91Ó°ÊÓ

The given expression is

(a) $\mathrm{δ}(x-a)=\mathrm{δ}(a-x)$.

(b) ${\mathrm{δ}}^{\text{'}}(x-a)=-{\mathrm{δ}}^{\text{'}}(a-x)$.

(c)$∫\mathrm{ϕ}\left(x\right)\mathrm{δ}\left(ax\right)dx=∫\mathrm{ϕ}\left(\frac{u}{a}\right)\mathrm{δ}\left(u\right)\frac{du}{a}$

(d)$\mathrm{δ}\left[\right(x-a\left)\right(x-b\left)\right]=\frac{1}{|a-b|}\left[\mathrm{δ}\right(x-a)+\mathrm{δ}(x-b\left)\right],(a≠b)$

(e) $\mathrm{δ}\left[f\right(x\left)\right]=\underset{i}{∑}\frac{\mathrm{δ}\left(x-x_i\right)}{\left|f^{\text{'}}\left(x\right)\right|}$

A function operator is a function that takes one (or more) functions as input and returns a function as output. In some ways, function operators are similar to functional

Let

$$\begin{gathered} u=-x \\ du=-dx \end{gathered}$$.

Now, let $u=a-x$.

Then$du=-dx$.

$$\begin{gathered} ∫\mathrm{ϕ}\left(x\right)\mathrm{δ}(a-x)dx=∫\mathrm{ϕ}(a-u)\mathrm{δ}\left(u\right)(-du) \\ =∫\mathrm{ϕ}(a-u)\mathrm{δ}\left(u\right)\left(du\right) \\ =∫-d^{\text{'}}(a-u)u\left(u\right)du \end{gathered}$$

Solve further

$$\begin{gathered} =-{∫}_0{\mathrm{ϕ}}^{\text{'}}(a-u)du \\ \mathrm{ϕ}\left(a\right)=-∫{\mathrm{ϕ}}^{\text{'}}\left(x\right)dx \\ =-∫{\mathrm{ϕ}}^{\text{'}}\left(x\right)u(x-a)dx \\ =∫\mathrm{ϕ}\left(x\right)\mathrm{δ}(x-a)dx \end{gathered}$$

That is,$\mathrm{δ}(x-a)=\mathrm{δ}(a-x)$ .

Solve the

${\mathrm{δ}}^{\text{'}}(-x)=-{\mathrm{δ}}^{\text{'}}\left(x\right)$and${\mathrm{δ}}^{\text{'}}(x-a)=-{\mathrm{δ}}^{\text{'}}(a-x).$

$$\begin{gathered} ∫\mathrm{ϕ}\left(x\right)x{\mathrm{δ}}^{\text{'}}(-x)dx=∫\mathrm{ϕ}(-u)(-u) \\ =∫\mathrm{ϕ}(-u)u{\mathrm{δ}}^{\text{'}}\left(u\right)du \\ =-∫\mathrm{ϕ}(-u)\mathrm{δ}\left(u\right)du \\ =∫\mathrm{ϕ}\left(x\right)\mathrm{δ}\left(x\right)dx \end{gathered}$$

Solve further

$$\begin{gathered} =∫\mathrm{ϕ}\left(x\right)\left(-x{\stackrel{˙}{\mathrm{δ}}}^{\text{'}}\left(x\right)\right)dx \\ =∫\mathrm{ϕ}\left(x\right)x\left(-{\mathrm{δ}}^{\text{'}}\left(x\right)\right)dx \end{gathered}$$

So,${\mathrm{δ}}^{\text{'}}(-x)=-{\mathrm{δ}}^{\text{'}}\left(x\right)$.

Now,

$$\begin{gathered} ∫\mathrm{ϕ}\left(x\right)\left[-{\mathrm{δ}}^{\text{'}}(a-x)\right]dx=-∫\mathrm{ϕ}\left(x\right){\mathrm{δ}}^{\text{'}}(a-x)dx \\ =-∫\mathrm{ϕ}(a-u){\mathrm{δ}}^{\text{'}}\left(u\right)(-du)(\text{Assume}u=a-x)=-∫\mathrm{ϕ}(a-u){\mathrm{δ}}^{\text{'}}\left(u\right)\left(du\right) \\ =-{\left.{(-1)}^1\left[-{\mathrm{ϕ}}^{\text{'}}(a-u)\right]\right|}_{x-0} \\ -{\mathrm{ϕ}}^{\text{'}}\left(a\right)=∫\mathrm{ϕ}\left(x\right){\mathrm{δ}}^{\text{'}}(x-a)dx \end{gathered}$$

$=∫\mathrm{ϕ}\left(x\right){\mathrm{δ}}^{\text{'}}(x-a)dx$

That is,${\mathrm{δ}}^{\text{'}}(x-a)=-{\mathrm{δ}}^{\text{'}}(a-x)$

If$a<0$, assume$a=-b,b>0$.

Then,

$$\begin{gathered} u=ax \\ u=-bx \\ du=-bdx \\ dx=\frac{dx}{-b} \end{gathered}$$.

After solve the equation

$$\begin{gathered} ∫\mathrm{ϕ}\left(x\right)\mathrm{δ}(-bx)dx=∫\mathrm{ϕ}\left(\frac{u}{-b}\right)\mathrm{δ} \\ =-\frac{1}{b}∫\mathrm{ϕ}\left(\frac{u}{-b}\right)\mathrm{δ}\left(u\right)du \\ =\frac{1}{b}∫\mathrm{ϕ}\left(\frac{u}{-b}\right)\mathrm{δ}\left(u\right)du \end{gathered}$$

By solve further

$$\begin{gathered} =\frac{1}{b}\mathrm{ϕ}\left(0\right) \\ =\frac{1}{|=|}\mathrm{ϕ}\left(0\right) \\ =\frac{1}{\left|a\right|}∫\mathrm{ϕ}\left(x\right)\mathrm{δ}\left(x\right)dx \\ =∫\mathrm{ϕ}\left(x\right)\left[\frac{1}{\left|a\right|}\mathrm{δ}\left(x\right)\right]dx \end{gathered}$$

If $a>0$, assume $u=ax$.

Then,

$$\begin{gathered} u=ax \\ du=adx \\ dx=\frac{du}{u} \end{gathered}$$

So,

$∫\mathrm{ϕ}\left(x\right)\mathrm{δ}\left(ax\right)dx=∫\mathrm{ϕ}\left(\frac{u}{a}\right)\mathrm{δ}\left(u\right)\frac{du}{a}$

Assume $a<b$.

The function $g\left(x\right)=(x-a)(x-b)$is upward parabola crossing -axis at $x=a$and $x=b$.

Then, it can be written as:

Now, differentiate this function to get,

$$\begin{gathered} u^{\text{'}}\left[\right(x-a\left)\right(x-b\left)\right](2x-a-b)=0-u^{\text{'}}(x-a)+u^{\text{'}}(x-b) \\ {a}^{\text{'}}\left[\right(x-a\left)\right(x-b\left)\right]=\frac{z^{\text{'}}(x-a)u^{\text{'}}(x-b)}{(x-b-b)} \\ \mathrm{δ}\left[\right(x-a\left)\right(x-b\left)\right]=\frac{-g(s-a)+d(x-b)}{[2x-b-b]} \end{gathered}$$

Thus,

width="360">δ[(x-a)(x-b)]=1|a-b|[δ(x-a)+δ(x-b)],(a≠b)

Assume $f\left(x\right)=a\left(x-x_1\right)\left(x-x_2\right)⋯\left(x-x_n\right)$with $x_1<x_2<…<x_n$.

Then,

By differentiating

$$\begin{gathered} f^{\text{'}}\left(x\right)u^{\text{'}}\left(f\right(x\left)\right)=Â\pm \left\{u^{\text{'}}\left(x-x_1\right)-u^{\text{'}}\left(x-x_2\right)+…-{(-1)}^n{u}^{\text{'}}\left(x-x_n\right)\right\} \\ {f}^{\text{'}}\left(x\right)\mathrm{δ}\left(f\right(x\left)\right)=Â\pm \left\{\mathrm{δ}\left(x-x_1\right)-\mathrm{δ}\left(x-x_2\right)+…-{(-1)}^n\mathrm{δ}\left(x-x_n\right)\right\} \\ \mathrm{δ}\left(f\right(x\left)\right)=Â\pm \frac{\left.âˆ\pounds z\left(x-x_1\right)d\left(x-x_2\right)+…-{(-1)}^{n}d\left(x-x_3\right)\right\}}{f^{\text{'}}\left(x\right)} \end{gathered}$$

Therefore,

That is, $\mathrm{δ}\left[f\right(x\left)\right]=\underset{i}{∑}\frac{\mathrm{δ}\left(x-x_i\right)}{\left|f^{\text{'}}\left(x\right)\right|}$if $f\left(x_i\right)=0$and $f^{\text{'}}\left(x_i\right)≠0$.