# Solving Stefan (moving-boundary) formulation of a diffusion problem using numerical and symbolic computing - Part 2

In the previous post, I quickly demonstrated the formulation of a moving-boundary problem for a 1D case, but in the end, we reached a problem related to the complexity of the derived equation. The complexity led to NotImplementedError, which means that the symbolic solver is not capable of solving that. As an alternative solution, in this post, we switch to a numerical solver to obtain the value of $$\alpha$$ through the following equation (see previous post for further explanation of this):

$\alpha=\frac{c_{0}-c_{\text {sat }}}{c_{\text {sol }}-c_{\text {sat }}} \sqrt{\frac{D}{\pi}} \frac{\exp \left(\frac{-\alpha^{2}}{D}\right)}{\operatorname{erfc}\left(\frac{-\alpha}{\sqrt{D}}\right)}$

To this end, we use Newton method, an iterative method to find the root of an equation. We reformulate the above equation and use the Newton method to find the value of the root, which will be $$\alpha$$ in this case.

Newton method works based on evaulation of the following iterative equation:

$x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$

We start by guessing the initial value of $$x_{n}$$, and then successively find a new value of $$x_{n+1}$$ and replace it with $$x_{n}$$ again. The desired function in our case can be defined based on the above expression for $$\alpha$$:

$f(\alpha) = \alpha - C \sqrt{\frac{D}{\pi}} \frac{\exp \left(\frac{-\alpha^{2}}{D}\right)}{\operatorname{erfc}\left(\frac{-\alpha}{\sqrt{D}}\right)}$

with $$C$$ being:

$C=\frac{c_{0}-c_{\text {sat }}}{c_{\text {sol }}-c_{\text {sat }}}$

Sympy can be used to easily calculate the derivative of $$f(\alpha)$$. Similar to the previous post, we define appropriate symbols in Sympy, but this time, we also assign some numerical values to them for facilitating further evaluations.

from sympy import init_session
init_session(quiet=True)

D_value = 0.00075
c_0 = 0
c_sol = 1735
c_sat = 134
C_value = (c_0 - c_sat)/(c_sol - c_sat)

a, D, C = symbols("alpha, D, C", positive=True)
func = a - C * sqrt(D/pi) * exp(-a**2/D) / erfc(-a/sqrt(D))
Eq(func, 0)


which produces the following output:

$\frac{C \sqrt{D} e^{- \frac{\alpha^{2}}{D}}}{\sqrt{\pi} \left(2 - \operatorname{erfc}{\left(\frac{\alpha}{\sqrt{D}} \right)}\right)} + \alpha = 0$

Now, we can simply calculate the derivitive of $$f(\alpha)$$:

dfunc = diff(func, a)
dfunc


which results to:

$\frac{2 C e^{- \frac{2 \alpha^{2}}{D}}}{\pi \left(2 - \operatorname{erfc}{\left(\frac{\alpha}{\sqrt{D}} \right)}\right)^{2}} + \frac{2 C \alpha e^{- \frac{\alpha^{2}}{D}}}{\sqrt{\pi} \sqrt{D} \left(2 - \operatorname{erfc}{\left(\frac{\alpha}{\sqrt{D}} \right)}\right)} + 1$

Although this looks a bit complicated, we don’t care about the complexity as Sympy will take care of the evaluation of this expression for obtaining $$f^{\prime}\left(\alpha\right)$$ in each iteration. Let’s define some functions for the corresponding terms in the Newton method equation:

func = func.subs([(D, D_value), (C, C_value)])
dfunc = dfunc.subs([(D, D_value), (C, C_value)])

def F(x):
return func.subs(a, x).evalf()

def dF(x):
return dfunc.subs(a, x).evalf()


And then, a pretyy simple declaration of the iterative Newton method:

def dx(f, x):
return abs(0-f(x))

def newtons_method(f, df, x0, e, verbose=False):
delta = dx(f, x0)
while delta > e:
x0 = x0 - f(x0)/df(x0)
delta = dx(f, x0)
if verbose:
print ('Delta is ', delta)

print ('>>>>> Root is at ', x0)
print ('>>>>> f(x) at root is ', f(x0))
return x0


In this definition, the code stops iterating when the error drops below a certain threshold (passed by e to the function).

Okay, everything is ready so far. We can start the iterations by calling newtons_method(F, dF, x0, 1e-7, True), in which x0 is the initial point to start with and the error threshold is $$10^{-7}$$. But, a better approach can be running the process for various start points and see if they all reach the same result or not:

x0s = [0, .5, 1]
for x0 in x0s:
newtons_method(F, dF, x0, 1e-7, True)


which generates this output, indicating that the value of $$\alpha$$ is $$-0.001366$$ with the selected values for the chemical parameters:

Delta is  8.68237762054906e-7
Delta is  3.82565589437956e-13
>>>>> Root is at  -0.00136691381199428
>>>>> f(x) at root is  3.82565589437956e-13
Delta is  0.00129321032032748
Delta is  8.68237762054906e-7
Delta is  3.82565589437956e-13
>>>>> Root is at  -0.00136691381199428
>>>>> f(x) at root is  3.82565589437956e-13
Delta is  0.00129321032032748
Delta is  8.68237762054906e-7
Delta is  3.82565589437956e-13
>>>>> Root is at  -0.00136691381199428
>>>>> f(x) at root is  3.82565589437956e-13


As you can see, the selected initial points all have converged to the same value, but of course the number of iterations is different to reach the root.