# Confirming a mathematical version (Lagrange solution and also geometry)

I am working with calculating stage layouts for alloys. These are plans for a product that show what stage, or mix of stages, a product will certainly exist in for a series of focus and also temperature levels (see this pdf presentation).

The critical action in attracting the borders that divide one stage from an additional on these layouts entails decreasing a free power function based on standard physical preservation restraints. I am mosting likely to omit the chemistry/physics and also wish that we can move on with the reduction making use of Lagrange multipliers.

The free power that is to be decreased is this:

$\widetilde{G}(x_1, x_2) = f^{(1)}G_{1}(x_1) + f^{(2)}G_{2}(x_2),$

based on:

$f^{(1)}x_1 + f^{(2)}x_2 = c_1,$

$f^{(1)} + f^{(2)} = 1.$

(as well as additionally that the $x_{i} > 0$ and also $f^{(i)} > 0$, for $i=1,2$.)

The Lagrange solution 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 reduction of $\widetilde{G}$ adheres to from locating the $x_{i}$'s that please $\nabla L = 0:$

$\frac{\partial L}{\partial x_{1}} = f^{(1)}G_{1}'(x_1) - \lambda_{1}f^{(1)} = 0$

$\frac{\partial L}{\partial x_2} = f^{(2)}G_{2}'(x_2) - \lambda_{1}f^{(2)} = 0$

$\frac{\partial L}{\partial f^{(1)}} = G_{1}(x_1) - \lambda_{1}x_{1} - \lambda_2 = 0$

$\frac{\partial L}{\partial f^{(2)}} = G_{2}(x_2) - \lambda_{1}x_{2} - \lambda_2 = 0$

which returns:

$(*) f^{(1)}\left[G_{1}'(x_1) - \lambda_1 \right] = 0$

$(**) f^{(2)}\left[G_{2}'(x_2) - \lambda_1 \right]= 0$

$(\***) G_{1}(x_1) - G_{2}(x_2) = \lambda_1 \left[ x_1 - x_2\right]$

Because $f^{(1)}$ and also $f^{(2)}$ are not to be absolutely no, from (*) and also (**) we have that

$G_{1}'(x_1) = G_{2}'(x_2) = \lambda_{1}.$

And also, an adjustment of formula (***) resembles

$\frac{G_{1}(x_1) -G_{2}(x_2)}{x_1 - x_2} = \lambda_{1}.$

Currently, consider $G_{i}$ as an also level polynomial (which it isn't, yet it's chart occasionally appears like one) in the aircraft. Allow the factors $x_1$ and also $x_2$ be areas along the x-axis that exist about listed below the minima of this contour. The restraints (*), (**), and also (***) define the problem that the line attracted in between $(x_1,G_{1}(x_1))$ and also $(x_2,G_{2}(x_2))$ create an usual tangent to the "wells" of the contour. It is these factors $x_1$ and also $x_2$, which stand for focus of pure parts in our alloy, that come to be mapped onto a stage layout. It is basically by duplicating this procedure for several temperature levels that we can map out the borders in the wanted stage layout.

The inquiry is: Considering this from a totally analytic geometry viewpoint, just how would certainly one acquire the "variational" strategy to locate an usual tangent line that we appear to have located making use of the above Lagrangian? (advising: I do not actually recognize just how to version points making use of variational approaches.)

And also, second of all: I have actually offered a version of a binary alloy, definition 2 variables to track standing for focus. I have been working with ternary alloys, where this free power $\widetilde{G}$ is a function of 3 variables (2 independent: $x_1,x_2,x_3$, where $x_3 = 1- x_1 - x_2$) and also is consequently a surface area over a Gibbs triangular. After that $\nabla L = 0$ generates partial by-products that no much longer "talk geometry" to me, although the remedy is an usual tangent aircraft (I have actually tried to identify an usual tangent aircraft. based totally in analytic geometry - entirely neglecting the Lagrangian - and also have actually thought of numerous relationships in between directional by-products ... How could directional by-products connect to the optimality problems state by the Lagrangian? )

MODIFY: Thanks Greg Graviton for learning this sub-optimal symbols and also mentioning numerous blunders in the declaration of the trouble. (Additionally, thanks for the excellent discussion below.)

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2019-05-07 01:38:08
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Concerning the physical definition, I take it that $f_1$ and also $f_2$ stand for the portions of both stages in the alloy (this indicates $f_1 + f_2 = 1$ ). I visualize $x_1$ and also $x_2$ to represent an intensional variable like stress, whose ordinary $f_1x_1 + f_2x_2 =: \bar x$ is held constant in the experiment. Currently, the alloy decreases the free power under these restraints.

To address your first inquiry : there is a geometric reason that the remedy is the usual tangent to $G_1$ and also $G_2$ when it comes to 2 measurements. Particularly, the portions $f_1$ and also $f_2$ are specifically the Barycentric coordinates of the ordinary $\bar x$ resting in between $x_1$ and also $x_2$ . Specifically, the value of the complete power $\tilde G$ is the elevation of the line attracted in between $(x_1,G_1(x_1))$ and also $(x_2,G_2(x_2))$ reviewed at $\bar x$ . Below's an illustration :

From this image, it is clear that if this line is not tangent to both $G_1$ and also $G_2$ , after that you can relocate a little bit to make sure that the value at $\bar x$ will certainly decrease.

To address your 2nd inquiry , the geometry conveniently includes greater measurements. As an example, for an alloy of 3 substances, one needs to take into consideration the triangular confined by the 3 factors $(x_1,G_1(x_1))$ , $(x_2,G_2(x_2))$ and also $(x_3,G_3(x_3))$ . The scenario is a little bit degenerate below, any kind of factor inside this triangular whose first coordinate is $\bar x$ stands for a legitimate value of $\tilde G$ . Of these, nature will certainly pick the tiniest one. Subsequently, the lower side of the triangular needs to be tangent to 2 of the specific free powers.

Similar thinking uses when the variable $x$ is not simply a number, yet, claim, a set of numbers, after that we're managing an aircraft tangent to 3 specific Gibbs features.

While not horribly valuable in this instance, there is additionally a really basic geometric analysis of the method of Lagrange multipliers. Particularly, take into consideration an objective function $f$ and also a holonomic restraint $g$ . After that, the Euler - Lagrange - formulas offer $\nabla f = \lambda \nabla g$ which suggests that $f$ adjustments just in instructions orthogonal to the surface area $g$ . Yet given that we're constrained to the surface area $g$ , this have to be an extremum. Wikipedia has a picture.

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2019-05-08 08:49:38
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