Allen Cahn

Note

Please view the examples related to the crack surface density functional in Allen-Cahn.

The Allen-Cahn equation is given by:

\[\frac{\partial \phi}{\partial t} = -M \mu\]

where \(\phi\) is the phase-field, \(M\) is the mobility, and \(\mu\) is the chemical potential.

Free energy \(\mu\)

In this case, the problem that needs to be solved is stationary and derives from an optimization principle. The phase field is given by the minimizer of the following energy-functional

\[W[\phi] = \int_\Omega \left( \frac{1}{l}f_{chem}(\phi) + \frac{l}{2} |\nabla \phi|^2 \right) dV\]

with

\[f_{chem}(\phi) = \frac{1}{4}(1-\phi^2)^2\]

Note

\[f'_{chem}(\phi) = \phi^3 - \phi\]

\[f''_{chem}(\phi) = 3\phi^2 - 1\]

The equilibrium equations can recovered from the optimality condition \(\delta W=0\). Next, we calculate the variations of the funcional, applying the Gateaux derivative.

\[\begin{split}\delta_\phi W & = \frac{d}{d \epsilon} W (\phi+\epsilon\delta_\phi) \bigg\rvert_{\epsilon=0} \\ & = \frac{d}{d \epsilon} \int \left( \frac{1}{l}f_{chem}(\phi +\epsilon \delta_\phi) + \frac{l}{2} |\nabla (\phi+\epsilon\delta_\phi)|^2 \right) dV \bigg\rvert_{\epsilon=0} \\ & = \int \left( \frac{1}{l}f'_{chem}(\phi+\epsilon \delta_\phi) \delta_\phi + \frac{l}{2} 2(\nabla (\phi+\epsilon\delta_\phi))\cdot \nabla \delta_\phi \right) dV \bigg\rvert_{\epsilon=0} \\ & = \int \left( \frac{1}{l}f'_{chem}(\phi+\epsilon \delta_\phi) \delta_\phi + l (\nabla (\phi+\epsilon\delta_\phi))\cdot \nabla \delta_\phi \right) dV \bigg\rvert_{\epsilon=0} \\ & = \int \left( \frac{1}{l}f'_{chem}(\phi) \delta_\phi + l \nabla \phi \cdot \nabla \delta_\phi \right) dV\end{split}\]

So, the weak form of the phase-field problem in the absence of external potential is given by:

\[\int_\Omega \left( \frac{1}{l} f'_{chem}(\phi)\delta\phi + l \nabla\phi \cdot \nabla \delta \phi \right) dV = 0\]

One dimension solution

Consider a broken bar with length \(L\), as depicted in fig , featuring a crack positioned at its center.w

For the one-dimensional scenario, the functional reads:

\[W_{1D}[\phi] = \int_\Omega \left( \frac{1}{l} f'_{chem}(\phi) + \frac{l}{2} \phi'^2 \right) dx\]

subject to the boundary conditions \(\phi(\pm \infty) = \pm 1\), \(\phi'(\pm\infty= = 0)\).

The problem can be reduced to a simple ordinary differential equation, with boundary conditions \(\phi(0) = 1\), and \(\phi'(A) = 0\):

\[\frac{1}{l}f'_{chem}(\phi) - l \phi''=0\]

The solution to this equation is given by:

\[\phi(x) = tanh\left(\frac{x}{l \sqrt{2}}\right)\]

\[\phi'(x) = \frac{1}{l\sqrt{2}} sech^2 \left( \frac{x}{l \sqrt{2}}\right)\]