TITLE: Validating a mathematical model (Lagrange formulation and geometry)
QUESTION [6 upvotes]: I am working on computing phase diagrams for alloys.  These are
blueprints for a material that show what phase, or combination of
phases, a material will exist in for a range of concentrations and
temperatures (see this
pdf presentation).  
The crucial step in drawing the boundaries that separate one phase
from another on these diagrams involves minimizing a free energy
function subject to basic physical conservation constraints.  I am
going to leave out the chemistry/physics and hope that we can move forward
with the minimization using Lagrange multipliers. 
The free energy that is to be minimized is this:
$\widetilde{G}(x_1, x_2) = f^{(1)}G_{1}(x_1) + f^{(2)}G_{2}(x_2),$
subject to:
$f^{(1)}x_1 + f^{(2)}x_2 = c_1,$
$f^{(1)} + f^{(2)} = 1. $
(and also that the $x_{i} > 0$ and $f^{(i)} > 0$, for $i=1,2$.)
The Lagrange formulation is:
$L(x_1,x_2,f^{(1)},f^{(2)},\lambda_1, \lambda_2, \lambda_3) =
f^{(1)}G_{1}(x_1) + f^{(2)}G_{2}(x_2)$  
$- \lambda_{1}(f^{(1)}x_1 + f^{(2)}x_2 - c_1)$
$- \lambda_{2}(f^{(1)} + f^{(2)} - 1) $
The minimization of $\widetilde{G}$ follows from finding the $x_{i}$'s  that satisfy $\nabla L = 0:$
$\frac{\partial L}{\partial x_{1}}   = f^{(1)}G_{1}'(x_1) - \lambda_{1}f^{(1)} = 0$
$\frac{\partial L}{\partial x_2}     = f^{(2)}G_{2}'(x_2) - \lambda_{1}f^{(2)} = 0$
$\frac{\partial L}{\partial f^{(1)}} = G_{1}(x_1) - \lambda_{1}x_{1} - \lambda_2 = 0$
$\frac{\partial L}{\partial f^{(2)}} = G_{2}(x_2) - \lambda_{1}x_{2} - \lambda_2 = 0$
which yields:
$(*) f^{(1)}\left[G_{1}'(x_1) - \lambda_1 \right] = 0$       
$(**) f^{(2)}\left[G_{2}'(x_2) - \lambda_1 \right]= 0 $       
$(\***) G_{1}(x_1) - G_{2}(x_2) = \lambda_1 \left[ x_1 - x_2\right]$ 
Because $f^{(1)}$ and $f^{(2)}$ are not to be zero, from (*) and (**) we have that 
$G_{1}'(x_1) = G_{2}'(x_2) = \lambda_{1}.$
And, a manipulation of equation (***) looks like 
$\frac{G_{1}(x_1) -G_{2}(x_2)}{x_1 - x_2} = \lambda_{1}.$
Now, think of $G_{i}$ as an even degree polynomial (which it isn't, but
it's graph sometimes resembles one) in the plane.  Let the points $x_1$
and $x_2$ be locations along the x-axis that lie roughly below the
minima of this curve.  The constraints (*),(**), and (***) describe the
condition that the line drawn between $(x_1,G_{1}(x_1))$ and $(x_2,G_{2}(x_2))$ form a common tangent
to the "wells" of the curve.  It is these points $x_1$ and $x_2$,
which represent concentrations of pure components in our alloy, that
become mapped onto a phase diagram.  It is essentially by repeating this procedure for many
temperatures that we can trace out the boundaries in the desired phase diagram.
The question is:  Looking at this from a purely analytic geometry
perspective, how would one derive the "variational" approach to find a common tangent line that we seem to have found using the above Lagrangian?  (warning: I don't really know how to
model things using variational methods.)  
And, secondly: I have presented a model of a binary alloy, meaning
two variables to keep track of representing concentrations.  I have
been working on ternary alloys, where this free energy $\widetilde{G}$
is a function of three variables (two independent: $x_1,x_2,x_3$,
where $x_3 = 1- x_1 - x_2$) and is therefore a surface over a Gibbs
triangle.  Then $\nabla L = 0$ produces partial derivatives that no
longer "speak geometry" to me, although the solution is a common tangent
plane.  (I have attempted to characterize a common tangent plane
based purely in analytic geometry - completely disregarding the
Lagrangian - and have come up with several relations between
directional derivatives... How might directional derivatives relate
to the optimality conditions set forth by the Lagrangian?)
EDIT:  Thank you Greg Graviton for wading through this sub-optimal notation and pointing out several mistakes in the statement of the problem.  (Also, thank you for the excellent discussion below.)

REPLY [4 votes]: Concerning the physical meaning, I take it that $f_1$ and $f_2$ represent the fractions of the two phases in the alloy (this implies $f_1 + f_2 = 1$). I imagine $x_1$ and $x_2$ to correspond to an intensional variable like pressure, whose average $f_1x_1 + f_2x_2 =: \bar x$ is held constant in the experiment. Now, the alloy minimizes the free energy under these constraints.
To answer your first question: there is a geometric reason why the solution is the common tangent to $G_1$ and $G_2$ in the case of two dimensions. Namely, the fractions $f_1$ and $f_2$ are exactly the Barycentric coordinates of the average $\bar x$ sitting between $x_1$ and $x_2$. In particular, the value of the total energy $\tilde G$ is the height of the line drawn between $(x_1,G_1(x_1))$ and $(x_2,G_2(x_2))$ evaluated at $\bar x$. Here's a sketch:

From this picture, it is clear that if this line is not tangent to both $G_1$ and $G_2$, then you can move it a bit so that the value at $\bar x$ will decrease.
To answer your second question, the geometry readily extends to higher dimensions. For instance, for an alloy of three compounds, one has to consider the triangle enclosed by the three points $(x_1,G_1(x_1))$, $(x_2,G_2(x_2))$ and $(x_3,G_3(x_3))$. The situation is a bit degenerate here, any point inside this triangle whose first coordinate is $\bar x$ represents a valid value of $\tilde G$. Of these, nature will choose the smallest one. Consequently, the lower side of the triangle has to be tangent to two of the individual free energies.

Similar reasoning applies when the variable $x$ is not just a number, but, say, a pair of numbers, then we're dealing with a plane tangent to three individual Gibbs functions.
While not terribly useful in this case, there is also a very general geometric interpretation of the method of Lagrange multipliers. Namely, consider a goal function $f$ and a holonomic constraint $g$. Then, the Euler-Lagrange-equations give $\nabla f = \lambda \nabla g$ which means that $f$ changes only in directions orthogonal to the surface $g$. But since we're confined to the surface $g$, this must be an extremum. Wikipedia has a picture.