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