Laplace Transforms

Laplace transforms are a widely used method for solving constant coefficient differential equations.

For inhomogeneous constant coefficient equations, we may use:

Laplace transforms are particularly useful when dealing with discontinuous forcing functions \( g(t) \).

The Laplace transform is an operator \( \mathcal{L} \) that maps a function \( f(t) \), defined for \( t \geq 0 \), to another function \( F(s) \). The key advantage of this method is that it converts differential equations into algebraic equations in \(s\)-space.

$$ \mathcal{L}(f(t)) = F(s) = \int_{0}^{\infty} e^{-st} f(t)\,dt $$

This is well-defined provided that:

A function \( f(t) \) is said to be of exponential order as \( t \to \infty \) if there exist constants \( a, A, K \) such that

$$ |f(t)| \leq K e^{at}, \quad t \geq A $$

In simple terms, \( f(t) \) does not grow faster than an exponential.

eg. Find the Laplace transform of \( \color{#00f0ff} f(t) = e^{-at} \)

$$ \mathcal{L}(e^{-at}) = \int_{0}^{\infty} e^{-st} e^{-at}\,dt $$ $$ = \int_{0}^{\infty} e^{-(s+a)t}\,dt $$ $$ = \left[ \frac{- \, e^{-(s+a)t}}{s+a} \right]_{0}^{\infty} $$ $$ = \frac{1}{s+a} $$

This basic result appears repeatedly and is one of the fundamental transform pairs.

Laplace Transform of a Derivative

Suppose \( f'(t) \) is piecewise continuous and of exponential order as \( t \to \infty \), and that \( f(t) \) is of exponential order

$$ \mathcal{L}(f'(t)) = s\mathcal{L}(f(t)) - f(0) $$

Therefore discontinuities of \( f'(t) \) do not affect \( \mathcal{L}(f'(t)) \)

Higher-Order Derivatives

Suppose that \( f, f', f'', \dots, f^{n-1} \) are continuous and that \( f^{n} \) is piecewise continuous on any interval \( [0,a] \). Suppose also that each of \( f, f', f'', \dots, f^{n-1}, f^{n} \) is of exponential order as \( t \to \infty \). Then

$$ \mathcal{L}(f^{n}(t)) = s^n \mathcal{L}(f(t)) - f^{n-1}(0) - s f^{n-2}(0) \; - \; ... \; - \; s^{n-1}f(0) $$

Equivalently,

$$ \mathcal{L}(f^{n}(t)) = s^n \mathcal{L}(f(t)) - \sum_{j=0}^{n-1} s^j f^{n-1-j}(0) $$

This formula allows us to incorporate initial conditions directly into the transformed equation.

Linearity of the Laplace Transform

The Laplace transform is a linear operator, i.e., suppose \( f(t) \) and \( g(t) \) have Laplace transforms \( F(s) \) and \( G(s) \), respectively. Then for any constants \( a \) and \( b \),

$$ \mathcal{L}(a f(t) + b g(t)) = aF(s) + bG(s) $$

This follows from the definition:

$$ \mathcal{L}(a f(t) + b g(t)) = \int_{0}^{\infty} (a f(t) + b g(t)) \, e^{-st}\,dt $$ $$ = a \int_{0}^{\infty} f(t) e^{-st}\,dt \, + \, b \int_{0}^{\infty} g(t) e^{-st}\,dt $$ $$ = aF(s) + bG(s) $$

Solving a Differential Equation Using Laplace Transforms

The general procedure is:

  1. Take the Laplace transform of the entire differential equation
  2. Use linearity and rules for transforming derivatives
  3. Input the initial conditions directly
  4. Solve the resulting equation for \( Y(s) \)
  5. Transform back from \( Y(s) \) to \( y(t) \)

eg. Use Laplace transforms to solve the IVP \( \quad \color{#00f0ff} y' + ay = 0, \; \; y(0) = y_0 \)

Take the Laplace transform of the differential equation:

$$ \mathcal{L}(y' + ay) = \mathcal{L}(y') + a\mathcal{L}(y) = 0 $$

Using the derivative formula

$$ s\mathcal{L}(y) - y(0) + a\mathcal{L}(y) = 0 $$ $$ (s + a)Y(s) - y_0 = 0 $$ $$ Y(s) = \frac{y_0}{s + a} $$ $$ \mathcal{L}(e^{-at}) = \frac{1}{s + a} $$

By linearity

$$ \mathcal{L}^{-1}(Y(s)) = y_0 \, \mathcal{L}^{-1}\!\left(\frac{1}{s + a}\right) $$ $$ y(t) = y_0 \, e^{-at} $$

eg. Use Laplace transforms to solve the IVP \( \quad \color{#00f0ff} y'' - 9y = 0, \; \; y(0) = 1, \; \; y'(0) = -1 \)

$$ \mathcal{L}(y'' - 9y) = \mathcal{L}(y'') - 9\mathcal{L}(y) = 0 $$

Using the derivative formula

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

therefore

$$ s^2Y(s) - y'(0) - sy(0) - 9Y(s) = 0 $$ $$ s^2Y(s) + 1 - s - 9Y(s) = 0 $$ $$ (s^2 - 9)Y(s) + 1 - s = 0 $$

Therefore

$$ Y(s) = \frac{s - 1}{s^2 - 9} $$

From this point, the problem reduces to expressing \( Y(s) \) in a form that matches known Laplace transforms.

$$ Y(s) = \frac{s - 1}{(s + 3)(s - 3)} $$

Use partial fractions

$$ \frac{s - 1}{(s + 3)(s - 3)} = \frac{A}{s + 3} + \frac{B}{s - 3} $$ $$ A(s - 3) + B(s + 3) = s - 1 $$ $$ A + B = 1 \quad -3A + 3B = -1 $$ $$ -3A + 3(1 - A) = -1 $$ $$ -6A = -4 \quad \Rightarrow \quad A = \frac{2}{3}, \quad B = \frac{1}{3} $$

So

$$ Y(s) = \frac{2}{3(s + 3)} + \frac{1}{3(s - 3)} $$

Taking the inverse Laplace transform

$$ y(t) = \frac{2}{3} \, e^{-3t} + \frac{1}{3} \, e^{3t} $$

There is a one-to-one correspondence between \( f(t) \) and \( F(s) \), so we can work meaningfully in \( s \)-space and then invert back to \( t \)-space. In practice, we use tables together with general results to identify how to invert a particular formula.

The following examples illustrate how different algebraic forms in \( s \)-space correspond to familiar time-domain solutions:

eg. Use Laplace transforms to solve the IVP \( \quad \color{#00f0ff} y'' - y' - 2y = 0, \; \; y(0) = 1, \; \; y'(0) = -1 \)

$$ \mathcal{L}(y'' - y' - 2y) = \mathcal{L}(y'') - \mathcal{L}(y') - 2\mathcal{L}(y) = 0 $$ $$ s^2Y(s) - y'(0) - sy(0) - (sY(s) - y(0)) - 2Y(s) = 0 $$ $$ s^2Y(s) + 1 - s - sY(s) + 1 - 2Y(s) = 0 $$ $$ (s^2 - s - 2)Y(s) = s - 2 $$

Therefore

$$ Y(s) = \frac{s - 2}{s^2 - s - 2} $$ $$ Y(s) = \frac{s - 2}{(s - 2)(s + 1)} $$

Use partial fractions:

$$ \frac{s - 2}{(s - 2)(s + 1)} = \frac{A}{s - 2} + \frac{B}{s + 1} $$ $$ A(s + 1) + B(s - 2) = s - 2 $$

Matching coefficients gives

$$ A + B = 1 $$ $$ A - 2B = -2 $$ $$ (1 - B) - 2B = -2 $$ $$ 1 - 3B = -2 $$ $$ B = 1, \quad A = 0 $$

So

$$ Y(s) = \frac{1}{s + 1} \quad \Rightarrow \quad y(t) = e^{-t} $$

eg. Use Laplace transforms to solve the IVP \( \quad \color{#00f0ff} y'' + 2y' + 10y = 0, \; \; y(0) = 1, \; \; y'(0) = -1 \)

$$ \mathcal{L}(y'' + 2y' + 10y) = \mathcal{L}(y'') + 2\mathcal{L}(y') + 10\mathcal{L}(y) = 0 $$ $$ s^2Y(s) - y'(0) - sy(0) + 2(sY(s) - y(0)) + 10Y(s) = 0 $$ $$ s^2Y(s) + 1 - s + 2sY(s) - 2 + 10Y(s) = 0 $$ $$ (s^2 + 2s + 10)Y(s) = s + 1 $$

Therefore,

$$ Y(s) = \frac{s + 1}{s^2 + 2s + 10} $$

Complete the square:

$$ Y(s) = \frac{s + 1}{(s + 1)^2 + 9} $$

Hence

$$ y(t) = e^{-t}\cos(3t) $$

eg. Same question, but with \( \color{#00f0ff} y'(0) = -4\)

Following the same steps,

$$ Y(s) = \frac{s - 2}{s^2 + 2s + 10} $$

Complete the square:

$$ Y(s) = \frac{s - 2}{(s + 1)^2 + 9} $$

Rewrite the numerator:

$$ Y(s) = \frac{s + 1}{(s + 1)^2 + 9} - \frac{3}{(s + 1)^2 + 9} $$

Therefore

$$ y(t) = e^{-t}\cos(3t) - e^{-t}\sin(3t) $$

eg. Use Laplace transforms to solve the IVP \( \quad \color{#00f0ff} y'' + 2y' + y = 0, \; \; y(0) = 1, \; \; y'(0) = -4 \)

$$ \mathcal{L}(y'') + 2\mathcal{L}(y') + \mathcal{L}(y) = 0 $$ $$ s^2Y(s) - y'(0) - sy(0) + 2(sY(s) - y(0)) + Y(s) = 0 $$ $$ s^2Y(s) + 4 - s + 2sY(s) - 2 + Y(s) = 0 $$ $$ (s^2 + 2s + 1)Y(s) = s - 2 $$ $$ Y(s) = \frac{s - 2}{s^2 + 2s + 1} $$ $$ Y(s) = \frac{s - 2}{(s + 1)^2} $$

Rewrite the numerator:

$$ Y(s) = \frac{(s + 1) - 3}{(s + 1)^2} $$ $$ Y(s) = \frac{1}{s + 1} - \frac{3}{(s + 1)^2} $$

Taking the inverse Laplace transform,

$$ y(t) = e^{-t} - 3t e^{-t} $$

eg. Use Laplace transforms to solve the IVP \( \quad \color{#00f0ff} y'' + 2y' + y = \cos(2t), \; \; y(0) = 1, \; \; y'(0) = -1 \)

$$ \mathcal{L}(y'') + 2\mathcal{L}(y') + \mathcal{L}(y) = \mathcal{L}(\cos(2t)) $$ $$ s^2Y(s) - y'(0) - sy(0) + 2(sY(s) - y(0)) + Y(s) = \frac{s}{s^2 + 4} $$ $$ s^2Y(s) + 1 - s + 2sY(s) - 2 + Y(s) = \frac{s}{s^2 + 4} $$ $$ (s^2 + 2s + 1)Y(s) = \frac{s}{s^2 + 4} + s + 1 $$ $$ (s + 1)^2Y(s) = \frac{s}{s^2 + 4} + s + 1 $$

Therefore,

$$ Y(s) = \frac{s}{(s^2 + 4)(s + 1)^2} + \frac{s + 1}{(s + 1)^2} $$ $$ Y(s) = \frac{s}{(s^2 + 4)(s + 1)^2} + \frac{1}{s + 1} $$

Decompose the first term:

$$ \frac{s}{(s^2 + 4)(s + 1)^2} = \frac{A}{s + 1} + \frac{B}{(s + 1)^2} + \frac{Cs + D}{s^2 + 4} $$ $$ s = A(s^2 + 4)(s + 1) + B(s^2 + 4) + (Cs + D)(s + 1)^2 $$

Expanding and matching coefficients gives

$$ A = \frac{3}{25}, \quad B = -\frac{1}{5}, \quad C = -\frac{3}{25}, \quad D = \frac{8}{25} $$ $$ \Rightarrow Y(s) = \frac{28}{25}\frac{1}{s + 1} - \frac{1}{5}\frac{1}{(s + 1)^2} - \frac{3}{25}\frac{s}{s^2 + 4} + \frac{4}{25}\frac{2}{s^2 + 4} $$

Using inverse Laplace transforms

$$ y(t) = \frac{28}{25} \, e^{-t} - \frac{1}{5}t e^{-t} - \frac{3}{25}\cos(2t) + \frac{4}{25}\sin(2t) $$