Improved approximations for some polymer extension models - Rheologica Acta
- ️Petrosyan, Rafayel
- ️Sat Oct 29 2016
Appendix 1
Here, we describe how the Eq. (1) has been obtained. We know the dependence of the relative extension on the normalized force for the FJC model Eq. (4), and we need to determine the dependence of the normalized force on the relative extension for this model.
$$ x(f)= \coth (f)-\frac{1}{f} $$
(4)
To achieve this goal, we adopted an approach that was previously used for deriving approximate formulas (Petrov and Schwille 2008; Petrov et al. 2012), namely, first determining the asymptotes of the desired dependence, summing those asymptotes and then reducing the relative error of this sum by fitting its absolute error with an appropriate function, and adding the resulting fitting function to that sum of asymptotes.
In order to find the asymptotic behavior for small extensions of the force-extension dependence, we write the Taylor series expansion of the right side of the Eq. (4) for the normalized force tending to 0:
$$ x(f)=\frac{f}{3}-\frac{f^3}{45}+O\left({f}^5\right) $$
(5)
Next, by taking only the first term in Eq. (5) for small relative extension, we will havef(x) ~ 3x.
For the normalized force tending to infinity from Eq. (4), due to the fact that the exponential terms grow much faster than the linear term, we will have the following:
$$ x(f)= \coth (f)-\frac{1}{f}=\frac{e^{2f}+1}{e^{2f}-1}-\frac{1}{f}\sim 1-\frac{1}{f} $$
(6)
Hence, for this limiting case, we have f(x) ~ 1/(1 − x). Next, we need to sum up these two limiting cases. However, care should be taken so that these asymptotes will not interfere with each other. For example, the term 1/(1 − x) equals to 1 when the relative extension is 0; this means that this term should be multiplied with some function that will make it 0 when the relative extension is 0 and it will still go to infinity as 1/(1 − x) when the relative extension tends to 1; clearly, the multiplier can be higher than the first power of the relative extension so that term 3x dominates when x tends to 0. We found that the relative error was minimal among the multipliers with an integer powers of x when this power was equal to 3. Now, we have the following approximation:
$$ f(x)=3x+\frac{x^3}{1-x} $$
(7)
Interestingly, Eq. (7) is identical to the expression proposed by Darabi and Itskov f(x) = x(x 2 − 3x + x)/(1 − x) (Darabi and Itskov 2015) which is identical to the expression found in the work by Gou, Lad, Ray, and Akhremitchev f(x) = 1/(1 − x) − (1 − x)2 (Guo et al. 2009). Next, the relative error of Eq. (7) was minimized by fitting its absolute error with the appropriate function and then adding that function with the resulting fitting parameters to the Eq. (7). As earlier here, too, care should be taken so that the fitting function will not interfere with the asymptotes obtained previously. In this case, we have found that the simple function of the form ax 2 sin (bx) is fitting well the absolute error of Eq. (7). Last error correction step can be repeated by fitting the absolute error of the last approximation with an appropriate function and then again adding it to that approximation. This high accuracy will be at the cost of the simplicity of the expression. Such accurate approximations, with more than one correction steps, have been obtained for the inverse Langevin function previously (Nguessong et al. 2014; Marchi and Arruda 2015).
Appendix 2
The obtainment of the Eq. (2) was relatively simpler, since in this case, the asymptotes have been already determined by Marko and Siggia (1995) f(x) = x − 0.25 + 0.25/(1 − x)2 and only the last step, namely, the determination of the appropriate function for fitting the absolute error of the Marko–Siggia approximation was required. In this case, we have found that the simple function of the form ax b is fitting well the absolute error of the Marko–Siggia approximation. Then, that term has been added to the Marko–Siggia approximation and the resulting approximation had significantly smaller relative error (Fig. 2).
Appendix 3
The derivation of the Eq. (3) was not that straightforward, since the determination of the asymptotes that are not interfering with each other was not trivial for this case. Note that when the relative extension tends to 0, the Taylor series expansion of the Marko–Siggia expression will give the following:
$$ f(x)=\frac{3x}{2}+\frac{3{x}^2}{4}+O\left({x}^3\right) $$
(8)
By taking only the first term in Eq. (8) for small normalized forces, we will have x(f) ~ 2f/3.
When the relative extension tends to 1, the normalized force tends to infinity asf(x) ~ 1/4(1 − x)2. Hence, for this limiting case, we have \( x(f)\sim 1-1/2\sqrt{f} \).
Let us understand why the Eq. (9) has been chosen for building the final approximation—Eq. (3).
$$ x(f)=\frac{4}{3}-\frac{4}{3\sqrt{f+1}}-\frac{10{e}^{\sqrt[4]{\frac{900}{f}}}}{\sqrt{f}{\left({e}^{\sqrt[4]{\frac{900}{f}}}-1\right)}^2} $$
(9)
When the normalized force tends to 0, the Taylor series expansion of the first two summands of the Eq. (9) will be given by Eq. (10).
$$ \frac{4}{3}-\frac{4}{3\sqrt{f+1}}=\frac{2f}{3}-\frac{f^2}{2}+O\left({f}^3\right) $$
(10)
Note that when the normalized force tends to 0, the third summand of Eq. (9) tends to 0 exponentially, i.e., much faster than the linear term 2f/3 and does not interfere with it.
When the normalized force tends to infinity, the Taylor series expansions of the first two summands and the third summand of Eq. (9) will be given by Eqs. (11) and (12), respectively.
$$ \frac{4}{3}-\frac{4}{3\sqrt{f+1}}=\frac{4}{3}-\frac{4}{3\sqrt{f}}+O\left({\left(\frac{1}{f}\right)}^{\frac{3}{2}}\right) $$
(11)
$$ \frac{10{e}^{\sqrt[4]{\frac{900}{f}}}}{\sqrt{f}{\left({e}^{\sqrt[4]{\frac{900}{f}}}-1\right)}^2}=\frac{1}{3}-\frac{5}{6\sqrt{f}}+O\left(\frac{1}{f}\right) $$
(12)
By considering only the first two terms on the right sides in Eqs. (11) and (12) and subtracting the Eq. (12) from the Eq. (11), we can obtain the desired limiting behavior for the relative extension when the normalized force tends to infinity.
The Eq. (9) can be further improved by fitting its absolute error with the function of the following form f a/(b + cf d) and then adding this function with the resulting fitting parameters to the Eq. (9). Thus, Eq. (3) can be obtained.