I am reading a document which claims that if we numerically integrate the following law with respect to time:

$$\dot{p}_c = \frac {1}{\lambda^*-\kappa^*}p_c \dot{\epsilon}_{vol}^c$$

We get the following result

$$p_{c,n+1} = p_{c,n} \exp\left[\frac{1}{\lambda^*-\kappa^*}\Delta\epsilon_{vol}^c\right]$$

where the subscripts $n$ and $n+1$ denote the value at $t = t_n$ and $t = t_n+\Delta t$

I have no idea of the intermediate steps used to make this claim. In any case I tried to think about the analytical integral with respect to time. In this case:

$$ \int \dot p_c dt = \int \frac {1}{\lambda^*-\kappa^*}p_c \dot{\epsilon}_{vol}^c dt$$

Should simply yield

$$p_c = \frac {1}{\lambda^*-\kappa^*}p_c \epsilon_{vol}^c$$ Should it not? Since the integral of the derivative of a function is simply that function. I have no clue how they're getting that exponential.

What are the intermediate steps to get to the numerically integrated expression. Why is the integral of the derivative simply not the function itself?


I now understand what I was overlooking in the original equation:

$p_c$ is a function of time and occurs on both sides of the original equation. Thus we have:

$$\int \frac{\dot{p}_c(t)}{p_c} dt= \int \frac{1}{\lambda^*+\kappa^*}\dot{\epsilon}_{vol}^c dt$$

using the u-substitution technique with $u = p_c(t)$ and $du = \dot{p}_c(t)dt$ we obtain $$\int \frac{\dot{p}_c(t)}{p_c} dt = \int \frac 1 u du $$

Then of course $$\int \frac 1 u du = \ln u = \ln p+c(t)$$

Thus we find:

$$\ln p_c(t) = \frac 1 {\lambda^*+\kappa^*}\epsilon_{vol}^c (t)$$

Rearranging we obtain:

$$p_c(t) = \exp\left[\frac 1 {\lambda^*+\kappa^*} \epsilon_{vol}^c(t)\right]$$

Which at least accounts for the exponential sign. So yes, the integral of the derivative of a function is that same function but it is important to know where the function occurs.

I'm still unsure as to where the numerical form $$p_{c,n+1} = p_{c,n} \exp\left[\frac{1}{\lambda^*-\kappa^*}\Delta\epsilon_{vol}^c\right]$$ comes from. If anyone could provide that explanation that would be great. Thanks.


If you integrate $$\dot p=kp$$ using that method, you need to incorporate the integration constant in some way, or do everything in definite integrals so that equals stay equals, for instance as $$ \int_{p_n}^{p_{n+1}}\frac{du}{u}=\int_{t_n}^{t_{n+1}}k\,ds $$ leading to (considering that no sign change takes place) $$ \ln|p_{n+1}|-\ln|p_n|=k\,\Delta t $$ which then transforms into the given equation $$ p_{n+1}=p_n\,\exp(k\,Δt). $$

  • $\begingroup$ great. Thank you very much. $\endgroup$ – user32882 Oct 21 '17 at 9:41

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