After the above preparation, we are in a position to prove our main theorem.

*Proof* To prove the theorem, we split the process into several steps. The first step is to construct the approximating solution sequence; in the second step, we prove the regularity of the solution we have obtained; and in the last step, we prove the uniqueness. To get the existence of a weak solution in the space {L}_{x}^{p}, it is natural to require {f}^{n} to be a Cauchy sequence in the strong topology of {L}_{x}^{p} and E(t,x) to be bounded in {L}_{x}^{{p}^{\prime}}, here \frac{1}{p}+\frac{1}{{p}^{\prime}}=1. To estimate the {L}^{{p}^{\prime}} norm of E(t,x), we will use the Hardy inequality.

Step 1: We construct an iterative solution sequence to approximate the solution of the original equation.

\{\begin{array}{l}\frac{\partial {f}^{n+1}}{\partial t}+v\cdot {\mathrm{\nabla}}_{x}{f}^{n+1}+{E}^{n}{|}_{\mathrm{\Omega}}\cdot {\mathrm{\nabla}}_{v}{f}^{n+1}-\sigma {\mathrm{\Delta}}_{v}{f}^{n+1}=0;\\ {E}^{n}(x,t)={\int}_{{\mathbb{R}}^{2}}\frac{x-y}{{|x-y|}^{2}}{\rho}^{n}(y,t)\phantom{\rule{0.2em}{0ex}}dy;\\ {\rho}^{n}(x,t)={\int}_{{\mathbb{R}}^{2}}{f}^{n}(x,v,t)\phantom{\rule{0.2em}{0ex}}dv;\\ {f}^{n+1}(x,v,0)={f}_{0}(x,v);\\ {\parallel {f}^{n+1}\parallel}_{{L}_{v}^{1}}{|}_{\partial \mathrm{\Omega}}=0.\end{array}

(3.1)

Firstly, on the one hand, to guarantee the weak convergence, we need the weak convergence of {E}^{n} in {L}^{{p}^{\prime}}, which in turn can be deduced by the boundedness of {E}^{n} in {L}^{{p}^{\prime}}. By the Hardy inequality [10–12], we have

{\parallel {E}^{n}\parallel}_{{L}^{{p}^{\prime}}}\le {\parallel \frac{1}{|x|}\underset{x}{\ast}{\rho}^{n}\parallel}_{{L}^{{p}^{\prime}}}\le C{\parallel {f}^{n}\parallel}_{{L}_{x}^{p}{L}_{v}^{1}},

where

\frac{1}{p}+\frac{1}{2}=1+\frac{1}{{p}^{\prime}}.

(3.2)

On the other hand, by Lemma 2.1 and Lemma 2.3, let {f}^{0}=0, we obtain a unique solution {f}^{n+1} to (3.1). To pass to the limit in the system of (1.1), we need a strong convergence. In order to get this, we compute the Cauchy sequence as follows:

\begin{array}{r}{\parallel {f}^{n+1}(x,v,t)-{f}^{n}(x,v,t)\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}\\ \phantom{\rule{1em}{0ex}}\le {\int}_{0}^{t}{\parallel ({E}^{n}(x,s)-{E}^{n-1}(x,s)){\parallel {\mathrm{\nabla}}_{v}{f}^{n}\parallel}_{{L}_{v}^{1}}\parallel}_{{L}_{x}^{p}}\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}\le {\int}_{0}^{t}{\parallel \frac{1}{|x|}\underset{x}{\ast}|{\rho}^{n}-{\rho}^{n-1}|{\parallel {\mathrm{\nabla}}_{v}{f}^{n}\parallel}_{{L}_{v}^{1}}\parallel}_{{L}_{x}^{p}}\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}\le {\int}_{0}^{t}{\int}_{{\mathbb{R}}_{y}^{2}}|{\rho}^{n}(y,s)-{\rho}^{n-1}(y,s)|\phantom{\rule{0.2em}{0ex}}dy\\ \phantom{\rule{2em}{0ex}}\cdot {\left({\int}_{{\mathbb{R}}_{x}^{2}}{\left(\frac{1}{{|x-y|}^{n-1}}{\parallel {\mathrm{\nabla}}_{v}{f}^{n}(x,v,s)\parallel}_{{L}_{v}^{1}}\right)}^{p}\phantom{\rule{0.2em}{0ex}}dx\right)}^{\frac{1}{p}}\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{1em}{0ex}}\le {\int}_{0}^{t}{\parallel {\rho}^{n}(\cdot ,s)-{\rho}^{n-1}(\cdot ,s)\parallel}_{{L}_{x}^{p}}{\parallel {\left(\frac{1}{{|x|}^{p}}\underset{x}{\ast}{\parallel {\mathrm{\nabla}}_{v}{f}^{n}(x,v,s)\parallel}_{{L}_{v}^{1}}^{p}\right)}^{\frac{1}{p}}\parallel}_{{L}_{x}^{{p}^{\prime}}}\phantom{\rule{0.2em}{0ex}}ds.\end{array}

Here

\frac{1}{p}+\frac{1}{{p}^{\prime}}=1.

(3.3)

Combining (3.2) with (3.3), we solve the indices p=\frac{4}{3} and {p}^{\prime}=4. We claim that

{\parallel {\left(\frac{1}{{|x|}^{p}}\underset{x}{\ast}{\parallel {\mathrm{\nabla}}_{v}{f}^{n}(x,v,s)\parallel}_{{L}_{v}^{1}}^{p}\right)}^{\frac{1}{p}}\parallel}_{{L}_{x}^{{p}^{\prime}}}\le C

(3.4)

uniformly in *n*. Thus,

{\parallel {f}^{n+1}(x,v,t)-{f}^{n}(x,v,t)\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}\le {\int}_{0}^{t}{\parallel {f}^{n}(\cdot ,s)-{f}^{n-1}(\cdot ,s)\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}.

(3.5)

By a process of induction, we obtain

{\parallel ({f}^{n+1}-{f}^{n})(x,v,t)\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}\le \frac{{C}^{n}{t}^{n}}{n!}\underset{t\in [0,T]}{max}{\parallel ({f}^{1}-{f}^{0})(x,v,t)\parallel}_{{L}_{x}^{p}{L}_{v}^{1}},

which implies {f}^{n} converges to some *f* in {L}_{\mathrm{loc}}^{\mathrm{\infty}}([0,T),{L}_{x}^{p}{L}_{v}^{1}).

Proof of the claim of (3.4).

A direct calculation by using the Hardy inequality yields

\begin{array}{r}{\parallel {\left(\frac{1}{{|x|}^{p}}\underset{x}{\ast}{\parallel {\mathrm{\nabla}}_{v}{f}^{n}(x,v,s)\parallel}_{{L}_{v}^{1}}^{p}\right)}^{\frac{1}{p}}\parallel}_{{L}_{x}^{{p}^{\prime}}}\\ \phantom{\rule{1em}{0ex}}\le C{\parallel {\parallel {\mathrm{\nabla}}_{v}{f}^{n}(x,v,s)\parallel}_{{L}_{v}^{1}}^{p}\parallel}_{{L}_{x}^{q}}\\ \phantom{\rule{1em}{0ex}}=C{\parallel {\mathrm{\nabla}}_{v}{f}^{n}(x,v,s)\parallel}_{{L}_{x}^{pq}{L}_{v}^{1}},\end{array}

here \frac{1}{q}+\frac{p}{2}=1+\frac{p}{{p}^{\prime}}, q>1.

Differentiating equation (3.1) by {D}_{x} and {D}_{v}, respectively, we get

\{\begin{array}{l}\frac{\partial ({D}_{x}{f}^{n+1})}{\partial t}+v\cdot {\mathrm{\nabla}}_{x}({D}_{x}{f}^{n+1})+{E}^{n}\cdot {\mathrm{\nabla}}_{v}({D}_{x}{f}^{n+1})-\sigma {\mathrm{\Delta}}_{v}({D}_{x}{f}^{n+1})=-{\mathrm{\nabla}}_{x}{E}^{n}\cdot {\mathrm{\nabla}}_{v}{f}^{n+1};\\ \frac{\partial ({D}_{v}{f}^{n+1})}{\partial t}+v\cdot {\mathrm{\nabla}}_{x}({D}_{v}{f}^{n+1})+{E}^{n}\cdot {\mathrm{\nabla}}_{v}({D}_{v}{f}^{n+1})-\sigma {\mathrm{\Delta}}_{v}({D}_{v}{f}^{n+1})=-{\mathrm{\nabla}}_{x}{f}^{n+1}.\end{array}

(3.6)

Applying Lemma 2.3 to (3.6) yields

\{\begin{array}{l}{\parallel {\mathrm{\nabla}}_{x}{f}^{n+1}\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}\le {\parallel {\mathrm{\nabla}}_{x}{f}_{0}\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}+{\int}_{0}^{t}{\parallel {\mathrm{\nabla}}_{x}{E}^{n}\cdot {\mathrm{\nabla}}_{v}{f}^{n+1}\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}\phantom{\rule{0.2em}{0ex}}ds;\\ {\parallel {\mathrm{\nabla}}_{v}{f}^{n+1}\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}\le {\parallel {\mathrm{\nabla}}_{v}{f}_{0}\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}+{\int}_{0}^{t}{\parallel {\mathrm{\nabla}}_{x}{f}^{n+1}\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}\phantom{\rule{0.2em}{0ex}}ds,\end{array}

(3.7)

where p>1. Since

{\parallel {\mathrm{\nabla}}_{x}{E}^{n}\cdot {\mathrm{\nabla}}_{v}{f}^{n+1}\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}\le {\parallel {\mathrm{\nabla}}_{x}{E}^{n}\parallel}_{{L}_{x}^{{p}_{1}}}{\parallel {\mathrm{\nabla}}_{v}{f}^{n+1}\parallel}_{{L}_{x}^{{p}_{2}}{L}_{v}^{1}},\phantom{\rule{2em}{0ex}}\frac{1}{{p}_{1}}+\frac{1}{{p}_{2}}=\frac{1}{p},

(3.8)

{\parallel {\mathrm{\nabla}}_{x}{E}^{n}\parallel}_{{L}_{x}^{{p}_{1}}}\le {\parallel \frac{1}{{|x|}^{N-1}}\underset{x}{\ast}{\mathrm{\nabla}}_{x}{\rho}^{n}\parallel}_{{L}_{x}^{{p}_{1}}}\le C{\parallel {\mathrm{\nabla}}_{x}{f}^{n}\parallel}_{{L}_{x}^{pq}{L}_{v}^{1}},\phantom{\rule{2em}{0ex}}\frac{1}{pq}+\frac{1}{2}=1+\frac{1}{{p}_{1}}.

(3.9)

We require {p}_{1}\le {p}^{\prime}, p<{p}^{\prime}, this is an easy thing. Note that {L}^{pq}(\mathrm{\Omega})\subset {L}^{p}(\mathrm{\Omega}) and {L}^{{p}_{2}}(\mathrm{\Omega})\subset {L}^{p}(\mathrm{\Omega}). Plugging (3.8) and (3.9) into (3.7) yields

\{\begin{array}{l}{\parallel {\mathrm{\nabla}}_{x}{f}^{n+1}\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}\le {\parallel {\mathrm{\nabla}}_{x}{f}_{0}\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}+{\int}_{0}^{t}{\parallel {\mathrm{\nabla}}_{x}{f}^{n}\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}{\parallel {\mathrm{\nabla}}_{v}{f}^{n+1}\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}\phantom{\rule{0.2em}{0ex}}ds;\\ {\parallel {\mathrm{\nabla}}_{v}{f}^{n+1}\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}\le {\parallel {\mathrm{\nabla}}_{v}{f}_{0}\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}+{\int}_{0}^{t}{\parallel {\mathrm{\nabla}}_{x}{f}^{n+1}\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}\phantom{\rule{0.2em}{0ex}}ds.\end{array}

Next, we only have to solve the Gronwall inequalities in the form

\{\begin{array}{l}f(t)\le {C}_{0}+{\int}_{0}^{t}f(s)g(s)\phantom{\rule{0.2em}{0ex}}ds,\\ g(t)\le {C}_{1}+{\int}_{0}^{t}f(s)\phantom{\rule{0.2em}{0ex}}ds.\end{array}

These inequalities only hold in finite time, we denote the maximal existence time by *T*, *i.e.*, for t\in [0,T), there exists \alpha (t)\in {L}^{\mathrm{\infty}}([0,T)) such that

{\parallel {\mathrm{\nabla}}_{v}{f}^{n}(t)\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}\le \alpha (t),\phantom{\rule{2em}{0ex}}{\parallel {\mathrm{\nabla}}_{x}{f}^{n}(t)\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}\le \alpha (t).

Therefore, the claim of (3.4) holds.

According to the standard weak convergence process, we conclude that *f* is a solution of the Cauchy problem of equations of (1.1).

Step 2: Regularity of the solution.

Denote {D}_{x} or {D}_{v} by *D*. Since {f}^{n}\to f in {L}_{x}^{p}{L}_{v}^{1}, which deduces D{f}^{n}\to Df in {\mathcal{D}}^{\prime}({\mathrm{\Omega}}_{x}\times {\mathbb{R}}_{v}^{2}), note that {\parallel D{f}^{n}\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}\le \text{const}, we have {\parallel Df\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}\le {\parallel D{f}^{n}\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}\le \text{const}.

By property (i) of Proposition A.3 in [5], we conclude that {f}^{n} is nonnegative. Moreover, {f}^{n}\to f, a.e. (x,v)\in {\mathrm{\Omega}}_{x}\times {\mathbb{R}}_{v}^{2}, since {f}^{n}\to f in {L}_{x}^{p}{L}_{v}^{1}, which implies that *f* is nonnegative.

Step 3: Uniqueness of the solution.

The uniqueness is a direct consequence of

{\parallel (f-\tilde{f})(x,v,t)\parallel}_{{L}_{x}^{p}{L}_{v}^{1}}\le C{\int}_{0}^{t}{\parallel (f-\tilde{f})(x,v,t)\parallel}_{{L}_{x}^{p}{L}_{v}^{1}},

(3.10)

which in turn is a result of a very similar process to (3.5).

Next we are going to deal with the boundary, *i.e.*, to show that the solution satisfied the boundary condition.

On the one hand, by Lemma 2.3, we have {\parallel {f}^{n}\parallel}_{{L}_{v}^{1}}\le {\parallel {f}_{0}\parallel}_{{L}_{v}^{1}} for *x* a.e.; on the other hand, {\parallel {f}^{n}(t,x,v)\parallel}_{{L}_{v}^{1}}\to {\parallel f(t,x,v)\parallel}_{{L}_{v}^{1}} for *x* a.e. Thus {\parallel f(t,x,v)\parallel}_{{L}_{v}^{1}}\le {\parallel {f}_{0}\parallel}_{{L}_{v}^{1}} for *x* a.e., note the assumption of {\parallel {f}_{0}\parallel}_{{L}_{v}^{1}}{|}_{\partial \mathrm{\Omega}}=0 in the sense of trace of {L}_{x}^{p}, we conclude in the sense of trace of {L}_{x}^{p}, since the boundary is {C}^{1} and {\parallel {f}^{n}\parallel}_{{L}_{v}^{1}}\in {W}_{x}^{1,p}:

{\parallel f(t,x,v)\parallel}_{{L}_{v}^{1}}{|}_{\partial \mathrm{\Omega}}=0.

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