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Ordinary Differential Equations from Physical Models

In engineering and physics, modeling physical phenomena is the means by which technological and natural phenomena are understood and predicted. A model is an abstraction of a physical system, often with simplifying assumptions, into a mathematical framework. Every model should be verifiable by an experiment that, to the greatest extent possible, satisfies the approximations that were used to obtain the model.

In the context of modeling, differential equations appear frequently. Learning how to model new and interesting systems is a learned skill--it is best to learn by following a few examples. Grain growth provides some interesting modeling examples that result in first-order ODES.

Grain Growth

In materials science and engineering, a grain usually refers a single element in an ensemble that comprises a polycrystal. In a single phase polycrystal, a grain is a contiguous region of material with the same crystallographic orientation. It is separated from other grains by grain boundaries where the crystallographic orientation changes abruptly.

A grain boundary contributes extra free energy to the entire system that is proportional to the grain boundary area. Thus, if the boundary can move to reduce the free energy it will.

Consider simple, uniformly curved, isolated two- and three-dimensional grains.

Figure 20-1: Illustration of a two-dimensional isolated circular grain and a three-dimensional isolated spherical grain. Because there is an extra energy in the system $ \Delta G_{2D} = 2 \pi R \gamma_{gb}$ and $ \Delta G_{3D} = 4 \pi R^2 \gamma _{gb}$ , there is a driving force to reduce the radius of the grain. A simple model for grain growth is that the velocity (normal to itself) of the grain boundary is $ v_{gb} = M_{gb} \gamma_{gb} \kappa$ where $ M_{gb}$ is the grain boundary mobility and $ \kappa$ is the mean curvature of the boundary. The normal velocity $ v_{gb}$ is towards the center of curvature.

A relevant question is ``how fast will a grain change its size assuming that grain boundary migration velocity is proportional to curvature?''

For the two-dimensional case, the rate of change of area can be formulated by considering the following illustration.

Figure 20-2: A segment of a grain boundary moving with normal velocity $ v_n$ will move a distance $ v_n \Delta t$ in a short time $ \Delta t$ . The motion will result in a change of area $ -\Delta A$ for the shrinking grain. Each segment, $ ds$ , of boundary contributes to the loss of area by $ \Delta A = -v_n \Delta t ds$ .

Because for a circle, the curvature is the same at each location on the grain boundary, the curvature is uniform and $ v_n = M_{gb} \kappa_{gb} \gamma_{gb} = M_{gb} \gamma_{gb}/R$ . Thus

$\displaystyle \ensuremath{\frac{d{A}}{d{t}}} = - M_{gb} \gamma_{gb} \frac{1}{R} 2 \pi R = - 2 \pi M_{gb} \gamma_{gb}$ (20-1)

Thus, the area of a circular grain changes at a constant rate, the rate of change of radius is:

$\displaystyle \ensuremath{\frac{d{A}}{d{t}}} = \ensuremath{\frac{d{\pi R^2}}{d{t}}} = 2 \pi R \ensuremath{\frac{d{R}}{d{t}}} = - 2 \pi M_{gb} \gamma_{gb}$ (20-2)

which is a first-order, separable ODE with solution:

$\displaystyle R^2(t) - R^2(t=0) = -2 M_{gb} \gamma_{gb} t$ (20-3)

For a spherical grain, the change in volume $ \Delta V$ due to the motion of a surface patch $ dS$ in a time $ \Delta t$ is $ \Delta V = v_n \Delta t   dS$ . The curvature of a sphere is

$\displaystyle \kappa_{sphere} = \left( \frac{1}{R} + \frac{1}{R} \right)$ (20-4)

Therefore the velocity of the interface is $ v_n = 2 M_{gb} \gamma_{gb}/R$ . The rate of change of volume due to the contributions of each surface patch is

$\displaystyle \ensuremath{\frac{d{V}}{d{t}}} = - M_{gb} \gamma_{gb} \frac{2}{R}...
... - 8 \pi M_{gb} \gamma_{gb} R = = -4 (6 \pi^2)^{1/3} M_{gb} \gamma_{gb} V^{1/3}$ (20-5)

which can be separated and integrated:

$\displaystyle V^{2/3}(t) - V^{2/3}(t=0) = -$constant$\displaystyle _1 t$ (20-6)


$\displaystyle R^2(t) - R^2(t=0) = -$constant$\displaystyle _2 t$ (20-7)

which is the same functional form as derived for two-dimensions.

The problem (and result) is more interesting if the grain doesn't have uniform curvature.

Figure 20-3: For a two-dimensional grain with non-uniform curvature, the local normal velocity (assumed to be proportional to local curvature) varies along the grain boundary. Because the motion is in the direction of the center of curvature, the velocity can be such that its motion increases the area of the interior grain for some regions of grain boundary and decreases the area in other regions.

However, it can still be shown that, even for an irregularly shaped two-dimensional grain, $ A(t) - A(t=0) = -($const$ ) t$ .

© W. Craig Carter 2003-, Massachusetts Institute of Technology