The only way I know how to explain the crux of the issue is from a technical perspective. It will involve some math, so bare with me.
V=I*R (Voltage is the product of current, Amps, and resistance, Ohms.) This can be rewritten as I=V/R and R=V/I.
P=VI (Power, Watts, is the product of voltage, Volts, and current, Amps.)
We will get back to this, but it is important to understand the basics upfront.
P is the important part as power is heat is vapor. It is what does work. It is also why DC's are popular, the low resistance and the large surface area of the two coils. To make a single low resistance coil, it is just a shorter section of wire than a standar resistance. A 1.5 Ohm dual coil is two 3.0 Ohm coils, so it is like having a 1.5 Ohm single coil with more wire surface in contact with the juice to create vapor. So, if we assume 3.7 Volts and 1.5 Ohm we can use the first formula I=V/R to see that we need 2.5 Amps to supply that DC.
Now, the problem is that most of the PV batteries cannot supply 2.5 Amps. The protection circuit cuts off the current supply, so the loaded voltage, the voltage that is measured at the DC to battery connection while it is operating, will actually be equal to V=IR or 1.5 Ohm * the maximum current. The next problem is that the majority of Li-Ion batteries, even in the 18650 size, have protection circuits much like the standard PV batteries and cannot supply the 2.5 Amps needed to run the DC to its full power. Only a very high quality protected 18650 can supply this current. An IMR, also known as safe chemistry, battery, like the AW IMR, can supply this current. So can a LiFEpo, which are more difficult to find in 18650, but is also a safe chemistry battery.
Now since P=VI and I=V/R then P=V*V/R, or power is equal to voltage squared divided by reistance. So, for a freshly charged Li-Ion battery, say an AW IMR for this example, then the power of a 1.5 Ohm DC is 11.8 Watts! That is some great vapor. Even when the charge has settled down to 3.7 Volts, the power is 9.1 Watts. (This is my usual range too.)
So IF you want to stick to 9.1 Watts, then a DC on a ProVari V2 that has a maximum current of 3.5 Amps at low voltage will work.
But if you were to get a 3.0 Ohm Single Coil, then ANY VV would work because you only need 1.8 Amps to make the same power at about 5.2 Volts.
The problem you are having with the LR DC's is that it takes so much current. The question you have to ask yourself is if you need to use DC's, why go with the expense of a VV when you could get the same thing from a good high drain 18650 battery, like IMR or LiFEpo controlled with a 5 amp button switch and a 510 connector in a box? This is how I started in the 9 Watt vaping realm. I got a box, switch and some battery springs from Radio Shack and mounted a 510 battery connecter and inserted a good 18650 battery. This was good enough to show me that higher power was the way I wanted to go.
When I went to VV mods, I did not use the low resistance coils I did with my homemade box. I use standard resistance at higher voltage to get the same and higher power. A well designed VV mod should have a current cutoff to help keep it safe. I just work within that cutoff and get all the power I need. That is why the VV box mod you have will not fire the DC's, you are over the current limit.
