Is there a geometrical explanation for a method using critical points in a multivariable context

107 Views Asked by At

I would like to describe you a technique that I found in a physics paper, for which I cannot give a mathematical interpretation.

Let $f(q,g,e,r)=\frac{r}{2}(g+q) + q^{\frac{p}{2}} e - \frac{T}{2} \ln{g} - \frac{\beta}{4} ( (g+q)^{p} - q^{p} - pgq^{p-1} ) - \frac{r}{2}$ where $T, \beta$ and $p$ are constants.

Then while we compute the stationary (critical) point, from the equation $0=\frac{df}{dr}$ we get that

\begin{equation} g + q -1 =0 \end{equation}, which implies that $g=1-q$.

Then put this formula in the initial formula for $f$. You will get $f=f(q,e) = q^{\frac{p}{2}}e - \frac{T}{2} \ln(1-q) -\frac{\beta}{4} ( 1 - (p-1)q^{p} - pq^{p-1})$

What geometrical explanation does it have ? Why in general would someone do this ? What do we gain ?

2

There are 2 best solutions below

1
On BEST ANSWER

One can get a good understanding of this issue in terms of shadows' limits i.e., limits of projections of a certain surface ; but in a first step, we will need to understand how envelopes of curves (or surfaces, etc.) are involved here.

Let us take an example.

Consider the following expression :

$$f(x,y,z):=(x-z)^2+y^2-z=0. \tag{1}$$

Let us interpret (1) as a family of circles :

$$(x-z)^2+y^2=\sqrt{z}^2$$

with center $(z,0)$ and radius $\sqrt{z}$ (see Fig. 1).

enter image description here

Fig. 1.

$z$ is now considered as a parameter.

If we differentiate (1) with respect to parameter $z$ and equate the result to $0$, we get :

$$-2(x-z)=1 \ \ \iff \ \ z=x+\frac12\tag{2}$$

If we plug (2) into (1), we get :

$$\frac14+y^2-(x+\frac12)=0$$

i.e.

$$y^2=x+\frac14 \ \ \iff \ \ y=\pm \sqrt{x+\frac14},$$

which is the equation of the envelope (the parabola in red on Fig. 1).

Why that ? Because it is known (https://en.wikipedia.org/wiki/Envelope_theorem) that the envelope of a family of curves $f(x,y,m)=0$ depending on a parameter $m$, under differentiability conditions that are fulfilled here, is obtained by eliminating $m$ from the two equations $$\begin{cases}f(x,y,m)&=&0\\\dfrac{\partial f}{\partial m}(x,y,m)&=&0\end{cases}\tag{3}$$

There is more to say. In fact, we can reinterpret Fig. 1 by taking $z$, not as any parameter but as a plain third coordinate ; in this way $f(x,y,z)=0$ is the implicit equation of a certain surface (here a paraboloid with oblique axis) ; the curve we have obtained is thus the boundary of the shadow of the projection on plane $xOy$ of this surface. This is clear from Fig. 2 which is an "elevation" of Fig. 1.

enter image description here

Fig. 2. : The paraboloid with equation (1) materialized by certain level lines (in black) and their projections onto the horizontal plane, giving back Fig. 1.

Remarks :

1) The concept of envelopes of curves and the associated technique (see (3)) for computing them can be extended in a straightforward manner to surfaces which are envelopes of families of surfaces.

2) A simple explanation for the case of straight lines envelopes I gave as an appendix to an answer : Loci of intersection of lines with positive intercepts??

3) Many examples of envelopes can be found on the net :

https://www.math24.net/envelope-family-curves-page-2/

https://people.duke.edu/~dgraham/handouts/EnvelopeTheorem.pdf

etc.

4) Red curve in Figure 1 can be interpreted as the moving "sound wave" of an aircraft.

5) In the example above, out of an equation with 3 variables $x,y,z$, the elimination process has generated an equation with 2 variables $x,y$. This kind of "dimension reduction by 1" is the most usual. In the example you give, one jumps from $4$ to $2$ variables : we are in a case of degeneracy.

6) See a slightly different explanation in gave to a cousin query : https://math.stackexchange.com/q/3095569

7) Related : https://www.math24.net/singular-solutions-differential-equations/

4
On

As far as I can tell, this is an instance of Implicit Function Theorem, where you are obtaining $z$ as a function of $x$ and $y$, say $z = g(x,y)$ so that the function $h(x,y) = f(x,y,g(x,y))$ becomes constant. The function $f$ needs to follow some basic properties for this to happen; essentially $f^{-1}(c)$ should locally look like the graph of a function.