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Assume $b,a \in \mathbb{R}$

$$\phi(x)=\begin{cases} b-\dfrac{b|x|}{a} & |x| <a \\ 0 & |x|>a \end{cases}$$

$$b-\dfrac{b|x|}{a}= \begin{cases} b-\dfrac{bx}{a} & x \geq 0\\ b+\dfrac{bx}{a} & x<0\end{cases}$$

How to find $$\phi(x+ct)+\phi(x-ct)$$ in terms of a four-condition piecewise function?

I did some algebra to obtain that

$$\phi(x+ct)=\begin{cases} b-\dfrac{b|x+ct|}{a} & x \in (-a-ct,a-ct)\\ 0 & x \notin (-a-ct,a-ct)\end{cases}$$

$$\phi(x-ct)=\begin{cases} b-\dfrac{b|x-ct|}{a} & x \in (ct-a,a+ct) \\ 0 & x \notin (ct-a,a+ct)\end{cases}$$

I find adding this piecewise functions together hard as a result of the constructive and destructive interference where there is overlap in their supports.

1 Answers1

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This may not be precisely what you want, but if you just want a closed form without worrying about piecewise components, then using the fact that the Heavyside step function can be expressed in terms of the absolute value:

$$H(x)=\frac{1}{2}\left(1+\frac{|x|}{x}\right)$$

We find that your function, $\phi$, can be expressed as

$$\phi(x)=\left(b-\frac{b|x|}{a}\right)H(x+a)H(a-x),$$

which then can be substituted into $\phi(x+ct)+\phi(x-ct)$. Although this is equivalent to your piecewise expression, given your bounds, it is undefined when $x=\pm a$. If you want it to be continuous, you can use the alternative definition $$H(x)=\left\lfloor\frac{1}{2}+\frac{1}{1+2^{-x}}\right\rfloor.$$

Graviton
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