Existence results for a class of nonlinear parabolic equations of generalized porous medium type with measure data

Abstract

In this paper, we study the existence of a special type of distributional solutions, the so-called “renormalized solutions” for some parabolic problems with unbounded term and general measure data. We obtain the a priori estimates and we establish the main result using “cut-off” functions.

Résumé

Dans ce papier, nous étudions l’existence d’un type spécifique de solutions de distributions, appelées “solutions renormalisées” pour certains problèmes paraboliques avec un terme non borné et une donnée mesure générale. Nous obtenons des estimations a priori et nous établirons le résultat à l’aide des fonctions isolées.

Appendix

As we have seen, the goal of this approach will be to pass to the limit using the equation solved by the truncations of $u_{n}$ (see Definition 3.1). The major advantage of this approach is that we can perform the passage to the limit without using the strong convergence of the truncations in $L^{p}(0,T;W^{1,p}_{0}(\Omega ))$ and the proof is based on the properties of truncations of renormalized solutions.

1.1 Some further properties and remarks

Let us stress a fact concerning the asymptotic reconstruction property of the singular part of the measure.

Proposition 6.1

Let $u_{n}$ be solution of (4.6), then

$$\begin{aligned} \underset{h\rightarrow \infty }{{\text {lim}}}\underset{n\rightarrow \infty }{{\text { lim sup}}}\int _{\lbrace h-1\le |b_{n}(x,u_{n})|\le h\rbrace }b_{s,n}(x,u_{n})a(t,x,u_{n},\nabla u_{n})\cdot \nabla u_{n}\psi \ dxdt =\int _{Q}\psi d\mu _{s} \end{aligned}$$

(6.1)

for every $\psi \in C^{\infty }_{c}([0,T]\times \Omega )$.

In the next, we prove the following Lemma, which is the key point to control singular sets where $\mu$ is concentrated and that will be developed in the proof of Proposition 6.1 (Fig. 6).

Fig. 6

The function $\Theta _{h}(s)$

Lemma 6.2

Let $u_{n}$ be solution of (4.6), $k>0$ and let $\psi _{\delta }$ be as in Lemma 5.1. Then

$$\begin{aligned} \int _{Q}\mu _{s}^{n}(k-|b_{n}(x,u_{n})|)^{+}\psi _{\delta }=\omega (n,\delta ). \end{aligned}$$

(6.2)

Proof

We multiply the Eq. (4.6) by $(k-|b_{n}(x,u_{n})|)^{+}\psi _{\delta }$ where $\psi _{\delta }$ is given by Lemma 5.1 and we integrate over Q to get

$$\begin{aligned} \begin{aligned}&-\int _{Q}\left( \int _{0}^{|b_{n}(x,u_{n})|}(k-s)^{+}ds\right) (\psi _{\delta })_{t}dxdt\\&\qquad +\int _{Q}a(t,x,u_{n},\nabla u_{n})\cdot \nabla \psi _{\delta }(k-|b_{n}(x,u_{n})|)^{+}dx dt\\&\quad =\int _{Q}a(t,x,u_{n},(b_{s,n}(x,u_{n}))^{-1}(\nabla T_{k}(b_{n}(x,u_{n}))\\&\qquad -\nabla _{x}b_{n}(x,u_{n}))\cdot \nabla T_{k}(b_{n}(x,u_{n}))\psi _{\delta }dx dt\\&\qquad +\int _{Q}(k-|b_{n}(x,u_{n})|)^{+}\psi _{\delta }d\mu _{n}+\int _{\Omega }\left( \int _{0}^{|b_{n}(x,u_{0}^{n})|}(k-s)^{+}ds\right) \psi _{\delta }(0)dx\\&\qquad +\int _{Q}(k-|b_{n}(x,u_{n})|)^{+}\psi _{\delta }d\mu _{n}+\int _{\Omega }\left( \int _{0}^{|b_{n}(x,u_{0}^{n})|}(k-s)^{+}ds\right) \psi _{\delta }(0)dx. \end{aligned} \end{aligned}$$

(6.3)

Now, using Proposition 4.1, observing that $\int _{0}^{|b_{n}(x,u_{n})|}(k-s)^{+}ds\in L^{p}(0,T;W^{1,p}_{0}(\Omega ))\cap L^{\infty }(Q)$ and that $\psi _{\delta }$ goes to zero in S, we get both

$$\begin{aligned} -\int _{Q}\left( \int _{0}^{|b_{n}(x,u_{n})|}(k-s)^{+}ds\right) (\psi _{\delta })_{t}=\omega (n,\delta ), \end{aligned}$$

(6.4)

and

$$\begin{aligned} \begin{aligned}&\int _{Q}a(t,x,u_{n},\nabla u_{n})\cdot \nabla \psi _{\delta }(k-|b_{n}(x,u_{n})|)^{+}dx dt\\&\quad =\int _{Q}a(t,x,u_{n},(b_{s,n}(x,u_{n}))^{-1}(\nabla T_{k}(b_{n}(x,u_{n}))-\nabla _{x}b_{n}(x,u_{n}))\\&\qquad \cdot \nabla \psi _{\delta }(k-|b_{n}(x,u_{n})|)^{+}=\omega (n,\delta ). \end{aligned} \end{aligned}$$

(6.5)

So that, dropping nonnegative terms in the right-hand side, we deduce (6.2). Let us also observe that, as a by-product, we also have the following property of the energy of the truncations near the singular set

$$\begin{aligned} \alpha \lambda \int _{Q}|\nabla u_{n}|^{p}dx dt\le \int _{Q} b_{s,n}(x,u_{n})a(t,x,u_{n},\nabla u_{n})\cdot \nabla u_{n}\psi _{\delta }dx dt\le \omega (n,\delta ). \end{aligned}$$

(6.6)

$\square$

Proof of Proposition 6.1

Let $\Theta _{h}(s)$ be a function defined by and let us take $\Theta _{h}(b_{n}(x,u_{n}))\Psi$ as test function in (4.6), where $\Psi \in C^{\infty }_{c}(Q)$, to obtain

$$\begin{aligned} \begin{aligned}&-\int _{Q}\left( \int _{0}^{|b_{n}(x,u_{n})|}\Theta _{h}(s)ds\right) \Psi _{t}dx dt\\&\qquad +\int _{\lbrace h-1\le |b_{n}(x,u_{n})|<h\rbrace }a(t,x,u_{n},\nabla u_{n})\cdot \nabla b_{n}(x,u_{n})\Psi dx dt\\&\qquad +\int _{Q}a(t,x,u_{n},\nabla u_{n})\cdot \nabla \Psi \Theta _{h}(b_{n}(x,u_{n}))dx dt\\&\quad =\int _{Q}\Psi \Theta _{h}(b_{n}(x,u_{n}))d\mu _{0}^{n}+\int _{Q}\Psi \Theta _{h}(b_{n}(x,u_{n}))d\mu _{s}^{n}\\&\qquad +\int _{\Omega }\left( \int _{0}^{|b_{n}(x,u_{0}^{n})|}\Theta (s)ds\right) \Psi (0)dx. \end{aligned} \end{aligned}$$

(6.7)

$\square$

Let us analyze the previous terms one by one. First of all, thanks to Proposition 4.1 we easily get

$$\begin{aligned} \left\{ \begin{aligned}&-\int _{Q}\left( \int _{0}^{|b_{n}(x,u_{n})|}\Theta _{h}(s)ds\right) \Psi _{t}dx dt=\omega (n,k),\\&\int _{Q}a(t,x,u_{n},\nabla u_{n})\cdot \nabla \Psi \Theta _{h}(b_{n}(x,u_{n}))dx dt=\omega (n,k). \end{aligned}\right. \end{aligned}$$

(6.8)

Similarly dropping the term at $t=0$, using the fact that $|\Theta _{h}(s)-1|\le (h-s)^{+}$ and Lemma 6.2, we have

$$\begin{aligned}&\int _{Q}\Psi \Theta _{h}(b_{n}(x,u_{n}))d\mu _{0}^{n}+\int _{Q}\Psi \Theta _{h}(b_{n}(x,u_{n}))d\mu _{s}^{n}\nonumber \\&\quad \le \left| \int _{\lbrace |b_{n}(x,u_{n})|\ge h-1\rbrace }\Psi d\mu _{0}^{n}\right| + \left| \int _{Q}\Psi d\mu _{s}^{n}\right| +\left| \int _{Q}\Psi (\Theta _{h}(b_{n}(x,u_{n})-1)d\mu _{s}^{n}\right| \nonumber \\&\quad \le \Vert \Psi \Vert _{L^{\infty }(Q)}\left( \left| \int _{\lbrace |b_{n}(x,u_{n})|\ge h-1\rbrace }d\mu _{0}^{n}\right| +\left| \int _{Q}d\mu _{s}^{n}\right| +\left| \int _{Q}(h-b_{n}(x,u_{n}))^{+}\psi _{\delta }d\mu _{s}^{n}\right| \right. \nonumber \\&\qquad \left. +\,\left| \int _{Q}(1-\psi _{\delta })d\mu _{s}^{n}\right| \right) \le \omega (n,k)+\omega (n,\delta )=\omega (n,k,\delta ). \end{aligned}$$

(6.9)

Finally, gathering together all these results we obtain

$$\begin{aligned} \underset{h\rightarrow \infty }{\text {lim}}\underset{n\rightarrow \infty }{\text {limsup}}\int _{\lbrace h-1\le |b_{n}(x,u_{n})|<h\rbrace } b_{s,n}(x,u_{n})a(t,x,u_{n},\nabla u_{n})\cdot \nabla u_{n}\Psi dx dt=\int _{Q}\Psi d\mu _{s}. \end{aligned}$$

1.2 Proof of lemma 2.12

Let us now consider the capacitary estimate of renormalized solutions: we want to prove that u satisfies (2.22) in Lemma 2.12, we still use the notations introduced in Sects. 2.3 and 2.1, in particular, for simplicity we consider the case of $\widetilde{a}(t,x,\zeta )=a(t,x,u(t,x),\zeta )=|\nabla \zeta |^{p-2}\zeta$ (i.e. $p-$Laplacian operator), so that $\widetilde{L}=L+|u|^{p-1}$, then the function $\widetilde{a}$ satisfies

$$\begin{aligned} |\widetilde{a}(t,x,\zeta )|\le \beta (\widetilde{L}+|\zeta |^{p-1})\quad \text {for a.e.}\quad (t,x)\in Q \;\text { and all }\zeta \in \mathbb {R}^{N} \end{aligned}$$

and (2.4)–(2.6) (without dependence in s). Hence, the problem (1.1) becomes

$$\begin{aligned}(\widetilde{\mathcal {P}})\quad {\left\{ \begin{array}{ll} b(x,\widetilde{u})_{t}-\text {div}(\widetilde{a}(t,x,\nabla \widetilde{u})) =\mu &{}\quad \text { in }\,\,Q=(0,T)\times \Omega ,\\ \widetilde{u}=0&{}\quad \text { in }\,\,(0,T)\times \partial \Omega ,\\ b(x,\widetilde{u})(0)=b(x,u_{0})&{}\quad \text { in }\,\,\Omega , \end{array}\right. }\end{aligned}$$

and consider also the condition $p>\frac{2N+1}{N+1}$, we assume in addition that $\mu \in \mathcal {M}_{b}(Q)$ and $b(x,u_{0})\in L^{1}(\Omega )$, hence, we have $b(x,\widetilde{u})\in L^{\infty }(0,T;L^{2}(\Omega ))\cap L^{p}(0,T;W^{1,p}_{0}(\Omega ))$. Actually, the proof will be split into three parts, in the first one we obtain the basic estimates.

Step. 1 Estimates on $T_{k}(b(x,\widetilde{u}))$ in the space $L^{\infty }(0,T;L^{2}(\Omega ))\cap L^{p}(0,T;W^{1,p}_{0}(\Omega ))$. For every $\tau \in \mathbb {R}$, let

$$\begin{aligned} \Theta _{k}(\tau )=\int _{0}^{s}T_{k}(\sigma )d\sigma . \end{aligned}$$

Taking $r\in [0,T]$ and choosing $v=T_{k}(b(x,\widetilde{u}))$ as test function in the weak formulation of $(\widetilde{\mathcal {P}})$ with $t=r$, we have

$$\begin{aligned}&\int _{\Omega }\Theta _{k}(b(x,\widetilde{u}))(r)dx+\int _{0}^{r}\int _{\Omega }\widetilde{a}(t,x,\nabla \widetilde{u})\cdot \nabla T_{k}(b(x,\widetilde{u}))dx dt\le k\Vert \mu \Vert _{\mathcal {M}_{b}(Q)}\\&\quad +\int _{\Omega }\Theta _{k}(b(x,u_{0}))dx. \end{aligned}$$

Observing that $\frac{T_{k}(s)^{2}}{2}\le \Theta _{k}(s)\le k|s|$, $\forall s\in \mathbb {R}$, we have

$$\begin{aligned}\begin{aligned}&\int _{\Omega }\frac{[T_{k}(b(x,\widetilde{u}))(r)]^{2}}{2}dx+\int _{0}^{r}\int _{\Omega }\widetilde{a}(t,x,\nabla \widetilde{u})\cdot \nabla _{x}b(x,\widetilde{u})\chi _{\{|b(x,\widetilde{u})|\le k\}}dx dt\\&\quad +\int _{0}^{r}\int _{\Omega }b_{s}(x,\widetilde{u})\widetilde{a}(t,x,\nabla \widetilde{u})\cdot \nabla \widetilde{u}\chi _{\lbrace |b(x,\widetilde{u}|\le k\rbrace }\le k(\Vert \mu \Vert _{\mathcal {M}_{b}(Q)}+\Vert b(x,u_{0})\Vert _{L^{1}(\Omega )}) \end{aligned} \end{aligned}$$

for any $r\in [0,T]$. In particular we deduce

$$\begin{aligned} \begin{aligned}&\int _{\Omega }\frac{[T_{k}(b(x,\widetilde{u}))(t)]^{2}}{2}dx +\alpha \int _{\lbrace |b(x,\widetilde{u})|\le k\rbrace }b_{s}(x,\widetilde{u})|\nabla \widetilde{u}|^{p}dx dt\\&\quad \le kM+\frac{\alpha }{2}\int _{\lbrace |b(x,\widetilde{u})|\le k\rbrace }b_{s}(x,\widetilde{u})|\nabla \widetilde{u}|^{p}dx dt+\frac{T}{p}(\Lambda +1)(\frac{2\beta p'}{\alpha \lambda })^{p-1}\Vert B\Vert _{L^{p}(\Omega )}^{p} \end{aligned} \end{aligned}$$

(6.10)

and we have

$$\begin{aligned} \int _{\Omega }\frac{[T_{k}(b(x,\widetilde{u}))(t)]^{2}}{2}dx +\frac{\alpha }{2}\int _{\lbrace |b(x,\widetilde{u})|\le k\rbrace }b_{s}(x,\widetilde{u})|\nabla \widetilde{u}|^{p}dx dt\le kM+C\Vert B\Vert _{L^{p}(\Omega )}^{p}. \end{aligned}$$

Then

$$\begin{aligned} \Vert T_{k}(b(x,\widetilde{u}))\Vert _{L^{\infty }(0,T;L^{2}(\Omega ))}^{2}\le CkM\text { and }\Vert T_{k}(b(x,\widetilde{u}))\Vert _{L^{p}(0,T;W^{1,p}_{0}(\Omega ))}^{p}\le CkM \end{aligned}$$

(6.11)

where

$$\begin{aligned} M=\Vert \mu \Vert _{\mathcal {M}_{b}(Q)}+\Vert b(x,u_{0})\Vert _{L^{2}(\Omega )}+\Vert B\Vert _{L^{p}(\Omega )}^{p}. \end{aligned}$$

(6.12)

Step. 2 Estimates in W. In order to deduce some estimates in W we use an idea from [32]. By standard results there exists a unique solution $z\in L^{\infty }(0,T;L^{2}(\Omega ))\cap L^{p}(0,T;W^{1,p}_{0}(\Omega ))$ of the backward problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -z_{t}-\Delta _{p}z=-2\Delta _{p}T_{k}(b(x,\widetilde{u}))&{}\quad \text { in }(0,T)\times \Omega ,\\ z=T_{k}(b(x,\widetilde{u}))&{}\quad \text { on }\lbrace T\rbrace \times \Omega ,\\ z=0&{}\quad \text { on }(0,T)\times \partial \Omega . \end{array}\right. } \end{aligned}$$

(6.13)

Let us multiply (6.13) by z and integrate between z and T. Using Young’s inequality, we obtain

$$\begin{aligned}&\int _{\Omega }\frac{[z(\tau )]^{2}}{2}dx+\frac{1}{2}\int _{0}^{T}\int _{\Omega }b_{s}(x,\widetilde{u})|\nabla z|^{p}dx dt\le \int _{\Omega }\frac{[T_{k}(b(x,\widetilde{u}))(T)]^{2}}{2}dx\\&\quad +\,C\int _{0}^{T}\int _{\Omega }|\nabla T_{k}(b(x,\widetilde{u}))|^{p}dx dt \end{aligned}$$

for every $\tau \in [0,T]$. Using (6.10) with $r=T$, we deduce

$$\begin{aligned}&\int _{\Omega }\frac{[z(\tau )]^{2}}{2}dx+\frac{1}{2}\int _{0}^{T}\int _{\Omega }b_{s}(x,\widetilde{u})|\nabla z|^{2}dx dt\\&\quad \le Ck\left( \Vert \mu \Vert _{\mathcal {M}_{b}(Q)}+\Vert b(x,u_{0})\Vert _{L^{1}(\Omega )}+\Vert B\Vert _{L^{p}(\Omega )}^{p}\right) \end{aligned}$$

for every $\tau \in [0,T]$. This implies

$$\begin{aligned} \Vert z\Vert _{L^{\infty }(0,T;L^{2}(\Omega ))}^{2}+\Vert z\Vert _{L^{p}(0,T;W^{1,p}_{0}(\Omega ))}^{p}\le CkM. \end{aligned}$$

(6.14)

Recall that $V=W^{1,p}_{0}(\Omega )\cap L^{2}(\Omega )$, thus

$$\begin{aligned} \Vert z\Vert _{L^{p}(0,T;V)}\le C\left( \Vert z\Vert _{L^{p}(0,T;W^{1,p}_{0}(\Omega ))}^{p}+\Vert z\Vert _{L^{p}(0,T;L^{2}(\Omega ))}^{p}\right) . \end{aligned}$$

We deduce from (6.14) that

$$\begin{aligned} \Vert z\Vert _{L^{p}(0,T;V)}\le C\left[ (kM)^{\frac{1}{p}}+(kM)^{\frac{1}{2}}\right] . \end{aligned}$$

(6.15)

Moreover, the equation in (6.13) implies

$$\begin{aligned} \Vert z_{t}\Vert _{L^{p'}(0,T;W^{-1,p'}(\Omega ))}\le \left( \Vert z\Vert _{L^{p}(0,T;W^{1,p}_{0}(\Omega ))}^{p-1}+\Vert T_{k}(b(x,\widetilde{u}))\Vert _{L^{p}(0,T;W^{1,p}_{0}(\Omega ))}^{p-1}\right) . \end{aligned}$$

Hence, using (6.11) and (6.14), we deduce

$$\begin{aligned} \Vert z_{t}\Vert _{L^{p'}(0,T;W^{-1,p'}(\Omega ))}\le C(kM)^{\frac{1}{p'}}. \end{aligned}$$

(6.16)

Combining (6.15) and (6.16), we conclude that

$$\begin{aligned} \Vert z\Vert _{W}\le C{\text {max}}\left\{ (kM)^{\frac{1}{p}},(kM)^{\frac{1}{p'}}\right\} , \end{aligned}$$

(6.17)

where M is defined in (6.12).

Step. 3 Proof completed for nonegative data. Let us assume that $\mu \ge 0$ and $b(x,u_{0})\ge 0$; hence, we have $b(x,\widetilde{u})_{t}-\Delta _{p}(b(x,\widetilde{u}))\ge 0$, and $b(x,\widetilde{u})\ge 0$ in Q. We claim that

$$\begin{aligned} T_{k}(b(x,\widetilde{u}))_{t}-\Delta _{p}T_{k}(b(x,\widetilde{u}))\ge 0. \end{aligned}$$

(6.18)

To prove (6.18), we consider $S_{k,\sigma }(s)$ the smooth approximation of $T_{k}(s)$ and its primitive $T_{k,\sigma }(s)$. Let $\varphi \in C^{\infty }_{c}(Q)$ be a nonnegative function and take $T'_{k,\sigma }(b(x,\widetilde{u}))\varphi$ as test function for $(\widetilde{\mathcal {P}})$. We obtain, using that $\mu \ge 0$ and that $T_{k,\sigma }(s)$ is concave for $s\ge 0$

$$\begin{aligned} -\int _{0}^{T}\varphi _{t}T_{k,\sigma }(b(x,\widetilde{u}))dt +\int _{Q}\widetilde{a}(t,x,\nabla \widetilde{u})\cdot \nabla \varphi S_{k,\sigma }(b(x,\widetilde{u}))dx dt\ge 0 \end{aligned}$$

which yields (6.18) as $\eta$ goes to 0. Combining (6.13) and (6.18), we obtain

$$\begin{aligned} -z_{t}-\Delta _{p}z\ge -(T_{k}(b(x,\widetilde{u})))_{t}-\Delta _{p}T_{k}(b(x,\widetilde{u})). \end{aligned}$$

(6.19)

Since both z and $T_k{(b(x,\widetilde{u}))}$ belong to $L^{p}(0,T;W^{1,p}_{0}(\Omega ))$, a standard comparison argument (multiply both sides of (6.19) by $(z-T_{k}(b(x,\widetilde{u})))^{-}$) allows us to conclude that $z\ge T_{k}(b(x,\widetilde{u}))$ a.e. in Q. In particular $z\ge k$ a.e. on $\lbrace b(x,\widetilde{u})>k\rbrace$. On the other hand, since $\widetilde{u}$ belongs to W, it has a unique $\text {cap}_{p}-$quasi continuous representative (still denoted by u), hence, the set $\lbrace u>k\rbrace$ is $\text {cap}_{p}-$quasi open and its capacity can be estimated with (2.11). Therefore, we get

$$\begin{aligned} \text {cap}(\lbrace |b(x,\widetilde{u})|>k\rbrace )\le \left\| \frac{z}{k}\right\| _{W} \end{aligned}$$

and using (6.17) we obtain the result (2.22).

Step. 4 Comparison with $\mu ^{+}$ and $\mu ^{-}$ when $\mu$ is a smooth function. Let us consider the case where $\mu \in C^{\infty }(\overline{Q})$, then $\mu ^{+}\in \mathcal {M}_{b}(Q)\cap L^{p'}(0,T;W^{-1,p'}(\Omega ))$ and we can consider the unique solution $v\in W$ of the problem

$$\begin{aligned} {\left\{ \begin{array}{ll} b(x,v)_{t}-\Delta _{p}v=\mu ^{+}&{}\quad \text { in }\,\,(0,T)\times \Omega ,\\ v=b(x,u_{0})^{+}&{}\quad \text { on }\,\,\lbrace 0\rbrace \times \Omega ,\\ v=0&{}\quad \text { on }\,\,(0,T)\times \partial \Omega . \end{array}\right. } \end{aligned}$$

(6.20)

By comparison principle, we have $v\ge \widetilde{u}$. Using Step. 3 we deduce that there exists a nonnegative function $z\in W$ such that

$$\begin{aligned} z\ge T_{k}(b(x,v))\ge T_{k}(b(x,\widetilde{u})) \end{aligned}$$

and

$$\begin{aligned} \Vert z\Vert _{W}\le C{\text {max}}\left\{ k^{\frac{1}{p}},k^{\frac{1}{p'}}\right\} \end{aligned}$$

where $C=C(\Vert \mu \Vert _{\mathcal {M}_{b}(Q)},\Vert b(x,u_{0})\Vert _{L^{1}(\Omega )},p)$. Similarly, using the solutions of (6.20) with data $-\mu ^{-}$ and $-b(x,u_{0})^{-}$ we deduce that there exists a nonnegative function $w\in W$ such that

$$\begin{aligned} T_{k}(b(x,\widetilde{u}))\ge -w ,\quad \Vert \widetilde{u}\Vert _{W}\le C{\text {max}}\left\{ k^{\frac{1}{p}},k^{\frac{1}{p'}}\right\} . \end{aligned}$$

We have thus proved that there exist two nonnegative function $z,w\in W$ such that

$$\begin{aligned} -w\le T_{k}(b(x,\widetilde{u}))\le z\text { and }\Vert z\Vert _{W}+\Vert w\Vert _{W}\le C{\text {max}}\left\{ k^{\frac{1}{p}},k^{\frac{1}{p'}}\right\} \end{aligned}$$

where C depends on $\Vert \mu \Vert _{\mathcal {M}_{b}(Q)}$, $\Vert b(x,u_{0})\Vert _{L^{1}(\Omega )}$ and p.

Step. 5 Proof completed. Let us fix $\Theta \in C^{\infty }_{c}(Q)$ and set $\widetilde{\mu }=\Theta \mu$. By standard properties of convolution [20, Lemma 2.25] and for a sequence of mollifiers $(\rho _{n})$, we have $\rho _{n}*\widetilde{\mu }\in C^{\infty }_{c}(Q)$ and

$$\begin{aligned} \rho _{n}*\widetilde{\mu }\rightarrow \widetilde{\mu }\text { strongly in }L^{p'}(0,T;W^{-1,p'}(\Omega )) \end{aligned}$$

and

$$\begin{aligned} \Vert \rho _{n}*\widetilde{\mu }\Vert _{\mathcal {M}_{b}(Q)}\le \Vert \widetilde{\mu }\Vert _{\mathcal {M}_{b}(Q)}\le \Vert \mu \Vert _{\mathcal {M}_{b}(Q)}. \end{aligned}$$

Take now $\lbrace \Theta _{j}\rbrace$ to be a sequence of $C^{\infty }_{c}(Q)$ functions such that $\Theta _{j}\rightarrow 1$ and consider the solutions $\widetilde{u}_{j,n}$ of the problem

$$\begin{aligned} {\left\{ \begin{array}{ll} b(x,\widetilde{u}_{j,n})_{t}-\Delta _{p}\widetilde{u}_{j,n}=\rho _{n}*(\Theta _{j}\mu ) &{}\quad \text { in }\,\,(0,T)\times \Omega ,\\ b(x,\widetilde{u}_{j,n})=b(x,u_{0})&{}\quad \text { on }\,\,\{0\}\times \Omega ,\\ \widetilde{u}_{j,n}=0&{}\quad \text { on }\,\,(0,T)\times \Omega . \end{array}\right. } \end{aligned}$$

(6.21)

As $n\rightarrow \infty$, the sequence $(\widetilde{u}_{j,n})$ converges in $L^{p}(0,T;W^{1,p}_{0}(\Omega ))$ to the solution $\widetilde{u}_{j}$ of (1.1) with $\Theta _{j}\mu$ as datum. Next, as $j\rightarrow +\infty$, as a consequence of a standard $L^{1}-$contraction argument

$$\begin{aligned} \widetilde{u}_{j}\rightarrow \widetilde{u}\quad \text { in }L^{\infty }(0,T;L^{1}(\Omega )). \end{aligned}$$

Indeed, subtracting equations (1.1) and (6.21), and taking $T_{k}(\widetilde{u}_{j,n}-\widetilde{u})$ as test function, we get (recall that both $\widetilde{u}_{j,n}$ and $\widetilde{u}$ belong to W)

$$\begin{aligned} \begin{aligned} \int _{\Omega }|\widetilde{u}_{j,n}-\widetilde{u}|(t)dx&\le C\Vert \rho _{n}*(\Theta _{j}\mu -\Theta _{j}\mu )\Vert _{L^{p'}(0,T;W^{-1,p'}(\Omega ))}\Vert T_{1}(\widetilde{u}_{j,n}-\widetilde{u})\Vert _{L^{p}(0,T;W^{1,p}_{0}(\Omega ))}\\&\quad + C\int _{\Omega }T_{1}(\widetilde{u}_{j,n}-\widetilde{u})(\Theta _{j}-1)d\mu \end{aligned} \end{aligned}$$

which yields

$$\begin{aligned} \begin{aligned} \Vert (\widetilde{u}_{j,n}-\widetilde{u})(t)\Vert _{L^{1}(\Omega )}&\le C\Vert \rho _{n}*(\Theta _{j}\mu )-\Theta _{j}\mu \Vert _{L^{p'}(0,T;W^{-1,p'}(\Omega ))}\Vert T_{1}(\widetilde{u}_{j,n}-\widetilde{u})\Vert _{L^{p}(0,T;W^{1,p}_{0}(\Omega ))}\\&\quad + C\Vert (1-\Theta _{j})\mu \Vert _{\mathcal {M}_{b}(Q)}. \end{aligned} \end{aligned}$$

Since for j fixed $\widetilde{u}_{j,n}$ is bounded in $L^{p}(0,T;W^{1,p}_{0}(\Omega ))$ as $n\rightarrow +\infty$, the first term in the right-hand side tends to 0, hence

$$\begin{aligned} \Vert (\widetilde{u}_{j}-\widetilde{u}\Vert (t)\Vert _{L^{1}(\Omega )}\le C\Vert (1-\Theta _{j})\mu \Vert _{\mathcal {M}_{b}(Q)}. \end{aligned}$$

Since the later term tends to zero as $j\rightarrow \infty$ by dominated convergence, we deduce the convergence of $\widetilde{u}_{j}$ to $\widetilde{u}$. By Step. 4 there exist a nonnegative functions $z_{j,n}$ and $w_{j,n}$ such that

$$\begin{aligned} -\,w_{j,n}\le T_{k}(b(x,\widetilde{u}_{j,n}))\le z_{j,n},\quad \Vert z_{j,n}\Vert _{W}+\Vert w_{j,n}\Vert _{W}\le C{\text {max}}\left\{ k^{\frac{1}{p}},k^{\frac{1}{p'}}\right\} \end{aligned}$$

where $C=C\left( \Vert \rho _{n}*(\Theta _{j}\mu ))\Vert _{\mathcal {M}_{b}(Q)},\Vert b(x,u_{0})\Vert _{L^{1}(\Omega )},p\right)$. Since

$$\begin{aligned} \Vert \rho _{n}*(\Theta _{j}\mu )\Vert _{\mathcal {M}_{b}(Q)}\le \Vert \mu \Vert _{\mathcal {M}_{b}(Q)}, \end{aligned}$$

the constant C can be chosen independent of n and j. The sequences $(z_{j,n})$ and $(w_{j,n})$ being bounded in W, they converge weakly up to subsequences to nonnegative functions $z,w\in W$ and almost everywhere in Q. Thus,

$$\begin{aligned} -w\le T_{k}(b(x,\widetilde{u}))\le z\text { a.e. in }Q,\quad \Vert z\Vert _{W}+\Vert w\Vert _{W}\le C{\text {max}}\left\{ k^{\frac{1}{p}},k^{\frac{1}{p'}}\right\} \end{aligned}$$

where $C=C(\Vert \mu \Vert _{\mathcal {M}_{b}(Q)},\Vert b(x,u_{0})\Vert _{L^{1}(\Omega )},p)$. Since $\widetilde{u}\in W$, it admits a uniquely defined $\text {cap}_{p}-$quasi continuous representative; hence, the sets $\lbrace \widetilde{u}>k\rbrace$ and $\lbrace \widetilde{u}<-k\rbrace$ are $\text {cap}_{p}-$quasi open. Using (2.11), we get

$$\begin{aligned}&\text {cap}_{p}(\lbrace |b(x,\widetilde{u})|>k)\rbrace )\le \text {cap}_{p}(\lbrace b(x,\widetilde{u})>k\rbrace )\\&\quad +\text {cap}_{p}(\lbrace |\widetilde{u}|<-k\rbrace )\le \left\| \frac{z}{k}\right\| _{W}+\left\| \frac{w}{k}\right\| _{W} \end{aligned}$$

which yields the result (2.22) for $u=\widetilde{u}$.