HEJ, HU ISSN 1418-7108 Manuscript no.: ANM-981030-A

Monotonicity and Lipschitz continuity

In many practical cases, the nonlinear behaviour of ferromagnetic materials cannot be neglected. Therefore, we extend the analysis of 2D monotone, Lipschitz continuous nonlinear problems (see [8]) to three dimensions. Let $\bar{\Omega}$ be decomposed into subdomains

$$\bar{\Omega} = \bigcup_{j=1}^{N_M}\bar{\Omega}_j, \quad ... ...\Omega_i \cap \Omega_j = \emptyset \quad \forall i \not= j$$

representing materials with different magnetic properties. We assume that the function $\nu \left( x, \vert{ \bf rot }{\bf u}\vert \right)$ in (8) depends on the position $x \in \Omega$, but $\nu$ is always the same function in one of the $\Omega_j$, i.e.,

$$\nu(x,z) = \nu^{(j)}(z) \quad \mbox{if} \quad x \in \Omega_j.$$ (50)

We suppose the following properties of the functions $\nu^{(j)}, \; j = 1, \ldots ,N_M$ which are justified by the physical model (cf. [8]):

(M0)

$\nu^{(j)}(z) = \nu^{(j)}_1 = \mbox{ const. } \quad \forall z \in [0,z^{(j)}_1],$

(M1)

$\nu^{(j)}(z) \ge \nu^{(j)}_1 \quad \forall z \ge 0,$

(M2)

$\nu^{(j)}(z)$ is a monotonic increasing function,

(M3)

$\lim_{z \rightarrow \infty} \nu^{(j)}(z) = \nu^{(j)}_{\infty},$ where $\nu^{(j)}_{\infty} = (\mu_0)^{-1}$ for ferromagnetic materials,

(M4)

there exists $\nu^{(j) \prime}(z) \quad \forall z \ge 0,$ and there exists a constant $M_1^{(j)}$ with
$\nu^{(j) \prime}(z) \le M_1^{(j)} \quad \forall z \ge 0,$

(M5)

there exists a constant $M^{(j)}$ with $\nu^{(j) \prime}(z) z + \nu^{(j)}(z) \le M^{(j)} \quad \forall z \ge 0.$

Let $\nu_1, M_1$ and $M$ be the global constants with

   

$$\nu_1 = \min_{j = 1, \ldots, N_M} \nu_1^{(j)}$$ (51)

$$M_1 = \max_{j = 1, \ldots, N_M} M_1^{(j)}$$ (52)

$$M = \max_{j = 1, \ldots, N_M} M^{(j)}.$$ (53)

We define the bivariate form $a : H^1(\Omega)^3 \times H^1(\Omega)^3 \longrightarrow {\bf R}$ which is nonlinear in its first argument, but linear in its second, by

$${a}({\bf u},{\bf v}) = \int_{\Omega} \nu(x,\vert{ \bf r... ... \bf rot }{\bf v} + \rho \div {\bf u} \div {\bf v} \;dx.$$ (54)

Then, the variational formulation for the problem (8), (9) can be written as:
(VF2) Find a pair $({\bf u},\lambda) \in {\bf Y}\times L^2_\star(\Omega)$ such that

  

$${a}({\bf u},{\bf v}) + b({\bf v}, \lambda) = \langle {\bf f},{\bf v} \rangle \quad \forall {\bf v} \in {\bf Y},$$ (55)

$$b({\bf u}, \mu) = 0 \quad \forall \mu \in {\bf M},$$ (56)

where the bilinear form $b$ is the identical as in (22). Now, we prove monotonicity and Lipschitz continuity of the form $a$ resp. the (nonlinear) operator $A : {\bf Y}\longrightarrow {\bf Y}^{\star}$ defined by

$$\langle A {\bf u}, {\bf v} \rangle = a( {\bf u}, {\bf v}) \quad \forall {\bf u},{\bf v} \in {\bf Y}.$$

Lemma 5   Assume (M1) and (M2). Then $A$ is strongly monotone, i.e., the inequality

$$\langle A {\bf u} - A {\bf v} , {\bf u} - {\bf v} \rangle... ...\vert _{\bf Y}^2 \quad \forall {\bf u}, {\bf v} \in {\bf Y}$$ (57)

holds.

Proof. The proof is similar to the proof of Lemma 2.1 in [8]. We consider vectors $s, t \in {{\bf R}}^3$, and assume that $\vert s\vert \ge \vert t\vert$ holds without loss of generality. Then, we get

$$\vert s\vert^2 - s \cdot t \ge \vert s\vert^2 - \vert s\vert\;\vert t\vert \ge 0$$

and from (M1) and (M2) it follows that

$$\bigl( \nu(\vert s\vert) s - \nu(\vert t\vert) t \bigr) \cdot (s-t) = \nu(\vert s\vert) \bigl(\vert s\vert^2-s \cdot t \bigr) + \nu(\vert t\vert) \bigl(\vert t\vert^2-s \cdot t \bigr)$$

$$\ge \nu(\vert t\vert) \bigl(\vert s\vert^2-s \cdot t \bigr) + \nu(\vert t\vert) \bigl(\vert t\vert^2-s \cdot t \bigr)$$

$$= \nu(\vert t\vert) \vert s-t\vert^2$$

$$\ge \nu_1 \vert s-t\vert^2$$

holds. Setting $s={ \bf rot }{\bf u}(x)$ and $t = { \bf rot }{\bf v}(x)$ and integrating we obtain

$$\int_{\Omega} \left( \nu(\vert{ \bf rot }{\bf u}\vert) {\... ...{\Omega} \vert{ \bf rot }({\bf u}-{\bf v})\vert^2 dx, \\$$

and with

$$\int_{\Omega} \rho \div ({\bf u}-{\bf v}) \div ({\bf u}-{\... ... = \rho \; \vert\vert\div ({\bf u}-{\bf v})\vert\vert _0^2$$ (58)

and Lemma 3 we get the estimate (57).

Lemma 6   Suppose that (M4) and (M5) hold. Then $A$ is Lipschitz-continuous, i.e., the inequality

$$\vert\vert A {\bf u} - A {\bf v} \vert\vert _{{\bf Y}^{\st... ...ert\vert _{\bf Y} \quad \forall {\bf u}, {\bf v} \in {\bf Y}$$ (59)

holds.

Proof. >From (M4) and (M5) we obtain

$$\nu^{\prime}(z) z + \nu(z) \le M \quad \forall z \ge 0,$$

and by the mean value theorem we get

$$\vert\nu(z_1) z_1 - \nu(z_2) z_2 \vert \le M \vert z_1 - z_2 \vert \quad \forall z_1, z_2 \ge 0.$$ (60)

Consider vectors $s, t \in {{\bf R}}^{3}$. The same calculation as in the proof of Lemma 2.2 in [8] yields

$$\bigl\vert \nu(\vert s\vert) s - \nu(\vert t\vert) t \bigr... ...\le M \vert s-t\vert \quad \forall s,t \in {{\bf R}}^{3}.$$ (61)

We set $s={ \bf rot }{\bf u}(x), t = { \bf rot }{\bf v}(x),$ multiply with $\vert{ \bf rot }{\bf w}(x)\vert$, and take the integral over $\Omega$. The result is

$$\int_{\Omega} \bigl\vert \nu(\vert{ \bf rot }{\bf u}\vert... ... - {\bf v})\vert \; \vert{ \bf rot }{\bf w}\vert \;dx ,$$

and with well-known estimates, an equation for the $\rho$ term, and Lemma 3 we obtain

$$\langle A {\bf u}- A {\bf v},{\bf w} \rangle \le \max(M,\r... ...\bf w} \in {\bf Y}\quad \forall {\bf u}, {\bf v} \in {\bf Y},$$

and the desired estimate (59).

HEJ, HU ISSN 1418-7108 Manuscript no.: ANM-981030-A