Why DE with \(V_0 \neq 0\) tracks LPM
The DE method with \(V_0 \neq 0\) agrees well with LPM because it ignores the low-voltage part of the waveform below the threshold \(V_0\). That is physically reasonable, since low voltage does not significantly contribute to leader development.
The DE method with \(V_0 \neq 0\) compares reasonably well with LPM, because \(V_0\) stops low-voltage portions from contributing unrealistically — only voltage above the threshold counts, which is physically reasonable since a voltage below a certain level does not produce leader development. It therefore stays useful for shorter, longer and oscillatory surges, provided the constants are properly calibrated.
The DE method with \(V_0 = 0\) is not even shown for the oscillatory comparison, because it gives invalid results when the voltage decays slowly or stays at a significant fraction of crest — treating all the voltage area as destructive overestimates the severity.
Why \(V_0 = 0\) fails for long surges
With \(V_0 = 0\), every part of the waveform contributes to the destructive effect, including the low-voltage tail. For long-duration waves this artificially raises the calculated destructive effect, so the method predicts a much lower \(E_{\max}\) and becomes overly conservative.
Why it is not even shown for the oscillatory surge
The oscillatory surge decays only to about 50% of its original value. With \(V_0 = 0\) the continuing oscillations would keep contributing to the destructive effect even when they may not be physically significant, so the \(V_0 = 0\) method is not valid for this case.