## On Patch by Patch Products - (RWLA,MCT,GT,AM Part III)

In my previous post, I described the concept of a "patchix" and of a special kind, the "patch." I described how to multiply a continuous function on $[0,1]$ by a patch(ix). Today I want to talk about how to multiply a patch by a patch and certain properties of it.

In my description in my informal paper, I basically said that in order to (right) multiply a patchix by a patchix, say $p(x,y)$ with $q(x,y)$, we would have to send $p(x,y) \rightsquigarrow p(1-y,t)$ and then integrate as:

$r(x,t) = \int_0^1 p(1-y,t) \cdot q(x,y) dy \rightsquigarrow r(x,y)$

If the patchix is furthermore special, so that both $\int_0^1 q(x,y) dx = \int_0^1 p(x,y) dx = u(y) = 1$, the uniform distribution on [0,1] (so that $u$ of any fiber is 1), then $p(x,y)$ and $q(x,y)$ are "patches." I want to show that, when we "patchix multiply" two patches we obtain another one. Here's why: assume then $p(x,y)$ and $q(x,y)$ are "patches." Then the resultant $r(x,t)$ is a patch too if $\int_0^1 r(x,t) dx = u(t) = 1$. Thus:

$\int_0^1 r(x,t) dx = \int_0^1 \int_0^1 p(1-y,t) \cdot q(x,y) dy dx$

If there is no issue with absolute convergence of the integrals (as there shouldn't be in a patch), by the Fubini theorem we can exchange the order of integration:

$\int_0^1 \int_0^1 p(1-y,t) \cdot q(x,y) dx dy = \int_0^1 p(1-y,t) \int_0^1 q(x,y) dx dy$

The inner integral evaluates to $u(y) = 1$ by hypothesis. Then we have $\int_0^1 p(1-y,t) dy = u(t) = 1$ because the transformation $x \rightsquigarrow 1-y$ changes the orientation so that the direction of integrating to 1 now changes to be in the $y$ direction ($dx \rightsquigarrow -dy$). Nicely, we have just proven closure of patches.

Because this strongly suggests that patches may form a group (as may patchixes with other properties), I want to attempt to show associativity, identity and inverses of patches in my next post (and of other patchixes with particular properties).

For now, I'm a little more interested in solving a concrete example by calculating self-powers. In my last post, I constructed the following patch:

$p(x,y) = x^2 y^3 + x \left( 2 - \frac{2 y^3}{3} \right)$

To calculate the second power, send $p(x,y) \rightsquigarrow p(1-y,t)$. I get, in expanded form, from my calculator:

$p(1-y,t) = t^3 y^2 - \frac{4 t^3 y}{3} + \frac{t^3}{3} - 2y + 2$

so that

$p_2(x,t) = \int_0^1 p(1-y,t) \cdot p(x,y) dy = \frac{29 x}{15} + \frac{t^3 x}{90} + \frac{x^2}{10} - \frac{t^3 x^2}{60}$

Last, let's send $p_2(x,t) \rightsquigarrow p_2(x,y) = \frac{29 x}{15} + \frac{y^3 x}{90} + \frac{x^2}{10} - \frac{y^3 x^2}{60}$

We can corroborate that this is a patch by integrating

$\int_0^1 p_2(x,y) dx = \int_0^1 \frac{29 x}{15} + \frac{y^3 x}{90} + \frac{x^2}{10} - \frac{y^3 x^2}{60} dx = 1$ which is indeed the case.

To calculate the third power, we can:

$p_3(x,t) = \int_0^1 p(1-y,t) \cdot p_2(x,y) dy = \frac{1741 x}{900} - \frac{t^3 x}{5400} + \frac{59 x^2}{600} + \frac{t^3 x^2}{3600}$

Then, send $p_3(x,t) \rightsquigarrow p_3(x,y) = \frac{1741 x}{900} - \frac{y^3 x}{5400} + \frac{59 x^2}{600} + \frac{y^3 x^2}{3600}$

Again, we can corroborate that this is a patch by

$\int_0^1 p_2(x,y) dx = \int_0^1 \frac{1741 x}{900} - \frac{y^3 x}{5400} + \frac{59 x^2}{600} + \frac{y^3 x^2}{3600} dx = 1$

which is the case.

Here's a countour 3D-plot of $p(x,y), p_2(x,y) \ldots p_7(x,y)$: in other words, the 7-step time evolution of the patch (the "brane").  By looking at the plot, you can probably begin to tell where I'm trying to get at:  the patch evolution shows how a fluid could evolve in time (its movement, oscillation), provided the appropriate first-patch generator can be found for a particular movement.  The fact that, if patches mirror Markov "thinking", a patch that will eventually settle to its long-term stable distribution means that this (patch) treatment, when applied to the physical world, takes into account some sort of "entropy," or loss of energy of the system.  Also some sort of "viscosity," is my belief.  The patch evolution catches nicely and inherently all relevant physical properties.  I will continue to explore this in my next post, I think.

The above image has been scaled differently for the different functions so that they can be better seen as they converge.  In my next post, I would like to expound on the evolution of the following first-patch: