Given a positive number λ, we define
$$\begin{aligned} G_{\lambda}(x,t)=\frac{1}{(T-t+\lambda)^{d/2}}e^{-\frac{\vert x-x_{0}\vert ^{2}}{4(T-t+ \lambda)}},\quad (x,t)\in\Omega\times(0,T). \end{aligned}$$
(2.1)
Then, for each \(t\in[0,T]\), we write
$$\begin{aligned}& H_{\lambda}(t)= \int_{\Omega}\bigl(\bigl\vert y(x,t)\bigr\vert ^{2}+ \bigl\vert z(x,t)\bigr\vert ^{2}\bigr)G_{\lambda}(x,t)\,dx, \end{aligned}$$
(2.2)
$$\begin{aligned}& D_{\lambda}(t)= \int_{\Omega}\bigl(\bigl\vert \nabla y(x,t)\bigr\vert ^{2}+\bigl\vert \nabla z(x,t)\bigr\vert ^{2}\bigr)G _{\lambda}(x,t)\,dx, \end{aligned}$$
(2.3)
and
$$\begin{aligned} N_{\lambda}(t)=\frac{2D_{\lambda}(t)}{H_{\lambda}(t)}, \end{aligned}$$
(2.4)
where \((y(x,t), z(x,t))\) are the solutions of equation (1.1). The function \(N_{\lambda}(t)\) was first discussed in [11], and it was called frequency function (see also [2, 12], and [9]). Throughout this section, we always work under the assumption \(H_{\lambda}(t)\neq0\). Next, we will discuss the properties for the functions \(G_{\lambda}(x,t)\), \(H_{\lambda}(t)\), \(D_{\lambda}(t)\) and \(N_{\lambda}(t)\). Now, we first fix a positive number r and a point \(x_{0}\) in the subset ω such that \(B_{r}\subset\omega\). \(B_{r}\) denotes the open ball, centered at the point \(x_{0}\) and of radius r, in \(\mathbb{R}^{d}\). Write \(m=\sup_{x\in\Omega} \vert x-x_{0}\vert ^{2}\). Lemma 2.1 is taken from [2, 6].
Lemma 2.1
For each
\(\lambda>0\), the function
\(G_{\lambda}\)
given in (2.1) holds the following four identities over
\(\mathbb{R}^{d}\times[0,T]\):
$$\begin{aligned}& \partial_{t}G_{\lambda}(x,t)+\triangle G_{\lambda}(x,t)=0, \end{aligned}$$
(2.5)
$$\begin{aligned}& \nabla G_{\lambda}(x,t)=\frac{-(x-x_{0})}{2(T-t+\lambda)}G_{\lambda }(x,t), \end{aligned}$$
(2.6)
$$\begin{aligned}& \partial_{i}^{2}G_{\lambda}(x,t)= \frac{-1}{2(T-t+\lambda)}G_{\lambda }(x,t)+\frac{\vert x_{i}-x_{0i}\vert ^{2}}{4(T-t+\lambda )^{2}}G_{\lambda}(x,t), \end{aligned}$$
(2.7)
and, for
\(i\neq j\),
$$\begin{aligned} \partial_{i}\partial_{j}G_{\lambda}(x,t)= \frac{(x_{i}-x_{0i})(x_{j}-x _{0j})}{4(T-t+\lambda)^{2}}G_{\lambda}(x,t). \end{aligned}$$
(2.8)
Lemma 2.2
For each
\(\lambda>0\), the following identity holds for
\(t\in(0,T)\):
$$\begin{aligned} \frac{d}{dt}\ln{H_{\lambda}(t)}=-N_{\lambda}(t)+ \frac{2}{H_{\lambda }(t)} \int_{\Omega} \bigl( y(\partial_{t}y-\triangle y)+z( \partial_{t}z- \triangle z) \bigr) G_{\lambda}\,dx. \end{aligned}$$
(2.9)
Proof
By a direct computation, we can obtain
$$\begin{aligned} H'_{\lambda}(t) =&2 \int_{\Omega}(y\partial_{t}y+z\partial_{t}z)G _{\lambda}\,dx+ \int_{\Omega}\bigl(y^{2}+z^{2}\bigr) \partial_{t}G_{\lambda}\,dx \\ =&2 \int_{\Omega}(y\partial_{t}y+z\partial_{t}z)G_{\lambda}\,dx- \int_{\Omega}\bigl(y^{2}+z^{2}\bigr)\Delta G_{\lambda}\,dx \\ =&2 \int_{\Omega}(y\partial_{t}y+z\partial_{t}z)G_{\lambda}\,dx- \int_{\Omega}\Delta\bigl(y^{2}+z^{2} \bigr)G_{\lambda}\,dx \\ =&2 \int_{\Omega}(y\partial_{t}y+z\partial_{t}z)G_{\lambda}\,dx-2 \int_{\Omega} \bigl( (\nabla y)^{2}+y\triangle y+(\nabla z)^{2}+z \triangle z \bigr) G_{\lambda}\,dx \\ =&2 \int_{\Omega} \bigl( y(\partial_{t}y-\triangle y)+z( \partial_{t}z- \triangle z) \bigr) G_{\lambda}\,dx-2 \int_{\Omega} \bigl( \vert \nabla y\vert ^{2}+\vert \nabla z\vert ^{2} \bigr) G_{\lambda}\,dx \\ =&2 \int_{\Omega} \bigl( y(\partial_{t}y-\triangle y)+z( \partial_{t}z- \triangle z) \bigr) G_{\lambda}\,dx-2D_{\lambda}(t). \end{aligned}$$
(2.10)
Therefore, for each \(t\in(0,T)\), we have
$$\begin{aligned} \frac{d}{dt}\ln{H_{\lambda}(t)}=\frac{H'_{\lambda}(t)}{H_{\lambda }(t)}=-N_{\lambda}(t)+ \frac{2}{H_{\lambda}(t)} \int_{\Omega} \bigl( y( \partial_{t}y-\triangle y)+z( \partial_{t}z-\triangle z) \bigr) G_{ \lambda}\,dx. \end{aligned}$$
(2.11)
This completes the proof of this lemma. □
Lemma 2.3
For each
\(\lambda>0\)
and
\(t\in(0,T)\), the functions
\(H_{\lambda}\)
and
\(D_{\lambda}\)
defined in (2.2) and (2.3), respectively, satisfy
$$\begin{aligned} 2H'_{\lambda}(t)D_{\lambda}(t) =&- \biggl[2 \int_{\Omega}y \biggl( \partial_{t}y-\frac{x-x _{0}}{2(T-t+\lambda)} \nabla y+\frac{1}{2}(\triangle y-\partial _{t}y) \biggr) G _{\lambda}\,dx \\ &{}+2 \int_{\Omega}z \biggl( \partial_{t}z- \frac{x-x_{0}}{2(T-t+\lambda)} \nabla z+\frac{1}{2}(\triangle z- \partial_{t}z) \biggr) G_{\lambda}\,dx \biggr]^{2} \\ &{}+ \biggl( \int_{\Omega}y(\triangle y-\partial_{t}y)G_{\lambda}\,dx+ \int_{\Omega}z(\triangle z-\partial_{t}z)G_{\lambda}\,dx \biggr) ^{2}. \end{aligned}$$
(2.12)
Proof
By (2.5), (2.6), we can rewrite \(H'_{\lambda}(t)\) as follows:
$$\begin{aligned} H'_{\lambda}(t) =&2 \int_{\Omega}(y\partial_{t}y+z\partial_{t}z)G _{\lambda}\,dx- \int_{\Omega}\bigl(y^{2}+z^{2}\bigr)\triangle G_{\lambda}\,dx \\ =&2 \int_{\Omega}(y\partial_{t}y+z\partial_{t}z)G_{\lambda}\,dx+ \int_{\Omega}\nabla\bigl(y^{2}+z^{2}\bigr) \nabla G_{\lambda}\,dx \\ =&2 \int_{\Omega}(y\partial_{t}y+z\partial_{t}z)G_{\lambda}\,dx- \int_{\Omega}(2y\nabla y+2z\nabla z)\frac{x-x_{0}}{2(T-t+\lambda)}G _{\lambda}\,dx \\ =&2 \int_{\Omega}y \biggl( \partial_{t}y- \frac{x-x_{0}}{2(T-t+\lambda)} \nabla y \biggr) G_{\lambda}\,dx+2 \int_{\Omega}z \biggl( \partial_{t}z-\frac{x-x_{0}}{2(T-t+\lambda)} \nabla z \biggr) G_{\lambda}\,dx \\ =&2 \int_{\Omega}y \biggl( \partial_{t}y- \frac{x-x_{0}}{2(T-t+\lambda)} \nabla y+\frac{1}{2}(\triangle y- \partial_{t}y) \biggr) G_{\lambda}\,dx \\ &{}+2 \int_{\Omega}z \biggl( \partial_{t}z- \frac{x-x_{0}}{2(T-t+\lambda)} \nabla z+\frac{1}{2}(\triangle z- \partial_{t}z) \biggr) G_{\lambda}\,dx \\ &{}- \int_{\Omega}y(\Delta y-\partial_{t}y)G_{\lambda}\,dx- \int_{\Omega }z(\Delta z-\partial_{t}z)G_{\lambda}\,dx. \end{aligned}$$
(2.13)
It follows from (2.6) that for each \(t \in(0,T)\)
$$\begin{aligned} D_{\lambda}(t) =& \int_{\Omega}\bigl(\vert \nabla y\vert ^{2}+\vert \nabla z\vert ^{2}\bigr)G_{ \lambda}\,dx \\ =& \int_{\Omega}\nabla y\nabla yG_{\lambda}\,dx+ \int_{\Omega}\nabla z\nabla zG_{\lambda}\,dx \\ =& \int_{\Omega}\operatorname{div}(y\nabla yG_{\lambda})\,dx- \int_{\Omega}y \operatorname{div}(\nabla yG_{\lambda})\,dx+ \int_{\Omega}\operatorname{div}(z \nabla zG_{\lambda})\,dx \\ &{}- \int_{\Omega}z\operatorname{div}(\nabla zG_{\lambda })\,dx \\ =&- \int_{\Omega}y\Delta yG_{\lambda}\,dx+ \int_{\Omega}y\nabla y\frac{x-x _{0}}{2(T-t+\lambda)}G_{\lambda}\,dx- \int_{\Omega}z\Delta zG_{\lambda }\,dx \\ &{}+ \int_{\Omega}z\nabla z\frac{x-x_{0}}{2(T-t+\lambda)}G_{\lambda }\,dx \\ =&- \int_{\Omega}y \biggl( \partial_{t}y- \frac{x-x_{0}}{2(T-t+\lambda)} \nabla y+\frac{1}{2}(\triangle y- \partial_{t}y) \biggr) G_{\lambda}\,dx-\frac{1}{2} \int_{\Omega }y(\triangle y-\partial_{t}y)G_{\lambda}\,dx \\ &{}- \int_{\Omega}z \biggl( \partial_{t}z-\frac{x-x_{0}}{2(T-t+\lambda)} \nabla z+\frac{1}{2}(\triangle z-\partial_{t}z) \biggr) G_{\lambda}\,dx \\ &{}- \frac{1}{2} \int_{\Omega}z(\triangle z-\partial_{t}z)G_{\lambda}\,dx. \end{aligned}$$
(2.14)
This, combining with (2.13), shows that
$$\begin{aligned} 2H'_{\lambda}(t)D_{\lambda}(t) =&- \biggl[2 \int_{\Omega}y \biggl( \partial_{t}y-\frac{x-x _{0}}{2(T-t+\lambda)} \nabla y+\frac{1}{2}(\triangle y-\partial _{t}y) \biggr) G _{\lambda}\,dx \\ &{}+2 \int_{\Omega}z \biggl( \partial_{t}z- \frac{x-x_{0}}{2(T-t+\lambda)} \nabla z+\frac{1}{2}(\triangle z- \partial_{t}z) \biggr) G_{\lambda}\,dx \biggr]^{2} \\ &{}+ \biggl( \int_{\Omega}y(\triangle y-\partial_{t}y)G_{\lambda}\,dx+ \int_{\Omega}z(\triangle z-\partial_{t}z)G_{\lambda}\,dx \biggr) ^{2}. \end{aligned}$$
(2.15)
This completes the proof of this lemma. □
Lemma 2.4
For each
\(\lambda>0\)
and
\(t\in(0,T)\), then we have
$$\begin{aligned} D'_{\lambda}(t) =&- \int_{\partial\Omega} \vert \nabla y\vert ^{2} \partial_{ \nu}G_{\lambda}\,d\sigma+2 \int_{\partial\Omega}\partial_{\nu}y( \nabla y\nabla G_{\lambda})\,d\sigma \\ &{}- \int_{\partial\Omega} \vert \nabla z\vert ^{2}\partial _{\nu}G_{\lambda}\,d\sigma+2 \int_{\partial\Omega}\partial_{\nu}z(\nabla z\nabla G_{ \lambda})\,d\sigma \\ &{}-2 \int_{\Omega} \biggl( \partial_{t}y-\frac{x-x_{0}}{2(T-t+\lambda)} \nabla y+\frac{1}{2}(\triangle y-\partial_{t}y) \biggr) ^{2}G_{\lambda }\,dx \\ &{}-2 \int_{\Omega} \biggl( \partial_{t}z-\frac{x-x_{0}}{2(T-t+\lambda)} \nabla z+\frac{1}{2}(\triangle z-\partial_{t}z) \biggr) ^{2}G_{\lambda }\,dx \\ &{}+\frac{1}{2} \int_{\Omega}(\triangle y-\partial_{t}y)^{2}G_{\lambda }\,dx+ \frac{1}{2} \int_{\Omega}(\triangle z-\partial_{t}z)^{2}G_{\lambda }\,dx +\frac{1}{(T-t+\lambda)}D_{\lambda}(t). \end{aligned}$$
(2.16)
Proof
By a direct computation, we can derive
$$\begin{aligned} D'_{\lambda}(t) =&2 \int_{\Omega}\nabla y\partial_{t}\nabla yG_{ \lambda}\,dx+ \int_{\Omega} \vert \nabla y\vert ^{2}\partial _{t}G_{\lambda}\,dx \\ &{}+2 \int_{\Omega}\nabla z\partial_{t}\nabla zG_{\lambda}\,dx+ \int_{\Omega} \vert \nabla z\vert ^{2} \partial_{t}G_{\lambda}\,dx \\ =&2 \int_{\Omega}\nabla y\partial_{t}\nabla yG_{\lambda}\,dx- \int_{\Omega} \vert \nabla y\vert ^{2}\triangle G_{\lambda}\,dx \\ &{}+2 \int_{\Omega}\nabla z\partial_{t}\nabla zG_{\lambda}\,dx- \int_{\Omega} \vert \nabla z\vert ^{2}\triangle G_{\lambda}\,dx \\ =&2 \int_{\Omega}\operatorname{div}(\nabla y\partial_{t}yG_{\lambda})\,dx-2 \int_{\Omega}\partial_{t}y\operatorname{div}(\nabla yG_{\lambda})\,dx- \int_{\Omega} \vert \nabla y\vert ^{2}\triangle G_{\lambda}\,dx \\ &{}+2 \int_{\Omega}\operatorname{div}(\nabla z\partial_{t}zG_{\lambda})\,dx-2 \int_{\Omega}\partial_{t}z\operatorname{div}(\nabla zG_{\lambda})\,dx- \int_{\Omega} \vert \nabla z\vert ^{2}\triangle G_{\lambda}\,dx \\ =&-2 \int_{\Omega}\partial_{t}y\triangle yG_{\lambda}\,dx-2 \int_{ \Omega}\partial_{t}y\nabla y\nabla G_{\lambda}\,dx- \int_{\Omega} \vert \nabla y\vert ^{2}\triangle G_{\lambda}\,dx \\ &{}-2 \int_{\Omega}\partial_{t}z\triangle zG_{\lambda}\,dx-2 \int_{ \Omega}\partial_{t}z\nabla z\nabla G_{\lambda}\,dx- \int_{\Omega} \vert \nabla z\vert ^{2}\triangle G_{\lambda}\,dx \\ =&-2 \int_{\Omega}\partial_{t}y\triangle yG_{\lambda}\,dx+2 \int_{ \Omega}\partial_{t}y\nabla y\frac{x-x_{0}}{2(T-t+\lambda)}G_{\lambda }\,dx- \int_{\Omega} \vert \nabla y\vert ^{2}\triangle G_{\lambda}\,dx \\ &{}-2 \int_{\Omega}\partial_{t}z\triangle zG_{\lambda}\,dx+2 \int_{ \Omega}\partial_{t}z\nabla z\frac{x-x_{0}}{2(T-t+\lambda)}G_{\lambda }\,dx- \int_{\Omega} \vert \nabla z\vert ^{2}\triangle G_{\lambda}\,dx. \end{aligned}$$
(2.17)
However,
$$\begin{aligned} \vert \nabla y\vert ^{2}\triangle G_{\lambda} =&\operatorname{div}\bigl(\vert \nabla y\vert ^{2} \nabla G_{\lambda}\bigr)-2\operatorname{div} \bigl( \nabla y(\nabla y\nabla G_{ \lambda}) \bigr) \\ &{}+2\triangle y\nabla y\nabla G_{\lambda}+2\sum _{i=1}^{d}\nabla y \partial_{i}y \partial_{i}\nabla G_{\lambda} \end{aligned}$$
(2.18)
and
$$\begin{aligned} \vert \nabla z\vert ^{2}\triangle G_{\lambda} =&\operatorname{div}\bigl(\vert \nabla z\vert ^{2} \nabla G_{\lambda}\bigr)-2\operatorname{div} \bigl( \nabla z(\nabla z\nabla G_{ \lambda}) \bigr) \\ &{}+2\triangle z\nabla z\nabla G_{\lambda}+2\sum _{i=1}^{d}\nabla z \partial_{i}z \partial_{i}\nabla G_{\lambda}. \end{aligned}$$
(2.19)
Now, we write
$$\begin{aligned} A = \int_{\partial\Omega} \vert \nabla y\vert ^{2} \partial_{\nu}G_{ \lambda}\,d\sigma-2 \int_{\partial\Omega}\partial_{\nu}y(\nabla y \nabla G_{\lambda})\,d\sigma \end{aligned}$$
(2.20)
and
$$\begin{aligned} B = \int_{\partial\Omega} \vert \nabla z\vert ^{2} \partial_{\nu}G_{ \lambda}\,d\sigma-2 \int_{\partial\Omega}\partial_{\nu}z(\nabla z \nabla G_{\lambda})\,d\sigma. \end{aligned}$$
(2.21)
Then, by (2.18), (2.19), (2.20), (2.21), we obtain
$$\begin{aligned} \int_{\Omega} \vert \nabla y\vert ^{2}\triangle G_{\lambda}\,dx =&A +2 \int_{\Omega}\triangle y\nabla y\nabla G_{\lambda}\,dx+2\sum _{i=1} ^{d} \int_{\Omega}\nabla y\partial_{i}y\partial_{i} \nabla G_{\lambda }\,dx \\ =&A -2 \int_{\Omega}\triangle y\nabla y\frac{x-x_{0}}{2(T-t+ \lambda)}G_{\lambda}\,dx \\ &{}+2\sum_{i=1}^{d} \int_{\Omega}\partial_{i}y\partial_{i}y \biggl( \frac {-1}{2(T-t+ \lambda)}G_{\lambda}+\frac{\vert x-x_{0}\vert ^{2}}{4(T-t+\lambda)^{2}}G_{ \lambda} \biggr) \,dx \\ &{}+2\sum_{i\neq j} \int_{\Omega}\partial_{j}y\partial_{i}y \frac{(x _{i}-x_{0i})(x_{j}-x_{0j})}{4(T-t+\lambda)^{2}}G_{\lambda}\,dx \\ =&A -2 \int_{\Omega}\triangle y\nabla y\frac{x-x_{0}}{2(T-t+ \lambda)}G_{\lambda}\,dx+ \int_{\Omega} \vert \nabla y\vert ^{2} \biggl( \frac{-1}{(T-t+ \lambda)} \biggr) G_{\lambda}\,dx \\ &{}+2 \int_{\Omega} \biggl( \frac{x-x_{0}}{2(T-t+\lambda)}\nabla y \biggr) ^{2}G_{\lambda}\,dx. \end{aligned}$$
(2.22)
In the same way, we get
$$\begin{aligned} \int_{\Omega} \vert \nabla z\vert ^{2}\triangle G_{\lambda}\,dx =&B -2 \int_{\Omega}\triangle z\nabla z\frac{x-x_{0}}{2(T-t+\lambda)}G_{ \lambda}\,dx+ \int_{\Omega} \vert \nabla z\vert ^{2} \biggl( \frac{-1}{(T-t+\lambda )} \biggr) G_{\lambda}\,dx \\ &{}+2 \int_{\Omega} \biggl( \frac{x-x_{0}}{2(T-t+\lambda)}\nabla z \biggr) ^{2}G_{\lambda}\,dx. \end{aligned}$$
(2.23)
Thus, we can rewrite (2.17) as
$$\begin{aligned} D'_{\lambda}(t) =&-A -B -2 \int_{\Omega} \biggl( \frac{x-x _{0}}{2(T-t+\lambda)}\nabla y \biggr) ^{2}G_{\lambda}\,dx-2 \int_{\Omega } \biggl( \frac{x-x_{0}}{2(T-t+\lambda)}\nabla z \biggr) ^{2}G_{\lambda}\,dx \\ &{}+2 \int_{\Omega}(\triangle y+\partial_{t}y) \frac{x-x_{0}}{2(T-t+ \lambda)}\nabla yG_{\lambda}\,dx+2 \int_{\Omega}(\triangle z+\partial _{t}z) \frac{x-x_{0}}{2(T-t+\lambda)}\nabla zG_{\lambda}\,dx \\ &{}-2 \int_{\Omega}\partial_{t}y\triangle yG_{\lambda}\,dx-2 \int_{ \Omega}\partial_{t}z\triangle zG_{\lambda}\,dx+ \frac{1}{(T-t+\lambda )}D_{\lambda}(t) \\ =&-A -B -2 \int_{\Omega} \biggl( \partial_{t}y-\frac{x-x _{0}}{2(T-t+\lambda)} \nabla y+\frac{1}{2}(\triangle y-\partial _{t}y) \biggr) ^{2}G_{\lambda}\,dx \\ &{}-2 \int_{\Omega} \biggl( \partial_{t}z-\frac{x-x_{0}}{2(T-t+\lambda)} \nabla z+\frac{1}{2}(\triangle z-\partial_{t}z) \biggr) ^{2}G_{\lambda }\,dx \\ &{}+2 \int_{\Omega}\frac{1}{4}(\triangle y+\partial_{t}y)^{2}G_{\lambda }\,dx-2 \int_{\Omega}\partial_{t}y\triangle yG_{\lambda}\,dx \\ &{}+2 \int_{\Omega}\frac{1}{4}(\triangle z+\partial_{t}z)^{2}G_{\lambda }\,dx-2 \int_{\Omega}\partial_{t}z\triangle zG_{\lambda}\,dx+ \frac{1}{(T-t+ \lambda)}D_{\lambda}(t) \\ =&-A -B -2 \int_{\Omega} \biggl( \partial_{t}y-\frac{x-x _{0}}{2(T-t+\lambda)} \nabla y+\frac{1}{2}(\triangle y-\partial _{t}y) \biggr) ^{2}G_{\lambda}\,dx \\ &{}-2 \int_{\Omega} \biggl( \partial_{t}z-\frac{x-x_{0}}{2(T-t+\lambda)} \nabla z+\frac{1}{2}(\triangle z-\partial_{t}z) \biggr) ^{2}G_{\lambda }\,dx \\ &{}+\frac{1}{2} \int_{\Omega}(\triangle y-\partial_{t}y)^{2}G_{\lambda }\,dx+ \frac{1}{2} \int_{\Omega}(\triangle z-\partial_{t}z)^{2}G_{\lambda }\,dx +\frac{1}{(T-t+\lambda)}D_{\lambda}. \end{aligned}$$
(2.24)
This completes the proof of this lemma. □
Lemma 2.5
For each
\(\lambda>0\)
and
\(t\in(0,T)\), it follows that
$$\begin{aligned} \frac{d}{dt}\bigl[(T-t+\lambda)N_{\lambda}(t)\bigr] \leqslant4M^{2}(T+\lambda ). \end{aligned}$$
(2.25)
Proof
Firstly, we compute \(N'_{\lambda}(t)\) as \(t\in(0,T)\). By (2.24) and (2.15), we derive that
$$\begin{aligned} N'_{\lambda}(t) =&2 \biggl( \frac{1}{H_{\lambda}(t)} \biggr) ^{2}\bigl[D'_{ \lambda}(t)H_{\lambda}(t)-D_{\lambda}(t)H'_{\lambda}(t) \bigr] \\ =&\frac{2}{H_{\lambda}(t)} \biggl\{ -A -B +\frac{1}{(T-t+ \lambda)}D_{\lambda}(t) \\ &{}-2 \int_{\Omega} \biggl( \partial_{t}y-\frac{x-x_{0}}{2(T-t+\lambda)} \nabla y+\frac{1}{2}(\triangle y-\partial_{t}y) \biggr) ^{2}G_{\lambda }\,dx+\frac{1}{2} \int_{\Omega}(\triangle y-\partial_{t}y)^{2}G_{\lambda }\,dx \\ &{}-2 \int_{\Omega} \biggl( \partial_{t}z-\frac{x-x_{0}}{2(T-t+\lambda)} \nabla z+\frac{1}{2}(\triangle z-\partial_{t}z) \biggr) ^{2}G_{\lambda }\,dx+\frac{1}{2} \int_{\Omega}(\triangle z-\partial_{t}z)^{2}G_{\lambda }\,dx \biggr\} \\ &{}- \biggl( \frac{1}{H_{\lambda}(t)} \biggr) ^{2} \biggl\{ - \biggl[2 \int_{\Omega}y \biggl( \partial_{t}y-\frac{x-x_{0}}{2(T-t+\lambda)} \nabla y+\frac{1}{2}(\triangle y-\partial_{t}y) \biggr) G_{\lambda}\,dx \\ &{}+2 \int_{\Omega}z \biggl( \partial_{t}z- \frac{x-x_{0}}{2(T-t+\lambda)} \nabla z+\frac{1}{2}(\triangle z- \partial_{t}z) \biggr) G_{\lambda}\,dx \biggr]^{2} \\ &{}+ \biggl[ \int_{\Omega}y(\triangle y-\partial_{t}y)G_{\lambda}\,dx+ \int_{\Omega}z(\triangle z-\partial_{t}z)G_{\lambda}\,dx \biggr] ^{2} \biggr\} \\ =&\frac{1}{(T-t+\lambda)}N_{\lambda}-2\frac{A +B }{H _{\lambda}(t)} \\ &{}+ \biggl( \frac{2}{H_{\lambda}(t)} \biggr) ^{2} \biggl\{ \biggl[ \int_{ \Omega}y \biggl( \partial_{t}y-\frac{x-x_{0}}{2(T-t+\lambda)} \nabla y+ \frac{1}{2}(\triangle y-\partial_{t}y) \biggr) G_{\lambda}\,dx \\ &{}+ \int_{\Omega}z \biggl( \partial_{t}z-\frac{x-x_{0}}{2(T-t+\lambda)} \nabla z+\frac{1}{2}(\triangle z-\partial_{t}z) \biggr) G_{\lambda}\,dx \biggr]^{2} \\ &{}- \int_{\Omega} \biggl( \partial_{t}y-\frac{x-x_{0}}{2(T-t+\lambda)} \nabla y+\frac{1}{2}(\triangle y-\partial_{t}y) \biggr) ^{2}G_{\lambda }\,dx \int_{\Omega}\bigl(y^{2}+z^{2} \bigr)G_{\lambda}\,dx \\ &{}- \int_{\Omega} \biggl( \partial_{t}z-\frac{x-x_{0}}{2(T-t+\lambda)} \nabla z+\frac{1}{2}(\triangle z-\partial_{t}z) \biggr) ^{2}G_{\lambda }\,dx \int_{\Omega}\bigl(y^{2}+z^{2} \bigr)G_{\lambda}\,dx \biggr\} \\ &{}- \biggl( \frac{1}{H_{\lambda}(t)} \biggr) ^{2} \biggl[ \int_{\Omega}y( \triangle y-\partial_{t}y)G_{\lambda}\,dx+ \int_{\Omega}z(\triangle z- \partial_{t}z)G_{\lambda}\,dx \biggr] ^{2} \\ &{}+\frac{1}{H_{\lambda}(t)} \biggl[ \int_{\Omega}(\triangle y-\partial _{t}y)^{2}G_{\lambda}\,dx+ \int_{\Omega}(\triangle z-\partial_{t}z)^{2}G _{\lambda}\,dx \biggr] . \end{aligned}$$
(2.26)
Let
$$\begin{aligned}& \alpha=\partial_{t}y-\frac{x-x_{0}}{2(T-t+\lambda)}\nabla y+ \frac{1}{2}( \triangle y-\partial_{t}y), \\& \beta=\partial_{t}z-\frac{x-x_{0}}{2(T-t+\lambda)}\nabla z+ \frac{1}{2}( \triangle z-\partial_{t}z). \end{aligned}$$
It follows from the Cauchy-Schwarz inequality that
$$\begin{aligned}& \int_{\Omega}\alpha^{2}G_{\lambda}\,dx \int_{\Omega}\bigl(y^{2}+z^{2}\bigr)G _{\lambda}\,dx+ \int_{\Omega}\beta^{2}G_{\lambda}\,dx \int_{\Omega}\bigl(y ^{2}+z^{2} \bigr)G_{\lambda}\,dx \\& \quad = \int_{\Omega}\bigl(\alpha^{2}+\beta^{2} \bigr)G_{\lambda}\,dx \int_{\Omega}\bigl(y ^{2}+z^{2} \bigr)G_{\lambda}\,dx \\& \quad \geq \biggl[ \int_{\Omega}y\alpha G_{\lambda}\,dx+ \int_{\Omega}z \beta G_{\lambda}\,dx \biggr] ^{2}. \end{aligned}$$
It, together with (2.26), shows that
$$\begin{aligned} \begin{aligned}[b] & N'_{\lambda}(t)-\frac{1}{(T-t+\lambda)}N_{\lambda}(t)+2 \frac{ A +B }{H_{\lambda}(t)} \\ &\quad{} -\frac{1}{H_{\lambda}(t)} \biggl[ \int _{\Omega}(\triangle y-\partial_{t}y)^{2}G_{\lambda}\,dx+ \int_{\Omega }(\triangle z-\partial_{t}z)^{2}G_{\lambda}\,dx \biggr] \leq0. \end{aligned} \end{aligned}$$
(2.27)
In what follows, we will discuss the properties of A and B. Since \(y=z=0\) on ∂Ω, we have \(\nabla y= \partial_{\nu}y\nu\), \(\nabla z=\partial_{\nu}z\nu\) on ∂Ω. If the domain Ω is convex, we can derive that \(((x-x_{0})\cdot\nu)\geq0\). According to (2.20) and (2.6), then
$$\begin{aligned} A =& \int_{\partial\Omega} \vert \nabla y\vert ^{2}\partial _{\nu}G_{ \lambda}\,d\sigma-2 \int_{\partial\Omega}\partial_{\nu}y(\nabla y \nabla G_{\lambda})\,d\sigma \\ =&-\frac{1}{2(T-t+\lambda)} \int_{\partial\Omega} \vert \nabla y\vert ^{2}\bigl((x-x _{0})\cdot\nu\bigr)G_{\lambda}\,d\sigma+\frac{1}{(T-t+\lambda)} \\ &{}\times \int_{\partial\Omega}\partial_{\nu}y\bigl((x-x_{0}) \nabla y\bigr)G _{\lambda}\,d\sigma \\ =&-\frac{1}{2(T-t+\lambda)} \int_{\partial\Omega} \vert \nabla y\vert ^{2}\bigl((x-x _{0})\cdot\nu\bigr)G_{\lambda}\,d\sigma+\frac{1}{(T-t+\lambda)} \\ &{}\times \int_{\partial\Omega} \vert \partial_{\nu}y \vert ^{2} \bigl((x-x_{0}) \cdot\nu\bigr)G_{\lambda}\,d\sigma \\ =&\frac{1}{2(T-t+\lambda)}\times \int_{\partial\Omega} \vert \partial _{\nu}y \vert ^{2}\bigl((x-x_{0})\cdot\nu\bigr)G_{\lambda}\,d\sigma \geq0. \end{aligned}$$
In the same way, we get
$$\begin{aligned} B =\frac{1}{2(T-t+\lambda)}\times \int_{\partial\Omega} \vert \partial_{\nu}z \vert ^{2} \bigl((x-x_{0})\cdot\nu\bigr)G_{\lambda }\,d\sigma\geq0. \end{aligned}$$
Combining with (2.27), we get
$$\begin{aligned} \begin{aligned}[b] & (T-t+\lambda)N'_{\lambda}(t)-N_{\lambda}(t) \\ &\quad{} - \frac{(T-t+\lambda)}{H _{\lambda}(t)} \biggl[ \int_{\Omega}(\triangle y-\partial_{t}y)^{2}G _{\lambda}\,dx+ \int_{\Omega}(\triangle z-\partial_{t}z)^{2}G_{\lambda }\,dx \biggr] \leq0. \end{aligned} \end{aligned}$$
(2.28)
Using equations (1.1) and (2.28) can be written as
$$\begin{aligned} (T-t+\lambda)N'_{\lambda}(t)-N_{\lambda}(t) \leq& \frac{(T-t+ \lambda)}{H_{\lambda}(t)} \biggl[ \int_{\Omega}(ay+bz)^{2}G_{\lambda }\,dx+ \int_{\Omega}(cy+dz)^{2}G_{\lambda}\,dx \biggr] \\ \leq&\frac{4M^{2}(T-t+\lambda)}{H_{\lambda}(t)} \int_{\Omega}\bigl(\vert y\vert ^{2}+\vert z\vert ^{2}\bigr)G _{\lambda}\,dx \\ \leq&4M^{2}(T+\lambda). \end{aligned}$$
Thus,
$$\begin{aligned} \frac{d}{dt} \bigl[ (T-t+\lambda)N_{\lambda}(t) \bigr] \leq4M^{2}(T+ \lambda), \quad \forall t \in(0,T). \end{aligned}$$
This completes the proof of this lemma. □
Let
$$\begin{aligned} \mathcal{K}_{M,y,z,T}=2\ln \biggl( \frac{\int_{\Omega} ( \vert y(x,0)\vert ^{2}+\vert z(x,0)\vert ^{2} ) \,dx}{ \int_{\Omega} ( \vert y(x,T)\vert ^{2}+\vert z(x,T)\vert ^{2} ) \,dx} \biggr) + \frac{m}{T}+8MT+4M^{2}T^{2}+\frac{d}{2}. \end{aligned}$$
Then we have the following lemma.
Lemma 2.6
For each
\(\lambda>0\), we have
$$\begin{aligned} \lambda N_{\lambda}(T)+\frac{d}{2}\leq \biggl( \frac{\lambda }{T}+1 \biggr) \mathcal{K}_{M,y,z,T}. \end{aligned}$$
(2.29)
Proof
Integrating (2.25) over \((t,T)\), we have
$$\begin{aligned} \lambda N_{\lambda}(T) \leq&(T-t+\lambda)N_{\lambda}(t)+4M^{2}(T+ \lambda) (T-t) \\ \leq&(T+\lambda)N_{\lambda}(t)+4M^{2}T(T+\lambda). \end{aligned}$$
Integrating the above over \((0,\frac{T}{2})\), we obtain
$$\begin{aligned} \frac{T}{2}\lambda N_{\lambda}(T)\leq(T+\lambda) \int_{0}^{ \frac{T}{2}}N_{\lambda}(t)\,dt+2M^{2}T^{2}(T+ \lambda). \end{aligned}$$
(2.30)
By integrating (2.11) over \((0,\frac{T}{2})\), we get
$$\begin{aligned} \int_{0}^{\frac{T}{2}}N_{\lambda}(t)\,dt =&- \int_{0}^{\frac{T}{2}}\frac{H'_{ \lambda}(t)}{H_{\lambda}(t)}\,dt+ \int_{0}^{\frac{T}{2}}\frac{2}{H_{ \lambda}(t)} \int_{\Omega} \bigl[ y(\partial_{t}y-\triangle y)+z( \partial_{t}z-\triangle z) \bigr] G_{\lambda}\,dx \,dt \\ =&-\ln\frac{H_{\lambda}(\frac{T}{2})}{H_{\lambda}(0)}+ \int_{0} ^{\frac{T}{2}}\frac{2}{H_{\lambda}(t)} \int_{\Omega} \bigl[ -y(ay+bz)-z(cy+dz) \bigr] G _{\lambda}\,dx \,dt \\ \leq&\ln\frac{H_{\lambda}(0)}{H_{\lambda}(\frac{T}{2})}+ \int_{0} ^{\frac{T}{2}}\frac{2}{H_{\lambda}(t)} \int_{\Omega}2M\bigl(y^{2}+z^{2}\bigr)G _{\lambda}\,dx \,dt \\ \leq&\ln\frac{H_{\lambda}(0)}{H_{\lambda}(\frac{T}{2})}+2MT. \end{aligned}$$
This, alone with (2.30), shows that
$$\begin{aligned} \frac{T}{2}\lambda N_{\lambda}(T)\leq(T+\lambda) \biggl[ \ln \frac{H _{\lambda}(0)}{H_{\lambda}(\frac{T}{2})}+2MT+2M^{2}T^{2} \biggr] . \end{aligned}$$
We have
$$\begin{aligned} \frac{H_{\lambda}(0)}{H_{\lambda}(\frac{T}{2})} =&\frac{\int_{ \Omega} ( \vert y(x,0)\vert ^{2}+\vert z(x,0)\vert ^{2} ) (T+\lambda)^{- \frac{d}{2}}\cdot e^{-\frac{\vert x-x_{0}\vert ^{2}}{4(T+\lambda)}}\,dx}{ \int_{\Omega} ( \vert y(x,\frac{T}{2})\vert ^{2}+\vert z(x,\frac{T}{2})\vert ^{2} ) ( \frac{T}{2}+\lambda)^{-\frac{d}{2}}\cdot e^{-\frac{\vert x-x_{0}\vert ^{2}}{4( \frac{T}{2}+\lambda)}}\,dx} \\ \leq&\frac{\int_{\Omega} ( \vert y(x,0)\vert ^{2}+\vert z(x,0)\vert ^{2} ) \,dx \cdot(\frac{T}{2}+\lambda)^{\frac{d}{2}}}{\int_{\Omega} (\vert y(x, \frac{T}{2})\vert ^{2}+\vert z(x,\frac{T}{2})\vert ^{2} ) \cdot e^{-\frac{\vert x-x _{0}\vert ^{2}}{4(\frac{T}{2}+\lambda)}}\,dx\cdot(T+\lambda )^{\frac{d}{2}}} \\ \leq&\frac{\int_{\Omega} ( \vert y(x,0)\vert ^{2}+\vert z(x,0)\vert ^{2} ) \,dx \cdot(\frac{T}{2}+\lambda)^{\frac{d}{2}}}{\int_{\Omega} (\vert y(x, \frac{T}{2})\vert ^{2}+\vert z(x,\frac{T}{2})\vert ^{2} ) \,dx\cdot(T+\lambda)^{ \frac{d}{2}}}\cdot e^{\frac{m}{4(\frac{T}{2}+\lambda)}} \\ \leq&e^{\frac{m}{2T}}\frac{\int_{\Omega} ( \vert y(x,0)\vert ^{2}+\vert z(x,0)\vert ^{2} ) \,dx}{ \int_{\Omega} ( \vert y(x,\frac{T}{2})\vert ^{2}+\vert z(x,\frac{T}{2})\vert ^{2} ) \,dx}. \end{aligned}$$
Therefore,
$$\begin{aligned} \frac{T}{2}\lambda N_{\lambda}(T) \leq&(T+\lambda) \biggl[ \frac{m}{2T}+ \ln \biggl( \frac{\int_{\Omega} ( \vert y(x,0)\vert ^{2}+\vert z(x,0)\vert ^{2} ) \,dx}{ \int_{\Omega} ( \vert y(x,\frac{T}{2})\vert ^{2}+\vert z(x,\frac{T}{2})\vert ^{2} ) \,dx} \biggr) +2MT+2M ^{2}T^{2} \biggr] . \end{aligned}$$
Using the energy estimate
$$\begin{aligned} \frac{\int_{\Omega} ( \vert y(x,T)\vert ^{2}+\vert z(x,T)\vert ^{2} ) \,dx}{ \int_{\Omega} ( \vert y(x,\frac{T}{2})\vert ^{2}+\vert z(x,\frac{T}{2})\vert ^{2} ) \,dx} \leq e^{2MT}, \end{aligned}$$
we obtain
$$\begin{aligned} \lambda N_{\lambda}(T) \leq& \biggl( \frac{\lambda}{T}+1 \biggr) \biggl[ 2 \ln \biggl( \frac{\int_{\Omega} ( \vert y(x,0)\vert ^{2}+\vert z(x,0)\vert ^{2} ) \,dx}{ \int_{\Omega} ( \vert y(x,T)\vert ^{2}+\vert z(x,T)\vert ^{2} ) \,dx} \biggr) + \frac{m}{T}+8MT+4M^{2}T^{2} \biggr] \\ =& \biggl( \frac{\lambda}{T}+1 \biggr) \biggl( \mathcal{K}_{M,y,z,T}- \frac{d}{2} \biggr) . \end{aligned}$$
Thus,
$$\begin{aligned} \lambda N_{\lambda}(T)+\frac{d}{2} \leq& \biggl( \frac{\lambda}{T}+1 \biggr) \mathcal{K}_{M,y,z,T}-\frac{\lambda}{T}\cdot \frac{d}{2} \\ \leq& \biggl( \frac{\lambda}{T}+1 \biggr) \mathcal{K}_{M,y,z,T}. \end{aligned}$$
This is (2.29). □
While most of the proof of the following lemma is similar to Lemma 3 of [2], we would rather give the proof in detail for the sake of completeness.
Lemma 2.7
For each
\(T>0\)
and
\(y_{0} \in L^{2}(\Omega)\), \(z_{0} \in L^{2}( \Omega)\), the solution
y
and
z, with
\(y(\cdot,0)=y_{0}(\cdot)\), \(z(\cdot,0)=z_{0}(\cdot)\), to (1.1) holds the estimate:
$$\begin{aligned}& \biggl[ 1-\frac{8\lambda}{r^{2}} \biggl( \frac{\lambda}{T}+1 \biggr) \mathcal{K}_{M,y,z,T} \biggr] \int_{\Omega} \vert x-x_{0}\vert ^{2} \bigl( \bigl\vert y(x,T)\bigr\vert ^{2}+\bigl\vert z(x,T)\bigr\vert ^{2} \bigr) e^{-\frac{\vert x-x_{0}\vert ^{2}}{4\lambda }}\,dx \\& \quad \leq8\lambda \biggl( \frac{\lambda}{T}+1 \biggr) \mathcal{K}_{M,y,z,T} \int_{B_{r}} \bigl( \bigl\vert y(x,T)\bigr\vert ^{2}+\bigl\vert z(x,T)\bigr\vert ^{2} \bigr) e ^{-\frac{\vert x-x_{0}\vert ^{2}}{4\lambda}}\,dx. \end{aligned}$$
Proof
We first point out the following fact: for any \(f \in H_{0}^{1}( \Omega)\) and for each \(\lambda>0\), we have \(0\leq\int_{\Omega} \vert \nabla ( f(x)\exp(-\frac{\vert x-x_{0}\vert ^{2}}{ 8\lambda}) ) \vert ^{2}\,dx\). By computing the right-hand term, we get
$$\begin{aligned} \int_{\Omega}\frac{\vert x-x_{0}\vert ^{2}}{8\lambda}\bigl\vert f(x)\bigr\vert ^{2}e^{-\frac{\vert x-x _{0}\vert ^{2}}{4\lambda}}\,dx \leq&2\lambda \int_{\Omega}\bigl\vert \nabla f(x)\bigr\vert ^{2}e ^{-\frac{\vert x-x_{0}\vert ^{2}}{4\lambda}}\,dx \\ &{}+\frac{d}{2} \int_{\Omega}\bigl\vert f(x)\bigr\vert ^{2}e^{-\frac {\vert x-x_{0}\vert ^{2}}{4 \lambda}}\,dx. \end{aligned}$$
(2.31)
It follows from (2.31) and (2.4) that
$$\begin{aligned}& \int_{\Omega} \vert x-x_{0}\vert ^{2} \bigl( \bigl\vert y(x,T)\bigr\vert ^{2}+\bigl\vert z(x,T)\bigr\vert ^{2} \bigr) e^{-\frac{\vert x-x_{0}\vert ^{2}}{4\lambda}}\,dx \\& \quad \leq 8\lambda \biggl(2\lambda \int_{\Omega} \bigl( \bigl\vert \nabla y(x,T)\bigr\vert ^{2}+\bigl\vert \nabla z(x,T)\bigr\vert ^{2} \bigr) e^{-\frac{\vert x-x_{0}\vert ^{2}}{4\lambda}}\,dx \\& \quad\quad{} +\frac{d}{2} \int_{\Omega} \bigl( \bigl\vert y(x,T)\bigr\vert ^{2}+\bigl\vert z(x,T)\bigr\vert ^{2} \bigr) e ^{-\frac{\vert x-x_{0}\vert ^{2}}{4\lambda}}\,dx \biggr) \\& \quad \leq 8\lambda \biggl( \lambda N_{\lambda}(T)+\frac{d}{2} \biggr) \int_{\Omega} \bigl( \bigl\vert y(x,T)\bigr\vert ^{2}+\bigl\vert z(x,T)\bigr\vert ^{2} \bigr) e^{-\frac{\vert x-x _{0}\vert ^{2}}{4\lambda}}\,dx \\& \quad \leq 8\lambda \biggl( \lambda N_{\lambda}(T)+\frac{d}{2} \biggr) \biggl[ \int_{B_{r}} \bigl( \bigl\vert y(x,T)\bigr\vert ^{2}+\bigl\vert z(x,T)\bigr\vert ^{2} \bigr) e^{-\frac{\vert x-x _{0}\vert ^{2}}{4\lambda}}\,dx \\& \quad\quad{} +\frac{1}{r^{2}} \int_{\Omega\setminus B_{r}}\vert x-x_{0}\vert ^{2} \bigl( \bigl\vert y(x,T)\bigr\vert ^{2}+\bigl\vert z(x,T)\bigr\vert ^{2} \bigr) e ^{-\frac{\vert x-x_{0}\vert ^{2}}{4\lambda}}\,dx \biggr]. \end{aligned}$$
(2.32)
It, combining with (2.29), shows that
$$\begin{aligned}& \int_{\Omega} \vert x-x_{0}\vert ^{2} \bigl( \bigl\vert y(x,T)\bigr\vert ^{2}+\bigl\vert z(x,T)\bigr\vert ^{2} \bigr) e^{-\frac{\vert x-x_{0}\vert ^{2}}{4\lambda}}\,dx \\& \quad \leq 8\lambda \biggl( \frac{\lambda}{T}+1 \biggr) \mathcal{K}_{M,y,z,T} \biggl[ \int_{B_{r}} \bigl( \bigl\vert y(x,T)\bigr\vert ^{2}+\bigl\vert z(x,T)\bigr\vert ^{2} \bigr) e^{-\frac{\vert x-x _{0}\vert ^{2}}{4\lambda}}\,dx \\& \quad\quad{} +\frac{1}{r^{2}} \int_{\Omega} \vert x-x_{0}\vert ^{2} \bigl( \bigl\vert y(x,T)\bigr\vert ^{2}+\bigl\vert z(x,T)\bigr\vert ^{2} \bigr) e ^{-\frac{\vert x-x_{0}\vert ^{2}}{4\lambda}}\,dx \biggr]. \end{aligned}$$
(2.33)
We get
$$\begin{aligned}& \biggl[ 1-\frac{8\lambda}{r^{2}} \biggl( \frac{\lambda}{T}+1 \biggr) \mathcal{K}_{M,y,z,T} \biggr] \int_{\Omega} \vert x-x_{0}\vert ^{2} \bigl( \bigl\vert y(x,T)\bigr\vert ^{2}+\bigl\vert z(x,T)\bigr\vert ^{2} \bigr) e^{-\frac{\vert x-x_{0}\vert ^{2}}{4\lambda }}\,dx \\& \quad \leq8\lambda \biggl( \frac{\lambda}{T}+1 \biggr) \mathcal{K}_{M,y,z,T} \int_{B_{r}} \bigl( \bigl\vert y(x,T)\bigr\vert ^{2}+\bigl\vert z(x,T)\bigr\vert ^{2} \bigr) e ^{-\frac{\vert x-x_{0}\vert ^{2}}{4\lambda}}\,dx. \end{aligned}$$
(2.34)
This completes the proof of the lemma. □