Greetings!

Its' exciting that chess world championship has just started! first game was a juggernaut, it took 7 hours. Just about first 40 moves completed, mistakes has been done by both sides, especially by Carlsen. He was way ahead but due to time trouble he couldn't find the correct moves. (he played too conservatively, he didnt want to do mistake)

Let's do some math,

Here is the question: How do you find the area ofe complex boundary?

$\underset{0}{\overset{L}{\int \; <!--\; integral\; -->}}\underset{q\left(x\right)}{\overset{p\left(x\right)}{\int \; <!--\; integral\; -->}}\mathrm{dx}\mathrm{dy}=?$

defined on a complex domain such as

It's the basic form of the problem I had to deal with when I tried to solve Schroidinger Eqn in 2D. Matrix element of spatial part of S. E is in the form above. This is standard change of variable of an integral in 2D

One way of solving s.E. eqn is to expand solution in terms of a complete basis. But it's not obvious what basis to use in a complex geometry Iike above. If we had a simple geometry like this:

its in this geometry to expand solution into a basis possible. Simple expansion in to sin and cos in × andy

I propose following transform:

And the jacobian of transformation

$\begin{array}{}v=\frac{y-q\left(x\right)}{p\left(x\right)-q\left(x\right)}=\frac{y-q\left(x\right)}{J\left(x\right)}\\ u=x\end{array}$

$\det \left(\begin{array}{cc}\frac{\partial \; <!--\; partial\; differential\; -->x}{\partial \; <!--\; partial\; differential\; -->u}& \frac{\partial \; <!--\; partial\; differential\; -->x}{\partial \; <!--\; partial\; differential\; -->v}\\ \frac{\partial \; <!--\; partial\; differential\; -->y}{\partial \; <!--\; partial\; differential\; -->u}& \frac{\partial \; <!--\; partial\; differential\; -->y}{\partial \; <!--\; partial\; differential\; -->v}\end{array}\right)=\left(\begin{array}{cc}1& 0\\ vj\text{'}+q\text{'}& J\end{array}\right)=J$

Therefore integral becomes

$\underset{R}{\int \; <!--\; integral\; -->}\int \; <!--\; integral\; -->\mathrm{dx}\mathrm{dy}=\underset{R^1}{\int \; <!--\; integral\; -->}\int \; <!--\; integral\; -->J·\; <!--\; middle\; dot\; -->dudv=\underset{0}{\overset{L}{\int \; <!--\; integral\; -->}}\underset{0}{\overset{1}{\int \; <!--\; integral\; -->}}J\left(u_1\right)·\; <!--\; middle\; dot\; -->dudv=\underset{0}{\overset{L}{\int \; <!--\; integral\; -->}}\underset{0}{\overset{1}{\int \; <!--\; integral\; -->}}p\left(u\right)-q\left(u\right)dudv$

For example,

$\begin{array}{}p\left(x\right)=1+x\\ Q\left(x\right)=0\\ L=2\end{array}$

$\begin{array}{}\underset{2}{\iint \; <!--\; double\; integral\; -->}\mathrm{dx}\mathrm{dy}=\underset{0}{\overset{2}{\int \; <!--\; integral\; -->}}\underset{0}{\overset{1}{\int \; <!--\; integral\; -->}}\left(1+u\right)dudv\\ A={\left(u+\frac{u^2}{2}\right)}_0^1\times \; <!--\; multiplication\; sign\; -->2=3\end{array}$

Indeed;

I will talk about solving PDE's by this method later blogs.