4 chapters down (technically), 3 chapters to go. (8.5)

Section 8.5 is entitled “Properties of Linear Systems.”

I have a feeling a lot of this section will be direct quotes from the book, considering that this section contains quite a few proofs.

“Suppose x₁ and x_2­ are solutions to the homogeneous linear system x’ = Ax. If C₁ and C₂ are any constants, then x = C₁x₁ + C₂x₂ is also a solution to [this system]” (362).

Using the properties of matrix multiplication, we can prove this:

Note that this theorem works if and only if the system is linear and homogeneous.

“Suppose that x₁, x₂, … , and x_k are all solutions to the homogeneous linear system x’ = Ax. Then any linear combination of x₁, x₂, … , and x_k is also a solution. Thus for any constants C₁, C₂, … , C_k, the function

is a solution to x’ = Ax” (363).

So if we have a system, and we want x to be a solution to that system such that x can be expressed as

In this case, x₁ and x_2­ are solutions to our system, and C₁ and C₂ are arbitrary constants. Our goal would be to find C₁ and C₂ which would make our expression of x satisfied for all t. If we know something about our solutions x₁ and x₂, like what their values would be at a specific time/point t, we could solve for C₁ and C₂ at that time/point. We have a handy exactness/uniqueness theorem we can use that would imply our equation would be satisfied for all t. Hooray!

However, something to note that we can only solve our constants provided the matrix

is nonsingular. The matrix will be nonsingular if x₁(t₀) and x₂(t₀) are linearly independent. In this case x­₁₁ and x₁₂ are the values for x₁ at our specific value of t, and x₂₁ and x₂₂ are the values for x₂ at our specific value of t. I suppose I could have made this a little more general, and expanded the rows to contain values x_1n and x_2n, but you get the idea. This works (just in case you’re a little foggy on matrix multiplication) because x(t₀) can be compacted to

The dimension of the first matrix is n × 2, and the dimension of the second matrix is 2 × 1. The multiplication is valid. Double hooray!

Let’s look at a proposition concerning this.

“Suppose y₁(t), y₂(t), … , and (t), … , and y_k(t) are solutions to the n-dimensional system y’ = Ay defined on the interval I = (α, β).

1. If the vectors y₁(t₀), y₂(t₀), … , and y_k(t₀) are linearly dependent for some t₀ ∈ I, then there are contestants C₁, C₂, … , and C_k, not all zero, such that C₁y₁(t) + C₂y₂(t) + … + C_ky_k(t) = 0 for all t ∈ I. In particular, y₁(t), y₂(t), … , and y_k(t) are linearly dependent for all t ∈ I.

2. If for some t₀ ∈ I the vectors y₁(t₀), y₂(t₀), .. , and y_k(t₀) are linearly independent, then y₁(t), y₂(t), … , and y_k(t) are linearly independent for all t ∈ I” (365).

The definition that arises from this is fairly straightforward. If there is one value of t that makes the linear system y’ = Ay linearly independent, then the set of all k solutions of the system is also linearly independent.

“Suppose y₁, …, and y_n are linearly independent solutions to the n-dimensional linear system y’(t) = Ay(t). Then any solution y can be expressed as a linear combination of y₁, …, and y_n. That is, there are constants C₁, …, and C­_n such that

for all t” (365).

If our homogeneous and linear system has a set of n linearly independent solutions, it’ll be called the fundamental set of solutions.

Our last theorem provides us with a way of finding general solutions to the homogeneous system y’ = Ay:

1. Find n linearly independent solutions y₁, y₂, …, y_n.

2. Show the n solutions are linearly independent.

3. Make the general solution

C₁, C₂, …, and C_n are arbitrary constants.

Elaborating on step 2, we only have to show linear independence for one value of t. A great way to do this is to use determinants and the Wronskian. I barely covered this in section 4.1 (which turned out to not be a thing in this class for a while). Hooray! My inability to follow the weekly schedule early on has finally paid off!

If for one value of t, the Wronskian is not equal to zero, then the n solutions are linearly independent.

That’s all for chapter 8! I know I’ll have the time, but if I have the motivation, then I’ll get started on chapter 9.

I’ll see you when I see you.