Question

In this problem we show how a general partial fraction expansion can be used to calculate many inverse Laplace transforms. Suppose that $$ F(s)=P(s) / Q(s) $$ where \(Q(s)\) is a polynomial of degree \(n\) with distinct zeros \(r_{1} \ldots r_{n}\) and \(P(s)\) is a polynomial of degree less than \(n .\) In this case it is possible to show that \(P(s) / Q(s)\) has a partial fraction cxpansion of the form $$ \frac{P(s)}{Q(s)}=\frac{A_{1}}{s-r_{1}}+\cdots+\frac{A_{n}}{s-r_{n}} $$ where the coefficients \(A_{1}, \ldots, A_{n}\) must be determined. \(\begin{array}{ll}{\text { (a) Show that }} & {} \\ {\qquad A_{k}=P\left(r_{k}\right) / Q^{\prime}\left(r_{k}\right),} & {k=1, \ldots, n}\end{array}\) Hint: One way to do this is to multiply Eq. (i) by \(s-r_{k}\) and then to take the limit as \(s \rightarrow r_{k}\) (b) Show that $$ \mathcal{L}^{-1}\\{F(s)\\}=\sum_{k=1}^{n} \frac{P\left(r_{k}\right)}{Q^{\prime}\left(r_{k}\right)} e^{r_{k} t} $$

Short answer

Based on the step-by-step solution provided above, write a short answer to this problem: To find the coefficients A_k of the partial fraction expansion for a given function F(s) expressed as a ratio of two polynomials P(s) and Q(s), we can use the formula \(A_{k}=\frac{P(r_{k})}{Q'(r_{k})}\). The inverse Laplace transform of F(s) can then be expressed as \(\mathcal{L}^{-1}\{F(s)\}=\sum_{k=1}^{n} \frac{P(r_{k})}{Q'(r_{k})} e^{r_{k} t}\).

Step-by-step solution

Part 1: Finding Coefficients A_k

We want to show that \(A_{k} = \frac{P(r_{k})}{Q'(r_{k})}\) for k = 1, ... , n. Using the hint given in the problem, let's multiply the partial fraction expansion equation by \((s-r_{k})\): $$ (s-r_{k}) \frac{P(s)}{Q(s)} = A_1 (s-r_{k})\frac{1}{s-r_1} + ... + A_k (s-r_{k})\frac{1}{s-r_k} + ... + A_n (s-r_{k})\frac{1}{s-r_n} $$ Now take the limit as s approaches \(r_k\): $$ \lim_{s \to r_k}(s-r_k) \frac{P(s)}{Q(s)} = \lim_{s \to r_k} \left[A_1 (s-r_k)\frac{1}{s-r_1} + ... + A_k (s-r_{k})\frac{1}{s-r_k} + ... + A_n (s-r_{k})\frac{1}{s-r_n}\right] $$ Outside of the k-th term, all other terms will go to 0 as s approaches \(r_k\). The k-th term becomes: $$ \lim_{s \to r_k} A_k (s-r_{k})\frac{1}{s-r_k} = A_k $$ By also taking the limit of the left side of the equation, we determine: $$ A_k = \lim_{s \to r_k} (s-r_k)\frac{P(s)}{Q(s)} $$ L'Hôpital's Rule can be applied, since both numerator and denominator approach 0: $$ A_k = \frac{P(r_k)}{Q'(r_k)} $$ Hence, we have proven that \(A_{k}=\frac{P(r_{k})}{Q'(r_{k})}\) for k = 1, ... , n.

Part 2: Inverse Laplace Transform of F(s)

Now that we found the coefficients \(A_k\), let's compute the inverse Laplace transform of \(F(s)\): $$ \mathcal{L}^{-1}\{F(s)\} = \mathcal{L}^{-1} \left\{ \sum_{k=1}^{n} \frac{A_{k}}{s-r_{k}} \right\} $$ Since the inverse Laplace transform is a linear operator, we can write it as a sum of the inverse Laplace transforms of the individual terms: $$ \mathcal{L}^{-1}\{F(s)\} = \sum_{k=1}^{n} \mathcal{L}^{-1}\left\{ \frac{A_{k}}{s-r_{k}} \right\} $$ Using the inverse Laplace transform formula for exponential functions, \(\mathcal{L}^{-1}\left\{ \frac{A_{k}}{s-r_{k}} \right\} = A_{k}e^{r_{k}t}\), we get: $$ \mathcal{L}^{-1}\{F(s)\} = \sum_{k=1}^{n} A_{k}e^{r_{k}t} $$ Substitute the expression for \(A_k\) we found in Part 1: $$ \mathcal{L}^{-1}\{F(s)\} = \sum_{k=1}^{n} \frac{P(r_{k})}{Q'(r_{k})} e^{r_{k}t} $$ Thus, we have shown that the inverse Laplace transform of F(s) can be expressed as \(\mathcal{L}^{-1}\{F(s)\}=\sum_{k=1}^{n} \frac{P(r_{k})}{Q'(r_{k})} e^{r_{k} t}\).

Key concepts

Partial Fraction Expansion

Partial fraction expansion is a powerful technique used in calculus and differential equations, particularly when working with complex rational expressions. This method decomposes a complicated rational expression into simpler fractions, making it more manageable to work with, especially for integration or taking inverse transforms in the context of Laplace transforms.

Consider a rational function where the numerator, denoted by P(s), is a polynomial of lower degree than the denominator polynomial, Q(s). If Q(s) has distinct roots r₁, r₂, ..., r_n, the expression can be rewritten as a sum of fractions with numerators A₁, A₂, ..., A_n and denominators corresponding to the polynomial's roots.

This approach simplifies the process of finding the inverse Laplace transform because it breaks down the complex rational function into individual components associated with simpler inverse transforms, typically exponential functions.

Laplace Transform

The Laplace transform is a widely used integral transform in mathematics with applications in engineering and physics, particularly within the field of differential equations. By transforming a function of time f(t) into a function of complex frequency F(s), it simplifies the analysis of systems governed by linear differential equations.

The Laplace transform has the remarkable property of transforming derivatives into algebraic terms, thereby converting differential equations into algebraic equations. This makes solving for the function f(t) more straightforward, with the inverse Laplace transform allowing one to recover f(t) from its transform F(s). Hence, the inverse Laplace transform is essential for returning from the frequency domain to the time domain, exemplified in our exercise by resolving the sum of exponentials.

Differential Equations

Differential equations play a central role in mathematics, physics, engineering, and many other sciences. They describe the relationship between a function and its derivatives, representing physical phenomena such as motion, heat, sound, and electrical currents.

Many differential equations cannot be solved explicitly, but methods like the Laplace transform provide a strategic workaround. By changing the domain from the time or spatial domain to the complex frequency domain, it turns the problem of solving a differential equation into an algebraic task. This is particularly useful in linear differential equations with constant coefficients, where the method leads to an equation that can be handled using algebraic techniques like partial fraction expansion.

Polynomial Functions

Polynomial functions form the basis of many mathematical expressions and are defined as sums of powers in a variable with coefficients. For example, a polynomial in s, such as P(s), might look like a₀ + a₁s + a₂s² + ... + a_nsⁿ, where the a_i are constants, and n determines the degree of the polynomial.

Significantly, the roots or zeros of a polynomial, often denoted as r₁, r₂, ..., r_n, are the values for which the polynomial equals zero and play a crucial role in techniques such as partial fraction expansion. Recognizing the structure of polynomial functions and their properties, including their roots, is instrumental when applying the Laplace transform to solve differential equations.