Q46P - Introduction to Electrodynamics

Question

 (a) Show that $\mathit{x}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\left(\delta \left(x\right)\right)\mathbf{=}\mathbf{-}\mathit{\delta }\left(x\right)$ 

[Hint: Use integration by parts.] 

 

(b) Let $\mathit{\theta }\left(x\right)$ be the step function:

 $\mathit{\theta }\left(x\right)\mathbf{=}\{\begin{array}{ll}1& \mathrm{if}\mathit{ }\mathit{x}>0\\ 0,& \mathrm{if}\mathit{ }\mathit{x}\le 0\end{array}$

Show that $\frac{\mathbf{d\theta }}{\mathbf{dx}}\mathbf{=}\mathit{\delta }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

Step-by-Step Solution

Answer

(a) The result $x\frac{d}{dx}\left(\delta \left(x\right)\right)=-\delta \left(x\right)$ has been proved.

 

(b) The result $\left(\frac{d\theta }{dx}\right)=\delta \left(x\right)$ has been proved. 

Step 1: Define Dirac delta function

The Dirac Delta function which is represented as $\mathit{\delta }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$, is defined as $\mathit{\theta }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{ll}\mathbf{1}& \mathbf{ }\mathbf{x}\mathbf{\ne }\mathbf{0}\\ \mathbf{0}\mathbf{,}& \mathbf{ }\mathbf{x}\mathbf{=}\mathbf{0}\end{array}$, .the Dirac delta function has the property  ${\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{\delta }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{d}\mathit{x}\mathbf{=}\mathit{f}\mathbf{\left(}\mathbf{0}\mathbf{\right)}$, where $\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ is a continuous containing $\mathit{x}\mathbf{=}\mathbf{0}$.

Step: 2 Prove x d d x ( δ ( x ) ) = - δ ( x )

(a)

Let function $u\left(x\right)=\delta \left(x\right)$ and $v\left(x\right)=x$. Differentiate $u\left(x\right)=\delta \left(x\right)$ and $v\left(x\right)=x$ with respect to $x$.

$\begin{array}{l}u\text{'}\left(x\right)=\frac{d}{dx}\delta \left(x\right)\\ v^{\text{'}}\left(x\right)=1\end{array}$ 

 

Substitute $\delta \left(x\right)$ for $u$ , $x$ for $v\left(x\right)$ into $\int udv=uv-\int vdu$ as,

$$\begin{gathered} {\int }_{-\infty }^{\infty }\delta \left(x\right)dx={\int }_{-\infty }^{\infty }c\left(x\right)\times 1dx \\ ={\overline{)x\delta \left(x\right)}}_{-\infty }^{\infty }-{\int }_{-\infty }^{\infty }x\frac{d}{dx}\left(\delta \left(x\right)\right)dx \end{gathered}$$                                 ….. (1)

 

Define $x\delta \left(x\right)$ as,

$x\delta \left(\mathrm{x}\right)=\{\begin{array}{ll}0\times \infty & \mathrm{x}=0\\ x=0& \mathrm{x}\ne 0\end{array}$ 

 

Thus the value of $x\delta \left(x\right)$ is 0 for all x.

 

Hence,  $x\delta {\overline{)\left(x\right)}}_{-\infty }^{\infty }=0$.

Now equation (1) can be rewritten as, 

$$\begin{gathered} {\int }_{-\infty }^{\infty }\delta \left(x\right)dx={\int }_{-\infty }^{\infty }\times \frac{d}{dx}\left(\delta \left(x\right)\right)dx \\ ={\int }_{-\infty }^{\infty }x\frac{d}{dx}\left(\delta \left(x\right)\right)dx \end{gathered}$$                                                      ……. (2)

 

Equation (2) can be rewritten as  $x\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{\delta }\left(\mathrm{x}\right)\right)=-\delta \left(\mathrm{x}\right)$

Step: 3 Prove ( d θ d x ) = δ ( x )

(b)

According to equation (1), ${\int }_{-\infty }^{\infty }\delta \left(x\right)dx={\overline{)x\delta \left(x\right)}}_{-\infty }^{\infty }-{\int }_{-\infty }^{\infty }\times \frac{d}{dx}\left(\delta \left(x\right)\right)dx$ . The second property of Dirac Delta function can be written as follows

$$\begin{gathered} {\int }_{-\infty }^{\infty }f\left(x\right)\left(\frac{d\theta }{dx}\right)dx=f\left(x\right)\theta {\overline{)\left(x\right)}}_{-\infty }^{\infty }-{\int }_{-\infty }^{\infty }\frac{df}{dx}\left(\theta \left(x\right)\right)dx \\ =f\left(x\right)\theta {\overline{)\left(x\right)}}_{-\infty }^{\infty }-\left[{\int }_{-\infty }^0\frac{df}{dx}\left(\theta \left(x\right)\right)dx+{\int }_0^{\infty }\frac{df}{dx}\left(\theta \left(x\right)\right)dx\right] \\ =f\left(x\right)\theta {\overline{)\left(x\right)}}_{-\infty }^{\infty }-\left[0+{\int }_0^{\infty }\frac{df}{dx}dx\right] \\ =f\left(x\right)(\infty )-{\int }_0^{\infty }\frac{df}{dx}dx \end{gathered}$$ 

 

 

Solve further as,

 $$\begin{gathered} {\int }_{-\infty }^{\infty }f\left(x\right)\left(\frac{d\theta }{dx}\right)dx=(f\left(\infty \right)-f\left(0\right)) \\ =f\left(0\right) \\ ={\int }_{-\infty }^{\infty }f\left(x\right)\delta \left(x\right)dx \end{gathered}$$

 

 

Thus, it can be written as $\left(\frac{d\theta }{dx}\right)=\delta x$.