text:
\section{Introduction and preliminaries} \label{sec1}
\label{sec1}

For given generalized stochastic processes of Kondratiev-type $F$ and $G$, and a bounded in space  generalized stochastic process of Kondratiev-type $Q$,  we consider Cauchy problems for  stochastic parabolic evolution equations
\begin{equation}
\label{general problem}
(\partial_t -\mathcal{L})U + Q \lozenge U =F, \quad U |_{t=0} = G,
\end{equation}
where $\mathcal{L}$ is an elliptic operator acting on the space variable. Since the  unknown generalized stochastic process $U$ is involved in product with another generalized stochastic proces, potential  $Q$, one must give sense to such product. Here, the Wick product denoted by  $\lozenge$ is used, see \cite{HOUZ1996}.

The special case, when $\mathcal{L}$ is the Laplacian, is the stochastic heat equation with random potential which, due to its various applications in biology, financial mathematics, aerodynamics, structural acoustics, has been widely studied, e.g. \cite{AHR1997,ANV2000,BDP1997,KL2017}.
Stochastic evolution equations with multiplicative noise are studied in \cite{LPSZ2015} and stochastic evolution problems with polynomial nonlinearities are studied in \cite{LPSZ2018}.

In this work we study problem \eqref{general problem}. Using the chaos expansion method from the white noise analysis developed in \cite{HOUZ1996} we prove the theorem
on existence of unique generalized stochastic process and provide estimates of coefficients in chaos expansion form of the solution.

To start with, we recall some basic notions from the white noise analysis, and for more details and proofs we refer to \cite{HOUZ1996}.
Denote by  $\mathcal{I} :=\mathbb N^m_0 $ the set of multi-indices having finite number of  nonzero components,
the zero vector by $\textbf{0}$,
and the length of a multi-index $\alpha = (\alpha_1,\alpha_2, \ldots,\alpha_m,0,0,\ldots)$, $\alpha_i \in \mathbb N_0$, $i=1, 2,\ldots, m$, $m\in\mathbb N$,  by  $|\alpha|= \sum_{i=1}^{m}\alpha_i$.
If $\alpha=(\alpha_1,\alpha_2,\dots)\in \mathcal I$ and $\beta=(\beta_1,\beta_2,\dots) \in \mathcal I$, then $\alpha \leq \beta$ if and only if $\alpha_k \leq \beta_k$ for all $k \in \mathbb{N}$.
The following lemma collects results needed later. For proofs and details see \cite{HOUZ1996,LPSZ2018,LPSZ2021sub}.
\begin{lem} \label{pomocna lema - ocene}
	Let $\alpha\in \mathcal{I}$ and $k\in \mathbb{N}_0$.

	(a) Let $k\leq |\alpha|$, and denote by $N(\alpha,k)$ the number of possibilities in which a multi-index $\alpha$ can be written as a sum of $k$ strictly smaller and nonzero multi-indices. Then $N(\alpha,k)\leq 2^{k|\alpha|}$.

	%
	(b) Define $(2\mathbb N)^\alpha: = \prod_{i=1}^{\infty} (2i)^{\alpha_i}$. Then $|\alpha| \leq (2\mathbb N)^\alpha$ and
	\begin{equation}\label{2N konv p>1}
	\sum_{\alpha\in \mathcal I} (2\mathbb N)^{-p\alpha} < \infty\quad  \Leftrightarrow \quad  p>1.
	\end{equation}

	Moreover, for every  $c>0$ there exists $s\geq 0$ such that $c^{|\alpha|}\leq (2\mathbb N)^{s\alpha}$ and   $c^\alpha\leq (2\mathbb N)^{s\alpha}.$
\end{lem}

For $\gamma\in \mathcal{I}$, the $\gamma$th {Fourier-Hermite polynomial} is defined by
$H_{\gamma}(\omega):=\prod_{k=1}^\infty h_{\gamma_k}(\langle\omega,
\xi_k\rangle)$, where $\xi_k$, $k\in \mathbb{N}$, is the Hermite function of order $k$,  and $h_k$, $k \in \mathbb N_0$, is the Hermite polynomial.
For a normed space $X$,  the tensor product
$X\otimes (S)_{-p}$ is the space of $X$-valued generalized stochastic processes of Kondratiev-type, and the space $X\otimes (S)_{-1}$ is the inductive limit of  spaces $X\otimes (S)_{-p }$, $p\geq 0$.
Every $X$-valued generalized stochastic process of Kondratiev-type, $F \in X\otimes (S)_{-1}$, can be  represented in {\em the chaos expansion form} $F(x,\omega) = \sum_{ \gamma\in \mathcal I}f_\gamma(x) H_\gamma(\omega)$,  $f_\gamma\in X$,  with
$\|F\|^2_{X\otimes (S)_{-p }}: = \sum_{ \gamma\in \mathcal{I}}\|f_\gamma\|^2_X (2\mathbb N)^{-p\gamma}$  finite for some $p\geq 0$. If it is finite for $p_0$, then it is finite for all $p\geq p_0$. The minimal such $p_0$ we call {\em the critical exponent}.
For $X$ a Banach space, $X\otimes(S)_{-p}$ is a Banach space for every $p\geq p_0$, and $X\otimes (S)_{-1}$ is a Frech\'et space.

The Wick product is introduced to overcome the multiplication problem for random variables in \cite{HOUZ1996} and it is generalized to the set of generalized stochastic processes in \cite{LPSZ2015}. Recall, if $F,G \in X \otimes (S)_{-p}$, $p\geq 0$, are generalized stochastic processes given in chaos expansion forms
$F= \sum_{\alpha\in \mathcal{I} }f_\alpha  H_\gamma$ and $G= \sum_{\beta\in \mathcal{I}} g_\beta  H_\beta$, then the Wick product $F \lozenge G,$ is  defined by
$F \lozenge G =\sum_{\gamma \in \mathcal I} \left( \sum_{\alpha + \beta = \gamma} f_\alpha g_\beta \right)  H_\gamma$.

We conclude the introductory section stating a theorem on the deterministic parabolic evolution problems, and a technical Lemma, both necessary for the later analysis. The proof of the theorem is similar to the proof of  Theorem 3 in \cite{GLO2021}, while  proof of the lemma is straightforward, thus both are omitted.

\begin{thm}\label{theorem estimates L infinity deterministic}
	Let the unbounded and closed operator $\mathcal L$ with a dense domain $D\subseteq L^2(\mathbb R^d)$,  be an infinitesimal generator of a $C_0$-semigroup $(T_t)_{t\geq 0}$ on $L^2(\mathbb R^d)$. Let the force term $f\in AC([0,T];L^2(\mathbb R^d))$, i.e., being differentiable a.e. on $[0,T]$ with $f' \in L^1(0,T;L^2(\mathbb R^d))$. Let  the initial condition $g\in D$,  and the potential $q \in L^{\infty} (\mathbb{R}^d)$. Then, the deterministic parabolic  initial value problem
	\[
	\left( \frac{\partial}{\partial t}  -  {\mathcal L} \right)  u(t, x) + {q (x)} \cdot  u(t, x)  = f(t, x ),
	\quad u(0, x) = g(x),
	\]
	has a unique bounded nonnegative solution $u\in AC([0,T];L^2(\mathbb R^d))$

	which satisfies
	\begin{equation*}
	\label{ee-perturbations}
	\|u(t,\cdot)\|_{L^2} \leq  M(t)  \left( \|g(\cdot)\|_{L^{2}} +\int_0^t \|f(s,\cdot)\|_{L^2} \,ds\right), \quad t\in (0,T],
	\end{equation*}
	where
	$M(t) : = M \exp {\left( \left(w + M \|q\|_{L^\infty}\right)t \right)},$ and $w \in \mathbb{R}$ and $M>0$ are the stability constants from the semigroup estimate $\|T_t\|_{L(L^2(\mathbb R^d))}\leq Me^{wt}$,  $t\geq 0.$
\end{thm}
\begin{lem} \label{Lema M(t)} Let $M(t)$ be as in Theorem \ref{theorem estimates L infinity deterministic} with the potential $q_{\mathbf{0}} \in L^{\infty} (\mathbb{R}^d)$. Then, the following estimates hold
	\begin{itemize}
		\item[(a)] 	$\tilde{M}(t) := \displaystyle \int_0^t M(s) \,ds  = \dfrac{M(t)-M}{w+M\|q_{\mathbf{0}} \|_{L^\infty}},$ \quad $\displaystyle \int_0^t s M(s) \,ds \leq t  \tilde{M}(t),$
		\item[(b)] $\displaystyle \int_0^t  M(s)  \tilde{M}(s)^n  \,ds \leq	 \  \tilde{M}(t)^{n+1},$ \quad  $\displaystyle \int_0^t s M(s) \tilde{M}(s)^n\,ds \leq  t \tilde{M}(t)^{n+1}$ for all $n \in \mathbb N.$
	\end{itemize}
\end{lem}

\section{Stochastic parabolic equations with random and space depending bounded potential}
\label{sec3}
Now we turn our attention to stochastic initial value problem  \eqref{general problem}, more precisely we consider
\begin{align}\label{Eq:ish2}
\left( \frac{\partial}{\partial t}  -  {\mathcal L} \right)  U(t, x,\omega) + {Q (x,\omega)} \lozenge  U(t, x,\omega)&  = F(t, x ,\omega), \quad t\in (0,T],\, x\in \mathbb R^d,\, \omega \in \Omega\\
U(0, x,\omega) &= G(x,\omega)\quad x\in \mathbb R^d,\, \omega \in \Omega .\nonumber
\end{align}

The main result  follows.
\begin{thm} \label{Thm0: stochastic weak sol 1}
	Let $\mathcal L$ be an unbounded closed operator with dense domain $D\subseteq L^2(\mathbb R^d)$, acting on the space component and  generating  a $C_0$-semigroup on $L^2(\mathbb R^d)$ satisfying $\|T_t\|_{L(L^2(\mathbb R^d))}\leq Me^{wt}$, and let $\tilde{M}(T)$ be as in Lemma \ref{Lema M(t)}.
	%
	Let the potential $Q \in L^\infty(\mathbb R^d) \otimes (S)_{-1}$ be a generalized stochastic process of Kondratiev-type with the critical exponent $p_1$ and the chaos expansion
	$Q(x, \omega) = \sum\limits_{\gamma\in \mathcal I} q_\gamma( x) \, H_\gamma(\omega)$, with $q_\mathbf{0}\in L^\infty(\mathbb R^d)$ such that $\tilde{M}(T) \|q_\mathbf{0}\|_{L^\infty} \not= 1$, and with $q_\gamma \in L^\infty(\mathbb R^d)$ such that  $\|q_{\gamma}\|_{L^\infty} \leq \|q_\mathbf{0}\|_{L^\infty}$,\\ for all $\gamma \in \mathcal{I}$.
	%
	Further, assume that  the force term $F\in AC([0,T];L^2(\mathbb R^d)) \otimes (S)_{-1}$ and  the initial condition $G\in D \otimes (S)_{-1}$ are  generalized stochastic processes of Kondratiev-type, with critical exponents $p_2$ and $p_3$, respectively.
	%
	Then,  there exists a unique  generalized stochastic process  \linebreak $U \in AC([0,T];D)  \otimes (S)_{-1} \subseteq AC([0,T];L^2 (\mathbb R^d)) \otimes (S)_{-1} $ satisfying the stochastic evolution initial value problem \eqref{Eq:ish2}.
	Moreover,  for all $t\in [0,T]$ the coefficients $u_\gamma$, $\gamma \in \mathcal{I}$ satisfy
	\begin{equation}
	\label{ocena koeficijenti}
	\norm{u_\gamma (t,\cdot)}{L^2}  \leq  M(t)\Bigg\{ a_{\gamma}(t)   + \sum\limits_{k=1}^{|\gamma|} \tilde{M}(t)^k \bigg( \sum_{\substack{0 \leq |\beta| \leq |\gamma|-k\\ \beta<\gamma}} a_\beta (t) \, \bigg(\sum_{\substack{\theta_1+\dots+\theta_k=\gamma-\beta\\\theta_i \not=\mathbf{0}, \, i=1,\dots, k}} \prod_{i=1}^{k} \|q_{\theta_i}\|_{L^\infty} \bigg)\bigg) \Bigg\},
	\end{equation}
	where $a_\gamma (t) :=  \|g_{\gamma}\|_{L^2} + t \| f_{\gamma}\|$, $\gamma\in\mathcal I$.
\end{thm}
\begin{proof}
Representing stochastic processes $Q$, $F$ and $G$ appearing in the problem \eqref{Eq:ish2} in their chaos expansion forms, assuming the solution $U$ in the form
$U(t, x, \omega) = \sum\limits_{\gamma\in \mathcal I} u_\gamma(t, x) H_\gamma(\omega)$,
and using the definition of the Wick product, we  formally obtain
\[\begin{split}
\sum_{\gamma\in \mathcal I} \left(\frac{\partial}{\partial t}  - \mathcal L \right) \, u_\gamma(t, x) H_\gamma(\omega) + \sum_{\gamma\in \mathcal I} \sum_{\alpha+ \beta=\gamma} q_\alpha(x) \, u_\beta(t, x) \, H_\gamma(\omega) &= \sum_{\gamma\in \mathcal I} f_\gamma(t, x)  \, H_\gamma(\omega),\\
\sum_{\gamma\in \mathcal I} u_\gamma(0,x) H_\gamma(\omega) &= \sum_{\gamma\in \mathcal I} g_\gamma(x) H_\gamma(\omega).
\end{split}\]
From the uniqueness of the chaos expansion representations, the problem is reduced to a triangular system of deterministic equations which can be solved recursively with respect to the length of $\gamma\in \mathcal I$.
For $|\gamma|=0$:
\begin{equation}
\label{deterministic system q(x): |gamma|=0}
\left(\frac{\partial}{\partial t}  - \mathcal L  \right) \, u_{\mathbf{0}}(t, x) + \, q_{\mathbf{0}} (x) \,  u_{\mathbf{0}} (t, x) =  f_{\mathbf{0}} (t, x), \quad \quad
u_{\mathbf{0}} (0, x) = g_{\mathbf{0}} (x).
\end{equation}
By assumptions,
$q_{\mathbf{0}} \in L^{\infty}(\mathbb R^d)$, $f_{\mathbf{0}}\in AC([0,T]; L^2(\mathbb R^d))$, and $g_{\mathbf{0}}\in D$. Theorem \ref{theorem estimates L infinity deterministic} implies a unique solution $u_{\mathbf{0}} \in AC([0,T],D)$ to  \eqref{deterministic system q(x): |gamma|=0} given by
$u_0(t,x) = S_t g_{\mathbf{0}}(x) + \int_0^t S_{t-s} f_{\mathbf{0}} (s,x) ds$, $t\in [0,T]$, $x\in \mathbb R^d$, where $(S_t)_{t \geq 0}$ is a $C_0$-semigroup on $L^2(\mathbb R^d)$ generated by $\mathcal L  - q_{\mathbf{0}}\text{Id}$,
satisfying
\begin{equation} \label{solution u_0 L2 estimate}
\|u_{\mathbf{0}}(t,\cdot)\|_{L^2} \leq M(t) \left( \|g_{\mathbf{0}} \|_{L^2 } + \int_0^t \| f_{\mathbf{0}} (s,\cdot)\|_{L^2} \,ds \right) \leq M(t) \left( \|g_{\mathbf{0}} \|_{L^2 } + t \| f_{\mathbf{0}}\|  \right)=M(t)a_{\mathbf{0}}(t).
\end{equation}

For $|\gamma|>1$ we have
$$	\left(\frac{\partial}{\partial t}  - \mathcal L  \right) \, u_{\gamma}(t, x) + \, q_{\mathbf{0}} (x) \,  u_{\gamma} (t, x) \, = 	\tilde{f}_{\gamma} (t, x),    \quad
u_{\gamma} (0, x) = g_{\gamma} (x)  ,$$
where $\tilde{f}_{\gamma} (t, x)  = f_{\gamma} (t, x)-\sum_{\substack{\alpha+\beta=\gamma\\ \alpha \not= \mathbf{0}}} q_\alpha(x) \, u_\beta (t, x)=f_{\gamma} (t, x)-\sum_{\mathbf{0}\leq \beta <\gamma} q_{\beta-\gamma}(x) \, u_\beta (t, x)$, with $u_{\mathbf{0}}$ and $ u_\beta$, $\beta<\gamma$ being the solutions  obtained in the previous steps.
Thus, we obtain a deterministic  problem of the same form as  for $|\gamma|=0$ satisfying, by assumptions, the conditions of  Theorem \ref{theorem estimates L infinity deterministic}  yielding a unique solution $u_{\gamma} \in AC([0,T],D)$ given by  $ u_{\gamma}(t,x) =  S_t g_{\gamma} (x) + \int_0^t S_{t-s} \tilde{f}_{\gamma} (t, x) ds$ and satisfying
\begin{equation}\label{estFromThm}
\norm{u_\gamma (t,\cdot)}{L^2}  \leq
M(t)\left( a_\gamma (t) +  \sum\limits_{\mathbf{0} \leq \beta < \gamma} \linf{q_{\gamma-\beta}} \int_0^t \ltwo{ u_{\beta} (s, \cdot)}\,ds\right).
\end{equation}

Next, by induction,  we will prove that $u_\gamma,$ $\gamma \in \mathcal{I}$, satisfy the estimate \eqref{ocena koeficijenti}. The estimate for $|\gamma|=0$ boils down to \eqref{solution u_0 L2 estimate}. We assume that the estimate \eqref{ocena koeficijenti} holds for every $\beta \in \mathcal{I}$ with $|\beta|\leq n$, i.e.,
\begin{equation} \label{opsta formula}
\norm{u_\beta (t,\cdot)}{L^2}  \leq  M(t)\Bigg\{ a_{\beta}(t)   + \sum\limits_{l=1}^{|\beta|} \tilde{M}(t)^l \bigg( \sum_{\substack{0 \leq |\alpha| \leq |\beta|-l\\ \alpha<\beta}} a_\alpha (t) \sum_{\substack{\theta_1+\dots+\theta_l=\beta-\alpha\\\theta_i \not=\mathbf{0}, \, i=1,\dots, l}} \prod_{i=1}^{l} \|q_{\theta_i}\|_{L^\infty} \bigg) \Bigg\},
\end{equation}
and want to show that \eqref{ocena koeficijenti} holds for $\gamma \in \mathcal{I}$ with $|\gamma|=n+1.$
Integrating  \eqref{opsta formula} and using Lemma \ref{Lema M(t)} we obtain
$$
\int_0^t \ltwo{ u_{\beta} (s, \cdot)}\,ds \leq a_\beta (t) \tilde{M}(t) + \sum\limits_{l=1}^{|\beta|} \tilde{M}(t)^{l+1} \Bigg( \sum_{\substack{0 \leq |\alpha| \leq |\beta|-l \\ \alpha<\beta}} a_\alpha (t) \sum_{\substack{\theta_1+\dots+\theta_l=\beta-\alpha \\ \theta_i \not=\mathbf{0}}} \prod_{i=1}^{l} \|q_{\theta_i}\|_{L^\infty} \Bigg).
$$
Starting from  \eqref{estFromThm} we obtain that $\norm{u_\gamma (t,\cdot)}{L^2}$ is bounded from above by $M(t)$ multiplied with
\begin{equation*}
a_\gamma (t)
+  \sum\limits_{\mathbf{0} \leq \beta < \gamma} \linf{q_{\gamma-\beta}} a_\beta (t) \tilde{M}(t)
+  \sum\limits_{\mathbf{0} < \beta < \gamma} \linf{q_{\gamma-\beta}} \sum\limits_{l=1}^{|\beta|} \tilde{M}(t)^{l+1} \bigg( \sum_{\substack{0 \leq |\alpha| \leq |\beta|-l \\ \alpha<\beta}} a_\alpha (t) \sum_{\substack{\theta_1+\dots+\theta_l=\beta-\alpha \\ \theta_i \not=\mathbf{0}}} \prod_{i=1}^{l} \|q_{\theta_i}\|_{L^\infty} \bigg).
\end{equation*}
We want to sum terms with respect to powers of $\tilde{M}(t)$. The first step is to change the order of the first two sums in the third term, where, while $\mathbf{0} < \beta<\gamma$, the length  $|\beta|$  varies from $1$ to $n$, thus
the third term becomes
\begin{eqnarray*}

	\sum_{l=1}^n \tilde{M}(t)^{l+1} \sum\limits_{\mathbf{0} < \beta < \gamma} \Bigg( \sum_{\substack{0 \leq |\alpha| \leq |\beta|-l \\ \alpha<\beta}} a_\alpha (t) \sum_{\substack{\theta_1+\dots+\theta_l=\beta-\alpha \\ \theta_i \not=\mathbf{0}}} \prod_{i=1}^{l} \|q_{\theta_i}\|_{L^\infty} \|q_{\gamma-\beta}\|_{L^\infty}\Bigg).
\end{eqnarray*}
Next we merge the sums over $\mathbf{0} < \beta < \gamma$ (which implies $|\beta| \leq |\gamma|-1$) and $\mathbf{0} \leq  \alpha<\beta$ with $ |\alpha| \leq |\beta|-l $ into the sum over $\mathbf{0} \leq \alpha <  \gamma$ with $ |\alpha| \leq |\gamma|-(l+1)$. Denoting $q_{\theta_{l+1}}:= q_{\gamma-\beta}$ we obtain
\begin{eqnarray*}

	&&\sum_{l=1}^n \tilde{M}(t)^{l+1} \sum_{\substack{\mathbf{0} \leq \alpha <\gamma \\ 0 \leq |\alpha| \leq |\gamma|-(l+1) }} a_\alpha (t) \sum_{\substack{\theta_1+\dots+\theta_{l+1}=\gamma-\alpha \\ \theta_i \not=\mathbf{0}}} \prod_{i=1}^{l+1} \|q_{\theta_i}\|_{L^\infty} \\
	&&= \sum_{k=2}^{|\gamma|} \tilde{M}(t)^{k} \sum_{\substack{  0 \leq |\alpha| \leq |\gamma|-k \\\alpha <\gamma }} a_\alpha (t) \sum_{\substack{\theta_1+\dots+\theta_{k}=\gamma-\alpha \\ \theta_i \not=\mathbf{0}}} \prod_{i=1}^{k} \|q_{\theta_i}\|_{L^\infty}.
\end{eqnarray*}
Therefore, $\norm{u_\gamma (t,\cdot)}{L^2} $ is bounded by
\begin{equation*}
M(t)\Bigg\{ a_\gamma (t)    + \sum\limits_{\mathbf{0} \leq \beta < \gamma} \linf{q_{\gamma-\beta}} a_\beta (t) \tilde{M}(t)+ \sum_{k=2}^{|\gamma|} \tilde{M}(t)^{k} \sum_{\substack{  0 \leq |\alpha| \leq |\gamma|-k \\\alpha <\gamma }} a_\alpha (t) \sum_{\substack{\theta_1+\dots+\theta_{k}=\gamma-\alpha \\ \theta_i \not=\mathbf{0}}} \prod_{i=1}^{k} \|q_{\theta_i}\|_{L^\infty} \Bigg\},
\end{equation*}
and thus \eqref{ocena koeficijenti} holds.
Next we show that the solution $U$ is a generalized stochastic process of Kondratiev-type, i.e., that the sum $\vertiii{U}^2  := \sum_{ \gamma\in \mathcal{I}}\|u_\gamma\|^2_{AC([0,T]; L^2(\mathbb R^d))} (2\mathbb N)^{-p\gamma}$  is finite for some critical exponent $p$ to be determined bellow.
Using  \eqref{ocena koeficijenti} and $(a_1+a_2+a_3)^2 \leq 3a_1^2 + 3a_2^2+3a_3^2$ we find that $\vertiii{U}^2 $ is bounded by
\begin{eqnarray*}

	&&3 M(T)^2\sum_{\gamma \in \mathcal{I}}\| g_{\gamma}\|_{L^2}^2 (2 \mathbb N)^{-p \gamma}  + 3T^2M(T)^2 \sum_{\gamma \in \mathcal{I}} \| f_{\gamma}\|^2 (2 \mathbb N)^{-p \gamma}\\
	& & +  \, 3 M(T)^2 \sum_{\gamma \in \mathcal{I}}\Bigg(\sum\limits_{k=1}^{|\gamma|} \tilde{M}(T)^k \bigg( \sum_{\substack{0 \leq |\beta| \leq |\gamma|-k \\ \beta<\gamma}} a_\beta (T) \bigg( \sum_{\substack{\theta_1+\dots+\theta_k=\gamma-\beta \\ \theta_i \not=\mathbf{0}}} \prod_{i=1}^{k} \|q_{\theta_i}\|_{L^\infty} \bigg)\bigg) \Bigg)^2(2 \mathbb N)^{-p \gamma} \\
	&& = : 3 M(T)^2 \left( S_1+ T^2 S_2 +S_3\right).
\end{eqnarray*}
Chosing $p \geq \max\{p_2,p_3\}$, according to assumptions on $G$ and $F$, we have
\begin{equation} \label{suma S1+T2 S2}
S_1+T^2 S_2
\leq  \sum_{\gamma \in \mathcal{I}}\| g_{\gamma}\|_{L^2}^2 (2 \mathbb N)^{-p_3 \gamma} + T^2 \sum_{\gamma \in \mathcal{I}}\| f_{\gamma}\|^2 (2 \mathbb N)^{-p_2 \gamma}
:=A < \infty.
\end{equation}
For $S_3$, using  $(\sum\limits_{k=1}^{|\gamma|} x_k)^2 \leq |\gamma| \sum\limits_{k=1}^{|\gamma|} x_k^2$ we obtain
\[
S_3 \leq  \sum_{\gamma \in \mathcal{I}}|\gamma|  \sum\limits_{k=1}^{|\gamma|}  \tilde{M}(T)^{2k} \bigg( \sum_{\substack{0 \leq |\beta| \leq |\gamma|-k \\ \beta<\gamma}} a_\beta (T) \bigg( \sum_{\substack{\theta_1+\dots+\theta_k=\gamma-\beta \\ \theta_i \not=\mathbf{0}}} \prod_{i=1}^{k} \|q_{\theta_i}\|_{L^\infty} \bigg)\bigg)^2 (2\mathbb N)^{-p\gamma}.
\]
Further, using $\left( \sum\limits_{ \alpha \in \mathcal{I}} |x_{\alpha} y_\alpha| \right)^2 \leq \left( \sum\limits_{\alpha \in \mathcal{I}}|x_{\alpha}|^2\right) \left(  \sum\limits_{ \alpha \in \mathcal{I}} |y_\alpha|^2 \right)$ and rearranging the powers of $(2\mathbb N)^{-p\gamma}$ we obtain
\begin{eqnarray*}

	S_3 \leq \sum_{\gamma \in \mathcal{I}} |\gamma|  \sum\limits_{k=1}^{|\gamma|}  \tilde{M}(T)^{2k} \sum_{\substack{0 \leq |\beta| \leq |\gamma|-k \\ \beta<\gamma}} (a_\beta (T) (2\mathbb N)^{-\frac{p\gamma}{6}})^2
	\sum_{\substack{0 \leq |\beta| \leq |\gamma|-k \\ \beta<\gamma}}
	\bigg( \sum_{\substack{\theta_1+\dots+\theta_k=\gamma-\beta \\ \theta_i \not=\mathbf{0}}} \prod_{i=1}^{k} \|q_{\theta_i}\|_{L^\infty} (2\mathbb N)^{-\frac{p\gamma}{6}}\bigg)^2  (2\mathbb N)^{-\frac{p\gamma}3}.
\end{eqnarray*}
Since $\beta<\gamma$ and $p \geq \max\{p_2,p_3\}$ (chosen in \eqref{suma S1+T2 S2}) we have
\begin{equation*}
\sum_{\substack{0 \leq |\beta| \leq |\gamma|-k \\ \beta<\gamma}} a_\beta (T)^2  (2\mathbb N)^{-\frac{p\gamma}3} \leq \sum_{\substack{0 \leq |\beta| \leq |\gamma|-k \\ \beta<\gamma}} a_\beta (T)^2  (2\mathbb N)^{-\frac{p\beta}3} \leq 2 A <\infty,
\end{equation*}
and by Lemma \ref{pomocna lema - ocene} we find
\[ S_3	 \leq  2A \sum_{\gamma \in \mathcal{I}} |\gamma|  \sum\limits_{k=1}^{|\gamma|}  \tilde{M}(T)^{2k}  \Bigg( \sum_{\substack{0 \leq |\beta| \leq |\gamma|-k \\ \beta<\gamma}} \|q_\textbf{0}\|^{2k}_{L^\infty}  \,  2^{2k|\gamma-\beta|}(2\mathbb N)^{-\frac{p(\gamma-\beta)}{3}}\Bigg) (2\mathbb N)^{-\frac{p\gamma}3}.
\]
By Lemma \ref{pomocna lema - ocene}, there exists $s\geq 0$ such that $2^{2k|\gamma-\beta|}\leq (2\mathbb N)^{\frac{s (\gamma-\beta)}3}$, and choosing $p$ to satisfy $p>s+3$ we have
\begin{eqnarray*}
	S_3
	&\leq & 2A \sum_{\gamma \in \mathcal{I}}  |\gamma|   \sum\limits_{k=1}^{|\gamma|}  \tilde{M}(T)^{2k}  \|q_\textbf{0}\|^{2k}_{L^\infty} \Bigg( \sum_{\substack{ |\gamma-\beta| \geq k \\ \beta<\gamma}} (2\mathbb N)^{-\frac{(p-s) (\gamma-\beta)}3}\Bigg) (2\mathbb N)^{-\frac{p\gamma}{3}}\\
	& \leq & 2 A C \sum_{\gamma \in \mathcal{I}}  |\gamma|   \sum\limits_{k=1}^{|\gamma|}  \tilde{M}(T)^{2k} \|q_\textbf{0}\|^{2k}_{L^\infty}   (2\mathbb N)^{-\frac{p\gamma}3},
\end{eqnarray*}
where $A$ is defined in \eqref{suma S1+T2 S2} and $C:=\sum_{\alpha \in \mathcal I} (2\mathbb N)^{-\frac{(p-s) \alpha}3}$ is finite by \eqref{2N konv p>1}. The assumption \mbox{$\tilde{M}(T) \|q_\textbf{0}\|_{L^\infty} \not= 1$} allows to
sum up the inner sum leading to
\begin{eqnarray*}

	S_3 \leq 2 A C \dfrac{\tilde{M}(T)^{2}\|q_\textbf{0}\|^{2}_{L^\infty}}{1-\tilde{M}(T)^{2}\|q_\textbf{0}\|^{2}_{L^\infty}} \left(  \sum_{\gamma \in \mathcal{I}}  |\gamma| (2\mathbb N)^{-\frac{p\gamma}3}  - \sum_{\gamma \in \mathcal{I}}  |\gamma| \tilde{M}(T)^{2|\gamma|}\|q_\textbf{0}\|^{2|\gamma|}_{L^\infty} (2\mathbb N)^{-\frac{p\gamma}3} \right).
\end{eqnarray*}

By Lemma \ref{pomocna lema - ocene},
for $\tilde{M} (T)^2 \|q_\textbf{0}\|^2 >0$ there exists $s_1>0$ so that
\begin{eqnarray} \label{s3 suma 2}
\sum_{\gamma \in \mathcal{I}}  |\gamma| \tilde{M}(T)^{2|\gamma|}\|q_\textbf{0}\|^{2|\gamma|}_{L^\infty} (2\mathbb N)^{-\frac{p\gamma}3}
\leq \sum_{\gamma \in \mathcal{I}} (2\mathbb N)^{s_1\gamma} (2\mathbb N)^{-(\frac{p}3-1)\gamma}.
\end{eqnarray}

Choosing $p>3s_1+6$, by Lemma \ref{pomocna lema - ocene} we have that  \eqref{s3 suma 2} is finite, and
\begin{equation*
\sum_{\gamma \in \mathcal{I}}  |\gamma| (2\mathbb N)^{-\frac{p\gamma}3} \leq  \sum_{\gamma \in \mathcal{I}}  (2\mathbb N)^\gamma (2\mathbb N)^{-\frac{p\gamma}3} =  \sum_{\gamma \in \mathcal{I}}   (2\mathbb N)^{-(\frac{p}3-1)\gamma} < \infty,
\end{equation*}
yielding that $S_3$ is finite.
Finally, $\vertiii{U}^2 < \infty$  for $p\geq \max\{p_2, p_3, s+3, 3s_1+6\}$, and  \linebreak $U \in AC([0,T]; D)  \otimes (S)_{-1}$. \\

The uniqueness of the solution $U$ follows from the uniqueness of its coefficients $u_\gamma$, $\gamma\in \mathcal I$ and the uniqueness of  the chaos expansion representation in the Fourier--Hermite basis of orthogonal stochastic polynomials.
\end{proof}

\section*{Acknowledgments}

Author Ljubica Oparnica is supported by the FWO Odysseus 1 grant no. G.0H94.18N: Analysis and Partial Differential Equations. Author Sne\v zana Gordi\'c is supported by Ministry of Education, Science and Technological Development of the Republic of Serbia, by Faculty of Education in Sombor.

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text:
Suspected 'honey oil' lab raided in Palmdale, 1 arrested
by Luis Meza • December 2, 2015
The bust happened Wednesday, Dec. 2, at a residence in the 5500 block of Laurel Avenue in Palmdale. [Photo by LUIS MEZA]
PALMDALE – A 29-year-old man was arrested Wednesday after deputies found a "honey oil" lab while serving a search warrant at a Palmdale home, authorities said.
Jimmy Echeverria was arrested around 7:45 a.m. on suspicion of manufacturing a controlled substance, according to LASD inmate records.
The bust happened at a residence in the 5500 block of Laurel Avenue in Palmdale.
[LUIS MEZA]
"The Sheriff's Department served a search warrant here this morning and discovered a marijuana extraction lab, and we came and processed the marijuana extraction lab," said Detective Scott Schulze of the Los Angeles Interagency Metropolitan Police Apprehension Crime Task Force (IMPACT).
"There were other narcotics found," Schulze added, but did not elaborate. The detective said he did not know the quantity or value of the drugs seized.
Officers spent several hours processing the scene on Wednesday. Several dozen cans of butane were recovered, as well as glass trays with residue from marijuana concentrates, according to deputies at the scene.
Manufacturing marijuana concentrates, often referred to as "honey oil", involves using highly flammable butane to extract Tetrahydrocannabinol (THC) from the cannabis plant, according to the Drug Enforcement Administration. Given the extremely volatile nature of heating butane and creating a gas, this process can result in violent explosions, according to the DEA. Read more here.
Echeverria was booked at the Palmdale Station jail Wednesday morning in lieu of $75,000 bail. He was released Wednesday evening after posting bond, according to LASD inmate records. The incident remains under investigation and no further information was immediately available.
Filed Under: Crime/ Safety
18 comments for "Suspected 'honey oil' lab raided in Palmdale, 1 arrested"
Ronald Reagan says
In the state of California any amount of Marijuana concentrate (Wax) Cultivation is Federal Felony punishable up to 3 years in prison and another 3 years for Possession. If this guy has evidence against him for Sales..then that's additional 4 years.
– That's total of 10 years in prison. Hope he can afford a good attorney….LOL
Federal Felony means your Medial Marijuana Recommendation wont save your freedom.
... says
Shut ur damn mouth
Flint says
The laws on hash and other concentrated forms of thc seem off. It seems like marijuana and oil etc… Should be citations or misdemeanors. If marijuana is legalized and taxed like alcohol, I assume that the government would make strict laws about growing and selling because that goes against the effort to derive tax dollars.
Matt says
You think your Namesake would be proud of incarcerating a man for a victimless crime? This guy didn't hurt, rob or extort anyone or vandalize any property. A crime without damage to persons or property is victimless. Therefore only a crime in the eyes of the government, who use their cronies to enforce the letter of the law, opposed to the spirit of the law. "Well the guy had FLAMMABLE chemicals in the house!" Ok, EVERYONE has at least half a dozen or so flammable chemicals in their house…yet I haven't seen any house on my block randomly explode for quite some time. Maybe if the sheriffs weren't so busy worrying about locking up a stoner, they'd have been more vigilant the night that woman was abducted and raped in Lancaster,…not a fair shot? Our Sheriffs' are so hell bent on enforcing laws that exist to exploit and generate revenue, they're hesitant to actually prevent or investigate real crime because there's 0 incentive to do so. Otherwise honey oil do-it-yourself-hobbyists wouldn't be on some Sheriffs blotter as a priority arrest.
Matt S. says
That's right, Reagan only used cocaine to fund
his dealings in South America and the middle east, not cannabis.
Yawn says
It's funny when ignorant people believe what's written, don't find out for themselves how anything works, then cast judgement. People scared of houses blowing up lol, keep drinking that kool-aid. Sheep are easy to shepard.
Tim Scott says
The number of people in prison because the fire department came and put out the remains of their home laboratories is not small enough to be ignored.
By the way, I'm no sheep so keep your pants on.
IF your meth lab explodes and sets the neighborhood on fire, that's negligence on your end, hands down. However if I fall asleep smoking a cigarette, play with a flare gun while drunk, throw fireworks at gasoline cans, do a Gene Simmons impression and trip etc… THESE CAN ALSO BURN DOWN NEIGHBORHOODS, that's when property is affected and that's when a real crime is committed. This isn't"Minority Report" and we get to condemn people for something they may or may not do in the future. It would be bad for this guy's bottom line if he burns his or any other home down, so I'm assuming there was some caution exercised, especially since (i believe) only the liquid is needed for making oil, not an open flame. If he was microbrewing some suds in his spare bathroom and/or distilling some sour mash in the backyard i doubt any of this self righteous indignation in behalf of people who yearn for some bygone age when"people knew their place" would be uttered.
I'm as much in favor of legalization as anyone, but I think you are making a mistake Matt. Legalization does not include "as long as marijuana is involved no laws apply." It doesn't matter WHAT you are manufacturing, you can't manufacture product in a house. People in houses in a residential area have a right to expect that their neighbors are also residents, and not exposing them to the hazards they would face if for some reason they chose to live in an industrial park.
Would it be better if there had been less of a big deal made out of WHAT this guy was manufacturing in his little home turned factory? Sure. But at the end of the day you can't turn a house into a manufacturing facility.
chris haze says
ive said once and ill say it again. Is this a joke? Plenty of better things the police of lancaster can be doing. this is ridiculous, what a joke of a police force.
Just Saying says
That would be true, if it happened in Lancaster.
AVHater says
Honey oil, AKA marijuana. Big [removed] deal. The War on Some Drugs is the never ending make work gravy train for cops and prison employees. LEO- do the world a favor and drive off a cliff.
Billandtimsbogusanswers says
Other drugs were found, is that ok.how about when the house blows up, is that ok.which illegal activities are ok.some addicts say meth, cocaine, should be ok.how about we just become a nation of no laws.
Yes, absolutely drugs are fine as long as the user minds their own business and doesn't harm another person. You probably wouldn't know a meth user if one bit you on the azz anyway.
Heartbroken says
You likely don't have a daughter you haven't seen or heard from in months. I don't know if she is dead or alive. All I can do is remember her the last time I saw her. All these boys glorify what there doing to get a li l lovin' , taking advantage is more like it. One thing leads to another, pot was no big deal, then speed, now she is doing heroin…my heart hurts every day as does the rest of our families…hospitalized from o.d….beat up…if u have a neice, daughter, aunt, mother missing or any other family member educate your friends to educate there kids and be involved more we are only given one life
An excellent demonstration of how pointless and unenforceable drug laws endanger us all. Let's make this illegal so the manufacturing process, involving large quantities of dangerously volatile fuel, will be moved into a house in a residential neighborhood and done by a guy with (perhaps) no particular skill or training, and with no supervision or regulation of any kind.
lol says
LASD doesnt have anything better to do then bust someone making honey oil? There are real criminals out there.
[removed] grimme


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