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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def kinematic_flux(vel, b, perturbation=False, axis=-1):
 r"""Compute the kinematic flux from two time series. Compute the kinematic flux from the time series of two variables `vel` and b. Note that to be a kinematic flux, at least one variable must be a component of velocity. Parameters vel : array_like A component of velocity b : array_like May be a component of velocity or a scalar variable (e.g. Temperature) perturbation : bool, optional `True` if the `vel` and `b` variables are perturbations. If `False`, perturbations will be calculated by removing the mean value from each variable. Defaults to `False`. Returns ------- array_like The corresponding kinematic flux Other Parameters axis : int, optional The index of the time axis, along which the calculations will be performed. Defaults to -1 Notes ----- A kinematic flux is computed as .. math:: \overline{u^{\prime} s^{\prime}} where at the prime notation denotes perturbation variables, and at least one variable is perturbation velocity. For example, the vertical kinematic momentum flux (two velocity components):
 .. math:: \overline{u^{\prime} w^{\prime}} or the vertical kinematic heat flux (one velocity component, and one scalar):
 .. math:: \overline{w^{\prime} T^{\prime}} If perturbation variables are passed into this function (i.e. `perturbation` is True), the kinematic flux is computed using the equation above. However, the equation above can be rewritten as .. math:: \overline{us} - \overline{u}~\overline{s} which is computationally more efficient. This is how the kinematic flux is computed in this function if `perturbation` is False. For more information on the subject, please see [Garratt1994]_. """

    kf = np.mean(vel * b, axis=axis)
    if not perturbation:
        kf -= np.mean(vel, axis=axis) * np.mean(b, axis=axis)
    return np.atleast_1d(kf)