Here is the equation I've been given. A(z,t) is the pulse propagation in an optical fiber. I'm solving the equation mathematically so that I can later use matlab to code the solution.

$$\dfrac{\partial A}{\partial z} = \frac{\alpha}{2}A - \beta_1 \dfrac{\partial A}{\partial t} -i \frac{\beta_2}{2} \dfrac{\partial ^2 A}{\partial t^2}$$

We're told explicitly to use the Fourier transform with respect to t, which I have little experience with as it is, and I certainly don't know how to take the left hand side with respect to a different variable.

Focusing on the right hand side and changing the left to 1, after taking the Fourier transform, I have this. $a(\omega)$ represents the Fourier transform of A(t).

$$a(\omega) = \frac{1}{\frac{\alpha}{2}-\beta_1(2i\omega\pi)+\frac{\beta_2}{2}(2\pi\omega)^2}$$

I got this conclusion using formulas from this website, swapping f for $\omega$.

My two questions: is that above formula correct, and how would I go about taking the Fourier transform of $\dfrac{\partial A}{\partial z}$ with respect to t?

The full problem description

EDIT: After some revision using help below as well as this website, I now have this equation.

$$ a_z (z, \omega)= \frac{\alpha}{2} a(z, \omega)- i\beta_1 \omega a(z, \omega) -i \frac{\beta_2}{2} \omega^2 a(z, \omega) $$

I'm writing it in this format where all the partials are expanded out so that I can get to an equation that can be solved for $a(z, \omega)$ which will be perfect for MATLAB. Just to be clear, the formula above is the original differential equation with Fourier transform respect to t done on both sides.

My lecture notes included the formula $F[f'(t)] = i\omega F(\omega)$ and I obtained the formula for $F[f''(t)]$ from the listed website.

Does this look right? I'm still not sure what to do about that $ a_z (z, \omega)$ but I've emailed the professor to see what he expects.

  • $\begingroup$ I think you mean how you find the fourier transform of $\frac{\partial A}{\partial t}$? $\endgroup$ – DaveNine Nov 7 '17 at 23:11
  • $\begingroup$ If you look at the image I linked as full problem description, you can see that there's no typo. The problem we're given has dA/dz as the first term, and it says to take the Fourier transform with respect to t on both sides. I have no idea what that means or how to go about it. $\endgroup$ – Roux Nov 8 '17 at 6:09
  • $\begingroup$ The Fourier transform doesn't discriminate among variables, the same definition I've in my answer works for the time domain as well as spatial. $\endgroup$ – DaveNine Nov 8 '17 at 6:11

The fourier transorm is typically taken with respect to the spacial variable, that is, the fourier transform of $f(x)$ is given by

$$\hat{f}(\xi) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}f(x)e^{-i\xi x}\, dx$$

For us, it should be easy to show, using things like integration by parts and other integral techniques, that

$\widehat{A_{tt}(z,t)} = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}A_{tt}(z,t)e^{-i\xi z}\, dz = \hat{A}_{tt}(\xi, t)$


$\widehat{A_{z}(z,t)} = i\xi \hat{A}(\xi,t)$

Using these, we can transform your equation into the following:

$$i\omega a(\omega,t) = \frac{\alpha}{2}a(\omega,t)-\beta_1 a_{t}(\omega,t)-i\frac{\beta_2}{2}a_{tt}(\omega,t)$$

Now, this is an ordinary differential equation in t, so we may solve for $a(\omega,t)$ by solving this equation. Then, once we have an expression for $a(\omega, t)$, we can use the inverse fourier transform to write down the solution to the PDE.

  • $\begingroup$ I do want to note that this is a formulation with respect to the spatial variable, $x$. There is absolutely no difference if you were to do it with respect to $t$ instead, it's just that you're going to be solving a different differential equation at the end there. $\endgroup$ – DaveNine Nov 7 '17 at 23:48
  • $\begingroup$ Where I'm getting lost is the flipping around of variables because I can't see the exact reason why. \widehat{A_{tt}(x,t)} goes from A as a function of x and t into a function of z and t within the integral and then finally becomes A as a function of \xi and t at the end. What's going on there? How do I follow that? Also, what do the arrows/hats on the top mean? $\endgroup$ – Roux Nov 8 '17 at 6:14
  • $\begingroup$ $\hat{f}(\xi, t)$ is the definition of the fourier transform of $f(x, t)$. The hat can be thought of as an operator. Another way to write this is with $\mathcal{F}(f(x,t))=\hat{f}(\xi, t)$ $\endgroup$ – DaveNine Nov 8 '17 at 6:18
  • $\begingroup$ So if I'm understanding you correctly, you're transforming the variable that you're taking the Fourier transform respect to into the \xi variable. $\endgroup$ – Roux Nov 8 '17 at 6:36
  • $\begingroup$ Also, I'm still getting stuck when trying to take the fourier transform of dA/dz. Using your top formula, I get the integral of dA/dz when respect to t. How do you represent that? What does that even mean exactly? $\endgroup$ – Roux Nov 8 '17 at 6:45

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