Proof. Since \(\alpha<d\), the Sobolev embedding theorem yields \[\label{Sobolev} \|v^{\frac{m}{2}}\|_{L_{x}^{\frac{\alpha d}{d-\alpha}}} \leq c \|\nabla v^{\frac{m}{2}}\|_{L_{x}^{\alpha}}.\tag{3}\] Let \(0 \leq \theta \leq 1\), and define \(r_1\) by \[\label{L:V00} \frac{1}{r_1} = (1-\theta) + \frac{2 \theta (d-\alpha)}{\alpha md}.\tag{4}\] By interpolation between \(L_{x}^{1}\) and \(L_{x}^{\frac{\alpha m d}{2(d-\alpha)}}\), we obtain \[\label{L:V01} \|v\|_{L^{r_1}_{x}} \leq \|v\|_{L^1_x}^{1-\theta} \|v\|^{\theta}_{L_{x}^{\frac{\alpha md}{2(d-\alpha)}}} = \|v\|_{L^1_x}^{1-\theta} \|v^{\frac{m}{2}}\|^{\frac{2\theta}{m}}_{L_{x}^{\frac{\alpha d}{d-\alpha}}} \leq c \|v\|_{L^1_x}^{1-\theta} \|\nabla v^{\frac{m}{2}}\|^{\frac{2\theta}{m}}_{L_{x}^{\alpha}}.\tag{5}\] If \(\theta =0\), then \(r_1 = 1\) and \(r_2 = \infty\), then \[\|v\|_{L^{1,\infty}_{x,t}} \leq \sup_{t\in(0,T)} \int_{\Omega} v \,dx.\] Assume that \(\theta \in (0, 1]\), and define \(r_2\) by \[\label{L:V000} \frac{1}{r_2} = \frac{2\theta}{\alpha m}.\tag{6}\] Taking the \(L_{t}^{r_2}\)-norm on both sides of 5 gives \[\label{L:V02} \|v\|_{L^{r_1,r_2}_{x,t}} \leq c\left( \int_{0}^{T} \|v\|_{L^1_x}^{(1-\theta)r_2} \|\nabla v^{\frac{m}{2}}\|^{\frac{2\theta r_2}{m}}_{L_{x}^{\alpha}} \,dt\right)^{\frac{1}{r_2}} \leq c\left(\sup_{t\in (0, T)}\|v\|_{L^1_x}\right)^{(1-\theta)} \|\nabla v^{\frac{m}{2}}\|_{L_{x,t}^{\alpha}}^{\frac{\alpha}{r_2}}.\tag{7}\] Since \(\theta = \frac{\alpha m}{2 r_2}\), this directly gives ?? . Finally, combining 4 and 6 with the condition \(\theta \in [0,1]\) yields the relations in ?? . ◻

Proof. We first note that, since \(d\ge2\) and \(m>1-\frac{1}{d}\), we have \(2-m\le m+1\). Hence, from \(u\in C([0,T];L^{m+1}(\Omega))\) and \(V\in L^\infty(\Omega_T)\), it follows that \(u^{2-m}|V|^2\in L^1(\Omega_T)\).

We justify the nonlinear test functions by a standard smooth approximation argument. More precisely, we first perform the following computations for smooth non-degenerate approximating solutions \(u_\kappa\) with zero lateral boundary data and zero initial data. For such solutions, all chain rules and integrations by parts below are classical, and the nonlinear test functions are admissible. The estimates obtained are independent of the approximation parameter \(\kappa\). Hence, passing to the limit \(\kappa\to0\), using the strong convergence of the approximations and the lower semicontinuity of the quadratic gradient terms, yields the estimates for the original regular solution. To simplify notation, we omit the subscript \(\kappa\) below.

Step 1: proof of ?? .

Fix \(\varepsilon>0\) and \(0<\delta<\varepsilon\). Define \[F_{\varepsilon,\delta}(s):= \begin{cases} 0, & 0\le s\le \delta,\\[1mm] \dfrac{s^{q-1}-\delta^{q-1}} {\varepsilon^{q-1}-\delta^{q-1}}, & \delta<s<\varepsilon,\\[3mm] 1, & s\ge \varepsilon, \end{cases} \quad \text{ and } \quad \Psi_{\varepsilon,\delta}(s) := \int_0^sF_{\varepsilon,\delta}(r)\,dr.\] Let \(\phi\in C^\infty([0,T])\) satisfy \(0\le \phi\le1\), \(-\phi_t\ge0\), \(\phi(T)=0\). We use \[\eta_1:=F_{\varepsilon,\delta}(u)\phi\] as a test function. Since \(u=0\) on \(\partial\Omega\times(0,T)\) and \(F_{\varepsilon,\delta}(0)=0\), we have \(\eta_1=0\) on \(\partial\Omega\times(0,T)\). Thus \(\eta_1\) is admissible.

Testing the equation by \(\eta_1\), we obtain \[I+II=III+IV,\] where \[I:= \int_{\Omega_T} u_tF_{\varepsilon,\delta}(u)\phi\,dxdt, \qquad II:= \int_{\Omega_T} \nabla u^m\cdot\nabla F_{\varepsilon,\delta}(u)\phi\,dxdt,\] \[III:= \int_{\Omega_T} \mu F_{\varepsilon,\delta}(u)\phi\,dxdt, \qquad IV:= \int_{\Omega_T} uV\cdot\nabla F_{\varepsilon,\delta}(u)\phi\,dxdt.\]

Since \(\partial_t\Psi_{\varepsilon,\delta}(u) = u_tF_{\varepsilon,\delta}(u)\), integrating by parts in time gives \[I = \int_{\Omega_T} \Psi_{\varepsilon,\delta}(u)(-\phi_t)\,dxdt -\int_\Omega \Psi_{\varepsilon,\delta}(u(x,0))\phi(0)\,dx + \int_\Omega \Psi_{\varepsilon,\delta}(u(x,T))\phi(T)\,dx.\] The terminal term is zero because \(\phi(T)=0\). The initial term is also zero, because \(u(\cdot,0)=0\) and \(\Psi_{\varepsilon,\delta}(0)=0\). Therefore \[\label{I01-smooth-u} I = \int_{\Omega_T} \Psi_{\varepsilon,\delta}(u)(-\phi_t)\,dxdt \ge0.\tag{8}\]

Moreover, since \(0\le F_{\varepsilon,\delta}\le1\) and \(0\le\phi\le1\), \[\label{III01-smooth-u} 0\le III\le \mu(\Omega_T).\tag{9}\]

Next, \[\nabla F_{\varepsilon,\delta}(u) = \frac{(q-1)u^{q-2}}{\varepsilon^{q-1}-\delta^{q-1}} \chi_{\{\delta<u<\varepsilon\}} \nabla u.\] Hence \[\begin{align} II &= \frac{m(q-1)}{\varepsilon^{q-1}-\delta^{q-1}} \int_{\{\delta<u<\varepsilon\}} u^{m+q-3} |\nabla u|^2\phi\,dxdt \\ &= \frac{4m(q-1)}{(m+q-1)^2(\varepsilon^{q-1}-\delta^{q-1})} \int_{\{\delta<u<\varepsilon\}} \left|\nabla u^{\frac{m+q-1}{2}}\right|^2\phi\,dxdt. \end{align}\] In particular, \(II\ge0\). Letting \(\delta\to0\) and using Fatou’s lemma, we get \[\label{II01-smooth-u} \liminf_{\delta\to0}II \ge \frac{4m(q-1)}{(m+q-1)^2\varepsilon^{q-1}} \int_{\{0<u<\varepsilon\}} \left|\nabla u^{\frac{m+q-1}{2}}\right|^2\phi\,dxdt.\tag{10}\]

For the drift term, Young’s inequality gives \[\begin{align} IV &= \frac{q-1}{\varepsilon^{q-1}-\delta^{q-1}} \int_{\{\delta<u<\varepsilon\}} u^{q-1}V\cdot\nabla u\,\phi\,dxdt \\ &\le \frac{1}{2} II + \frac{q-1}{2m(\varepsilon^{q-1}-\delta^{q-1})} \int_{\{\delta<u<\varepsilon\}} u^{q+1-m}|V|^2\phi\,dxdt. \end{align}\] On \(\{\delta<u<\varepsilon\}\), we have \(u^{q+1-m} = u^{2-m}u^{q-1} \le \varepsilon^{q-1}u^{2-m}\). Therefore \[\label{IV01-smooth-u} IV \le \frac{1}{2} II + \frac{(q-1)\varepsilon^{q-1}}{2m(\varepsilon^{q-1}-\delta^{q-1})} \int_{\{\delta<u<\varepsilon\}} u^{2-m}|V|^2\phi\,dxdt.\tag{11}\]

Combining 8 –11 , and then letting \(\delta\to0\), we obtain \[\label{comp1-smooth-u} \begin{align} &\int_{\Omega_T} \left[ \int_0^u \min\left\{1,\frac{s^{q-1}}{\varepsilon^{q-1}}\right\}\,ds \right] (-\phi_t)\,dxdt + \frac{2m(q-1)}{(m+q-1)^2\varepsilon^{q-1}} \int_{\{0<u<\varepsilon\}} \left|\nabla u^{\frac{m+q-1}{2}}\right|^2\phi\,dxdt \\ &\qquad\le \mu(\Omega_T) + \frac{q-1}{2m} \int_{\{0<u<\varepsilon\}} u^{2-m}|V|^2\phi\,dxdt. \end{align}\tag{12}\]

Letting \(\varepsilon\to0\), the last term on the right-hand side tends to zero because \(u^{2-m}|V|^2\in L^1(\Omega_T)\). Also, \[\int_0^u \min\left\{1,\frac{s^{q-1}}{\varepsilon^{q-1}}\right\}\,ds \longrightarrow u \qquad\text{in }L^1(\Omega_T).\] Thus \[\int_{\Omega_T}u(-\phi_t)\,dxdt \le \mu(\Omega_T)\] for every smooth nonincreasing \(\phi\) satisfying \(0\le\phi\le1\) and \(\phi(T)=0\).

Now fix \(\tau\in(0,T)\). Choose nonnegative functions \(\rho_j\in C_c^\infty(0,T)\) such that \[\int_0^T\rho_j(t)\,dt=1, \qquad \rho_j\rightharpoonup\delta_\tau\] in the sense of measures, and set \[\phi_j(t):=\int_t^T\rho_j(s)\,ds.\] Then \(0\le\phi_j\le1\), \(-(\phi_j)_t=\rho_j\), and \(\phi_j(T)=0\). Since \(u\in C([0,T];L^{m+1}(\Omega))\), the function \[t\mapsto\int_\Omega u(x,t)\,dx\] is continuous. Passing to the limit \(j\to\infty\) gives \[\int_\Omega u(x,\tau)\,dx \le \mu(\Omega_T).\] Taking the supremum over \(\tau\in(0,T)\) proves ?? .

Step 2: proof of ?? .

Let \(A>0\) and \(\xi>1\). We again fix \(\varepsilon>0\) and \(0<\delta<\varepsilon\), and use the test function \[\eta_2:= \frac{F_{\varepsilon,\delta}(u)\phi}{\left(A^{q-1}+u^{q-1}\right)^{\xi-1}},\] where \(\phi\in C^\infty([0,T])\) satisfies \(0\le\phi\le1\), \(-\phi_t\ge0\), and \(\phi(T)=0\). Since \(F_{\varepsilon,\delta}(0)=0\), the function \(\eta_2\) vanishes on \(\partial\Omega\times(0,T)\), and hence it is admissible.

Testing the equation by \(\eta_2\) gives \[I+II_1+II_2=III+IV_1+IV_2,\] where \[I:= \int_{\Omega_T} u_t \frac{F_{\varepsilon,\delta}(u)\phi}{\left(A^{q-1}+u^{q-1}\right)^{\xi-1}} \,dxdt, \qquad II_1:= \int_{\Omega_T} \frac{ \nabla u^m\cdot\nabla F_{\varepsilon,\delta}(u) }{\left(A^{q-1}+u^{q-1}\right)^{\xi-1}} \phi\,dxdt,\] \[II_2:= (1-\xi) \int_{\Omega_T} \frac{ F_{\varepsilon,\delta}(u) \nabla u^m\cdot\nabla u^{q-1} }{\left(A^{q-1}+u^{q-1}\right)^\xi} \phi\,dxdt, \qquad III:= \int_{\Omega_T} \mu \frac{F_{\varepsilon,\delta}(u)\phi}{\left(A^{q-1}+u^{q-1}\right)^{\xi-1}} \,dxdt,\] \[IV_1:= \int_{\Omega_T} \frac{ uV\cdot\nabla F_{\varepsilon,\delta}(u) }{\left(A^{q-1}+u^{q-1}\right)^{\xi-1}} \phi\,dxdt, \qquad IV_2:= (1-\xi) \int_{\Omega_T} \frac{ F_{\varepsilon,\delta}(u) uV\cdot\nabla u^{q-1} }{\left(A^{q-1}+u^{q-1}\right)^\xi} \phi\,dxdt.\]

Since \(F_{\varepsilon,\delta}(u)=0\) on \(\{u\le\delta\}\), all terms involving \(\nabla u^{q-1}\) are supported in \(\{u>\delta\}\). Hence no singularity occurs at \(u=0\). Moreover, \[\nabla u^m\cdot\nabla u^{q-1} = m(q-1)u^{m+q-3}|\nabla u|^2 \ge0 \qquad\text{on }\{u>\delta\}.\] Since \(1-\xi<0\), we have \[II_2\le0.\] Therefore \[\label{basic-smooth-u} -II_2 \le |I|+II_1+III+|IV_1|+|IV_2|.\tag{13}\]

Define \[\Theta_{\varepsilon,\delta}(s) := \int_0^s \frac{F_{\varepsilon,\delta}(r)}{\left(A^{q-1}+r^{q-1}\right)^{\xi-1}} \,dr.\] Then \[I = \int_{\Omega_T} \partial_t\Theta_{\varepsilon,\delta}(u)\phi\,dxdt.\] As in Step 1, the initial and terminal boundary terms vanish, and hence \[I = \int_{\Omega_T} \Theta_{\varepsilon,\delta}(u)(-\phi_t)\,dxdt.\] Moreover, \[0\le \Theta_{\varepsilon,\delta}(u) \le A^{(q-1)(1-\xi)}u.\] Using ?? , we get \[\label{I2-smooth-u} |I| \le A^{(q-1)(1-\xi)}\mu(\Omega_T).\tag{14}\] Similarly, \[0\le \frac{F_{\varepsilon,\delta}(u)}{\left(A^{q-1}+u^{q-1}\right)^{\xi-1}} \le A^{(q-1)(1-\xi)},\] so \[\label{III2-smooth-u} 0\le III \le A^{(q-1)(1-\xi)}\mu(\Omega_T).\tag{15}\]

For \(II_1\), we have \[\begin{align} II_1 &= \frac{m(q-1)}{\varepsilon^{q-1}-\delta^{q-1}} \int_{\{\delta<u<\varepsilon\}} \frac{ u^{m+q-3}|\nabla u|^2\phi }{\left(A^{q-1}+u^{q-1}\right)^{\xi-1}} \,dxdt \\ &\le A^{(q-1)(1-\xi)} \frac{m(q-1)}{\varepsilon^{q-1}-\delta^{q-1}} \int_{\{\delta<u<\varepsilon\}} u^{m+q-3}|\nabla u|^2\phi \,dxdt. \end{align}\] The last integral is precisely the diffusion term from Step 1. Hence, using 12 before letting \(\varepsilon\to0\), we obtain \[\label{II1-smooth-u} \begin{align} \limsup_{\delta\to0}II_1 &\le 2A^{(q-1)(1-\xi)} \left[ \mu(\Omega_T) + \frac{q-1}{2m} \int_{\{0<u<\varepsilon\}} u^{2-m}|V|^2\phi\,dxdt \right]. \end{align}\tag{16}\]

Next, by Fatou’s lemma, \[\label{II2-smooth-u} \liminf_{\varepsilon\to0} \liminf_{\delta\to0} (-II_2) \ge \frac{4m(q-1)(\xi-1)}{(m+q-1)^2} \int_{\Omega_T} \frac{ \left|\nabla u^{\frac{m+q-1}{2}}\right|^2\phi }{\left(A^{q-1}+u^{q-1}\right)^\xi} \,dxdt.\tag{17}\]

For \(IV_1\), Young’s inequality gives \[\begin{align} |IV_1| &= \left| \frac{q-1}{\varepsilon^{q-1}-\delta^{q-1}} \int_{\{\delta<u<\varepsilon\}} \frac{ u^{q-1}V\cdot\nabla u\,\phi }{\left(A^{q-1}+u^{q-1}\right)^{\xi-1}} \,dxdt \right| \\ &\le II_1 + \frac{q-1}{4m(\varepsilon^{q-1}-\delta^{q-1})} \int_{\{\delta<u<\varepsilon\}} \frac{ u^{q+1-m}|V|^2\phi }{\left(A^{q-1}+u^{q-1}\right)^{\xi-1}} \,dxdt. \end{align}\] Since \(\left(A^{q-1}+u^{q-1}\right)^{1-\xi} \le A^{(q-1)(1-\xi)}\) and \(u^{q+1-m} = u^{2-m}u^{q-1} \le \varepsilon^{q-1}u^{2-m}\) on \(\{\delta<u<\varepsilon\}\), we infer \[\label{IV1-smooth-u} \limsup_{\delta\to0}|IV_1| \le \limsup_{\delta\to0}II_1 + o_\varepsilon(1),\tag{18}\] where \(o_\varepsilon(1)\to0\) as \(\varepsilon\to0\).

For \(IV_2\), Young’s inequality yields \[|IV_2| \le \frac{1}{4}(-II_2) + \frac{(q-1)(\xi-1)}{m} \int_{\Omega_T} \frac{ F_{\varepsilon,\delta}(u)u^{q+1-m}|V|^2\phi }{\left(A^{q-1}+u^{q-1}\right)^\xi} \,dxdt.\] Using \(\left(A^{q-1}+u^{q-1}\right)^\xi \ge A^{(q-1)(\xi-1)}u^{q-1}\) on \(\{u>0\}\), we obtain \[\frac{u^{q+1-m}}{\left(A^{q-1}+u^{q-1}\right)^\xi} \le A^{(q-1)(1-\xi)}u^{2-m}.\] Therefore \[\label{IV2-smooth-u} |IV_2| \le \frac{1}{4}(-II_2) + \frac{(q-1)(\xi-1)A^{(q-1)(1-\xi)}}{m} \int_{\Omega_T}u^{2-m}|V|^2\phi\,dxdt.\tag{19}\]

Combining 13 –19 , then letting \(\delta\to0\) and \(\varepsilon\to0\), and using \[\int_{\{0<u<\varepsilon\}} u^{2-m}|V|^2\phi\,dxdt \to0 \qquad\text{as }\varepsilon\to0,\] we get \[\int_{\Omega_T} \frac{ \left|\nabla u^{\frac{m+q-1}{2}}\right|^2\phi }{\left(A^{q-1}+u^{q-1}\right)^\xi} \,dxdt \le \frac{(m+q-1)^2A^{(q-1)(1-\xi)}}{m(q-1)(\xi-1)} \left[ 2\mu(\Omega_T) + \frac{(q-1)(\xi-1)}{3m} \int_{\Omega_T}u^{2-m}|V|^2\phi\,dxdt \right].\]

Finally, choose \(\{\phi_k\}_{k=1}^\infty\subset C^\infty([0,T])\) such that \[0\le\phi_k\le1,\qquad -(\phi_k)_t\ge0,\qquad \phi_k(T)=0,\] \[\phi_k(t)=1 \quad\text{for }0\le t\le T-\frac{1}{k}, \qquad \phi_k\nearrow1 \quad\text{pointwise in }(0,T).\] Applying the previous estimate with \(\phi=\phi_k\) and then letting \(k\to\infty\), the monotone convergence theorem gives ?? , which completes the proof. ◻

Proof. We argue as in the proof of Lemma 4, using the same approximation argument, notation, and truncated test functions. It remains only to observe that the drift term vanishes.

Let \(\Phi:[0,\infty)\to\mathbb{R}\) be Lipschitz with \(\Phi(0)=0\), and let \[\eta=\Phi(u)\phi(t),\] where \(\phi\in C^\infty([0,T])\), \(0\le\phi\le1\), \(-\phi_t\ge0\), and \(\phi(T)=0\). Define \[H_\Phi(s):=\int_0^s r\Phi'(r)\,dr.\] Then \[\nabla H_\Phi(u)=u\Phi'(u)\nabla u.\] Since \(u=0\) on \(\partial\Omega\times(0,T)\), we have \(H_\Phi(u)=0\) on the lateral boundary. Hence, using \(\nabla\cdot V=0\), \[\label{divfree-cancel} \int_{\Omega_T}uV\cdot\nabla\eta\,dxdt = \int_{\Omega_T}\phi V\cdot\nabla H_\Phi(u)\,dxdt = -\int_{\Omega_T}\phi(\nabla\cdot V)H_\Phi(u)\,dxdt =0.\tag{20}\]

For ?? , we take \[\eta_1=F_{\varepsilon,\delta}(u)\phi\] as in Lemma 4. By 20 , the drift term vanishes, and the identity in Step 1 of Lemma 4 reduces to \[I+II=III.\] The estimates of \(I,II,\) and \(III\) are the same as in Lemma 4, except that no drift contribution appears. In particular, \[I\ge0,\qquad 0\le III\le\mu(\Omega_T),\] and hence \[I+II\le\mu(\Omega_T).\] Letting \(\delta\to0\), then \(\varepsilon\to0\), and finally approximating a Dirac mass at any \(\tau\in(0,T)\) by \(-\phi_t\), exactly as in Lemma 4, gives \[\int_\Omega u(x,\tau)\,dx\le\mu(\Omega_T).\] Taking the supremum over \(\tau\in(0,T)\) proves ?? .

For ?? , we take \[\eta_2 = \frac{F_{\varepsilon,\delta}(u)\phi}{\left(A^{q-1}+u^{q-1}\right)^{\xi-1}}.\] Again, by 20 , the drift terms vanish. Therefore the identity in Step 2 of Lemma 4 reduces to \[I+II_1+II_2=III.\] Since \(II_2\le0\) and \(III\ge0\), we have \[-II_2\le I+II_1.\] As in Lemma 4, \[I\le A^{(q-1)(1-\xi)}\mu(\Omega_T).\] Moreover, \[II_1 \le A^{(q-1)(1-\xi)} \int_{\Omega_T}\nabla u^m\cdot\nabla F_{\varepsilon,\delta}(u)\phi\,dxdt.\] The last integral is precisely the diffusion term \(II\) from Step 1. Since the divergence-free Step 1 gives \(II\le\mu(\Omega_T)\), we obtain \[II_1\le A^{(q-1)(1-\xi)}\mu(\Omega_T).\] Thus \[-II_2\le2A^{(q-1)(1-\xi)}\mu(\Omega_T).\] Letting \(\delta\to0\) and \(\varepsilon\to0\), Fatou’s lemma gives, exactly as in Lemma 4, \[\frac{4m(q-1)(\xi-1)}{(m+q-1)^2} \int_{\Omega_T} \frac{\left|\nabla u^{\frac{m+q-1}{2}}\right|^2\phi}{\left(A^{q-1}+u^{q-1}\right)^\xi} \,dxdt \le 2A^{(q-1)(1-\xi)}\mu(\Omega_T).\] Therefore, \[\int_{\Omega_T} \frac{\left|\nabla u^{\frac{m+q-1}{2}}\right|^2\phi}{\left(A^{q-1}+u^{q-1}\right)^\xi} \,dxdt \le \frac{(m+q-1)^2A^{(q-1)(1-\xi)}}{2m(q-1)(\xi-1)} \mu(\Omega_T).\] Finally, taking \(\phi_k\nearrow1\) as in Lemma 4 and using the monotone convergence theorem gives the same estimate without \(\phi\). Hence ?? follows. Note that since the drift term is canceled, the restriction \(1-\frac{1}{d}<m\le2\) used in Lemma 4 is not needed here. Thus the result holds for all \(m>0\). ◻

Proof. The estimate ?? follows from the estimate ?? . Now we prove ?? for \(1\le \alpha < \frac{2(2+md)}{2+d(m+q-1)}\). The estimate for \(\alpha<1\), whenever needed, follows from Hölder’s inequality after proving the estimate for a larger exponent. First, it follows that by applying the Hölder inequality, \[\begin{align} \fint_{\Omega_T} \left|\nabla u^{\frac{m+q-1}{2}}\right|^{\alpha} \,dxdt &= \fint_{\Omega_T} \left( \frac{\left|\nabla u^{\frac{m+q-1}{2}}\right|^2}{\left(A^{q-1}+u^{q-1}\right)^\xi} \right)^{\frac{\alpha}{2}} \left(A^{q-1}+u^{q-1}\right)^{\frac{\xi\alpha}{2}} \,dxdt\\ &\le \left( \fint_{\Omega_T} \frac{\left|\nabla u^{\frac{m+q-1}{2}}\right|^2}{\left(A^{q-1}+u^{q-1}\right)^\xi} \,dxdt \right)^{\frac{\alpha}{2}} \underbrace{\left( \fint_{\Omega_T} \left(A^{q-1}+u^{q-1}\right)^{\frac{\xi\alpha}{2-\alpha}} \,dxdt \right)^{\frac{2-\alpha}{2}}}_{=:I}. \end{align}\] Then we have \[I \le c(\alpha) A^{\frac{\xi\alpha(q-1)}{2}} + \left( \fint_{\Omega_T} u^{\frac{\xi\alpha(q-1)}{2-\alpha}} \,dxdt \right)^{\frac{2-\alpha}{2}} = c(\alpha) A^{\frac{\xi\alpha(q-1)}{2}},\] by choosing the constant \(A\) with \[\label{est03-01} A^{\frac{\xi\alpha(q-1)}{2}} = \left( \fint_{\Omega_T} u^{\frac{\xi\alpha(q-1)}{2-\alpha}} \,dxdt \right)^{\frac{2-\alpha}{2}}.\tag{21}\] We combine the above inequalities and ?? to discover \[\label{est03-02} \fint_{\Omega_T} \left|\nabla u^{\frac{m+q-1}{2}}\right|^{\alpha} \,dxdt \le \left[\frac{(m+q-1)^2}{m(q-1)(\xi-1)}\right]^{\frac{\alpha}{2}} A^{\frac{\alpha(q-1)}{2}} \left[ \frac{2\mu(\Omega_{T})}{|\Omega_T|} + \frac{(q-1)(\xi-1)}{3m}\fint_{\Omega_T} u^{2-m} |V|^2 \,dxdt\right]^{\frac{\alpha}{2}}.\tag{22}\] On the other hand, Lemma 1 and ?? imply \[\label{est03-03} \begin{align} \fint_{\Omega_T} \left(u^{\frac{m+q-1}{2}}\right)^{\alpha\frac{d+\frac{2}{m+q-1}}{d}} \,dxdt &\le c \fint_{\Omega_T} \left|\nabla u^{\frac{m+q-1}{2}}\right|^{\alpha} \,dxdt \left(\sup_{t \in (0, T)} \int_{\Omega} \left(u^{\frac{m+q-1}{2}}\right)^{\frac{2}{m+q-1}} \,dx \right)^{\frac{\alpha}{d}}\\ &\le c [\mu(\Omega_{T})]^{\frac{\alpha}{d}} \fint_{\Omega_T} \left|\nabla u^{\frac{m+q-1}{2}}\right|^{\alpha} \,dxdt, \end{align}\tag{23}\] where \(c=c(m,d,q,\alpha)\ge 1\). Now we choose \(\xi >1\) such that \[\frac{\xi\alpha(q-1)}{2-\alpha} = \frac{m+q-1}{2}\alpha\frac{d+\frac{2}{m+q-1}}{d},\] that is, \[\begin{gather} \xi = \frac{(2-\alpha)\{2+d(m+q-1)\}}{2d(q-1)} > 1 \iff \alpha < 2 - \frac{2d(q-1)}{2+d(m+q-1)} = \frac{2(2+md)}{2+d(m+q-1)},\\ \xi-1=\frac{2(2+md)-\alpha\{2+d(m+q-1)\}}{2d(q-1)},\\ \frac{(m+q-1)^2}{m(q-1)(\xi-1)} = \frac{2d(m+q-1)^2}{m[2(2+md)-\alpha\{2+d(m+q-1)\}]}. \end{gather}\] Inserting 23 into 21 , we find \[A^{\frac{\alpha(q-1)}{2}} \le c [\mu(\Omega_{T})]^{\frac{\alpha(q-1)}{2+d(m+q-1)}} \left( \fint_{\Omega_T} \left|\nabla u^{\frac{m+q-1}{2}}\right|^\alpha\,dxdt \right)^{\frac{d(q-1)}{2+d(m+q-1)}}.\] Combining the above estimates, we obtain \[\label{est03-04} \begin{align} &\left(\fint_{\Omega_T} \left|\nabla u^{\frac{m+q-1}{2}}\right|^{\alpha} \,dxdt\right)^{1-\frac{d(q-1)}{2+d(m+q-1)}} \\ &\qquad \le c \left[\frac{2d(m+q-1)^2}{m[2(2+md)-\alpha\{2+d(m+q-1)\}]}\right]^{\frac{\alpha}{2}} [\mu(\Omega_{T})]^{\frac{\alpha(q-1)}{2+d(m+q-1)}} \left[ \frac{2\mu(\Omega_{T})}{|\Omega_T|}\right]^{\frac{\alpha}{2}}\\ &\qquad \quad + c [\mu(\Omega_{T})]^{\frac{\alpha(q-1)}{2+d(m+q-1)}} \left[\frac{(m+q-1)^2}{m^2}\fint_{\Omega_T} u^{2-m} |V|^2 \,dxdt\right]^{\frac{\alpha}{2}}, \end{align}\tag{24}\] where \(c=c(m,d,q,\alpha)\ge 1\). This implies the estimate ?? . ◻

Proof. Now we handle the term of \(V\) in the estimate ?? . Let us set \[\mathcal{V}= \int_{\Omega_T} u^{2-m} |V|^2 \,dxdt.\]

If \(m=2\), then \(\mathcal{V}= \|V\|^{2}_{L_{x,t}^{2,2}}\). Therefore, the estimate ?? directly becomes \[\label{E95V9522} \left(\fint_{\Omega_T} \left|\nabla u^{\frac{m}{2}}\right|^{\alpha} \,dxdt\right)^{\frac{1}{\alpha}} \le c \left[ \frac{\mu(\Omega_{T})}{|\Omega_T|}\right]^{\frac{1}{2}}+ c \left(\fint_{\Omega_T} |V|^2 \,dxdt\right)^{\frac{1}{2}}.\tag{25}\]

Now let \(1-\frac{1}{d}< m <2\). Applying Hölder inequalities in space with \(q_1 >2\) and time with \(q_2 >2\), we obtain \[\mathcal{V} \leq \|V\|_{L_{x,t}^{q_1, q_2}}^{2} \|u\|_{L_{x,t}^{r_1, r_2}}^{2-m}, \quad r_1 = \frac{(2-m)q_1}{q_1 - 2}, \quad r_2 = \frac{(2-m)q_2}{q_2 - 2}.\] Requiring the pair \((r_1, r_2)\) to satisfy ?? in Lemma 2 is equivalent to \[\label{V95alpha} \begin{gather} V \in L_{x,t}^{q_1,q_2} \;\text{ where } \frac{d}{q_1} + \frac{\alpha (2+md) - 2d}{2q_2} = \frac{\alpha(2+md)+2d(m-2)}{4}, \\ \text{ for } \begin{cases} \frac{m-1}{2} \leq \frac{1}{q_1} \leq \frac{1}{2}-\frac{(2-m)(d-\alpha)}{\alpha md}, \; \frac{1}{2}-\frac{2-m}{\alpha m} \leq \frac{1}{q_2} \leq \frac{1}{2} &\text{if } 1\leq m < 2, \;\frac{2d}{2+md} \leq \alpha < 2,\\ 0\leq \frac{1}{q_1} \leq \frac{\alpha(2+md)+2d(m-2)}{4d}, \; 0 \leq \frac{1}{q_2} \leq \frac{1}{2}-\frac{d(1-m)}{\alpha(2+md)-2d} & \text{if } 1-\frac{1}{d}< m < 1, \;\frac{2d(2-m)}{2+md} \leq \alpha < 2. \end{cases} \end{gather}\tag{26}\] Then, combining ?? in Lemma 2 and ?? in Lemma 4 yields \[\label{V:est:02} \mathcal{V} \leq c\|V\|_{L_{x,t}^{q_1, q_2}}^{2} \left[\mu(\Omega_{T})\right]^{\frac{2(2-m) q_2 - \alpha m (q_2 -2)}{2q_2}} \left(\int_{\Omega_T} \left|\nabla u^{\frac{m}{2}}\right|^{\alpha}\,dxdt\right)^{1-\frac{2}{q_2}}.\tag{27}\]

We now derive ?? for each case.

\(\bullet\) (Porous medium case) If \(m=2\), the desired estimate ?? follows immediately from 25 by applying Hölder inequalities for any \(q_1, q_2 \geq 2\).

Let \(1\leq m <2\) and \(V \in L_{x,t}^{\frac{2}{m-1}, 2}\). In this setting, \(V\) belongs to the scaling-invariant class \(\mathcal{S}_{m}^{(q_1, q_2)}\) (with the convention that \(\frac{2}{m-1}=\infty\) if \(m=1\)). Applying Hölder inequality in space with \((2-m)+(m-1)=1\) and using ?? , we have \[\label{V:est:01} \mathcal{V} \leq \left(\sup_{t\in (0,T)} \int_{\Omega} u \,dx\right)^{2-m} \|V\|_{L_{x,t}^{\frac{2}{m-1}, 2}}^{2} \leq [\mu(\Omega_T)]^{2-m} \cdot \|V\|_{L_{x,t}^{\frac{2}{m-1}, 2}}^{2}.\tag{28}\] Note that 28 is also valid for \(m=1\). This gives ?? with \(\sigma_1 = 1\) and \(\sigma_{2}=\frac{2-m}{2}\) since \(q_2 =2\).

If \(1\leq m <2\) and \(V\) satisfies 26 , inserting 27 into ?? implies \[\label{V:est:03} \left(\fint_{\Omega_T} \left|\nabla u^{\frac{m}{2}}\right|^{\alpha} \,dxdt\right)^{\frac{1}{\alpha}} \le c \left[ \frac{\mu(\Omega_{T})}{|\Omega_T|}\right]^{\frac{1}{2}} + c \|V\|_{L_{x,t}^{q_1, q_2}} \frac{\left[\mu(\Omega_{T})\right]^{\frac{2(2-m)q_2 - \alpha m (q_2 -2)}{4q_2}}}{|\Omega_{T}|^{\frac{1}{q_2}}} \left(\fint_{\Omega_T} \left|\nabla u^{\frac{m}{2}}\right|^{\alpha}\,dxdt\right)^{\frac{q_2 -2}{2q_2}}.\tag{29}\] By applying Young’s inequality with \(\frac{\alpha(q_2 -2)}{2q_2} + \frac{(2-\alpha)q_2 + 2\alpha}{2q_2}=1\), we obtain \[\label{V:est:04}\begin{align} \left(\fint_{\Omega_T} \left|\nabla u^{\frac{m}{2}}\right|^{\alpha} \,dxdt\right)^{\frac{1}{\alpha}} &\le c \left[ \frac{\mu(\Omega_{T})}{|\Omega_T|}\right]^{\frac{1}{2}} + \frac{1}{2} \left(\fint_{\Omega_T} \left|\nabla u^{\frac{m}{2}}\right|^{\alpha} \,dxdt\right)^{\frac{1}{\alpha}} + c\|V\|_{L_{x,t}^{q_1, q_2}}^{\frac{2q_2}{(2-\alpha)q_2 + 2\alpha}} \frac{\left[\mu(\Omega_{T})\right]^{\frac{2(2-m)q_2-\alpha m(q_2-2)}{2\left[(2-\alpha)q_2 + 2\alpha\right]}}}{|\Omega_{T}|^{\frac{2}{(2-\alpha)q_2 + 2\alpha}}}, \end{align}\tag{30}\] which implies ?? . Since ?? is obtained for any \(\alpha \in (0,2)\), the condition 26 is valid for \(1\leq m <2\) and \(\frac{2d}{2+md} \leq \alpha < 2\). As \(\alpha \to 2\), 26 corresponds to \[V \in \mathfrak{S}_{m}^{(q_1,q_2)} \;\text{ where } \;\frac{md}{(2-m)+d(m-1)} < q_1 \leq \frac{2}{m-1}, \; 2 \leq q_2 < \frac{m}{m-1}.\]

If \(V\) satisfies ?? , then \((q_1,q_2)\) either equals \(\left( \frac{2}{m - 1}, 2 \right)\) (where 28 or 25 applies), or satisfies 26 (where 29 applies), or lies in the subcritical regime of 26 . In the subcritical case, there exists a pair \((\tilde{q}_1, \tilde{q}_2)\) satisfying 26 such that \(\tilde{q}_1 \leq q_1\) and \(\tilde{q}_2 \leq q_2\). Therefore, we have \[\label{subSalpha} \|V\|_{L^{\tilde{q}_1, \tilde{q}_2}_{x,t}} \leq |\Omega|^{\frac{1}{\tilde{q}_1} - \frac{1}{q_1}} T^{\frac{1}{\tilde{q}_2} - \frac{1}{q_2}} \|V\|_{L^{q_1, q_2}_{x,t}},\tag{31}\] which again leads to 29 . Thus ?? is valid for all \(V\) satisfying ?? .

\(\bullet\) (Fast diffusion case) Assume that \(1-\frac{1}{d}<m<1\). By the same reasoning as above, the class 26 as \(\alpha \to 2\) corresponds to ?? . Consequently, for any \(V\) satisfying ?? , the estimate ?? holds.

\(\bullet\) (Divergence-free case) Since both ?? and ?? are independent of \(V\), the estimate ?? from Lemma 6 contains no drift integral term. Therefore, it directly yields ?? . ◻

Proof. We proceed under the condition ?? , which is established in Lemma 4 and Theorem 3.

To show that \(\partial_t u \in L^{1}(0,T; W^{-1,1}(\Omega))\), it suffices to bound the right-hand side of the weak formulation \[\label{compact01} \int_{\Omega_T} \partial_t u \,\varphi \,dxdt = -\int_{\Omega_T}\nabla u^m\cdot\nabla\varphi\,dxdt + \int_{\Omega_T}uV\cdot\nabla\varphi\,dxdt + \int_{\Omega_T}\mu\varphi\,dxdt\tag{32}\] for every \(\varphi \in C^{\infty}(\overline{\Omega} \times [0,T])\) such that \(\varphi = 0\) on \(\partial \Omega \times (0,T)\) and \(\varphi(\cdot, T)=0\).

First, it is straightforward to see that \[\int_{\Omega_T} \mu \varphi \,dxdt \leq \|\mu\|_{L^1(\Omega_T)}\|\varphi\|_{L^{\infty}(\Omega_T)} < \infty.\]

Next, using the identity \(|\nabla u^m| = 2 u^{\frac{m}{2}}|\nabla u^{\frac{m}{2}}|\), we obtain \[\int_{\Omega_T} |\nabla u^m| |\nabla\varphi| \,dxdt \leq 2\|\nabla \varphi\|_{L^{\infty}(\Omega_{T})} \int_{\Omega_T} u^{\frac{m}{2}}|\nabla u^{\frac{m}{2}}| \,dxdt.\] By Hölder’s inequality, it follows that \[\begin{align} \int_{\Omega_T} u^{\frac{m}{2}}|\nabla u^{\frac{m}{2}}| \,dxdt &\leq \left(\int_{\Omega_T} |\nabla u^{\frac{m}{2}}|^{\alpha} \,dxdt\right)^{\frac{1}{\alpha}} \left(\int_{\Omega_T} u^{\frac{\alpha m}{2(\alpha -1)}} \,dxdt\right)^{1-\frac{1}{\alpha}}. \end{align}\] Consequently, the second integral is finite provided that the exponent satisfies \[\frac{\alpha m}{2(\alpha -1)} \leq \begin{cases} 1, & \text{ for } (1-\frac{2}{d})_{+}<m < 1 \text{ and } \alpha \in [\frac{2}{2-m},2), \\ \frac{\alpha (2+md)}{2d}, & \text{ for } m \geq 1 \text{ and } \alpha \in [\frac{2(1+md)}{2+md}, 2). \end{cases}\] Indeed, by Lemma 1, the assumption ?? ensures that \(u \in L_{x,t}^{\frac{\alpha(2+md)}{2d}}\), and in the singular case, we directly use the first condition in ?? .

Lastly, applying Hölder’s inequality with \(q_1, q_2 > 1\) yields \[\label{compact02} \int_{\Omega_T} u |V| |\nabla \varphi| \,dxdt \leq \|\nabla \varphi\|_{L^{\infty}} \|V\|_{L_{x,t}^{q_1,q_2}} \|u\|_{L_{x,t}^{r_1,r_2}},\tag{33}\] where \(r_1 = \frac{q_1}{q_1 -1}\) and \(r_2 = \frac{q_2}{q_2 -1}\). Note that the bound for the case \(V\in L_{x,t}^{\infty,1}\) in ?? is immediate since \(u\in L_{x,t}^{1,\infty}(\Omega_{T})\). Furthermore, Lemma 2 combined with ?? guarantees that \(\|u\|_{L_{x,t}^{r_1,r_2}} < \infty\) whenever the pair \((r_1, r_2)\) satisfies ?? , which corresponds to the pair \((q_1, q_2)\) in ?? .

Combining the above estimates in 32 , we conclude that \[\partial_t u \in L^{1}(0,T; W^{-1,1}(\Omega)),\] which completes the proof. ◻

Proof of Theorem 1. Let \(\mu\) be a nonnegative finite measure on \(\Omega_T\). Then there exists a sequence of nonnegative functions \(\{\mu_n\}\subset L^\infty(\Omega_T)\) such that \[\mu_n(\Omega_T):=\int_{\Omega_T}\mu_n\,dxdt \le \mu(\Omega_T)\] and \[\mu_n \rightharpoonup \mu \qquad \text{weakly in the sense of measures.}\] Likewise, we choose a sequence \(\{V_n\}\subset C^\infty(\Omega_{T})\cap L^\infty(\Omega_T)\) such that \[V_n \to V \quad \text{strongly in } L_{x,t}^{q_1,q_2}(\Omega_T),\] and \[\sup_n \|V_n\|_{L_{x,t}^{q_1,q_2}(\Omega_T)}<\infty.\] When one of the exponents \(q_1,q_2\) is infinite, this approximation is understood in finite lower exponents which are still admissible for the estimates below, together with the corresponding uniform endpoint bound. In particular, the endpoint \((q_1,q_2)=(\infty,2)\) for \(m=1\) is handled by approximating in \(L_{x,t}^{q,2}\) with a fixed sufficiently large finite \(q\), while retaining the uniform \(L_{x,t}^{\infty,2}\) bound.

For each \(n\), let \(u_n\) be a regular solution in the sense of Definition 3 to the regularized problem \[\label{reg-PME} \partial_t u_n-\Delta u_n^m+\nabla\cdot(u_nV_n)=\mu_n\tag{34}\] with zero initial and boundary data. Moreover, for the case of measure-valued initial data, a similar argument applies, see Remark 6.

Choose \(\bar\alpha\in(1,2)\) sufficiently close to \(2\) so that \[1<\kappa<\frac{\bar\alpha(2+md)}{2(d+1)}\] for some \(\kappa>1\). This is possible because \(m>1-\frac{1}{d}\). Applying the estimates from Section 2 and Theorem 3, we obtain \[\sup_{t\in(0,T)}\int_\Omega u_n(x,t)\,dx \le \mu_n(\Omega_T) \le \mu(\Omega_T),\] and \[\sup_n \int_{\Omega_T}\left|\nabla u_n^{\frac{m}{2}}\right|^{\bar\alpha}\,dxdt \le C.\] Moreover, Lemma 1 yields \[\sup_n \|u_n\|_{L^r(\Omega_T)}\le C, \qquad r:=\frac{\bar\alpha(2+md)}{2d}.\]

Next, Hölder’s inequality gives \[\begin{align} \int_{\Omega_T} |\nabla u_n|^\kappa\,dxdt &= c(m)\int_{\Omega_T} \left|\nabla u_n^{\frac{m}{2}}\right|^\kappa u_n^{\frac{\kappa(2-m)}{2}} \,dxdt \\ &\le c(m) \left(\int_{\Omega_T}\left|\nabla u_n^{\frac{m}{2}}\right|^{\bar\alpha}\,dxdt\right)^{\frac{\kappa}{\bar\alpha}} \left(\int_{\Omega_T}u_n^{\frac{\bar\alpha \kappa(2-m)}{2(\bar\alpha-\kappa)}}\,dxdt\right)^{1-\frac{\kappa}{\bar\alpha}}. \end{align}\] Since \[\frac{\bar\alpha \kappa(2-m)}{2(\bar\alpha-\kappa)} \le \frac{\bar\alpha(2+md)}{2d}=r \qquad \Longleftrightarrow \qquad \kappa\le \frac{\bar\alpha(2+md)}{2(d+1)},\] the right-hand side is uniformly bounded in \(n\). Hence \[\sup_n \|\nabla u_n\|_{L^\kappa(\Omega_T)}<\infty.\] Because \(u_n=0\) on \(\partial\Omega\times(0,T)\), Poincaré’s inequality implies \[\sup_n \|u_n\|_{L^\kappa(0,T;W_0^{1,\kappa}(\Omega))}<\infty.\]

By Lemma 7, \[\sup_n \|\partial_t u_n\|_{L^1(0,T;W^{-1,1}(\Omega))}<\infty.\] We now apply Lemma 3 with \[X_0=W_0^{1,\kappa}(\Omega), \qquad X=L^1(\Omega), \qquad X_1=W^{-1,1}(\Omega).\] Since \[W_0^{1,\kappa}(\Omega)\Subset L^1(\Omega) \qquad \text{and} \qquad L^1(\Omega)\hookrightarrow W^{-1,1}(\Omega),\] we conclude that, up to a subsequence, \[u_n \to u \quad \text{strongly in } L^1(\Omega_T)\] and almost everywhere in \(\Omega_T\).

On the other hand, after passing to a further subsequence, \[\nabla u_n^{\frac{m}{2}}\rightharpoonup w \quad \text{weakly in } L^{\bar\alpha}(\Omega_T;\mathbb{R}^d),\] and \[u_n \rightharpoonup u \quad \text{weakly in } L^r(\Omega_T).\] By interpolation between the strong convergence in \(L^1(\Omega_T)\) and the uniform bound in \(L^r(\Omega_T)\), \[u_n \to u \quad \text{strongly in } L^\gamma(\Omega_T) \quad \text{for every } 1\le \gamma<r.\] In particular, \[u_n^{\frac{m}{2}}\to u^{\frac{m}{2}} \quad \text{strongly in } L^{\frac{\bar\alpha}{\bar\alpha-1}}(\Omega_T).\]

To identify the weak limit, let \(\eta\in C_c^\infty(\Omega_T)\). Then \[\int_{\Omega_T} w \eta\,dxdt = \lim_{n\to\infty} \int_{\Omega_T}\nabla u_n^{\frac{m}{2}}\eta\,dxdt = -\lim_{n\to\infty} \int_{\Omega_T}u_n^{\frac{m}{2}}\,\nabla\eta\,dxdt.\] Since \(u_n^{\frac{m}{2}}\to u^{\frac{m}{2}}\) strongly in \(L^{\frac{\bar\alpha}{\bar\alpha-1}}(\Omega_T)\), we obtain \[\int_{\Omega_T} w\eta\,dxdt = -\int_{\Omega_T}u^{\frac{m}{2}}\,\nabla\eta\,dxdt.\] Hence \[w=\nabla u^{\frac{m}{2}} \qquad \text{in the sense of distributions.}\]

It remains to pass to the limit in the weak formulation for 34 . The time derivative term and the measure term are standard. For the diffusion term, we write \[\nabla u_n^m = 2u_n^{\frac{m}{2}}\nabla u_n^{\frac{m}{2}},\] and use the strong convergence of \(u_n^{m/2}\) together with the weak convergence of \(\nabla u_n^{m/2}\) to deduce \[\nabla u_n^m \rightharpoonup \nabla u^m \quad \text{weakly in } L^1(\Omega_T;\mathbb{R}^d).\] For the drift term, let \[r_1=\frac{q_1}{q_1-1}, \qquad r_2=\frac{q_2}{q_2-1}.\] Since \((r_1,r_2)\) satisfies ?? , Lemma 2, together with the uniform bounds above, yields \[\sup_n \|u_n\|_{L_{x,t}^{r_1,r_2}(\Omega_T)}<\infty.\] Hence, up to a subsequence, \[u_n \rightharpoonup u \quad \text{weakly in } L_{x,t}^{r_1,r_2}(\Omega_T).\] Since \(V_n\to V\) strongly in \(L_{x,t}^{q_1,q_2}(\Omega_T)\), it follows that \[u_nV_n \rightharpoonup uV \quad \text{weakly in } L^1(\Omega_T;\mathbb{R}^d).\] Passing to the limit in the weak formulation, we conclude that \(u\) is a weak solution of 1 in the sense of Definition 1. ◻

Proof of Theorem 2. We note that a naive mollification of a divergence-free vector field on \(\Omega\) does not in general remain divergence-free on the whole domain \(\Omega\). Therefore, instead of working directly on \(\Omega\), we use an interior exhaustion.

Let \(\{\Omega_n\}_{n=1}^\infty\) be a sequence of bounded smooth domains such that \[\overline{\Omega_n}\subset \Omega_{n+1}\Subset \Omega, \qquad \bigcup_{n=1}^\infty \Omega_n=\Omega.\] Set \(\Omega_{n,T}:=\Omega_n\times(0,T).\) Let \(\mu\) be a nonnegative finite measure on \(\Omega_T\). Then there exists a sequence of nonnegative functions \(\{\mu_n\}\subset C_c^\infty(\Omega_{n,T})\) such that \[\mu_n(\Omega_{n,T})=\int_{\Omega_{n,T}}\mu_n\,dxdt \le \mu(\Omega_T)\] and \(\mu_n \rightharpoonup \mu\) weakly in the sense of measures on \(\Omega_T\). Let \(\widetilde{V}\) denote the extension of \(V\) by zero outside \(\Omega_T\). Choose \(\varepsilon_n>0\) so that \(\varepsilon_n < \frac{1}{2} \operatorname{dist}(\Omega_n,\partial\Omega),\) and let \(\rho_{\varepsilon_n}^x\) and \(\eta_{\varepsilon_n}^t\) be standard mollifiers in space and time. Define \[V_n:=\eta_{\varepsilon_n}^t *_t \left(\rho_{\varepsilon_n}^x *_x \widetilde{V}\right) \qquad \text{in } \Omega_{n,T}.\] Then \(V_n\in C^\infty(\Omega_{n,T})\cap L^\infty(\Omega_{n,T}),\) and \(V_n \to V\) strongly in the ambient space corresponding to either ?? or ?? , locally in \(\Omega_T\). Moreover, \(\nabla\cdot V_n=0\) in \(\Omega_{n,T}\) for every \(n\).

For each \(n\), let \(u_n\) be a regular solution in \(\Omega_{n,T}\) of \[\label{reg-PME95n} \partial_t u_n-\Delta u_n^m+\nabla\cdot(u_nV_n)=\mu_n \qquad \text{in } \Omega_{n,T},\tag{35}\] with zero initial data and zero lateral boundary data on \(\partial\Omega_n\times(0,T).\) We extend \(u_n\) by zero from \(\Omega_{n,T}\) to \(\Omega_T\), and still denote the extension by \(u_n\).

Fix \(\bar\alpha\in(1,2)\) such that \[\label{alpha-star-divfree} \bar\alpha> \max\left\{ \frac{2d}{2+md}, \, \frac{2}{2-m} \chi_{\{m<1\}}, \, \frac{2(1+md)}{2+md} \chi_{\{m\ge1\}} \right\}.\tag{36}\] This is possible since \(m>(1-\frac{2}{d})_+\). By Theorem 3 (iii), applied on \(\Omega_{n,T}\), we obtain \[\label{divfree-est-1} \sup_{t\in(0,T)}\int_{\Omega_n} u_n(x,t)\,dx \le \mu_n(\Omega_{n,T}) \le \mu(\Omega_T),\tag{37}\] and \[\label{divfree-est-2} \sup_n \int_{\Omega_{n,T}} \left|\nabla u_n^{\frac{m}{2}}\right|^{\bar\alpha}\,dxdt \le C.\tag{38}\] Since \(\bar\alpha>\frac{2d}{2+md}\), Lemma 1 yields \[\label{divfree-est-3} \sup_n \|u_n\|_{L^p(\Omega_{n,T})}\le C, \qquad p:=\frac{\bar\alpha(2+md)}{2d}>1.\tag{39}\]

We now prove compactness. The argument is divided according to the diffusion range.

Case 1: \(1\le m\le2\). Choose \(\kappa\) such that \[1<\kappa<\frac{\bar\alpha(2+md)}{2(d+1)}.\] Exactly as in the proof of Theorem 1, Hölder’s inequality gives \[\sup_n \|\nabla u_n\|_{L^\kappa_{\rm loc}(\Omega_T)}<\infty.\] Hence, for every \(U\Subset\Omega\) and every \(0<t_1<t_2<T\), \[\sup_n \|u_n\|_{L^\kappa(t_1,t_2;W^{1,\kappa}(U))}<\infty.\] Moreover, by Lemma 7, \[\sup_n \|\partial_t u_n\|_{L^1(t_1,t_2;W^{-1,1}(U))}<\infty.\] Therefore, by Lemma 3, after passing to a subsequence, \[u_n\to u \quad \text{strongly in }L^1_{\rm loc}(\Omega_T)\] and almost everywhere in \(\Omega_T\).

Case 2: \(\left(1-\frac{2}{d}\right)_+<m<1\). For \(k\in\mathbb{N}\), define \[w_n^{(k)}:=T_k(u_n):=\min\{u_n,k\}.\] Since \(0<m<1\), we have \[|\nabla w_n^{(k)}| = \frac{2}{m}\left(w_n^{(k)}\right)^{1-\frac{m}{2}} \left|\nabla \left(w_n^{(k)}\right)^{\frac{m}{2}}\right| \le C(k,m)\left|\nabla u_n^{\frac{m}{2}}\right|.\] Hence, for every \(U\Subset\Omega\) and every \(0<t_1<t_2<T\), \[\label{divfree-trunc-grad} \sup_n \|w_n^{(k)}\|_{L^{\bar\alpha}(t_1,t_2;W^{1,\bar\alpha}(U))}\le C(k,U,t_1,t_2).\tag{40}\] At this point we use the truncation theory from Sections 3 and 4.3 in [@BDG15]. In the present divergence-free setting, the drift term disappears in the truncation estimates exactly as in the proof of Lemma 5. Therefore the arguments of Lemmas 3.1–3.4 and Section 4.3.1–4.3.2 in [@BDG15] apply to the truncations \(T_k(u_n)\) and yield the local time-derivative estimate \[\label{divfree-trunc-time} \sup_n \|\partial_t w_n^{(k)}\|_{L^1(t_1,t_2;W^{-1,1}(U))} \le C(k,U,W,t_1,t_2, \|V\|_{L^{1}(\Omega_T)})\tag{41}\] for every \(U\Subset W\Subset\Omega\) and every \(0<t_1<t_2<T\). Consequently, for every fixed \(k\in\mathbb{N}\), Lemma 3 implies that, after passing to a subsequence, \[T_k(u_n)\to T_k(u) \quad \text{strongly in }L^1_{\rm loc}(\Omega_T).\] Repeating the diagonal construction exactly as in Section 4.3.2 of [@BDG15], we conclude that \[u_n\to u \quad \text{strongly in }L^1_{\rm loc}(\Omega_T)\] and almost everywhere in \(\Omega_T\).

Case 3: \(m>2\). In this case, the direct estimate of \(\nabla u_n\) used in Case 1 is not available because \(u_n^{1-\frac{m}{2}}\) is singular near the zero set of \(u_n\). We therefore appeal to the compactness theory for the porous medium equation with measure data in [@BDG13]. As in Case 2, the divergence-free condition allows us to rewrite the drift contribution in the local truncation estimates as a lower-order cutoff term. This term is controlled locally by the integrability of \(V_n\), and therefore it does not affect the compactness argument.

Consequently, the local compactness method of [@BDG13] applies to the sequence \(\{u_n\}\) and gives, up to a subsequence, \[u_n\to u \quad \text{strongly in }L^1_{\rm loc}(\Omega_T)\] and almost everywhere in \(\Omega_T\).

In all cases, we have now obtained \[u_n\to u \quad \text{strongly in }L^1_{\rm loc}(\Omega_T)\] and almost everywhere in \(\Omega_T\). Interpolating with 39 , we also obtain \[\label{divfree-strong-gamma} u_n\to u \quad \text{strongly in }L^\gamma(\Omega_T) \quad \text{for every }1\le \gamma<p.\tag{42}\]

From 38 , after passing to a further subsequence if necessary, we may assume that \[\nabla u_n^{\frac{m}{2}}\rightharpoonup z \quad \text{weakly in }L^{\bar\alpha}_{\rm loc}(\Omega_T;\mathbb{R}^d).\] Since 36 implies \(\frac{m\bar\alpha}{2(\bar\alpha-1)}<p,\) the strong convergence 42 yields \[u_n^{\frac{m}{2}}\to u^{\frac{m}{2}} \quad \text{strongly in }L^{\frac{\bar\alpha}{\bar\alpha-1}}_{\rm loc}(\Omega_T).\] Let \(\eta\in C_c^\infty(\Omega_T)\). Then \[\int_{\Omega_T} z\eta\,dxdt = \lim_{n\to\infty}\int_{\Omega_T}\nabla u_n^{\frac{m}{2}}\eta\,dxdt = -\lim_{n\to\infty}\int_{\Omega_T}u_n^{\frac{m}{2}}\,\nabla\eta\,dxdt = -\int_{\Omega_T}u^{\frac{m}{2}}\,\nabla\eta\,dxdt.\] Hence \(z=\nabla u^{\frac{m}{2}}\) in the sense of distributions.

Let \(\varphi \in C^{\infty} \left(\overline{\Omega} \times [0, T]\right)\) which vanishes on \(\partial \Omega \times (0,T)\) and \(\varphi(\cdot,T)=0\). We cannot use \(\varphi\) directly as a test function in the approximate problem on \(\Omega_n\), since \(\varphi\) does not necessarily vanish on \(\partial\Omega_n\). By using a suitable spatial cutoff function, we can construct a sequence of \(\{\phi_j\} \subset C_c^\infty(\Omega\times[0,T))\) with \(\phi_j(\cdot,T)=0\) such that \(\varphi_j\to\varphi\) in \(C^1_{\rm loc}(\Omega_T)\), and \(\varphi_j\), \((\varphi_j)_t\), \(\nabla\varphi_j\) are uniformly bounded. Then for each fixed \(j\), the function \(\varphi_j\) is admissible in the weak formulation for \(u_n\) on \(\Omega_{n,T}\) for all sufficiently large \(n\). Testing the approximate equation by \(\varphi_j\) and passing to the limit \(n\to\infty\), exactly as in the proof of Theorem 1, gives \[\int_{\Omega_T} \left\{ -u(\varphi_j)_t+\nabla u^m\cdot\nabla\varphi_j -uV\cdot\nabla\varphi_j \right\}\,dxdt = \int_{\Omega_T}\varphi_j\,d\mu .\] Finally, letting \(j\to\infty\), we obtain that \(u\) is a weak solution of 1 in \(\Omega_T\). This completes the proof. ◻

Proof. We only explain the core a priori estimates for smooth solutions. The standard regularization and compactness argument producing a weak solution pair is exactly the same as in Section 3, and is therefore omitted.

We set \(\nu:=\|\rho_0\|_{L^1(\Omega)}+\mu(\Omega_T)\). Since \(v\) is divergence-free, Theorem 2, together with Remark 6, applies to the equation 44 . In particular, for every \(0<\alpha<2\), \[\label{app-rho-general} \sup_{t\in(0,T)}\int_\Omega \rho(\cdot,t)\,dx + \int_{\Omega_T}\left|\nabla \rho^{\frac{m}{2}}\right|^\alpha\,dxdt \le C(\nu),\tag{46}\] where the constant is independent of the drift \(v\).

We now estimate the fluid component.

\(\bullet\) (2d case) Assume that \(d=2\) and \(m>0\). Choose \(\bar\alpha\in \left( \max\left\{1,\frac{2}{m+1}\right\},2 \right).\) Then 46 yields \[\label{app-2d-rho} \sup_{t\in(0,T)}\int_\Omega \rho(\cdot,t)\,dx + \int_{\Omega_T}\left|\nabla \rho^{\frac{m}{2}}\right|^{\bar\alpha}\,dxdt \le C(\nu).\tag{47}\]

Testing 45 against \(v\), and using \(\operatorname{div}v=0\), we obtain \[\frac{1}{2}\frac{d}{dt}\|v\|_{L^2(\Omega)}^2+\|\nabla v\|_{L^2(\Omega)}^2 = -\int_\Omega \rho\,\nabla\phi\cdot v\,dx \le \|\nabla\phi\|_{L^\infty}\|\rho\|_{L^q}\|v\|_{L^p},\] where \(\frac{1}{p}+\frac{1}{q}=1\). By the two-dimensional Gagliardo–Nirenberg inequality, \[\|v\|_{L^p} \le C\|v\|_{L^2}^{\frac{2}{p}}\|\nabla v\|_{L^2}^{1-\frac{2}{p}}.\] Moreover, by interpolation between \(L^1(\Omega)\) and the Sobolev exponent for \(\rho^{m/2}\), \[\|\rho\|_{L^q} \le C \|\rho\|_{L^1}^{\theta} \|\nabla \rho^{\frac{m}{2}}\|_{L^{\bar\alpha}}^{\frac{2}{m}(1-\theta)}, \qquad \theta=\frac{(m+q)\bar\alpha-2q}{q(\bar\alpha(m+1)-2)}.\] Substituting these bounds into the energy inequality for \(v\) and applying Young’s inequality, we obtain \[\frac{d}{dt}\|v\|_{L^2}^2+\|\nabla v\|_{L^2}^2 \le \varepsilon \|\nabla \rho^{\frac{m}{2}}\|_{L^{\bar\alpha}}^{\bar\alpha} + C_\varepsilon\left(1+\|v\|_{L^2}^2\right),\] provided that \(p>2\) is chosen sufficiently large. Since 47 controls \(\|\nabla \rho^{\frac{m}{2}}\|_{L^{\bar\alpha}(\Omega_T)}\), Gronwall’s inequality yields \[\label{app-2d-v} \sup_{t\in(0,T)}\|v(\cdot,t)\|_{L^2(\Omega)}^2 + \int_{\Omega_T}|\nabla v|^2\,dxdt \le C.\tag{48}\] Combining 47 and 48 , and then using 46 again for arbitrary \(0<\alpha<2\), we conclude that \[\sup_{t\in(0,T)}\int_\Omega \left(\rho(\cdot,t)+|v(\cdot,t)|^2\right)\,dx + \int_{\Omega_T}\left(\left|\nabla \rho^{\frac{m}{2}}\right|^\alpha+|\nabla v|^2\right)\,dxdt \le C\] for every \(0<\alpha<2\).

\(\bullet\) (3d case) Assume that \(d=3\) and \(m>\frac{2}{3}\). Choose \(\bar\alpha\in \left( \max\left\{1,\frac{8}{3m+2}\right\},2 \right).\) Then 46 gives \[\label{app-3d-rho} \sup_{t\in(0,T)}\int_\Omega \rho(\cdot,t)\,dx + \int_{\Omega_T}\left|\nabla \rho^{\frac{m}{2}}\right|^{\bar\alpha}\,dxdt \le C(\nu).\tag{49}\]

Testing 45 against \(v\), and using \(\operatorname{div}v=0\), we obtain \[\frac{1}{2}\frac{d}{dt}\|v\|_{L^2(\Omega)}^2+\|\nabla v\|_{L^2(\Omega)}^2 = -\int_\Omega \rho\,\nabla\phi\cdot v\,dx \lesssim \|\rho\|_{L^{6/5}}\|v\|_{L^6} \le C\|\rho\|_{L^{6/5}}^2+\frac{1}{2}\|\nabla v\|_{L^2}^2.\] Hence, we get \[\label{app-3d-v1} \frac{1}{2}\frac{d}{dt}\|v\|_{L^2}^2+\|\nabla v\|_{L^2}^2 \lesssim \|\rho\|_{L^{6/5}}^2.\tag{50}\]

We next estimate \(\|\rho\|_{L^{6/5}}\). By interpolation, together with the uniform \(L^1\)-bound on \(\rho\), \[\|\rho\|_{L^{6/5}}^2 \le \|\rho\|_{L^1}^{2\theta} \|\rho\|_{L^{\frac{3m\bar\alpha}{2(3-\bar\alpha)}}}^{2(1-\theta)} \lesssim \left\|\rho^{\frac{m}{2}}\right\|_{L^{\frac{3\bar\alpha}{3-\bar\alpha}}}^{\frac{4(1-\theta)}{m}} \lesssim \left\|\nabla \rho^{\frac{m}{2}}\right\|_{L^{\bar\alpha}}^{\frac{4(1-\theta)}{m}},\] where \(\displaystyle \theta=\frac{(5m+4)\bar\alpha-12}{2(\bar\alpha(3m+2)-6)}\). A direct computation gives \[\frac{4(1-\theta)}{m} = \frac{2\bar\alpha}{\bar\alpha(3m+2)-6}.\] Therefore, we have \[\label{app-3d-v2} \frac{d}{dt}\|v\|_{L^2}^2+\|\nabla v\|_{L^2}^2 \lesssim \left\|\nabla \rho^{\frac{m}{2}}\right\|_{L^{\bar\alpha}}^{\frac{2\bar\alpha}{\bar\alpha(3m+2)-6}}.\tag{51}\] Since \(\bar\alpha>\frac{8}{3m+2},\) it follows that \[\frac{2\bar\alpha}{\bar\alpha(3m+2)-6}\le \bar\alpha.\] Combining 51 with 49 , we conclude that \[\label{app-3d-v} \sup_{t\in(0,T)}\|v(\cdot,t)\|_{L^2(\Omega)}^2 + \int_{\Omega_T}|\nabla v|^2\,dxdt \le C.\tag{52}\] Together with 46 , this yields \[\sup_{t\in(0,T)}\int_\Omega \left(\rho(\cdot,t)+|v(\cdot,t)|^2\right)\,dx + \int_{\Omega_T}\left(\left|\nabla \rho^{\frac{m}{2}}\right|^\alpha+|\nabla v|^2\right)\,dxdt \le C\] for every \(0<\alpha<2\).

Finally, the above estimates are stable under the standard regularization procedure, and the compactness argument from Section 3 yields a weak solution pair \((\rho,v)\) in the sense of Definition 4. This completes the proof. ◻