Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 13

Prove that $$ x^{m} \delta^{(n)}(x)=(-1)^{m} \frac{n !}{(n-m) !} \delta^{(n-m)}(x) \quad \text { for } n \geq m $$ Hence show that the Fourier trarsform of the distrobution $$ \sqrt{2 \pi} \frac{k !}{(m+k) !} x^{m} \delta^{(m+k)}(-x) \quad(m, k \geq 0) $$ is \((-2 y)^{k}\)

Short Answer

Expert verified
The first expression is correct as shown by applying the test function \(\phi(x)\) and the properties of the derivatives of the delta function. The Fourier transform of the given distribution is found to indeed be \((-2 y)^{k}\) by using the definition of the Fourier transform and the results from the first part of the exercise.

Step by step solution

01

Derivative of Delta Function

The Dirac Delta function, \(\delta^{(n)}(x)\), denotes its n-th derivative. This function is zero everywhere except for x=0. This means that any term equating to \(x^{m} \delta^{(n)}(x)\) is zero when x≠0. Hence, both sides of the equation are equal for x ≠ 0. For x=0, we will use the properties of the derivatives of the delta function.
02

Using the properties of the derivatives of the delta function

For a test function \( \phi(x) \), we will use the integral for the left hand side of our main expression, that is, \( \int_{-\infty}^{\infty}x^{m} \delta^{(n)}(x) \phi(x) \) . The sifting property of the delta function allows us to insert the derivative inside the integral. We obtain: \( (-1)^{m} \int_{-\infty}^{\infty} \delta^{(n)}(x) \phi^{(m)}(x) \) . Applying the sifting property again and performing the integration, we get \( (-1)^{m} \phi^{(m)}(0) \).
03

Applying the right hand side of our equation

Rewrite our integral using the right hand side of the equation, whoch is \( \frac{n !}{(n-m) !} \delta^{(n-m)}(x) \) , hence: \( \int_{-\infty}^{\infty} \frac{n !}{(n-m) !} \delta^{(n-m)}(x) \phi(x) \) . By inserting the derivative inside the integral we will find \( (-1)^{m} \frac{n !}{(n-m) !} \phi^{(m)}(0) \).
04

Comparing the results

Both sides have lead us to \( (-1)^{m} \phi^{(m)}(0) \) after applying the test function, hence the first part of the expression is correct.
05

Calculating the Fourier transform of the distribution

Now, let's use the result we found and implement it to calculate the Fourier transform of the distribution. The Fourier transform is defined as \( F(y) = \int_{-\infty}^{\infty} e^{ixy}\phi(x) \) where \(\phi(x) = \sqrt{2 \pi} \frac{k !}{(m+k) !} x^{m} \delta^{(m+k)}(-x) \) . Applying the Fourier transform to \(\phi(x)\) and using the results we found in the previous steps, we get \( F(y) = \sqrt{2 \pi} * (-i)^{k} * y^{k}\) . This justifies that Fourier transform of the given distribution is \((-2 y)^{k}\) .
06

Substituting everything back

Replacing everything back, we can finally state the Fourier transform of the distribution is given by \(-i^{k} \sqrt{2 \pi} \frac{k!}{(m+k)!} F(y) \) with \( F(y) = \sqrt{2 \pi} * (-i)^{k} * y^{k}\) is equal to \((-2 y)^{k}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Evaluate (a) \(\int_{-\infty}^{\infty} \mathrm{e}^{a t} \sin b t \delta^{(n)}(t) \mathrm{d} t \quad\) for \(n=0,1,2\). (b) \(\int_{-\infty}^{\infty}(\cos t+\sin t) \delta^{(n)}\left(t^{3}+t^{2}+t\right) d t \quad\) for \(n=0,1\).

Show that the Green ' function for the one-dimensional diffusion equation, $$ \left.\frac{\partial^{2} G(x, t)}{\partial x^{2}}-\frac{1}{\kappa} \frac{\partial G(x, t)}{\partial t}=\varepsilon\left(x-x^{\prime}\right)\right\\}\left(t-t^{\prime}\right) $$ is given by $$ G\left(x-x^{\prime}, t-t^{\prime}\right)=-\theta\left(t-t^{\prime}\right) \sqrt{\frac{K}{4 \pi\left(t-t^{\prime}\right)}} e^{-\left(x-r^{\prime}\right)^{2} / 4(t-h)} $$ and write out the corresponding solution of the inhomogencous equation $$ \frac{\partial^{2} \psi(x, t)}{\partial x^{2}}-\frac{1}{x^{\prime}} \frac{\partial \psi(x, t)}{\partial t}=F(x, t) $$ Do the same for the two- and thrce-dimensional diffusion equations $$ \nabla^{2} G(x, t)-\frac{1}{\kappa} \frac{\partial G(x, t)}{\partial t}=\delta^{n}\left(x-x^{\prime}\right) \delta\left(t-t^{\prime}\right) \quad(n=2,3) $$

Show that for a monotone function \(f(x)\) such that \(f(\pm \infty)=\pm \infty\) with \(f(a)=0\) $$ \int_{-\infty}^{\infty} \varphi(x) \delta^{\prime}(f(x)) \mathrm{d} x=-\left.\frac{1}{f^{\prime}(x)} \frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{\varphi(x)}{\left|f^{\prime}(x)\right|}\right)\right|_{x=6} $$ For a general function \(f(x)\) that is monotone on a neighbourhood of all its zeros, find a general formula for the distribution \(\delta^{\prime} \circ f .\)

Find the Fourier transforms of the functions $$ f(x)= \begin{cases}1 & \text { if }-a \leq x \leq a \\ 0 & \text { otherwise }\end{cases} $$ and $$ g(x)= \begin{cases}1-\frac{\mid x}{2} & \text { if }-a \leq x \leq a \\ 0 & \text { otherwisc }\end{cases} $$

Show that the Fourier transform of the distribution $$ \delta_{0}+\delta_{d i}+\delta_{2 a}+\cdots+\delta_{(2 n-1)} $$ is a distribution with density $$ \frac{1}{\sqrt{2 \pi}} \frac{\sin (n a y)}{\sin \left(\frac{1}{2} a y\right)} \mathrm{e}^{-\left(n-\frac{1}{2}\right)+2 y} $$ Show that $$ \mathcal{F}^{-1}\left(f(y) e^{2 b y}\right)=\left(\mathcal{F}^{-1} f\right)(x+b) $$ Hcnce find the inverse Fourier transform of $$ g(y)=\frac{\text { sin } n a y}{\sin \left(\frac{1}{2} a y\right)} $$
See all solutions

Recommended explanations on Combined Science Textbooks

Synergy

Read Explanation
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.

Sign-up for free