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Klein-Gordon equation is ${鈭�}^{2}u=\left(\frac{1}{v^2}\right)\frac{{鈭�}^{2}u}{鈭�t^2}+{位}^{2}u$.

Any disturbance or energy transfer from one location to another is referred to as a wave. Each wave is guided by a mathematical formula. A wave can be either standing or stationary.

Consider the Klein Gordon equation.

${鈭�}^{2}u=\left(\frac{1}{v^2}\right)\frac{{鈭�}^{2}u}{鈭�t^2}+{位}^{2}u$

Express it in polar coordinates.

${鈭�}^{2}u=\frac{{鈭�}^{2}u}{鈭�r^2}+\frac{1}{r}\left(\frac{鈭�u}{鈭�r}\right)+\frac{1}{r^2}\left(\frac{{鈭�}^{2}u}{鈭�{胃}^2}\right)$ 鈥�.. (1)

Separate the spatial and temporal variables to write the above equation in the form of $u=F\left(r,胃\right)T\left(t\right)$.

Assume the separation constant.$-k^2$

$\left(\frac{{鈭�}^{2}F}{鈭�r^2}\right)+\frac{1}{r}\left(\frac{鈭�F}{鈭�r}\right)+\frac{1}{r^2}\left(\frac{{鈭�}^{2}F}{鈭�{胃}^2}\right)+\left(k^2-{位}^2\right)F=0$ 鈥�.. (2)

$\frac{d^{2}T}{d{t}^2}+k^2{v}^{2}T=0$ 鈥�.. (3)

Solve equation (3) to get $T=\left\{\begin{array}{l}Cos\left(kvt\right)\\ Sin\left(kvt\right)\end{array}\right.$.

Separate spatial variable from $F\left(r,胃\right)$and express in terms of $F\left(r,胃\right)=R\left(r\right)螛\left(胃\right)$.

$\frac{1}{螛}\left(\frac{{鈭�}^2螛}{鈭�{胃}^2}\right)=-n^2$

Solve the above equation to get$螛=\left\{\begin{array}{l}\cos \left(n胃\right)\\ sin\left(n胃\right)\end{array}\right.$ .

Write equation (2) for r.

$r^2\left(\frac{{鈭�}^{2}R}{鈭�r^2}\right)+r\left(\frac{鈭�R}{鈭�r}\right)-m^{2}R+\left(k^2-{位}^2\right)r^{2}R=0$

Solve the equation by changing the variables $z={\left(k^2-{位}^2\right)}^{\frac{1}{2}}r$.

This gives us the Bessel鈥檚 differential equation and the solution can be written in the form of $R=Z_p\left(z\right)$.

Change to the previous variable.

$R=Z_p\left(\sqrt{k^2-{位}^{2}r}\right)$

Consider the requirement of finite u at r=0.

The general solution is$u=J_p\left(\sqrt{k^2-{位}^{2}r}\right)\left(\begin{array}{c}\cos \left(n胃\right)\\ sin\left(n胃\right)\end{array}\right)\left(\begin{array}{c}\cos \left(kvt\right)\\ \sin \left(kvt\right)\end{array}\right)$ .

Consider the requirement of u=0.

At r=R the equation is:

$\sqrt{k^2-{位}^2}R=k_{nm}$ 鈥�.. (4)

Square both sides in equation (4) and solve for $k^2$.

$k^2={\left(\frac{k_{nm}}{R}\right)}^2+{位}^2$ 鈥�.. (5)

As $J_p\left(k_{nm}\right)=0$, $k_{nm}$is a solution of $J_p\left(z\right)$.

Replace R=a in equation (5) and find the characteristic frequencies of oscillation.

$$\begin{gathered} 谓=\frac{蝇}{2蟺} \\ =\frac{kv}{2蟺} \\ =\frac{v}{2蟺}\sqrt{{\left(\frac{k_{nm}}{a}\right)}^2+{位}^2} \end{gathered}$$

Hence the characteristic frequencies of oscillation is $v=\frac{v}{2蟺}\sqrt{{\left(\frac{k_{nm}}{a}\right)}^2+{位}^2}$.