# Optimum design with lagrangian finite elements: design of an electromagnet

Americo Marrocco, Olivier Pironneau

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## To cite this version:

Americo Marrocco, Olivier Pironneau. Optimum design with lagrangian finite elements: design of an electromagnet. [Research Report] IRIA-RR-251, IRIA. 1977, pp.58. hal-04716425

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OPTIMUM DESIGN
WITH LAGRANGIAN
FINITE ELEMENTS :
DESIGN OF
AN ELECTROMAGNET

Americo MARROCCO
Olivier PIRONNEAU

Rapport de Recherche N° 251

Août 1977
OPTIMUM DESIGN
WITH LAGRANGIAN FINITE ELEMENTS:
DESIGN OF AN ELECTROMAGNET

Americo MARROCCO and Olivier PIRONNEAU
IRIA/LABORIA

Résumé :

Pour illustrer la méthode d'optimum design, on optimise l'entrefer d'un électroaimant à l'aide
des techniques du contrôle optimal, l'approximation étant faite par éléments finis triangulaires
de degré 1.

Abstract :

As an illustration of the method an electromagnet with constant interpolar field is designed by
the techniques of optimal control discretized with Lagrangian triangular finite elements of degree 1.
# I. INTRODUCTION

In this paper we shall discuss some aspects related to the design of systems for which a good numerical simulation by partial differential equations can be done. The practical situation is the following : some engineers and numerical analysts have spend a non neglectible amount of time at making an efficient numerical simulation of a system. This numerical scheme is now currently used by the engineers in the design of new products. There, the "know how" of the engineer is used : he knows that if such and such parameters are changed such behavior is induced. The numerical analyst raise the following question : can this be done automatically ? In general, the answer is no; indeed dialogues with the engineers reveal that the (often subconscious) criteria for optimal design are many and that the constraints are hundreds. Also very often the criteria are non differentiable and the constraints are thought of only once the unsuitable answer is given. However the numerical analyst can be of great help to the engineer intuition. With respect to one differentiable criteria, he can :

- 1. Give the optimum design under certain constraints

- 2. Tell the direction of changes to be done to improve the design.

The first can be done by solving an "optimum design" problem ; the second involves the computation of the gradient of the criteria with respect to the variation of the domain of the PDE.

Optimum design problems have been studied theoretically by several authors (see PIRONNEAU [1] and the bibliography there in, for example). By changes of variables bringing the variable domains into a fixed domain one may convert the problem into an optimum control problem where the control variable is in the coefficient of the PDE. This method allowed BEGIS-GLOWINSKI [2] and MORICE [3] to device very satisfactory methods for optimum design problems ; with two drawbacks however :

- 1. It is difficult to take into account geometrical constraints

- 2. A completely new study of the (new) state equation is necessary.

Therefore we propose a method which operates in the physical plan and which can be a natural extension of the numerical study of the state equation. The method is the discrete analogue of the continuous case in PIRONNEAU [4].
To fix the idea we have illustrate the method on an electromagnetic system : the
design of an electromagnet. The problem is also interesting in itself for the de-
sign of very large electromagnets and for tape recorder heads. The state equation
is a non linear elliptic equation studied numerically in GLOWINSKI-MARROCCO [5],
[6]. We have look for electromagnet with constant interpolar field and electro-
magnet with minimal electromagnetic losses, we hope that these design examples
are sufficient for the reader to be able to adapt the technique to other PDE's
and other criteria.

The computation are done with triangular Lagrangian piecewise linear elements. The
nodes are allowed to vary on prescribed curves (as suggested by YVON to avoid in-
tercrossing but it is not an obligation). The gradient of the criteria with res-
pect to the coordinates of the moving nodes is computed (as an answer to the se-
cond question above) and the criteria is then minimized by the method of steepest
descent. The results are excellent : a few iterations are sufficient for a very
good precision.

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## II. STATEMENT OF THE MAGNETOSTATIC PROBLEM.

### I.1. Preliminaries.

Let us consider the Maxwell's equation for magnetostatic

$$ (2.1) \qquad \nabla \times \vec{H} = \vec{j} $$

$$ (2.2) \qquad \vec{B} = \mu \vec{H} $$

$$ (2.3) \qquad \nabla \cdot \vec{B} = 0 $$

where

* $\vec{H}$ is the **magnetizing force** (or magnetic field intensity)

* $\vec{j}$ is the **current density vector** (and (2.1), is the well known Maxwell-Ampère relation)

* $\vec{B}$ is the **flux density** (or magnetic induction)

* $\mu$ is the **magnetic permeability** (a scalar function for isotropic material)

Equation (2.2) gives the relation between magnetizing force $\vec{H}$ and flux density $\vec{B}$. This relation is linear in air, where we have $\mu = \mu_0 = 4\pi 10^{-7}$ MKSA, but it becomes non linear in iron because the magnetic permeability $\mu$ is a function of $\vec{H}$ (or $\vec{B}$). The behavior of $\mu$ versus $\vec{H}$ or $\vec{B}$ can be extracted from the characteristic curve (see fig. (II.1)). The hysteresis effect is not considered, only the saturation effect is taken into consideration. With relation (2.3) it is classical [see DURAND [7] for example] to introduce the potential vector $\vec{A}$ such that

$$ (2.4) \qquad \vec{B} = \nabla \times \vec{A} $$

Using this last relation, equations (2.1), (2.2), (2.3) may be written in the following way

$$ (2.5) \qquad \nabla \times (\nu \nabla \times \vec{A}) = \vec{j} $$

where $\nu$ is the **magnetic reluctivity** given by $\nu = \frac{1}{\mu}$

In our two-dimensional problem, potential vector $\vec{A}$ and current density vector $\vec{j}$ have a particular structure.

$$ (2.6) \qquad \left\{ \begin{array}{l} \vec{A} = (0, 0, A_3) \\ \vec{j} = (0, 0, j_3) \end{array} \right. $$

and $A_3$, $j_3$ are only $x_1$ and $x_2$ dependant.

In this particular case, equation (2.5) can be written :
$$ (2.7) \qquad - \sum_{i=1}^{2} \frac{\partial}{\partial x_i} \left( \nu \frac{\partial A_3}{\partial x_i} \right) = j_3 $$

or

$$ (2.7 \text{ bis}) \qquad - \nabla \cdot (\nu \nabla A_3) = j_3 $$

and equation (2.7) is the equation we have to study. Equation (2.5) or (2.7) is theoretically given for the whole space but in practice, for computation, we take a bounded domain $\Omega$. Then the potential vector $\vec{A}$ must satisfy suitable properties on the boundaries of $\Omega$.

For our computational example we have taken an electro-magnet (see fig. (II.2)). This example is extracted from [DURAND [7]] but here we consider the two-dimensional cartesian problem. For computations it is sufficient to consider a quarter of the electro-magnet because of symmetry and anti-symmetry conditions.

The boundaries conditions for the "potential vector $A_3$" are shown on figure II.3. On $\overline{DC}$ and $\overline{CB}$ we take $A_3 = 0$, i.e. no-flux line cross these boundaries. On AB we have electric anti-symmetric condition but related with geometric symmetry this leads to $A_3 = 0$. Then no flux lines cross the boundary $\overline{AB}$.

On $\overline{DA}$ we have natural symmetry condition, which is given by the Newmann condition

$$ \frac{\partial A_3}{\partial n} = 0 $$

(In our practical case $\frac{\partial A_3}{\partial x_1} = 0$, and the flux lines cross the boundary AB orthogonally).

In the following we denote by

$$ \Gamma_0 = \overline{DC} \cup \overline{CB} \cup \overline{BA} $$

$$ \Gamma_1 = \overline{DA} $$

## II.2. Mathematical model.

The domain $\Omega$ is a rectangle in $\mathbb{R}^2$ and the boundary $\Gamma$ is separated in two parts $\Gamma_0$ and $\Gamma_1$.

The magnetic field determination in $\Omega$ is equivalent to the solution of partial differential equation

$$ (2.8) \qquad \left\{ \begin{array}{ll} - \frac{\partial}{\partial x_1} \left( \nu(x, A) \frac{\partial A}{\partial x_1} \right) - \frac{\partial}{\partial x_2} \left( \nu(x, A) \frac{\partial A}{\partial x_2} \right) = j & \text{in } \Omega \\ A = 0 & \text{on } \Gamma_0 \\ \frac{\partial A}{\partial n} = 0 & \text{on } \Gamma_2 \end{array} \right. $$

where $A$ is function of $x_1, x_2$. The reluctivity $\nu$ is a function of the space va-
riable x = (x₁, x₂) and also function of |**B**| in the iron.
In two dimensional cartesian case we have

$$
(2.9) \qquad |\vec{B}| = |\nabla \times \vec{A}| = |\nabla A|
$$

if **A** = (0,0,A), so that **v** is also function of **A**.

We can write the relectivity v as

$$
(2.10) \qquad \nu = \nu_o \nu_r
$$

where $v_o = \frac{1}{\mu_o} = \frac{1}{4\Pi 10^{-7}}$ MKSA is the reactivity in air and $v_r$ the relative reactivity with respect to air.

In iron, $v_r$ is a non linear function of $|\vec{B}|$ ; but it is easy to see that it is
more convenient to express $v_r$ as a function of $|\vec{B}|^2 = |\nabla A|^2 = |\nabla \times \vec{A}|^2$ .

Following this remark, we can write (2.8) more precisely as

$$
\begin{equation}
\begin{aligned}
& \left\{
  \begin{array}{ll}
    -\frac{\partial}{\partial x_1} [v_r(x, |\nabla A|^2) \frac{\partial A}{\partial x_1}] - \frac{\partial}{\partial x_2} [v_r(x, |\nabla A|^2) \frac{\partial A}{\partial x_1}] = \mu_o j & \text{in } \Omega \\
    A = 0 & \text{on } \Gamma_o \\
    \frac{\partial A}{\partial n} = 0 & \text{on } \Gamma_1
  \end{array}
\right.
\end{aligned}
\tag{2.11}
\end{equation}
$$

The first equation in (2.11) can be simply written

$$
(2.12) \quad - \nabla \cdot [v_r(x, |\nabla A|^2) \nabla A] = \mu_o j \quad \text{in } \Omega
$$

It is easy to see [GLOWINSKI-MARROCCO [5], [6]] that (2.11) (or (2.12)) is the
Euler-equation of an optimisation problem in a suitable functional space.

Let us denote by $\Psi$ the function such that

$$
\begin{equation}
\left\{
\begin{array}{@{}l@{}}
  \displaystyle \frac{\partial \Psi}{\partial |\vec{B}|^2}(x, |\vec{B}|^2) = v_r(x, |\vec{B}|^2) \\[1em]
  \Psi(x, 0) = 0
\end{array}
\right.
\end{equation}
$$

Then the energy functional of the optimisation problem is given by

$$
(2.14) \quad \mathcal{F}(A) = \frac{1}{2} \int_{\Omega} \Psi(x, |\nabla A|^2) dx - \int_{\Omega} \mu_o j A \, dx
$$

In iron, the *relative reactivity* $v_r$, function of $|\vec{B}|^2$ is approximated by an ana-
lytical function of the family defined by
$$ (2.15) \qquad \tilde{v}_{\epsilon, \alpha, c, T}(x) = \epsilon + (c-\epsilon) \frac{x^\alpha}{x^\alpha + T} $$

The parameters $\epsilon, \alpha, c, T$ are chosen to fit the data extracted from characteristic curve.

A typical set of values for these parameters is

$$ \epsilon = 5.163619 \ 10^{-4} $$

$$ c = 0.175775 $$

$$ \alpha = 5.419241 $$

$$ T = 8.758756 \ 10^3 $$

In (2.13) we replace obviously $v_r$ by our analytical approximation (see fig. [II.4])
We can now have a rigorous formulation for our problem.

Let us define the functional space $V$ by

$$ (2.16) \qquad \nu = \{ v | v \in L^2(\Omega), \frac{\partial v}{\partial x_i} \in L^2(\Omega), v|_{\Gamma_o} = 0 \} $$

It is easy to see that with the norm

$$ (2.17) \qquad v \to \|v\|_V = \int_{\Omega} |\nabla v|^2 dx $$

$V$ is an Hilbert space.

The optimisation problem can be precisely formulated by

$$ \text{Find } A \in \mathcal{V} \text{ such that} \\ \mathfrak{F}(A) \le \mathfrak{F}(v) \quad \text{for every } v \in \mathcal{V} \\ \text{where } \mathfrak{F} \text{ is given by} \\ \mathfrak{F}(v) = \frac{1}{2} \int_{\Omega} \Psi(x, |\nabla v|^2) dx - \int_{\Omega} \mu_o j.v \, dx \\ \text{and } \Psi \text{ given by (2.13).} $$

## II.3. Fundamentals results.

### Theorem 2.1.

The optimization problem (2.18) has a unique solution.

This result comes directly from the following properties

- $\mathfrak{F}$ is strongly continuous on $\mathcal{V}$

- $\mathfrak{F}$ is strictly convex

- $\mathfrak{F}(v) \to \infty$ when $\|v\|_V \to \infty$

which are derived easily from reluctivity properties (2.15). We can also prove
that A, solution of the optimization problem (2.18) is also the unique solution
of (2.11) in v.

An other form of the optimality condition is given by the variational formulation

$$
(2.19) \quad \left\{
\begin{array}{@{}l@{}}
\displaystyle \int_{\Omega'} v_r(x, |\nabla A|^2) \nabla A \cdot \nabla v \, dx - \int_{\Omega_0^o} \mu_o j.v \, dx = 0 \quad \forall v \in \mathcal{V} \\[1em]
\displaystyle \qquad \hat{A} \in \mathcal{V}
\end{array}
\right.
$$

II.4. Finite element approximation

The triangulation $\mathcal{J}_h$ will be a set of triangles $T$ such that

$$
(2.20) \quad \bigcup_{T \in \mathcal{J}_h} T \subset \bar{\Omega}
$$

of $T_1$ and $T_2 \in \mathcal{J}_h$ we have

$$
(2.21) \quad \left\{
\begin{array}{@{}l@{\quad}l@{}}
  & T_1 \cap T_2 = \emptyset \\[1em]
\text{or } & T_1 \text{ and } T_2 \text{ have a common side} \\
\text{or } & T_1 \text{ and } T_2 \text{ have a common vertex}
\end{array}
\right.
$$

As usual h is the largest edges 'lenght of J_h
We define

$$
(2.22) \quad \Omega_h = \bigcup_{T \in \mathcal{T}_h} T
$$

$$
(2.23) \quad \bar{\Theta}_h = \{ P \mid P \in \Omega_h, P \text{ vertex of } T, T \in \mathcal{T}_h \}
$$

$$
(2.24) \quad \Gamma_{oh} = \partial \Omega_h \cap \Gamma_o
$$

$$
(2.25) \quad \Gamma_{1h} = \partial \Omega_h \cap \Gamma_1
$$

$$
(2.26) \quad \gamma_{oh} = \{ P \mid P \in \Gamma_{oh}, P \text{ vertex of } T, T \in \mathcal{T}_h \}
$$

$$
(2.27) \quad \theta_h = \bar{\theta}_h - \gamma_{oh}
$$

Remark 2.1.

With the triangulation used, the boundaries between air and iron, air and copper,
never cross a triangle and are unions of edges of triangles which belong to J_h.
(see fig. [II.5] for an initial triangulation retained).

The functional space $\mathcal{V}$ (2.16) is approximated by

$$
(2.28) \quad \nu_{oh} = \{ v_h | v_h \in C^0(\bar{\Omega}_h), d^o v_h \le 1 \text{ on } T, T \in \mathcal{T}_h, v_h = 0 \text{ on } \Gamma_{oh} \}
$$
and we can express the approximated problem by

$$
\begin{array}{l}
\text{Find } A_h \in \mathcal{V}_{oh} \text{ such that} \\
(2.29) \qquad \mathfrak{F}(A_h) \le \mathfrak{F}(v_h) \quad \forall v_h \in \mathcal{V}_{oh} \\
\text{where } \mathfrak{F} \text{ is given by (2.14)}
\end{array}
$$

Some remark about notation (2.22), (2.27).

We can number from 1 to $\overline{N(h)}$, the vertices of the triangulation $\mathcal{J}_h$, and an other
notation for $\bar{\Theta}_h$ (2.23) is

$$
(2.30) \quad \bar{\Theta}_h = \{ x_k \mid x_k \text{ vertex (or node) of } \mathcal{J}_h \} \\
\phantom{(2.30) \quad \bar{\Theta}_h = \{ x_k \mid} \kappa = 1,2,\dots \bar{N}(h)
$$

The vertices which belong to θh (2.27) may be numbered from 1 to N(h), and N(h)
is the dimension of the space Th.

A function $v_h \in \mathcal{V}_{oh}$ is completely determined by the values it takes on $\theta_h$, we
shall denote

$$
(2.31) \quad v_k = v_h(x_k) , \quad \forall x_k \in \Theta_k \quad (\text{or for } k = 1,2,\dots,N(h))
$$

Let us define $\omega_j \in \mathcal{V}_{oh}$ by

$$
(2.32) \quad \omega_j(x_i) = \delta_{ij} \quad \text{for } i,j = 1,2,\dots,N(h)
$$

Then {$\omega_j$} $\sum_{j=1}^{N(h)}$ is a basis of $\mathcal{V}_{oh}$ and $A_h$, discrete potential vector may be explicitly plicited as

$$
(2.33) \quad A_h(x) = \sum_{i=1}^{N(h)} A_i \omega_i(x)
$$

and $A_h$ is solution of the non linear system

$$
(2.34) \quad \left\{
\begin{array}{@{}l@{}}
  \displaystyle \int_{\Omega} v_r(x, |\nabla A_h|^2) \nabla A_h \cdot \nabla \omega_j \, dx - \int_{\Omega} \mu_{0j} \omega_j \, dx = 0 \\
  \quad j = 1, 2, \dots, N(h)
\end{array}
\right.
$$

An other expression for (2.34), used to explicit the algorithm is

$$
(2.35) \quad \frac{\partial \mathfrak{F}}{\partial v_i}(A_h) = 0 \quad i = 1, 2, \dots, N(h)
$$
### III. STATEMENT OF THE OPTIMUM DESIGN PROBLEM - (CONTINUOUS CASE).

Ω is an open set of $R^N$ ($N = 2$ in our case), and j and $E_d$ are two given function of $L^2(Ω)$.

Let F be the ferrous region, C the copper region and $G = Ω - F \cup C$.

The magnetic potential vector $A_F$ (which depends on the shape of ferrous region F), is the solution of the non linear elliptic variational equality.

$$ (3.1) \qquad \left\{ \begin{array}{l} \displaystyle \int_{\Omega} \nu_r (x, |\nabla A_F|^2) \nabla A_F \cdot \nabla \omega \, dx = \int_{\Omega} j_1 \cdot \omega \, dx \quad \forall \omega \in \nu \\[2ex] A_F \in \nu \end{array} \right. $$

In (3.1), $j_1$ denote simply $\mu_0$ and the magnetic relative reluctivity $\nu_r$ is some known smooth function of $|\nabla A_F|^2$ in F (satisfying the coercivity condition) and is equal to 1 elsewhere.

$\nu$ is the Hilbert space defined in (2.16).

If D is some open subset of G we wish to solve the problem

$$ (3.2) \qquad \begin{array}{ll} \min & J(F) \\ F & \in \mathfrak{F} \end{array} $$

where

$$ (3.3) \qquad J(F) = \int_D |\nabla A_F - E_d|^2 \, dx $$

and $\mathfrak{F}$ is the set of admissible shapes for the ferrous region. For (3.2) to be well defined we must have

$$ (3.4) \qquad \mathfrak{F} = \{ F / F \subset \Omega, F \cap C = \emptyset \} $$

Strictly speaking (3.2) is an optimal control problem for the distributed parameter system (3.1), where the control appears in the coefficient of the P.D.E. as a discontinuity of $\nu_r$.

Although not much can be said on the existence and uniqueness of solution of (3.2), the tools used by LIONS [8], CHAVENT et all [9] [10], TARTAR [11], seem to indicate that if no constraints are imposed on the regularity of the boundary of F, minimizing sequences for (3.2) may not converge to any solution of (3.1) for some F.

Furthermore it should be pointed out that problem (3.2) is very similar to an optimum design problem (optimal control problem where the unknown is a boundary of the domain). Optimum design problems with Neumann condition on the unknown
boundary are also known for not having a solution if no regularity is assumed for
the boundary (PIRONNEAU-SAGUEZ [12]). However if $\mathfrak{J}$ contains only sets with uni-
formely Lipschitz boundaries then these problems have at least one solution
(CHENAIS [13])

Now let us outline how such problems can be solved by gradient techniques. Since
our purpose is to analyse in detail the discrete case, we shall only recall the
essence of the method only to make the analysis of the discrete case clearer.
More details can be found on a similar problem in PIRONNEAU [4]

Let $F'$ be obtained from $F$ by

$$
(3.5) \qquad \partial F' = \{x + \alpha(x) \mid x \in \partial F\}
$$

where $\alpha$ is a given vector valued function (usually on takes $\vec{\alpha}(x) = \beta(x) \vec{n}(x)$
where $\vec{n}$ is the normal to $\partial F$).

We must be able to evaluate the first order term in $\alpha$ of the variation

$$
\partial J = J(F') - J(F)
$$

For the sake of clarity let us assume that $D$ does not depend upon $\alpha$.

From (3.3) with $\delta A_F = A_{F'} - A_F$

$$
(3.6) \qquad \partial J = 2 \int_D (\nabla A_F - E_d) \delta A_F \, dx + o(\delta A_F)
$$

and from (3.1) with $\delta v_r = v_r(x, |\nabla A_F|^2) - v_r(x, |\nabla A_F|^2)$

$$
(3.7) \qquad \left\{
\begin{array}{l}
\displaystyle \int_{\Omega} \left[ \nabla v_r(x, |\nabla A_F|^2) \nabla \delta A_F \cdot \nabla \omega + \delta v_r \nabla A_F \cdot \nabla \omega \right] \, dx = o(\delta A_F) \\[1em]
\forall \omega \in \mathcal{V}
\end{array}
\right.
$$

Now see (fig. III.1)

$$
\text{in } F' \cap F \qquad \delta v_r = 2 \mathring{v}_r (x, |\nabla A_F|^2) \nabla A_F \cdot \nabla \delta A_F
$$

$$
\text{in } G' \cap G \qquad \delta v_r = 0
$$

$$
\text{in } F' \cap G \qquad \delta v_r = v_r(x, |\nabla A_F|^2) - 1
$$

$$
\text{in } F \cap G' \qquad \delta v_r = 1 - v_r(x, |\nabla A_F|^2)
$$

where

$$
\mathring{v}_r(x, |\nabla A|^2) = \frac{\partial}{\partial |\nabla A|^2} [v_r(x, |\nabla A|^2)]
$$

Therefore (3.7) becomes
$$
(3.8) \quad \left\{
\begin{array}{@{}l@{}}
\displaystyle \int_{\Omega} [v_r(x, |\nabla A_F|^2) \nabla \delta_{A_F} \cdot \nabla \omega + 2 \dot{v}_r(x, |\nabla A_F|^2) \nabla_{AF} \cdot \nabla \delta_{AF} \nabla_{AF} \cdot \nabla \omega] \, dx \\
\displaystyle \quad + \int_{F' \cap G} [v_r(x, |\nabla A_F|^2) - 1] \nabla_{AF} \cdot \nabla \omega \, dx + \\
\displaystyle \quad + \int_{F \cap G'} [1 - v_r(x, |\nabla A_F|^2)] \nabla_{AF} \cdot \nabla \omega \, dx = o(\delta_{AF}) \\
\forall \omega \in \nu
\end{array}
\right.
$$

If $\alpha(x) \ll 1$ the last two integrals in (3.8) can be approximated by the surface
integral

$$
(3.9) \quad \int_{\partial F} [v_r(x, |\nabla A_F|^2) - 1] \nabla_{AF} \nabla \omega \alpha(x) \cdot n(x) \, dS(x) + o(\alpha)
$$

Therefore if we define $P_F$ as the solution in $\nu$ of

$$
(3.10) \quad \left\{
\begin{array}{@{}l@{}}
\displaystyle \int_{\Omega} [v_r(x, |\nabla A_F|^2) \nabla P_F \cdot \nabla \omega + 2 \dot{v}_r(x, |\nabla A_F|^2) \nabla_{AF} \cdot \nabla P_F \nabla_{AF} \cdot \nabla \omega] \, dx = \\
\displaystyle \quad 2 \int_D (\nabla A_F - E_d) \cdot \nabla \omega \, dx, \quad \forall \omega \in \nu
\end{array}
\right.
$$

Then by letting $\omega = \delta_{AF}$ in (3.10) and $\omega = P_F$ in (3.8) we find from (3.6), (3.8),
(3.9) that

$$
(3.11) \quad \delta J = \int_{\partial F} [v_r(x, |\nabla A_F|^2) - 1] \nabla_{AF} \cdot \nabla P_F \alpha(x) n(x) \, dS(x) + o(\alpha)
$$

Now it appears from (3.11) that by choosing $\alpha(x)$ such that

$$
(3.12) \quad \alpha(x) n(x) = - \rho [v_r(x, |\nabla A_F|^2) - 1] \nabla_{AF} \cdot \nabla P_F \Big|_{\partial F}
$$

(where $\rho$ is a small positive parameter), we can decrease the criterion $J$ by

$$
(3.13) \quad \delta J = - \rho \int_{\partial F} [v_r(x, |\nabla A_F|^2) - 1]^2 [\nabla_{AF} \cdot \nabla P_F]^2 dS
$$

This is the essence of the gradient method below

Algorithm 1.

0. Choose $F_o$, $\rho$ and set $i = 0$

1. Compute $A_{F_i}$, $P_{F_i}$ from (3.1), (3.10)

2. Compute $\beta_i = - (v_r(x, |\nabla A_{F_i}|^2) - 1) \nabla_{AF_i} \cdot \nabla P_{F_i} \Big|_{\partial F_i}$
$$
3. \text{ Set } \partial F_{i+1} = \{x + \rho \beta_i(x) n(x) \mid x \in \partial F_i\} \\
\text{Set } i = i+1 \text{ and go back to 1}
$$

It can be shown that so long as all ∂Fᵢ's are smooth algorithm 1 generates sequences which have accumulation point satisfying β = 0. But β = 0 implies
∇A_F∇P_F = 0 on ∂F, which in turn implies, in general, from (3.10), that P_F = 0
and that F is a local minimum of (3.2).

It seemed at first that if the discretization grid of any numerical method was
refined enough one could simply discretize (3.12) to obtain a method for solving
(3.2). It turned out that the ∂Fᵢ's become more and more irregular and we had
to analyse the discrete case separately. It is the object of the next paragraph.
IV. OPTIMUM DESIGN VIA TRIANGULAR LINEAR FINITE ELEMENTS.

In order to implement algorithm 1 we must obtain the discrete analogue of (3.11)

It is well known in optimal control theory that one should obtain the optimality conditions for the discretized problem rather than discretize directly (3.17).

For the approximation by finite element method we keep the same notations as in (II), let us recall simply that the discrete vector potential $A_h$ is solution of the non linear system

$$ (4.1) \qquad \int_{\Omega} v_r(x, |\nabla A_h|^2) \nabla A_h \cdot \nabla \omega_j \, dx = \int_{\Omega} j_1 \cdot \omega_j \, dx \quad j = 1, 2, \dots, n $$

As in the continuous case we search for $\Omega$ such that $\nabla A$ is as close to $E_d$ as possible in a subdomain $D$ of $\Omega$.

Therefore we define

$$ (4.2) \qquad J_h(\pi_h) = \int_D |\nabla A_h - E_d|^2 \, dx $$

and we wish to find the coordinates of the nodes of $\pi_h$ such that $J_h(\pi_h)$ is minimum. Thus if $\chi$ is the set of admissible position of the nodes

$X = \{x_k\}$, $k = 1, 2, \dots, N(h)$ then our problem is

$$ (4.3) \qquad \min_{X \in \mathcal{X}} J_h(\pi_h) $$

Problem (4.3) is a constrained minimization problem in $R^{N \times N(h)}$. In order to apply a gradient or a conjugate gradient method we must compute

$$ (4.4) \qquad \partial_{\alpha_k} J_h(\pi_h) = \lim_{\|\alpha_k\| \to 0} \frac{[J_h(\pi_h') - J_h(\pi_h)] / \|\alpha_k\|}{k = 1, 2, \dots, N(h)} $$

where $\pi_h'$ is the triangulation obtained from $\pi_h$ by moving the $k$th node in position $x_k + \alpha_k$.

It is convenient to assume that

$$ (4.5) \qquad D = \bigcup_{j \in J} T'_j $$

Note that $D$ may or may not depend upon $\alpha$.

Therefore, if $E_d$ is piecewise constant on $J_h$

$$ (4.6) \qquad \delta J_h = J_h(\pi_h') - J_h(\pi_h) = \sum_j \int_{T'_j} |\nabla A'_h - E_d|^2 dx - \int_{T_j} |\nabla A_h - E_\alpha|^2 dx $$
where $A'_h$ is computed on $\pi'_h$ by (4.1) and where $T'_j$ are the triangles of $\pi'_h$.

If $C$ and $C'$ are two sets of $\mathbb{R}^N$, we shall denote

$$ (4.7) \qquad \left\{ \begin{array}{l} C \setminus C' = C - C' \cap C \\ C \cap C = C' - C' \cap C \end{array} \right. $$

Then (4.6) can be rewritten as

$$ (4.8) \quad \delta J_h = \sum_{j \in J} \int_{T_j \cap T'_j} (|\nabla A'_h - E_d|^2 - |\nabla A_h - E_d|^2) dx + \int_{T'_j \setminus T_j} |\nabla A'_h - E_d|^2 dx \\ - \int_{T_j \setminus T'_j} |\nabla A_h - E_d|^2 dx $$

Define (see 2.33) for notation $A_i$)

$$ (4.9) \qquad \delta A_i = A'_i - A_i \quad i = 1, 2, \dots, N(h) $$

and

$$ (4.10) \qquad \delta \omega_i(x) = \omega'_i(x) - \omega_i(x) \quad i = 1, 2, \dots, N(h) $$

where $\omega'_i$ is defined on $\pi'_h$ by (2.32). Then (2.33) implies that

$$ (4.11) \qquad A'_h - A_h = \sum_{i=1}^{N(h)} \delta A_i \omega_i + A_i \delta \omega_i + \delta A_i \delta \omega_i $$

We shall show below (Lemma. 4.1.) the key relation

$$ (4.12) \qquad \left\{ \begin{array}{l} \delta \omega_i(x) = - \omega_k(x) \nabla \omega_i \circ \omega_k \quad \forall x \in T_j \cap T'_j, \forall i, j \text{ such that} \\ \phantom{\delta \omega_i(x) =} x_i \text{ and } x_k \text{ are nodes of } T_j \\ = 0 \quad \forall x \in T_j \cap T'_j, \forall i, j \text{ such that} \\ \phantom{= 0} x_k \notin T_j \end{array} \right. $$

Therefore $\delta A_i \delta \omega_i$ is a higher order term and the first integral of (4.8) is equal to

$$ (4.13) \quad 2 \sum_{j \in J} \int_{T_j \cap T'_j} (\nabla A_h - E_d) \left[ \sum_{i=1}^{N(h)} \delta A_i \nabla \omega_i - \sum_{i \in I_{jk}} A_i (\nabla \omega_i \circ \alpha_k) \nabla \omega_k \right] dx + o(\alpha_h) $$

where $I_{jk}$ is the set of indices of nodes of $T_j$ if $x_k \in T_j$ and $I_{jk} = \emptyset$ if $x_k \notin T_j$.
We shall also show later (on Lemma 4.2) that the rest of (4.8) is equal to

$$
(4.14) \quad \int_{\Omega} |\nabla A_h - E_d|^2 \nabla \omega_k \cdot \alpha_k \, dx + o(\alpha_k)
$$

Therefore if we denote by

$$
(4.15) \quad \tilde{\delta}_{A_i} = \sum_{i=1}^{N(h)} \delta_{A_i} \omega_i
$$

and by $\tilde{\alpha}_k(\cdot)$ the function of $(L^2(\Omega))^N$ which is piece wise constant on $T_j$ and
such that

$$
(4.16) \quad \left\{
\begin{array}{ll}
\tilde{\alpha}_k(x) = \alpha_k & \text{if } x \text{ and } x_k \in T_j \\
& \\
& = 0 \quad \text{if } x \in T_j \text{ but } x_k \notin T_j,
\end{array}
\right.
$$

then (4.8) becomes

$$
(4.17) \quad \delta_{J_h} = \int_D \left[ 2(\nabla A_h - E_d) \left[ \tilde{\delta A} - \nabla A_h \cdot \tilde{\alpha}_k \nabla \omega_k \right] + |\nabla A_h - E_d|^2 \nabla \omega_k \cdot \alpha_k \right] dx + o(\alpha_k)
$$

To evaluate the first term we go back to (4.15), (4.9) and to the definition of
$A_i$ (2.33); (4.11)

From (4.1)

$$
(4.18) \quad \left\{
\begin{array}{@{}l@{}}
\displaystyle \sum_j \int_{T_j} v_r(x, |\nabla A'_h|^2) \nabla A'_h \cdot \nabla \omega'_1 \, dx - \int_{T_j} v_r(x, |\nabla A_h|^2) \nabla A_h \cdot \nabla \omega_1 \, dx \\
\displaystyle \qquad = \int_{\Omega} j_1 (\omega'_1 - \omega_1) \, dx \quad 1=1,2,\dots,N(h)
\end{array}
\right.
$$

As above from (4.11) it is easy to show that

$$
(4.19) \quad \left\{
\begin{aligned}
& \int_{T'_j \cap T_j} [v_r(x, |\nabla A'_h|^2) \nabla A'_h \cdot \nabla \omega'_1 - v_r(x, |\nabla A_h|^2) \nabla A_h \cdot \nabla \omega_1] \, dx = \\
& 2 \int_{T'_j \cap T_j} v_r(x, |\nabla A_h|^2) \nabla A_h \cdot \nabla (\delta_{A_i} \omega_i + A_i \, \delta \omega_i) \nabla A_h \cdot \nabla \omega_1 \, dx \\
& + \int_{T'_j \cap T_j} v_r(x, |\nabla A_h|^2) [\nabla (\delta_{A_i} \omega_i + A_i \, \delta \omega_i) \cdot \nabla \omega_1 + \nabla A_h \cdot \nabla \delta \omega_1] \, dx + o(\alpha_k)
\end{aligned}
\right.
$$

From lemma below

$$
(4.20) \quad \int_{T'_j \setminus T_j} v'_r(x, |\nabla A'_h|^2) \nabla A'_h \cdot \nabla \omega'_1 \, dx - \int_{T_j \setminus T'_j} v_r(x, |\nabla A_h|^2) \nabla A_h \cdot \nabla \omega_1 \, dx =
$$
$$ \int_{\Omega} v_r(x, |\nabla A_h|^2) (\nabla A_h \cdot \nabla \omega_l) (\nabla \omega_k \cdot \tilde{\alpha}_k) \, dx + o(\alpha_k) $$

In our example the copper region ($j_1 \neq 0$) is given, therefore the right member of (4.18) is zero. (It is not more difficult to optimize the copper region; (4.18) can be evaluated from (4.12) but if $j_1$ is not piecewise constant, it contains a boundary integral (see Lemma)).

Now (4.15) and (4.12) is used in (4.19) and together with (4.20) and (4.18) we find that

$$ (4.21) \left\{ \begin{array}{l} \sum_j \int_{T' \cap T_j} 2 \dot{v}_r(x, |\nabla A_h|^2) (\nabla A_h \cdot \nabla \tilde{\delta}A - \nabla A_h \nabla \omega_k \nabla A_h \cdot \tilde{\alpha}_k) \nabla A_h \cdot \nabla \omega_l + \\ \sum_j \int_{T' \cap T_j} v_r(x, |\nabla A_h|^2) \{[\nabla \tilde{\delta}A - \nabla A_h \cdot \tilde{\alpha}_k \nabla \omega_k] \omega_l - \nabla A_h \cdot \nabla \omega_k \nabla \omega_l \cdot \tilde{\alpha}_k\} \, dx \\ + \int_{\Omega} v_r(x, |\nabla A_h|^2) \nabla A_h \cdot \nabla \omega_l \nabla \omega_k \tilde{\alpha}_k \, dx = o(\alpha_k) \end{array} \right. $$

We rewrite (4.21) as

$$ (4.22) \left\{ \begin{array}{l} \int_{\Omega} [2 \dot{v}_r(x, |\nabla A_h|^2) \nabla A_h \cdot \nabla \tilde{\delta}A \nabla A_h \cdot \nabla \omega_l + v_r(x, |\nabla A_h|^2) \nabla \tilde{\delta}A \cdot \nabla \omega_l] \, dx = \\ \int_{\Omega} (2 \dot{v}_r(x, |\nabla A_h|^2) \nabla A_h \cdot \nabla \omega_k \nabla A_h \cdot \tilde{\alpha}_k \nabla A_h \cdot \nabla \omega_l + v_r(x, |\nabla A_h|^2) \nabla A_h \cdot \tilde{\alpha}_k \nabla \omega_k \nabla \omega_l) \, dx \\ + \int_{\Omega} v_r(x, |\nabla A_h|^2) [\nabla A_h \cdot \nabla \omega_k \nabla \omega_l \cdot \tilde{\alpha}_k - \nabla A_h \cdot \nabla \omega_l \nabla \omega_k \cdot \tilde{\alpha}_k] \, dx + o(\alpha_k) \end{array} \right. $$

So to evaluate the first term in (4.17) we must introduce the adjoint state $P_h$ solution of

$$ P_h \in \mathcal{V}_{oh} $$

$$ (4.23) \left\{ \begin{array}{l} \int_{\Omega} 2 \dot{v}_r(x, |\nabla A_h|^2) \nabla A_h \cdot \nabla \omega_l \cdot \nabla A_h \cdot \nabla P_h + v_r(x, |\nabla A_h|^2) \nabla \omega_l \cdot \nabla P_h \, dx = \\ \int_{\Omega} 2(\nabla A_h - E_d) \cdot \nabla \omega_l \, dx \quad 1 = 1, 2, \dots, N(h) \end{array} \right. $$

Then from (4.22) we find that

$$ (4.24) \int_{\Omega} 2(\nabla A_h - E_d) \nabla \tilde{\delta}A \, dx = \int_{\Omega} 2 \dot{v}_r(x, |\nabla A_h|^2) \nabla A_h \cdot \nabla \omega_k \nabla A_h \cdot \tilde{\alpha}_k \nabla A_h \cdot \nabla P_h \, dx $$
$$
\begin{align*}
&+ \int_{\Omega} v_r(x, |\nabla A_h|^2) \nabla A_h \cdot \tilde{\alpha}_k \nabla \omega_k \cdot \nabla P_h \, dx \\
&+ \int_{\Omega} v_r(x, |\nabla A_h|^2) \nabla A_h \nabla \omega_k \nabla P_h \tilde{\alpha}_k \, dx \\
&- \int_{\Omega} v_r(x, |\nabla A_h|^2) \nabla A_h \nabla P_h \nabla \omega_k \tilde{\alpha}_k \, dx + o(\alpha_k)
\end{align*}
$$

So $\partial_{\alpha_k} J_h(\mathfrak{J}_h)$ can be evaluated from (4.17) and (4.24). It remains to prove
the following two lemmas.

**Lemma 4.1.**

If the triangulation $\mathfrak{R}'_h$ is obtained from $\mathfrak{J}_h$ by moving the $k^{th}$ node from $x_k$ into
$x_k + \alpha_k$ then

$$
\begin{align*}
\omega'_i(x) - \omega_i(x) &= - \omega_k(x) \nabla \omega_i \alpha_k \\
\forall x \in T_j \cap T'_j, \quad \forall i,j \text{ such that } x_i, x_k \in T_j
\end{align*}
$$

Proof.

If $x_i, x_l, x_k$ are the 3 nodes of $T_j$

$$
(4.25) \qquad \omega_i'(x) = \frac{(x-x_l) \times (x_l-x_k)}{(x_i-x_l) \times (x_l-x_k)}
$$

$$
\text{where } a \times b = a_1^b b_2 - a_2^b b_1
$$

Therefore

$$
\omega'_{i}(x) = \frac{(x-x_{1}) \times (x_{1}-x_{k}-\alpha_{k})}{(x_{i}-x_{1}) \times (x_{1}-x_{k}-\alpha_{k})}
$$

$$
(4.26) \quad \omega'_{i}(x) = \omega_{i}(x) - \frac{(x-x_{1}) \times \alpha_{k}}{(x_{i}-x_{1}) \times (x_{1}-x_{k})} + \frac{(x-x_{1}) \times (x_{1}-x_{k})}{[(x_{i}-x_{1}) \times (x_{1}-x_{k})]^{2}} (x_{i}-x_{1}) \times \alpha_{k}
$$

from which it is seen that $\omega'_i - \omega_i$ is linear in $x$, takes for values 0 at $x = x_1$,
0 at $x = x_i$ and

$$
- \frac{(x_k - x_l) \times \alpha_k}{(x_i - x_l) \times (x_l - x_k)} = \nabla \omega_i \cdot \alpha_k \quad \text{at } x = x_k
$$

Therefore

$$
\omega'_{i}(x) - \omega_{i}(x) = - \omega_{k}(x) \nabla \omega_{i}(x) \cdot \alpha_{k}
$$

The same formula can be shown for $i = k$
### Remark 4.1

Lemma (4.1) is also true if higher order finite elements are used. Indeed in that case $\omega_i$ is a polynomial function of the coordinates of degree L

$$ \omega_i(x) = \sum_{|1| \le L} a_i^1 x_1^{1_1} \cdots x_N^{1_N} $$

and the coefficient $a_i^1$ are found from the linear system

$$ \omega_i(x_j) = \delta_{ij} $$

Therefore if $x_k$ varies by $\alpha_k$, $a_i^1$ varies by $\delta a_i^1$ with

$$ \begin{aligned} \sum_{|1| \le L} \delta a_i^1 x_j^{1_1} \cdots x_N^{1_N} &= - \sum a_i^1 \frac{\partial}{\partial x_k} (x_j^{1_1} \cdots x_N^{1_N}) \alpha_k + o(\alpha_k) \\ &= - \nabla \omega_i(x_j) \alpha_k \delta_{kj} \end{aligned} $$

Therefore $a_k^1$ and $\delta a_i^1$ are solution of the same linear system where the right members differ only from a multiplicative constant $-\nabla \omega_i(x_k) \alpha_k$
Hence

$$ (4.27) \qquad \delta \omega_i(x) = - \omega_k(x) \nabla \omega_i(x_k) \alpha_k $$

### Lemma 4.2.

Let $f$ be a piece wise constant function on $\Omega = \bigcup_j T_j$. Then if $x_k$ moves into $x_k + \alpha_k$

$$ (4.28) \qquad \sum_j \int_{T_j \setminus T_k} f \, dx - \int_{T_j \setminus T_k} f \, dx = \int_{\Omega} f \nabla \omega_k \cdot \alpha_k \, dx + o(\alpha_k) $$

#### Proof.

We shall prove (4.28) on an example. It is straight forward to extend the proof to the general case but the notation are cumbersome

$T'_1 \setminus T_1$ and $T_1 \setminus T'_1$ are shown on (fig. IV.1.)

Then

$$ (4.29) \qquad 2 \int_{T'_1 \setminus T_1} f \, dx - 2 \int_{T_1 \setminus T'_1} f \, dx = \int_{x_k^1 x_2^1} f|_{T_1} \alpha_k n^1 ds + \int_{x_3^1 x_k^1} f|_{T_1} \alpha_k n^1 ds $$
where $n^1$ is the (exterior) normal to $T_j$. Similarly

$$ (4.30) \qquad 2 \int_{T'_2 \setminus T_2} f \, dx - 2 \int_{T_2 \setminus T'_2} f \, dx = \int_{x_3^1 x_k^1} f|_{T_2} \alpha_k n^2 ds + \int_{x_k^1 x_1^1} f|_{T_2} \alpha_k n^2 ds $$
$$
(4.31) \quad 2 \int_{T'_3 \setminus T_3} f \, dx - 2 \int_{T_3 \setminus T'_3} f \, dx = \int_{x_1^+ x_k^+} f|_{T_3} \alpha_k n^3 ds + \int_{x_k^+ x_2^+} f|_{T_3} \alpha_k n^3 ds
$$

In (4.29), (4.30), (4.31) segment x_j x_1 are not orientated.

Now Green's formula applied on $T_1$, for example, gives

$$
(4.32) \quad \int_{T_1} f \nabla \omega_k \cdot \alpha_k \, dx = \int_{\partial T_1} f|_{T_1} \omega_k \alpha_k n_1 \, ds = \frac{1}{2} \int_{x_3^+ x_k^+ + x_k x_2^+} f|_{T_1} \alpha_k n_1^1 \, ds
$$

Therefore

$$
2[(4.32) \text{ on } T_1 + (4.32) \text{ on } T_2 + (4.32) \text{ on } T_3] \text{ gives } (4.29) + (4.30) + (4.31)
$$

**Remark 4.2.**

If f is not piece wise constant the lemma (4.2) does not hold. One can however
express the left member of (4.28) in terms of α_k by means of boundary integrals.
We summarize the results in the following proposition

**Proposition 4.1.**

Assuming that $E_d$ is piece wise constant on $\mathcal{J}_h$ and that $j_1|_{T_1} = 0$ for all $j$ such that $x_k \in T_j$, if $\mathcal{J}'_h$ is obtained from $\mathcal{J}_h$ by moving the $k^{th}$ node from the position $x_k$ into the position $x_k + \alpha_k$. Then, $J_h(\mathcal{J}_h)$ given by (4.2) varies of

$$
(4.33) \quad \left\{
\begin{aligned}
J_h(\tau'_h) - J_h(\tau_h) &= \int_{\Omega'} [2 \cdot \dot{v}_r \nabla_A \nabla \omega_k \nabla_A \alpha_k \nabla_A \nabla P_h + v_r (\nabla_A \alpha_k \nabla \omega_k \nabla P_h \\
&\quad + \nabla_A \nabla \omega_k \nabla P_h \alpha_k - \nabla_A \nabla P_h \nabla \omega_k \alpha_k)] \, dx \\
&\quad + \int_D [|\nabla_A - E_d|^2 \nabla \omega_k \alpha_k - 2(\nabla_A - E_d) \nabla \omega_k \nabla_A \alpha_k] \, dx + o(\alpha_k)
\end{aligned}
\right.
$$

Where $P_h$ is the solution of (4.23), and $\dot{v}_r$, $v_r$ are always functions of $x$ and
$|A_h|^2$ as previously.

Remark 4.3.

Clearly from (4.4)

$$
\begin{equation}
\begin{aligned}
(4.34) \quad \partial_{\alpha_k} J_h(\tau_h) = & \int_{\Omega'} [2 \cdot \dot{v}_r \nabla_A \nabla \omega_k \nabla_A D_k \nabla_A \nabla P_h + v_r \nabla_A D_k \nabla \omega_k \nabla P_h \\
& \phantom{\int_{\Omega'} [2 \cdot \dot{v}_r \nabla_A \nabla \omega_k \nabla_A D_k \nabla_A \nabla P_h} + v_r \nabla_A \nabla \omega_k \nabla P_h D_k - v_r \nabla_A \nabla P_h \nabla \omega_k D_k] \, dx \\
& + \int_D [|\nabla_A - E_d|^2 \nabla \omega_k D_k - 2(|\nabla_A - E_d) \nabla \omega_k \nabla_A D_k] \, dx
\end{aligned}
\end{equation}
$$
where

$$ (4.35) \qquad D_k = \alpha_k / \|\alpha_k\| $$

Now it is straight to solve (4.3) by any optimization technique. To illustrate the method let us write the gradient method with fixed stepsize for the solution of (4.3) in the case of an electromagnet. Then we wish to find the shape of the air-gap which gives a constant given field $E_d$. In that case $\chi$ is the set of coordinates of the nodes of all the proper triangulation of $\Omega$, with the same topological properties (same number of triangles, same number of nodes). To remove the constraints on $\{x_k\}_{k=1}^n$ we found that it is feasible to assume that all interior nodes are constructed from the boundary nodes of the unknown boundary from continuous mappings :

$$ (4.36) \qquad x_k = X_k(x_1, x_2, \dots, x_s), \quad k = s+1, \dots, n $$

It will be seen later that it is not necessary to know $X_k$ explicitly, only $\frac{\partial X_k}{\partial x_j}$ are required.

Furthermore, in order to avoid the nodes of the unknown boundary to cross each other we may constrained them to remain on prescribed curves. It should be kept in mind that the solution may depend upon those curves if they have not been property choosen. In any case :

$$ (4.37) \qquad x_j = X_j(t_j) \quad j = 1, \dots, s \quad t_j \in \mathbb{R} $$

Now (4.3) is an unconstrained optimization problem in $t \in \mathbb{R}^s$ and from (4.33), (4.36), (4.40) and the composite function theorem

$$ (4.38) \qquad \frac{\partial J_h}{\partial t_j} = \sum_{k=s+1}^{n} \partial_j J_h \alpha_k^j + \partial_j J_h \frac{\partial X_j}{\partial t_j} $$

with

$$ (4.39) \qquad \alpha_k^j = \partial_{x_j} X_k \frac{\partial X_j}{\partial t_j} \quad k = s+1, \dots, n $$

Thus we obtain the following algorithm

**Algorithm 2. (Gradient method with fixed step size)**
0. Choose $t^0$ and $\rho$, set $i = 0$, compute the triangulation from (4.36), (4.37)

1. Compute $A_h$ and $P_h$ from (4.1) and (4.23)

2. Set $\alpha_k^j = \partial_x_j X_k \frac{\partial X_j}{\partial t_k}$, $k = s+1, \dots, n$

$$ \alpha_k^k = \frac{\partial X_k}{\partial t_k}, \quad k = 1, 2, \dots, s $$

and compute

$$ \partial_{\alpha_k^j} J_h \text{ from (4.34), (4.35), and compute } \frac{\partial J_h}{\partial t_j} \text{ according to (4.38).} $$

3. Set $h_j = -\frac{\partial J_h}{\partial t_j}$, $j = 1, 2, \dots, s$, Set $t_j^{i+1} = t_j^i + \rho h_j$

Set $i = i+1$ compute the new triangulation from (4.36), (4.37) and go back to 1.

### Remark 4.4.

In practice it is better to optimize the step size $\rho$ at each iteration, and conjugate the direction $h$ eventually (see POLAK [14] for example).

For our computational example, relation such a type (4.36) or (4.37) became very simple. We give on fig. IV.2. an initial triangulation for the region of interest in our optimum design problem.

The triangulation $\mathcal{J}_h$ is parametrized in the following way.

The nodes □ are fixed nodes

The nodes * may be called, **principal moving nodes** (nodes at the boundary between iron and air, in the air-gap) and the nodes O, **associated moving nodes**.

The coordinates of these associated nodes are simply related with fixed nodes and principal moving nodes.

All the "moving nodes" must move on the prescribed curves (which are in our case, straight lines $\Delta_j$). We know directly the $\frac{\partial X_j}{\partial x_j}$ from the parametrized construction. We give on fig. IV.3. an admissible shape for the triangulation.
## V. NUMERICAL EXPERIMENTS.

The numerical experiments have been made on the electro-magnet showed in fig.

### II.3.

The triangulation, according with remark 2.1, is composed by 267 triangles and 158 nodes. The number of unknows is 124.

### V.1. Description of the program and algorithms used.

The optimum design can be composed by the following modules

<figure><img src="image_5.png" /></figure>

A "module" for solving the direct problem (or the magnetostatic problem). We give again the formulation of the magnetostatic problem.

$$ (5.1) \qquad \begin{cases} \text{Find } A_h \in \mathcal{V}_{oh} \text{ such that} \\ \int_{\Omega} v_r (x, |\nabla A_h|^2) \nabla A_h \cdot \nabla v_h \, dx - \int_{\Omega} \mu_o j v_h \, dx = 0 \quad \forall v_h \in \mathcal{V}_{oh} \\ \text{or Find } A_h \in \mathcal{V}_{oh} \text{ such that} \\ (5.2) \qquad \qquad \mathfrak{F}(A_h) \le \mathfrak{F}(v_h) \quad \forall v_h \in \mathcal{V}_{oh} \end{cases} $$

The algorithm used to solve (5.2) (or (5.1)) is the E.G.S.N. algorithm given in GLOWINSKI-MARROCCO [5], which can be briefly explicited by

$$ (5.3) \qquad A_h^o \text{ given in } \mathcal{V}_{oh} \quad [A_h^o \equiv 0 \text{ for the first time (global iterations)} \\ \text{we can write} \qquad A_h = (A_1, A_2, A_3, \dots, A_N(h) \text{ and after initialized with the previous state}]. $$

and the algorithm formula is given by

$$ (5.4) \quad \left\{ \begin{array}{ll} A_i^{n+1} = A_i^n - \omega \frac{\frac{\partial \mathfrak{F}}{\partial v_i} (A_1^{n+1}, A_2^{n+1}, \dots, A_{i-1}^{n+1}, A_i^n, \dots, A_N^n)}{\frac{\partial^2 \mathfrak{F}}{\partial v_i^2} (A_1^{n+1}, A_2^{n+1}, \dots, A_{i-1}^{n+1}, A_i^n, \dots, A_N^n)} & \\ \\ \text{for } i = 1, 2, \dots, N(h) & \\ \\ \omega \text{ is the relaxation parameter, } 0 < \omega < 2. & \end{array} \right. $$

The process described by (5.3), (5.4) is stopped when
$$ (5.5) \qquad \frac{N(h) \sum_{i=1} |A_i^{n+1} - A_i^n|}{\sum_{i=1} |A_i^{n+1}|} \le \eta $$

(for a given value of $\eta$. In computational experiment, the value taken for $\eta$ is $10^{-5}$)

(B) A "module" for solving the adjoint-state problem, solution of which is needed to compute the descent direction.

The adjoint state $P_h$ is given by the solution of the following variational linear equation:

$$ (5.6) \qquad \left\{ \begin{array}{l} \text{Find } P_h \in \mathcal{V}_{oh} \text{ such that} \\[1em] \qquad \int_{\Omega} \nu_r(x, |\nabla A_h|^2) \nabla P_h \cdot \nabla \nu_h \, dx + 2 \int_{\Omega} \dot{\nu}_r(x, |\nabla A_h|^2) \nabla A_h \cdot \nu_h \nabla P_h \cdot \nabla A_h \, dx \\[1em] \qquad = 2 \int_D \nabla \nu_h (\nabla A_h - E_d) \, dx \\[1em] \qquad \text{for every } \nu_h \in \mathcal{V}_{oh} \end{array} \right. $$

In (5.6), $A_h$ is given, and is the solution of the magnetostatic problem. $D$ is the domain ($D \subset \Omega$), in which is computed the criterion, and

$$ (5.7) \qquad \dot{\nu}_r(x, |\nabla A_h|^2) = \frac{\partial}{\partial |\nabla A_h|^2} \nu_r(x, |\nabla A_h|^2) $$

For this linear problem (5.6) we use a direct method for solution - Cholesky-Band-Algorithm.

To use efficiently the band structure, the nodes of the triangulation must be suitably numbered, or be renumbered in order to reduce the bandwidth of the matrix, using for example the reverse Cuthill-Mc Kee algorithm.

(C) A "module" for the descent direction computation, when we know $A_h$ solution of the magnetostatic problem and $P_h$ solution of the adjoint state problem. We must take into account in the formula if the criterion domain is variable or not.

For each moving node we can compute the real vector or the projected vector if
a displacement direction is imposed ; as in our computational experiments.

(D) A "module" minimizing the criterion functional when we know a descent direction. This is a classical optimization algorithm, but we must evidently include geometric constraints in it.

In our computational example, these geometric constraints may be explicit as :
if $M = (M_{x_1}, M_{x_2})$ is a moving point, the constraint is

$$
(5.8) \qquad a \le M_{x_1} \le b
$$

We used the "gradient method with optimal choice of step length ρ" with eventually projection. We have tested the conjugate gradient method so.

(E) To analyse very quickly the computational results it is convenient to
have a "drawing module" for the plotting of flux lines (or other characteris-
tics) related with a given geometry.

V.2. Numerical experiments with FIXED CRITERION DOMAIN.

We can see on fig. (V.1.), the domain in which is evaluated the criterion

$$
(5.9) \qquad J = \int_D |\nabla A_h - E_d|^2 dx
$$

and we can see also the initial iron geometry retained (but an other initial
geometry may be chosen).

The initial state is given in fig. (V.2), which corresponds to current density
$5.10^6$ MKSA. Even with this current density the magnetostatic problem is non
linear.

In the first example we take

$$
(5.10) \qquad \left\{ \begin{array}{l} E_d = (0, 1.3) \\ \\ B_x = 1.3 \text{ teslas} \\ \\ B_y = 0 \end{array} \right.
$$

With these parameters, the wish is to have a uniform flux density $\vec{B}$ in D such
that

$$
(5.11) \quad \left\{
\begin{array}{l}
B_x = 1.3 \text{ teslas} \\
B_y = 0
\end{array}
\right.
$$

The starting value of the criterion is

$$
(5.12) \qquad J_o = .61160 \ 10^{-5}
$$
and after 9 iterations in the optimization process (D) the criterion is

$$ (5.13) \qquad J_9 = .35450 \ 10^{-5} $$

We can see on fig. (V.3, V.4, V.5, V.6), the flux lines at iteration 1,2,5,9, and in fig. (V.7) the variation of the criterion (5.9).

We can remark that the variation of J (5.9) is relatively important in the first two or three steps.

For our second example, we take

$$ (5.14) \qquad E_d = (0, 1.6) $$

The initial value of the criterion is

$$ (5.15) \qquad J_o = .51906 \ 10^{-4} $$

and the final value is

$$ (5.16) \qquad J_5 = .17862 \ 10^{-4} $$

On fig. (V.8, V.9, V.10), we give the flux lines at steps 1,2,5.

In the two following examples, we take a current density $20.10^6$ MKSA, and the initial state (flux lines) is given by fig (V.11).

With this current density value, the flux density in the air-gap (region of interest) is more large than 1,3 or 1,6 (see on table V.1, the initial flux distribution).

In the third example, we take

$$ (5.17) \qquad E_d = (0, 1.3) $$

The initial value of the criterion is

$$ (5.18) \qquad J_o = .10030 \ 10^{-3} $$

and the final value is

$$ (5.19) \qquad J_9 = .46190 \ 10^{-6} $$
(at the first step the value is $J_1 = .60574 \ 10^{-6}$).

On fig. (V.12, V.13, V.14, V.15), we give the flux distribution at step 1,2,5,9.

We can see on **table V.2** the final flux distribution in region D.

For the **fourth example**, we take

$$ (5.20) \qquad E_d = (0,1.6) $$

The initial value of the criterion is

$$ (5.21) \qquad J_o = .25427 \ 10^{-4} $$

and the final value is

$$ (5.22) \qquad J_8 = .18863 \ 10^{-5} $$

The flux lines are plotted for the steps 1,2,5,8, on fig. (V.16, V.17, V.18, V.19).

### V.3. Numerical experiments with MOVING CRITERION DOMAIN.

We can see on fig. (V.20) the initial domain in which is the criterion evaluated.

As the boundary between iron and air (in the air-gap) can move, the set D is a moving domain.

#### First example.

The current density $j = 5 \ 10^6$ MKSA. The initial state is always given by fig (V.2). For the criterion we take

$$ (5.23) \qquad E_d = (0,1.3) $$

The initial value of the criterion is

$$ (5.24) \qquad J_o = .24089 \ 10^{-4} $$

and after 20 iterations the criterion value is

$$ (5.25) \qquad J_{20} = .65713 \ 10^{-5} $$

As in the previous example, the decrease is fast in the first step but slow after.
At iteration 6 we have

$$ (5.26) \qquad J_6 = .68109 \ 10^{-5} $$

In fig. (V.21, V.22, V.23, V.24, V.25, V.26) we give the successive state of the problem, precisely at iteration 1, 2, 5, 10, 15, 20.

The fig. (V.27) give the variation of the criterion along the iteration.

## Second example

We keep the same current density, but we take

$$ (5.27) \qquad E_d = (0,1.6) $$

for the criterion

The initial value of the criterion is

$$ (5.28) \qquad J_0 = .91754 \ 10^{-4} $$

and after 20 iterations the criterion value is

$$ (5.29) \qquad J_{20} = .31068 \ 10^{-4} $$

The same remark as in first example holds; at iteration 5 we have

$$ J_5 = .32056 \ 10^{-4} $$

The fig. (V.28, V.33) give the flux repartition, and the air-gap geometry at iteration 1, 2, 5, 10, 15, 20, and the variation of the criterion along the iteration is given in fig. (V.27).

In the two following example we take the value of current density

$$ (5.30) \qquad j = 20.10^6 \text{ MKSA} $$

and the initial state in term of flux distribution can be seen in fig. V.11. For the

## Third example

We take

$$ (5.31) \qquad E_d = (0,1.3) $$
The initial value of the criterion is

$$ (5.32) \qquad J_o = .27241 \ 10^{-3} $$

and we can reduce this value to

$$ J_{56} = .67852 \ 10^{-6} $$

but after 56 iterations. The total time is approximately 35 mn on CII IRIS 80.
For this example the result is obtained with 3 runs, in the following way.
First run, as the previous example.

We perform 16 iterations, and the final criterion value is

$$ (5.33) \qquad J_{16} = .11360 \ 10^{-4} $$

and if we look at the air-gap geometry, we observe an "oscillating boundary".
We substitute to this oscillating boundary a straight boundary and we obtain a
new initial geometry given in fig. (V.47).

The new criterion with this substitution is then

$$ (5.34) \qquad J_{16} = .97890 \ 10^{-5} $$

We obtain so manually a better value for the criterion. In the second run, we
perform 20 iterations with these new initial geometry and we obtain for the cri-
terion

$$ (5.35) \qquad J_{36} = .18465 \ 10^{-5} $$

We make a new "manual fitting" of the oscillating boundary as previously and we
obtain a new geometry given in fig. (V.48).

The new value of the criterion is then

$$ (5.36) \qquad J_{36} = .36073 \ 10^{-5} $$

and after 20 new iterations we obtain (3<sup>d</sup> run)

$$ (5.37) \qquad J_{56} = .67852 \ 10^{-6} $$

And we can continue the process ! For this process an interactive program with
display would be very useful.

We give the flux lines for the iterations 1, 5, 10, 16, 21, 26, 31, 36, 36 bis, 41, 46, 51, 56 on fig. (V.34-V.46).

The variation of the criterion along this iteration is given in fig (V.49).

In the Fourth example, we take the value

$$ (5.38) \qquad E_d = (0,1.6) $$

The initial criterion is

$$ (5.39) \qquad J_o = .99395 \ 10^{-4} $$

We can see on table V.3 the values of the flux density B in the region of interest. These are the values with the initial geometry.

After 20 iterations, the criterion is

$$ (5.40) \qquad J_{20} = .16558 \ 10^{-5} $$

We can see on table V.4, the values obtained after optimization.

We can see on fig. (V.50, V.54), the state at iteration 1, 2, 5, 10, 15.

In the fifth example, we considered a problem which may be related with the problem of "minimizing the losses in an electrical machine".

The density current is $j = 5.10^6$ MKSA, and in the domain showed in fig. (V.55), we want to have no flux line; this is equivalent to have a constant vector potential in this region or $\nabla A = 0$; then we take for the criterion parameter

$$ (5.41) \qquad E_d = (0,0) $$

The initial value of the criterion is

$$ J_o = .13839 \ 10^{-2} $$

and after 8 steps, the criterion value is

$$ J_8 = .97475 \ 10^{-3} $$

The geometric constraints (5.8) are saturated, for all the nodes, except the first. The algorithm tends to reduce the size of the air-gap. The initial magnetostatic state is given by fig. (V.2) and the final state in fig. (V.56).
## VI. VARIOUS REMARKS AND CONCLUSION.

1. Of course other iterative schemes can be used to solve the non linear state equation (see GLOWINSKI-MARROCCO [6], [15] for other methods) however the EGSN method used here seems well adapted since the state vector changes very little after a few iterations.

2. The computing times indicated (about 8 min on CII IRIS 80 is 40 seconds on IBM 370-168) are good except for the third example where we have tested if any substantial improvement could be made by letting the program run a very long time.

3. An interactive system with visualization would be very appropriate to such studies; it would then be possible to adapt the triangulations if the geometries change very much.

Thus, once the state equation is mastered, it takes only an additional "automatic" triangulation subroutine to be able to optimize the design. However it should be pointed out that the topology of the answer must be provided. If a "hole" is missing the method will not find it.

In conclusion let us also say that if the automatic triangulation subroutine is found to be too expensive to produce, one may compute the gradient vector only. Engineers find it to be a very valuable information for the improvement of their design.
REFERENCES

[1] O. PIRONNEAU. Variational method for the numerical solution of free boundary problems and optimum design problems. In "Control Theory of Systems governed by P.D.E. A. AZIS ed. academic press 1977.

[2] D. BEGIS, R. GLOWINSKI. Application de la méthode des éléments finis à l'ap-
proximation d'un problème de domaine optimal. Applied Math.
and Optimization. Vol. 2, n° 2, 1975.

[3] P. MORICE. Une méthode d'optimisation de forme de domaine. Proc. Congrès IFIP-IRIA 1974, Springer-Verlag.

[4] O. PIRONNEAU. Sur les problèmes d'optimisation de structure en mécanique des fluides. Thèse de Doctorat, Paris VI, 1976.

[5] R. GLOWINSKI, A. MARROCCO . Analyse numérique du champ magnétique d'un alter-
nateur par éléments finis et surrelaxation ponctuelle non li-
néaire, Computer methods in applied mechanics and engineering
Vol. 3, n° 1, Jan. 1974.

[6] R. GLOWINSKI, A. MARROCCO . Finite element approximation and iterative methods of solution for 2.D non linear magnetostatic problems . Proceedings of COMPUMAG Conference OXFORD 1976, pp. 112-125.

[7] E. DURAND. *Magnetostatique*, Masson et Cie, Paris 1968.

[8] J.L. LIONS . *Sur quelques questions d'analyse, de mécanique et de contrôle optimal*, Les presses de l'Université de Montréal, 1976.

[9] G. CHAVENT . Analyse fonctionnelle et identification de coefficients répartis dans les équations aux dérivées partielles, Thèse de Doctorat, Paris.

[10] A. BAMBERGER, G. CHAVENT, P. LAILLY . Etude mathématique et numérique d'un problème inverse pour l'équation des ondes à une dimension. Rapport interne n° 14, Ecole Polytechnique, Jan. 1977.
[11] L. TARTAR. Cours au Collège de France, Jan. 1977.

[12] O. PIRONNEAU, G. SAGUEZ. Assymptotic behaviour of solutions of P.D.E. with respect to the domain. IRIA Laboria report n° 218, 1977.

[13] D. CHENAIS. On the existence of a solution in a domain identification problem, J. of Math. Anal and Appli. 52,2, 1975.

[14] E. POLAK. Computational method in optimization, Academic Press, 1971.

[15] R. GLOWINSKI, A. MARROCCO. Numerical solution of two-dimensional magnetostatic problems by augmented Lagrangian methods, to appear in Computer Methods in applied mechanics and engineering.
<figure><img src="image_6.png" /><figcaption>FIG. II.1. IRON CHARACTERISTIC CURVE</figcaption></figure>

<figure><img src="image_7.png" /><figcaption>FIG. II.2. ELECTRO MAGNET CROSS SECTION</figcaption></figure>
<figure><img src="image_8.png" /><figcaption>FIG. II.3. COMPUTATIONAL DOMAIN</figcaption></figure>

<figure><img src="image_9.png" /><figcaption>LABORIA- A. MAROCCO

FIG. II.4. ANALYTICAL APPROXIMATION OF RELUCTIVITY</figcaption></figure>
<figure><img src="image_10.png" /></figure>

NOMBRE DE NOEUDS 158

NOMBRE D'ÉLÉMENTS 267

<figure><img src="image_11.png" /><figcaption>FIG. III.1.</figcaption></figure>

<figure><img src="image_12.png" /><figcaption>FIG. IV.1. TRANSFORMATION OF THE TRIANGULATION BY
MOVING THE NODE xₖ</figcaption></figure>
<figure><img src="image_13.png" /><figcaption>FIG. IV.2. TRIANGULATION OF THE REGION OF INTEREST FOR
OPTIMUM DESIGN PROBLEM</figcaption></figure>

□ NODES ARE FIXED, *, ○ NODES ARE THE MOVING ONES

<figure><img src="image_14.png" /><figcaption>FIG. IV.3. TRANSFORMATION OF THE TRIANGULATION BY
MOVING THE NODES</figcaption></figure>

<figure><img src="image_15.png" /><figcaption>FIG. V.1. INITIAL MAGNET GEOMETRY AND PART OF Ω IN WHICH IS THE CRITERION EVALUATED (D)</figcaption></figure>
<figure><img src="image_16.png" /></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0.0105E+0
DENSITE DE COURANT MKSA 0.500E+7

<figure><img src="image_17.png" /><figcaption>FIG. V.2. INITIAL MAGNETOSTATIC STATE FOR
$d = -5.10^{-6}$ MKSA</figcaption></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0.0105E+0
DENSITE DE COURANT MKSA 0.500E+7
ETAT DESIRE BIX= 0.
BY= 1.300

<figure><img src="image_18.png" /><figcaption>FIG. V.3. STEP 1 CRITERION . 40428 $10^{-5}$
(INITIAL VALUE . 61160 $10^{-5}$)</figcaption></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0.0105E+0
DENSITE DE COURANT MKSA 0.500E+7
ETAT DESIRE BIX= 0.
BY= 1.300

<caption>FIG. V.5 STEP 5 CRITERION . 35466 $10^{-5}$</caption>

MAGNETOSTATIC STATE AND AIR GAP

SHAPE FOR

$$J = S 10^6 \text{ MKSA}$$

$$E_d = (0,13)$$

<figure><img src="image_19.png" /></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0.0105E+0
DENSITE DE COURANT MKSA 0.500E+7
ETAT DESIRE BIX= 0.
BY= 1.300

<figure><img src="image_20.png" /><figcaption>FIG. V.4. STEP 2 CRITERION . 36334 $10^{-5}$</figcaption></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0.0105E+0
DENSITE DE COURANT MKSA 0.500E+7
ETAT DESIRE BIX= 0.
BY= 1.300

<caption>FIG. V.6 STEP 9 CRITERION . 35450 $10^{-5}$</caption>
<figure><img src="image_21.png" /><figcaption>FIG. V.7. VARIATION OF THE CRITERION VALUE (D FIXED)</figcaption></figure>

1. $J = 5 \cdot 10^6 \text{ AND } E_d = (0,1.3)$

2. $J = 5 \cdot 10^6 \text{ AND } E_d = (0,1.6)$

3. $J = 20 \cdot 10^6 \text{ AND } E_d = (0,1.3)$

4. $J = 20 \cdot 10^6 \text{ AND } E_d = (0,1.6)$

<figure><img src="image_22.png" /></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0.0 0.108E+0
DENSITE DE COURANT MKSA 0.500E+7
ETAT DESIRE BX= 0.
BY= 1.600

<caption>FIG. V.9. STEP 2 CRITERION. 18088 $10^{-4}$</caption>

MAGNETOSTATIC STATE AND AIR

GAP SHAPE FOR

$$J = 5 \cdot 10^6 \text{ MKSA}$$

$$E_d = (0,1.6)$$

<figure><img src="image_23.png" /></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0.0 0.108E+0
DENSITE DE COURANT MKSA 0.500E+7
ETAT DESIRE BX= 0.
BY= 1.600

<figure><img src="image_24.png" /><figcaption>FIG. V.8. STEP 1 CRITERION. 19963 $10^{-4}$
(INITIAL VALUE . 51906 $10^{-4}$)</figcaption></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0.0 0.108E+0
DENSITE DE COURANT MKSA 0.500E+7
ETAT DESIRE BX= 0.
BY= 1.600

<caption>FIG. V.10. STEP 5 CRITERION. 17862 $10^{-4}$</caption>
<figure><img src="image_25.png" /></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0.0144E+0
DENSITE DE COURANT MKSA 0.200E+8

<figure><img src="image_26.png" /><figcaption>FIG. V.11. INITIAL MAGNETOSTATIC STATE
FOR J = 20 10⁶ MKSA</figcaption></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0.0144E+0
DENSITE DE COURANT MKSA 0.200E+8
ETAT DESIRE BIX= 0.
BY= 1.300

<figure><img src="image_27.png" /><figcaption>FIG. V.12. STEP 1 CRITERION . 60574 10⁻⁶
(INITIAL VALUE . 10030 10⁻³)</figcaption></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0.0144E+0
DENSITE DE COURANT MKSA 0.200E+8
ETAT DESIRE BIX= 0.
BY= 1.300

<caption>FIG. V.14. STEP 5 CRITERION . 46671 10⁻⁶</caption>

MAGNETOSTATIC STATE AND AIR GAP SHAPE
FOR

$$J = 20 10^6 \text{ MKSA}$$

$$E_d = (0,1.3)$$

<figure><img src="image_28.png" /></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0.0144E+0
DENSITE DE COURANT MKSA 0.200E+8
ETAT DESIRE BIX= 0.
BY= 1.300

<figure><img src="image_29.png" /><figcaption>FIG. V.13. STEP 2 CRITERION . 53566 10⁻⁶</figcaption></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0.0144E+0
DENSITE DE COURANT MKSA 0.200E+8
ETAT DESIRE BIX= 0.
BY= 1.300

<caption>FIG. V.15. STEP 9 CRITERION . 46190 10⁻⁶</caption>
MAGNETOSTATIC STATE AND AIR GAP SHAPE FOR

$$J = 20 \times 10^{-6} \text{ MASA}$$

$$E_d = (0,1, 6)$$

<figure><img src="image_30.png" /><figcaption>FIG. V.16. STEP 1 CRITERION. $22023 \times 10^{-5}$
(INITIAL VALUE $25427 \times 10^{-4}$)</figcaption></figure>

LABORIA- A.MAROCCO

<figure><img src="image_31.png" /><figcaption>FIG. V.17. STEP 2 CRITERION. $19971 \times 10^{-5}$</figcaption></figure>

LABORIA- A.MAROCCO

<figure><img src="image_32.png" /><figcaption>FIG. V.18. STEP 5 CRITERION. $18921 \times 10^{-5}$</figcaption></figure>

LABORIA- A.MAROCCO

<figure><img src="image_33.png" /><figcaption>FIG. V.19. STEP 8 CRITERION $18863 \times 10^{-5}$</figcaption></figure>

LABORIA- A.MAROCCO
<figure><img src="image_34.png" /></figure>

ZONE D EVALUATION DU CRITERE

FIG. V.20. INITIAL MAGNET GEOMETRY AND PART OF Ω IN WHICH IS THE CRITERION EVALUATED (VARIABLE D)

MAGNETOSTATIC STATE AND AIR GAP SHAPE FOR

J = 5 10⁶ MKSA AND E_d = (0,1.3)

<figure><img src="image_35.png" /></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0.104E+0
DENSITE DE COURANT MKSA 0.500E+7
ETAT DESIRE BIX= 0.
BY= 1.300

LABORIA- A.MARROCCO

FIG. V.21. STEP 1 CRITERION . 16334 10⁻⁴
(INITIAL VALUE . 24089 10⁻⁴)

<figure><img src="image_36.png" /></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0.104E+0
DENSITE DE COURANT MKSA 0.500E+7
ETAT DESIRE BIX= 0.
BY= 1.300

LABORIA- A.MARROCCO

FIG. V.22. STEP 2 CRITERION . 12204 10⁻⁴
<figure><img src="image_37.png" /><figcaption>FIG. V.23. STEP 5 CRITERION . 75190 $10^{-5}$</figcaption></figure>

<figure><img src="image_38.png" /><figcaption>FIG. V.24. STEP 10 CRITERION . 66455 $10^{-1}$</figcaption></figure>

<figure><img src="image_39.png" /><figcaption>FIG. V.25. STEP 15 CRITERION . 66217 $10^{-5}$</figcaption></figure>

<figure><img src="image_40.png" /><figcaption>FIG. V.26. STEP 20 CRITERION . 65713 $10^{-5}$</figcaption></figure>
<figure><img src="image_41.png" /><figcaption>FIG. V.27. VARIATION OF THE CRITERION VALUE</figcaption></figure>

$$1. J = 5 \cdot 10^6 \quad \text{AND} \quad E_d = (0,1.3)$$

$$2. J = 5 \cdot 10^6 \quad \text{AND} \quad E_d = (0,1.6)$$

MAGNETOSTATIC STATE AND AIR GAP SHAPE FOR

$$J = 5 \cdot 10^6 \quad \text{AND} \quad E_d = (0,1.6)$$

<figure><img src="image_42.png" /></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0.010E+0
DENSITE DE COURANT MKSA 0.500E+7
ETAT DESIRE BIX= 0.
BY= 1.600

LABORIA- A.MARROCCO

FIG. V.28. STEP 1 CRITERION . 49405 $10^{-4}$
(INITIAL VALUE 091754 $10^{-4}$)

<figure><img src="image_43.png" /></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0.010E+0
DENSITE DE COURANT MKSA 0.500E+7
ETAT DESIRE BIX= 0.
BY= 1.600

LABORIA- A.MARROCCO

FIG. V.29. STEP 2 CRITERION . 40805 $10^{-4}$
<figure><img src="image_44.png" /></figure>

LABORIA- A.MARROCCO

<figure><img src="image_45.png" /><figcaption>FIG. V.30. STEP 5 CRITERION . 32056 $10^{-4}$</figcaption></figure>

LABORIA- A.MARROCCO

<figure><img src="image_46.png" /><figcaption>FIG. V.31. STEP 10 CRITERION . 31550 $10^{-4}$</figcaption></figure>

LABORIA- A.MARROCCO

<figure><img src="image_47.png" /><figcaption>FIG. V.32. STEP 15 CRITERION . 31292 $10^{-4}$</figcaption></figure>

LABORIA- A.MARROCCO

<caption>FIG. V.33. STEP 20 CRITERION . 31068 $10^{-4}$</caption>
<figure><img src="image_48.png" /></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0. 0.145E+0
DENSITE DE COURANT MKSA 0.200E+8
ETAT DESIRE BIX= 0.
BY= 1.300

<figure><img src="image_49.png" /><figcaption>FIG. V.34. STEP 1 CRITERION . 12162 10⁻³
(INITIAL VALUE . 27241 10⁻³)</figcaption></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0. 0.145E+0
DENSITE DE COURANT MKSA 0.200E+8
ETAT DESIRE BIX= 0.
BY= 1.300

<figure><img src="image_50.png" /><figcaption>FIG. V.35. STEP 5 CRITERION . 48866 10⁻⁴</figcaption></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0. 0.145E+0
DENSITE DE COURANT MKSA 0.200E+8
ETAT DESIRE BIX= 0.
BY= 1.300

<caption>FIG. V. 37 STEP 16 CRITERION . 97890 10⁻⁵
(BEFORE "FITTING" : . 11360 10⁻⁴)</caption>

MAGNETOSTATIC STATE AND AIR GAP SHAPE FOR
J = 20 10⁶ MKSA
E_d = (0,1.3)

<figure><img src="image_51.png" /></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0. 0.145E+0
DENSITE DE COURANT MKSA 0.200E+8
ETAT DESIRE BIX= 0.
BY= 1.300

<figure><img src="image_52.png" /><figcaption>FIG. V.36. STEP 10 CRITERION .22615 10⁻⁴</figcaption></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0. 0.145E+0
DENSITE DE COURANT MKSA 0.200E+8
ETAT DESIRE BIX= 0.
BY= 1.300

<caption>FIG. V.38. STEP 21 CRITERION .64430 10⁻¹</caption>
<figure><img src="image_53.png" /></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0. 0.145E+0
DENSITE DE COURANT MKSA 0.200E+8
ETAT DESIRE BIX= 0.
BY= 1.300

<caption>FIG. V.39. STEP 26 CRITERION . 41371 10⁻⁵</caption>

MAGNETOSTATIC STATE AND AIR GAP SHAPE FOR
J = 20 10⁻⁶ MKSA
E₉ = (0,1.3)

<figure><img src="image_54.png" /></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0. 0.145E+0
DENSITE DE COURANT MKSA 0.200E+8
ETAT DESIRE BIX= 0.
BY= 1.300

<figure><img src="image_55.png" /><figcaption>FIG. V.40 STEP 31 CRITERION . 27266 10⁻⁵</figcaption></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0. 0.145E+0
DENSITE DE COURANT MKSA 0.200E+8
ETAT DESIRE BIX= 0.
BY= 1.300

<figure><img src="image_56.png" /><figcaption>FIG. V.41. STEP 36 CRITERION . 18465 10⁻⁵</figcaption></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0. 0.145E+0
DENSITE DE COURANT MKSA 0.200E+8
ETAT DESIRE BIX= 0.
BY= 1.300

<figure><img src="image_57.png" /><figcaption>FIG. V.42. STEP 36 CRITERION . 36073 10⁻⁵</figcaption></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0. 0.145E+0
DENSITE DE COURANT MKSA 0.200E+8
ETAT DESIRE BIX= 0.
BY= 1.300

<caption>FIG. V.43. STEP 41 CRITERION . 13949 10⁻⁵</caption>
<figure><img src="image_58.png" /></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0. 0.145E+0
DENSITE DE COURANT MKSA 0.200E+8
ETAT DESIRE BI= 0.
BY= 1.300

<figure><img src="image_59.png" /><figcaption>FIG. V.44. STEP 46 CRITERION . 99631 10⁻⁶</figcaption></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0. 0.145E+0
DENSITE DE COURANT MKSA 0.200E+8
ETAT DESIRE BI= 0.
BY= 1.300

<figure><img src="image_60.png" /><figcaption>FIG. V.45. STEP 51 CRITERION . 79043 10⁻⁶</figcaption></figure>

NOMBRE DE NOEUDS 158
NOMBRE D ELEMENTS 267

<caption>FIG. V.47. NEW TRIANGULATION AT STEP 16</caption>

MAGNETOSTATIC STATE AND AIR GAP SHAPE FOR

$$J = 20 \ 10^6 \text{ MKSA}$$

$$E_d = (0,1.3)$$

<figure><img src="image_61.png" /></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0. 0.145E+0
DENSITE DE COURANT MKSA 0.200E+8
ETAT DESIRE BI= 0.
BY= 1.300

<figure><img src="image_62.png" /><figcaption>FIG. V.46. STEP 56. CRITERION . 67852 10⁻⁶</figcaption></figure>

NOMBRE DE NOEUDS 158
NOMBRE D ELEMENTS 267

<caption>FIG.V. 48 NEW TRIANGULATION AT STEP 36</caption>
<figure><img src="image_63.png" /><figcaption>FIG. V.49.</figcaption></figure>

VARIATION OF THE CRITERION

$$1. J = 20 \ 10^6, E_d = (0,1.3)$$

$$2. J = 20 \ 10^6, E_d = (0,1.6)$$

MAGNETOSTATIC STATE AND AIR GAP SHAPE FOR

$$J = 20 \ 10^6 \text{ MKSA}, E_d = (0,1.6)$$

<figure><img src="image_64.png" /></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0. 0.146E+0
DENSITE DE COURANT MKSA 0.200E+8
ETAT DESIRE BI= 0.
BY= 1.600

FIG. V.50. STEP 1 CRITERION . 16112 10⁻⁴
(INITIAL VALUE . 99395 10⁻⁴)

<figure><img src="image_65.png" /></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0. 0.146E+0
DENSITE DE COURANT MKSA 0.200E+8
ETAT DESIRE BI= 0.
BY= 1.600

FIG. V.51. STEP 2 CRITERION . 69781 10⁻⁵
<figure><img src="image_66.png" /></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0. 0.146E+0
DENSITE DE COURANT MKSA 0.200E+8
ETAT DESIRE BX= 0.
BY= 1.600

<figure><img src="image_67.png" /><figcaption>FIG.V.52.STEP 5 CRITERION . 22778 10⁻⁵</figcaption></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0. 0.146E+0
DENSITE DE COURANT MKSA 0.200E+8
ETAT DESIRE BX= 0.
BY= 1.600

<figure><img src="image_68.png" /><figcaption>FIG.V.53. STEP 10 CRITERION . 17220 10⁻⁵</figcaption></figure>

ZONE D EVALUATION DU CRITERE

<figure><img src="image_69.png" /><figcaption>FIG. V.55. SUBSET D FOR THE LAST EXAMPLE</figcaption></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0. 0.146E+0
DENSITE DE COURANT MKSA 0.200E+8
ETAT DESIRE BX= 0.
BY= 1.600

<figure><img src="image_70.png" /><figcaption>FIG. V.54. STEP 15 CRITERION . 16688 10⁻⁵</figcaption></figure>

PAS ENTRE 2 LIGNES 0.600E-2
VALEURS EXTREMES DE LA FONCTION
0. 0.110E+0
DENSITE DE COURANT MKSA 0.500E+7
ETAT DESIRE BX= 0.
BY= 0.

<caption>FIG. V.56. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 8 CRITERION . 97475⁻⁵
(INITIAL VALUE . 13839 10⁻²²)</caption>
<table><caption>TABLE V.1 - Initial value of flux density $\vec{B}$ in the subset $D$ (fixed)</caption><thead><tr><th>TRIANGLE</th><th>I</th><th>MODULE</th><th>I</th><th>COMP.X</th><th>I</th><th>COMP.Y</th><th>I</th></tr></thead><tbody><tr><td>240</td><td>I</td><td>1.60867</td><td>I</td><td>.20220</td><td>I</td><td>1.59592</td><td>I</td></tr><tr><td>241</td><td>I</td><td>1.64087</td><td>I</td><td>.16562</td><td>I</td><td>1.63249</td><td>I</td></tr><tr><td>242</td><td>I</td><td>1.60118</td><td>I</td><td>.12910</td><td>I</td><td>1.59597</td><td>I</td></tr><tr><td>243</td><td>I</td><td>1.56817</td><td>I</td><td>.16568</td><td>I</td><td>1.55959</td><td>I</td></tr><tr><td>244</td><td>I</td><td>1.80034</td><td>I</td><td>.12910</td><td>I</td><td>1.80172</td><td>I</td></tr><tr><td>245</td><td>I</td><td>1.86110</td><td>I</td><td>.10560</td><td>I</td><td>1.85810</td><td>I</td></tr><tr><td>246</td><td>I</td><td>1.80563</td><td>I</td><td>.08216</td><td>I</td><td>1.80131</td><td>I</td></tr><tr><td>247</td><td>I</td><td>1.74862</td><td>I</td><td>.10566</td><td>I</td><td>1.74543</td><td>I</td></tr><tr><td>248</td><td>I</td><td>1.86461</td><td>I</td><td>.08216</td><td>I</td><td>1.86280</td><td>I</td></tr><tr><td>249</td><td>I</td><td>1.90283</td><td>I</td><td>.06595</td><td>I</td><td>1.90169</td><td>I</td></tr><tr><td>250</td><td>I</td><td>1.86530</td><td>I</td><td>.04969</td><td>I</td><td>1.86264</td><td>I</td></tr><tr><td>251</td><td>I</td><td>1.82495</td><td>I</td><td>.06590</td><td>I</td><td>1.82376</td><td>I</td></tr><tr><td>252</td><td>I</td><td>1.86919</td><td>I</td><td>.04969</td><td>I</td><td>1.89855</td><td>I</td></tr><tr><td>253</td><td>I</td><td>1.92586</td><td>I</td><td>.03951</td><td>I</td><td>1.92346</td><td>I</td></tr><tr><td>254</td><td>I</td><td>1.89842</td><td>I</td><td>.02879</td><td>I</td><td>1.89820</td><td>I</td></tr><tr><td>255</td><td>I</td><td>1.87570</td><td>I</td><td>.03917</td><td>I</td><td>1.87330</td><td>I</td></tr><tr><td>256</td><td>I</td><td>1.91882</td><td>I</td><td>.02879</td><td>I</td><td>1.91861</td><td>I</td></tr><tr><td>257</td><td>I</td><td>1.93446</td><td>I</td><td>.02224</td><td>I</td><td>1.93433</td><td>I</td></tr><tr><td>258</td><td>I</td><td>1.91842</td><td>I</td><td>.01553</td><td>I</td><td>1.91836</td><td>I</td></tr><tr><td>259</td><td>I</td><td>1.90276</td><td>I</td><td>.02213</td><td>I</td><td>1.90263</td><td>I</td></tr><tr><td>260</td><td>I</td><td>1.92936</td><td>I</td><td>.01553</td><td>I</td><td>1.92935</td><td>I</td></tr><tr><td>261</td><td>I</td><td>1.93979</td><td>I</td><td>.01123</td><td>I</td><td>1.93974</td><td>I</td></tr><tr><td>262</td><td>I</td><td>1.92918</td><td>I</td><td>.00682</td><td>I</td><td>1.92916</td><td>I</td></tr><tr><td>263</td><td>I</td><td>1.91875</td><td>I</td><td>.01117</td><td>I</td><td>1.91871</td><td>I</td></tr><tr><td>264</td><td>I</td><td>1.93597</td><td>I</td><td>.00682</td><td>I</td><td>1.93396</td><td>I</td></tr><tr><td>265</td><td>I</td><td>1.94213</td><td>I</td><td>.00342</td><td>I</td><td>1.94213</td><td>I</td></tr><tr><td>266</td><td>I</td><td>1.93394</td><td>I</td><td>.00000</td><td>I</td><td>1.93394</td><td>I</td></tr><tr><td>267</td><td>I</td><td>1.92577</td><td>I</td><td>.00340</td><td>I</td><td>1.92577</td><td>I</td></tr></tbody></table>

<table><thead><tr><th>TRIANGLE</th><th>I</th><th>MODULE</th><th>I</th><th>COMP.X</th><th>I</th><th>COMP.Y</th><th>I</th></tr></thead><tbody><tr><td>240</td><td>I</td><td>1.29559</td><td>I</td><td>.09501</td><td>I</td><td>1.29210</td><td>I</td></tr><tr><td>241</td><td>I</td><td>1.33829</td><td>I</td><td>.04975</td><td>I</td><td>1.33756</td><td>I</td></tr><tr><td>242</td><td>I</td><td>1.29213</td><td>I</td><td>.00452</td><td>I</td><td>1.29213</td><td>I</td></tr><tr><td>243</td><td>I</td><td>1.24786</td><td>I</td><td>.04978</td><td>I</td><td>1.24687</td><td>I</td></tr><tr><td>244</td><td>I</td><td>1.29889</td><td>I</td><td>.00452</td><td>I</td><td>1.29889</td><td>I</td></tr><tr><td>245</td><td>I</td><td>1.30013</td><td>I</td><td>.00400</td><td>I</td><td>1.30012</td><td>I</td></tr><tr><td>246</td><td>I</td><td>1.29891</td><td>I</td><td>.00350</td><td>I</td><td>1.29891</td><td>I</td></tr><tr><td>247</td><td>I</td><td>1.29769</td><td>I</td><td>.00402</td><td>I</td><td>1.29768</td><td>I</td></tr><tr><td>248</td><td>I</td><td>1.30522</td><td>I</td><td>.00350</td><td>I</td><td>1.30522</td><td>I</td></tr><tr><td>249</td><td>I</td><td>1.30800</td><td>I</td><td>.00151</td><td>I</td><td>1.30800</td><td>I</td></tr><tr><td>250</td><td>I</td><td>1.30516</td><td>I</td><td>.00051</td><td>I</td><td>1.30516</td><td>I</td></tr><tr><td>251</td><td>I</td><td>1.29833</td><td>I</td><td>.00149</td><td>I</td><td>1.29833</td><td>I</td></tr><tr><td>252</td><td>I</td><td>1.30159</td><td>I</td><td>.00051</td><td>I</td><td>1.30159</td><td>I</td></tr><tr><td>253</td><td>I</td><td>1.30537</td><td>I</td><td>.00125</td><td>I</td><td>1.30537</td><td>I</td></tr><tr><td>254</td><td>I</td><td>1.30147</td><td>I</td><td>.00204</td><td>I</td><td>1.30147</td><td>I</td></tr><tr><td>255</td><td>I</td><td>1.29969</td><td>I</td><td>.00156</td><td>I</td><td>1.29969</td><td>I</td></tr><tr><td>256</td><td>I</td><td>1.29861</td><td>I</td><td>.00204</td><td>I</td><td>1.29861</td><td>I</td></tr><tr><td>257</td><td>I</td><td>1.29745</td><td>I</td><td>.00156</td><td>I</td><td>1.29745</td><td>I</td></tr><tr><td>258</td><td>I</td><td>1.29854</td><td>I</td><td>.00111</td><td>I</td><td>1.29854</td><td>I</td></tr><tr><td>259</td><td>I</td><td>1.29769</td><td>I</td><td>.00111</td><td>I</td><td>1.29769</td><td>I</td></tr><tr><td>260</td><td>I</td><td>1.29767</td><td>I</td><td>.00066</td><td>I</td><td>1.29667</td><td>I</td></tr><tr><td>261</td><td>I</td><td>1.29662</td><td>I</td><td>.00066</td><td>I</td><td>1.29665</td><td>I</td></tr><tr><td>262</td><td>I</td><td>1.29765</td><td>I</td><td>.00023</td><td>I</td><td>1.29765</td><td>I</td></tr><tr><td>263</td><td>I</td><td>1.29872</td><td>I</td><td>.00068</td><td>I</td><td>1.29872</td><td>I</td></tr><tr><td>264</td><td>I</td><td>1.29779</td><td>I</td><td>.00023</td><td>I</td><td>1.29780</td><td>I</td></tr><tr><td>265</td><td>I</td><td>1.29751</td><td>I</td><td>.00011</td><td>I</td><td>1.29751</td><td>I</td></tr><tr><td>266</td><td>I</td><td>1.29773</td><td>I</td><td>.00000</td><td>I</td><td>1.29770</td><td>I</td></tr><tr><td>267</td><td>I</td><td>1.29807</td><td>I</td><td>.00012</td><td>I</td><td>1.29807</td><td>I</td></tr></tbody></table>

$$ \text{FONC} = .4613850759E-06 $$

<caption>TABLE V.2 - Value of flux density $\vec{B}$ after optimization for $E_d$ (0,1.3)</caption>
<table><thead><tr><th>TRIANGLE</th><th>I</th><th>MODULE</th><th>I</th><th>COMP.X</th><th>I</th><th>COMP.Y</th><th>I</th></tr></thead><tbody><tr><td>214</td><td>I</td><td>1.98593</td><td>I</td><td>.76513</td><td></td><td>1.83267</td><td>I</td></tr><tr><td>215</td><td>I</td><td>1.69740</td><td>I</td><td>.46491</td><td></td><td>1.63249</td><td>I</td></tr><tr><td>216</td><td>I</td><td>1.43631</td><td>I</td><td>.48798</td><td></td><td>1.38694</td><td>I</td></tr><tr><td>217</td><td>I</td><td>1.01781</td><td>I</td><td>.29273</td><td></td><td>1.99646</td><td>I</td></tr><tr><td>218</td><td>I</td><td>1.17542</td><td>I</td><td>.22294</td><td></td><td>1.16397</td><td>I</td></tr><tr><td>219</td><td>I</td><td>1.90910</td><td>I</td><td>.12450</td><td></td><td>1.99522</td><td>I</td></tr><tr><td>220</td><td>I</td><td>1.05305</td><td>I</td><td>.10613</td><td></td><td>1.05031</td><td>I</td></tr><tr><td>221</td><td>I</td><td>1.77904</td><td>I</td><td>.05804</td><td></td><td>1.97810</td><td>I</td></tr><tr><td>222</td><td>I</td><td>1.90958</td><td>I</td><td>.05287</td><td></td><td>1.99888</td><td>I</td></tr><tr><td>223</td><td>I</td><td>1.96607</td><td>I</td><td>.02812</td><td></td><td>1.96587</td><td>I</td></tr><tr><td>224</td><td>I</td><td>1.77708</td><td>I</td><td>.02628</td><td></td><td>1.97691</td><td>I</td></tr><tr><td>225</td><td>I</td><td>1.95935</td><td>I</td><td>.01161</td><td></td><td>1.95932</td><td>I</td></tr><tr><td>226</td><td>I</td><td>1.96848</td><td>I</td><td>.01085</td><td></td><td>1.96845</td><td>I</td></tr><tr><td>227</td><td>I</td><td>1.95661</td><td>I</td><td>.00000</td><td></td><td>1.95661</td><td>I</td></tr><tr><td>228</td><td>I</td><td>1.95065</td><td>I</td><td>.01207</td><td></td><td>1.95661</td><td>I</td></tr><tr><td>229</td><td>I</td><td>1.94213</td><td>I</td><td>.00000</td><td></td><td>1.94213</td><td>I</td></tr><tr><td>230</td><td>I</td><td>1.93979</td><td>I</td><td>.11348</td><td></td><td>1.93974</td><td>I</td></tr><tr><td>231</td><td>I</td><td>1.95952</td><td>I</td><td>.02815</td><td></td><td>1.95932</td><td>I</td></tr><tr><td>232</td><td>I</td><td>1.93461</td><td>I</td><td>.03232</td><td></td><td>1.93434</td><td>I</td></tr><tr><td>233</td><td>I</td><td>1.96653</td><td>I</td><td>.05333</td><td></td><td>1.96586</td><td>I</td></tr><tr><td>234</td><td>I</td><td>1.92453</td><td>I</td><td>.06394</td><td></td><td>1.92347</td><td>I</td></tr><tr><td>235</td><td>I</td><td>1.98052</td><td>I</td><td>.09587</td><td></td><td>1.97820</td><td>I</td></tr><tr><td>236</td><td>I</td><td>1.90554</td><td>I</td><td>.12137</td><td></td><td>1.90168</td><td>I</td></tr><tr><td>237</td><td>I</td><td>1.00222</td><td>I</td><td>.16813</td><td></td><td>1.99522</td><td>I</td></tr><tr><td>238</td><td>I</td><td>1.87167</td><td>I</td><td>.22527</td><td></td><td>1.85807</td><td>I</td></tr><tr><td>239</td><td>I</td><td>1.01639</td><td>I</td><td>.28293</td><td></td><td>1.99645</td><td>I</td></tr><tr><td>240</td><td>I</td><td>1.60867</td><td>I</td><td>.20220</td><td></td><td>1.59597</td><td>I</td></tr><tr><td>241</td><td>I</td><td>1.64087</td><td>I</td><td>.16562</td><td></td><td>1.63249</td><td>I</td></tr><tr><td>242</td><td>I</td><td>1.60113</td><td>I</td><td>.12910</td><td></td><td>1.59597</td><td>I</td></tr><tr><td>243</td><td>I</td><td>1.56817</td><td>I</td><td>.16568</td><td></td><td>1.55930</td><td>I</td></tr><tr><td>244</td><td>I</td><td>1.80634</td><td>I</td><td>.12910</td><td></td><td>1.80177</td><td>I</td></tr><tr><td>245</td><td>I</td><td>1.86710</td><td>I</td><td>.10560</td><td></td><td>1.85810</td><td>I</td></tr><tr><td>246</td><td>I</td><td>1.80368</td><td>I</td><td>.08216</td><td></td><td>1.80184</td><td>I</td></tr><tr><td>247</td><td>I</td><td>1.74862</td><td>I</td><td>.10566</td><td></td><td>1.74547</td><td>I</td></tr><tr><td>248</td><td>I</td><td>1.86461</td><td>I</td><td>.08216</td><td></td><td>1.86280</td><td>I</td></tr><tr><td>249</td><td>I</td><td>1.90283</td><td>I</td><td>.06595</td><td></td><td>1.90169</td><td>I</td></tr><tr><td>250</td><td>I</td><td>1.86530</td><td>I</td><td>.04969</td><td></td><td>1.86264</td><td>I</td></tr><tr><td>251</td><td>I</td><td>1.87495</td><td>I</td><td>.06590</td><td></td><td>1.82376</td><td>I</td></tr><tr><td>252</td><td>I</td><td>1.86919</td><td>I</td><td>.04969</td><td></td><td>1.89855</td><td>I</td></tr><tr><td>253</td><td>I</td><td>1.92586</td><td>I</td><td>.03931</td><td></td><td>1.92346</td><td>I</td></tr><tr><td>254</td><td>I</td><td>1.89842</td><td>I</td><td>.02879</td><td></td><td>1.89820</td><td>I</td></tr><tr><td>255</td><td>I</td><td>1.87570</td><td>I</td><td>.03917</td><td></td><td>1.87350</td><td>I</td></tr><tr><td>256</td><td>I</td><td>1.91882</td><td>I</td><td>.02879</td><td></td><td>1.91861</td><td>I</td></tr><tr><td>257</td><td>I</td><td>1.93446</td><td>I</td><td>.02224</td><td></td><td>1.93433</td><td>I</td></tr><tr><td>258</td><td>I</td><td>1.91842</td><td>I</td><td>.01558</td><td></td><td>1.91836</td><td>I</td></tr><tr><td>259</td><td>I</td><td>1.90276</td><td>I</td><td>.02213</td><td></td><td>1.90263</td><td>I</td></tr><tr><td>260</td><td>I</td><td>1.92936</td><td>I</td><td>.01558</td><td></td><td>1.92930</td><td>I</td></tr><tr><td>261</td><td>I</td><td>1.92978</td><td>I</td><td>.01123</td><td></td><td>1.92974</td><td>I</td></tr><tr><td>262</td><td>I</td><td>1.92913</td><td>I</td><td>.00682</td><td></td><td>1.92916</td><td>I</td></tr><tr><td>263</td><td>I</td><td>1.91875</td><td>I</td><td>.01117</td><td></td><td>1.91871</td><td>I</td></tr></tbody></table>

<table><caption>TABLE V.3 - Initial value of flux density $\vec{B}$ in the subset $\mathcal{D}$ (variable)</caption><thead><tr><th>TRIANGLE</th><th>I</th><th>MODULE</th><th>I</th><th>COMP.X</th><th>I</th><th>COMP.Y</th><th>I</th></tr></thead><tbody><tr><td>264</td><td>I</td><td>1.93597</td><td>I</td><td>.00682</td><td></td><td>1.93596</td><td>I</td></tr><tr><td>265</td><td>I</td><td>1.94213</td><td>I</td><td>.00342</td><td></td><td>1.94213</td><td>I</td></tr><tr><td>266</td><td>I</td><td>1.93594</td><td>I</td><td>.00000</td><td></td><td>1.93594</td><td>I</td></tr><tr><td>267</td><td>I</td><td>1.92577</td><td>I</td><td>.00340</td><td></td><td>1.92577</td><td>I</td></tr></tbody></table>
<table><thead><tr><th>TRIANGLE</th><th>MODULE</th><th>COMP.X</th><th>COMP.Y</th></tr></thead><tbody><tr><td>214</td><td>1.62270</td><td>0.1918</td><td>1.62259</td></tr><tr><td>215</td><td>1.62455</td><td>0.24436</td><td>1.62395</td></tr><tr><td>216</td><td>1.67175</td><td>0.0551</td><td>1.67174</td></tr><tr><td>217</td><td>1.60611</td><td>0.0504</td><td>1.60610</td></tr><tr><td>218</td><td>1.60174</td><td>0.00324</td><td>1.60174</td></tr><tr><td>219</td><td>1.60147</td><td>0.00325</td><td>1.60147</td></tr><tr><td>220</td><td>1.59330</td><td>0.00053</td><td>1.59330</td></tr><tr><td>221</td><td>1.60161</td><td>0.00124</td><td>1.60161</td></tr><tr><td>222</td><td>1.60503</td><td>0.00039</td><td>1.60503</td></tr><tr><td>223</td><td>1.60101</td><td>0.00133</td><td>1.60101</td></tr><tr><td>224</td><td>1.59750</td><td>0.00076</td><td>1.59750</td></tr><tr><td>225</td><td>1.59971</td><td>0.00004</td><td>1.59971</td></tr><tr><td>226</td><td>1.59967</td><td>0.00004</td><td>1.59967</td></tr><tr><td>227</td><td>1.59960</td><td>0.00000</td><td>1.59960</td></tr><tr><td>228</td><td>1.59956</td><td>0.00028</td><td>1.59956</td></tr><tr><td>229</td><td>1.59895</td><td>0.00000</td><td>1.59895</td></tr><tr><td>230</td><td>1.59903</td><td>0.00032</td><td>1.59903</td></tr><tr><td>231</td><td>1.59979</td><td>0.00058</td><td>1.59979</td></tr><tr><td>232</td><td>1.59924</td><td>0.00067</td><td>1.59924</td></tr><tr><td>233</td><td>1.60067</td><td>0.00109</td><td>1.60067</td></tr><tr><td>234</td><td>1.59999</td><td>0.00126</td><td>1.59999</td></tr><tr><td>235</td><td>1.60166</td><td>0.00155</td><td>1.60166</td></tr><tr><td>236</td><td>1.60211</td><td>0.00140</td><td>1.60211</td></tr><tr><td>237</td><td>1.60714</td><td>0.00205</td><td>1.60714</td></tr><tr><td>238</td><td>1.60882</td><td>0.00135</td><td>1.60882</td></tr><tr><td>239</td><td>1.66794</td><td>0.02238</td><td>1.66779</td></tr><tr><td>240</td><td>1.56164</td><td>0.16491</td><td>1.56291</td></tr><tr><td>241</td><td>1.62661</td><td>0.9391</td><td>1.62390</td></tr><tr><td>242</td><td>1.55308</td><td>0.2292</td><td>1.55291</td></tr><tr><td>243</td><td>1.43489</td><td>0.9391</td><td>1.48194</td></tr><tr><td>244</td><td>1.59249</td><td>0.2292</td><td>1.59237</td></tr><tr><td>245</td><td>1.60892</td><td>0.1603</td><td>1.60884</td></tr><tr><td>246</td><td>1.59229</td><td>0.0013</td><td>1.59226</td></tr><tr><td>247</td><td>1.57582</td><td>0.1602</td><td>1.57574</td></tr><tr><td>248</td><td>1.59576</td><td>0.0013</td><td>1.59573</td></tr><tr><td>249</td><td>1.60212</td><td>0.00647</td><td>1.60211</td></tr><tr><td>250</td><td>1.59575</td><td>0.00382</td><td>1.59575</td></tr><tr><td>251</td><td>1.58939</td><td>0.00648</td><td>1.58937</td></tr><tr><td>252</td><td>1.59735</td><td>0.00382</td><td>1.59735</td></tr><tr><td>253</td><td>1.59999</td><td>0.00272</td><td>1.59999</td></tr><tr><td>254</td><td>1.59739</td><td>0.00165</td><td>1.59739</td></tr><tr><td>255</td><td>1.59476</td><td>0.00274</td><td>1.59476</td></tr><tr><td>256</td><td>1.59811</td><td>0.00165</td><td>1.59811</td></tr><tr><td>257</td><td>1.59924</td><td>0.00117</td><td>1.59924</td></tr><tr><td>258</td><td>1.59814</td><td>0.00072</td><td>1.59814</td></tr><tr><td>259</td><td>1.59701</td><td>0.00119</td><td>1.59701</td></tr><tr><td>260</td><td>1.59843</td><td>0.00072</td><td>1.59843</td></tr><tr><td>261</td><td>1.59903</td><td>0.00049</td><td>1.59903</td></tr><tr><td>262</td><td>1.59850</td><td>0.00027</td><td>1.59850</td></tr><tr><td>263</td><td>1.59795</td><td>0.00050</td><td>1.59795</td></tr></tbody></table>

<table><thead><tr><th>TRIANGLE</th><th>MODULE</th><th>COMP.X</th><th>COMP.Y</th></tr></thead><tbody><tr><td>264</td><td>1.59862</td><td>0.00027</td><td>1.59862</td></tr><tr><td>265</td><td>1.59895</td><td>0.00013</td><td>1.59895</td></tr><tr><td>266</td><td>1.59863</td><td>0.00000</td><td>1.59863</td></tr><tr><td>267</td><td>1.59839</td><td>0.00014</td><td>1.59839</td></tr></tbody></table>

$$ \text{RMC} = .16537118307E-05 $$

<caption>TABLE V.4 - Value of flux density $\vec{B}$ after optimization for $E_c$ (0,1.6)</caption>
- Fig. II.1. IRON CHARACTERISTIC CURVE

II.2. ELECTRO-MAGNET CROSS SECTION

II.3. COMPUTATIONAL DOMAIN

II.4. ANALYTICAL APPROXIMATION OF THE RELATIVE RELUCTIVITY

II.5. ONE INITIAL TRIANGULATION OF Ω

- Fig. III.1. -

- Fig. IV.1. TRANSFORMATION OF THE TRIANGULATION BY MOVING THE NODE X<sub>K</sub>

IV.2. TRIANGULATION OF THE REGION OF INTEREST FOR OPTIMUM DESIGN PROBLEM
□ NODES ARE FIXED

*, ○ NODES ARE THE MOVING ONES (PRINCIPAL AND ASSOCIATED NODES)

IV.3. TRANSFORMATION OF THE TRIANGULATION BY MOVING THE NODES

- Fig. V.1. INITIAL MAGNET GEOMETRY AND PART OF Ω IN WHICH IS THE CRITERION EVALUATED (D)

V.2. INITIAL MAGNETOSTATIC STATE FOR J = 5.10<sup>6</sup> MKSA

V.3. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 1

J = 5 10⁶ MKSA, E<sub>d</sub> = (0,1.3)
CRITERION VALUE . 40428 10⁻⁵ (INITIAL . 61160 10⁻⁵)

V.4. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 2

J = 5 10⁶ MKSA , E<sub>d</sub> = (0,1.3)
CRITERION VALUE . 36334 10⁻⁵

V.5. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 5

J = 5 10⁶ MKSA , E<sub>d</sub> = (0,1.3)
CRITERION VALUE . 35466 10⁻⁵

V.6. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 9

J = 5. 10⁶ MKSA , E<sub>d</sub> = (0,1.3)
CRITERION VALUE . 35450 10⁻⁵

V.7. VARIATION OF THE CRITERION VALUE (D FIXED)

CURVE 1. FOR J = 5.10⁶ AND E<sub>d</sub> = (0,1.3)

CURVE 2. FOR J = 5 10⁶ AND E<sub>d</sub> = (0,1.6)

CURVE 3. FOR J = 20 10⁶ AND E<sub>d</sub> = (0,1.3)

CURVE 4. FOR J = 20 10⁶ AND E<sub>d</sub> = (0,1.6)

V.8. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 1

J = 5. 10⁶ MKSA , E<sub>d</sub> = (0,1.6)

CRITERION VALUE . 19963 10⁻⁴ (INITIAL . 51906 10⁻⁴)
V.9. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 2
J = 5 10⁶ MKSA, Ed = (0,1.6)
CRITERION VALUE .18088 10⁻⁴

V.10. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 5
J = 5 10⁶ MKSA, Ed = (0,1.6)
CRITERION VALUE .17862 10⁻⁴

V.11. INITIAL MAGNETOSTATIC STATE FOR J = 20 10⁶ MKSA

V.12. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 1
J = 20 10⁶ MKSA, Ed = (0,1.3)
CRITERION VALUE .60574 10⁻⁶ (INITIAL .10030 10⁻³)

V.13. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 2
J = 20 10⁶ MKSA, Ed = (0,1.3)
CRITERION VALUE .53566 10⁻⁶

V.14. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 5
J = 20 10⁶ MKSA, Ed = (0,1.3)
CRITERION VALUE .46671 10⁻⁶

V.15. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 9
J = 20 10⁶ MKSA, Ed = (0,1.3)
CRITERION VALUE .46190 10⁻⁶

V.16. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 1
J = 20 10⁶ MKSA, Ed = (0,1.6)
CRITERION VALUE .22023 10⁻⁵ (INITIAL .25427 10⁻⁴)

V.17. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 2
J = 20 10⁶ MKSA, Ed = (0,1.6)
CRITERION VALUE .18921 10⁻⁵

V.18. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 5
J = 20 10⁶ MKSA, Ed = (0,1.6)
CRITERION VALUE .18921 10⁻⁵

V.19. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 8
J = 20 10⁶ MKSA, Ed = (0,1.6)
CRITERION VALUE .18863 10⁻⁵

V.20. INITIAL MAGNET GEOMETRY AND PART OF Ω IN WHICH IS THE CRITERION
EVALUATED (D)

V.21. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 1
J = 5 10⁶ MKSA, Ed = (0,1.3)
CRITERION VALUE .16334 10⁻⁴ (INITIAL .24089 10⁻⁴)
V.22. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 2
J = 5 10⁶ MKSA, Ed = (0,1.3)
CRITERION VALUE . 12204 10⁻⁴

V.23. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 5
J = 5 10⁶ MKSA, Ed = (0,1.3)
CRITERION VALUE . 75190 10⁻⁵

V.24. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 10
J = 5 10⁶ MKSA, Ed = (0,1.3)
CRITERION VALUE . 66455 10⁻⁵

V.25. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 15
J = 5 10⁶ MKSA, Ed = (0,1.3)
CRITERION VALUE . 66217 10⁻⁵

V.26. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 20
J = 5 10⁶ MKSA, Ed = (0,1.3)
CRITERION VALUE . 65713 10⁻⁵

V.27. VARIATION OF THE CRITERION VALUE
CURVE 1 FOR J = 5 10⁶ AND Ed = (0,1.3)
CURVE 2 FOR J = 5 10⁶ AND Ed = (0,1.6)

V.28. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 1
J = 5 10⁶ MKSA, Ed = (0,1.6)
CRITERION VALUE . 49405 10⁻⁴ (INITIAL . 91754 10⁻⁴)

V.29. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 2
J = 5 10⁶ MKSA, Ed = (0,1.6)
CRITERION VALUE . 40805 10⁻⁴

V.30. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 5
J = 5 10⁶ MKSA, Ed = (0,1.6)
CRITERION VALUE . 32056 10⁻⁴

V.31. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 10
J = 5 10⁶ MKSA, Ed = (0,1.6)
CRITERION VALUE . 31550 10⁻⁴

V.32. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 15
J = 5 10⁶ MKSA, Ed = (0,1.6)
CRITERION VALUE . 31292 10⁻⁴

V.33. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 20
J = 5 10⁶ MKSA, Ed = (0,1.6)
CRITERION VALUE . 31068 10⁻⁴
V.34. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 1
J = 20 10⁶ MKSA, Ed = (0,1.3)

CRITERION VALUE . 12162 10⁻³ (INITIAL . 27241 10⁻³)

V.35. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 5
J = 20 10⁶ MKSA, Ed = (0,1.3)

CRITERION VALUE . 48866 10⁻⁴

V.36. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 10
J = 20 10⁶ MKSA, Ed = (0,1.3)

CRITERION VALUE . 22615 10⁻⁴

V.37. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 16
J = 20 10⁶ MKSA, Ed = (0,1.3)

CRITERION VALUE . 97890 10⁻⁵ (BEFORE "FITTING" . 11360 10⁻⁴)

V.38. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 21
J = 20 10⁶ MKSA, Ed = (0,1.3)

CRITERION VALUE . 64430 10⁻⁵

V.39. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 26
J = 20 10⁶ MKSA, Ed = (0,1.3)

CRITERION VALUE . 41371 10⁻⁵

V.40. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 31
J = 20 10⁶ MKSA, Ed = (0,1.3)

CRITERION VALUE . 18465 10⁻⁵

V.41. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 36
J = 20 10⁶ MKSA, Ed = (0,1.3)

CRITERION VALUE . 18465 10⁻⁵

V.42. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 36
J = 20 10⁶ MKSA, Ed = (0,1.3)

CRITERION VALUE . 36073 10⁻⁵ (AFTER "FITTING")

V.43. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 41
J = 20 10⁶ MKSA, Ed = (0,1.3)

CRITERION VALUE . 13949 10⁻⁵

V.44. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 46
J = 20 10⁶ MKSA, Ed = (0,1.3)

CRITERION VALUE . 99631 10⁻⁶

V.45. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 51
J = 20 10⁶ MKSA, Ed = (0,1.3)

CRITERION VALUE . 79043 10⁻⁶
V.46. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 56
J = 20 10⁶ MKSA, Ed = (0,1.3)
CRITERION VALUE . 67852 10⁻⁶

V.47. NEW TRIANGULATION AT STEP 16

V.48. NEW TRIANGULATION AT STEP 36

V.49. VARIATION OF THE CRITERION VALUE
CURVE 1 FOR J = 20 10⁶ AND Ed = (0,1,3)
CURVE 2 FOR J = 20 10⁶ AND Ed = (0,1.6)

V.50. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 1
J = 20 10⁶ MKSA, Ed = (0,1.6)
CRITERION VALUE . 16112 10⁻⁴ (INITIAL . 99395 10⁻⁴)

V.51. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 2
J = 20 10⁶ MKSA, Ed = (0,1.6)
CRITERION VALUE . 69781 10⁻⁵

V.52. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 5
J = 20 10⁶ MKSA, Ed = (0,1.6)
CRITERION VALUE . 22378 10⁻⁵

V.53. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 10
J = 20 10⁶ MKSA, Ed = (0,1.6)
CRITERION VALUE . 17220 10⁻⁵

V.54. MAGNETOSTATIC STATE AND AIR GAP SHAPE AT STEP 15
J = 20 10⁵ MKSA, Ed = (0,1.6)
CRITERION VALUE . 16688 10⁻⁵

V.55. PART OF Ω IN WHICH IS THE CRITERION EVALUATED FOR THE
FIFTH EXAMPLE

V.56. FINAL MAGNETOSTATIC STATE AND AIR GAP SHAPE (STEP 8)
J = 5 10⁶ MKSA, Ed = (0,0)
CRITERION VALUE . 97475 10⁻³ (INITIAL . 13839 10⁻²)