What are the Eigenvalues and Stability of a Second Order Differential Equation?

[Kamekui]

Homework Statement

Consider the second order equation x''+ ax' + bx = 0:
(a) Convert the equation to a 2 x 2 system.
(b) Compute the eigenvalues.
(c) In circuit and spring problems, both constants are nonnegative. Assume that they
are actually positive, and show that the eigenvalues have a negative real part, and conclude
that the trivial solution is asymptotically stable.

Homework Equations

The Attempt at a Solution

(a). Let x1=x, x2=x'

x1'=x2, x2'=-ax'-b''


{{0,1},{-b,-a}}=A



(b) det(A-λI)=
\begin{bmatrix}
-λ & 1 \\
-b & -a-λ\\

\end{bmatrix}

= x²+aλ+b

→λ=-a/2+-√(a²-4b)/2

-a/2 is real and negative... but I just don't feel like I'm going about getting the solution the correct way.

(c) No idea...

 

[Mark44]

Homework Statement

Consider the second order equation x''+ ax' + bx = 0:
(a) Convert the equation to a 2 x 2 system.
(b) Compute the eigenvalues.
(c) In circuit and spring problems, both constants are nonnegative. Assume that they
are actually positive, and show that the eigenvalues have a negative real part, and conclude
that the trivial solution is asymptotically stable.

Homework Equations

The Attempt at a Solution

(a). Let x1=x, x2=x'

x1'=x2, x2'=-ax'-b''

Typo above. The 2nd equation should be x₂' = bx₁ - ax₂.

{{0,1},{-b,-a}}=A



(b) det(A-λI)=
\begin{bmatrix}
-λ & 1 \\
-b & -a-λ\\

\end{bmatrix}

= x²+aλ+b

Another typo above. The equation you're solving is λ² + aλ + b = 0.

→λ=-a/2+-√(a²-4b)/2

-a/2 is real and negative... but I just don't feel like I'm going about getting the solution the correct way.

Other than the typos I pointed out, I think you're on the right track. Based on what you have here, and not knowing the relative sizes of a and b, I don't see any way of showing that the part under the radical is either positive or negative. Is there some information you didn't include?

 

[Kamekui]

Unfortunately, no. I directly copied and pasted the directions i here. The only thing that makes sense to me beyond the given directions is that 4b>a^2 so we would get a definite Real and Complex part.

 

[Kamekui]

Alright, I went back and worked out part (b) to this:

det(A-λI)=λ²+aλ+b

→λ=(1/2)*(-a±√(a²-4b))

Now, there are three possible cases:

√(a²-4b)=0
√(a²-4b)>0
√(a²-4b)<0

Case 1: If √(a²-4b))=0 then trivially, λ= -a/2
By assumption, a ε ℝ and a>0 → -a/2<0


Case 2: If √(a²-4b)>0, then sub-cases arise:

(i). Consider λ=(1/2)*(-a-√(a²-4b))
Claim (1/2)*(-a±√(a²-4b))<0
Proof Assume (1/2)*(-a±√(a²-4b))>0
-a/2 > √(a²-4b)/2

By assumption √(a²-4b)>0 and -a/2<0, so it follows
-a/2 >/ √(a²-4b)/2

Therefore a contradiction has been reached and it follows that:
λ=(1/2)*(-a±√(a²-4b))<0


(ii.) Consider λ=(1/2)*(-a+√(a²-4b))
Claim: (1/2)*(-a+√(a²-4b))<0
Proof: Assume (1/2)*(-a+√(a²-4b))>0

√(a²-4b)/2 > a/2
(a²-4b)/4 > a²/4

By assumption b>0 → -4b<0 → a²-4b< a²

Therefore a contradiction has been reached and it follows that:
λ=(1/2)*(-a+√(a²-4b))<0



Case 3: If √(a²-4b)<0, then
λ=(1/2)*(-a±√(a²-4b))

Since √(a²-4b)<0, we can re-write this as
i√(4b-a²)

So λ=(1/2)*(-a±i√(4b-a²))

and we can break the solution into real and complex parts. Therefore,
the real part of the λ is -a/2<0


So this takes care of part 2; however, I'm not sure how to go about part (c), any help appreciated (Sorry about not using latex, I tried and it didnt go well..)

 

[Mark44]

Alright, I went back and worked out part (b) to this:

det(A-λI)=λ²+aλ+b

→λ=(1/2)*(-a±√(a²-4b))

Now, there are three possible cases:

√(a²-4b)=0
√(a²-4b)>0
√(a²-4b)<0

You need to lose the square root symbols. The three cases are the possible values of the discriminant, the quantity inside the radical.

Case 1: If √(a²-4b))=0 then trivially, λ= -a/2
By assumption, a ε ℝ and a>0 → -a/2<0

You don't need to say all of this.
If a²-4b = 0, then λ= -a/2.
Since by assumption a > 0, then -a/2 < 0.
.

Case 2: If √(a²-4b)>0, then sub-cases arise:

(i). Consider λ=(1/2)*(-a-√(a²-4b))
Claim (1/2)*(-a±√(a²-4b))<0
Proof Assume (1/2)*(-a±√(a²-4b))>0
-a/2 > √(a²-4b)/2

By assumption √(a²-4b)>0 and -a/2<0, so it follows
-a/2 >/ √(a²-4b)/2

Therefore a contradiction has been reached and it follows that:
λ=(1/2)*(-a±√(a²-4b))<0


(ii.) Consider λ=(1/2)*(-a+√(a²-4b))
Claim: (1/2)*(-a+√(a²-4b))<0
Proof: Assume (1/2)*(-a+√(a²-4b))>0

√(a²-4b)/2 > a/2
(a²-4b)/4 > a²/4

By assumption b>0 → -4b<0 → a²-4b< a²

Therefore a contradiction has been reached and it follows that:
λ=(1/2)*(-a+√(a²-4b))<0



Case 3: If √(a²-4b)<0, then
λ=(1/2)*(-a±√(a²-4b))

This -- √(a²-4b)<0 -- makes no sense. If the quantity under the radical is >= 0, its square root will be >= 0.

If the quantity under the radical is < 0, then you have an imaginary number, which cannot be compared to zero using < or >.

Since √(a²-4b)<0, we can re-write this as
i√(4b-a²)

So λ=(1/2)*(-a±i√(4b-a²))

and we can break the solution into real and complex parts. Therefore,
the real part of the λ is -a/2<0


So this takes care of part 2; however, I'm not sure how to go about part (c), any help appreciated (Sorry about not using latex, I tried and it didnt go well..)

For the c part, I'm not exactly sure what is being asked, but the systems involved are physical systems. For a damped spring-mass system, could you have an undriven system in which the mass started oscillating in cycles of greater and greater amplitude? This is connected to whether the eigenvalues are real or complex.

Similarly, for the inductor-capacitor system with resistance, could you have an undriven system in which the voltage oscillated higher and higher?

 

[Kamekui]

Ok, this is where I am getting confused with you in part (c): In circuit and spring problems, both constants are nonnegative. Assume that they
are actually positive, and show that the eigenvalues have a negative real part, and conclude that the trivial solution is asymptotically stable.

I see your point about Case 1,2,3.

As for part (c), we need to take the eigenvalues we found earlier, consider the different possibilities (i.e. case 1,2,3) and find the matrix exponential and show that it behaves in an asymptotically stable manner at the origin.

Now, the possible eigenvalues would be: λ=(1/2)*(-a±√(a²-4b))

Plug λ into A-λI to obtain a new matrix (call it Z). Row-reduce Z to obtain an eigenvector. Repeat this until you have two linearly independent eigenvectors. Use these eigenvectors to form a new matrix (call it T) such that T^-1AT=J (Jordan Form)

Break J into its diagonal(D) and nilpotent(N) parts such that J= D+N

e^At=Te^JtT^-1, so
Te^Dte^NtT^-1

If this is done correctly I should get solutions that are asymptotically stable. Does this process seem correct? (I wanted your opinion before I work all that out)

 

[Ray Vickson]

Ok, this is where I am getting confused with you in part (c): In circuit and spring problems, both constants are nonnegative. Assume that they
are actually positive, and show that the eigenvalues have a negative real part, and conclude that the trivial solution is asymptotically stable.

I see your point about Case 1,2,3.

As for part (c), we need to take the eigenvalues we found earlier, consider the different possibilities (i.e. case 1,2,3) and find the matrix exponential and show that it behaves in an asymptotically stable manner at the origin.

Now, the possible eigenvalues would be: λ=(1/2)*(-a±√(a²-4b))

Plug λ into A-λI to obtain a new matrix (call it Z). Row-reduce Z to obtain an eigenvector. Repeat this until you have two linearly independent eigenvectors. Use these eigenvectors to form a new matrix (call it T) such that T^-1AT=J (Jordan Form)

Break J into its diagonal(D) and nilpotent(N) parts such that J= D+N

e^At=Te^JtT^-1, so
Te^Dte^NtT^-1

If this is done correctly I should get solutions that are asymptotically stable. Does this process seem correct? (I wanted your opinion before I work all that out)

From the expressions you have for the eigenvalues, and for a, b >0, it is easy to see that both eigenvalues have negative real parts. Look at three cases: (1) a^2 - 4b > 0. Then both eigenvalues are real, and you need to show both are < 0. (2) a^2 - 4b = 0: easy. (3) a^2 - 4b < 0. Now the eigenvalues are complex, but what are their real parts?

RGV

 

[Kamekui]

Guys thank you for all the help! I was just sitting here and I finally saw the solution, I was just over thinking this problem big time.

I determined the Jordan normal form for A, found the matrix exponential for it. The trivial solution for the matrix exponential displays asymptotically stable behavior. Confirmed with the instructor today. Again, thanks for all the help!