32E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Determine the Laplace transform of each of the following functions:

$\left(a\right)f_1\left(t\right)=\left\{\begin{array}{l}t, t=1,2,3,....,\\ e^t, t\ne 1,2,3,....\end{array}\right.$

$\left(b\right)f_2\left(t\right)=\left\{\begin{array}{l}{e}^t, t\ne 5,8,\\ 6, t=5,\\ 0, t=8,\end{array}\right.$

$\left(c\right)f_3\left(t\right)=e^t.$

Which of the preceding functions is the inverse Laplace transform of $\frac{1}{\left(s-1\right)}?$

Step-by-Step Solution

Answer

$$\begin{gathered} \mathcal{L}\left\{f_1\left(t\right)\right\}\left(s\right)=\mathcal{L}\left\{f_2\left(t\right)\right\}\left(s\right)=\mathcal{L}\left\{f_3\left(t\right)\right\}\left(s\right)=\frac{1}{s-1} \\ {\mathcal{L}}^{-1}\left\{\frac{1}{s-1}\right\}\left(t\right)=f_3\left(t\right)=e^t \end{gathered}$$

Step 1: Define Inverse Laplace transform

Given a function $F\left(s\right)$ ,if there is a function $f\left(t\right)$  that is continuous on

$[0,\infty )$ and satisfies $\mathcal{L}\left\{f\right\}=F$,then we say that $f\left(t\right)$ is the inverse Laplace transform of $F\left(s\right)$ and employ the notation $f={\mathcal{L}}^{-1}\left\{F\right\}$ 

Non-repeated Linear Factors

If  $Q\left(s\right)$ can be factored into a product of distinct linear factors,

$Q\left(s\right)=\left(s-r_1\right)\left(s-r_2\right).....\left(s-r_n\right)$

where the  $r_i$'s are all distinct real numbers, then the partial fraction expansion has the form

$\frac{P\left(s\right)}{Q\left(s\right)}=\frac{A_1}{s-r_1}+\frac{A_2}{s-r_2}+.....+\frac{A_{n_{}}}{s-r_n}$

where the $A_i$'s are real numbers. There are various ways of determining the constants $A_1.....A_n$. In the next example, we demonstrate two such methods.2.

Repeated Linear Factors

If  $s-r$ is a factor of $Q\left(s\right)$ and  ${\left(s-r\right)}^m$ is the highest power of  $s-r$ that divides $Q\left(s\right)$, then the portion of the partial fraction expansion of $P\left(s\right)/Q\left(s\right)$ that corresponds to the term  ${\left(s-r\right)}^m$ is

$\frac{A_1}{s-r}+\frac{A_2}{{\left(s-r\right)}^2}+........+\frac{A_m}{{\left(s-r\right)}^m}$

where the $A_i$'s are real numbers.

Quadratic Factors

If ${\left(s-\alpha \right)}^2+{\beta }^2$ is a quadratic factor of  $Q\left(s\right)$  that cannot be reduced to linear factors with real coefficients and  is the highest power of ${\left(s-\alpha \right)}^2+{\beta }^2$ that divides $Q\left(s\right)$, then the portion of the partial fraction expansion that corresponds to ${\left(s-\alpha \right)}^2+{\beta }^2$ is

$\frac{C_{1}s+D_1}{{\left(s-\alpha \right)}^2+{\beta }^2}+\frac{C_{2}s+D_2}{\left[{\left(s-\alpha \right)}^2+{\beta }^2\right]}+......+\frac{C_{m}s+D_m}{{\left[{\left(s-\alpha \right)}^2+{\beta }^2\right]}^m}$

it is more convenient to express  $C_{i}s+D_i$ in the form  $A_i\left(s-\alpha \right)+\beta B_i$ when we look up the Laplace transforms. So let's agree to write this portion of the partial fraction expansion in the equivalent form

$\frac{A_1\left(s-\alpha \right)+\beta B_1}{{\left(s-\alpha \right)}^2+{\beta }^2}+\frac{A_2\left(s-\alpha \right)+\beta B_2}{{\left[{\left(s-\alpha \right)}^2+{\beta }^2\right]}^2}+.......+\frac{A_m\left(s-\alpha \right)+\beta B_m}{{\left[{\left(s-\alpha \right)}^2+{\beta }^2\right]}^m}$

Step 2: Find the factor of the denominator

a. b. c. Since the functions $f_1\left\{t\right\},f_2\left\{t\right\},$ and $f_3\left\{t\right\}$ differ at finite number of points and Laplace transform is a definite integral which doesn't depend on values of functions at finite number of points, we have that all three functions has the same Laplace transform.

Let's calculate it:

$$\begin{gathered} \mathcal{L}\left\{f_1\left(t\right)\right\}\left(s\right)=\mathcal{L}\left\{f_2\left(t\right)\right\}\left(s\right) \\ =\mathcal{L}\left\{f_3\left(t\right)\right\}\left(s\right)=\mathcal{L}\left\{e^t\right\}\left(s\right) \\ =\frac{1}{s-1} \\ Therefore, \\ \mathcal{L}\left\{f_1\left(t\right)\right\}\left(s\right)=\mathcal{L}\left\{f_2\left(t\right)\right\}\left(s\right)=\mathcal{L}\left\{f_3\left(t\right)\right\}\left(s\right)=\frac{1}{s-1} \end{gathered}$$

Since the inverse Laplace transform must be continuous function on $[0,\infty )$$f_1\left(t\right)$ and $f_2\left(t\right)$  have discontinuities at some points we have that

${\mathcal{L}}^{-1}\left\{\frac{1}{s-1}\right\}\left(t\right)=f_3\left(t\right)=e^t$