^{1}

^{*}

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

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

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

A α : R 3 + 1 → g by 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 η α β : = diag ( − 1 , 1 , 1 , 1 ) ; for more details, see [

The curvature of a connection F α β by

F α β : = ∂ α A β − ∂ β A α + [ A α , A β ]

Here [,] denotes the Lie bracket of g . It appears in calculations whenever we commute covariant derivatives [

∂ α F α β + [ A α , F α β ] = 0

We can expand this as

□ A β − ∂ β ( ∂ α A β ) + [ A α , ∂ α A β ] − [ A α , ∂ β A α ] + [ A α , [ A α , A β ] ] = 0

where □ : = − ∂ t 2 + Δ , α , β = 0 , 1 , 2 , 3.

The Cauchy problem for Yang-mills equation is not well-posed because of gauge invariance (see [

∂ t ( div A ) + [ A i , ∂ t A i ] = 0 (1)

and

□ A j − ∂ j ( div A ) + + [ A i , ∂ i A j ] − [ A i , ∂ j A i ] + [ A i , [ A i , A j ] ] = 0 (2)

where i , j = 1 , 2 , 3 .

The local well-posedness of the Equations (1) and (2) have already proved in [

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

A i = ∂ x i + a i i = 1 , 2 , 3. (3)

where a i = a i ( x 1 , x 2 , x 3 ) .

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

f = ∑ α 1 + α 2 + α 3 ≤ 2 c α 1 α 2 α 3 x 1 α 1 x 2 α 2 x 3 α 3

α 1 , α 2 , α 3 = 0 , 1 , 2. c α 1 α 2 α 3 as coefficient and c α 1 α 2 α 3 ∈ R . So we calculate the equation. The first calculation can be

( □ A j ) f = ( Δ − ∂ t 2 ) ( ∂ x j + a j ) f = [ Δ ∂ x j + ( Δ a j ) + a j Δ − ∂ t 2 ∂ x j − ∂ 2 a j ∂ t 2 − a j ∂ t 2 ] f = [ ( Δ a j ) + a j Δ ] f = ( Δ a j ) f + 2 a j ( c 200 + c 020 + c 002 )

Divergence terms can be

[ ∂ j ( div A ) ] f = [ ∂ x j ( ∂ x 1 ∂ x 1 + a 1 ∂ x 1 + ∂ a 1 ∂ x 1 + ∂ x 2 ∂ x 2 + a 2 ∂ x 2 + ∂ a 2 ∂ x 2 + ∂ x 3 ∂ x 3 + a 3 ∂ x 3 + ∂ a 3 ∂ x 3 ) ] f = [ a 1 ∂ x j ∂ x 1 + ∂ a 1 ∂ x j ∂ x 1 + ∂ 2 a 1 ∂ x j ∂ x 1 + ∂ a 1 ∂ x 1 ∂ x j + ∂ a 2 ∂ x j ∂ x 2 + a 2 ∂ x j ∂ x 2 + ∂ 2 a 2 ∂ x j ∂ x 2 + ∂ a 2 ∂ x 2 ∂ x j + ∂ a 3 ∂ x j ∂ x 3 + a 3 ∂ x j ∂ x 3 + ∂ 2 a 3 ∂ x j ∂ x 3 + ∂ a 3 ∂ x 3 ∂ x j ] f = ( ∂ a 1 ∂ x j + a 1 ∂ x j ) ( c 100 + c 110 x 2 + c 101 x 3 + 2 c 200 x 1 ) + ( ∂ a 2 ∂ x j + a 2 ∂ x j ) ⋅ ( c 010 + c 110 x 1 + c 011 x 3 + 2 c 020 x 2 ) + ( ∂ a 3 ∂ x j + a 3 ∂ x j ) ( c 001 + c 101 x 1 + c 011 x 2 + 2 c 002 x 3 ) + ( ∂ 2 a 1 ∂ x j ∂ x 1 + ∂ a 1 ∂ x 1 ∂ x j + ∂ 2 a 2 ∂ x j ∂ x 2 + ∂ a 2 ∂ x 2 ∂ x j + ∂ 2 a 3 ∂ x j ∂ x 3 + ∂ a 3 ∂ x 3 ∂ x j ) f

Finally, the sections of Lie bracket can be

[ A i , ∂ i A j ] f = [ ( A i ⋅ ∂ i A j − ∂ i A j ⋅ A i ) ] f = [ ( ∂ x i + a i ) ( ∂ x j ∂ x i + ∂ a j ∂ x i + a j ∂ x i ) − ( ∂ x j ∂ x i + ∂ a j ∂ x i + a j ∂ x i ) ( ∂ x i + a i ) ] f = ( ∂ 2 a j ∂ x i ∂ x i + ∂ a j ∂ x i ∂ x i − ∂ 2 a i ∂ x j ∂ x i − ∂ a i ∂ x i ∂ x j − ∂ a i ∂ x j ∂ x i − a j ∂ a i ∂ x i ) f = ( ∂ 2 a j ∂ x i ∂ x i − a j ∂ a i ∂ x i − ∂ a i ∂ x i ∂ x j ) f + ( ∂ a j ∂ x 1 + ∂ a 1 ∂ x 1 ) ( c 100 + c 110 x 2 + c 101 x 3 + 2 c 200 x 1 ) = ( ∂ a j ∂ x 2 + ∂ a 2 ∂ x 2 ) ( c 010 + c 110 x 1 + c 011 x 3 + 2 c 020 x 2 ) + ( ∂ a j ∂ x 3 + ∂ a 3 ∂ x 3 ) ( c 001 + c 101 x 1 + c 011 x 2 + 2 c 002 x 3 )

[ A i , ∂ j A i ] f = [ ( A i ⋅ ∂ j A i − ∂ j A i ⋅ A i ) ] f = [ ( ∂ x i + a i ) ( ∂ x j ∂ x i + ∂ a i ∂ x j + a i ∂ x j ) − ( ∂ x j ∂ x i + ∂ a i ∂ x j + a i ∂ x j ) ( ∂ x i + a i ) ] f = − ( ∂ a i ∂ x j ∂ x i + a i ∂ a i ∂ x j ) f = − [ ∂ a 1 ∂ x j ( c 100 + c 110 x 2 + c 101 x 3 + 2 c 200 x 1 ) + ∂ a 2 ∂ x j ( c 010 + c 110 x 1 + c 011 x 3 + 2 c 020 x 2 ) + ∂ a 3 ∂ x j ( c 001 + c 101 x 1 + c 011 x 2 + 2 c 002 x 3 ) + ( a 1 ∂ a 1 ∂ x j + a 2 ∂ a 2 ∂ x j + a 3 ∂ a 3 ∂ x j ) f ]

[ A i , [ A i , A j ] ] f = [ A i , A i A j − A j A i ] f = [ A i , ( ∂ x i + a i ) ( ∂ x j + a j ) − ( ∂ x j + a j ) ( ∂ x i + a i ) ] f = [ A i , ∂ a j ∂ x i − ∂ a i ∂ x j ] f = [ ( ∂ x i + a i ) ( ∂ a j ∂ x i − ∂ a i ∂ x j ) − ( ∂ a j ∂ x i − ∂ a i ∂ x j ) ( ∂ x i + a i ) ] f = ( ∂ 2 a j ∂ x i ∂ x i − ∂ 2 a i ∂ x j ∂ x i ) f

Combining the above calculations we have

2 a j ( c 200 + c 020 + c 002 ) + [ Δ a j − ∂ 2 a j ∂ t 2 + 2 ∂ 2 a j ∂ x i 2 − 2 ( ∂ 2 a 1 ∂ x j ∂ x 1 + ∂ 2 a 2 ∂ x j ∂ x 2 + ∂ 2 a 3 ∂ x j ∂ x 3 ) + a j ( ∂ a 1 ∂ x 1 + ∂ a 2 ∂ x 2 + ∂ a 3 ∂ x 3 ) + a 1 ∂ a 1 ∂ x j + a 2 ∂ a 2 ∂ x j + a 3 ∂ a 3 ∂ x j ] f + ∂ a j ∂ x 1 ( c 100 + c 110 x 2 + c 101 x 3 + 2 c 200 x 1 ) + ∂ a j ∂ x 2 ( c 010 + c 110 x 1 + c 011 x 3 + 2 c 020 x 2 ) + ∂ a j ∂ x 3 ( c 001 + c 101 x 1 + c 011 x 2 + 2 c 002 x 3 ) − ( ∂ a 1 ∂ x 1 ∂ x j + ∂ a 2 ∂ x 2 ∂ x j + ∂ a 3 ∂ x 3 ∂ x j ) f = 0

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

Δ a j − ∂ 2 a j ∂ t 2 + 2 ∂ 2 a j ∂ x i 2 − 2 ( ∂ 2 a 1 ∂ x j ∂ x 1 + ∂ 2 a 2 ∂ x j ∂ x 2 + ∂ 2 a 3 ∂ x j ∂ x 3 ) + a j ( ∂ a 1 ∂ x 1 + ∂ a 2 ∂ x 2 + ∂ a 3 ∂ x 3 ) + a 1 ∂ a 1 ∂ x j + a 2 ∂ a 2 ∂ x j + a 3 ∂ a 3 ∂ x j = 0 (4)

First, we consider j = 1 .

The constant coefficient equation is

2 a 1 ( c 200 + c 020 + c 002 ) + Δ a 1 − ∂ 2 a 1 ∂ t 2 + 2 ∂ 2 a 1 ∂ x i 2 − 2 ( ∂ 2 a 1 ∂ x 1 ∂ x 1 + ∂ 2 a 2 ∂ x 1 ∂ x 2 + ∂ 2 a 3 ∂ x 1 ∂ x 3 ) + a 1 ( ∂ a 1 ∂ x 1 + ∂ a 2 ∂ x 2 + ∂ a 3 ∂ x 3 ) + a 1 ∂ a 1 ∂ x 1 + a 2 ∂ a 2 ∂ x 1 + a 3 ∂ a 3 ∂ x 1 + ∂ a 1 ∂ x 1 c 100 + ∂ a 2 ∂ x 2 c 010 + ∂ a 3 ∂ x 3 c 001 − ( ∂ a 1 ∂ x 1 + ∂ a 2 ∂ x 2 + ∂ a 3 ∂ x 3 ) c 100 = 0

The coefficient equation of x 1 is

∂ a 1 ∂ x 2 c 110 + ∂ a 1 ∂ x 3 c 101 − 2 ∂ a 2 ∂ x 2 c 200 − 2 ∂ a 3 ∂ x 3 c 200 = 0 (5)

The coefficient equation of x 2 is

2 ∂ a 1 ∂ x 2 c 020 + ∂ a 1 ∂ x 3 c 011 − ∂ a 2 ∂ x 2 c 110 − ∂ a 3 ∂ x 3 c 110 = 0 (6)

The coefficient equation of x 3 is

∂ a 1 ∂ x 2 c 011 + 2 ∂ a 1 ∂ x 3 c 002 − ∂ a 2 ∂ x 2 c 101 − ∂ a 3 ∂ x 3 c 101 = 0 (7)

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

2 a 1 ( c 200 + c 020 + c 002 ) + ∂ a 1 ∂ x 2 c 010 + ∂ a 1 ∂ x 3 c 001 − ∂ a 2 ∂ x 2 c 100 − ∂ a 3 ∂ x 3 c 100 = 0 (8)

( 6 ) × c 110 × c 101 − ( 7 ) × c 200 × c 101 − ( 8 ) × c 200 × c 110 we have

∂ a 1 ∂ x 2 ( c 110 c 110 c 101 − 2 c 020 c 200 c 101 − c 011 c 200 c 110 ) + ∂ a 1 ∂ x 3 ( c 101 c 110 c 101 − c 011 c 200 c 101 − 2 c 020 c 011 c 200 ) = 0 (9)

Deformation by (6), we have

∂ a 2 ∂ x 2 + ∂ a 3 ∂ x 3 = 2 ∂ a 1 ∂ x 2 c 020 + ∂ a 1 ∂ x 3 c 011 / c 110 (10)

Simulaneous (8) and (10), we have

2 a 1 ( c 200 + c 020 + c 002 ) + ∂ a 1 ∂ x 2 ( c 010 − 2 c 020 c 100 c 110 ) + ∂ a 1 ∂ x 3 ( c 001 − c 011 c 100 c 110 ) = 0 (11)

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

a 1 = C 1 [ x 1 ] [ − ( c 101 c 110 c 101 − c 011 c 200 c 101 − 2 c 020 c 011 c 200 ) x 2 + ( c 110 c 110 c 101 − 2 c 020 c 200 c 101 − c 011 c 200 c 110 ) x 3 c 110 c 110 c 101 − 2 c 020 c 200 c 101 − c 011 c 200 c 110 ]

where C 1 is a constant, [ ] denotes the arbitrary combination of functions represented as independent variables in square brackets. For example, [ x ] is represented as x sin x or e x ln x cos x and so on.

Next, from (11) we can obtain

a 1 = C 2 e 2 ( c 200 + c 020 + c 002 ) x 2 c 010 − 2 c 020 c 100 / c 110 [ x 1 ] [ − ( c 001 − c 011 c 100 / c 110 ) x 2 + ( c 010 − 2 c 020 c 100 / c 110 ) x 3 c 010 − 2 c 020 c 100 / c 110 ]

where C 2 is a constant.

We can observe the above a 1 and the general properties of two surfaces, a 1 is irrelevant to the x 2 and x 3 , so a 1 = a 1 ( x 1 ) .

Because of a 1 = a 1 ( x 1 ) , we take a 1 into the (11) can be obtain

2 a 1 ( c 200 + c 020 + c 002 ) = 0

By two surfaces we can obtain

a 1 = 0

Similarly, we can prove that j = 2 , 3 , we have

a 2 , a 3 = 0

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

{ A 1 = ∂ x 1 A 2 = ∂ x 2 A 3 = ∂ x 3

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 i = a i x 1 + b i x 2 + c i x 3 + d i

where i = 1 , 2 , 3 , A i is satisfied the (1) because of not contain time t. Then we just need to bring A i into (2). We have

Δ A j − ∂ j ( div A ) + ∑ i = 1 3 ( ∂ j A i A i − A j ∂ i A i ) = 0 (12)

Equation (12) is composed of three equations. First we consider the case of j = 1 . So the constant coefficient equation is

b 2 d 2 − b 2 d 1 + a 3 d 3 − c 3 d 1 = 0

The coefficient equation of x 1 is

a 2 b 2 − a 1 b 2 + a 3 a 3 − a 1 c 3 = 0

The coefficient equation of x 2 is

b 2 b 2 − b 1 b 2 + a 3 b 3 − b 1 c 3 = 0

The coefficient equation of x 3 is

b 2 c 2 − b 2 c 1 + a 3 c 3 − c 1 c 3 = 0

When j = 2 , the relationship of the coefficients are

{ b 1 d 1 − a 1 d 2 + c 3 d 3 − c 3 d 2 = 0 a 1 b 1 − a 1 a 2 + a 3 c 3 − a 2 c 3 = 0 b 1 b 1 − a 1 b 2 + b 3 c 3 − b 2 c 3 = 0 b 1 c 1 − a 1 c 2 + c 3 c 3 − c 2 c 3 = 0

When j = 3 , the relationship of the coefficients are

{ c 1 d 1 − a 1 d 3 + c 2 d 2 − b 2 d 3 = 0 a 1 c 1 − a 1 a 3 + a 2 c 2 − a 2 b 3 = 0 b 1 c 1 − a 1 b 3 + b 2 c 2 − b 2 b 3 = 0 c 1 c 1 − a 1 c 3 + c 2 c 2 − b 2 c 3 = 0

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

a 1 = a 2 = a 3 = b 1 = b 2 = b 3 = c 1 = c 2 = c 3

d 1 = d 2 = d 3

Therefore

A i = a x 1 + a x 2 + a x 3 + b (13)

where a , b ∈ R i = 1 , 2 , 3 .

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

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 1 = ∑ α 1 + α 2 + α 3 ≤ 2 a α 1 α 2 α 3 x 1 α 1 x 2 α 2 x 3 α 3 A 2 = ∑ β 1 + β 2 + β 3 ≤ 2 b β 1 β 2 β 3 x 1 β 1 x 2 β 2 x 3 β 3 A 3 = ∑ γ 1 + γ 2 + γ 3 ≤ 2 c γ 1 γ 2 γ 3 x 1 γ 1 x 2 γ 2 x 3 γ 3

where α i , β i , γ i ∈ ℕ i = 1 , 2 , 3 , a α 1 α 2 α 3 , b β 1 β 2 β 3 , c γ 1 γ 2 γ 3 ∈ R are coefficients. So

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

A 1 , A 2 , A 3 into (12), we have

Δ A j − ∂ x j x 1 ( ∑ α 1 + α 2 + α 3 ≤ 2 a α 1 α 2 α 3 x 1 α 1 x 2 α 2 x 3 α 3 ) − ∂ x j x 2 ( ∑ β 1 + β 2 + β 3 ≤ 2 b β 1 β 2 β 3 x 1 β 1 x 2 β 2 x 3 β 3 ) − ∂ x j x 3 ( ∑ γ 1 + γ 2 + γ 3 ≤ 2 c γ 1 γ 2 γ 3 x 1 γ 1 x 2 γ 2 x 3 γ 3 ) + [ ∂ x j ( ∑ α 1 + α 2 + α 3 ≤ 2 a α 1 α 2 α 3 x 1 α 1 x 2 α 2 x 3 α 3 ) ] ⋅ ∑ α 1 + α 2 + α 3 ≤ 2 a α 1 α 2 α 3 x 1 α 1 x 2 α 2 x 3 α 3 − A j ⋅ [ ∂ x 1 ( ∑ α 1 + α 2 + α 3 ≤ 2 a α 1 α 2 α 3 x 1 α 1 x 2 α 2 x 3 α 3 ) ] + [ ∂ x j ( ∑ β 1 + β 2 + β 3 ≤ 2 b β 1 β 2 β 3 x 1 β 1 x 2 β 2 x 3 β 3 ) ] ⋅ ∑ β 1 + β 2 + β 3 ≤ 2 b β 1 β 2 β 3 x 1 β 1 x 2 β 2 x 3 β 3

− A j ⋅ [ ∂ x 2 ( ∑ β 1 + β 2 + β 3 ≤ 2 b β 1 β 2 β 3 x 1 β 1 x 2 β 2 x 3 β 3 ) ] + [ ∂ x j ( ∑ γ 1 + γ 2 + γ 3 ≤ 2 c γ 1 γ 2 γ 3 x 1 γ 1 x 2 γ 2 x 3 γ 3 ) ] ⋅ ∑ γ 1 + γ 2 + γ 3 ≤ 2 c γ 1 γ 2 γ 3 x 1 γ 1 x 2 γ 2 x 3 γ 3 − A j ⋅ [ ∂ x 3 ( ∑ γ 1 + γ 2 + γ 3 ≤ 2 c γ 1 γ 2 γ 3 x 1 γ 1 x 2 γ 2 x 3 γ 3 ) ] = 0

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

a 200 = a 020 = a 002 = a 110 = a 101 = a 011 = 0

b 200 = b 020 = b 002 = b 110 = b 101 = b 011 = 0

c 200 = c 020 = c 002 = c 110 = c 101 = c 011 = 0

a 100 = a 010 = a 011 = b 100 = b 010 = b 001 = c 100 = c 010 = c 001 ∈ ℝ

a 000 = b 000 = c 000 ∈ ℝ

So the solution of the equation can be written

A i = a x 1 + a x 2 + a x 3 + b (14)

where a , b ∈ ℝ i = 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.

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

{ A 1 = ∑ α 1 + α 2 + α 3 ≤ n a α 1 α 2 α 3 x 1 α 1 x 2 α 2 x 3 α 3 A 2 = ∑ β 1 + β 2 + β 3 ≤ n b β 1 β 2 β 3 x 1 β 1 x 2 β 2 x 3 β 3 A 3 = ∑ γ 1 + γ 2 + γ 3 ≤ n c γ 1 γ 2 γ 3 x 1 γ 1 x 2 γ 2 x 3 γ 3 (15)

where α i , β i , γ i ∈ ℕ i = 1 , 2 , 3 , a α , b β , c γ ∈ R are coefficients.

In the front two sections, it is easy for us to conclude that when n = 1 , 2 the solutions are the same. So we will use mathematical induction to prove that when n ≥ 2 the solution is also (14).

First, we assume that when n ( n ≥ 2 ) the solution of the equation is

A i = a x 1 + a x 2 + a x 3 + b

where a , b ∈ ℝ i = 1 , 2 , 3.

Now when n + 1 , we have

Δ A j − ∂ x j x 1 ( ∑ α 1 + α 2 + α 3 ≤ n + 1 a α 1 α 2 α 3 x 1 α 1 x 2 α 2 x 3 α 3 ) − ∂ x j x 2 ( ∑ β 1 + β 2 + β 3 ≤ n + 1 b β 1 β 2 β 3 x 1 β 1 x 2 β 2 x 3 β 3 ) − ∂ x j x 3 ( ∑ γ 1 + γ 2 + γ 3 ≤ n + 1 c γ 1 γ 2 γ 3 x 1 γ 1 x 2 γ 2 x 3 γ 3 ) + [ ∂ x j ( ∑ α 1 + α 2 + α 3 ≤ 2 a α 1 α 2 α 3 x 1 α 1 x 2 α 2 x 3 α 3 ) ] ⋅ ∑ α 1 + α 2 + α 3 ≤ n + 1 a α 1 α 2 α 3 x 1 α 1 x 2 α 2 x 3 α 3

− A j ⋅ [ ∂ x 1 ( ∑ α 1 + α 2 + α 3 ≤ n + 1 a α 1 α 2 α 3 x 1 α 1 x 2 α 2 x 3 α 3 ) ] + [ ∂ x j ( ∑ β 1 + β 2 + β 3 ≤ 2 b β 1 β 2 β 3 x 1 β 1 x 2 β 2 x 3 β 3 ) ] ⋅ ∑ β 1 + β 2 + β 3 ≤ n + 1 b β 1 β 2 β 3 x 1 β 1 x 2 β 2 x 3 β 3 − A j ⋅ [ ∂ x 2 ( ∑ β 1 + β 2 + β 3 ≤ n + 1 b β 1 β 2 β 3 x 1 β 1 x 2 β 2 x 3 β 3 ) ] + [ ∂ x j ( ∑ γ 1 + γ 2 + γ 3 ≤ 2 c γ 1 γ 2 γ 3 x 1 γ 1 x 2 γ 2 x 3 γ 3 ) ] ⋅ ∑ γ 1 + γ 2 + γ 3 ≤ n + 1 c γ 1 γ 2 γ 3 x 1 γ 1 x 2 γ 2 x 3 γ 3 − A j ⋅ [ ∂ x 3 ( ∑ γ 1 + γ 2 + γ 3 ≤ n + 1 c γ 1 γ 2 γ 3 x 1 γ 1 x 2 γ 2 x 3 γ 3 ) ] = 0

To further simplify (15), we have

{ A 1 = ∑ α 1 + α 2 + α 3 ≤ n a α 1 α 2 α 3 x 1 α 1 x 2 α 2 x 3 α 3 + ∑ α 1 + α 2 + α 3 = n + 1 a α 1 α 2 α 3 x 1 α 1 x 2 α 2 x 3 α 3 A 2 = ∑ β 1 + β 2 + β 3 ≤ n b β 1 β 2 β 3 x 1 β 1 x 2 β 2 x 3 β 3 + ∑ β 1 + β 2 + β 3 = n + 1 b β 1 β 2 β 3 x 1 β 1 x 2 β 2 x 3 β 3 A 3 = ∑ γ 1 + γ 2 + γ 3 ≤ n c γ 1 γ 2 γ 3 x 1 γ 1 x 2 γ 2 x 3 γ 3 + ∑ γ 1 + γ 2 + γ 3 = n + 1 c γ 1 γ 2 γ 3 x 1 γ 1 x 2 γ 2 x 3 γ 3

To bring into the equation, we have

where J j is

{ J 1 = ∑ α 1 + α 2 + α 3 = n + 1 a α 1 α 2 α 3 x 1 α 1 x 2 α 2 x 3 α 3 J 2 = ∑ β 1 + β 2 + β 3 = n + 1 b β 1 β 2 β 3 x 1 β 1 x 2 β 2 x 3 β 3 J 3 = ∑ γ 1 + γ 2 + γ 3 = n + 1 c γ 1 γ 2 γ 3 x 1 γ 1 x 2 γ 2 x 3 γ 3

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

A i = a x 1 + a x 2 + a x 3 + b

And the number of more than n + 1 of the items in the n-sub surfaces is always equal to zero.

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

A i = a x 1 + a x 2 + a x 3 + b

where a , b ∈ ℝ i = 1 , 2 , 3.

Zhu, P. and Ding, L.Y. (2017) The Solution of Yang-Mills Equations on the Surface. Applied Mathematics, 8, 35-43. http://dx.doi.org/10.4236/am.2017.81004