User:Jclaer/space

Fitting the Hubbert curve
Over the long haul populations grow and decay. M. King Hubbert chose a simple equation to describe the growth and decay of civilization's dependence on petroleum. (See for example, Beyond Oil: The view from Hubbert's Peak, by Ken Deffeyes. ) 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 diminish as does the number of remaining fish. Hubbert's equation models both growth and decay with a single equation of three parameters to be chosen from the data. His conclusion 50 years ago was that our oil production would be peaking about now (2008). It is.

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.

τ is the year at which production peaks.

ω is an inverse decay time (imaginary frequency).

Three forms of Hubbert's equation
The Hubbert's equation can be expressed in three 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 at $$t=-\infty$$ and $$t=+\infty$$. As production begins and Q is small the solution to dQ/dt=ωQ exhibits exponential growth at a rate ω. As production ends at $$ Q \approx Q_\infty$$ the non-linear equation reduces to exponential decay. To prove this fact change variables from Q to q where $$ Q=Q_\infty - q$$ and see an equation for q exponentially decaying hence the production P=dQ/dt decaying, decaying at the same rate ω.

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 ω = P/Q. Knowing everything but $$Q_\infty$$ we can solve for it with equation (2).

The third form of Hubbert's equation is the one best known. It looks like a Guassian, but it isn't. (A Guassian 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)} \!$$

Examining the form of this equation we see it is symmetric about the point t=τ. Asymptotically it decreases (or increases) exponentially towards its maximum value at t=τ. 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 τ. 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).

In real life there is no reason for the decay rate to match the growth rate. It's just a consequence of this simplified theory. It's a challenge for the coming few decades to replace the two same ω's in equation (3) by two different ones, and then show how to find them.

Final algebra check
We haven't yet shown that equation (3) is indeed the solution to equation (1). That requires only using the rule for the derivative of a quotient, d(u/v) = (1/v)du - u/v2dv.