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The Solution of Yang-Mills Equations on the Surface

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^{s}if the norm is sufficiently. Here, we construct a solution on the quadric that is independent of the time. And we also construct a solution of the polynomial form. In the process of solving, the polynomial is used to solve the problem before solving.

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1. Introduction and Preliminaries

This paper is concerned with the solution of the Yang-Mills equation.

We shall denote -valued tensors define on Minkowski space-time

by bold character, where ranges over 0, 1, 2, 3. We use the usual summation conventions on, and raise and lower indices with respect to the Minkowski metric; for more details, see [1] [2] [3] . Given an arbitrary -valued tensor.

The curvature of a connection by

Here [,] denotes the Lie bracket of. It appears in calculations whenever we commute covariant derivatives [4] [5] , or more precisely that

We can expand this as

where

The Cauchy problem for Yang-mills equation is not well-posed because of gauge invariance (see [6] [7] ). However, if one fixes the connection to lie in the temporal gauge, the Yang-Mills equations become essentially hyperbolic [8] [9] , and simplify to

(1)

and

(2)

where.

The local well-posedness of the Equations (1) and (2) have already proved in [10] . Here in not described in detail. This paper will show that the solution of operator and polynomial type.

2. Exact Solution of Equation

Below we will construct the exact solution of the equation on the general quadric that denotes by

(3)

where.

We bring (3) to Equation (2), because the equation is used in the two general surfaces, we define the general quadric by

as coefficient and. So we calculate the equation. The first calculation can be

Divergence terms can be

Finally, the sections of Lie bracket can be

Combining the above calculations we have

We will use the properties of polynomials to list the coefficient equations in order to solve the (3). For the cross terms and square terms coefficient, we have

(4)

First, we consider.

The constant coefficient equation is

The coefficient equation of is

(5)

The coefficient equation of is

(6)

The coefficient equation of is

(7)

Because of the (4), the coefficient equation of constant can be

(8)

we have

(9)

Deformation by (6), we have

(10)

Simulaneous (8) and (10), we have

(11)

First, for (9) we can use mathematica to get

where is a constant, denotes the arbitrary combination of functions represented as independent variables in square brackets. For example, is represented as or and so on.

Next, from (11) we can obtain

where is a constant.

We can observe the above and the general properties of two surfaces, is irrelevant to the and, so.

Because of, we take into the (11) can be obtain

By two surfaces we can obtain

Similarly, we can prove that, we have

In summary, when the Equation (2) is acting on the quadric, we have

3. Polynomial Solutions

3.1. First Order Polynomial Solution

Below we construct a polynomial solution. First, the constant must satisfy the equation so that all constant are the solutions of the Equation (1) and (2). Then we define the solution of a polynomial form on a surface by

where, is satisfied the (1) because of not contain time t. Then we just need to bring into (2). We have

(12)

Equation (12) is composed of three equations. First we consider the case of. So the constant coefficient equation is

The coefficient equation of is

The coefficient equation of is

The coefficient equation of is

When, the relationship of the coefficients are

When, the relationship of the coefficients are

There exist 12 equations. By solving the above equations, we can obtain

Therefore

(13)

where.

In summary, the solution of the polynomial form of Yang-Mills equation is expressed in the form of (13).

3.2. The Quadratic Polynomial Solution

In this section, we mainly discuss the solution of the quadratic polynomial form of the Yang-Mills equation on the two surfaces. We define by

where, are coefficients. So

must satisfy the Equation (1), therefore, it just needs to take

into (12), we have

There exist 30 equations and 30 unknowns. Solving the equations we can obtain the following results

So the solution of the equation can be written

(14)

where

In summary, the solution of the quadratic polynomial form of Yang-Mills equation is (14). It obvious that (13) is equal to (14). So we conjecture that the solution of n-degree polynomial on n-sub surface is also (14). In the next section, we will proof the hypothesis.

3.3. Solution of N-Degree Polynomial

In this section, we mainly use mathematical induction to prove the hypothesis. We define that by

(15)

where, are coefficients.

In the front two sections, it is easy for us to conclude that when the solutions are the same. So we will use mathematical induction to prove that when the solution is also (14).

First, we assume that when the solution of the equation is

where

Now when, we have

To further simplify (15), we have

To bring into the equation, we have

where is

On the number of in the above equation is either less than, or more than. When the number of is less than, the solution of the equation is

And the number of more than of the items in the n-sub surfaces is always equal to zero.

4. Conclusion

In summary, we can get the solution of the polynomial type of Yang-Mills equation by mathematical induction is

where

Cite this paper

*Applied Mathematics*,

**8**, 35-43. doi: 10.4236/am.2017.81004.

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