User:Jclaer/Hubbert

Over the long haul populations grow and decay. To describe the growth and decay of civilization's dependence on nuclear and fossil fuels, M. King Hubbert , chose a simple equation.

Introduce bacteria to food and their population will grow exponentially until there no longer is food. As we remove all the fish from the ocean our regular catch will be proportional to the number of remaining fish. Hubbert's equation models both exponential growth and decay with a single equation of three parameters to be chosen from the data. He claimed 52 years ago that worldwide oil production would be peaking about now (2008). It is.

Hubbert's math has four different expressions which we examine before showing they are mathematically equivalent.

Basic definitions
We define:

t is time in years

Q(t)  is cumulative production in billion barrels at year t.

$$Q_\infty$$ is the ultimate recoverable resource.

P(t)=dQ/dt is production in billion barrels/year at year t.

? is the year at which production peaks.

? is an inverse decay time (imaginary frequency).

The four forms of Hubbert's equation
The Hubbert's equation can be expressed in four forms. First, the differential form


 * $$\frac{dQ}{dt}=P=\omega\ Q\left(1 - \frac{Q}{Q_\infty}\right) \qquad \mbox{(1)} \!$$

This equation is non-linear in Q but it reduces to familiar linear equations near the beginning and $$Q \approx 0$$ and near the end at $$Q \approx Q_{\infty}$$. As production begins and $$Q/Q_\infty$$ is small, equation (1) reduces to dQ/dt=?Q which displays exponential growth at a rate ?. As production ends near $$ Q \approx Q_\infty$$ the non-linear equation reduces to exponential decay. To prove this fact change variables from Q to q where $$ Q=Q_\infty - q$$ and evaluate the result at small q.

Dividing equation (1) by Q we get the second form of the Hubbert equation sometimes called the Hubbert Linearization.


 * $$\frac{P}{Q}= \omega\ \left(1 - \frac{Q}{Q_\infty}\right) \qquad \mbox{(2)} \!$$

The important thing about this equation is that it is linear in the two variables Q and P/Q. If you have historical measurements of Pi and Qi, you can plot these points in the (Q,P/Q)-plane and hope for them to reasonably fit a straight line. Fitting the best line to the scattered points we can read the axis intercepts. At Q=0 with equation (2) we can read off the value of the growth/decay parameter $$\omega = (P/Q)_{\rm intercept}$$. For world oil, according to Deffeyes it is 5.3%/year. At the other intercept, P/Q=0 we must have $$Q=Q_{\infty}$$. Again, according to Deffeyes, $$Q_{\infty}$$ is two trillion barrels.

The third form of Hubbert's equation is the one best known. It looks like a Gaussian, but it isn't. (A Gaussian decays much faster.)  The current production P=dQ/dt is



P(t) = Q_\infty \ \omega \ {       1 \over (e^{-(\omega/2) (\tau-t)} + e^{(\omega/2) (\tau-t)})^2 } \qquad \mbox{(3)} \!$$

This is the equation of a blob, also known as Hubbert's pimple, symmetric about the point t=?. Asymptotically it decreases (or increases) exponentially towards its maximum value at the center at t=?. The function resembles a gaussian but exponential decay is much weaker than gaussian decay. Exponential growth is common in ecological systems which may also decay exponentially as resources are depleted or predator numbers grow exponentially.

All that remains is to figure out ?. The Hubbert curve is symmetrical and reaches its maximum when half the oil is gone. That happens when $$Q=Q_\infty/2$$. In the case of USA production which has passed its peak we can find the year that Q reached that value (about 1973). There is some debate about what year world production peaks, but general agreement is that it is about now (2008). Under Hubbert assumptions the decline curve is a mirror of the rise curve. That means we start down gently over the next decade, but about 25 years from now we hit the inflection point and see a 5%/year decline every year thereafter.

In real life there is no reason for the decay rate to match the growth rate. The decay could be faster because of horizontal drilling. The decay could be slower because we tax to conserve or successfully invest in technologies. As liquid oil depleats, society is switching to mining tar sands.

The Hubbert equation, in all its forms, follows as a consequence of the definition of the logistic function Q(t). It ranges from 0 in the past to $$Q_\infty$$ in the future.


 * $$Q(t) = {Q_\infty \over 1 + e^{\omega (\tau-t)}} \qquad \mbox{(4)} \!$$

Verification the four forms are equivalent
If you buy the idea that your data scatter in (Q,P/Q)-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).



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



P(t) = Q_\infty \ \omega \ {       1 \over (e^{-(\omega/2) (\tau-t)} + e^{(\omega/2) (\tau-t)})^2 } \qquad \mbox{(6)} \!$$ which is equation (3).

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


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


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


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


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

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