1. Introduction and PreliminariesThis 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 [1] [2] [3] . Given an arbitrary g -valued tensor F α β : R 3 + 1 → g .
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 [4] [5] , or more precisely that
∂ 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 [6] [7] ). However, if one fixes the connection to lie in the temporal gauge A 0 = 0 , the Yang-Mills equations become essentially hyperbolic [8] [9] , and simplify to
∂ 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 [10] . Here in not described in detail. This paper will show that the solution of operator and polynomial type.
2. Exact Solution of EquationBelow 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
3. Polynomial Solutions3.1. First Order Polynomial SolutionBelow 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).
3.2. The Quadratic Polynomial SolutionIn 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.
3.3. Solution of N-Degree PolynomialIn 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.