The word "Linear": Definitions and Theorems

Section 9.5 The word “Linear”: Definitions and Theorems

In the next section, Section 9.6 we will give, without proof, several important theorems about linear differential equations. But before we get to the theorems, you will need to understand what is meant by the word linear so that you can understand the content of these theorems. Before you read this section, you should also make sure you know the definitions and notation in Section 9.1 and Section 4.3.

Homogeneous Linear Differential Equations.

You have often used the fact that the derivative operator acting on functions is linear:

\begin{equation} \frac{d}{dx} \left(\alpha f\right)= \alpha \left(\frac{d}{dx} f\right)\tag{9.5.1} \end{equation}

\begin{equation} \frac{d}{dx}\left(f+g\right)=\left(\frac{d}{dx}f\right) + \left(\frac{d}{dx}g\right)\tag{9.5.2} \end{equation}

e.g. For example, how do you calculate \(\frac{d}{dx} \left(3x^2 + \cos{x}\right)\text{?}\)

By a straightforward extension, the linear differential operator \(\LL\) defined by

\begin{equation} \LL\equiv\frac{d^n}{dx^n}+a_{n-1}(x)\frac{d^{n-1}}{dx^{n-1}} +\dots+a_0(x)\tag{9.5.3} \end{equation}

is indeed linear. The important feature here is that all of the (\(n\)th order) derivatives are to the first power and not inside of any other special functions. (Do not be confused by the notation for \(n\)th order derivatives, which looks like the notation for the \(n\)th power.) Most differential operators in physics ARE linear, so this should look very comfortable and familiar. As counterexamples, the following strange-looking differential operators are NOT linear:

\begin{equation} \left(\frac{d}{dx} f\right)^2\text{,}\tag{9.5.4} \end{equation}

\begin{equation} \sin\left(\frac{d}{dx} f\right)\text{.}\tag{9.5.5} \end{equation}

Inhomogeneous Linear Differential Equations.

As with Example 1 in Section 4.3 for straight lines through the origin and not through the origin, if you take a homogeneous linear differential operator \(\LL\) as in example 1 above and add to it an inhomogeneous term \(b(x)\text{,}\) the resulting operator which takes \(y\) to \(\LL y - b(x)\) is NOT linear. If \(y_p\) and \(y_q\) are both solutions of

\begin{equation} \LL y=b(x)\text{,}\tag{9.5.6} \end{equation}

then

\begin{equation} \LL \left(y_p+y_q\right)=2b(x)\ne b(x)\text{.}\tag{9.5.7} \end{equation}

You can NOT add two solutions of an inhomogeneous differential equation and get another solution. However, you CAN add ANY solution of the homogeneous equation to a solution of the inhomogeneous equation to get another solution of the inhomogeneous equation.