Book cover for Advanced Engineering Mathematics

Advanced Engineering Mathematics

Dennis G. Zill, Michael R. Cullen

ISBN #9780763740955

3rd Edition

4,310 Questions

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17,647 Students Helped

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Summary
Learning Objectives
Key Concepts
Example Problems
Explanations
Common Mistakes

Summary

This textbook section covers the integral transform methods including the Laplace and Fourier transforms, emphasizing their utility in solving differential equations and boundary?value problems. Central ideas include the definition and properties of the error function and complementary error functions, the convolution theorem and translation theorem, and the use of Fourier integrals for function representation. Also highlighted are numerical aspects like the Fast Fourier Transform (FFT) and its matrix properties. The material reinforces how these transforms simplify the process of solving complex PDEs by reducing them to algebraic forms in the frequency or transform domain.

Learning Objectives

1

Explain the definition and properties of the error function (erf) and its complementary function (erfc), and how they emerge in Laplace transform solutions.

2

Apply the Laplace transform method to solve differential and partial differential equations using techniques like the convolution theorem and the translation theorem.

3

Understand and utilize Fourier integrals and Fourier transforms (both sine and cosine) for expressing functions and solving boundary?value problems.

4

Analyze and solve problems involving fast Fourier transform (FFT) matrices and their properties in numerical approaches.

5

Integrate theoretical knowledge with practical examples to solve PDEs and boundary value problems using integral transform methods.

Key Concepts

CONCEPT

DEFINITION

Laplace Transform

An integral transform defined by L{f(t)} = ∫₀∞ e^(−st)f(t) dt used to convert differential equations into algebraic equations.

Error Function (erf)

A special function defined as erf(t) = (2/√π) ∫₀ᵗ e^(−u²) du, commonly arising in solutions of diffusion and heat equations.

Complementary Error Function (erfc)

Defined by erfc(x) = 1 − erf(x), used to express solutions and transform properties in PDEs and probability theory.

Fourier Transform

A tool to decompose functions into sinusoids. The Fourier transform of f(x) is F(α) = ∫₋∞∞ f(x)e^(−iαx) dx and is useful in solving differential equations and analyzing signals.

Fourier Integral

Representation of a function via an integral involving sine and cosine functions, allowing reconstruction from its frequency components.

Convolution Theorem

A theorem stating that the Laplace or Fourier transform of a convolution of two functions is the product of their individual transforms.

Translation Theorem

A property that relates a shift in time (or space) to the multiplication by an exponential factor in the transform domain.

Fast Fourier Transform (FFT)

An efficient algorithm to compute the discrete Fourier transform (DFT) and its inverse, important for numerical solutions of PDEs.

Example Problems

Example 1

(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 $$\begin{aligned} \mathscr{L}\{\operatorname{erf}(\sqrt{t})\} &=\frac{1}{\sqrt{\pi}} \mathscr{L}\left\{\int_{0}^{t} \frac{e^{-\tau}}{\sqrt{\tau}} d \tau\right\}=\frac{1}{\sqrt{\pi}} \mathscr{L}\{1\} \mathscr{L}\left\{t^{-1 / 2} e^{-t}\right\}=\left.\frac{1}{\sqrt{\pi}} \frac{1}{s} \mathscr{L}\left\{t^{-1 / 2}\right\}\right|_{s \rightarrow s+1} \\&=\frac{1}{\sqrt{\pi}} \frac{1}{s} \frac{\sqrt{\pi}}{\sqrt{s+1}}=\frac{1}{s \sqrt{s+1}}\end{aligned}$$

Example 2

since erfc $(\sqrt{t})=1-\operatorname{erf}(\sqrt{t})$ we have $$\mathscr{L}\{\operatorname{erfc}(\sqrt{t})\}=\mathscr{L}\{1\}-\mathscr{L}\{\operatorname{erf}(\sqrt{t})\}=\frac{1}{s}-\frac{1}{s \sqrt{s+1}}=\frac{1}{s}\left[1-\frac{1}{\sqrt{s+1}}\right].$$

Example 3

By the first translation theorem, $$\mathscr{L}\left\{e^{t} \operatorname{erf}(\sqrt{t})\right\}=\left.\mathscr{L}\{\operatorname{erf}(\sqrt{t})\}\right|_{s \rightarrow s-1}=\left.\frac{1}{s \sqrt{s+1}}\right|_{s \rightarrow s-1}=\frac{1}{\sqrt{s}(s-1)}.$$

Example 4

By the first translation theorem and the result of Problem 2, $$\begin{aligned}\mathscr{L}\left\{e^{t} \operatorname{erfc}(\sqrt{t})\right\} &=\left.\mathscr{L}\{\operatorname{erfc}(\sqrt{t})\}\right|_{s \rightarrow s-1}=\left.\left(\frac{1}{s}-\frac{1}{s \sqrt{s+1}}\right)\right|_{s \rightarrow s-1}=\frac{1}{s-1}-\frac{1}{\sqrt{s}(s-1)} \\&=\frac{\sqrt{s}-1}{\sqrt{s}(s-1)}=\frac{\sqrt{s}-1}{\sqrt{s}(\sqrt{s}+1)(\sqrt{s}-1)}=\frac{1}{\sqrt{s}(\sqrt{s}+1)}.\end{aligned}$$

Example 5

From entry 3 in Table 15.1 and the first translation theorem we have $$\mathscr{L}\left\{e^{-G t / C} \operatorname{erf}\left(\frac{x}{2} \sqrt{\frac{R C}{t}}\right)\right\}=\mathscr{L}\left\{e^{-G t / C}\left[1-\operatorname{erfc}\left(\frac{x}{2} \sqrt{\frac{R C}{t}}\right)\right]\right\}$$ $$\begin{array}{l} =\mathscr{L}\left\{e^{-G t / C}\right\}-\mathscr{L}\left\{e^{-G t / C} \operatorname{erfc}\left(\frac{x}{2} \sqrt{\frac{R C}{t}}\right)\right\} \\=\frac{1}{s+G / C}-\left.\frac{e^{-x \sqrt{R C}} \sqrt{s}}{s}\right|_{s \rightarrow s+G / C} \\=\frac{1}{s+G / C}-\frac{e^{-x \sqrt{R C} \sqrt{s+G / C}}}{s+G / C}=\frac{C}{C s+G}\left(1-e^{x \sqrt{R C s+R G}}\right).\end{array}$$
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Step-by-Step Explanations

QUESTION

How can one derive the Laplace transform of erf(√t) using a substitution such as τ = u²?

STEP-BY-STEP ANSWER:

Step 1: Begin with the definition: erf(√t) = (2/√π) ∫₀^(√t) e^(−u²) du.
Step 2: Use the substitution τ = u²; then dτ = 2u du, so that u = √τ and du = dτ/(2√τ).
Step 3: Substitute into the integral to express erf(√t) in terms of τ: erf(√t) = (2/√π) ∫₀^(t) e^(−τ) dτ/(2√τ) = (1/√π) ∫₀^(t) e^(−τ)/√τ dτ.
Step 4: Recognize that the Laplace transform of t^(−1/2) is known, and apply the convolution theorem combined with the first translation theorem to obtain the transform result.
Final Answer: The Laplace transform yields an expression of the form 1/(s√(s+1)) or similar, depending on translation shifts.

Laplace Transform of the Error Function

QUESTION

How is a function f(x) represented as a Fourier integral using sine and cosine transforms?

STEP-BY-STEP ANSWER:

Step 1: Write the Fourier integral representation as f(x) = (1/π) ∫₀∞ [A(α) cos(αx) + B(α) sin(αx)] dα.
Step 2: Determine the coefficients A(α) and B(α) by computing the integrals A(α) = ∫ f(x) cos(αx) dx and B(α) = ∫ f(x) sin(αx) dx under proper limits.
Step 3: Substitute these coefficients back into the representation and use the properties of sine and cosine to reconstruct the function.
Final Answer: f(x) is expressed as an integral combination of cos(αx) and sin(αx) weighted by A(α)/B(α), facilitating analysis in the frequency domain.

Fourier Integral Representation

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Common Mistakes

  • Misapplying substitution rules in the derivation of the error function’s Laplace transform, particularly handling the differential conversion (du vs. d?).
  • Confusing the error function (erf) with its complement (erfc), and misinterpreting their respective limit values.
  • Overlooking the importance of conditions such as symmetry when deciding to use Fourier sine or cosine transforms.
  • Forgetting to correctly apply the convolution theorem, especially when handling shifted transforms via the translation theorem.
  • Errors in sign conventions while expanding series or evaluating integrals in transform methods.