The standard equation for the doppler effect from a frame of reference with constant velocity is
$f_0$ initial frequency;
$v$ is the velocity of waves in the medium;
$v_r$ is the velocity of the receiver relative to the medium;
$v_s$ is the velocity of the source relative to the medium;
$$f = \left(\frac{v + v_r}{v + v_s}\right)f_0$$
Now I'd like to know what's the equivalent for an accelerating frame, I have
$$f = \left(\frac{v'\text{dt} + v_r}{v'\text{dt} + v_s}\right)f_0$$
But it just doesn't seem correct.
EDIT
Mea Culpa – I was thinking of the signal being emitted from an accelerating train B as being the frame of reference.
So, second attempt.
1)You are standing at a platform.
2)The wind is stationary.
3)A train is accelerating towards you – blowing its whistle.
What is the formula for the frequency received at the station? Where the standard formula is
$$f = \left(\frac{v }{v + v_s}\right)f_0$$
rewriting for an accelerating frame of reference for the source
$$f = \left(\frac{v }{v + (dv_s/dt)dt}\right)f_0$$
but I don't think this goes anywhere, and yes a non-linear solution is probably the answer but it's like a three legged black dog with rabies – it's better than no dog at all.
Best Answer
In the classical Doppler effect, you should think of $v_s$ and $v_r$ to be the velocities of the source and receiver relative to the medium, at the time when the signal was sent and received. In an accelerated reference frame, it is essentially the medium that is moving. Remember that the Doppler shift concerns the frequency sent by the source compared to the frequency seen by the receiver. So it does not depend on which reference frame you, as a third party, is in, but only depends on the intrinsic wave speed of the medium $v$ (which we assume to be constant, when measured in a frame in which the medium is at rest), and the velocities relative to the medium of the source and receivers.
With that said, that means you can compute relative to the unaccelerated reference frame by doing a coordinate change.
Suppose you are in a reference frame in which the medium itself is moving with variable velocity $\tilde{v}(t)$, and you measure the source and receivers to be moving with velocity $\tilde{v}_s(t)$ and $\tilde{v}_r(t)$. Then the Doppler effect can be computed by looking at the reference frame in which the medium is not moving (so the corresponding velocities are $v_s(t) = \tilde{v}_s(t) - \tilde{v}(t)$ and similarly $\tilde{v}_r(t)$). If in this reference frame $v_s$ and $v_r$ are not constant in time, then given a time $t_r$ when the signal was received, you will have to compute (using that the wave travels with constant speed $v$ and using the trajectory of the receiver versus the source) the time $t_s$ at which the signal was sent, and plug those values into your Doppler shift expression (with $v_s \to v_s(t_s)$ and $v_r \to v_r(t_r)$).
If the wave speed itself $v$ depends on the macroscopic of the entire medium, then you will have to solve the very complicated system for elastodynamics/acoustic in the medium, for which the final answer would depend on how you model the medium.
Edit for the new clarification
In the particular case, assume you, the receiver at the platform is measuring the signal at time $t_0 = 0$. And choose coordinate system so that you are at the origin. Let $x(t)$ denote the position of the train that is approaching the station. (or leaving the station, you just need to adjust the sign appropriately.) The signal you observe will have been emitted by the train at time $t_s < 0$, when $|x(t_s)| = v \cdot (0-t_s)$ where $v$ is the speed of sound in air. (For the time being we'll assume you have a non-supersonic train, so there is a unique time $t_s$ that satisfies the requirements.) Then the expression for the doppler shift would be
$$ f = \left( \frac{v}{v + \dot{x}(t_s)}\right) f_0 $$
Unfortunately, solving for $t_s$ is generally hard. But we can get a good estimate on $f$ if we assume that our train is reasonably close to the station when you are making the observation, and if the train is not accelerating too much, by using Newton's method.
To the zero'th order you'd get $t_s = 0$. To the first order in Newton's method, you can solve and get $t_s ~ x(0) / (v - \dot{x}(0))$. Then we can approximate the velocity $\dot{x}(t_s) \sim \dot{x}(0) + \ddot{x}(0) t_s$ which leads to a first order correction to the doppler shift formula
$$ f \sim \left( \frac{v}{v+ \dot{x}(0) + \frac{\ddot{x}(0) x(0)}{v - \dot{x}(0)}}\right) f_0 $$