# Controlling a nonlinear Fokker-Planck equation via inputs with nonlocal
action

Ştefana-Lucia Aniţa
“Octav Mayer” Institute of Mathematics of the Romanian Academy Iaşi 700506
E-mail<EMAIL_ADDRESS>
##  1 Formulation of the problem

Consider the following optimal control problem:

$\underset{\zeta\in{\cal M}}{\text{Minimize}}\
\int_{0}^{T}\int_{\mathbb{R}^{d}}G(t,x)\rho^{\zeta}(t,x)dx\
dt+\int_{\mathbb{R}^{d}}G_{T}(x)\rho^{\zeta}(T,x)dx+\int_{\mathbb{R}^{d}}Q(x,\zeta(x))dx,$
$None$

where $T\in(0,+\infty)$, $d\in\mathbb{N},\ d\geq 3$, and $\rho^{\zeta}$ is the
weak solution to the following nonlinear Fokker-Planck equation:

$\left\\{\begin{array}[]{ll}\displaystyle{{\partial\rho}\over{\partial
t}}(t,x)=-\nabla\cdot(u(x)b(\rho(t,x))\rho(t,x))+\Delta\beta(\rho(t,x)),&t\in[0,T],\
x\in\mathbb{R}^{d}\\\ {}&{}\\\
\rho(0,x)=\rho_{0}(x),&x\in\mathbb{R}^{d}.\end{array}\right.$ (1)

Here $u(x)=-\nabla\left(\int_{\mathbb{R}^{d}}{\cal
G}_{R}(x-x^{\prime})\zeta(x^{\prime})dx^{\prime}\right)={\cal K}(\zeta)(x)$,

$\zeta\in{\cal M}=\\{\theta\in L^{\infty}(\mathbb{R}^{d});\
0\leq\theta(x)\leq\tilde{M}_{0}\ \mbox{\rm a.e. }x\in\mathbb{R}^{d},\
\theta(x)=0\ \mbox{\rm a.e. }|x|_{d}>R_{0}\\};$

$\tilde{M}_{0}$, $R$, $R_{0}$ are positive constants, and we assume that
${\cal G}_{R}\in C_{0}^{2}(\mathbb{R}^{d})$, ${\cal G}_{R}(x)>0$ if
$|x|_{d}<R$ and ${\cal G}_{R}(x)=0$ if $|x|_{d}\geq R$. We have that ${\cal
K}:L^{\infty}(\mathbb{R}^{d})\longrightarrow L^{\infty}(\mathbb{R}^{d})^{d}$
is defined by

${\cal K}(\zeta)(x)=-\nabla({\cal G}_{R}*\zeta)(x)$

(where ${\cal G}_{R}*\zeta$ is the convolution product of ${\cal G}_{R}$ and
$\zeta$).

The weak solution to (1) may be viewed as the probabilistic density of a
population while $\zeta(x)$ is the density at $x\in\mathbb{R}^{d}$ of a second
population (or of another entity, which may be a substance or a signal) which
produces a stimulus to the first population. This second population is time-
independent, imobile, located in $\overline{B(0_{d};R_{0})}$ $(R_{0}>0$), and
it repels the individuals of the first population which are at a distance less
than $R$. It means that $\zeta$ is an input (control) with nonlocal action.
This action, denoted by $u$, is expressed mathematically in terms of the so-
caled “generalized gradient” (nonlocal gradient) with kernel ${\cal G}_{R}$.

Actually, $u(x)=-\nabla({\cal G}_{R}*\zeta)(x)={\cal K}(\zeta)(x)$ describes
the nonlocal action (effect) of $\zeta$ (the second population) towards the
individuals of the first population located at $x\in\mathbb{R}^{d}$. The term
$-\nabla\cdot(u(x)b(\rho(t,x))\rho(t,x))=\nabla\cdot(\nabla({\cal
G}_{R}*\zeta)(x)b(\rho(t,x))\rho(t,x))$ describes a cross-diffusion (see e.g.
[19, 20]), and leads to a particular form of Keller-Segel’s models (see e.g.
[17, 18]).

Problem $(P)$ asks to optimally displace a population via the repeller action
produced by a second population of density $\zeta$. It is a natural
requirement (due to logistic constraints) to place this second population in a
bounded domain (here we have considered it to be a ball) and to have bounded
density.

Notice that the case of a local repeller action at $x\in\mathbb{R}^{d}$ is a
limit case of the nonlocal one, where

${\cal G}_{R}=\delta$

($\delta$ is the Dirac distribution) and

$u(x)=-\nabla\zeta(x).$

If the second population attracts the first one, then we have to take
$u(x)=\nabla({\cal G}_{R}*\zeta)(x)$ instead (and $u(x)=\nabla\zeta(x)$ in the
case of local action).

Assume that the following hypotheses hold:

  * (H1)

$\beta\in C^{2}(\mathbb{R})$, $\beta(0)=0$ and $\exists
0<\gamma_{0}<\gamma_{1}$ such that

$\gamma_{0}\leq\beta^{\prime}(r)\leq\gamma_{1},\ \forall
r\in\mathbb{R},\quad\mbox{\rm and }\Psi\in C^{2}(\mathbb{R}),\ \mbox{\rm where
}\Psi(r)={{\beta(r)}\over r};$

  * (H2)

$b\in C^{1}(\mathbb{R})$, $b$ is bounded, $b(r)\geq 0$, $\forall r\geq 0$ and

$|b^{*}(r)-b^{*}(\tilde{r})|\leq\alpha|\beta(r)-\beta(\tilde{r})|,\quad\forall
r,\tilde{r}\in\mathbb{R},$

where $\alpha>0$ and $b^{*}(r)=b(r)r$;

  * (H3)

$\rho_{0}\in L^{1}(\mathbb{R}^{d})\cap L^{\infty}(\mathbb{R}^{d})$,
$\rho_{0}(x)\geq 0$ a.e. $x\in\mathbb{R}^{d}$,
$\int_{\mathbb{R}^{d}}\rho_{0}(x)dx=1$;

  * (H4)

$G\in C_{b}([0,T]\times\mathbb{R}^{d})\cap L^{2}((0,T)\times\mathbb{R}^{d})$,
$G_{T}\in C_{b}(\mathbb{R}^{d})\cap H^{1}(\mathbb{R}^{d})$,
$Q:\mathbb{R}^{d}\times\mathbb{R}\longrightarrow\mathbb{R}$,
$Q|_{B(0_{d};R_{0})\times[0,\tilde{M}_{0}]}\in
C_{b}^{0,1}(B(0_{d};R_{0})\times[0,\tilde{M}_{0}])$,

$0\leq G(t,x),\ \forall(t,x)\in[0,T]\times\mathbb{R}^{d},\quad 0\leq
G_{T}(x),\ \forall x\in\mathbb{R}^{d},$ $0\leq Q(x,z),\
\forall(x,z)\in\mathbb{R}^{d}\times[0,\tilde{M}_{0}],\quad Q(x,z)=0,\ \forall
x\in\mathbb{R}^{d},|x|_{d}>R_{0},\forall z\in[0,\tilde{M}_{0}],$

and the mapings $z\mapsto Q(x,z)$ are convex with respect to
$z\in[0,\tilde{M}_{0}]$ for any $x\in\mathbb{R}^{d}$.

The following notations $\nabla=\left({{\partial}\over{\partial
x_{1}}},{{\partial}\over{\partial x_{2}}},\cdots,{{\partial}\over{\partial
x_{d}}}\right)$, $\Delta=\sum_{i=1}^{d}{{\partial^{2}}\over{\partial
x_{i}^{2}}}$ will be used. Sometimes we will use the simplified notation
$\|\cdot\|_{L^{k}}$ (instead of $\|\cdot\|_{L^{k}(\mathbb{R}^{d})}$ or
$\|\cdot\|_{L^{k}(\mathbb{R}^{d};\mathbb{R}^{d})}$) for the usual norm of
$L^{k}(\mathbb{R}^{d})$ or of $L^{k}(\mathbb{R}^{d};\mathbb{R}^{d})$.

For some recent results concerning the optimal control of Fokker-Planck
equations we refer to [2, 4, 5, 9, 22]. The difficulty of our present study is
mainly due to the nonlinearities in the Fokker-Planck equation (1).

In section 2 we will recall that there exists a unique weak solution to the
nonlinear Fokker-Planck equation (1) (see [9]) and that our optimal control
problem is deeply related to a stochastic optimal control problem $(P_{S})$
related to a McKean-Vlasov equation. In section 3 one proves the existence of
an optimal control for problem $(P)$. The existence of an optimal control is
established and necessary optimality conditions are derived for a penalized
optimal control problem $(P_{h})$ related to a backward Euler scheme for the
nonlinear Fokker-Planck equation (with a constant discretization step $h$) in
section 4. Using the optimality conditions for $(P_{h})$ one derives the
necessary optimality conditions for problem $(P)$ in section 5. Section 6
contains final remarks and comments.

##  2 The nonlinear Fokker-Planck equation and its relationship to a McKean-
Vlasov equation

For any $\zeta\in{\cal M}$ we consider the equation (1) as the following
Cauchy problem in $L^{1}(\mathbb{R}^{d})$:

$\left\\{\begin{array}[]{ll}\displaystyle{{d\rho}\over{dt}}(t)+A^{\zeta}\rho(t)=0,\quad
t\in[0,T]\\\ {}\\\ \rho(0)=\rho_{0},\end{array}\right.$ (2)

where $A^{\zeta}:D(A^{\zeta})\subset L^{1}(\mathbb{R}^{d})\longrightarrow
L^{1}(\mathbb{R}^{d})$ is an m-accretive operator in $L^{1}(\mathbb{R}^{d})$
given by

$D(A^{\zeta})=\\{\theta\in L^{1}(\mathbb{R}^{d});\
-\Delta\beta(\theta)+\nabla\cdot({\cal K}(\zeta)b(\theta)\theta)\in
L^{1}(\mathbb{R}^{d})\\},$
$A^{\zeta}(\theta)=-\Delta\beta(\theta)+\nabla\cdot({\cal
K}(\zeta)b(\theta)\theta)\quad\mbox{\rm in }{\cal
D}^{\prime}(\mathbb{R}^{d}),\quad\theta\in D(A^{\zeta})$

(see Proposition 2.3 in [14]). Moreover,
$\overline{D(A^{\zeta})}=L^{1}(\mathbb{R}^{d})$ and $(I+\lambda
A^{\zeta})^{-1}({\cal P})\subset{\cal P},\ \forall\lambda>0$, where ${\cal P}$
is the set of all probability densities on $\mathbb{R}^{d}$.

Definition 2.1. Let $\zeta\in{\cal M}$. A function $\rho^{\zeta}\in
C([0,T];L^{1}(\mathbb{R}^{d}))$ is called a weak/mild solution to (1)/(2) if

$\rho^{\zeta}(t)=\underset{h\rightarrow
0+}{\text{lim}}\tilde{\rho}^{\zeta}_{h}(t)\ \mbox{\rm in
}L^{1}(\mathbb{R}^{d}),\ \mbox{\rm uniformly for }t\in[0,T],$

where for any $h>0$, $\tilde{\rho}^{\zeta}_{h}:[0,T]\rightarrow
L^{1}(\mathbb{R}^{d})$ is given by

$\tilde{\rho}^{\zeta}_{h}(t)=\rho_{h}^{\zeta,i}\quad\mbox{\rm if
}t\in[ih,(i+1)h),t\leq T,\ i=0,1,...,N=\left[{T\over h}\right]$

and $\rho^{\zeta,i+1}_{h}$ is the solution to

$\left\\{\begin{array}[]{ll}\rho^{\zeta,0}_{h}=\rho_{0},\quad\rho^{\zeta,i}_{h}\in
L^{1}(\mathbb{R}^{d}),&i=1,...,N,\\\ {}&{}\\\
\rho^{\zeta,i+1}_{h}-h\Delta\beta(\rho^{\zeta,i+1}_{h})+h\nabla\cdot({\cal
K}(\zeta)b(\rho^{\zeta,i+1}_{h})\rho^{\zeta,i+1}_{h})=\rho^{\zeta,i}_{h}\
\mbox{\rm in }{\cal
D}^{\prime}(\mathbb{R}^{d}),&i=0,1,...,N-1.\end{array}\right.$ (3)

The existence and uniqueness of such a weak/mild solution $\rho^{\zeta}$ to
(1)/(2) follow via Theorem 2.1 in [14] (which is based on Crandall-Liggett’s
existence theorem (see [7])). Moreover, by Theorem 2.1 in [14] we get that
$\rho^{\zeta}$ is also a distributional solution to (1). Furthermore, we have
that $\rho^{\zeta}(t)\in{\cal P},\ \forall t\in[0,T]$ and $\rho^{\zeta}\in
L^{\infty}((0,T)\times\mathbb{R}^{d})$.

Remark 2.1. If we consider $N\in\mathbb{N}^{*}$ and $h={T\over N}$, then we
get that

$\rho^{\zeta}(t)=\underset{h\rightarrow
0+}{\text{lim}}\rho^{\zeta}_{h}(t)=\underset{N\rightarrow+\infty}{\text{lim}}\rho^{\zeta}_{h}(t)\
\mbox{\rm in }L^{1}(\mathbb{R}^{d}),\ \mbox{\rm uniformly for }t\in[0,T],$

where $\rho^{\zeta}_{h}:[0,T]\rightarrow L^{1}(\mathbb{R}^{d})$ is given by

$\rho^{\zeta}_{h}(t)=\left\\{\begin{array}[]{ll}\rho_{0},&\mbox{\rm if
}t=0,\\\ {}&{}\\\ \rho^{\zeta,i+1}_{h},&\mbox{\rm if }t\in(ih,(i+1)h],\quad
i=0,1,...,N-1.\end{array}\right.$

Indeed, if $t=ih$ ($i\in\\{0,1,...,N\\}$), then
$\rho_{h}^{\zeta}(t)=\rho_{h}^{\zeta,i}=\tilde{\rho}_{h}^{\zeta}(t)$ and so

$\|\rho_{h}^{\zeta}(t)-\rho^{\zeta}(t)\|_{L^{1}(\mathbb{R}^{d})}=\|\tilde{\rho}_{h}^{\zeta}(t)-\rho^{\zeta}(t)\|_{L^{1}(\mathbb{R}^{d})}.$

On the other hand, if $t\in(ih,(i+1)h),\ i\in\\{0,1,...,N-1\\},$ then
$\rho_{h}^{\zeta}(t)=\rho_{h}^{\zeta}((i+1)h)$ and so

$\|\rho_{h}^{\zeta}(t)-\rho^{\zeta}(t)\|_{L^{1}(\mathbb{R}^{d})}=\|\rho_{h}^{\zeta}((i+1)h)-\rho^{\zeta}(t)\|_{L^{1}(\mathbb{R}^{d})}=\|\tilde{\rho}_{h}^{\zeta}((i+1)h)-\rho^{\zeta}(t)\|_{L^{1}(\mathbb{R}^{d})}$
$\leq\|\tilde{\rho}_{h}^{\zeta}((i+1)h)-\rho^{\zeta}((i+1)h)\|_{L^{1}(\mathbb{R}^{d})}+\|\rho^{\zeta}((i+1)h)-\rho^{\zeta}(t)\|_{L^{1}(\mathbb{R}^{d})}.$

Taking into account the convergence of $\tilde{\rho}^{\zeta}_{h}(t)$ to
$\rho^{\zeta}(t)$ in $L^{1}(\mathbb{R}^{d})$, uniformly for $t\in[0,T]$ and
the fact that $\rho^{\zeta}\in C([0,T];L^{1}(\mathbb{R}^{d}))$, we get the
conclusion.

Consider now the following controlled McKean-Vlasov SDE on $\mathbb{R}^{d}$:

$\left\\{\begin{array}[]{ll}dX(t)=f\left(X(t),{\cal K}(\zeta)(X(t)),{{d{\cal
L}_{X(t)}}\over{dx}}(X(t))\right)dt+\sigma\left(X(t),{{d{\cal
L}_{X(t)}}\over{dx}}(X(t))\right)dW(t),\quad t\in[0,T]\\\ {}\\\
X(0)=X_{0},\end{array}\right.$ (4)

where ${\cal L}_{Y}$ is the law/distribution of the random variable $Y$ and
${{d{\cal L}_{Y}}\over{dx}}$ is its density (when it exists) with respect to
the Lebesgue measure $dx$, and

$f(x,u,r)=ub(r),\quad\sigma(x,r)=\left({{2\beta(r)}\over r}\right)^{1\over
2}.$

Since for any $\zeta\in{\cal M}$, $\rho^{\zeta}$ is a distributional solution
to the nonlinear Fokker-Planck solution, which is also $t$-narrowly
continuous, then by superposition’s principle (see [21, 26, 24]) we conclude
that there exists a unique (in law) probabilistically weak solution
$(X^{\zeta}(t))_{t\in[0,T]}$ to (4) (which has $\rho^{\zeta}(t)$ as a
probability density). Actually, the weak solution $X^{\zeta}$ corresponds to a
probability space $(\Omega,{\cal F},\mathbb{P})$ with normal filtration
$({\cal F}_{t})_{t\in[0,T]}$ and $(W(t))_{t\in[0,T]}$ is an
$\mathbb{R}^{d}-({\cal F}_{t})_{t\in[0,T]}$ Brownian motion.

Hence,

$\mathbb{E}\left[\int_{0}^{T}G(t,X^{\zeta}(t))dt\right]+\mathbb{E}[G_{T}(X^{\zeta}(T))]=\int_{0}^{T}\int_{\mathbb{R}^{d}}G(t,x)\rho^{\zeta}(t,x)dx\
dt+\int_{\mathbb{R}^{d}}G_{T}(x)\rho^{\zeta}(T,x)dx.$

It follows that the next stochastic optimal control problem

$\underset{\zeta\in{\cal M}}{\text{Minimize}}\
\mathbb{E}\left[\int_{0}^{T}G(t,X^{\zeta}(t))dt\right]+\mathbb{E}[G_{T}(X^{\zeta}(T))]+\int_{\mathbb{R}^{d}}Q(x,\zeta(x))dx,$
$None$

is equivalent to the deterministic optimal control problem $(P)$.

We mention that lately there is a high interest in McKean-Vlasov SDEs and in
stochastic optimal control problems. Some recent and important results
concerning the McKean-Vlasov can be found in [9, 12, 13, 16]. As concerns
problem $(P_{S})$ (which is a stochastic optimal control problem related to
McKean-Vlasov SDE) we see that it is deeply related to an optimal control
problem $(P)$ for a nonlinear Fokker-Planck equation. A similar approach,
reducing the investigation of a stochastic optimal control problem to a
deterministic optimal control problem, has been recently used for the case of
feedback controllers independent of time in [1, 3, 8] and in [2, 22] for
controllers dependent of time as well, for optimal control problems with other
cost functionals and related to stochastic differential equations with drift
and diffusion independent of the probabilistic density. For other approaches
for stochastic optimal control problems see [10, 11, 15] while for standard
results concerning the control of stochastic differential equations we refer
to the monographs [23, 25].

##  3 Existence of an optimal control for problem (P)

Let $m^{*}=\inf_{\zeta\in{\cal M}}I(\zeta)\in[0,+\infty)$, where

$I(\zeta)=\int_{0}^{T}\int_{\mathbb{R}^{d}}G(t,x)\rho^{\zeta}(t,x)dx\
dt+\int_{\mathbb{R}^{d}}G_{T}(x)\rho^{\zeta}(T,x)dx+\int_{\mathbb{R}^{d}}Q(x,\zeta(x))dx.$

For any $\zeta\in{\cal M}$ we have that ${\cal K}(\zeta)\in
C_{0}^{1}(\mathbb{R}^{d};\mathbb{R}^{d})$,

$\begin{array}[]{ll}supp\ {\cal
K}(\zeta)\subset\overline{B(0_{d};R+R_{0})},\\\ \displaystyle|{\cal
K}(\zeta)(x)|_{d}\leq\tilde{M}_{0}\int_{\mathbb{R}^{d}}|\nabla{\cal
G}_{R}(-x^{\prime})|_{d}dx^{\prime}=M,\quad\forall x\in\mathbb{R}^{d},\\\
|\nabla{\cal K}(\zeta)(x)|_{d\times d}\leq M_{1},\quad\forall
x\in\mathbb{R}^{d},\end{array}$

where $M_{1}$ is a constant independent of $\zeta\in{\cal M}$.

For any sequence $\\{\zeta_{k}\\}_{k\in\mathbb{N}^{*}}\subset{\cal M}$, there
exists a subsequence $\\{\zeta_{k_{l}}\\}_{l\in\mathbb{N}^{*}}$ and
$\zeta^{*}\in{\cal M}$ such that

$\zeta_{k_{l}}\longrightarrow\zeta^{*}\quad\mbox{\rm weakly * in
}L^{\infty}(\mathbb{R}^{d}).$

This implies that

${\cal K}(\zeta_{k_{l}})(x)=-\int_{\mathbb{R}^{d}}\nabla{\cal
G}_{R}(x-x^{\prime})\zeta_{k_{l}}(x^{\prime})dx^{\prime}\longrightarrow-\int_{\mathbb{R}^{d}}\nabla{\cal
G}_{R}(x-x^{\prime})\zeta^{*}(x^{\prime})dx^{\prime}={\cal
K}(\zeta^{*})(x),\quad\forall x\in\mathbb{R}^{d},$

and that

$\nabla{\cal K}(\zeta_{k_{l}})(x)\longrightarrow\nabla{\cal
K}(\zeta^{*})(x),\quad\forall x\in\mathbb{R}^{d}.$

Let us notice that by Arzelà’s compactness theorem we may infer that there
exists a subsequence $\\{{\cal K}(\zeta_{k_{l_{r}}})\\}_{r\in\mathbb{N}^{*}}$
such that

${\cal K}({\zeta_{k_{l_{r}}}})\longrightarrow\tilde{u},\quad\mbox{\rm
uniformly for }x\in\overline{B(0_{d};R+R_{0})}.$

Since ${\cal K}(\zeta_{k_{l_{r}}})(x)=0,\ \forall|x|_{d}>R+R_{0}$, we get that

${\cal K}({\zeta_{k_{l_{r}}}})\longrightarrow\tilde{u},\quad\mbox{\rm
uniformly for }x\in\mathbb{R}^{d}$

(where $\tilde{u}$ has been extended by the value $0_{d}$ outside
$\overline{B(0_{d};R+R_{0})}$). Since ${\cal
K}({\zeta_{k_{l_{r}}}})\longrightarrow{\cal K}({\zeta^{*}})$, $\forall
x\in\mathbb{R}^{d}$, we conclude that

${\cal K}({\zeta_{k_{l_{r}}}})\longrightarrow{\cal
K}({\zeta^{*}}),\quad\mbox{\rm uniformly for }x\in\mathbb{R}^{d}$

We postpone for the time being proof of the next auxiliary result:

Lemma 3.1. If $\\{\zeta_{n}\\}_{n\in\mathbb{N}^{*}}\subset{\cal M}$,
$\zeta_{n}\longrightarrow\zeta\in{\cal M}$ weakly * in
$L^{\infty}(\mathbb{R}^{d})$,

${\cal K}(\zeta_{n})=-\nabla({\cal
G}_{R}*\zeta_{n})\longrightarrow-\nabla({\cal G}_{R}*\zeta)={\cal
K}(\zeta),\quad\mbox{\it uniformly for }x\in\mathbb{R}^{d},$

then for any $h>0$ sufficiently small and for any $f\in L^{1}(\mathbb{R}^{d})$
we have that

$y_{n}\longrightarrow y\quad\mbox{\it in }L^{1}(\mathbb{R}^{d}),$

where $y_{n},\ y\in L^{1}(\mathbb{R}^{d})$,

$y_{n}-h\Delta\beta(y_{n})+h\nabla\cdot({\cal
K}(\zeta_{n})b^{*}(y_{n}))=f\quad\mbox{\it in }{\cal
D}^{\prime}(\mathbb{R}^{d})$ (5)

and

$y-h\Delta\beta(y)+h\nabla\cdot({\cal K}(\zeta)b^{*}(y))=f\quad\mbox{\it in
}{\cal D}^{\prime}(\mathbb{R}^{d}).$ (6)

Remark 3.1. By Lemma 2.5 in [14] we have that equations (5) and (6) have
unique solutions, respectively. Moreover, if one uses more accurate notations:
$y_{n}(f)$ for the solution to (5), and $y(f)$ for the solution to (6), then

$\|y_{n}(f)-y_{n}(g)\|_{L^{1}(\mathbb{R}^{d})}\leq\|f-g\|_{L^{1}(\mathbb{R}^{d})},$

for any $f,\ g\in L^{1}(\mathbb{R}^{d})$. By Lemma 3.1 one obtains that

$\|y(f)-y(g)\|_{L^{1}(\mathbb{R}^{d})}\leq\|f-g\|_{L^{1}(\mathbb{R}^{d})},\quad
f,\ g\in L^{1}(\mathbb{R}^{d}).$

Using now Trotter-Kato’s theorem (Theorem 2.1-p.241, [6]) we may conclude that

Lemma 3.2. If $\\{\zeta_{n}\\}_{n\in\mathbb{N}^{*}}\subset{\cal M}$,
$\zeta_{n}\longrightarrow\zeta\in{\cal M}$ weakly * in
$L^{\infty}(\mathbb{R}^{d})$, ${\cal K}(\zeta_{n})\longrightarrow{\cal
K}(\zeta),$ uniformly for $x\in\mathbb{R}^{d},$ then

$\rho^{\zeta_{n}}\longrightarrow\rho^{\zeta}\quad\mbox{\it in
}C([0,T];L^{1}(\mathbb{R}^{d})).$

Theorem 3.3. Problem $(P)$ has at least one optimal control.

Proof. Let $\\{\zeta_{k}\\}_{k\in\mathbb{N}^{*}}\subset{\cal M}$ such that

$m^{*}\leq I(\zeta_{k})<m^{*}+{1\over k},\quad k\in\mathbb{N}^{*}.$

Since $\\{\zeta_{k}\\}_{k\in\mathbb{N}^{*}}\subset{\cal M}$ it follows that
there exists a subsequence (also denoted by $\\{\zeta_{k}\\}$) such that

$\zeta_{k}\longrightarrow\zeta^{*}\in{\cal M}\ \mbox{\rm weakly * in
}L^{\infty}(\mathbb{R}^{d}),\quad{\cal K}(\zeta_{k})\longrightarrow{\cal
K}(\zeta^{*}),\ \mbox{\rm uniformly for }x\in\mathbb{R}^{d}.$

Using the convexity of $Q$ with respect to $\zeta$, we obtain that

$I(\zeta_{k})\geq\int_{0}^{T}\int_{\mathbb{R}^{d}}G(t,x)\rho^{\zeta_{k}}(t,x)dx\
dt+\int_{\mathbb{R}^{d}}G_{T}(x)\rho^{\zeta_{k}}(T,x)dx$
$+\int_{\mathbb{R}^{d}}Q(x,\zeta^{*}(x))dx+\int_{\mathbb{R}^{d}}{{\partial
Q}\over{\partial z}}(x,\zeta^{*}(x))(\zeta_{k}(x)-\zeta^{*}(x))dx.$

The last inequality implies via Lemma 3.2 that

$m^{*}\geq\liminf_{k\rightarrow+\infty}I(\zeta_{k})\geq I(\zeta^{*}),$

and so $\zeta^{*}$ is an optimal control for problem $(P)$.

Proof of Lemma 3.1. Consider for the beginning the function $\Phi$ defined in
the proof of Lemma 3.3 in [16]. Actually, we recall that this particular
function satisfies $\Phi\in C^{2}(\mathbb{R}^{d})$, $1\leq\Phi(x)$, $\forall
x\in\mathbb{R}^{d}$ and

$\lim_{|x|_{d}\rightarrow+\infty}\Phi(x)=+\infty,\quad\nabla\Phi\in
L^{\infty}(\mathbb{R}^{d};\mathbb{R}^{d}),\quad\Delta\Phi\in
L^{\infty}(\mathbb{R}^{d}).$

  * Case 1.

Let $f\in L^{1}(\mathbb{R}^{d})\cap L^{\infty}(\mathbb{R}^{d})$ such that
$\int_{\mathbb{R}^{d}}|f(x)|\Phi(x)dx<+\infty$.

From the proof of Lemma 2.5 in [14] we have that $y_{n}$, the unique solution
to (5), satisfies $\beta(y_{n})\in H^{1}(\mathbb{R}^{d})$.

By equation (5) we get that

$\int_{\mathbb{R}^{d}}y_{n}\beta(y_{n})dx+h\int_{\mathbb{R}^{d}}|\nabla\beta(y_{n})|_{d}^{2}dx-h\int_{\mathbb{R}^{d}}{\cal
K}(\zeta_{n})b(y_{n})y_{n}\cdot\nabla\beta(y_{n})dx=\int_{\mathbb{R}^{d}}f\beta(y_{n})dx.$

Notice that $\beta(y_{n})=\beta(y_{n})-\beta(0)=\beta^{\prime}(\xi_{n})y_{n}$
and $\nabla\beta(y_{n})=\beta^{\prime}(y_{n})\nabla y_{n}$ (where $\xi_{n}(x)$
is an intermediate point). It follows that for $h>0$ sufficiently small,
$\\{\beta(y_{n})\\}_{n\in\mathbb{N}^{*}}$ is bounded in
$H^{1}(\mathbb{R}^{d})$ and $\\{y_{n}\\}_{n\in\mathbb{N}^{*}}$ is bounded in
$H^{1}(\mathbb{R}^{d})$ as well.

It follows that there exists a subsequence $y_{n_{k}}\longrightarrow\tilde{y}$
in $L^{2}(B(0_{d};1))$, there exists a subsequence
$y_{n_{k_{l}}}\longrightarrow\tilde{y}$ in $L^{2}(B(0_{d};2))$, …

Taking the diagonal subsequence (also denoted by
$\\{y_{n}\\}_{n\in\mathbb{N}^{*}}$) we get that
$y_{n}\longrightarrow\tilde{y}$ in $L^{2}_{loc}(\mathbb{R}^{d})$. Repeating
the argument in Lemma 3.3 in [16] we obtain that there exists a nonnegative
constant $C$ such that

$\int_{\mathbb{R}^{d}}|y_{n}(x)|\Phi(x)dx\leq C,\quad\forall
n\in\mathbb{N}^{*}.$

Using Fatou’s lemma we obtain
$\displaystyle\int_{\mathbb{R}^{d}}|\tilde{y}(x)|\Phi(x)dx\leq C$.

Since
$\displaystyle\int_{\mathbb{R}^{d}}[y_{n}\varphi-h\beta(y_{n})\Delta\varphi-h{\cal
K}(\zeta_{n})b^{*}(y_{n})\cdot\nabla\varphi]dx=\int_{\mathbb{R}^{d}}f\varphi
dx$, $\forall\varphi\in C^{\infty}_{0}(\mathbb{R}^{d})$, we get that

$\int_{\mathbb{R}^{d}}[\tilde{y}\varphi-h\beta(\tilde{y})\Delta\varphi-h{\cal
K}(\zeta)b^{*}(\tilde{y})\cdot\nabla\varphi]dx=\int_{\mathbb{R}^{d}}f\varphi
dx.$

So, $\tilde{y}$ satisfies (6) in distributional sense.

On the other hand, by Lemma 2.5 in [14] we have that
$\displaystyle\int_{\mathbb{R}^{d}}|y_{n}|dx\leq\int_{\mathbb{R}^{d}}|f|dx$
and consequently
$\displaystyle\int_{\mathbb{R}^{d}}|\tilde{y}|dx\leq\int_{\mathbb{R}^{d}}|f|dx$.

We may infer that $\tilde{y}\in L^{1}(\mathbb{R}^{d}),\ \beta(\tilde{y})\in
L^{1}(\mathbb{R}^{d})$ (and so $-\Delta\beta(\tilde{y})+\nabla\cdot({\cal
K}(\zeta)b^{*}(\tilde{y}))\in L^{1}(\mathbb{R}^{d})$). So, $\tilde{y}$ is
solution to (6) and by Lemma 2.5 in [14] we conclude that $\tilde{y}=y$.

Let $\tilde{\Phi}_{\cal R}=\min\\{\Phi(x);|x|_{d}\geq{\cal R}\\}$, ${\cal
R}>0$. This yields

$\int_{\mathbb{R}^{d}}|y_{n}-y|dx=\int_{B(0_{d};{\cal
R})}|y_{n}-y|dx+\int_{|x|_{d}\geq{\cal R}}|y_{n}-y|dx.$

Since

$\displaystyle\int_{|x|_{d}\geq{\cal R}}|y_{n}|\tilde{\Phi}_{\cal
R}dx\leq\int_{|x|_{d}\geq{\cal
R}}|y_{n}|\Phi(x)dx\leq\int_{\mathbb{R}^{d}}|y_{n}|\Phi(x)dx\leq C,$

it follows that

$\int_{|x|_{d}\geq{\cal R}}|y_{n}|dx\leq{{C}\over{\tilde{\Phi}_{\cal R}}}\
\mbox{\rm and so }\int_{|x|_{d}\geq{\cal
R}}|y|dx\leq{{C}\over{\tilde{\Phi}_{\cal R}}}.$

On the other hand, since $\tilde{\Phi}_{\cal R}\rightarrow+\infty$ as ${\cal
R}\rightarrow+\infty$, it follows that for any $\varepsilon>0$, there exists
for ${\cal R}>0$ sufficiently large such that

$\int_{|x|_{d}\geq{\cal R}}|y_{n}-y|dx\leq\int_{|x|_{d}\geq{\cal
R}}(|y_{n}|+|y|)dx\leq 2\cdot{{\varepsilon}\over 2}=\varepsilon,$

and so

$\limsup_{n\rightarrow+\infty}\int_{\mathbb{R}^{d}}|y_{n}-y|dx\leq\varepsilon,\
\forall\varepsilon>0,\quad\mbox{\rm and }y_{n}\longrightarrow y\quad\mbox{\rm
in }L^{1}(\mathbb{R}^{d}).$

Remark 3.2. The conclusion holds not only for a subsequence of
$\\{y_{n}\\}_{n\in\mathbb{N}^{*}}$, but for the sequence itself. We argue by
contradiction.

Assume that there exists a subsequence $\\{y_{n_{l}}\\}_{l\in\mathbb{N}^{*}}$
such that $\|y_{n_{l}}-y\|_{L^{1}(\mathbb{R}^{d})}\rightarrow\theta_{0}>0$.
The first part of the proof allows us to conclude that on a subsequence
$\\{y_{n_{l_{q}}}\\}_{q\in\mathbb{N}^{*}}$ we have that
$y_{n_{l_{q}}}\longrightarrow y$ in ${L^{1}(\mathbb{R}^{d})}$ and so
$\|y_{n_{l_{q}}}-y\|_{L^{1}(\mathbb{R}^{d})}\longrightarrow 0$; absurd.

  * Case 2.

Let $f\in L^{1}(\mathbb{R}^{d})$ and $\\{f_{k}\\}_{k\in\mathbb{N}^{*}}$
satisfying the hypotheses of $f$ in case 1 and satisfying
$f_{k}\longrightarrow f$ in $L^{1}(\mathbb{R}^{d})$. Let

    * $\bullet$

$y^{k}_{n}$ solution to (6) with $\zeta:=\zeta_{n}$, $f:=f_{k}$;

    * $\bullet$

$y_{n}$ solution to (6) with $\zeta:=\zeta_{n}$;

    * $\bullet$

$y^{k}$ solution to (6) with $f:=f_{k}$.

We have that

$\|y_{n}-y\|_{L^{1}(\mathbb{R}^{d})}\leq\|y_{n}-y^{k}_{n}\|_{L^{1}(\mathbb{R}^{d})}+\|y^{k}_{n}-y^{k}\|_{L^{1}(\mathbb{R}^{d})}+\|y^{k}-y\|_{L^{1}(\mathbb{R}^{d})}.$

Since

$\|y_{n}-y^{k}_{n}\|_{L^{1}(\mathbb{R}^{d})}\leq\|f-f_{k}\|_{L^{1}(\mathbb{R}^{d})},\quad\|y^{k}-y\|_{L^{1}(\mathbb{R}^{d})}\leq\|f-f_{k}\|_{L^{1}(\mathbb{R}^{d})},$

we conclude that

$\|y_{n}-y\|_{L^{1}(\mathbb{R}^{d})}\leq
2\|f-f_{k}\|_{L^{1}(\mathbb{R}^{d})}+\|y^{k}_{n}-y^{k}\|_{L^{1}(\mathbb{R}^{d})}.$

For a fixed $k$ we have that $y^{k}_{n}-y^{k}\longrightarrow 0$ in
$L^{1}(\mathbb{R}^{d})$.

Hence $\limsup_{n\rightarrow+\infty}\|y_{n}-y\|_{L^{1}(\mathbb{R}^{d})}\leq
2\|f-f_{k}\|_{L^{1}(\mathbb{R}^{d})}$, $\forall k\in\mathbb{N}^{*}$, and so

$\limsup_{n\rightarrow+\infty}\|y_{n}-y\|_{L^{1}(\mathbb{R}^{d})}\leq 0\
\mbox{\rm and }y_{n}\longrightarrow y\ \mbox{\rm in }L^{1}(\mathbb{R}^{d}).$

##  4 A penalized optimal control problem ${\bf(P_{h})}$

For any $N\in\mathbb{N}^{*}$ we consider $h={T\over N}$ and the following
optimal control problem:

$\underset{\zeta\in{\cal M}}{\text{Minimize}}\
\int_{0}^{T}\int_{\mathbb{R}^{d}}G(t,x)\rho^{\zeta}_{h}(t,x)dx\
dt+\int_{\mathbb{R}^{d}}G_{T}(x)\rho^{\zeta}_{h}(T,x)dx$ $None$
$+\int_{\mathbb{R}^{d}}Q(x,\zeta(x))dx+{1\over
2}\int_{\mathbb{R}^{d}}|\zeta(x)-\zeta^{*}(x)|^{2}dx,$

where $\rho^{\zeta}_{h}$ and $\rho^{\zeta,i}_{h}$ were defined in section 2,
and $\zeta^{*}$ is an optimal control for problem $(P)$.

Remark 4.1. The following equality holds

$\int_{0}^{T}\int_{\mathbb{R}^{d}}G(t,x)\rho^{\zeta}_{h}(t,x)dx\
dt+\int_{\mathbb{R}^{d}}G_{T}(x)\rho^{\zeta}_{h}(T,x)dx$
$=\sum_{i=1}^{N}\int_{\mathbb{R}^{d}}\tilde{G}^{i}(x)\rho^{\zeta,i}_{h}(x)dx+\int_{\mathbb{R}^{d}}G_{T}(x)\rho^{\zeta,N}_{h}(x)dx,$

where

$\tilde{G}^{i}(x)=\int_{(i-1)h}^{ih}G(t,x)dt,\quad i=1,2,...,N.$

For the remainder of the section we consider $h={T\over N}$ to be sufficiently
small.

###  4.1 Existence of an optimal control for $\bf(P_{h})$

Lemma 4.1. If $\\{\zeta_{n}\\}_{n\in\mathbb{N}^{*}}\subset{\cal M}$,
$\zeta_{n}\longrightarrow\zeta\in{\cal M}$ weakly * in
$L^{\infty}(\mathbb{R}^{d})$, ${\cal K}(\zeta_{n})\longrightarrow{\cal
K}(\zeta)$, uniformly for $x\in\mathbb{R}^{d},$ and if
$\\{f_{n}\\}_{n\in\mathbb{N}^{*}}\subset L^{1}(\mathbb{R}^{d})$,
$f_{n}\longrightarrow f$ in $L^{1}(\mathbb{R}^{d})$, then

$y^{n}_{n}\longrightarrow y\quad\mbox{\it in }L^{1}(\mathbb{R}^{d}),$

where $y_{n}^{k},\ y_{n}$ where introduced in section 3 (in the proof of Lemma
3.1).

Proof. By Lemma 2.5 in [14] we get that

$\|y^{n}_{n}-y\|_{L^{1}(\mathbb{R}^{d})}\leq\|y^{n}_{n}-y_{n}\|_{L^{1}(\mathbb{R}^{d})}+\|y_{n}-y\|_{L^{1}(\mathbb{R}^{d})}\leq\|f_{n}-f\|_{L^{1}(\mathbb{R}^{d})}+\|y_{n}-y\|_{L^{1}(\mathbb{R}^{d})}.$

Using the convergence of $\\{f_{n}\\}_{n\in\mathbb{N}^{*}}$ and that
$y_{n}\longrightarrow y$ in $L^{1}(\mathbb{R}^{d})$ (by Lemma 3.1) we get the
conclusion.

Lemma 4.2. If $\\{\zeta_{n}\\}_{n\in\mathbb{N}^{*}}\subset{\cal M}$,
$\zeta_{n}\longrightarrow\zeta\in{\cal M}$ weakly * in
$L^{\infty}(\mathbb{R}^{d})$, ${\cal K}(\zeta_{n})\longrightarrow{\cal
K}(\zeta)$, uniformly for $x\in\mathbb{R}^{d},$ and if
$(\rho^{\zeta_{n},i}_{h})_{i=0,1,...,N}$ and
$(\rho^{\zeta,i}_{h})_{i=0,1,...,N}$ are defined by $(\ref{eq2.7})$, then

$\rho^{\zeta_{n},i}_{h}\longrightarrow\rho^{\zeta,i}_{h}\quad\mbox{\it in
}L^{1}(\mathbb{R}^{d}),\quad\forall i\in\\{0,1,...,N\\}.$

Proof. This follows by finite induction and Lemma 4.1.

Theorem 4.3. Problem $(P_{h})$ has at least one optimal control
$\zeta^{*}_{h}$.

Proof. Denote by $I_{h}$ the cost functional associated to problem $(P_{h})$.
Let $m^{*}_{h}=\underset{\zeta\in{\cal
M}}{\text{inf}}I_{h}(\zeta)\in[0,+\infty),$ and let
$\\{\zeta_{n}\\}_{n\in\mathbb{N}^{*}}\subset{\cal M}$,
$\zeta_{n}\longrightarrow\zeta^{*}_{h}\in{\cal M}$ weakly * in
$L^{\infty}(\mathbb{R}^{d})$,

${\cal K}(\zeta_{n})\longrightarrow{\cal K}(\zeta_{h}^{*}),\quad\mbox{\rm
uniformly for }x\in\mathbb{R}^{d},$

such that:

$m^{*}_{h}\leq I_{h}(\zeta_{n})<m^{*}_{h}+{1\over n},\quad
n\in\mathbb{N}^{*}.$

By Lemma 4.2 we have that

$\int_{\mathbb{R}^{d}}G_{T}\rho^{\zeta_{n},N}_{h}dx\longrightarrow\int_{\mathbb{R}^{d}}G_{T}\rho^{\zeta^{*}_{h},N}_{h}dx,$

and

$\sum_{i=1}^{N}\int_{\mathbb{R}^{d}}\tilde{G}^{i}\rho^{\zeta_{n},i}_{h}dx\longrightarrow\sum_{i=1}^{N}\int_{\mathbb{R}^{d}}\tilde{G}^{i}\rho^{\zeta^{*}_{h},i}_{h}dx.$

On the other hand (from the convexity of $Q$ with respect to $\zeta$)

$\int_{\mathbb{R}^{d}}Q(x,\zeta_{n}(x))dx+{1\over
2}\int_{\mathbb{R}^{d}}|\zeta_{n}(x)-\zeta^{*}(x)|^{2}dx\geq\int_{\mathbb{R}^{d}}Q(x,\zeta_{h}^{*}(x))dx$
$+\int_{\mathbb{R}^{d}}{{\partial Q}\over{\partial
z}}(x,\zeta_{h}^{*}(x))(\zeta_{n}(x)-\zeta_{h}^{*}(x))dx.$

This implies that

$m^{*}_{h}\geq\lim_{n\rightarrow+\infty}I_{h}(\zeta_{n})=\liminf_{n\rightarrow+\infty}I_{h}(\zeta_{n})\geq
I(\zeta_{h}^{*})\geq m_{h}^{*}.$

We conclude that $\zeta^{*}_{h}$ is an optimal control for problem $(P_{h})$.

###  4.2 Necessary optimality conditions for $\bf(P_{h})$

Let $\zeta_{h}^{*}$ be an optimal control for problem $(P_{h})$. Let $\xi\in
L^{\infty}(\mathbb{R}^{d})$ be such that $\zeta_{h}^{*}+\varepsilon\xi\in{\cal
M}$, for any $\varepsilon>0$ sufficiently small.

For any $\zeta\in{\cal M}$, let $z^{\zeta,i+1}$ be the weak solution to

$\left\\{\begin{array}[]{ll}z^{\zeta,i+1}-h\Delta(\beta^{\prime}(\rho_{h}^{\zeta,i+1})z^{\zeta,i+1})+h\nabla\cdot({\cal
K}(\zeta)(b^{*})^{\prime}(\rho_{h}^{\zeta,i+1})z^{\zeta,i+1})+h\nabla\cdot({\cal
K}(\xi)b^{*}(\rho_{h}^{\zeta.i+1}))=z^{\zeta,i},\\\ \hskip
284.52756pti=0,1,...,N-1,\\\ z^{\zeta,0}=0.\end{array}\right.$ (7)

Actually, $z^{\zeta,i+1}$ is a weak solution to the equation in (7) if
$\tilde{z}^{i+1}\in H^{1}(\mathbb{R}^{d})$ is the unique weak solution to

${1\over{\beta^{\prime}(\rho_{h}^{\zeta,i+1})}}\tilde{z}^{i+1}-h\Delta\tilde{z}^{i+1}+h\nabla\cdot\left({\cal
K}(\zeta){{(b^{*})^{\prime}(\rho_{h}^{\zeta,i+1})}\over{\beta^{\prime}(\rho_{h}^{\zeta,i+1})}}\tilde{z}^{i+1}\right)=-h\nabla\cdot({\cal
K}(\xi)b^{*}(\rho_{h}^{\zeta.i+1}))+{1\over{\beta^{\prime}(\rho_{h}^{\zeta,i})}}\tilde{z}^{i}$

and $\tilde{z}^{0}=0$. Here
$\tilde{z}^{i}=\beta^{\prime}(\rho_{h}^{\zeta,i})z^{\zeta,i}$. The existence
and uniqueness of a weak solution to this equation follows via Lax-Milgram’s
lemma (for sufficiently small $h$).

We postpone the proof of the following lemma.

Lemma 4.4. The following convergences hold for any $i\in\\{1,2,...,N\\}$:

  * (i)

$\rho_{h}^{\zeta^{*}_{h}+\varepsilon\xi,i}\longrightarrow\rho_{h}^{\zeta^{*}_{h},i}\quad\mbox{\it
in }L^{2}(\mathbb{R}^{d}),\ \mbox{\it as }\varepsilon\longrightarrow 0+$;

  * (ii)

${{\rho_{h}^{\zeta^{*}_{h}+\varepsilon\xi,i}-\rho_{h}^{\zeta^{*}_{h},i}}\over{\varepsilon}}\longrightarrow
z^{\zeta^{*}_{h},i}\quad\mbox{\it in }L^{2}(\mathbb{R}^{d}),\ \mbox{\it as
}\varepsilon\longrightarrow 0+$ (on a subsequence).

Theorem 4.5. If $p_{h}^{\zeta^{*}_{h},i+1}$ is the unique weak solution to

$\left\\{\begin{array}[]{ll}p^{\zeta,i}-h\beta^{\prime}(\rho_{h}^{\zeta,i+1})\Delta
p^{\zeta,i}-h{\cal K}(\zeta)(b^{*})^{\prime}(\rho_{h}^{\zeta,i+1})\cdot\nabla
p^{\zeta,i}+\tilde{G}^{i+1}=p^{\zeta,i+1},\\\ \hskip
284.52756pti=N-1,...,1,0,\\\ p^{\zeta,N}=-G_{T},\end{array}\right.$ (8)

corresponding to $\zeta:=\zeta^{*}_{h}$, then

$\int_{\mathbb{R}^{d}}{\cal
K}(\xi)\cdot[\sum_{i=1}^{N}[hb^{*}(\rho_{h}^{\zeta^{*}_{h},i})\nabla
p_{h}^{\zeta^{*}_{h},i-1}]dx-\int_{\mathbb{R}^{d}}\xi[{{\partial
Q}\over{\partial
z}}(x,\zeta_{h}^{*}(x))+(\zeta^{*}_{h}(x)-\zeta^{*}(x))]dx\leq 0,$ (9)

for any $\xi\in L^{\infty}(\mathbb{R}^{d})$ such that
$\zeta_{h}^{*}+\varepsilon\xi\in{\cal M}$, for any $\varepsilon>0$
sufficiently small.

We say that $p^{\zeta,i}\in H^{1}(\mathbb{R}^{d})$ is a weak solution to the
equation in (8) if it is a weak solution to

${1\over{\beta^{\prime}(\rho_{h}^{\zeta,i+1})}}p^{\zeta,i}-h\Delta
p^{\zeta,i}-h{\cal
K}(\zeta){{(b^{*})^{\prime}(\rho_{h}^{\zeta,i+1})}\over{\beta^{\prime}(\rho_{h}^{\zeta,i+1})}}\cdot\nabla
p^{\zeta,i}+{1\over{\beta^{\prime}(\rho_{h}^{\zeta,i+1})}}\tilde{G}^{i+1}={1\over{\beta^{\prime}(\rho_{h}^{\zeta,i+1})}}p^{\zeta,i+1}.$

The existence and uniqueness of a weak solution follows via Lax-Milgram’s
lemma (for sufficiently small $h$).

Remark 4.2. We have that $\zeta_{h}^{*}(x)=0$ a.e. $x\in Ext(B(0_{d};R_{0}))$.
On the other hand for almost any $x\in B(0_{d};R_{0})$ we have that

$\zeta_{h}^{*}(x)=\left\\{\begin{array}[]{ll}0,&\mbox{\rm if
}\displaystyle\int_{\mathbb{R}^{d}}\nabla{\cal
G}_{R}(x^{\prime}-x)\cdot\sum_{i=1}^{N}[hb^{*}(\rho_{h}^{\zeta^{*}_{h},i})\nabla
p_{h}^{\zeta^{*}_{h},i-1}](x^{\prime})dx^{\prime}\\\ {}&\displaystyle\ \ \ \ \
+{{\partial Q}\over{\partial
z}}(x,\zeta_{h}^{*}(x))+(\zeta^{*}_{h}(x)-\zeta^{*}(x))>0\\\ {}&{}\\\
\tilde{M}_{0},&\mbox{\rm if }\displaystyle\int_{\mathbb{R}^{d}}\nabla{\cal
G}_{R}(x^{\prime}-x)\cdot\sum_{i=1}^{N}[hb^{*}(\rho_{h}^{\zeta^{*}_{h},i})\nabla
p_{h}^{\zeta^{*}_{h},i-1}](x^{\prime})dx^{\prime}\\\ {}&\ \ \ \ \
\displaystyle+{{\partial Q}\over{\partial
z}}(x,\zeta_{h}^{*}(x))+(\zeta^{*}_{h}(x)-\zeta^{*}(x))<0.\end{array}\right.$

Proof of Theorem 4.5. Since $\zeta^{*}_{h}$ is an optimal control for problem
$(P_{h})$, then $I_{h}(\zeta^{*}_{h})\leq
I_{h}(\zeta^{*}_{h}+\varepsilon\xi)$, for any $\varepsilon>0$ sufficiently
small. After an easy evaluation and using the Lemma 4.4 we get that

$0\leq\sum_{i=1}^{N}\int_{\mathbb{R}^{d}}\tilde{G}^{i}z^{\zeta^{*}_{h},i}dx+\int_{\mathbb{R}^{d}}G_{T}z^{\zeta^{*}_{h},N}dx+\int_{\mathbb{R}^{d}}[{{\partial
Q}\over{\partial
z}}(x,\zeta_{h}^{*}(x))+(\zeta_{h}^{*}(x)-\zeta^{*}(x))]\xi(x)dx.$ (10)

We multiply (8) by $z^{\zeta^{*}_{h},i+1}$ and integrate over
$\mathbb{R}^{d}$. After some evaluation and taking into account (7) we obtain
that

$\int_{\mathbb{R}^{d}}\tilde{G}^{i+1}z^{\zeta^{*}_{h},i+1}dx=\int_{\mathbb{R}^{d}}p_{h}^{\zeta^{*}_{h},i+1}z^{\zeta^{*}_{h},i+1}dx+\int_{\mathbb{R}^{d}}p_{h}^{\zeta^{*}_{h},i}[-z^{\zeta^{*}_{h},i}+h\nabla\cdot({\cal
K}(\xi)b^{*}(\rho_{h}^{\zeta^{*}_{h},i+1}))]dx.$

By summation and using that $z^{\zeta^{*}_{h},0}=0$ we obtain that

$\sum_{i=1}^{N}\int_{\mathbb{R}^{d}}\tilde{G}^{i}z^{\zeta^{*}_{h},i}dx=-\int_{\mathbb{R}^{d}}G_{T}z^{\zeta^{*}_{h},N}dx-\sum_{i=1}^{N}\int_{\mathbb{R}^{d}}hb^{*}(\rho_{h}^{\zeta^{*}_{h},i}){\cal
K}(\xi)\cdot\nabla p_{h}^{\zeta^{*}_{h},i-1}dx$

and using (10) we get (9).

Proof of Lemma 4.4. By Lemma 3.1 in [16] it follows that there exists $C>0$
such that

$\|\rho_{h}^{\zeta^{*}_{h}+\varepsilon\xi,i}\|_{L^{\infty}(\mathbb{R}^{d})}\leq
C,\quad\forall 0\leq\varepsilon\leq\varepsilon_{0},\ i=0,1,...,N.$

Since
$\rho_{h}^{\zeta^{*}_{h}+\varepsilon\xi,i}\longrightarrow\rho_{h}^{\zeta^{*}_{h},i}\
\mbox{\rm in }L^{1}(\mathbb{R}^{d}),\ i=0,1,...,N$, as
$\varepsilon\longrightarrow 0+$ (by Lemma 4.2) we conclude that (i) is
satisfied.

Let $y^{\varepsilon,i}=\beta(\rho_{h}^{\zeta^{*}_{h}+\varepsilon\xi,i}),\quad
y^{i}=\beta(\rho_{h}^{\zeta^{*}_{h},i})$,
$\eta^{i}=\beta^{\prime}(\rho_{h}^{\zeta^{*}_{h},i})z^{\zeta^{*}_{h},i}$,
$\eta_{\varepsilon}^{i}=y^{\varepsilon,i}-y^{i}$,
$\tilde{\eta}_{\varepsilon}^{i}={{\eta_{\varepsilon}^{i}}\over{\varepsilon}}$.

We have that $\zeta_{\varepsilon}^{i+1}$ is the weak solution to

$\beta^{-1}(y^{\varepsilon,i+1})-\beta^{-1}(y^{i+1})-h\Delta\eta_{\varepsilon}^{i+1}+h\nabla\cdot({\cal
K}(\zeta_{h}^{*})(b^{*}(\beta^{-1}(y^{\varepsilon,i+1}))-b^{*}(\beta^{-1}(y^{i+1}))))$
$+\varepsilon h\nabla\cdot({\cal
K}(\xi)b^{*}(\beta^{-1}(y^{\varepsilon,i+1})))=\beta^{-1}(y^{\varepsilon,i})-\beta^{-1}(y^{i}).$

If we denote by $w_{\varepsilon}^{i}=\tilde{\eta}_{\varepsilon}^{i}-\eta^{i}$,
then we have that $w_{\varepsilon}^{i+1}$ is the unique weak solution to

${{\beta^{-1}(y^{\varepsilon,i+1})-\beta^{-1}(y^{i+1})}\over{\varepsilon}}-{1\over{\beta^{\prime}(\beta^{-1}(y^{i+1}))}}\eta^{i+1}-h\Delta
w_{\varepsilon}^{i+1}$ $+h\nabla\cdot\left({\cal
K}(\zeta_{h}^{*})({{b^{*}(\beta^{-1}(y^{\varepsilon,i+1}))-b^{*}(\beta^{-1}(y^{i+1}))}\over{\varepsilon}}-{{(b^{*})^{\prime}(\beta^{-1}(y^{i+1}))}\over{\beta^{\prime}(\beta^{-1}(y^{i+1}))}}\eta^{i+1})\right)$
$={{\beta^{-1}(y^{\varepsilon,i})-\beta^{-1}(y^{i})}\over{\varepsilon}}-{1\over{\beta^{\prime}(\beta^{-1}(y^{i}))}}\eta^{i}.$

If we apply the Lagrange’s mean value theorem we get after some evaluation and
by finite induction that for a subsequence (also indexed by $\varepsilon$):

$w_{\varepsilon}^{i}\longrightarrow 0\quad\mbox{\rm in
}H^{1}(\mathbb{R}^{d}),\quad\forall i=0,1,...,N,$

and so

${{\beta(\rho_{h}^{\zeta^{*}_{h}+\varepsilon\xi,i})-\beta(\rho_{h}^{\zeta^{*}_{h},i})}\over{\varepsilon}}\longrightarrow\beta^{\prime}(\rho_{h}^{\zeta^{*}_{h},i})z^{\zeta^{*}_{h},i}\quad\mbox{\rm
in }L^{2}(\mathbb{R}^{d}),\quad\mbox{\rm as }\varepsilon\longrightarrow 0+,\
\forall i=0,1,...,N.$

If we apply once more the Lagrange’s mean value theorem we get that on a
subsequence (ii) holds.

Remark 4.3. Note that for any $\zeta\in{\cal M}$, (3) is a backward Euler
scheme for (1). Moreover, for any $\zeta\in{\cal M}$ we get that

$\rho_{h}^{\zeta}(t)\longrightarrow\rho^{\zeta}(t)\quad\mbox{\rm in
}L^{1}(\mathbb{R}^{d}),\ \mbox{\rm as }h\longrightarrow 0,$

uniformly for $t\in[0,T]$.

##  5 Relationship between problems $\bf(P)$ and $\bf(P_{h})$. Optimality
conditions for $\bf(P)$

For any $h=\displaystyle{T\over N}$, $N\in\mathbb{N}^{*}$, we consider
$\zeta^{*}_{h}\in{\cal M}$ an optimal control for problem $(P_{h})$.

Lemma 5.1. The sequence $\\{\rho_{h}^{\zeta}\\}$ is bounded in
$L^{\infty}(0,T;L^{2}(\mathbb{R}^{d}))$, uniformly for $\zeta\in{\cal M}$ and
$\\{\nabla\rho_{h}^{\zeta}\\}$ is bounded in
$L^{2}(0,T;L^{2}(\mathbb{R}^{d})^{d})$, uniformly for $\zeta\in{\cal M}$, for
$h>0$ sufficiently small.

Proof. By (3) we obtain that for any $\zeta\in{\cal M}$:

$\int_{\mathbb{R}^{d}}|\rho_{h}^{\zeta,i+1}|^{2}dx+h\int_{\mathbb{R}^{d}}\nabla\beta(\rho_{h}^{\zeta,i+1})\cdot\nabla\rho_{h}^{\zeta,i+1}dx=\int_{\mathbb{R}^{d}}\rho_{h}^{\zeta,i}\rho_{h}^{\zeta,i+1}dx+h\int_{\mathbb{R}^{d}}{\cal
K}(\zeta)b^{*}(\rho_{h}^{\zeta,i+1})\cdot\nabla\rho_{h}^{\zeta,i+1}dx,$

and so

${1\over
2}\int_{\mathbb{R}^{d}}|\rho_{h}^{\zeta,i+1}|^{2}dx+h\gamma_{0}\int_{\mathbb{R}^{d}}|\nabla\rho_{h}^{\zeta,i+1}|_{d}^{2}dx\leq{1\over
2}\int_{\mathbb{R}^{d}}|\rho_{h}^{\zeta,i}|^{2}dx+h\tilde{M}_{1}\int_{\mathbb{R}^{d}}|\rho_{h}^{\zeta,i+1}||\nabla\rho_{h}^{\zeta,i+1}|_{d}dx$

where $\tilde{M}_{1}$ is a nonnegative constant. We immediately obtain that
there exists a positive constant $M_{2}$ such that

${1\over
2}\int_{\mathbb{R}^{d}}|\rho_{h}^{\zeta,i+1}|^{2}dx+{{h\gamma_{0}}\over
2}\int_{\mathbb{R}^{d}}|\nabla\rho_{h}^{\zeta,i+1}|_{d}^{2}dx\leq{1\over
2}\int_{\mathbb{R}^{d}}|\rho_{h}^{\zeta,i}|^{2}dx+{{hM_{2}}\over
2}\int_{\mathbb{R}^{d}}|\rho_{h}^{\zeta,i+1}|^{2}dx,$

and consequently for $h>0$ sufficiently small

$(1-hM_{2})\int_{\mathbb{R}^{d}}|\rho_{h}^{\zeta,i+1}|^{2}dx+h\gamma_{0}\int_{\mathbb{R}^{d}}|\nabla\rho_{h}^{\zeta,i+1}|_{d}^{2}dx\leq\int_{\mathbb{R}^{d}}|\rho_{h}^{\zeta,i}|^{2}dx.$

After an easy evaluation we get that

$(1-hM_{2})^{N}\|\rho_{h}^{\zeta}(t)\|^{2}_{L^{2}}+\gamma_{0}(1-hM_{2})^{N}\int_{0}^{t}\int_{\mathbb{R}^{d}}|\nabla\rho_{h}^{\zeta}(s,x)|^{2}_{d}dx\
ds\leq\|\rho_{0}\|^{2}_{L^{2}},$

for any $t\in[0,T],\ \zeta\in{\cal M}.$ Since $h={T\over N}$ we get the
conclusion of the lemma.

We postpone the proof ot the following result.

Lemma 5.2. If for a sequence $\\{\zeta_{h}\\}_{h>0}\subset{\cal M}$ we have
that $\zeta_{h}\longrightarrow\zeta\in{\cal M}$ weakly * in
$L^{\infty}(\mathbb{R}^{d}),$

${\cal K}(\zeta_{h})\longrightarrow{\cal K}(\zeta)\ \mbox{\it uniformly in
}\mathbb{R}^{d},\quad\nabla{\cal K}(\zeta_{h})(x)\longrightarrow\nabla{\cal
K}(\zeta)(x),\ \forall x\in\mathbb{R}^{d},$

as $h\rightarrow 0$, then

$\rho_{h}^{\zeta_{h}}(t)\longrightarrow\rho^{\zeta}(t)\ in\
L^{1}(\mathbb{R}^{d}),\ \mbox{\it uniformly for }t\in[0,T].$

Theorem 5.3 (The relationship between $\bf(P_{h})$ and $\bf(P)$). The
following convergences hold

  * (j)

$\zeta_{h}^{*}\longrightarrow\zeta^{*}$ in $L^{2}(\mathbb{R}^{d})$;

  * (jj)

$I_{h}(\zeta_{h}^{*})\longrightarrow I(\zeta^{*})$;

  * (jjj)

$I(\zeta_{h}^{*})\longrightarrow I(\zeta^{*})$, as $h\rightarrow 0$.

Proof. Since $\\{\zeta^{*}_{h}\\}\subset{\cal M}$ it follows that there exists
a subsequence (also denoted by $\\{\zeta^{*}_{h}\\}$) such that
$\zeta^{*}_{h}\longrightarrow\tilde{\zeta}\in{\cal M}$ weakly * in
$L^{\infty}(\mathbb{R}^{d})$ and

$\rho_{h}^{\zeta^{*}_{h}}(t)\longrightarrow\rho^{\tilde{\zeta}}(t)\quad\mbox{\rm
in }L^{1}(\mathbb{R}^{d}),\ \mbox{\rm uniformly for }t\in[0,T].$

This yields

$I_{h}(\zeta^{*})\geq
I_{h}(\zeta^{*}_{h})\geq\int_{0}^{T}\int_{\mathbb{R}^{d}}G(t,x)\rho_{h}^{\zeta^{*}_{h}}(t,x)dx\
dt+\int_{\mathbb{R}^{d}}G_{T}(x)\rho_{h}^{\zeta^{*}_{h}}(T,x)dx$
$+\int_{\mathbb{R}^{d}}Q(x,\tilde{\zeta}(x))dx+\int_{\mathbb{R}^{d}}{{\partial
Q}\over{\partial
z}}(x,\tilde{\zeta}(x))(\zeta^{*}_{h}(x)-\tilde{\zeta}(x))dx+{1\over
2}\int_{\mathbb{R}^{d}}|\zeta^{*}_{h}(x)-\zeta^{*}(x)|_{d}^{2}dx.$

By Crandall-Liggett’s theorem we have that
$\rho_{h}^{\zeta^{*}}(t)\longrightarrow\rho^{\zeta^{*}}(t)$ in
$L^{1}(\mathbb{R}^{d})$, uniformly for $t\in[0,T]$, and so we get that

$I_{h}(\zeta^{*})\longrightarrow I(\zeta^{*}),$

and

$I(\zeta^{*})\geq\limsup_{h\rightarrow
0}I_{h}(\zeta^{*}_{h})\geq\liminf_{h\rightarrow 0}I_{h}(\zeta^{*}_{h})\geq
I(\tilde{\zeta})+{1\over
2}\int_{\mathbb{R}^{d}}|\tilde{\zeta}(x)-\zeta^{*}(x)|^{2}_{d}dx\geq
I(\zeta^{*}).$

Since $\zeta^{*}$ is an optimal control for problem $(P)$ we may infer that
$\tilde{\zeta}(x)=\zeta^{*}(x)$ a.e. $x\in\mathbb{R}^{d}$ and that on a
subsequence (j) and (jj) hold.

Arguing by contradiction we get that (j) and (jj) hold in general (not only on
a subsequence).

On the other hand, using that
$\rho^{\zeta^{*}_{h}}(t)\longrightarrow\rho^{\zeta^{*}}(t)$ in
$L^{1}(\mathbb{R}^{d})$, uniformly for $t\in[0,T]$ (by Lemma 3.2), we get that
(jjj) holds.

Theorem 5.4 (Necessary optimality conditions for $\bf(P)$). If $p\in
W^{1,2}([0,T];L^{2}(\mathbb{R}^{d}))\cap L^{2}(0,T;H^{2}(\mathbb{R}^{d}))$
satisfies

$\left\\{\begin{array}[]{ll}\displaystyle{{dp}\over{dt}}(t)=-\beta^{\prime}(\rho^{\zeta^{*}}(t))\Delta
p(t)-{\cal K}(\zeta^{*})(b^{*})^{\prime}(\rho^{\zeta^{*}}(t))\cdot\nabla
p(t)+G(t),\quad t\in[0,T]\\\ p(T)=-G_{T},\end{array}\right.$ (11)

then

$\int_{\mathbb{R}^{d}}\int_{0}^{T}{\cal K}(\xi)\cdot
b^{*}(\rho^{\zeta^{*}})\nabla p\ dt\ dx-\int_{\mathbb{R}^{d}}\xi(x){{\partial
Q}\over{\partial z}}(x,\zeta^{*}(x))dx\leq 0,$ (12)

for any $\xi\in L^{\infty}(\mathbb{R}^{d})$ such that
$\zeta^{*}+\varepsilon\xi\in{\cal M}$, for any $\varepsilon>0$ sufficiently
small.

Proof. We shall use the results obtained for problem $(P_{h})$ in the previous
section.

For the sake of clarity we shall renote by $\zeta_{h}$ (instead of
$\zeta^{*}_{h}$) an optimal control for $(P_{h})$, and assume that $G$ is
time-independent. Let

$p_{h}^{\zeta_{h}}(t,x)=\left\\{\begin{array}[]{ll}-G_{T}(x),&\mbox{\rm if
}t=T\\\ p_{h}^{\zeta_{h},i-1}(x),&\mbox{\rm if }t\in[(i-1)h,ih),\
i=1,2,...,N.\end{array}\right.$

Multiplying (8) by $-\Delta p_{h}^{\zeta_{h},i}$ we get that

$\int_{\mathbb{R}^{d}}|\nabla
p_{h}^{\zeta_{h},i}|^{2}_{d}dx+h\int_{\mathbb{R}^{d}}\beta^{\prime}(\rho_{h}^{\zeta_{h},i+1})|\Delta
p_{h}^{\zeta_{h},i}|^{2}dx=\int_{\mathbb{R}^{d}}\nabla
p_{h}^{\zeta_{h},i}\cdot\nabla p_{h}^{\zeta_{h},i+1}dx$
$+h\int_{\mathbb{R}^{d}}G(x,\zeta_{h}(x))\Delta
p_{h}^{\zeta,i}dx-h\int_{\mathbb{R}^{d}}{\cal
K}(\zeta_{h})(b^{*})^{\prime}(\rho_{h}^{\zeta_{h},i+1})\cdot\nabla
p_{h}^{\zeta_{h},i}\Delta p_{h}^{\zeta_{h},i}dx,$

and so there exist two positive constants $M_{1}^{0},\ M_{2}^{0}$ such that

${1\over 2}\int_{\mathbb{R}^{d}}|\nabla
p_{h}^{\zeta_{h},i}|^{2}_{d}dx+{{h\gamma_{0}}\over
2}\int_{\mathbb{R}^{d}}|\Delta p_{h}^{\zeta_{h},i}|^{2}dx\leq{1\over
2}\int_{\mathbb{R}^{d}}|\nabla p_{h}^{\zeta_{h},i+1}|^{2}_{d}dx$
$+{{hM_{1}^{0}}\over 2}\int_{\mathbb{R}^{d}}|\nabla
p_{h}^{\zeta_{h},i}|^{2}_{d}dx+{{hM_{2}^{0}}\over
2}\int_{\mathbb{R}^{d}}G(x)^{2}dx.$

It follows that

$(1-hM_{1}^{0})\int_{\mathbb{R}^{d}}|\nabla
p_{h}^{\zeta_{h},i}|^{2}_{d}dx+h\gamma_{0}\int_{\mathbb{R}^{d}}|\Delta
p_{h}^{\zeta_{h},i}|^{2}dx\leq\int_{\mathbb{R}^{d}}|\nabla
p_{h}^{\zeta_{h},i+1}|^{2}_{d}dx$
$+hM_{2}^{0}\int_{\mathbb{R}^{d}}G(x)^{2}dx.$

After an easy evaluation we have that for $h>0$ sufficiently small

$(1-hM_{1}^{0})^{N}\int_{\mathbb{R}^{d}}|\nabla
p_{h}^{\zeta_{h},i}|^{2}_{d}dx+h\gamma_{0}(1-hM_{1}^{0})^{N}\sum_{i=0}^{N-1}\int_{\mathbb{R}^{d}}|\Delta
p_{h}^{\zeta_{h},i}|^{2}dx\leq\int_{\mathbb{R}^{d}}|\nabla G_{T}|^{2}_{d}dx$
$+hM_{2}^{0}[1+(1-hM_{1}^{0})+(1-hM_{1}^{0})^{2}+\cdots]\int_{\mathbb{R}^{d}}G(x)^{2}dx.$

This implies that there exists a nonnegative constant $\tilde{M}$ such that
for any $t\in[0,T]$:

$\int_{\mathbb{R}^{d}}|\nabla
p_{h}^{\zeta_{h}}(t,x)|^{2}_{d}dx+\int_{0}^{T}\int_{\mathbb{R}^{d}}|\Delta
p_{h}^{\zeta_{h}}(t,x)|^{2}dx\ dt\leq\tilde{M}.$

So, $\\{\nabla p_{h}^{\zeta_{h}}\\}$ is bounded in
$L^{\infty}(0,T;L^{2}(\mathbb{R}^{d})^{d})$ and $\\{\Delta
p_{h}^{\zeta_{h}}\\}$ is bounded in $L^{2}(0,T;L^{2}(\mathbb{R}^{d}))$ (for
$h>0$ sufficiently small).

If we multiply now (8) by $p_{h}^{\zeta_{h},i}$ and integrate over
$\mathbb{R}^{d}$ we obtain in the same manner that $\\{p_{h}^{\zeta_{h}}\\}$
is bounded in $L^{\infty}(0,T;L^{2}(\mathbb{R}^{d}))$.

We may infer that there exists a subsequence (also denoted by
$\\{p_{h}^{\zeta_{h}}\\}$) and a function $p$ such that

$\left\\{\begin{array}[]{ll}p_{h}^{\zeta_{h}}\longrightarrow p&\mbox{\rm
weakly in }L^{2}(0,T;L^{2}(\mathbb{R}^{d})),\\\ \nabla
p_{h}^{\zeta_{h}}\longrightarrow\nabla p&\mbox{\rm weakly in
}L^{2}(0,T;L^{2}(\mathbb{R}^{d})^{d}),\\\ \Delta
p_{h}^{\zeta_{h}}\longrightarrow\Delta p&\mbox{\rm weakly in
}L^{2}(0,T;L^{2}(\mathbb{R}^{d})).\end{array}\right.$

By (8) we have that

$p_{h}^{\zeta_{h},i}(x)=-G_{T}(x)+\int_{ih}^{T}\beta^{\prime}(\rho_{h}^{\zeta_{h}}(s,x))\Delta
p_{h}^{\zeta_{h}}(s,x)ds+\int_{ih}^{T}{\cal
K}(\zeta_{h})(x)(b^{*})^{\prime}(\rho_{h}^{\zeta_{h}}(s,x))\cdot\nabla
p_{h}^{\zeta_{h}}(s,x)ds$ $-(T-ih)G(x)\quad\mbox{\rm in
}L^{2}(\mathbb{R}^{d}).$

Since $\zeta_{h}\longrightarrow\zeta^{*}$ a.e. $x\in\mathbb{R}^{d}$ (on a
subsequence), we may conclude that for any $t\in[0,T]$:

$p(t)=-G_{T}+\int_{t}^{T}\beta^{\prime}(\rho^{\zeta^{*}}(s))\Delta
p(s)ds+\int_{t}^{T}{\cal
K}(\zeta^{*})(b^{*})^{\prime}(\rho^{\zeta^{*}}(s))\cdot\nabla p(s)ds$
$-\int_{t}^{T}G(x)ds\quad\mbox{\rm in }L^{2}(\mathbb{R}^{d}).$

We may conclude that $p\in W^{1,2}([0,T];L^{2}(\mathbb{R}^{d}))\cap
L^{2}(0,T;H^{2}(\mathbb{R}^{d}))$ and that $p$ satisfies (11). The uniqueness
of the solution to (11) follows in a standard manner.

By (9) we have that

$\int_{\mathbb{R}^{d}}\int_{0}^{T}{\cal K}(\xi)\cdot
b^{*}(\rho_{h}^{\zeta^{*}_{h}})\nabla p_{h}^{\zeta^{*}_{h}}dt\
dx-\int_{\mathbb{R}^{d}}\xi[{{\partial Q}\over{\partial
z}}(x,\zeta_{h}^{*}(x))+(\zeta^{*}_{h}(x)-\zeta^{*}(x))]dx\leq 0,$

for any $\xi\in L^{\infty}(\mathbb{R}^{d})$ such that
$\zeta_{h}^{*}+\varepsilon\xi\in{\cal M}$, for any $\varepsilon>0$
sufficiently small, which immediately implies (12).

Proof of Lemma 5.2.  We have that

$\rho_{h}^{\zeta_{h},i+1}-h\Delta\beta(\rho_{h}^{\zeta_{h},i+1})+h\nabla\cdot({\cal
K}(\zeta_{h})b^{*}(\rho_{h}^{\zeta_{h},i+1}))=\rho^{\zeta_{h},i}_{h}$

and

$\rho_{h}^{\zeta,i+1}-h\Delta\beta(\rho_{h}^{\zeta,i+1})+h\nabla\cdot({\cal
K}(\zeta_{h})b^{*}(\rho_{h}^{\zeta,i+1}))=\rho^{\zeta,i}_{h}+h\nabla\cdot(({\cal
K}(\zeta_{h})-{\cal K}(\zeta))b^{*}(\rho_{h}^{\zeta,i+1})).$

Using the properties of $A^{\zeta_{h}}$, $A^{\zeta}$ we get that

$\|\rho_{h}^{\zeta_{h},i+1}-\rho_{h}^{\zeta,i+1}\|_{L^{1}}\leq\|\rho^{\zeta_{h},i}_{h}-\rho^{\zeta,i}_{h}-h\nabla\cdot(({\cal
K}(\zeta_{h})-{\cal K}(\zeta))b^{*}(\rho_{h}^{\zeta,i+1}))\|_{L^{1}}$
$\leq\|\rho^{\zeta_{h},i}_{h}-\rho^{\zeta,i}_{h}\|_{L^{1}}+h\|\nabla\cdot({\cal
K}(\zeta_{h})-{\cal K}(\zeta))b^{*}(\rho_{h}^{\zeta,i+1})\|_{L^{1}}+h\|({\cal
K}(\zeta_{h})-{\cal
K}(\zeta)(b^{*})^{\prime}(\rho_{h}^{\zeta,i+1})\cdot\nabla\rho_{h}^{\zeta,i+1}\|_{L^{1}}.$

By summation we obtain that for any $t\in[0,T]$:

$\|\rho_{h}^{\zeta_{h}}(t)-\rho_{h}^{\zeta}(t)\|_{L^{1}}\leq\||\nabla\cdot({\cal
K}(\zeta_{h})-{\cal
K}(\zeta))|\int_{0}^{T}|b^{*}(\rho_{h}^{\zeta}(s))|ds\|_{L^{1}}$ $+\||{\cal
K}(\zeta_{h})-{\cal
K}(\zeta)|_{d}\int_{0}^{T}|(b^{*})^{\prime}(\rho_{h}^{\zeta}(s))\cdot\nabla\rho_{h}^{\zeta}(s)|_{d}ds\|_{L^{1}}$
$\leq\tilde{M}_{1}\|{\cal K}(\zeta_{h})-{\cal
K}(\zeta)\|_{L^{2}}+\tilde{M}_{2}\|\nabla\cdot({\cal K}(\zeta_{h})-{\cal
K}(\zeta))\|_{L^{2}},$

where $\tilde{M}_{1},\ \tilde{M}_{2}$ are positive constants. Using now Lemma
5.1 we may infer that

$\rho_{h}^{\zeta_{h}}(t)-\rho_{h}^{\zeta}(t)\longrightarrow 0\quad\mbox{\rm in
}L^{1}(\mathbb{R}^{d}),\ \mbox{\rm uniformly for }t\in[0,T].$

On the other hand

$\rho_{h}^{\zeta}(t)-\rho^{\zeta}(t)\longrightarrow 0\quad\mbox{\rm in
}L^{1}(\mathbb{R}^{d}),\ \mbox{\rm uniformly for }t\in[0,T]$

via Crandall-Liggett theorem. We may infer that

$\rho_{h}^{\zeta_{h}}(t)\longrightarrow\rho^{\zeta}(t)\quad\mbox{\rm in
}L^{1}(\mathbb{R}^{d}),\ \mbox{\rm uniformly for }t\in[0,T].$

Remark 5.1. By (12) we get that for almost any $x\in B(0_{d};R_{0})$ we have
that

$\zeta^{*}(x)=\left\\{\begin{array}[]{ll}0,&\mbox{\rm if
}\displaystyle\int_{\mathbb{R}^{d}}\nabla{\cal
G}_{R}(x^{\prime}-x)\cdot\int_{0}^{T}[b^{*}(\rho^{\zeta^{*}})\nabla
p](t,x^{\prime})dt\ dx^{\prime}+{{\partial Q}\over{\partial
z}}(x,\zeta^{*}(x))>0\\\ {}&{}\\\ \tilde{M}_{0},&\mbox{\rm if
}\displaystyle\int_{\mathbb{R}^{d}}\nabla{\cal
G}_{R}(x^{\prime}-x)\cdot\int_{0}^{T}[b^{*}(\rho^{\zeta^{*}})\nabla
p](t,x^{\prime})dt\ dx^{\prime}+{{\partial Q}\over{\partial
z}}(x,\zeta^{*}(x))<0.\end{array}\right.$

##  6 Final comments

Problem $(P)$ has an obvious meaning if we consider $G$ and $G_{T}$ to be
approximations of the characteristic functions of two compact subsets
$K\subset[0,T]\times\mathbb{R}^{d}$ and $K_{T}\subset\mathbb{R}^{d}$.

Remark 6.1. The results in the present paper can be easily extended to the
case of

$\zeta\in{\cal M}_{0}=\\{\theta\in L^{\infty}(\mathbb{R}^{d});\
0\leq\theta(x)\leq\tilde{M}_{0}\ \mbox{\rm a.e. }x\in\mathbb{R}^{d},\
\theta(x)=0\ \mbox{\rm a.e. }x\in Ext(D)\\},$

where $D$ is a bounded and open subset of $\mathbb{R}^{d}$.

Remark 6.2. Due to Lemma 5.1 it follows that the results in the previous
sections 3-5 hold if we consider for $G$ the weaker hypotheses

$G\in L^{2}((0,T)\times\mathbb{R}^{d}),\quad 0\leq G(t,x)\ \mbox{\rm a.e.
}(t,x)\in(0,T)\times\mathbb{R}^{d}.$

Let us briefly discuss the situation when $Q\equiv 0$ (i.e. the cost of the
control $\zeta$ is negligible). Problem $(P)$ becomes

$\underset{\zeta\in{\cal M}}{\text{Minimize}}\
\int_{0}^{T}\int_{\mathbb{R}^{d}}G(t,x)\rho^{\zeta}(t,x)dx\
dt+\int_{\mathbb{R}^{d}}G_{T}(x)\rho^{\zeta}(T,x)dx,$ $None$

where $\rho^{\zeta}$ is the weak solution to the following nonlinear Fokker-
Planck equation (1). For any $h={T\over N}$ (where $N\in\mathbb{N}^{*}$)
sufficiently small we consider the following approximating optimal control
problem

$\underset{\zeta\in{\cal M}}{\text{Minimize}}\
\int_{0}^{T}\int_{\mathbb{R}^{d}}G(t,x)\rho^{\zeta}_{h}(t,x)dx\
dt+\int_{\mathbb{R}^{d}}G_{T}(x)\rho^{\zeta}_{h}(T,x)dx,$ $None$

where $\rho^{\zeta}_{h}$ and $\rho^{\zeta,i}_{h}$ were defined in section 2.

Since problem $(\tilde{P})$ is a particular case of problem $(P)$, it follows
that there exists at least one optimal control for $(\tilde{P})$. The
existence of an optimal control $\tilde{\zeta}_{h}^{*}$ for problem
$(\tilde{P_{h}})$ is proven in the same way as for problem $(P_{h})$.

The following results follow in an analogous manner to Theorem 5.3 and Theorem
4.5.

Theorem 6.1 (The relationship between $\bf(\tilde{P}_{h})$ and
$\bf(\tilde{P})$). If $\tilde{\zeta}^{*}$ is a weak * accumulation point in
$L^{\infty}(\mathbb{R}^{d})$ for $\\{\tilde{\zeta}_{h}^{*}\\}$ (and denote the
convergent subsequence by $\\{\tilde{\zeta}_{h}^{*}\\}$ as well), then

  * (l)

$\tilde{\zeta}^{*}$ is an optimal control for problem $(\tilde{P})$;

  * (ll)

$\tilde{I}_{h}(\tilde{\zeta}_{h}^{*})\longrightarrow\tilde{I}(\tilde{\zeta}^{*})$;

  * (lll)

$\tilde{I}(\tilde{\zeta}_{h}^{*})\longrightarrow\tilde{I}(\tilde{\zeta}^{*})$,
as $h\rightarrow 0$.

Here $\tilde{I}$ and $\tilde{I}_{h}$ are the cost functionals for problems
$(\tilde{P})$ and $(\tilde{P}_{h})$, respectively.

If we replace the assumption $G_{T}\in C_{b}(\mathbb{R}^{d})\cap
H^{1}(\mathbb{R}^{d})$ by the weaker one

$G_{T}\in C_{b}(\mathbb{R}^{d})\cap L^{2}(\mathbb{R}^{d}),$

then the next result holds

Theorem 6.2. If $p_{h}^{\tilde{\zeta}^{*}_{h},i+1}$ is the unique weak
solution to $(\ref{eq4.1})$ corresponding to $\zeta:=\tilde{\zeta}^{*}_{h}$,
then

$\int_{\mathbb{R}^{d}}{\cal
K}(\xi)\cdot[\sum_{i=1}^{N}[hb^{*}(\rho_{h}^{\tilde{\zeta}^{*}_{h},i})\nabla
p_{h}^{\tilde{\zeta}^{*}_{h},i-1}]dx\leq 0,$

for any $\xi\in L^{\infty}(\mathbb{R}^{d})$ such that
$\tilde{\zeta}_{h}^{*}+\varepsilon\xi\in{\cal M}$, for any $\varepsilon>0$
sufficiently small.

Remark 6.3. We can adapt the proofs in the previous sections to investigate
the more general optimal control problem

$\underset{\zeta\in{\cal M}}{\text{Minimize}}\
\int_{0}^{T}\int_{\mathbb{R}^{d}}G(t,x,\rho^{\zeta}(t,x))dx\
dt+\int_{\mathbb{R}^{d}}G_{T}(x,\rho^{\zeta}(T,x))dx+\int_{\mathbb{R}^{d}}Q(x,\zeta(x))dx,$
$None$

where $\rho^{\zeta}$ is the weak solution to (1) if appropriate assumptions on
$G,\ G_{T}$ are considered.

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