Eigenvalues and vectors of a 4 by 4 matrix

[Ron Burgundypants]

Homework Statement

Coupled Harmonic Oscillators. In this series of exercises you are asked
to generalize the material on harmonic oscillators in Section 6.2 to the
case where the oscillators are coupled. Suppose there are two masses m1
and m2 attached to springs and walls as shown in Figure 6.10. The springs
connecting mj to the walls both have spring constants k1, while the spring
connecting m1 and m2 has spring constant k2. This coupling means that
the motion of either mass affects the behavior of the other.
Let xj denote the displacement of each mass from its rest position, and
assume that both masses are equal to 1. The differential equations for
these coupled oscillators are then given by

x₁'' = -k(₁ + k₂)x₁ + k₂x₂
x₂'' = k₂x₁ - (k₁ + k₂)x₂

These equations are derived as follows. If m1 is moved to the right
(x1 > 0), the left spring is stretched and exerts a restorative force on
m1 given by -k1x1. Meanwhile, the central spring is compressed, so it
exerts a restorative force on m1 given by -k2x1. If the right spring is
stretched, then the central spring is compressed and exerts a restorative
force on m1 given by k2x2 (since x2 < 0). The forces on m2 are similar.

(a) Write these equations as a first-order linear system.
(b) Determine the eigenvalues and eigenvectors of the corresponding
matrix.
(c) Find the general solution.
(d) Let ω1 = √k1 and ω2 = √k1 + 2k2. What can be said about the
periodicity of solutions relative to the ωj? Prove this.

Homework Equations

(A - λI) = Ax

The Attempt at a Solution

Part a is ok but I'm stuck for the rest

For part a we set y₁ = dx₁/dt and therefore dy₁/dt = d²x/dt²
Then y₂= dx₂/dt so we have a similar case of dy₂/dt = d²x/dt²

I think this is enough to satisfy part a) of the question

So then we put them into matrix form to try and find the eigenvalues and vectors and get the following by rows. Columns are x1,y1, x2, y2 respectively.

dx₁/dt = 0 1 0 0
dy₁/dt = -(k1+k2) 0 k2 0
dx₂/dt = 0 0 0 1
dy₂/dt = k2 0 -(k1+k2) 0

So there is a diagonal of zeros meaning we have zero trace. By using the formula (A - λI) = Ax. An attempt was made to try and get the eigenvalues. I subtract λ across each term in the diagonal, multiply out the expression and I end up with the following expression.

k1²λ + k1k2λ + k2²λ. I don't know how to deal with this expression. I'm used to seeing things in the form λ² + λ - c.

So any helpful tips?

 

[Mark44]

Homework Statement

Coupled Harmonic Oscillators. In this series of exercises you are asked
to generalize the material on harmonic oscillators in Section 6.2 to the
case where the oscillators are coupled. Suppose there are two masses m1
and m2 attached to springs and walls as shown in Figure 6.10. The springs
connecting mj to the walls both have spring constants k1, while the spring
connecting m1 and m2 has spring constant k2. This coupling means that
the motion of either mass affects the behavior of the other.
Let xj denote the displacement of each mass from its rest position, and
assume that both masses are equal to 1. The differential equations for
these coupled oscillators are then given by

x₁'' = -k(₁ + k₂)x₁ + k₂x₂
x₂'' = k₂x₁ - (k₁ + k₂)x₂

These equations are derived as follows. If m1 is moved to the right
(x1 > 0), the left spring is stretched and exerts a restorative force on
m1 given by -k1x1. Meanwhile, the central spring is compressed, so it
exerts a restorative force on m1 given by -k2x1. If the right spring is
stretched, then the central spring is compressed and exerts a restorative
force on m1 given by k2x2 (since x2 < 0). The forces on m2 are similar.

(a) Write these equations as a first-order linear system.
(b) Determine the eigenvalues and eigenvectors of the corresponding
matrix.
(c) Find the general solution.
(d) Let ω1 = √k1 and ω2 = √k1 + 2k2. What can be said about the
periodicity of solutions relative to the ωj? Prove this.

Homework Equations

(A - λI) = Ax

The Attempt at a Solution

Part a is ok but I'm stuck for the rest

For part a we set y₁ = dx₁/dt and therefore dy₁/dt = d²x/dt²
Then y₂= dx₂/dt so we have a similar case of dy₂/dt = d²x/dt²

I think this is enough to satisfy part a) of the question

So then we put them into matrix form to try and find the eigenvalues and vectors and get the following by rows. Columns are x1,y1, x2, y2 respectively.

dx₁/dt = 0 1 0 0
dy₁/dt = -(k1+k2) 0 k2 0
dx₂/dt = 0 0 0 1
dy₂/dt = k2 0 -(k1+k2) 0

So there is a diagonal of zeros meaning we have zero trace. By using the formula (A - λI) = Ax. An attempt was made to try and get the eigenvalues. I subtract λ across each term in the diagonal, multiply out the expression and I end up with the following expression.

k1²λ + k1k2λ + k2²λ. I don't know how to deal with this expression. I'm used to seeing things in the form λ² + λ - c.

So any helpful tips?

Let's call your matrix A. Write the matrix $A - \lambda I$, and then take its determinant, which you set equal to zero. That should give you an equation that you can solve for $\lambda$.

The idea is that if $\lambda$ is an eigenvalue (with associated eigenvector $\vec x$, then by definition, $A\vec x = \lambda \vec x$. Equivalently, $(A - \lambda I)\vec x = \vec 0$, which for nonzero vectors $\vec x$, means that $|A - \lambda I| = 0$.

 

[Ron Burgundypants]

Ok maybe I wasn't clear enough. Thanks for your response. I'm aware this is how you find the eigenvalue. My question is how do i solve the final expression to give me some eigenvalues. I also just realized I made a mistake...

My actual final expression is λ⁴ - 2λ³k2 + λ²(k2²+k2 - k1) + λ(k1k2 - k2²) = 0

Getting eigenvalues form this? Yikes...

 

[PeroK]

Ok maybe I wasn't clear enough. Thanks for your response. I'm aware this is how you find the eigenvalue. My question is how do i solve the final expression to give me some eigenvalues. I also just realized I made a mistake...

My actual final expression is λ⁴ - 2λ³k2 + λ²(k2²+k2 - k1) + λ(k1k2 - k2²) = 0

Getting eigenvalues form this? Yikes...

The first eigenvalue is obvious; the second requires a bit of guesswork; and then you have a quadratic.