# Hubbert model

• Jan 21st 2010, 06:58 PM
xzotica
Hubbert model
Sorry i am new here. But i dont understand how hubbert derived his model. from dQ/dt to Q using integration.

Integral of dQ/dt=kQ(1-Q/Qt) = Q = Qt/(1+e^(tm-t))
• Jan 21st 2010, 11:02 PM
CaptainBlack
Quote:

Originally Posted by xzotica
Sorry i am new here. But i dont understand how hubbert derived his model. from dQ/dt to Q using integration.

Integral of dQ/dt=kQ(1-Q/Qt) = Q = Qt/(1+e^(tm-t))

Hubberts model represents the production rate of a resource by the logistic equation:

$\displaystyle \frac{dQ}{dt}=k Q \left(1-\frac{Q}{Q_{max}}\right)$

where $\displaystyle Q(t)$ is the total production up to time $\displaystyle t$, $\displaystyle k$ is a constant and $\displaystyle Q_{max}$ is the total available resource.

This is an first order ordinary differential equation of variables seperable type. So:

$\displaystyle \int_0^{\tau} \frac{1}{k Q \left(1-\frac{Q}{Q_{max}}\right)}\ dQ$ $\displaystyle =\int_0^{\tau} \;dt$

CB
• Jan 22nd 2010, 01:17 AM
xzotica
can you please give me step by step on that integration? I dont undertstand how it became:

Q= Qmax/(1+e^k(tm-t))
• Jan 22nd 2010, 02:36 AM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
Hubberts model represents the production rate of a resource by the logistic equation:

$\displaystyle \frac{dQ}{dt}=k Q \left(1-\frac{Q}{Q_{max}}\right)$

where $\displaystyle Q(t)$ is the total production up to time $\displaystyle t$, $\displaystyle k$ is a constant and $\displaystyle Q_{max}$ is the total available resource.

This is an first order ordinary differential equation of variables seperable type. So:

$\displaystyle \int_0^{\tau} \frac{1}{k Q \left(1-\frac{Q}{Q_{max}}\right)}\ dQ$ $\displaystyle =\int_0^{\tau} \;dt$

CB

Quote:

Originally Posted by xzotica
can you please give me step by step on that integration? I dont undertstand how it became:

Q= Qmax/(1+e^k(tm-t))

First we integrate the right hand side:

$\displaystyle \int_0^{\tau} \frac{1}{k Q \left(1-\frac{Q}{Q_{max}}\right)}\ dQ$ $\displaystyle = \tau$

Now we resolve the left hand side into partial fractions and move the $\displaystyle k$ to a more convienient position:

$\displaystyle \int_0^{\tau} \frac{1}{(Q_{max}-Q)}+ \frac{1}{Q}\ dQ$ $\displaystyle = k\tau$

Now we can integrate the left hand side (note $\displaystyle Q(t)$ and $\displaystyle (Q_{max}-Q(t))$ are both positive):

$\displaystyle -\ln(Q_{max}-Q)+\ln(Q) + K$ $\displaystyle = k\tau$

where $\displaystyle K$ is a constant depending on the quantity of the resource used at time $\displaystyle 0$ so:

$\displaystyle \frac{KQ}{Q_{max}-Q}=e^{k\tau}$

Which after some jiggery pokery gives:

$\displaystyle Q(\tau)=\frac{Q_{max}}{1+Ke^{-k\tau}}$

Which might not be what you posted but it is what the Wikipedia article has.

CB
• Jan 23rd 2010, 12:02 AM
CaptainBlack
Since the OP has disappeared I might as well add a comment on the Hubbert model.

As a predictive model it is almost completely without value as there is no reason to suppose that resource recovery should follow the logistic equation. Every logistic like model with positive exponents attached to the two major terms has all the properties that the basic logistic equation has and so without calibration are equally valid. This would be unimportant if they all predicted the same time and value for peak production. Then there is the question of what other models have the required properties of an initial growth period with a limit to the total possible recovery.

To make this valid you need calibration or validation data and then to show that the simple logistic model performs better that its rivals. Even then you will need a theoretical basis for justifying the assumption that the same parameters are applicable to other resources etc.

CB