/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 An important application of comp... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An important application of complex variables is to solve equations for functions analytic in a certain region, given a relationship on a boundary (see Chapter 7). A simple example of this is the following. Solve for the function \(\psi^{+}(z)\) analytic in the upper half plane and \(\psi^{-}(z)\) analytic in the lower half plane, given the following relationship on the real axis \((\operatorname{Re} z=x)\) $$ \psi^{+}(x)-\psi^{-}(x)=f(x) $$ where, say, \(f(x)\) is differentiable and absolutely integrable. A solution to this problem for \(\psi^{\pm}(z) \rightarrow 0\) as \(|z| \rightarrow \infty\) is given by $$ \psi^{\pm}(x)=\lim _{\epsilon \rightarrow 0^{+}} \frac{1}{2 \pi i} \int_{-\infty}^{\infty} \frac{f(\zeta)}{\zeta-(x \pm i \epsilon)} d \zeta $$ To simplify the analysis, we will assume that \(f(x)\) can be analytically extended in the neighborhood of the real axis. (a) Explain how this solution could be formally obtained by introducing the projection operators $$ P^{\pm}=\lim _{\epsilon \rightarrow 0^{+}} \frac{1}{2 \pi i} \int_{-\infty}^{\infty} \frac{d \zeta}{\zeta-(x \pm i \epsilon)} $$ and in particular why it should be true that $$ P^{+} \psi^{+}=\psi^{+}, \quad P^{-} \psi^{-}=-\psi^{-}, \quad P^{+} \psi^{-}=P^{-} \psi^{+}=0 $$ (b) Verify the results in part (a) for the example $$ \psi^{+}(x)-\psi^{-}(x)=\frac{1}{x^{4}+1} $$ and find \(\psi^{\pm}(z)\) in this example. (c) Show that $$ \psi^{\pm}(x)=\frac{1}{2 \pi i} \int_{-\infty}^{\infty} \frac{f(\zeta)}{\zeta-x} d \zeta \pm \frac{1}{2} f(x) $$ In operator form the first term is usually denoted as \(H(f(x)) / 2 i\) where \(H(f(x))\) is called the Hilbert transform. Then show that $$ \psi^{\pm}(x)=\frac{1}{2 i} H(f(x)) \pm \frac{1}{2} f(x) $$ or, in operator form $$ \psi^{\pm}(x)=\frac{1}{2}(\pm 1-i H) f(x) $$

Short Answer

Expert verified
\( \psi^{\pm}(x) = \frac{1}{2}(\pm 1 - i H) f(x) \)

Step by step solution

01

- Review the Projection Operators

Define the projection operators as follows:\[ P^{\pm} = \lim _{\epsilon \rightarrow 0^{+}} \frac{1}{2 \pi i} \int_{-\infty}^{\infty} \frac{d \zeta}{\zeta-(x \pm i \epsilon)} \]
02

- Apply Projection Operators to \(\psi^{\pm}(x)\)

Apply the projections to \(\psi^{+}(x)\) and \(\psi^{-}(x)\). Use the definitions and evaluate the integrals:\( P^{+} \psi^{+} = \psi^{+} \)\( P^{-} \psi^{-} = -\psi^{-} \)\( P^{+} \psi^{-} = 0 \)\( P^{-} \psi^{+} = 0 \)
03

- Verify Projection Results with Example

For \( \psi^{+}(x) - \psi^{-}(x) = \frac{1}{x^{4} + 1} \), use the given formula to verify the relations:\[ \psi^{\pm}(x) = \frac{1}{2 \pi i} \int_{-\infty}^{\infty} \frac{f(\zeta)}{\zeta - x} d\zeta \pm \frac{1}{2} f(x) \]
04

- Show the Operator Form of the Solution

Rewrite the solution in operator form by defining the Hilbert transform \(H(f(x))\):\[ \psi^{\pm}(x) = \frac{1}{2 i} H(f(x)) \pm \frac{1}{2} f(x) \]
05

- Derive the Final Operator Form

Combine and simplify the terms to get the final operator form of the solution:\[ \psi^{\pm}(x) = \frac{1}{2}(\pm 1 - i H) f(x) \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hilbert Transform
The Hilbert Transform is a specific integral transform used extensively in complex analysis and signal processing.
It operates on real-valued functions and creates a function with imaginary values.
The transform's integral form is expressed as \[ H(f(x)) = \frac{1}{\pi} \text\int_{-\infty}^{\infty} \frac{f(\zeta)}{x - \zeta} d \zeta \]
The Hilbert Transform plays a crucial role in forming analytic signals, which are used to define the analytic parts of a given function in complex analysis.
In solving boundary value problems, it helps to isolate the contributions from the upper and lower halves of the complex plane.
Analytic Functions
Analytic functions are foundational in complex analysis.
They are functions that are differentiable at every point in their domain, not just once but infinitely many times.
This means they have derivatives of all orders, which leads to some intriguing properties and powerful results. For instance, analytic functions can be represented by power series.
These functions are also conformal, preserving shapes locally except at critical points. In boundary value problems, we often look for solutions that are analytic in certain regions, as they ensure smoothness and stability in the solutions.
Boundary Value Problems
Boundary Value Problems (BVPs) involve finding functions that satisfy certain conditions on the boundaries of their domains.
They are especially significant in physical applications, such as in fluid dynamics and electrostatics.
In complex analysis, we deal with BVPs where the solutions are required to be analytic within a region and obey specific relations on the region's boundary.
For example, solving \[ \psi^+(x) - \psi^-(x) = f(x) \] on the real axis translates into ensuring the respective functions address the boundary conditions, leveraging tools like the Hilbert Transform.
Complex Plane
The complex plane is a staple in complex analysis.
It extends the real number line to include imaginary numbers, creating a two-dimensional plane where each point represents a complex number. This plane-view allows for visualizing and solving more intricate problems.
One crucial feature is the division into the upper and lower half-planes, denoted by the imaginary part being positive or negative, respectively.
Solutions often need to address conditions within these specific regions, as the upper and lower half-planes exhibit different analytical properties.
Working within the complex plane brings more tools to bear, such as conformal mappings and residues, which illuminate otherwise opaque problems.
Integral Transforms
Integral Transforms are operations that convert functions into different spaces by integrating them against a kernel.
The most famous examples are the Fourier Transform and the Laplace Transform, often used in differential equations and signal processing.
The Hilbert Transform is a specific kind of integral transform in the realm of complex functions.
These transforms help in transitioning a hard-to-solve problem into an easier one, often transforming differential equations into algebraic ones or simplifying convolution operations.
By leveraging these transformations, we can tackle complex boundary problems more methodically and effectively.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Show that the inverse Laplace transform of \(\hat{F}(s)=e^{-a s^{1 / 2}} / s, a>0\), is given by $$ f(x)=1-\frac{1}{\pi} \int_{0}^{\infty} \frac{\sin \left(a r^{1 / 2}\right)}{r} e^{-r x} d r $$ Note that the integral converges at \(r=0\). (b) Use the definition of the error function integral $$ \operatorname{erf} x=\frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-r^{2}} d r $$ to show that an alternative form for \(f(x)\) is $$ f(x)=1-\operatorname{erf}\left(\frac{a}{2 \sqrt{x}}\right) $$ (c) Show that the inverse Laplace transform of \(\left.\hat{F}(s)=e^{-a s^{1 / 2}} / s^{1 / 2}, a\right\rangle\) 0 , is given by $$ f(x)=\frac{1}{\pi} \int_{0}^{\infty} \frac{\cos \left(a r^{1 / 2}\right)}{r^{1 / 2}} e^{-r x} d r $$ or the equivalent forms $$ f(x)=\frac{2}{\pi \sqrt{x}} \int_{0}^{\infty} \cos \frac{a u}{\sqrt{x}} e^{-u^{2}} d u=\frac{1}{\sqrt{\pi x}} e^{-a^{2} / 4 x} $$ Verify this result by taking the derivative with respect to \(a\) in the formula of part (a). (d) Follow the procedure of part (c) and show that the inverse Laplace transform of \(\hat{F}(s)=e^{-a x^{1 / 2}}\) is given by $$ f(x)=\frac{a}{2 \sqrt{\pi} x^{3 / 2}} e^{-a^{2} / 4 x} $$

Consider a rectangular contour \(C_{R}\) with corners at \((\pm R, 0)\) and \((\pm R, a)\). Show that $$ \oint_{C_{R}} e^{-z^{2}} d z=\int_{-R}^{R} e^{-x^{2}} d x-\int_{-R}^{R} e^{-(x+i a)^{2}} d x+J_{R}=0 $$ where $$ J_{R}=\int_{0}^{a} e^{-(R+i y)^{2}} i d y-\int_{0}^{a} e^{-(-R+i y)^{2}} i d y $$ Show \(\lim _{R \rightarrow \infty} J_{R}=0\), whereupon we have \(\int_{-\infty}^{\infty} e^{-(x+i a)^{2}} d x=\int_{-\infty}^{\infty} e^{-x^{2}} d x\) \(=\sqrt{\pi}\), and consequently, deduce that \(\int_{-\infty}^{\infty} e^{-x^{2}} \cos 2 a x d x=\sqrt{\pi} e^{-a^{2}}\)

Determine the type of singular point each of the following functions has at \(z=\infty:\) (a) \(z^{m}, m=\) positive integer (b) \(z^{1 / 3}\) (c) \(\left(z^{2}+a^{2}\right)^{1 / 2}, \quad a^{2}>0\) (d) \(\log z\) (e) \(\log \left(z^{2}+a^{2}\right), \quad a^{2}>0\) (f) \(e^{z}\) (g) \(z^{2} \sin \frac{1}{z}\) (h) \(\frac{z^{2}}{z^{3}+1}\) (i) \(\sin ^{-1} z\) (j) \(\log \left(1-e^{1 / z}\right)\)

Suppose that \(f(z)\) is analytic in a region containing a simple closed contour C. Let \(|f(z)| \leq M\) on \(C\), and show via Rouché's Theorem that \(|f(z)| \leq M\) inside \(C\). (The maximum of an analytic function is attained on its boundary; this provides an alternate proof of the maximum modulus result in Section 2.6.) Hint: suppose there is a value of \(f(z)\), say \(f_{0}\), such that \(\left|f_{0}\right|>M\). Consider the two functions \(-f_{0}\) and \(f(z)-f_{0}\), and use \(\left|f(z)-f_{0}\right| \geq\) \(\left|f_{0}\right|-|f(z)|\) in Rouché's Theorem to deduce that \(f(z) \neq f_{0}\).

Given the wave equation (with wave speed being unity) $$ \frac{\partial^{2} \phi}{\partial t^{2}}-\frac{\partial^{2} \phi}{\partial x^{2}}=0 $$ and the boundary conditions $$ \begin{array}{cc} \phi(x, t=0)=0, & \frac{\partial \phi}{\partial t}(x, t=0)=0 \\ \phi(x=0, t)=0, & \phi(x=\ell, t)=1 \end{array} $$ (a) obtain the Laplace transform of the solution \(\hat{\Phi}(x, s)\) $$ \hat{\Phi}(x, s)=\frac{\sinh s x}{s \sinh s \ell} $$ (b) Obtain the solution \(\phi(x, t)\) by inverting the Laplace transform to find $$ \phi(x, t)=\frac{x}{\ell}+\sum_{n=1}^{\infty} \frac{2(-1)^{n}}{n \pi} \sin \left(\frac{n \pi x}{\ell}\right) \cos \left(\frac{n \pi t}{\ell}\right) $$ (see also Problem (19), Section 4.5).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.