# Well-posedness of martingale problem for SBM
with interacting branching

Lina Ji111Department of Mathematics, Southern University of Science and
Technology, Shenzhen, China. Email<EMAIL_ADDRESS>Jie
Xiong222Department of Mathematics & National Center for Applied Mathematics
(Shenzhen), Southern University of Science and Technology, Shenzhen, China.
Email<EMAIL_ADDRESS>and Xu Yang333School of Mathematics and
Information Science, North Minzu University, Yinchuan, China. Email:
<EMAIL_ADDRESS>
###### Abstract

In this paper a martingale problem for super-Brownian motion with interactive
branching is derived. The uniqueness of the solution to the martingale problem
is obtained by using the pathwise uniqueness of the solution to a
corresponding system of SPDEs with proper boundary conditions. The existence
of the solution to the martingale problem and the Hölder continuity of the
density process are also studied.

Keywords. super-Brownian motion; interacting branching; function-valued
process; stochastic partial differential equation.

AMS Mathematics subject classification (2010): 60H15; 60J68.

##  1 Introduction and main results

Let $M(\mathbb{R})$ be the collection of all finite Borel measures on
$\mathbb{R}.$ Let $C_{b}^{n}(\mathbb{R})$ be the collection of all bounded
continuous functions on $\mathbb{R}$ with bounded derivatives up to $n$th
order. We consider a $M(\mathbb{R})$-valued process $(X_{t})_{t\geq 0}$
satisfying the following martingale problem (MP): for $\phi\in
C_{b}^{2}(\mathbb{R}),$ the process

$\displaystyle
M_{t}(\phi)\equiv\left<X_{t},\phi\right>-\left<X_{0},\phi\right>-\int^{t}_{0}\left<X_{s},\frac{1}{2}\phi^{\prime\prime}\right>ds$
(1.1)

is a continuous martingale with

$\displaystyle\left<M(\phi)\right>_{t}=\int_{0}^{t}\left<X_{s},{\gamma}(\mu_{s},\cdot)\phi^{2}\right>ds,$
(1.2)

where $\gamma$ is the interacting branching rate depending on the density
process. The notation $\left<\nu,\phi\right>$ denotes the integral of the
function $\phi$ with respect to the measure $\nu.$ In this paper we assume
$X_{0}(dx)=\mu_{0}(x)dx$ with $\mu_{0}\in L^{2}(\mathbb{R})$, where
$L^{2}(\mathbb{R})$ is the space of functions $f$ such that
$\int_{\mathbb{R}}f(x)^{2}dx<\infty.$ When $\gamma$ is a constant, the process
$(X_{t})_{t\geq 0}$ is a super-Brownian motion. In this case the well-
posedness of the MP (1.1, 1.2) was established by the nonlinear partial
differential equation satisfied by its log-Laplace transform. Moreover, a new
approach for the well-posedness of the MP was suggested by Xiong [21], in
which a relationship between the super-Brownian motion and a stochastic
partial differential equation (SPDE) satisfied by its corresponding
distribution-function-valued process was established. The weak uniqueness of
the solution to the MP (1.1, 1.2) was also obtained by the strong uniqueness
of the solution to the corresponding SPDE in [21]. See He et. al [9] for the
case of super-Lévy process.

While the superprocesses with interaction are more natural since the branching
and spatial motion for many species depend on the processes themselves. When
the spatial motion is interactive, the well-posedness of the martingale
problem was studied by Donnelley and Kurtz [4], see also Perkins [15, Theorem
V.5.1] and Li et. al [12]. Uniqueness for the historical superprocesses with
certain interaction was investigated by Perkins [15]. Furthermore, the
superprocesses with interactive immigration were studied by [2, 7, 17], see
also Li [11, Section 10]. The well-posedness of the martingale problem for the
interactive immigration process was solved by Mytnik and Xiong [14]. See also
[22] for the well-posedness of the martingale problem for a superprocess with
location-dependent branching, interactive immigration mechanism and spatial
motion.

However, the hard case for the superprocess with interactive branching was
rarely investigated. We are interested in the case of
$\gamma(\mu_{s},x)=\gamma(\mu_{s}(x)),$ i.e., the interactive branching
mechanism depending on the density process $(\mu_{t}(x))_{t\geq
0,x\in\mathbb{R}}.$ However, in this case the well-posedness of the MP (1.1,
1.2) is still an open problem. The weak uniqueness of the solution to the MP
(1.1, 1.2) is very difficult to prove. As a first step, throughout this paper
we assume $\gamma$ satisfies the following condition:

######  Condition 1.1.

Fixed integer $n\geq 0.$ For
$-\infty=a_{0}<a_{1}<\cdots<a_{n}<a_{n+1}=\infty$, let

$\displaystyle{\gamma}(\mu_{s},x)=\begin{cases}g_{i}^{2}(\mu_{s}(a_{i+1})),&a_{i}\leq
x<a_{i+1},\ i=0,\cdots,n-1,\\\
g_{n}^{2}\left(\int_{a_{n}}^{\infty}\mu_{s}(y)dy\right),&a_{n}\leq
x<\infty,\end{cases}$

where $g_{i},\ i=0,\cdots,n$ are positive continuous bounded functions from
$\mathbb{R}_{+}$ to $\mathbb{R}_{+}$.

The existence and Hölder continuity of the density process
$(\mu_{t}(x))_{t\geq 0,x\in\mathbb{R}}$ are investigated in this paper, which
satisfies the following SPDE:

$\displaystyle\frac{\partial}{\partial
t}\mu_{t}(x)=\frac{1}{2}\Delta\mu_{t}(x)+\sqrt{\mu_{t}(x)\gamma(\mu_{t},x)}\dot{W}(t,x),$
(1.3)

where $\Delta$ denotes the one-dimensional Laplacian operator and
$\\{W(t,x):t\geq 0,x\in\mathbb{R}\\}$ is a time-space Gaussian white noise
based on the Lebesgue measure and the dot denotes the derivative in
distribution sense. The existence of the solution to the MP (1.1, 1.2) is
given by showing the existence of the solution to (1.3). Moreover, we show the
weak uniqueness of the solution to the MP (1.1, 1.2) in Theorem 1.4. The main
idea is to relate the MP with a system of SPDEs, which is satisfied by a
sequence of corresponding function-valued processes on intervals. The weak
uniqueness of solution to the MP follows from the pathwise uniqueness of the
solution to the system of SPDEs, see Section 3. Throughout this paper we
always assume that all random variables defined on the same filtered
probability space $(\Omega,\mathscr{F},\mathscr{F}_{t},\mathbb{P}).$ Let
$\mathbb{E}$ denote the corresponding expectation.

The main results of this paper without special explanation are stated as
below.

######  Theorem 1.2.

(Existence) There exists a $M(\mathbb{R})$-valued continuous process
$(X_{t})_{t\geq 0}$ satisfying the MP (1.1, 1.2). Moreover, $X_{t}(dx)$ is
absolutely continuous respect to $dx$ with density $\mu_{t}(x)$ satisfying
(1.3).

######  Theorem 1.3.

(Joint Hölder continuity) Suppose that $(\mu_{t}(x))_{t\geq 0,x\in\mathbb{R}}$
satisfies (1.3) and $\mu_{0}$ is Hölder continuous with exponent
$0<\lambda<\frac{1}{2}$. Then
$[0,T]\times\mathbb{R}\ni(t,x){\rightarrow}\mu_{t}(x)$ is Hölder continuous
with exponent $\lambda_{1}/2$ in time variable and with exponent $\lambda_{2}$
in space variable, where $\lambda_{1},\lambda_{2}\in(0,1/2).$ Namely, there
exists a random variable $K\geq 0$ only depending on $\lambda_{1}$ and
$\lambda_{2}$ such that

$\displaystyle|\mu_{t}(x)-\mu_{r}(y)|\leq
K(|t-r|^{\lambda_{1}/2}+|x-y|^{\lambda_{2}}),\qquad t,r\in[0,T],\
x,y\in\mathbb{R}.$

######  Theorem 1.4.

(Uniqueness) Assume $g_{n}$ is $\beta$-Hölder continuous with
$\frac{1}{2}\leq\beta\leq 1$, i.e.,

$\displaystyle|g_{n}(x)-g_{n}(y)|\leq K|x-y|^{\beta},\qquad x,y\geq 0$ (1.4)

for some constant $K.$ Then the weak uniqueness of the solution to the MP
(1.1, 1.2) holds.

######  Remark 1.5.

Taking $n=0$ in Condition 1.1, the branching rate depends on the total mass
process, i.e., the quadratic variation process of the martingale defined by
(1.1) is

$\displaystyle\langle M(\phi)\rangle_{t}=\int_{0}^{t}\gamma(\langle
X_{s},1\rangle)\langle X_{s},\phi^{2}\rangle ds.$ (1.5)

The well-posedness of the MP (1.1, 1.5) is a corollary of Theorem 1.2 and 1.4.

We introduce some notation. Let $\mathcal{X}_{0}$ be the Hilbert space
consisting of all functions $f$ such that

$\displaystyle\|f\|_{0}^{2}:=\int_{\mathbb{R}}f(x)^{2}e^{-|x|}dx<\infty.$

We denote the corresponding inner product by $\left<\cdot,\cdot\right>_{0}.$
Let $\mathscr{B}(\mathbb{R})$ (resp. $\mathscr{B}(a_{1},a_{2})$) denote the
Borel $\sigma$-algebra on $\mathbb{R}$ (resp. $(a_{1},a_{2})$). Let
$B(\mathbb{R})^{+}$ (resp. $C_{b}(\mathbb{R})^{+}$) be the collection of all
bounded positive (resp. bounded positive continuous) functions on
$\mathbb{R}.$ Let $B[a_{1},a_{2}]$ be the Banach space of bounded measurable
functions on $[a_{1},a_{2}]$. For $f,g\in B[a_{1},a_{2}]$ let
$\left<f,g\right>=\int_{a_{1}}^{a_{2}}f(x)g(x){\mbox{\rm d}}x$. Define
$C_{b}[a_{1},a_{2}]$ be the set of bounded continuous functions on
$[a_{1},a_{2}]$. For any integer $n\geq 0,$ let $C_{b}^{n}[a_{1},a_{2}]$ be
the subset of $C_{b}[a_{1},a_{2}]$ of functions with bounded continuous
derivatives up to the $n$th order. Let $C_{c}^{n}(a_{1},a_{2})$ denote the
subset of $C_{b}^{n}[a_{1},a_{2}]$ of functions with compact supports in
$(a_{1},a_{2}).$

The rest of the paper is organized as follows. In Section 2, we give the
proofs of Theorem 1.2 and 1.3. The weak uniqueness of the solution to the MP
(1.1, 1.2), i.e., the proof of Theorem 1.4, is given in Section 3. Throughout
the paper we use $\nabla$ to be the first order spatial differential operator,
use $K$ to denote a non-negative constant whose value may change from line to
line. In the integrals, we make convention that, for $a\leq b\in\mathbb{R},$

$\displaystyle\int_{a}^{b}=\int_{(a,b]}\quad\text{and}\quad\int_{a}^{\infty}=\int_{(a,\infty)}.$

##  2 Proofs of Theorem 1.2 and 1.3

In this section, we give the proofs of Theorem 1.2 and 1.3. The existence of
the solution to MP (1.1, 1.2) is obtained by the existence of the
corresponding density process. Moreover, the Hölder continuity of density
process $(\mu_{t}(x))_{t\geq 0,x\in\mathbb{R}}$ is given by a standard
argument.

######  Lemma 2.1.

The martingale defined in (1.1,1.2) induces a $(\mathscr{F}_{t})$-martingale
measure $\\{M_{t}(B):t\geq 0,B\in\mathscr{B}(\mathbb{R})\\}$ satisfying

$\displaystyle M_{t}(\phi)=\int_{0}^{t}\int_{\mathbb{R}}\phi(x)M(ds,dx),\qquad
t\geq 0,\ \phi\in C_{b}^{2}(\mathbb{R}),$

where $M(ds,dx)$ is an orthogonal martingale measure on
$\mathbb{R}_{+}\times\mathbb{R}$ with covariance measure
$ds\int_{\mathbb{R}}\big{[}\gamma(X_{s},z)\delta_{z}(dx)\delta_{z}(dy)\big{]}X_{s}(dz)$.

###### Proof.

Notice that $\gamma$ is bounded by Condition 1.1. By the MP (1.1, 1.2), one
can see that $\mathbb{E}[\left<X_{t},1\right>]=\left<X_{0},1\right><\infty.$
For each $n\geq 1$ we define the measure $\Gamma_{n}\in M(\mathbb{R})$ by

$\displaystyle\Gamma_{n}(\phi)=\mathbb{E}\Big{[}\int_{0}^{n}ds\int_{\mathbb{R}}\gamma(\mu_{s},x)\phi(x)X_{s}(dx)\Big{]}$

with $\phi\in B(\mathbb{R})^{+}.$ Then $\Gamma_{n}(\phi)\leq
K\int_{0}^{n}\mathbb{E}[\left<X_{s},1\right>]ds$ is bounded for each $n\geq
1.$ The rest proof follows by changing $c(z)=\gamma(\mu_{s},z)$ and
$H(z,{\mbox{\rm d}}\nu)=0$ in the proof of Li [11, Theorem 7.25]. We omit it
here. ∎

Let $T_{t}(x,dy)$ be the semigroup generated by $\Delta/2,$ which is
absolutely continuous with respect to Lebesgue measure ${\mbox{\rm d}}y$ with
density $p_{t}(x,y)$ satisfying

$p_{t}(x,y)=p_{t}(x-y)=\frac{1}{\sqrt{2\pi t}}{\mbox{\rm
e}}^{-|x-y|^{2}/(2t)}.$

######  Lemma 2.2.

Suppose that $(\mu_{t}(x))_{t\geq 0,x\in\mathbb{R}}$ is a solution to (1.3)
with $\mu_{0}\in L^{2}(\mathbb{R})$. Then for every $t\geq 0,$ we have
$\mathbb{E}[\langle\mu_{t},1\rangle]<\infty.$

###### Proof.

By (1.3) we have

$\displaystyle\langle\mu_{t},1\rangle=\langle\mu_{0},1\rangle+\int_{0}^{t}\int_{\mathbb{R}}\sqrt{\mu_{s}(x)\gamma(\mu_{s},x)}W(ds,dx).$

Notice that $\sqrt{x}\leq x+1$ for any $x\geq 0.$ By Hölder’s inequality one
can check that

$\displaystyle\mathbb{E}[\langle\mu_{t},1\rangle]\\!\\!$
$\displaystyle\leq\\!\\!$
$\displaystyle\langle\mu_{0},1\rangle+\mathbb{E}\left|\int_{0}^{t}\int_{\mathbb{R}}\sqrt{\mu_{s}(x)\gamma(\mu_{s},x)}W(ds,dx)\right|$
$\displaystyle\leq\\!\\!$
$\displaystyle\langle\mu_{0},1\rangle+K\left[\mathbb{E}\left(\int_{0}^{t}\int_{\mathbb{R}}\mu_{s}(x)dsdx\right)\right]^{1/2}$
$\displaystyle\leq\\!\\!$ $\displaystyle
K+K\int_{0}^{t}\mathbb{E}\left[\langle\mu_{s},1\rangle\right]ds.$

The result follows by Gronwall’s inequality. ∎

Similar with Li [11, Theorem 7.26, Theorem 7.28], we have the following
result.

######  Proposition 2.3.

Suppose that $(X_{t})_{t\geq 0}$ is a solution to the MP (1.1, 1.2). Then for
each $t\geq 0$ the random measure $X_{t}(dx)$ is absolutely continuous respect
to $dx$ with density $\mu_{t}(x)$ satisfying (1.3). Conversely, assume that
$(\mu_{t}(x))_{t\geq 0,x\in\mathbb{R}}$ is a solution to (1.3) with
$\mu_{0}\in L^{2}(\mathbb{R})$. Then $(X_{t})_{t\geq 0}$ satisfies the MP
(1.1, 1.2).

###### Proof.

Suppose that $(X_{t})_{t\geq 0}$ satisfies the MP (1.1, 1.2). By Lemma 2.1 and
the proof of [11, Theorem 7.26], we have

$\displaystyle\left<X_{t},\phi\right>=\left<X_{0},T_{t}\phi\right>+\int_{0}^{t}\int_{\mathbb{R}}T_{t-s}\phi(x)M(ds,dx),$

where $M(ds,dx)$ is an orthogonal martingale measure defined in Lemma 2.1.
Recall that $\gamma$ is bounded by Condition 1.1. Let $\phi\in
C_{b}(\mathbb{R})^{+}$ and set $p_{t}(x,z)=0$ for all $t\leq 0.$ For any
$n\geq 1$ we have

$\displaystyle\mathbb{E}\Big{[}\int_{\mathbb{R}}\phi(z)dz\int_{0}^{n}\int_{\mathbb{R}}p_{t-s}(x-z)^{2}{\gamma(\mu_{s},x)}X_{s}(dx)ds\Big{]}$
$\displaystyle\qquad\leq
K\mathbb{E}\Big{[}\int_{\mathbb{R}}\phi(z)dz\int_{0}^{t}\int_{\mathbb{R}}p_{t-s}(x-z)^{2}X_{s}(dx)ds\Big{]}$
$\displaystyle\qquad\leq
K\sup_{z\in\mathbb{R}}\phi(z)\int_{0}^{t}\frac{\mathbb{E}\left[\left<X_{s},1\right>\right]}{\sqrt{2\pi(t-s)}}{\mbox{\rm
d}}s=K\sup_{z\in\mathbb{R}}\phi(z)\sqrt{t}<\infty.$

Then by stochastic Fubini’s theorem (e.g., see Li [11, Theorem 7.24]), we get
$\left<X_{t},\phi\right>=\int_{\mathbb{R}}\mu_{t}(x)\phi(x)dx$ with

$\displaystyle\mu_{t}(x)=\int_{\mathbb{R}}p_{t}(x-z)\mu_{0}(z){\mbox{\rm
d}}z+\int_{0}^{t}\int_{\mathbb{R}}p_{t-s}(x-z)M(ds,dz).$

By El Karoui and Méléard [5, Theorem III-6], on some extension of the
probability space one can define a white noise $W(ds,dz)$ on
$\mathbb{R}_{+}\times\mathbb{R}$ based on $dsdz$ such that the following
holds:

$\displaystyle\mu_{t}(x)=\left<\mu_{0},p_{t}(x-\cdot)\right>+\int_{0}^{t}\int_{\mathbb{R}}\sqrt{\mu_{s}(z)\gamma(\mu_{s},z)}p_{t-s}(x-z)W(ds,dz),$
(2.1)

which implies (1.3).

Conversely, suppose that $\mu_{t}(x)$ satisfies (1.3) with $\mu_{0}\in
L^{2}(\mathbb{R})$ and denote $X_{t}(dx)=\mu_{t}(x)dx$. It follows from Lemma
2.2 that $X_{t}\in M(\mathbb{R})$ almost surely for every $t\geq 0$. For any
$\phi\in C_{b}^{2}(\mathbb{R})$ one can check that

$\displaystyle\left<X_{t},\phi\right>\\!\\!$ $\displaystyle=\\!\\!$
$\displaystyle\int_{\mathbb{R}}\mu_{t}(x)\phi(x)dx$ $\displaystyle=\\!\\!$
$\displaystyle\int_{\mathbb{R}}\mu_{0}(x)\phi(x)dx+\frac{1}{2}\int_{0}^{t}\int_{\mathbb{R}}\mu_{s}(x)\phi^{\prime\prime}(x)dxds$
$\displaystyle+\int_{0}^{t}\int_{\mathbb{R}}\phi(x)\sqrt{\mu_{s}(x)\gamma(\mu_{s},x)}W(ds,dx)$
$\displaystyle=\\!\\!$
$\displaystyle\left<X_{0},\phi\right>+\frac{1}{2}\int_{0}^{t}\left<X_{s},\phi^{\prime\prime}\right>ds+M_{t}(\phi)$

with $\left<M(\phi)\right>_{t}$ satisfying (1.2), which completes the proof. ∎

Now we show the existence of the solution to (1.3). For any $T>0,$ let $m\geq
1$ and $t_{k}=kT/m$ with $k=0,1,\cdots,m.$ Define a sequence of approximation
by

$\displaystyle\mu_{t}^{m}(x)=\mu_{0}(x)+\frac{1}{2}\int_{0}^{t}\Delta\mu_{s}^{m}(x)ds+\sum_{k=1}^{m}\sqrt{\gamma(\mu_{t_{k-1}}^{m},x)}\int_{t_{k-1}\wedge
t}^{t_{k}\wedge t}G_{m}(\mu_{s}^{m}(x))\dot{W}(s,x)ds,$ (2.2)

where

$\displaystyle G_{m}(x)=\int_{\mathbb{R}}p_{m^{-1}}(x-y)(\sqrt{|y|}\wedge
m)dy$

is a Lipschitz function on $[0,\infty)$ for fixed $m\geq 1$ and
$\lim_{m\rightarrow\infty}G_{m}(x)=\sqrt{x}$ for all $x\geq 0.$ For all $m\geq
1$ and $x\geq 0,$ one can check that there is a constant $K>0$ such that

$\displaystyle G_{m}(x)\\!\\!$ $\displaystyle\leq\\!\\!$
$\displaystyle\int_{\mathbb{R}}p_{m^{-1}}(x-y)(1+|y|)dy=1+\mathbb{E}[|B_{m^{-1}}^{x}|]\leq
1+{\left[\mathbb{E}\left(|B_{m^{-1}}^{x}|^{2}\right)\right]^{1/2}}$ (2.3)
$\displaystyle\leq\\!\\!$ $\displaystyle 1+(1+x^{2})^{1/2}\leq K(x+1)$ (2.4)

where $(B_{t}^{x})_{t\geq 0}$ is a Brownian motion with initial value $x.$ For
any $k=1,\cdots,m,$ conditioned on $\mathscr{F}_{t_{k-1}},$ it is well known
that there is a strong unique non-negative solution
$\\{\mu_{t}^{m}(x):t\in[t_{k-1},t_{k}),x\in\mathbb{R}\\}$ to

$\displaystyle\mu_{t}^{m}(x)=\mu_{t_{k-1}}^{m}(x)+\frac{1}{2}\int_{t_{k-1}}^{t}\Delta\mu_{s}^{m}(x)ds+\sqrt{\gamma(\mu_{t_{k-1}}^{m},x)}\int_{t_{k-1}}^{t}G_{m}(\mu_{s}^{m}(x))\dot{W}(s,x)ds,$

see [1, Theorem 5.1]. Therefore, the process $(\mu_{t}^{m}(x))_{t\geq
0,x\in\mathbb{R}}$ is the unique strong solution to (2.2).

Let $J(x)=\int_{\mathbb{R}}e^{-|y|}\rho(x-y)dy,$ where $\rho$ is the mollifier
given by $\rho(x)=C\exp\\{-1/(1-x^{2})\\}1_{\\{|x|<1\\}},$ and $C$ is a
constant such that $\int_{\mathbb{R}}\rho(x)dx=1.$ Then for any $n\geq 1,$
there are constants $c_{n}$ and $C_{n}$ such that

$\displaystyle c_{n}e^{-|x|}\leq J^{(n)}(x)\leq C_{n}e^{-|x|},\qquad\forall\
x\in\mathbb{R},$ (2.5)

see (2.1) of Mitoma [13]. We may and will replace $e^{-|x|}$ by $J(x)$ in the
definition of the space $\mathcal{X}_{0},$ and define
$\left<f,g\right>_{0}=\int_{\mathbb{R}}f(x)g(x)J(x)dx$ for any $f,g\in
C_{b}^{2}(\mathbb{R})$.

######  Lemma 2.4.

Assume that $\mu_{0}$ satisfies
$\int_{\mathbb{R}}\mu_{0}(x)^{2p}J(x)dx<\infty$ for some $p\geq 1$. Then for
every $T>0$, we have

$\displaystyle\sup_{0\leq t\leq T,m\geq
1}\mathbb{E}\left[\int_{\mathbb{R}}\mu_{t}^{m}(x)^{2p}J(x)dx\right]<\infty.$

###### Proof.

Using the convolution form, the solution $\mu_{t}^{m}(x)$ to (2.2) can be
represented as

$\displaystyle\mu_{t}^{m}(x)\\!\\!$ $\displaystyle=\\!\\!$
$\displaystyle\left<\mu_{0},p_{t}(x-\cdot)\right>$ (2.6)
$\displaystyle+\sum_{k=1}^{m}\int_{t_{k-1}\wedge t}^{t_{k}\wedge
t}\int_{\mathbb{R}}\sqrt{\gamma(\mu_{t_{k-1}}^{m},z)}G_{m}(\mu_{s}^{m}(z))p_{t-s}(x-z)W(ds,dz).$

By Hölder’s inequality we have

$\displaystyle\left|\int_{0}^{t}ds\int_{\mathbb{R}}\mu_{s}^{m}(z)^{2}p_{t-s}(x-z)^{2}dz\right|^{p}$
$\displaystyle\qquad\leq\left|\int_{0}^{t}\frac{1}{\sqrt{2\pi(t-s)}}ds\int_{\mathbb{R}}\mu_{s}^{m}(z)^{2}p_{t-s}(x-z)dz\right|^{p}$
$\displaystyle\qquad\leq
K\int_{0}^{t}\frac{1}{\sqrt{2\pi(t-s)}}ds\left|\int_{\mathbb{R}}\mu_{s}^{m}(z)^{2}p_{t-s}(x-z)dz\right|^{p}\cdot\left|\int_{0}^{t}\frac{1}{\sqrt{2\pi(t-s)}}ds\right|^{p/q}$
$\displaystyle\qquad\leq
K\int_{0}^{t}\frac{1}{\sqrt{2\pi(t-s)}}ds\int_{\mathbb{R}}\mu_{s}^{m}(z)^{2p}p_{t-s}(x-z)dz$

with $1/p+1/q=1$ and $p,q\geq 1.$ By the above inequality, (2.3) and
Burkholder-Davis-Gundy’s inequality one can see that

$\displaystyle\mathbb{E}\left[\int_{\mathbb{R}}J(x)dx\left|\sum_{k=1}^{m}\int_{t_{k-1}\wedge
t}^{t_{k}\wedge
t}\int_{\mathbb{R}}\sqrt{\gamma(\mu_{t_{k-1}}^{m},z)}G_{m}(\mu_{s}^{m}(z))p_{t-s}(x-z)W(ds,dz)\right|^{2p}\right]$
$\displaystyle\qquad\leq
K\mathbb{E}\left[\int_{\mathbb{R}}J(x)dx\left|\int_{0}^{t}\int_{\mathbb{R}}G_{m}(\mu_{s}^{m}(z))^{2}p_{t-s}(x-z)^{2}dsdz\right|^{p}\right]$
$\displaystyle\qquad\leq
K\mathbb{E}\left[\int_{\mathbb{R}}J(x)dx\left|\int_{0}^{t}ds\int_{\mathbb{R}}(\mu_{s}^{m}(z)^{2}+1)p_{t-s}(x-z)^{2}dz\right|^{p}\right]$
$\displaystyle\qquad{\leq
K+K\mathbb{E}\left[\int_{0}^{t}\frac{1}{\sqrt{2\pi(t-s)}}ds\int_{\mathbb{R}}J(x)dx\int_{\mathbb{R}}\mu_{s}^{m}(z)^{2p}p_{t-s}(x-z)dz\right]}$
$\displaystyle\qquad{\leq
K+K\mathbb{E}\left[\int_{0}^{t}\frac{1}{\sqrt{2\pi(t-s)}}ds\int_{\mathbb{R}}\mu_{s}^{m}(z)^{2p}dz\int_{\mathbb{R}}J(u+z)p_{t-s}(u)du\right]}$
$\displaystyle\qquad{\leq
K+K\int_{0}^{t}\frac{1}{\sqrt{t-s}}\mathbb{E}\left[\int_{\mathbb{R}}\mu_{s}^{m}(z)^{2p}J(z)dz\right]ds.}$

By (2.5) we have $J(u+z)\leq J(z)e^{|u|}$ which implies the last inequality,
since

$\displaystyle\int_{\mathbb{R}}e^{u}p_{t-s}(u)du\leq\int_{\mathbb{R}}e^{T|u|}p_{1}(u)du=\mathbb{E}\left[e^{T|B_{1}|}\right]\leq
K$

for $0\leq s\leq t\leq T,$ where $B_{t}$ is a standard Brownian motion.
Moreover, we have

$\displaystyle\int_{\mathbb{R}}\left<\mu_{0},p_{t}(x-\cdot)\right>^{2p}J(x)dx\\!\\!$
$\displaystyle=\\!\\!$
$\displaystyle\int_{\mathbb{R}}J(x)dx\left[\int_{\mathbb{R}}p_{t}(x-z)\mu_{0}(z)dz\right]^{2p}$
$\displaystyle\leq\\!\\!$
$\displaystyle\int_{\mathbb{R}}J(x)dx\int_{\mathbb{R}}p_{t}(x-z)\mu_{0}(z)^{2p}dz$
$\displaystyle\leq\\!\\!$ $\displaystyle
K\int_{\mathbb{R}}\mu_{0}(z)^{2p}J(z)dz<\infty.$

Combining the above inequality, we have

$\displaystyle\mathbb{E}\left[\int_{\mathbb{R}}\mu_{t}^{m}(x)^{2p}J(x)dx\right]\leq
K+K\int_{0}^{t}\frac{1}{\sqrt{t-s}}\mathbb{E}\left[\int_{\mathbb{R}}\mu_{s}^{m}(x)^{2p}J(x)dx\right]ds.$

Iterating the above once, one can check that

$\displaystyle\mathbb{E}\left[\int_{\mathbb{R}}\mu_{t}^{m}(x)^{2p}J(x)dx\right]\\!\\!$
$\displaystyle\leq\\!\\!$ $\displaystyle
K+K\int_{0}^{t}\mathbb{E}\left[\int_{\mathbb{R}}\mu_{r}^{m}(x)^{2p}J(x)dx\right]dr\int_{r}^{t}\frac{1}{\sqrt{(t-s)(s-r)}}ds$
$\displaystyle\leq\\!\\!$ $\displaystyle
K+K\int_{0}^{t}\mathbb{E}\left[\int_{\mathbb{R}}\mu_{r}^{m}(x)^{2p}J(x)dx\right]dr.$

The result follows by Gronwall’s inequality. ∎

We proceed to proving the tightness of $(\mu^{m}_{\cdot})$ in
$C([0,T]\times\mathbb{R}).$ Denote

$\displaystyle\nu_{t}^{m}(x)=\sum_{k=1}^{m}\int_{t_{k-1}\wedge t}^{t_{k}\wedge
t}\int_{\mathbb{R}}\sqrt{\gamma(\mu_{t_{k-1}}^{m},z)}G_{m}(\mu_{s}^{m}(z))p_{t-s}(x-z)W(ds,dz).$
(2.7)

######  Lemma 2.5.

Assume that $\mu_{0}$ satisfies
$\int_{\mathbb{R}}\mu_{0}(x)^{2p}J(x)dx<\infty$ for some $p\geq 1$. For fixed
$0<\alpha<1$ and $T>0,$ there is a constant $K$ such that

$\displaystyle\mathbb{E}[|\nu_{t}^{m}(x)-\nu_{r}^{m}(x)|^{2p}]\leq
K(t-r)^{\alpha p/2}$

for all $0<r<t\leq T.$

###### Proof.

Recall that $\gamma$ is bounded by Condition 1.1. By (2.3), (2.7) and
Burkholder-Davis-Gundy’s inequality one can check that

$\displaystyle\mathbb{E}\left[|\nu_{t}^{m}(x)-\nu^{m}_{r}(x)|^{2p}\right]\\!\\!$
$\displaystyle\leq\\!\\!$ $\displaystyle
K\mathbb{E}\left[\left|\int_{0}^{r}\int_{\mathbb{R}}\left[p_{r-s}(x-z)-p_{t-s}(x-z)\right]^{2}\left[\mu_{s}^{m}(z)+1\right]dsdz\right|^{p}\right]$
$\displaystyle+K\mathbb{E}\left[\Big{|}\int_{r}^{t}\int_{\mathbb{R}}[\mu_{s}^{m}(z)+1]p_{t-s}(x-z)^{2}dsdz\Big{|}^{p}\right]$
$\displaystyle=\\!\\!$ $\displaystyle I_{1}^{m}(t,r)+I_{2}^{m}(t,r).$

By [16, Lemma III4.5], for any $0<s<r<t\leq T$ and $\delta\in(0,1/4),$ there
is a constant $K=K(T)>0$ such that

$\displaystyle\left[p_{r-s}(x-z)-p_{t-s}(x-z)\right]^{2}$ (2.8)
$\displaystyle\qquad\leq
K(t-r)^{\delta}(r-s)^{-3\delta/2}\left[p_{r-s}(x-z)^{2-\delta}+p_{t-s}(x-z)^{2-\delta}\right].$
(2.9)

By [21, Lemma 1.4.4], we have

$\displaystyle\int_{0}^{r}\int_{\mathbb{R}}[p_{r-s}(x-z)-p_{t-s}(x-z)]^{2}dsdz\leq
K|t-r|^{1/2}$

and $\int_{r}^{t}\int_{\mathbb{R}}p_{t-s}(x-z)^{2}dsdz\leq K|t-r|^{1/2}.$ From
(2.8) and Lemma 2.4 one can obtain that

$\displaystyle\mathbb{E}\left[\int_{0}^{r}ds\int_{\mathbb{R}}[p_{r-s}(x-z)-p_{t-s}(x-z)]^{2}|\mu_{s}^{m}(z)|^{p}dz\right]$
$\displaystyle\quad\leq
K(t-r)^{\delta}\int_{0}^{r}(r-s)^{-3\delta/2}ds\mathbb{E}\left[\int_{\mathbb{R}}[p_{r-s}(x-z)^{2-\delta}+p_{t-s}(x-z)^{2-\delta}]|\mu_{s}^{m}(z)|^{p}dz\right]$
$\displaystyle\quad\leq
K(t-r)^{\delta}\int_{0}^{r}(r-s)^{-3\delta/2}ds\left(\int_{\mathbb{R}}[p_{r-s}(x-z)^{4-2\delta}+p_{t-s}(x-z)^{4-2\delta}]e^{|z|}dz\right)^{1/2}$
$\displaystyle\qquad\cdot\mathbb{E}\left[\left(\int_{\mathbb{R}}|\mu_{s}^{m}(z)|^{2p}J(z)dz\right)^{1/2}\right]$
$\displaystyle\quad\leq
K(t-r)^{\delta}\left[\int_{0}^{r}(r-s)^{-(3+4\delta)/4}ds+\int_{0}^{r}(r-s)^{-3\delta/2}(t-s)^{-(3-2\delta)/4}ds\right]$
$\displaystyle\quad\leq K(t-r)^{\delta}.$

By Hölder’s inequality, it implies that

$\displaystyle I_{1}^{m}(t,r)\\!\\!$ $\displaystyle\leq\\!\\!$ $\displaystyle
K\mathbb{E}\left[\int_{0}^{r}ds\int_{\mathbb{R}}[p_{r-s}(x-z)-p_{t-s}(x-z)]^{2}|\mu_{s}^{m}(z)|^{p}dz\right]$
$\displaystyle\quad\cdot\left[\int_{0}^{r}\int_{\mathbb{R}}[p_{r-s}(x-z)-p_{t-s}(x-z)]^{2}dsdz\right]^{p-1}+K(t-r)^{p/2}$
$\displaystyle\leq\\!\\!$ $\displaystyle
K(t-r)^{\delta+(p-1)/2}+K(t-r)^{p/2}.$ (2.11)

Similarly, by Hölder’s inequality and Lemma 2.4 one can see that

$\displaystyle\mathbb{E}\left[\int_{r}^{t}\int_{\mathbb{R}}|\mu_{s}^{m}(z)|^{p}p_{t-s}(x-z)^{2}dsdz\right]$
$\displaystyle\qquad\leq\mathbb{E}\left[\int_{r}^{t}ds\left[\int_{\mathbb{R}}|\mu_{s}^{m}(z)|^{2p}J(z)dz\right]^{1/2}\left[\int_{\mathbb{R}}p_{t-s}(x-z)^{4}e^{|z|}dz\right]^{1/2}\right]$
$\displaystyle\qquad\leq K\int_{r}^{t}(t-s)^{-3/4}ds=K(t-r)^{1/4},$

which implies that

$\displaystyle I_{2}^{m}(t,r)\\!\\!$ $\displaystyle\leq\\!\\!$ $\displaystyle
K\mathbb{E}\left[\int_{r}^{t}\int_{\mathbb{R}}|\mu_{s}^{m}(z)|^{p}p_{t-s}(x-z)^{2}dsdz\right]\left[\int_{r}^{t}\int_{\mathbb{R}}p_{t-s}(x-z)^{2}dsdz\right]^{p-1}$
(2.12) $\displaystyle+K(t-r)^{p/2}$ $\displaystyle\leq\\!\\!$ $\displaystyle
K(t-r)^{1/4+(p-1)/2}+K(t-r)^{p/2}.$

The result follows. ∎

Similar with above, we have the following result.

######  Lemma 2.6.

Assume that $\mu_{0}$ satisfies
$\int_{\mathbb{R}}\mu_{0}(x)^{2p}J(x)dx<\infty$ for some $p\geq 1$. For fixed
$0<\beta<1,$ there is a constant $K$ such that

$\displaystyle\mathbb{E}\left[|\nu_{t}^{m}(x)-\nu_{t}^{m}(y)|^{2p}\right]\leq
K|x-y|^{\beta p}$

for any $x,y\in\mathbb{R}$.

Proof of Theorem 1.2. By (2.6) one can check that

$\displaystyle\mu_{t}^{m}(x)\\!\\!$ $\displaystyle=\\!\\!$
$\displaystyle\left<\mu_{0},p_{t}(x-\cdot)\right>+\nu_{t}^{m}(x).$

By Kolmogorov’s criteria (see [10, Corollary 16.9]), Lemma 2.5 and Lemma 2.6,
for each fixed $T,K>0,$ the sequence of laws of
$\\{\mu_{t}^{m}(x):(t,x)\in[0,T]\times[-K,K]\\}$ on $C([0,T]\times[-K,K])$ is
tight, and hence, has a convergent subsequence. By the standard diagonlization
argument, there exists a subsequence $(\mu_{t}^{m_{k}},W_{t}^{m_{k}})$ which
converges to $(\mu_{t},W_{t})$ in law on $C([0,T]\times[-K,K])$ for each $K$
and $T.$ Therefore, $(\mu_{t}^{m_{k}},W_{t}^{n_{k}})_{t\geq 0}$ converges in
law as $k\rightarrow\infty.$ Applying Skorokhod’s representation (e.g. [6,
Theorem 1.8]), on another probability space, there are continuous processes
$(\hat{\mu}_{t}^{m_{k}},\hat{W}_{t}^{m_{k}})_{t\geq 0}$ and
$(\hat{\mu}_{t},\hat{W}_{t})_{t\geq 0}$ with the same distribution as
$(\mu_{t}^{m_{k}},W_{t}^{m_{k}})$ and $(\mu_{t},W_{t}),$ respectively.
Moreover,

$\displaystyle(\hat{\mu}_{t}^{m_{k}},\hat{W}_{t}^{m_{k}})_{t\geq
0}\rightarrow(\hat{\mu}_{t},\hat{W}_{t})_{t\geq 0}$

almost surely as $k\rightarrow\infty$. Recall that there is a unique strong
non-negative solution to (2.2). For any $f\in C_{b}^{2}(\mathbb{R})$ with
$|f|+|f^{\prime\prime}|\leq KJ$, we have

$\displaystyle\left<\hat{\mu}_{t}^{m_{k}},f\right>\\!\\!$
$\displaystyle=\\!\\!$
$\displaystyle\left<\mu_{0},f\right>+\frac{1}{2}\int_{0}^{t}\left<\hat{\mu}_{s}^{m_{k}},f^{\prime\prime}\right>ds$
$\displaystyle+\sum_{j=1}^{m_{k}}\int_{t_{j-1}\wedge t}^{t_{j}\wedge
t}\int_{\mathbb{R}}\sqrt{\gamma(\hat{\mu}_{t_{j-1}}^{m_{k}},z)}G_{m_{k}}(\hat{\mu}_{s}^{m_{k}}(z))f(z)\hat{W}^{m_{k}}(ds,dz).$

It follows from Lemma 2.4 that for every $t>0$, we have

$\displaystyle\lim_{k\rightarrow\infty}\mathbb{E}\left[\int_{\mathbb{R}}|\hat{\mu}_{t}^{m_{k}}(x)-\hat{\mu}_{t}(x)|^{2}J(x)dx\right]=0$
(2.13)

and

$\displaystyle\sup_{0\leq t\leq T,k\geq
1}\mathbb{E}\left[\int_{\mathbb{R}}[\hat{\mu}_{t}(x)+\hat{\mu}_{t}^{m_{k}}(x)]^{2}J(x)dx\right]<\infty,$
(2.14)

since $\mu_{0}(x)\in L^{2}(\mathbb{R}).$ Then by the above and dominated
convergence theorem, we have

$\displaystyle\mathbb{E}\left[\int_{0}^{t}\langle|\hat{\mu}_{s}^{m_{k}}-\hat{\mu}_{s}|,f^{\prime\prime}\rangle
ds\right]\leq\mathbb{E}\left[\int_{0}^{t}ds\int_{\mathbb{R}}|\hat{\mu}_{s}^{m_{k}}(x)-\hat{\mu}_{s}(x)|^{2}J(x)dx\right]^{1/2}\rightarrow
0$

as $k\rightarrow\infty.$ Recall that $\gamma$ satisfies Condition 1.1. Thus,

$\displaystyle\mathbb{E}\left[\left|\int_{0}^{t}\int_{\mathbb{R}}\sqrt{\gamma(\hat{\mu}_{s},z)\hat{\mu}_{s}(z)}f(z)\hat{W}^{m_{k}}(ds,dz)\right|^{2}\right]$
$\displaystyle\quad\leq
K\mathbb{E}\left[\int_{0}^{t}\int_{\mathbb{R}}\hat{\mu}_{s}(z)J(z)dsdz\right]\leq
K\left\\{\mathbb{E}\left[\int_{0}^{t}\int_{\mathbb{R}}\hat{\mu}_{s}(z)^{2}J(z)dsdz\right]\right\\}^{1/2}<\infty.$

It thus follows from [23, Lemma 2.4] that

$\displaystyle\left<\hat{\mu}_{t},f\right>\\!\\!$ $\displaystyle=\\!\\!$
$\displaystyle\left<\mu_{0},f\right>+\frac{1}{2}\int_{0}^{t}\left<\hat{\mu}_{s},f^{\prime\prime}\right>ds+\int_{0}^{t}\int_{\mathbb{R}}\sqrt{\gamma(\hat{\mu}_{s},z)\hat{\mu}_{s}(z)}f(z)\hat{W}(ds,dz).$

That completes the existence of solution to (1.3). By Proposition 2.3 one can
see that $X_{t}(dx)=\mu_{t}(x)dx$ satisfies the MP (1.1, 1.2), which implies
the conclusion. ∎

Proof of Theorem 1.3. Assume that $(\mu_{t}(x))_{t\geq 0,x\in\mathbb{R}}$ is a
solution to SPDE (1.3). Then it also satisfies (2.1) by Proposition 2.3. For
any $0<r\leq t$ and $p\geq 1$, we have

$\displaystyle\mathbb{E}\big{[}|\mu_{t}(x)-\mu_{r}(x)|^{2p}\big{]}\\!\\!$
$\displaystyle\leq\\!\\!$ $\displaystyle
K\Big{|}\int_{\mathbb{R}}[p_{t}(x-z)-p_{r}(x-z)]\mu_{0}(z)dz\Big{|}^{2p}$
$\displaystyle+K\mathbb{E}\left[\Big{|}\int_{r}^{t}\int_{\mathbb{R}}p_{t-s}(x-z)\sqrt{\mu_{s}(z)}W(ds,dz)\Big{|}^{2p}\right]$
$\displaystyle+K\mathbb{E}\left[\Big{|}\int_{0}^{r}\int_{\mathbb{R}}[p_{r-s}(x-z)-p_{t-s}(x-z)]\sqrt{\mu_{s}(z)}W(ds,dz)\Big{|}^{2p}\right]$
$\displaystyle=:\\!\\!$ $\displaystyle K(I_{1}+I_{2}+I_{3}).$

Recall that $\mu_{0}$ is Hölder continuous with exponent
$\lambda<\frac{1}{2}$. Similar with the proof of [8, Theorem 1.1], by Hölder’s
inequality one can see that

$\displaystyle I_{1}\\!\\!$ $\displaystyle=\\!\\!$
$\displaystyle\Big{|}\int_{\mathbb{R}}[p_{t}(x-z)-p_{r}(x-z)]\mu_{0}(z)dz\Big{|}^{2p}$
$\displaystyle=\\!\\!$
$\displaystyle\Big{|}\mathbb{E}\big{[}\mu_{0}(x+B_{t})-\mu_{0}(x+B_{r})\big{]}\Big{|}^{2p}\leq
K\mathbb{E}\left[|B_{t-r}|^{2p\lambda}\right]$ $\displaystyle\leq\\!\\!$
$\displaystyle
K\int_{\mathbb{R}}|x|^{2p\lambda}\frac{1}{\sqrt{2\pi(t-r)}}e^{-\frac{x^{2}}{2(t-r)}}dx\leq
K(t-r)^{\lambda p},$

where $(B_{t})_{t\geq 0}$ is a standard Brownian motion. Moreover, similar
with (2) and (2.12) one can see that

$\displaystyle I_{2}\leq
K\mathbb{E}\left[\Big{|}\int_{r}^{t}\int_{\mathbb{R}}p_{t-s}(x-z)^{2}\mu_{s}(z)dsdz\Big{|}^{p}\right]\leq
K(t-r)^{1/4+(p-1)/2},$

and

$\displaystyle I_{3}\\!\\!$ $\displaystyle\leq\\!\\!$ $\displaystyle
K\mathbb{E}\left[\Big{|}\int_{0}^{r}\int_{\mathbb{R}}[p_{r-s}(x-z)-p_{t-s}(x-z)]^{2}\mu_{s}(z)dsdz\Big{|}^{p}\right]\leq
K(t-r)^{\delta+(p-1)/2}$

with $\delta\in(0,1/4).$ Then there exists $\alpha\in(0,1)$ such that

$\displaystyle\mathbb{E}\big{[}|\mu_{t}(x)-\mu_{r}(x)|^{2p}\big{]}\\!\\!$
$\displaystyle\leq\\!\\!$ $\displaystyle K(t-r)^{\lambda
p}+K(t-r)^{1/4+(p-1)/2}+K(t-r)^{\delta+(p-1)/2}$ (2.15)
$\displaystyle\leq\\!\\!$ $\displaystyle K(t-r)^{\alpha p/2}.$

Similarly, for any $x,y\in\mathbb{R}$ we have

$\displaystyle\mathbb{E}\left[|\mu_{t}(x)-\mu_{t}(y)|^{2p}\right]\leq
K|x-y|^{2p\lambda}+K|x-y|^{\beta p}$ (2.16)

with $\beta\in(0,1).$ For $t,r\in[0,T]$ and $x,y\in\mathbb{R},$ notice that

$\displaystyle\mathbb{E}\left[|\mu_{t}(x)-\mu_{r}(y)|^{2p}\right]\\!\\!$
$\displaystyle\leq\\!\\!$ $\displaystyle
K\mathbb{E}\left[|\mu_{t}(x)-\mu_{t}(x)|^{2p}\right]+K\mathbb{E}\left[|\mu_{t}(x)-\mu_{t}(y)|^{2p}\right].$

Combining with the above, (2.15) and (2.16), the result follows from
Kolmogorov’s continuity criteria (see e.g. [19, Corollary 1.2]). ∎

##  3 Proof of Theorem 1.4

In this section we show the weak uniqueness of the solution to the MP (1.1,
1.2) under Condition 1.1. The main idea of uniqueness is to relate the MP
(1.1, 1.2) with a system of SPDEs, which is satisfied by a sequence of
corresponding function-valued processes $(u_{t}^{i})_{t\geq 0},\
i=0,1,\cdots,n.$ The weak uniqueness of the solution to the MP (1.1, 1.2)
follows from the pathwise uniqueness of the solution to the system of SPDEs.

For a solution $(X_{t})_{t\geq 0}$ to the MP (1.1, 1.2), we define the
function-valued processes as

$\displaystyle u_{t}^{i}(x)=X_{t}((a_{i},x]),\qquad a_{i}\leq x<a_{i+1},\
i=0,\cdots,n.$ (3.1)

In Proposition 3.1 we show that (3.1) is a solution to the following system of
SPDEs:

$\displaystyle\begin{split}u_{t}^{i}(x)&=u_{0}^{i}(x)+\int_{0}^{t}\frac{1}{2}\Delta
u_{s}^{i}(x)ds+\int_{0}^{t}\int_{0}^{u_{s}^{i}(x)}g_{i}(\nabla
u^{i}_{s}(a_{i+1}))W_{i}(ds,dz),\\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad x\in[a_{i},a_{i+1}),\
i=0,\cdots,n-1;\end{split}$ (3.2)
$\displaystyle\begin{split}u_{t}^{n}(x)&=u_{0}^{n}(x)+\int_{0}^{t}\frac{1}{2}\Delta
u_{s}^{n}(x)ds+\int_{0}^{t}\int_{0}^{u_{s}^{n}(x)}g_{n}(u_{s}^{n}(\infty))W_{n}(ds,dz),\\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
x\in[a_{n},\infty).\end{split}$ (3.3)
$\displaystyle\begin{split}&u^{i}(a_{i})=0,\;\;\;i=0,1,\cdots,n,\\\ &\nabla
u^{i-1}_{t}(a_{i})=\nabla u^{i}_{t}(a_{i}),\;\;\;i=1,\cdots,n.\end{split}$
(3.4)

where $u_{s}^{n}(\infty):=\lim_{x\rightarrow\infty}u_{s}^{n}(x)$ and
$W_{i}({\mbox{\rm d}}s,{\mbox{\rm d}}z),\ i=0,\cdots,n$ are independent time-
space white noises on $\mathbb{R}_{+}\times\mathbb{R}_{+}$ with intensity
${\mbox{\rm d}}s{\mbox{\rm d}}z$. The pathwise uniqueness of the solution to
the above system of SPDEs is obtained in Proposition 3.3.

The system of SPDEs (3.2, 3.3) can be understood in the following form: for
any $\phi_{i}\in C_{b}^{2}[a_{i},a_{i+1}]$ with
$\phi_{i}(a_{i})=\phi^{\prime}_{i}(a_{i+1})=0,\ i=0,1,\cdots,n-1,$ and
$\phi_{n}\in C_{b}^{2}[a_{n},\infty)$ with
$\phi_{n}(a_{n})=\phi_{n}(\infty)=0$ (given
$\phi_{n}(\infty):=\lim_{x\rightarrow\infty}\phi_{n}(x)$), we have

$\displaystyle\begin{split}\langle u_{t}^{i},\phi_{i}\rangle&=\langle
u_{0}^{i},\phi_{i}\rangle+\frac{1}{2}\int_{0}^{t}\big{[}\langle
u_{s}^{i},\phi_{i}^{\prime\prime}\rangle+\phi_{i}(a_{i+1})\nabla
u^{i}_{s}(a_{i+1})\big{]}ds\\\
&\qquad+\int_{0}^{t}\int_{0}^{\infty}\int_{a_{i}}^{a_{i+1}}1_{\\{z\leq
u^{i}_{s}(x)\\}}\phi_{i}(x)dxg_{i}(\nabla
u^{i}_{s}(a_{i+1}))W_{i}(ds,dz);\end{split}$ (3.5)
$\displaystyle\begin{split}\langle u_{t}^{n},\phi_{n}\rangle&=\langle
u_{0}^{n},\phi_{n}\rangle+\frac{1}{2}\int_{0}^{t}\langle
u_{s}^{n},\phi_{n}^{\prime\prime}\rangle ds\\\
&\qquad+\int_{0}^{t}\int_{0}^{\infty}\int_{a_{n}}^{\infty}1_{\\{z\leq
u^{n}_{s}(x)\\}}\phi_{n}(x)dxg_{n}\big{(}u_{s}^{n}(\infty)\big{)}W_{n}(ds,dz).\end{split}$
(3.6)

######  Proposition 3.1.

Suppose that $(X_{t})_{t\geq 0}$ is a solution to the MP (1.1, 1.2). Then
$\\{u_{t}^{i}:t\geq 0,x\in[a_{i},a_{i+1})\\},i=0,1,\cdots,n$ defined as (3.1),
solves the group of SPDEs (3.2,3.3) with boundary condition (LABEL:0.7).

###### Proof.

For any $\phi_{i}\in C^{3}_{c}(a_{i},a_{i+1})$ with $i=0,1,\cdots,n-1$, by
integration by parts, we have

$\displaystyle\left<u^{i}_{t},\phi_{i}^{\prime}\right>\\!\\!$
$\displaystyle=\\!\\!$
$\displaystyle-\left<X_{t},\phi_{i}\right>=-M_{t}(\phi_{i})-\left<X_{0},\phi_{i}\right>-\frac{1}{2}\int^{t}_{0}\left<X_{s},\phi_{i}^{\prime\prime}\right>ds$
$\displaystyle=\\!\\!$ $\displaystyle-
M_{t}(\phi_{i})+\left<u^{i}_{0},\phi_{i}^{\prime}\right>+\frac{1}{2}\int^{t}_{0}\left<u^{i}_{s},\phi_{i}^{\prime\prime\prime}\right>ds.$

Thus

$-M_{t}(\phi_{i})=\left<u^{i}_{t},\phi_{i}^{\prime}\right>-\left<u^{i}_{0},\phi_{i}^{\prime}\right>-\frac{1}{2}\int^{t}_{0}\left<u^{i}_{s},(\phi_{i}^{\prime})^{\prime\prime}\right>ds$
(3.7)

is a continuous martingale. By Lemma 2.1 we have

$\displaystyle-
M_{t}(\phi_{i})=\int_{0}^{t}\int_{\mathbb{R}}\phi_{i}(x)M(ds,dx)$

with

$\displaystyle\left<-M(\phi_{i})\right>_{t}\\!\\!$ $\displaystyle=\\!\\!$
$\displaystyle\int^{t}_{0}g_{i}(\nabla
u^{i}_{s}(a_{i+1}))^{2}ds\int_{\mathbb{R}}\phi_{i}(x)^{2}X_{s}(dx)$
$\displaystyle=\\!\\!$ $\displaystyle\int^{t}_{0}g_{i}(\nabla
u^{i}_{s}(a_{i+1}))^{2}ds\int_{0}^{u_{s}^{i}(a_{i+1})}\phi_{i}(u_{s}^{i}(y)^{-1})^{2}dy$
$\displaystyle=\\!\\!$
$\displaystyle\int^{t}_{0}\int^{\infty}_{0}\left(\int_{a_{i}}^{a_{i+1}}1_{\\{y\leq
u^{i}_{s}(x)\\}}\phi_{i}^{\prime}(x)dx\right)^{2}g_{i}(\nabla
u^{i}_{s}(a_{i+1}))^{2}dsdy,$

where $u_{s}^{i}(y)^{-1}$ denotes the generalized inverse of the nondecreasing
function $u_{s}^{i},$ that is,

$\displaystyle u_{s}^{i}(y)^{-1}=\sup\\{x\in[a_{i},a_{i+1}):u_{s}^{i}(x)\leq
y\\}.$

Moreover, for $\phi_{n}\in C_{c}^{3}(a_{n},\infty)$, one can see that

$-M_{t}(\phi_{n})=\left<u^{n}_{t},\phi_{n}^{\prime}\right>-\left<u^{n}_{0},\phi_{n}^{\prime}\right>-\frac{1}{2}\int^{t}_{0}\left<u^{n}_{s},\phi_{n}^{\prime\prime\prime}\right>ds$
(3.8)

is a continuous martingale with

$\displaystyle\left<-M(\phi_{n})\right>_{t}\\!\\!$ $\displaystyle=\\!\\!$
$\displaystyle\int^{t}_{0}g_{n}(u_{s}^{n}(\infty))^{2}ds\int_{\mathbb{R}}\phi_{n}(x)^{2}X_{s}(dx)$
$\displaystyle=\\!\\!$
$\displaystyle\int^{t}_{0}\int^{\infty}_{0}\left(\int_{a_{n}}^{\infty}1_{\\{y\leq
u^{n}_{s}(x)\\}}\phi_{n}^{\prime}(x)dx\right)^{2}g_{n}(u_{s}^{n}(\infty))^{2}dsdy.$

Similar with Lemma 2.1, the family $\\{-M_{t}(\phi_{i}):t\geq 0,\phi_{i}\in
C_{c}^{3}(a_{i},a_{i+1})\\}$ determines a martingale measure
$\\{M_{t}(B):t\geq 0,B\in\mathscr{B}(a_{i},a_{i+1})\\}.$ Moreover, for
$\phi_{i}\in C_{c}^{3}(a_{i},a_{i+1})$ and $\phi_{j}\in
C_{c}^{3}(a_{j},a_{j+1})$ with $i\neq j,$ we have

$\displaystyle\left<-M(\phi_{i}),-M(\phi_{j})\right>_{t}\\!\\!$
$\displaystyle=\\!\\!$ $\displaystyle\int_{0}^{t}{\mbox{\rm
d}}s\int_{\mathbb{R}}\gamma(X_{s},z)X_{s}(dz)\int_{\mathbb{R}}\phi_{i}(x)\delta_{z}(dx)\int_{\mathbb{R}}\phi_{j}(y)\delta_{z}(dy)$
$\displaystyle=\\!\\!$ $\displaystyle\int_{0}^{t}{\mbox{\rm
d}}s\int_{\mathbb{R}}\gamma(X_{s},z)\phi_{i}(z)\phi_{j}(z)X_{s}(dz)=0.$

By El Karoui and Méléard [5, Theorem III-7, Corollary III-8], on some
extension of the probability space on can define a sequence independent
Gaussian white noise $W_{i}(ds,du),i=0,\cdots,n$ on $(0,\infty)^{2}$ based on
$dsdu$ such that

$\displaystyle-M_{t}(\phi_{i})\\!\\!$ $\displaystyle=\\!\\!$
$\displaystyle\int^{t}_{0}\int^{\infty}_{0}\int_{a_{i}}^{a_{i+1}}1_{\\{z\leq
u^{i}_{s}(x)\\}}\phi_{i}^{\prime}(x)\sqrt{\gamma(X_{s},x)}dxW_{i}(ds,dz)$
(3.9) $\displaystyle=\\!\\!$
$\displaystyle\int_{a_{i}}^{a_{i+1}}\phi_{i}^{\prime}(x)dx\int^{t}_{0}\int^{u^{i}_{s}(x)}_{0}g_{i}(\nabla
u^{i}_{s}(a_{i+1}))W_{i}(ds,dz)$

for any $\phi_{i}\in C_{c}^{3}(a_{i},a_{i+1}),i=0,\cdots,n-1,$ and

$\displaystyle-M_{t}(\phi_{n})\\!\\!$ $\displaystyle=\\!\\!$
$\displaystyle\int_{a_{n}}^{\infty}\phi_{n}^{\prime}(x)dx\int^{t}_{0}\int^{u^{n}_{s}(x)}_{0}g_{n}(u^{n}_{s}(\infty))W_{n}(ds,dz)$

for any $\phi_{n}\in C_{c}^{3}(a_{n},\infty).$ Then (3.5) and (3.6) follow
from the approximation method and Lemma 4.2 in Appendix. ∎

######  Lemma 3.2.

Suppose that $(u_{t}^{n})_{t\geq 0}$ satisfies (3.3). Then
$(u_{t}^{n}(\infty))_{t\geq 0}$ satisfies

$\displaystyle
u_{t}^{n}(\infty)=u_{0}^{n}(\infty)+\int_{0}^{t}\int_{0}^{u_{s}^{n}(\infty)}g_{n}(u_{s}^{n}(\infty))W_{n}(ds,dz).$
(3.10)

###### Proof.

Recall that $p_{t}(x)=\frac{1}{\sqrt{2\pi}t}{\mbox{\rm e}}^{-x^{2}/(2t)}$ and
let

$\displaystyle q_{t}^{x}(y):=p_{t}(x+a_{n}-y)-p_{t}(x-a_{n}+y)$

for $t>0$ and $x,y\geq a_{n}$. Then (3.3) can be written into the following
mild form:

$\displaystyle u_{t}^{n}(x)\\!\\!$ $\displaystyle=\\!\\!$
$\displaystyle\left<u_{0}^{n},q_{t}^{x}\right>$ (3.11)
$\displaystyle+\int_{0}^{t}\int_{0}^{\infty}\Big{[}\int_{a_{n}}^{\infty}1_{\\{z\leq
u_{s}^{n}(y)\\}}q_{t-s}^{x}(y){\mbox{\rm
d}}y\Big{]}g_{n}(u_{s}^{n}(\infty))W_{n}(ds,dz)$

for any $x\geq a_{n}.$ By a change of variable, we have

$\displaystyle\left<u_{0}^{n},q_{t}^{x}\right>\\!\\!$ $\displaystyle=\\!\\!$
$\displaystyle\int_{a_{n}}^{\infty}u_{0}^{n}(y)[p_{t}(x+a_{n}-y)-p_{t}(x-a_{n}+y)]dy$
$\displaystyle=\\!\\!$
$\displaystyle\int_{-\infty}^{x}u_{0}^{n}(x+a_{n}-z)p_{t}(z)dz-\int_{x}^{\infty}u_{0}^{n}(z-x+a_{n})p_{t}(z)dz$
$\displaystyle\to\\!\\!$ $\displaystyle
u_{0}^{n}(\infty)\int_{-\infty}^{\infty}p_{t}(z)dz=u_{0}^{n}(\infty)$

and

$\displaystyle\int_{a_{n}}^{\infty}1_{\\{z\leq
u_{s}^{n}(y)\\}}q_{t-s}^{x}(y)dy\to 1_{\\{z\leq u_{s}^{n}(\infty)\\}}$

as $x\to\infty,$ which ends the proof. ∎

######  Proposition 3.3.

(Pathwise uniqueness) Suppose that $(u_{t})_{t\geq 0}$ and
$(\tilde{u}_{t})_{t\geq 0}$ are two solutions to (3.2, 3.3) satisfying the
boundary conditions (LABEL:0.7). If $u_{0}(x)=\tilde{u}_{0}(x)$ for all
$x\in\mathbb{R},$ then $\mathbb{P}\\{u_{t}(x)=\tilde{u}_{t}(x)\ \text{for
all}\ t\geq 0\ \text{and}\ x\in\mathbb{R}\\}=1.$

###### Proof.

The pathwise uniqueness of the solution to (3.10) holds by [3, Theorem 2.1].
Moreover, the pathwise uniqueness of the solution $\\{u_{t}^{n}(x):t\geq
0,x\in[a_{n},\infty)\\}$ holds for (3.3) by [24, Theorem 1.4]. That implies
the strong uniqueness of $(\nabla u_{t}^{n}(a_{n}))_{t\geq 0}.$ By the proof
of [24, Theorem 1.4] and induction method, one can get the pathwise uniqueness
of the solution to (3.2) for $i=0,1,\cdots,n-1.$ The proof ends here. ∎

Proof of Theorem 1.4 The result is a direct conclusion of Proposition 3.3. ∎

##  4 Appendix

In this section we give some results about the process $(u_{t}(x))_{t\geq
0,x\in[0,1]}$ satisfying the following SPDE:

$\displaystyle u_{t}(x)=u_{0}(x)+\int_{0}^{t}\frac{1}{2}\Delta
u_{s}(x)ds+\int_{0}^{t}\int_{0}^{u_{s}(x)}g(\nabla u_{s}(1))W(ds,dz),$

where $g$ is a positive continuous bounded function from $\mathbb{R}_{+}$ to
$\mathbb{R}_{+}$, and $W(ds,dz)$ is a time-space Gaussian white noise with
density $dsdz$. Let $\Phi\in C_{c}^{2}(0,1)$ satisfies $0\leq\Phi\leq 2$ and
$\int_{0}^{1}\Phi(x){\mbox{\rm d}}x=1$. For $k\geq 1$ and $x\in[0,1]$ let

$\displaystyle h_{k}(x):=\int_{0}^{kx}\Phi(z){\mbox{\rm
d}}z\cdot\int_{x^{k}}^{1}\Phi(z){\mbox{\rm d}}z.$ (4.1)

Then $h_{k}\in C_{c}^{2}(0,1)$ for all $k\geq 1$.

######  Lemma 4.1.

Suppose that $f\in C[0,1]$ with $f^{\prime}(1)$ and $f^{\prime}(0)$ exist.
Then

$\displaystyle\lim_{k\to\infty}\left<f,h_{k}^{\prime}\right>=f(0)-f(1),\quad\lim_{k\to\infty}\left<f,h_{k}^{\prime\prime}\right>=f^{\prime}(1)-f^{\prime}(0).$

###### Proof.

Observe that for each $n\geq 1$,

$\displaystyle h_{k}^{\prime}(x)=k\Phi(kx)\int_{x^{k}}^{1}\Phi(z)dz-
kx^{k-1}\Phi(x^{k})\int_{0}^{kx}\Phi(z)dz,\qquad x\in[0,1].$

Then by change of variables and dominated convergence, as $k\to\infty$,

$\displaystyle\left<f,h_{k}^{\prime}\right>\\!\\!$ $\displaystyle=\\!\\!$
$\displaystyle\int_{0}^{1}f(x)k\Phi(kx)\Big{[}\int_{x^{k}}^{1}\Phi(z)dz\Big{]}dx-\int_{0}^{1}f(x)kx^{k-1}\Phi(x^{k})\Big{[}\int_{0}^{kx}\Phi(z)dz\Big{]}dx$
$\displaystyle=\\!\\!$
$\displaystyle\int_{0}^{1}f(y/k)\Phi(y)\Big{[}\int_{(y/k)^{k}}^{1}\Phi(z)dz\Big{]}dy-\int_{0}^{1}f(y^{1/k})\Phi(y)\Big{[}\int_{0}^{ky^{1/k}}\Phi(z)dz\Big{]}dy$

converges to $f(0)-f(1)$, which gives the first assertion.

In the following we prove the second assertion. Observe that

$\displaystyle h_{k}^{\prime\prime}(x)\\!\\!$ $\displaystyle=\\!\\!$
$\displaystyle k^{2}\Phi^{\prime}(kx)\int_{x^{k}}^{1}\Phi(z){\mbox{\rm
d}}z-2k^{2}x^{k-1}\Phi(kx)\Phi(x^{k})$ (4.2)
$\displaystyle-\big{[}k(k-1)x^{k-2}\Phi(x^{k})+k^{2}x^{2k-2}\Phi^{\prime}(x^{k})\big{]}\int_{0}^{kx}\Phi(z)dz$
$\displaystyle=:\\!\\!$ $\displaystyle M_{1,k}(x)-2M_{2,k}(x)-M_{3,k}(x).$

By change of variables and dominated convergence again, as $k\to\infty$,

$\displaystyle\int_{0}^{1}[f(x)-f(0)]M_{1,k}(x)dx\\!\\!$
$\displaystyle=\\!\\!$
$\displaystyle\int_{0}^{1}k[f(y/k)-f(0)]\Phi^{\prime}(y)\Big{[}\int_{y^{k}k^{-k}}^{1}\Phi(z)dz\Big{]}dy$
(4.3) $\displaystyle\to\\!\\!$ $\displaystyle
f^{\prime}(0)\int_{0}^{1}y\Phi^{\prime}(y)dy=-f^{\prime}(0)$

and

$\displaystyle\int_{0}^{1}[f(x)-f(1)]M_{2,k}(x)dx\\!\\!$
$\displaystyle=\\!\\!$
$\displaystyle\int_{0}^{1}\frac{f(y^{1/k})-f(1)}{y^{1/k}-1}k(y^{1/k}-1)\Phi(y^{1/k})\Phi(y)dy$
(4.4) $\displaystyle\to\\!\\!$ $\displaystyle 0.$

Similarly, as $k\to\infty$,

$\displaystyle\int_{0}^{1}[f(x)-f(1)]M_{3,k}(x)dx$ (4.5)
$\displaystyle\quad=\int_{0}^{1}\frac{f(y^{1/k})-f(1)}{y^{1/k}-1}k(y^{1/k}-1)\big{[}k^{-1}(k-1)y^{-1/k}\Phi(y)$
$\displaystyle\quad\qquad+y^{(k-1)/k}\Phi^{\prime}(y)\big{]}\cdot\Big{[}\int_{0}^{ky^{1/k}}\Phi(z)dz\Big{]}dy$
$\displaystyle\quad\to f^{\prime}(1)\int_{0}^{1}\ln
y[\Phi(y)+y\Phi^{\prime}(y)]dy=-f^{\prime}(1).$

Applying integration by parts and the fact $0\leq\Phi\leq 2$ and
$\mbox{supp}(\Phi)\subset(0,1)$,

$\displaystyle\int_{0}^{1}M_{1,k}(x)dx\\!\\!$ $\displaystyle=\\!\\!$
$\displaystyle\int_{0}^{1}\Big{(}\int_{0}^{kx}\Phi(z)dz\Big{)}^{\prime\prime}\cdot\Big{(}\int_{x^{k}}^{1}\Phi(z)dz\Big{)}dx=\int_{0}^{1}M_{2,k}(x)dx$
$\displaystyle=\\!\\!$ $\displaystyle
k\int_{0}^{1}\Phi(ky^{1/k})\Phi(y)dy=k\int_{0}^{k^{-k}}\Phi(ky^{1/k})\Phi(y)dy\leq
4k^{1-k}$

and

$\displaystyle\int_{0}^{1}M_{3,k}(x)dx\\!\\!$ $\displaystyle=\\!\\!$
$\displaystyle-\int_{0}^{1}\Big{(}\int_{x^{k}}^{1}\Phi(z)dz\Big{)}^{\prime\prime}\cdot\Big{(}\int_{0}^{kx}\Phi(z)dz\Big{)}dx$
$\displaystyle=\\!\\!$ $\displaystyle\int_{0}^{1}M_{2,k}(x)dx\leq 4k^{1-k}.$

Then combining (4.2) with (4.3)-(4.5) one completes the proof. ∎

######  Lemma 4.2.

Suppose that for each $\phi\in C_{c}^{2}(0,1)$, $(u_{t})_{t\geq 0}$ satisfies

$\displaystyle\left<u_{t},\phi\right>\\!\\!$ $\displaystyle=\\!\\!$
$\displaystyle\left<u_{0},\phi\right>+\frac{1}{2}\int_{0}^{t}\left<u_{s},\phi^{\prime\prime}\right>ds$
(4.6) $\displaystyle+\int_{0}^{t}\int_{0}^{\infty}g(\nabla
u_{s}(1))\Big{[}\int_{0}^{1}1_{\\{z\leq u_{s}(x)\\}}\phi(x)dx\Big{]}W(ds,dz).$

Then for each $\phi\in C_{b}^{2}[0,1]$,

$\displaystyle\left<u_{t},\phi\right>\\!\\!$ $\displaystyle=\\!\\!$
$\displaystyle\left<u_{0},\phi\right>+\frac{1}{2}\int_{0}^{t}\big{[}\left<u_{s},\phi^{\prime\prime}\right>+F_{s}(\phi)]ds$
(4.7) $\displaystyle+\int_{0}^{t}\int_{0}^{\infty}g(\nabla
u_{s}(1))\Big{[}\int_{0}^{1}1_{\\{z\leq u_{s}(x)\\}}\phi(x){\mbox{\rm
d}}x\Big{]}W(ds,dz),$

where

$\displaystyle F_{s}(\phi):=[\phi(1)\nabla u_{s}(1)-\phi(0)\nabla
u_{s}(0)]-[u_{s}(1)\phi^{\prime}(1)-u_{s}(0)\phi^{\prime}(0)].$ (4.8)

###### Proof.

Recall $h_{k}$ in (4.1). For $m\geq 1$ define stopping time $\tau_{m}$ by

$\displaystyle\tau_{m}:=\inf\Big{\\{}t\geq 0:\sup_{x\in[0,1]}|u_{t}(x)|\geq
m\Big{\\}}$

with the convention $\inf\emptyset=\infty$. Then
$\lim_{m\to\infty}\tau_{m}=\infty$ almost surely. It follows from (4.6) that

$\displaystyle\left<u_{t\wedge\tau_{m}},\phi h_{k}\right>\\!\\!$
$\displaystyle=\\!\\!$
$\displaystyle\int_{0}^{t\wedge\tau_{m}}\int_{0}^{\infty}g(\nabla
u_{s}(1))\Big{[}\int_{0}^{1}1_{\\{z\leq
u_{s}(x)\\}}\phi(x)h_{k}(x)dx\Big{]}W(ds,dz)$ (4.9)
$\displaystyle+\left<u_{0},\phi
h_{k}\right>+\frac{1}{2}\int_{0}^{t\wedge\tau_{m}}\left<u_{s},(\phi
h_{k})^{\prime\prime}\right>ds.$

Notice that

$\displaystyle\left<u_{s},(\phi
h_{k})^{\prime\prime}\right>=\left<u_{s},\phi^{\prime\prime}h_{k}\right>+2\left<u_{s},\phi^{\prime}h^{\prime}_{k}\right>+\left<u_{s},\phi
h^{\prime\prime}_{k}\right>.$

It follows from Lemma 4.1 that

$\displaystyle\lim_{k\to\infty}\left<u_{s},(\phi
h_{k})^{\prime\prime}\right>\\!\\!$ $\displaystyle=\\!\\!$
$\displaystyle\left<u_{s},\phi^{\prime\prime}\right>+[u_{s}(0)\phi^{\prime}(0)-u_{s}(1)\phi^{\prime}(1)]-[\phi(0)\nabla
u_{s}(0)-\phi(1)\nabla u_{s}(1)]$ $\displaystyle=\\!\\!$
$\displaystyle\left<u_{s},\phi^{\prime\prime}\right>+F_{s}(\phi).$

Thus letting $k\to\infty$ in (4.9) we obtain

$\displaystyle\left<u_{t\wedge\tau_{m}},\phi\right>\\!\\!$
$\displaystyle=\\!\\!$
$\displaystyle\left<u_{0},\phi\right>+\frac{1}{2}\int_{0}^{t\wedge\tau_{m}}\big{[}\left<u_{s},\phi^{\prime\prime}\right>+F_{s}(\phi)\big{]}ds$
$\displaystyle+\int_{0}^{t\wedge\tau_{m}}\int_{0}^{\infty}g(\nabla
u_{s}(1))\Big{[}\int_{0}^{1}1_{\\{z\leq u_{s}(x)\\}}\phi(x)dx\Big{]}W(ds,dz).$

Letting $m\to\infty$ the result holds. ∎

Acknowledgements The research of L. Ji was supported in part by the fellowship
of China Postdoctoral Science Foundation 2020M68194; the research of J. Xiong
was supported in part by NSFC grants 61873325, 11831010 and SUSTech fund
Y01286120; the research of X. Yang was supported in part by NSFC grants
11771018 and 12061004.

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timings and the adaptive POD using a large magnetic multi-material cleaver.
This model, and the included contour plots, show the importance of using
prismatic layers to capture the thin electromagnetic skin depths, and the
importance of our new MPT calculation formulation.

The procedures presented in this work are expected to be invaluable for
constructing large dictionaries of MPT characterisations of complex in-
homogeneous real-world metallic objects. An open dataset of these
characterisations will be released in the near future.

##  10 Acknowledgements

The authors would like to thank Prof. Peyton, Prof. Lionheart and Dr Davidson
for their helpful discussions and comments on polarizability tensors. The
authors are grateful for the financial support received from the Engineering
and Physical Science Research Council (EPSRC, U.K.) through the research grant
EP/V009028/1.

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