Heaviside's Expansion Formula. Let P(s) and Q(s) be polynomials with the degree of P(s) less than the degree of Q(s) . Letrole="math" localid="1664358171256" Q(s) =s-r1 s-r2...s-rnrole="math" localid="1664359866810" , L-1PQt=∑i=1nPriPrierit

40E - Fundamentals Of Differential Equations And Boundary Value Problems

40E

Question

Heaviside's Expansion Formula. Let $P (s)$ and $Q (s)$ be polynomials with the degree of $P (s)$ less than the degree of $Q (s)$ . Let

$Q (s) = (s - r_{1}) (s - r_{2}) . . . (s - r_{n})$ $, L^{- 1} \{\frac{P}{Q}\} (t) = \sum_{i = 1}^{n} \frac{P (r_{i})}{P (r_{i})} e^{r_{i} t}$

Step-by-Step Solution

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Answer

Using partial fractions we find that $L^{- 1} \{\frac{P (s)}{Q (s)}\} (t) = \sum_{i = 1}^{n} \frac{P (r_{i})}{P (r_{i})} e^{r_{i} t}$ and then similarly as in problem 38 we find that $A_{i} = \frac{P (r_{i})}{Q (r_{i})}$ from where it follows the wanted result.

1Step 1: Define Inverse Laplace transform

Given a function , if there is a function that is continuous on

and satisfies,then we say that is the inverse Laplace transform of and employ the notation

Non-repeated Linear Factors

If can be factored into a product of distinct linear factors,

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

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

Repeated Linear Factors

If is a factor of and is the highest power of that divides , then the portion of the partial fraction expansion of that corresponds to the term is

where the 's are real numbers.

Quadratic Factors

If is a quadratic factor of that cannot be reduced to linear factors with real coefficients and is the highest power of that divides , then the portion of the partial fraction expansion that corresponds to is

it is more convenient to express in the form when we look up the Laplace transforms. So let's agree to write this portion of the partial fraction expansion in the equivalent form

2Step 2: Find the factor of the denominator

Since

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

Then, using partial fractions we get

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

from where it follows

$L^{- 1} \{\frac{P (s)}{Q (s)}\} (t) = {\sum L^{- 1}}_{i = 1}^{n} \{\frac{A_{i}}{s - r_{i}}\} = \sum_{i = 1}^{n} A_{i} e^{r_{i} t}$

Multiplying equation 1 with $s - r_{i}$ we get

$\frac{s - r_{i} P (s)}{Q (s)} = A_{i} + \sum_{i \neq j, j = 1}^{n} \frac{(s - r_{i}) A_{j}}{s - r_{j}}$

3Step 3: Applying the limit

Applying $\lim_{s \to r_{j}}$ gives

$\lim_{s \to r_{i}} \frac{(s - r_{i}) P (s)}{Q (s)} = \lim_{s \to r} (A_{i} + \sum_{i \neq j, j = 1}^{n} \frac{(s - r_{i}) A_{j}}{s - r_{j}}) = A_{i} + 0 + . . . + 0 = A_{i}$

from where it follows that

$A_{i} = \lim_{s \to r} \frac{P (s)}{\frac{Q (s)}{(s - r_{i})}} = \frac{P (s)}{\lim_{s \to r} \frac{Q (s)}{(s - r_{i})}} = \frac{P (r_{i})}{Q (r_{i})}$

Therefore,

$L^{- 1} \{\frac{P}{Q}\} (t) = \sum_{i = 1}^{n} \frac{P (r_{i})}{P (r_{i})} e^{r_{i} t}$

39E

43E

Other exercises in this chapter

38E

Residue Computation. Let PsQs be a rational function with degP<degQ and suppose s-r is a non-repeated linear factor of Qs . Pr

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39E

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43E

Use the residue formulas derived in Problems 38 and 42 to determine the partial fraction expansion forF(s)=6s2+28s2-2s+5(s+2)

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Q13E

Use the Laplace transform table and the linearity of the Laplace transform to determine the following transforms.\(L\left\{ {6{e^{ - 3t}} - {t^2} + 2t - 8} \rig

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