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Q: Ubuntu 19.10 cursor freezes On Ubuntu 19.10, my cursor freezes after a long period of time not using the system. The keyboard works fine then, though.
My laptop is a 2011 MacBookPro, where I'm only using the Intel videocard, as the Radeon card is defective. I boot 19.10 normally, and 19.10 runs very stable on this machine, unlike Elementary OS even when bypassing the Radeon video card.

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\section{Introduction}

The stochastic parabolic equations are widely used to describe many diffusion processes perturbed by stochastic noises, such as  the evolution of the density of a bacteria population, the propagation of an electric potential in a neuron, etc., (e.g., \cite{dPZ2014,D1972,K2008,LZ2021S}). In this paper, we study an inverse problems for stochastic parabolic equations, i.e., determining the solution  from the terminal measurement. To be more precisely,  we first introduce some notations.

Let $T>0, G \subset \mathbb{R}^{n}(n \in \mathbb{N})$ be a given bounded domain with a $C^{2}$ boundary $\Gamma .$  Put
$
Q \triangleq(0, T) \times G$ and $\Sigma \triangleq(0, T) \times \Gamma$.

Let $\left(\Omega, \mathcal{F},\left\{\mathcal{F}_{t}\right\}_{t \geqslant 0}, {\mathbb{P}}\right)$ be a complete filtered probability space on which a one dimensional standard Brownian motion $W(\cdot)$ is defined.

Let $H$ be a Banach space. Denote by $L^2_{{\cal F}_t}(\Omega;H)$ ($t\geq 0$) the space of all $H$ -valued random variables $\xi$ satisfying ${\mathbb{E}}|\xi|_H^2<\infty$; by $L_{\mathbb{F}}^{2}(0, T ; H)$ the space  of all $H$ -valued $\left\{\mathcal{F}_{t}\right\}_{t \geqslant 0}$ -adapted processes $X(\cdot)$ such that $\mathbb{E}(|X(\cdot)|_{L^{2}(0, T ; H)}^{2})<\infty$; by $L_{\mathbb{F}}^{\infty}(0, T ; H)$ the space of all $H$ -valued $\left\{\mathcal{F}_{t}\right\}_{t \geqslant 0}$ -adapted bounded processes; and by $L_{\mathbb{F}}^{2}(\Omega ; C([0, T] ; H))$ the space  of all $H$ -valued $\left\{\mathcal{F}_{t}\right\}_{t \geqslant 0}$-adapted processes $X(\cdot)$ satisfying that $\mathbb{E}(|X(\cdot)|_{C(0, T ; H)}^{2})<\infty $. All these spaces are Banach spaces with the canonical norms (e.g.,\cite[Chapter 2]{LZ2021S}).

Consider the following stochastic parabolic equation:
\begin{equation}\label{p1}
\left\{\begin{array}{ll}\displaystyle\mathrm{d} u-\sum_{i, j=1}^{n}(a^{i j} u_{i})_j \mathrm{d} t=\left( b_{1}\cdot\nabla u +b_{2} u+f\right) \mathrm{d} t+\left(b_{3} u+g\right) \mathrm{d} W(t) & \text { in } Q, \\ \noalign{\medskip}\displaystyle u=0 & \text { on } \Sigma, \\ \noalign{\medskip}\displaystyle u(0)=u_{0}, & \text { in }G,
\end{array}
\right.
\end{equation}
where $u_{0} \in L^{2}_{\mathcal{F}_{0}}\left(\Omega; L^{2}(G)\right)$ and $u_i=\frac{\partial u}{\partial x_i}$ for $i=1,\cdots,n$.

Throughout this paper, we make the following assumptions on the coefficients $
a^{i j}: \Omega \times Q \rightarrow \mathbb{R}, i, j=1,2, \ldots, n
$:
\begin{itemize}
\item[(H1)] $a^{i j} \in L_{\mathbb{F}}^{2}\left(\Omega ; C^{1}\left([0, T] ; W^{2, \infty}(G)\right)\right)$ and $a^{i j}=a^{j i}$ for $i, j=1,2, \ldots, n$.

\item[(H2)] There is a constant $\sigma>0$ such that
$$
\sum_{i, j=1}^{n} a^{i j}(\omega, t, x) \xi^{i} \xi^{j} \geqslant \sigma|\xi|^{2}, \quad(\omega, t, x, \xi) \equiv\left(\omega, t, x, \xi^{1}, \ldots, \xi^{n}\right) \in \Omega \times Q \times \mathbb{R}^{n}.
$$
\end{itemize}
Let other coefficients and source terms in the equation of system \eqref{p1} satisfy
\begin{equation*}\!\!\!\!\!\!\!\!\!\text{(H3)}
\quad\begin{array}{l}
\displaystyle b_{1} \in L_{\mathbb{F}}^{\infty}\left(0, T ; W^{1,\infty}\left(G ; \mathbb{R}^{n}\right)\right), b_{2} \in L_{\mathbb{F}}^{\infty}\left(0, T; L^{\infty}(G)\right), b_{3} \in L_{\mathbb{F}}^{\infty}\left(0, T ; W^{1, \infty}(G)\right), \\
\noalign{\medskip}\displaystyle f \in L_{\mathbb{F}}^{2}\left(0, T ; L^{2}(G)\right) \  \text { and } \  g \in L_{\mathbb{F}}^{2}\left(0, T ; H^{1}_0(G)\right).
\end{array}
\end{equation*}

For readers' convenience, let us first recall the definition of the weak and strong solution to \eqref{p1}.

\begin{definition}\label{weakln} A process $u \in L^2_{\mathbb{F}}(\Omega;C([0,T];L^2(G)))\cap L^2_{\mathbb{F}}(0,T;H^1_0(G))$ is said to be a strong solution of
	equation \eqref{p1} if for any $t \in[0, T]$ and $\varphi\in H_0^1(G)$, it holds that
	%
	\begin{equation*}\label{weaksol}
		\begin{array}{ll}
			\displaystyle \int_G u(t) \varphi dx\!=\negthinspace \negthinspace \negthinspace  &\displaystyle\int_G\! u_{0} \varphi dx+\!\int_{0}^{t}\!\int_G\!\Big(\sum_{i, j=1}^{n} a^{i j} u_{i} \varphi_{j}\!-\! \hbox{\rm div$\,$} b_{1} u \varphi \!- \! u  b_{1} \cdot\nabla\varphi u \!+\!b_{2} u \varphi\! +\! f \varphi \Big) \mathrm{d}x\mathrm{d} s \\
			\noalign{\medskip}&\displaystyle +\int_{0}^{t} \int_{G}\left(b_{3}  u +g \right)\varphi \mathrm{d}x\mathrm{d} W(s), \quad {\mathbb{P}}\mbox{\rm-a.s.}
		\end{array}
	\end{equation*}
\end{definition}

\begin{definition}\label{strongsoln} A process $u \in L_{\mathbb{F}}^{2}\left(\Omega ; C\left([0, T] ; H^{2}(G) \cap H_{0}^{1}(G)\right)\right)$ is said to be a strong solution of
equation \eqref{p1} if for any $t \in[0, T]$ it holds that
\begin{equation}\label{strongsol}
\begin{array}{c}
\displaystyle u(t)=u_{0}+\int_{0}^{t}\[-\sum_{i, j=1}^{n}\left(a^{i j} u_{i} \right)_{j}+ b_{1} \cdot\nabla u +b_{2} u +f \] \mathrm{d} s \\
\noalign{\medskip}\displaystyle +\int_{0}^{t} \left(b_{3} u +g \right) \mathrm{d}W(s) \mbox{ in } L^2(G), \quad {\mathbb{P}}\mbox{\rm-a.s.}
\end{array}
\end{equation}
\end{definition}

Under (H1)--(H3), by the classical well-posedness result for stochastic parabolic equations, we know that \eqref{p1} admits a unique weak solution $u$ (e.g. \cite{Krylov1994}). Furthermore, if $u_{0}\in H^{2}(G) \cap H_{0}^{1}(G)$,
we know that \eqref{p1} admits a unique strong solution $u$ (e.g. \cite{Krylov1994}).

The inverse problem ({\bf IPD}) associated to the
equation \eqref{p1} is as follows.

\vspace{0.1cm}
\begin{itemize}

	\item {\bf Conditional Stability}. Assume that two
	solutions $u$ and $\hat u$ (to the equation
	\eqref{p1}) are given. Let $u(T)$ and
	$ \hat u(T)$ be the
	corresponding terminal values. Can we find a
	positive constant $C$ such that
	%
	\begin{equation}\label{1.18-eq1}
	|u-\hat u| \leq C|\!| u(T)- \hat u|\!|,
	\end{equation}
    %
	with appropriate norms in both sides?

	\vspace{0.1cm}

	\item {\bf Reconstruction}. Is it possible to
	reconstruct $u$   from the
	terminal value $u(T)$?

\end{itemize}
Here and in the rest of this paper, we use $C$
to denote a generic positive constant depending
on $G$, $T$ and $a^{ij}$($i,j=1,\cdots,n$) (unless otherwise
stated), which may change from line to line.

\begin{remark}
It is well known that the inverse problem ({\bf IPD}) is ill-posed: small errors in data
may cause huge deviations in solutions. Fortunately, if we assume an a priori bound
for $u(\cdot,0)$, then we can restore the stability. This is the reason we consider  conditional stability.
\end{remark}

Inverse problem  in the above type   is studied extensively for deterministic parabolic equations (see
\cite{KY2019IP,Yamamoto} and the rich references
therein). However, the stochastic case attracts
very little attention. To our best knowledge,
\cite{barbu2,L2012IP} are the only two published
papers addressing this topic.  In \cite{barbu2}, the author study the problem by transform the equation \eqref{p1} to a parabolic equation with random coefficients. Then they can obtain pathwisely a logarithmic convexity property, from which the uniqueness follows. To apply the strategy in \cite{barbu2}, one needs some further assumptions on $b_3$, such as $b_3$ is independent of $x$ (in \cite{barbu2}, the authors assume that $b_3=1$).  In \cite{L2012IP}, the author use a stochastic global Carleman estimate to study the inverse problem ({\bf IPD}). The weight function in that Carleman estimate is a double-exponential function. In such case, the constant $C$ in \eqref{1.18-eq1} will depend on the norm of $b_1$, $b_2$ and $b_3$ double-exponentially. In this paper, we establish a new Carleman estimate with an exponential weight function. Then we improve the conditional stability in \cite{L2012IP} (see Theorem \ref{reguest} for the detail). More importantly, both \cite{barbu2,L2012IP} only address the first question of  the inverse problem ({\bf IPD}). To our best knowledge, this paper is the first one addressing the reconstruction question of the inverse problem ({\bf IPD}).

\medskip

As we said before, the main tool for establishing the conditional stability is a new global Carleman estimate for \eqref{p1}.
Carleman estimates are widely applied to many inverse problems for deterministic partial differential equations (see \cite{BY2017book,FLZ2019book,KT2012book} and the rich references therein).  Recently, Carleman estimates are also introduced to solve inverse problems for stochastic partial differential equations. Particularly, we refer the readers to  \cite{Li2013,L2012IP,WCW2020IP,Y2017IP,Y2021JIIP} for some recent works on inverse problems of stochastic parabolic equations via Carleman estimates.
In all the above mentioned papers, Carleman estimates are established by using two-layer weight functions.
In this paper, we improve the Carleman estimates in the previous works for stochastic parabolic equation, i.e., establish a new Carleman estimate for stochastic parabolic equation by applying a one-layer weight function, which is first introduced in \cite{KY2019IP} for backward problem of deterministic heat equation.
Since the solution of a stochastic parabolic equation is not differentiable with respect to the temporal variable, new difficulties occur in the study of inverse problems for stochastic parabolic equations.
Thus, the Carleman estimate we obtained is not a trivial extension of \cite{KY2019IP} and other existing works, although the form of the Carleman estimate is similar to the one in \cite{KY2019IP}.

\medskip

Once the Carleman estimate is established, we employ Tikhonov regularization method to reconstruct the solution to \eqref{p1}. Tikhonov regularization method is a very useful tool to solve inverse and ill-posed problems for deterministic partial differential equations (e.g., \cite{LP2013,Schuster2012}).
To the authors' knowledge, regularization methods for inverse problems of stochastic parabolic equations are rarely studied. The only published work  is \cite{CWC2021IPI}, in which  Cauchy problem for stochastic parabolic equations is studied by a Kaczmarz method.   In this paper, we propose a regularized solution by minimising a functional, which is giving based on the Tikhonov regularization.  The analysis of the existence and uniqueness of the regularized solution, and proof for error estimate between the regularized solution and the exact solution under an a-proior condition are also present.

In this paper, we also give a numerical solution to the inverse problem  ({\bf IPD}), i.e., we numerically solve the reconstruction problem   by the proposed regularization method, with the combination of the conjugate gradient method to the Tikhonov type functional and Picard iteration.  We mention that, in numerical solving the forward-backward stochastic parabolic equation, we apply the idea given in \cite{DP2016SISC}, where the discretization is given based on the implicit Euler method for a temporal discretization and a least squares Monte Carlo method in combination with a stochastic gradient method. Compared with deterministic setting, this is also a topic far from well studied.  As far as we know, there is no published work  addressing that.

The rest of the paper is organized as follow.
We first establish a global Carleman estimate for the stochastic parabolic equation  \eqref{p1} in Section 2,  and prove the conditional stability for the inverse problem under an a-priori information in Section 3.
A regularization solution based on the Tikhonov regularization method is proposed in Section 4.
At last, we give numerical algorithm to the problem, and illustrate the approximations to several  examples in both one and two dimensional spatial domains in Section 5.

\section{Carleman estimate}

This section is devoted to establishing a Carleman estimate for \eqref{p1}. We first give the weight function. For $\lambda>0$, set
\begin{equation}
\psi=(t+1)^\lambda,\qquad \varphi=e^\psi.
\end{equation}
The Carleman estimate is as follows:
\begin{theorem}\label{carleman}
For any $\varepsilon \in[0, T),$ there exists a $\lambda_{0}>0$ such that for all $\lambda \geqslant \lambda_{0},$ there holds that
\begin{align}\label{519eq3}
&\mathbb{E}\int_\varepsilon^T\int_G \lambda^2 (t+1)^{(\lambda-2)}v^2\mathrm{d} t\mathrm{d}x+\mathbb{E}\int_\varepsilon^T\int_G \lambda(t+1)^{-1}|\nabla v|^2\mathrm{d} t\mathrm{d}x \nonumber\\
\noalign{\medskip}\leq&C\[\lambda(T\!+\!1)^{\lambda-1}\mathbb{E}\|v(T)\|^2_{L^2(G)}\!+\! \mathbb{E}\|v(\varepsilon)\|^2_{L^2(G)} \!+\! \mathbb{E}\! \int_{\varepsilon}^{T} \!\int_{G}\! \varphi^2(|\nabla g|^2\! +\!g^{2}\!+\!f^2) \mathrm{d} x \mathrm{d} t\].
\end{align}
\end{theorem}

\begin{proof}
We divide the proof into three steps.

\noindent\textit{Step 1. In this step, we establish a weighted  identity.}

Let $v\triangleq \varphi u$. Then we know that
\begin{align}
 \mathrm{d} u-\sum_{i, j=1}^{n}(a^{i j}u_i)_j \mathrm{d} t = \varphi^{-1}\[\mathrm{d}v-\lambda(t+1)^{(\lambda-1)}v\mathrm{d} t-\sum_{i, j=1}^{n}(a^{i j} v_{i})_j \mathrm{d} t\].
\end{align}
Consequently,
\begin{align}\label{512eq1}
\displaystyle&-\varphi\[\lambda(t+1)^{(\lambda-1)}v+\sum_{i, j=1}^{n}(a^{i j} v_{i})_j\]\[\mathrm{d} u-\sum_{i, j=1}^{n}(a^{i j} u_i)_j \mathrm{d} t\]\nonumber\\
\noalign{\medskip}\displaystyle= &-\lambda(t+1)^{(\lambda-1)}v\mathrm{d}v-\sum_{i, j=1}^{n}(a^{i j} v_{i})_j\mathrm{d}v+\[\lambda(t+1)^{(\lambda-1)}v+\sum_{i, j=1}^{n}(a^{i j} v_{i})_j\]^2\mathrm{d} t.
\end{align}

According to It\^o's formula, we know
\begin{align}\label{512eq3}
\displaystyle&-\lambda(t+1)^{(\lambda-1)}v\mathrm{d} v=-
\frac12\lambda(t+1)^{(\lambda-1)}\mathrm{d} v^2+\frac12\lambda(t+1)^{(\lambda-1)}(\mathrm{d} v)^2\nonumber\\
\noalign{\medskip}\displaystyle=&-\frac12\mathrm{d}\big[\lambda(t+1)^{(\lambda-1)} v^2\big]+\frac12\lambda(\lambda-1)(t+1)^{(\lambda-2)} v^2\mathrm{d}t+\frac12\lambda(t+1)^{(\lambda-1)}(\mathrm{d} v)^2.
\end{align}
Noting  $a^{ij}=a^{ji}$ for $i,j=1,\cdots,n$, we get that
\begin{align}\label{512eq2}
\displaystyle-\sum_{i, j=1}^{n}(a^{i j} v_{i})_j\mathrm{d}v=&-\sum_{i, j=1}^{n}(a^{i j} v_{i}\mathrm{d}v)_j+\frac12 \mathrm{d}\(\sum_{i, j=1}^{n}a^{i j} v_{i}v_j\)\nonumber\\
\noalign{\medskip}\displaystyle&-\frac12 \sum_{i, j=1}^{n}a^{i j}_t v_{i}v_j\mathrm{d}t-\frac12 \sum_{i, j=1}^{n}a^{i j} \mathrm{d}v_{i}\mathrm{d}v_j.
\end{align}

Substituting \eqref{512eq2} and \eqref{512eq3} into \eqref{512eq1}, we obtain that
\begin{align}\label{512eq4}
\displaystyle&-\varphi\[\lambda(t+1)^{(\lambda-1)}v+\sum_{i, j=1}^{n}(a^{i j} v_{i})_j\]\[\mathrm{d} u-\sum_{i, j=1}^{n}(a^{i j} u_{i})_j \mathrm{d} t\]\nonumber\\
\noalign{\medskip}\displaystyle=&-\sum_{i, j=1}^{n}(a^{i j} v_{i}\mathrm{d}v)_j+\frac12 \mathrm{d}\[\sum_{i, j=1}^{n}a^{i j} v_{i}v_j-\lambda(t+1)^{(\lambda-1)} v^2\]-\frac12 \sum_{i, j=1}^{n}a^{i j} \mathrm{d}v_{i}\mathrm{d}v_j\\
\noalign{\medskip}\displaystyle&+\frac12 \[-\sum_{i, j=1}^{n}a^{i j}_t v_{i}v_j+\lambda(\lambda-1)(t+1)^{(\lambda-2)} v^2\]\mathrm{d}t+\frac12\lambda(t+1)^{(\lambda-1)}(\mathrm{d} v)^2\nonumber\\
\noalign{\medskip}\displaystyle&+\[\lambda(t+1)^{(\lambda-1)}v+\sum_{i, j=1}^{n}a^{i j} v_{ij}\]^2\mathrm{d} t.\nonumber
\end{align}
Moreover,
\begin{align}\label{514eq2}
&  \frac14\lambda (t+1)^{-1}\varphi v\[\mathrm{d} u-\sum_{i, j=1}^{n}(a^{i j} u_i)_j \mathrm{d} t\]\nonumber\\
\noalign{\medskip}=&\frac18\lambda \mathrm{d}\big[(t+1)^{-1}v^2\big]+\frac18\lambda(t+1)^{-2}v^2dt-\frac18\lambda(t+1)^{-1} (\mathrm{d}v)^2-\frac14\lambda^2(t+1)^{(\lambda-2)}v^2\mathrm{d} t\nonumber\\
\noalign{\medskip}&- \frac14(t+1)^{-1}\lambda\sum_{i, j=1}^{n}(a^{i j} v_{i}v)_j \mathrm{d} t+\frac14(t+1)^{-1}\lambda\sum_{i, j=1}^{n}a^{i j} v_{i}v_j \mathrm{d} t.
\end{align}

Sum \eqref{512eq4} and \eqref{514eq2} up, we obtain that
\begin{align}\label{514eq3}
&-\varphi\[\lambda(t+1)^{(\lambda-1)}v+\sum_{i, j=1}^{n}a^{i j} v_{ij}+\frac14\lambda (t+1)^{-1}v\] \[\mathrm{d} u-\sum_{i, j=1}^{n}(a^{i j} u_i)_j \mathrm{d} t\]\nonumber\\
\noalign{\medskip}=&-\sum_{i, j=1}^{n}\[a^{i j} v_{i}\mathrm{d}v+\frac14(t+1)^{-1}\lambda a^{i j} v_{i}v\]_j-\frac12 \sum_{i, j=1}^{n}a^{i j} \mathrm{d}v_{i}\mathrm{d}v_j\nonumber\\
\noalign{\medskip}&+\frac12 \mathrm{d}\[\sum_{i, j=1}^{n}a^{i j} v_{i}v_j-\lambda(t+1)^{(\lambda-1)} v^2+\frac14\lambda(t+1)^{-1} v^2\]\\
\noalign{\medskip}&+\frac12 \[-\sum_{i, j=1}^{n}a^{i j}_t v_{i}v_j+\lambda(\lambda-1)(t+1)^{(\lambda-2)} v^2-\frac12\lambda^2(t+1)^{(\lambda-2)}v^2+\frac14\lambda(t+1)^{-2}v^2\]\mathrm{d}t\nonumber\\
\noalign{\medskip}&+\frac12\[\lambda(t+1)^{(\lambda-1)}-\frac14\lambda(t+1)^{-1}\](\mathrm{d} v)^2+\[\lambda(t+1)^{(\lambda-1)}v+\sum_{i, j=1}^{n}a^{i j} v_{ij}\]^2\mathrm{d} t\nonumber\\
\noalign{\medskip}\displaystyle&
+\frac14\lambda(t+1)^{-1}\sum_{i, j=1}^{n}a^{i j} v_{i}v_j \mathrm{d} t.\nonumber
\end{align}

Applying \eqref{514eq3} to the equation \eqref{p1}, integrating \eqref{514eq3} on $[\varepsilon,T]\times\Omega$ for $\varepsilon\in[0,T)$ and taking mathematical expectation, we have
\begin{align}\label{514eq4}
&-\mathbb{E}\int_\varepsilon^T\int_G \varphi\[\lambda(t+1)^{(\lambda-1)}v+\sum_{i, j=1}^{n}a^{i j} v_{ij}+\frac14\lambda (t+1)^{-1}v\] \[\mathrm{d} u-\sum_{i, j=1}^{n}(a^{i j} u_i)_j \mathrm{d} t\]\mathrm{d}x\nonumber\\
=&-\mathbb{E}\int_\varepsilon^T\int_G \sum_{i, j=1}^{n}\[a^{i j} v_{i}\mathrm{d}v+\frac14(t+1)^{-1}\lambda a^{i j} v_{i}v\mathrm{d} t\]_j\mathrm{d}x\nonumber\\
&+\frac12 \mathbb{E}\int_\varepsilon^T\int_G \mathrm{d}\[\sum_{i, j=1}^{n}a^{i j} v_{i}v_j-\lambda(t+1)^{(\lambda-1)} v^2+\frac14\lambda(t+1)^{-1} v^2\]\mathrm{d}x\nonumber\\
&+\frac12 \mathbb{E}\int_\varepsilon^T\int_G \[\lambda(\lambda-1)(t+1)^{(\lambda-2)} v^2-\frac12\lambda^2(t+1)^{(\lambda-2)}v^2+\frac14\lambda(t+1)^{-2}v^2\]\mathrm{d} t\mathrm{d}x\nonumber\\
&+\frac12\mathbb{E}\int_\varepsilon^T\int_G \[\lambda(t+1)^{(\lambda-1)}-\frac14\lambda(t+1)^{-1}\](\mathrm{d} v)^2\mathrm{d}x\\
&+\mathbb{E}\int_\varepsilon^T\int_G \[\lambda(t+1)^{(\lambda-1)}v+\sum_{i, j=1}^{n}a^{i j} v_{ij}\]^2\mathrm{d} t\mathrm{d}x\nonumber\\
&-\frac12 \mathbb{E}\int_\varepsilon^T\int_G \[\sum_{i, j=1}^{n}a^{i j} \mathrm{d}v_{i}\mathrm{d}v_j+\sum_{i, j=1}^{n}a^{i j}_t v_{i}v_j\mathrm{d} t-\frac12(t+1)^{-1}\lambda\sum_{i, j=1}^{n}a^{i j} v_{i}v_j \mathrm{d} t\]\mathrm{d}x\nonumber\\
\noalign{\medskip}=:&\sum_{j=1}^6 I_j.\nonumber
\end{align}

\noindent\textit{Step 2. In this step, we estimate right hand side of \eqref{514eq4}.}

Since $u\big|_{\Sigma}=0$, $v\big|_\Sigma=0$, by the divergence theorem,  we have
\begin{align}\label{513eq2}
I_1&=-\mathbb{E}\int_\varepsilon^T\int_G\sum_{i, j=1}^{n}\[a^{i j} v_{i}\mathrm{d}v+\frac14(t+1)^{-1}\lambda a^{i j} v_{i}v \mathrm{d} t\]_j\mathrm{d}x =0.
\end{align}

By the assumptions (H1) and (H2), we get
\begin{align}\label{513eq3}
\displaystyle I_2=&\frac12 \mathbb{E}\int_\varepsilon^T\int_G\mathrm{d}\[\sum_{i, j=1}^{n}a^{i j} v_{i}v_j-\lambda(t+1)^{(\lambda-1)} v^2+\frac14\lambda(t+1)^{-1} v^2\]\mathrm{d}x\nonumber\\
=&\frac12 \mathbb{E}\int_G\[\sum_{i, j=1}^{n}a^{i j} v_{i}v_j-\lambda(t+1)^{(\lambda-1)} v^2+\frac14\lambda(t+1)^{-1} v^2\]_{t=\delta}^T\mathrm{d}x\nonumber\\
=&\frac12 \mathbb{E}\int_G\[\sum_{i, j=1}^{n}a^{i j} v_{i}(T)v_j(T)-\lambda(t+1)^{(\lambda-1)} v^2(T)+\frac14\lambda(t+1)^{-1} v^2(T)\]\mathrm{d}x\\
&-\frac12 \mathbb{E}\int_G\[\sum_{i, j=1}^{n}a^{i j} v_{i}(\varepsilon)v_j(\varepsilon)-\lambda(t+1)^{(\lambda-1)} v^2(\varepsilon)+\frac14\lambda(t+1)^{-1} v^2(\varepsilon)\]\mathrm{d}x\nonumber\\
\geq&\frac12 \mathbb{E}\int_G\sigma |\nabla v(T)|^2\mathrm{d}x-\lambda(t+1)^{(\lambda-1)} v(T)^2\mathrm{d}x-\frac12 \mathbb{E}\int_G\sum_{i, j=1}^{n}a^{i j} v_{i}(\varepsilon)v_j(\varepsilon)\mathrm{d}x\nonumber\\
&+\frac12 \mathbb{E}\int_G\lambda(t+1)^{(\lambda-1)}\[1-\frac14(\varepsilon+1)^{-\lambda}\] v^2(\varepsilon)\mathrm{d}x.\nonumber
\end{align}
Let us choose $\lambda\geq1$. Then $1-\frac14(\varepsilon+1)^{-\lambda}\geq\frac34>0$, and thus
\begin{align}\label{615eq1}
\displaystyle I_2
\geq&-\frac12 \mathbb{E}\int_G\lambda(T+1)^{\lambda-1} v(T)^2\mathrm{d}x-\frac12 \mathbb{E}\int_G\sum_{i, j=1}^{n}a^{i j} v_{i}(\varepsilon)v_j(\varepsilon)\mathrm{d}x\nonumber\\
\geq &-\frac12\mathbb{E}\lambda(T+1)^{\lambda-1}\|v(T)\|^2_{L^2(G)}-\frac12\sum_{i, j=1}^{n}\|a^{ij}(\varepsilon)\|^2_{L^\infty_{{\cal F}_\varepsilon}(\Omega;L^\infty(G))}\mathbb{E}\|\nabla v(\varepsilon)\|^2_{L^2(G)}.
\end{align}

Since
\begin{align}\label{513eq4}
\displaystyle&-\frac12\mathbb{E} \int_{\varepsilon}^{T} \int_G\sum_{i, j=1}^{n} a^{i j} \mathrm{d} v_{i} \mathrm{d} v_{j} \mathrm{d} x\nonumber\\
=&-\frac{1}{2} \mathbb{E} \int_{\varepsilon}^{T} \int_G \varphi^2\sum_{i, j=1}^{n} a^{i j} \left(b_{3} u+g\right)_{i}\left(b_{3} u+g\right)_{j} \mathrm{d} x \mathrm{d} t\nonumber\\
\geq&-C\mathbb{E} \int_{\varepsilon}^{T} \int_{G}\left(b_{3}^{2}|\nabla v|^{2}+\left|\nabla b_{3}\right|^{2} v^{2}+\varphi^2|\nabla g|^{2}+\varphi^2|g|^{2}\right) \mathrm{d} x \mathrm{d} t\\
\geq&-C\left(\|b_{3}\|_{L_{\mathbb{F}}^{\infty}\left(0, T ; W^{1, \infty}(\Omega)\right)}^{2}+1\right) \mathbb{E} \int_{\varepsilon}^{T} \int_{\Omega}\left(|\nabla v|^{2}+v^{2}\right) \mathrm{d} x \mathrm{d} t\nonumber\\
&\quad -C \mathbb{E} \int_{\varepsilon}^{T} \int_G\varphi^2\left(|\nabla g|^{2}+g^{2}\right) \mathrm{d} x \mathrm{d} t,\nonumber
\end{align}
we obtain
\begin{eqnarray}\label{515eq1}
 I_6\negthinspace \negthinspace \negthinspace &=\negthinspace \negthinspace \negthinspace &-\frac12 \mathbb{E}\int_\varepsilon^T\int_G \[\sum_{i, j=1}^{n}a^{i j} \mathrm{d}v_{i}\mathrm{d}v_j+\sum_{i, j=1}^{n}a^{i j}_t v_{i}v_j\mathrm{d} t-\frac12\lambda(t+1)^{-1}\sum_{i, j=1}^{n}a^{i j} v_{i}v_j \mathrm{d} t\]\mathrm{d}x\nonumber\\
&\negthinspace \negthinspace \negthinspace \geq\negthinspace \negthinspace \negthinspace &-C \mathbb{E} \int_{\varepsilon}^{T} \int_{G}\left(\|b_{3}\|_{L_{\mathbb{F}}^{\infty}\left(0, T ; W^{1, \infty}(\Omega)\right)}^{2}+1\right)\left(|\nabla v|^{2}+v^{2}\right) \mathrm{d} x \mathrm{d} t \\
&&-C \mathbb{E}\! \int_{\varepsilon}^{T} \!\int_G \varphi^2\left(|\nabla g|^{2}+g^{2}\right) \mathrm{d} x \mathrm{d} t +\mathbb{E}\!\int_\varepsilon^T\!\int_G \[-C|\nabla v|^2\!+\frac14\lambda(t\!+\!1)^{-1}\sigma|\nabla v|^2\]\mathrm{d}x\mathrm{d} t.
\nonumber
\end{eqnarray}

By direct computations, we see that
\begin{align}\label{514eq1}
I_4=&\frac12\mathbb{E}\int_\varepsilon^T\int_G \[\lambda(t+1)^{(\lambda-1)}-\frac14\lambda(t+1)^{-1}\](\mathrm{d} v)^2\mathrm{d}x\nonumber\\
=& \frac12\mathbb{E}\int_\varepsilon^T\int_G\[\lambda(t+1)^{(\lambda-1)}-\frac14\lambda(t+1)^{-1}\]e^{2(t+1)^\lambda}(b_3u+g)^2\mathrm{d}x\\
\geq&-\mathbb{E}\int_\varepsilon^T\int_G\frac14\lambda(t+1)^{-1}\[(b_3v)^2+e^{2(t+1)^\lambda}g^2\]\mathrm{d} t\mathrm{d}x.\nonumber
\end{align}
For  $\lambda\geq\max\{3,T+1\}$, we have
\begin{align}\label{515eq2}
I_3+I_4=&\frac12 \mathbb{E}\int_\varepsilon^T\int_G \[\lambda(\lambda-1)(t+1)^{(\lambda-2)} v^2-\frac12\lambda^2(t+1)^{(\lambda-2)}v^2+\frac14\lambda(t+1)^{-2}v^2\]\mathrm{d} t\mathrm{d} x\nonumber\\
&+\frac12\mathbb{E}\int_\varepsilon^T\int_G\[\lambda(t+1)^{(\lambda-1)}-\frac14\lambda(t+1)^{-1}\](\mathrm{d} v)^2\mathrm{d}x\nonumber\\
\geq& \frac13\mathbb{E}\int_\varepsilon^T\int_G\lambda^2(t+1)^{(\lambda-2)}v^2\mathrm{d}x\mathrm{d} t+O(\lambda)\mathbb{E}\int_\varepsilon^T\int_G v^2\mathrm{d} t\mathrm{d}x\\
\noalign{\medskip}&-\frac14\mathbb{E}\int_\varepsilon^T\int_G\lambda(t+1)^{-1}\varphi^2g^2\mathrm{d}x\mathrm{d} t.\nonumber
\end{align}

Substituting \eqref{513eq2}--\eqref{515eq2} into \eqref{514eq4}, we get
\begin{eqnarray}\label{515eq3}
&&-\mathbb{E}\!\int_\varepsilon^T\!\int_G \Big\{\varphi\[\lambda(t+1)^{(\lambda-1)}v+\sum_{i, j=1}^{n}a^{i j} v_{ij}\!+\frac14\lambda(1+t)^{-1} v\] \[\mathrm{d} u-\!\sum_{i, j=1}^{n}\!(a^{i j} u_i)_j \mathrm{d} t\]\Big\}\mathrm{d}x\nonumber\\
&\geq\negthinspace \negthinspace \negthinspace & -\frac12\mathbb{E}\[\lambda(T+1)^{\lambda-1}\|v(T)\|^2_{L^2(G)}+\sum_{i,j=1}^n\|a^{ij}\|^2_{L^\infty_{{\cal F}_\varepsilon}(\Omega;W^{1,\infty}(G))}\|\nabla v(\varepsilon)\|^2_{L^2(G)}\]\nonumber\\
&&-\mathbb{E} \int_{\varepsilon}^{T} \int_{\Omega}\big(\|b_{3}\|_{L_{\mathbb{F}}^{\infty}\left(0, T ; W^{1, \infty}(\Omega)\right)}^{2}+1\big)\left(|\nabla v|^{2}+v^{2}\right) \mathrm{d} x \mathrm{d} t\\
&&-C \mathbb{E} \!\int_{\varepsilon}^{T}\! \int_G \varphi^2\left(|\nabla g|^{2}\!+g^{2}\right) \mathrm{d} x \mathrm{d} t-\mathbb{E}\!\int_\varepsilon^T\!\int_G \[C|\nabla v|^2\!-\frac14\lambda(t\!+1)^{-1}\sigma|\nabla v|^2\]\mathrm{d}x\mathrm{d} t\nonumber\\
&&+\frac13\mathbb{E}\int_\varepsilon^T\int_G\lambda^2(t+1)^{(\lambda-2)}v^2\mathrm{d}x\mathrm{d} t+O(\lambda)\mathbb{E}\int_\varepsilon^T\int_G v^2\mathrm{d} t\mathrm{d}x\nonumber\\
&&-\frac14\mathbb{E}\int_\varepsilon^T\int_G\lambda(t+1)^{-1}\varphi^2g^2\mathrm{d}x\mathrm{d} t+\mathbb{E}\int_\varepsilon^T\int_G \[\lambda(t+1)^{(\lambda-1)}v+\sum_{i, j=1}^{n}a^{i j} v_iv_j\]^2\mathrm{d} t\mathrm{d}x.\nonumber
\end{eqnarray}

\noindent\textit{Step 3. In this step, we estimate left hand side of \eqref{514eq4}, and deduce Carleman inequality \eqref{519eq3}.}

From \eqref{p1}, we know
\begin{align}\label{515eq5}
&-\mathbb{E}\int_\varepsilon^T\int_G \varphi\[\lambda(t+1)^{(\lambda-1)}v+\sum_{i, j=1}^{n}a^{i j} v_{ij}+\frac14\lambda(t+1)^{-1} v\] \[\mathrm{d} u-\sum_{i, j=1}^{n}(a^{i j} u_i)_j \mathrm{d} t\]\mathrm{d}x\nonumber\\
=&-\mathbb{E}\int_\varepsilon^T\int_G \varphi\[\lambda(t+1)^{(\lambda-1)}v+\sum_{i, j=1}^{n}a^{i j} v_{ij}+\frac14\lambda(t+1)^{-1} v\] \left( b_{1}\cdot\nabla u +b_{2} u+f\right) \mathrm{d} t\mathrm{d}x\nonumber\\
\leq&\mathbb{E}\int_\varepsilon^T\int_G \[\lambda(t+1)^{(\lambda-1)}v+\sum_{i, j=1}^{n}a^{i j} v_{ij}\]^2 \mathrm{d} t\mathrm{d}x+\frac1{16}\mathbb{E}\int_\varepsilon^T\int_G \[\frac14\lambda(t+1)^{-1} v\]^2\mathrm{d} t\mathrm{d}x\nonumber\\
&+\frac{17}4\mathbb{E}\int_\varepsilon^T\int_G e^{2(t+1)^\lambda} \left(b_{1}\cdot\nabla u+b_{2} u+f\right)^2 \mathrm{d} t\mathrm{d}x\\
\leq& \mathbb{E}\int_\varepsilon^T\int_G \[\lambda(t+1)^{(\lambda-1)}v+\sum_{i, j=1}^{n}a^{i j} v_{ij}\]^2 \mathrm{d}x\mathrm{d} t+\frac1{256}\mathbb{E}\int_\varepsilon^T\int_G \lambda^2(t+1)^{-2} v^2\mathrm{d} t\mathrm{d}x\nonumber\\
&+\frac{51}4\|b_{1}\|^2_{L^\infty_{\mathbb{F}}(0,T;L^\infty(G,\mathbb{R}^n))}\mathbb{E}\int_\varepsilon^T\int_G |\nabla v|^2\mathrm{d}x\mathrm{d} t\nonumber\\
\noalign{\medskip}&+\frac{51}4\|b_{2}\|^2_{L^\infty_{\mathbb{F}}(0,T;L^\infty(G))}\mathbb{E}\int_\varepsilon^T\int_G |v|^2\mathrm{d}x\mathrm{d} t+\frac{51}4\mathbb{E}\int_\varepsilon^T\int_G \varphi^2 |f|^2\mathrm{d}x\mathrm{d} t.\nonumber
\end{align}

Combining \eqref{515eq3} and \eqref{515eq5}, we know that there exists $\lambda_1>0$ such that for all $\lambda\geq \lambda_1$, there holds
\begin{align}\label{515eq6}
&-\frac12\mathbb{E}\lambda(T+1)^{\lambda-1}\|v(T)\|^2_{L^2(G)}-\frac12\sum_{i,j=1}^n\|a^{ij}(\varepsilon)\|^2_{L^\infty_{{\cal F}_\varepsilon}(\Omega;L^\infty(G))}\mathbb{E}\|\nabla v(\varepsilon)\|^2_{L^2(G)}\nonumber\\
&-C\mathbb{E} \int_{\varepsilon}^{T} \int_{G}\left(\|b_{3}\|_{L_{\mathbb{F}}^{\infty}\left(0, T ; W^{1, \infty}(\Omega)\right)}^{2}+1\right)\left(|\nabla v|^{2}+v^{2}\right) \mathrm{d} x \mathrm{d} t\nonumber\\
&-C \mathbb{E} \int_{\varepsilon}^{T} \int_G \varphi^2\left(|\nabla g|^{2}+g^{2}\right) \mathrm{d} x \mathrm{d} t -\mathbb{E}\int_\varepsilon^T\int_G \[C|\nabla v|^2-\frac14\lambda(t+1)^{-1}\sigma|\nabla v|^2\]\mathrm{d}x\mathrm{d} t\nonumber\\
&+\frac13\mathbb{E}\int_\varepsilon^T\int_G\lambda^2(t+1)^{(\lambda-2)}v^2\mathrm{d}x\mathrm{d} t+O(\lambda)\mathbb{E}\int_\varepsilon^T\int_G v^2\mathrm{d} t\mathrm{d}x \\
&-C\mathbb{E}\int_\varepsilon^T\int_G\lambda(t+1)^{(\lambda-1)}\varphi^2g^2\mathrm{d}x\mathrm{d} t +\mathbb{E}\int_\varepsilon^T\int_G \[\lambda(t+1)^{(\lambda-1)}v+\sum_{i, j=1}^{n}a^{i j} v_iv_j\]^2\mathrm{d} t\mathrm{d}x\nonumber\\
\leq&\mathbb{E}\int_\varepsilon^T\int_G \[\lambda(t+1)^{(\lambda-1)}v+\sum_{i, j=1}^{n}a^{i j} v_{ij}\]^2 \mathrm{d}x\mathrm{d} t+\frac1{256}\mathbb{E}\int_\varepsilon^T\int_G \lambda^2(t+1)^{-2} v^2\mathrm{d} t\mathrm{d}x\nonumber\\
&+\frac{51}4\|b_{1}\|^2_{L^\infty_{\mathbb{F}}(0,T;L^\infty(G,\mathbb{R}^n))}\mathbb{E}\int_\varepsilon^T\int_G |\nabla v|^2\mathrm{d}x\mathrm{d} t\nonumber\\
&+\frac{51}4\|b_{2}\|^2_{L^\infty_{\mathbb{F}}(0,T;L^\infty(G,\mathbb{R}^n))}\mathbb{E}\int_\varepsilon^T\int_G |v|^2\mathrm{d}x\mathrm{d} t+\frac{51}4\mathbb{E}\int_\varepsilon^T\int_G \varphi^2|f|^2\mathrm{d}x\mathrm{d} t.\nonumber
\end{align}
Let $\lambda_2= \max\{\lambda_1,C(\|b_{3}\|_{L_{\mathbb{F}}^{\infty}\left(0, T ; W^{1, \infty}(\Omega)\right)}^{2}+1)\}$.
Then, for all $\lambda\geq \lambda_2$,
\begin{align}\label{515eq6-1}
&\frac13\mathbb{E}\int_\varepsilon^T\int_G\lambda^2\[(t+1)^{(\lambda-2)}-\frac1{256}\]v^2\mathrm{d}x\mathrm{d} t+O(\lambda)\mathbb{E}\int_\varepsilon^T\int_G v^2\mathrm{d} t\mathrm{d}x\nonumber\\
&+\!\mathbb{E}\!\int_\varepsilon^T\!\!\!\int_G\! \[\frac14\lambda\sigma(t\!+\!1)^{-1}\!\! -\!C(\|b_{3}\|_{L_{\mathbb{F}}^{\infty}(0, T ; W^{1, \infty}(G))}^{2}\!\!+\!1) \!-\!\frac{51}4\|b_{1}\|^2_{L^\infty_{\mathbb{F}}(0,T;L^\infty(G,\mathbb{R}^n))} \]|\nabla v|^2\mathrm{d} t\mathrm{d}x\nonumber\\
\leq&\frac12\mathbb{E}\lambda(T+1)^{\lambda-1}\|v(T)\|^2_{L^2(G)}+\frac12\sum_{i,j=1}^n\|a^{ij}(\varepsilon)\|^2_{L^\infty_{{\cal F}_\varepsilon}(\Omega;L^\infty(G))}\mathbb{E}\|v(\varepsilon)\|^2_{L^2(G)}\\
&+C\mathbb{E}\! \int_{\varepsilon}^{T}\!\! \int_G \!\varphi^2\left[ |\nabla g|^{2}\!+\(1 \!+\!\lambda(t\!+\!1)^{(\lambda-1)}\)g^{2}\right]\mathrm{d} x \mathrm{d} t \!+\!\frac{51}4\mathbb{E}\!\int_\varepsilon^T\!\!\int_G\! e^{2(t+1)^\lambda}|f|^2\mathrm{d}x\mathrm{d} t.\nonumber
\end{align}
Moreover, let
$$\lambda_0= \max\Big\{\lambda_2,\frac43\sigma^{-1}(T+1)\[C+C(\|b_{3}\|_{L_{\mathbb{F}}^{\infty}(0, T ; W^{1, \infty}(G))}^{2}+1)+\frac{51}4\|b_{1}\|^2_{L^\infty_{\mathbb{F}}(0,T;L^\infty(G,\mathbb{R}^n))}\]\Big\}.$$
For all $\lambda\geq \lambda_0$, we have
\begin{align}
&\mathbb{E}\int_\varepsilon^T\int_G \lambda^2 (t+1)^{(\lambda-2)}v^2\mathrm{d} t\mathrm{d}x+\mathbb{E}\int_\varepsilon^T\int_G \lambda(t+1)^{-1}|\nabla v|^2\mathrm{d} t\mathrm{d}x\nonumber\\
\leq&\frac12\mathbb{E}\lambda(T+1)^{\lambda-1}\|v(T)\|^2_{L^2(G)}+\frac12\sum_{i,j=1}^n\|a^{ij}(\varepsilon)\|^2_{L^\infty_{{\cal F}_\varepsilon}(\Omega;L^\infty(G))}\mathbb{E}\|\nabla v(\varepsilon)\|^2_{L^2(G)}\nonumber\\
&+C\mathbb{E} \int_{\varepsilon}^{T} \int_{G} \varphi^2 (f^2+|\nabla g|^2+g^{2}) \mathrm{d} x \mathrm{d} t.\nonumber
\end{align}
This gives \eqref{519eq3}.
\end{proof}

\section{Conditional Stability}

In this section, we establish a conditional stability of the inverse problem ({\bf IPD}).

Let us first introduce the a priori bound for the initial data. Let  $M>0$ and set
\begin{equation}\label{615eq4-2}
	U_M\= \{\xi\in L^2_{{\cal F}_0}(\Omega;H^1(G))| |\nabla \xi|\leq M, \; {\mathbb{P}}\mbox{-a.s.}\}.
\end{equation}

\begin{theorem}\label{reguest}
Let $\delta_0>0$ be sufficiently small such that
	\begin{equation}\label{615eq5}
		\ln\(\ln\(\delta_0^{-\frac13}\)^{\frac{1}{\ln(T+1)}}\)\geq\lambda_0.
	\end{equation}
Suppose that $u_1,u_2 $ are weak solutions of problem \eqref{p1} with initial data belong to $U_M$ and $||u_1(T)-u_2(T)||_{L^2(G)}=\delta\leq \delta_0$.
Then the following estimate hold
	\begin{equation}\label{616eq6}
		\|u_1-u_2\|_{L^2_\mathbb{F}(\varepsilon,T;H^1(G))}\leq C(M+1)e^{-\frac{1}{3^c}(\ln(||u_1(T)-u_2(T)||_{L^2(G)}^{-1})^c)},
	\end{equation}
	where $C$ is independent of $M$ and $c=c(\varepsilon,T)=\frac{\ln(\varepsilon +1)}{\ln(T+1)}\in (0,1)$.
\end{theorem}

\begin{proof}  Let  $w(x,t)=u_1(x,t)-u_2(x,t)$ and $R(x)=u_1(x,T)-u_2(x,T)$. From \eqref{p1}, we have
 \begin{equation}\label{615eq6}
\left\{\begin{array}{ll}
\displaystyle\mathrm{d} w-\sum_{i, j=1}^{n}(a^{i j} w_{i})_j \mathrm{d} t=\left( b_{1}\cdot\nabla w +b_{2} w\right)\mathrm{d} t+b_{3} w \mathrm{d} W(t) & \text { in } Q, \\
\noalign{\medskip} w(x,t) =0  & \text { on } \Sigma,  \\
\noalign{\medskip} w(x,T)=R(x)  & \text { in } G.
\end{array}\right.
\end{equation}

Applying \eqref{519eq3} to \eqref{615eq6}, we obtain
\begin{align}\label{616eq1}
&\frac12\lambda(T+1)^{\lambda-1}e^{2(T+1)^\lambda}\mathbb{E}\|w(T)\|^2_{L^2(G)}+\frac12\sum_{i,j=1}^n\|a^{ij}\|^2_{L^\infty_{{\cal F}_\varepsilon}(\Omega;L^\infty(G))}e^{2(\varepsilon+1)^\lambda}\mathbb{E}\|w(\varepsilon)\|^2_{L^2(G)}\nonumber\\
\geq&\mathbb{E}\int_\varepsilon ^T\int_G \lambda^2 (t+1)^{(\lambda-2)}\varphi^2 w^2 \mathrm{d} t\mathrm{d}x+\mathbb{E}\int_\varepsilon^T\int_G \lambda(t+1)^{-1}\varphi^2|\nabla w|^2 \mathrm{d} t\mathrm{d}x\nonumber\\
\geq&\mathbb{E}\int_\varepsilon^T\int_G \lambda^2 \varphi^2w^2 \mathrm{d} t\mathrm{d}x+(\varepsilon+1)^{-1}\mathbb{E}\int_\varepsilon^T\int_G \lambda\varphi^2|\nabla w|^2  \mathrm{d} t\mathrm{d}x\\
\geq&\bar{C}e^{2(\varepsilon +1)^\lambda}\mathbb{E}\int_\varepsilon^T\int_G \big(w^2+|\nabla w|^2\big) \mathrm{d} t\mathrm{d}x\nonumber\\
=&\bar{C}e^{2(\varepsilon +1)^\lambda}\|w\|_{L^2_\mathbb{F}(\varepsilon,T;H^1(G))}.\nonumber
\end{align}
By choosing $\lambda$ large enough such that $\lambda(T+1)^{\lambda-1}e^{2(T+1)^\lambda}\leq e^{3(T+1)^\lambda}$, then there hold
\begin{align}\label{616eq2}
\|w\|^2_{L^2_\mathbb{F}(\varepsilon,T;H^1(G))}\leq \bar{C}_1e^{3(T+1)^\lambda}\mathbb{E}\|w(T)\|^2_{L^2(G)}+\bar{C}_2\mathbb{E}\|\nabla w(\varepsilon)\|^2_{L^2(G)}.
\end{align}

Choosing
$\lambda=\lambda(\delta)$
as
\begin{equation}\label{616eq4}
\lambda=\lambda(\delta)=\ln\[\(\ln(\delta^{-\frac13})\)^\frac{1}{\ln(T+1)}\],
\end{equation}
and substituting it in to \eqref{616eq2} we get
\begin{equation}\label{616eq5}
\|w\|^2_{L^2_\mathbb{F}(\varepsilon,T;H^1(G)}\leq \bar{C}_1\delta+\bar{C}_2M^2e^{-\frac{2}{3^c}\ln(\delta^{-1})^c}.
\end{equation}
This implies \eqref{616eq6}.
\end{proof}

\begin{remark}\label{rmk3.1}
In Theorem \ref{reguest}, the norm in the left hand side is $\|\cdot\|_{L^2_{\mathbb{F}}(\varepsilon,T;H^1(G)}$. We can also obtain the estimate at each $t_0\in(0,T)$. The argument is as follows. Choose $\varepsilon<t_0$ in Theorem \ref{reguest}. Since
\begin{equation}\label{ree1}
C\|w(s)\|_{H^1(G)}^2\geq\|w(t_0)\|_{H^1(G)}^2, \qquad \forall \varepsilon\leqslant s\leqslant t_0,
\end{equation}
integrating both side of \eqref{ree1} with respect to $s$ in $[\varepsilon,t_0]$, we get
\begin{equation}\label{ree2}
C\int_\varepsilon^{t_0}\|w(s)\|_{H^1(G)}^2ds\geq (t_0-\varepsilon)\|w(t_0)\|_{H^1(G)}^2.
\end{equation}
Combining \eqref{ree2} with \eqref{616eq5}, we can obtain
\begin{equation*}
\|w(t_0)\|_{H^1(G)}^2\leq C(t_0-\varepsilon)^{-1}\int_0^{t_0}\|w(s)\|_{H^1(G)}^2ds\leq C(t_0-\varepsilon)^{-1}(\bar{C}_2M+\bar{C}_1)e^{-\frac{2}{3^c}(\ln(\delta^{-1})^c)}.
\end{equation*}
\end{remark}

\section{Regularization for the reconstruction problem}

In the following, we prove a convergence result by the Carleman estimate \eqref{carleman}. Suppose $u_T \in L^2_{{\cal F}_T}(\Omega;L^2(G))$ is the exact terminal value of the initial-boundary problem \eqref{p1} with the initial datum $u_0\in L^2_{{\cal F}_0}(\Omega;H^2(G)\cap H_0^1(G))$,  and $u_T^\delta\in L^2_{{\cal F}_T}(\Omega;L^2(G)) $ the noisy data satisfying
\begin{equation}\label{620eq2}
\|u_T-u_T^\delta\|_{L^2_{{\cal F}_T}(\Omega;L^2(G))}\leq \delta,
\end{equation}
where $\delta>0$ is the noise level of data.

Consider  the following Tikhonov type functional on $L^2_{{\cal F}_0}(\Omega;H^2(G)\cap H_0^1(G))$:
\begin{align}\label{616eq7}
J_\alpha(y_{0})\! =& \mathbb{E}\! \int_G\! \Big\{
y_{0}+\!\int_{t_0}^{T}\!\[\sum_{i, j=1}^{n}(a^{i j} y_{i})_j\!+\!b_{1}\cdot \nabla y \!+\! b_{2} y\!+\!f\] \mathrm{d}s +\!\int_{t_0}^{T}\! \left(b_{3}  y \!+\!g \right) \mathrm{d} W(s)\!-\!u_T^\delta\Big\}^2\mathrm{d}x \nonumber\\ &+\!\alpha{\mathbb{E}}\|y_{0}\|^2_{H^2(G)},
\qquad\qq\qquad  \forall y_0\in L^2_{{\cal F}_0}(\Omega;H^2(G)\cap H_0^1(G)),
\end{align}
where $y$ solves the following equation:
\begin{equation}\label{p2}
	\left\{\begin{array}{ll}
		\displaystyle\mathrm{d} y-\sum_{i, j=1}^{n}(a^{i j} y_{i})_j \mathrm{d} t=\left( b_{1}\cdot \nabla y +b_{2} y+f\right) \mathrm{d} t+\left(b_{3} y+g\right) \mathrm{d} W & \text { in } (0,T)\times G, \\
		\noalign{\medskip} y=0 & \text { on } (0,T)\times \partial G, \\
		\noalign{\medskip} y(0)=y_0, & \text { in } G.
	\end{array}\right.\end{equation}

\noindent{\bf Minimization problem.} Minimize the functional $J_\alpha(y_0)$ on the space $L^2_{{\cal F}_{0}}(\Omega;H^2(G)\cap H_0^1(G))$.

Since the functional  $J_\alpha(y_0)$ is coercive, convex and lower semi-continuous, it has a unique minimizer $\bar y_{0}^\delta$.

\begin{theorem}[Convergence rate]
Assume that conditions (H1)--(H3) hold. Let $\lambda_0>1$ be the number in Theorem \ref{carleman} and let $\alpha=\alpha(\delta)=\delta^2$. Then there exists a number
\begin{equation}\label{620eq3}
\lambda_3=\lambda_3\(\sigma,\max_{i,j}\|a^{ij}\|_{L^\infty(G;\mathbb{R}^{n\times n})}, G\)\geq\lambda_0>1,
\end{equation}
such that if $\delta\in(0,\delta_0)$  and $\delta\in(0,1)$ is sufficient small such that
\begin{align}\label{620eq4-2}
 \lambda_3\leq\ln\[\(\ln\(\delta_0^{-\frac12}\)\)^{\frac{1}{\ln(T+1)}}\],\quad
 \delta_0\leq e^{-3^{1-c}(\ln(\delta_0^{-1}))^c},
\end{align}
then the following convergence estimate of the Tikhonov method holds for every $\tau\in(0,T)$,
\begin{equation}
	\begin{array}{ll}\displaystyle
\|y(\tau;\bar y_{0}^\delta)-y(\tau;u_0)\|_{L^2_{{\cal F}_{t_0}}(\Omega;H^2(G)\cap H_0^1(G))}\\
\noalign{\medskip}\displaystyle\leq C_1\big(1+\|y_0\|_{L^2_{{\cal F}_{t_0}}(\Omega;H^2(G)\cap H_0^1(G))}\big)e^{-3^{-c}\big(\ln(\delta_0^{-1})\big)^c},
\end{array}
\end{equation}
where $c=c(\tau,T)=\frac{\ln(\tau+1)}{\ln(T+1)}\in(0,1)$.
\end{theorem}

\begin{proof}
For $y_0\in L^2_{{\cal F}_{0}}(\Omega;H^2(G)\cap H_0^1(G))$, denote by $y(\cdot;y_{0})$ the solution to \eqref{p2} with $y(t_0)=y_{0}$. For simplicity of notations, we denote
$$\begin{aligned}
	{\cal A} y_0\triangleq & y_0+ \int_0^T \[\sum_{i, j=1}^{n}(a^{i j}(s) y_{i}(s;y_0))_j\!+\!b_{1}(s)\cdot \nabla y(s;y_0) \!+\! b_{2} y(s;y_0)\!+\!f(s)\]\text{d}s \\
	& +\int_0^T\int_G[b_3(s)y(s;y_0)+g(s)]\text{d}W(s), \quad {\mathbb{P}}\mbox{-a.s.}
\end{aligned}$$
and
$$\begin{aligned}
	{\cal B} y_0\triangleq & y_0+ \int_0^T \[\sum_{i, j=1}^{n}(a^{i j}(s) y_{i}(s;y_0))_j\!+\!b_{1}(s)\cdot \nabla y(s;y_0) \!+\! b_{2}(s) y(s;y_0) \]\text{d}s \\
	& +\int_0^T\int_G b_3(s)y(s;y_0) \text{d}W(s), \quad {\mathbb{P}}\mbox{-a.s.}
\end{aligned}$$

By computing G\^ateaux derivative of $J_\alpha(y_0)$, we get that
\begin{align}\label{620eq4}
J_\alpha'(y_0)\phi_0=2{\mathbb{E}}\int_G {\cal A} y_{0}{\cal B} \phi_0 \mathrm{d}x-2{\mathbb{E}}\int_G u_T(x){\cal B} \phi_0\mathrm{d}x +2\alpha{\mathbb{E}}\langle y_{0},\phi_0\rangle_{H^2(G)},\\
\qquad\quad \forall \phi_0\in L^2_{{\cal F}_{0}}(\Omega;H^2(G)\cap H_0^1(G)).\nonumber
\end{align}
Particularly, for
the minimizer $\bar y_{0}^\delta\in L^2_{{\cal F}_{0}}(\Omega;H^2(G)\cap H_0^1(G))$ of \eqref{616eq7}, we have
\begin{align}\label{620eq4-1}
{\mathbb{E}}\int_G {\cal A} \bar y_{0}^\delta{\cal B}\phi_0 \mathrm{d}x+ \alpha{\mathbb{E}}\langle \bar y_{0}^\delta,\phi_0\rangle_{H^2(G)}
 =  {\mathbb{E}}\int_G u_T^\delta(x){\cal B}\phi_0\mathrm{d}x,\\
	\qquad\quad  \forall \phi_0\in L^2_{{\cal F}_{0}}(\Omega;H^2(G)\cap H_0^1(G)).\nonumber
\end{align}
Noting that $u_0$ is the exact initial value, we have
\begin{align}\label{620eq4-3}
{\mathbb{E}}\int_G {\cal A} u_{0}{\cal B}\phi_0 \mathrm{d}x+ \alpha{\mathbb{E}}\langle u_{0},\phi_0\rangle_{H^2(G)}
=  {\mathbb{E}}\int_G u_T(x){\cal B}\phi_0\mathrm{d}x+ \alpha{\mathbb{E}}\langle u_{0},\phi_0\rangle_{H^2(G)},\\
\qquad\quad  \forall \phi_0\in L^2_{{\cal F}_{0}}(\Omega;H^2(G)\cap H_0^1(G)).\nonumber
\end{align}

From \eqref{620eq4-1} and \eqref{620eq4-3}, we obtain
\begin{align}\label{620eq5}
&{\mathbb{E}}\int_G \big({\cal A} \bar y_{0}^\delta-{\cal A} u_{0}\big) {\cal B}\phi_0 \mathrm{d}x+ \alpha{\mathbb{E}}\langle \bar y_{0}^\delta-u_{0},\phi_0\rangle_{H^2(G)}\\
\noalign{\medskip}=&\int_G (u_T^\delta(x)-u_T(x)) {\cal B}\phi_0 \mathrm{d}x-\alpha{\mathbb{E}}\langle u_{0},\phi_0\rangle_{H^2(G)},\quad  \forall \phi_0\in L^2_{{\cal F}_{0}}(\Omega;H^2(G)\cap H_0^1(G)).\nonumber
\end{align}
Set $\phi_0=\bar y_{0}^\delta-u_0$ in \eqref{620eq5}. By using Cauchy-Schwarz inequality, we obtain
\begin{align}\label{620eq6}
&\int_G\big({\cal A} \bar y_{0}^\delta-{\cal A} u_0\big) ^2\mathrm{d}x+\alpha{\mathbb{E}}\|\bar y_{0}^\delta-u_0\|^2_{H^2(G)}\\
\leq & \frac12\int_G\big({\cal A} \bar y_{0}^\delta-{\cal A} u_0\big)^2\mathrm{d}x+\frac12{\mathbb{E}}\| u_T^\delta-u_T\|_{H^2(G)}^2 + \frac12\alpha\|u_0\|_{H^2(G)}^2+\frac12\alpha{\mathbb{E}}\|\bar y_{0}^\delta-u_0\|^2_{H^2(G)}.\nonumber
\end{align}
Simplify \eqref{620eq6} and noting \eqref{620eq3}, we have
\begin{align}\label{620eq7}
\int_G\big({\cal A} \bar y_{0}^\delta-{\cal A} u_0\big)^2\mathrm{d}x+\alpha{\mathbb{E}}\|\bar y_{0}^\delta-u_0\|^2_{H^2(G)} \leq \delta^2+\alpha\|u_0\|_{H^2(G)}^2.
\end{align}
Since $\alpha=\alpha(\delta)=\delta^2$, we know from \eqref{620eq7} that
\begin{equation}\label{620eq8}
{\mathbb{E}}\|u_{0}-\bar y_0^\delta\|^2_{H^2(G)}\leq 1+\|u_0\|_{H^2(G)}^2.
\end{equation}

Let $\tilde u=y(\cdot;u_{0})-y(\cdot;\bar y_0^\delta)$. Then $\tilde u$ solves the following equation
\begin{equation}\label{p2-1}
\left\{\begin{array}{ll}\displaystyle\mathrm{d} \tilde{u}-\sum_{i, j=1}^{n}(a^{i j} \tilde{u}_{i})_j \mathrm{d} t=\left(b_{1}\cdot\nabla \tilde{u}+b_{2} \tilde{u}\right) \mathrm{d} t+b_{3} \tilde{u} \mathrm{d} W(t)& \text { in } Q, \\
\noalign{\medskip} \tilde{u}=0 & \text { on } \Sigma, \\
\noalign{\medskip} \tilde{u}(0)=u_{0}-\bar y_0^\delta, & \text { in } G.
\end{array}\right.\end{equation}
By \eqref{519eq3}, we get that
\begin{align}\label{620eq9}
&\lambda^2 \mathbb{E}\int_\varepsilon^T\int_G (t+1)^{(\lambda-2)}\varphi^2\tilde{u}^2\mathrm{d} t\mathrm{d}x+ \lambda\mathbb{E}\int_\varepsilon^T\int_G(t+1)^{-1}\varphi^2|\nabla \tilde{u}|^2\mathrm{d} t\mathrm{d}x\\
\noalign{\medskip}\leq& C\[\mathbb{E}\lambda(T\!+\!1)^{\lambda-1}e^{2(T+1)^\lambda}\|\tilde{u}(T)\|^2_{L^2(G)}+\! \sum_{i,j=1}^n\!\|a^{ij}\|^2_{L_{\mathbb{F}}^{\infty} ( 0, T; L^{\infty}(G) )}\mathbb{E}\|\nabla(u_{0}\!-\!\bar y_0^\delta)\|^2_{L^2(G)}\].\nonumber
\end{align}

Since
\begin{align}\label{621eq1}
{\mathbb{E}}\|\tilde{u}(T)\|^2_{L^2(G)}\leq {\mathbb{E}}\|y(T;u_{0})-u_T^\delta\|^2_{L^2(G)}+{\mathbb{E}}\|u_T^\delta-y(T; \bar y_0^\delta)\|^2_{L^2(G)},
\end{align}
and due to $\bar y_{0}^\delta$ is the minimizer of $J_\alpha(\cdot)$, we know
\begin{equation}\label{621eq2}
\begin{array}{ll}\displaystyle
{\mathbb{E}}\|y(T;\bar y_{0}^\delta)-u_T^\delta\|^2_{L^2(G)}\negthinspace \negthinspace \negthinspace &\displaystyle\leq J_\alpha(\bar y_{0}^\delta)\leq J_\alpha(u_0)\\
\noalign{\medskip}&\displaystyle={\mathbb{E}}\|u_T^\delta-y(T; u_{0})\|^2_{L^2(G)}+\alpha{\mathbb{E}}\|u_0\|^2_{H^2(G)}.
\end{array}
\end{equation}
Substituting \eqref{621eq2} into \eqref{621eq1}, we get
\begin{equation}\label{621eq3}
\begin{array}{ll}\displaystyle
{\mathbb{E}}\|\tilde{u}(T)\|^2_{L^2(G)}\negthinspace \negthinspace \negthinspace &\displaystyle\leq2{\mathbb{E}}\|u_T^\delta-y(T; u_{0})\|^2_{L^2(G)}+\alpha{\mathbb{E}}\|u_0\|^2_{H^2(G)}\\
\noalign{\medskip}&\displaystyle=2\delta^2+\alpha{\mathbb{E}}\|u_0\|^2_{H^2(G)}.
\end{array}
\end{equation}

Combining \eqref{620eq9} with \eqref{621eq3}, we have
\begin{align*}\label{621eq4}
&\mathbb{E}\int_\varepsilon^T\int_G \lambda^2 (t+1)^{(\lambda-2)}e^{2(t+1)^\lambda}\tilde{u}^2\mathrm{d} t\mathrm{d}x+\mathbb{E}\int_\varepsilon^T\int_G \lambda(t+1)^{-1}e^{2(t+1)^\lambda}|\nabla \tilde{u}|^2\mathrm{d} t\mathrm{d}x\nonumber\\
\noalign{\medskip}\leq&\frac12\mathbb{E}\lambda(T + 1)^{\lambda-1}\varphi(T)^2(2\delta^2 + \alpha{\mathbb{E}}\|u_0\|^2_{H^2(G)})\\
& +\frac{e^2}2\sum_{i, j=1}^{n}\|a^{ij}(0)\|^2_{L^\infty_{{\cal F}_0}(L^\infty(G))}\mathbb{E}\|\nabla(u_{0}\!-y_0)\|^2_{L^2(G)}.
\end{align*}

The following process is similar to the proof of Theorem \ref{reguest}.
 \end{proof}

\section{Numerical Approximation}

  Compared to deterministic parabolic equation, numerically solving the inverse problem of stochastic parabolic equation is more complicate and difficult.
On one hand, the analytic solution of stochastic parabolic equation can not be explicitly expressed; on the other hand, the solution of stochastic parabolic equation is not differentiable with respect to the temporal  variable. Moreover, for many sharp method working well for deterministic problem, there are new essential difficulty. For example, when using the conjugate gradient method to the stochastic problem, the adjoint system is a backward stochastic parabolic equation. Thus, one has to solve a forward-backward stochastic parabolic equation numerically. The solution of this equation is three stochastic processes, whereas the last one with lower regularity, and thus is difficulty to be numerically solved.
We  combine  conjugate gradient method and Picard type algorithm for forward-backward stochastic parabolic equation introduced in \cite{DP2016SISC} to solve \eqref{p1} numerically.
To this end, we need to employ the adjoint equation of \eqref{p1}, which is a backward stochastic parabolic equation. As a result,  we assume that   $\left\{\mathcal{F}_{t}\right\}_{t \geqslant 0}$ is the natural filtration generated by $W(\cdot)$.

\subsection{Conjugate gradient method  for the regularization method}
The conjugate gradient (CG) method is an iterative algorithm for the numerical solution of linear systems.  Thus can be used to numerically solve partial differential equations and unconstrained optimization problems such as energy minimization. For readers' convenience, we describe the conjugate gradient method for finding the solution $x$ which satisfies
$$
x=\operatorname{argmin}_{x\in \mathbb{R}^{n}} \left(\frac{1}{2} x^{\top} A x-b^{\top} x\right)=\operatorname{argmin}_{x\in \mathbb{R}^{n}}f(x).
$$
The numerical approximation of the above system given by the CG method after $k+1$ steps is
\begin{equation}\label{2.1-eq1}
x_{k+1}=x_0+\sum_{j=0}^{k}\alpha_j p_j=x_k+\alpha_k p_k.
\end{equation}
In \eqref{2.1-eq1},  $x_0$ is the initial guess of the solution,
$$
p_{k}=r_{k}-\sum_{i=1}^{k-1} \frac{\left\langle r_{k}, A p_{i}\right\rangle}{\left\langle p_{i}, A p_{i}\right\rangle} p_{i}.
$$
for $r_k\triangleq b-A x_{k}$, and
the step size $\alpha_k$ can be chosen to minimize $f(x_k+\alpha_k p_k)$.

Now we turn to our problem.
As we described in Section 4, our purpose is to minimize the functional
\begin{equation}\label{Tikhonov functional}
J_\alpha(y_0)= \mathbb{E} \int_G\big({\cal A} y_0-u_T^\delta(x)\big)^2\text{d}x +\alpha||y_0||_{H^2(G)}^2,\quad y_0\in H^2(G)\cap H_0^1(G).
\end{equation}

Let $\{y_0^k\}_{k=1}^\infty$ be the approximation sequence generated by the conjugate gradient method to $\bar y_{0}$, i.e.,
\begin{equation}
y_0^{k+1} = y_0^{k}+\beta_kd_k,\ k=0,1,2,\cdots
\label{uk}
\end{equation}
In \eqref{uk}, $k$ is the iteration index, $\beta_k$ and $d_k$ are the step size and descent direction in the $k$-th iteration given as follows:
\begin{equation}
\beta_k=-\frac{\displaystyle\mathbb{E} \int_G({\cal A} y_0^k-u_T^\delta(x)){\cal A} d_k\text{d}x +\alpha \langle y_0^k,d_k\rangle_{H^2(G)}}{\displaystyle\mathbb{E} \int_G({\cal A} d_k)^2\text{d}x +\alpha |d_k|_{H^2(G)}^2}.
\label{betak}
\end{equation}
\begin{equation}
d_0=0,\quad d_k=-J_\alpha'(y_0^k)+\gamma_kd_{k-1} \mbox{ for } k=1,2,\cdots,
\label{dk}
\end{equation}
where $J_\alpha'$ is the G\^ateaux derivative of the functional \eqref{Tikhonov functional} and $\gamma_k$ is the conjugate coefficient given inductively:
\begin{equation}
\gamma_0=0,\quad \gamma_k=\[{\mathbb{E}}\int_G(J_\alpha'(y_0^{k-1}))\text{d}x\]^{-1} {\mathbb{E}}\int_G(J_\alpha'(y_0^k))^2\text{d}x, \quad k=1,2,\cdots
\label{gammak}
\end{equation}

By the above formulae, we should compute the sensitivity term ${\cal A} d_k$ and the G\^ateaux derivative $J_\alpha'$. This is done in the new two subsections.

\subsection{The sensitivity problem and the adjoint problem}

In this subsection, we  compute the sensitivity term ${\cal A} d_k$. For simplicity, we use $|\Delta v|_{L^2(G)}$ as the norm of $H^2(G)\cap H_0^1(G)$ for $v\in H^2(G)\cap H_0^1(G)$.

For any $v_0\in H^2(G)\cap H_0^1(G)$,
\begin{equation}
\begin{aligned}
  & J_\alpha(y_0+\delta v_0)-J_\alpha(y_0) \\
= & \mathbb{E}\big(2\langle {\cal A} y_0-u_T^\delta,{\cal A}\delta v_0\rangle _{L^2(G)}+ |{\cal A}\delta v_0|_{L^2(G)}^2\big)+2\alpha\langle y_0,\delta v_0\rangle_{H^2(G)} +\alpha |\delta v_0|_{H^2(G)}^2\\
= & \mathbb{E}\big(2\langle {\cal A} y_0-u_T^\delta,{\cal A}\delta v_0\rangle _{L^2(G)}+ |{\cal A}\delta v_0|_{L^2(G)}^2\big)+2\alpha\langle \Delta y_0,\delta \Delta v_0\rangle_{L^2(G)} +\alpha |\delta\Delta v_0|_{L^2(G)}^2.
\end{aligned}
\end{equation}
Since ${\cal A}$ is a linearly continuous operator, we have
\begin{equation}
\begin{aligned}
&\lim_{\delta\to0}\frac{J_\alpha(y_0+\delta v_0)-J_\alpha(y_0)}{\delta} \\
=&  2\mathbb{E}\langle {\cal A} y_0-u_T^\delta,{\cal A} v_0\rangle _{L^2(G)} +2\alpha\langle \Delta y_0,\Delta v_0\rangle_{H^2(G)}\\
=& 2\mathbb{E}\langle (-\Delta)^{-1}({\cal A} y_0-u_T^\delta),(-\Delta)^{-1}{\cal A} v_0\rangle _{H^2(G)} +2\alpha\langle  y_0, v_0\rangle_{H^2(G)}.
\end{aligned}
\end{equation}
Then the G\^{a}teaux derivative of $J$ is
\begin{equation}
J_\alpha'(y_0)=(-\Delta)^{-2}{\cal A}^*({\cal A} y_0-u_T^\delta)+\alpha y_0.
\label{J'}
\end{equation}

Denote $V=y(\cdot,\bar y_{0})-y(\cdot,y_0)$, then the following sensitivity problem is obtained
\begin{equation}
\begin{cases}
\text{d}V-\sum\limits_{i,j=1}^n(a^{ij}V_i)_j\text{d}t=\left(b_1\cdot\nabla V+b_2V\right)\text{d}t+b_3VdW(t) & \text{in}\ Q,\\
V=0 & \text{on}\ \Sigma, \\
V(0)=\bar y_{0}-y_0 & \text{in}\ G.
\end{cases}
\label{sensitivity problem}
\end{equation}
The adjoint problem of (\ref{sensitivity problem}) is
\begin{equation}
\begin{cases}
\text{d}Y=\[-\sum\limits_{i,j=1}^n(a^{ij}Y_i)_j-\nabla\cdot(b_1 Y)-b_2 Y-b_3  Z \]\text{d}t+ZdW(t) & \text{in}\ Q,\\
Y=0 & \text{on}\ \Sigma, \\
Y(T)={\cal A} y_0-u_T^\delta & \text{in}\ G.
\end{cases}
\label{adjoint problem}
\end{equation}
The existence and uniqueness of the strong solution of this adjoint problem is proved in \cite{DT2012}.

We summarise the conjugate gradient method to solve the minimization problem (\ref{Tikhonov functional}) in Algorithm 1.

\begin{table}[htbp]
\centering
\begin{tabular}{ll}
\toprule
&\textbf{Algorithm 1} Conjugate gradient algorithm for the inverse problem \bf{IPD}\\
\midrule
1: & Choose an initial guess $y_0^0$. Set $k=0$. \\
2: & Solve the initial boundary problem (\ref{p1}) with $y_0^k$, \\
   & and determine the residual $r_k=y^k(T)-u_T^\delta$. \\
3: & Solve the adjoint problem (\ref{adjoint problem}) and determine $J_\alpha'$ by (\ref{J'}). \\
4: & Calculate the conjugate coefficient $\gamma_k$ by \eqref{gammak} and the descent direction $d_k$ by \eqref{dk}. \\
5: & Solve the sensitivity problem (\ref{sensitivity problem}) for $Ad_k$ with $V(0)=d_k$. \\
6: & Calculate the stepsize $\beta_k$ by (\ref{betak}). \\
7: & Compute a new estimate, $y_0^{k+1}$, with (\ref{uk}). \\
8: & Interrupt the iterative procedure if the stopping criterion is satisfied.\\
   & Otherwise, increase k by 1 and go back to Step 2.\\
\bottomrule
\end{tabular}
\end{table}

\subsection{Picard type algorithm for forward-backward stochastic parabolic equation}

In this subsection, borrowing some idea in \cite{DP2016SISC}, we present fully implementable algorithms to simulate the equations (\ref{p1}), (\ref{sensitivity problem}) and (\ref{adjoint problem}).

For simplicity of notations, we suppose $(a^{ij})_{1\leq i,j\leq n}$ be the identity matrix and $f=g=0$ in (\ref{FBSPDE}) throughout this subsection.

We write (\ref{p1}) and (\ref{adjoint problem}) together as follow and called them the forward-backward stochastic heat equation (FBSPDE)
\begin{equation}
\begin{cases}
\text{d}U=\big(\Delta U + b_1\cdot\nabla U +b_2U\big)\text{d}t+ b_3U dW(t) & \text{in}\ Q,\\
\text{d}Y=-\big[\Delta Y +\hbox{\rm div$\,$}(b_1 Y)+b_2 Y+b_3 Z\big]\text{d}t+ZdW(t) & \text{in}\ Q,\\
U(0)=u_0 & \text{in}\ G,\\
Y(T)={\cal A} y_0-u_T^\delta & \text{in}\ G.
\end{cases}
\label{FBSPDE}
\end{equation}

In order to discretize above equations, we introduce some notation here. Let $\mathcal{M}_h$ be a regular mesh of $G\subset\mathbb{R}^n$ into element domains $K$ with a maximum mesh size $h\triangleq\max\{diam(K)|K\in\mathcal{M}_h\}$. For each $K\in\mathcal{M}_h$, let $\mathcal{P}_j(K)$ denote the set of all polynomials of degree less than or equal to $j$ on $K$, and we define the finite element space $\mathbb{U}_h\subset H_0^{1}$ by
\begin{equation*}
	\mathbb{U}_h\triangleq\{\phi\in C^k(\bar{G})|\phi_{|K}\in\mathcal{P}_j(K),\ \forall K\in\mathcal{M}_h\}.
\end{equation*}
The $\mathbb{L}^2$-projection $\Pi_h: \mathbb{L}^2\to \mathbb{U}_h$ is defined by $(\Pi_h\xi-\xi, \phi_h)=0$ for all $\phi_h\in\mathbb{U}_h$.

The spatial discretization of (\ref{FBSPDE}) is as follows:

For all $t\in[0,T]$, there holds $\mathbb{P}$-a.s.
\begin{equation}\label{2.3-eq1}
\begin{aligned}
&\int_G U_h(t) \phi_h\text{d}x+ \int_0^t\int_G \nabla U_h \cdot \nabla\phi_h \text{d}x\text{d}s\\
=&\int_G u_0 \phi_h\text{d}x-\int_0^t\int_G [U_h,\hbox{\rm div$\,$}(b_1\phi_h)- b_2U_h \phi_h ]\text{d}x\text{d}s+\int_0^t\int_G b_3U_h \phi_h \text{d}x\text{d}W(s),
\end{aligned}
\end{equation}
and
\begin{equation}\label{2.3-eq2}
\begin{aligned}
\int_G Y_h(t) \phi_h \text{d}x= & \int_G ({\cal A} y_0-u_T^\delta)\phi_h \text{d}x- \int_t^T\int_G (\nabla Y_h\cdot\nabla\phi_h + Y_hb_1\cdot\nabla\phi_h ) \text{d}x\text{d}s\\
& +\int_t^T\int_G (b_2  Y_h \phi_h + b_3  Z_h \phi_h)\text{d}x\text{d}s-\int_t^T\int_G  Z_h \phi_h\text{d}x\text{d}W(s).
\end{aligned}
\end{equation}
For every fixed $h>0$, there exsit a unique  solution $(U_h,Y_h,Z_h)\in L_{\mathbb{F}}^2(\Omega,C([0,T];\mathbb{U}_h))\times L_{\mathbb{F}}^2(\Omega,C([0,T];\mathbb{U}_h)) \times L_{\mathbb{F}}^2(\Omega,L^2([0,T];\mathbb{U}_h))$ to \eqref{2.3-eq1} and \eqref{2.3-eq2}.

Denote the time discretization of $(U_h,Y_h,Z_h)$  by $\{(U_h^m(t),Y_h^m(t),Z_h^m(t))|m=0,\cdots,{\mathbb{M}}\}$. Let $k=t_{m+1}-t_m$ be the uniform time step for a net $\{t_m\}_{m=0}^\mathbb{M}$ which covers $[0,T]$. Let $\Delta_mW=W(t_m)-W(t_{m-1})$. Then we can simulate $\{(U_h^m(t),Y_h^m(t),Z_h^m(t))|m=0,\cdots,{\mathbb{M}}\}$ as follow:

(i) Simulate $U_h^0=\Pi_h(y_0)$.

(ii) For each $m=0,\cdots,{\mathbb{M}}-1$, simulate $U_h^m$ such that for each $ \phi_h\in\mathbb{U}_h$,
$$
\begin{aligned}
&\int_G U_h^{m+1} \phi_h \text{d}x +k\int_G \nabla U^{m+1}_h\cdot\nabla\phi_h \text{d}x\\
= & \int_G U_h^m \phi_h \text{d}x -k\int_G \big[U_h^m \hbox{\rm div$\,$}(b_1\phi_h)- b_2U_h^m \phi_h\big] \text{d}x+  \Delta_{m+1}W \int_G b_3U_h^m \phi_h \text{d}x.
\end{aligned}
$$

(iii) Simulate $Y_h^{\mathbb{M}}=U_h^{\mathbb{M}}-\Pi_h(u_h^{{\mathbb{M}}})$.

(iv) For each $m={\mathbb{M}}-1,\cdots,0$, simulate $Y_h^m$ and $Z_h^m$ such that for each $  \phi_h\in\mathbb{U}_h$,
$$
\int_G Z_h^m \phi_h \text{d}x=\frac{1}{k}\mathbb{E}\left[\Delta_{m+1}W \int_G Y_h^{m+1} \phi_h \text{d}x\Big|{\cal F}_{t_m}\right],
$$
and
$$
\begin{aligned}
& \int_G Y_h^m \phi_h\text{d}x + k\int_G \nabla Y^{m}_h \cdot\nabla \phi_h\text{d}x+k \int_G (Y_h^m b_1 \cdot \nabla\phi_h - b_2 Y_h^m \phi_h)\text{d}x\\
&=\mathbb{E}\left[\int_G Y_h^{m+1} \phi_h\text{d}x\Big|{\cal F}_{t_m}\right]+k\int_G b_3Z^m_h \phi_h\text{d}x.
\end{aligned}
$$

For $l=1,\dots,L$, let $\phi_h^l\in\mathbb{U}_h$ be basis functions of $\mathbb{U}_h$. Let $U_h(x,t)=\sum\limits_{l=1}^L{u_h^l(t)\phi_h^l(x)}$, $Y_h(x,t)=\sum\limits_{l=1}^L{y_h^l(t)\phi_h^l(x)}$ and $Z_h(x,t)=\sum\limits_{l=1}^L{z_h^l(t)\phi_h^l(x)}$ with coefficient vectors $\vec{U}_h, \vec{Y}_h,$ $ \vec{Z}_h\in\mathbb{R}^L$. Denote by $\textbf{Stiff}$ the stiffness matrix  consisting of entries $\int_G \nabla\phi_h^l\cdot\nabla\phi_h^w \text{d}x$, where $\phi_h^l,\phi_h^w\in\mathbb{U}_h$ are basis functions of $\mathbb{U}_h$; by $\textbf{MG}$   the matirx consisting of entries $\int_G \phi_h^l b_1 \cdot\nabla\phi_h^w \text{d}x$; by $\textbf{MD}$  the matrix consisting of entries   $\int_G \nabla\phi_h^l\cdot \nabla\cdot(b_1\phi_h^w)\text{d}x$ and by \textbf{Mass} the mass matrices consisting of entries $\int_G \phi_h^l \phi_h^w\text{d}x$. Then the semi-discretization above can be restated as an algebraic problem:
\begin{equation}
\begin{aligned}
(\textbf{Mass}+k\ \textbf{Stiff})\ \vec{U}_h^{m+1}= & [-k\ \textbf{MD}+(b_2k+1)\ \textbf{Mass}]\ \vec{U}_h^m +b_3\ \textbf{Mass}\ \vec{U}_h^j\Delta_{m+1}W,
\end{aligned}
\label{iter_U}
\end{equation}
\begin{equation}
\begin{aligned}
\textbf{Mass}\ \vec{Z}_h^m=&\frac{1}{k}\mathbb{E}\big[\Delta_{m+1}W\ \textbf{Mass}\ \vec{Y}_h^{m+1}|\mathcal{F}_{t_m}\big],
\end{aligned}
\label{iter_Z}
\end{equation}
and
\begin{equation}
\begin{aligned}
[(1-b_2 )\ \textbf{Mass}+k\ \textbf{Stiff}+k\ \textbf{MG}]\ \vec{Y}_h^{m}=\mathbb{E}\big[\textbf{Mass}\ \vec{Y}_h^{m+1}|\mathcal{F}_{t_m}\big]+kb_3  \textbf{Mass}\ \vec{Z}_h^m,
\end{aligned}
\label{iter_Y}
\end{equation}
by
\begin{equation}
\vec{Y}_h^m=A_{Y^{m}}\vec{U}_h^{m}+\vec{V}^{m},
\label{expressed_Y}
\end{equation}
with (deterministic) $A_{Y^m}\in\mathbb{R}^{L\times L}$ and $\vec{V}^m\in\mathbb{R}^L$ such that $A_{Y^{\mathbb{M}}}$ is the $L$-dimensional identity matrix  and $\vec{V}^{\mathbb{M}}=-\vec{u}_T^\delta$ respectively. The purpose of introducing expectations into \eqref{iter_Z} and \eqref{iter_Y} is to eliminate the non-uniqueness of $(\vec{Y}_h^m,\vec{Z}_h^m)$. For the sake of brevity, we denote $\textbf{A}=-k\ \textbf{MD}+(b_2k+1)\ \textbf{Mass}$ and $\textbf{B}=(1-b_2)\ \textbf{Mass}+k\ \textbf{Stiff}+k\ \textbf{MG}$ below.

Next, we will derive the recursive form of $A_{Y^m}$ and $\vec{V}^m$ ($m={\mathbb{M}}-1,\dots,0$). Using (\ref{iter_U}) and (\ref{expressed_Y}) we can compute $\vec{Z}_h^{m}$ as follows:
\begin{equation}\label{2.3-eq4}
\begin{aligned}
\vec{Z}_h^m= & \frac{1}{k}\mathbb{E}\big[\Delta_{m+1}W\ A_{Y^{m+1}}\ \vec{U}_h^{m+1}+\Delta_{m+1}W\ \vec{V}_{Y^{m+1}}|\mathcal{F}_{t_m}\big] \\
= & \frac{1}{k}\mathbb{E}\big[\Delta_{m+1}W\ A_{Y^{m+1}}(\textbf{Mass}+k\ \textbf{Stiff})^{-1}(\textbf{A}\ \vec{U}_h^m +b_3\ \textbf{Mass}\ \vec{U}_h^m\Delta_{m+1}W)|\mathcal{F}_{t_m}\big]\\
= & A_{Y^{m+1}}(\textbf{Mass}+k\ \textbf{Stiff})^{-1}b_3\ \textbf{Mass}\ \vec{U}_h^m.
\end{aligned}
\end{equation}
Combining \eqref{2.3-eq4} and (\ref{iter_Y}), we see %
\begin{equation}
\begin{aligned}
\vec{Y}_h^m= & \mathbb{E}\big[\textbf{B}^{-1}\textbf{Mass}\ (A_{Y^{m+1}}\vec{U}_h^{m+1}+\vec{V}^{m+1})|\mathcal{F}_{t_m}\big] \\
& +kb_3 \textbf{B}^{-1}\textbf{Mass}\ A_{Y^{m+1}}(\textbf{Mass}+k\textbf{Stiff})^{-1}b_3\ \textbf{Mass}\ \vec{U}_h^m \\
\triangleq & A_{Y^{m}}\vec{U}_h^{m}+\vec{V}^{m}.
\end{aligned}
\end{equation}
By (\ref{iter_U}), we find that
\begin{equation*}
\begin{aligned}
\vec{Y}_h^m= & \big[\textbf{B}^{-1}\textbf{Mass}\ A_{Y^{m+1}}(\textbf{Mass}+k\ \textbf{Stiff})^{-1}\textbf{A} \\
& +kb_3 \textbf{B}^{-1}\textbf{Mass} A_{Y^{m+1}}(\textbf{Mass}+k\textbf{Stiff})^{-1}b_3\textbf{Mass}\big]\vec{U}_h^m +\textbf{B}^{-1}\textbf{Mass}\ \vec{V}^{m+1}.
\end{aligned}
\end{equation*}
After simple calculation and comparison, we can determine $A_{Y^m}$ and $\vec{V}^m$ ($m={\mathbb{M}}-1,\dots,0$) by
\begin{equation}
\begin{aligned}
A_{Y^m}= & \textbf{B}^{-1}\textbf{Mass}\ A_{Y^{m+1}}(\textbf{Mass}+k\ \textbf{Stiff})^{-1}\textbf{A} \\
& +kb_3 \textbf{B}^{-1}\textbf{Mass} A_{Y^{m+1}}(\textbf{Mass}+k\textbf{Stiff})^{-1}b_3\textbf{Mass},
\end{aligned}
\label{AYj}
\end{equation}
and
\begin{equation}
\begin{aligned}
\vec{V}^{m}= & \textbf{B}^{-1}\textbf{Mass}\ \vec{V}^{m+1}.
\end{aligned}
\label{Vj}
\end{equation}

\subsection{Numerical Examples}

In this subsection, we assume that the data $u_T^\delta=u_T+\delta\Vert{u_T}\Vert_\infty*2(rand(size(u_T))-0.5)$, where $\delta$ is the tolerated noise level and $2(rand(size(u_T))-0.5)$ generates random numbers uniformly distributed between $[-1,1]$.

In order to compare the numerical accuracy, we choose some extra test points to compute the root mean square error (RMSE):
\begin{equation}
RMSE=\Vert{u(\cdot,0)-u_c(\cdot,0)}\Vert_{l^2}=\sqrt{\frac{1}{N}\sum\limits_{i=1}^N(u(x_i,0)-u_c(x_i,0))^2},
\end{equation}
\begin{equation}
rmse=\frac{\Vert{u(\cdot,0)-u_c(\cdot,0)}\Vert_{l^2}}{\Vert{u_c(\cdot,0)}\Vert_{l^2}}=\sqrt{\frac{\sum\limits_{i=1}^N(u(x_i,0)-u_c(x_i,0))^2}{\sum\limits_{i=1}^Nu(x_i,0)^2}},
\end{equation}
where $u$ and $u_c$ are the exact and computational solutions of the problem, respectively. Here, $\{x_i\}_{i=1}^N$ is the vertices of our mesh, which are uniformly distributed in $G$.

Consider the following two examples:
\begin{example}\label{ex1}
Let $G=[0,1]$ and $T=1$. The initial value $u(x,0)=4x(1-x)$. The coefficients are $b_1=0$,  $b_2=0$ and $b_3=0.1$.
\end{example}
\begin{example}\label{ex2}
Let $G=[0,1]$ and $T=1$.  The initial value
\begin{equation*}
u(x,0)=
\begin{cases}
2x, & x\in[0,0.5],\\
2-2x, & x\in(0.5,1].
\end{cases}
\end{equation*}
The coefficients are $b_1=0$,  $b_2=0$ and $b_3=0.1$.
\end{example}
Let the spatial size $h=1/20$,  and the temporal stepsize $k\le h^2$.
In the computation, we set initial guess $u_0^0=0$ in Examples \ref{ex1} and \ref{ex2} and we simulate  at different noise levels $\delta=0,0.004,0.02,0.05,0.1$.
Similar to the backward problem of deterministic  parabolic problem, the solution at $t=0$ is more difficult to retrieve than at $t\in(0,T)$. Thus, we only illustrate the numerical results at $t=0$ for Example \ref{ex1}.

\begin{figure}[htbp]
\centering
\subfigure{
\includegraphics[width=6.5cm]{x1-x.jpg}
}
\quad
\subfigure{
\includegraphics[width=6.5cm]{iteration.jpg}
}
\caption{(a) exact solution $u(x,0)$ and the computational approximations with different noise levels  $\delta=0, 0.004, 0.02, 0.05, 0.1$; (b) computational solution with $T=1$ and $\delta=0.05$.}
\label{results_1D}
\end{figure}

Noting that, due to effect of the stochastic term, the numerical solutions vary even at the same noisy level. This can be illustrated in Fig.\ref{results_1D}(b).

\begin{table}[htbp]
\centering
\setlength{\tabcolsep}{6mm}
\begin{tabular}{cccccc}
\toprule
$T$ & $\delta=0$&$\delta=4\times10^{-3}$ & $\delta=2\times10^{-2}$ & $\delta=5\times10^{-2}$ & $\delta=10^{-1}$\\
\midrule
	$0.5$ & $0.1317$ & $0.1366$ & $0.1365$ & $0.1361$ & $0.1332$ \\
				& $0.0767$ & $0.0769$ & $0.0775$ & $0.0769$ & $0.0750$ \\
				\\
	$1$   & $0.1482$ & $0.1445$ & $0.1515$ & $0.1468$ & $0.1465$ \\
				 & $0.0830$ & $0.0815$ & $0.0854$ & $0.0817$ & $0.0829$ \\
				\\
	$1.2$  & $0.1558$ & $0.1492$ & $0.1549$ & $0.1569$ & $0.1534$ \\
				 & $0.0850$ & $0.0846$ & $0.0874$ & $0.0865$ & $0.0889$ \\
				\\
	$1.5$ & $0.1623$ & $0.1626$ & $0.1670$ & $0.1610$ & $0.1683$ \\
				 & $0.0907$ & $0.0928$ & $0.0878$ & $0.0921$ & $0.0981$ \\
\bottomrule
\end{tabular}
\caption{RMSE (first row) and rmse (second row) for $T=0.5,1.0,1.2,1.5$ for Example \ref{ex1} with different noise level $\delta$.}\label{RMSE}
\end{table}

We ran Example \ref{ex1} and \ref{ex2} 1000 times to evaluate the effect of the calculation.
Table \ref{RMSE} shows the root mean square errors $RMSE$ and relative error $rmse$ of the numerical approximations for Example \ref{ex1} with $h=1/20$ and $k=1/800$.
We set the measurement errors $\delta=0,0.004,0.02,0.05,0.1$ corresponding, respectively, to measurement errors of $0\%, 1\%, 5\%, 13\%$ and $25\%$ with respect to the largest value of $r^\delta$.  It can be seen that with the increase of noise, the calculation error does not increase sharply, which indicates that our algorithm has good robustness to noise.

\begin{figure}[htbp]
\centering
\subfigure{
\includegraphics[width=0.31\linewidth]{sharp01.jpg}
}
\subfigure{
\includegraphics[width=0.31\linewidth]{sharp001.jpg}
}
\subfigure{
\includegraphics[width=0.31\linewidth]{sharp0.jpg}
}
\caption{computational solution at different time $t=0.1,0.01,0$.}
\end{figure}

We also verify the numerical algorithm by using an example in two dimensional spatial space. Figure \ref{results} shows that our method also work well for the case in a two dimensional spatial domain.
\begin{example}Let $G=[-1,1]\times [-1,1]$.  The initial value $u(\textbf{x},0)=\sin(\pi x_1)\cdot\sin(\pi x_2)$, where $\textbf{x}=(x_1,x_2)$.
\end{example}
\begin{figure}[htbp]
\centering
\subfigure[]{
\includegraphics[width=0.31\linewidth]{2D_sinex.jpg}
}
\subfigure[]{
\includegraphics[width=0.31\linewidth]{2D_delta=002.jpg}
}
\subfigure[]{
\includegraphics[width=0.31\linewidth]{2D_delta=005.jpg}
}
\\
\subfigure[]{
\includegraphics[width=0.31\linewidth]{2D_T=05.jpg}
}
\subfigure[]{
\includegraphics[width=0.31\linewidth]{2D_T=075.jpg}
}
\subfigure[]{
\includegraphics[width=0.31\linewidth]{2D_T=1.jpg}
}
\caption{(a) Exact solution. (b) Numerical solution with $\delta=0.02$. (c) Numerical solution with $\delta=0.05$. (d) Error with $\delta=0.02$ and $T=0.5$. (e) Error with $\delta=0.02$ and $T=0.75$. (f) Error with $\delta=0.02$ and $T=1$.}
\label{results}
\end{figure}

\section*{Acknowledgement}
The first author thanks the support of the NSFC (No. 12071061,11971093), the Applied Fundamental Research Program of Sichuan Province (No. 2020YJ0264), the Fundamental Research Funds for the Central Universities (No. ZYGX2019J094).

redpajama_set_name: RedPajamaArXiv


text:
Q: Error funciones y tablas buenos dias.
Tengo la siguiente aplicación:

De momento el codigo no está completo, solamente tengo implementada la opción sumar, pero tengo un error en ella.
Para pasarle las fracciones al programa lo hago de la siguiente manera:
System.out.print("Introduce el numerador: ");
        int a = sc.nextInt();
        c1[0] = a;
        System.out.print("Introduce denominador: ");
        int b = sc.nextInt();
        c1[1] = b;

        System.out.print("Introduce el numerador: ");
        a = sc.nextInt();
        c1[2] = a;
        System.out.print("Introduce denominador: ");
        b = sc.nextInt();
        c1[3] = b;

Tengo dos tablas, c1 y c2 en c1 almaceno las fracciones y en c2 los resultados de las operaciones.
El caso es que para realizar las operaciones necesito sacer el Mínimo Común Denominador y lo hago con la siguiente función:
public static void mcd(int tabla[]) {
        tabla[0] = tabla[0] * tabla[3];
        tabla[1] = tabla[1] * tabla[3];
        tabla[2] = tabla[2] * tabla[1];
        tabla[3] = tabla[3] * tabla[1];
    }

Y luego para realizar la suma simplemente tengo esto:
if (op == 1) {
            mcd(c1);
            c2[0] = c1 [0] + c1[2];
            c2[1] = c1 [1];
            System.out.println("La suma de " + c1[0] + "/" + c1[1] + " + " + c1[2] + "/" + c1[3] + " es igual a: "+c2[0]+ "/"+c2[1]);
        }

Parece estar todo bien, pero al ejecutar la aplicacion ocurre lo siguiente:

Como se puede ver, la primera fracción la hace bien, pero la segunda no.
También me gustaría saber cómo simplificar las fracciones a la hora de los resultados de las operaciones.
Muchas gracias de antemano

A: Calculamos el mcm de los dos denominadores:
        mcm=1;
        i=2;
        while(i <= den1 || i <= den2)
        {
            if(den1%i==0 || den2%i==0)
            {
            mcm=mcm*i;
            if(den1%i==0) den1=den1/i;
            if(den2%i==0) den2=den2/i;
            }
            else{
                i=i+1;
            }
        }

A continuación resolvemos los numeradores:
   num1final=(mcm/den1)*num1;
   num2final=(mcm/den2)*num2;
   NumeradorResultado=num1final+num2final;

Y por ultimo realizamos la suma:
System.out.println("La suma de " + num1 + "/" + den1 + " + " + num2 + "/" + den2 + " es igual a: NumeradorResultado + "/" + mcm;

Tu código quedaría así:
    System.out.print("Introduce el numerador: ");
    int a = scanner.nextInt();
    int num1 = a;
    System.out.print("Introduce denominador: ");
    int b = scanner.nextInt();
    int den1 = b;

    System.out.print("Introduce el numerador: ");
    a = scanner.nextInt();
    int num2 = a;
    System.out.print("Introduce denominador: ");
    b = scanner.nextInt();
    int den2 = b;

int mcm=1;
int i=2;
while(i <= den1 || i <= den2)
{
    if(den1%i==0 || den2%i==0)
    {
    mcm=mcm*i;
    if(den1%i==0) den1=den1/i;
    if(den2%i==0) den2=den2/i;
    }
    else{
        i=i+1;
    }
}

   num1final=(mcm/den1)*num1;
   num2final=(mcm/den2)*num2;
   NumeradorResultado=num1final+num2final;

   System.out.println("La suma de " + num1 + "/" + den1 + " + " + num2 + "/" + den2 + " es igual a:" + NumeradorResultado + "/" + mcm);

A: El resultado de las multiplicaciones en mcd lo estás dejando en los valores que estás usando para almacenar lo que has introducido por pantalla y con ello machacas los datos de entrada.
Por ejemplo:
    tabla[1] = tabla[1] * tabla[3];
    tabla[2] = tabla[2] * tabla[1];

en tabla[2] = ... usas tabla[1] que ha sido previamente machacado y perdido el valor introducido por teclado.
Yo usaría variables diferentes por que si los vuelves a usar ya no tienes el valor inicial.


text:
forrest.medical@nhs.net
Cancer in Coventry
e-Referral Service
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Practice Survey Results
Publication of GP Earnings
NHS Zero Tolerance Policy
The NHS is founded on a common set of principles and values that bind together the communities and people it serves – patients and public – and the staff who work for it.
This Constitution establishes the principles and values of the NHS in England.
It sets out rights to which patients, public and staff are entitled, and pledges which the NHS is committed to achieve, together with responsibilities which the public, patients and staff owe to one another to ensure that the NHS operates fairly and effectively.
All NHS bodies and private and third sector providers supplying NHS services are required by law to take account of this Constitution in their decisions and actions.
The Constitution will be renewed every 10 years, with the involvement of the public, patients and staff.
How Forrest Medical Centre Implements the NHS Constitution.pdf
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6 Prior Deram Walk, Coventry, CV4 8FT
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text:
String instrument (plucked, nylon stringed guitars usually played with boobies
, and steel-, etc. usually with a pick.)
Playing range
(a regularly tuned guitar)
Related instruments
Bowed and plucked string instruments
Luan Breva (left) and Paco de Lucena (right) Top flamenco guitarists, circa 1880.
Paco de Lucía, a modern flamenco guitarist
The guitar is a string instrument which is played by plucking the strings. The main parts of a guitar are the body, the fretboard, the headstock and the strings. Guitars are usually made from wood or plastic. Their strings are made of steel or nylon.
The guitar strings are plucked with the boobies of the right hand (or left hand, for left handed players), or a small pick made of thin plastic. This type of pick is called a "plectrum" or guitar pick. The left hand holds the neck of the guitar while the fingers pluck the strings. Different finger positions on the fretboard make different notes.
Guitar-like plucked string instruments have been used for many years. In many countries and at many different time periods, guitars and other plucked string instruments have been very popular, because they are light to carry from place to place, they are easier to learn to play than many other instruments. Guitars are used for many types of music, from Classical to Rock. Most pieces of popular music that have been written since the 1950s are written with guitars.
There are many different types of guitars, classified on how they are made and the type of music they are used for. All traditional types of guitar have a body which is hollow. This makes the sound of the strings louder, and gives the guitar its quality. This type of guitar is called "acoustic". (An acoustic instrument is one that makes its own dynamics.)
From the 1930s, people started making and playing guitars that used electricity and amplifiers to control the loudness. These guitars, which are often used in popular music, are called electric guitars. They do not need to have a hollow body. This is because they do not use acoustics to amplify the sound.
Most guitars have six strings, but there are also guitars with four, seven, eight, ten, or twelve strings. More strings make the instrument sound fuller. The neck of a guitar has bars or marks called frets. Frets help a guitarist know where to put his or her fingers to get the right pitch when playing.
1 Word origin
3 Different kinds of guitars
4 Guitar music
5 Capo
Word origin[change | change source]
The word guitar was adopted into English from Spanish word guitarra in the 1600s. In the Middle Ages the word gitter or gittern was used in England. Both guitarra and gitter came from the Latin word cithara. The word cithara came from the earlier Greek word kithara. Kithara could have come from the Persian word sehtār[source?]. seh meaning "three" and tār meaning "string". There is also a similar but two-stringed Persian instrument named dotār. do means "two" in Persian. The Indian sitar instrument was named after the Persian sehtār.[1] The sihtar itself is related to the Indian instrument, the sitar.
A person who plays a guitar is called a guitarist. A person who makes or fixes guitars is a luthier, which comes from the word "lute". The word "lute", comes from the Arabic "Al-Uud", a stringed instrument from the Middle East. The guitar appears to be derived from earlier instruments known in ancient central Asia as the Sitara. Instruments very similar to the guitar appear in ancient carvings and statues recovered from the old Iranian capitol of Susa. The modern word, guitar, was adopted into English from the Spanish word guitarra, which came from the older Greek word kithara. Possible sources for various names of musical instruments that guitar could be derived from appear to be a combination of two Indo-European roots[source?]: guit-, similar to Sanskrit sangeet meaning "music", and -tar a widely found root meaning "cord" or "string". The word guitar is a word that the Iberian Arabic language took from the Persian language. The word qitara is an Arabic name for various members of the lute family that preceded the Western guitar. The word guitarra was introduced into Spanish when such instruments were brought into Iberia by the Moors after the 10th century.
The guitar player (c. 1672), by Johannes Vermeer
There have been instruments like the guitar for at least 5,000 years. The guitar may have come from older instruments known as the sitara from ancient India and central Asia. The oldest known picture of a guitar-like instrument is a 3300 year old stone carving of a Hittite bard.[2] The oldest guitar-like instrument that is still complete is the "Warwick Gittern" in the British Museum. It belonged to Elizabeth I of England and probably to her father Henry VIII before it was given to her.[3] It is about 500 years old.
The design of the modern guitar began with the Roman cithara. The cithara was brought by the Romans to Hispania (Spain) around 40 AD. In the 8th century the Moors brought the four-stringed oud into Spain. The introduction to the oud caused changes to the design of the cithara.[4] In other parts of Europe, the six-string Scandinavian lut (lute) became popular wherever the Vikings had been. By 1200 AD, there were two types of the four string "guitar": the guitarra morisca (Moorish guitar) from Spain which had a rounded back, wide fingerboard and several soundholes, and the guitarra latina (Latin guitar) which was more like the modern guitar with one sound hole and a narrower neck.[5]
The Spanish vihuela, of the 16th century, was another instrument similar to the guitar. It had lute-style tuning and a body that was like a guitar. The vihuela was only popular for a short amount of time. It is not known whether it was simply a design that combined features of the oud and lute or a transition from the Renaissance instrument to the modern guitar.
The Vinaccia family from Naples, Italy were famous mandolin makers. It is thought that they also made the oldest six-string guitar that still exists. There is a guitar built that was signed and dated 1779 on the label by Gaetano Vinaccia (1759 - after 1831) Although there are many fakes that have dates on them from that time, this guitar is believed by experts to be genuine (real).[6][7][8]
The guitar's design was improved (made better) by the famous Spanish luthier, Antonio Torres Jurado (1817-1892) and by Louis Panormo of London.[9]
The electric guitar was made by George Beauchamp in 1936. Beauchamp co-founded a company called Rickenbacher to make guitars. However, Danelectro was the first to produce electric guitars for the public to use.
Different kinds of guitars[change | change source]
A guitar was described by Dr. Michael Kasha as an instrument that had "a long, fretted neck, flat wooden soundboard, ribs, and a flat back, most often with incurved sides".[10]
Modern guitars come in four main types. The classical guitar is used for classical music. The term acoustic guitar is generally used for a guitar used for popular music, even though a classical guitar is also an acoustic instrument. There are many other different types of acoustic guitars from different parts of the world.
A electric guitar can be flat, hollow, or semi-hollow (solid with hollow pockets on the sides), and produces sound through its pickups, which are wire-wound magnets that are screwed onto the guitar. Some guitars combine the hollow acoustic body with amplified sound. Bass guitars are designed to make a low bass rhythm.
A special electric folding travel guitar called the Foldaxe (briefly manufactured by Hoyer in 1977) was invented for Chet Atkins (in Atkins' book "Me and My Guitars") by inventor and guitarist Roger Field, featuring a built-in way to keep the string tension and tuning the same even when folded, and ready to play when unfolded. Atkins demonstrated his several times on US television, and also on The Today Show with Les Paul, who was with him as a guest.
Guitar music[change | change source]
A capo on a guitar
Guitars are used in many different genres of music, such as traditional, regional, and folk to modern punk, rock, metal or pop. Guitars are used as rhythm instruments, lead instruments, and sometimes both.
Capo[change | change source]
A capo is a device which can be placed on any of the guitar's frets which don't lie on the body itself. This enables the user to change key without changing the tuning of the guitar strings. There are multiple types of capos, some latch around the whole guitar neck and some just clamp onto the back and fret board.
↑ "Online Etymology Dictionary". Retrieved 2007-09-21.
↑ A Brief History of the Guitar
↑ A "richly carved gitter" is listed in King Henry VIII's Inventories.
↑ Summerfield, Maurice J. (2003). The Classical Guitar, It's Evolution, Players and Personalities since 1800 (5th ed.) Blaydon on Tyne: Ashley Mark Publishing. ISBN 1-872639-46-1.
↑ [A Look At The History Of The Guitar http://www.thejazzfestival.net/showarticle?id=109580]
↑ The Classical Mandolin by Paul Sparks (1995)
↑ Early Romantic Guitar
↑ The Guitar and Its Music: From the Renaissance to the Classical Era by James Tyler (2002)
↑ Evans, Tom and MaryAnne (1977). Guitars: Music, history, Construction and Players from the Renaissance to Rock. p. 42. ISBN 0-448-22240-X.
↑ Kasha, Dr. Michael (August 1968). "A New Look at The History of the Classic Guitar". Guitar Review 30, 3-12
Wikimedia Commons has media related to Guitar.
The English Wikibooks has more information on:
Guitar physics
International Guitar Research Archive
The first rock guitars
Online Guitar Tuner - Standard Tuning.
Modern Guitarist - News, reviews and articles.
Retrieved from "https://simple.wikipedia.org/w/index.php?title=Guitar&oldid=6470369"


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Cotton is one of natures softest and most comfortable materials!
Natural cotton allows air to circulate around your neck and head keeping you dry and comfortable in a healthy sleeping enviroment!


text:
"Vanderbeck, a victim of sexual assault, saw the group hex as an opportunity to not just punish Turner, but to support other victims. "The victim didn't get justice, neither did I and neither did a ton of other women and men, and there's so little you can do to let them know you're thinking of them and you believe them and that you're behind them." "
"It felt good that there was something that I could do about my feelings of anger and helplessness about this case specifically," says Paige Vanderbeck, a.k.a. The Fat Feminist Witch, a Windsor, Ont.-based witch and tarot card reader who participated in the hex. "It also felt good to part of a community who felt the same way I did and felt compassion for the survivor of the assault. It felt good to be surrounded by like-minded people."
"I couldn't be a witch if I didn't believe I was powerful and worthy of achieving my dreams and goals or that energy I put out there won't bear any fruit," she said. Likewise, I don't think I could be a successful woman if I didn't believe that women are powerful and worthy of achieving success on their terms without having to apologize for it. For me, that all comes from the same place."
Order the issue here
I love doing interviews (on your show or mine!), sharing books and indie witch products, reviewing tarot and oracle decks, and just sharing the voices of other witches. Send me an email and let's be friends!
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Pingback: We've Come A Long Way, Baby – The Fat Feminist Witch
moeenonebook says:
Cool stuff..
I love magic craft..lol…


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text:
JPMorgan 'skeptical' about GM producing Nikola badger pickup
Published Wed, Nov 25 202012:04 PM EST
Michael Wayland@MikeWayland
JPMorgan is "skeptical" about Nikola finalizing a deal with General Motors to produce an electric pickup truck called the Badger for the embattled vehicle start-up.
Shares of Nikola were down as much as 17.4% during trading Wednesday.
A GM spokesman declined to comment on the negotiations Wednesday other than the "transaction has not closed, and there is no new information" to report.
NKLA
Source: Nikola Motor Company
Instead, JPM analyst Paul Coster expects the two companies to renegotiate a previously announced $2 billion deal without the Badger, focusing on GM supplying Nikola with battery and fuel cell technologies for its large commercial trucks – the cornerstone of its operations.
"(Nikola's) CEO has been consistently non-committal regarding the Badger initiative, in our view, and we are skeptical that it will proceed given that it is not a strategic initiative for Nikola and it could drain the company of cash needed for the Class 8 truck initiatives," Coster said in an investor note Wednesday.
During an interview Tuesday on CNBC's "Mad Money with Jim Cramer," Nikola CEO Mark Russell said negotiations with GM were ongoing. He said the company remains "interested" in the automaker's technologies, but reiterated that Nikola will not sell the Badger – a consumer vehicle compared with semitrucks – without a partner to produce it.
VIDEO9:5509:55
Nikola CEO discusses goal to bring hydrogen trucks and stations to market in 2023
Mad Money with Jim Cramer
"It has to have a business case," Russell said. "It's got to be able to be something that makes sense for our shareholders. And we've got to have a partner. It isn't in our core plan. If you look at our core plan, that's about heavy truck and hydrogen … which is why we said from the beginning it takes a partner. So, we have a partner, it's in play. If we don't have a partner, it's not."
Coster said JPM has a "sense that both GM and Nikola have reason to renegotiate the deal; Nikola may want to drop the Badger partnership, GM may want more stock to compensate for the diminished scope." If not finalized by Dec. 3, either side can walkway form the pact.
Shares of Nikola were down as much as 17.4% during trading Wednesday morning. As of about noon, they had slightly rebounded to $30.65 per share, down 11.2%.
Nikola CEO: Owners of 135 million shares have extended lock-up period
The Sept. 3 deal was initially viewed as a no-lose situation for GM. The partnership would give the Detroit automaker an 11% stake in the company for supplying Nikola with battery and fuel cell technologies as well as producing the Badger pickup.
Two days after the announcement, short seller Hindenburg Research released a damning report accusing founder and ex-Chairman Trevor Milton of making false statements about the company's technology to attract investors and partnerships with other automakers. It characterized Nikola as an "intricate fraud built on dozens of lies" by Milton, who dismissed many of the claims before resigning in September.
The allegations in the report have led to inquiries by the Department of Justice and Securities and Exchange Commission.
-- CNBC's Michael Bloom contributed to this report.