Question Wave Equation Show that \[ $u(x, t)=\frac{1}{2}[f(x-c t)+f(x+c t)]$ \] is a solution to the one-dimensional wave equation $\frac{\partial^{2} u}{\partial t^{2}}=c^{2} \frac{\partial^{2} u}{\partial x^{2}} .$ (This equation describes the small transverse vibration of an elastic string such as those on certain musical instruments.)
\[
$u(x, t)=\frac{1}{2}[f(x-c t)+f(x+c t)]$
\]
is a solution to the one-dimensional wave equation $\frac{\partial^{2} u}{\partial t^{2}}=c^{2} \frac{\partial^{2} u}{\partial x^{2}} .$
(This equation describes the small transverse vibration of an elastic string such as those on certain musical instruments.)
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u(x, t)=\frac{1}{2}[f(x-c t)+f(x+c t)]Let r=x-c t and s=x+c t.\begin{array}{l}\text { Then } u(r, s)=\frac{1}{2}[f(r)+f(s)] \\\frac{\partial u}{\partial t}=\frac{\partial u}{\partial r} \frac{\partial r}{\partial t}+\frac{\partial u}{\partial s} \frac{\partial s}{\partial t}=\frac{1}{2} \frac{d f}{d r}(-c)+\frac{1}{2} \frac{d f}{d s}(c) \\\frac{\partial^{2} u}{\partial t^{2}}=\frac{1}{2} \frac{d^{2} f}{d r^{2}}(-c)^{2}+\frac{1}{2} \frac{d^{2} f}{d s^{2}}(c)^{2}=\frac{c^{2}}{2}\left[\frac{d^{2} f}{d r^{2}}+\frac{d^{2} f}{d s^{2}}\right] \\\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r} \frac{\partial r}{\partial x}+\frac{\partial u}{\partial s} \frac{\partial s}{\partial x}=\frac{1}{2} \frac{d f}{d r}(1)+\frac{1}{2} \frac{d f}{d s}(1) \\\frac{\partial^{2} u}{\partial x^{2}}=\frac{1}{2} \frac{d^{2} f}{2 d r^{2}}(1)^{2}+\frac{1}{2} \frac{d^{2} f}{d s^{2}}(1)^{2}=\frac{1}{2}\left[\frac{d^{2} f}{d r^{2}}+\frac{d^{2} f}{d s^{2}}\right]\end{array}So, \frac{\partial^{2} u}{\partial t^{2}}=c^{2} \frac{\partial^{2} u}{\partial x^{2}} ...