Subsection 7.1.2 D’Alembert’s Solution of Vibrating String
The equation of vibrating string is given by
\begin{equation}
\frac{\partial^{2} y}{\partial x^{2}} = \frac{1}{v^{2}}\frac{\partial^{2} y}{\partial t^{2}} \tag{7.1.25}
\end{equation}
The solution of equation (7.1.25) may be obtained by changing the variable \(x\) and \(t\) in the other parameters as \(p=x+vt\text{,}\) and \(q=x-vt\text{,}\) so that
\begin{equation*}
\frac{\partial p}{\partial x}=\frac{\partial q}{\partial x}=1\quad \text{and}\quad \frac{\partial p}{\partial t} =v, \frac{\partial q}{\partial t} =-v, \quad \text{now,}\quad y=f(p,q)
\end{equation*}
or,
\begin{equation*}
\frac{\partial y}{\partial x} = \frac{\partial y}{\partial p} \frac{\partial p}{\partial x}+\frac{\partial y}{\partial q} \frac{\partial q}{\partial x}
\end{equation*}
at constant \(t\text{.}\) Or,
\begin{equation}
\frac{\partial y}{\partial x} = \frac{\partial y}{\partial p}+\frac{\partial y}{\partial q}\tag{7.1.26}
\end{equation}
In operator form
\begin{equation}
\frac{\partial}{\partial x} \equiv \frac{\partial}{\partial p}+\frac{\partial}{\partial q}\tag{7.1.27}
\end{equation}
and
\begin{equation*}
\frac{\partial^{2}y}{\partial x^{2}}=\frac{\partial}{\partial x} \left(\frac{\partial y}{\partial x}\right)= \left(\frac{\partial}{\partial p}+\frac{\partial}{\partial q}\right)\left(\frac{\partial y}{\partial p}+\frac{\partial y}{\partial q}\right)
\end{equation*}
\begin{equation}
= \frac{\partial^{2}y}{\partial p^{2}}+2\frac{\partial^{2}y}{\partial p \partial q}+\frac{\partial^{2}y}{\partial q^{2}}\tag{7.1.28}
\end{equation}
also
\begin{equation*}
\frac{\partial y}{\partial t}=\frac{\partial y}{\partial p}\frac{\partial p}{\partial t}+\frac{\partial y}{\partial q}\frac{\partial q}{\partial t}
\end{equation*}
at constant \(x\text{.}\) Or,
\begin{equation*}
\frac{\partial y}{\partial t}=v\left[\frac{\partial y}{\partial p}-\frac{\partial y}{\partial q}\right]
\end{equation*}
i.e.,
\begin{equation*}
\frac{\partial}{\partial t}=v\left[\frac{\partial}{\partial p}-\frac{\partial}{\partial q}\right]
\end{equation*}
and
\begin{equation*}
\frac{\partial^{2}y}{\partial t^{2}}=\frac{\partial}{\partial t} \left(\frac{\partial y}{\partial t}\right)= v^{2}\left(\frac{\partial}{\partial p}-\frac{\partial}{\partial q}\right)\left(\frac{\partial y}{\partial p}-\frac{\partial y}{\partial q}\right)
\end{equation*}
\begin{equation}
= v^{2}\frac{\partial^{2}y}{\partial p^{2}}+2\frac{\partial^{2}y}{\partial p \partial q}+\frac{\partial^{2}y}{\partial q^{2}} \tag{7.1.29}
\end{equation}
\begin{equation*}
\frac{\partial^{2}y}{\partial p\partial q}=0.
\end{equation*}
Integrating this expression with respect to \(q\text{,}\) we get -
\begin{equation*}
\int\frac{\partial}{\partial q}\left(\frac{\partial y}{\partial p}\right)\,dq=\int 0.\,dq
\end{equation*}
or,
\begin{equation*}
\frac{\partial y}{\partial p} = constant = f(p)\quad \text{(say)}.
\end{equation*}
Integrating it again w. r. t. ’p’, we get -
\begin{equation*}
y=\int f(p) \,dp + constant. =\phi(p) + \psi(q) \quad \text{(say)}.
\end{equation*}
\begin{equation}
\therefore \quad y(x,t) = \phi(x+vt) + \psi(x-vt)\tag{7.1.30}
\end{equation}
This is a D’ Alembert’s solution of one - dimensional wave equation (7.1.25). Physically equation (7.1.30) represents two plane waves travelling in opposite directions with the same velocity and same period. Verification of equation (7.1.30) with the help of boundary conditions
\begin{equation}
y(0,t) = 0 = y(l,t) \tag{7.1.31}
\end{equation}
and the initial conditions
\begin{equation}
y(x,0) =f(x)\tag{7.1.32}
\end{equation}
and
\begin{equation}
\left(\frac{\partial y}{\partial t}\right)_{t=0} =f'(x)\tag{7.1.33}
\end{equation}
from equation (7.1.30), we have
\begin{equation}
\frac{\partial y}{\partial t} =v\phi'(x+vt)-v\psi'(x-vt)\tag{7.1.34}
\end{equation}
\begin{equation}
y(x,0) =f(x)=\phi(x)+\psi(x)\tag{7.1.35}
\end{equation}
\begin{equation}
\frac{\partial y}{\partial t} =f'(x) = v[\phi'(x)-\psi'(x)] \tag{7.1.36}
\end{equation}
assuming \(f'(x)=0\text{,}\) we have from equation (7.1.36),
\begin{equation*}
v[\phi'(x)-\psi'(x)]=0
\end{equation*}
or,
\begin{equation*}
\phi'(x) = \psi'x)
\end{equation*}
which on integrating gives
\begin{equation}
\phi(x) =\psi(x)+\lambda \tag{7.1.37}
\end{equation}
\begin{equation}
\phi(x)=\frac{1}{2}[f(x)+\lambda]\tag{7.1.38}
\end{equation}
and
\begin{equation}
\psi(x)=\frac{1}{2}[f(x)-\lambda]\tag{7.1.39}
\end{equation}
\begin{equation}
y(x,t) = \frac{1}{2}[f(x+vt)+f(x-vt)]\tag{7.1.40}
\end{equation}
which reduces to
\begin{equation}
y(0,t) = \frac{1}{2}[f(vt)+f(-vt)]\tag{7.1.41}
\end{equation}
and
\begin{equation}
y(l,t) = \frac{1}{2}[f(l+vt)+f(l-vt)]\tag{7.1.42}
\end{equation}
