Let us first give the definition of weak solutions to problem (1).
Definition 1
Let \(\mathbf{m}_{0}\in\mathbb{H}^{1}(\Omega)\) with \(|\mathbf{m}_{0}|=1\) a.e., we say that a three dimensional vector \(\mathbf{m}=(m_{1},m_{2},m_{3})\) is a weak solution of problem (1) if
-
for all \(T>0\), \(\mathbf{m}\in\mathbb{H}^{1}(Q)\), \(\partial _{t}\mathbf{m}\in\mathbb{L}^{2}(Q)\), and \(|\mathbf{m}|=1\) a.e. in Q;
-
for all \({\boldsymbol{\phi}}\in\mathcal {C}^{\infty }(\overline{Q})\) with \(\boldsymbol{\phi}(\cdot,0)=\boldsymbol{\phi} (\cdot,T)=0\), we have
$$\begin{aligned}& \int_{Q}\partial _{t}\mathbf{m}\cdot{\boldsymbol{ \phi}} \,\mathrm {d}x\,\mathrm{d}t-\alpha \int _{Q}\mathbf{m}\times\partial _{t}\mathbf{m} \cdot{\boldsymbol{\phi}} \,\mathrm{d}x\,\mathrm{d}t \\& \quad =\beta \int_{Q}\mathbf{m}\times\nabla\mathbf{m}\cdot\nabla { \boldsymbol{\phi}} \,\mathrm{d}x\,\mathrm{d}t+\lambda \int_{Q}\mathbf{m}\times\Delta \mathbf{m}\cdot\mathbf{m} \times{\boldsymbol{\phi}} \,\mathrm{d}x\,\mathrm{d}t; \end{aligned}$$
(6)
-
\(\mathbf{m}(x,0)=\mathbf{m}_{0}(x)\) in the trace sense;
-
for all \(t\in(0,T)\), there holds
$$\begin{aligned}& \frac{\beta+\alpha\lambda}{2} \int_{\Omega}\bigl\vert \nabla\mathbf {m}(t)\bigr\vert ^{2} \,\mathrm{d}x +\alpha \int_{0}^{t} \int_{\Omega} |\partial _{t}\mathbf {m}|^{2} \,\mathrm{d} x\,\mathrm{d}t \\& \quad \leq \frac{\beta+\alpha\lambda}{2} \int_{\Omega}|\nabla\mathbf{m} _{0}|^{2} \, \mathrm{d}x. \end{aligned}$$
(7)
Remark 2
We will show in Section 2.2 that \(\mathbf{m}\times\Delta \mathbf{m}\) makes sense in \(\mathbb{L}^{2}(Q)\), and for this reason, it will be clear that (6) makes sense.
To prove the global existence of weak solutions of the problem (1) we proceed as in [6–8].
2.1 The penalty problem
Let \(\varepsilon>0\). We introduce the following penalty problem.
For initial datum \(\mathbf{m}_{0}\in\mathbb{H}^{1}(\Omega)\), and for each positive number T, find a vector field \(\mathbf{m}_{\varepsilon}\) in Q such as to satisfy the equation
$$ \partial _{t}\mathbf{m}^{\varepsilon} \times \mathbf{m}^{\varepsilon }+\beta\Delta\mathbf{m} ^{\varepsilon}+\lambda \mathbf{m}^{\varepsilon}\times\Delta\mathbf{m} ^{\varepsilon} -\alpha\partial _{t}\mathbf{m}^{\varepsilon} -\frac{1}{\varepsilon } \bigl(\bigl\vert \mathbf{m} ^{\varepsilon}\bigr\vert ^{2}-1 \bigr) \mathbf{m}^{\varepsilon}=0 $$
(8)
subject to the initial and boundary conditions
$$ \left \{ \textstyle\begin{array}{l} \mathbf{m}^{\varepsilon}(\cdot,0)=\mathbf{m}_{0},\qquad |\mathbf{m}_{0}|=1 \quad \mbox{in } \Omega, \\ \partial_{\nu}\mathbf{m}^{\varepsilon}=0 \quad \mbox{on } (0,T)\times \partial\Omega. \end{array}\displaystyle \right . $$
(9)
The last term of equation (8) was introduced at the end to represent the constraint \(|\mathbf{m}|=1\).
We have the following result.
Theorem 1
For each fixed positive
ε, there is a weak solution
\(\mathbf{m} ^{\varepsilon}\)
of Problem (8)-(9) such that
$$\begin{aligned}& \int_{Q}\partial _{t}\mathbf{m}^{\varepsilon} \times\mathbf {m}^{\varepsilon}\cdot {\boldsymbol{\varphi}} \,\mathrm{d}x\, \mathrm{d}t -\beta \int _{Q}\nabla\mathbf{m}^{\varepsilon }\cdot\nabla{ \boldsymbol{\varphi}} \,\mathrm{d}x\,\mathrm{d}t -\lambda \int_{Q}\mathbf{m}^{\varepsilon}\times\nabla\mathbf {m}^{\varepsilon }\cdot\nabla{\boldsymbol{\varphi}} \,\mathrm{d}x\,\mathrm{d}t \\& \quad {}-\alpha \int_{Q}\partial _{t}\mathbf{m}^{\varepsilon} \cdot {\boldsymbol{\varphi}} \,\mathrm{d}x\,\mathrm{d}t -\frac{1}{\varepsilon} \int_{Q} \bigl(\bigl\vert \mathbf{m} ^{\varepsilon}\bigr\vert ^{2}-1 \bigr)\mathbf{m}^{\varepsilon}\cdot{\boldsymbol{ \varphi}} \,\mathrm{d}x\,\mathrm{d}t =0 \end{aligned}$$
for any
φ
in
\(\mathbb{H}^{1}(Q)\). Moreover, the following energy estimate holds:
$$\begin{aligned}& \frac{\beta+\alpha\lambda}{2} \int_{\Omega}\bigl\vert \nabla\mathbf{m} ^{\varepsilon}\bigr\vert ^{2}(t) \,\mathrm{d}x +\alpha \int_{0}^{t} \int_{\Omega} \bigl\vert \partial _{t} \mathbf{m}^{\varepsilon} \bigr\vert ^{2} \,\mathrm{d}x\,\mathrm{d}t \\& \quad {}+\frac{1}{4\varepsilon}\biggl(1+\frac{\alpha\lambda}{\beta}\biggr) \int _{\Omega}\bigl(\bigl\vert \mathbf{m}^{\varepsilon}\bigr\vert ^{2}-1\bigr)^{2}(t) \,\mathrm{d}x \leq \frac{\beta+\alpha\lambda}{2} \int_{\Omega} \vert \nabla\mathbf{m} _{0}\vert ^{2} \,\mathrm{d}x \end{aligned}$$
(10)
for all
\(t\in(0,T)\).
Proof
To show the existence of a solution for the penalized problem using the method of Faedo-Galerkin and since \(\mathbb{H}^{1}(\Omega)\) is a separable Hilbert space we can approximate m by \(\mathbf {m}^{\varepsilon,N}\). We set
$$ \mathbf{m}^{\varepsilon,N}(x,t)=\sum_{i=1}^{N} \mathbf{a}_{i}(t)f_{i}(x), $$
where \(\lbrace f_{i}\rbrace_{i\in\mathbb{N}}\) is an orthonormal basis of \(L^{2}(\Omega)\) and orthogonal in \(H^{1}(\Omega)\) consisting of eigenfunctions of −Δ, i.e.,
$$ \left \{ \textstyle\begin{array}{l} -\Delta f_{i}=\lambda_{i}f_{i}, \quad i=1,2,\ldots, \\ \partial_{\nu}f_{i}=0 \quad \mbox{on } \partial\Omega, \end{array}\displaystyle \right . $$
(11)
where \(\mathbf{a}_{i}(t)\) are \(\mathbb{R}^{3}\)-valued vectors.
We obtain the following approximate problem:
$$\begin{aligned}& \partial _{t}\mathbf{m}^{\varepsilon,N} \times\mathbf {m}^{\varepsilon,N}+\beta \Delta\mathbf{m}^{\varepsilon,N}+\lambda \mathbf{m}^{\varepsilon ,N}\times\Delta \mathbf{m}^{\varepsilon,N} \\& \quad {}-\alpha\partial _{t}\mathbf {m}^{\varepsilon,N} - \frac{1}{\varepsilon} \bigl(\bigl\vert \mathbf{m}^{\varepsilon,N}\bigr\vert ^{2}-1 \bigr)\mathbf{m} ^{\varepsilon,N}=0, \end{aligned}$$
(12)
on Q with the following initial and boundary conditions:
$$\begin{aligned}& \mathbf{m}^{\varepsilon,N}(\cdot,0)=\mathbf{m}^{N}(\cdot,0)\quad \mbox{in } \Omega, \\& \partial_{\nu}\mathbf{m}^{\varepsilon,N}=0 \quad \mbox{on } \partial \Omega, \end{aligned}$$
and
$$ \int_{\Omega}\mathbf{m}^{N}(\cdot,0)f_{i} \, \mathrm{d}x = \int_{\Omega }\mathbf{m} _{0}f_{i} \, \mathrm{d}x. $$
Multiplying equation (12) by \(f_{i}\) and integrating over Ω, we get an ordinary differential system. In fact, we have
$$ \partial _{t}\mathbf{m}^{\varepsilon,N} \times\mathbf {m}^{\varepsilon,N}-\alpha\partial _{t}\mathbf{m}^{\varepsilon,N} = \mathbb{M}\bigl(\mathbf{m}^{\varepsilon ,N}\bigr)\partial _{t}\mathbf{m} ^{\varepsilon,N}, $$
where
$$\mathbb{M}\bigl(\mathbf{m}^{\varepsilon,N}\bigr)= \begin{pmatrix} -\alpha& m^{\varepsilon,N}_{3} & -m^{\varepsilon,N}_{2} \\ -m^{\varepsilon,N}_{3} & -\alpha& m^{\varepsilon,N}_{1} \\ m^{\varepsilon,N}_{2} & -m^{\varepsilon,N}_{1} & -\alpha \end{pmatrix} . $$
We can write equation (12) in the form
$$ \mathbb{M}\bigl(\mathbf{m}^{\varepsilon,N}\bigr)\partial _{t}\mathbf {m}^{\varepsilon,N} =-\beta \Delta\mathbf{m}^{\varepsilon,N}-\lambda \mathbf{m}^{\varepsilon ,N}\times\Delta \mathbf{m}^{\varepsilon,N}+\frac{1}{\varepsilon} \bigl(\bigl\vert \mathbf {m}^{\varepsilon ,N}\bigr\vert ^{2}-1 \bigr) \mathbf{m}^{\varepsilon,N}. $$
Note that
$$ \det\mathbb{M}\bigl(\mathbf{m}^{\varepsilon,N}\bigr)=-\alpha\bigl(\alpha ^{2}+\bigl\vert \mathbf{m} ^{\varepsilon,N}\bigr\vert ^{2} \bigr)\neq0. $$
Hence \(\mathbb{M}(\mathbf{m}^{\varepsilon,N})\) is invertible.
The resulting system is then locally Lipschitz. There exists a unique local solution for the approximate problem that can extend on \([0, T]\) using an a priori estimate.
To get bounds on the solutions, we multiply equation (12) by \(\partial _{t}\mathbf{m}^{\varepsilon,N} \) and integrate over Ω to obtain
$$\begin{aligned}& \frac{\beta}{2}\frac{\mathrm{d}}{\mathrm{d}t } \int_{\Omega}\bigl\vert \nabla \mathbf{m}^{\varepsilon,N}\bigr\vert ^{2} \,\mathrm{d}x -\lambda \int_{\Omega }\mathbf{m} ^{\varepsilon,N}\times\Delta \mathbf{m}^{\varepsilon,N}\cdot \partial _{t}\mathbf{m} ^{\varepsilon,N} \,\mathrm{d}x \\& \quad {}+\alpha \int_{\Omega}\bigl\vert \partial _{t} \mathbf{m}^{\varepsilon,N} \bigr\vert ^{2} \,\mathrm{d}x + \frac{1}{4\varepsilon}\frac{\mathrm{d}}{\mathrm{d}t } \int_{\Omega}\bigl(\bigl\vert \mathbf{m}^{\varepsilon ,N}\bigr\vert ^{2}-1\bigr)^{2} \,\mathrm{d}x =0. \end{aligned}$$
(13)
Multiply again equation (12) with \(\Delta\mathbf {m}^{\varepsilon ,N}\) and integrate over Ω; we get
$$\begin{aligned}& \int_{\Omega}\partial _{t}\mathbf{m}^{\varepsilon,N} \times \mathbf{m} ^{\varepsilon,N}\cdot\Delta\mathbf{m}^{\varepsilon,N} \, \mathrm{d}x +\beta \int_{\Omega}\bigl\vert \Delta\mathbf{m}^{\varepsilon,N}\bigr\vert ^{2} \,\mathrm{d}x \\& \quad {}+\frac{\alpha}{2}\frac{\mathrm{d}}{\mathrm{d}t } \int_{\Omega}\bigl\vert \nabla \mathbf{m}^{\varepsilon,N}\bigr\vert ^{2} \,\mathrm{d}x - \frac {1}{\varepsilon} \int _{\Omega}\bigl(\bigl\vert \mathbf{m}^{\varepsilon,N}\bigr\vert ^{2}-1\bigr)\cdot\Delta\mathbf {m}^{\varepsilon ,N} \,\mathrm{d}x =0. \end{aligned}$$
(14)
Multiplying (14) by λ and taking the sum with (13)
$$\begin{aligned}& \frac{\beta}{2}\frac{\mathrm{d}}{\mathrm{d}t } \int_{\Omega}\bigl\vert \nabla \mathbf{m}^{\varepsilon,N}\bigr\vert ^{2} \,\mathrm{d}x +\alpha \int_{\Omega }\bigl\vert \mathbf{m} ^{\varepsilon,N}\bigr\vert ^{2} \,\mathrm{d}x +\frac{1}{4\varepsilon}\frac{\mathrm{d}}{\mathrm{d}t } \int _{\Omega }\bigl(\bigl\vert \mathbf{m}^{\varepsilon,N}\bigr\vert ^{2}-1\bigr)^{2} \,\mathrm{d}x \\& \qquad {}+\lambda\beta \int_{\Omega}\bigl\vert \Delta\mathbf{m}^{\varepsilon ,N}\bigr\vert ^{2} \,\mathrm{d}x +\frac{\lambda\alpha}{2}\frac{\mathrm{d}}{\mathrm {d}t } \int _{\Omega}\bigl\vert \nabla\mathbf{m}^{\varepsilon,N}\bigr\vert ^{2} \,\mathrm{d}x \\& \quad =\frac{\lambda}{\varepsilon} \int_{\Omega}\bigl(\bigl\vert \mathbf {m}^{\varepsilon ,N}\bigr\vert ^{2}-1\bigr)\cdot\Delta\mathbf{m}^{\varepsilon,N} \,\mathrm{d}x. \end{aligned}$$
(15)
On the other hand, the Young inequality gives
$$\begin{aligned}& \frac{\lambda}{\varepsilon} \int_{\Omega}\bigl(\bigl\vert \mathbf {m}^{\varepsilon ,N}\bigr\vert ^{2}-1\bigr)\mathbf{m}^{\varepsilon,N}\cdot\Delta\mathbf {m}^{\varepsilon,N} \,\mathrm{d}x \\& \quad \leq\frac{\lambda}{2d\varepsilon^{2}} \int_{\Omega}\bigl(\bigl\vert \mathbf{m} ^{\varepsilon,N}\bigr\vert ^{2}-1\bigr)^{2}\bigl\vert \mathbf{m}^{\varepsilon,N} \bigr\vert ^{2} \,\mathrm{d}x +\frac{\lambda d}{2} \int_{\Omega}\bigl\vert \Delta\mathbf{m}^{\varepsilon ,N}\bigr\vert ^{2} \,\mathrm{d}x \end{aligned}$$
(16)
for any constant \(d>0\).
We multiply (12) by \((|\mathbf{m}^{\varepsilon ,N}|^{2}-1)\mathbf{m} ^{\varepsilon,N}\) and integrate over Ω to get
$$\begin{aligned}& \beta \int_{\Omega}\bigl(\bigl\vert \mathbf{m}^{\varepsilon,N}\bigr\vert ^{2}-1\bigr)\mathbf{m} ^{\varepsilon,N}\cdot\Delta \mathbf{m}^{\varepsilon,N} \,\mathrm{d}x - \frac{\alpha}{4}\frac{\mathrm{d}}{\mathrm{d}t } \int_{\Omega }\bigl(\bigl\vert \mathbf{m} ^{\varepsilon,N}\bigr\vert ^{2}-1\bigr)^{2} \,\mathrm{d}x \\& \quad {}-\frac{1}{\varepsilon} \int_{\Omega}\bigl(\bigl\vert \mathbf{m}^{\varepsilon ,N}\bigr\vert ^{2}-1\bigr)^{2}\bigl\vert \mathbf{m}^{\varepsilon,N} \bigr\vert ^{2} \,\mathrm{d}x =0. \end{aligned}$$
Hence
$$\begin{aligned}& \frac{\lambda}{\varepsilon} \int_{\Omega}\bigl(\bigl\vert \mathbf {m}^{\varepsilon ,N}\bigr\vert ^{2}-1\bigr)\mathbf{m}^{\varepsilon,N}\cdot\Delta\mathbf {m}^{\varepsilon,N} \,\mathrm{d}x \\& \quad =\frac{\alpha\lambda}{4\beta\varepsilon}\frac{\mathrm {d}}{\mathrm{d}t } \int_{\Omega}\bigl(\bigl\vert \mathbf{m}^{\varepsilon,N}\bigr\vert ^{2}-1\bigr)^{2} \,\mathrm{d}x + \frac{\lambda}{\beta\varepsilon^{2}} \int_{\Omega}\bigl(\bigl\vert \mathbf{m} ^{\varepsilon,N}\bigr\vert ^{2}-1\bigr)^{2}\bigl\vert \mathbf{m}^{\varepsilon,N} \bigr\vert ^{2} \,\mathrm{d}x. \end{aligned}$$
Therefore
$$\begin{aligned}& \frac{\alpha\lambda}{4\beta\varepsilon}\frac{\mathrm {d}}{\mathrm{d}t } \int_{\Omega}\bigl(\bigl\vert \mathbf{m}^{\varepsilon,N}\bigr\vert ^{2}-1\bigr)^{2} \,\mathrm{d}x + \frac{\lambda}{\beta\varepsilon^{2}} \int_{\Omega}\bigl(\bigl\vert \mathbf{m} ^{\varepsilon,N}\bigr\vert ^{2}-1\bigr)^{2}\bigl\vert \mathbf{m}^{\varepsilon,N} \bigr\vert ^{2} \,\mathrm{d}x \\& \quad \leq\frac{\lambda}{2d\varepsilon^{2}} \int_{\Omega}\bigl(\bigl\vert \mathbf{m} ^{\varepsilon,N}\bigr\vert ^{2}-1\bigr)^{2}\bigl\vert \mathbf{m}^{\varepsilon,N} \bigr\vert ^{2} \,\mathrm{d}x + \frac{\lambda d}{2} \int_{\Omega}\bigl\vert \Delta\mathbf{m}^{\varepsilon ,N}\bigr\vert ^{2} \,\mathrm{d}x. \end{aligned}$$
That is,
$$\begin{aligned}& \frac{\alpha\lambda}{4\beta\varepsilon}\frac{\mathrm {d}}{\mathrm{d}t } \int_{\Omega}\bigl(\bigl\vert \mathbf{m}^{\varepsilon,N}\bigr\vert ^{2}-1\bigr)^{2} \,\mathrm{d}x + \frac{\lambda}{\varepsilon^{2}} \biggl(\frac{1}{\beta}-\frac {1}{2d}\biggr) \int_{\Omega}\bigl(\bigl\vert \mathbf{m}^{\varepsilon ,N}\bigr\vert ^{2}-1\bigr)^{2}\bigl\vert \mathbf{m} ^{\varepsilon,N}\bigr\vert ^{2} \,\mathrm{d}x \\& \quad \leq\frac{\lambda d}{2} \int_{\Omega}\bigl\vert \Delta\mathbf {m}^{\varepsilon ,N}\bigr\vert ^{2} \,\mathrm{d}x. \end{aligned}$$
So, for \(d>\frac{\beta}{2}\),
$$\begin{aligned}& \frac{\lambda}{2d\beta\varepsilon^{2}} \int_{\Omega}\bigl(\bigl\vert \mathbf{m} ^{\varepsilon,N}\bigr\vert ^{2}-1\bigr)^{2}\bigl\vert \mathbf{m}^{\varepsilon,N} \bigr\vert ^{2} \,\mathrm{d}x \\& \quad \leq\frac{\lambda d}{2(2d-\beta)} \int_{\Omega}\bigl\vert \Delta\mathbf{m} ^{\varepsilon,N}\bigr\vert ^{2} \,\mathrm{d}x -\frac{\alpha\lambda}{4\beta \varepsilon(2d-\beta)}\frac{\mathrm{d}}{\mathrm{d}t } \int_{\Omega}\bigl(\bigl\vert \mathbf{m}^{\varepsilon ,N}\bigr\vert ^{2}-1\bigr)^{2} \,\mathrm{d}x. \end{aligned}$$
Therefore from (16), we have
$$\begin{aligned}& \frac{\lambda}{\varepsilon} \int_{\Omega}\bigl(\bigl\vert \mathbf {m}^{\varepsilon ,N}\bigr\vert ^{2}-1\bigr)\mathbf{m}^{\varepsilon,N}\cdot\Delta\mathbf {m}^{\varepsilon,N} \,\mathrm{d}x \\& \quad \leq\frac{\lambda d}{2}\biggl(1+\frac{\beta}{2d-\beta}\biggr) \int_{\Omega }\bigl\vert \Delta\mathbf{m}^{\varepsilon,N}\bigr\vert ^{2} \,\mathrm{d}x - \frac{\alpha\lambda}{4\varepsilon(2d-\varepsilon)}\frac {\mathrm{d}}{\mathrm{d}t } \int_{\Omega}\bigl(\bigl\vert \mathbf{m}^{\varepsilon ,N}\bigr\vert ^{2}-1\bigr)^{2} \,\mathrm{d}x \end{aligned}$$
and then from (15)
$$\begin{aligned}& \frac{\beta}{2}\frac{\mathrm{d}}{\mathrm{d}t } \int_{\Omega}\bigl\vert \nabla \mathbf{m}^{\varepsilon,N}\bigr\vert ^{2} \,\mathrm{d}x +\alpha \int_{\Omega }\bigl\vert \mathbf{m} ^{\varepsilon,N}\bigr\vert ^{2} \,\mathrm{d}x +\frac{1}{4\varepsilon}\frac{\mathrm{d}}{\mathrm{d}t } \int _{\Omega }\bigl(\bigl\vert \mathbf{m}^{\varepsilon,N}\bigr\vert ^{2}-1\bigr)^{2} \,\mathrm{d}x \\& \quad {}+\lambda\beta \int_{\Omega}\bigl\vert \Delta\mathbf{m}^{\varepsilon ,N}\bigr\vert ^{2} \,\mathrm{d}x +{}\frac{\lambda\alpha}{2}\frac{\mathrm{d}}{\mathrm{d}t } \int _{\Omega}\bigl\vert \nabla\mathbf{m}^{\varepsilon,N}\bigr\vert ^{2} \,\mathrm{d}x \leq\frac {\lambda d^{2}}{2d-\beta} \int_{\Omega}\bigl\vert \Delta\mathbf{m}^{\varepsilon ,N}\bigr\vert ^{2} \,\mathrm{d}x. \end{aligned}$$
That is,
$$\begin{aligned}& \frac{\beta+\alpha\lambda}{2}\frac{\mathrm{d}}{\mathrm{d}t } \int _{\Omega}\bigl\vert \nabla\mathbf{m}^{\varepsilon,N}\bigr\vert ^{2} \,\mathrm{d}x +\alpha \int _{\Omega}\bigl\vert \partial _{t} \mathbf{m}^{\varepsilon,N} \bigr\vert ^{2} \,\mathrm{d}x +\lambda \biggl(\beta -\frac{d^{2}}{2d-\beta}\biggr) \int_{\Omega}\bigl\vert \Delta\mathbf {m}^{\varepsilon ,N}\bigr\vert ^{2} \,\mathrm{d}x \\& \quad {}+\frac{1}{4\varepsilon}\biggl(1+\frac{\alpha\lambda}{2d-\beta }\biggr)\frac{\mathrm{d}}{\mathrm{d}t } \int_{\Omega}\bigl(\bigl\vert \mathbf {m}^{\varepsilon ,N}\bigr\vert ^{2}-1\bigr)^{2} \,\mathrm{d}x \leq0. \end{aligned}$$
Taking \(d=\beta\) we get \(\beta-\frac{d^{2}}{2d-\beta}=0\) and therefore
$$\begin{aligned}& \frac{\beta+\alpha\lambda}{2}\frac{\mathrm{d}}{\mathrm{d}t } \int _{\Omega}\bigl\vert \nabla\mathbf{m}^{\varepsilon,N}\bigr\vert ^{2} \,\mathrm{d}x +\alpha \int _{\Omega}\bigl\vert \partial _{t} \mathbf{m}^{\varepsilon,N} \bigr\vert ^{2} \,\mathrm{d}x \\& \quad {}+\frac{1}{4\varepsilon}\biggl(1+\frac{\alpha\lambda}{\beta}\biggr)\frac {\mathrm{d}}{\mathrm{d}t } \int_{\Omega}\bigl(\bigl\vert \mathbf{m}^{\varepsilon ,N}\bigr\vert ^{2}-1\bigr)^{2} \,\mathrm{d} x \leq0. \end{aligned}$$
We integrate from 0 to t to get
$$\begin{aligned}& \frac{\beta+\alpha\lambda}{2} \int_{\Omega}\bigl\vert \nabla\mathbf{m} ^{\varepsilon,N}\bigr\vert ^{2}(t) \,\mathrm{d}x +\alpha \int_{0}^{t} \int _{\Omega} \bigl\vert \partial _{t} \mathbf{m}^{\varepsilon,N} \bigr\vert ^{2} \,\mathrm{d}x\,\mathrm {d}t \\& \qquad {}+\frac{1}{4\varepsilon}\biggl(1+\frac{\alpha\lambda}{\beta}\biggr) \int _{\Omega}\bigl(\bigl\vert \mathbf{m}^{\varepsilon,N}\bigr\vert ^{2}-1\bigr)^{2}(t) \,\mathrm{d}x \\& \quad \leq\frac{\beta+\alpha\lambda}{2} \int_{\Omega}\bigl\vert \nabla\mathbf{m} ^{N}\bigr\vert ^{2}(0) \,\mathrm{d}x +\frac{1}{4\varepsilon}\biggl(1+ \frac{\alpha \lambda }{\beta}\biggr) \int_{\Omega}\bigl(\bigl\vert \mathbf{m}^{N}\bigr\vert ^{2}-1\bigr)^{2}(0) \,\mathrm{d}x \end{aligned}$$
(17)
for all \(t\in(0,T)\).
The right-hand side is uniformly bounded. Indeed \(\mathbb {H}^{1}(\Omega )\hookrightarrow\mathbb{L}^{4}(\Omega)\) with continuous embedding, therefore
$$\begin{aligned} \int_{\Omega}\bigl(\bigl\vert \mathbf{m}^{\varepsilon,N}\bigr\vert ^{2}-1\bigr)^{2}(0) \,\mathrm{d}x =& \int _{\Omega}\bigl\vert \mathbf{m}^{N}(0)\bigr\vert ^{4} \,\mathrm{d}x -2 \int_{\Omega }\bigl\vert \mathbf{m} ^{N}(0)\bigr\vert ^{2} \,\mathrm{d}x +\operatorname{meas}(\Omega) \\ \leq& \bigl\Vert \mathbf{m}^{N}(0)\bigr\Vert ^{4}_{\mathbb{L}^{4}(\Omega)}+\operatorname {meas}(\Omega) \\ \leq& C_{1}\bigl\Vert \mathbf{m}^{N}(0)\bigr\Vert ^{4}_{\mathbb{H}^{1}(\Omega)}+C_{2}, \end{aligned}$$
where \(C_{1}\) et \(C_{2}\) are two constants independent of ε and N.
Furthermore, note that \(\mathbf{m}^{\varepsilon,N}(0)=\mathbf {m}^{N}(0)\), and since \(\mathbf{m}^{N}(0)\) has the same components as \(\mathbf{m}_{0}\) in the basis \(\lbrace f_{i}\rbrace_{i\in\mathbb{N}}\) and \(\mathbf{m}_{0}\in \mathbb{H} ^{1}(\Omega)\), \(\|\mathbf{m}_{0}\|_{\mathbb{H}^{1}(\Omega)}\leq C_{3}\) with \(C_{3}>0\) is a constant independent of ε and N. Hence
$$ \bigl\Vert \mathbf{m}^{N}(0)\bigr\Vert _{\mathbb{H}^{1}(\Omega)}\leq C_{3}. $$
Therefore
$$ \bigl\Vert \nabla\mathbf{m}^{N}(0)\bigr\Vert _{\mathbb{L}^{2}(\Omega)}\leq C_{3}. $$
Thus for ε fixed, we have
$$\begin{aligned}& \bigl(\bigl|\mathbf{m}^{\varepsilon,N}\bigr|^{2}-1\bigr)_{N}\quad \mbox{is bounded in } L^{\infty }\bigl(0,T,L^{2}(\Omega) \bigr), \\& \bigl(\nabla\mathbf{m}^{\varepsilon,N}\bigr)_{N} \quad \mbox{is bounded in } L^{\infty }\bigl(0,T,\mathbb{L}^{2}(\Omega)\bigr). \end{aligned}$$
By the Young inequality
$$ \int_{\Omega}\bigl\vert \mathbf{m}^{\varepsilon,N}\bigr\vert ^{2} \,\mathrm{d}x \leq C+ \int _{\Omega}\bigl(\bigl\vert \mathbf{m}^{\varepsilon,N}\bigr\vert ^{2}-1\bigr)^{2} \,\mathrm{d}x, $$
where C is a constant which does not depend on N. Therefore
$$\begin{aligned}& \bigl(\mathbf{m}^{\varepsilon,N}\bigr)_{N}\quad \mbox{is bounded in } L^{\infty }\bigl(0,T,\mathbb{H} ^{1}(\Omega)\bigr), \\& \bigl(\partial _{t}\mathbf{m}^{\varepsilon,N} \bigr)_{N} \quad \mbox{is bounded in } L^{2}\bigl(0,T,\mathbb{L} ^{2}( \Omega)\bigr):=\mathbb{L}^{2}(Q). \end{aligned}$$
Then we have the following convergences for a subsequence further noted \(\mathbf{m}^{\varepsilon,N}\) for any \(1< p<\infty\):
$$\begin{aligned}& \mathbf{m}^{\varepsilon,N}\rightharpoonup\mathbf{m}^{\varepsilon} \quad \mbox{weakly in } L^{p}\bigl(0,T,\mathbb{H}^{1}(\Omega) \bigr), \end{aligned}$$
(18)
$$\begin{aligned}& \mathbf{m}^{\varepsilon,N}\longrightarrow\mathbf{m}^{\varepsilon} \quad \mbox{strongly in } L^{2}\bigl(0,T,\mathbb{L}^{2}( \Omega)\bigr) \mbox{ and a.e.}, \end{aligned}$$
(19)
$$\begin{aligned}& \partial _{t}\mathbf{m}^{\varepsilon,N} \rightharpoonup \partial _{t}\mathbf{m} ^{\varepsilon} \quad \mbox{weakly in } L^{2}\bigl(0,T,\mathbb{L}^{2}(\Omega)\bigr), \end{aligned}$$
(20)
$$\begin{aligned}& \bigl\vert \mathbf{m}^{\varepsilon,N}\bigr\vert ^{2}-1\rightharpoonup\zeta\quad \mbox{weakly in } L^{p} \bigl(0,T,L^{2}(\Omega)\bigr). \end{aligned}$$
(21)
The convergence (19) is a consequence of (18) and the compactness embedding of \(L^{2}(0,T,\mathbb{H}^{1}(\Omega))\) in \(L^{2}(0,T,\mathbb{L} ^{2}(\Omega))\). On the other hand \(\zeta=|\mathbf{m}^{\varepsilon }|^{2}-1\). This is provided by the following lemma.
Lemma 2
Let Θ be a bounded open subset of
\(\mathbb{R}^{d}_{x}\times \mathbb{R}_{t}\), \(h_{n}\), and
h
are functions of
\(L^{q}(\Theta)\)
with
\(1< q<\infty\)
such as
\(\|h_{n}\|_{L^{q}(\Theta)}\leq C\), \(h_{n}\longrightarrow h\)
a.e. in Θ; then
\(h_{n}\rightharpoonup h\)
weakly in
\(L^{q}(\Theta)\).
The proof of Lemma 2 can be found in [17]. In our case \(\Theta=Q\), \(h_{N}=|\mathbf{m}^{\varepsilon,N}|^{2}-1\), \(h=|\mathbf{m} ^{\varepsilon}|^{2}-1\), and \(q=2\), and from (19) \(|\mathbf{m} ^{\varepsilon,N}|^{2}-1\longrightarrow|\mathbf{m}^{\varepsilon}|^{2}-1\) a.e., and we have in particular \(|\mathbf{m}^{\varepsilon,N}|^{2}-1 \in L^{2}(\Theta)\), \(|\mathbf{m}^{\varepsilon}|^{2}-1 \in L^{2}(\Theta )\), and \(\||\mathbf{m}^{\varepsilon}|^{2}-1\|_{L^{2}(\Theta)}\leq C\).
Now, we pass to the limit as \(N\rightarrow\infty\). Multiplying the equation (12) by \({\boldsymbol{\varphi}} \in{\mathcal{C}}^{\infty }(\overline{Q})\) and integrating on Q,
$$\begin{aligned}& \int_{Q}\partial _{t}\mathbf{m}^{\varepsilon,N} \times\mathbf {m}^{\varepsilon ,N}\cdot{\boldsymbol{\varphi}} \,\mathrm{d}x\, \mathrm{d}t -\beta \int _{Q}\nabla\mathbf{m} ^{\varepsilon,N}\cdot\nabla{ \boldsymbol{\varphi}} \,\mathrm {d}x\,\mathrm{d}t -\alpha \int_{Q}\partial _{t}\mathbf{m}^{\varepsilon,N} \cdot {\boldsymbol{\varphi}} \,\mathrm{d}x\,\mathrm{d}t \\& \quad {}-\lambda \int_{Q}\mathbf {m}^{\varepsilon,N}\times \nabla \mathbf{m}^{\varepsilon,N}\cdot\nabla{\boldsymbol{\varphi} } \,\mathrm{d}x\, \mathrm{d}t - \frac{1}{\varepsilon} \int_{Q} \bigl(\bigl\vert \mathbf{m}^{\varepsilon ,N}\bigr\vert ^{2}-1 \bigr)\mathbf{m}^{\varepsilon,N}\cdot{\boldsymbol{\varphi} } \,\mathrm{d}x\,\mathrm{d}t =0. \end{aligned}$$
(22)
We have
$$ \partial _{t}\mathbf{m}^{\varepsilon,N} \rightharpoonup\partial _{t}\mathbf{m}^{\varepsilon}\quad \mbox{weakly in } \mathbb{L}^{2}(Q) $$
and
$$ \mathbf{m}^{\varepsilon,N}\longrightarrow\mathbf{m}^{\varepsilon}\quad \mbox{strongly in } L^{2}\bigl(0,T,\mathbb{L}^{2}(\Omega) \bigr). $$
Thus
$$ \int_{Q}\partial _{t}\mathbf{m}^{\varepsilon,N} \times\mathbf {m}^{\varepsilon,N}\cdot {\boldsymbol{\varphi}} \,\mathrm{d}x\, \mathrm{d}t \longrightarrow \int _{Q}\partial _{t}\mathbf{m}^{\varepsilon} \times\mathbf{m}^{\varepsilon}\cdot {\boldsymbol{\varphi}} \,\mathrm{d}x\, \mathrm{d}t. $$
On the other hand
$$ \nabla\mathbf{m}^{\varepsilon,N}\rightharpoonup\nabla\mathbf {m}^{\varepsilon} \quad \mbox{weakly in } \mathbb{L}^{2}(Q). $$
Therefore
$$ \int_{Q}\nabla\mathbf{m}^{\varepsilon,N}\cdot\nabla { \boldsymbol{\varphi}} \,\mathrm{d}x\,\mathrm{d}t \longrightarrow \int_{Q}\nabla\mathbf{m}^{\varepsilon}\cdot \nabla { \boldsymbol{\varphi}} \,\mathrm{d}x\,\mathrm{d}t $$
and
$$ \int_{Q}\mathbf{m}^{\varepsilon,N}\times\nabla\mathbf {m}^{\varepsilon,N}\cdot \nabla{\boldsymbol{\varphi}} \,\mathrm{d}x\,\mathrm{d}t \longrightarrow- \int_{Q}\mathbf{m}^{\varepsilon}\times\nabla \mathbf{m} ^{\varepsilon}\cdot\nabla{\boldsymbol{\varphi}} \,\mathrm {d}x\,\mathrm{d}t, $$
and from (20)
$$ \int_{Q}\partial _{t}\mathbf{m}^{\varepsilon,N} \cdot{\boldsymbol{\varphi}} \,\mathrm{d} x\,\mathrm{d}t \longrightarrow \int_{Q}\partial_{t}\mathbf {m}^{\varepsilon}\cdot {\boldsymbol{\varphi}} \,\mathrm{d}x\,\mathrm{d}t. $$
Taking into account (21), we have
$$ \int_{Q} \bigl(\bigl\vert \mathbf{m}^{\varepsilon,N}\bigr\vert ^{2}-1 \bigr)\mathbf {m}^{\varepsilon ,N}\cdot{\boldsymbol{ \varphi}} \,\mathrm{d}x\,\mathrm{d}t \longrightarrow \int_{Q} \bigl(\bigl\vert \mathbf{m}^{\varepsilon}\bigr\vert ^{2}-1 \bigr)\mathbf{m}^{\varepsilon}\cdot {\boldsymbol{ \varphi} } \,\mathrm{d}x\,\mathrm{d}t. $$
Using the previous convergences and passing to the limit (\(N\rightarrow \infty\)) in (22), we get
$$\begin{aligned}& \int_{Q}\partial _{t}\mathbf{m}^{\varepsilon} \times\mathbf {m}^{\varepsilon}\cdot {\boldsymbol{\varphi}} \,\mathrm{d}x\, \mathrm{d}t -\beta \int _{Q}\nabla\mathbf{m}^{\varepsilon }\cdot\nabla{ \boldsymbol{\varphi}} \,\mathrm{d}x\,\mathrm{d}t -\lambda \int_{Q}\mathbf{m} ^{\varepsilon}\times\nabla \mathbf{m}^{\varepsilon}\cdot\nabla {\boldsymbol{\varphi}} \,\mathrm{d}x\, \mathrm{d}t \\& \quad {}- \alpha \int_{Q}\partial _{t}\mathbf{m}^{\varepsilon} \cdot {\boldsymbol{\varphi}} \,\mathrm{d}x\,\mathrm{d}t -\frac{1}{\varepsilon} \int_{Q} \bigl(\bigl\vert \mathbf{m} ^{\varepsilon}\bigr\vert ^{2}-1 \bigr)\mathbf{m}^{\varepsilon}\cdot{\boldsymbol{ \varphi}} \,\mathrm{d}x\,\mathrm{d}t =0 \end{aligned}$$
(23)
for all φ in \({\mathcal{C}}^{\infty}(\overline {Q})\), and this relation holds for all \({\boldsymbol{\varphi}}\in\mathbb {H}^{1}(Q)\) by a density argument. Inequality (10) follows from (17), and Theorem 1 is now completely proved. □
2.2 Convergence of the approximate solutions
To pass to the limit as \(\varepsilon\rightarrow0\), we need the first estimate (17) and the following lemma.
Lemma 3
If
\(\mathbf{m}^{\varepsilon}\)
satisfies (23) then
\(|\mathbf{m} ^{\varepsilon}|\leq\)1 a.e. in
Q.
Proof
Noting that
$$ \int_{\Omega}\mathbf{g}\cdot{\boldsymbol{\varphi}} \,\mathrm {d}x = \int_{\lbrace |\mathbf{m}^{\varepsilon}|^{2}>1\rbrace}\mathbf{g}\cdot {\boldsymbol{\varphi}} \,\mathrm{d}x + \int_{\lbrace|\mathbf{m}^{\varepsilon}|^{2}\leq1\rbrace }\mathbf{g}\cdot{\boldsymbol{\varphi}} \,\mathrm{d}x $$
for all g, φ in \(\mathbb {L}^{2}(\Omega)\).
So if we choose
$$ {\boldsymbol{\varphi}}= \bigl(\max\bigl(\bigl\vert \mathbf{m}^{\varepsilon } \bigr\vert ^{2},1\bigr)-1 \bigr)\mathbf{m} ^{\varepsilon} $$
we have
$$ \left \{ \textstyle\begin{array}{l@{\quad}l} {\boldsymbol{\varphi}}=0, & \mbox{if } |\mathbf{m}^{\varepsilon }|^{2}\leq1, \\ {\boldsymbol{\varphi}}=(|\mathbf{m}^{\varepsilon}|^{2}-1)\mathbf {m}^{\varepsilon},& \mbox{if } |\mathbf{m}^{\varepsilon}|^{2}>1. \end{array}\displaystyle \right . $$
Let \(A=\lbrace|\mathbf{m}^{\varepsilon}|^{2}>1\rbrace\), then (23) becomes
$$\begin{aligned}& - \beta \int_{0}^{T} \int_{A}\nabla\mathbf{m}^{\varepsilon}\cdot \nabla \bigl( \bigl(\bigl|\mathbf{m}^{\varepsilon}\bigr|^{2}-1\bigr)\mathbf{m}^{\varepsilon} \bigr) \,\mathrm{d} x\,\mathrm{d}t - \alpha \int_{0}^{T} \int_{A}\bigl(\partial _{t}\mathbf{m}^{\varepsilon} \cdot\mathbf{m} ^{\varepsilon}\bigr) \bigl(\bigl|\mathbf{m}^{\varepsilon}\bigr|^{2}-1 \bigr) \,\mathrm {d}x\,\mathrm{d}t \\& \quad {}- \frac{1}{\varepsilon} \int_{0}^{T} \int_{A}\bigl(\bigl|\mathbf {m}^{\varepsilon }\bigr|^{2}-1 \bigr)^{2}\bigl|\mathbf{m}^{\varepsilon}\bigr|^{2} \,\mathrm{d}x\, \mathrm{d}t =0. \end{aligned}$$
That is,
$$\begin{aligned}& - \frac{\beta}{2} \int_{0}^{T} \int_{A}\bigl\vert \nabla\bigl(\bigl\vert \mathbf {m}^{\varepsilon }\bigr\vert ^{2}-1\bigr)\bigr\vert ^{2} \,\mathrm{d}x\,\mathrm{d}t -\beta \int_{0}^{T} \int_{A}\bigl(\bigl\vert \mathbf{m}^{\varepsilon }\bigr\vert ^{2}-1\bigr)\bigl\vert \nabla \mathbf{m}^{\varepsilon}\bigr\vert ^{2} \,\mathrm{d}x\,\mathrm{d}t \\& \quad {}- \frac{\alpha}{4} \int_{0}^{T}\frac{\mathrm{d}}{\mathrm{d}t } \int _{A}\bigl(\bigl\vert \mathbf{m}^{\varepsilon}\bigr\vert ^{2}-1\bigr)^{2} \,\mathrm{d}x\,\mathrm{d}t - \frac {1}{\varepsilon} \int_{0}^{T} \int_{A}\bigl(\bigl\vert \mathbf{m}^{\varepsilon }\bigr\vert ^{2}-1\bigr)^{2}\bigl\vert \mathbf{m}^{\varepsilon} \bigr\vert ^{2} \,\mathrm{d}x\,\mathrm{d}t =0. \end{aligned}$$
Then
$$ \frac{\alpha}{4} \int_{0}^{T}\frac{\mathrm{d}}{\mathrm{d}t } \int _{A}\bigl(\bigl\vert \mathbf{m}^{\varepsilon}\bigr\vert ^{2}-1\bigr)^{2} \,\mathrm{d}x\,\mathrm{d}t \leq0. $$
Therefore
$$ \int_{A}\bigl(\bigl\vert \mathbf{m}^{\varepsilon}\bigr\vert ^{2}-1\bigr)^{2}(T) \,\mathrm{d}x \leq \int _{A}\bigl(\bigl\vert \mathbf{m}^{\varepsilon}\bigr\vert ^{2}-1\bigr)^{2}(0) \,\mathrm{d}x, $$
and as \(| \mathbf{m}^{\varepsilon}(0)|=1\), we get
$$ \int_{A}\bigl(\bigl\vert \mathbf{m}^{\varepsilon}\bigr\vert ^{2}-1\bigr)^{2}(T) \,\mathrm{d}x \leq0, $$
which implies that \(|\mathbf{m}^{\varepsilon}|\leq\)1 a.e. on Q. □
Now we will look for an estimate of the term \(\mathbf{m}^{\varepsilon }\times \Delta\mathbf{m}^{\varepsilon}\). Going back to (17) and taking into account the previous convergences in N, we get
$$\begin{aligned}& \frac{\beta+\alpha\lambda}{2} \int_{\Omega}\bigl\vert \nabla\mathbf{m} ^{\varepsilon}\bigr\vert ^{2}(t) \,\mathrm{d}x +\alpha \int_{0}^{t} \int_{\Omega} \bigl\vert \partial _{t} \mathbf{m}^{\varepsilon} \bigr\vert ^{2} \,\mathrm{d}x\,\mathrm{d}t \\& \quad {}+ \frac{1}{4\varepsilon}\biggl(1+\frac{\alpha\lambda}{\beta}\biggr) \int _{\Omega}\bigl(\bigl\vert \mathbf{m}^{\varepsilon}\bigr\vert ^{2}-1\bigr)^{2}(t) \,\mathrm{d}x \leq \frac{\beta+\alpha\lambda}{2} \int_{\Omega} \vert \nabla\mathbf{m} _{0}\vert ^{2} \,\mathrm{d}x \end{aligned}$$
(24)
for all \(t\in(0,T)\).
Thus
$$\begin{aligned}& \bigl(\partial _{t}\mathbf{m}^{\varepsilon} \bigr)_{\varepsilon} \quad \mbox{is bounded in } \mathbb{L}^{2}(Q), \\& \bigl(\bigl\vert \mathbf{m}^{\varepsilon}\bigr\vert ^{2}-1 \bigr)_{\varepsilon} \quad \mbox{is bounded in } L^{\infty}\bigl(0,T, L^{2}(\Omega)\bigr), \\& \bigl(\mathbf{m}^{\varepsilon}\bigr)_{\varepsilon}\quad \mbox{is bounded in } L^{\infty }\bigl(0,T, \mathbb{H}^{1}(\Omega)\bigr). \end{aligned}$$
Multiplying equation (8) by \(\mathbf{m}^{\varepsilon}\times \partial _{t}\mathbf{m}^{\varepsilon} \) and integrating over Ω, we get
$$\begin{aligned} \begin{aligned}[b] &-\int_{\Omega}\bigl\vert \mathbf{m}^{\varepsilon}\times\partial _{t}\mathbf {m}^{\varepsilon } \bigr\vert ^{2} \, \mathrm{d}x +\beta \int_{\Omega}\Delta\mathbf {m}^{\varepsilon}\cdot \mathbf{m}^{\varepsilon}\times\partial _{t}\mathbf{m}^{\varepsilon} \,\mathrm{d}x \\ &\quad {}+\lambda \int_{\Omega}\mathbf{m}^{\varepsilon}\times\Delta \mathbf{m} ^{\varepsilon}\cdot\mathbf{m}^{\varepsilon}\times\partial _{t} \mathbf{m}^{\varepsilon } \,\mathrm{d}x =0. \end{aligned} \end{aligned}$$
(25)
Multiply this time equation (8) by \(\mathbf{m}^{\varepsilon }\times \Delta\mathbf{m}^{\varepsilon}\) and integrating over Ω we obtain
$$\begin{aligned}& -\alpha \int_{\Omega}\mathbf{m}^{\varepsilon}\times\Delta \mathbf{m} ^{\varepsilon}\cdot\partial _{t}\mathbf{m}^{\varepsilon} \,\mathrm {d}x - \int_{\Omega}\mathbf{m}^{\varepsilon}\times\Delta\mathbf {m}^{\varepsilon }\cdot\mathbf{m}^{\varepsilon}\times\partial _{t} \mathbf {m}^{\varepsilon} \,\mathrm{d}x \\& \quad {}+\lambda \int_{\Omega}\bigl\vert \mathbf{m}^{\varepsilon}\times\Delta \mathbf{m} ^{\varepsilon}\bigr\vert ^{2} \,\mathrm{d}x =0. \end{aligned}$$
(26)
Multiplying equation (26) by λ and taking the sum with (25), we get
$$ - \int_{\Omega}\bigl\vert \mathbf{m}^{\varepsilon}\times\partial _{t}\mathbf {m}^{\varepsilon } \bigr\vert ^{2} \, \mathrm{d}x +(\beta+\alpha\lambda) \int_{\Omega}\Delta \mathbf{m} ^{\varepsilon}\cdot \mathbf{m}^{\varepsilon}\times\partial _{t}\mathbf{m}^{\varepsilon } \,\mathrm{d}x +\lambda^{2} \int_{\Omega}\bigl\vert \mathbf{m}^{\varepsilon}\times\Delta \mathbf{m} ^{\varepsilon}\bigr\vert ^{2} \,\mathrm{d}x =0. $$
Then
$$ \lambda^{2} \int_{\Omega}\bigl\vert \mathbf{m}^{\varepsilon}\times\Delta \mathbf{m} ^{\varepsilon}\bigr\vert ^{2} \,\mathrm{d}x = \int_{\Omega}\bigl\vert \mathbf {m}^{\varepsilon }\times\partial _{t}\mathbf{m}^{\varepsilon} \bigr\vert ^{2} \, \mathrm{d}x -(\beta+\alpha \lambda) \int_{\Omega}\Delta\mathbf{m}^{\varepsilon}\cdot\mathbf{m} ^{\varepsilon}\times\partial _{t}\mathbf{m}^{\varepsilon} \, \mathrm{d}x. $$
(27)
Multiplying (8) by \(\partial_{t}\mathbf{m}^{\varepsilon}\), integrating over Ω, replacing \(\int_{\Omega}\Delta\mathbf{m} ^{\varepsilon}\cdot\mathbf{m}^{\varepsilon}\times\partial _{t}\mathbf{m}^{\varepsilon } \,\mathrm{d}x\) by its value in (27) and using Lemma 3, we have
$$\begin{aligned}& \lambda^{2} \int_{\Omega}\bigl\vert \mathbf{m}^{\varepsilon}\times\Delta \mathbf{m} ^{\varepsilon}\bigr\vert ^{2} \,\mathrm{d}x \\& \quad = \int_{\Omega}\bigl\vert \mathbf {m}^{\varepsilon }\times\partial _{t}\mathbf{m}^{\varepsilon} \bigr\vert ^{2} \, \mathrm{d}x +\frac{\alpha(\beta +\alpha\lambda)}{\lambda} \int_{\Omega}\bigl\vert \partial _{t}\mathbf {m}^{\varepsilon} \bigr\vert ^{2} \,\mathrm{d}x \\& \qquad {}+ \frac{\beta(\beta+\alpha\lambda)}{2\lambda}\frac{\mathrm{d} }{\mathrm{d}t } \int_{\Omega}\bigl\vert \nabla\mathbf{m}^{\varepsilon}\bigr\vert ^{2} \,\mathrm{d}x + \frac{(\beta+\alpha\lambda)}{4\varepsilon\lambda}\frac{\mathrm{d} }{\mathrm{d}t } \int_{\Omega}\bigl(\bigl\vert \mathbf{m}^{\varepsilon }\bigr\vert ^{2}-1\bigr)^{2} \,\mathrm{d}x \\& \quad \leq \int_{\Omega}\bigl\vert \mathbf{m}^{\varepsilon}\bigr\vert ^{2}\bigl\vert \partial _{t}\mathbf{m} ^{\varepsilon} \bigr\vert ^{2} \,\mathrm{d}x +\frac{\alpha(\beta+\alpha \lambda )}{\lambda} \int_{\Omega}\bigl\vert \partial _{t} \mathbf{m}^{\varepsilon} \bigr\vert ^{2} \,\mathrm{d}x + \frac{\beta(\beta+\alpha\lambda)}{2\lambda}\frac{\mathrm{d} }{\mathrm{d}t } \int_{\Omega}\bigl\vert \nabla\mathbf{m}^{\varepsilon}\bigr\vert ^{2} \,\mathrm{d}x \\& \qquad {}+ \frac{(\beta+\alpha\lambda)}{4\varepsilon\lambda}\frac {\mathrm{d}}{\mathrm{d}t } \int_{\Omega}\bigl(\bigl\vert \mathbf{m}^{\varepsilon }\bigr\vert ^{2}-1\bigr)^{2} \,\mathrm{d}x \\& \quad \leq \biggl(1+\frac{\alpha(\beta+\alpha\lambda)}{\lambda}\biggr) \int _{\Omega}\bigl\vert \partial _{t} \mathbf{m}^{\varepsilon} \bigr\vert ^{2} \,\mathrm{d}x + \frac{\beta(\beta+\alpha\lambda)}{2\lambda}\frac{\mathrm{d} }{\mathrm{d}t } \int_{\Omega}\bigl\vert \nabla\mathbf{m}^{\varepsilon}\bigr\vert ^{2} \,\mathrm{d}x \\& \qquad {}+ \frac{(\beta+\alpha\lambda)}{4\varepsilon\lambda}\frac {\mathrm{d}}{\mathrm{d}t } \int_{\Omega}\bigl(\bigl\vert \mathbf{m}^{\varepsilon }\bigr\vert ^{2}-1\bigr)^{2} \,\mathrm{d}x. \end{aligned}$$
We integrate from 0 to t, and using (24), we get
$$ \lambda^{2} \int_{0}^{t} \int_{\Omega}\bigl\vert \mathbf{m}^{\varepsilon }\times \Delta \mathbf{m}^{\varepsilon}\bigr\vert ^{2} \,\mathrm{d}x\,\mathrm{d}t \leq C, $$
(28)
where C is a constant independent of ε. Hence
$$ \bigl(\mathbf{m}^{\varepsilon}\times\Delta\mathbf{m}^{\varepsilon } \bigr)_{\varepsilon}\quad \mbox{is bounded in } \mathbb{L}^{2}(Q). $$
(29)
Up to a subsequence, we have the following convergences for \(1< p<\infty\):
$$\begin{aligned}& \mathbf{m}^{\varepsilon}\rightharpoonup\mathbf{m}\quad \mbox{weakly in } L^{p}\bigl(0,T, \mathbb{H} ^{1}(\Omega)\bigr), \\& \partial _{t}\mathbf{m}^{\varepsilon} \rightharpoonup\partial _{t}\mathbf{m}\quad \mbox{weakly in } \mathbb{L}^{2}(Q), \\& \bigl\vert \mathbf{m}^{\varepsilon}\bigr\vert ^{2}-1 \longrightarrow0 \quad \mbox{strongly in } L^{2}\bigl(0,T,L^{2}(\Omega)\bigr) \mbox{ and } \vert \mathbf{m}\vert =1 \mbox{ a.e.}, \\& \mathbf{m}^{\varepsilon}\times\Delta\mathbf{m}^{\varepsilon } \rightharpoonup \chi\quad \mbox{weakly in } \mathbb{L}^{2}(Q). \end{aligned}$$
(30)
By the compactness embedding of \(\mathbb{H}^{1}(Q)\) into \(\mathbb {L}^{q}(Q)\) with \(2\leq q<6\), we have
$$ \mathbf{m}^{\varepsilon}\longrightarrow\mathbf{m}\quad \mbox{strongly in } \mathbb{L} ^{2}(Q)\mbox{ and in } \mathbb{L}^{4}(Q). $$
(31)
In the following, we show
$$ \chi=\mathbf{m}\times\Delta\mathbf{m}\in\mathbb{L}^{2}(Q). $$
(32)
Letting \({\boldsymbol{\varphi}}\in\mathbb{H}^{1}(Q)\), using the Green formula,
$$ \int_{Q}\mathbf{m}^{\varepsilon}\times\Delta\mathbf {m}^{\varepsilon}\cdot{\boldsymbol{\varphi}} \,\mathrm{d}x\,\mathrm{d}t=- \int_{Q}\mathbf{m}^{\varepsilon }\times\nabla \mathbf{m}^{\varepsilon}\cdot\nabla{\boldsymbol{\varphi}} \,\mathrm {d}x \, \mathrm{d}t . $$
By the previous convergences,
$$\begin{aligned} \int_{Q}\mathbf{m}^{\varepsilon}\times\nabla\mathbf {m}^{\varepsilon}\cdot \nabla{\boldsymbol{\varphi}} \,\mathrm{d}x \,\mathrm {d}t \longrightarrow& \int_{Q}\mathbf{m} \times\nabla\mathbf{m}\cdot\nabla{ \boldsymbol{\varphi}} \,\mathrm {d}x \,\mathrm{d}t \\ =&- \int_{Q}\mathbf{m}\times\Delta\mathbf{m}\cdot{\boldsymbol{ \varphi}} \,\mathrm{d}x\,\mathrm{d}t, \end{aligned}$$
and therefore (32) is proved. In particular, we have
$$ \mathbf{m}^{\varepsilon}\times\Delta\mathbf{m}^{\varepsilon }\rightharpoonup \mathbf{m} \times\Delta\mathbf{m} \quad \mbox{weakly in } \mathbb{L}^{2}(Q). $$
Now going back to (23) and taking \({\boldsymbol{\varphi} }=\mathbf{m} ^{\varepsilon}\times{\boldsymbol{\phi}}\) with \({\boldsymbol{\phi} }\in{\mathcal{C}}^{\infty}(\overline{Q})\), we get
$$\begin{aligned}& \int_{Q}\partial _{t}\mathbf{m}^{\varepsilon} \times\mathbf {m}^{\varepsilon}\cdot \mathbf{m}^{\varepsilon}\times{\boldsymbol{ \phi}} \,\mathrm {d}x\,\mathrm{d}t +\beta \int _{Q}\mathbf{m}^{\varepsilon}\times\nabla \mathbf{m}^{\varepsilon }\cdot\nabla {\boldsymbol{\phi}} \,\mathrm{d}x\, \mathrm{d}t \\& \quad {}+\lambda \int_{Q}\mathbf{m}^{\varepsilon}\times\Delta\mathbf{m} ^{\varepsilon}\cdot\mathbf{m}^{\varepsilon}\times{\boldsymbol{\phi}} \, \mathrm{d} x\,\mathrm{d}t -\alpha \int_{Q}\partial _{t}\mathbf{m}^{\varepsilon} \cdot\mathbf{m} ^{\varepsilon}\times{\boldsymbol{\phi}} \,\mathrm{d}x\, \mathrm{d}t =0. \end{aligned}$$
(33)
For the first term of (33), we set \(D_{\varepsilon}=\int _{Q}\partial _{t}\mathbf{m}^{\varepsilon} \times\mathbf {m}^{\varepsilon}\cdot\mathbf{m} ^{\varepsilon}\times{\boldsymbol{\phi}} \,\mathrm{d}x\,\mathrm{d}t \). We have
$$ D_{\varepsilon}= \int_{Q}\bigl(\mathbf{m}^{\varepsilon}\cdot {\boldsymbol{ \phi}}\bigr)\mathbf{m} ^{\varepsilon}\cdot\partial _{t} \mathbf{m}^{\varepsilon} \,\mathrm {d}x\,\mathrm{d}t- \int _{Q}\bigl\vert \mathbf{m}^{\varepsilon}\bigr\vert ^{2}\partial _{t}\mathbf {m}^{\varepsilon} \cdot{ \boldsymbol{\phi} } \,\mathrm{d}x\,\mathrm{d}t. $$
On the one hand
$$\begin{aligned} \int_{Q}\bigl\vert \mathbf{m}^{\varepsilon}\bigr\vert ^{2}\partial _{t}\mathbf {m}^{\varepsilon} \cdot { \boldsymbol{\phi}} \,\mathrm{d}x\,\mathrm{d}t =& \int_{Q}\bigl(\bigl\vert \mathbf {m}^{\varepsilon}\bigr\vert ^{2}-1\bigr)\partial _{t}\mathbf{m}^{\varepsilon} \cdot{\boldsymbol{\phi}} \,\mathrm {d}x\,\mathrm{d}t \\ &{}+ \int_{Q}\partial _{t}\mathbf{m}^{\varepsilon} \cdot{\boldsymbol{\phi}} \,\mathrm {d}x\,\mathrm{d}t \\ \longrightarrow& \int_{Q}\partial _{t}\mathbf{m}\cdot{\boldsymbol{ \phi}} \,\mathrm{d}x\,\mathrm{d}t. \end{aligned}$$
On the other hand
$$\begin{aligned} \int_{Q}\bigl(\mathbf{m}^{\varepsilon}\cdot{\boldsymbol{ \phi}}\bigr)\mathbf {m}^{\varepsilon}\cdot \partial _{t} \mathbf{m}^{\varepsilon} \,\mathrm{d}x\,\mathrm{d}t =& \frac{1}{2} \int _{Q}\partial _{t}\bigl(\bigl\vert \mathbf{m}^{\varepsilon}\bigr\vert ^{2}-1\bigr) \mathbf {m}^{\varepsilon}\cdot{\boldsymbol{\phi}} \,\mathrm{d}x\,\mathrm{d}t \\ =& \frac{1}{2} \biggl[ \int_{\Omega}\bigl(\bigl\vert \mathbf{m}^{\varepsilon }\bigr\vert ^{2}-1\bigr)\mathbf{m} ^{\varepsilon}\cdot{\boldsymbol{\phi}} \,\mathrm{d}x \biggr]_{0}^{T} \\ &{} -\frac{1}{2} \int_{Q}\bigl(\bigl\vert \mathbf{m}^{\varepsilon}\bigr\vert ^{2}-1\bigr)\partial _{t}\bigl(\mathbf{m} ^{\varepsilon}\cdot{\boldsymbol{\phi}}\bigr) \,\mathrm{d}x\,\mathrm{d}t. \end{aligned}$$
We choose ϕ so that \({\boldsymbol{\phi}}=0\) in \(t=0\) and \(t=T\); then
$$ \biggl[ \int_{\Omega}\bigl(\bigl\vert \mathbf{m}^{\varepsilon}\bigr\vert ^{2}-1\bigr)\mathbf {m}^{\varepsilon }\cdot{\boldsymbol{\phi}} \,\mathrm{d}x \biggr]_{0}^{T}=0. $$
Therefore
$$\begin{aligned} \int_{Q}\bigl(\mathbf{m}^{\varepsilon}\cdot{\boldsymbol{ \phi}}\bigr)\mathbf {m}^{\varepsilon}\cdot \partial _{t} \mathbf{m}^{\varepsilon} \,\mathrm{d}x\,\mathrm{d}t =& -\frac{1}{2} \int _{Q}\bigl(\bigl\vert \mathbf{m}^{\varepsilon}\bigr\vert ^{2}-1\bigr)\partial _{t}\bigl(\mathbf {m}^{\varepsilon}\cdot{\boldsymbol{\phi}}\bigr) \,\mathrm{d}x\,\mathrm{d}t \\ =& -\frac{1}{2} \int_{Q}\bigl(\bigl\vert \mathbf{m}^{\varepsilon}\bigr\vert ^{2}-1\bigr)\partial _{t}\mathbf{m} ^{\varepsilon} \cdot{\boldsymbol{\phi}} \,\mathrm{d}x\,\mathrm{d}t \\ & -\frac{1}{2} \int_{Q}\bigl(\bigl\vert \mathbf{m}^{\varepsilon}\bigr\vert ^{2}-1\bigr)\mathbf{m} ^{\varepsilon}\cdot\partial _{t}{\boldsymbol{\phi}} \,\mathrm {d}x\,\mathrm{d}t \longrightarrow0. \end{aligned}$$
Hence
$$ D_{\varepsilon}\longrightarrow- \int_{Q}\partial _{t}\mathbf{m}\cdot {\boldsymbol{ \phi}} \,\mathrm{d}x\,\mathrm{d}t. $$
For the second term of (33), we have
$$ \beta \int_{Q}\mathbf{m}^{\varepsilon}\times\nabla\mathbf {m}^{\varepsilon }\cdot\nabla{\boldsymbol{\phi}} \,\mathrm{d}x\,\mathrm{d}t \longrightarrow \beta \int_{Q}\mathbf{m}\times\nabla\mathbf{m}\cdot\nabla{ \boldsymbol{\phi}} \,\mathrm{d}x\,\mathrm{d}t. $$
For the third term of (33), we get
$$ \lambda \int_{Q}\mathbf{m}^{\varepsilon}\times\Delta\mathbf {m}^{\varepsilon }\cdot\mathbf{m}^{\varepsilon}\times{\boldsymbol{\phi}} \, \mathrm {d}x\,\mathrm{d}t \longrightarrow\lambda \int_{Q}\mathbf{m}\times\Delta\mathbf {m}\cdot\mathbf{m}\times {\boldsymbol{\phi}} \,\mathrm{d}x\,\mathrm{d}t. $$
For the last term of (33), we have
$$ \int_{Q}\partial _{t}\mathbf{m}^{\varepsilon} \cdot\mathbf {m}^{\varepsilon}\times {\boldsymbol{\phi}} \,\mathrm{d}x\, \mathrm{d}t \longrightarrow \int _{Q}\partial _{t}\mathbf{m}\cdot \mathbf{m} \times{\boldsymbol{\phi}} \,\mathrm{d}x\,\mathrm{d}t. $$
Letting ε tend to 0 in (33), we get
$$\begin{aligned}& - \int_{Q}\partial _{t}\mathbf{m}\cdot{\boldsymbol{ \phi}} \,\mathrm {d}x\,\mathrm{d}t +\beta \int _{Q}\mathbf{m}\times\nabla\mathbf{m}\cdot\nabla{ \boldsymbol{\phi} } \,\mathrm{d}x\,\mathrm{d}t \\& \quad {}+\lambda \int_{Q}\mathbf{m}\times\Delta\mathbf{m}\cdot\mathbf {m} \times{\boldsymbol{\phi} } \,\mathrm{d}x\,\mathrm{d}t -\alpha \int_{Q}\partial _{t}\mathbf{m}\cdot\mathbf{m} \times {\boldsymbol{\phi}} \,\mathrm{d} x\,\mathrm{d}t =0 \end{aligned}$$
for all \({\boldsymbol{\phi}}\in\mathcal{C}^{\infty}(\overline{Q})\). Inequality (7) follows from (24). We proved the following.
Theorem 2
Let
\(\mathbf{m}_{0}\in\mathbb{H}^{1}(\Omega)\)
with
\(|\mathbf{m}_{0}|=1\)
a.e., then there exists a global weak solution of the problem (1) in the sense of Definition
1.