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Question

Prove Theorem 7, page 368, on the linearity of the inverse transform. [Hint: Show that the right-hand side of equation (3) is a continuous function on $[0, \infty)$ whose Laplace transform is $[F_{1} (s) + F_{2} (s)]$

Step-by-Step Solution

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Answer

It is proved that:

$L^{- 1} \{F_{1} + F_{2}\} = L^{- 1} \{F_{1}\} + L^{- 1} \{F_{2}\} L^{- 1} \{c F\} = c L^{- 1} F$

1Step 1: Define Inverse Laplace transform

Given a function F(s) ,if there is a function f(t) that is continuous on $[0, \infty)$

and satisfies $L \{f\} = F$ then we say that f(t) is the inverse Laplace transform of F(s) and employ the notation $f = L^{- 1} \{F\}$ .

Non-repeated Linear Factors

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

$Q (s) = (s - r_{1}) (s - r_{2}) . . . (s - r_{n})$

where the $r_{i}$ 's are all distinct real numbers, then the partial fraction expansion has the form

$\frac{P (s)}{Q (s)} = \frac{A_{1}}{s - r_{1}} + \frac{A_{2}}{s - r_{2}} + . . . + \frac{A_{n}}{s - r_{n}}$

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

$\frac{A_{1}}{s - r} + \frac{A_{2}}{{(s - r_{1})}^{2}} + . . . + \frac{A_{m}}{{(s - r)}^{m}}$

where the $A_{i}$ 's are real numbers.

Quadratic Factors

If ${(s - α)}^{2} + β^{2}$ is a quadratic factor of $Q (s)$ that cannot be reduced to linear factors with real coefficients and m is the highest power of ${(s - α)}^{2} + β$ that divides Q(s) , then the portion of the partial fraction expansion that corresponds to " width="9" height="19" style="max-width: none;" > ${(s - α)}^{2} + β^{2}$ is

$\frac{C_{1} S + D_{1}}{{(s - α)}^{2} + β^{2}} + \frac{C_{2} S + D_{2}}{{[{(s - α)}^{2} + β^{2}]}^{2}} + . . . + \frac{C_{m} S + D_{m}}{{[{(s - α)}^{2} + β^{2}]}^{m}}$

it is more convenient to express $C_{i} s + D_{i}$ in the form $A_{i} (s - α) + β B_{i}$ when we look up the Laplace transforms. So let's agree to write this portion of the partial fraction expansion in the equivalent form

$\frac{A_{1} (s - α) + β B_{1}}{{(S - α)}^{2} + β^{2}} + \frac{A_{2} (s - α) + β B_{2}}{{[{(S - α)}^{2} + β^{2}]}^{2}} + . . . + \frac{A_{M} (s - α) + β B_{M}}{{[{(S - α)}^{2} + β^{2}]}^{m}}$

2Step 2: Find the factor of the denominator

Since $L^{- 1} \{F_{1}\}$ and $L^{- 1} \{F_{2}\}$ are both continuous functions on $[0, \infty)$ we have that their sum also is continuous function on $[0, \infty)$