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

   
Equivalence with magnetostatics

Next, we have to prove that the solution $({\bf u},\lambda)$ of the variational problem (23), (24) represents a solution of the magnetostatic problem, too. Indeed, in (23) we have added the terms $\int_{\Omega}\rho \div {\bf u} \div {\bf v} \;dx$ and $- \int_{\Omega} \lambda \div {\bf v} \; dx$ to the formulation of (15). We show that for the solution $({\bf u},\lambda)$ the sum of both terms vanishes. We recall the following theorem from [7].

THEOREM 3   Assume that $\Omega$ is bounded, Lipschitz-continuous and simply-connected. A function ${\bf v}\in L^2(\Omega)^3$ satisfies

$${ \bf rot }{\bf v} = {\bf0} \quad in \; \Omega$$

iff there exists a unique function $\dot{p} \in H^1(\Omega)\backslash {\bf R}$ such that

$${\bf u} = { \rm grad }\dot{p}.$$

Proof. See [7], Theorem 2.9.
In the next theorem, we will assume that the electrical current density ${\bf f}$ is divergence-free. Indeed, since electrical charges cannot appear or disappear, this assumption represents the physical behaviour.

THEOREM 4   Assume that $\Omega$ is bounded, Lipschitz-continuous and simply-connected. Assume further that ${\bf f} \in H(\div ;\Omega)$ is divergence-free, i.e.,

$$\div {\bf f} = 0.$$ (34)

Then, the unique solution $({\bf u},\lambda)$ of

  

$$\int_{\Omega} \nu { \bf rot }{\bf u} \cdot { \bf rot }{\bf v}... ...} \rho \div {\bf u} \div {\bf v} \;dx - \int_{\Omega} \lambda \div {\bf v} \;dx = \int_{\Omega} {\bf f} \; {\bf v} \;dx \quad \forall {\bf v}\in{\bf Y}$$ (35)

$$- \int_{\Omega} \mu \div {\bf u} \; dx = 0 \quad \forall \mu\in{\bf M}$$ (36)

fulfills

$$\int_{\Omega} \nu { \bf rot }{\bf u} \cdot { \bf rot ... ...ega} {\bf f} \; {\bf v} \;dx \quad \forall {\bf v}\in{\bf Y}.$$ (37)

Proof. Let us choose an arbitrary $\phi \in L^2(\Omega)$. Then it is well-known that the Dirichlet problem

 

$$\Delta p = \phi \quad {\rm in}\; \Omega$$ (38)

$$p = 0 \quad {\rm on}\; \partial\Omega$$ (39)

has a unique weak solution $p\in H^1_0(\Omega)$. >From Theorem 3 we deduce that there exists a ${\bf v}\in L^2(\Omega)^3$ with

$${ \bf rot }{\bf v} = 0$$ (40)

and

$${\bf v} = { \rm grad }p.$$ (41)

Next, from (39) we get that the tangential derivatives of $p$ vanish, i.e. that the tangential components of ${\bf v}$ are zero. Thus, ${\bf v}\in H_0({ \bf rot };\Omega)$ holds. Further, we can determine the divergence of ${\bf v}$ as follows

$$\div {\bf v} = \div\left({ \rm grad }p \right)$$

$$= \Delta p$$

$$= \phi.$$ (42)

Consequently, ${\bf v} \in H(\div ;\Omega)$ and ${\bf v} \in {\bf Y}$. We can apply ${\bf v}$ as a test function in (35), and we get with (40)

$$\int_{\Omega} \rho \div {\bf u} \div {\bf v} \;dx - \int_{\Omega} \lambda \div {\bf v} \;dx = \int_{\Omega} {\bf f} \; {\bf v} \;dx \quad \forall {\bf v}\in{\bf Y},$$ (43)

$$\int_{\Omega} \left( \rho \div {\bf u} - \lambda \right) \phi \;dx = \int_{\Omega} {\bf f} \; { \rm grad }p \;dx,$$ (44)

$$\int_{\Omega} \left( \rho \div {\bf u} - \lambda \right) \phi \;dx = - \int_{\Omega} \div {\bf f} p \;dx + \int_{\partial\Omega} {\bf f} \; p \;{\rm n} \;ds.$$ (45)

With the assumptions (39) and (34) on $p$ and ${\bf f}$, the right-hand side is zero. Therefore,

$$\int_{\Omega} \left( \rho \div {\bf u} - \lambda \right) \phi \;dx = 0$$ (46)

holds, and setting $\phi := \div {\bf v} \in L^2(\Omega)$ for any ${\bf v} \in {\bf Y}$, we arrive at (37).
Thus, with (15) we get

$$\int_{\Omega} (\nu { \bf rot }{\bf u}) \cdot ({ \bf rot\... ...ega} {\bf f} \; {\bf v} \;dx \quad \forall {\bf v}\in{\bf Y},$$ (47)

and we conclude the relation

$${ \bf rot }\left( \nu { \bf rot }{\bf u} \right) = {\bf f}$$ (48)

in the weak sense. We remark that we do not require that $\bf {f}{\rm n} = 0$ on $\partial\Omega$. Further, setting $\phi := \lambda \in L^2(\Omega)_\star$, we get from (46) and (36) that

$$\begin{eqnarray*} \int_{\Omega} \left( \rho \div {\bf u} - \lambda \right) \lambda \;dx &=& 0,\\ \int_{\Omega} \lambda^2 \;dx &=& 0, \end{eqnarray*}$$

and $\lambda=0$. Therefore, we can omit the $\lambda$-term in (35), and solve the coercive problem:
(VF1$^\star$) Find ${\bf u} \in {\bf Y}$ such that

$$\int_{\Omega} \nu { \bf rot }{\bf u} \cdot { \bf rot }{... ...ega} {\bf f} \; {\bf v} \;dx \quad \forall {\bf v}\in{\bf Y}.$$ (49)

The unique solution ${\bf u}$ of (VF1$^\star$) coincides with the ${\bf u}$ in the solution of (VF1). We may discretize and solve the problem in the formulation (VF1$^\star$) instead of (VF1).

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