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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

\begin{eqnarray*}
\frac{\partial}{\partial x} y_r\left(t-\frac{x}{c}\right)
&=&...
...c}\right)
&=& \frac{1}{c}{\dot y}_l\left(t+ \frac{x}{c}\right).
\end{eqnarray*}

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

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

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

\begin{eqnarray*}
y''_r&=& \left(-\frac{1}{c}\right)^2 {\ddot y}_r= \frac{1}{c^2...
...eft(\frac{1}{c}\right)^2 {\ddot y}_l= \frac{1}{c^2} {\ddot y}_l.
\end{eqnarray*}

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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Previous: Traveling-Wave Solution
Next: Use of the Chain Rule

written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.