Chapter 12: Problem 7

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

Expert verified

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 \(\delta'(x^{2} - 1) = \frac{1}{4} (\delta'(x-1) - \delta'(x+1) + \delta(x-1) + \delta(x+1))\).