The Solution of Yang-Mills Equations on the Surface

We show that Yang-Mills equation in 3 dimensions is local well-posedness in s H 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.


Introduction and Preliminaries
This paper is concerned with the solution of the Yang-Mills equation.
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 0 0 A = , the Yang-Mills equations become essentially hyperbolic [8] [9], and simplify to and ( ) div , , , , 0 where , 1, 2, 3 i j = .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.

Exact Solution of Equation
Below we will construct the exact solution of the equation on the general quadric that denotes by , , i i a a x x x = .We bring (3) to Equation (2), because the equation is used in the two general surfaces, we define the general quadric by , , 0,1, 2.c α α α α α α = as coefficient and 1 2 3 c R α α α ∈ .So we calculate the equation.The first calculation can be

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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 First, we consider 1 j = .
The constant coefficient equation is ( ) The coefficient equation of The coefficient equation of Because of the (4), the coefficient equation of constant can be ( ) Deformation by (6), we have where 2 C is a constant.
We can observe the above 1 a and the general properties of two surfaces, 1 a is irrelevant to the 2 x and 3 x , so ( ) Because of ( ) a a x = , we take 1 a into the (11) can be obtain ( ) By two surfaces we can obtain , 0 a a = In summary, when the Equation ( 2) is acting on the quadric, we have

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 A is satisfied the (1) because of not contain time t.Then we just need to bring i A into (2).We have ( ) ( ) Equation ( 12) is composed of three equations.First we consider the case of 1 j = .So the constant coefficient equation is When 3 j = , the relationship of the coefficients are There exist 12 equations.By solving the above equations, we can obtain where , 1, 2, 3 In summary, the solution of the polynomial form of Yang-Mills equation is expressed in the form of (13).

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 , , A A A must satisfy the Equation ( 1), therefore, it just needs to take , , A A A 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 where , 1, 2, 3.
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.

Solution of N-Degree Polynomial
In this section, we mainly use mathematical induction to prove the hypothesis.
We define that by where , , 1, 2, 3 In the front two sections, it is easy for us to conclude that when 1, 2 n = the solutions are the same.So we will use mathematical induction to prove that when 2 n ≥ the solution is also (14).
First, we assume that when ( ) the solution of the equation is To further simplify (15), we have To bring into the equation, we have ∑ where j J is  And the number of more than 1 n + of the items in the n-sub surfaces is al- ways equal to zero.

Conclusion
In summary, we can get the solution of the polynomial type of Yang-Mills equation by mathematical induction is bold character A α , where α ranges over 0, 1, 2, 3. We use the usual summation conventions on α , and raise and lower indices with respect to the Minkowski metric

1 C
11)First, for(9) we can use mathematica to get is a constant, [ ] denotes the arbitrary combination of functions represented as independent variables in square brackets.For example, [ ] b b c b b c c a c c c b c

∑
On the number of x in the above equation is either less than n , or more than 1 n + .When the number of x is less than n , the solution of the equation is