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Duality Principles and Numerical Procedures for a Large Class of Non-convex Models in the Calculus of Variations

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14 August 2023

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14 August 2023

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Abstract
This article develops duality principles and numerical results for a large class of non-convex variational models. The main results are based on fundamental tools of convex analysis, duality theory and calculus of variations. More specifically the approach is established for a class of non-convex functionals similar as those found in some models in phase transition. Finally, in the last section we present a concerning numerical example and the respective software.
Keywords: 
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1. Introduction

In this section we establish a dual formulation for a large class of models in non-convex optimization.
The main duality principle is applied to double well models similar as those found in the phase transition theory.
Such results are based on the works of J.J. Telega and W.R. Bielski [2,3,15,16] and on a D.C. optimization approach developed in Toland [17].
About the other references, details on the Sobolev spaces involved are found in [1]. Related results on convex analysis and duality theory are addressed in [5,7,8,10,14].
Finally, in this text we adopt the standard Einstein convention of summing up repeated indices, unless otherwise indicated.
In order to clarify the notation, here we introduce the definition of topological dual space.
Definition 1.1
(Topological dual spaces). Let U be a Banach space. We shall define its dual topological space, as the set of all linear continuous functionals defined on U. We suppose such a dual space of U, may be represented by another Banach space U * , through a bilinear form · , · U : U × U * R (here we are referring to standard representations of dual spaces of Sobolev and Lebesgue spaces). Thus, given f : U R linear and continuous, we assume the existence of a unique u * U * such that
f ( u ) = u , u * U , u U .
The norm of f , denoted by f U * , is defined as
f U * = sup u U { | u , u * U | : u U 1 } u * U * .
At this point we start to describe the primal and dual variational formulations.

2. A general duality principle non-convex optimization

In this section we present a duality principle applicable to a model in phase transition.
This case corresponds to the vectorial one in the calculus of variations.
Let Ω R n be an open, bounded, connected set with a regular (Lipschitzian) boundary denoted by Ω .
Consider a functional J : V R where
J ( u ) = F ( u 1 , , u N ) + G ( u 1 , , u N ) u i , f i L 2 ,
and where
V = { u = ( u 1 , , u N ) W 1 , p ( Ω ; R N ) : u = u 0 on Ω } ,
f L 2 ( Ω ; R N ) , and 1 < p < + .
We assume there exists α R such that
α = inf u V J ( u ) .
Moreover, suppose F and G are Fréchet differentiable but not necessarily convex. A global optimum point may not be attained for J so that the problem of finding a global minimum for J may not be a solution.
Anyway, one question remains, how the minimizing sequences behave close the infimum of J.
We intend to use duality theory to approximately solve such a global optimization problem.
Denoting V 0 = W 0 1 , p ( Ω ; R N ) , Y 1 = Y 1 * = L 2 ( Ω ; R N × n ) , Y 2 = Y 2 * = L 2 ( Ω ; R N × n ) , Y 3 = Y 3 * = L 2 ( Ω ; R N ) , at this point we define, F 1 : V × V 0 R , G 1 : V R , G 2 : V R , G 3 : V 0 R and G 4 : V R , by
F 1 ( u , ϕ ) = F ( u 1 + ϕ 1 , , u N + ϕ N ) + K 2 Ω u j · u j d x + K 2 2 Ω ϕ j · ϕ j d x
and
G 1 ( u 1 , , u n ) = G ( u 1 , , u N ) + K 1 2 Ω u j u j d x u i , f i L 2 ,
G 2 ( u 1 , , u N ) = K 1 2 Ω u j · u j d x ,
G 3 ( ϕ 1 , , ϕ N ) = K 2 2 Ω ϕ j · ϕ j d x ,
and
G 4 ( u 1 , , u N ) = K 1 2 Ω u j u j d x .
Define now J 1 : V × V 0 R ,
J 1 ( u , ϕ ) = F ( u + ϕ ) + G ( u ) u i , f i L 2 .
Observe that
J 1 ( u , ϕ ) = F 1 ( u , ϕ ) + G 1 ( u ) G 2 ( u ) G 3 ( ϕ ) G 4 ( u ) F 1 ( u , ϕ ) + G 1 ( u ) u , z 1 * L 2 ϕ , z 2 * L 2 u , z 3 * L 2 + sup v 1 Y 1 { v 1 , z 1 * L 2 G 2 ( v 1 ) } + sup v 2 Y 2 { v 2 , z 2 * L 2 G 3 ( v 2 ) } + sup u V { u , z 3 * L 2 G 4 ( u ) } = F 1 ( u , ϕ ) + G 1 ( u ) u , z 1 * L 2 ϕ , z 2 * L 2 u , z 3 * L 2 + G 2 * ( z 1 * ) + G 3 * ( z 2 * ) + G 4 * ( z 3 * ) = J 1 * ( u , ϕ , z * ) ,
u V , ϕ V 0 , z * = ( z 1 * , z 2 * , z 3 * ) Y * = Y 1 * × Y 2 * × Y 3 * .
From the general results in [17], we may infer that
inf ( u , ϕ ) V × V 0 J ( u , ϕ ) = inf ( u , ϕ , z * ) V × V 0 × Y * J 1 * ( u , ϕ , z * ) .
On the other hand
inf u V J ( u ) inf ( u , ϕ ) V × V 0 J 1 ( u , ϕ ) .
From these last two results we may obtain
inf u V J ( u ) inf ( u , ϕ , z * ) V × V 0 × Y * J 1 * ( u , ϕ , z * ) .
Moreover, from standards results on convex analysis, we may have
inf u V J 1 * ( u , ϕ , z * ) = inf u V { F 1 ( u , ϕ ) + G 1 ( u ) u , z 1 * L 2 ϕ , z 2 * L 2 u , z 3 * L 2 + G 2 * ( z 1 * ) + G 3 * ( z 2 * ) + G 4 * ( z 3 * ) } = sup ( v 1 * , v 2 * ) C * { F 1 * ( v 1 * + z 1 * , ϕ ) G 1 * ( v 2 * + z 3 * ) ϕ , z 2 * L 2 + G 2 * ( z 1 * ) + G 3 * ( z 2 * ) + G 4 * ( z 3 * ) } ,
where
C * = { v * = ( v 1 * , v 2 * ) Y 1 * × Y 3 * : div ( v 1 * ) i + ( v 2 * ) i = 0 , i { 1 , , N } } ,
F 1 * ( v 1 * + z 1 * , ϕ ) = sup v 1 Y 1 { v 1 , z 1 * + v 1 * L 2 F 1 ( v 1 , ϕ ) } ,
and
G 1 * ( v 2 * + z 2 * ) = sup u V { u , v 2 * + z 2 * L 2 G 1 ( u ) } .
Thus, defining
J 2 * ( ϕ , z * , v * ) = F 1 * ( v 1 * + z 1 * , ϕ ) G 1 * ( v 2 * + z 3 * ) ϕ , z 2 * L 2 + G 2 * ( z 1 * ) + G 3 * ( z 2 * ) + G 4 * ( z 3 * ) ,
we have got
inf u V J ( u ) inf ( u , ϕ ) V × V 0 J 1 ( u , ϕ ) = inf ( u , ϕ , z * ) V × V 0 × Y * J 1 * ( u , ϕ , z * ) = inf z * Y * inf ϕ V 0 sup v * C * J 2 * ( ϕ , z * , v * ) .
Finally, observe that
inf u V J ( u ) inf z * Y * inf ϕ V 0 sup v * C * J 2 * ( ϕ , z * , v * ) sup v * C * inf ( z * , ϕ ) Y * × V 0 J 2 * ( ϕ , z * , v * ) .
This last variational formulation corresponds to a concave relaxed formulation in v * concerning the original primal formulation.

4. A convex dual variational formulation for a third similar model

In this section we present another duality principle for a third related model in phase transition.
Let Ω = [ 0 , 1 ] R and consider a functional J : V R where
J ( u ) = 1 2 Ω min { ( u 1 ) 2 , ( u + 1 ) 2 } d x + 1 2 Ω u 2 d x u , f L 2 ,
and where
V = { u W 1 , 2 ( Ω ) : u ( 0 ) = 0 and u ( 1 ) = 1 / 2 }
and f L 2 ( Ω ) .
A global optimum point is not attained for J so that the problem of finding a global minimum for J has no solution.
Anyway, one question remains, how the minimizing sequences behave close to the infimum of J.
We intend to use the duality theory to solve such a global optimization problem in an appropriate sense to be specified.
At this point we define, F : V R and G : V R by
F ( u ) = 1 2 Ω min { ( u 1 ) 2 , ( u + 1 ) 2 } d x = 1 2 Ω ( u ) 2 d x Ω | u | d x + 1 / 2 F 1 ( u ) ,
and
G ( u ) = 1 2 Ω u 2 d x u , f L 2 .
Denoting Y = Y * = L 2 ( Ω ) we also define the polar functional F 1 * : Y * R and G * : Y * R by
F 1 * ( v * ) = sup v Y { v , v * L 2 F 1 ( v ) } = 1 2 Ω ( v * ) 2 d x + Ω | v * | d x ,
and
G * ( ( v * ) ) = sup u V { u , v * L 2 G ( u ) } = 1 2 Ω ( ( v * ) + f ) 2 d x 1 2 v * ( 1 ) .
Observe this is the scalar case of the calculus of variations, so that from the standard results on convex analysis, we have
inf u V J ( u ) = max v * Y * { F 1 * ( v * ) G * ( ( v * ) ) } .
Indeed, from the direct method of the calculus of variations, the maximum for the dual formulation is attained at some v ^ * Y * .
Moreover, the corresponding solution u 0 V is obtained from the equation
u 0 = G ( ( v ^ * ) ) ( v * ) = ( v ^ * ) + f .
Finally, the Euler-Lagrange equations for the dual problem stands for
( v * ) + f v * sign ( v * ) = 0 , in Ω , ( v * ) ( 0 ) + f ( 0 ) = 0 , ( v * ) ( 1 ) + f ( 1 ) = 1 / 2 ,
where sign ( v * ( x ) ) = 1 if v * ( x ) > 0 , sign ( v * ( x ) ) = 1 , if v * ( x ) < 0 and
1 sign ( v * ( x ) ) 1 ,
if v * ( x ) = 0 .
We have computed the solutions v * and corresponding solutions u 0 V for the cases in which f ( x ) = 0 and f ( x ) = sin ( π x ) / 2 .
For the solution u 0 ( x ) for the case in which f ( x ) = 0 , please see Figure 3.
For the solution u 0 ( x ) for the case in which f ( x ) = sin ( π x ) / 2 , please see Figure 4.
Remark 4.1.
Observe that such solutions u 0 obtained are not the global solutions for the related primal optimization problems. Indeed, such solutions reflect the average behavior of weak cluster points for concerning minimizing sequences.

4.1. The algorithm through which we have obtained the numerical results

In this subsection we present the software in MATLAB through which we have obtained the last numerical results.
This algorithm is for solving the concerning Euler-Lagrange equations for the dual problem, that is, for solving the equation
( v * ) + f v * sign ( v * ) = 0 , in Ω , ( v * ) ( 0 ) = 0 , ( v * ) ( 1 ) = 1 / 2 .
Here the concerning software in MATLAB. We emphasize to have used the smooth approximation
| v * | ( v * ) 2 + e 1 ,
where a small value for e 1 is specified in the next lines.
*************************************
  • clear all
  • m 8 = 800 ; (number of nodes)
  • d = 1 / m 8 ;
  • e 1 = 0.00001 ;
  • f o r i = 1 : m 8
    y o ( i , 1 ) = 0.01 ;
    y 1 ( i , 1 ) = sin ( π * i / m 8 ) / 2 ;
    e n d ;
  • f o r i = 1 : m 8 1
    d y 1 ( i , 1 ) = ( y 1 ( i + 1 , 1 ) y 1 ( i , 1 ) ) / d ;
    e n d ;
  • f o r k = 1 : 3000 (we have fixed the number of iterations)
    i = 1 ;
    h 3 = 1 / v o ( i , 1 ) 2 + e 1 ;
    m 12 = 1 + d 2 * h 3 + d 2 ;
    m 50 ( i ) = 1 / m 12 ;
    z ( i ) = m 50 ( i ) * ( d y 1 ( i , 1 ) * d 2 ) ;
  • f o r i = 2 : m 8 1
    h 3 = 1 / v o ( i , 1 ) 2 + e 1 ;
    m 12 = 2 + h 3 * d 2 + d 2 m 50 ( i 1 ) ;
    m 50 ( i ) = 1 / m 12 ;
    z ( i ) = m 50 ( i ) * ( z ( i 1 ) + d y 1 ( i , 1 ) * d 2 ) ;
    e n d ;
  • v ( m 8 , 1 ) = ( d / 2 + z ( m 8 1 ) ) / ( 1 m 50 ( m 8 1 ) ) ;
  • f o r i = 1 : m 8 1
    v ( m 8 i , 1 ) = m 50 ( m 8 i ) * v ( m 8 i + 1 ) + z ( m 8 i ) ;
    e n d ;
  • v ( m 8 / 2 , 1 )
  • v o = v ;
    e n d ;
  • f o r i = 1 : m 8 1
    u ( i , 1 ) = ( v ( i + 1 , 1 ) v ( i , 1 ) ) / d + y 1 ( i , 1 ) ;
    e n d ;
  • f o r i = 1 : m 8 1
    x ( i ) = i * d ;
    e n d ;
    p l o t ( x , u ( : , 1 ) )
********************************

6. An exact convex dual variational formulation for a non-convex primal one

In this section we develop a convex dual variational formulation suitable to compute a critical point for the corresponding primal one.
Let Ω R 2 be an open, bounded, connected set with a regular (Lipschitzian) boundary denoted by Ω .
Consider a functional J : V R where
J ( u ) = F ( u x , u y ) u , f L 2 ,
V = W 0 1 , 2 ( Ω ) and f L 2 ( Ω ) .
Here we denote Y = Y * = L 2 ( Ω ) and Y 1 = Y 1 * = L 2 ( Ω ) × L 2 ( Ω ) .
Defining
V 1 = { u V : u 1 , K 1 }
for some appropriate K 1 > 0 , suppose also F is twice Fréchet differentiable and
det 2 F ( u x , u y ) v 1 v 2 0 ,
u V 1 .
Define now F 1 : V R and F 2 : V R by
F 1 ( u x , u y ) = F ( u x , u y ) + ε 2 Ω u x 2 d x + ε 2 Ω u y 2 d x ,
and
F 2 ( u x , u y ) = ε 2 Ω u x 2 d x + ε 2 Ω u y 2 d x ,
where here we denote d x = d x 1 d x 2 .
Moreover, we define the respective Legendre transform functionals F 1 * and F 2 * as
F 1 * ( v * ) = v 1 , v 1 * L 2 + v 2 , v 2 * L 2 F 1 ( v 1 , v 2 ) ,
where v 1 , v 2 Y are such that
v 1 * = F 1 ( v 1 , v 2 ) v 1 ,
v 2 * = F 1 ( v 1 , v 2 ) v 2 ,
and
F 2 * ( v * ) = v 1 , v 1 * + f 1 L 2 + v 2 , v 2 * L 2 F 2 ( v 1 , v 2 ) ,
where v 1 , v 2 Y are such that
v 1 * + f 1 = F 2 ( v 1 , v 2 ) v 1 ,
v 2 * = F 2 ( v 1 , v 2 ) v 2 .
Here f 1 is any function such that
( f 1 ) x = f , in Ω .
Furthermore, we define
J * ( v * ) = F 1 * ( v * ) + F 2 * ( v * ) = F 1 * ( v * ) + 1 2 ε Ω ( v 1 * + f 1 ) 2 d x + 1 2 ε Ω ( v 2 * ) 2 d x .
Observe that through the target conditions
v 1 * + f 1 = ε u x ,
v 2 * = ε u y ,
we may obtain the compatibility condition
( v 1 * + f 1 ) y ( v 2 * ) x = 0 .
Define now
A * = { v * = ( v 1 * , v 2 * ) B r ( 0 , 0 ) Y 1 * : ( v 1 * + f 1 ) y ( v 2 * ) x = 0 , in Ω } ,
for some appropriate r > 0 such that J * is convex in B r ( 0 , 0 ) .
Consider the problem of minimizing J * subject to v * A * .
Assuming r > 0 is large enough so that the restriction in r is not active, at this point we define the associated Lagrangian
J 1 * ( v * , φ ) = J * ( v * ) + φ , ( v 1 * + f ) y ( v 2 * ) x L 2 ,
where φ is an appropriate Lagrange multiplier.
Therefore
J 1 * ( v * ) = F 1 * ( v * ) + 1 2 ε Ω ( v 1 * + f 1 ) 2 d x + 1 2 ε Ω ( v 2 * ) 2 d x + φ , ( v 1 * + f ) y ( v 2 * ) x L 2 .
The optimal point in question will be a solution of the corresponding Euler-Lagrange equations for J 1 * .
From the variation of J 1 * in v 1 * we obtain
F 1 * ( v * ) v 1 * + v 1 * + f ε φ y = 0 .
From the variation of J 1 * in v 2 * we obtain
F 1 * ( v * ) v 2 * + v 2 * ε + φ x = 0 .
From the variation of J 1 * in φ we have
( v 1 * + f ) y ( v 2 * ) x = 0 .
From this last equation, we may obtain u V such that
v 1 * + f = ε u x ,
and
v 2 * = ε u y .
From this and the previous extremal equations indicated we have
F 1 * ( v * ) v 1 * + u x φ y = 0 ,
and
F 1 * ( v * ) v 2 * + u y + φ x = 0 .
so that
v 1 * + f = F 1 ( u x φ y , u y + φ x ) v 1 ,
and
v 2 * = F 1 ( u x φ y , u y + φ x ) v 2 .
From this and equation (26) and (27) we have
ε F 1 * ( v * ) v 1 * x ε F 1 * ( v * ) v 2 * y + ( v 1 * + f 1 ) x + ( v 2 * ) y = ε u x x ε u y y + ( v 1 * ) x + ( v 2 * ) y + f = 0 .
Replacing the expressions of v 1 * and v 2 * into this last equation, we have
ε u x x ε u y y + F 1 ( u x φ y , u y + φ x ) v 1 x + F 1 ( u x φ y , u y + φ x ) v 2 y + f = 0 ,
so that
F ( u x φ y , u y + φ x ) v 1 x + F ( u x φ y , u y + φ x ) v 2 y + f = 0 , in Ω .
Observe that if
2 φ = 0
then there exists u ^ such that u and φ are also such that
u x φ y = u ^ x
and
u y + φ x = u ^ y .
The boundary conditions for φ must be such that u ^ W 0 1 , 2 .
From this and equation (29) we obtain
δ J ( u ^ ) = 0 .
Summarizing, we may obtain a solution u ^ W 0 1 , 2 of equation δ J ( u ^ ) = 0 by minimizing J * on A * .
Finally, observe that clearly J * is convex in an appropriate large ball B r ( 0 , 0 ) for some appropriate r > 0

10. A duality principle for a general vectorial case in the calculus of variations

In this section we develop a duality principle for a general vectorial case in variational optimization.
Let Ω R 3 be an open, bounded and connected set with a regular (Lipschitzian) boundary denoted by Ω . Let J : V R be a functional where
J ( u ) = G ( u 1 , , u N ) u , f L 2 ,
where
V = W 0 1 , 2 ( Ω ; R N )
and
f = ( f 1 , , f N ) L 2 ( Ω ; R N ) .
Here we have denoted u = ( u 1 , , u N ) V and
u , f L 2 = u i , f i L 2 ,
so that we may also denote
J ( u ) = G ( u ) u , f L 2 .
Assume
G ( u ) = Ω g ( u ) d x
where g : R 3 N R is a differentiable function such that
g ( y ) +
as | y | . Moreover, suppose there exists α R such that
α = inf u V J ( u ) .
It is well known that
α = inf u V J ( u ) = inf u V J * * ( u ) = inf u V { ( G ) * * ( u ) u , f L 2 } .
Under some mild hypotheses, from convexity, we have that
inf u V { ( G ) * * ( u ) u , f L 2 } = sup v * A * { ( G ) * ( d i v v * ) } = ( G ) * ( f ) ,
where
A * = { v * Y = Y * = L 2 ( Ω ; R 3 N ) : d i v v * + f = 0 } .
Now observe that the restriction v = u for some u V is equivalent to the restriction
curl v i = 0 , in Ω
where v = { v i } = { v i j } j = 1 3 , i { 1 , , N } , with appropriate boundary conditions, so that with an appropriate Lagrange multiplier ϕ = { ϕ i } , we obtain
( G ) * ( d i v v * ) = sup u V { u , d i v v * L 2 G ( u ) } = sup u V { u , v * L 2 G ( u ) } inf ϕ Y * sup v Y { v , v * L 2 G ( v ) + ϕ , curl v L 2 = inf ϕ Y * G * ( v * + curl ϕ ) .
where we have denoted
curl v = { curl v i }
and
curl ϕ = { curl ϕ i } .
Joining the pieces, we have got
inf u V J ( u ) = inf u V { G ( u ) u , f L 2 } sup ( v * , ϕ ) A * × Y * { G * ( v * + curl ϕ ) } ,
where we recall that Y = Y * = L 2 ( Ω ; R 3 N ) .
We emphasize such a dual formulation in ( v * , ϕ ) is convex (in fact concave).

11. A note on the Galerkin Functional

Let Ω R 3 be an open, bounded and connected set with a regular (Lipschitzian) boundary denoted by Ω .
Consider the functional J : V R where
J ( u ) = γ 2 Ω u · u d x + α 4 Ω u 4 d x β 2 Ω u 2 d x u , f L 2
Here V = W 0 1 , 2 ( Ω ) , γ > 0 , α > 0 , β > 0 .
We denote also
Y = Y * = L 2 ( Ω ) .
At this point we define
A + = { u V : u f 0 , in Ω } ,
V 2 = { u V : u K 3 } ,
for some appropriate real constant K 3 > 0 and
V 1 = A + V 2 .
Observe that
J ( u ) = γ 2 u + α u 3 β f ,
so that we define the Galerkin functional J 1 : V R by
J 1 ( u ) = 1 2 J ( u ) 2 2 = 1 2 Ω ( γ 2 u + α u 3 β u f ) 2 d x .
From this, we get
2 J 1 ( u ) u 2 = ( γ u + α u 3 β u f ) 6 α u + ( γ 2 + 3 α u 2 β ) 2 .
Define now
φ 2 = ( γ 2 u + α u 3 β u f ) 2 .
At this point, for an appropriate small real constant ε 1 > 0 and bounded constant operator M 1 > ε 1 , we set the intended non-active restriction
3 α | u | | M 1 + γ 2 + β | ,
and define
B 1 = { u V 1 : 3 α | u | | M 1 + γ 2 + β | } .
Observe that since for u V 1 we have u f 0 in Ω so that if u 1 , u 2 V 1 then
sign ( u 1 ) = sign ( u 2 ) , in Ω ,
we may infer that B 1 is a convex set.
Furthermore, if u B 1 , then
3 α | u | | M 1 + γ 2 + β | ,
so that
3 α u 2 M 1 + γ 2 + β ,
and hence
δ 2 J ( u ) = γ 2 + 3 α u 2 β M 1 > ε 1 > 0 .
For a small parameter ε > 0 we define the intended non-active restriction
φ 2 ε , in Ω ,
and define
B 2 = { u V 1 : φ 2 ε , in Ω } .
Observe that for α > 0 and β > 0 sufficiently large φ 2 is convex in V 1 (positive definite Hessian) so that B 2 is a convex set. Assuming 0 < ε ε 1 1 , define B 3 = B 1 B 2 , which is a convex set.
Summarizing, if u B 3 , then
δ 2 J 1 ( u ) 0 .
With such results in mind, we define the following convex optimization problem for finding a critical point of J.
Minimize
J 1 ( u ) = 1 2 J ( u ) 2 2 = 1 2 Ω ( γ 2 u + α u 3 β u f ) 2 d x ,
subject to
u B 3 .
Observe that a critical point u 0 B 3 of J 1 , from such a concerning convexity of J 1 on the convex set B 1 , is also such that
J ( u 0 ) = min u B 3 J 1 ( u ) .
Finally, we may also define the convex optimization problem of minimizing
J 3 ( u ) = K 1 J 1 ( u ) + J ( u ) = K 1 2 Ω ( γ 2 u + α u 3 β u f ) 2 d x + γ 2 Ω u · u d x + α 4 Ω u 4 d x β 2 Ω u 2 d x u , f L 2 ,
subject to
u B 3 .
Here K 1 > 0 is a large real constant.
Such a functional J 3 is also convex on B 3 so that a critical point u 0 B 3 of J is also a critical point of J 3 , and thus
J 3 ( u 0 ) = min u B 3 J 3 ( u ) .

12. A note on the Legendre-Galerkin functional

Let Ω R 3 be an open, bounded and connected set with a regular (Lipschitzian) boundary denoted by Ω .
Consider the functional J : V R where
J ( u ) = γ 2 Ω u · u d x + α 4 Ω u 4 d x β 2 Ω u 2 d x u , f L 2
Here V = W 0 1 , 2 ( Ω ) , γ > 0 , α > 0 , β > 0 .
We denote also
Y = Y * = L 2 ( Ω )
and F 1 : V R , F 2 : V R and F 3 : V R by
F 1 ( u ) = γ 2 Ω u · u d x ,
F 2 ( u ) = α 4 Ω u 4 d x ,
F 3 ( u ) = β 2 Ω u 2 d x .
Moreover, we define F 1 * , F 2 * , F 3 * : Y * R by
F 1 * ( v 1 * ) = sup u V { u , v 1 * L 2 F 1 ( u ) } = 1 2 Ω ( v 1 * ) 2 γ 2 d x ,
F 2 * ( v 2 * ) = sup u V { u , v 2 * L 2 F 2 ( u ) } = 3 4 Ω ( v 2 * ) 4 / 3 α 1 / 3 d x ,
F 3 * ( v 3 * ) = sup u V { u , v 3 * L 2 F 3 ( u ) } = 1 2 β Ω ( v 3 * ) 2 d x .
Observe now that these three last suprema are attained through the equations,
v 1 * = F 1 ( u ) u = γ 2 u ,
v 2 * = F 2 ( u ) u = α u 3
v 3 * = F 3 ( u ) u = β u .
From such results, at a critical point, we obtain the following compatibility conditions
u = v 1 * γ 2 = v 2 * β 1 / 3 = v 3 * β .
From such relations we have
v 1 * γ 2 = v 3 * β ,
and
v 2 * = α v 3 * β 3 ,
so that
v 1 * = γ 2 v 3 * β ,
and
v 2 * = α v 3 * β 3 .
Moreover, we define the functional F 4 * : Y * R , by
F 4 * ( v * ) = sup u V { u , v 1 * + v 2 * v 3 * L 2 u , f L 2 } .
Therefore
F 4 * ( v * ) = 0 , if v 1 * + v 2 * v 3 * f = 0 , in Ω , + , otherwise .
Hence, a critical point of J corresponds to the solution of the following system of equations
v 1 * = γ 2 v 3 * β ,
v 2 * = α v 3 * β 3 ,
and
v 1 * + v 2 * v 3 * f = 0 , in Ω .
From this last equation we may obtain
v 1 * = v 2 * + v 3 * + f ,
so that the final equations to be solved are
v 2 * + v 3 * + f + γ 2 v 3 * β = 0
and
v 2 * α v 3 * β 3 = 0 , in Ω ,
with the boundary conditions
u = v 3 * β = 0 , on Ω .
With such results in mind, we define the Legendre-Galerkin functional J * : [ Y * ] 2 R , where
J * ( v * ) = 1 2 Ω v 2 * + v 3 * + f + γ 2 v 3 * β 2 d x + 1 2 Ω v 2 * α v 3 * β 3 2 d x .
At this point, defining
φ = v 2 * α v 3 * β 3 ,
we obtain
2 J * ( v * ) ( v 2 * ) 2 = 2 ;
2 J * ( v * ) ( v 3 * ) 2 = 1 γ 2 β 2 + 9 α 2 ( v 3 * ) 4 β 6 + O ( φ ) ,
2 J * ( v * ) v 2 * v 3 * = 3 α ( v 3 * ) 2 β 3 + 1 γ 2 β .
From such results we may infer that
det 2 J * ( v * ) v 2 * v 3 * = 2 J * ( v * ) ( v 2 * ) 2 2 J * ( v * ) ( v 3 * ) 2 2 J * ( v * ) v 2 * v 3 * 2 = 1 γ 2 β + 3 α ( v 3 * ) 2 β 3 2 + O ( φ )
Observe that a critical point φ = 0 so that δ 2 J * ( v * ) > 0 at a neighborhood of any critical point.
At this point we define
A + = v * = ( v 2 * , v 3 * ) [ Y * ] 2 : v 3 * β f 0 , in Ω ,
D * = { v * = ( v 2 * , v 3 * ) [ Y * ] 2 : v * K } ,
for an appropriate real constant K > 0 .
Define now E * = A + D * ,
C 1 * = { v * = ( v 2 * , v 3 * ) E * : φ 2 ε , in Ω } ,
for a small real constant ε > 0 ,
C 2 * = v * = ( v 2 * , v 3 * ) E * : 1 γ 2 β + 3 α ( v 3 * ) 2 β 3 ε 1 ,
and
C * = C 1 * C 2 * .
Similarly as done in the previous section, we may prove that C * is a convex set.
Furthermore, for 0 < ε ε 1 1 , we have that J * is convex on C * .
Summarizing, we may define the following convex optimization problem to obtain a critical point of the primal functional J,
Minimize J * ( v 2 * , v 3 * ) subject to v * = ( v 2 * , v 3 * ) C * .
We call J * the Legendre-Galerkin functional associated to J.

12.1. Numerical examples

We have obtained numerical solutions for two one-dimensional examples.
  • For γ = 1.0 , α = 3.0 , β = 30.0 , f 10 , in Ω = [ 0 , 1 ] .
    For the respective solution please see Figure 7.
  • For γ = 0.01 , α = 3.0 , β = 30.0 , f 10 , in Ω = [ 0 , 1 ] .
    For the respective solution please see Figure 8.

13. A general concave dual variational formulation for global optimization

Let Ω R 3 be an open, bounded and connected set a regular (Lipschitzian) boundary denoted by Ω .
Consider a functional J : V R where
J ( u ) = G ( u ) u , f L 2 , u V .
Here V = W 0 1 , 2 ( Ω ) , f L 2 ( Ω ) and we also denote Y = Y * = L 2 ( Ω ) .
Assume there exists α R such that
α = inf u V J ( u ) .
Furthermore, suppose G is three times Fréchet differentiable and there exists K > 0 such that
2 G ( u ) u 2 + K > 0 , u V .
Define now J 1 : V × Y R where,
J 1 ( u , v ) = G 1 ( u , v ) + F ( u ) ,
where
G 1 ( u , v ) = G ( v ) ε 2 Ω v 2 d x + K 2 Ω ( v u ) 2 d x ,
and
F ( u ) = ε 2 Ω u 2 d x u , f L 2 .
Moreover, we define the polar functionals G 1 * : Y * × V R and F * : Y * R , where
G 1 * ( v * , u ) = sup v Y { v , v * L 2 G 1 ( u , v ) } = G K ε * ( v * + K u ) + K 2 Ω u 2 d x ,
G K ε * ( v * + K u ) = sup v Y v , v * L 2 G ( v ) K 2 Ω v 2 d x + ε 2 Ω v 2 d x ,
and
F * ( v * ) = sup u V { u , v * L 2 F ( u ) } = 1 2 ε Ω ( v * f ) 2 d x .
At this point we define the functional J 2 * : Y * × V R by
J 2 * ( v * , u ) = G K ε * ( v * + K u ) + K 2 Ω u 2 d x F * ( v * ) .
With such results in mind we define
V 1 = { u V : u K 3 } ,
and
D * = { v * Y * : v * K 4 } ,
for appropriated real constants K 3 > 0 and K 4 > 0 .
Moreover, we define also the penalized functional J 3 * : Y * × V R where
J 3 * ( v * , u ) = J 2 * ( v * , u ) K 1 2 Ω v * G ( u ) u + ε u 2 d x .
Finally, we remark that for ε > 0 sufficiently small and K 1 > 0 sufficiently large, J 3 * is concave in D * × V 1 around a concerning critical point. We recall that a critical point
v * G ( u ) u + ε u = 0 , in Ω .

15. One more dual variational formulation

In this section we develop one more dual variational formulation for a related model.
Let Ω = [ 0 , 1 ] R and consider the functional J : V R defined by
J ( u ) = 1 2 Ω ( ( u ) 2 1 ) 2 d x + 1 2 Ω u 2 d x u , f L 2 ,
where
V = { u W 1 , 4 ( Ω ) : u ( 0 ) = 0 and u ( 1 ) = 1 / 2 } .
We define also the relaxed functional J 1 : V × V 0 R , already including a concerning restriction and corresponding non-negative Lagrange multiplier Λ 2 , where
J 1 ( u , v , Λ ) = 1 2 Ω ( ( u + v ) 2 1 ) 2 d x + 1 2 Ω u 2 d x u , f L 2 + Λ 2 , ( v ) 2 K L 2 .
where
V 0 = { v W 0 1 , 4 ( Ω ) : ( v ) 2 K 0 in Ω } .
Observe that
1 2 Ω ( ( u + v ) 2 1 ) 2 d x + 1 2 Ω u 2 d x u , f L 2 + Λ 2 , ( v ) 2 K L 2 = v 0 * , ( u + v ) 2 1 L 2 + 1 2 Ω ( ( u + v ) 2 1 ) 2 d x + v 0 * , ( u + v ) 2 1 L 2 + Λ 2 , ( v ) 2 K L 2 u , v 1 * L 2 v , v 2 * L 2 + u , v 1 * L 2 + v , v 2 * L 2 + 1 2 Ω u 2 d x u , f L 2 inf w Y v 0 * , w L 2 + 1 2 Ω ( w ) 2 d x inf ( v 1 , v 2 ) Y × Y v 0 * , ( v 1 + v 2 ) 2 1 L 2 + Λ 2 , ( v 2 ) 2 K L 2 v 1 , v 1 * L 2 v 2 , v 2 * L 2 + inf ( u , v ) V × V 0 u , v 1 * L 2 + v , v 2 * L 2 + 1 2 Ω u 2 d x u , f L 2 = 1 2 Ω ( v 0 * ) 2 d x Ω v 0 * d x 1 4 Ω ( v 1 * ) 2 v 0 * d x 1 2 Ω ( v 1 * v 2 * ) 2 2 Λ 2 d x 1 2 Ω ( ( v 1 * ) + f ) 2 d x 1 2 Ω K Λ 2 d x + v 1 * ( 1 ) u ( 1 ) .
Here, we highlight v 2 * = c R in Ω , for some real constant c.
Hence, denoting
J 1 * ( v * , Λ ) = 1 2 Ω ( v 0 * ) 2 d x Ω v 0 * d x 1 4 Ω ( v 1 * ) 2 v 0 * d x 1 2 Ω ( v 1 * v 2 * ) 2 2 Λ 2 d x 1 2 Ω ( ( v 1 * ) + f ) 2 d x 1 2 Ω K Λ 2 d x + v 1 * ( 1 ) u ( 1 )
and
J 2 ( u , v ) = 1 2 Ω ( ( u + v ) 2 1 ) 2 d x + 1 2 Ω u 2 d x u , f L 2 ,
we have obtained
inf ( u , v ) V × V 0 J 2 ( u , v ) } sup ( v * , Λ ) A * × [ Y * ] × R × Y * J 1 * ( v * , Λ ) .
Finally, for
A * = { v 0 * Y * : v 0 * ε in Ω }
we emphasize J 1 * is concave on A * × [ Y * ] × R × Y * .
Here ε > 0 is a small regularizing real constant.
Remark 15.1.
The constraint ( v ) 2 K 0 , in Ω is included to restrict the action of v on the region where the primal functional is non-convex, through an appropriate constant K > 0 .

17. A model in superconductivity through an eigenvalue approach

In this section we intend to model superconductivity through a two phase eigenvalue approach.
Let Ω = [ 0 , 5 ] R be a straight wire corresponding to a one-dimensional super-conducting sample.
Consider the functional J : V × V × R R where
J ( u , v , E ) = γ 1 2 Ω u · u d x + α 1 2 Ω | u | 4 d x ω 2 2 Ω | u | 2 d x + γ 2 2 Ω v · v d x + α 2 2 Ω | v | 4 d x ω 1 2 2 K 3 2 Ω | v | 2 d x E 2 Ω ( | u | 2 + | v | 2 ) d x m T .
Here, in atomic units, m T is the total electronic charge, V = W 0 1 , 2 ( Ω ) and we set α 1 = 10 4 corresponding to higher self-interacting energy which is related to a normal phase. We also set α 2 = 10 1 corresponding to a lower self-interacting energy which is related to a super-conducting phase and respective super-currents.
Moreover, we set γ 1 = γ 2 = 1 , and initially ω = 1.8 which is gradually decreased to ω = 1.0 .
Furthermore, we define
| ϕ N | 2 = | u | 2 | u | 2 + | v | 2
and
| ϕ S | 2 = | v | 2 | u | 2 + | v | 2
where ϕ N corresponds to a normal phase and ϕ S to a super-conducting one.
At this point we observe that the temperature T = T ( x , t ) is proportional the frequency ω / ( 2 π ) of vibration for the normal phase.
We start the process with ω = 1.8 which in atomic units corresponds to a higher temperature and gradually decreases it to the value ω = 1.0
Between ω = 1.2 and ω = 1.0 the system changes from an almost total normal phase to an almost total super-conducting phase, as expected.
We highlight that the temperature is proportional to the vibrational kinetics energy
E 1 ( t ) = 1 2 Ω | u | 2 r N ( x , t ) t · r N ( x , t ) t d x
so that for
r N ( x , t ) = e i ω t w 5 ( x )
and for a suitable vectorial function w 5 , we have
T E 1 ω 2
so that we may model the decreasing of temperature T through the decreasing of ω 2 .
For ω = 1.8 , for the corresponding normal phase ϕ N and super-conducting phase ϕ S , please se Figure 11 and Figure 12, respectively.
For ω = 1.0 , for the corresponding normal phase ϕ N and super-conducting phase ϕ S , please se Figure 13 and Figure 14, respectively.
Finally, we have set ω 1 / K 3 1 which for large ω 1 corresponds to the super-currents.

18. A simplified qualitative many body model for the hydrogen nuclear fusion

In this section we develop a qualitative simple model for the hydrogen nuclear fusion.
Let Ω = [ 0 , L ] 3 R 3 be a box in which is confined a gas comprised by an amount of ionized deuterium and tritium isotopes of hydrogen.
Though a suitable increasing in temperature, we intend to develop the following nuclear reaction
Deuterium + + Tritium + Hellion + + + Neutron ( energetic ) .
We recall that the ionized Deuterium atom comprises a proton and a neutron and the ionized Tritium atom comprises a proton and two neutrons.
Under certain conditions and at a suitable high temperature the ionized Deuterium and Tritium atoms react chemically resulting in an ionized Hellion atom, comprised by two protons and two neutrons and resulting also in one more single energetic neutron. We emphasize the higher kinetics neutron energy level has many potential practical applications, including its conversion in electric energy.
At this point we denote by m D , m T , m H e and m N the masses of the ionized Deuterium, Tritium and Hellion atoms, and the single neutron, respectively.
Therefore, we have the following mass relation
m D + m T = m H e + m N .
To simplify our analysis, in such a chemical reaction, denoting the total masses of ionized Deuterium, Tritium, Hellion and single Neutrons by ( m D ) T , ( m T ) T , ( m H e ) T and ( m N ) T we assume there is a real constant c > 0 such that
( m D ) T = c m D , ( m T ) T = c m T , ( m H e ) T = c m H e , ( m N ) T = c m N .
With such statements and definitions in mind, we define the following functional J, where
J ( ϕ , r ) = J ( ϕ D , ϕ T , ϕ H e , ϕ N , r ) = G ( ϕ ) + F ( ϕ ) + E c ( ϕ , r ) ,
where, in a simplified many body context,
| ϕ D ( x , y ) | 2 = | ϕ p D ( y ) | 2 + | ϕ N D ( x , y ) | 2 | ϕ p D ( y ) | 2 1 m p ,
| ϕ T ( x , y ) | 2 = | ϕ p T ( y ) | 2 + ( | ϕ N 1 T ( x , y ) | 2 + | ϕ N 2 T ( x , y ) | 2 ) | ϕ p T ( y ) | 2 1 m p ,
| ϕ H e ( x , y ) | 2 = | ϕ 2 P H e ( y ) | 2 + ( | ϕ N 1 H e ( x , y ) | 2 + | ϕ N 2 H e ( x , y ) | 2 ) | ϕ 2 P H e ( y ) | 2 1 2 m p ,
ϕ N = ϕ N ( x ) .
Here x , y Ω R 3 refers to the particle densities.
Furthermore, we assume γ p D > 0 , γ p T > 0 , γ N D > 0 , γ N 1 T > 0 , γ N 2 T > 0 , γ 2 p H e > 0 , γ N 1 H e > 0 ,   γ N 2 H e > 0 , γ N > 0 , and α D > 0 , α T > 0 , α H e > 0 , α N > 0 , α D T > 0 , α H e N > 0 , so that
G ( ϕ ) = γ p D 2 Ω ( ϕ p D ) · ( ϕ p D ) d y + γ N D 2 Ω ( ϕ N D ) · ( ϕ N D ) d x d y γ p T 2 Ω ( ϕ p T ) · ( ϕ p T ) d y + γ N 1 T 2 Ω ( ϕ N 1 T ) · ( ϕ N 1 T ) d x d y + γ N 2 T 2 Ω ( ϕ N 2 T ) · ( ϕ N 2 T ) d x d y + γ 2 p H e 2 Ω ( ϕ 2 p H e ) · ( ϕ 2 p H e ) d y + γ N 1 H e 2 Ω ( ϕ N 1 H e ) · ( ϕ N 1 H e ) d x d y + γ N 2 H e 2 Ω ( ϕ N 2 H e ) · ( ϕ N 2 H e ) d x d y + γ N 2 Ω ( ϕ N ) · ( ϕ N ) d x ,
and,
F ( ϕ ) = α D 2 Ω | ϕ D ( x ξ 1 , y ξ 2 ) | 2 | ϕ D ( ξ 1 , ξ 2 ) | 2 | ( x , y ) ( ξ 1 , ξ 2 ) | d x d y d ξ 1 d ξ 2 + α T 2 Ω | ϕ T ( x ξ 1 , y ξ 2 ) | 2 | ϕ T ( ξ 1 , ξ 2 ) | 2 | ( x , y ) ( ξ 1 , ξ 2 ) | d x d y d ξ 1 d ξ 2 + α D T 2 Ω | ϕ D ( x ξ 1 , y ξ 2 ) | 2 | ϕ T ( ξ 1 , ξ 2 ) | 2 | ( x , y ) ( ξ 1 , ξ 2 ) | d x d y d ξ 1 d ξ 2 + α H e 2 Ω | ϕ H e ( x ξ 1 , y ξ 2 ) | 2 | ϕ H e ( ξ 1 , ξ 2 ) | 2 | ( x , y ) ( ξ 1 , ξ 2 ) | d x d y d ξ 1 d ξ 2 + α N 2 0 t f Ω | ϕ N ( x ξ ) | 2 | ϕ N ( ξ ) | 2 | x ξ | d x d ξ + j = 1 2 α H e N 2 Ω | ϕ H e ( x 1 ξ 1 , y ξ 2 ) | 2 | ϕ N ( ξ j ) | 2 | ( x , y ) ( ξ 1 , ξ 2 ) | d x d y d ξ 1 d ξ 2
and the kinetics energy is expressed by
E c ( ϕ , r ) = 1 2 Ω | ϕ D | 2 r D t · r D t d x d y + 1 2 Ω | ϕ T | 2 r T t · r T t d x d y + 1 2 Ω | ϕ H e | 2 r H e t · r H e t d x d y + 1 2 Ω | ϕ N | 2 r N t · r N t d x d y ,
where we also assume
r D e i ω t w 5 ( x , y ) ,
r T e i ω t w 6 ( x , y ) ,
so that considering such a vibrational motion, the temperature T is proportional to ω 2 , that is
T ω 2 .
Therefore, an increasing in T corresponds to a proportional increasing in ω 2 .
Summarizing, we have supposed
E c ( ϕ , r ) 1 2 ω 2 Ω | ϕ D | 2 + | ϕ T | 2 d x C 1 + 1 2 ω 1 2 Ω | ϕ N | 2 d x C 2 ,
so that we represent the increasing in T through an increasing in ω 2 .
Moreover, we denote by m N the mass of a single neutron and by m p the mass of a single proton.
Thus, denoting also by λ 1 , λ 2 the proportion of non-reacted and reacted masses respectively, we have the following constraints.
  • Ω | ϕ N D ( x , y ) | 2 d x = m N ,
  • Ω | ϕ N 1 T ( x , y ) | 2 d x = m N ,
  • Ω | ϕ N 2 T ( x , y ) | 2 d x = m N ,
  • Ω | ϕ N 1 H e ( x , y ) | 2 d x = m N ,
  • Ω | ϕ N 2 H e ( x , y ) | 2 d x = m N ,
  • Ω | ϕ p D ( y ) | 2 d y = λ 1 c m p ,
  • Ω | ϕ p T ( y ) | 2 d y = λ 1 c m p ,
  • Ω | ϕ 2 P H e ( y ) | 2 d y = λ 2 ( 2 c m p ) ,
Similar constraints are valid corresponding to the charge of a single proton.
We have also the following complementing constraints,
  • Ω | ϕ D | 2 d x d y = λ 1 ( m D ) T ,
  • Ω | ϕ T | 2 d x d y = λ 1 ( m T ) T ,
  • Ω | ϕ H e | 2 d x d y = λ 2 ( m H e ) T ,
  • Ω | ϕ N | 2 d x d y = λ 2 ( m N ) T ,
  • λ 1 + λ 2 = 1 .
With such results and statements in mind and simplifying the interacting terms, we re-define the functional J now denoting it by J 1 , here already including the Lagrange multipliers concerning the constraints, where
J 1 ( ϕ , ω , E , λ ) = γ p D 2 Ω ( ϕ p D ) · ( ϕ p D ) d y + γ N D 2 Ω ( ϕ N D ) · ( ϕ N D ) d x d y γ p T 2 Ω ( ϕ p T ) · ( ϕ p T ) d y + γ N 1 T 2 Ω ( ϕ N 1 T ) · ( ϕ N 1 T ) d x d y + γ N 2 T 2 Ω ( ϕ N 2 T ) · ( ϕ N 2 T ) d x d y + γ 2 p H e 2 Ω ( ϕ 2 p H e ) · ( ϕ 2 p H e ) d y + γ N 1 H e 2 Ω ( ϕ N 1 H e ) · ( ϕ N 1 H e ) d x d y + γ N 2 H e 2 Ω ( ϕ N 2 H e ) · ( ϕ N 2 H e ) d x d y + γ N 2 Ω ( ϕ N ) · ( ϕ N ) d x + α D 2 Ω | ϕ D | 4 d x + α T 2 Ω | ϕ T | 4 d x + α H e 2 Ω | ϕ H e | 4 d x + α N 2 Ω | ϕ N | 4 d x ω 2 Ω ( | ϕ D | 2 + | ϕ T | 2 ) d x ω 1 2 Ω | ϕ N | 2 d x + J A u x ,
where the functional J A u x stands for
J A u x = Ω ( E N D ) 5 ( y ) Ω | ϕ N D ( x , y ) | 2 d x m N d y Ω ( E N 1 T ) 6 ( y ) Ω | ϕ N 1 T ( x , y ) | 2 d x m N d y Ω ( E N 2 T ) 7 ( y ) Ω | ϕ N 2 T ( x , y ) | 2 d x m N d y Ω ( E N 1 H e ) 8 ( y ) Ω | ϕ N 1 H e ( x , y ) | 2 d x m N d y Ω ( E N 2 H e ) 9 ( y ) Ω | ϕ N 2 H e ( x , y ) | 2 d x m N d y ( E D ) 2 Ω | ϕ p D ( y ) | 2 d y λ 1 c m p ( E T ) 3 Ω | ϕ p T ( y ) | 2 d y λ 1 c m p ( E H e ) 3 Ω | ϕ 2 P H e ( x , y ) | 2 d y λ 2 2 c m p E 5 Ω | ϕ D | 2 d x d y λ 1 ( m D ) T E 6 Ω | ϕ T | 2 d x d y λ 1 ( m T ) T E 7 Ω | ϕ H e | 2 d x d y λ 2 ( m H e ) T E 8 Ω | ϕ N | 2 d x d y λ 2 ( m N ) T E 9 ( λ 1 + λ 2 1 ) .
Remark 18.1.
In order to obtain consistent results it is necessary to set
( α N , α H e ) ( α D , α T ) .
In such a case, a higher temperature corresponding to a large ω 2 , though such a nuclear reaction, will result in a small λ 1 and a higher kinetics energy for the neutron field, corresponding to a large ω 1 2 and λ 2 closer to 1.

19. A more detailed mathematical description of the hydrogen nuclear fusion

In this section we develop in more details another model for the hydrogen nuclear fusion.
Let Ω R 3 be an open, bounded and connected set with a regular (Lipschitzian) boundary denoted by Ω .
Here such a set Ω stands for a control volume in which an ionized gas (plasma) flows. Such a gas comprises ionized Deuterium and Tritium atoms intended, through a suitable higher temperature, to chemically react resulting in atoms of Hellion and a field of single energetic Neutrons.
Symbolically such a reaction stands for
Deuterium + + Tritium + Hellion + + + Neutron ( energetic ) .
We recall that the ionized Deuterium atom is comprised by a proton and a neutron and the ionized Tritium atom is comprised by a proton and two neutrons.
Moreover, the ionized Hellion atom is comprised by two protons and two neutrons.
As previously mentioned, resulting from such a chemical reaction up surges also an energetic neutron which the higher kinetics energy has a great variety of applications, including its conversion in electric energy.
We highlight the model here presented includes electric and magnetic fields and the corresponding potential ones.
Denoting by t the time on the interval [ 0 , t f ] , at this point we define the following density functions:
  • For the Deuterium field
    | ϕ D ( x , y , t ) | 2 = | ϕ p D ( y , t ) | 2 + | ϕ N D ( x , y , t ) | 2 | ϕ p D ( y , t ) | 2 1 m p ,
  • For the Tritium field
    | ϕ T ( x , y , t ) | 2 = | ϕ p T ( y , t ) | 2 + ( | ϕ N 1 T ( x , y , t ) | 2 + | ϕ N 2 T ( x , y , t ) | 2 ) | ϕ p T ( y , t ) | 2 1 m p ,
  • For the Hellion field
    | ϕ H e ( x , y , t ) | 2 = | ϕ 2 p H e ( y , t ) | 2 + ( | ϕ N 1 H e ( x , y , t ) | 2 + | ϕ N 2 H e ( x , y , t ) | 2 ) | ϕ 2 p H e ( y , t ) | 2 1 2 m p ,
  • For the Neutron field
    ϕ N = ϕ N ( x , t ) ,
  • For the electronic field resulting from the ionization
    ϕ e = ϕ e ( x , y , t ) .
Furthermore, we define also the related densities
  • ρ D ( y , t ) = Ω | ϕ D ( x , y , t ) | 2 d x ,
  • ρ T ( y , t ) = Ω | ϕ T ( x , y , t ) | 2 d x ,
    ρ H e ( y , t ) = Ω | ϕ H e ( x , y , t ) | 2 d x ,
    ρ N ( x , t ) = | ϕ N ( x , t ) | 2 ,
    ρ e ( y , t ) = Ω | ϕ e ( x , y , t ) | 2 d x .
For the chemical reaction in question we consider that one unit of mass of fractional proportion α D of ionized Deuterium and α T of ionized Tritium results in one unit of mass of fractional proportion α H e of ionized Hellion and α N of neutrons.
Symbolic, this stands for
1 = α D + α T = α H e + α N .
Concerning the control volume Ω in question and related surface control Ω , we assume such a volume has an initial (fot t = 0 ) amount of ionized Deuterium of ( m D ) 0 and an initial amount of ionized Tritium of ( m T ) 0 . The initial amount of ionized Hellion and single neutrons are supposed to be zero.
On the other hand, about the surface control Ω , we assume there is a part Ω 1 Ω for which is allowed the entrance and exit of Deuterium and Tritium ionized atoms.
We assume also there is another part Ω 2 Ω such that Ω 1 Ω 2 = for which is allowed only the exit of ionized Hellion atoms and neutrons, but not their entrance.
In Ω 2 is allowed the exit only (not the entrance) of ionized Deuterium and Tritium atoms.
Indeed, we assume the following relations for the masses:
  • ( m H e , N ) T ( t ) = m H e , N ( t ) + 0 t Ω 2 ( ρ H e ( x , τ ) + ρ N ( x , τ ) ) u · n d S d τ ,
  • m H e , N ( t ) = m H e ( t ) + m N ( t ) ,
  • m H e ( t ) = Ω ρ H e ( x , t ) d x ,
  • m N ( t ) = Ω ρ N ( x , t ) d x ,
  • ( m D ) ( t ) = ( m D ) 0 0 t Ω 1 Ω 2 ( ρ D ( x , τ ) ) u · n d S d τ α D ( m H e , N ) T ( t ) ,
  • ( m T ) ( t ) = ( m T ) 0 0 t Ω 1 Ω 2 ( ρ T ( x , τ ) ) u · n d S d τ α T ( m H e , N ) T ( t ) ,
  • ( m e ) T ( t ) = m e ( t ) + 0 t Ω 2 ( ρ e ( x , τ ) ) u · n d S d τ ,
  • m e ( t ) = Ω ρ e ( x , t ) d x .
  • m e ( t ) = Ω | ϕ p D ( x , t ) | 2 d x m e m p + Ω | ϕ p T ( x , t ) | 2 d x m e m p + Ω | ϕ 2 p H e ( x , t ) | 2 d x m e m p .
Here n denotes the outward normal vectorial fields to the concerning surfaces.
Having clarified such masses relations, we define the functional
J ( ϕ , ρ , r , u , E , A , B )
where
J = G ( u ) + F ( ϕ ) + E c ( ϕ , r ) + F 1 + F 2 + F 3 ,
and where we assume γ p D > 0 , γ p T > 0 , γ N D > 0 , γ N 1 T > 0 , γ N 2 T > 0 , γ 2 p H e > 0 , γ N 1 H e > 0 ,   γ N 2 H e > 0 , γ N > 0 , γ e > 0 and α D > 0 , α T > 0 , α H e > 0 , α N > 0 , α D T > 0 , α H e N > 0 , α e , e > 0 , α H e , e < 0 so that
G ( ϕ ) = γ p D 2 0 t f Ω ( ϕ p D ) · ( ϕ p D ) d y d t + γ N D 2 0 t f Ω ( ϕ N D ) · ( ϕ N D ) d x d y d t γ p T 2 0 t f Ω ( ϕ p T ) · ( ϕ p T ) d y d t + γ N 1 T 2 0 t f Ω ( ϕ N 1 T ) · ( ϕ N 1 T ) d x d y d t + γ N 2 T 2 0 t f Ω ( ϕ N 2 T ) · ( ϕ N 2 T ) d x d y d t + γ 2 p H e 2 0 t f Ω ( ϕ 2 p H e ) · ( ϕ 2 p H e ) d y d t + γ N 1 H e 2 0 t f Ω ( ϕ N 1 H e ) · ( ϕ N 1 H e ) d x d y d t + γ N 2 H e 2 0 t f Ω ( ϕ N 2 H e ) · ( ϕ N 2 H e ) d x d y d t + γ N 2 0 t f Ω ( ϕ N ) · ( ϕ N ) d x d t + γ e 2 0 t f Ω ( ϕ e ) · ( ϕ e ) d x d y d t ,
and
F ( ϕ ) = α D 2 0 t f Ω | ϕ D ( x ξ 1 , y ξ 2 , t ) | 2 | ϕ D ( ξ 1 , ξ 2 , t ) | 2 | ( x , y ) ( ξ 1 , ξ 2 ) | d x d y d ξ 1 d ξ 2 d t + α T 2 0 t f Ω | ϕ T ( x ξ 1 , y ξ 2 , t ) | 2 | ϕ T ( ξ 1 , ξ 2 , t ) | 2 | ( x , y ) ( ξ 1 , ξ 2 ) | d x d y d ξ 1 d ξ 2 d t + α D T 2 0 t f Ω | ϕ D ( x ξ 1 , y ξ 2 , t ) | 2 | ϕ T ( ξ 1 , ξ 2 , t ) | 2 | ( x , y ) ( ξ 1 , ξ 2 ) | d x d y d ξ 1 d ξ 2 d t + α H e 2 0 t f Ω | ϕ H e ( x ξ 1 , y ξ 2 , t ) | 2 | ϕ H e ( ξ 1 , ξ 2 , t ) | 2 | ( x , y ) ( ξ 1 , ξ 2 ) | d x d y d ξ 1 d ξ 2 d t + α N 2 0 t f Ω | ϕ N ( x ξ , t ) | 2 | ϕ N ( ξ ) | 2 | x ξ , t | d x d ξ d t + j = 1 2 α H e N 2 0 t f Ω | ϕ H e ( x 1 ξ 1 , y ξ 2 , t ) | 2 | ϕ N ( ξ j , t ) | 2 | ( x , y ) ( ξ 1 , ξ 2 ) | d x d y d ξ 1 d ξ 2 d t + α H e , e 2 0 t f Ω | ϕ H e ( x ξ 1 , y ξ 2 , t ) | 2 | ϕ e ( ξ 1 , ξ 2 , t ) | 2 | ( x , y ) ( ξ 1 , ξ 2 ) | d x d y d ξ 1 d ξ 2 d t + α e , e 2 0 t f Ω | ϕ e ( x ξ 1 , y ξ 2 , t ) | 2 | ϕ e ( ξ 1 , ξ 2 , t ) | 2 | ( x , y ) ( ξ 1 , ξ 2 ) | d x d y d ξ 1 d ξ 2 d t
and the internal kinetics energy is expressed by
E c ( ϕ , r ) = 1 2 0 t f Ω | ϕ D | 2 r D t · r D t d x d y d t + 1 2 0 t f Ω | ϕ T | 2 r T t · r T t d x d y d t + 1 2 0 t f Ω | ϕ H e | 2 r H e t · r H e t d x d y d t + 1 2 0 t f Ω | ϕ N | 2 r N t · r N t d x d y d t + 1 2 0 t f Ω | ϕ e | 2 r e t · r e t d x d y d t ,
Here it is worth highlighting we have approximated the initially discrete set of indices s of particles as a continuous positive real variable s.
Moreover,
F 1 = 1 4 π 0 t f curl A B 0 2 d t ,
F 2 = 0 t f Ω E i n d · K p | ϕ p D | 2 u + r D t d x d y d t + 0 t f Ω E i n d · K p | ϕ p T | 2 u + r T t d x d y d t + 0 t f Ω E i n d · K p | ϕ 2 p H e | 2 u + r H e t d x d y d t + 0 t f Ω E i n d · K e | ϕ e | 2 u + r e t d x d y d t ,
where K p and K e are appropriate real constants related to the respective charges.
Here u = ( u 1 , u 2 , u 3 ) is the fluid velocity field and
r D , r T , r H e , r N , r e
are fields of displacements for the corresponding atom fields.
Also A denotes the magnetic potential, B 0 an external magnetic field and B is the total magnetic field.
Moreover, E i n d is an induced electric field.
Finally,
F 3 = C D 2 0 t f Ω ( x , y ) r D · ( x , y ) r D d x d y d t + C T 2 0 t f Ω ( x , y ) r T · ( x , y ) r T d x d y d t C H e 2 0 t f Ω ( x , y ) r H e · ( x , y ) r H e d x d y d t + C N 2 0 t f Ω ( x , y ) r N · ( x , y ) r N d x d y d t C e 2 0 t f Ω ( x , y ) r e · ( x , y ) r e d x d y d t ,
for appropriate real positive constants C D C T , C H e , C N , C e .
Such a functional J is subject to the following constraints:
  • The momentum conservation equation for the fluid motion
    ρ u k t + u j u k x j = ρ f k P x k + τ k j , j + ( F E ) k + ( F M ) k ,
    k { 1 , 2 , 3 } .
    Here ρ = ρ D + ρ T + ρ H e + ρ N + ρ e is the total density and P is the fluid pressure field.
    Furthermore,
    τ i j = μ u i x j + u j x i 2 3 δ i j k = 1 3 u k x k ,
    i , j { 1 , 2 , 3 } ,
    F E = { ( F E ) k } = K p ( | ϕ p D | 2 + | ϕ p T | 2 + | ϕ 2 p H e | 2 ) + K e Ω | ϕ e | 2 d x E ,
    and
    F M = { ( F M ) k } = K p | ϕ p D | 2 u + r D t | ϕ p T | 2 u + r T t + | ϕ 2 p H e | 2 u + r H e t + K e | ϕ e | 2 u + r e t × B .
  • Mass conservation equation:
    ρ t + div ( ρ u ) = 0 .
  • Energy equation
    ρ D e D t + P ( div u ) = Q t div q ,
    where we assume the Fourier law
    q = K T ,
    where T = T ( x , t ) is the scalar field of temperature.
    Also,
    e = ρ 2 u · u + ρ D 2 r D t · r D t + ρ T 2 r T t · r T t + ρ H e 2 r H e t · r H e t + ρ N 2 r N t · r N t + ρ e 2 r e t · r e t
    and
    D e D t = e t + u j e x j .
  • P = F 7 ( ρ , T ) ,
    for an appropriate scalar function F 7 .
  • Mass relations
    (a)
    m D ( t ) = Ω ρ D ( x , t ) d x ,
    (b)
    m T ( t ) = Ω ρ T ( x , t ) d x ,
    (c)
    m H e ( t ) = Ω ρ H e ( x , t ) d x ,
    (d)
    m N ( t ) = Ω ρ N ( x , t ) d x ,
    (e)
    m e ( t ) = Ω ρ e ( x , t ) d x ,
    where,
    (a)
    ( m H e , N ) T ( t ) = m H e , N ( t ) + 0 t Ω 2 ( ρ H e ( x , τ ) ) u · n d S d τ ,
    (b)
    m H e , N ( t ) = m H e ( t ) + m N ( t ) ,
    (c)
    ( m D ) ( t ) = ( m D ) 0 0 t Ω 1 Ω 2 ( ρ D ( x , τ ) ) u · n d S d τ α D ( m H e , N ) T ( t ) ,
    (d)
    ( m T ) ( t ) = ( m T ) 0 0 t Ω 1 Ω 2 ( ρ T ( x , τ ) ) u · n d S d τ α T ( m H e , N ) T ( t ) ,
    (e)
    ( m e ) T ( t ) = m e ( t ) + 0 t Ω 2 ( ρ T ( x , τ ) ) u · n d S d τ .
    (f)
    m e ( t ) = Ω | ϕ p D ( x , t ) | 2 d x m e m p + Ω | ϕ p T ( x , t ) | 2 d x m e m p + Ω | ϕ 2 p H e ( x , t ) | 2 d x m e m p .
  • Other mass constraints
    (a)
    Ω | ϕ N D ( x , y , t ) | 2 d x = m N ,
    (b)
    Ω | ϕ N 1 T ( x , y , t ) | 2 d x = m N ,
    (c)
    Ω | ϕ N 2 T ( x , y , t ) | 2 d x = m N ,
    (d)
    Ω | ϕ N 1 H e ( x , y , t ) | 2 d x = m N ,
    (e)
    Ω | ϕ N 2 H e ( x , y , t ) | 2 d x = m N .
  • For the induced electric field, we must have
    curl E i n d + 1 c curl K ^ p | ϕ p D | 2 u + r D t + K ^ p | ϕ p T | 2 u + r T t + K ^ p | ϕ 2 p H e | 2 u + r H e t + K ^ e Ω | ϕ e ( x , y , t ) | 2 u ( y , t ) + r e ( x , y , t ) t d x × curl A B 0 1 c t curl A B 0 = 0 ,
    where K ^ p and K ^ e are appropriate real constants related to the respective charges.
  • A Maxwell equation:
    div B = 0 ,
    where
    B = B 0 curl A .
  • Another Maxwell equation:
    div E = 4 π K p ( | ϕ p D | 2 + | ϕ p T | 2 + | ϕ 2 p H e | 2 ) + K e Ω | ϕ e ( x , y , t ) | 2 d x ,
    where the total electric field E stands for
    E = E i n d + E ρ ,
    and where generically denoting
    F ( ϕ ) = Ω f 5 ( ϕ , x , ξ ) d x d ξ ,
    we have also
    E ρ = Ω f 5 ( ϕ , x , ξ ) x k d ξ .
At this point we generically denote
h 1 , h 2 L 2 = 0 t f Ω h 1 h 2 d x d y d t .
Thus, already including the Lagrange multipliers concerning the restrictions indicated, the extended functional J 3 stands for
J 3 = J 3 ( ϕ , u , r , P , A , B , E , Λ , E ) = G ( ϕ ) + F ( ϕ ) + E c ( ϕ , r ) + F 1 + F 2 + F 3 + Λ k , ρ u k t + u j u k x j ρ f k + P x k τ k j , j ( F E ) k ( F M ) k L 2 + Λ 4 , ρ t + div ( ρ u ) L 2 + J A u x 1 + J A u x 2 + J A u x 3 + J A u x 4 ,
where,
J A u x 1 = Λ 5 , ρ D e D t + P ( div u ) Q t + div q L 2 + Λ 6 , P F 7 ( ρ , T ) L 2 ,
J A u x 2 = Λ 7 , m D ( t ) Ω ρ D ( x , t ) d x L 2 + Λ 8 , m T ( t ) Ω ρ T ( x , t ) d x L 2 Λ 9 , m H e ( t ) Ω ρ H e ( x , t ) d x L 2 Λ 10 , m N ( t ) Ω ρ N ( x , t ) d x L 2 Λ 11 , m e ( t ) Ω ρ e ( x , t ) d x L 2 ,
J A u x 3 = 0 t f Ω ( E N D ) 5 ( y , t ) Ω | ϕ N D ( x , y , t ) | 2 d x m N d y d t 0 t f Ω ( E N 1 T ) 6 ( y , t ) Ω | ϕ N 1 T ( x , y , t ) | 2 d x m N d y d t 0 t f Ω ( E N 2 T ) 7 ( y , t ) Ω | ϕ N 2 T ( x , y , t ) | 2 d x m N d y d t 0 t f Ω ( E N 1 H e ) 8 ( y , t ) Ω | ϕ N 1 H e ( x , y , t ) | 2 d x m N d y d t 0 t f Ω ( E N 2 H e ) 9 ( y , t ) Ω | ϕ N 2 H e ( x , y , t ) | 2 d x m N d y d t ,
J A u x 4 = Λ 12 , curl E i n d + 1 c curl K ^ p | ϕ p D | 2 u + r D t + K ^ p | ϕ p T | 2 u + r T t + K ^ p | ϕ 2 p H e | 2 u + r H e t + K ^ e Ω | ϕ e ( x , y , t ) | 2 u ( y , t ) + r e ( x , y , t ) t d x × curl A B 0 1 c t curl A B 0 L 2 + Λ 13 , div B L 2 + Λ 14 , div E 4 π K p ( | ϕ p D | 2 + | ϕ p T | 2 + | ϕ 2 p H e | 2 ) + K e Ω | ϕ e ( x , y , t ) | 2 d x L 2 .
Here we recall the following definitions and relations:
  • For the Deuterium field
    | ϕ D ( x , y , t ) | 2 = | ϕ p D ( y , t ) | 2 + | ϕ N D ( x , y , t ) | 2 | ϕ p D ( y , t ) | 2 1 m p ,
  • For the Tritium field
    | ϕ D ( x , y , t ) | 2 = | ϕ p D ( y , t ) | 2 + ( | ϕ N 1 D ( x , y , t ) | 2 + | ϕ N 2 D ( x , y , t ) | 2 ) | ϕ p D ( y , t ) | 2 1 m p ,
  • For the Hellion field
    | ϕ H e ( x , y , t ) | 2 = | ϕ 2 p H e ( y , t ) | 2 + ( | ϕ N 1 H e ( x , y , t ) | 2 + | ϕ N 2 H e ( x , y , t ) | 2 ) | ϕ 2 p H e ( y , t ) | 2 1 2 m p ,
  • For the Neutron field
    ϕ N = ϕ N ( x , t ) ,
  • For the electronic field resulting from the ionization
    ϕ e = ϕ e ( x , y , t ) .
  • ρ D ( y , t ) = Ω | ϕ D ( x , y , t ) | 2 d x ,
  • ρ T ( y , t ) = Ω | ϕ T ( x , y , t ) | 2 d x ,
    ρ H e ( y , t ) = Ω | ϕ H e ( x , y , t ) | 2 d x ,
    ρ N ( x , t ) = | ϕ N ( x , t ) | 2 ,
    ρ e ( y , t ) = Ω | ϕ e ( x , y , t ) | 2 d x .
Also,
ρ = ρ D + ρ T + ρ H e + ρ N + ρ e ,
  • ( m H e , N ) T ( t ) = m H e , N ( t ) + 0 t Ω 2 ( ρ H e ( x , τ ) + ρ N ( x , τ ) ) u · n d S d τ ,
  • m H e , N ( t ) = m H e ( t ) + m N ( t ) ,
  • m H e ( t ) = Ω ρ H e ( x , t ) d x ,
  • m N ( t ) = Ω ρ N ( x , t ) d x ,
  • ( m D ) ( t ) = ( m D ) 0 0 t Ω 1 Ω 2 ( ρ D ( x , τ ) ) u · n d S d τ α D ( m H e , N ) T ( t ) ,
  • ( m T ) ( t ) = ( m T ) 0 0 t Ω 1 Ω 2 ( ρ T ( x , τ ) ) u · n d S d τ α T ( m H e , N ) T ( t ) ,
  • ( m e ) T ( t ) = m e ( t ) 0 t Ω 2 ( ρ e ( x , τ ) ) u · n d S d τ ,
  • m e ( t ) = Ω ρ e ( x , t ) d x .
  • m e ( t ) = Ω | ϕ p D ( x , t ) | 2 d x m e m p + Ω | ϕ p T ( x , t ) | 2 d x m e m p + Ω | ϕ 2 p H e ( x , t ) | 2 d x m e m p .
Finally,
E = E i n d + E ρ ,
and where generically denoting
F ( ϕ ) = Ω f 5 ( ϕ , x , ξ ) d x d ξ ,
we have also
E ρ = Ω f 5 ( ϕ , x , ξ ) x k d ξ .
and,
B = B 0 curl A .

20. A final mathematical description of the hydrogen nuclear fusion

In this section we develop in even more details another model for the hydrogen nuclear fusion.
Let Ω R 3 be an open, bounded and connected set with a regular (Lipschitzian) boundary denoted by Ω .
Here such a set Ω stands for a control volume in which an ionized gas (plasma) flows. Such a gas comprises ionized Deuterium and Tritium atoms intended, through a suitable higher temperature, to chemically react resulting in atoms of Hellion and a field of single energetic Neutrons.
Symbolically such a reaction stands for
Deuterium + + Tritium + Hellion + + + Neutron ( energetic ) .
We recall that the ionized Deuterium atom is comprised by a proton and a neutron and the ionized Tritium atom is comprised by a proton and two neutrons.
Moreover, the ionized Hellion atom is comprised by two protons and two neutrons.
As previously mentioned, resulting from such a chemical reaction up surges also an energetic neutron which the higher kinetics energy has a great variety of applications, including its conversion in electric energy.
We highlight the model here presented includes electric and magnetic fields and the corresponding potential ones.
Denoting by t the time on the interval [ 0 , t f ] , at this point we define the following density functions:
  • For a single Deuterium atom indexed by s:
    | ϕ D ( x , y , t , s ) | 2 = | ϕ p D ( y , t , s ) | 2 + | ϕ N D ( x , y , t , s ) | 2 | ϕ p D ( y , t , s ) | 2 1 m p ,
  • For a single Tritium atom indexed by s:
    | ϕ T ( x , y , t , s ) | 2 = | ϕ p T ( y , t , s ) | 2 + ( | ϕ N 1 T ( x , y , t , s ) | 2 + | ϕ N 2 T ( x , y , t , s ) | 2 ) | ϕ p T ( y , t , s ) | 2 1 m p ,
  • For a single Hellion atom indexed by s:
    | ϕ H e ( x , y , t , s ) | 2 = | ϕ 2 p H e ( y , t , s ) | 2 + ( | ϕ N 1 H e ( x , y , t , s ) | 2 + | ϕ N 2 H e ( x , y , t , s ) | 2 ) | ϕ 2 p H e ( y , t , s ) | 2 1 2 m p ,
  • For the Neutron field:
    ϕ N = ϕ N ( x , t , s ) ,
  • For the electronic field resulting from the ionization
    ϕ e = ϕ e ( x , y , t , s ) .
Furthermore, we define also the related densities
  • ρ D ( y , t ) = 0 N D ( t ) Ω | ϕ D ( x , y , t , s ) | 2 d x d s ,
  • ρ T ( y , t ) = 0 N T ( t ) Ω | ϕ T ( x , y , t , s ) | 2 d x d s ,
    ρ H e ( y , t ) = 0 N H e ( t ) Ω | ϕ H e ( x , y , t , s ) | 2 d x d s ,
    ρ N ( x , t ) = 0 N N ( t ) | ϕ N ( x , t , s ) | 2 d s ,
    ρ e ( y , t ) = 0 N e ( t ) Ω | ϕ e ( x , y , t , s ) | 2 d x d s .
For the chemical reaction in question we consider that one unit of mass of fractional proportion α D of ionized Deuterium and α T of ionized Tritium results in one unit of mass of fractional proportion α H e of ionized Hellion and α N of neutrons.
Symbolically, this stands for
1 = α D + α T = α H e + α N .
Concerning the control volume Ω in question and related surface control Ω , we assume such a volume has an initial (fot t = 0 ) amount of ionized Deuterium of ( m D ) 0 and an initial amount of ionized Tritium of ( m T ) 0 . The initial amount of ionized Hellion and single neutrons are supposed to be zero.
On the other hand, about the surface control Ω , we assume there is a part Ω 1 Ω for which is allowed the entrance and exit of Deuterium and Tritium ionized atoms.
We assume also there is another part Ω 2 Ω such that Ω 1 Ω 2 = for which is allowed only the exit of ionized Hellion atoms and neutrons, but not their entrance.
In Ω 2 is allowed the exit only (not the entrance) of ionized Deuterium and Tritium atoms.
Indeed, we assume the following relations for the masses:
  • ( m H e , N ) T ( t ) = m H e , N ( t ) + 0 t Ω 2 ( ρ H e ( x , τ ) + ρ N ( x , τ ) ) u · n d S d τ ,
  • m H e , N ( t ) = m H e ( t ) + m N ( t ) ,
  • m H e ( t ) = Ω ρ H e ( x , t ) d x ,
  • m N ( t ) = Ω ρ N ( x , t ) d x ,
  • ( m D ) ( t ) = ( m D ) 0 0 t Ω 1 Ω 2 ( ρ D ( x , τ ) ) u · n d S d τ α D ( m H e , N ) T ( t ) ,
  • ( m T ) ( t ) = ( m T ) 0 0 t Ω 1 Ω 2 ( ρ T ( x , τ ) ) u · n d S d τ α T ( m H e , N ) T ( t ) ,
  • ( m e ) T ( t ) = m e ( t ) + 0 t Ω 2 ( ρ e ( x , τ ) ) u · n d S d τ ,
  • m e ( t ) = Ω ρ e ( x , t ) d x .
  • m e ( t ) = 0 N D ( t ) Ω | ϕ p D ( y , t , s ) | 2 d y d s m e m p + 0 N T ( t ) Ω | ϕ p T ( y , t , s ) | 2 d y d s m e m p + 0 N p ( t ) Ω | ϕ 2 p H e ( y , t , s ) | 2 d y d s m e m p .
Here n denotes the outward normal vectorial fields to the concerning surfaces.
Having clarified such masses relations, denoting by N D ( t ) N T ( t ) , N H e ( t ) , N N ( t ) , N e ( t ) the respective indexed number of particles at time t, we define the functional
J ( ϕ , ρ , r , u , E , A , B , { N D , N T , N H e , N N , N e } )
where
J = G ( u ) + F ( ϕ ) + E c ( ϕ , r ) + F 1 + F 2 + F 3 + F 4 ,
and where we assume γ p D > 0 , γ p T > 0 , γ N D > 0 , γ N 1 T > 0 , γ N 2 T > 0 , γ 2 p H e > 0 , γ N 1 H e > 0 ,   γ N 2 H e > 0 , γ N > 0 , γ e > 0 and α D > 0 , α T > 0 , α H e > 0 , α N > 0 , α D T > 0 , α H e N > 0 , α e , e > 0 , α H e , e < 0 so that
G ( ϕ ) = γ p D 2 0 t f 0 N D ( t ) Ω ( ϕ p D ) · ( ϕ p D ) d y d s d t + γ N D 2 0 t f 0 N D ( t ) Ω ( ϕ N D ) · ( ϕ N D ) d x d y d s d t γ p T 2 0 t f 0 N T ( t ) Ω ( ϕ p T ) · ( ϕ p T ) d y d s d t + γ N 1 T 2 0 t f 0 N T ( t ) Ω ( ϕ N 1 T ) · ( ϕ N 1 T ) d x d y d s d t + γ N 2 T 2 0 t f 0 N T ( t ) Ω ( ϕ N 2 T ) · ( ϕ N 2 T ) d x d y d s d t + γ 2 p H e 2 0 t f 0 N H e ( t ) Ω ( ϕ 2 p H e ) · ( ϕ 2 p H e ) d y d s d t + γ N 1 H e 2 0 t f 0 N H e ( t ) Ω ( ϕ N 1 H e ) · ( ϕ N 1 H e ) d x d y d s d t + γ N 2 H e 2 0 t f 0 N H e ( t ) Ω ( ϕ N 2 H e ) · ( ϕ N 2 H e ) d x d y d s d t + γ N 2 0 t f 0 N N ( t ) Ω ( ϕ N ) · ( ϕ N ) d x d s d t + γ e 2 0 t f 0 N e ( t ) Ω ( ϕ e ) · ( ϕ e ) d x d y d s d t ,
and
F ( ϕ ) = α D 2 0 t f 0 N D ( t ) 0 N D ( t ) Ω | ϕ D ( x ξ 1 , y ξ 2 , t , s s 1 ) | 2 | ϕ D ( ξ 1 , ξ 2 , t , s 1 ) | 2 | ( x , y ) ( ξ 1 , ξ 2 ) | d x d y d ξ 1 d ξ 2 d s d s 1 d t + α T 2 0 t f 0 N T ( t ) 0 N T ( t ) Ω | ϕ T ( x ξ 1 , y ξ 2 , t , s s 1 ) | 2 | ϕ T ( ξ 1 , ξ 2 , t , s 1 ) | 2 | ( x , y ) ( ξ 1 , ξ 2 ) | d x d y d ξ 1 d ξ 2 d s d s 1 d t + α D T 2 0 t f 0 N D ( t ) 0 N T ( t ) Ω | ϕ D ( x ξ 1 , y ξ 2 , t , s s 1 ) | 2 | ϕ T ( ξ 1 , ξ 2 , t , s 1 ) | 2 | ( x , y ) ( ξ 1 , ξ 2 ) | d x d y d ξ 1 d ξ 2 d t + α H e 2 0 t f 0 N H e ( t ) 0 N H e ( t ) Ω | ϕ H e ( x ξ 1 , y ξ 2 , t , s s 1 ) | 2 | ϕ H e ( ξ 1 , ξ 2 , t ) | 2 | ( x , y ) ( ξ 1 , ξ 2 , s 1 ) | d x d y d ξ 1 d ξ 2 d s d s 1 d t + α N 2 0 t f 0 N N ( t ) 0 N N ( t ) Ω | ϕ N ( x ξ , t , s s 1 ) | 2 | ϕ N ( ξ , t , s 1 ) | 2 | x ξ | d x d ξ d s d s 1 d t + j = 1 2 α H e N 2 0 t f 0 N H e ( t ) 0 N D ( t ) Ω | ϕ H e ( x 1 ξ 1 , y ξ 2 , t ) | 2 | ϕ N ( ξ j , t ) | 2 | ( x , y ) ( ξ 1 , ξ 2 ) | d x d y d ξ 1 d ξ 2 d s d s 1 d t + α H e , e 2 0 t f 0 N H e ( t ) 0 N e ( t ) Ω | ϕ H e ( x ξ 1 , y ξ 2 , t , s s 1 ) | 2 | ϕ e ( ξ 1 , ξ 2 , t , s 1 ) | 2 | ( x , y ) ( ξ 1 , ξ 2 ) | d x d y d ξ 1 d ξ 2 d s d s 1 d t + α e , e 2 0 t f 0 N e ( t ) 0 N e ( t ) Ω | ϕ e ( x ξ 1 , y ξ 2 , t , s s 1 ) | 2 | ϕ e ( ξ 1 , ξ 2 , t , s 1 ) | 2 | ( x , y ) ( ξ 1 , ξ 2 ) | d x d y d ξ 1 d ξ 2 d s d s 1 d t
and the internal kinetics energy is expressed by
E c ( ϕ , r ) = 1 2 0 t f 0 N D ( t ) Ω | ϕ D | 2 r D t · r D t d x d y d s d t + 1 2 0 t f 0 N T ( t ) Ω | ϕ T | 2 r T t · r T t d x d y d s d t + 1 2 0 t f 0 N H e ( t ) Ω | ϕ H e | 2 r H e t · r H e t d x d y d s d t + 1 2 0 t f 0 N N ( t ) Ω | ϕ N | 2 r N t · r N t d x d y d s d t + 1 2 0 t f 0 N e ( t ) Ω | ϕ e | 2 r e t · r e t d x d y d s d t ,
Moreover,
F 1 = 1 4 π 0 t f curl A B 0 2 d t ,
F 2 = 0 t f 0 N D ( t ) Ω E i n d · K p | ϕ p D | 2 u + r D t d x d y d s d t + 0 t f 0 N T ( t ) Ω E i n d · K p | ϕ p T | 2 u + r T t d x d y d s d t + 0 t f 0 N H e ( t ) Ω E i n d · K p | ϕ 2 p H e | 2 u + r H e t d x d y d s d t + 0 t f 0 N e ( t ) Ω E i n d · K e | ϕ e | 2 u + r e t d x d y d s d t ,
where K p and K e are appropriate real constants related to the respective charges.
Here u = ( u 1 , u 2 , u 3 ) is the fluid velocity field and
r D , r T , r H e , r N , r e
are fields of displacements for the corresponding particle fields.
Also A denotes the magnetic potential, B 0 an external magnetic field and B is the total magnetic field.
Moreover, E i n d is an induced electric field.
Also,
F 3 = C D 2 0 t f 0 N D ( t ) Ω ( x , y ) r D · ( x , y ) r D d x d y d s d t + C T 2 0 t f 0 N T ( t ) Ω ( x , y ) r T · ( x , y ) r T d x d y d s d t + C H e 2 0 t f 0 N H e ( t ) Ω ( x , y ) r H e · ( x , y ) r H e d x d y d s d t + C N 2 0 t f 0 N N ( t ) Ω ( x , y ) r N · ( x , y ) r N d x d y d s d t C e 2 0 t f 0 N e ( t ) Ω ( x , y ) r e · ( x , y ) r e d x d y d s d t ,
for appropriate real positive constants C D C T , C H e , C N , C e .
Finally,
F 4 = ε D 2 0 t f N D ( t ) t 2 d t + ε T 2 0 t f N D ( t ) t 2 d t + ε N 2 0 t f N N ( t ) t 2 d t + ε H e 2 0 t f N H e ( t ) t 2 d t + ε e 2 0 t f N e ( t ) t 2 d t ,
where ε D , ε T , ε N , ε H e , ε e are small real positive constants.
Such a functional J is subject to the following constraints:
  • The momentum conservation equation for the fluid motion
    ρ u k t + u j u k x j = ρ f k P x k + τ k j , j + ( F E ) k + ( F M ) k ,
    k { 1 , 2 , 3 } .
    Here ρ = ρ D + ρ T + ρ H e + ρ N + ρ e is the total density and P is the fluid pressure field.
    Furthermore,
    τ i j = μ u i x j + u j x i 2 3 δ i j k = 1 3 u k x k ,
    i , j { 1 , 2 , 3 } ,
    F E = { ( F E ) k } = K p 0 N D ( t ) | ϕ p D | 2 d s + 0 N T ( t ) | ϕ p T | 2 d s + 0 N H e ( t ) | ϕ 2 p H e | 2 d s + K e 0 N e ( t ) | ϕ e | 2 d s E ,
    and
    F M = { ( F M ) k } = K p 0 N D ( t ) | ϕ p D | 2 u + r D t d s 0 N T ( t ) | ϕ p T | 2 u + r T t d s + 0 N H e ( t ) | ϕ 2 p H e | 2 u + r H e t d s + K e 0 N e ( t ) | ϕ e | 2 u + r e t d s × B .
  • Mass conservation equation:
    ρ t + div ( ρ u ) = 0 .
  • Energy equation
    ρ D e D t + P ( div u ) = Q t div q ,
    where we assume the Fourier law
    q = K T ,
    where T = T ( x , t ) is the scalar field of temperature.
    Also,
    e = ρ 2 u · u + ρ D 2 r D t · r D t + ρ T 2 r T t · r T t + ρ H e 2 r H e t · r H e t + ρ N 2 r N t · r N t + ρ e 2 r e t · r e t
    and
    D e D t = e t + u j e x j .
  • P = F 7 ( ρ , T ) ,
    for an appropriate scalar function F 7 .
  • Mass relations
    (a)
    m D ( t ) = Ω ρ D ( x , t ) d x ,
    (b)
    m T ( t ) = Ω ρ T ( x , t ) d x ,
    (c)
    m H e ( t ) = Ω ρ H e ( x , t ) d x ,
    (d)
    m N ( t ) = Ω ρ N ( x , t ) d x ,
    (e)
    m e ( t ) = Ω ρ e ( x , t ) d x ,
    where,
    (a)
    ( m H e , N ) T ( t ) = m H e , N ( t ) + 0 t Ω 2 ( ρ H e ( x , τ ) ) u · n d S d τ ,
    (b)
    m H e , N ( t ) = m H e ( t ) + m N ( t ) ,
    (c)
    ( m D ) ( t ) = ( m D ) 0 0 t Ω 1 Ω 2 ( ρ D ( x , τ ) ) u · n d S d τ α D ( m H e , N ) T ( t ) ,
    (d)
    ( m T ) ( t ) = ( m T ) 0 0 t Ω 1 Ω 2 ( ρ T ( x , τ ) ) u · n d S d τ α T ( m H e , N ) T ( t ) ,
    (e)
    ( m e ) T ( t ) = m e ( t ) + 0 t Ω 2 ( ρ T ( x , τ ) ) u · n d S d τ .
    (f)
    m e ( t ) = 0 N D ( t ) Ω | ϕ p D ( y , t , s ) | 2 d y d y d s m e m p + 0 N T ( t ) Ω | ϕ p T ( y , t , s ) | 2 d y d s m e m p + 0 N p ( t ) Ω | ϕ 2 p H e ( y , t , s ) | 2 d y d s m e m p .
  • Other mass constraints
    (a)
    Ω | ϕ N D ( x , y , t , s ) | 2 d x = m N ,
    (b)
    Ω | ϕ N 1 T ( x , y , t , s ) | 2 d x = m N ,
    (c)
    Ω | ϕ N 2 T ( x , y , t , s ) | 2 d x = m N ,
    (d)
    Ω | ϕ N 1 H e ( x , y , t , s ) | 2 d x = m N ,
    (e)
    Ω | ϕ N 2 H e ( x , y , t , s ) | 2 d x = m N ,
    (f)
    Ω | ϕ p D ( x , t , s ) | 2 d x = m p ,
    (g)
    Ω | ϕ p T ( x , t , s ) | 2 d x = m p ,
    (h)
    Ω | ϕ 2 p H e ( x , t , s ) | 2 d x = 2 m p ,
  • m D ( t ) = m p N D ( t ) + m N N D ( t )
    m T ( t ) = m p N T ( t ) + m N N T ( t ) ,
    m H e ( t ) = 2 m p N H e ( t ) + 2 m N N H e ( t ) ,
    m e ( t ) = m e N D ( t ) + m e N T ( t ) + 2 m e N H e ( t ) .
  • For the induced electric field, we must have
    curl E i n d + 1 c curl K ^ p 0 N D ( t ) | ϕ p D | 2 u + r D t d s + K ^ p 0 N T ( t ) | ϕ p T | 2 u + r T t d s + K ^ p 0 N H e ( t ) | ϕ 2 p H e | 2 u + r H e t d s + K ^ e 0 N e ( t ) Ω | ϕ e ( x , y , t , s ) | 2 u ( y , t ) + r e ( x , y , t ) t d x d s × curl A B 0 1 c t curl A B 0 = 0 ,
    where K ^ p and K ^ e are appropriate real constants related to the respective charges.
  • A Maxwell equation:
    div B = 0 ,
    where
    B = B 0 curl A .
  • Another Maxwell equation:
    div E = 4 π K p 0 N D ( t ) | ϕ p D | 2 d s + 0 N T ( t ) | ϕ p T | 2 d s + 0 N H e ( t ) | ϕ 2 p H e | 2 d s + K e 0 N e ( t ) Ω | ϕ e ( x , y , t , s ) | 2 d x d s ,
    where the total electric field E stands for
    E = E i n d + E ρ ,
    and where generically denoting
    F ( ϕ ) = Ω f 5 ( ϕ , x , t ξ , s ) d x d ξ d s ,
    we have also
    E ρ = Ω f 5 ( ϕ , x , t , ξ , s ) x k d ξ d s .
At this point we generically denote
h 1 , h 2 L 2 = 0 t f Ω h 1 h 2 d x d y d t .
Thus, already including the Lagrange multipliers concerning the restrictions indicated, the extended functional J 3 stands for
J 3 = J 3 ( ϕ , u , r , P , A , B , E , Λ , E , { N D , N T , N H e , N N , N e } ) = G ( ϕ ) + F ( ϕ ) + E c ( ϕ , r ) + F 1 + F 2 + F 3 + F 4 + Λ k , ρ u k t + u j u k x j ρ f k + P x k τ k j , j ( F E ) k ( F M ) k L 2 + Λ 4 , ρ t + div ( ρ u ) L 2 + J A u x 1 + J A u x 2 + J A u x 3 + J A u x 4 + J A u x 5 ,
where,
J A u x 1 = Λ 5 , ρ D e D t + P ( div u ) Q t + div q L 2 + Λ 6 , P F 7 ( ρ , T ) L 2 ,
J A u x 2 = Λ 7 , m D ( t ) Ω ρ D ( x , t ) d x L 2 + Λ 8 , m T ( t ) Ω ρ T ( x , t ) d x L 2 Λ 9 , m H e ( t ) Ω ρ H e ( x , t ) d x L 2 Λ 10 , m N ( t ) Ω ρ N ( x , t ) d x L 2 Λ 11 , m e ( t ) Ω ρ e ( x , t ) d x L 2 ,
J A u x 3 = 0 t f Ω ( E N D ) 5 ( y , t , s ) Ω | ϕ N D ( x , y , t , s ) | 2 d x m N d y d t 0 t f Ω ( E N 1 T ) 6 ( y , t , s ) Ω | ϕ N 1 T ( x , y , t , s ) | 2 d x m N d y d t 0 t f Ω ( E N 2 T ) 7 ( y , t , s ) Ω | ϕ N 2 T ( x , y , t , s ) | 2 d x m N d y d t 0 t f Ω ( E N 1 H e ) 8 ( y , t , s ) Ω | ϕ N 1 H e ( x , y , t , s ) | 2 d x m N d y d t 0 t f Ω ( E N 2 H e ) 9 ( y , t , s ) Ω | ϕ N 2 H e ( x , y , t , s ) | 2 d x m N d y d t , 0 t f Ω ( E p D ) ( t , s ) Ω | ϕ p D ( y , t , s ) | 2 d y m p d s d t , 0 t f Ω ( E p T ) ( t , s ) Ω | ϕ p T ( y , t , s ) | 2 d y m p d s d t , 0 t f Ω ( E 2 p H e ) ( t , s ) Ω | ϕ 2 p H e ( y , t , s ) | 2 d y 2 m p d s d t ,
J A u x 4 = Λ 12 , curl E i n d + 1 c curl K ^ p 0 N D ( t ) | ϕ p D | 2 u + r D t d s + K ^ p 0 N T ( t ) | ϕ p T | 2 u + r T t d s + K ^ p 0 N H e ( t ) | ϕ 2 p H e | 2 u + r H e t d s + K ^ e 0 N e ( t ) Ω | ϕ e ( x , y , t , s ) | 2 u ( y , t ) + r e ( x , y , t , s ) t d x d s × curl A B 0 1 c t curl A B 0 L 2 + Λ 13 , div B L 2 + Λ 14 , div E 4 π K p 0 N D ( t ) | ϕ p D | 2 d s + 0 N T ( t ) | ϕ p T | 2 d s + 0 N H e ( t ) | ϕ 2 p H e | 2 d s + K e Ω | ϕ e | 2 d x d s L 2 .
J A u x 5 = Λ 15 , m D ( t ) ( m p N D ( t ) + m N N D ( t ) ) L 2 + Λ 16 , m T ( t ) ( m p N T ( t ) + m N N T ( t ) ) L 2 + Λ 17 , m H e ( t ) ( 2 m p N H e ( t ) + 2 m N N H e ( t ) ) L 2 + Λ 18 , m e ( t ) ( m e N D ( t ) + m e N T ( t ) + 2 m e N H e ( t ) ) L 2 .
Here we recall the following definitions and relations:
  • For the Deuterium field
    | ϕ D ( x , y , t , s ) | 2 = | ϕ p D ( y , t , s ) | 2 + | ϕ N D ( x , y , t , s ) | 2 | ϕ p D ( y , t , s ) | 2 1 m p ,
  • For the Tritium field
    | ϕ T ( x , y , t , s ) | 2 = | ϕ p T ( y , t , s ) | 2 + ( | ϕ N 1 T ( x , y , t , s ) | 2 + | ϕ N 2 T ( x , y , t , s ) | 2 ) | ϕ p D ( y , t , s ) | 2 1 m p ,
  • For the Hellion field
    | ϕ H e ( x , y , t , s ) | 2 = | ϕ 2 p H e ( y , t , s ) | 2 + ( | ϕ N 1 H e ( x , y , t , s ) | 2 + | ϕ N 2 H e ( x , y , t , s ) | 2 ) | ϕ 2 p H e ( y , t , s ) | 2 1 2 m p ,
  • For the Neutron field
    ϕ N = ϕ N ( x , t , s ) ,
  • For the electronic field resulting from the ionization
    ϕ e = ϕ e ( x , y , t , s ) .
  • ρ D ( y , t ) = 0 N D ( t ) Ω | ϕ D ( x , y , t , s ) | 2 d x d s ,
  • ρ T ( y , t ) = 0 N T ( t ) Ω | ϕ T ( x , y , t , s ) | 2 d x d s ,
    ρ H e ( y , t ) = 0 N H e ( t ) Ω | ϕ H e ( x , y , t , s ) | 2 d x d s ,
    ρ N ( x , t ) = 0 N N ( t ) | ϕ N ( x , t , s ) | 2 d s ,
    ρ e ( y , t ) = 0 N e ( t ) Ω | ϕ e ( x , y , t , s ) | 2 d x d s .
Also,
ρ = ρ D + ρ T + ρ H e + ρ N + ρ e ,
  • ( m H e , N ) T ( t ) = m H e , N ( t ) + 0 t Ω 2 ( ρ H e ( x , τ ) + ρ N ( x , τ ) ) u · n d S d τ ,
  • m H e , N ( t ) = m H e ( t ) + m N ( t ) ,
  • m H e ( t ) = Ω ρ H e ( x , t ) d x ,
  • m N ( t ) = Ω ρ N ( x , t ) d x ,
  • ( m D ) ( t ) = ( m D ) 0 0 t Ω 1 Ω 2 ( ρ D ( x , τ ) ) u · n d S d τ α D ( m H e , N ) T ( t ) ,
  • ( m T ) ( t ) = ( m T ) 0 0 t Ω 1 Ω 2 ( ρ T ( x , τ ) ) u · n d S d τ α T ( m H e , N ) T ( t ) ,
  • ( m e ) T ( t ) = m e ( t ) 0 t Ω 2 ( ρ e ( x , τ ) ) u · n d S d τ ,
  • m e ( t ) = Ω ρ e ( x , t ) d x .
  • m e ( t ) = 0 N D ( t ) Ω | ϕ p D ( y , t , s ) | 2 d y d y d s m e m p + 0 N T ( t ) Ω | ϕ p T ( y , t , s ) | 2 d y d s m e m p + 0 N p ( t ) Ω | ϕ 2 p H e ( y , t , s ) | 2 d y d s m e m p .
Finally,
E = E i n d + E ρ ,
and where generically denoting
F ( ϕ ) = Ω f 5 ( ϕ , x , t , ξ , s ) d x d ξ d s ,
we have also
E ρ = Ω f 5 ( ϕ , x , t , ξ , s ) x k d ξ d s .
and,
B = B 0 curl A .

21. A qualitative modeling for a general phase transition process

In this section we develop a general qualitative modeling for a phase transition process.
Let Ω R 3 be an open, bounded and connected set with a regular (Lipschitzian) boundary denoted by Ω .
Such a set Ω is supposed to a be a fixed volume in which an amount of mass of a substance A with a density function u will develop phase a transition for another phase with corresponding density function v . The total mass m T is suppose to be kept constant throughout such a process.
We model such transition in phase through a functional J : V × V R where
J ( u , v ) = γ 1 2 Ω u · u d x + α 1 2 Ω u 4 d x γ 2 2 Ω v · v d x + α 2 2 Ω v 4 d x 1 2 Ω ω 2 ( u 2 + v 2 ) d x E 2 Ω ( u 2 + v 2 ) d x m T .
Here γ 1 > 0 , γ 2 > 0 , α 1 > 0 , α 2 > 0 and V = W 1 , 2 ( Ω ) .
The phases corresponding to u and v are connected through a Lagrange multiplier E, which represents the chemical potential of the chemical process in question.
We assume the temperature is directly proportional to the internal kinetics E C energy where
E C = 1 2 Ω u 2 r u t · r u t d x .
For a internal vibrational motion, we assume approximately
r u e i ω t w 5 ( x ) ,
for an appropriate frequency ω and vectorial function w 5 .
Thus, the temperature T = T ( x , t ) is indeed proportional to ω 2 , that is, symbolically, we may write
T E 1 ω 2 .
Therefore, we start with the system with a phase corresponding to u 1 and v 0 at ω = 1 . Gradually increasing the temperature to a corresponding ω = 15 , we obtain a transition to a phase corresponding to u 0 and v 1 .
At this point, we also define the index normalized corresponding densities
ϕ u = u 2 u 2 + v 2
and
ϕ v = v 2 u 2 + v 2 .
Finally, we have obtained some numerical results for the following parameters:
Ω = [ 0 , 1 ] R , γ 1 = γ 2 = 1 , α = 0.1 , α 2 = 10 3 .
  • We start with ω = 1 corresponding to ϕ u 1 and ϕ v 0 in Ω .
    For the corresponding solutions ϕ u and ϕ v , please see Figure 15 and Figure 16, respectively.
  • We end the process with ω = 15 corresponding to ϕ u 0 and ϕ v 1 in Ω .
    For the corresponding solutions ϕ u and ϕ v , please see Figure 17 and Figure 18, respectively.

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Figure 1. solution u 0 ( x ) for the case f ( x ) = 0 .
Figure 1. solution u 0 ( x ) for the case f ( x ) = 0 .
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Figure 2. solution u 0 ( x ) for the case f ( x ) = sin ( π x ) / 2 .
Figure 2. solution u 0 ( x ) for the case f ( x ) = sin ( π x ) / 2 .
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Figure 3. solution u 0 ( x ) for the case f ( x ) = 0 .
Figure 3. solution u 0 ( x ) for the case f ( x ) = 0 .
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Figure 4. solution u 0 ( x ) for the case f ( x ) = sin ( π x ) / 2 .
Figure 4. solution u 0 ( x ) for the case f ( x ) = sin ( π x ) / 2 .
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Figure 5. Density t ( x , y ) for the Case A.
Figure 5. Density t ( x , y ) for the Case A.
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Figure 6. Density t ( x , y ) for the Case B.
Figure 6. Density t ( x , y ) for the Case B.
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Figure 7. Solution u ( x ) = v 3 * ( x ) / β for the example 1.
Figure 7. Solution u ( x ) = v 3 * ( x ) / β for the example 1.
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Figure 8. Solution u ( x ) = v 3 * ( x ) / β for the example 2.
Figure 8. Solution u ( x ) = v 3 * ( x ) / β for the example 2.
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Figure 9. Solution u 0 ( x ) for the example A.
Figure 9. Solution u 0 ( x ) for the example A.
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Figure 10. Solution u 0 ( x ) for the example B.
Figure 10. Solution u 0 ( x ) for the example B.
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Figure 11. Solution ϕ N ( x ) for the ω = 1.8 .
Figure 11. Solution ϕ N ( x ) for the ω = 1.8 .
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Figure 12. Solution ϕ S ( x ) for the ω = 1.8 .
Figure 12. Solution ϕ S ( x ) for the ω = 1.8 .
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Figure 13. Solution ϕ N ( x ) for the ω = 1.0 .
Figure 13. Solution ϕ N ( x ) for the ω = 1.0 .
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Figure 14. Solution ϕ S ( x ) for the ω = 1.0 .
Figure 14. Solution ϕ S ( x ) for the ω = 1.0 .
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Figure 15. Solution ϕ u ( x ) for ω = 1 .
Figure 15. Solution ϕ u ( x ) for ω = 1 .
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Figure 16. Solution ϕ v ( x ) for ω = 1 .
Figure 16. Solution ϕ v ( x ) for ω = 1 .
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Figure 17. Solution ϕ u ( x ) for ω = 15 .
Figure 17. Solution ϕ u ( x ) for ω = 15 .
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Figure 18. Solution ϕ v ( x ) for ω = 15 .
Figure 18. Solution ϕ v ( x ) for ω = 15 .
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