Question

Why differential equations have infinite eigenvalues?

Answer 1

solutions to the system,

→
x
′
=
A
→
x
where the eigenvalues are repeated eigenvalues. Since we are going to be working with systems in which
A
is a
2
×
2
matrix we will make that assumption from the start. So, the system will have a double eigenvalue,
λ
.

This presents us with a problem. We want two linearly independent solutions so that we can form a general solution. However, with a double eigenvalue we will have only one,

→
x
1
=
→
η
e
λ
t
So, we need to come up with a second solution. Recall that when we looked at the double root case with the second order differential equations we ran into a similar problem. In that section we simply added a
t
to the solution and were able to get a second solution. Let’s see if the same thing will work in this case as well. We’ll see if

→
x
=
t
e
λ
t
→
η
will also be a solution.

To check all we need to do is plug into the system. Don’t forget to product rule the proposed solution when you differentiate!

→
η
e
λ
t
+
λ
→
η
t
e
λ
t
=
A
→
η
t
e
λ
t
Now, we got two functions here on the left side, an exponential by itself and an exponential times a
t
. So, in order for our guess to be a solution we will need to require,

A
→
η
=
λ
→
η
⇒
(
A
−
λ
I
)
→
η
=
→
0
→
η
=
→
0
The first requirement isn’t a problem since this just says that
λ
is an eigenvalue and it’s eigenvector is
→
η
. We already knew this however so there’s nothing new there. The second however is a problem. Since
→
η
is an eigenvector we know that it can’t be zero, yet in order to satisfy the second condition it would have to be.

So, our guess was incorrect. The problem seems to be that there is a lone term with just an exponential in it so let’s see if we can’t fix up our guess to correct that. Let’s try the following guess.

→
x
=
t
e
λ
t
→
η
+
e
λ
t
→
ρ
where
→
ρ
is an unknown vector that we’ll need to determine.

As with the first guess let’s plug this into the system and see what we get.

→
η
e
λ
t
+
λ
→
η
t
e
λ
t
+
λ
→
ρ
e
λ
t
=
A
(
→
η
t
e
λ
t
+
→
ρ
e
λ
t
)
(
→
η
+
λ
→
ρ
)
e
λ
t
+
λ
→
η
t
e
λ
t
=
A
→
η
t
e
λ
t
+
A
→
ρ
e
λ
t
Now set coefficients equal again,

λ
→
η
=
A
→
η
⇒
(
A
−
λ
I
)
→
η
=
→
0
→
η
+
λ
→
ρ
=
A
→
ρ
⇒
(
A
−
λ
I
)
→
ρ
=
→
η
As with our first guess the first equation tells us nothing that we didn’t already know. This time the second equation is not a problem. All the second equation tells us is that
→
ρ
must be a solution to this equation.

It looks like our second guess worked. Therefore,

→
x
2
=
t
e
λ
t
→
η
+
e
λ
t
→
ρ
will be a solution to the system provided
→
ρ
is a solution to

(
A
−
λ
I
)
→
ρ
=
→
η
Also, this solution and the first solution are linearly independent and so they form a fundamental set of solutions and so the general solution in the double eigenvalue case is,

→
x
=
c
1
e
λ
t
→
η
+
c
2
(
t
e
λ
t
→
η
+
e
λ
t
→
ρ
I hope this will help you