KindiakMath

Who Cares about Laplace Transforms?

Consider the Heaviside step function, commonly denoted U(\cdot) or H(\cdot), defined by

U(t) \equiv \mathbb I_{[0,\infty)}(t) = \begin{cases} 1, & t \geq 0, \\ 0, & t < 0. \end{cases}

How do we find the general solution for the differential equation below?

\displaystyle \frac{\mathrm d^2 y}{\mathrm dt^2} - 5 \frac{\mathrm dy}{\mathrm dt} + 6y = U(t)

We can write this in \mathcal D-notation to obtain the equation

(\mathcal D^2 - 5\mathcal D + 6)(y) = U(t),

so perhaps we can find y_{\mathrm C} = C_1 e^{2t} + C_2 e^{3t} and

\displaystyle y_{\mathrm P} = \frac{1}{\mathcal D^2 - 5\mathcal D + 6}(U(t)).

Except that y_{\mathrm P} isn’t immediately possible. Such conundrums motivate the Laplace transform as a powerful problem-solving tool.

Definition 1. Let f be a function. For any s, the Laplace transform of f, denoted \mathcal L\{f\}, is defined by

\displaystyle \mathcal L\{f\}(s) := \int_0^{\infty} f(t)e^{-st}\, \mathrm dt \equiv \lim_{T \to \infty} \int_0^T f(t)e^{-st}\, \mathrm dt,

whenever the right-hand side is well-defined. Defining F = \mathcal L\{ f\}, we sometimes abuse notation and denote

F = F(s) \equiv \mathcal L\{f(t)\} = \mathcal L\{ f \}.

Example 1. For any s > 0, \mathcal L\{1\} = 1/s.

Proof. By the definition of Laplace transforms,

\displaystyle \begin{aligned} \mathcal L\{1\} &= \int_0^{\infty} 1 \cdot e^{-st}\, \mathrm dt = \int_0^{\infty} e^{-st}\, \mathrm dt\\ &= \frac 1{-s}\left[ e^{-st} \right]_0^\infty = \frac{1}{-s}(0 - 1) = \frac 1s. \end{aligned}

We will explore more commonly used Laplace transforms and connect them to commonly used solving techniques in future posts. Let’s first verify a familiar linearity property for the Laplace transforms that justifies the technique of “splitting”.

Theorem 1. Let f,g be functions such that \mathcal L\{f\} and \mathcal L\{g\} exist, and c be a constant. Then \mathcal L\{f+g\} = \mathcal L\{f\} + \mathcal L\{g\} and \mathcal L\{cf\} = c \mathcal L\{f\} both exist.

Proof. We prove the result for \mathcal L\{f+g\} = \mathcal L\{f\} + \mathcal L\{g\}. By definition of the Laplace transform,

\displaystyle \begin{aligned} \mathcal L\{f(t)\} &= \lim_{T \to \infty} \int_0^T f(t)e^{-st}\, \mathrm dt \end{aligned}

By the linearity of taking integrals and taking limits,

\displaystyle \begin{aligned} L\{(f+g)(t)\} &= \lim_{T \to \infty} \int_0^T (f(t) + g(t))e^{-st}\, \mathrm dt\\ &= \lim_{T \to \infty} \left(\int_0^T f(t)e^{-st}\, \mathrm dt + \int_0^T g(t)e^{-st}\, \mathrm dt \right) \\ &= \lim_{T \to \infty} \int_0^T f(t) e^{-st}\, \mathrm dt + \lim_{T \to \infty} \int_0^T g(t) e^{-st}\, \mathrm dt \\ &= \mathcal L\{f(t)\} + \mathcal L\{g(t)\}. \end{aligned}

Coupled with linearity, we also need the Laplace transform of derivatives.

Theorem 2. Let f be a differentiable function such that F(s) = \mathcal L\{f(t)\} exists, f' is integrable on any [0, t] and f(t)e^{-st} \to 0 as t \to \infty. Then \mathcal L\{f'(t)\} exists and is given by

\displaystyle \mathcal L\{f'(t)\} = s F(s) - f(0).

Proof. Integrating by parts,

\displaystyle \begin{aligned} \mathcal L\{f'(t)\} &= \int_0^\infty f'(t) e^{-st}\, \mathrm dt \\ &= [ \underbrace{f(t)}_{\mathrm I} \underbrace{e^{-st}}_{\mathrm S} ]_0^\infty - \int_0^\infty \underbrace{f(t)}_{\mathrm I} \cdot \underbrace{-se^{-st}}_{\mathrm D}\, \mathrm dt \\ &= (0 - f(0) \cdot 1) + s \int_0^\infty f(t)e^{-st}\, \mathrm dt\\&= - f(0) + s \mathcal L\{f(t)\} \\ &= sF(s) - f(0). \end{aligned}

Corollary 1. Let f be a twice-differentiable function such that \mathcal L\{f(t)\} and \mathcal L\{f'(t)\} exist, f'' is integrable on any [0, t], and f(t)e^{-st} \to 0, f'(t)e^{-st} \to 0 as t \to \infty. Then \mathcal L\{f''(t)\} exists and is given by

\displaystyle \mathcal L\{f''(t)\} = s^2 F(s) - sf(0) - f'(0).

Proof. Applying the result in Theorem 2 to f'' = (f')',

\begin{aligned} \mathcal L\{f''(t)\} &= s \cdot \mathcal L\{f'(t)\} - f'(0)\\ &= s \cdot (s \cdot \mathcal L\{f(t)\} - f(0)) - f'(0) \\ &= s^2 F(s) - sf(0) - f'(0). \end{aligned}

These results give us the main power of the Laplace transform, which we will illustrate in the motivating example.

Example 2. Given the initial value problem

\displaystyle y'' - 5 y' + 6y = U(t),\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) \}.

By linearity,

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

By Corollary 1,

\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 Theorem 2,

\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 a result that we will discuss in the future, \mathcal L \{ U(t) \} = 1/s. Substituting all of these results,

\displaystyle s^2 Y(s) - 5sY(s) + 6Y(s) = \frac 1s.

By basic algebruh,

\displaystyle Y(s) = \frac{1}{s(s-2)(s-3)}.

Now, we recall that Y(s) = \mathcal L \{y(t)\}. How do we recover y(t)? Effectively, we would take the inverse Laplace transform

y(t) = \mathcal L^{-1} \{Y(s)\},

and employ several more tools at our disposal, like partial fractions, and more often than not, the shift theorems in the context of Laplace transforms. We will discuss these ideas more in a future post.

—Joel Kindiak, 5 Feb 25, 0041H

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