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Is the following statement correct and if yes does it need to satisfy specific requirement to be correct: $${ d \over dt} \sum_{j=1}^{N(t)} f(t,j) = \sum_{j=1}^{N(t)} {df(t,j) \over dt} + f(t,N(t)) {dN(t) \over dt}$$

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What is the nature of $N(t)$? As I understand (as it is the upper limit of a sumation), it maps reals into naturals, so it cannot be continuous unless it is constant, but it can be locally continuous. Is it? – Carlos Eugenio Thompson Pinzón Oct 16 '13 at 22:11
$N(t)$ is integer and it is a big number. During a time period of interest, let's say, $t=[0,T]$, it changes between $10^6$ to $2 \times 10^6$. – Narj Oct 16 '13 at 22:22
Without presuming to speak for @Carlos, I believe his point was that if $N(t)$ is integer then it's either a constant map, or it's discontinuous (hence not differentiable on $[0,T]$. More context should therefore be provided. – Jonathan Y. Oct 16 '13 at 22:35
Thank you both. I think I see your point now. $N(t)$ is not a constant map and it seems to be discontinuous. Here is more context: $N(t)$ is the number of cells in a medium under a specific experiment. $N$ grows as a result of cell-division. Since N is a big number I have to work with average values. Suppose that $f(t,j)$ is amount of one chemical agent at time t in cell j, then its average is, $ \overline f = { { \sum_{j=1}^{N(t)}f(t,j)}\over{N(t)}}$. – Narj Oct 16 '13 at 22:54
I know the expression for ${{df(t,j)}\over{dt}}$ from some information about the intracellular kinetics of cells. I need to connect ${{d\overline f}\over{dt}}$ with ${{df(t,j)}\over{dt}}$. Is this any more clear? – Narj Oct 16 '13 at 22:55
up vote 1 down vote accepted

There are two approaches for this problem: one is to define: $$f_j(t)=\left\{\begin{array}{ll}f(t,j)&j\le N(t)\\0&j>N(t)\end{array}\right.$$ Then $$\sum_{j=1}^{N(t)}f(t,j)=\sum_{j=1}^{\infty}f_j(t)$$ and continue as you already know. $$\frac d{dt}\sum_{j=1}^{N(t)}f(t,j)=\sum_{j=1}^{\infty}f_j'(t)$$ Beware where each $f_j$ is non-continues.

The other approach is to have all points $\{t_i\}_{i\in I}$ in which $N(t)$ take a jump, meaning $N(t_i^-)\ne N(t_i^+)$. If $\{t_i\}_{i\in I}$ is an well ordered set, you can have: $0<t_1<t_2<\cdots<t_k<\cdots<T$, ($t_0=0$) and we define $N_i=N(t)$ for $t_{i-1}<t<t_i$. Note that $N_i$ should be well defined as there are no jumps in $N$ between $t_{i-1}$ and $t_i$.

So now you divide in this function defined by parts: $$\left.\frac d{dt}\sum_{j=1}^{N(t)}f(t,j)\right|_{t_{k-q}<t<t_k}=\sum_{j=1}^{N_k}\frac{df(t,j)}{dt}$$

The derivative at each $t_k$ does not exist unless both limits exists and are equal.

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This is helpful. Thank you. I think the first approach is more suitable for me. About the second case, however, the discontinuity in $N$ happens very often and also it depends on the resolution of experimental measurements, while I am only interpolating some measurements to obtain the whole profile of $N$. So, perhaps I can't use the second approach. I just need to understand one more thing. In second equation, is there any reason for replacing $t$ with $x$ in the functions under summation? – Narj Oct 17 '13 at 9:21
No reason. Just careless editing. Corrected. – Carlos Eugenio Thompson Pinzón Oct 17 '13 at 10:18

Since $N(t)$ is an integer, $N'(t)$ only makes sense at points $t_0$ where $N(t)=N(t_0)$ for all $t$ in some neighbourhood of $t_0$; in such a case, $N'(t_0)=0$. On any point where $N(t)$ changes value, its derivative is not defined.

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