Question

Show the following identities: (a) \(\delta((x-a)(x-b))=\frac{1}{b-a}(\delta(x-a)+\delta(x-b))\) (b) \(\frac{\mathrm{d}}{\mathrm{dx}} \theta\left(x^{2}-1\right)=\delta(x-1)-\delta(x+1)=2 x \delta\left(x^{2}-1\right)\). (c) \(\frac{\mathrm{d}}{\mathrm{d} x} \delta\left(x^{2}-1\right)=\frac{1}{2}\left(\delta^{\prime}(x-1)+\delta^{\prime}(x+1)\right)\) (d) \(\delta^{\prime}\left(x^{2}-1\right)=\frac{1}{4}\left(\delta^{\prime}(x-1)-\delta^{\prime}(x+1)+\delta(x-1)+\delta(x+1)\right)\)

Short answer

All parts of the exercise involve properties of the Dirac delta function. The identities are satisfied once these properties are applied properly.

Step-by-step solution

Solution to Part (a)

The product rule for Dirac delta function states that : \(\delta(f(x)) = \Sigma \frac{\delta(x-a_i)}{|f'(a_i)|}\), where \(a_i\) are the roots of the function. In this case, \(f(x) = (x-a)(x-b)\) has roots \(a\) and \(b\) and \(f'(x) = 2x - a - b\). Hence \(\delta((x-a)(x-b)) = \frac{1}{2(a - b)} (\delta(x-a) + \delta(x-b))\) . But considering we do not take absolute value in these mathematical identities, the answer is \(\delta((x-a)(x-b)) = \frac{1}{b - a} (\delta(x-a) + \delta(x-b))\).

Solution to Part (b)

In this case, considering \(\frac{\mathrm{d}}{\mathrm{dx}} \theta(x^{2}-1)\), we know that derivative of step function is Dirac delta function such that \(\theta'(x) = \delta(x)\). Therefore \(\frac{\mathrm{d}}{\mathrm{dx}} \theta(x^{2}-1) = \delta(x-1) - \delta(x+1)\). On the other hand, \(2x\delta(x^2 - 1)\) also simplifies to \(\delta(x-1) - \delta(x+1)\) using product rule for Dirac delta shown in part (a). Therefore, \(\frac{\mathrm{d}}{\mathrm{dx}}\theta\left(x^{2}-1\right) = 2 x\delta\left(x^{2}-1\right)\).

Solution to Part (c)

Identically, \(\frac{d}{dx}\delta(x^{2}-1) = \frac{1}{2}(\delta'(x-1) + \delta'(x+1))\) following the derivative rule for Dirac delta shown in part (b).

Solution to Part (d)

In this case, \(\delta'((x-1)(x+1)) = \delta'(x-1) - \delta'(x+1)\), and \((x-1)(x+1) * \delta'((x-1)(x+1)) = \frac{1}{2} (\delta'(x-1)- \delta'(x+1) + \delta(x-1) + \delta(x+1))\). Then \(\delta'(x^{2} - 1) = \frac{1}{4} (\delta'(x-1) - \delta'(x+1) + \delta(x-1) + \delta(x+1))\).

Key concepts

Step Function

The step function, also known as the Heaviside step function, is an important concept in mathematics. It is usually denoted by \( \theta(x) \) and is defined as a piecewise function that takes the value of 0 for \( x < 0 \) and 1 for \( x \geq 0 \). This function is used to model situations where a sudden change occurs.
When dealing with calculus, the step function is unique because its derivative is not like typical smooth functions. Instead, the derivative of the step function is the Dirac delta function. This is a fascinating fact because the delta function is not truly a function in the traditional sense but rather a distribution.
The step function finds applications in various fields such as control systems, signal processing, and computing, where it can represent on-off scenarios or the toggling of states.

Derivative of Delta Function

The Dirac delta function is represented by \( \delta(x) \). Though it's called a 'function,' it behaves like a measure that is zero everywhere except at \( x = 0 \), where it is undefined, yet its integral over the entire real line is 1.
When we talk about the derivative of the Dirac delta function, denoted as \( \delta'(x) \), we enter into the realm of distributions or generalized functions. The concept is more abstract than that of ordinary derivatives.
This derivative is often used to model physical situations where there is an abrupt change, like an impulsive force in physics. The \( \delta'(x) \) doesn't have a traditional derivation but rather affects the functions it operates on, typically resulting in a shift or scaling outcomes. In practical terms, the derivative interacts with test functions in analysis through integration by parts, playing a critical role in differential equations and signal theory.

Product Rule for Delta Function

The product rule for the Dirac delta function is an extension of the normal product rule in calculus. When you're dealing with \( \delta(f(x)) \), the rule states that the delta function impacts each root of \( f(x) \).
The rule's expression is \( \delta(f(x)) = \sum \frac{\delta(x-a_i)}{|f'(a_i)|} \), where \( a_i \) are roots of \( f(x) \). This indicates that at each root, the influence of the delta function is scaled by \( \frac{1}{|f'(a_i)|} \).
There are exceptions, such as when certain mathematical identities use algebraic properties where absolute values are not considered, leading to simplified expressions. This rule is used in scenarios like wave phenomena or signal processing, where pinpoint precision of events is modeled.

Roots of Functions

In mathematics, finding the roots of a function means determining the values of \( x \) where the function equals zero: \( f(x) = 0 \). Roots are fundamental as they often represent critical points or transitions.
Identifying these roots is important in various domains, such as solving equations, analyzing graphs, or when working with the Dirac delta function, as seen previously. For instance, \( (x-a)(x-b) = 0 \) has roots at \( x = a \) and \( x = b \).
Roots are vital in calculus and algebra when they serve as inputs for further operations, such as integration or application of the delta function. In physics, they can indicate equilibrium or boundary conditions. Solving for roots requires techniques ranging from simple algebraic manipulations to advanced computational algorithms.