# Analytical Solution for a 1D equilibrium interface

Composition $$c$$$for 2D simulation domain Order parameter $$\eta$$$ along $$y=0$$$Composition $$c$$$ along $$y=0$$$Click here to see the input file for this example: kks_example_noflux.i In the KKS model, an analytical solution exists for the order parameter $$\eta$$$ and composition $$c$$$through a 1D equilibrium interface: \eta_0(x) = \frac{1}{2} \left[1-\tanh{\left( \frac{\sqrt{w}}{\sqrt{2} \epsilon} x\right)} \right] c_0(x) = h(\eta_0(x))c_S^e + [1-h(\eta_0(x))]c_L^e where we use the switching function $$h(\eta) = \eta^3(6\eta^2-15\eta+10)$$$, the gradient energy coefficient $$\epsilon^2 = 1$$$and the barrier function height $$w=1$$$. (Note that there is a typo in Eq. (49) of the paper, $$\epsilon$$$should be in the denominator of the argument to the $$\tanh$$$ function, not $$\sqrt{\epsilon}$$$.) In the following example, we equilibrate a flat interface between a solid phase ($$\eta = 1$$$, $$f^S = (c_S-c_S^e)^2$$$, $$c_S^e = 0.9$$$) and a liquid phase ($$\eta = 0$$$, $$f^L = (c_L-c_L^e)^2$$$, $$c_L^e = 0.1$$$) in a 2D simulation. The vector postprocessor LineValueSampler is used to obtain the values of $$\eta$$$ and $$c$$$along $$y=0$$$, the results are output to a CSV file, and plotted together with the 1D analytical solution. (We will use no-flux boundary conditions, so a boundary conditions block is not required in the input file.)

# Verification against analytical solution

$$L^2$$$error for order parameter $$\eta$$$

To perform a more quantitative comparison of the simulation results to the analytical solution, we will calculate the $$L^2$$$norm of the difference between the simulation result and the analytical solution. The $$L^2$$$ norm is defined for this case as
\|u - u_h\|_{L^2} = \left( \int_\Omega (u - u_h)^2 d\Omega \right)^{1/2}
where $$u$$$is the analytical solution, $$u_h$$$ is the equilibrium solution from simulation, and $$\Omega$$$is the domain. The $$L^2$$$ norm can be obtained in the MOOSE framework using the ElementL2Error postprocessor. It can be shown from the properties of the finite element method that for the linear Lagrange elements used in the split formulation,
\|u - u_h\|_{L^2} \le Ch^2
where $$h$$$is the characteristic element size (for 2D square elements, $$h$$$ is the length of one side) and $$C$$$is a constant specific to the problem. Thus, as the mesh is refined, $$L^2$$$ error should decrease (at least) quadratically with $$h$$$. (In performing this comparison between the analytical solution and simulation results, if a no-flux boundary condition is used in the simulation, the order parameter and composition profiles may shift slightly in the $$+x$$$ or $$-x$$$direction, even if the analytical solution is used as an initial condition. This makes it difficult to compare to an analytical solution centered at $$x=0$$$. Therefore, we simulate the $$x \ge 0$$$half of the domain and use a Dirichlet boundary condition of $$\eta=0.5$$$ and $$c=0.5$$$at $$x=0$$$, which prevents the interface from moving. For the verification, the size of the domain was also reduced to lower the computational cost.)
To verify that $$L^2$$$error decreases quadratically with $$h$$$, see the plot showing $$L^2$$$error versus $$h$$$ for $$\eta$$\$ in this problem. As expected, on a log-log scale, the points are fit well by a straight line. The slope was determined to be 1.995, in good agreement with the expected value of 2.