Example 7.2.1.
Find the solution of the wave equation
\begin{equation*}
\frac{\partial^{2} y}{\partial t^{2}} = v^{2}\frac{\partial^{2} y}{\partial x^{2}}
\end{equation*}
such that \(y=p_{o}\cos pt, (p_{o}\) is a constant) when \(x=1\) and \(y=0\) when \(x=0\text{.}\)
Solution.
The wave equation is
\begin{equation}
\frac{\partial^{2} y}{\partial t^{2}} = v^{2}\frac{\partial^{2} y}{\partial x^{2}} \tag{7.2.1}
\end{equation}
Let us assume its solution as
\begin{equation}
y(x,t) = X(x) T(t) \tag{7.2.2}
\end{equation}
so that
\begin{equation*}
\frac{\partial y}{\partial t} =X\frac{\partial T}{\partial t}
\end{equation*}
and
\begin{equation*}
\frac{\partial^{2} y}{\partial t^{2}}=X\frac{\partial^{2} T}{\partial t^{2}}
\end{equation*}
also
\begin{equation*}
\frac{\partial y}{\partial x} =T\frac{\partial X}{\partial x}
\end{equation*}
and
\begin{equation*}
\frac{\partial^{2} y}{\partial x^{2}}=T\frac{\partial^{2} X}{\partial x^{2}}
\end{equation*}
Substituting these values in eqn. (7.2.1)(i), we get -
\begin{equation*}
X\frac{\partial^{2} T}{\partial t^{2}} = Tv^{2}\frac{\partial^{2} X}{\partial x^{2}}
\end{equation*}
or,
\begin{equation*}
\frac{1}{X}\frac{\partial^{2} X}{\partial x^{2}} =\frac{1}{ v^{2}T} \frac{\partial^{2} T}{\partial t^{2}} =-k^{2} \text{(say)}
\end{equation*}
or, and
\begin{equation}
\frac{\partial^{2} X}{\partial x^{2}}+k^{2}X = 0 \quad \text{and} \quad \frac{\partial^{2} T}{\partial t^{2}}+k^{2}v^{2}T = 0 \tag{7.2.3}
\end{equation}
Now, auxiliary equations are - \(m^{2}+k^{2}=0\text{,}\) or,
\begin{equation*}
m=\pm ik
\end{equation*}
and \(m^{2}+k^{2}v^{2}=0\) or,
\begin{equation*}
m=\pm ikv\text{.}
\end{equation*}
\begin{equation}
X=c_{1}\cos kx+c_{2}\sin kx \quad \text{and} \quad T=c_{3}\cos kvt+c_{4}\sin kvt \tag{7.2.4}
\end{equation}
Thus
\begin{equation}
y(x,t)=(c_{1}\cos kx+c_{2}\sin kx)(c_{3}\cos kvt+c_{4}\sin kvt)\tag{7.2.5}
\end{equation}
\begin{equation*}
0=c_{1}(c_{3}\cos kvt+c_{4}\sin kvt) \Rightarrow c_{1} =0.
\end{equation*}
Hence, eqn. (7.2.3)(iii) reduces to
\begin{equation*}
y=c_{2}\sin kx(c_{3}\cos kvt+c_{4}\sin kvt)
\end{equation*}
or,
\begin{equation}
y=c_{2}c_{3}\cos kvt.\sin kx+c_{2}c_{4}\sin kvt.\sin kx\tag{7.2.6}
\end{equation}
put \(y=p_{o}\cos pt\) when \(x=1\text{,}\)
\begin{equation*}
\therefore \quad p_{o}\cos pt = c_{2}c_{3}\cos kvt.\sin kl+c_{2}c_{4}\sin kvt.\sin kl
\end{equation*}
equating the coefficients of \(\sin\) and \(\cos\) on both sides, we get -
\begin{equation*}
p_{o} =c_{2}c_{3}\sin kl \quad \Rightarrow c_{2}c_{3} = \frac{p_{o}}{\sin kl},
\end{equation*}
\begin{equation*}
0=c_{2}c_{4}\sin kl \quad \Rightarrow c_{2}c_{2}c_{4} =0,
\end{equation*}
and
\begin{equation*}
p=kv \quad \Rightarrow \frac{p}{v}=k
\end{equation*}
Hence, eqn. (7.2.6) becomes -
\begin{equation*}
y=\frac{p_{o}}{\sin kl}\cos pt.\sin \frac{p}{v}x = \frac{p_{o}}{\sin (\frac{p}{v})l}\cos pt.\sin (\frac{p}{v})x
\end{equation*}
