Oscillations in the Oral Glucose Tolerance Test (OGTT)

by Reinaldo Baretti Machín   ( UPR- Humacao)

e-mail     reibaretti2004@yahoo.com

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References:

1. The human body's response to glucose and three physical models American Journal of Physics -- July 1987 -- Volume 55, Issue 7, pp. 641-645

2. E. Ackerman et. al, Phys. Med. Biol. , 9 ,203, (1964)

3. Classical Mechanics  , by Herbert Goldstein

4. Classical Mechanics: 2nd Edition (Dover Books on Physics) , H.C. Corben and Philip Stehle

A normal glucose response curve to an oral glucose test (OGTT) looks like a damped oscillator (see fig 1.) around baseline values of 90-100 mg/dL. It is not exactly a damped oscillator since the first semi period is less than 2 hours while the second semi period is more than two hour long. Fig 1. Oscillations of the blood glucose level during an OGTT.

Reproduced from http://themedicalbiochemistrypage.org/diabetes.php

Our aim in this note is to model the normal curve using the analogy of a damped oscillator.

A damped oscillator is described by the differential equation (DE) ,  , ,

d2x/dt2 + 2β dx/dt + (ω0 )2 x = 0             ,    (  1 )

where  x ~length , β ~ 1/time    , (ω0 )2 ~ 1/time 2 . For a particular solution two initial conditions have to be specified ,  x(0)  , (dx/dt) = v(0)  .

We change the dependent variable from x to the glucose concentration g(t) ~ mg/dL ~miligrams/deciliter and write

d2g/dt2 + 2β dg/dt + (ω0 )2 g = 0                                      (  2   )

The solution is

g(t) = A exp(-βt) sin ( ω't ) + g(0)                                   ( 3   )

Substitution of  ( 3 ) in  (2)  gives

exp(-βt) (ω'2 - ω022 ) cos(ω't) = 0                         ,    (  4  )

implying that the modified frequency ω' is

ω'2 = ω02 - β2    ~rad/ time 2                              .    (  5  )

From the normal glucose curve ( Fig. 1) we have the initial value g(0) = 90 mg/dL . To find the parameters A , β ,ω ' we will examine characteristics of the curve. The period is approximately  τ  ≈ 4 hours = 240 min . Hence the angular frequency

ω' = 2π/τ = (2 π/240) rad /min                                ,  ( 6  )

Taking the first derivative

dg/dt = -β A exp(-βt) sin ( ω't ) + ω' A exp(-βt) cos(ω't )            .        (  7  )

The estimation of β will depend on whether  we take the first maxima  at t = 1hour  or the first minima at t= 2.5 hour.

We see that about t = t1 = 1 hour = 60 min , g(t) has reached a relative maxima. Equating  dg/dt = 0 yields

β = ω' cot (ω't1 ) = 2.62E-2   cot ( 2.62E-2 *60)                       ,   (  8 )

= - 3.15E-5  /min     ,

a negative value which is unacceptable for  β . However using the firts minima at t = 2.5 hr = 150 min yields

β = ω' cot (ω't1 ) = 2.62E-2   cot ( 2.62E-2 *150 )

= 2.60E-2/min .                                                                .  (9)

To get the  amplitude A we see that the maxima at t= 60 min is about 125 mg/dL . Solving (3) for A

A ≈  {gmax - g(0)}/[ exp(2.60E-2*60) sin(2.62E-2*60)  ]

= {125-90}/[ exp(2.60E-2*60) sin(2.62E-2*60)  ]

= 166 mg/dL                                                                                    . (10)

Hence our approximation for g(t) is

g(t) ≈ 166 exp(-2.60E-2 t) sin (2.62E-2 t) + 90 ~ mg/dL   , t ~min        . ( 11 ) Fig. 2 Damped oscillator of eq ( 11 ) , compare with normal response in Fig. 1.

Conclusions: The glucose oscillations can be simulated with a damped oscillator solution by adjusting four parameters  , the base line g0 ~mg/dL , the frequency ω'~rad/min , the damping parameter β ~1/min and the amplitude A ~mg/dL .

End of Problem