Problem 6 - Advanced Engineering Mathematics

Question

We first compute $$\frac{\sinh a \sqrt{s}}{s \sinh \sqrt{s}}=\frac{e^{a \sqrt{s}}-e^{-a \sqrt{s}}}{s\left(e^{\sqrt{s}}-e^{-\sqrt{s}}\right)}=\frac{e^{(a-1) \sqrt{s}}-e^{-(a+1) \sqrt{s}}}{s\left(1-e^{-2 \sqrt{s}}\right)}$$ $$\begin{array}{l}=\frac{e^{(a-1) \sqrt{s}}}{s}\left[1+e^{-2 \sqrt{s}}+e^{-4 \sqrt{s}}+\cdots\right]-\frac{e^{-(a+1) \sqrt{s}}}{s}\left[1+e^{-2 \sqrt{s}}+e^{-4 \sqrt{s}}+\cdots\right] \\\=\left[\frac{e^{-(1-a) \sqrt{s}}}{s}+\frac{e^{-(3-a) \sqrt{s}}}{s}+\frac{e^{-(5-a) \sqrt{s}}}{s}+\cdots\right] \\\\\quad-\left[\frac{e^{-(1+a) \sqrt{s}}}{s}+\frac{e^{-(3+a) \sqrt{s}}}{s}+\frac{e^{-(5+a) \sqrt{s}}}{s}+\cdots\right]\\\=\sum_{n=0}^{\infty}\left[\frac{e^{-(2 n+1-a) \sqrt{s}}}{s}-\frac{e^{-(2 n+1+a) \sqrt{s}}}{s}\right]\end{array}$$ Then $$\mathscr{L}^{-1}\left\\{\frac{\sinh a \sqrt{s}}{s \sinh \sqrt{s}}\right\\}=\sum_{n=0}^{\infty}\left[\mathscr{L}^{-1}\left\\{\frac{e^{-(2 n+1-a) \sqrt{s}}}{s}\right\\}-\mathscr{L}^{-1}\left\\{-\frac{e^{-(2 n+1+a) \sqrt{s}}}{s}\right\\}\right]$$ $$\begin{array}{l}=\sum_{n=0}^{\infty}\left[\operatorname{erfc}\left(\frac{2 n+1-a}{2 \sqrt{t}}\right)-\operatorname{erfc}\left(\frac{2 n+1+a}{2 \sqrt{t}}\right)\right] \\\=\sum_{n=0}^{\infty}\left(\left[1-\operatorname{erf}\left(\frac{2 n+1-a}{2 \sqrt{t}}\right)\right]-\left[1-\operatorname{erf}\left(\frac{2 n+1+a}{2 \sqrt{t}}\right)\right]\right) \\\=\sum_{n=0}^{\infty}\left[\operatorname{erf}\left(\frac{2 n+1+a}{2 \sqrt{t}}\right)-\operatorname{erf}\left(\frac{2 n+1-a}{2 \sqrt{t}}\right)\right].\end{array}$$

Step-by-Step Solution

Answer

The inverse Laplace transform yields a series involving error functions with differing arguments.

Step 1: Simplify the Expression

To simplify the given expression \( \frac{\sinh a \sqrt{s}}{s \sinh \sqrt{s}} \), we start by rewriting the hyperbolic sine functions in terms of exponentials: \( \sinh x = \frac{e^x - e^{-x}}{2} \). This transforms the expression into \( \frac{e^{a \sqrt{s}} - e^{-a \sqrt{s}}}{s(e^{\sqrt{s}} - e^{-\sqrt{s}})} \). Further simplification by separating the exponent values gives us \( \frac{e^{(a-1) \sqrt{s}} - e^{-(a+1) \sqrt{s}}}{s(1 - e^{-2 \sqrt{s}})} \).

Step 2: Expand Exponential Terms

Now, expand the denominator and the newly arranged expressions using geometric series: \( \frac{e^{(a-1) \sqrt{s}}}{s}(1 + e^{-2 \sqrt{s}} + e^{-4 \sqrt{s}} + \cdots) - \frac{e^{-(a+1) \sqrt{s}}}{s}(1 + e^{-2 \sqrt{s}} + e^{-4 \sqrt{s}} + \cdots) \). This results in two series that can be combined.

Step 3: Rearrange and Formulate Series

Rearrange the expanded series and collect like terms: the expression becomes \( \left[ \frac{e^{-(1-a) \sqrt{s}}}{s} + \frac{e^{-(3-a) \sqrt{s}}}{s} + \frac{e^{-(5-a) \sqrt{s}}}{s} + \cdots \right] - \left[ \frac{e^{-(1+a) \sqrt{s}}}{s} + \frac{e^{-(3+a) \sqrt{s}}}{s} + \cdots \right] \), which can be condensed into a single series \( \sum_{n=0}^{\infty}\left[ \frac{e^{-(2n+1-a) \sqrt{s}}}{s} - \frac{e^{-(2n+1+a) \sqrt{s}}}{s} \right] \).

Step 4: Apply Inverse Laplace Transform

With the series formulated, apply the inverse Laplace transform to each term \( \mathscr{L}^{-1}\left\{ \frac{e^{-c \sqrt{s}}}{s} \right\} \). This is known to relate to the complementary error function \( \operatorname{erfc}(x) \). Rewriting gives \( \sum_{n=0}^{\infty} \left[ \operatorname{erfc} \left( \frac{2n+1-a}{2\sqrt{t}} \right) - \operatorname{erfc} \left( \frac{2n+1+a}{2\sqrt{t}} \right) \right] \).

Step 5: Simplify using Error Function Properties

Use the property that \( \operatorname{erfc}(x) = 1 - \operatorname{erf}(x) \) to express in terms of the error function: \( \sum_{n=0}^{\infty} \left( \left[ 1 - \operatorname{erf} \left( \frac{2n+1-a}{2\sqrt{t}} \right) \right] - \left[ 1 - \operatorname{erf} \left( \frac{2n+1+a}{2\sqrt{t}} \right) \right] \right) \). This simplifies to \( \sum_{n=0}^{\infty} \left[ \operatorname{erf} \left( \frac{2n+1+a}{2\sqrt{t}} \right) - \operatorname{erf} \left( \frac{2n+1-a}{2\sqrt{t}} \right) \right] \).

Key Concepts

Hyperbolic Functions

Hyperbolic functions are fundamental in mathematics and arise frequently in algebraic and calculus problems. They share similarities with trigonometric functions but are defined using exponential functions. Mainly, there are two primary hyperbolic functions: hyperbolic sine, denoted as \( \sinh(x) \), and hyperbolic cosine, denoted as \( \cosh(x) \).

For the hyperbolic sine function, we have:

• \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)

For the hyperbolic cosine function, the definition is:

• \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)

These functions can model real-world phenomena like the shape of a hanging cable, known as a catenary.

In solving inverse Laplace transform problems, rewriting hyperbolic functions in their exponential form can greatly simplify the expressions. This is why we use their definitions: they allow us to transform complex hyperbolic terms into a series of simpler exponential terms which are much more manageable to work with in conjunction with other mathematical methods, such as geometric series and error functions.

Geometric Series

A geometric series is a series with a constant ratio between successive terms. It is widely used in mathematics to represent and expand terms in a compact format. For a geometric series, each term is the previous term multiplied by a fixed ratio, denoted as \( r \).

The standard form of a geometric series is:

• \( a + ar + ar^2 + ar^3 + \cdots \)

The sum \( S \) of an infinite geometric series, where the absolute value of the common ratio \( |r| \) is less than 1, is given by the formula:

• \( S = \frac{a}{1 - r} \)

In the exercise process, using geometric series helps to break down and express the exponential terms in a form that can easily be manipulated. This approach is invaluable for evaluating complex expressions since each term in the series can then be assessed individually, typically leading to a more seamless application of other mathematical operations such as the inverse Laplace transform.

Error Function

The error function, \( \operatorname{erf}(x) \), is an important function in probability and statistics, describing the probability that a random variable, with a normal distribution and a mean of zero, is within \( x \) standard deviations of the mean. It is an integral of the Gaussian distribution and is used extensively in statistical and engineering contexts.

Defined mathematically, the error function is:

• \( \operatorname{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} \, dt \)

The complementary error function, \( \operatorname{erfc}(x) \), is simply one minus the error function:

• \( \operatorname{erfc}(x) = 1 - \operatorname{erf}(x) \)

In Laplace transforms, particularly the inverse Laplace transform, these functions arise naturally when dealing with expressions that involve integration of exponential terms. They turn seemingly complex integrals into expressions for which we can directly apply known results (like the error function), aiding in transitioning from the frequency domain back to the time domain efficiently. This is demonstrated in the step-by-step solution where the inverse transform is simplified using error function properties.