Bessel Functions and the Wave Equation

TL;DR

This paper presents a method for solving the n-dimensional wave equation by leveraging elementary properties of Bessel functions, providing a new analytical approach.

Contribution

It introduces a novel solution technique for the wave equation using Bessel functions, which simplifies the problem with elementary properties.

Findings

Explicit solutions for the wave equation in multiple dimensions.

Demonstration of the effectiveness of Bessel functions in wave analysis.

Potential applications in physics and engineering problems.

Abstract

We solve the Cauchy problem for the $n$-dimensional wave equation using elementary properties of the Bessel functions.

Equations

$$D_{x_k{x}_k}^2=\frac{{\partial }^2}{\partial x_k^2},1\le k\le n,$$

$$D_{tt}^{2}u(x,t)={\nabla }^{2}u(x,t),x\in {\mathbb{R}}^n,t>0,$$

$$u(x,0)=\phi (x)\text{and}{D}_{t}u(x,0)=\psi (x),x\in {\mathbb{R}}^n,$$

$$D_{tt}^2\hat{u}(\xi ,t)+|\xi {|}^2\hat{u}(\xi ,t)=0,\xi \in {\mathbb{R}}^n,t>0,$$

$$\hat{u}(\xi ,0)=\hat{\phi }(\xi )\text{and}{D}_t\hat{u}(\xi ,0)=\hat{\psi }(\xi ),\xi \in {\mathbb{R}}^n.$$

$$\hat{u}(\xi ,t)=\hat{\phi }(\xi )\cos (t|\xi |)+\hat{\psi }(\xi )\frac{\sin (t|\xi |)}{|\xi |},\xi \in {\mathbb{R}}^n,t>0.$$

$$u(x,t)=\frac{1}{(2\pi {)}^n}{\int }_{{\mathbb{R}}^n}\hat{\phi }(\xi )$$

$$+\frac{1}{(2\pi {)}^n}{\int }_{{\mathbb{R}}^n}\hat{\psi }(\xi )\frac{\sin (t|\xi |)}{|\xi |}{e}^{i\xi \cdot x}d\xi .$$

$$\frac{\sin(R|\xi|)}{|\xi|}=c_{n}\,\Big{(}\frac{1}{R}\,\frac{\partial}{\partial R}\Big{)}^{(n-3)/2}\Big{(}\frac{1}{\,\omega_{n}\,R}\int_{\partial B(0,R)}e^{-ix\cdot\xi}\,d\sigma(x)\Big{)}\,,$$

$$\frac{\sin(R|\xi|)}{|\xi|}=d_{n}\,\Big{(}\frac{1}{R}\,\frac{\partial}{\partial R}\Big{)}^{(n-2)/2}\Big{(}\frac{1}{\,v_{n}}\int_{B(0,R)}\frac{1}{\sqrt{R^{2}-|x|^{2}}}\,e^{-ix\cdot\xi}\,dx\Big{)}\,,$$

$$x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+(x^2-{\nu }^2)y=0.$$

$$\frac{d}{dx}(x^{\nu }{J}_{\nu }(x))=x^{\nu }{J}_{\nu -1}(x),$$

$$J_{\nu }(x)=\frac{(x/2{)}^{\nu }}{\Gamma (\nu +1/2)\Gamma (1/2)}{\int }_{-1}^1(1-s^2{)}^{\nu -1/2}{e}^{ixs}ds,$$

$$J_{1/2}(x)=\frac{\sqrt{2}}{\sqrt{\pi }}\frac{1}{x^{1/2}}\sin (x),$$

$$\frac{1}{{\omega }_{n-1}R}{\int }_{\partial B(0,R)}{e}^{-ix\cdot \xi }d\sigma (x)=R^{n-2}{\int }_{-1}^1{e}^{iR|\xi |s}(1-s^2{)}^{(n-3)/2}ds,$$

$$\frac{1}{{\omega }_{n}R}{\int }_{\partial B(0,R)}{e}^{-ix\cdot \xi }d\sigma (x)$$

$$=c_n\frac{1}{|\xi {|}^{n-2}}(R|\xi |{)}^{(n-2)/2}{J}_{(n-2)/2}(R|\xi |).$$

$$\frac{\partial}{\partial R}\Big{(}\frac{1}{\omega_{n}R}\,\int_{\partial B(0,R)}e^{-ix\cdot\xi}\,d\sigma(x)\Big{)}=c_{n}\,\frac{1}{|\xi|^{n-2}}|\xi|\,\,(R|\xi|)^{(n-2)/2}J_{(n-4)/2}(R|\xi|)\,,$$

$$\frac{1}{R}\frac{\partial}{\partial R}\Big{(}\frac{1}{\omega_{n}R}\,\int_{\partial B(0,R)}e^{-ix\cdot\xi}\,d\sigma(x)\Big{)}=c_{n}\,\frac{1}{|\xi|^{n-4}}\,(R|\xi|)^{(n-4)/2}J_{(n-4)/2}(R|\xi|)\,.$$

$$\displaystyle\Big{(}\frac{1}{R}\frac{\partial}{\partial R}\Big{)}^{(n-3)/2}\Big{(}\frac{1}{\omega_{n}R}\,$$

$$=c_n\frac{1}{|\xi |}(R|\xi |{)}^{1/2}{J}_{1/2}(R|\xi |)$$

$$=c_n\frac{1}{|\xi |}(R|\xi |{)}^{1/2}\frac{\sin (R|\xi |)}{(R|\xi |{)}^{1/2}}$$

$$=c_n\frac{\sin (R|\xi |)}{|\xi |}.$$

$${\int }_0^{\infty }\sin (R\rho )J_0(t\rho )d\rho =\frac{1}{\sqrt{R^2-t^2}}H(R-t),R,t>0,$$

$$\Big{(}\frac{1}{R}\frac{\partial}{\partial R}\Big{)}\Big{(}\int_{0}^{\infty}\sin(R\rho)\,\rho^{\nu-1}J_{\nu-1}(t\,\rho)\,d\rho\Big{)}=\frac{1}{t}\int_{0}^{\infty}\sin(R\rho)\,\rho^{\nu}J_{\nu}(t\rho)\,d\rho\,,$$

$$\Big{(}\frac{1}{R}\frac{\partial}{\partial R}\Big{)}^{j}\Big{(}\int_{0}^{\infty}\sin(R\rho)\,\rho^{\nu-j}J_{\nu-j}(t\,\rho)\,d\rho\Big{)}=\frac{1}{t^{j}}\int_{0}^{\infty}\sin(R\rho)\,\rho^{\nu}J_{\nu}(t\rho)\,d\rho\,.$$

$$\displaystyle\frac{\partial}{\partial R}\Big{(}\int_{0}^{\infty}\sin(R\rho)\,\rho^{\nu-1}J_{\nu-1}(t\rho)\,d\rho\Big{)}$$

$$=-\frac{1}{t^{\nu }}{\int }_0^{\infty }\cos (R\rho )(t\rho {)}^{\nu }{J}_{\nu -1}(t\rho )d\rho ,$$

$$\frac{-1}{t^{\nu+1}}\int_{0}^{\infty}\cos(R\rho)\,\frac{\partial}{\partial\rho}\big{(}\big{(}t\rho)^{\nu}J_{\nu}(t\rho)\big{)}\,d\rho=\frac{R}{t}\int_{0}^{\infty}\sin(R\rho)\,\rho^{\nu}J_{\nu}(t\rho)\,d\rho\,,$$

$$\hat{f}(|\xi |)=d_n\frac{1}{|\xi {|}^{(n-2)/2}}{\int }_0^{\infty }{\rho }^{n/2}f(\rho )J_{(n-2)/2}(|\xi |\rho )d\rho .$$

$${\int }_{{\mathbb{R}}^n}\frac{\sin (R|\xi |)}{|\xi |}{e}^{-ix\cdot \xi }d\xi =d_n\frac{1}{|x{|}^{(n-2)/2}}{\int }_0^{\infty }{\rho }^{n/2}\frac{\sin (R\rho )}{\rho }{J}_{(n-2)/2}(|x|\rho )d\rho .$$

$${\int }_{{\mathbb{R}}^n}\frac{\sin (R|\xi |)}{|\xi |}{e}^{-ix\cdot \xi }d\xi =d_n\frac{1}{|x{|}^{(k-1)}}{\int }_0^{\infty }\sin (R\rho ){\rho }^{k-1}{J}_{k-1}(|x|\rho )d\rho ,$$

$${\int }_{{\mathbb{R}}^n}\frac{\sin (R|\xi |)}{|\xi |}{e}^{-ix\cdot \xi }d\xi$$

Full text

With $\nabla^{2}=D^{2}_{x_{1}x_{1}}+\cdots+D^{2}_{x_{n}x_{n}}$ the Laplacian in ${\mathbb{R}}^{n}$, where

[TABLE]

and $D_{t}$ and $D_{tt}$ indicating the first and second order derivatives with respect to the variable $t\in{\mathbb{R}}$, respectively, the wave equation in the upper half-space ${\mathbb{R}}^{n+1}_{+}$ is given by

[TABLE]

and the Cauchy problem for this equation consists of finding $u(x,t)$ that satisfies (1) subject to the initial conditions

[TABLE]

where for simplicity we shall take $\varphi$ and $\psi$ in $\mathcal{S}({\mathbb{R}}^{n})$.

Applying the Fourier transform to (1) in the space variables, considering $t$ as a parameter, it readily follows that $\widehat{\nabla^{2}u}(\xi,t)=-|\xi|^{2}\,\widehat{u}(\xi,t)$, and so $\widehat{u}$ satisfies

[TABLE]

subject to

[TABLE]

For each fixed $\xi\in{\mathbb{R}}^{n}$ this resulting ordinary differential equation in $t$ is the simple harmonic oscillator equation with constant angular frequency $|\xi|$, and so

[TABLE]

Hence, the Fourier inversion formula gives for $(x,t)\in{\mathbb{R}}^{n+1}_{+}$,

[TABLE]

Since the first integral in (2) can be obtained from the second by differentiating with respect to $t$, we will concentrate on the latter. The idea is to interpret $\sin(|\xi|t)/|\xi|$ as the Fourier transform of a tempered distribution, and the key ingredient for this are the following representation formulas established in [1].

[TABLE]

where $R>0$, $\omega_{n}$ is the surface measure of the unit ball in ${\mathbb{R}}^{n}$, and $c_{n}^{-1}=(n-2)(n-4)\cdots 1$.

On the other hand, if $n$ is an even integer greater than or equal to $2$,

[TABLE]

where $R>0$, $d_{n}^{-1}={n(n-2)(n-4)\cdots 2}$, and $v_{n}$ is the volume of the unit ball in ${\mathbb{R}}^{n}$.

The purpose of this note is to establish (3) and (4) using elementary properties of Bessel functions. $J_{\nu}(x)$, the Bessel function of order $\nu$, is defined as the solution of the second order linear equation

[TABLE]

Several basic properties of the Bessel functions follow readily from their power series expression [2]. They include the recurrence formula

[TABLE]

the integral representation of Poisson type

[TABLE]

and the identity

[TABLE]

for $x>0$.

We will consider the odd dimensional case first. The dimensional constant $c_{n}$ may vary from appearance to appearance until it is finally determined at the end of the proof. To begin recall that for $n\geq 3$, as established in (18) in [1],

[TABLE]

which combined with (6) above with $\nu-1/2=(n-3)/2$ there, i.e., $\nu=(n-2)/2$, gives

[TABLE]

Now, by (5) we obtain that

[TABLE]

or

[TABLE]

Thus, applying the above reasoning $(n-3)/2$ times, (7) gives

[TABLE]

The value of $c_{n}$ is readily obtained as in [1], and (3) has been established.

To consider the case $n$ even, one generally proceeds at this point by a reasoning akin to Hadamard’s method of descent, i.e., the desired result for the wave equation in even dimension $n$ is derived from the result in odd dimension $n+1$, as is done for instance in [1] for the representation formulas. On the other hand, Bessel functions provide the desired result for the wave equation in even dimensions directly, by a method akin to ascent: the result for the wave equation for dimension $n=2$ is obtained explicitly, and for even dimension $n+2$ is obtained from the result in even dimension $n$.

We will first prove a preliminary result. The dimensional constant $d_{n}$ may vary from appearance to appearance until it is finally determined at the end of the proof.

[TABLE]

where $H$ denotes the Heavyside function.

Furthermore, for $\nu\geq 1$,

[TABLE]

and, consequently, for $1\leq j\leq\nu$,

[TABLE]

Now,

[TABLE]

which, by (1), equals

[TABLE]

which proves (9).

(10) follows by repeated applications of (9), and we have finished. ∎

Finally, recall that the Fourier transform of a radial function $f$ on ${\mathbb{R}}^{n}$ is given by the expression [2],

[TABLE]

In particular, we have

[TABLE]

Let now $n=2k$ be an even integer. Then by (11),

[TABLE]

and, therefore, (10) with $\nu=j=k-1$ there yields

[TABLE]

Thus by the Fourier inversion formula,

[TABLE]

The constant $d_{n}$ is readily determined as in [1], and we have finished.