Question 11.5.1: Find the solution of the two-dimensional wave equation (3) i......

Assuming that u(r, t) = R(r)T(t), and substituting for u(r, t) in equation (3), we obtain

\frac{R^{\prime\prime}+(1/r)R^{\prime}}{R}=\frac{1}{a^{2}}\frac{T^{\prime\prime}}{T}=-\lambda^{2}. (6)

We have anticipated that the separation constant must be negative by writing it as −λ² with λ > 0.^{12}

Then equation (6) yields the following two ordinary differential equations:

r^{2}R^{\prime\prime}+r R^{\prime}+\lambda^{2}r^{2}R=0, (7)

T^{\prime\prime}+\lambda^{2}a^{2}T=0. (8)

Thus, from equation (8),

T(t)=k_{1}\sin(\lambda a t)+k_{2}\cos(\lambda a t). (9)

Introducing the new independent variable \xi = λr into equation (7), we obtain

\xi^{2}\frac{d^{2}R}{d\xi^{2}}+\xi\frac{d R}{d\xi}+\xi^{2}R=0, (10)

which is Bessel’s equation of order zero. Thus

R\left(\frac{\xi}{\lambda}\right)=c_{1}J_{0}(\xi)+c_{2}Y_{0}(\xi), (11)

where J_{0} and Y_{0} are Bessel functions of the first and second kinds, respectively, of order zero (see Section 11.4). In terms of r, we have

R(r)=c_{1}J_{0}(\lambda r)+c_{2}Y_{0}(\lambda r). (12)

The boundedness condition on u(r, t) requires that R remain bounded as r\to0^{+}. Since Y_{0}(\lambda r)\rightarrow-\infty as r\to0^{+}, we must choose c_2 = 0. The boundary condition (4) then requires that

J_{0}(\lambda)=0. (13)

Consequently, the allowable values of the separation constant are obtained from the roots of the transcendental equation (13). Recall from Section 11.4 that J_{0}(\lambda) has an infinite set of discrete positive zeros, which we denote by \lambda_{1},\lambda_{2},\lambda_{3},\,\dots\,,\lambda_{n},\,\dots\,, ordered in increasing magnitude. Further, the functions J_{0}(\lambda_{n}r) are the eigenfunctions of a singular Sturm-Liouville problem and can be used as the basis of a series expansion for the given function f . The fundamental solutions of this problem, which satisfy the partial differential equation (3), the boundary condition (4), and the boundedness condition, are

u_{n}(r,t)=J_{0}(\lambda_{n}r)\sin(\lambda_{n}a t),\quad n=1,2,\,\dots\,, (14)

\nu_{n}(r,t)=J_{0}(\lambda_{n}r)\cos(\lambda_{n}a t),\quad n=1,2,\,\dots\,, (15)

Next we assume that u(r, t) can be expressed as an infinite linear combination of the fundamental solutions (14), (15):

u(r,t)=\sum_{n=1}^{\infty}(k_{n}u_{n}(r,t)+c_{n}\nu_{n}(r,t))\\=\sum_{n=1}^{\infty}(k_{n}J_{0}(\lambda_{n}r)\;\sin(\lambda_{n}a t)+c_{n}J_{0}(\lambda_{n}r)\;\cos(\lambda_{n}a t)). (16)

The initial conditions require that

u(r,0)=\sum_{n=1}^{\infty}c_{n}J_{0}(\lambda_{n}r)=f(r), (17)

u_{t}(r,0)=\sum_{n=1}^{\infty}\lambda_{n}a k_{n}J_{0}(\lambda_{n}r)=0. (18)

From equation (24) of Section 11.4,

c_{m}=\frac{\displaystyle\int_{0}^{1}x f(x)\,J_{0}\Bigl(\sqrt{\lambda_{m}}x\Bigr)d x}{\displaystyle\int_{0}^{1}x J_{0}^{2}\Bigl(\sqrt{\lambda_{m}}x\Bigr)d x}, (24)

we obtain

k_{n}=0,\quad c_{n}=\frac{\displaystyle\int_{0}^{1}r f(r)J_{0}(\lambda_{n}r)\,d r}{\displaystyle\int_{0}^{1}r(J_{0}(\lambda_{n}r))^{2}\,d r},\quad n=1,2,\ \dots\ . (19)

Thus the solution of the partial differential equation (3) satisfying the boundary condition (4) and the initial conditions (5) is given by

u(r,t)=\sum_{n=1}^{\infty}c_{n}J_{0}(\lambda_{n}r)\;\cos(\lambda_{n}a t) (20)

with the coefficients c_{n} defined by equation (19).

^{12}By denoting the separation constant by −λ², rather than simply by −λ, we avoid the appearance of numerous radical signs in the following discussion.