Solution for the simple harmonic oscillator problem via unilateral Fourier transform

Abstract

The solution of the simple harmonic oscillator problem is properly determined by means of the unilateral Fourier transform.

Keywords:
Fourier transform; Simple harmonic oscillator; Unilateral Fourier transform

The differential equation for the simple harmonic oscillator (SHO)

(1) $$\frac{d^{2}x(t)}{d{t}^2}+{\omega }^{2}x(t)=0,\omega >0$$

works as an excellent pedagogical tool for illustrating in a simple way several techniques for solving second-order differential equations such as power series expansion, and also Laplace transform (see, e.g. [¹]) and Fourier series expansion [²] (see also [³]). Here, the differential equation for the SHO is approached by unilateral Fourier transform.

Let us begin with a brief description of the unilateral Fourier transform and a few of its properties. The direct Fourier sine and cosine transforms of $f(\xi )$ are denoted by ${ℱ}_s\{f(\xi )\}=F_s(k)$ and ${ℱ}_c\{f(\xi )\}=F_c(k)$, respectively, and are defined by the integrals (see, e.g. [¹])

(2) $$\begin{array}{cc}\hfill F_s(k)& ={ℱ}_s\{f(\xi )\}=\sqrt{{\displaystyle \frac{2}{\pi }}}{\displaystyle {\int }_0^{\mathrm{\infty }}}\mathit{d}\xi f(\xi )\sin k\xi ,\hfill \\ \hfill F_c(k)& ={ℱ}_{c}f(\xi )=\sqrt{{\displaystyle \frac{2}{\pi }}}{\displaystyle {\int }_0^{\mathrm{\infty }}}\mathit{d}\xi f(\xi )\cos k\xi .\hfill \end{array}$$

The original function $f(\xi )$ can be recovered by the inverse unilateral Fourier transforms ${ℱ}_s^{-1}\{F_s(k)\}$ and ${ℱ}_c^{-1}\{F_c(k)\}$ expressed as

(3) $$\begin{array}{cc}\hfill f(\xi )& ={ℱ}_s^{-1}\{F_s(k)\}=\sqrt{{\displaystyle \frac{2}{\pi }}}{\displaystyle {\int }_0^{\mathrm{\infty }}}\mathit{d}k{F}_s(k)\sin k\xi ,\hfill \\ \hfill f(\xi )& ={ℱ}_c^{-1}\{F_c(k)\}=\sqrt{{\displaystyle \frac{2}{\pi }}}{\displaystyle {\int }_0^{\mathrm{\infty }}}\mathit{d}k{F}_c(k)\cos k\xi .\hfill \end{array}$$

We now observe that $f(\xi )$ retrieved by $F_s(k)$ must satisfy the homogeneous Dirichlet boundary condition at the origin, whereas $f(\xi )$ retrieved by $F_c(k)$ must satisfy the homogeneous Neumann boundary condition at the origin:

(4) $$\begin{array}{cc}\hfill F_s(k)& \Rightarrow {f(\xi )|}_{\xi =0}=0,\hfill \\ \hfill F_c(k)& \Rightarrow {{\displaystyle \frac{df(\xi )}{d\xi }}|}_{\xi =0}=0.\hfill \end{array}$$

Those boundary conditions are often overlooked in the literature [⁴, ⁵, ⁶, ⁷, ⁸, ⁹, ¹⁰, ¹¹, ¹², ¹³] (see [¹⁴] for a merciless criticism). The unilateral Fourier transforms have the following derivative properties

(5) $$\begin{array}{cc}\hfill {ℱ}_s\left\{{\displaystyle \frac{d^{2}f(\xi )}{d{\xi }^2}}\right\}& =-k^2{F}_s(k),\hfill \\ \hfill {ℱ}_c\left\{{\displaystyle \frac{d^{2}f(\xi )}{d{\xi }^2}}\right\}& =-k^2{F}_c(k),\hfill \end{array}$$

where the proper boundary conditions have already been used.

We are now ready to address the SHO delineated by the homogeneous Dirichlet boundary condition (${x(t)|}_{t=0}=0$) and the sine Fourier transform, or the homogeneous Neumann condition (${dx(t)/dt|}_{t=0}=0$) and the cosine Fourier transform. Using (2), with $X(k)$ denoting the unilateral Fourier transform of $x(t)$, one obtains

(6) $$(k^2-{\omega }^2)X(k)=0.$$

The solution $X(k)=0$ is certainly valid for $k\ne \pm \omega$. The complete solution for all $k$ can be found by using the property $z\delta (z)=0$ (see, e.g. [¹]), where $\delta (z)$ is the Dirac delta symbol. Remembering that $0\le k<\mathrm{\infty }$, one can write the solution as

(7) $$X(k)=\sqrt{\frac{\pi }{2}}\delta (k-\omega )\times \{\begin{array}{cc}{N}_s,\hfill & \text{for}{ℱ}_s\{x(t)\}\hfill \\ N_c,\hfill & \text{for}{ℱ}_c\{x(t)\},\hfill \end{array}$$

where $N_s$ and $N_c$ are arbitrary constants. Using the property (see, e.g. [¹])

(8) $${\int }_0^{\mathrm{\infty }}\mathit{d}kF(k)\delta (k-k_0)=F(k_0),k_0>0,$$

it is not difficult to see that each inverse Fourier transform yields one of the two linearly independent solutions of our problem:

(9) $$x(t)=\{\begin{array}{cc}{N}_s\sin \omega t,\hfill & \text{for}{ℱ}_s\{x(t)\}\hfill \\ N_c\cos \omega t,\hfill & \text{for}{ℱ}_c\{x(t)\}.\hfill \end{array}$$

In conclusion, the complete solution of the SHO can be approached with simplicity via the unilateral Fourier transform method. To the best of author’s knowledge, the SHO was never approached in this way.

Acknowledgement

Grant 09126/2019-3, Conselho Nacional de Desenvolvimento Científico e Tecnológico (CNPq), Brazil.

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Publication Dates

• Publication in this collection
  03 Feb 2023
• Date of issue
  2023

History

• Received
  28 Nov 2022
• Accepted
  07 Jan 2023