The global existence of the smoothing solution for the Navier-Stokes equations
aa r X i v : . [ m a t h . A P ] J u l The global existence of the smoothing solutionfor the Navier-Stokes equations
Wang Jianfeng
Department of Mathematics, Nanjing University, Nanjing, 210093,P. R. China.
Abstract.
This paper discussed the global existence of the smoothing solution for the Navier-Stokes equations. At first, we construct the theory of the linear equations which is about theunknown four variables functions with constant coefficients. Secondly, we use this theoryto convert the Navier-Stokes equations into the simultaneous of the first order linear partialdifferential equations with constant coefficients and the quadratic equations. Thirdly, we usethe Fourier transformation to convert the first order linear partial differential equations withconstant coefficients into the linear equations, and we get the explicit general solution of it.At last, we convert the quadratic equations into the integral equations or the question to findthe fixed-point of a continuous mapping. We use the theories about the Poisson’s equation,the heat-conduct equation, the Schauder fixed-point theorem to prove that the fixed-point isexist, hence the smoothing solution for the Navier-Stokes equations is globally exist.
Keywords. smoothing solution, Poisson’s equation, heat-conduct equation, the Schauderfixed-point theorem, globally exist.
We consider the dynamical equations for an viscous and incompressible fluid just as follows. divu = 0 (1.1) ∂u∂t − µ ∆ u + X k =1 u k ∂u∂x k + 1 ρ gradp = F (1.2)where u = ( u , u , u ) is the velocity vector, it is the macro motion velocity of the material particle. µ is the dynamic viscosity coefficient, we assume it is a constant. ρ is the density of the materialparticle, it is the mass of per volume fluid. Because the fluid is incompressible, we can assume ρ is a constant, too. p is intensity of the pressure, it is the pressure on per area fluid, the directionis perpendicular to such area. F = ( F , F , F ) is the density of the body force, it is the externalforce of per unit mass. u = ( u , u , u ) and p, F are all functions on the variables of the time t and position x = ( x , x , x ) . We assume τ = 1 ρ , τ is called the specific volume of the fluid, it isjust the volume of per unit mass. These equations are the second order partial differential equationsabout four unknown functions: u = ( u , u , u ) , p , and they are completed. These equations arecalled the Navier-Stokes equations. 1n order to discuss it more conveniently, we often rewrite the Navier-Stokes equations as follows. ∂u ∂x + ∂u ∂x + ∂u ∂x = 0 (1.3) ∂u j ∂t − µ ( ∂ u j ∂x + ∂ u j ∂x + ∂ u j ∂x ) + X k =1 u k ∂u j ∂x k + τ ∂p∂x j = F j , j = 1 , , . (1.4)And we assume as follows. Assumption 1.1 (1)We only discuss the Navier-Stokes equations on the region as follows, t ∈ [0 , T ] , T is given, ( x , x , x ) T ∈ K , K is a bounded and closed set in R , and when t isnot in [0 , T ] , or ( x , x , x ) T is not in K , u j ≡ , p ≡ , F j ≡ , j = 1 , , . (2)In the region K ′ = [0 , T ] × K , u j ∈ C , p ∈ C , F j ∈ C , j = 1 , , . (3)The boundary of K , ∂K satisfies the exterior ball condition, ∀ ( x , x , x ) T ∈ ∂K , there existsa ball B ρ ( y ) ⊂ R \ K , such that B ρ ( y ) ∩ K = ( x , x , x ) T , where y = ( y , y , y ) T , B ρ ( y ) = { z ∈ R | | z − y | ≤ ρ , ρ > } , K is the interior of K .(4) K and F j , j = 1 , , , satisfy the following, max ≤ j ≤ { | F − ( α Tj Y ) | , | F − ( α Tj Y ) | , | F − ( α Tj Y ) F − ( α Tj Y ) |} ≤ θ (1 − θ ) M T, M ( K )(2 θ + M T, M ( K )) , where < θ < , and F − ( α Tj Y ) , F − ( α Tj Y ) , M T, , M ( K ) will be defined in the section 3. After that we will show that the smoothing solution for the Navier-Stokes equations is globally existin the region K ′ . Theorem 2.1
We consider the linear equations as follows. AX = β , where A = ( a ij ) m × s ∈ R m × s is a constant matrix, X = ( X ( x , x , x , t ) , X ( x , x , x , t ) , · · · , X s ( x , x , x , t )) T is the unknown s dimensional four variables functional vector, β = ( b ( x , x , x , t ) , b ( x , x , x , t ) , · · · , b m ( x , x , x , t )) T is the known m dimensional four variables functional vector.We can get the following conclusions.(1)A necessary and sufficient condition for the existence of the solution of this equations is ∀ ( x , x , x , t ) T ∈ R , rank ( A, β ( x , x , x , t )) = rank ( A ) , where β ( x , x , x , t ) is the value of β , when ( x , x , x , t ) T is given.(2)If the solution of this equations is exist, rank ( A ) = r, r < s , X is a particular solution of thisequations, and the constant linearly independent solutions of the equations AX = 0 are η , η , · · · , η s − r , we denote A ( η ) = [ η , η , · · · , η s − r ] , then its general solution can be expressedas X = X + A ( η ) Z , where Z is the arbitrary s − r dimensional four variables functional vector. heorem 2.2 If u , u , u ∈ C , p ∈ C , F , F , F ∈ C , the Navier-Stokes equations can beconverted into the simultaneous of the first order linear partial differential equations with constantcoefficients and the quadratic equations, just as the following. ∂ ( HZ + h ) ∂t = H Z + h ∂ ( HZ + h ) ∂x i = H i Z , i = 1 , , . (2.1) e T Z = ( e T Z )( − α T Z ) , e T Z = ( e T Z )( e T Z ) , e T Z = ( e T Z )( e T Z ) ,e T Z = ( e T Z )( e T Z ) , e T Z = ( e T Z )( e T Z ) , e T Z = ( e T Z )( e T Z ) ,e T Z = ( e T Z )( e T Z ) , e T Z = ( e T Z )( e T Z ) , e T Z = ( e T Z )( e T Z ) , (2.2) where H = ( − α , − α , − α , − α , e , e , e , e , e , e , e , e , e , e , e , e ) T ,H = ( e , e , e , e , e , e , − α , e , e , e , − α , e , e , e , − α , e ) T ,H = ( e , e , e , e , e , e , − α , e , e , e , e , e , e , e , e , e ) T ,H = ( e , e , e , e , e , e , e , e , e , e , e , e , e , e , e , e ) T ,H = ( e , e , e , e , e , e , e , e , e , e , e , e , e , e , e , e ) T ,α = (0 , , , , , , , , , , ) T ,α = (0 , τ, , − µ, − µ, − µ, , , , , ) T ,α = (0 , τ, , − µ, − µ, − µ, , , , , ) T ,α = (0 , τ, , − µ, − µ, − µ, , , , , T , and e i is the ith dimensional unit coordinate vector, ≤ i ≤ , h = (0 , F , , F , , F , T ,h = (0 , F , F , F , ) T , Z is an unknown dimensional functional vector. Moreover ( e , e , e , e ) T Z = ( u , u , u , p ) T is the solution of the Navier-Stokes equations. Theorem 2.3
We consider the first order linear partial differential equations with constant coef-ficients and the quadratic equations in the theorem 2.2, moreover when t is not in [0 , T ] , or ( x , x , x ) T is not in K , Z ≡ , then we can get the following conclusions.(1)We can use the Fourier transformation to convert (2.1) into the linear equations as follows. iξ H − H iξ H − H iξ H − H iξ H − H F ( Z ) = F ( h ) − iξ F ( h ) − iξ F ( h ) − iξ F ( h ) − iξ F ( h ) , (2.3) where F ( Z ) = Z R Ze − iξ t − i P j =1 ξ j x j dtdx dx dx ,F ( h ) = Z R he − iξ t − i P j =1 ξ j x j dtdx dx dx ,F ( h ) = Z R h e − iξ t − i P j =1 ξ j x j dtdx dx dx . We can get the explicit general solution of it, F ( Z ) = Y + A ( η ) Z , or Z = F − ( Y + A ( η ) Z ) , here Y = ( iξ y , iξ y , y , iξ y , iξ y , iξ y , y , iξ y , iξ y , iξ y , y ,iξ y , iξ y , iξ y , iξ y , y , ( iξ ) y , ( iξ ) y , ( iξ ) y , ( iξ ) y , iξ iξ y ,iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , ( iξ ) y , ( iξ ) y , ( iξ ) y , ( iξ ) y ,iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , ( iξ ) y , ( iξ ) y , ( iξ ) y , ( iξ ) y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , T × ) T , and y = iξ ( iξ F ( F ) + iξ F ( F ) + iξ F ( F )) a bτ − F ( F ) aτ ,y = iξ ( iξ F ( F ) + iξ F ( F ) + iξ F ( F )) a bτ − F ( F ) aτ ,y = iξ ( iξ F ( F ) + iξ F ( F ) + iξ F ( F )) a bτ − F ( F ) aτ ,y = iξ F ( F ) + iξ F ( F ) + iξ F ( F ) abτ , where a = µ [( iξ ) + ( iξ ) + ( iξ ) ] − iξ τ = 0 , b = ( iξ ) + ( iξ ) + ( iξ ) a = 0 , when ( ξ , ξ , ξ ) = 0 . A ( η ) = ( η , η , η , η , η , η , η , η , η ) , and η j = ( iξ y , iξ y , y , iξ y , iξ y , iξ y , y , iξ y , iξ y , iξ y , y ,iξ y , iξ y , iξ y , iξ y , y , ( iξ ) y , ( iξ ) y , ( iξ ) y , ( iξ ) y , iξ iξ y ,iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , ( iξ ) y , ( iξ ) y , ( iξ ) y , ( iξ ) y ,iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , ( iξ ) y , ( iξ ) y , ( iξ ) y , ( iξ ) y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , e Tj ) T , here e j is the jth dimensional unit coordinate vector, ≤ j ≤ , moreover when j = 1 , , , y = ξ a bτ + 1 aτ , y = − iξ iξ a bτ , y = − iξ iξ a bτ , y = − iξ abτ , when j = 4 , , , y = ξ a bτ + 1 aτ , y = − iξ iξ a bτ , y = − iξ iξ a bτ , y = − iξ abτ , when j = 7 , , , y = ξ a bτ + 1 aτ , y = − iξ iξ a bτ , y = − iξ iξ a bτ , y = − iξ abτ .Z = ( Z j ) × , moreover we assume F − ( Y + A ( η ) Z I { ( ξ , ξ , ξ ) =0 } ) = F − ( Y + A ( η ) Z ) , and Z ∈ Ω , where Ω = { Z | H [ F − ( Y + A ( η ) Z )] = H [ F − ( Y + A ( η ) Z )] I K ′ ∈ C ( K ′ ) . } . Z = F [ F − ( e T ( Y + A ( η ) Z )) F − ( − α T ( Y + A ( η ) Z ))] = f ( Z ) ,Z = F [ F − ( e T ( Y + A ( η ) Z )) F − ( e T ( Y + A ( η ) Z ))] = f ( Z ) ,Z = F [ F − ( e T ( Y + A ( η ) Z )) F − ( e T ( Y + A ( η ) Z ))] = f ( Z ) ,Z = F [ F − ( e T ( Y + A ( η ) Z )) F − ( e T ( Y + A ( η ) Z ))] = f ( Z ) ,Z = F [ F − ( e T ( Y + A ( η ) Z )) F − ( e T ( Y + A ( η ) Z ))] = f ( Z ) ,Z = F [ F − ( e T ( Y + A ( η ) Z )) F − ( e T ( Y + A ( η ) Z ))] = f ( Z ) ,Z = F [ F − ( e T ( Y + A ( η ) Z )) F − ( e T ( Y + A ( η ) Z ))] = f ( Z ) ,Z = F [ F − ( e T ( Y + A ( η ) Z )) F − ( e T ( Y + A ( η ) Z ))] = f ( Z ) ,Z = F [ F − ( e T ( Y + A ( η ) Z )) F − ( e T ( Y + A ( η ) Z ))] = f ( Z ) . (2.4) This is also the question to find the fixed-points of f ( Z ) , where f ( Z ) = ( f j ( Z )) × . We canuse the theories about the Poisson’s equation, the heat-conduct equation, the Schauder fixed-pointtheorem to prove that the fixed-point is exist in the region K ′ , and the smoothing solution for theNavier-Stokes equations is globally exist in the region K ′ . Proof of theorem2-1 . (1)Necessity. Because the solution of AX = β is exist, we assume theparticular solution is X , then ∀ ( x , x , x , t ) T ∈ R , we know that AX ( x , x , x , t ) = β ( x , x , x , t )satisfies, X ( x , x , x , t ) and β ( x , x , x , t )are values of X and β , when ( x , x , x , t ) T is given. Hence rank ( A, β ( x , x , x , t )) = rank ( A ) , ∀ ( x , x , x , t ) T ∈ R . Sufficiency. Because ∀ ( x , x , x , t ) T ∈ R , we can get rank ( A, β ( x , x , x , t )) = rank ( A ) , then the solution of AX = β ( x , x , x , t )is exist. We assume the particular solution is X ( x , x , x , t ) , then ∀ ( x , x , x , t ) T ∈ R ,we let X = ( X ( x , x , x , t ) , X ( x , x , x , t ) , · · · , X s ( x , x , x , t )) T = X ( x , x , x , t ) , and we can get X satisfies AX = β .(2)When the solution of AX = β is exist, we assume X is a particular solution, X is the arbitrarysolution of this equations, then ∀ ( x , x , x , t ) T ∈ R , we can get A ( X ( x , x , x , t ) − X ( x , x , x , t )) = 0 , C , C , · · · , C s − r such that X ( x , x , x , t ) − X ( x , x , x , t ) = C η + C η + · · · + C s − r η s − r , we let Z ( x , x , x , t ) = ( C , C , · · · , C s − r ) T , then X ( x , x , x , t ) = X ( x , x , x , t ) + A ( η ) Z ( x , x , x , t ) . Because ( x , x , x , t ) T is arbitrary, we can learn that X = X + A ( η ) Z . On the other hand A [ X + A ( η ) Z ] = β + 0 Z = β . This is to say ∀ Z , X + A ( η ) Z is the solution of the equation AX = β . Hence X + A ( η ) Z is the general solution of the equation AX = β . Proof of theorem2-2 . We can rewrite the Navier-Stokes equations as following. AX = β , where A = , , , , , , , , , , , , , , , , , , , τ, , − µ, − µ, − µ, , , , , , , , , , τ, , − µ, − µ, − µ, , , , , , , , , , τ, , − µ, − µ, − µ, , , , , × n is the row vector which is made up of n zeroes, X includes u , u , u , p and all their first orderpartial derivative, and all the second order partial derivative of u , u , u , and all the productswhich they are in the Navier-Stokes equations, X = ( ∂u ∂x , ∂u ∂t , ∂u ∂t , ∂u ∂t , ∂u ∂x , ∂u ∂x , u , ∂u ∂x , ∂u ∂x , ∂u ∂x , u , ∂u ∂x , ∂u ∂x , ∂u ∂x , u ,∂p∂t , ∂p∂x , ∂p∂x , ∂p∂x , p , ∂ u ∂t , ∂ u ∂x , ∂ u ∂x , ∂ u ∂x , ∂ u ∂t∂x , ∂ u ∂t∂x , ∂ u ∂t∂x , u ∂x ∂x , ∂ u ∂x ∂x , ∂ u ∂x ∂x , ∂ u ∂t , ∂ u ∂x , ∂ u ∂x , ∂ u ∂x , ∂ u ∂t∂x , ∂ u ∂t∂x , ∂ u ∂t∂x ,∂ u ∂x ∂x , ∂ u ∂x ∂x , ∂ u ∂x ∂x , ∂ u ∂t , ∂ u ∂x , ∂ u ∂x , ∂ u ∂x , ∂ u ∂t∂x , ∂ u ∂t∂x , ∂ u ∂t∂x ,∂ u ∂x ∂x , ∂ u ∂x ∂x , ∂ u ∂x ∂x , u ∂u ∂x , u ∂u ∂x , u ∂u ∂x , u ∂u ∂x , u ∂u ∂x , u ∂u ∂x ,u ∂u ∂x , u ∂u ∂x , u ∂u ∂x ) T , β = (0 , F , F , F ) . We can see that A = ( E , A ) , A is a 4 ×
55 matrix. We assume A = ( α , α , α , α ) T , where α = (0 , , , , , , , , , , ) T ,α = (0 , τ, , − µ, − µ, − µ, , , , , ) T ,α = (0 , τ, , − µ, − µ, − µ, , , , , ) T ,α = (0 , τ, , − µ, − µ, − µ, , , , , T . We let X = (0 , F , F , F , ) T , A ( η ) = − A E ! × , then AX = β , AA ( η ) = 0 , andthe columns of A ( η ) are linear independent, so according to (2) in the theorem 2.2, we can get thegeneral solution of AX = β as follows, X = X + A ( η ) Z .
We write the 59 components of X in details. ∂u ∂x = − α T Z , ∂u ∂t = F − α T Z , ∂u ∂t = F − α T Z , ∂u ∂t = F − α T Z , ∂u ∂x = e T Z ,∂u ∂x = e T Z , u = e T Z , ∂u ∂x = e T Z , ∂u ∂x = e T Z , ∂u ∂x = e T Z , u = e T Z ,∂u ∂x = e T Z , ∂u ∂x = e T Z , ∂u ∂x = e T Z , u = e T Z , ∂p∂t = e T Z , ∂p∂x = e T Z ,∂p∂x = e T Z , ∂p∂x = e T Z , p = e T Z , ∂ u ∂t = e T Z , ∂ u ∂x = e T Z , ∂ u ∂x = e T Z ,∂ u ∂x = e T Z , ∂ u ∂t∂x = e T Z , ∂ u ∂t∂x = e T Z , ∂ u ∂t∂x = e T Z , ∂ u ∂x ∂x = e T Z ,∂ u ∂x ∂x = e T Z , ∂ u ∂x ∂x = e T Z , ∂ u ∂t = e T Z , ∂ u ∂x = e T Z , ∂ u ∂x = e T Z ,∂ u ∂x = e T Z , ∂ u ∂t∂x = e T Z , ∂ u ∂t∂x = e T Z , ∂ u ∂t∂x = e T Z , ∂ u ∂x ∂x = e T Z ,∂ u ∂x ∂x = e T Z , ∂ u ∂x ∂x = e T Z , ∂ u ∂t = e T Z , ∂ u ∂x = e T Z , ∂ u ∂x = e T Z , u ∂x = e T Z , ∂ u ∂t∂x = e T Z , ∂ u ∂t∂x = e T Z , ∂ u ∂t∂x = e T Z , ∂ u ∂x ∂x = e T Z ,∂ u ∂x ∂x = e T Z , ∂ u ∂x ∂x = e T Z , u ∂u ∂x = e T Z , u ∂u ∂x = e T Z , u ∂u ∂x = e T Z ,u ∂u ∂x = e T Z , u ∂u ∂x = e T Z , u ∂u ∂x = e T Z , u ∂u ∂x = e T Z , u ∂u ∂x = e T Z ,u ∂u ∂x = e T Z , here e i is the ith
55 dimensional unit coordinate vector, 1 ≤ i ≤ . There are 9 quadratic items,hence we can get 9 quadratic constraints which Z need to satisfy just as following: e T Z = ( e T Z )( − α T Z ) , e T Z = ( e T Z )( e T Z ) , e T Z = ( e T Z )( e T Z ) ,e T Z = ( e T Z )( e T Z ) , e T Z = ( e T Z )( e T Z ) , e T Z = ( e T Z )( e T Z ) ,e T Z = ( e T Z )( e T Z ) , e T Z = ( e T Z )( e T Z ) , e T Z = ( e T Z )( e T Z ) . (3.1)Because u , u , u ∈ C , p ∈ C , we can get 46 differential constraints which Z need to satisfyjust as following: ∂ ( e T Z ) ∂x = − α T Z , ∂ ( e T Z ) ∂t = F − α T Z , ∂ ( e T Z ) ∂t = F − α T Z , ∂ ( e T Z ) ∂t = F − α T Z ,∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = e T Z ,∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂t = e T Z , ∂ ( e T Z ) ∂x = e T Z ,∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂t = e T Z , ∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = e T Z ,∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂t∂x = e T Z , ∂ ( e T Z ) ∂t∂x = e T Z , ∂ ( e T Z ) ∂t∂x = e T Z , ∂ ( e T Z ) ∂x ∂x = e T Z ,∂ ( e T Z ) ∂x ∂x = e T Z , ∂ ( e T Z ) ∂x ∂x = e T Z , ∂ ( e T Z ) ∂t = e T Z , ∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = e T Z ,∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂t∂x = e T Z , ∂ ( e T Z ) ∂t∂x = e T Z , ∂ ( e T Z ) ∂t∂x = e T Z , ∂ ( e T Z ) ∂x ∂x = e T Z ,∂ ( e T Z ) ∂x ∂x = e T Z , ∂ ( e T Z ) ∂x ∂x = e T Z , ∂ ( e T Z ) ∂t = e T Z , ∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = e T Z ,∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂t∂x = e T Z , ∂ ( e T Z ) ∂t∂x = e T Z , ∂ ( e T Z ) ∂t∂x = e T Z , ∂ ( e T Z ) ∂x ∂x = e T Z ,∂ ( e T Z ) ∂x ∂x = e T Z , ∂ ( e T Z ) ∂x ∂x = e T Z .
Because the second order partial derivative can be taken as the partial derivative of the first orderpartial derivative, and u , u , u ∈ C , we can learn that their second order mixed partial derivatives8re equal, the above differential constraints are equivalent to 64 first order differential constraints asfollows: ∂ ( e T Z ) ∂x = − α T Z , ∂ ( e T Z ) ∂t = F − α T Z , ∂ ( e T Z ) ∂t = F − α T Z , ∂ ( e T Z ) ∂t = F − α T Z ,∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = e T Z ,∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂t = e T Z , ∂ ( e T Z ) ∂x = e T Z ,∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = e T Z , ∂ ( F − α T Z ) ∂t = e T Z , ∂ ( − α T Z ) ∂x = e T Z ,∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = e T Z , ∂ ( F − α T Z ) ∂x = ∂ ( − α T Z ) ∂t = e T Z ,∂ ( F − α T Z ) ∂x = ∂ ( e T Z ) ∂t = e T Z , ∂ ( F − α T Z ) ∂x = ∂ ( e T Z ) ∂t = e T Z ,∂ ( − α T Z ) ∂x = ∂ ( e T Z ) ∂x = e T Z , ∂ ( − α T Z ) ∂x = ∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = ∂ ( e T Z ) ∂x = e T Z ,∂ ( F − α T Z ) ∂t = e T Z , ∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = e T Z ,∂ ( F − α T Z ) ∂x = ∂ ( e T Z ) ∂t = e T Z , ∂ ( F − α T Z ) ∂x = ∂ ( e T Z ) ∂t = e T Z ,∂ ( F − α T Z ) ∂x = ∂ ( e T Z ) ∂t = e T Z , ∂ ( e T Z ) ∂x = ∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = ∂ ( e T Z ) ∂x = e T Z ,∂ ( e T Z ) ∂x = ∂ ( e T Z ) ∂x = e T Z , ∂ ( F − α T Z ) ∂t = e T Z , ∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = e T Z ,∂ ( e T Z ) ∂x = e T Z , ∂ ( F − α T Z ) ∂x = ∂ ( e T Z ) ∂t = e T Z , ∂ ( F − α T Z ) ∂x = ∂ ( e T Z ) ∂t = e T Z ,∂ ( F − α T Z ) ∂x = ∂ ( e T Z ) ∂t = e T Z , ∂ ( e T Z ) ∂x = ∂ ( e T Z ) ∂x = e T Z , ∂ ( e T Z ) ∂x = ∂ ( e T Z ) ∂x = e T Z ,∂ ( e T Z ) ∂x = ∂ ( e T Z ) ∂x = e T Z .
We write them into the equations, we assume U = ( ∂u ∂x , ∂u ∂t , ∂u ∂t , ∂u ∂t , ∂u ∂x , ∂u ∂x , u , ∂u ∂x , ∂u ∂x , ∂u ∂x , u , ∂u ∂x , ∂u ∂x , ∂u ∂x , u , p ) T ,H = ( − α , − α , − α , − α , e , e , e , e , e , e , e , e , e , e , e , e ) T ,h = (0 , F , F , F , ) T , then we can get U = HZ + h , moreover we assume H = ( e , e , e , e , e , e , − α , e , e , e , − α , e , e , e , − α , e ) T ,H = ( e , e , e , e , e , e , − α , e , e , e , e , e , e , e , e , e ) T ,H = ( e , e , e , e , e , e , e , e , e , e , e , e , e , e , e , e ) T ,H = ( e , e , e , e , e , e , e , e , e , e , e , e , e , e , e , e ) T , = (0 , F , , F , , F , T , then we can get ∂U∂t = H Z + h ∂U∂x i = H i Z , i = 1 , , , (3.2)or ∂ ( HZ + h ) ∂t = H Z + h ∂ ( HZ + h ) ∂x i = H i Z , i = 1 , , . (3.3)Now we have converted the Navier-Stokes equations into the simultaneous of the first order linearpartial differential equations with constant coefficients (3.3) and 9 quadratic polynomial equations(3.1).And we get a necessary condition for the existence of the solution of the Navier-Stokes equations asfollows, ∃
55 dimensional four variables functional vector Z , such that Z satisfies the first order linearpartial differential equations with constant coefficients (3.3) and 9 quadratic polynomial equations(3.1).In fact, this condition is also sufficient. If Z satisfies the above first order linear partial differentialequations with constant coefficients (3.3) and 9 quadratic polynomial equations (3.1), we let( u , u , u , p ) T = ( e , e , e , e ) T Z , then from (3.3) we can get U = HZ + h , where U = ( ∂u ∂x , ∂u ∂t , ∂u ∂t , ∂u ∂t , ∂u ∂x , ∂u ∂x , u , ∂u ∂x , ∂u ∂x , ∂u ∂x , u , ∂u ∂x , ∂u ∂x , ∂u ∂x , u , p ) T , again from (3.3) we can learn that ∂U∂t = H Z + h , ∂U∂x i = H i Z , i = 1 , , X = X + A ( η ) Z , where X includes u , u , u , p and all their first order partial derivative,and all the second order partial derivative of u , u , u , and all the products which they are in theNavier-Stokes equations, then we can get AX = β , this is just the Navier-Stokes equations.Hence we can get the corollary as follows. Corollary 3.1
If we assume u , u , u ∈ C , p ∈ C , F , F , F ∈ C , then a necessaryand sufficient condition for the existence of the solution for the Navier-Stokes equations is that ∃ dimensional four variables functional vector Z , such that Z satisfies the first order linear partialdifferential equations with constant coefficients (3.3) and quadratic equations (3.1). Under this circumstance, ( u , u , u , p ) T = ( e , e , e , e ) T Z is the solution of the Navier-Stokesequations, here e i is the ith
55 dimensional unit coordinate vector, 1 ≤ i ≤
55 .
Proof of theorem2-3 . (1) Under the assumption (1.1) and from the theorem 2.2, we can learn thatwhen t is not in [0 , T ] , or ( x , x , x ) is not in K , Z ≡ Z can do the Fourier trans-formation with t , x , x , x , we use the Fourier transformation to convert (2.1) into the linear10quations as follows. iξ H − H iξ H − H iξ H − H iξ H − H F ( Z ) = F ( h ) − iξ F ( h ) − iξ F ( h ) − iξ F ( h ) − iξ F ( h ) , (3.4)where F ( Z ) = Z R Ze − iξ t − i P j =1 ξ j x j dtdx dx dx ,F ( h ) = Z R he − iξ t − i P j =1 ξ j x j dtdx dx dx ,F ( h ) = Z R h e − iξ t − i P j =1 ξ j x j dtdx dx dx . We assume that B = iξ H − H iξ H − H iξ H − H iξ H − H × , G = F ( h ) − iξ F ( h ) − iξ F ( h ) − iξ F ( h ) − iξ F ( h ) × , next we solve the linear equations BY = G , where Y = ( y j ) × . We can get rank ( B ) = 46 , when( ξ , ξ , ξ ) = 0 , moreover the first 46 columns are linear independent.We write out all the rows of the matrix B. − iξ α T − e T , − iξ α T − e T , − iξ α T − e T , − iξ α T − e T , − iξ α T − e T , − iξ α T − e T , − iξ α T − e T , − iξ α T − e T , − iξ α T − e T , − iξ α T − e T , − iξ α T − e T , − iξ α T − e T , − iξ α T − e T , − iξ α T − e T , − iξ α T − e T , − iξ α T − e T ,iξ e T − e T , iξ e T − e T , iξ e T − e T , iξ e T − e T ,iξ e T − e T , iξ e T − e T , iξ e T − e T , iξ e T − e T ,iξ e T + α T , iξ e T + α T , iξ e T − e T , iξ e T − e T ,iξ e T − e T , iξ e T − e T , iξ e T − e T , iξ e T − e T ,iξ e T − e T , iξ e T − e T , iξ e T − e T , iξ e T − e T ,iξ e T − e T , iξ e T − e T , iξ e T − e T , iξ e T − e T ,iξ e T + α T , iξ e T − e T , iξ e T − e T , iξ e T − e T ,iξ e T − e T , iξ e T − e T , iξ e T − e T , iξ e T − e T ,iξ e T − e T , iξ e T − e T , iξ e T − e T , iξ e T − e T ,iξ e T − e T , iξ e T − e T , iξ e T − e T , iξ e T − e T ,iξ e T + α T , iξ e T − e T , iξ e T − e T , iξ e T − e T ,iξ e T − e T , iξ e T − e T , iξ e T − e T , iξ e T − e T , e i is the ith
55 dimensional unit coordinate vector, 1 ≤ i ≤
55 , and α = (0 , , , , , , , , , , ) T ,α = (0 , τ, , − µ, − µ, − µ, , , , , ) T ,α = (0 , τ, , − µ, − µ, − µ, , , , , ) T ,α = (0 , τ, , − µ, − µ, − µ, , , , , T . First we let y j − = 0 , ≤ j ≤ BY = 0 is only 0 , when( ξ , ξ , ξ ) = 0 , hence the first 46 columns are linear independent, and rank ( B ) ≥
46 .From the 7th row we can get y = iξ y , y = iξ y , y + y = − iξ y , and from the first, thesecond, the 5th , the 6th rows, we can get y = ( iξ ) y , y = ( iξ ) y , y = ( iξ ) y , y = ay , where a = µ [( iξ ) + ( iξ ) + ( iξ ) ] − iξ τ = 0 , when ( ξ , ξ , ξ ) = 0 , moreover y = ( iξ ) y , y = iξ iξ y , y = iξ iξ y , y = iξ iξ y ,y = iξ iξ y , y = iξ iξ y , y = iξ iξ y .From the 11th row we can get y = iξ y , y = iξ y , y = iξ y , and from the 8th, the third, the9th , the 10th rows, we can get y = ( iξ ) y , y = ( iξ ) y , y = ( iξ ) y , y = ay , moreover y = ( iξ ) y , y = iξ iξ y , y = iξ iξ y , y = iξ iξ y , y = iξ iξ y , y = iξ iξ y , y = iξ iξ y .From the 15th row we can get y = iξ y , y = iξ y , y = iξ y , and from the 12th, the 13th,the 14th , the 4th rows, we can get y = ( iξ ) y , y = ( iξ ) y , y = ( iξ ) y , y = ay , moreover y = ( iξ ) y , y = iξ iξ y , y = iξ iξ y , y = iξ iξ y , y = iξ iξ y , y = iξ iξ y , y = iξ iξ y .And from the last row, we can get y = iξ y , y = iξ y , y = iξ y , y = iξ y .Because y = ay , y = ay , y = ay , we can get y = iξ a y , y = iξ a y , y = iξ a y . From ( y + y ) = y + y , we can get − iξ y = iξ y + iξ y . This is equal to the following.[ ( iξ ) a + ( iξ ) a + ( iξ ) a ] y = 0 . Hence y = 0 , when ( ξ , ξ , ξ ) = 0 , and we can get y = 0 , y = 0 , y = 0 , the solution of BY = 0 is only 0 , the first 46 columns are linear independent, and rank ( B ) ≥
46 .Next we will in turn let y j − = 1 , y k = 0 , ≤ k ≤ , k = 47 + j − , ≤ j ≤ BY = 0 , we assume they are η j , ≤ j ≤ rank ( B ) ≤
46 , hence rank ( B ) = 46 , when ( ξ , ξ , ξ ) = 0 .If we let y j − = 1 , y k = 0 , ≤ k ≤ , k = 47 + j − , ≤ j ≤ α T Y will change into α T Y + 1 , and y = iξ y = ay − τ , then we can get y = iξ a y + 1 aτ , and y = iξ a y , y = iξ a y . Hence we can get the following, by + iξ aτ = 0 , or y = − iξ abτ . y , y , y and η j , ≤ j ≤ y j − = 1 , y k = 0 , ≤ k ≤ , k = 47 + j − , ≤ j ≤ α T Y will change into α T Y + 1 , and y = iξ y = ay − τ , then we can get y = iξ a y + 1 aτ , and y = iξ a y , y = iξ a y . Hence we can get the following, by + iξ aτ = 0 , or y = − iξ abτ . Moreover we can work out y , y , y and η j , ≤ j ≤ y j − = 1 , y k = 0 , ≤ k ≤ , k = 47 + j − , ≤ j ≤ α T Y will change into α T Y + 1 , and y = iξ y = ay − τ , then we can get y = iξ a y + 1 aτ ,and y = iξ a y , y = iξ a y . Hence we can get the following, by + iξ aτ = 0 , or y = − iξ abτ . Moreover we can work out y , y , y and η j , ≤ j ≤ η j , ≤ j ≤ η j = ( iξ y , iξ y , y , iξ y , iξ y , iξ y , y , iξ y , iξ y , iξ y , y ,iξ y , iξ y , iξ y , iξ y , y , ( iξ ) y , ( iξ ) y , ( iξ ) y , ( iξ ) y , iξ iξ y ,iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , ( iξ ) y , ( iξ ) y , ( iξ ) y , ( iξ ) y ,iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , ( iξ ) y , ( iξ ) y , ( iξ ) y , ( iξ ) y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , e Tj ) T , here e j is the jth ≤ j ≤ j = 1 , , y = ξ a bτ + 1 aτ , y = − iξ iξ a bτ , y = − iξ iξ a bτ , y = − iξ abτ , where b = ( iξ ) + ( iξ ) + ( iξ ) a = 0 , when ( ξ , ξ , ξ ) = 0 , when j = 4 , , y = ξ a bτ + 1 aτ , y = − iξ iξ a bτ , y = − iξ iξ a bτ , y = − iξ abτ , when j = 7 , , y = ξ a bτ + 1 aτ , y = − iξ iξ a bτ , y = − iξ iξ a bτ , y = − iξ abτ . Finally we work out a particular solution Y of BY = G . We can see that G = (0 , − iξ F ( F ) , − iξ F ( F ) , − iξ F ( F ) , , F ( F ) , , F ( F ) , , F ( F ) , , ,iξ F ( F ) , − iξ F ( F ) , − iξ F ( F ) , , , iξ F ( F ) , − iξ F ( F ) , − iξ F ( F ) , , , iξ F ( F ) , − iξ F ( F ) , − iξ F ( F ) , ) T ,
13f we let y j − = 0 , ≤ j ≤ y = iξ y = ay + F ( F ) τ , y = iξ y = ay + F ( F ) τ , and y = iξ y = ay + F ( F ) τ . Hence we can get the following, by = iξ F ( F ) + iξ F ( F ) + iξ F ( F ) aτ , or y = iξ F ( F ) + iξ F ( F ) + iξ F ( F ) abτ . Moreover we can work out y , y , y and Y .We write Y as follows. Y = ( iξ y , iξ y , y , iξ y , iξ y , iξ y , y , iξ y , iξ y , iξ y , y ,iξ y , iξ y , iξ y , iξ y , y , ( iξ ) y , ( iξ ) y , ( iξ ) y , ( iξ ) y , iξ iξ y ,iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , ( iξ ) y , ( iξ ) y , ( iξ ) y , ( iξ ) y ,iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , ( iξ ) y , ( iξ ) y , ( iξ ) y , ( iξ ) y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , T × ) T , where y = iξ ( iξ F ( F ) + iξ F ( F ) + iξ F ( F )) a bτ − F ( F ) aτ ,y = iξ ( iξ F ( F ) + iξ F ( F ) + iξ F ( F )) a bτ − F ( F ) aτ ,y = iξ ( iξ F ( F ) + iξ F ( F ) + iξ F ( F )) a bτ − F ( F ) aτ ,y = iξ F ( F ) + iξ F ( F ) + iξ F ( F ) abτ . And we can test such Y satisfies BY = G .If we assume A ( η ) = ( η , η , η , η , η , η , η , η , η ) , then we can get the explicit generalsolutions of (3.4) as follows, F ( Z ) = Y + A ( η ) Z , or Z = F − ( Y + A ( η ) Z ) , where Z = ( Z j ) × , moreover we assume F − ( Y + A ( η ) Z I { ( ξ , ξ , ξ ) =0 } ) = F − ( Y + A ( η ) Z ) , and Z ∈ Ω . Because Z need to satisfy HZ = HZI K ′ ∈ C ( K ′ ) , we can get Ω = { Z | H [ F − ( Y + A ( η ) Z )] = H [ F − ( Y + A ( η ) Z )] I K ′ ∈ C ( K ′ ) . } , where H = ( − α , − α , − α , − α , e , e , e , e , e , e , e , e , e , e , e , e ) T . F ( e T j − Z ) = Z j , ≤ j ≤ Z = F [ F − ( e T ( Y + A ( η ) Z )) F − ( − α T ( Y + A ( η ) Z ))] = f ( Z ) ,Z = F [ F − ( e T ( Y + A ( η ) Z )) F − ( e T ( Y + A ( η ) Z ))] = f ( Z ) ,Z = F [ F − ( e T ( Y + A ( η ) Z )) F − ( e T ( Y + A ( η ) Z ))] = f ( Z ) ,Z = F [ F − ( e T ( Y + A ( η ) Z )) F − ( e T ( Y + A ( η ) Z ))] = f ( Z ) ,Z = F [ F − ( e T ( Y + A ( η ) Z )) F − ( e T ( Y + A ( η ) Z ))] = f ( Z ) ,Z = F [ F − ( e T ( Y + A ( η ) Z )) F − ( e T ( Y + A ( η ) Z ))] = f ( Z ) ,Z = F [ F − ( e T ( Y + A ( η ) Z )) F − ( e T ( Y + A ( η ) Z ))] = f ( Z ) ,Z = F [ F − ( e T ( Y + A ( η ) Z )) F − ( e T ( Y + A ( η ) Z ))] = f ( Z ) ,Z = F [ F − ( e T ( Y + A ( η ) Z )) F − ( e T ( Y + A ( η ) Z ))] = f ( Z ) . (3.5)This is also the question to find the fixed-points of f ( Z ) , where f ( Z ) = ( f j ( Z )) × . But we cannot see immediately f ( Z ) ∈ Ω , even if Z ∈ Ω . We need to attain this point first. We assumeΩ as follows. Ω = { Z |∃ h = h I K ′ ∈ C ( K ′ ) , such that Z = F ( h ) . } , where h = ( h j ) × , F ( h ) = Z R h e − iξ t − i P j =1 ξ j x j dtdx dx dx . We will prove that Ω ⊂ Ω and f ( Z ) ∈ Ω , if Z ∈ Ω . We look at a lemma as follows. Lemma 3.1 ∀ h = h I K ′ ∈ C ( K ′ ) , and h satisfies the H¨older condition, ∃ h = h I K ′ ∈ C [0 , T ] T C ∞ ( K ) , such that F ( h ) = a bτ F ( h ) , where a = µ [( iξ ) + ( iξ ) + ( iξ ) ] − iξ τ , b = ( iξ ) + ( iξ ) + ( iξ ) a , and h satisfies the H¨older condition means that ∃ C > , < α < , such that | h ( x t ) − h ( x ′ t ) | = max ≤ j ≤ | h j ( x t ) − h j ( x ′ t ) | ≤ C | x t − x ′ t | α , ∀ x t , x ′ t ∈ K ′ . Proof of lemma3-1 . We see that a bτ = ( µ [( iξ ) + ( iξ ) + ( iξ ) ] − iξ )(( iξ ) + ( iξ ) + ( iξ ) ) ,hence this lemma is equivalent to ∀ h = h I K ′ ∈ C ( K ′ ) , and h satisfies the H¨older condition, ∃ h = h I K ′ ∈ C [0 , T ] T C ∞ ( K ) , such that µ X j =1 3 X k =1 ∂ h ∂x j ∂x k − X j =1 ∂ h ∂t∂x j = h , a.e. (3.6)If we let v = µ △ h − ∂h /∂t , then we convert (3.6) into follows, △ v = h ,µ △ h − ∂h ∂t = v . (3.7)15e see that (3.7) are the simultaneous of Poisson’s equation and the heat-conduct equation, theyare all classical mathematical-physics equations and h satisfies the H¨older condition, the boundaryof K , ∂K satisfies the exterior ball condition, we know their solutions are all exist. We can write h as follows. v ( t, M ) = − π Z K h ( t, M ) r MM dx dx dx , where M = ( x , x , x ) , M = ( x , x , x ) ,r MM = p ( x − x ) + ( x − x ) + ( x − x ) ,h ( t, M ) = ( 12 √ πµ ) Z t Z R v ( τ , y , y , y )( √ t − τ ) e − ( x − y x − y x − y µ ( t − τ dy dy dy dτ . We assume the measure of the boundary of K is 0 , then we can get that h satisfies (3.6) . Nextwe will prove h = h I K ′ ∈ C [0 , T ] T C ∞ ( K ) .First we can see v is continuous on the region K ′ , and from ∂ k h ( t, M ) ∂x a ∂x a ∂x a = ( 12 √ πµ ) Z t Z R v ( τ , y , y , y )( √ t − τ ) ∂ k e − ( x − y x − y x − y µ ( t − τ ∂x a ∂x a ∂x a dy dy dy dτ , where k = a + a + a , a , a , a are all nonnegative integral numbers, we know h = h I K ′ ∈ C ∞ ( K ) . And because ∂h ∂t = µ △ h − v , we know h = h I K ′ ∈ C [0 , T ] T C ∞ ( K ) .We know h satisfies the H¨older condition if h = h I K ′ ∈ C ( K ′ ) , hence we can get the corollaryas follows. Corollary 3.2 ∀ h = h I K ′ ∈ C ( K ′ ) , ∃ h = h I K ′ ∈ C [0 , T ] T C ∞ ( K ) , such that F ( h ) = a bτ F ( h ) , where a = µ [( iξ ) + ( iξ ) + ( iξ ) ] − iξ τ , b = ( iξ ) + ( iξ ) + ( iξ ) a . From the corollary 3.2 and the assumption 1.1 , we can get ∃ F j = F j I K ′ ∈ C [0 , T ] T C ∞ ( K ) ,such that F ( F j ) = a bτ F ( F j ) , 1 ≤ j ≤ F j ( t, M ) = ( 12 √ πµ ) Z t Z R F jv ( τ , y , y , y )( √ t − τ ) e − ( x − y x − y x − y µ ( t − τ dy dy dy dτ ,F jv ( t, M ) = − π Z K F j ( t, M ) r MM dx dx dx , ≤ j ≤ . Next we see H [ F − ( Y )] , where H = ( − α , − α , − α , − α , e , e , e , e , e , e , e , e , e , e , e , e ) T ,α = (0 , , , , , , , , , , ) T ,α = (0 , τ, , − µ, − µ, − µ, , , , , ) T ,α = (0 , τ, , − µ, − µ, − µ, , , , , ) T ,α = (0 , τ, , − µ, − µ, − µ, , , , , T , e i is the ith
55 dimensional unit coordinate vector, 1 ≤ i ≤
55 , Y = ( iξ y , iξ y , y , iξ y , iξ y , iξ y , y , iξ y , iξ y , iξ y , y ,iξ y , iξ y , iξ y , iξ y , y , ( iξ ) y , ( iξ ) y , ( iξ ) y , ( iξ ) y , iξ iξ y ,iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , ( iξ ) y , ( iξ ) y , ( iξ ) y , ( iξ ) y ,iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , ( iξ ) y , ( iξ ) y , ( iξ ) y , ( iξ ) y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , T × ) T , and y = iξ ( iξ F ( F ) + iξ F ( F ) + iξ F ( F )) a bτ − F ( F ) aτ = iξ ( iξ F ( F ) + iξ F ( F ) + iξ F ( F )) − abF ( F ) ,y = iξ ( iξ F ( F ) + iξ F ( F ) + iξ F ( F )) a bτ − F ( F ) aτ = iξ ( iξ F ( F ) + iξ F ( F ) + iξ F ( F )) − abF ( F ) ,y = iξ ( iξ F ( F ) + iξ F ( F ) + iξ F ( F )) a bτ − F ( F ) aτ = iξ ( iξ F ( F ) + iξ F ( F ) + iξ F ( F )) − abF ( F ) ,y = iξ F ( F ) + iξ F ( F ) + iξ F ( F ) abτ = a ( iξ F ( F ) + iξ F ( F ) + iξ F ( F )) . We assume H [ F − ( Y )] = W ( F , F , F ) , where W ( F , F , F ) = ( w j ) × , and we canget the following. w = − ( e T + e T ) F − ( Y ) = − F − ( iξ y + iξ y ) ,w = − [ τ e T − µ ( e T + e T + e T ) + e T + e T + e T ] F − ( Y )= − F − { τ iξ y − µ [( iξ ) + ( iξ ) + ( iξ ) ] y } ,w = − [ τ e T − µ ( e T + e T + e T ) + e T + e T + e T ] F − ( Y )= − F − { τ iξ y − µ [( iξ ) + ( iξ ) + ( iξ ) ] y } ,w = − [ τ e T − µ ( e T + e T + e T ) + e T + e T + e T ] F − ( Y )= − F − { τ iξ y − µ [( iξ ) + ( iξ ) + ( iξ ) ] y } ,w = e T F − ( Y ) = F − ( iξ y ) , w = e T F − ( Y ) = F − ( iξ y ) ,w = e T F − ( Y ) = F − ( y ) , w = e T F − ( Y ) = F − ( iξ y ) ,w = e T F − ( Y ) = F − ( iξ y ) , w = e T F − ( Y ) = F − ( iξ y ) ,w = e T F − ( Y ) = F − ( y ) , w = e T F − ( Y ) = F − ( iξ y ) ,w = e T F − ( Y ) = F − ( iξ y ) , w = e T F − ( Y ) = F − ( iξ y ) ,w = e T F − ( Y ) = F − ( y ) , w = e T F − ( Y ) = F − ( y ) . Because F j = F j I K ′ ∈ C [0 , T ] T C ∞ ( K ) , 1 ≤ j ≤ F − ( iξ F ( F j )) = F − ( F ( ∂F j ∂t )) = ∂F j ∂t , ≤ j ≤ , F − ( iξ F ( F j )) = ∂F j ∂x , F − ( iξ F ( F j )) = ∂F j ∂x , F − ( iξ F ( F j )) = ∂F j ∂x , ≤ j ≤ . Hence we can get the following, F − ( y ) = ∂ F ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x − ∂ F ∂x ,F − ( y ) = ∂ F ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x − ∂ F ∂x ,F − ( y ) = ∂ F ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x − ∂ F ∂x ,F − ( y ) = 1 τ [ µ △ ( ∂F ∂x + ∂F ∂x + ∂F ∂x ) − ∂ F ∂t∂x − ∂ F ∂t∂x − ∂ F ∂t∂x ] . And we can get the following, w = − ∂F − ( y ) ∂x − ∂F − ( y ) ∂x = ∂ F ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x ∂x − ∂ F ∂x ∂x ,w = − ∂τ F − ( y ) ∂x + µ △ F − ( y ) , w = − ∂τ F − ( y ) ∂x + µ △ F − ( y ) ,w = − ∂τ F − ( y ) ∂x + µ △ F − ( y ) ,w = ∂F − ( y ) ∂x = ∂ F ∂x ∂x + ∂ F ∂x ∂x ∂x − ∂ F ∂x − ∂ F ∂x ∂x ,w = ∂F − ( y ) ∂x = ∂ F ∂x ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x ∂x − ∂ F ∂x ,w = F − ( y ) = ∂ F ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x − ∂ F ∂x ,w = ∂F − ( y ) ∂x = ∂ F ∂x ∂x + ∂ F ∂x ∂x ∂x − ∂ F ∂x − ∂ F ∂x ∂x ,w = ∂F − ( y ) ∂x = ∂ F ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x ∂x − ∂ F ∂x ∂x ,w = ∂F − ( y ) ∂x = ∂ F ∂x ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x ∂x − ∂ F ∂x ,w = F − ( y ) = ∂ F ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x − ∂ F ∂x ,w = ∂F − ( y ) ∂x = ∂ F ∂x ∂x + ∂ F ∂x ∂x ∂x − ∂ F ∂x − ∂ F ∂x ∂x ,w = ∂F − ( y ) ∂x = ∂ F ∂x ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x ∂x − ∂ F ∂x ,w = ∂F − ( y ) ∂x = ∂ F ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x ∂x − ∂ F ∂x ∂x , = F − ( y ) = ∂ F ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x − ∂ F ∂x ,w = F − ( y ) = 1 τ [ µ △ ( ∂F ∂x + ∂F ∂x + ∂F ∂x ) − ∂ F ∂t∂x − ∂ F ∂t∂x − ∂ F ∂t∂x ] . And we can see that W is the function to do the partial derivation with F , F , F no morethan the fourth order and their linear combination, moreover no more than the first order with thevariable t . Hence H [ F − ( Y )] = H [ F − ( Y )] I K ′ ∈ C ( K ′ ) .If Z ∈ Ω , then there exists h = h I K ′ ∈ C ( K ′ ) , such that Z = F ( h ) . Again from thecorollary 3.2 , we can get ∃ h = h I K ′ ∈ C [0 , T ] T C ∞ ( K ) , such that F ( h ) = a bτ F ( h ) . Nextwe see H [ F − ( A ( η ) Z )] , where A ( η ) = ( η , η , η , η , η , η , η , η , η ) , η j = ( iξ y , iξ y , y , iξ y , iξ y , iξ y , y , iξ y , iξ y , iξ y , y ,iξ y , iξ y , iξ y , iξ y , y , ( iξ ) y , ( iξ ) y , ( iξ ) y , ( iξ ) y , iξ iξ y ,iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , ( iξ ) y , ( iξ ) y , ( iξ ) y , ( iξ ) y ,iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , ( iξ ) y , ( iξ ) y , ( iξ ) y , ( iξ ) y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , iξ iξ y , e Tj ) T , here e j is the jth ≤ j ≤ j = 1 , , y = 1 a bτ ( ξ + ab ) , y = 1 a bτ ( − iξ iξ ) , y = 1 a bτ ( − iξ iξ ) , y = 1 a bτ ( − iξ a ) , where b = ( iξ ) + ( iξ ) + ( iξ ) a = 0 , and ( ξ , ξ , ξ ) = (0 , ,
0) ,when j = 4 , , y = 1 a bτ ( ξ + ab ) , y = 1 a bτ ( − iξ iξ ) , y = 1 a bτ ( − iξ iξ ) , y = 1 a bτ ( − iξ a ) , when j = 7 , , y = 1 a bτ ( ξ + ab ) , y = 1 a bτ ( − iξ iξ ) , y = 1 a bτ ( − iξ iξ ) , y = 1 a bτ ( − iξ a ) . Hence we can get H [ F − ( A ( η ) Z )] = H [ F − ( A ( η ) F ( h ))]= H [ F − ( A ( η ) a bτ F ( h ))] = H { F − [( a bτ A ( η )) F ( h )] } , and we can see that a bτ A ( η ) is a polynomial matrix. We assume H { F − [( a bτ A ( η )) F ( h )] } = W ( h ) , where W ( h ) = ( w j ) × , and we assume h = ( h j ) × , then we can get the following. w = − ( e T + e T ) F − [( a bτ A ( η )) F ( h )]= − F − { iξ [ − iξ iξ ( F ( h ) + F ( h ) + F ( h )) + ( ξ + ab )( F ( h ) + F ( h ) + F ( h )) − iξ iξ ( F ( h ) + F ( h ) + F ( h ))] + iξ [ − iξ iξ ( F ( h ) + F ( h ) + F ( h )) − iξ iξ ( F ( h ) + F ( h ) + F ( h )) + ( ξ + ab )( F ( h ) + F ( h ) + F ( h ))] } , = − [ τ e T − µ ( e T + e T + e T ) + e T + e T + e T ] F − [( a bτ A ( η )) F ( h )]= − F − { [ τ iξ ( − iξ a ) − µ (( iξ ) + ( iξ ) + ( iξ ) )( ξ + ab ) + a bτ ]( F ( h ) + F ( h ) + F ( h ))+[ τ iξ ( − iξ a ) − µ (( iξ ) + ( iξ ) + ( iξ ) )( − iξ iξ )]( F ( h ) + F ( h ) + F ( h ))+[ τ iξ ( − iξ a ) − µ (( iξ ) + ( iξ ) + ( iξ ) )( − iξ iξ )]( F ( h ) + F ( h ) + F ( h )) } ,w = − [ τ e T − µ ( e T + e T + e T ) + e T + e T + e T ] F − [( a bτ A ( η )) F ( h )]= − F − { [ τ iξ ( − iξ a ) − µ (( iξ ) + ( iξ ) + ( iξ ) )( − iξ iξ )]( F ( h ) + F ( h ) + F ( h ))+[ τ iξ ( − iξ a ) − µ (( iξ ) + ( iξ ) + ( iξ ) )( ξ + ab ) + a bτ ]( F ( h ) + F ( h ) + F ( h ))+[ τ iξ ( − iξ a ) − µ (( iξ ) + ( iξ ) + ( iξ ) )( − iξ iξ )]( F ( h ) + F ( h ) + F ( h )) } ,w = − [ τ e T − µ ( e T + e T + e T ) + e T + e T + e T ] F − [( a bτ A ( η )) F ( h )]= − F − { [ τ iξ ( − iξ a ) − µ (( iξ ) + ( iξ ) + ( iξ ) )( − iξ iξ )]( F ( h ) + F ( h ) + F ( h ))+[ τ iξ ( − iξ a ) − µ (( iξ ) + ( iξ ) + ( iξ ) )( − iξ iξ )]( F ( h ) + F ( h ) + F ( h ))+[ τ iξ ( − iξ a ) − µ (( iξ ) + ( iξ ) + ( iξ ) )( ξ + ab ) + a bτ ]( F ( h ) + F ( h ) + F ( h )) } ,w = e T F − [( a bτ A ( η )) F ( h )] = F − { iξ [( ξ + ab )( F ( h ) + F ( h ) + F ( h )) − iξ iξ ( F ( h ) + F ( h ) + F ( h )) − iξ iξ ( F ( h ) + F ( h ) + F ( h ))] } ,w = e T F − [( a bτ A ( η )) F ( h )] = F − { iξ [( ξ + ab )( F ( h ) + F ( h ) + F ( h )) − iξ iξ ( F ( h ) + F ( h ) + F ( h )) − iξ iξ ( F ( h ) + F ( h ) + F ( h ))] } ,w = e T F − [( a bτ A ( η )) F ( h )] = F − { ( ξ + ab )( F ( h ) + F ( h ) + F ( h )) − iξ iξ ( F ( h ) + F ( h ) + F ( h )) − iξ iξ ( F ( h ) + F ( h ) + F ( h )) } ,w = e T F − [( a bτ A ( η )) F ( h )] = F − { iξ [ − iξ iξ ( F ( h ) + F ( h ) + F ( h ))+( ξ + ab )( F ( h ) + F ( h ) + F ( h )) − iξ iξ ( F ( h ) + F ( h ) + F ( h ))] } ,w = e T F − [( a bτ A ( η )) F ( h )] = F − { iξ [ − iξ iξ ( F ( h ) + F ( h ) + F ( h ))+( ξ + ab )( F ( h ) + F ( h ) + F ( h )) − iξ iξ ( F ( h ) + F ( h ) + F ( h ))] } ,w = e T F − [( a bτ A ( η )) F ( h )] = F − { iξ [ − iξ iξ ( F ( h ) + F ( h ) + F ( h ))+( ξ + ab )( F ( h ) + F ( h ) + F ( h )) − iξ iξ ( F ( h ) + F ( h ) + F ( h ))] } ,w = e T F − [( a bτ A ( η )) F ( h )] = F − {− iξ iξ ( F ( h ) + F ( h ) + F ( h ))+( ξ + ab )( F ( h ) + F ( h ) + F ( h )) − iξ iξ ( F ( h ) + F ( h ) + F ( h )) } ,w = e T F − [( a bτ A ( η )) F ( h )] = F − { iξ [ − iξ iξ ( F ( h ) + F ( h ) + F ( h )) − iξ iξ ( F ( h ) + F ( h ) + F ( h )) + ( ξ + ab )( F ( h ) + F ( h ) + F ( h ))] } ,w = e T F − [( a bτ A ( η )) F ( h )] = F − { iξ [ − iξ iξ ( F ( h ) + F ( h ) + F ( h )) − iξ iξ ( F ( h ) + F ( h ) + F ( h )) + ( ξ + ab )( F ( h ) + F ( h ) + F ( h ))] } ,w = e T F − [( a bτ A ( η )) F ( h )] = F − { iξ [ − iξ iξ ( F ( h ) + F ( h ) + F ( h )) − iξ iξ ( F ( h ) + F ( h ) + F ( h )) + ( ξ + ab )( F ( h ) + F ( h ) + F ( h ))] } ,w = e T F − [( a bτ A ( η )) F ( h )] = F − {− iξ iξ ( F ( h ) + F ( h ) + F ( h )) − iξ iξ ( F ( h ) + F ( h ) + F ( h )) + ( ξ + ab )( F ( h ) + F ( h ) + F ( h )) } ,w = e T F − [( a bτ A ( η )) F ( h )] = F − {− iξ a ( F ( h ) + F ( h ) + F ( h )) − iξ a ( F ( h ) + F ( h ) + F ( h )) − iξ a ( F ( h ) + F ( h ) + F ( h )) } . h = h I K ′ ∈ C [0 , T ] T C ∞ ( K ) , we can get F − ( iξ F ( h )) = F − ( F ( ∂h ∂t )) = ∂h ∂t , for the same reason we can get F − ( iξ F ( h )) = ∂h ∂x , F − ( iξ F ( h )) = ∂h ∂x , F − ( iξ F ( h )) = ∂h ∂x . If we assume h = h + h + h , h = h + h + h , h = h + h + h , then we can getthe following. w = ∂ h ∂x ∂x + ∂ h ∂x ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x , w = ∂ h ∂t∂x + ∂ h ∂t∂x − ∂ h ∂t∂x ∂x − ∂ h ∂t∂x ∂x ,w = ∂ h ∂t∂x + ∂ h ∂t∂x − ∂ h ∂t∂x ∂x − ∂ h ∂t∂x ∂x , w = ∂ h ∂t∂x + ∂ h ∂t∂x − ∂ h ∂t∂x ∂x − ∂ h ∂t∂x ∂x ,w = ∂ h ∂x + ∂ h ∂x ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x ∂x , w = ∂ h ∂x ∂x + ∂ h ∂x − ∂ h ∂x ∂x ∂x − ∂ h ∂x ∂x ,w = ∂ h ∂x + ∂ h ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x , w = ∂ h ∂x + ∂ h ∂x ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x ∂x ,w = ∂ h ∂x ∂x + ∂ h ∂x ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x , w = ∂ h ∂x ∂x + ∂ h ∂x − ∂ h ∂x ∂x ∂x − ∂ h ∂x ∂x ,w = ∂ h ∂x + ∂ h ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x , w = ∂ h ∂x + ∂ h ∂x ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x ∂x ,w = ∂ h ∂x ∂x + ∂ h ∂x ∂x − ∂ h ∂x ∂x ∂x − ∂ h ∂x ∂x , w = ∂ h ∂x ∂x + ∂ h ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x ,w = ∂ h ∂x + ∂ h ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x ,w = 1 τ ( ∂ h ∂t∂x + ∂ h ∂t∂x + ∂ h ∂t∂x ) − µτ ( ∂ △ h ∂x + ∂ △ h ∂x + ∂ △ h ∂x ) . We can see that W is the function to do the partial derivation with the components of h no morethan the third order and their linear combination, moreover no more than the first order with thevariable t . Hence H [ F − ( A ( η ) Z )] = H [ F − ( A ( η ) Z )] I K ′ ∈ C ( K ′ ) .Now we can get H [ F − ( Y + A ( η ) Z )] = H [ F − ( Y + A ( η ) Z )] I K ′ ∈ C ( K ′ ) , if Z ∈ Ω , thismeans that Ω ⊂ Ω , moreover the components of F − ( f ( Z )) are all in the H [ F − ( Y + A ( η ) Z )] , hence f ( Z ) ∈ Ω , if Z ∈ Ω . And we can get if Z ∈ Ω and Z = f ( Z ) , then Z ∈ Ω .Secondly we prove the fixed-point of f ( Z ) is exist, where Z ∈ Ω . In order to discuss moreconveniently, we assume f j ( Z ) = F [ F − ( α Tj ( Y + A ( η ) Z )) F − ( α Tj ( Y + A ( η ) Z ))] , ≤ j ≤ α = e , α = e , α = e , α = e , α = e , α = e , α = e , α = e ,α = e , α = − α , α = e , α = e , α = e , α = e , α = e , α = e ,α = e , α = e , and α = e + e , e i is the ith
55 dimensional unit coordinate vector,1 ≤ i ≤
55 . 21nd we assume the set Ω C as follows.Ω C = { h | h ∈ C ( K ′ ) , ∃ M > , such that | h | ≤ M , moreover ∃ C > , < α < , such that ∀ x t , x ′ t ∈ K ′ , | h ( x t ) − h ( x ′ t ) | ≤ C | x t − x ′ t | α . } , where h = ( h j ) × , | h | = max ≤ j ≤ | h j | , | h j | = max x t ∈ K ′ | h j ( x t ) | , x t = ( t, x , x , x ) T . From the Arzela-Ascoli theorem, we know Ω C is a compact set. And we assume M in the Ω C satisfiesthe following.max ≤ j ≤ { | F − ( α Tj Y ) | , | F − ( α Tj Y ) | , | F − ( α Tj Y ) F − ( α Tj Y ) |} ≤ θM , where 0 < θ < , | F − ( α Tj Y ) | = max x t ∈ K ′ | F − ( α Tj Y )( x t ) | , | F − ( α Tj Y ) | = max x t ∈ K ′ | F − ( α Tj Y )( x t ) | , | F − ( α Tj Y ) F − ( α Tj Y ) | = max x t ∈ K ′ | F − ( α Tj Y )( x t ) F − ( α Tj Y )( x t ) | , ≤ j ≤ . We assume ∀ x t , x ′ t ∈ K ′ , ∀ j , ≤ j ≤ , ∃ C > | F − ( α Tj Y )( x t ) − F − ( α Tj Y )( x ′ t ) | ≤ C | x t − x ′ t | α , | F − ( α Tj Y )( x t ) − F − ( α Tj Y )( x ′ t ) | ≤ C | x t − x ′ t | α , | F − ( α Tj Y )( x t ) F − ( α Tj Y )( x t ) − F − ( α Tj Y )( x ′ t ) F − ( α Tj Y )( x ′ t ) | ≤ C | x t − x ′ t | α . Next we assume g ( h ) = F − [ f ( F ( h ))], and g j ( h ) = F − [ f j ( F ( h ))] , ≤ j ≤ h ∈ Ω C .We will prove that the fixed-point of g ( h ) is exist. Moreover we can get h ∈ C ( K ′ ) , if h = g ( h )and h ∈ Ω C .We only need to show | g j ( h ) | ≤ M , moreover ∀ x t , x ′ t ∈ K ′ , we can get | g j ( x t ) − g j ( x ′ t ) | ≤ C | x t − x ′ t | α , ≤ j ≤ g j ( h ) = F − [ α Tj ( Y + A ( η ) F ( h ))] F − [ α Tj ( Y + A ( η ) F ( h ))] , = F − ( α Tj Y ) F − ( α Tj Y ) + F − ( α Tj Y ) F − ( α Tj A ( η ) F ( h )) + F − ( α Tj Y ) F − ( α Tj A ( η ) F ( h )) + F − ( α Tj A ( η ) F ( h )) F − ( α Tj A ( η ) F ( h )) , and from the lemma 3.1 , we can get ∃ h = h I K ′ ∈ C [0 , T ] T C ∞ ( K ) , such that g j ( h ) = F − ( α Tj Y ) F − ( α Tj Y ) + F − ( α Tj Y ) F − ( α Tj A ( η ) a bτ F ( h )) + F − ( α Tj Y ) F − ( α Tj A ( η ) a bτ F ( h )) + F − ( α Tj A ( η ) a bτ F ( h )) F − ( α Tj A ( η ) a bτ F ( h ))= F − ( α Tj Y ) F − ( α Tj Y ) + F − ( α Tj Y ) W j ( h ) + F − ( α Tj Y ) W j ( h ) + W j ( h ) W j ( h ) , where W j ( h ) = F − ( α Tj ( a bτ A ( η )) F ( h )) , W j ( h ) = F − ( α Tj ( a bτ A ( η )) F ( h )) , ≤ j ≤ , and α = e , α = e , α = e , α = e , α = e , α = e , α = e , α = e ,α = e , α = − α , α = e , α = e , α = e , α = e , α = e , α = e , = e , α = e , and α = e + e , e i is the ith
55 dimensional unit coordinate vector,1 ≤ i ≤
55 . And we can get the following. F − ( α T Y ) = w = ∂ F ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x − ∂ F ∂x ,F − ( α T Y ) = w = ∂ F ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x ∂x − ∂ F ∂x ∂x ,F − ( α T Y ) = w = ∂ F ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x − ∂ F ∂x ,F − ( α T Y ) = w = ∂ F ∂x ∂x + ∂ F ∂x ∂x ∂x − ∂ F ∂x − ∂ F ∂x ∂x ,F − ( α T Y ) = w = ∂ F ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x − ∂ F ∂x ,F − ( α T Y ) = w = ∂ F ∂x ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x ∂x − ∂ F ∂x ,F − ( α T Y ) = w = ∂ F ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x − ∂ F ∂x ,F − ( α T Y ) = w = ∂ F ∂x ∂x + ∂ F ∂x ∂x ∂x − ∂ F ∂x − ∂ F ∂x ∂x ,F − ( α T Y ) = w = ∂ F ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x − ∂ F ∂x ,F − ( α T Y ) = w = ∂ F ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x ∂x − ∂ F ∂x ∂x ,F − ( α T Y ) = w = ∂ F ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x − ∂ F ∂x ,F − ( α T Y ) = w = ∂ F ∂x ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x ∂x − ∂ F ∂x ,F − ( α T Y ) = w = ∂ F ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x − ∂ F ∂x ,F − ( α T Y ) = w = ∂ F ∂x ∂x + ∂ F ∂x ∂x ∂x − ∂ F ∂x − ∂ F ∂x ∂x ,F − ( α T Y ) = w = ∂ F ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x − ∂ F ∂x ,F − ( α T Y ) = w = ∂ F ∂x ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x ∂x − ∂ F ∂x ,F − ( α T Y ) = w = ∂ F ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x − ∂ F ∂x ,F − ( α T Y ) = w = ∂ F ∂x ∂x + ∂ F ∂x ∂x − ∂ F ∂x ∂x − ∂ F ∂x ∂x , j ( t, M ) = ( 12 √ πµ ) Z t Z R F jv ( τ , y , y , y )( √ t − τ ) e − ( x − y x − y x − y µ ( t − τ dy dy dy dτ ,F jv ( t, M ) = − π Z K F j ( t, M ) r MM dx dx dx , ≤ j ≤ ,W ( h ) = w = ∂ h ∂x + ∂ h ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x ,W ( h ) = w = ∂ h ∂x ∂x + ∂ h ∂x ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x ,W ( h ) = w = ∂ h ∂x + ∂ h ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x ,W ( h ) = w = ∂ h ∂x + ∂ h ∂x ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x ∂x ,W ( h ) = w = ∂ h ∂x + ∂ h ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x ,W ( h ) = w = ∂ h ∂x ∂x + ∂ h ∂x − ∂ h ∂x ∂x ∂x − ∂ h ∂x ∂x ,W ( h ) = w = ∂ h ∂x + ∂ h ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x ,W ( h ) = w = ∂ h ∂x + ∂ h ∂x ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x ∂x ,W ( h ) = w = ∂ h ∂x + ∂ h ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x ,W ( h ) = w = ∂ h ∂x ∂x + ∂ h ∂x ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x ,W ( h ) = w = ∂ h ∂x + ∂ h ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x ,W ( h ) = w = ∂ h ∂x ∂x + ∂ h ∂x − ∂ h ∂x ∂x ∂x − ∂ h ∂x ∂x ,W ( h ) = w = ∂ h ∂x + ∂ h ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x ,W ( h ) = w = ∂ h ∂x + ∂ h ∂x ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x ∂x ,W ( h ) = w = ∂ h ∂x + ∂ h ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x ,W ( h ) = w = ∂ h ∂x ∂x + ∂ h ∂x ∂x − ∂ h ∂x ∂x ∂x − ∂ h ∂x ∂x ,W ( h ) = w = ∂ h ∂x + ∂ h ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x ,W ( h ) = w = ∂ h ∂x ∂x + ∂ h ∂x − ∂ h ∂x ∂x − ∂ h ∂x ∂x , h = ( h j ) × , and h = h + h + h , h = h + h + h , h = h + h + h .Hence we can get W j ( h ) , W j ( h ) , ≤ j ≤ h no more than the third order and their linear combination, moreover only dothe partial derivation with the variables x , x , x . From the lemma 3.1 , we know v ( t, M ) = − π Z K h ( t, M ) r MM dx dx dx , where M = ( x , x , x ) , M = ( x , x , x ) ,r MM = p ( x − x ) + ( x − x ) + ( x − x ) ,h ( t, M ) = ( 12 √ πµ ) Z t Z R v ( τ , y , y , y )( √ t − τ ) e − ( x − y x − y x − y µ ( t − τ dy dy dy dτ . And we can get the lemma as follows.
Lemma 3.2 ∃ M T, > , and M T, is independent with M , K , ∀ h ∈ Ω C , we can get thefollowing, max ≤ i ≤ {| h | , | ∂h ∂t | , | ∂h ∂x i | , | ∂ ∂h ∂x i ∂t |} ≤ M M T, M ( K ) , where h = ( h m ) × , h = h ( t, M ) = ( h m ( t, M )) × , | h | = max ≤ m ≤ max ( t, M ) ∈ K ′ | h m ( t, M ) | , | ∂h ∂t | = max ≤ m ≤ max ( t, M ) ∈ K ′ | ∂h m ( t, M ) ∂t | , | ∂h ∂x i | = max ≤ m ≤ max ( t, M ) ∈ K ′ | ∂h m ( t, M ) ∂x i | , | ∂ ∂h ∂x i ∂t | = max ≤ m ≤ max ( t, M ) ∈ K ′ | ∂ ∂h m ( t, M ) ∂x i ∂t | , M ( K ) = max M ∈ K π Z K r MM dx dx dx . Proof of lemma3-2 . We denote h m = h m ( t, M ) , ≤ m ≤ v ( t, M ) = ( v m ) × ,where v m = v m ( t, M ) , then from the lemma 3.1 , we can get ∀ m , ≤ m ≤ v m ( t, M ) = − π Z K h m ( t, M ) r MM dx dx dx , where M = ( x , x , x ) , M = ( x , x , x ) ,r MM = p ( x − x ) + ( x − x ) + ( x − x ) ,h m ( t, M ) = ( 12 √ πµ ) Z t Z R v m ( τ , y , y , y )( √ t − τ ) e − ( x − y x − y x − y µ ( t − τ dy dy dy dτ . if h ∈ Ω C , then we can get | v m | ≤ | h m | π Z K r MM dx dx dx ≤ M M ( K ) , moreover | h m | ≤ ( 12 √ πµ ) Z t Z R M M ( K )( √ t − τ ) e − ( x − y x − y x − y µ ( t − τ dy dy dy dτ ≤ T M M ( K ) , where M ( K ) = max M ∈ K π Z K r MM dx dx dx . ∂h m ∂t = lim τ → t ( 12 √ πµ ) Z R v m ( τ , y , y , y )( √ t − τ ) e − ( x − y x − y x − y µ ( t − τ dy dy dy +( 12 √ πµ ) Z t Z R ∂ ( v m ( τ , y , y , y )( √ t − τ ) e − ( x − y x − y x − y µ ( t − τ ) ∂t dy dy dy dτ . If we let y j = x j + 2 p µ ( t − τ ) θ j , j = 1 , , ∂h m ∂t = lim τ → t ( 1 √ π ) Z R v m ( τ , x j + 2( p µ ( t − τ )) θ j , j = 1 , , , ) e − ( θ + θ + θ ) dθ dθ dθ +( 12 √ πµ ) Z t Z R − v m ( τ , y , y , y )( √ t − τ ) e − ( x − y x − y x − y µ ( t − τ dy dy dy dτ +( 12 √ πµ ) Z t Z R v m ( τ , y , y , y )4 µ ( √ t − τ ) X i =1 ( x i − y i ) e − P i =1( xi − yi )24 µ ( t − τ dy dy dy dτ . If we let y j = x j + 2 √ µ ( p ( t − τ )) θ j , j = 1 , , , to the second integral, and we let y j = x j + 2 √ µ ( p ( t − τ )) θ j , j = 1 , , , to the third integral, then we can get as follows, ∂h m ∂t = v m ( t, x , x , x ) + ( 1 √ π ) Z t Z R − v (1) m ( t − τ ) e − ( t − τ ) ( θ + θ + θ ) dθ dθ dθ dτ +( 1 √ π ) Z t Z R v (2) m ( t − τ ) ( θ + θ + θ ) e − ( t − τ ) ( θ + θ + θ ) dθ dθ dθ dτ , where v (1) m = v m ( τ , x j +2 √ µ ( p ( t − τ )) θ j , j = 1 , , , ) , v (2) m = v m ( τ , x j +2 √ µ ( p ( t − τ )) θ j , j = 1 , , , ) . Hence we can get | ∂h m ∂t | ≤ M M ( K ) + ( 1 √ π ) Z t Z R M M ( K )( t − τ ) e − ( t − τ ) ( θ + θ + θ ) dθ dθ dθ dτ +( 1 √ π ) Z t Z R M M ( K )( t − τ ) ( θ + θ + θ ) e − ( t − τ ) ( θ + θ + θ ) dθ dθ dθ dτ . We assume ϕ ( t, τ ) , ϕ ( t, τ ) as follows, ϕ ( t, τ ) = Z R ( t − τ ) e − ( t − τ ) ( θ + θ + θ ) dθ dθ dθ ,ϕ ( t, τ ) = Z R ( t − τ ) ( θ + θ + θ ) e − ( t − τ ) ( θ + θ + θ ) dθ dθ dθ . We can see that ϕ ( t, τ ) , ϕ ( t, τ ) are continuous about t, τ on the region 0 ≤ τ ≤ t , ≤ t ≤ T , hence they are bounded on the region 0 ≤ τ ≤ t , ≤ t ≤ T . We assume there exist M ′ > | ϕ ( t, τ ) | ≤ M ′ , | ϕ ( t, τ ) | ≤ M ′ , where 0 ≤ τ ≤ t , ≤ t ≤ T .
Hence we can get | ∂h m ∂t | ≤ M M ( K ) + ( 1 √ π ) T M M ( K ) M ′ . ∂h m ∂x i = ( 12 √ πµ ) Z t Z R v m ( τ , y , y , y )4 µ ( √ t − τ ) ( − x i − y i ) e − P j =1( xj − yj )24 µ ( t − τ dy dy dy dτ . If we let y j = x j + 2 √ µ ( p ( t − τ )) θ j , j = 1 , , , we can get the follows, ∂h m ∂x i = ( 1 √ π ) Z t Z R √ µ v (1) m ( √ t − τ ) θ i e − ( t − τ ) ( θ + θ + θ ) dθ dθ dθ dτ . We assume ϕ ( t, τ ) as follows, ϕ ( t, τ ) = Z R ( √ t − τ ) θ i e − ( t − τ ) ( θ + θ + θ ) dθ dθ dθ , We can also see that ϕ ( t, τ ) is continuous about t, τ on the region 0 ≤ τ ≤ t , ≤ t ≤ T , henceit is bounded on the region 0 ≤ τ ≤ t , ≤ t ≤ T . We assume there exist M ′′ > | ϕ ( t, τ ) | ≤ M ′′ , where 0 ≤ τ ≤ t , ≤ t ≤ T .
Hence we can get | ∂h m ∂x i | ≤ ( 1 √ π ) T √ µ M M ( K ) M ′′ . At last we see from ∂ ∂h m ∂x i ∂t = lim τ → t ( 12 √ πµ ) Z R v m ( τ , y , y , y )4 µ ( √ t − τ ) ( − x i − y i ) e − P j =1( xj − yj )24 µ ( t − τ dy dy dy +( 12 √ πµ ) [ Z t Z R v m ( τ , y , y , y )4 µ ( √ t − τ ) (5)( x i − y i ) e − P j =1( xj − yj )24 µ ( t − τ dy dy dy dτ + Z t Z R v m ( τ , y , y , y )16 µ ( √ t − τ ) ( − x i − y i ) X j =1 ( x j − y j ) e − P i =1( xi − yi )24 µ ( t − τ dy dy dy dτ ] . If we let y j = x j + 2 √ µ ( p ( t − τ )) θ j , j = 1 , , , to the first integral, and we let y j = x j + 2 √ µ ( p ( t − τ )) θ j , j = 1 , , , to the second integral, y j = x j + 2 √ µ ( p ( t − τ )) θ j , j = 1 , , , to the third integral, then we can get as follows, ∂ ∂h m ∂x i ∂t = lim τ → t ( 1 √ π ) Z R √ µ v (1) m ( √ t − τ ) θ i e − ( t − τ ) ( θ + θ + θ ) dθ dθ dθ +( 1 √ π ) Z t Z R v (2) m √ µ ( √ t − τ ) (5) θ i e − ( t − τ ) P j =1 θ j dy dy dy dτ +( 1 √ π ) Z t Z R v (3) m √ µ ( √ t − τ ) θ i ( θ + θ + θ ) e − ( t − τ ) P j =1 θ j dy dy dy dτ , where v (3) m = v m ( τ , x j + 2 √ µ ( p ( t − τ )) θ j , j = 1 , , , ) .
27e assume ϕ ( t, τ ) , ϕ ( t, τ ) as follows, ϕ ( t, τ ) = Z R ( √ t − τ ) θ i e − ( t − τ ) ( θ + θ + θ ) dθ dθ dθ ,ϕ ( t, τ ) = Z R ( √ t − τ ) θ i ( θ + θ + θ ) e − ( t − τ ) ( θ + θ + θ ) dθ dθ dθ . We can see that ϕ ( t, τ ) , ϕ ( t, τ ) are continuous about t, τ on the region 0 ≤ τ ≤ t , ≤ t ≤ T , hence they are bounded on the region 0 ≤ τ ≤ t , ≤ t ≤ T . We assume there exist M ′′′ > | ϕ ( t, τ ) | ≤ M ′′′ , | ϕ ( t, τ ) | ≤ M ′′′ , where 0 ≤ τ ≤ t , ≤ t ≤ T .
Hence we can get | ∂ ∂h m ∂x i ∂t | ≤ ( 1 √ π ) M M ( K ) √ µ M ′′ + ( 1 √ π ) √ µ T M M ( K ) M ′′′ . If we let M T, = max { T , √ π ) T M ′ , ( 1 √ π ) T √ µ M ′′ , ( 1 √ π ) √ µ M ′′ + ( 1 √ π ) √ µ T M ′′′ } , then we can get max ≤ i ≤ {| h | , | ∂h ∂t | , | ∂h ∂x i | , | ∂ ∂h ∂x i ∂t |} ≤ M M T, M ( K ) . Hence the lemma is true.