Hubbert math

Verification the four forms are equivalent

If you buy the idea that your data scatter in $(Q_i,P_i/Q_i)$-space is a straight line, then you have bought equation (2). If you buy any one of equations (1),(2),(3), or (4), then you have bought them all because they are mathematically equivalent. Starting from the definition (4) using the rule from calculus that $d(1/v)/dt= -(dv/dt)/v^2$ yields equation (3).

$$\frac{dQ}{dt} = P(t) = Q_\infty \omega\ \frac{ e^{\omega (\tau-t)} }{ (1 + e^{\omega (\tau-t)})^2 }$$ (5)

$$P(t) = Q_\infty \omega \ \frac{ 1 }{ (e^{-(\omega/2) (\tau-t)} + e^{(\omega/2) (\tau-t)})^2 }$$ (6)

which is equation (3).

Equation (4) allows us to eliminate the denominator in equation (5) getting equation (2)

$$P/Q = (Q/Q_\infty) \omega e^{\omega (\tau-t)}$$ (7)

$$P/Q = (Q/Q_\infty) \omega ((1+ e^{\omega (\tau-t)}) -1)$$ (8)

$$P/Q = (Q/Q_\infty) \omega (Q_\infty/Q-1)$$ (9)

$$P/Q = \omega (1-Q/Q_\infty)$$ (10)

which is equation (2). Multiplying both sides by $Q$ gives equation (1).

Hubbert math