Subsection 7.1.4 Solution of Two Dimensional Wave Equation
The two - dimensional wave equation is given as
\begin{equation}
\frac{\partial^{2} u}{\partial t^{2}}=v^{2}\left[\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}}\right]\tag{7.1.45}
\end{equation}
The equation can be solved by the method of separation of variables for two cases, one, when the membrane is a rectangle and second, when it is a circle.
Subsubsection 7.1.4.1 Rectangular Membrane
To solve equation (7.1.45), let
\begin{equation*}
\left(\frac{\partial u}{\partial t}\right)_{t=0} =g(x,y)
\end{equation*}
be the initial velocity, \(u(x,y,0) =f(x,y)\) be the initial displacement, and the following are the boundary conditions
\begin{equation}
u(0,y,t)=u(a,y,t) =u(x,0,t) =u(x,b,t)=0 \tag{7.1.46}
\end{equation}
where \(a\) and \(b\) are sides of the membrane along X and Y axes, respectively, as shown in figure Figure 7.1.4.
Let
\begin{equation}
u(x,y,t) =X(x) Y(y) T(t)\tag{7.1.47}
\end{equation}
\begin{equation*}
XY \frac{\partial^{2} T}{\partial t^{2}}=v^{2}\left[YT \frac{\partial^{2} X}{\partial x^{2}}+XT \frac{\partial^{2} Y}{\partial y^{2}}\right]
\end{equation*}
or,
\begin{equation}
\frac{1}{X}\frac{\partial^{2} X}{\partial x^{2}}+\frac{1}{Y}\frac{\partial^{2} Y}{\partial y^{2}}=\frac{1}{v^{2}T}\frac{\partial^{2} T}{\partial t^{2}} = \lambda (say)\tag{7.1.48}
\end{equation}
Case - I when constant \(\lambda\) is zero.
\begin{equation}
\left.\begin{aligned}
\frac{1}{X}\frac{\partial^{2} X}{\partial x^{2}}=0 & \Rightarrow X=c_{1}x+c_{2}\\
\frac{1}{Y}\frac{\partial^{2} Y}{\partial y^{2}} = 0 & \Rightarrow Y=c_{3}y+c_{4}\\
\text{and} \quad \frac{1}{v^{2}T}\frac{\partial^{2} T}{\partial t^{2}} = 0 & \Rightarrow T=c_{5}t+c_{6}
\end{aligned}\right\}\tag{7.1.49}
\end{equation}
Case - II when constant \(\lambda \) is positive.
\begin{equation}
\left.\begin{aligned}
\frac{1}{X}\frac{\partial^{2} X}{\partial x^{2}}=k^{2}_{1} & \Rightarrow X=c_{1}e^{k_{1}x}+c_{2}e^{-k_{1}x}\\
\frac{1}{Y}\frac{\partial^{2} Y}{\partial y^{2}} = k^{2}_{2} & \Rightarrow Y=c_{3}e^{k_{2}y}+c_{4}e^{-k_{2}y}\\
\text{and} \quad \frac{1}{v^{2}T}\frac{\partial^{2} T}{\partial t^{2}} = k^{2} & \Rightarrow T=c_{5}e^{vkt}+c_{6}e^{-vkt}
\end{aligned}\right\}\tag{7.1.50}
\end{equation}
where \(k^{2}=k^{2}_{1}+k^{2}_{2}\)
Case - III when constant \(\lambda \) is negative.
\begin{equation}
\left.\begin{aligned}
\frac{1}{X}\frac{\partial^{2} X}{\partial x^{2}}=-k^{2}_{1} & \Rightarrow X=c_{1}\cos k_{1}x+c_{2}\sin k_{1}x\\
\frac{1}{Y}\frac{\partial^{2} Y}{\partial y^{2}} = -k^{2}_{2} & \Rightarrow Y=c_{3}\cos k_{2}y+c_{4}\sin k_{2}y\\
\text{and} \quad \frac{1}{v^{2}T}\frac{\partial^{2} T}{\partial t^{2}} =- k^{2} & \Rightarrow T=c_{5}\cos vkt+c_{6}\sin vkt
\end{aligned}\right\}\tag{7.1.51}
\end{equation}
where \(k^{2}=k^{2}_{1}+k^{2}_{2}\text{.}\)
Considering the case I and II with the given boundary conditions give no solution, hence only case III will be able to give the solution. From case III, we have
\begin{equation*}
u(0,y,t) =0
\end{equation*}
\begin{equation*}
=(c_{1}+0)(c_{3}\cos k_{2}y+c_{4}\sin k_{2}y)( c_{5}\cos vkt+c_{6}\sin vkt) \quad \Rightarrow c_{1} =0.
\end{equation*}
or,
\begin{equation*}
u(x,0,t) =0
\end{equation*}
\begin{equation*}
=(c_{1}\cos k_{1}x+c_{2}\sin k_{1}x)(c_{3}+0)( c_{5}\cos vkt+c_{6}\sin vkt)\quad \Rightarrow c_{3} =0.
\end{equation*}
or,
\begin{equation*}
u(a,y,t) =0
\end{equation*}
\begin{equation*}
=(c_{1}\cos k_{1}a+c_{2}\sin k_{1}a)(c_{3}\cos k_{2}y+c_{4}\sin k_{2}y)(c_{5}\cos vkt+c_{6}\sin vkt)
\end{equation*}
\begin{equation*}
=(c_{2}\sin k_{1}a)(c_{4}\sin k_{2}y)(c_{5}\cos vkt+c_{6}\sin vkt) \Rightarrow \sin k_{1}a =0
\end{equation*}
\([\because c_{2}\neq 0 ].\) i.e., \(k_{1}a=m\pi \) and \(k_{1}=\frac{m\pi}{a},\) \([m=1,2,3,\cdots]\text{.}\)
and
\begin{equation*}
u(x,b,t)=0=X(x)(c_{3}\cos k_{2}b+c_{4}\sin k_{2}b)T(t)\quad \Rightarrow \sin k_{2}b =0
\end{equation*}
\([ \because c_{4}\neq 0 ].\) or,
\begin{equation*}
k_{2}b =n\pi, k_{2} =\frac{n\pi}{b}, \quad [n = 1,2,3,\cdots]
\end{equation*}
Hence the appropriate solution of two-dimensional wave equation can be taken as
\begin{equation*}
u(x,y,t) =\sum\limits_{m=1}^{\infty}\sum\limits_{n=1}^{\infty} u_{mn}(x,y,t)
\end{equation*}
\begin{equation*}
= \sum\limits_{m=1}^{\infty}\sum\limits_{n=1}^{\infty} \left(A_{mn}\cos vk_{mn}t+B_{mn}\sin vk_{mn}t\right)\sin \frac{m\pi x}{a}\sin \frac{n\pi y}{b}
\end{equation*}
where,
\begin{equation*}
k^{2} = k_{mn}^{2}=\pi^{2}\left(\frac{m^{2}}{a^{2}}+ \frac{n^{2}}{b^{2}}\right).
\end{equation*}
and
\begin{equation*}
c_{2}c_{4}c_{5} =A_{mn}, \quad c_{2}c_{4}c_{6} = B_{mn}
\end{equation*}
are another constants.
\begin{equation}
=\sum\limits_{m=1}^{\infty}\sum\limits_{n=1}^{\infty} \left(A_{mn}\cos \lambda_{mn}t+B_{mn}\sin \lambda_{mn}t\right)\sin \frac{m\pi x}{a}\sin \frac{n\pi y}{b},\tag{7.1.52}
\end{equation}
where \(\lambda_{mn}=vk_{mn}. \) These functions are called eigen function or characteristics functions and the numbers \(\lambda_{mn}\) are called the eigen values of rectangular vibrating membrane. The frequency of \(u_{mn}\) is (\(\lambda_{mn}/2\pi\)).
Now imploying the initial conditions in equation (7.1.52), we get-
\begin{equation}
u(x,y,0) = f(x,y)=\sum\limits_{m=1}^{\infty}\sum\limits_{n=1}^{\infty} A_{mn}\sin \frac{m\pi x}{a}\sin \frac{n\pi y}{b}\tag{7.1.53}
\end{equation}
This series is known as double Fourier series of \(f(x,y)\)
\begin{equation*}
\therefore A_{mn} = \frac{2}{a}\cdot\frac{2}{b}\int\limits_{x=0}^{a} \int\limits_{y=0}^{b}f(x,y) \sin \frac{m\pi x}{a}\sin \frac{n\pi y}{b}\,dx\,dy
\end{equation*}
\begin{equation}
=\frac{4}{ab}\int\limits_{x=0}^{a} \int\limits_{y=0}^{b}f(x,y) \sin \frac{m\pi x}{a}\sin \frac{n\pi y}{b}\,dx\,dy \tag{7.1.54}
\end{equation}
Now using equation (7.1.52) and \(\left(\frac{\partial u}{\partial t}\right)_{t=0}=g(x,y)\text{,}\) we get -
\begin{equation}
g(x,y) = \sum\limits_{m=1}^{\infty}\sum\limits_{n=1}^{\infty}\lambda_{mn} B_{mn}\sin \frac{m\pi x}{a}\sin \frac{n\pi y}{b}\tag{7.1.55}
\end{equation}
This is again a double Fourier sine series.
\begin{equation*}
\therefore \lambda_{mn} B_{mn} = \frac{4}{ab}\int\limits_{x=0}^{a} \int\limits_{y=0}^{b}g(x,y) \sin \frac{m\pi x}{a}\sin \frac{n\pi y}{b}\,dx\,dy
\end{equation*}
or,
\begin{equation}
B_{mn}=\frac{4}{ab\lambda_{mn}}\int\limits_{x=0}^{a} \int\limits_{y=0}^{b}g(x,y) \sin \frac{m\pi x}{a}\sin \frac{n\pi y}{b}\,dx\,dy\tag{7.1.56}
\end{equation}
Hence the solution of the wave equation (7.1.45) is given by equation (7.1.52) with the coefficients \(A_{mn}\) and \(B_{mn}\) as given by equations (7.1.54) and (7.1.56), respectively.
Special Case. when initial velocity vanishes, i.e.,
\begin{equation*}
\left(\frac{\partial u}{\partial t}\right)_{t=0}=g(x,y) =0,
\end{equation*}
which implies \(B_{mn} =0.\) Therefore, the solution of wave equation is given by
\begin{equation}
u(x,y,t) = \sum\limits_{m=1}^{\infty}\sum\limits_{n=1}^{\infty}A_{mn}\cos\lambda_{mn}t\sin \frac{m\pi x}{a}\sin \frac{n\pi y}{b}\tag{7.1.57}
\end{equation}
where
\begin{equation*}
A_{mn}= \frac{4}{ab}\int\limits_{x=0}^{a} \int\limits_{y=0}^{b}f(x,y) \sin\frac{m\pi x}{a}\sin \frac{n\pi y}{b}\,dx\,dy.
\end{equation*}
Subsubsection 7.1.4.2 Circular Membrane
Consider the vibration of the circular membrane of radius \(a\text{.}\) The vibrations of a circular membrane are governed by the two-dimensional wave equation
\begin{equation*}
\frac{\partial^{2} u}{\partial t^{2}}=v^{2}\left[\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial y^{2}}\right].
\end{equation*}
Using polar coordinates as \(x=r\cos\theta, y=r\sin\theta,\) the wave equation is transformed into
\begin{equation}
\frac{\partial^{2} u}{\partial t^{2}}=v^{2}\left[\frac{\partial^{2}u}{\partial r^{2}}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2} u}{\partial \theta^{2}}\right]\tag{7.1.58}
\end{equation}
(see Appendix B)
The boundary conditions are \(u(a,0,t)=0\text{,}\) \(-\pi \leq \theta \leq \pi \text{,}\) \(t \geq 0\text{;}\) the initial conditions are
\begin{equation*}
\left(\frac{\partial u}{\partial t}\right)_{t=0} =g(r,\theta),
\end{equation*}
the initial velocity; \(u(r,\theta,0)=f(r,\theta),\) the initial displacement; where \(0\leq r \leq a \) and \(-\pi \leq \theta \leq \pi.\) Let
\begin{equation}
u(r,0,t)=R(r)\Theta(\theta)T(t) \tag{7.1.59}
\end{equation}
be the solution of equation (7.1.58). Hence with help of equation (7.1.59), equation (7.1.58) becomes -
\begin{equation*}
R\Theta\frac{\partial^{2} T}{\partial t^{2}}=v^{2}\left[\Theta T\frac{\partial^{2}R}{\partial r^{2}}+\frac{1}{r}\Theta T\frac{\partial R}{\partial r}+\frac{1}{r^{2}}RT\frac{\partial^{2} \Theta}{\partial \theta^{2}}\right]
\end{equation*}
or,
\begin{equation*}
\frac{1}{v^{2}T}\frac{\partial^{2} T}{\partial t^{2}}
\end{equation*}
\begin{equation}
=\left(\frac{1}{R}\frac{\partial^{2} R}{\partial r^{2}}+\frac{1}{rR}\frac{\partial R}{\partial r}\right)+\frac{1}{r^{2}\Theta}\frac{\partial^{2} \Theta}{\partial \theta^{2}} =-k^{2} \quad (say)\tag{7.1.60}
\end{equation}
now taking
\begin{equation*}
\frac{1}{\Theta}\frac{\partial^{2} \Theta}{\partial \theta^{2}} =-m^{2}
\end{equation*}
or,
\begin{equation}
\frac{\partial^{2} \Theta}{\partial \theta^{2}} +m^{2}\Theta =0\tag{7.1.61}
\end{equation}
we get-
\begin{equation*}
\frac{1}{R}\frac{\partial^{2} R}{\partial r^{2}}+\frac{1}{rR}\frac{\partial R}{\partial r}+\frac{1}{r^{2}}(-m^{2})=-k^{2}
\end{equation*}
or,
\begin{equation}
\frac{\partial^{2} R}{\partial r^{2}}+\frac{1}{r}\frac{\partial R}{\partial r}+\left(k^{2}-\frac{m^{2}}{r^{2}}\right)R=0\tag{7.1.62}
\end{equation}
Also, we have -
\begin{equation*}
\frac{1}{v^{2}T}\frac{\partial^{2} T}{\partial t^{2}}=-k^{2}
\end{equation*}
i.e.,
\begin{equation}
\frac{\partial^{2} T}{\partial t^{2}}+v^{2} k^{2}T=0\tag{7.1.63}
\end{equation}
Hence equations (7.1.61), (7.1.62) and (7.1.63) are the ordinary differential equation for \(\Theta, R\text{,}\) and \(T\text{,}\) respectively. The solution of equation (7.1.61) is of the form
\begin{equation}
\Theta =De^{\pm im\theta} \text{and} \quad m=0,1,2\cdots \tag{7.1.64}
\end{equation}
Now taking \(s=kr\text{,}\) we get -
\begin{equation*}
\frac{\partial R}{\partial r} = \frac{\partial R}{\partial s}\frac{\partial s}{\partial r} = k\frac{\partial R}{\partial s}
\end{equation*}
and
\begin{equation*}
\frac{\partial^{2} R}{\partial r^{2}} =k^{2}\frac{\partial^{2} R}{\partial s^{2}}.
\end{equation*}
therefore equation (7.1.62) gives
\begin{equation*}
k^{2}\frac{\partial^{2} R}{\partial s^{2}}+\frac{1}{r}k\frac{\partial R}{\partial s}+\left(k^{2}-\frac{m^{2}}{r^{2}}\right)R =0
\end{equation*}
\begin{equation}
\frac{\partial^{2} R}{\partial r^{2}}=\frac{\partial^{2} R}{\partial s^{2}}+\frac{1}{s}\frac{\partial R}{\partial s}+\left(1-\frac{m^{2}}{s^{2}}\right)R = 0\tag{7.1.65}
\end{equation}
which is Bessel’s equation and hence its general solution is
\begin{equation}
R=c_{1}J_{m}(s)+c_{2}J_{-m}(s)=c_{1}J_{m}(kr)+c_{2}J_{-m}(kr)\tag{7.1.66}
\end{equation}
But the deflection of the membrane is finite when \(J_{-m}\to \infty\) as \(r\to 0\text{,}\) so we can not make use of \(J_{-m} \) and must take \(c_{2}=0 \text{.}\) Thus equation (7.1.66) reduces to
\begin{equation}
R=c_{1}J_{m}(kr)\tag{7.1.67}
\end{equation}
Now using the boundary condition, we have
\begin{equation*}
u(a,\theta,t) =R(a)\Theta(\theta)T(t)=0.
\end{equation*}
which implies \(R(a)=0\text{;}\) otherwise if \(\Theta(\theta)=0\) or, \(T(t)=0\) causes \(u=0\text{.}\)
\begin{equation}
\therefore R(a) =c_{1}J_{m}(ka)=0 \Rightarrow J_{m}(ka) =0\tag{7.1.68}
\end{equation}
assuming \(k_{1}{,} k_{2}{,} k_{3}{,}\cdots \) be the positive roots of equation (7.1.67), the general solution of equation (7.1.63) is
\begin{equation}
T=A\cos vkt+B\sin vkt\tag{7.1.69}
\end{equation}
\begin{equation}
\left\{A\cos vkt+B\sin vkt\right\}e^{\pm im\theta}J_{m}(kr)\tag{7.1.70}
\end{equation}
Trying superposition and using distinct values of constants A and B for each choice of \(m\) and \(n\text{,}\) the general solution of equation (7.1.58) may be taken as
\begin{equation*}
u(r,\theta,t)=\sum\limits_{m=1}^{\infty}\sum\limits_{n=1}^{\infty}\left\{A_{mn}\cos (vk_{mn}t)\right.
\end{equation*}
\begin{equation}
\left.+B_{mn}\sin (vk_{mn}t)\right\}e^{\pm im\theta}J_{m}(k_{mn}r)\tag{7.1.71}
\end{equation}
which satisfies the boundary condition \(u(a,\theta,t)=0\text{.}\) In this case equation (7.1.58) is radially symmetrical i.e., the solution is independent of \(\theta\text{,}\) we have from equation (7.1.71) by putting \(m=0\text{.}\)
\begin{equation}
u(r,t)=\sum\limits_{n=1}^{\infty}\left\{A_{n}\cos (vk_{n}t)+ B_{n}\sin (vk_{n}t)\right\}J_{o}(k_{n}r) \tag{7.1.72}
\end{equation}
where \(k_{1}, k_{2}, k_{3},\cdots \) are the positive roots of \(J_{o}(ka) =0\text{.}\)
Now from equation (7.1.72) and the boundary condition when \(t=0\text{,}\) we have
\begin{equation*}
u(r,0)= \sum\limits_{n=1}^{\infty}A_{n}J_{o}(k_{n}r)=f(r)
\end{equation*}
\([\because f(r,\theta)\) becomes \(f(r)\) when it is independent of \(\theta\)]. Therefore \(A'_{n}s\) must be the coefficient of Fourier - Bessel series, which represents \(f(r)\) in terms of \(J_{o}(k_{n}r) \) i.e.,
\begin{equation}
A_{n}=\frac{2}{a^{2}J_{1}(k_{n}a)}\int\limits_{0}^{a}rf(r)J_{o}(k_{n}r)\,dr\tag{7.1.73}
\end{equation}
Also, from
\begin{equation*}
\left(\frac{\partial u}{\partial t}\right)_{t=0}=g(r,\theta),
\end{equation*}
or,
\begin{equation*}
u(r,0)=\sum\limits_{n=1}^{\infty}vk_{n}B_{n}J_{o}(k_{n}r)=g(r),
\end{equation*}
\(g\) being independent of \(\theta\text{.}\) Hence,
\begin{equation}
B_{n}=\frac{2}{vk_{n}a^{2}J^{2}_{1}(k_{n}a)}\int\limits_{0}^{a}rg(r)J_{o}(k_{n}r)\,dr\tag{7.1.74}
\end{equation}
Subsubsection 7.1.4.3 Three Dimensional Wave Equation
The three - dimensional wave equation is given by
\begin{equation}
\frac{\partial^{2} u}{\partial t^{2}}=v^{2}\left[\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}+\frac{\partial^{2} u}{\partial z^{2}}\right] =v^{2}\nabla^{2}u.\tag{7.1.75}
\end{equation}
where
\begin{equation*}
\nabla^{2} = \frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}}.
\end{equation*}
The solution of which can be obtained under the conditions
- \begin{equation*} \frac{\partial u}{\partial x}=0; \quad \text{when} \quad x=0,x=a. \end{equation*}
- \begin{equation*} \frac{\partial u}{\partial y}=0; \quad \text{when} \quad y=0,y=a \end{equation*}
- \begin{equation*} \frac{\partial u}{\partial z}=0; \quad \text{when} \quad z=0,z=a \end{equation*}
- \begin{equation*} u\neq 0 \quad \text{at} \quad t=0. \end{equation*}
Let
\begin{equation}
u(x,y,t)=X(x)Y(y)T(t)\tag{7.1.76}
\end{equation}
be the solution of equation (7.1.75). Hence, making use of equation (7.1.76), equation (7.1.75) becomes -
\begin{equation}
\frac{1}{X}\frac{\partial^{2} X}{\partial x^{2}}+\frac{1}{Y}\frac{\partial^{2} Y}{\partial y^{2}}+\frac{1}{Z}\frac{\partial^{2} Z}{\partial z^{2}}=\frac{1}{v^{2}T}\frac{\partial^{2} T}{\partial t^{2}}\tag{7.1.77}
\end{equation}
here variables have been separated. Hence each of the four terms in this equation must be constant say
\begin{equation*}
\frac{1}{X}\frac{\partial^{2} X}{\partial x^{2}}=-\lambda_{1}^{2}; \quad \frac{1}{Y}\frac{\partial^{2} Y}{\partial y^{2}}=-\lambda_{2}^{2};
\end{equation*}
\begin{equation*}
\frac{1}{Z}\frac{\partial^{2} Z}{\partial z^{2}}=-\lambda_{3}^{2};
\end{equation*}
and
\begin{equation*}
\frac{1}{v^{2}T}\frac{\partial^{2} T}{\partial t^{2}}= -\lambda^{2}
\end{equation*}
so that
\begin{equation*}
\lambda^{2} = \lambda_{1}^{2}+\lambda_{2}^{2}+\lambda_{3}^{2}
\end{equation*}
Now, we get -
\begin{equation*}
X=A_{1}\cos (\lambda_{1}x+\alpha_{1}),\quad Y=A_{2}\cos (\lambda_{2}y+\alpha_{2}),
\end{equation*}
\begin{equation*}
Z=A_{3}\cos (\lambda_{3}z+\alpha_{3}),
\end{equation*}
and
\begin{equation*}
T=B\cos (\lambda vt+\alpha)
\end{equation*}
where \(A_{1},A_{2},A_{3},B,\alpha_{1}, \alpha_{2} , \alpha_{3}, \alpha\) all are arbitrary constants. Hence the solution of equation, (7.1.75) is
\begin{equation*}
u=A\cos (\lambda_{1}x+\alpha_{1})
\end{equation*}
\begin{equation}
\cos (\lambda_{2}y+\alpha_{2})\cos (\lambda_{3}z+\alpha_{3})\cos (\lambda vt+\alpha) \tag{7.1.78}
\end{equation}
we can also write the solution ofequation (7.1.75) in the form
\begin{equation}
u=Ce^{\pm i(\lambda_{1}x+\lambda_{2}y+\lambda_{3}z+\lambda vt)}\tag{7.1.79}
\end{equation}
where
\begin{equation*}
\lambda^{2}= \lambda^{2}_{1}+\lambda^{2}_{2}+\lambda^{2}_{3}
\end{equation*}
we have
\begin{equation*}
e^{\pm i \lambda_{1}x} =A_{1} \cos\lambda_{1}x \pm A_{2} \sin\lambda_{1}x
\end{equation*}
or,
\begin{equation*}
\frac{\partial u}{\partial x}=0,
\end{equation*}
when \(x=0\) that implies \(A_{2}=0, \) etc.
Then
\begin{equation}
u=C \cos\lambda_{1}x\cos\lambda_{2}y\cos\lambda_{3}z\cos\lambda vt\tag{7.1.80}
\end{equation}
The result (7.1.80) may be deduced from equation (7.1.77), since, when \(x=0, \sin\alpha_{1}x =0\) etc. According to boundary condition i., we have -
\begin{equation*}
\sin\lambda_{1}a\cos\lambda_{2}y\cos\lambda_{3}z\cos\lambda vt=0,
\end{equation*}
i.e., \(\sin\lambda_{1}a=0 \) or, \(\sin\lambda_{1}a =N_{1}\pi\) and \(\lambda_{1}=\frac{N_{1}\pi}{a}, N_{1}\) being an integer. Similarly,
\begin{equation*}
\lambda_{2}=\frac{N_{2}\pi}{a}, \quad \lambda_{3}=\frac{N_{3}\pi}{a},
\end{equation*}
and
\begin{equation*}
\lambda =\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}+\lambda_{3}^{2}}
\end{equation*}
\begin{equation*}
=\frac{\pi}{a}\sqrt{N_{1}^{2}+N_{2}^{2}+N_{3}^{2}}
\end{equation*}
now, equation (7.1.80) reduces to
\begin{equation*}
u=C\cos \frac{N_{1}\pi x}{a} \cos \frac{N_{2}\pi y}{a}\cos \frac{N_{3}\pi z}{a} \cos\{\sqrt{N_{1}^{2}+N_{2}^{2}+N_{3}^{2}}\frac{\pi vt}{a}\}
\end{equation*}
Hence the general solution of equation (7.1.75) is
\begin{equation*}
u=\sum_{N_{1}=1}^{\infty}\sum_{N_{2}=1}^{\infty}\sum_{N_{3}=1}^{\infty}C_{N_{1}N_{2}N_{3}}\cos( \frac{N_{1}\pi x}{a})
\end{equation*}
\begin{equation}
\cos (\frac{N_{2}\pi y}{a}) \cos (\frac{N_{3}\pi z}{a}) \cos\left\{\sqrt{N_{1}^{2}+N_{2}^{2}+N_{3}^{2}}\frac{\pi vt}{a}\right\}\tag{7.1.81}
\end{equation}
