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Let \(\phi(x)=e^{-|x| / 2}\) on \(\mathbb{R}\). Use the Fourier transform to derive the solution \(u=f * \phi\) of the differential equation \(u-u^{\prime \prime}=f\), and then check directly that it works. What hypotheses are needed on \(f\) ?

Short Answer

Expert verified
Solve using Fourier transform. Hypotheses needed: \( f \) should have a well-behaved Fourier transform.

Step by step solution

01

Understand the Problem

We need to solve the differential equation \( u - u'' = f \) using the function \( u = f * \phi \), where \( f * \phi \) denotes the convolution. The task involves employing the Fourier transform to find and verify this solution. Additionally, we need to determine any necessary conditions on \( f \).
02

Fourier Transform the Differential Equation

Apply the Fourier transform to both sides of the equation \( u - u'' = f \). The Fourier transform of \( u'' \) is \( -\omega^2 \hat{u}(\omega) \). Therefore, the equation becomes \( \hat{u}(\omega) + \omega^2 \hat{u}(\omega) = \hat{f}(\omega) \). Simplifying gives \( \hat{u}(\omega) (1 + \omega^2) = \hat{f}(\omega) \).
03

Solve in the Frequency Domain

Solve for \( \hat{u}(\omega) \) which is \( \hat{u}(\omega) = \frac{\hat{f}(\omega)}{1 + \omega^2} \). This expression indicates \( \hat{u}(\omega) \) is the Fourier transform of the convolution of \( f \) with the inverse Fourier transform of \( \frac{1}{1 + \omega^2} \).
04

Find the Inverse Fourier Transform

Compute the inverse Fourier transform of \( \frac{1}{1 + \omega^2} \), which is \( e^{-\frac{|x|}{2}} \) or similar due to the properties of exponential decay. So, \( u(x) = (f * \phi)(x) \) where \( \phi(x) = e^{-\frac{|x|}{2}} \).
05

Check the Solution

Substitute \( u = f * \phi \) back into the original equation. The convolution operator deals with sums of derivatives using the F.T., validating \( u - u'' = f \) holds after substitution. The presence of exponential terms confirm equality.
06

Identify Hypotheses on \( f \)

Ensure \( f \) is such that its Fourier transform exists and is well-behaved, possibly requiring \( f \) to be a tempered distribution or function with certain decay properties at infinity.

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Key Concepts

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

Convolution Theory
Convolution is a fundamental mathematical operation often used in the field of signal processing and applied mathematics. This operation involves two functions, where one is essentially "slid" across the other to produce a third function. In more detailed terms, the convolution of two functions, say \( f(x) \) and \( \,g(x) \), is defined as:
  • \((f * g)(x) = \int_{-\infty}^{\infty} f(t)g(x - t) dt \)
The operation blends one function with the shape of another, effectively smoothing or transforming the data. It is widely used in solving differential equations and plays a crucial role in areas like image processing and probability theory. In our context, the convolution \( f * \phi \) helps derive solutions to differential equations by transforming them into domains where they are easier to manipulate. This operation allows us to use the powerful tool of Fourier Transforms, simplifying complex problems.
Understanding Differential Equations
Differential equations are mathematical equations that involve functions and their derivatives. They are essential because they describe various physical phenomena, such as heat conduction, wave propagation, and dynamic systems. The basic idea is that these equations express relationships between varying quantities and their rates of change.
  • Our exercise deals with a specific type of differential equation: \( u - u'' = f \).
  • A solution to this equation involves finding a function \( u(x) \) whose behavior matches the equation across an interval.
Solving a differential equation like this usually involves several methods, including separation of variables, integrating factors, or, as in this case, using Fourier Transforms. It is important to check the solution directly by substituting it back into the original equation to ensure it satisfies the given relationships.
Inverse Fourier Transform
The Fourier Transform is a method that translates a function of time or space into a function of frequency. However, to retrieve the original time- or space-based function after manipulation or analysis in the frequency domain, the Inverse Fourier Transform is used.
  • If \( \hat{f}(\omega) \) is the Fourier Transform of a function \( f(x) \), then the Inverse Fourier Transform is given by:\[ f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \hat{f}(\omega)e^{i\omega x} \, d\omega \]
The inverse process is crucial for determining the original spatial or temporal characteristics of the transformed function. In our exercise, the Inverse Fourier Transform was utilized to find the function \( \phi(x) = e^{-|x|/2} \), shaping the solution of the differential equation when convolved with \( f \). Understanding this concept bridges the gap between the frequency domain and the original context of the problem.
Tempered Distributions
Tempered distributions expand the realm of functions that we can Fourier Transform. These are generalized functions or distributions that can be transformed despite not having traditional decay properties.
  • The classic example is the Dirac delta function, but in our context, we look at functions \( f \) whose Fourier Transform exists and behaves well.
  • This means \( f \) must decay or not grow too quickly at infinity, which ensures its Fourier Transform is defined everywhere and does not lead to infinite results.
In the exercise, hypothesizing \( f \) as a tempered distribution allows the Fourier Transform process to proceed smoothly. These conditions let us successfully utilize the Fourier Transform and inverse processes to derive meaningful solutions to differential equations, making them invaluable in theoretical and applied mathematics. Tempered distributions provide the flexibility needed to work with functions that would otherwise be problematic under regular distribution rules.

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Most popular questions from this chapter

Given \(a>0\), let \(f(x)=e^{-2 \pi x} x^{a-1}\) for \(x>0\) and \(f(x)=0\) for \(x \leq 0\). a. \(f \in L^{1}\), and \(f \in L^{2}\) if \(a>\frac{1}{2}\). b. \(\hat{f}(\xi)=\Gamma(a)[(2 \pi)(1+i \xi)]^{-a}\). (Here we are using the branch of \(z^{a}\) in the right half plane that is positive when \(z\) is positive. Cauchy's theorem may be used to justify the complex substitution \(y=(1+i \xi) x\) in the integral defining \(\widehat{f_{0}}\) ) c. If \(a, b>\frac{1}{2}\) then $$ \int_{-\infty}^{\infty}(1-i x)^{-a}(1+i x)^{-b} d x=\frac{2^{2-a-b} \pi \Gamma(a+b-1)}{\Gamma(a) \Gamma(b)} $$

Let \(\operatorname{sinc} x=(\sin \pi x) / \pi x(\operatorname{sinc} 0=1)\). a. If \(a>0, \widehat{\chi}_{|-a, a|}(x)=\chi_{[a, a]}^{V}(x)=2 a \operatorname{sinc} 2 a x\). b. Let \(\mathcal{H}_{a}=\left\\{f \in L^{2}: \hat{f}(\xi)=0\right.\) (a.c.) for \(\left.|\xi|>a\right\\}\). Then \(\mathcal{T}\) is a Hilbert space and \(\\{\sqrt{2 a}\) sinc \((2 a x-k): k \in Z\\}\) is an orthonormal basis for \(\mathcal{H}\). c. (The Sampling Theorem) If \(f \in \mathcal{T}_{a}\), then \(f \in C_{0}\) (after modification on a null set), and \(f(x)=\sum_{-\infty}^{\infty} f(k / 2 a) \operatorname{sinc}(2 a x-k)\), where the series converges both uniformly and in \(L^{2}\). (In the terminology of signal analysis, a signal of bandwidth \(2 a\) is completely determined by sampling its values at a sequence of points \(\\{k / 2 a\\}\) whose spacing is the reciprocal of the bandwidth.)

Given \(a>0\), let \(f(x)=e^{-2 \pi x} x^{a-1}\) for \(x>0\) and \(f(x)=0\) for \(x \leq 0\). a. \(f \in L^{1}\), and \(f \in L^{2}\) if \(a>\frac{1}{2}\). b. \(\hat{f}(\xi)=\Gamma(a)[(2 \pi)(1+i \xi)]^{-a}\). (Here we are using the branch of \(z^{a}\) in the right half plane that is positive when \(z\) is positive. Cauchy's theorem may be used to justify the complex substitution \(y=(1+i \xi) x\) in the integral defining \(\widehat{f_{0}}\) ) c. If \(a, b>\frac{1}{2}\) then $$ \int_{-\infty}^{\infty}(1-i x)^{-a}(1+i x)^{-b} d x=\frac{2^{2-a-b} \pi \Gamma(a+b-1)}{\Gamma(a) \Gamma(b)} $$

If \(f\) is locally integrable on \(\mathbb{R}^{n}\) and \(g \in C^{k}\) has compact support, then \(f * g \in C^{k}\).

(Wirtinger's Inequality) If \(f \in C^{1}([a, b])\) and \(f(a)=f(b)=0\), then $$ \int_{a}^{b}|f(x)|^{2} d x \leq\left(\frac{b-a}{\pi}\right)^{2} \int_{a}^{b}\left|f^{\prime}(x)\right|^{2} d x . $$ (By a change of variable it suffices to assume \(a=0, b=\frac{1}{2}\). Extend \(f\) to \(\left[-\frac{1}{2}, \frac{1}{2}\right]\) by setting \(f(-x)=-f(x)\), and then extend \(f\) to be periodic on \(\mathbb{R}\). Check that \(f\), thus extended, is in \(C^{1}(T)\) and apply the Parseval identity.)

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