9.8, a kind-of review over things we sort-of talked about in the past

Section 9.8 is entitled “Higher Order Linear Equations.”

This is another one of those sections that is filled with the wonderful gifts of theorems and definitions and propositions. It’s also quite long. It’s also a section that is kind of a throwback to previous sections. I’ve only skimmed it at this point, but I saw familiar words and terms such as “linear independent”, “uniqueness and existence”, and “Wronskian.” Of course, this is super exciting for me because I tend to be better at summarizing these sections. Obviously the reason for this is that I’m familiar with these terms, and the only difference between these terms from the past and in the present is that the context is slightly different. In any event, it’s not like the context is super different than before.

We’ll start out by talking about what a linear equation is; it’s of the form

Back in section 8.1, we found out how to replace higher-order equations with a single-order equation. We did this by replacing the variables in the equation with new ones, e.g.

This system is a system of n equations and n unknown functions x₁, x₂,…, x_n. Because of how we defined our variables, this is equivalent to our original equation involving y. The function y(t) will be a solution to our original equation if the following vector-valued function is a solution to the system of equations:

The matrix notation of our system would be x’ = Ax + f. In this case, the matrix A would be 

Also, f(t) = (0, 0, … , F(t))^T. So our original linear equation would be homogeneous if F(t) = 0.

Also, the initial value problem for a higher-order equation would look like

Here is a theorem concerning existence and uniqueness:

“Suppose the coefficients of the equation

are continuous functions of t in an interval (α, β). Then, for any t₀ ∈ (α, β), and for any constants y₀, y₁,…, y_n-1, [the equation], together with the initial conditions

has a unique solution defined for all t ∈ (α, β)” (434).

So y(t) is a solution to the homogeneous and higher-order equation

This is true if and only if x(t) = (y(t), y’(t),…, y^(n-1))^T is a solution to x’ = Ax. A is still defined above. A theorem concerning this notion: “Suppose that y₁(t), y₂(t),…, and y_k(t) are all solutions to [the equation above]. Then any linear combination of these functions is also a solution” (435).

Onto a definition concerning linear independence and linear dependence: “Suppose the functions y₁(t), y₂(t),…, and y_n(t) are all defined on the interval (α, β). The functions are linearly dependent if there are constants c₁, c₂,…, and c_n, not all equal to 0, such that

for all t ∈ (α, β). The functions are linearly independent if they are not linearly dependent” (435).

No seriously, that’s the wise definition of linear independence. Now you know something important, specifically concerning the relationship between linear independence and dependence, specifically that they are not the same thing.

So it’s easy to know if two or three functions are linearly independent or not. But what if you have a lot more than two or three functions? Why, you look at the Wronskian, of course! :D

Let’s define the Wronskian of y₁, y₂,…, and y_n is the Wronskian of x₁, x₂,…, and x_n:

A proposition about this: “The solutions y­₁(t), y₂(t),…, y_n(t) to [the homogeneous, higher-order equation I presented a while ago] are linearly independent if and only if the corresponding solutions to the system [x’ = Ax] are linearly independent. This, in turn, is equivalent to W(t­₀) ≠ 0 for some t₀” (437).

A theorem concerning the structure of the general solution: “Suppose that y₁(t), y₂(t),…, and y_n(t) are linearly independent solutions to [the homogeneous, higher-order equation]. Then every solution to [this higher-order equation] is a linear combination of y₁(t), y₂(t),…, and y_n(t)” (438).

The fundamental set of solutions is the set of n solutions y₁(t), y₂(t),…, and y_n(t) that are linearly independent. Then the general solution is a linear combination

Remember when we looked for exponential solutions to solve systems? Suppose we have a solution to the system x’ = Ax, i.e. x = (y, y’,…, y^(n-1))^T, in which y is a solution to

The characteristic polynomial is denoted as p(λ) and is to the differential equation above. The following equation is called the characteristic equation:

The roots of the characteristic equation for the differential equation above are equal to the eigenvalues of the matrix A in the system x’ = Ax. If the characteristic equation has n distinct roots, then the exponential solutions to the system are linearly independent.

A theorem for you concerning real roots:

“If λ is a real root to the characteristic polynomial of algebraic multiplicity q, then

are q linearly independent solutions” (441).

And now the corresponding theorem concerning complex roots:

“If λ = α + iβ is a complex root of the characteristic polynomial with multiplicity q, then so is λ = α – iβ. In addition,

are 2q linearly independent solutions” (442).

I’m not a tremendously big fan of ending these summary things with direct quote (mostly because the high school English teacher that lives within my soul likes to remind me that starting or ending with a quote from someone else is a big no-no in the world of high school English papers; however, the college kid that lives within my soul likes to remind me that this is not the world of high school English papers, and I haven’t had to write a formal paper yet in my college career) but I’m doing so anyway since this is the end of the summary. This means I have 9.9 to summarize and then I’m done with summarizing chapter 9.

This also means I have two more chapters (4 and 5) and 12 more sections to summarize before I’m free from summarizing these things. I am slightly exciting but also dreading this day, considering these summaries take a couple of hours of boredom from my 12 credit semester and turn them into semi-useful math hours.

I don’t know if I’ll have 9.9 up by tomorrow. Actually, I can assure you that unless I get it up tonight (which is looking less and less likely as these minutes whiz by me), 9.9 will be up on Tuesday. Then I’ll be moving onto (or backwards to) chapter 4!

I’ll see you when I see you.