ヘヴィサイドの階段関数の負数・和・差

ヘヴィサイドの階段関数の負数・和・差

(1)

\[ H_{a}\left(-x\right)=-H_{a}\left(x\right)+1+\left(2a-1\right)\delta_{0,x} \]

(2)

\[ H_{a}\left(x\right)+H_{b}\left(x\right)=2H_{\frac{a+b}{2}}\left(x\right) \]

(3)

\[ H_{a}\left(x\right)-H_{b}\left(x\right)=\left(a-b\right)\delta_{0,x} \]

(4)

\[ H_{c}\left(x\right)-H_{c}\left(y\right)=H_{c}\left(-y\right)-H_{c}\left(-x\right)+\left(2c-1\right)\left(\delta_{x,0}-\delta_{y,0}\right) \]

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\(H\left(x\right)\)はヘヴィサイドの階段関数
\(\delta_{ij}\)はクロネッカーのデルタ

(1)

\begin{align*} H_{a}\left(-x\right) & =H_{\frac{1}{2}}\left(-x\right)+\left(a-\frac{1}{2}\right)\delta_{0,-x}\\ & =\frac{\sgn\left(-x\right)+1}{2}+\left(a-\frac{1}{2}\right)\delta_{0,-x}\\ & =-\frac{\sgn\left(x\right)+1}{2}+1+\left(a-\frac{1}{2}\right)\delta_{0,-x}\\ & =-H_{\frac{1}{2}}\left(x\right)+1+\left(a-\frac{1}{2}\right)\delta_{0,-x}\\ & =-\left(H_{\frac{1}{2}}\left(x\right)+\left(a-\frac{1}{2}\right)\delta_{0,x}\right)+1+2\left(a-\frac{1}{2}\right)\delta_{0,x}\\ & =-H_{a}\left(x\right)+1+\left(2a-1\right)\delta_{0,x} \end{align*}

(2)

\begin{align*} H_{a}\left(x\right)+H_{b}\left(x\right) & =H_{a}\left(x\right)+H_{a}\left(x\right)+\left(b-a\right)\delta_{0,x}\\ & =2\left(H_{a}\left(x\right)+\frac{b-a}{2}\delta_{0,x}\right)\\ & =2\left(H_{a}\left(x\right)+\left(\frac{b+a}{2}-a\right)\delta_{0,x}\right)\\ & =2H_{\frac{a+b}{2}}\left(x\right) \end{align*}

(3)

\begin{align*} H_{a}\left(x\right)-H_{b}\left(x\right) & =H_{a}\left(x\right)-\left(H_{a}\left(x\right)+\left(b-a\right)\delta_{0,x}\right)\\ & =\left(a-b\right)\delta_{0,x} \end{align*}

(4)

(1)より、
\begin{align*} H_{a}\left(x\right)-H_{a}\left(y\right) & =1-H_{a}\left(-x\right)+\left(2a-1\right)\delta_{x,0}-\left\{ 1-H_{a}\left(-y\right)+\left(2a-1\right)\delta_{y,0}\right\} \\ & =H_{a}\left(-y\right)-H_{a}\left(-x\right)+\left(2a-1\right)\left(\delta_{x,0}-\delta_{y,0}\right) \end{align*} となるので与式は成り立つ。

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ヘヴィサイドの階段関数の負数・和・差
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