Resource depletion model Functions
Resource usage starts as free growth:
current growth of usage is proportional to current value of usage
See Math Details at end for functional discussion, and derivations.
We need a function that
• starts growth (x to far left of graph) as an exponential,
• peaks,
• returns to exponential drop toward zero on far right.
Reciprocal Cantenary Model
F(t) = 1/[catenary curve] =
Diagram: f(t) = 2F(t).
Asymptotic Form
t ≪ 0: F(t) =
t ≫ 0: F(t) =
2
et  e t
2
t
e e
t
2
1
= t t
e e
cosh(t)
Heavy blue dotted – pure exponential growth. Thin red, locaton of inflection
et
et
=
2 et
e2 t  1
e –t
2 e –t
=
1  e –2 t
e –t
t
… t = –10, Denominator e2(–10) + 1 ≈ 1
–t
… t = +10, Denominator 1 + e–2(10) ≈ 1
  2 e
t –
  2 e
t 
To use the reciprocal of cosh x, need multiplier of 0.5 for early pure exp growth.
Peak
F’(t) = – sinh(t) = 0 at peak sinh (tPK) = 0
Inflection
2
2
F”(t) = 2 sinh t – cosh t = 0 at inflection point, t = T.
cosh2 (t)

tPK = 0
cosh3 t
T=
1
2
Function value:
ln
 2 1
 2 1
, T ≈ –0.8817, +0.8817
F(T) = 1/cosh(T) = 2/(eT + e–T) = 1/ 2
 F(T) = 2 /2 ≈ 0.707…
Living in initial growth period, it is hard to see where usage data
diverges from exponential free growth because of data “noise.”
Must get well away from exp curve to see end of free growth.
Near-Peak non-exponential … Time when usage falls below exponential growth:
Value of t (call it t*) where exponential growth just exceeds usage F(t) by fraction A …
Recall: F = t 2 t . The “2e±t ” term is the asyptotic form of F.
t < 0 … 2et – F ≥ A*F
t > 0 … 2e–t – F ≥ A*F
e e
Solution:
t < 0 … t* = + 1 ln A
2
t > 0 … t* = – 1
2
ln A
The t values where the free growth exponential is A greater than the curve
is given in the table (right)
A
t*
t*
t<0
t>0
10 %
–1.15
1.15
20 %
–0.81
0.81
30 %
–0.60
0.60
50 %
–0.35
0.35
Resource depletion model Functions
Hubbert’s Model:
Derivative Of The Logistics Curve
See Math Details at end for solutions to derivations.
t ≪ 0: L (t) = et
Logistics Curve
1
L(t) =
L(t) =
1  e t
Asymptotic behavior
e t/2
… t = – 10, 1 + e– (–10)
≈e10
L( – 10) = e–10
t ≫ 0: L (t) = 1 – e–t … t = +10, e– t < 1
asymtote is Taylor series
2 cosh(t/2)
The logistics starts correctly, Hubbert chose it’s deriviative as the usage model
Hubbert’s Function for Usage Model
F(t) =
d
e– t
L(t) =
dt
(1  e t )2
or
F(t) =
1
4 cosh2 (t/2)
Diagram: f(t) = 4F(t). Heavy blue dotted – pure exponential growth
Thin red, locaton of inflection.
Peak where F’(t) = 0.
F’(t) = –
sinh t/2
4 cosh3 t/2
peak at sinh(t/2) = 0, t0 = 0
Inflection Point where F”(T) =0
F”(t) =
1 2 – cosh(t)
8
4
f(t) = 4·F(t) so peak is f = 1.0
so T is where cosh T = 2
cosh (t/2)
T = ln (2 ±
3 ) = ± 1.317 these points are marked by the thin red lines
Inflection value of the utilization model fuction
f(t) = 4 e–t/(1+e—t)2 =
f(1.317) = 0.667
Non-Exponential Portion … Time when usage falls below exponential growth:
Value of t (call it t*) where exponential growth exceeds usage F(t) by fraction A.
2
where F =
. A is 0.1 for 10%, 0.4 for 40%, etc.
e t  e t
Solution for t* where exponential curve is exactly A*100% bigger than F
Solution is
t* = – ln( A  1 – 1)
the graph demonstrates both + t*, and – t*.
2et – F ≥ A*F,
A
t*
10 %
3.02
20 %
2.35
30 %
1.96
50 %
1.49
Resource depletion model Functions
t
t
t
t
cosh x = e  e , sinh x = e  e
Math Details
2

identities:
2
d
cosh(at) = a sinh(at).
dt
d
sinh(at) = a cosh(at).
dt
cosh x – sinh x = e–x
cosh2x – sinh2x = 1
cosh 2x = 1 +2 sinh2x
Reciprocal Catenary Model
F(t) =
2 et
e2 t  1


d
F(t)
dt
d2
dt 2
1
–1
=
sinh (t) = – sinh(t)
2
dt cosh(t)
cosh (t)
cosh2 (t)
= d
2
d sinh(t)
1
= –
= 2 sinh t – 1
3
dt cosh2 (t)
dt cosh(t)
cosh(t)
cosh t
Solve for t0: 2 sinh2 t – cosh2t = 0

2
2
= 2 sinh t – cosh t
F(t) = d
t
–t

t
cosh2t = 2 sinh2t
–t
(e + e ) = ± 2 (e – e )
t=
1
2
2 1
ln

(± 2 + 1)e
≈ 0.8817… , t =
2 1

1
2
ln
–t
cosh3 t
cosh t = ± 2 sinh t
= (± 2 – 1)et
– 2 1
– 2 1

e2t =
 2 1
 2 1
≈ – 0.8817… + and – acceptable in
argument.
Solve for t*: 2e+t –

≥ A
e+t* (et*+ e–t*) = A+1
Solve for t: 2e–t –

2
e t  e t
2
e t  e t

≥ A
e–t* (et*+ e–t*) = A+1

2
e t  e t
Solve at boundry  e+t* = (A+1)
(e2t*+1) = A+1
2
e t  e t

e2t* = A

t* =
1
2
Solve at boundry  e–t* = (A+1)
e–2t*+1 = A+1

e–2t* = A

1
et*  e– t*

e+t* =
A 1
et*  e– t*
ln A
1
et*  e– t*
t* = – 1 ln A
2

e–t* =
A 1
et*  e– t*
Resource depletion model Functions
Hubbert Function
Logistics Function:
L(t) =
1
1  e t

=
1
e t/2 (e t/2  e t/2 )
d
d
1
L(t) =
dt
dt 1  e  t
=
e– t
(1  e t )2
=
=
=

e t/2
=
2 cosh(t/2)
d
(1 + e–t)–1 = –(1 + e–t)–2 (–e–t) = e–t (1 + e–t)–2
dt
e t
=
e 2t/2 (e  t/2  e t/2 )2
1
(e  t/2  e t/2 )2
1
4 cosh2 (t/2)
d
e t/2
d
d t/2
L(t) =
=1
e (cosh(t/2))–1
2 dt
dt
dt 2 cosh(t/2)
=
1 1 et/2) (cosh t/2)–1 + et/2(–cosh–2 t/2) 1
2 2
2
t/2
cosh t/2 – sinh t/2 =
= e
4
cosh2 t/2
1
4 cosh2 (t/2)
sinh(t/2) =
e t/2
4
1
cosh t/2
– sinh t/2
cosh2 t/2
… cosh x – sinh x = e–x
,
Hubbert’s model
F(t) =
1
4 cosh2 (t/2)
or F(t) =
e– t
(1  e t )2
sinh t/2

d
F(t)
dt
=
1
4

d
F(t)
dt
=
–t
–t
– e– t – 2 e– t
(–e–t) = e–t 2e
– 1e
=–
(1  e t )2
(1  e t )3
(1  e t )3 (1  e t )3
(–2 cosh–3t/2)
1
2
sinh (t/2) = –
4 cosh3 t/2
= – sinh t/2
4 cosh3 t/2

d2
dt 2
F(t) = –
1 1 cosh t/2
4
2 cosh3 t/2
cosh4 t/2
2
1 1 – 2 sinh (t/2)
8
4
cosh (t/2)
2
–
cosh(t)
1
+
8
cosh4 (t/2)
x
–x
1
2
sinh (t/2) = –
= –
… cosh2x – sinh2x = 1
=
… cosh 2x =1 +2 sinh2x
Solve for x: a = cosh x = (e + e )/2

– 3 sinh t/2

let u = ex, get 2au = 1 + u2
2a = (ex + e–x)






3
 t/2 
 t/2
e
2 e

2


2
2
1 cosh (t/2) – 3 sinh (t/2)
8
cosh4 (t/2)
2a = e–x (1 + e2x)
0 = u2 + 2au + 1
 t/2
 t/2
e
2 e
2

u = ex = a ±

2a ex = 1 + e2x
a2  1
x = ln a ±
a2  1
for a = 2: x = ln(4 ± 31/2) ≈ ± 1.317 (this solves for inverse cosh a)
Solve for t*: e–t –

A 1
e– t*
e– t
e– t
–t*
≥
A
Solve
at
boundry
 e
=
(A+1)
1=
(1  et*)2
(1  et*)2
(1  e t )2
(1  e t )2
(1+ e–t*)2 = (A+1)

e–t* =
A 1
–1

t* = – ln( A  1 – 1)