Problem 1 - Advanced Engineering Mathematics

Question

(a) The result follows by letting \(\tau=u^{2}\) or \(u=\sqrt{\tau}\) in erf \((\sqrt{t})=\frac{2}{\sqrt{\pi}} \int_{0}^{\sqrt{t}} e^{-u^{2}} d u.\) (b) Using \(\mathscr{L}\left\\{t^{-1 / 2}\right\\}=\frac{\sqrt{\pi}}{s^{1 / 2}}\) and the first translation theorem, it follows from the convolution theorem that $$\begin{aligned} \mathscr{L}\\{\operatorname{erf}(\sqrt{t})\\} &=\frac{1}{\sqrt{\pi}} \mathscr{L}\left\\{\int_{0}^{t} \frac{e^{-\tau}}{\sqrt{\tau}} d \tau\right\\}=\frac{1}{\sqrt{\pi}} \mathscr{L}\\{1\\} \mathscr{L}\left\\{t^{-1 / 2} e^{-t}\right\\}=\left.\frac{1}{\sqrt{\pi}} \frac{1}{s} \mathscr{L}\left\\{t^{-1 / 2}\right\\}\right|_{s \rightarrow s+1} \\\&=\frac{1}{\sqrt{\pi}} \frac{1}{s} \frac{\sqrt{\pi}}{\sqrt{s+1}}=\frac{1}{s \sqrt{s+1}}\end{aligned}$$

Step-by-Step Solution

Answer

The Laplace Transform of \( \mathrm{erf}(\sqrt{t}) \) is \( \frac{1}{s \sqrt{s+1}} \).

Step 1: Substitute Variables in the Integral

We start with the error function expression: \[ ext{erf}(\sqrt{t}) = rac{2}{\sqrt{\pi}} \int_{0}^{\sqrt{t}} e^{-u^2} du \] We are asked to perform a substitution where \( \tau = u^2 \) or \( u = \sqrt{\tau} \). This means \( du = \frac{1}{2\sqrt{\tau}} d\tau \). Substitute these into the integral.

Step 2: Transform the Integral Limits

Given \( \tau = u^2 \), the limits of integration change from \( u = 0 \) to \( \sqrt{t} \). For \( \tau \), these become from \( 0 \) to \( t \). The new integral is: \[ rac{2}{\sqrt{\pi}} \int_{0}^{t} e^{-\tau} \frac{1}{2\sqrt{\tau}} \, d\tau \] which simplifies to \[ \frac{1}{\sqrt{\pi}} \int_{0}^{t} \frac{e^{-\tau}}{\sqrt{\tau}} \, d\tau \]

Step 3: Apply Laplace Transform Properties

Using the convolution theorem and given expression \( \mathscr{L}\{t^{-1/2}\} = \frac{\sqrt{\pi}}{s^{1/2}} \), we compute the Laplace Transform of the error function. The first translation theorem implies a shift in the Laplace domain, denoted by substituting \( s \rightarrow s+1 \).

Step 4: Find the Transformed Function

Compute the Laplace Transform for \( \mathscr{L}\{1\} \and \mathscr{L}\{t^{-1/2} e^{-t}\} = \frac{1}{s} \cdot \frac{\sqrt{\pi}}{\sqrt{s+1}} \).

Step 5: Combine the Results

The final calculation is \[ \frac{1}{\sqrt{\pi}} \cdot \frac{1}{s} \cdot \frac{\sqrt{\pi}}{\sqrt{s+1}} = \frac{1}{s\sqrt{s+1}} \] which is the Laplace Transform of the error function \( \mathrm{erf}(\sqrt{t}) \).

Key Concepts

Error Function

The error function, often denoted as \(\text{erf}(x)\), is a special function encountered in probability, statistics, and partial differential equations. It is used to describe the probability of a random variable falling within a certain range under a normal distribution. The error function is defined as:

• \(\text{erf}(x) = \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-u^2} \, du\)

This formula represents the integral of the Gaussian function over the range from 0 to \(x\). By substituting variables, such as letting \(\tau = u^2\), the error function can be evaluated more easily over different limits. In the context of the Laplace Transform, the error function can be transformed to a simpler expression that is easier to analyze mathematically.
One interesting property of the error function is that it is an odd function, meaning \(\text{erf}(-x) = -\text{erf}(x)\). This symmetry makes it a crucial component in solving problems related to heat equations and diffusion processes.

Convolution Theorem

The convolution theorem is a fundamental principle that connects the process of convolution in the time domain with multiplication in the Laplace domain. This theorem is especially useful for simplifying the calculation of Laplace Transforms of products of functions.
The convolution of two functions \(f(t)\) and \(g(t)\) is defined as:

• \((f * g)(t) = \int_{0}^{t} f(\tau)g(t-\tau) \, d\tau\)

In Laplace Transform terms, the convolution theorem states:

• \(\mathscr{L}\{f * g\}(s) = \mathscr{L}\{f\}(s) \cdot \mathscr{L}\{g\}(s)\)

This allows us to calculate the Laplace Transform of a complicated expression by combining simpler transforms. In the context of the problem, it was used to find the Laplace Transform of the expression involving the error function \(\text{erf}(\sqrt{t})\). By transforming both components separately and then applying the convolution theorem, the solution simplifies to an expression easier to handle mathematically.

Translation Theorem

The translation theorem in the Laplace Transform reveals how a shift in the time domain translates into the frequency domain. It is used when the function we are transforming is multiplied by an exponential function \(e^{at}\).
The first translation theorem, or shifting theorem, is given by:

• If \(F(s) = \mathscr{L}\{f(t)\}\), then \(\mathscr{L}\{e^{at}f(t)\} = F(s-a)\)

This shows how adding an exponential shift to a function affects its Laplace Transform by relocating it along the \(s\)-axis. In the exercise, this theorem was used by replacing \(s\) with \(s+1\) to account for the exponential decay factor \(e^{-t}\) in the error function. This adjustment simplifies calculating the Laplace Transform of such shifted functions by merely substituting \(s+a\) instead of recalculating the entire Transform.

Integral Substitution

Integral substitution is a technique to simplify the evaluation of integrals by changing variables. This approach makes an integral easier to handle or match a known form.

• Begin with the substitution: Choose a new variable \(\tau\) that relates to \(u\) via \(\tau = u^2\).
• Adjust the differential: Derive \(du\) in terms of \(\tau\), so \(du = \frac{1}{2\sqrt{\tau}}d\tau\).
• Transform the limits: Change the original limits of integration from the \(u\)-domain to the \(\tau\)-domain. For example, when \(u\) moves from 0 to \(\sqrt{t}\), \(\tau\) varies from 0 to \(t\).

After applying substitution, integrals such as \(\int e^{-u^2} \, du\) can be transformed into simpler integrals like \(\int \frac{e^{-\tau}}{\sqrt{\tau}} \, d\tau\). This aids in reducing complex expressions and conditions that emerge in problems like calculating the Laplace Transform of the error function.