) 0 − 6eu(eu − 1) ) t− u
t| − eu(u +eu − 1) eu − eu (2u + 3eu − 3) | || t− u || du
0 ( u u u u ) ( ( 2 ) )
e (e − 1) 0 e (3e − 2) ln (t− u) + 1
This gives a perfectly good description of a particular solution. Thus the general solution is of the
form
Here c is an arbitrary vector in F^{n}. Note how this is essentially a return to the notion of the variation of
constants formula presented earlier.
Appendix A Integrals
The integral is needed in order to consider the basic questions of differential equations. Consider the
following initial value problem, differential equation and initial condition.
A′(x) = ex2,A (0) = 0.
So what is the solution to this initial value problem and does it even have a solution? More generally, for
which functions, f does there exist a solution to the initial value problem, y^{′}
(x)
= f
(x)
,y
(0)
= y_{0}? The
solution to these sorts of questions depend on the integral. Since this is usually not done well in beginning
calculus courses, I will give a presentation of the theory of the integral. I assume the reader is familiar
with the usual techniques for finding antiderivatives and integrals such as partial fractions, integration by
parts and integration by substitution. These topics are usually done very well in beginning calculus
courses.
The integral depends on completeness of ℝ.
Definition A.0.1One of the equivalent definitions of completeness of ℝ is that if S is anynonempty subset of ℝ which is bounded above, then there exists a least upper bound for S and ifS is bounded below, then there exists a greatest lower bound for S.The least upper bound of S isdenoted as sup
(S )
or sometimes as l.u.b.
(S )
while the greatest lower bound of S is denoted as
inf
(S)
sometimes as g.l.b.
(S )
. If there is no upper bound for S we say sup
(S )
= ∞. If there is nolower bound, we say inf
(S )
= −∞.
The words mean exactly what they say. sup
(S)
is a number with the property that s ≤ sup
(S)
for all
s ∈ S and out of all such “upper bounds” it is the smallest. inf
(S)
has the property that
inf
(S)
≤ s for all s ∈ S and if l ≤ s for all s ∈ S, then l ≤ inf
(S)
. In words, it is the largest lower
bound and sup
(S)
is the smallest upper bound. Here the meaning of small and large are as
follows. To say that x is smaller than y means that x ≤ y which also says that y is larger than
x.
A consequence of this axiom is the nested interval lemma, Lemma A.0.2.
Lemma A.0.2Let I_{k} =
[ak,bk]
and supposethat for all k = 1,2,
⋅⋅⋅
,
Ik ⊇ Ik+1.
Then there exists a point, c ∈ ℝ which is an element of every I_{k}. If
lim bk − ak = 0
k→ ∞
then there is exactly one point in all of these intervals.
Proof:Since I_{k}⊇ I_{k+1}, this implies
ak ≤ ak+1,bk ≥ bk+1. (1.1)
(1.1)
Consequently, letting k ≤ l,
l l l k
a ≤ a ≤ b ≤ b . (1.2)
(1.2)
Thus
{ l } { l } k
c ≡ sup a : l = 1,2,⋅⋅⋅ = sup a : l = k,k+ 1,⋅⋅⋅ ≤ b
because b^{k} is an upper bound for all the a^{l}. Then c ≥ a^{l} for all l. In other words x ≥ a^{k} for all k. Also
c ≤ b^{k} for all k. Therefore, c ∈
[ k k]
a ,b
for all k.
If the length of these intervals converges to 0, then there can be at most one point in their intersection
since otherwise, you would have two different points c,d and the length of the k^{th} interval would then be at
least as large as
|d− c|
but both of these points would need to be in intervals having smaller length than
this which can’t happen. ■
Corollary A.0.3Suppose
{xn}
is a sequence contained in
[a,b]
. Then there exists x ∈
[a,b]
anda subsequence
{xnk}
such that lim_{k→∞}x_{nk} = x.
Proof:Consider a sequence of closed intervals contained in
[a,b]
,I_{1},I_{2},
⋅⋅⋅
where I_{k+1} is one half of
I_{k} and each I_{k} contains x_{n} for infinitely many values of n. Thus I_{1} =
[a,b]
,I_{2} is either
[a, a+b]
2
or
[a+b,b]
2
, depending on which one contains x_{n} for infinitely many values of n. If both intervals have this
property, just pick one. Let x_{nk}∈ I_{k} and let x_{nk+1}∈ I_{k+1} with n_{k+1}> n_{k}. This is possible to do
because each I_{k} contains x_{n} for infinitely many values of n. Then by the nested interval lemma,
there exists a unique point x contained in all of these intervals and
|x − x |
nk
<
(b− a)
∕2^{k}.
■
This has an obvious corollary in the case that
[a,b]
is replaced with Q = ∏_{i=1}^{n}
[a,b ]
i i
the Cartesian
product of closed intervals.
Corollary A.0.4Suppose Q = ∏_{i=1}^{n}
[a ,b]
i i
and let
{x }
k
be a sequence which is contained in Q.Then there exists x ∈ Q and a subsequence
{x }
kj
such that lim_{j→∞}x_{kj} = x.
Proof: Let x_{k} =
( )
x1k,x2k,⋅⋅⋅,xnk
. By Corollary A.0.3 there exists a subsequence still denoted by
{xk}
such that lim_{k→∞}x_{k}^{1} = x_{1}∈
[a1,b1]
. Then by the same corollary, there is a further
subsequence, still denoted as
{xk}
such that in addition to this, x_{k}^{2}→ x_{2}∈
[a2,b2]
. Continuing
this way, taking subsequences, we eventually obtain a subsequence, still denoted as
{xk}
such
that
kli→m∞ xik = xi ∈ [ai,bi]
for each i ≤ n. However, from the way we measure distance in ℝ^{n} as
(∑ )1 ∕2
|x − y| = (xi − yi)2
i
The above limits say exactly that lim_{k→∞}
|xk − x|
= 0 and that x ∈ Q. ■
The next corollary is the extreme value theorem from calculus.
Corollary A.0.5If f :
[a,b]
→ ℝ is continuous, then there exists x_{M}∈
[a,b]
such that
f (xM ) = sup{f (x) : x ∈ [a,b]}
and there exists x_{m}∈
[a,b]
such that
f (xm) = inf{f (x) : x ∈ [a,b]}
Proof:From the definition of inf
{f (x) : x ∈ [a,b]}
, there exists x_{n}∈
[a,b]
such that
f (xn) ≤ inf{f (x) : x ∈ [a,b]}+ 1∕n
That is, lim_{n→∞}f
(x )
n
= inf
{f (x) : x ∈ [a,b]}
. This is called a minimizing sequence. Therefore, there is a
subsequence
{x }
nk
which converges to x ∈
[a,b]
. By continuity of f it follows that
inf{f (x ) : x ∈ [a,b]} = lim f (xnk) = f (x)
k→ ∞
The case where f achieves its maximum is similar. You just use a maximizing sequence. ■
You might write down a corresponding generalization in case
Proof: The Cauchy sequence is contained in some closed interval
[a,b]
thanks to Theorem ??. By
Corollary A.0.3, there is a subsequence of the Cauchy sequence which converges to some x ∈
[a,b]
. Then by
Theorem ??, the original Cauchy sequence converges to x. ■
Actually, the convergence of every Cauchy sequence is equivalent to completeness and so it gives
another way of defining completeness in contexts where no order is available.
The Riemann integral pertains to bounded functions which are defined on a bounded interval. Let
[a,b]
be a closed interval. A set of points in
[a,b]
,
{x0,⋅⋅⋅,xn}
is a partition if
a = x0 < x1 < ⋅⋅⋅ < xn = b.
Such partitions are denoted by P or Q.
Definition A.0.7A function f :
[a,b]
→ ℝ is bounded if the set of values of f is contained in someinterval. Thus
sup{f (x) : x ∈ [a,b]} < ∞, inf{f (x) : x ∈ [a,b]} > − ∞
. Letting P denote a partition,
∥P ∥ ≡ max {|xi+1 − xi| : i = 0,⋅⋅⋅,n− 1}.
A Riemann sum for a bounded f corresponding to a partition P =
{x ,⋅⋅⋅,x }
0 n
is a sum of theform
∑ ∑n
f ≡ f (yi)(xi − xi−1)
P i=1
where y_{i}∈
[xi−1,xi]
. Then there are really many different Riemann sums corresponding to a givenpartition, depending on which y_{i}is chosen.
For example, suppose f is a function with positive values. The above Riemann sum involves adding
areas of rectangles. Here is a picture:
PICT
The area under the curve is close to the sum of the areas of these rectangles and one would imagine that
this would become an increasingly good approximation if you included more and narrower
rectangles.
Definition A.0.8A bounded function defined on an interval
[a,b]
is Riemann integrablemeans that there exists a number I such that for every ε > 0, there exists a δ > 0 such thatwhenever
∥P∥
< δ, and∑_{P}f is some Riemann sum corresponding to this partition, it followsthat
| |
||∑ ||
|| f − I|| < ε
P
This is written as
lim ∑ f = I
∥P ∥→0 P
and when this number exists, it is denoted by
∫
b
I = a f (x) dx
One of the big theorems is on the existence of the integral whenever f is a continuous function. This
requires a technical lemma which follows.
Lemma A.0.9Let f :
[a,b]
→ ℝ be continuous. Then for every ε > 0 there exists a δ > 0 suchthat if
|x− y|
< δ,x,y ∈
[a,b]
, it follows that
|f (x)− f (y)|
< ε.
Proof:If not, then there exists ε > 0 and x_{n},y_{n},