4

So I have a Hermite-Gaussian profile given by the initial value problem $$ iu_z + u_{xx}=0,\quad (x,z)\in\mathbb{R}\times\mathbb{R}_+ $$ where the condition initially is given by this $$ u_0(x) = H_n\left(\sqrt{2b}x\right)e^{-bx^2} $$ where $H_n(x)$ is the Hermite polynomial of nth order

The problem is to find $u(x,z)$ and compute various moments of $x$ to determine the behavior of the profile with propagation distance.

I know that the Fourier Transform of this Hermite polynomial is given by thanks to this answer $$ \mathscr{F}(H_n(\sqrt{2b}x)e^{-bx^2}) =(-i)^n\sqrt{\frac{2\pi}{2b}}H_n\left(\frac{\xi}{\sqrt{2b}}\right)e^{-\xi^2/4b} $$ I took the Fourier Transform of the initial equation and got a new equation with two variables $\xi_1, \xi_2$ , namely $$ \mathscr{F}(iu_z + u_{xx}) = [i(i\xi_1) + i^2\xi^2_2]\widehat u = -(\xi_1+\xi_2^2)\widehat u = 0 $$

I can't for the life of me understand how to extend this initial condition to two variables (x,z) to find the inverse Fourier Transform. Any help is appreciated.

  • 1
    Hint: Take the Fourier transform of the equation itself – Ninad Munshi Jan 15 '23 at 10:59
  • @NinadMunshi I did. I got stuck after that. I know I need to take the inverse FT to get to the answer, but I can't figure out where the initial condition comes in – Aadrsh Rishi Jan 15 '23 at 11:06
  • Just writing here to make sure you get notified – I'm not sure whether you're still getting notifications for my comments under my answer, since you deleted the comment that I was responding to. – joriki Jan 15 '23 at 19:58

2 Answers2

4

(Partial answer)

You only need to take the Fourier transform with respect to $x$, such that your differential equation becomes $i\hat{u}_z-\xi^2\hat{u}=0$, hence $\hat{u}(z,\xi) = A(\xi)e^{-iz\xi^2}$. The initial condition gives then $$ A(\xi) = \hat{u}(0,\xi) = \hat{u}_0(\xi) = (-i)^n\sqrt{\frac{\pi}{b}}H_n\left(\frac{\xi}{\sqrt{2b}}\right)e^{-\xi^2/4b}. $$ I let you taking the inverse Fourier transform $-$ N.B. : the fact that Hermite functions are eigenfunctions of the Fourier transform will help (see here).


Addendum. I don't know in which context your differential equation arises (it looks like Schrödinger equation for a free particle), but it is to be noticed that it can take the form of the standard 1D heat equation, i.e. $u_t = u_{xx}$, after a Wick rotation $z \mapsto -it$.

Abezhiko
  • 14,205
  • Thanks for your elegant answer, I'm trying to tackle the Inverse Fourier Transform right now, and I hope to be able to complete the solution soon. As for your addendum, this is a Fourier Optics course. This proble specifically deals with Hermite-Gaussian modes. – Aadrsh Rishi Jan 15 '23 at 12:17
  • Hey, I'm struggling with the Inverse Fourier as well. I got $$ u(x,z) = \mathscr{F}^{-1}(\hat{u}(\xi,z)) = \int(-i)^n\sqrt{\frac{2\pi}{2a}}H_n\left(\frac{\xi}{\sqrt{2a}}\right)e^{-\xi^2/4a}e^{-i\xi^2z}e^{-2\pi ix\xi}d\xi $$ I'm not really sure how to use hermite functions as eigenfunctions here. Can you please expand on this a little. Sorry if this asking a lot, but I'm really struggling with this course :( – Aadrsh Rishi Jan 15 '23 at 14:36
  • @AadrshRishi Oh I see, I remember gaussian beams and stuff from a quantum optics course long ago; closed forms exist (https://en.wikipedia.org/wiki/Gaussian_beam#Hermite-Gaussian_modes), but they are complicated to get (quite a nightmare to my memory too), so that it is usually easier to stay in the Fourier space in order to carry further manipulations. I'm afraid I can't be of any help :/ – Abezhiko Jan 15 '23 at 18:18
  • Well, that's fine. Thanks for all your help. – Aadrsh Rishi Jan 15 '23 at 18:40
2

@Abezhiko’s answer is great and solves the problem efficiently. Just to provide another perspective on the problem: You can view applying the Fourier transform as a form of separation of variables. If the transform does separate the variables, then you can write the solution in the form of a single integral where each component is a product of a Fourier component in $x$ and a Fourier component in $z$, with each product individually solving the differential equation. So

$$ u(x,z) = \int\mathrm d\xi\,\hat u(\xi)\mathrm e^{\mathrm i\xi x}\mathrm e^{\mathrm i\xi_z(\xi)z}\;, $$

where the function $\xi_z(\xi)$ is determined by the condition that $\mathrm e^{\mathrm i\xi x}\mathrm e^{\mathrm i\xi_z(\xi)z}$ solves the differential equation. Your derivation shows $\xi^2+\xi_z(\xi)=0$, so $\xi_z(\xi)=-\xi^2$. Thus the general solution has the form

$$ u(x,z) = \int\mathrm d\xi\,\hat u(\xi)\mathrm e^{\mathrm i\xi x}\mathrm e^{-\mathrm i\xi^2z}\;, $$

and since you know the Fourier transform of $u(x,0)$, you know $\hat u(\xi)$ and can obtain $u(x,z)$ by performing the inverse transform. This is of course the same inverse transform that you need to perform in @Abezhiko’s answer.

joriki
  • 242,601
  • @AadrshRishi: a) You're missing a factor $\mathrm e^{\mathrm i\xi x}$. b) In your comment under Abezhiko’s answer, you have a factor of $2\pi$ in the exponent. Make sure to be consistent in including or not including those factors. In your derivation of $\xi_1+\xi_2^2=0$ in the question, you didn't include them (there would be one of them for the first derivative and two for the second, so they wouldn't cancel). – joriki Jan 15 '23 at 19:55
  • c) To compute this integral, my approach would be to complete the square in the exponent, shift $\xi$ so the exponential is a pure square (which implicitly shifts the integration contour in the complex plane, which is valid because the integrand vanishes at infinity), express the shifted Hermite polynomial as a linear comination of unshifted Hermite polynomials (see Wikipedia) and then use the orthogonality of the Hermite polynomials to find that only one term in the sum survives. – joriki Jan 15 '23 at 19:56
  • Let me know if you encounter any obstacles in that (I haven't carried it out in detail, so I'm not 100% sure it works like that, but I think it should). About $z$: Yes, in the integral over $\xi$, you need to treat $z$ as a constant. – joriki Jan 15 '23 at 19:56
  • I see now that you removed your comment – I guess that means you sorted it out yourself? – joriki Jan 15 '23 at 19:56
  • Hey @joriki, sorry I wasn't able to see your comments. I had an early class today so I called it an early night. I have made some progress where I used the convolution theorem to approach the IFT, but I got stuck nonetheless. I've made a separate question for this here https://math.stackexchange.com/questions/4619364/please-help-me-calculate-this-integral-for-an-inverse-fourier-transform – Aadrsh Rishi Jan 16 '23 at 06:36
  • @AadrshRishi: And? Did you try the approach I suggested? – joriki Jan 16 '23 at 08:12
  • No I didn't sadly. I'm trying to solve my approach with by parts. If that doesn;t pan out,I'm planning on giving your approach a go – Aadrsh Rishi Jan 16 '23 at 08:14
  • 2
    Hey, I solved the problem. It worked with the convolution approach that I was using, although I had to cheese it out a little. I wrote the first 4 Hermite Polynomials, solved those individual integrals and then worked out the pattern for the general solution. I got $$ u(x,z) = \frac{H_n(\sqrt{2a}x)e^{\frac{-ax^{2}}{4iaz+1}}}{(4iaz+1)^{n+\frac{1}{2}}} $$ I think this is correct, atleast solves the initial condition – Aadrsh Rishi Jan 16 '23 at 16:51
  • @AadrshRishi: OK. I'm surprised it looks so complicted. You could write this up as an answer to your other question and accept it so that the question doesn't remain unanswered. – joriki Jan 16 '23 at 16:52
  • 1
    Yes, I'll do that definitely once I make my submission. Thanks a lot for your help though, I couldn't have done it without you. – Aadrsh Rishi Jan 16 '23 at 16:54