# Guided Self-Accelerating Airy Beams--A Mini-Review
## 2 Results and Discussion
### Harmonic Potential Guidance
#### 2.2.1 One-Dimensional Airy Beams  
In the one-dimensional (1D) case, the paraxial propagation of a beam in a linear medium with an external harmonic potential is described by the following equation [63; 64]:  
\[i\frac{\partial\psi}{\partial z}+\frac{1}{2}\frac{\partial^{2}\psi}{\partial x ^{2}}-\frac{1}{2}\alpha^{2}x^{2}\psi=0, \tag{14}\]  
where \(\alpha\) determines the width of the harmonic potential. The Fourier transformation (FT) of Equation (14) leads to the corresponding equation in the inverse space:  
\[i\frac{\partial\hat{\psi}}{\partial z}+\frac{1}{2}\alpha^{2}\frac{\partial^{2 }\hat{\psi}}{\partial k^{2}}-\frac{1}{2}k^{2}\hat{\psi}=0. \tag{15}\]  
Clearly, if \(\alpha=1\), Equations (14) and (15) have the same form, so that both equations have the same solutions but expressed in real (\(x\)) and inverse (\(k\)) spaces [63], respectively. As before, we are interested in the behavior of Airy beams. The propagation of a truncated Airy beam in the harmonic potential is shown in Figure 3.  
Generally, the solution of Equation (14) can be written as [63; 64; 65; 66; 67; 68; 69]:  
\[\psi(x,\,z)=\int_{-\infty}^{+\infty}\psi(\xi,\,0)\sqrt{\mathcal{H}(x,\,\xi,\,z )d\xi}, \tag{16}\]  
where  
\[\mathcal{H}(x,\,\xi,\,z)=-\frac{i}{2\pi}\alpha\csc\left(\alpha z\right)\exp \left\{i\alpha\cot\left(\alpha z\right)\left[x^{2}+\xi^{2}-2x\xi\sec\left( \alpha z\right)\right]\right\} \tag{17}\]  
Figure 3: (Color online) Propagation of a finite energy Airy beam in a harmonic potential. (**a1,a2**) Real space; (**b1,b2**) Inverse space. Periodic inversion and an automatic Fourier transform of the beam are evident. The parameters are: \(a=0.1\), \(\alpha=1\) (**a1, b1**) and \(\alpha=0.5\) (**a2, b2**). Reproduced with permission from [63], Copyright Elsevier, 2015.  
is associated with the corresponding kernel. Combining Equations (16) and (17), after some algebra one arrives at  
\[\psi(x,\,z)=f(x,\,z)\int_{-\infty}^{+\infty}\left[\psi(\xi,\,0)\exp\left(ib\xi^{2 }\right)\right]\exp(-iK\xi)d\xi, \tag{18}\]  
where \(b=\frac{\xi}{2}\cot\left(\alpha z\right),\,K=\alpha x\csc\left(\alpha z\right),\) and  
\[f(x,\,z)=\sqrt{-\frac{i}{2\pi}\frac{K}{x}}\exp\left(ibx^{2}\right).\]  
One can see that the integral in Equation (18) is a Fourier transform of \(\varphi(x,0)\exp(ibx^{2})\). In other words, the propagation of a beam in a harmonic potential is equivalent to an automatic FT, that is, to the periodic change from the beam to the FT of the beam with a parabolic chirp and back. It is worth mentioning that the same formula also represents a fractional Fourier transform of the initial beam [67; 70; 71], the \"degree\" of which is proportional to the propagation distance. By choosing the input as \(\psi(x,0)=Ai(x)\exp(ax)\), the solution in Equation (18) can be found using the following steps:  
(i) Find the Fourier transforms of \(\psi(x,\,0)=\text{Ai}(x)\exp(ax)\) and \(\exp(ibx^{2})\), which can be written as [2; 9; 63; 64]:  
\[\hat{\psi}(k)=\exp\left(-ak^{2}\right)\exp\left[\frac{a^{3}}{3}+\frac{i}{3} \left(k^{3}-3a^{2}k\right)\right], \tag{19}\]  
and  
\[\sqrt{i\frac{\pi}{b}}\exp\left(-\frac{i}{4b}k^{2}\right), \tag{20}\]  
respectively. (ii) Perform the convolution of the two Fourier transforms in Equations (19) and (20), and using the definition [72]  
\[i\frac{\partial\hat{\psi}}{\partial z}+\frac{1}{2}\alpha^{2}\frac{\partial^{2 }\hat{\psi}}{\partial k^{2}}-\frac{1}{2}k^{2}\hat{\psi}=0. \tag{15}\]0  
find the inverse Fourier transform:  
\[i\frac{\partial\hat{\psi}}{\partial z}+\frac{1}{2}\alpha^{2}\frac{\partial^{2 }\hat{\psi}}{\partial k^{2}}-\frac{1}{2}k^{2}\hat{\psi}=0. \tag{15}\]1 \[i\frac{\partial\hat{\psi}}{\partial z}+\frac{1}{2}\alpha^{2}\frac{\partial^{2 }\hat{\psi}}{\partial k^{2}}-\frac{1}{2}k^{2}\hat{\psi}=0. \tag{15}\]2  
From this expression, one obtains the accelerating trajectory of the initial beam:  
\[i\frac{\partial\hat{\psi}}{\partial z}+\frac{1}{2}\alpha^{2}\frac{\partial^{2 }\hat{\psi}}{\partial k^{2}}-\frac{1}{2}k^{2}\hat{\psi}=0. \tag{15}\]3  
with the period \(D=2\pi/\alpha\).