KindiakMath

The Step Functions

Recall that we are interested in solving initial value problems of the form

ay'' + by' + cy = Q(t),\quad y(0) = y_0, \quad y'(0) = y_0',

where y_0, y_0' are pre-determined constants. Initial value problems are basically differential equations, with some starting conditions (i.e. specified values of y(0) and y'(0). It is called an initial value problem since we are specifying conditions at the initial moment, namely, t = 0.

Many a time, the function Q(x) takes the form of a step function.

Definition 1. Given distinct real numbers a_0, a_1,\dots, a_n, a function f is a step function if there exist t_1 <\dots < t_n such that

f(t) = \begin{cases} a_0, & t < t_1, \\ a_i, & t_i \leq t < t_{i+1},\quad 1 \leq i \leq {n-1}, \\ a_{n}, & t \geq t_n,\\ \end{cases}

Example 1. The unit step function at 0 is a step function, since

U(t) = \begin{cases} 0, & t < 0, \\ 1, & t \geq 0. \end{cases}

There are several generalisations of the idea described in Definition 1. Namely, in each interval, the function need not be constant (i.e. we can replace each a_i with f_i(t) and the function will still make sense). Nevertheless, we will stick with constants a_i for simplicity.

The unit step function is itself of considerable interest. Firstly, we can shift it to obtain specific functions that match our problem. Next, we can truncate the unit step functions so that its integral is trivial to compute (it’s just the area of a rectangle).

Lemma 1. For any set K, define the indicator function by

\mathbb I_K(t) := \begin{cases} 1, & t \in K, \\ 0, & t \notin K. \end{cases}

Then for any real a < b,

  • \mathbb I_{(-\infty, a)} = 1 - U_a,
  • \mathbb I_{[a, b)} = U_a - U_b,
  • \mathbb I_{[b,\infty)} =U_b.

Here, we denote U_a(t) \equiv U(t-a) for brevity.

In particular, \displaystyle \int_{-\infty}^\infty \mathbb I_{[a, b)}(t)\, \mathrm dt = b-a.

Proof. We prove \mathbb I_{[a, b)} = U_a - U_b and leave the rest as exercises.

We need to prove that

U(t- a) - U(t-b) = \begin{cases}0, & t < a, \\ 1, & a \leq t < b, \\ 0, & t \geq b.\end{cases}

For t < a, since a < b, we have t < b, so that

U(t- a) - U(t-b) = 0 - 0 = 0.

For a \leq t < b,

U(t- a) - U(t-b) = 1 - 0 = 1.

For t \geq b, since a < b, we have t \geq a, so that

U(t-a) - U(t-b) = 1 - 1 = 0.

Therefore,

U(t- a) - U(t-b) = \begin{cases}0, & t < a, \\ 1, & a \leq t < b, \\ 0, & t \geq b,\end{cases}

as required.

Its Laplace transform is trivial to compute.

Theorem 1. For any real a, \mathcal L\{U_a(t)\} \equiv \mathcal L\{U(t-a)\} = e^{-as}/s. In particular,

\displaystyle \mathcal L\{U(t)\} = \frac 1s = \mathcal L\{1\}.

Proof. Plugging in the definition of the Laplace transform,

\displaystyle \begin{aligned}\mathcal L\{U(t-a)\} &= \int_0^\infty U(t-a) e^{-st}\, \mathrm dt \\ &= \int_0^a U(t-a) e^{-st}\, \mathrm dt + \int_a^\infty U(t-a) e^{-st}\, \mathrm dt \\ &= \int_0^a 0 \cdot e^{-st}\, \mathrm dt + \int_a^\infty 1\cdot e^{-st}\, \mathrm dt \\ &= \int_a^\infty e^{-st}\, \mathrm dt = \left[\frac{e^{-st}}{-s}\right]_a^\infty \\ &= 0 - \frac{e^{-as}}{-s} = \frac{e^{-as}}{s}.\end{aligned}

Finally, we can scale these functions accordingly to obtain functions that match our problem. When we combine these functions, we obtain a step function f. The converse is true as well: every step function can be constructed by “combining” a bunch of unit step functions together.

Theorem 2. Let f be a defined by

f(t) = \begin{cases} a_0, & t < t_1, \\ a_i, & t_{i} \leq t < t_{i+1},\quad 1 \leq i \leq n-1, \\ a_n, & t \geq t_n,\\ \end{cases}

For each 1 \leq i \leq n, define the i-th jump by \Delta a_i := a_i - a_{i-1}. Then

\displaystyle f = a_0 + \Delta a_1 U_{t_i} + \cdots + \Delta a_n U_{t_n}.

Proof. We will illustrate the case n = 2 for simplicity. The general case follows by careful bookkeeping.

Suppose

f(t) = \begin{cases} a_0, & t < t_1, \\ a_1, & t_1 \leq t < t_2, \\ a_2, & t \geq t_2. \end{cases}

Then adding the various components,

\displaystyle \begin{aligned} f &= a_0(1 - U_{t_1}) + a_1 (U_{t_1} - U_{t_2}) + a_2U_{t_2} \\ &= a_0 - a_0 U_{t_1} + a_1 U_{t_1} - a_1 U_{t_2} + a_2 U_{t_2} \\ &= a_0 + (a_1 - a_0) U_{t_1} + (a_2 - a_1) U_{t_2} \\ &= a_0 + \Delta a_1 U_{t_1} + \Delta a_2 U_{t_2}. \end{aligned}

These step functions can be used to define a powerful notion of integration known as Lebesgue integration. Of more immediate interest, its greatest utility in the context of differential equations is that its Laplace transform becomes trivial to compute.

Corollary 1. Let f be a defined by

f(t) = \begin{cases} a_0, & t < t_1, \\ a_i, & t_{i} \leq t < t_{i+1},\quad 1 \leq i \leq n-1, \\ a_n, & t \geq t_n,\\ \end{cases}

Then

\displaystyle \begin{aligned} \mathcal L\{f(t)\} &= a_0 \cdot \frac 1s + \sum_{i=1}^n \Delta a_i \cdot \frac{e^{-st_i}}{s}. \end{aligned}

Finally, given a step function f, how do we compute the following Laplace transform?

\mathcal L\{f^2\} \equiv \mathcal L\{f^2(t)\} \equiv \mathcal L\{(f(t))^2\}

We will consider f = 5U_2 - 7U_3 for simplicity. A direct computation yields

f^2 = (5U_1 - 7U_2)^2 = 25U_1^2 - 70U_1U_2 + 49U_2^2.

But how do we evaluate the products of unit step functions?

Theorem 3. For any real a \leq b, U_a U_b = U_b.

Proof. We observe that

U_a \cdot U_b = \mathbb I_{[a,\infty)} \cdot \mathbb I_{[b,\infty)} = \mathbb I_{[a,\infty) \cap [b,\infty)} = \mathbb I_{[b,\infty)} = U_b.

We can then compute the calculation.

Example 2. Let f = 5U_2 - 7U_3 be a step function. Then

f^2 = 25U_2 - 21U_3

so that \displaystyle \mathcal L\{(f(t))^2\} = 25 \cdot \frac{e^{-2s}}{s} - 21 \cdot \frac{e^{-3s}}{s}.

To answer the question we started with: how do we solve initial value problems involving step functions? By taking Laplace transforms.

Example 3. Given the initial value problem

\displaystyle y'' - 5 y' + 6y = U(t-2),\quad y(0)=y'(0) = 0

and denoting Y(s) \equiv \mathcal L\{y(t)\} \equiv \mathcal L\{y\}(s), evaluate Y(s).

Solution. Taking Laplace transforms on all sides,

\displaystyle \mathcal L\{y'' - 5 y' + 6y\} = \mathcal L\{ U(t-2) \}.

By linearity,

\displaystyle \mathcal L\{y''\} - 5 \mathcal L\{y'\} + 6 \mathcal L\{y\} = \mathcal L\{ U(t-2) \}.

By the Laplace transform of the second derivative,

\displaystyle \begin{aligned} \mathcal L\{y''\} &= s^2 Y(s) - s y(0) - y'(0) \\ &= s^2 Y(s) - 0 - 0 = s^2 Y(s). \end{aligned}

By the Laplace transform of the first derivative,

\displaystyle \begin{aligned} \mathcal L\{y'\} &= s Y(s) - y(0) \\ &= s Y(s) - 0 = s Y(s). \end{aligned}

By notation, \mathcal L\{y\} = Y(s). By Theorem 1,

\displaystyle \mathcal L \{ U(t-2) \} = \frac{e^{-2s}}s.

Substituting all of these results,

\displaystyle s^2 Y(s) - 5sY(s) + 6Y(s) = \frac {e^{-2s}}s.

By basic algebruh,

\displaystyle Y(s) = \frac{e^{-2s}}{s(s-2)(s-3)}.

From this result, we still face another pressing problem. We will need to compute

\displaystyle y(t) = \mathcal L^{-1}\{Y(s)\} = \mathcal L^{-1} \left\{ e^{-2s} \cdot \frac{1}{s(s-2)(s-3)} \right\}.

How do we actually evaluate the expression on the right? This is where we will employ the help of inverse Laplace transforms and the shift theorems, in our next discussion.

—Joel Kindiak, 10 Feb 25, 1151H

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