It's not my field but I'll try to give my 2 cents ...
Unfortunately it's a long time I don't use Mathematica but I'll try to provide some hints that are implementation-independent.
First of all, I guess you are interested in computing \frac{M_{t}}{M_{\infty}} (Left Member of Eq 3 on page 320) that should be the normalized mass.
I suppose you know values of the R, L, D Params.
I think you should decice up to what level of precision you want the solution, this is to say up to which value of n you want to truncate the summations: let's call it n'.
You can see that increasing n' (and so getting higher value solutions to Eq 4) will give you smaller and smaller terms to the summations in (Eq 3) and so you have basically to decide which is the threshold value beyond which the contributions become neglectible, for your study.
This way you could easily solve (Eq 4) getting the n' roots and use them to compute the n' terms of the summation in Eq 3 getting and approximated solution with desired precision.
Thanks Brian. I have just started to learn Mathematica to model my experiments. It is almost problem to me to solve. I am grateful to you. The value of "R" equals to L*H/D and beta*Tan(beta), and both should be satisfied. I am not sure how to correlate both and incorporate into the eqn. 3.In addition to that I would like to manipulate "H" and "D" value to fit my experimental data set, aiming to find out the value of "H" and "D" of experimental data. Is it possible to superimpose the simulated curve on experimental data and manipulate "H" and "D". Furthermore, the axis will be normalised mass verses t^(1/2).Thanks again for your help.
See the attached file. I have added some additional code to make the calculations dimensionless. and also show how to plot the function versus t^1/2. This problem has 3 dimensional parameters L, H and D, and the combination produces R a dimensionless quantity called a Sherwood number (equivalent to a Nusselt number in heat transfer). Since the LHS of the equation is dimensionless, the RHS should also be dimensionless. Thus it is important to male time dimensionless. The obvious parameters to use are D and L.
In the attached file I select a pair of values for for D and H from Table 2. This allows me to find R and then make time dimensionless. The value of L( thickness of the polymer sheet) is given in the beginning of the paper.
When you plot the dimensionless eqn 2 versus tau^1/2 you get the sigmoid curve.
Note I had a typo in the Eq. 2 in the previous PDF which I fixed in this version
Mohammed, Sorry I for got to answer you last question about fitting the data to the model. The answer is yes it is trivial task using mathematica and Mathematica's FindFit function. In this case the dimensionless group R is the fitting parameter and the data should be plotted versus tau^1/2 not t^1/2
Thanks Brian. I have been trying to write the codes you provided but could not successful. May I have the "mathematica file"? In addition to that, from my experiment I get "mass uptake" as a function of "time". My final goal to find the diffusion coefficient "D" and the surface mass transfer "H". Thanks again for your kind help.
Sorry for asking again. If is it needed to plot mass ratio Vs tau^(1/2), here it is assumed that tau is a dimensionless quantities equals to t* D/H. However, in the final graph, it seems that tau and t are the similar, if not before fitting, how do i convert my data as i dont know the values of those. Another request, could you please provide some additional code for the Findfit function. Thanks.
The point is that the equation in the original article shows t as a dimensional quantity. By making t dimensionless with L and D to get tau, you now have a UNIVERSAL function on the RHS that depends only on the dimensionless Sherwood number. Suppose you have real data Mratio vs time. One would then rescale time with the values of D and L for your experiment so that you have Mratio versus Tau and then fit that data to the universal function to determine the Sherwood number.
In principle you could work with a dimensional time but then when it comes to fitting you would have to fit 3 parameters R, D and L, and recall R depends on D and H. So you would be fitting against D,H, and L again 3 parameters.
But recall that the eigenvalues depend on a highly nonlinear equation and separate errors in D H and L may cause problems.
Rather than rewrite the code in terms of tau I simply set D=L=1 so there was no explicit dependence on D and L on the RHS, which would be the case if t is made dimensionless with D and L. Since H has units of time one could also use H and L to make t dimensionless but the theory shows that D and L are the appropriate variables to make t dimensionless.
Brian, I went through the mathematica notebook you provided. However, I am still in confusion how to get the D,H value from the Sherwood Number. Thanks again.
Mohammad, The procedure I have outlined in the notebook, reduces the problem to determining a single dimensionless parameter: R (Sherwood number). Thus in order to extract out one of the parameters ( H,D,L) , two of them must be known.
It is possible to redesign the optimization problem to find multiple parameters, say two, or even three, but the optimization problem becomes more complex and there are likely many local minima in the 3-D parameter space and finding a global minimum is more challenging. What I tried to do in the notebooks is to give you a sense of how one would go about programming such a task. FindFit can find global minima, but its success all depends on the initial guess for the parameters- continuation methods or homotropy may be a strategy to navigate from one known global minimum to another.
On your other question, rescaling time with D and L is simply a definition for tau. But in order to extract a real time from tau, you will require knowing both L and D.
I would guess L is usually known. If you do an experiment with a known gas where D in the film is known, then you can extract H using the technique described above., Or you may know H for your system or able to estimate it. Note: H depends on the external mass transfer environment.