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Traveling-Wave Partial Derivatives

Because we have defined our traveling-wave components $ y_r(t-x/c)$ and $ y_l(t+x/c)$ as having arguments in units of time, the partial derivatives with respect to time $ t$ are identical to simple derivatives of these functions. Let $ {\dot y}_r$ and $ {\dot y}_l$ denote the (partial) derivatives with respect to time of $ y_r$ and $ y_l$ , respectively. In contrast, the partial derivatives with respect to $ x$ are

\frac{\partial}{\partial x} y_r\left(t-\frac{x}{c}\right)
&=& -\frac{1}{c}{\dot y}_r\left(t- \frac{x}{c}\right)\\ [10pt]
\frac{\partial}{\partial x} y_l\left(t+\frac{x}{c}\right)
&=& \frac{1}{c}{\dot y}_l\left(t+ \frac{x}{c}\right).

Denoting the spatial partial derivatives by $ y'_r$ and $ y'_l$ , respectively, we can write more succinctly

y'_r&=& -\frac{1}{c}{\dot y}_r\\ [5pt]
y'_l&=& \frac{1}{c}{\dot y}_l,

where this argument-free notation assumes the same $ t$ and $ x$ for all terms in each equation, and the subscript $ l$ or $ r$ determines whether the omitted argument is $ t + x/c$ or $ t - x/c$ .

Now we can see that the second partial derivatives in $ x$ are

y''_r&=& \left(-\frac{1}{c}\right)^2 {\ddot y}_r= \frac{1}{c^2} {\ddot y}_r\\ [5pt]
y''_l&=& \left(\frac{1}{c}\right)^2 {\ddot y}_l= \frac{1}{c^2} {\ddot y}_l.

These relations, together with the fact that partial differention is a linear operator, establish that

$\displaystyle y(t,x) = y_r\left(t-\frac{x}{c}\right) + y_l\left(t+\frac{x}{c}\right). \protect$

obeys the ideal wave equation $ {\ddot y}= c^2y''$ for all twice-differentiable functions $ y_r$ and $ y_l$ .

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``Physical Audio Signal Processing'', by Julius O. Smith III, W3K Publishing, 2010, ISBN 978-0-9745607-2-4
Copyright © 2023-08-20 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University