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

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.
\end{displaymath} (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]):
$ \nu^{(j)}(z) = \nu^{(j)}_1 = \mbox{ const. }
\quad \forall z \in [0,z^{(j)}_1], $
$ \nu^{(j)}(z) \ge \nu^{(j)}_1 \quad \forall z \ge 0, $
$ \nu^{(j)}(z) $ is a monotonic increasing function,
$ \lim_{z \rightarrow \infty} \nu^{(j)}(z) =
\nu^{(j)}_{\infty}, $ where $\nu^{(j)}_{\infty} = (\mu_0)^{-1} $ for ferromagnetic materials,
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, $
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
$\displaystyle \nu_1$ $\textstyle =$ $\displaystyle \min_{j = 1, \ldots, N_M} \nu_1^{(j)}$ (51)
$\displaystyle M_1$ $\textstyle =$ $\displaystyle \max_{j = 1, \ldots, N_M} M_1^{(j)}$ (52)
$\displaystyle M$ $\textstyle =$ $\displaystyle \max_{j = 1, \ldots, N_M} M^{(j)}.$ (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.
\end{displaymath} (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
$\displaystyle {a}({\bf u},{\bf v}) + b({\bf v}, \lambda)$ $\textstyle =$ $\displaystyle \langle {\bf f},{\bf v} \rangle
\quad \forall {\bf v} \in {\bf Y},$ (55)
$\displaystyle b({\bf u}, \mu)$ $\textstyle =$ $\displaystyle 0 \quad \forall \mu \in {\bf M},$ (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}
\end{displaymath} (57)


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
$\displaystyle \bigl( \nu(\vert s\vert) s - \nu(\vert t\vert) t \bigr) \cdot (s-t)$ $\textstyle =$ $\displaystyle \nu(\vert s\vert) \bigl(\vert s\vert^2-s \cdot t \bigr) +
\nu(\vert t\vert) \bigl(\vert t\vert^2-s \cdot t \bigr)$  
  $\textstyle \ge$ $\displaystyle \nu(\vert t\vert) \bigl(\vert s\vert^2-s \cdot t \bigr) +
\nu(\vert t\vert) \bigl(\vert t\vert^2-s \cdot t \bigr)$  
  $\textstyle =$ $\displaystyle \nu(\vert t\vert) \vert s-t\vert^2$  
  $\textstyle \ge$ $\displaystyle \nu_1 \vert s-t\vert^2$  

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) {\...
\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
\end{displaymath} (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}
\end{displaymath} (59)


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

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

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.
\end{displaymath} (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...
M \vert s-t\vert \quad \forall s,t
\in {{\bf R}}^{3}.
\end{displaymath} (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
Frontpage previous next