Graphing solutions of the plucked string equation (Strauss PDE)

Solution 1:

$\def\abs#1{\left|#1\right|}\def\paren#1{\left(#1\right)}$First, in both editions of the book, $φ$ is defined as$$ φ(x) = \begin{cases} b - \dfrac{b|x|}{\color{red}{a}}; & |x| < a\\ 0; & |x| > a \end{cases}. $$ For the graph of $u(\,·\,, t)$ at $t = \dfrac{a}{2c}$, since the solution $u(x, t) = \dfrac{1}{2} (φ(x + ct) + φ(x - ct))$ contains piecewise terms, it is natural to divide the real line in intervals to get simpler expressions of $u$ on each interval. To see where to cut the real line, note that$$ φ(x) = \frac{b}{2a} (|x + a| + |x - a| - 2|x|), $$ thus\begin{gather*} u\paren{ x, \frac{a}{2c} } = \frac{1}{2} \paren{ φ\paren{ x + \frac{a}{2} } + φ\paren{ x - \frac{a}{2} } }\\ = \frac{b}{2a} \paren{ \abs{ x + \frac{3a}{2} } - \abs{ x + \frac{a}{2} } - \abs{ x - \frac{a}{2} } + \abs{ x - \frac{3a}{2} } }. \end{gather*} The critical points of the absolute terms above are $\pm\dfrac{a}{2}$ and $\pm\dfrac{3a}{2}$, and the segments in the graph of $u(\,·\,, t)$ change slopes exactly at these points, so these are the points to cut.