5.3 Linear Systems - ESE 2030 📏

1Reading¶

Material related to this page, as well as additional exercises, can be found in ALA 7.4.

2Learning Objectives¶

By the end of this page, you should know:

how linear systems are defined as linear functions
examples of general linear systems: derivative and evaluation operators
how the superposition principle is applied to general linear systems

3Linear Systems¶

We recover our familiar matrix-vector linear system $A\vv u = \vv f$ if $U = \mathbb{R}^n$ and $V = \mathbb{R}^m$ , and $L(\vv u) = A\vv u$ . However, we can express much more interesting problems in this framework.

Example 1

Consider a typical initial value problem

u'' + u' - 2u = f(t), \quad u(0) = 1, \quad u'(0) = -1

(2)

for some unknown scalar function $u(t)$ . First, we rewrite each equation in terms of derivative operators and evaluation operators:

\begin{align*} L_1(u) &= u'' + u' - 2u = D^2(u) + D(u) - 2u = (D^2 + D - 2)(u) = f(t) \\ L_2(u) &= u(0) = E_0(u) = 1 \\ L_3(u) &= u'(0) = E_0(D(u)) = -1 \end{align*}

(3)

If we then define the linear operator $M(u)$ and RHS $\vv f$ as

M(u) = \begin{bmatrix} L_1(u) \\ L_2(u) \\ L_3(u) \end{bmatrix} \quad \text{and} \quad \vv f = \begin{bmatrix} f(t) \\ 1 \\ -1 \end{bmatrix}

(4)

we can pose the initial value problem as a linear system $M(u) = \vv f$ . In (4), what are the domain $U$ and codomain $V$ of the operator $M: U \to V$ ?

The reason for introducing this extra layer of abstraction is that it lets us port over ideas from systems of linear equations. For example, the superposition principle holds here too!

We’ll focus on solutions to homogeneous linear systems now, but if you’re interested, Section 7.4 of ALA covers the general setting. The superposition principle here says that if a homogeneous linear system $L(\vv z) = \vv 0$ , for $L: U \to V$ a linear function, with two solutions $\vv z_1$ and $\vv z_2$ satisfying $L(\vv z_1) = \vv 0$ and $L(\vv z_2) = \vv 0$ , then any linear combination $c\vv z_1 + d\vv z_2$ is also a solution. This follows immediately from the linearity of $L$ :

L(c\vv z_1 + d\vv z_2) = cL(\vv z_1) + dL(\vv z_2) = c\vv 0 + d\vv 0 = 0.

(5)

Example 2

Consider the $2^{nd}$ order linear differential operator

L = D^2 - 2D - 3

(7)

which maps a function $u(t)$ to the function

\begin{align*} L(u) &= (D^2 - 2D - 3)(u) = D^2u - 2Du - 3u \\ &= u'' - 2u' - 3u. \end{align*}

(8)

The associated homogeneous system then encodes a homogeneous, linear, constant coefficient $2^{nd}$ order differential equation:

L(u) = u'' - 2u' - 3u = 0. \quad (\text{ODE})

(9)

Therefore if we can characterize the kernel of $L$ , we will have a general solution to this ODE.

Using techniques you would have seen in Math 1400, you can check that two linearly independent solutions (within the domain $C^2[0,1]$ ) to (ODE) are

u_1(t) = e^{3t} \quad \text{and} \quad u_2(t) = e^{-t}.

(10)

According to the superposition principle, every linear combination

u(t) = c_1u_1(t) + c_2u_2(t) = c_1e^{3t} + c_2e^{-t}

(11)

is also a solution (try some values of $c_1$ and $c_2$ and check!). In fact, we can show that dim Ker $L = 2$ , and so any solution to (ODE) takes the form (11).