Can you explain what is a second order exponential decay function: $$ y(x) = y_0+A_{1}e^{-\frac{x}{t_1}}+A_{2}e^{-\frac{x}{t_2}} $$

(the $t_i$, $A_i$, and $y_0$ are constants and, presumably, the "decay constants" $t_i$ are positive)?

  • Qualitatively, what is the difference between "first order" and "second order"? (A first order exponential function has the form $y(t)=y_0 + A_1 e^{-\frac{x}{t}}$.)

  • How can we estimate $t_1$ and $t_2$ from data?

  • 2
    $\begingroup$ This is also known as a biexponential function: see google.com/search?btnG=1&pws=0&q=biexponential+model $\endgroup$
    – StasK
    Commented Oct 16, 2012 at 20:23
  • $\begingroup$ Please consult; en.wikipedia.org/wiki/Exponential_decay In order to find decay constants, fit the data with 2nd order exponential decay NLfit (Non-linear curve fit) using originlab 8.5 or higher. Good Luck $\endgroup$
    – user29402
    Commented Aug 21, 2013 at 17:56
  • $\begingroup$ You can model this type of function in JMP software.(non linear regression) It will come up with best estimates for all parameters $\endgroup$
    – user55901
    Commented Sep 16, 2014 at 17:26
  • $\begingroup$ Your second question has subsequently been answered at stats.stackexchange.com/questions/260042. It remains only to respond to the first "qualitative" question. $\endgroup$
    – whuber
    Commented Jan 29, 2021 at 16:12

1 Answer 1


BTW the link is broken. If one takes $y_0=0$, and $x$ to be time, $t$, the equation refers to a particular solution to the second order linear differential equation with constant coefficients using maximum initial functional value (at $t=0$). The reason for the simplifying assumption of $y_0=0$ is to formulate the answer as a density function, as a density function requires a finite area under its support. It is possible in some circumstances to have a system with a non-zero $y_0$. Said second order differential equation enter image description here was used to model venous concentration following intravenous injection of exogenous creatinine in dogs, in a so-called two compartment model 1.

  • $t_1$ is $\frac{1}{\ln2}$ times the "half-life" associated with the "first compartment" and $t_2$ is $\frac{1}{\ln2}$ times the so-called "terminal half-life." Actually, this is not exactly true as half-life of the plasma concentration is for all models, with the exception of a monoexponential, a function of time and not a collection of constants 2: enter image description here Note that in the monoexponential case, and only in the monoexponential case, the equation above reduces to a constant half-life.
  • The so-called one-compartment model, or a monoexponential corresponds to the second OP question. It takes the form of the solution to a first order differential equation:$$\frac{dC(t)}{dt}=-k\;C(t)\;\;,$$ where $C(t)$ is a monoexponential model of venous concentration subsequent to a bolus venous injection. Again, this is not actually true for injected substances in organisms as the samples are drawn remote to the injection site, such that the initial concentration at $t=0$ is not a maximum value, but is actually zero. This model, and the "second order" one above, would more accurately describe one or two leaky gas cylinders whose internal pressures equilibrate much more quickly than they leak to each other or to an external vacuum. Similar to the above, the name "first order" is inherited from the "first order" differential equation.

1.) Sapirstein, L. A.; Vidt, D. G.; Mandel, M. J. & Hanusek, G. Volumes of distribution and clearances of intravenously injected creatinine in the dog. American Journal of Physiology. 1955, 181, 330-336

2.) Wesolowski, C. A.; Wesolowski, M. J.; Babyn, P. S. & Wanasundara, S. N. Time varying apparent volume of distribution and drug half-lives following intravenous bolus injections. PLoS ONE, (2016), 11, e0158798


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.