How can speed of geostrophic wind change when forces are always perpendicular to speed?

Question

Equation for acceleration of an air parcel is

$$D \vec U/Dt = -2 \vec \omega \times \vec U -\frac{1}{\rho} \vec \nabla p + \vec g$$

Now, for geostrophic wind, U is parallel to the contour lines of constant p.

But if this is the case, then all forces above (beside g) are perpendicular to U. So how can U change when the lines of constant pressure get closer: in this case velocity should rise because larger gradient of p. In the moment I regard this as a discrepancy. How to resolve it?

EDIT:

This is what I mean in a picture:

As the the blue air parcel moves from left to right on a line of constant pressure it becomes slower (red arrow), although the local change of speed is zero (stationary flow). It means, there must be a force on it which directs along its direction of flow (in fact its energy is decreased, because it is slower). The question is: where is this force coming from, since both pressure force and Coriolis force are perpendicular to the parcel's speed in each moment.

Answer 1

You're confusing time-dependent and time independent equations: The time-dependent model tells you how $\vec U$ will evolve, it doesn't have to evolve into a steady-state. The steady, time-independent model tells you which $\vec U$ fulfills that equation as a criterion.
Therefore that $\vec U$ usually gets a special name, the geostrophic velocity $\vec U_g$, and the time-dependent equation also can be reformulated into a form of $\vec U =\vec U' -\vec U_g$.

So when the current $\vec U_g$ changes, we can see how $d_t \vec U$ generates non-perpendicular terms until $\vec U = \vec U_g$ and balance is restored.

In the moment I regard this as a discrepancy.

I don't presume you regard the existence of satellites as fantasy, to which your argument would equally apply? For a circular orbit, the centrifugal force is perpendicular to the velocity, so it's all a hoax?