Help understanding this pulsing LED circuit
Essentially this circuit is a relaxation oscillator built using an op-amp or comparator and relying on hysteresis to oscillate.

If you ignore the two 2.2mF capacitors, the circuit would still work.

In terms of the rest of the circuit, the three 10k resistors connected to the "+" input of the op-amp act firstly to form a bias around mid supply as you have only a single supply rail, and secondly to add some hysteresis to the circuit. When the output changes, the bias voltage will also change as a result.

The magic happens then with the interaction of this hysteresis with the 10k resistor (I will call R1) and 100uF capacitor (I will call C1) combination. R1 and C1 act as an R-C circuit, which in couple with the hysteresis provides the oscillation.

At t=0

Lets say C1 is discharged so the "-" terminal is at 0V, and the "+" terminal is biased to either ~3V or ~6V depending on the output voltage. We can see in this condition that we have positive feedback forcing the output high (the "+" terminal is way above the "-" terminal). With the output high, our "+" terminal is therefore biased to 6V.

At t=RC

Over the last RC time period, C1 has been slowly charging up through R1. After one RC time constant the voltage across C1 and hence at the "-" terminal is charged to about 6V (actually it takes a little bit longer, but I'm going to ignore that for simplicity).

When this condition is reached, the "-" terminal becomes higher than the "+" terminal. This causes the output to start swinging down towards 0. Due to the feedback resistor in the bias circuit, this also causes the bias voltage to begin dropping as well. This positive feedback causes the output to slam down to 0V, and the bias voltage to drop to 3V. C1 starts to discharge

At t=2RC

C1 has been discharging through R1 because the output is low, and has now discharged to ~3V. When this happens the voltage at the "-" terminal is now below the voltage at the "+" terminal. 

This causes the output to start swinging down towards 9V. Due to the feedback resistor in the bias circuit, this also causes the bias voltage to begin rising as well. This positive feedback causes the output to slam up to 9V, and the bias voltage to rise back to 6V. C1 starts charging again.



The process then continues forming the required oscillation. 

I believe the two 2.2mF capacitors are added to give the oscillator a kick to start running, potentially to overcome the pull-down effect of the LEDs. However I could be wrong about that.