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Sidharth Ghoshal
Sidharth Ghoshal
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In one functional variableFinding Square roots:

Then $L^2 = \Omega$. In particular this produces a system of equations $A^2 = 0_n$, $AB + BA = I_n$, $AC+CA=0_n$, $BC+CB=-I_n$ and $C^2 = -f'*I_n$. Where $0_n$ and $I_n$ refer to the 0 and identity matrix over $n \times n$ matrices.

Then $L^2 = \Omega$. In particular this produces a system of equations $A^2 = 0_n$, $AB + BA = I_n$, $AC+CA=0_n$, $BC+CB=-I_n$ and $C^2 = -f'*I_n$. Where $0_n$ and $I_n$ refer to the 0 and identity matrix over $n \times n$ matrices.

I'm not sure when but there definitely exists some $n$ for which this has a solution. And if you wanted to, you could go looking for it, but even with $2\times 2$ matrices this becomes a system of 10 quadratic algebraic equations with 12 unknowns (so very likely solvable but very tedious to solve)

I'm not sure when but there definitely exists anThe interaction of $n$ for which this has a solution. And if you wanted to, you could go looking for it, but even$\frac{\partial}{\partial t}$ with $2\times 2$ matrices$f(t)$ makes this becomes a systemlot more complicated but still probably solvable. Instead of 10$10$ quadratic algebraic equations with 12in $12$ numerical unknowns. You instead have (so very likely solvable but very tedious to solve)$10$ second order PDEs in $12$ functional variable unknowns.

Your particular instance is $q=4$ all of order $1$.

Connection to Diracs Equation:

It's not obvious to me what the relationship between factoring the euler lagrange equations and dirac's equation is. Let me explain why: 1. Dirac's Equations is a square root of a PDE operator (That accepts a function and outputs a function).

The euler lagrange equations are superoperators (that accept an operator and produce an operator).

Your question is in some sense asking what's the relationship between a particular function and a number. Philosophically one might ask what is the relationship between $e^x$ and $e$, but it would be very hard to find coherent and useful ideas explaining the relationship between $e^x$ and say the number $3$.

Similarly one might ask does knowing about the relationships between $\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}, \frac{d}{dx}$ allow you to know anything about $x^{\sqrt{2}}, x^2 $ the latter being a function $g$ s.t. $g^2 = g(g(x)) = x^2$.

You're question in particular lives on the next rung of this tower. You are not comparing a number and a function. You are not comparing a function and an operator. You are comparing a linear operator (dirac's equation) to a superlinear operator (square root of Euler lagrange).

In one functional variable:

Then $L^2 = \Omega$. In particular this produces a system of equations $A^2 = 0_n$, $AB + BA = I_n$, $AC+CA=0_n$, $BC+CB=-I_n$ and $C^2 = -f'*I_n$. Where $0_n$ and $I_n$ refer to the 0 and identity matrix over $n \times n$ matrices.

I'm not sure when but there definitely exists an $n$ for which this has a solution. And if you wanted to, you could go looking for it, but even with $2\times 2$ matrices this becomes a system of 10 quadratic algebraic equations with 12 unknowns (so very likely solvable but very tedious to solve)

Your particular instance is $q=4$ all of order $1$.

Finding Square roots:

Then $L^2 = \Omega$. In particular this produces a system of equations $A^2 = 0_n$, $AB + BA = I_n$, $AC+CA=0_n$, $BC+CB=-I_n$ and $C^2 = -f'*I_n$. Where $0_n$ and $I_n$ refer to the 0 and identity matrix over $n \times n$ matrices.

I'm not sure when but there definitely exists some $n$ for which this has a solution. And if you wanted to, you could go looking for it, but even with $2\times 2$ matrices this becomes a system of 10 quadratic algebraic equations with 12 unknowns (so very likely solvable but very tedious to solve)

The interaction of $\frac{\partial}{\partial t}$ with $f(t)$ makes this a lot more complicated but still probably solvable. Instead of $10$ quadratic equations in $12$ numerical unknowns. You instead have $10$ second order PDEs in $12$ functional variable unknowns.

Your particular instance is $q=4$ all of order $1$.

Connection to Diracs Equation:

It's not obvious to me what the relationship between factoring the euler lagrange equations and dirac's equation is. Let me explain why: 1. Dirac's Equations is a square root of a PDE operator (That accepts a function and outputs a function).

The euler lagrange equations are superoperators (that accept an operator and produce an operator).

Your question is in some sense asking what's the relationship between a particular function and a number. Philosophically one might ask what is the relationship between $e^x$ and $e$, but it would be very hard to find coherent and useful ideas explaining the relationship between $e^x$ and say the number $3$.

Similarly one might ask does knowing about the relationships between $\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}, \frac{d}{dx}$ allow you to know anything about $x^{\sqrt{2}}, x^2 $ the latter being a function $g$ s.t. $g^2 = g(g(x)) = x^2$.

You're question in particular lives on the next rung of this tower. You are not comparing a number and a function. You are not comparing a function and an operator. You are comparing a linear operator (dirac's equation) to a superlinear operator (square root of Euler lagrange).

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Sidharth Ghoshal
Sidharth Ghoshal
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In one functional variable:

These square roots do exist however the solutions can get very complex quickly.

First we recall the general form of the LHS of the Euler-Lagrange equations with standard boundary conditions of an operator depending on variables $t,f,f',...f^{(n)}$

$$ \frac{\partial}{\partial t} - \frac{d}{dt}\left( \frac{\partial }{\partial f} \right) + \frac{d^2}{dt^2}\left( \frac{\partial }{\partial f'} \right) - ... $$

If our operator depends ONLY on $t$ then this is just:

$$ \frac{\partial }{\partial t} $$

And therefore this has an operator square root via the fractional derivative:

Now we wade a bit deeper into the water and consider an operator that depends exclusively on $t,f$ and no higher order terms. The Euler lagrange equations in this case are of the form

$$ \frac{\partial}{\partial t} - \frac{d}{dt}\left( \frac{\partial }{\partial f} \right) $$

Which can be expanded out as:

$$ \Omega = \frac{\partial}{\partial t} - \frac{\partial^2 }{\partial f \partial t} - \frac{\partial^2}{\partial f^2}f' $$

So this is a linear superoperator. It takes a non linear operator (where an operator itself a function that takes in a function and outputs a function) and produces another non linear operator. That also obeys given two operators $O_1, O_2$ we have that $\Omega[O_1 + O_2] = \Omega[O_1] + \Omega[O_2]$.

So one can try to find a matrix of operators $A,B,C$ s.t. given

$$ L = A*I + B \frac{\partial}{\partial t} + C \frac{\partial }{\partial f} $$

Then $L^2 = \Omega$. In particular this produces a system of equations $A^2 = 0_n$, $AB + BA = I_n$, $AC+CA=0_n$, $BC+CB=-I_n$ and $C^2 = -f'*I_n$. Where $0_n$ and $I_n$ refer to the 0 and identity matrix over $n \times n$ matrices.

I'm not sure when but there definitely exists an $n$ for which this has a solution. And if you wanted to, you could go looking for it, but even with $2\times 2$ matrices this becomes a system of 10 quadratic algebraic equations with 12 unknowns (so very likely solvable but very tedious to solve)

Considering higher order euler lagrange dependencies similarly follows. With multiple variables the problem gets even more complex and in general an interesting mathematical question could be given an operator depending on $q$ functional variables $f_1 ... f_q$ of differential order $c_1 ... c_q$ what is the minimum $n$ so that its square root can be taken of the Euler lagrange equations over the $n \times n$ real matrices.

Your particular instance is $q=4$ all of order $1$.