Standing wave and energy flux Here is a problem I have been asked that I do not know the answer. Consider two ideal wave generators (it can be sound generator or whatever) separated by a distance L and facing each other. At t=0 they start to emit a wave. Basic calculus show a standing wave is created. So the energy flux through any surface on the path of the wave is in average zero. In the same time, the generators continuously feed energy into the system. Question: where does the energy  emitted by the generators go ? 
 A: The standing wave relies on the travelling waves being reflected at each end of your distance, and this reflection is never 100%. In addition to this you get viscous losses because the propagation of the wave requires shear motion in the fluid i.e. the non-zero velocity gradients.
In the case of your loudspeakers, hi-fi loudspeakers are specifically designed not to reflect sound, so without an energy input your standing wave would be lost very quickly. I would guess this energy loss is far greater than viscous losses in the air.
In the case of water waves, most of us must have discovered that it's easy to set up a zero mode standing wave in a bath (usually to our mother's dismay) and this wave will persist for quite a while. In this case the main energy loss would be viscous losses where the water moves over the walls of the bath and viscous losses in the bulk. I would guess the biggest velocity gradients are at the walls so most of the energy would be lost there.
Response to comment: if you split your standing wave into it's left moving and right moving components there is only no net energy flux if the two components have the same amplitude. However the energy losses due to viscosity mean that the amplitude of the left moving wave decreases with distance to the right, and conversely for the right moving wave. This means there is a net energy flux.
A: I agree with you that the the issue can be specified theoretically without general losses.
The error in your reasoning is that you assume an actuator has a dynamic nature independent of the wave medium, so that you'd always pour energy into each actuator and thus a paradox ensues since, as you note, a standing wave pattern in a medium has a zero energy transport.
In the limit of a perfect medium where you create the standing wave (or any other wave in general), for example with two actuators and an ideal wire, you get a coupling of the two actuators which disallows you to lose energy in them in total, that is, any voltage drop due to driving current in speaker/actuator A will present as a voltage gain in speaker/actuator B (acting as a microphone) and vice versa. More specifically in the standing-wave situation both ends will lose all driving impedance they had before.
That is, without losses, the setup in total is just an ideal transmission line.
In the setup with one actuator and one perfectly reflecting wall, the sole actuator will lose all its impedance and be infinitely easy to drive.
There are related questions that usually arise about energy conservation and "complete" destructive interference, with the accompanied confusion of thousands of forum posts :)
