Singular diffusion equations based on the convex–nonconvex variation
Based on the new variational model, the following diffusion equation is proposed:
$$\begin{aligned} &\frac{\partial u}{\partial t}=\mu _{1}\operatorname{div} \biggl( \frac{\nabla u}{\sqrt{1+ \vert \nabla u \vert ^{2}}} \biggr)+\mu _{2} \operatorname{div} \biggl( \frac{\nabla u}{1+ \vert \nabla u \vert ^{2}} \biggr)- \lambda (u-f), \end{aligned}$$
(3)
$$\begin{aligned} &u(x,0)=f,\quad x\in \Omega , \end{aligned}$$
(4)
$$\begin{aligned} &\frac{\partial u}{\partial \vec{n}}=0,\quad (x,t)\in \partial \Omega \times (0,T). \end{aligned}$$
(5)
For this special equation, what we obtain in this paper will reveal another aspect for the existence, namely the existence of a discontinuous solution. Note that the equation is strongly degenerate at the discontinuous points of such a solution. On the other hand, the new equation can be considered as a perturbation of Perona–Malik model [23]. Such a perturbation is not the usual viscous one, for example, Δu or \(\Delta ^{2} u\), which has standard regularity effects. The perturbation has no hazard for the equation to permit the existence of discontinuous solutions, which has particular meaning: with the new perturbation, the new model is still an anisotropic diffusion equation. That is to say, inside the regions where the magnitude of the gradient of u is weak, the new equation acts as Gaussian smoothing, resulting in isotropic smoothing; near the region’s boundaries where the magnitude of the gradient is large, the regularization is “stopped” and the edges are preserved.
Let
$$\begin{aligned}& Z(X)=\nabla \varphi (X)=\mu _{1}\frac{X}{\sqrt{1+ \vert X \vert ^{2}}}+\mu _{2} \frac{X}{1+ \vert X \vert ^{2}} \end{aligned}$$
and
$$\begin{aligned}& \varphi (X)=\mu _{1}\sqrt{1+ \vert X \vert ^{2}}+ \frac{\mu _{2}}{2}\ln \bigl(1+ \vert X \vert ^{2} \bigr), \end{aligned}$$
for \(X\in \mathbb{R}^{N}\). Therefore, the new diffusion equation can be rewritten as
$$\begin{aligned} &\frac{\partial u}{\partial t}=\operatorname{div} \bigl(Z(\nabla u) \bigr)- \lambda (u-f), \quad (x,t)\in \Omega \times (0,T), \\ &u(x,0)=f, \quad x\in \Omega , \\ &\frac{\partial u}{\partial \vec{n}}=0,\quad (x,t)\in \partial \Omega \times (0,T). \end{aligned}$$
Let \(\varphi ^{**}\) denote the convexification of φ, namely,
$$ \varphi ^{**}(X)=\sup \bigl\{ l(X):l\leq \varphi ,l \textrm{ is convex} \bigr\} , $$
and
$$ Z(X)=\nabla \varphi (X),\quad\quad Z^{**}(X)=\nabla \varphi ^{**}(X), \quad X\in \mathbb{R}^{N}. $$
Since \(\varphi \in C^{1}(\mathbb{R}^{N})\), \(\varphi ^{**}\in C^{1}(\mathbb{R}^{N})\) is convex.
Definition 1
A Young measure solution to problem (3)–(5) is a function
$$ u\in L^{\infty } \bigl((0,T);\operatorname{BV}(\Omega ) \bigr)\cap L^{\infty }(Q_{T}),\quad \frac{\partial u}{\partial t}\in L^{2}(Q_{T}) $$
and there exists a \(W^{1,1}(Q_{T})\)-gradient Young measure \(\nu =(\nu _{x,t})_{(x,t)\in Q_{T}}\) on \(\mathbb{R}^{N}\) such that
$$\begin{aligned} & \iint _{Q_{T}} \bigl(\langle \nu ,Z\rangle \cdot \nabla \zeta +u_{t} \zeta +\lambda (u-f_{p})\zeta \bigr)\,dx\,dt=0, \\ & \quad \forall \zeta \in C^{\infty }( \overline{Q}_{T})\textrm{ with }\zeta (x,0)=\zeta (x,T)=0, \end{aligned}$$
(6)
$$\begin{aligned} & \nabla u(x,t)=\langle \nu _{x,t},{\mathrm{id}}\rangle \quad \text{a.e. }(x,t) \in Q_{T}, \end{aligned}$$
(7)
$$\begin{aligned} &\langle \nu _{x,t},Z\cdot {\mathrm{id}}\rangle \leq \langle \nu _{x,t},Z \rangle \cdot \langle \nu _{x,t},{\mathrm{id}}\rangle \quad \text{a.e. }(x,t) \in Q_{T}, \end{aligned}$$
(8)
$$\begin{aligned} &\operatorname{supp}\nu _{x,t}\subset \bigl\{ X\in \mathbb{R}^{N}:\varphi (X)=\varphi ^{**}(X) \bigr\} \quad \text{a.e. }(x,t)\in Q_{T}, \end{aligned}$$
(9)
where id is the identity mapping in \(\mathbb{R}^{N}\),
$$ \langle \nu ,Z\rangle = \int _{\mathbb{R}^{N}}Z(X)\,d\nu (X),\quad\quad \langle \nu ,{\mathrm{id}}\rangle = \int _{\mathbb{R}^{N}}X\,d\nu (X),\quad\quad \langle \nu ,Z\cdot {\mathrm{id}}\rangle = \int _{\mathbb{R}^{N}}Z(X)\,d\nu (X), $$
and
$$\begin{aligned}& u(x,0)=f(x),\quad x\in \Omega , \end{aligned}$$
(10)
in the sense of trace.
Theorem 2
Let \(f\in \operatorname{BV}(\Omega )\cap L^{\infty }(\Omega )\). Then problem (3)–(5) admits at least one Young measure solution.
Preliminaries
We use \(C_{0}(\mathbb{R}^{d})\) to denote the closure of the set of continuous functions on \(\mathbb{R}^{d}\) with compact supports. The dual of \(C_{0}(\mathbb{R}^{d})\) can be identified with the space \(\mathcal{M}(\mathbb{R}^{d})\) of signed Radon measures with finite mass via the pairing
$$ \langle \nu ,f \rangle = \int _{\mathbb{R}^{d}}f\,d\nu _{x}. $$
Let \(D\subset \mathbb{R}^{n}\) be a measurable set of finite measure. A map \(\nu :D\to \mathcal{M}(\mathbb{R}^{d})\) is called weakly* measurable if the functions \(x\mapsto \int _{\mathbb{R}^{d}}f\,d\nu _{x}\) are measurable for all \(f\in C_{0}(\mathbb{R}^{d})\), where \(\nu _{x}=\mu (x)\).
For \(p\geq 1\), define
$$ \mathcal{E}_{0}^{p} \bigl(\mathbb{R}^{d} \bigr)= \biggl\{ \varphi \in C \bigl(\mathbb{R}^{d} \bigr): \lim _{ \vert X \vert \to +\infty }\frac{ \vert \varphi (X) \vert }{1+ \vert X \vert ^{p}} \text{ exists} \biggr\} . $$
As noted in [11], the space \(\mathcal{E}_{0}^{p}(\mathbb{R}^{d})\) is a separable Banach space with the norm
$$ \Vert \psi \Vert _{\mathcal{E}_{0}^{p}}=\sup_{X\in \mathbb{R}^{d}} \frac{ \vert \psi (X) \vert }{1+ \vert X \vert ^{p}}. $$
We define
$$ \mathcal{E}^{p} \bigl(\mathbb{R}^{d} \bigr)= \biggl\{ \varphi \in C \bigl(\mathbb{R}^{d} \bigr):\sup_{ \vert X \vert \in \mathbb{R}^{d}} \frac{ \vert \varphi (X) \vert }{1+ \vert X \vert ^{p}}< +\infty \biggr\} , $$
which is an inseparable space with the above norm.
Definition 2
Let \(p\geq 1\). A Young measure \(\nu =(\nu _{x})_{x\in D}\) on \(\mathbb{R}^{d}\) is called a \(W^{1,p}\)-gradient Young measure if
-
(i)
\(x\in D\mapsto \int _{\mathbb{R}^{d}}f \,d\nu _{x}\in \mathbb{R}\) is a Lebesgue measurable function for all f bounded and continuous on \(\mathbb{R}^{d}\);
-
(ii)
There is a sequence of functions \(\{u^{k}\}_{k=1}^{\infty }\subset W^{1,p}(D)\) for which the representation formula
$$\begin{aligned}& \lim_{k\to \infty } \int _{E}\psi \bigl(\nabla u^{k}(x) \bigr)\,dx= \int _{E} \langle \nu _{x},\psi \rangle \,dx \end{aligned}$$
(11)
holds for all measurable \(E\subset D\) and all \(\psi \in \mathcal{E}_{0}^{p}(\mathbb{R}^{d})\), where \(\langle \nu _{x},\psi \rangle =\int _{\mathbb{R}^{d}}\psi \,d\nu _{x}\).
We also call ν the \(W^{1,p}(D)\)-gradient Young measure generated by \(\{\nabla u^{k}\}_{k=1}^{\infty }\) and \(\{\nabla u^{k}\}_{k=1}^{\infty }\) the \(W^{1,p}(D)\)-gradient generating sequence of ν. In addition, the representation formula (11) also holds for \(\psi \in \mathcal{E}^{p}(\mathbb{R}^{d})\). By the fundamental theorem for Young measure, we see that
$$ \bigl\Vert \nu (x) \bigr\Vert _{\mathcal{M}(\mathbb{R}^{d})}=1 \quad \text{a.e. }x\in D. $$
Definition 3
Let \(\{z^{k}\}_{k=1}^{\infty }\subset L^{1}(D)\) and \(z\in L^{1}(D)\). We say that \(\{z^{k}\}_{k=1}^{\infty }\) converges to z in the bitting sense if there is a decreasing sequence of subsets \(E_{j+1}\subset E_{j}\) of D with \(\lim_{j\to \infty }\operatorname{meas}(E_{j})=0\) such that \(\{z^{k}\}_{k=1}^{\infty }\) converges weakly to z in \(L^{1}(D\backslash E_{j})\) for all j.
Definition 4
Let \(p\geq 1\). A Young measure \(\nu =(\nu _{x})_{x\in D}\) on \(\mathbb{R}^{d}\) is called a \(W^{1,p}(D)\)-bitting Young measure if there is a sequence \(\{z^{k}\}_{k=1}^{\infty }\subset L^{p}(D)\) and \(z\in L^{1}(D)\) such that \(\{\vert z^{k}\vert ^{p}\}_{k=1}^{\infty }\) converges to z and \(\{\psi (z^{k}(x))\}_{k=1}^{\infty }\) converges to \(\langle \nu _{x},\psi \rangle \) in the bitting sense for all \(\psi \in \mathcal{E}_{0}^{p}(\mathbb{R}^{d})\) (or \(\mathcal{E}^{p}(\mathbb{R}^{d})\)).
We also call ν the \(W^{1,p}(D)\)-bitting Young measure generated by \(\{z^{k}\}_{k=1}^{\infty }\) and \(\{z^{k}\}_{k=1}^{\infty }\) the \(W^{1,p}(D)\)-bitting generating sequence of ν. By the fundamental theorem for Young measure, we see that
$$ \bigl\Vert \nu (x) \bigr\Vert _{\mathcal{M}(\mathbb{R}^{d})}=1 \quad \text{a.e. }x\in D. $$
Kinderlehrer and Pedregal [37] showed a property which characterizes \(W^{1,p}\)-gradient Young measures as described in the following lemma.
Lemma 1
Let \(\nu =(\nu _{x})_{x\in D}\) be a Young measure on \(\mathbb{R}^{d}\). Then \(\nu =(\nu _{x})_{x\in D}\) is a \(W^{1,p}(D)\)-gradient Young measure if and only if
-
(i)
There exists \(u\in W^{1,p}(D)\) such that
$$ \nabla u(x)= \int _{\mathbb{R}^{d}}A\,d\nu _{x}(x)\quad \textit{a.e. }x\in D; $$
-
(ii)
Jensen’s inequality
$$ \psi \bigl(\nabla u(x) \bigr)\leq \int _{\mathbb{R}^{d}}\psi (A)\,d\nu _{x}(A) $$
holds for all \(\psi \in \mathcal{E}^{p}(\mathbb{R}^{d})\) continuous, quasiconvex, and bounded below;
-
(iii)
The function
$$ x\mapsto \int _{\mathbb{R}^{d}} \vert A \vert ^{p}\,d\nu _{x}(A) $$
is in \(L^{1}(D)\).
We give the following two lemmas. The proofs can be found in [38, 39].
Lemma 2
Suppose \(f\in \mathcal{E}^{p}(\mathbb{R}^{d})\), for some \(p\geq 1\), is quasiconvex and bounded below and let \(\{u^{k}\}_{k=1}^{\infty }\) converge weakly to u in \(W^{1,p}(D)\). Then
-
(i)
For all measurable \(E\subset D\),
$$ \int _{E}f(\nabla u)\,dx\leq \liminf_{k\to \infty } \int _{E}f \bigl(\nabla u^{k} \bigr)\,dx; $$
-
(ii)
If
$$ \lim_{k\to \infty } \int _{D}f \bigl(\nabla u^{k} \bigr)\,dx= \int _{D}f(\nabla u)\,dx, $$
then \(\{f(\nabla u^{k})\}_{k=1}^{\infty }\) are weakly sequentially precompact in \(L^{1}(D)\) and the sequence converges weakly to \(f(\nabla u)\).
Lemma 3
Let f and \(\{u^{k}\}_{k=1}^{\infty }\) be as in Lemma 2 (ii) and assume in addition that
$$ \bigl(c \vert X \vert ^{p}-1 \bigr)^{+}\leq f(X)\leq C \vert X \vert ^{p}+1 $$
for \(0< c\leq C\). Let v be generated by the gradients \(\{\nabla u\}_{k=1}^{\infty }\). Then ν is a \(W^{1,p}(D)\)-gradient Young measure.
We now state and prove a result for the sequences of gradient-generated Young measures [40].
Lemma 4
Let \(1\leq p<2\). Suppose that \(\{\nu ^{\alpha }=(\nu _{x}^{\alpha })_{x\in D}\}_{\alpha >0}\) is a family of \(W^{1,p}(D)\)-gradient Young measures and each is generated by \(\{\nabla u^{\alpha ,m}\}_{m=1}^{\infty }\), where \(u^{\alpha ,m}\) is in \(W^{1,p}(D)\) uniformly bounded in α and m. Then there exist a subsequence of \(\{\nu ^{\alpha }\}_{\alpha >0}\), denoted by \(\{\nu ^{\alpha _{i}}\}_{i=1}^{\infty }\), and a \(W^{1,p}(D)\)-gradient Young measure ν such that
-
(i)
\(\{\nu ^{\alpha _{i}}\}_{i=1}^{\infty }\) converges weakly* to ν in \(L^{\infty }(D;\mathcal{M}(\mathbb{R}^{d}))\), namely, \(\{\langle \nu ^{\alpha _{i}},\psi \rangle \}_{i=1}^{\infty }\) converges weakly* to \(\langle \nu ,\psi \rangle \) in \(L^{\infty }(D)\) for all \(\psi \in C_{0}(\mathbb{R} ^{d})\);
-
(ii)
For \(1\leq q< p\), \(\{\nu ^{\alpha _{i}}\}_{i=1}^{\infty }\) converges weakly to ν in \(L^{1}(D;(\mathcal{E}_{0}^{q}(\mathbb{R}^{d}))')\), namely, \(\{\langle \nu ^{\alpha _{i}},\psi \rangle \}_{i=1}^{\infty }\) converges weakly to \(\langle \nu ,\psi \rangle \) in \(L^{\infty }(D)\) for all \(\psi \in \mathcal{E}_{0}^{q}(\mathbb{R}^{d})\);
-
(iii)
\(\{\nu ^{\alpha _{i}}\}_{i=1}^{\infty }\) converges to ν in \(L^{1}(D;(\mathcal{E}_{0}^{q}(\mathbb{R}^{d}))')\) in the bitting sense, namely, \(\psi \in \mathcal{E}_{0}^{p}(\mathbb{R}^{d})\), \(\{\langle \nu ^{\alpha _{i}},\psi \rangle \}_{i=1}^{\infty }\) converges to \(\langle \nu ,\psi \rangle \) in the bitting sense.
Existence of solution to the approximation problem
Since equation (3) is degenerate, singular, and forward–backward, some necessary approximations are required for discussing the existence of solutions. Our approximations will be divided into two steps. For this purpose, we need to approximate the initial datum f. By the density properties of BV functions in [7], there exists some subsequence \(\{f_{p}\}\subset C_{0}^{\infty }(\Omega )\) such that \(\Vert f_{p}\Vert _{L^{\infty }(\Omega )}\) and \(\Vert \nabla f_{p}\Vert _{L^{1}(\Omega )}\) are uniformly bounded in p, and \(\{f_{p}\}_{0< p<1}\) converges to f in \(L^{1}(\Omega )\).
As the first step, we consider the following evolution problem:
$$\begin{aligned} &\frac{\partial u}{\partial t}=\operatorname{div} \bigl(Z_{p}( \nabla u) \bigr)-\lambda (u-f_{p}), \quad (x,t)\in \Omega \times (0,T), \end{aligned}$$
(12)
$$\begin{aligned} &u(x,0)=f_{p}, \quad x\in \Omega , \end{aligned}$$
(13)
$$\begin{aligned} &\frac{\partial u}{\partial \vec{n}}=0,\quad (x,t)\in \partial \Omega \times (0,T), \end{aligned}$$
(14)
where
$$\begin{aligned}& Z_{p}(X)=\nabla \varphi _{p}(X)=\mu _{1} \frac{ \vert X \vert ^{2(p-1)}X}{\sqrt{1+\frac{1}{1+\delta } \vert X \vert ^{2p}}} +{\mu _{2}} \frac{X}{1+ \vert X \vert ^{2}} \end{aligned}$$
and
$$\begin{aligned}& \varphi _{p}(X)=\mu _{1}\sqrt{1+\frac{1}{p} \vert X \vert ^{2p}}+ \frac{\mu _{2}}{2}\ln \bigl(1+ \vert X \vert ^{2} \bigr). \end{aligned}$$
Let \(\varphi ^{**}_{p}\) denote the convexification of \(\varphi _{p}\), namely,
$$ \varphi ^{**}_{p}(X)=\sup \bigl\{ f(X):f\leq \varphi _{p},f \textrm{ is convex} \bigr\} $$
and
$$ Z_{p}^{**}=\nabla \varphi ^{**}_{p}. $$
Since \(\varphi _{p}\in C^{1}(\mathbb{R}^{N})\), \(\varphi ^{**}\in C^{1}(\mathbb{R}^{N})\) is convex. In addition,
$$\begin{aligned}& Z_{p}(X)\cdot {\mathrm{id}}(X)\geq Z(X)\cdot {\mathrm{id}}(X),\quad \forall X\in \bigl\{ X\in \mathbb{R}^{N}: \vert X \vert \geq C_{0} \bigr\} , \\& \bigl\{ X\in \mathbb{R}^{N}:\varphi _{p}(X)=\varphi ^{**}(X) \bigr\} \subset \bigl\{ X \in \mathbb{R}^{N}:\varphi (X)=\varphi ^{**}(X) \bigr\} . \end{aligned}$$
Definition 5
A Young measure solution to problem (12)–(14) is a function
$$ u\in L^{\infty } \bigl((0,T);W^{1,p}(\Omega ) \bigr)\cap L^{\infty }(Q_{T}),\quad \frac{\partial u}{\partial t}\in L^{2}(Q_{T}) $$
and there exists a \(W^{1,p}(Q_{T})\)-gradient Young measure \(\nu =(\nu _{x,t})_{(x,t)\in Q_{T}}\) on \(\mathbb{R}^{N}\) such that
$$\begin{aligned} & \iint _{Q_{T}} \bigl(\langle \nu ,Z_{p}\rangle \cdot \nabla \zeta +u_{t} \zeta +\lambda (u-f_{p})\zeta \bigr)\,dx \,dt=0, \\ & \quad \forall \zeta \in C^{\infty }( \overline{Q}_{T})\textrm{ with }\zeta (x,0)=\zeta (x,T)=0, \end{aligned}$$
(15)
$$\begin{aligned} & \nabla u(x,t)=\langle \nu _{x,t},{\mathrm{id}}\rangle \quad \text{a.e. }(x,t) \in Q_{T}, \end{aligned}$$
(16)
$$\begin{aligned} &\langle \nu _{x,t},Z_{p}\cdot {\mathrm{id}}\rangle =\langle \nu _{x,t},Z_{p} \rangle \cdot \langle \nu _{x,t},{\mathrm{id}}\rangle \quad \text{a.e. }(x,t) \in Q_{T}, \end{aligned}$$
(17)
$$\begin{aligned} &\operatorname{supp}\nu _{x,t}\subset \bigl\{ X\in \mathbb{R}^{N}:\varphi _{p}(X)= \varphi ^{**}_{p}(X) \bigr\} \quad \text{a.e. }(x,t)\in Q_{T}, \end{aligned}$$
(18)
and
$$\begin{aligned}& u(x,0)=f_{p}(x),\quad x\in \Omega , \end{aligned}$$
(19)
in the sense of trace.
Theorem 3
Let \(f_{p}\in W^{1,p}(\Omega )\cap L^{\infty }(\Omega )\). Then problem (12)–(14) admits at least one Young measure solution.
The following existence proof follows by the ideas due to Kinderlehrer and Pedregal [39], Demoulini [38], and Yin and Wang [40].
In order to obtain the theorem above, the following functionals defined on \(W^{1,p}(\Omega )\) are considered:
$$ \mathcal{F}_{h} \bigl(v;u^{h,j-1} \bigr)= \int _{\Omega }\varphi _{p}(\nabla v)\,dx+ \frac{1}{2h} \int _{\Omega } \bigl(v-u^{h,j-1} \bigr)^{2}\,dx + \frac{\lambda }{2} \int _{\Omega } \vert v-f_{p} \vert ^{2} \,dx,\quad v\in W^{1,p}(\Omega ), $$
and
$$ \mathcal{F}_{h}^{**} \bigl(v;u^{h,j-1} \bigr)= \int _{\Omega }\varphi _{p}^{**}(\nabla v)\,dx+ \frac{1}{2h} \int _{\Omega } \bigl(v-u^{h,j-1} \bigr)^{2}\,dx + \frac{\lambda }{2} \int _{\Omega } \vert v-f_{p} \vert ^{2} \,dx,\quad v\in W^{1,p}(\Omega ), $$
where \(0< h<1\), \(u^{h,0}=f_{p}\), j is an integer and \(1\leq j\leq T/h+1\).
Lemma 5
There exists \(u^{h,j}\in W^{1,p}(\Omega )\cap L^{\infty }(\Omega )\) such that \(u^{h,j}\) is a minimum of \(\mathcal{F}_{h}^{**}(v;u^{h,j-1})\) and
$$ \inf \bigl\{ \mathcal{F}_{h} \bigl(v;u^{h,j-1} \bigr);v\in W^{1,p}(\Omega ) \bigr\} =\min \bigl\{ \mathcal{F}^{**}_{h} \bigl(v;u^{h,j-1} \bigr);v\in W^{1,p}(\Omega ) \bigr\} =\mathcal{F}^{**}_{h} \bigl(u^{h,j};u^{h,j-1} \bigr). $$
Moreover,
$$ \bigl\Vert u^{h,j} \bigr\Vert _{W^{1,p}(\Omega )}+ \bigl\Vert u^{h,j} \bigr\Vert _{L^{\infty }(\Omega )} \leq M_{1} $$
and
$$ \mathcal{F}^{**}_{h} \bigl(u^{h,j};u^{h,j-1} \bigr)\leq \Lambda \vert \nabla f_{p} \vert ^{p}+1, $$
where Λ only depends of \(\mu _{1}\) and \(\mu _{2}\).
Proof
By the relaxation theorem (cf. [43]), we get that
$$\begin{aligned}& \mathcal{F}^{**}_{h} \bigl(v;u^{h,j-1} \bigr)\leq \mathcal{F}_{h} \bigl(v;u^{h,j-1} \bigr),\quad \forall v\in W^{1,p}(\Omega ), \\& \inf \bigl\{ \mathcal{F}_{h} \bigl(v;u^{h,j-1} \bigr);v\in W^{1,p}(\Omega ) \bigr\} =\inf \bigl\{ \mathcal{F}^{**}_{h} \bigl(v;u^{h,j-1} \bigr);v\in W^{1,p}(\Omega ) \bigr\} . \end{aligned}$$
Let \(\{u^{h,j,k}\}_{k=1}^{\infty }\subset W^{1,1+\delta }(\Omega )\) be a minimizing sequence of \(\mathcal{F}_{h}\) and \(\mathcal{F}_{h}^{**}\). Then
$$\begin{aligned}& \lim_{k\to \infty } \int _{\Omega }\varphi _{p}^{**} \bigl(\nabla u^{h,j,k} \bigr)\,dx = \lim_{k\to \infty } \int _{\Omega }\varphi _{p} \bigl(\nabla u^{h,j,k} \bigr)\,dx \end{aligned}$$
(20)
and, for k sufficiently large,
$$\begin{aligned}& \bigl\Vert u^{h,j,k} \bigr\Vert _{L^{\infty }(\Omega )}\leq \bigl\Vert u^{h,j-1} \bigr\Vert _{L^{\infty }( \Omega )}+2^{-j}, \\& \begin{aligned} & \int _{\Omega }\varphi _{p}^{**} \bigl(\nabla u^{h,j,k} \bigr)\,dx + \frac{\lambda }{2} \int _{\Omega } \bigl(u^{h,j,k}-f^{\delta } \bigr)^{2}\,dx \\ &\quad \leq \mathcal{F}^{**}_{h} \bigl(u^{h,j,k};u^{h,j-1} \bigr)\leq \mathcal{F}^{**}_{h} \bigl(u^{h,j-1};u^{h,j-1} \bigr)+2^{-j} \\ &\quad = \int _{\Omega }\varphi _{p}^{**} \bigl(\nabla u^{h,j-1} \bigr)\,dx + \frac{\lambda }{2} \int _{\Omega } \bigl(u^{h,j-1}-f^{\delta } \bigr)^{2}\,dx+2^{-j}. \end{aligned} \end{aligned}$$
From the growth condition, we see that \(\{u^{h,j,k}\}_{k=1}^{\infty }\) is bounded in \(W^{1,p}(\Omega )\cap L^{\infty }(\Omega )\), and therefore
$$ \bigl\Vert u^{h,j,k} \bigr\Vert _{W^{1,p}(\Omega )}+ \bigl\Vert u^{h,j,k} \bigr\Vert _{L^{\infty }(\Omega )} \leq M_{1}, $$
where \(M_{1}\) is a constant independent of p, h, j, and k. Hence there exist \(u^{h,j}\in W^{1,p}(\Omega )\cap L^{\infty }(\Omega )\) and a subsequence of \(\{u^{h,j,k}\}_{k=1}^{\infty }\), denoted the same, such that
$$\begin{aligned}& u^{h,j,k}\to u^{h,j},\quad \textrm{weakly in }W^{1,p}( \Omega ), \\& u^{h,j,k}\to u^{h,j},\quad \textrm{strongly in }L^{p}( \Omega ), \\& \mathcal{F}^{**}_{h} \bigl(u^{h,j};u^{h,j-1} \bigr)=\inf \bigl\{ {\mathcal{F}^{**}_{h} \bigl(v;u^{h,j-1} \bigr);v \in W^{1,p}(\Omega )} \bigr\} , \\& \bigl\Vert u^{h,j} \bigr\Vert _{W^{1,p}(\Omega )}+ \bigl\Vert u^{h,j} \bigr\Vert _{L^{\infty }(\Omega )} \leq M_{1}, \end{aligned}$$
and \(\{u^{h,j,k}\}_{k=1}^{\infty }\subset L^{\infty }(\Omega )\) yields
$$\begin{aligned}& \lim_{k\to \infty } \int _{\Omega } \bigl(u^{h,j,k}-u^{h,j-1} \bigr)^{2}\,dx = \int _{\Omega } \bigl(u^{h,j}-u^{h,j-1} \bigr)^{2}\,dx \end{aligned}$$
(21)
and
$$\begin{aligned}& \lim_{k\to \infty } \int _{\Omega } \bigl(u^{h,j,k}-f_{p} \bigr)^{2}\,dx = \int _{\Omega } \bigl(u^{h,j}-f_{p} \bigr)^{2}\,dx. \end{aligned}$$
(22)
Thus \(u^{h,j}\) is a minimum of \(\mathcal{F}_{h}^{**}(v;u^{h,j-1})\) and
$$\begin{aligned} &\mathcal{F}_{h}^{**} \bigl(u^{h,j};u^{h,j} \bigr)-\mathcal{F}_{h}^{**} \bigl(u^{h,j-1};u^{h,j-1} \bigr) \\ &\quad = \int _{\Omega } \bigl(\varphi ^{**}_{p} \bigl( \nabla u^{h,j} \bigr)-\varphi ^{**}_{p} \bigl( \nabla u^{h,j-1} \bigr) \bigr)\,dx+\frac{\lambda }{2} \int _{\Omega } \bigl( \bigl(u^{h,j}-f_{p} \bigr)^{2}- \bigl(u^{h,j-1}-f_{p} \bigr)^{2} \bigr)\,dx \\ &\quad \leq \int _{\Omega }Z^{**}_{p} \bigl(\nabla u^{h,j} \bigr)\cdot \bigl(\nabla u^{h,j}- \nabla u^{h,j-1} \bigr)\,dx+\frac{\lambda }{2} \int _{\Omega } \bigl(u^{h,j}+u^{h,j-1}-2f_{p} \bigr) \bigl(u^{h,j}-u^{h,j-1} \bigr)\,dx \\ &\quad \leq - \int _{\Omega }\operatorname {div}\bigl(Z^{**}_{p} \bigl( \nabla u^{h,j} \bigr) \bigr) \bigl( u^{h,j}- u^{h,j-1} \bigr)\,dx+\frac{\lambda }{2} \int _{\Omega } \bigl(u^{h,j}+u^{h,j-1}-2f_{p} \bigr) \bigl(u^{h,j}-u^{h,j-1} \bigr)\,dx. \end{aligned}$$
Note that
$$ \frac{u^{h,j}-u^{h,j-1}}{h}=\operatorname {div}\bigl(Z^{**}_{p} \bigl(\nabla u^{h,j} \bigr) \bigr)- \lambda \bigl(u^{h,j}-f^{\delta } \bigr), $$
and then
$$\begin{aligned} \mathcal{F}_{h}^{**} \bigl(u^{h,j};u^{h,j} \bigr)-\mathcal{F}_{h}^{**} \bigl(u^{h,j-1};u^{h,j-1} \bigr)&\leq - \int _{\Omega }\frac{(u^{h,j}-u^{h,j-1})^{2}}{h}- \frac{\lambda }{2} \int _{\Omega } \bigl(u^{h,j}-u^{h,j-1} \bigr)^{2}\,dx\leq 0. \end{aligned}$$
Thus
$$\begin{aligned}& \mathcal{F}_{h}^{**} \bigl(u^{h,j};u^{h,j-1} \bigr)\leq \mathcal{F}_{h}^{**} \bigl(u^{h,j-1};u^{h,j-1} \bigr) \leq \mathcal{F}_{h}^{**} \bigl(u^{h,j-2};u^{h,j-2} \bigr)\leq \cdots \leq \mathcal{F}_{h}^{**} \bigl(u^{h,0};u^{h,0} \bigr). \end{aligned}$$
Hence
$$\begin{aligned}& \mathcal{F}_{h}^{**} \bigl(u^{h,j};u^{h,j-1} \bigr)\leq \int _{\Omega }\varphi ^{**}_{p}( \nabla f_{p})\leq \Lambda \vert \nabla f_{p} \vert ^{p}+1. \end{aligned}$$
□
Let \(\nu ^{h,j}=(v_{x}^{h,j})_{x\in \Omega }\) be the Young measure generated by \(\{\nabla u^{h,j,k}\}_{k=1}^{\infty }\) in the proof of Lemma 5. By Lemma 3, \(\nu ^{h,j}\) is a \(W^{1,p}\)-gradient Young measure. Then
$$\begin{aligned} \int _{\Omega } \bigl\langle \nu ^{h,j},\varphi ^{**}_{\delta } \bigr\rangle \,dx&= \lim_{k\to \infty } \int _{\Omega }\varphi ^{**}_{p} \bigl(\nabla u^{h,j,k} \bigr)\,dx \\ &=\lim_{k\to \infty } \int _{\Omega }\varphi _{p} \bigl(\nabla u^{h,j,k} \bigr)\,dx = \int _{\Omega } \bigl\langle \nu ^{h,j},\varphi _{p} \bigr\rangle \,dx. \end{aligned}$$
Noticing that \(\varphi ^{**}_{p}\leq \varphi _{p}\), we see that
$$\begin{aligned}& \operatorname{supp}\nu _{x}^{h,j}\subset \bigl\{ X\in R^{N}:\varphi _{p}(X)=\varphi ^{**}_{p}(X) \bigr\} ,\quad \text{a.e. }x\in \Omega . \end{aligned}$$
(23)
Thus,
$$\begin{aligned}& \bigl\langle \nu ^{h,j},Z_{p} \bigr\rangle = \bigl\langle \nu ^{h,j},Z_{p}^{**} \bigr\rangle \quad \text{a.e. in } \Omega , \\& \nabla u^{h,j}= \bigl\langle \nu ^{h,j},{\mathrm{id}}\bigr\rangle \quad \text{a.e. in }\Omega . \end{aligned}$$
Let \(\chi ^{h,j}\) be the indicator function of \([hj,h(j+1))\) and
$$\begin{aligned}& \beta ^{h,j}(t)= \textstyle\begin{cases} \frac{t}{h}-j, & t\in [hj,h(j+1)), \\ 0,&\textrm{otherwise}. \end{cases}\displaystyle \end{aligned}$$
Define
$$\begin{aligned}& u^{h}(x,t)=\sum_{0\leq j\leq T/h}\chi ^{h,j}(t) \bigl\{ u^{h,j}(x)+\beta ^{h,j}(t) \bigl(u^{h,j+1}(x)-u^{h,j}(x) \bigr) \bigr\} , \end{aligned}$$
(24)
$$\begin{aligned}& w^{h}(x,t)=\sum_{0\leq j\leq T/h}\chi ^{h,j}(t)u^{h,j}(x), \end{aligned}$$
(25)
$$\begin{aligned}& \nu ^{h}= \bigl(\nu ^{h}_{x,t} \bigr)_{(x,t)\in Q_{T}}=\sum_{0\leq j\leq T/h} \chi ^{h,j}(t)\nu _{x}^{h,j}. \end{aligned}$$
(26)
Then
$$\begin{aligned}& u^{h}(x,t)\in L^{\infty } \bigl((0,T);W^{1,p}(\Omega ) \bigr)\cap L^{\infty }(Q_{T}), \\& \nu ^{h}\in L^{1} \bigl(Q_{T}; \bigl(\mathcal{E}_{0}^{p} \bigl(\mathbb{R}^{N} \bigr) \bigr)' \bigr)\cap L^{\infty } \bigl((0,T); \bigl(\mathcal{E}_{0}^{p} \bigl(\mathbb{R}^{N} \bigr) \bigr)' \bigr), \\& u^{h}(x,0)=f_{p}(x),w^{h}(x,0)=f_{p}(x),u^{h}(x,T)=w^{h}(x,T) \quad \text{a.e. }x\in \Omega , \\& \sup_{0\leq t\leq T} \bigl\Vert u^{h} \bigr\Vert _{W^{1,p}(\Omega )}+ \bigl\Vert u^{h} \bigr\Vert _{L^{\infty }(Q_{T})}\leq M_{1}, \\& \sup_{0\leq t\leq T} \bigl\Vert w^{h} \bigr\Vert _{W^{1,p}(\Omega )}+ \bigl\Vert w^{h} \bigr\Vert _{L^{\infty }(Q_{T})}\leq M_{1}. \end{aligned}$$
Based on the facts above, we can obtain the following lemma.
Lemma 6
The functions \(u^{h}\), \(w^{h}\), and the Young measure \(\nu ^{h}\) defined above satisfy
$$\begin{aligned}& \iint _{Q_{T}} \bigl\langle \nu ^{h},Z_{p} \bigr\rangle \cdot \nabla \zeta \,dx\,dt+ \iint _{Q_{T}}\frac{\partial u^{h}}{\partial t}\zeta \,dx\,dt+\lambda \iint _{Q_{T}} \bigl(w^{h}-f_{p} \bigr)\zeta \,dx\,dt=0, \end{aligned}$$
(27)
for \(\zeta \in C^{\infty }(\overline{Q}_{T})\), with \(\zeta (x,0)=\zeta (x,T)=0\). Moreover,
$$\begin{aligned}& \sup_{0\leq t\leq T} \bigl\Vert u^{h} \bigr\Vert _{W^{1,p}(\Omega )}+ \bigl\Vert u^{h} \bigr\Vert _{L^{\infty }(Q_{T})}+ \biggl\Vert \frac{\partial u^{h}}{\partial t} \biggr\Vert _{L^{2}(Q_{T})} + \bigl\Vert \bigl\langle \nu ^{h}_{x,t},Z_{p} \bigr\rangle \bigr\Vert _{L^{(p/(p-1))}(Q_{T})} \leq M, \\& \sup_{0\leq t\leq T} \bigl\Vert w^{h} \bigr\Vert _{W^{1,p}(\Omega )}+ \bigl\Vert w^{h} \bigr\Vert _{L^{\infty }(Q_{T})}\leq M, \end{aligned}$$
where M are independent of p and h.
Proof
Let \(\xi \in C_{0}^{\infty }(\Omega )\), \(-1<{\epsilon }<1\). Then there exists \(C>0\) such that
$$ \varphi _{p}^{**}(X+{\epsilon }\nabla \xi )\leq \Lambda \bigl(1+ \vert X \vert ^{p} \bigr). $$
By Lemma 2, we can see that
$$\begin{aligned} \mathcal{F}^{**}_{h} \bigl(u^{h,j};u^{h,j-1} \bigr)&\leq \mathcal{F}_{h}^{**} \bigl(u^{h,j}+{ \epsilon }\xi ;u^{h,j-1} \bigr) \\ &= \int _{\Omega }\varphi _{p}^{**} \bigl(\nabla u^{h,j}+{\epsilon }\nabla \xi \bigr)\,dx +\frac{1}{2h} \int _{\Omega } \bigl(u^{h,j}+{\epsilon }\xi -u^{h,j-1} \bigr)^{2}\,dx \\ &\quad{} +\frac{\lambda }{2} \int _{\Omega } \bigl(u^{h,j}+{\epsilon }\xi -f^{\delta } \bigr)^{2}\,dx \\ &\leq \lim_{k\to \infty } \int \varphi _{p}^{**} \bigl(\nabla u^{h,j,k}+{ \epsilon }\nabla \xi \bigr)\,dx+\frac{1}{2h} \int _{\Omega } \bigl(u^{h,j}+{ \epsilon }\xi -u^{h,j-1} \bigr)^{2}\,dx \\ &\quad{}+\frac{\lambda }{2} \int _{\Omega } \bigl(u^{h,j}+{ \epsilon }\xi -f^{\delta } \bigr)^{2}\,dx \\ &= \int _{\Omega } \int _{\mathbb{R}^{N}}\varphi _{p}^{**}(X+{\epsilon } \nabla \xi )\,d\nu _{x}^{h,j}(X)\,dx+\frac{1}{2h} \int _{\Omega } \bigl(u^{h,j}+{ \epsilon }\xi -u^{h,j-1} \bigr)^{2}\,dx \\ &\quad{}+\frac{\lambda }{2} \int _{\Omega } \bigl(u^{h,j}+{ \epsilon }\xi -f^{\delta } \bigr)^{2}\,dx, \end{aligned}$$
which implies the equilibrium equation
$$\begin{aligned}& \int _{\Omega } \bigl\langle \nu ^{h,j},Z^{**}_{p} \bigr\rangle \cdot \nabla \xi \,dx+ \frac{1}{h} \int _{\Omega } \bigl(u^{h,j}-u^{h,j-1} \bigr)\xi \,dx+\lambda \int _{\Omega } \bigl(u^{h,j}-f_{p} \bigr)\xi \,dx=0, \end{aligned}$$
\(\forall \xi \in C_{0}^{\infty }(\Omega )\). At the minimizer \(u^{h,j}\), the Gâteaux derivative of \(\mathcal{F}^{**}_{h}\) is zero, and we obtain
$$ \int _{\Omega }Z_{p}^{**} \bigl(\nabla u^{h,j}(x) \bigr)\cdot \nabla \xi \,dx+ \frac{1}{h} \int _{\Omega } \bigl(u^{h,j}-u^{h,j-1} \bigr)\xi \,dx+\lambda \int _{\Omega } \bigl(u^{h,j}-f_{p} \bigr)\xi \,dx=0, $$
\(\forall \xi \in C_{0}^{\infty }(\Omega )\). Thus
$$\begin{aligned}& Z_{p} \bigl(\nabla u^{h,j}(x) \bigr)= \bigl\langle \nu _{x}^{h,j},Z_{p} \bigr\rangle = \bigl\langle \nu _{x}^{h,j},Z^{**}_{p} \bigr\rangle =Z^{**}_{p} \bigl(\nabla u^{h,j}(x) \bigr), \quad \text{a.e. }x\in \Omega , \end{aligned}$$
(28)
$$\begin{aligned}& \int _{\Omega } \bigl\langle \nu ^{h,j},Z_{p} \bigr\rangle \cdot \nabla \xi \,dx+ \frac{1}{h} \int _{\Omega } \bigl(u^{h,j}-u^{h,j-1} \bigr)\xi \,dx+\lambda \int _{\Omega } \bigl(u^{h,j}-f_{p} \bigr)\xi \,dx=0, \end{aligned}$$
(29)
and, by Lemma 5, we get the estimate
$$\begin{aligned} \bigl\Vert \bigl\langle \nu ^{h,j},Z_{p} \bigr\rangle \bigr\Vert _{L^{p/(p-1)}(\Omega )}&= \bigl\Vert Z_{p}^{**} \bigl( \nabla u^{h,j} \bigr) \bigr\Vert _{L^{p/(p-1)}(\Omega )}\leq \Lambda \biggl( \int _{\Omega } \bigl\vert \nabla u^{h,j} \bigr\vert ^{p}\,dx \biggr)^{(p-1)/p} \\ &\leq \Lambda \bigl\Vert u^{h,j} \bigr\Vert ^{\delta }_{W^{1,p}(\Omega )} \leq \Lambda (M_{1}+1). \end{aligned}$$
From (24)–(26) and (29),
$$ \int _{\Omega } \bigl\langle \nu ^{h},Z_{p} \bigr\rangle \cdot \nabla \xi \,dx+ \frac{\partial u^{h}}{\partial t}\xi \,dx+\lambda \int _{\Omega } \bigl(w^{h}-f_{p} \bigr) \xi \,dx=0,\quad \forall \xi \in C^{\infty }(\overline{\Omega }), \text{a.e. }t \in [0,T], $$
which implies that
$$\begin{aligned}& \iint _{Q_{T}} \bigl\langle \nu ^{h},Z_{p} \bigr\rangle \cdot \nabla \zeta \,dx\,dt+ \iint _{Q_{T}}\frac{\partial u^{h}}{\partial t}\zeta \,dx\,dt+\lambda \iint _{Q_{T}} \bigl(w^{h}-f_{p} \bigr)\zeta \,dx\,dt=0, \end{aligned}$$
(30)
for \(\zeta \in C^{\infty }(\overline{Q}_{T})\), with \(\zeta (x,0)=\zeta (x,T)=0\). From the direct calculation, we see that
$$\begin{aligned}& \bigl\Vert \bigl\langle \nu ^{h}_{x,t},Z_{p} \bigr\rangle \bigr\Vert _{L^{p/(p-1)}(Q_{T})}\leq T \sup_{0\leq j\leq T/h} \bigl\Vert \bigl\langle \nu ^{h,j}_{x},Z_{p} \bigr\rangle \bigr\Vert _{L^{p/(p-1)}( \Omega )}\leq T(M_{1}+1). \end{aligned}$$
Because
$$\begin{aligned}& \frac{\partial u^{h}(x,t)}{\partial t}=\frac{1}{h}\sum_{0\leq j \leq T/h}\chi ^{h,j}(t) \bigl(u^{h,j+1}(x)-u^{h,j}(x) \bigr)\in L^{\infty } \bigl((0,T);W^{1,p}( \Omega ) \bigr)\cap L^{\infty }(Q_{T}), \end{aligned}$$
\(\partial _{t} u^{h}\) can be chosen as the test function in (27), and therefore
$$\begin{aligned}& \iint _{Q_{T}} \bigl\langle \nu ^{h},Z_{p} \bigr\rangle \cdot \nabla \frac{\partial u^{h}}{\partial t} \,dx\,dt+ \iint _{Q_{T}} \biggl( \frac{\partial u^{h}}{\partial t} \biggr)^{2} \,dx\,dt \\& \quad{} +\lambda \iint _{Q_{T}} \bigl\vert w^{h}-f_{p} \bigr\vert ^{p-2} \bigl(w^{h}-f_{p} \bigr) \frac{\partial u^{h}}{\partial t} \,dx\,dt=0. \end{aligned}$$
Then
$$\begin{aligned} & \iint _{Q_{T}} \biggl(\frac{\partial u^{h}}{\partial t} \biggr)^{2} \,dx+ \lambda \int _{\Omega } \bigl\vert u^{h}(x,T)-f_{p} \bigr\vert ^{2}\,dx \\ &\quad =- \iint _{Q_{T}} \bigl\langle \nu ^{h},Z_{p} \bigr\rangle \cdot \nabla \frac{\partial u^{h}}{\partial t} \,dx\,dt +\lambda \int _{\Omega } \bigl(u^{h}(x,T)-f_{p} \bigr)f_{p}\,dx \\ &\quad =-\sum_{0\leq j\leq \frac{T}{h}} \iint _{Q_{T}}\chi ^{h,j}(t) \bigl\langle \nu ^{h},Z_{p} \bigr\rangle \cdot \frac{\partial (\nabla u^{h})}{\partial t}\,dx\,dt \\ &\quad\quad {} +\lambda \int _{\Omega } \bigl(u^{h}(x,T)-f_{p} \bigr)f_{p}\,dx \\ &\quad =-\sum_{0\leq j\leq \frac{T}{h}} \iint _{Q_{T}}\chi ^{h,j}(t) \frac{\partial (\langle \nu ^{h},Z_{p} \rangle \cdot \nabla u^{h})}{\partial t}\,dx\,dt \\ &\quad\quad {} +\lambda \int _{\Omega } \bigl(u^{h}(x,T)-f_{p} \bigr)f_{p}\,dx \\ &\quad =- \int _{\Omega } \bigl\langle \nu ^{h}_{x,T},Z_{p} \bigr\rangle \cdot \nabla u^{h}(x,T)\,dx+ \int _{\Omega } \bigl\langle \nu ^{h}_{x,0},Z_{p} \bigr\rangle \cdot \nabla u^{h}(x,0)\,dx \\ &\quad\quad {} +\lambda \int _{\Omega } \bigl(u^{h}(x,T)-f_{p} \bigr)f_{p}\,dx \\ &\quad \leq \bigl\Vert \bigl\langle \nu ^{h}_{x,T},Z_{p} \bigr\rangle \bigr\Vert _{L^{p/(p-1)}(\Omega )} \bigl\Vert \nabla u^{h}(x,T) \bigr\Vert _{L^{p}(\Omega )}+ \bigl\Vert \bigl\langle \nu ^{h}_{x,0},Z_{p} \bigr\rangle \bigr\Vert _{L^{p/(p-1)}(\Omega )} \Vert \nabla f_{p} \Vert _{L^{p}(\Omega )} \\ &\qquad {} +\frac{\lambda }{2} \int _{\Omega } \bigl\vert u^{h}(x,T)-f_{p} \bigr\vert ^{2}\,dx+ \frac{\lambda }{2} \int _{\Omega }f_{p}^{2}\,dx, \end{aligned}$$
and therefore
$$\begin{aligned} & \iint _{Q_{T}} \biggl(\frac{\partial u^{h}}{\partial t} \biggr)^{2} \,dx+ \frac{\lambda }{2} \int _{\Omega } \bigl\vert u^{h}(x,T)-f_{p} \bigr\vert ^{2}\,dx \\ &\quad \leq \bigl\Vert \bigl\langle \nu ^{h}_{x,T},Z_{p} \bigr\rangle \bigr\Vert _{L^{p/(p-1)}(\Omega )} \bigl\Vert \nabla u^{h}(x,T) \bigr\Vert _{L^{p}(\Omega )} \\ &\qquad {} + \bigl\Vert \bigl\langle \nu ^{h}_{x,0},Z_{p} \bigr\rangle \bigr\Vert _{L^{p/(p-1)}( \Omega )} \Vert \nabla f_{p} \Vert _{L^{p}(\Omega )}+\frac{\lambda }{2} \int _{\Omega }f_{p}^{2}\,dx, \\ &\quad \leq 2\Lambda M_{1}(M_{1}+1)+\frac{\lambda }{2} \int _{\Omega }f_{p}^{2}\,dx, \end{aligned}$$
which implies
$$\begin{aligned}& \biggl\Vert \frac{\partial u^{h}}{\partial t} \biggr\Vert _{L^{2}(Q_{T})} \leq \bigl(2 \Lambda +\lambda ^{1/2}+1 \bigr) (M_{1}+1). \end{aligned}$$
□
Let
$$ v^{h,k}(x,t)=\sum_{0\leq j\leq \frac{T}{h}}\chi ^{h,j}(t)u^{h,j,k}(x), $$
for \(k\geq 1\). By Lemma 5, \(\{v^{h,k}\}_{k=1}^{\infty }\subset L^{\infty }((0,T);W^{1,p}(\Omega )) \cap L^{\infty }(Q_{T})\), \(\{v^{h,k}\}_{k=1}^{\infty }\) is the \(W^{1,p}(Q_{T})\)-gradient generating sequence of \(\nu ^{h}\) and
$$\begin{aligned}& \sup_{0\leq t\leq T} \bigl\Vert v^{h,k} \bigr\Vert _{W^{1,p}(\Omega )}+ \bigl\Vert v^{h,k} \bigr\Vert _{L^{\infty }(Q_{T})}\leq M_{1}. \end{aligned}$$
(31)
When \(h\to 0\) and \(k\to \infty \), we have the following lemma.
Lemma 7
There exist \(u,w\in L^{\infty }((0,T);W^{1,p}(\Omega ))\cap L^{\infty }(Q_{T})\), \(\nu \in L^{\infty }((0,T);(\mathcal{E}^{p}(\mathbb{R}^{N}))')\), \(\frac{\partial u}{\partial t}\in L^{2}(Q_{T})\), and subsequences of \(\{u^{h}\}_{0< h<1}\), \(\{w^{h}\}_{0< h<1}\), \(\{v^{h,k}\}_{0< h<1,k=1}^{\infty }\), and \(\{\nu ^{h}\}_{0< h<1}\), denoted by \(\{u^{h_{m}}\}_{m=1}^{\infty }\), \(\{w^{h_{m}}\}_{m=1}^{\infty }\), \(\{v^{m}=v^{h_{m},k_{m}}\}_{m=1}^{\infty }\), and \(\{\nu ^{h_{m}}\}_{m=1}^{\infty }\), respectively, such that
$$\begin{aligned}& u^{h_{m}}\to u, \quad \textrm{weakly in } L^{\infty } \bigl((0,T);W^{1,p}( \Omega ) \bigr)\cap L^{\infty }(Q_{T}) \textrm{ and strongly in } L^{p}(Q_{T}), \\& \frac{\partial u^{h_{m}}}{\partial t} \to \frac{\partial u}{\partial t}, \quad \textrm{weakly in } L^{2}(Q_{T}), \\& w^{h_{m}}\to u, \quad \textrm{weakly in } L^{\infty } \bigl((0,T);W^{1,p}( \Omega ) \bigr)\cap L^{\infty }(Q_{T}) \textrm{ and strongly in } L^{p}(Q_{T}), \\& v^{m}\to v, \quad \textrm{weakly in } L^{\infty } \bigl((0,T);W^{1,p}(\Omega ) \bigr) \cap L^{\infty }(Q_{T}) \textrm{ and strongly in } L^{p}(Q_{T}), \\& \nu ^{h_{m}} \to \nu \quad \textrm{weakly in }L^{1} \bigl(Q_{T}; \bigl(\mathcal{E}^{p} \bigl( \mathbb{R}^{N} \bigr) \bigr)' \bigr), \textrm{ weakly* in }L^{\infty } \bigl(Q_{T}; \mathcal{M} \bigl(\mathbb{R}^{N} \bigr) \bigr), \\& \textrm{and in } L^{\infty } \bigl(Q_{T}; \bigl(\mathcal{E}^{p} \bigl(\mathbb{R}^{N} \bigr) \bigr)' \bigr) \textrm{ in the bitting sense} \end{aligned}$$
and
$$\begin{aligned}& u(x,t)=v(x,t),\quad \textit{a.e. }(x,t)\in Q_{T}, \\& u(x,0)=f_{p}(x),\quad x\in \Omega . \\& \nabla u=\nabla w=\nabla v=\langle \nu ,{\mathrm{id}}\rangle . \end{aligned}$$
Proof
From Lemma 6, we see that \(\{u^{h}\}_{0< h<1}\) and \(\{w^{h}\}_{0< h<1}\) are bounded in \(L^{\infty }((0,T); W^{1,p}(\Omega ))\cap L^{\infty }(Q_{T})\), \(\{\partial _{t} u^{h}\}_{0< h<1}\) is bounded in \(L^{2}(Q_{T})\), \(\{\langle \nu ^{h},Z_{p} \rangle \}_{0< h<1}\) is bounded in \(L^{p/(p-1)}(Q_{T})\), \(\{\nu ^{h}\}_{0< h<1}\) is bounded in \(L^{\infty }((0,T);(\mathcal{E}_{0}^{p}(\mathbb{R}^{N}))')\), and the bounds are independent of δ and h. By weak compactness, we can obtain the convergence of the subsequences \(\{u^{h_{m}}\}_{m=1}^{\infty }\), \(\{w^{h_{m}}\}_{m=1}^{\infty }\), and \(\{\nu ^{h_{m}}\}_{m=1}^{\infty }\). From (26), we can see that (12) holds, namely
$$ u(x,0)=f_{p}(x),\quad x\in \Omega . $$
From (31), by Lemma 4 and the weak sequential precompactness of \(L^{\infty }((0,T); W^{1,p}(\Omega ))\cap L^{\infty }(Q_{T})\), there exist \(w\in L^{\infty }((0,T); W^{1,p}(\Omega ))\cap L^{\infty }(Q_{T})\) and a diagonal subsequence of \(\{v^{h_{m},k}\}_{m=1}^{\infty }\), denoted by \(\{v^{m}=w^{h_{m},k_{m}}\}_{m=1}^{\infty }\), such that \(\{v^{m}\}_{m=1}^{\infty }\) converges to w weakly in \(L^{\infty }((0,T); W^{1,p}(\Omega ))\cap L^{\infty }(Q_{T})\) and strongly in \(L^{p}(Q_{T})\),
$$ \sup_{1\leq j\leq T/h_{m}} \bigl\Vert u^{h_{m},j,k_{m}}-u^{h_{m},j} \bigr\Vert _{L^{2}( \Omega )}\leq h_{m}, $$
\(\{v^{m}\}_{m=1}^{\infty }\subset L^{\infty }((0,T); W^{1,p}(\Omega )) \cap L^{\infty }(Q_{T})\),
$$\begin{aligned}& \sup_{0\leq t\leq T} \bigl\Vert w^{m} \bigr\Vert _{W^{1,p}(\Omega )}+ \bigl\Vert w^{m} \bigr\Vert _{L^{\infty }(Q_{T})}\leq M_{1}, \end{aligned}$$
and \(\{\nabla v^{m}\}_{m=1}^{\infty }\) is the \(W^{1,1}(Q_{T})\)-bitting generating sequence of ν, and then
$$ \nabla v=\langle \nu ,{\mathrm{id}}\rangle . $$
Now let us prove that
$$ u(x,t)=v(x,t),\quad \text{a.e. } (x,t)\in Q_{T}. $$
Since
$$\begin{aligned}& u^{h}(x,t)=\sum_{0\leq j\leq T/h}\chi ^{h,j}(t) \bigl\{ u^{h,j}(x)+\beta ^{h,j}(t) \bigl(u^{h,j+1}(x)-u^{h,j}(x) \bigr) \bigr\} , \\& v^{h,k}(x,t)=\sum_{0\leq j\leq T/h}\chi ^{h,j}(t)u^{h,j,k}(x), \end{aligned}$$
by Lemma 5, we have
$$\begin{aligned} & \iint _{Q_{T}} \bigl\vert u^{h}(x,t)-v^{h,k}(x,t) \bigr\vert ^{2}\,dx\,dt \\ &\quad \quad {}+ 2\sum_{0\leq j\leq \frac{T}{h}} \biggl( \iint _{Q_{T}} \bigl\vert \chi ^{h,j}(t) \lambda ^{h,j}(t) \bigl(u^{h,j+1}-u^{h,j}(x) \bigr) \bigr\vert ^{2}\,dx\,dt \\ &\qquad {} + \iint _{Q_{T}} \bigl\vert \chi ^{h,j}(t) \bigl(u^{h,j}(x)-u^{h,j,k}(x) \bigr) \bigr\vert ^{2} \,dx\,dt \biggr) \\ &\quad \leq 2h \sum_{0\leq j\leq \frac{T}{h}} \biggl( \int _{\Omega } \bigl\vert u^{h,j+1}(x)-u^{h,j}(x) \bigr\vert ^{2}\,dx + \int _{\Omega } \bigl\vert u^{h,j}(x)-u^{h,j,k}(x) \bigr\vert ^{2}\,dx \biggr) \\ &\quad \leq h^{2}\sum_{0\leq j\leq \frac{T}{h}} \bigl(\mathcal{F}_{h} \bigl(u^{h,j+1};u^{h,j} \bigr)+ \mathcal{F}_{h} \bigl(u^{h,j,k};u^{h,j} \bigr) \bigr) \\ &\quad \leq 8hT\sum_{0\leq j\leq \frac{T}{h}}\mathcal{F}_{h} \bigl(u^{h,j+1};u^{h,j} \bigr)+ \frac{4hT}{k} \\ &\quad \leq 8hT\Lambda \int _{\Omega } \vert \nabla f_{p} \vert \,dx+4hT \\ &\quad \rightarrow 0 \quad (h\to 0). \end{aligned}$$
Since \(\{u^{h_{m}}\}_{m=1}^{\infty }\) converges to u in \(L^{p}(Q_{T})\) and \(\{v^{h_{m},k_{m}}\}_{m=1}^{\infty }\) converges to w in \(L^{p}(Q_{T})\), we see that
$$ u(x,t)=v(x,t),\quad \text{a.e. }(x,t)\in Q_{T}. $$
Hence
$$ \nabla u=\nabla v=\langle \nu ,{\mathrm{id}}\rangle , $$
which implies (7). Similar to the above arguments, we also obtain that \(\{v^{h_{m}}\}_{m=1}^{\infty }\) converges to u in \(L^{p}(Q_{T})\). □
Proof of Theorem 3
From Lemmas 6 and 7, we can obtain
$$\begin{aligned}& \iint _{Q_{T}}\langle \nu ,Z_{p} \rangle \cdot \nabla \zeta \,dx\,dt+ \iint _{Q_{T}}\frac{\partial u}{\partial t}\zeta \,dx\,dt+\lambda \iint _{Q_{T}}(u-f_{p}) \zeta \,dx\,dt=0, \end{aligned}$$
(32)
for \(\zeta \in C^{\infty }(\overline{Q}_{T})\), with \(\zeta (x,0)=\zeta (x,T)=0\). Therefore, if we prove (8), we will obtain the weak solution of the problem (3)–(5). Let \(\{u^{h,j,k}\}_{k=1}^{\infty }\subset W^{1,p}(\Omega )\) be a minimizing sequence of \(\mathcal{F}_{h}\) in the proof of Lemma 5. For all \(\xi \in W^{1,p}(\Omega )\), we see that
$$\begin{aligned} &\lim_{k\to \infty } \int _{\Omega }Z^{**}_{p} \bigl(\nabla u^{h,j,k} \bigr)\cdot \nabla \xi \,dx+\frac{1}{h} \int _{\Omega } \bigl(u^{h,j,k}-u^{h,j-1} \bigr)\xi \,dx+ \lambda \int _{\Omega } \bigl(u^{h,j,k}-f_{p} \bigr)\xi \,dx \\ &\quad = \int _{\Omega } \bigl\langle \nu ^{h,j},Z^{**}_{p} \bigr\rangle \cdot \nabla \xi \,dx+\frac{1}{h} \int _{\Omega } \bigl(u^{h,j}-u^{h,j-1} \bigr)\xi \,dx+ \lambda \int _{\Omega } \bigl(u^{h,j}-f_{p} \bigr)\xi \,dx=0 \end{aligned}$$
and
$$\begin{aligned} &\lim_{k\to \infty } \biggl\vert \int _{\Omega } \bigl(Z^{**}_{p} \bigl(\nabla u^{h,j,k} \bigr)- \bigl\langle \nu ^{h,j},Z^{**}_{p} \bigr\rangle \bigr)\cdot \bigl(\nabla u^{h,j,k}- \nabla u^{h,j} \bigr) \,dx \biggr\vert \\ &\quad = \lim_{k\to \infty } \biggl\vert \int _{\Omega }\frac{u^{h,j,k}-u^{h,j}}{h} \bigl(u^{h,j,k}-u^{h,j} \bigr)\,dx \\ &\qquad {} +\lambda \int _{\Omega } \bigl( \bigl(u^{h,j,k}-f_{p} \bigr)- \bigl(u^{h,j}-f_{p} \bigr) \bigr) \bigl(u^{h,j,k}-u^{h,j} \bigr)\,dx \biggr\vert \\ &\quad \leq 1/h\lim_{k\to \infty } \bigl\Vert u^{h,j,k}-u^{h,j} \bigr\Vert ^{2}_{L^{2}( \Omega )} \\ &\qquad {} +\lambda \lim_{k\to \infty } \biggl( \int _{\Omega } \bigl\vert \bigl\vert u^{h,j,k}-f_{p} \bigr\vert ^{2}- \bigl\vert u^{h,j}-f_{p} \bigr\vert ^{2} \bigr\vert \,dx \biggr)^{1/2} \biggl( \int _{\Omega } \bigl\vert u^{h,j,k}-u^{h,j} \bigr\vert ^{2}\,dx \biggr)^{1/2} \\ &\quad =0. \end{aligned}$$
Since \(Z_{p}^{**}(\nabla u^{h,j,k})\cdot \nabla u^{h,j,k}\) converges weakly to \(\langle \nu ^{h,j},Z^{**}_{p}\cdot {\mathrm{id}}\rangle \) in \(L^{1}(\Omega )\), \(Z_{p}^{**}(\nabla u^{h,j,k})\) converges weakly to \(\langle \nu ^{h,j},Z^{**}_{p} \rangle \) in \(L^{p/(p-1)}(\Omega )\), \(\nabla u^{h,j,k}\) converges weakly to \(\langle \nu ^{h,j},{\mathrm{id}}\rangle \) in \(L^{p}(\Omega )\) as k tends to infinity, we get that
$$\begin{aligned}& \bigl\langle \nu _{x}^{h,j},Z_{p}^{**} \cdot {\mathrm{id}}\bigr\rangle = \bigl\langle \nu _{x}^{h,j},Z^{**}_{p} \bigr\rangle \cdot \bigl\langle \nu _{x}^{h,j},{\mathrm{id}}\bigr\rangle \quad \text{a.e. }x\in \Omega . \end{aligned}$$
Thus (23) implies that
$$\begin{aligned}& \bigl\langle \nu _{x}^{h,j},Z_{p}\cdot {\mathrm{id}}\bigr\rangle = \bigl\langle \nu _{x}^{h,j},Z_{p} \bigr\rangle \cdot \bigl\langle \nu _{x}^{h,j},{\mathrm{id}}\bigr\rangle \quad \text{a.e. }x\in \Omega . \end{aligned}$$
By the definition of \(\nu ^{h}\) in (26), we see that
$$\begin{aligned}& \bigl\langle \nu _{x,t}^{h},Z_{p}\cdot {\mathrm{id}}\bigr\rangle = \bigl\langle \nu _{x,t}^{h},Z_{p} \bigr\rangle \cdot \bigl\langle \nu _{x,t}^{h},{\mathrm{id}}\bigr\rangle \quad \text{a.e. }x\in Q_{T}. \end{aligned}$$
From Lemma 7, we can obtain that \(\langle \nu _{x,t}^{h_{m}},Z_{p}\cdot {\mathrm{id}}\rangle \) converges weakly to \(\langle \nu _{x,t},Z_{p}\cdot {\mathrm{id}}\rangle \) in the biting sense, \(\langle \nu _{x,t}^{h_{m}},Z_{p} \rangle \) converges weakly to \(\langle \nu _{x,t},Z_{p} \rangle \) in \(L^{p/(p-1)}(Q_{T})\), \(\langle \nu _{x,t}^{h_{m}},{\mathrm{id}}\rangle \) converges weakly to \(\langle \nu _{x,t},{\mathrm{id}}\rangle \) in \(L^{p}(Q_{T})\) as m tends to infinity. Thus for all \(\eta \in C^{\infty }(\overline{Q}_{T})\), with \(\eta (x,0)=\eta (x,T)=0\), (17) and (19) imply that
$$\begin{aligned} & \biggl\vert \iint _{Q_{T}} \bigl( \bigl\langle \nu ^{h_{m}},Z_{p} \bigr\rangle -\langle \nu ,Z_{p} \rangle \bigr)\cdot \bigl\langle \nu ^{h_{m}},{\mathrm{id}}\bigr\rangle \eta \,dx\,dt \biggr\vert \\ &\quad \leq \biggl\vert \iint _{Q_{T}} \bigl( \bigl\langle \nu ^{h_{m}},Z_{p} \bigr\rangle - \langle \nu ,Z_{p} \rangle \bigr) \cdot \nabla \bigl(w^{h_{m}}\eta \bigr)\,dx\,dt \biggr\vert \\ &\quad\quad {} + \biggl\vert \iint _{Q_{T}}w^{h_{m}} \bigl( \bigl\langle \nu ^{h_{m}},Z_{p} \bigr\rangle -\langle \nu ,Z_{p} \rangle \bigr) \cdot \nabla \eta \,dx\,dt \biggr\vert \\ &\quad \leq \biggl\vert \iint _{Q_{T}} \biggl( \frac{\partial u^{h_{m}}}{\partial t}-\frac{\partial u}{\partial t} \biggr)w^{h_{m}}\eta \,dx\,dt \biggr\vert +\lambda \biggl\vert \iint _{Q_{T}} \bigl( \bigl(w^{h_{m}}-f_{p} \bigr) -(u-f_{p}) \bigr)w^{h_{m}}\eta \,dx\,dt \biggr\vert \\ &\qquad {} + \biggl\vert \iint _{Q_{T}}w^{h_{m}} \bigl( \bigl\langle \nu ^{h_{m}},Z_{p} \bigr\rangle -\langle \nu ,Z_{p} \rangle \bigr) \cdot \nabla \eta \,dx\,dt \biggr\vert \\ &\quad \leq \biggl\vert \iint _{Q_{T}} \biggl( \frac{\partial u^{h_{m}}}{\partial t}-\frac{\partial u}{\partial t} \biggr)u\eta \,dx\,dt \biggr\vert + \biggl\vert \iint _{Q_{T}} \biggl( \frac{\partial u^{h_{m}}}{\partial t}-\frac{\partial u}{\partial t} \biggr) \bigl(w^{h_{m}}-u \bigr)\eta \,dx\,dt \biggr\vert \\ &\qquad {} +\lambda \biggl\vert \iint _{Q_{T}} \bigl( \bigl(w^{h_{m}}-f_{p} \bigr)-(u-f_{p}) \bigr)w^{h_{m}} \eta \,dx\,dt \biggr\vert \\ &\qquad {} + \biggl\vert \iint _{Q_{T}}u \bigl( \bigl\langle \nu ^{h_{m}},Z_{p} \bigr\rangle - \langle \nu ,Z_{p} \rangle \bigr) \cdot \nabla \eta \,dx \,dt \biggr\vert \\ &\qquad {} + \biggl\vert \iint _{Q_{T}} \bigl(w^{h_{m}}-u \bigr) \bigl( \bigl\langle \nu ^{h_{m}},Z_{p} \bigr\rangle -\langle \nu ,Z_{p} \rangle \bigr) \cdot \nabla \eta \,dx\,dt \biggr\vert \\ &\quad \rightarrow 0\quad (m\to \infty ). \end{aligned}$$
Hence,
$$\begin{aligned} & \biggl\vert \iint _{Q_{T}} \bigl( \bigl\langle \nu ^{h_{m}},Z_{p} \bigr\rangle \cdot \bigl\langle \nu ^{h_{m}},{\mathrm{id}}\bigr\rangle -\langle \nu ,Z_{p} \rangle \cdot \langle \nu ,{\mathrm{id}}\rangle \bigr)\eta \,dx\,dt \biggr\vert \\ &\quad \leq \biggl\vert \iint _{Q_{T}} \bigl( \bigl\langle \nu ^{h_{m}},Z_{p} \bigr\rangle -\langle \nu ,Z_{p} \rangle \bigr)\cdot \bigl\langle \nu ^{h_{m}}, {\mathrm{id}}\bigr\rangle \eta \,dx\,dt \biggr\vert \\ &\quad \quad {} + \biggl\vert \iint _{Q_{T}}\langle \nu ,Z_{p} \rangle \cdot \bigl( \bigl\langle \nu ^{h_{m}},{\mathrm{id}}\bigr\rangle -\langle \nu ,{\mathrm{id}}\rangle \bigr)\eta \,dx\,dt \biggr\vert \\ &\quad \rightarrow 0\quad (m\to \infty ). \end{aligned}$$
So \(\langle \nu _{x,t}^{h_{m}},Z_{p} \rangle \cdot \langle \nu ^{h_{m}}_{x,t}, {\mathrm{id}}\rangle \) weakly converges to \(\langle \nu _{x,t},Z_{p} \rangle \cdot \langle \nu _{x,t},{\mathrm{id}}\rangle \) in \(L^{1}(Q_{T})\). Since \(\langle \nu _{x,t}^{h_{m}},Z_{p}\cdot {\mathrm{id}}\rangle \) converges to \(\langle \nu _{x,t},Z_{p}\cdot {\mathrm{id}}\rangle \) in the bitting sense, we obtain that
$$\begin{aligned}& \langle \nu _{x,t},Z_{p}\cdot {\mathrm{id}}\rangle =\langle \nu _{x,t},Z_{p} \rangle \cdot \langle \nu _{x,t},{\mathrm{id}}\rangle \quad \text{a.e. }(x,t) \in Q_{T}\setminus E_{j}, \forall j\geq 1. \end{aligned}$$
Since \(\langle \nu _{x,t},Z_{p}\cdot {\mathrm{id}}\rangle \in L^{1}(Q_{T})\),
$$\begin{aligned}& \langle \nu _{x,t},Z_{p}\cdot {\mathrm{id}}\rangle =\langle \nu _{x,t},Z_{p} \rangle \cdot \langle \nu _{x,t},{\mathrm{id}}\rangle \quad \text{a.e. }(x,t) \in Q_{T}, \end{aligned}$$
which implies (8). Hence, u is the desired Young measure solution of problem (3)–(5). The proof of Theorem 3 is complete. □
Remark 2
Let u be the Young measure solution of problem (12)–(14) obtained in the proof of Theorem 3. Then from the proof we see that there exists a constant M depending only on \(\Vert f_{p}\Vert _{W^{1,p}(\Omega )}\), \(\Vert f_{p}\Vert _{L^{\infty }(\Omega )}\), Λ, and \(\operatorname{meas}(\Omega )\), but independent of p and T, such that
$$\begin{aligned}& \Vert u \Vert _{L^{\infty }((0,T);W^{1,p}(\Omega ))}+ \Vert u \Vert _{L^{\infty }(Q_{T})} \leq M, \\& \biggl\Vert \frac{\partial u}{\partial t} \biggr\Vert _{L^{2}(Q_{T})}\leq M, \end{aligned}$$
namely, \(u\in L^{\infty }(\mathbb{R}^{+}; W^{1,p}(\Omega ))\cap L^{\infty }(Q_{\infty })\), \(\partial u/\partial t\in L^{2}(Q_{\infty })\), where \(Q_{\infty }=\Omega \times \mathbb{R}^{+}\).
Existence of solution to problem (3)–(5)
In this subsection, we consider the limit case of problem (3)–(5), namely, \(p\to 1\).
Proof of Theorem 2
Let \(u_{p}\) be the Young measure solution of problem (12)–(14) with the initial data \(f_{p}\) with respect to the \(W^{1,p}(Q_{T})\)-gradient Young measures \(\nu ^{p}\) generated by the sequence \(\{\nabla w^{p,k}\}_{k=1}^{\infty }\), which we obtained in the proof of Theorem 3. We see that
$$\begin{aligned}& u_{p}\in L^{\infty } \bigl((0,T);W^{1,p}(\Omega ) \bigr) \cap L^{\infty }(Q_{T}), \quad \frac{\partial u_{p}}{\partial t}\in L^{2}(Q_{T}), \\& \nu _{p}\in L^{\infty } \bigl((0,T); \bigl(\mathcal{E}_{0}^{p} \bigl(\mathbb{R}^{N} \bigr) \bigr)' \bigr),\quad \langle \nu _{p},Z_{p} \rangle \in L^{p/(p-1)}(Q_{T}), \\& w^{p,k}\in L^{\infty } \bigl((0,T);W^{1,p}(\Omega ) \bigr) \cap L^{\infty }(Q_{T}), \end{aligned}$$
and there exists a constant \(M_{0}\) depending only on \(\Vert f_{p}\Vert _{W^{1,p(\Omega )}}\), \(\Vert f_{p}\Vert _{L^{\infty }(\Omega )}\), Λ, and \(\operatorname{meas}(\Omega )\), but independent of p, such that
$$\begin{aligned}& \Vert u_{p} \Vert _{L^{\infty }((0,T);W^{1,p}(\Omega ))}+ \Vert u_{p} \Vert _{L^{\infty }(Q_{T})} \leq M_{0},\quad\quad \biggl\Vert \frac{\partial u_{p}}{\partial t} \biggr\Vert _{L^{2}(Q_{T})} \leq M_{0}, \\& \Vert \nu _{p} \Vert _{L^{\infty }((0,T);(\mathcal{E}^{p}(\mathbb{R}^{N}))')}\leq M_{0}, \quad \quad \bigl\Vert \langle \nu _{p},Z_{p} \rangle \bigr\Vert _{L^{p/(p-1)}(Q_{T})}\leq M_{0}, \\& \sup_{0\leq t\leq T} \bigl\Vert w^{p,k} \bigr\Vert _{W^{1,p}(\Omega )}+ \bigl\Vert w^{p,k} \bigr\Vert _{L^{\infty }(Q_{T})}\leq M_{0}. \end{aligned}$$
So there exist \(u\in L^{\infty }((0,T);\operatorname{BV}(\Omega ))\cap L^{\infty }(Q_{T})\) with \(\partial u/\partial t\in L^{2}(Q_{T})\) and a subsequence of \(\{u_{p}\}_{1< p<2}\), denoted by \(\{u_{p_{m}}\}_{m=1}^{\infty }\), such that
$$\begin{aligned}& u_{p_{m}}\to u,\quad \textrm{strongly in } L^{1}(Q_{T}), \\& \frac{\partial u_{p_{m}}}{\partial t}\to \frac{\partial u}{\partial t},\quad \textrm{weakly in } L^{2}(Q_{T}). \end{aligned}$$
By Lemma 4, there exist a \(W^{1,1}(Q_{T})\)-gradient Young measure \(\nu \in L^{\infty }((0,T);(\mathcal{E}_{0}^{1}(\mathbb{R}^{N}))')\) and a subsequence of \(\{\nu ^{p_{m}}\}_{m=1}^{\infty }\), denoted the same, such that
$$\begin{aligned}& \nu ^{p_{m}}\to \nu ,\quad \textrm{weakly* in } L^{\infty } \bigl(Q_{T}; \mathcal{M} \bigl(\mathbb{R}^{N} \bigr) \bigr), \\& \nu ^{p_{m}}\to \nu ,\quad \textrm{in } \mathcal{E}_{0}^{1} \bigl(\mathbb{R}^{d} \bigr) \textrm{ in the biting sense}, \end{aligned}$$
which implies that there is a decreasing sequence of subsets \(E_{j+1}\subset E_{j}\) of \(Q_{T}\) with \(\lim_{j\to \infty }\operatorname{meas}(E_{j})=0\) such that \(\langle \nu ^{p_{m}},\psi \rangle \) converges weakly to \(\langle \nu ,\psi \rangle \) in \(L^{1}(Q_{T}\backslash E_{j})\) for all \(\psi \in \mathcal{E}_{0}^{1}(\mathbb{R}^{d})\) and all \(j\geq 1\). By (18) we get that
$$ \operatorname{supp}\nu _{x,t}^{p}\subset \bigl\{ x\in \mathbb{R}^{N}: \varphi _{p}(X)= \varphi _{p}^{**}(X) \bigr\} \subset \bigl\{ X\in \mathbb{R}^{N}:\varphi (X)= \varphi ^{**}(X) \bigr\} $$
which implies (9). By Lemma 4, there exist \(w\in L^{\infty }((0,T);\operatorname{BV}(\Omega ))\) and a subsequence of \(\{w^{p_{m},k}\}_{m,k=1}^{\infty }\), denoted by \(\{w^{k}\}_{k=1}^{\infty }\), such that \(\{w^{k}\}_{k=1}^{\infty }\) converges to w in \(L^{1}(Q_{T})\) and \(\{\nabla w^{k}\}_{k=1}^{\infty }\) is the \(W^{1,1}(Q_{T})\)-biting generating sequence of ν, namely, there is a decreasing sequence of subsets \(G_{j+1}\subset G_{j}\) of \(Q_{T}\) with \(\lim_{j\to \infty }\operatorname{meas}(G_{j})=0\) such that \(\langle \nu ^{p_{m}},\psi \rangle \) converges weakly to \(\langle \nu ,\psi \rangle \) in \(L^{1}(Q_{T}\backslash G_{j})\) for all \(\psi \in \mathcal{E}_{0}^{1}(\mathbb{R}^{d})\) and all \(j\geq 1\).
To prove (6), we first prove that \(\{\langle \nu ^{p_{m}},Z_{p_{m}} \rangle \}_{m=1}^{\infty }\) converges weakly to \(\langle \nu ,Z \rangle \) in \(L^{1}(Q_{T})\). For \(i\geq 1\), define
$$\begin{aligned}& \theta ^{i}(A)= \textstyle\begin{cases} 1,& \vert A \vert \leq i, \\ i+1- \vert A \vert ,& i\leq \vert A \vert \leq i+1, \\ 0,& \vert A \vert \geq i+1. \end{cases}\displaystyle \end{aligned}$$
Let \(\eta \in L^{\infty }(Q_{T};\mathbb{R}^{N})\). Then
$$\begin{aligned}& \begin{aligned} & \biggl\vert \iint _{Q_{T}} \bigl( \bigl\langle \nu ^{p_{m}},Z_{p} \bigr\rangle - \bigl\langle \nu ^{p_{m}},Z \bigr\rangle \bigr)\cdot \eta \,dx\,dt \biggr\vert \\ &\quad \leq \Vert \eta \Vert _{L^{\infty }(Q_{T})} \iint _{Q_{T}} \bigl\vert \bigl\langle \nu _{p}, \bigl(1- \theta ^{i} \bigr) (Z_{p}-Z) \bigr\rangle \bigr\vert \,dx\,dt + \biggl\vert \iint _{Q_{T}} \bigl\langle \nu ^{p},\theta ^{i}(Z_{p}-Z) \bigr\rangle \cdot \eta \,dx\,dt \biggr\vert \\ &\quad =I+\mathit{II}, \end{aligned} \\& \begin{aligned} I&\leq \Vert \eta \Vert _{L^{\infty }(Q_{T})}\lim _{k\to \infty } \iint _{\{(x,t): \vert \nabla w^{p,k} \vert \geq i\}} \bigl\vert Z_{p} \bigl(\nabla w^{p,k} \bigr)-Z \bigl(\nabla w^{p,k} \bigr) \bigr\vert \,dx \,dt \\ &\leq 2\Lambda \Vert \eta \Vert _{L^{\infty }(Q_{T})}\lim_{k\to \infty } \iint _{ \{(x,t): \vert \nabla w^{p,k} \vert \geq i\}} \bigl\vert \nabla w^{p,k} \bigr\vert ^{p-1}\,dx\,dt. \end{aligned} \end{aligned}$$
Noticing that
$$ \lim_{i\to \infty }\operatorname{meas} \bigl\{ (x,t): \bigl\vert \nabla w^{p,k} \bigr\vert \geq i \bigr\} =0, $$
we see that I tends uniformly to 0 in p as \(i\to \infty \). For II, we get that
$$\begin{aligned} \mathit{II}&\leq \Vert \eta \Vert _{L^{\infty }(Q_{T})} \bigl\Vert \theta ^{i}(Z_{p}-Z) \bigr\Vert _{L^{\infty }(Q_{T})} \iint _{Q_{T}} \bigl\langle \nu ^{p},1 \bigr\rangle \,dx \,dt \\ &= \Vert \eta \Vert _{L^{\infty }(Q_{T})} \bigl\Vert \theta ^{i}(Z_{p}-Z) \bigr\Vert _{L^{\infty }(Q_{T})} \operatorname{meas}(Q_{T}). \end{aligned}$$
So
$$ \lim_{p\to 1^{+}} \iint _{Q_{T}} \bigl\langle \nu ^{p},Z_{p}-Z \bigr\rangle \cdot \eta \,dx\,dt=0. $$
Therefore,
$$\begin{aligned} & \biggl\vert \iint _{Q_{T}} \bigl\langle \nu ^{p_{m}},Z_{p_{m}} \bigr\rangle \cdot \eta \,dx\,dt- \iint _{Q_{T}}\langle \nu ,Z \rangle \cdot \eta \,dx\,dt \biggr\vert \\ &\quad \leq \biggl\vert \iint _{Q_{T}} \bigl( \bigl\langle \nu ^{p_{m}},Z_{p_{m}} \bigr\rangle - \bigl\langle \nu ^{p_{m}},Z \bigr\rangle \bigr)\cdot \eta \,dx\,dt \biggr\vert + \biggl\vert \iint _{Q_{T}} \bigl( \bigl\langle \nu ^{p_{m}},Z \bigr\rangle -\langle \nu ,Z \rangle \bigr)\cdot \eta \,dx\,dt \biggr\vert \\ &\quad \to 0\quad (m\to \infty ). \end{aligned}$$
So \(\{\langle \nu ^{p_{m}},Z_{p_{m}} \rangle \}_{m=1}^{\infty }\) converges weakly to \(\langle \nu ,Z \rangle \) in \(L^{1}(Q_{T})\). Thus (6) holds, namely,
$$\begin{aligned} & \iint _{Q_{T}} \bigl(\langle \nu ,Z_{p}\rangle \cdot \nabla \zeta +u_{t} \zeta -\lambda (u-f_{p})\zeta \bigr)\,dx \,dt=0, \\ &\quad \forall \zeta \in C^{\infty }(\overline{Q}_{T}) \textrm{ with }\zeta (x,0)=\zeta (x,T)=0. \end{aligned}$$
Since \(\{w^{k}\}_{k=1}^{\infty }\) converges to w in \(L^{1}(Q_{T})\), we get that for all \(\eta \in C_{0}^{\infty }(Q_{T}\backslash G_{j};\mathbb{R}^{N})\),
$$\begin{aligned} &\lim_{k\to \infty } \biggl( \iint _{Q_{T}\backslash G_{j}}\eta \cdot \nabla w^{k}\,dx\,dt- \iint _{Q_{T}\backslash G_{j}}\eta \cdot \nabla w \,dx\,dt \biggr) \\ &\quad =\lim_{k\to \infty } \iint _{Q_{T}\backslash G_{j}} \bigl(w-w^{k} \bigr) \operatorname {div}(\eta )\,dx \,dt=0. \end{aligned}$$
Thus for all \(j\geq 1\),
$$ \nabla w=\langle \nu ,{\mathrm{id}}\rangle \quad \textrm{in }Q_{T} \backslash G_{j} $$
in the sense of measure. Let \(G=\bigcap_{j=1}^{\infty }G_{j}\). Then \(\operatorname{meas}(G)=0\) and
$$ \nabla w=\langle \nu ,{\mathrm{id}}\rangle \quad \textrm{in }Q_{T} \backslash G $$
in the sense of measure. Similar to the proof of Theorem 3, we get that
$$ u(x,t)=w(x,t)\quad \text{a.e. }(x,t)\in Q_{T}. $$
Thus (7) holds, namely,
$$ \nabla u(x,t)=\langle \nu _{x,t},{\mathrm{id}}\rangle \quad \text{a.e. }(x,t) \in Q_{T} $$
in the sense of measure.
We now prove (8). For all \(\eta \in C_{0}^{\infty }(Q_{T})\), we get that
$$\begin{aligned} & \biggl\vert \iint _{Q_{T}} \bigl( \bigl\langle \nu ^{h_{m}},Z_{p_{m}} \bigr\rangle - \langle \nu ,Z \rangle \bigr)\cdot \bigl\langle \nu ^{p_{m}},{\mathrm{id}}\bigr\rangle \eta \,dx\,dt \biggr\vert \\ &\quad \leq \biggl\vert \iint _{Q_{T}} \bigl( \bigl\langle \nu ^{p_{m}},Z_{p_{m}} \bigr\rangle - \langle \nu ,Z \rangle \bigr) \cdot \nabla (u_{p_{m}}\eta ) \,dx\,dt \biggr\vert \\ &\quad\quad {} + \biggl\vert \iint _{Q_{T}}u_{p_{m}} \bigl( \bigl\langle \nu ^{p_{m}},Z_{p_{m}} \bigr\rangle -\langle \nu ,Z \rangle \bigr) \cdot \nabla \eta \,dx\,dt \biggr\vert \\ &\quad \leq \biggl\vert \iint _{Q_{T}} \biggl( \frac{\partial u_{p_{m}}}{\partial t}-\frac{\partial u}{\partial t} \biggr)u_{p_{m}}\eta \,dx\,dt \biggr\vert +\lambda \biggl\vert \iint _{Q_{T}} \bigl((u_{p_{m}}-f_{p_{m}}) -(u-f) \bigr)u_{p_{m}}\eta \,dx\,dt \biggr\vert \\ &\quad \quad {} + \biggl\vert \iint _{Q_{T}}u_{p_{m}} \bigl( \bigl\langle \nu ^{p_{m}},Z_{p_{m}} \bigr\rangle -\langle \nu ,Z \rangle \bigr) \cdot \nabla \eta \,dx\,dt \biggr\vert \\ & \quad \leq \biggl\vert \iint _{Q_{T}} \biggl( \frac{\partial u_{p_{m}}}{\partial t}-\frac{\partial u}{\partial t} \biggr)u\eta \,dx\,dt \biggr\vert + \biggl\vert \iint _{Q_{T}} \biggl( \frac{\partial u_{p_{m}}}{\partial t}-\frac{\partial u}{\partial t} \biggr) (u_{p_{m}}-u)\eta \,dx\,dt \biggr\vert \\ &\quad \quad {} +\lambda \biggl\vert \iint _{Q_{T}}(u_{p_{m}}-u)u_{p_{m}}\eta \,dx\,dt \biggr\vert +\lambda \biggl\vert \iint _{Q_{T}}(f_{p_{m}}-f)u_{p_{m}}\eta \,dx\,dt \biggr\vert \\ &\quad \quad {} + \biggl\vert \iint _{Q_{T}}u \bigl( \bigl\langle \nu ^{p_{m}},Z_{p_{m}} \bigr\rangle -\langle \nu ,Z \rangle \bigr) \cdot \nabla \eta \,dx\,dt \biggr\vert \\ &\quad \quad {} + \biggl\vert \iint _{Q_{T}} \bigl(u^{p_{m}}-u \bigr) \bigl( \bigl\langle \nu ^{h_{m}},Z_{p_{m}} \bigr\rangle -\langle \nu ,Z \rangle \bigr) \cdot \nabla \eta \,dx\,dt \biggr\vert \\ &\quad \rightarrow 0\quad (m\to \infty ). \end{aligned}$$
Since \(\langle \nu _{x,t}^{p_{m}},{\mathrm{id}}\rangle \) converges to \(\langle \nu _{x,t},{\mathrm{id}}\rangle \) in the biting sense, we see that for all \(j\geq 1\) and all \(\eta \in C_{0}^{\infty }(Q_{T}\backslash E_{j})\),
$$\begin{aligned} & \biggl\vert \iint _{Q_{T}\backslash E_{j}} \bigl( \bigl\langle \nu ^{p_{m}},Z_{p_{m}} \bigr\rangle \cdot \bigl\langle \nu ^{p_{m}},{\mathrm{id}}\bigr\rangle -\langle \nu ,Z \rangle \cdot \langle \nu ,{\mathrm{id}}\rangle \bigr)\eta \,dx\,dt \biggr\vert \\ &\quad \leq \biggl\vert \iint _{Q_{T}\backslash E_{j}} \bigl( \bigl\langle \nu ^{p_{m}},Z_{p_{m}} \bigr\rangle -\langle \nu ,Z \rangle \bigr)\cdot \bigl\langle \nu ^{p_{m}},{\mathrm{id}}\bigr\rangle )\eta \,dx\,dt \biggr\vert \\ &\qquad {} + \biggl\vert \iint _{Q_{T}\backslash E_{j}} \bigl( \bigl\langle \nu ^{p_{m}}, {\mathrm{id}}\bigr\rangle -\langle \nu ,{\mathrm{id}}\rangle \bigr)\cdot \langle \nu ,Z \rangle )\eta \,dx\,dt \biggr\vert \to 0 \quad (m\to \infty ). \end{aligned}$$
Thus for all \(j\geq 1\),
$$\begin{aligned}& \lim_{m\to \infty } \bigl\langle \nu _{x,t}^{p_{m}},Z_{p_{m}} \cdot {\mathrm{id}}\bigr\rangle =\langle \nu _{x,t},Z \rangle \cdot \langle \nu _{x,t}, {\mathrm{id}}\rangle \quad \text{a.e. }(x,t)\in Q_{T} \backslash E_{j}. \end{aligned}$$
Fix \(i\geq C_{0}\). Let \(0\leq \eta \in C_{0}^{\infty }(Q_{T}\backslash E_{j})\). Then
$$\begin{aligned} & \iint _{Q_{T}} \bigl\langle \nu ^{p_{m}},Z_{p_{m}} \cdot {\mathrm{id}}\bigr\rangle \eta \,dx\,dt- \iint _{Q_{T}}\langle \nu ,Z\cdot {\mathrm{id}}\rangle \eta \,dx\,dt \\ &\quad = \biggl( \iint _{Q_{T}} \bigl\langle \nu ^{p_{m}},\theta ^{i}(Z_{p_{m}} \cdot {\mathrm{id}}) \bigr\rangle \eta \,dx\,dt- \iint _{Q_{T}} \bigl\langle \nu ,\theta ^{i}(Z \cdot {\mathrm{id}}) \bigr\rangle \eta \,dx\,dt \biggr) \\ &\qquad {} + \biggl( \iint _{Q_{T}} \bigl\langle \nu ^{p_{m}}, \bigl(1-\theta ^{i} \bigr) (Z_{p_{m}} \cdot {\mathrm{id}}) \bigr\rangle \eta \,dx \,dt- \iint _{Q_{T}} \bigl\langle \nu , \bigl(1- \theta ^{i} \bigr) (Z\cdot {\mathrm{id}}) \bigr\rangle \eta \,dx\,dt \biggr) \\ &\quad =I+\mathit{II}. \end{aligned}$$
Because \(\theta ^{i}(Z_{p_{m}}\cdot {\mathrm{id}})\) converge uniformly to \(\theta ^{i}(Z\cdot {\mathrm{id}})\) in \(\mathbb{R}^{N}\), we see that
$$\begin{aligned} & \biggl\vert \iint _{Q_{T}} \bigl\langle \nu ^{p_{m}},\theta ^{i}(Z_{p_{m}} \cdot {\mathrm{id}}) \bigr\rangle \eta \,dx\,dt- \iint _{Q_{T}} \bigl\langle \nu ^{p_{m}}, \theta ^{i}(Z\cdot {\mathrm{id}}) \bigr\rangle \eta \,dx\,dt \biggr\vert \\ &\quad \leq \Vert \eta \Vert _{L^{\infty }(Q_{T})} \bigl\Vert \theta ^{i}(Z_{p_{m}}\cdot {\mathrm{id}}) -\theta ^{i}(Z\cdot {\mathrm{id}}) \bigr\Vert _{L^{\infty }(Q_{T})} \iint _{Q_{T}} \bigl\langle \nu ^{p_{m}},1 \bigr\rangle \,dx\,dt \\ &\quad = \Vert \eta \Vert _{L^{\infty }(Q_{T})} \bigl\Vert \theta ^{i}(Z_{p_{m}}\cdot {\mathrm{id}})-\theta ^{i}(Z\cdot {\mathrm{id}}) \bigr\Vert _{L^{\infty }(Q_{T})} \operatorname{meas}(Q_{T})\to 0 \quad (m\to \infty ). \end{aligned}$$
Hence
$$\begin{aligned} \lim_{m\to \infty } I&=\lim_{m\to \infty } \biggl( \iint _{Q_{T}} \bigl\langle \nu ^{p_{m}},\theta ^{i}(Z_{p_{m}}\cdot {\mathrm{id}}) \bigr\rangle \eta \,dx\,dt- \iint _{Q_{T}} \bigl\langle \nu ^{p_{m}},\theta ^{i}(Z\cdot {\mathrm{id}}) \bigr\rangle \eta \,dx\,dt \biggr) \\ &\quad {} +\lim_{m\to \infty } \biggl( \iint _{Q_{T}} \bigl\langle \nu ^{p_{m}}, \theta ^{i}(Z\cdot {\mathrm{id}}) \bigr\rangle \eta \,dx\,dt- \iint _{Q_{T}} \bigl\langle \nu ,\theta ^{i}(Z\cdot {\mathrm{id}}) \bigr\rangle \eta \,dx\,dt \biggr) \\ &=0. \end{aligned}$$
From \((1-\theta ^{i})(Z_{p_{m}}\cdot {\mathrm{id}})\geq (1-\theta ^{i})(Z\cdot {\mathrm{id}})\), we get that
$$\begin{aligned} \lim_{m\to \infty }\mathit{II}&\geq \lim_{m\to \infty } \biggl( \iint _{Q_{T}} \bigl\langle \nu ^{p_{m}}, \bigl(1-\theta ^{i} \bigr) (Z\cdot {\mathrm{id}}) \bigr\rangle \eta \,dx\,dt - \iint _{Q_{T}} \bigl\langle \nu , \bigl(1-\theta ^{i} \bigr) (Z\cdot {\mathrm{id}}) \bigr\rangle \eta \,dx\,dt \biggr) \\ &=0. \end{aligned}$$
Thus
$$ \lim_{m\to \infty } \iint _{Q_{T}} \bigl\langle \nu ^{p_{m}},Z_{p_{m}} \cdot {\mathrm{id}}\bigr\rangle \eta \,dx\,dt\geq \iint _{Q_{T}}\langle \nu ,Z\cdot {\mathrm{id}}\rangle \eta \,dx\,dt. $$
By the arbitrariness of \(0\leq \eta \in C_{0}^{\infty }(Q_{T}\backslash E_{j})\), we see that
$$ \lim_{m\to \infty } \bigl\langle \nu _{x,t}^{p_{m}},Z_{p_{m}} \cdot {\mathrm{id}}\bigr\rangle \geq \langle \nu _{x,t},Z\cdot {\mathrm{id}}\rangle , \quad \text{a.e. }(x,t)\in Q_{T}\backslash E_{j}. $$
Thus for all \(j\geq 1\),
$$ \langle \nu _{x,t},Z\cdot {\mathrm{id}}\rangle \leq \langle \nu _{x,t},Z \rangle \cdot \langle \nu _{x,t},{\mathrm{id}}\rangle \quad \text{a.e. }(x,t) \in Q_{T}\backslash E_{j}. $$
Since \(\langle \nu _{x,t},Z\cdot {\mathrm{id}}\rangle \), \(\langle \nu _{x,t},Z \rangle \cdot \langle \nu _{x,t},{\mathrm{id}}\rangle \in L^{1}(Q_{T})\), we see that
$$ \langle \nu _{x,t},Z\cdot {\mathrm{id}}\rangle \leq \langle \nu _{x,t},Z \rangle \cdot \langle \nu _{x,y},{\mathrm{id}}\rangle \quad \text{a.e. }(x,t) \in Q_{T}. $$
So (8) holds, and therefore u is the Young measure solution of the problem (3)–(5). The proof of the theorem is complete. □
Remark 3
Note that if the initial data \(f\equiv C\), and then f is the Young measure of the problem (3)–(5).