Nonstandard Martingales, Markov Chains and the Heat Equation

TL;DR

This paper introduces a nonstandard martingale derived from a discrete Markov chain, providing a novel approach to solving the heat equation with both smooth and non-smooth initial conditions.

Contribution

It develops a nonstandard martingale framework that connects discrete Markov chains to solutions of the heat equation, extending classical methods.

Findings

Nonstandard martingale construction from Markov chains.

Solution to heat equation with non-smooth initial conditions.

Classical solution recovered for smooth initial conditions.

Abstract

We construct a nonstandard martingale from a discrete Markov chain. This is shown to be useful for solving the heat equation with a non smooth initial condition. We show that the nonstandard solution to the heat equation with a smooth initial condition specialises to the classical solution.

Full text

One of the most fundamental results in the theory of Markov chains is the following;

$p_{ij}^{(n)}\rightarrow\pi_{j}$*, for all states $i,j$, as $n\rightarrow\infty$

$\overline{p}_{(i,j)(k,l)}=p_{ik}p_{jl}$

and initial distribution $\mu_{(i,j)}=\lambda_{i}\pi_{j}$. A simple calculation shows that;

$\overline{p}^{(n)}_{(i,j)(k,l)}=p_{ik}^{(n)}p_{jl}^{(n)}$ for fixed states $i,j,k,l$

As $P$ is irreducible and aperiodic, we have that $min(p_{ik}^{(n)},p_{jl}^{(n)})>0$, for sufficiently large $n$. Hence, for such $n$, $\overline{p}^{(n)}_{(i,j)(k,l)}>0$ and $\overline{P}$ is irreducible. A similar straightforward calculation gives that the distribution $\mu_{(i,j)}$ is invariant for $\overline{P}$. By well known results, this implies that $\overline{P}$ is positive recurrent. Fix a state $b$, and let $S=inf\{n\geq 1:X_{n}=Y_{n}=b\}$. Then $S$ is the first passage time in the system $\{W_{n}\}_{n\geq 0}$ to $(b,b)$, and $P(S<\infty)=1$ follows by known results, and the fact that $\overline{P}$ is irreducible and recurrent. Clearly $P(S<\infty)\leq P(T<\infty)$, so $(*)$ follows. We now calculate;

$P(X_{n}=j)=P(X_{n}=j,n\geq T)+P(X_{n}=j,n<T)$

$=P(Y_{n}=j,n\geq T)+P(X_{n}=j,n<T)$

by definition of $T$ and the fact that $\{X_{n}\}_{n\geq 0}$ and $\{Y_{n}\}_{n\geq 0}$ have the same transition matrix. Then;

$P(X_{n}=j)=P(Y_{n}=j,n\geq T)+P(Y_{n}=j,n<T)$

$\indent\ \ \ \ \ \ \ \ \ \ \ -P(Y_{n}=j,n<T)+P(X_{n}=j,n<T)$

$=P(Y_{n}=j)-P(Y_{n}=j,n<T)+P(X_{n}=j,n<T)$

$=\pi_{j}-P(Y_{n}=j,n<T)+P(X_{n}=j,n<T)$ $(**)$

We have that $P(Y_{n}=j,n<T)\leq P(n<T)$ and $P(n<T)\rightarrow P(T=\infty)=0$ as $n\rightarrow\infty$, using $(*)$. Similarly, $P(X_{n}=j,n<T)\rightarrow 0$ as $n\rightarrow\infty$. It follows that $P(X_{n}=j)\rightarrow\pi_{j}$, using $(**)$, as required. The final claim is a consequence of the fact that $p_{ij}^{(n)}=P(X_{n}=j)$ where the initial distribution of $X_{0}$ is the dirac function $\delta_{i}$.

∎

We now establish a rate of convergence result.

$|p_{ij}^{(n)}-\pi_{j}|\leq(1-\rho)^{{n\over m}-1}$*, for all states $i,j$

$P(X_{n}=j)=\pi_{j}-P(Y_{n}=j,n<T)+P(X_{n}=j,n<T)$

Hence;

$|p_{ij}^{(n)}-\pi_{j}|\leq P(n<T)$

As $P$ is irreducible and aperiodic, we have that $p_{kl}^{(n)}>0$ for all sufficiently large $n$, and all states $k,l$. As $P$ is finite, there exists an $m\geq 1$ such that $p_{kl}^{(m)}>0$ for all $k,l$. In particular, there exists $\rho\in(0,1)$ such that $p_{kl}^{(m)}>\rho$. We have that;

$P_{k,l}(T\leq m)\geq\sum_{u}\overline{p}^{(m)}_{(k,l),(u,u)}=\sum_{u}p_{ku}^{(m)}p_{lu}^{(m)}\geq\rho\sum_{u}p_{ku}^{(m)}=\rho$

$P_{k,l}(T>m)\leq(1-\rho)$

$P(T>m)=\sum_{(k,l)}P_{(k,l)}(T>m)\delta_{ik}\pi_{l}\leq(1-\rho)$

Moreover;

$P(T>n)\leq P(T>[{n\over m}]m)$

We claim that, for $k\geq 1$, $P(T>(k+1)m|T>km)\leq 1-\rho$. We have that;

$P(T>(k+1)m|T>km)$

$=\sum_{i_{km}\neq j_{km},i_{km-1}\neq j_{km-1},\ldots,i,j}P(T>(k+1)m|W_{km}=(i_{km},j_{km}),W_{km-1}=(i_{km-1},j_{km-1}),\ldots,W_{0}=(i,j))P(W_{km}=(i_{km},j_{km}),W_{km-1}=(i_{km-1},j_{km-1}),\ldots,W_{0}=(i,j)|T>km)$

$=\sum_{i_{km}\neq j_{km},i_{km-1}\neq j_{km-1},\ldots,i,j}P(T>(k+1)m|W_{km}=(i_{km},j_{km}))P(W_{km}=(i_{km},j_{km}),W_{km-1}=(i_{km-1},j_{km-1}),\ldots,W_{0}=(i,j)|T>km)$

$\leq(1-\rho)\sum_{i_{km}\neq j_{km},i_{km-1}\neq j_{km-1},\ldots,i,j}P(W_{km}=(i_{km},j_{km}),W_{km-1}=(i_{km-1},j_{km-1}),\ldots,W_{0}=(i,j)|T>km)$

$=(1-\rho)$

Inductively, we have that;

$P(T>km)=P(T>km,T>(k-1)m)$

$=P(T>km|T>(k-1)m)P(T>(k-1)m)$

$\leq P(T>km|T>(k-1)m)(1-\rho)^{k-1}$

$\leq(1-\rho)^{k}$

It follows that $|p_{ij}^{(n)}-\pi_{j}|\leq(1-\rho)^{[{n\over m}]}\leq(1-\rho)^{{n\over m}-1}$

∎

$|p_{ij}^{(n)}-{1\over N}|\leq({3^{N-1}-1\over 3^{N-1}})^{{n\over N-1}-1}=\epsilon_{n}$

*If $N$ is odd, we can choose $m={N-1\over 2}$ and $\rho={1\over 3^{{N-1\over 2}}}$ in Lemma 0.2. It follows that;

$|p_{ij}^{(n)}-{1\over N}|\leq({3^{{N-1\over 2}}-1\over 3^{{N-1\over 2}}})^{{2n\over N-1}-1}=\delta_{n}$

*Moreover, for any initial probability distribution $\pi_{0}$, letting $\pi_{j}^{n}=P(X_{n}=j)$, we have that;

$|\pi_{j}^{n}-{1\over N}|\leq\epsilon_{n}$*, $0\leq j\leq N-1$ $(*)$

*and, similarly, with $\delta_{n}$ replacing $\epsilon_{n}$, when $N$ is odd. For any initial distribution $\lambda_{0}$ of positive numbers, with sum $K$, letting $\lambda_{n}=\lambda_{0}P^{n}$, we have that;

$|\lambda_{j}^{n}-{K\over N}|\leq K\epsilon_{n}$*, $0\leq j\leq N-1$

and, similarly, with $\delta_{n}$ replacing $\epsilon_{n}$ when $N$ is odd.

$\pi_{j}^{n}=\pi_{0}^{0}p_{0j}^{(n)}+\pi_{1}^{0}p_{1j}^{(n)}+\ldots+\pi_{N-1}^{0}p_{N-1,j}^{(n)}$

$=(\pi_{0}^{0}+\ldots+\pi_{N-1}^{0})({1\over N})+\pi_{0}^{0}\epsilon_{n}^{0}+\ldots+\pi_{N-1}^{0}\epsilon_{n}^{N-1}$

$={1\over N}+\epsilon_{n}^{\prime}$

where $\epsilon_{n}^{j}\leq\epsilon_{n}$, for $0\leq j\leq N-1$ and $\epsilon_{n}^{\prime}\leq\epsilon_{n}$. The case $N$ odd is the same. The final claim follows by observing that $\lambda_{n}=\lambda_{0}P^{n}=K\pi_{0}P^{n}$ for an initial probability distribution $\pi_{0}$. Then multiply the result $(*)$ through by $K$, similarly for $N$ odd.

∎

$n\geq(\eta-1)(1+{log(\epsilon)\over log(3^{\eta-1}-1)-log(3^{\eta-1})}$* $(*)$

*we have for any initial probability distribution $\pi_{0}$, that;

$\pi_{j}^{n}\simeq{1\over\eta}$* for $0\leq j\leq\eta-1$ $(**)$

*We obtain the same result, if $\eta$ is odd, $\epsilon$ is an infinitesimal and;

$n\geq{\eta-1\over 2}(1+{log(\epsilon)\over log(3^{{\eta-1\over 2}}-1)-log(3^{{\eta-1\over 2}})})$* $(***)$

*for any initial probability distribution $\pi_{0}$. If $\lambda_{0}$ is a nonstandard distribution with sum $\lambda$, possibly infinite, then if $\epsilon$ is an infinitesimal with $\lambda\epsilon\simeq 0$, and $n$ satisfies $(*)$, we obtain that;

$\pi_{j}^{n}\simeq{\lambda\over\eta}$* for $0\leq j\leq\eta-1$ $(****)$

and, the same result holds when $\eta$ is odd and $n$ satisfies $(***)$.

$Prob_{N}=\{f\in Seq_{1}:(\forall_{m\geq N+1}f(m)=0)\wedge(\forall_{1\leq m\leq N}f(m)\geq 0)\wedge\sum_{1\leq m\leq N}f(m)=1\}$.

encode probability vectors of length $N$. Let $G:\mathcal{N}\rightarrow Seq_{2}$ be defined by;

$G(N,1,1)=G(N,1,2)=G(N,1,N)={1\over 3}$, $(\forall_{m\neq 1,2,N}G(N,1,m)=0$

$(\forall_{1\leq k\leq N-1}\forall_{2\leq m\leq N}G(N,k+1,1)=G(N,k,N)$, $G(N,k+1,m)=G(N,k,m-1)$

$\forall_{k\geq N+1}\forall_{m\geq N+1}G(N,k,m)=0$

$G$ encodes the transition matrices for the given Markov chain with $N$ states. Let $H:\mathcal{N}^{2}\rightarrow Seq_{2}$ be defined by;

$(\forall_{1\leq i,j\leq N})H(1,N,i,j)=G(N,i,j)$

$(\forall_{i,j>N})H(1,N,i,j)=0$

$(\forall_{1\leq i,j\leq N})H(n,N,i,j)=\sum_{1\leq k\leq N}H(n-1,i,k)G(N,k,j)$

$(\forall_{i,j>N})H(n,N,i,j)=0$

$H$ encodes the powers $G(N)^{n}$ of the transition matrices. We define maps $L(N,n):Prob_{N}\rightarrow Prob_{N}$ by;

$(\forall_{1\leq j\leq N})L(N,n)(f)(j)=\sum_{1\leq k\leq N}f(k)H(n,N,k,j)$

$(\forall_{j>N})L(N,n)(f)(j)=0$

$L(N,n)(f)$ encodes the probability vectors $\pi^{n}$ for an initial distribution $\pi_{0}$ represented by $f$.

By a simple rearrangement, we have that the bound in $|\pi_{j}^{n}-{1\over N}|$, from lemma 0.3, can be formulated in first order logic as;

$\forall N\in\mathcal{N}\forall\pi\in Prob_{N}\forall\epsilon\in\mathcal{R}_{>0}\forall{n\in\mathcal{N}}(n\geq(N-1)(1+{log(\epsilon)\over log(3^{N-1}-1)-log(3^{N-1})})\rightarrow(|L(n,N)(\pi)(j)-{1\over N}|\leq\epsilon,0\leq j\leq N-1)$

By transfer, we obtain a corresponding result, quantifying over ${{}^{*}\mathcal{N}}$. Taking $\epsilon$ to be an infinitesimal and $\eta$ to be an infinite natural number, we obtain the first result. Observe that by construction of $G,H,L$, the nonstandard Markov chain with $\eta$ states evolves by the usual nonstandard matrix multiplication by the transition matrix, of the initial probability distribution. The remaining claims are similar and left to the reader.

∎

*We let $\overline{\Omega_{\eta}}=\{x\in{{}^{*}\mathcal{R}}:0\leq x<1\}$

*and $\overline{\mathcal{T}_{\nu}}=\{t\in{{}^{*}\mathcal{R}}:0\leq x\leq 1\}$

*We let $\mathcal{C}_{\eta}$ consist of internal unions of the intervals $[{i\over\eta},{i+1\over\eta})$, for $0\leq i\leq\eta-1$, and let $\mathcal{D}_{\nu}$ consist of internal unions of $[{i\over\nu},{i+1\over\nu})$ and $\{1\}$, for $0\leq i\leq\nu-1$.

*We define counting measures $\mu_{\eta}$ and $\lambda_{\nu}$ on $\mathcal{C}_{\eta}$ and $\mathcal{D}_{\nu}$ respectively, by setting $\mu_{\eta}([{i\over\eta},{i+1\over\eta}))={1\over\eta}$, $\lambda_{\nu}([{i\over\nu},{i+1\over\nu}))={1\over\nu}$, for $0\leq i\leq\eta-1$, $0\leq i\leq\nu-1$ respectively, and $\lambda_{\nu}(\{1\})=0$.

We let $(\overline{\Omega_{\eta}},\mathcal{C}_{\eta},\mu_{\eta})$ and $(\overline{\mathcal{T}}_{\nu},\mathcal{D}_{\nu},\lambda_{\nu})$ be the resulting ${*}$-finite measure spaces, in the sense of [2]. We let $(\overline{\Omega_{\eta}}\times\overline{\mathcal{T}}_{\nu},\mathcal{C}_{\eta}\times\mathcal{D}_{\nu},\mu_{\eta}\times\lambda_{\nu})$ denote the corresponding product space.

$F({i\over\eta},{j\over\nu})=(\pi_{f}K^{j})(i)$*, for $0\leq i\leq\eta-1$, $0\leq j\leq\nu$

$F(x,y)=F({[x\eta]\over\eta},{[y\nu]\over\nu})$*, $(x,y)\in\overline{\Omega_{\eta}}\times\overline{\mathcal{T}_{\nu}}$

where $\pi_{f}$ is the nonstandard distribution vector corresponding to $f$, $K$ is the transition matrix of the above Markov chain with $\eta$ states, and $K^{j}$ denotes a nonstandard power.

∎

*(i). $\mathcal{E}_{\eta,0}=\mathcal{E}_{\eta}$

*(ii). $\mathcal{E}_{\eta,i}\subseteq\mathcal{E}_{\eta,j}$, if $i\geq j$.

*We say that $F:\overline{\Omega_{\eta}}\times\overline{\mathcal{T}_{\nu}}\rightarrow{{}^{*}\mathcal{R}}$ is adapted to the filtration if $F$ is measurable with respect to $\mathcal{E}_{\eta}\times\mathcal{D}_{\nu}$ and $F_{i\over\nu}:\overline{\Omega_{\eta}}\rightarrow{{}^{*}\mathcal{R}}$ is measurable with respect to $\mathcal{E}_{\eta,i}$, for $0\leq i\leq\nu$.

*If $f:\overline{\Omega_{\eta}}\rightarrow{{}^{*}\mathcal{R}}$ is measurable with respect to $\mathcal{E}_{\eta,j}$ and $0\leq j\leq i\leq\nu$, we define the conditional expectation $E_{\eta}(f|\mathcal{E}_{\eta,i})$ to be the unique $g:\overline{\Omega_{\eta}}\rightarrow{{}^{*}\mathcal{R}}$ such that $g$ is measurable with respect to $\mathcal{E}_{\eta,i}$ and;

$\int_{U}gd\gamma_{\eta}=\int_{U}fd\gamma_{\eta}$

*for all $U\in\mathcal{E}_{\eta,i}$. We say that $F:\overline{\Omega_{\eta}}\times\overline{\mathcal{T}_{\nu}}\rightarrow{{}^{*}\mathcal{R}}$ is a reverse martingale if;

*(i). $F$ is adapted to the reverse filtration on $\overline{\Omega_{\eta}}$

*(ii). $E_{\eta}(F_{j\over\nu}|\mathcal{E}_{\eta,i})=F_{i\over\nu}$ for $0\leq j\leq i\leq\nu$

*Given $F:\overline{\Omega_{\eta}}\times\overline{\mathcal{T}_{\nu}}\rightarrow{{}^{*}\mathcal{R}}$ measurable with respect to $\mathcal{E}_{\eta}\times\mathcal{D}_{\nu}$, we define the cumulative density function; $P:{{}^{*}\mathcal{R}}\times[0,1]\rightarrow{{}^{*}\mathcal{R}}$ by;

$P(x,t)=\gamma_{\eta}(F_{t}\leq x)$

*We say that $F_{1}$ and $F_{2}$ are equivalent in distribution, if their respective cumulative density functions $P_{1}$ and $P_{2}$ coincide.

Base case; the points $\{({i\over\eta},1):0\leq i\leq\eta-1\}$ are special points of the first kind.

Inductive hypothesis; Suppose the special points of the first and second kind on $\overline{\Omega_{\eta}}\times[{\nu-i\over\nu},1]$ have been defined for $0\leq i\leq\nu-1$

Then a point $(x,{\nu-i-1\over\nu})$ is special of the first kind if $(x,{\nu-i\over\nu})$ is special of the first or second kind, in addition, the points $(x+{k\over 3^{i+1}\eta},{\nu-i-1\over\nu})$, $1\leq k\leq 2$, are special of the second kind, when $(x,{\nu-i\over\nu})$ is special of the first kind.

We call a point of the form $({i\over\eta},{j\over\nu})$ for $0\leq i\leq\eta-1$, $0\leq j\leq\nu$ original. We associate original points to special points inductively as follows;

Base case; We associate the original point $({i\over\eta},1)$ to the special point $({i\over\eta},1)$, for $0\leq i\leq\eta-1$

Inductive hypothesis: Suppose we have associated original points to special points on $\overline{\Omega_{\eta}}\times[{\nu-i\over\nu},1]$,for $0\leq i\leq\nu-1$.

Then if $(x,{\nu-i-1\over\nu})$ is special of the first kind, and $({i_{x}\over\eta},{\nu-i\over\nu})$ is associated to $(x,{\nu-i\over\nu})$, we associate $({i_{x}\over\eta},{\nu-i-1\over\nu})$ to $(x,{\nu-i-1\over\nu})$. We match original points to special points of the second kind, by associating $({i_{x}-1\over\eta},{\nu-i-1\over\nu})$ to $(x+{1\over 3^{i+1}\eta},{\nu-i-1\over\nu})$ and $({i_{x}+1\over\eta},{\nu-i-1\over\nu})$ to $(x+{2\over 3^{i+1}\eta},{\nu-i-1\over\nu})$, where $(x,{\nu-i-1\over\nu})$ is special of the first kind. We adopt the convention that ${-1\over\eta}={\eta-1\over\eta}$ and ${\eta\over\eta}={0\over\eta}$.

We define $\overline{F}$ on $\overline{\Omega_{\eta}}\times\overline{\mathcal{T}_{\nu}}$ by setting $\overline{F}(z)=F(\Gamma(z))$, $(*)$, where $z$ is a special point of $\overline{\Omega_{\eta}}\times\overline{\mathcal{T}_{\nu}}$ and $\Gamma(z)$ is the associated original point. We then set $\overline{F}(x,y)=\overline{F}({[x3^{\nu-[\nu y]}\eta]\over 3^{\nu-[\nu y]}\eta},{[\nu y]\over\nu})$, $(**)$.

We claim that $\overline{F}$ is a reverse martingale. By the definition $(*)$, $\overline{F}$ is adapted to the reverse filtration, $(i)$. To verify $(ii)$ of the definition of a reverse martingale in Definition 0.9, by the tower law for conditional expectation, it is sufficient to prove that $E_{\eta}(\overline{F}_{i\over\nu}|\mathcal{E}_{i+1})=\overline{F}_{i+1\over\nu}$, for $0\leq i\leq\nu-1$. We have that;

$E_{\eta}(\overline{F}_{i\over\nu}|\mathcal{E}_{i+1})({j\over 3^{\nu-i-1}\eta})$

$=3^{\nu-i-1}\eta\int_{[{j\over 3^{\nu-i-1}\eta},{j+1\over 3^{\nu-i-1}\eta})}E_{\eta}(\overline{F}_{i\over\nu}|\mathcal{E}_{i+1})d\gamma_{\eta}$

$=3^{\nu-i-1}\eta\int_{[{j\over 3^{\nu-i-1}\eta},{j+1\over 3^{\nu-i-1}\eta})}\overline{F}_{i\over\nu}d\gamma_{\eta}$

$={3^{\nu-i-1}\eta\over 3^{\nu-i}\eta}(\sum_{k=0}^{2}\overline{F}_{i\over\nu}({3j+k\over 3^{\nu-i}\eta}))$

$={1\over 3}(F_{i\over\nu}({i_{x}\over\eta})+F_{i\over\nu}({i_{x}-1\over\eta})+F_{i\over\nu}({i_{x}+1\over\eta}))$

$=F_{i+1\over\nu}({i_{x}\over\eta})=\overline{F}_{i+1\over\nu}({j\over 3^{\nu-i-1}\eta})$

where $({i_{x}\over\eta},{i+1\over\nu})$ is associated to the special point $({j\over 3^{\nu-i-1}\eta},{i+1\over\nu})=(x,{i+1\over\nu})$

We now claim that if $({i\over\eta},{\nu-j\over\nu})$ is an original point, then it is associated to $3^{j}$ special points of the form $(y,{\nu-j\over\nu})$. We prove this by induction.

Base Case; $j=0$, then the original points $\{({i\over\eta},1):0\leq i\leq\eta-1\}$ are in bijection with the same special points.

Inductive hypothesis; Suppose that, for $0\leq i\leq\eta-1$, each original point $({i\over\eta},{\nu-j\over\nu})$ is associated to $3^{j}$ special points of the form $(y,{\nu-j\over\nu})$.

If the original point $({i\over\eta},{\nu-j-1\over\nu})$ is associated to a special point $(x,{\nu-j-1\over\nu})$, there are $3$ cases, either $({i\over\eta},{\nu-j\over\nu})$ is associated to $(x,{\nu-j\over\nu})$, or, $({i-1\over\eta},{\nu-j\over\nu})$ is associated to $(x-{1\over 3^{j+1}\eta},{\nu-j\over\nu})$, or $({i+1\over\eta},{\nu-j\over\nu})$ is associated to $(x-{2\over 3^{j+1}\eta},{\nu-j\over\nu})$. These cases are disjoint and all occur, so we obtain a total of $3.3^{j}=3^{j+1}$ assignments.

Using this result and the definition of $\overline{F}$, for a given $\alpha\in{{}^{*}\mathcal{R}}$;

$3^{\nu-[\nu t]}{{}^{*}Card}(\{i:0\leq i\leq\eta-1,F({i\over\eta},{[\nu t]\over\nu})=\alpha\})$

$={{}^{*}Card}(\{i:0\leq i\leq 3^{\nu-[\nu t]}\eta-1,\overline{F}({i\over 3^{\nu-[\nu t]}\eta},{[\nu t]\over\nu})=\alpha\})$

We now calculate;

$\gamma_{\eta}(F_{t}=\alpha)$

$={1\over\eta}{{}^{*}Card}(\{i:0\leq i\leq\eta-1,F({i\over\eta},{[\nu t]\over\nu})=\alpha\})$

$={1\over 3^{\nu-[\nu t]}\eta}{{}^{*}Card}(\{i:0\leq i\leq 3^{\nu-[\nu t]}\eta-1,\overline{F}({i\over 3^{\nu-[\nu t]}\eta},{[\nu t]\over\nu})=\alpha\})$

$=\gamma_{\eta}(\overline{F}_{t}=\alpha)$

Using the measurability of $F_{[\nu t]\over\nu}$ and $\overline{F}_{[\nu t]\over\nu}$ with respect to the algebras $\mathcal{C}_{\eta}$ and $\mathcal{E}_{\eta,\nu-[\nu t]}$ respectively.

∎

We make some considerations in connection with the heat equation.

$\mathcal{F}(h)(m)={1\over 2\pi}\int_{-\pi}^{\pi}h(x)e^{-imx}dx$

*If $g\in C(T)$, we define its Fourier transform in space by;

$\mathcal{F}(g)(m,t)={1\over 2\pi}\int_{-\pi}^{\pi}g(x,t)e^{-imx}dx$**

${\partial{G}\over\partial t}={\partial^{2}G\over\partial x^{2}}$* $(*)$

*on $T^{0}$.

$\mathcal{F}({\partial{G}\over\partial t}-{\partial^{2}G\over\partial x^{2}})(m,t)=0$ $(t>0,m\in\mathcal{Z})$

Differentiating under the integral sign, we have that;

$\mathcal{F}({\partial{G}\over\partial t})={\partial\mathcal{F}(G)\over\partial t}(m,t)$, for $t>0,m\in\mathcal{Z}$

Integrating by parts and using the fact that $G_{t}\in C^{\infty}([-\pi,\pi])$, for $t>0$, we have that;

$\mathcal{F}{\partial^{2}G\over\partial x^{2}}=-m^{2}\mathcal{F}(G)(m,t)$, for $t>0,m\in\mathcal{Z}$

We thus obtain the sequence of ordinary differential equations, indexed by $m\in\mathcal{Z}$;

${\partial\mathcal{F}(G)\over\partial t}+m^{2}\mathcal{F}(G)(m,t)=0$ $(t>0)$

with initial condition, given by;

$\mathcal{F}(G)(m,0)=\mathcal{F}(g)(m)$

By Picard’s Theorem, this has the unique solution, given by;

$\mathcal{F}(G)(m,t)=e^{-m^{2}t}\mathcal{F}(g)(m)$ $(t\geq 0)$

As $G_{t}\in C^{\infty}([-\pi,\pi])$, its Fourier series converges absolutely to $G_{t}$ and, in particular, $G_{t}$ is determined by its Fourier coefficients, for $t>0$. It follows that $G$ is a unique solution.

If $g\in C^{\infty}([-\pi,\pi])$, its Fourier series converges absolutely to $g$, hence, the series;

$\sum_{m\in\mathcal{Z}}e^{-m^{2}t}\mathcal{F}(g)(m)e^{imx}$

are absolutely convergent for $t>0$. It follows that $G$ defined by;

$G(x,t)=\sum_{m\in\mathcal{Z}}e^{-m^{2}t}\mathcal{F}(g)(m)e^{imx}$

is a solution of the required form. ∎

*We let $([-\pi,\pi],\mathfrak{D},\mu)$ denote the interval $[-\pi,\pi]$, with the completion $\mathfrak{D}$ of the Borel field, and $\mu$ the restriction of Lebesgue measure. We let $(\mathcal{R}_{\geq 0}\cup\{+\infty\},\mathfrak{C},\lambda)$ denote the extended real half line, with the completion $\mathfrak{C}$ of the extended Borel field, and $\lambda$ the extension of Lebesgue measure, with $\lambda(+\infty)=\infty$, see….

*We let $(\overline{\mathcal{V}_{\eta}}\times\overline{\mathcal{T}_{\nu}},\mathcal{D}_{\eta}\times\mathcal{C}_{\nu},\mu_{\eta}\times\lambda_{\nu})$ be the associated product space and $(\overline{\mathcal{V}_{\eta}}\times\overline{\mathcal{T}_{\nu}},L(\mathcal{D}_{\eta}\times\mathcal{C}_{\eta}),L(\mu_{\eta}\times\lambda_{\nu}))$ be the corresponding Loeb space. $(\overline{\mathcal{V}_{\eta}}\times\overline{\mathcal{T}_{\nu}},L(\mathcal{D}_{\eta})\times L(\mathcal{C}_{\nu}),L(\mu_{\eta})\times L(\lambda_{\nu}))$ is the complete product of the Loeb spaces $(\overline{\mathcal{V}_{\eta}},L(\mathcal{D}_{\eta}),L(\mu_{\eta}))$ and $(\overline{\mathcal{T}_{\nu}},L(\mathcal{C}_{\nu}),L(\lambda_{\nu}))$. Similarly, $([-\pi,\pi]\times(\mathcal{R}_{\geq 0}\cup\{+\infty\},\mathfrak{D}\times\mathfrak{C},\mu\times\lambda)$ is the complete product of $([-\pi,\pi],\mathfrak{D},\mu)$ and $(\mathcal{R}_{\geq 0}\cup\{+\infty\},\mathfrak{C},\lambda)$.

*We let $({{}^{*}\mathcal{R}},{{}^{*}\mathfrak{E}})$ denote the hyperreals, with the transfer of the Borel field $\mathfrak{C}$ on $\mathcal{R}$. A function $f:(\overline{\mathcal{V}_{\eta}},\mathcal{D}_{\eta})\rightarrow({{}^{*}\mathcal{R}},{{}^{*}\mathfrak{E}})$ is measurable, if $f^{-1}:{{}^{*}\mathfrak{E}}\rightarrow\mathcal{D}_{\eta}$. The same definition holds for ${\mathcal{T}_{\nu}}$. Similarly, $f:(\overline{\mathcal{V}_{\eta}}\times\overline{\mathcal{T}_{\nu}},\mathcal{D}_{\eta}\times\mathcal{C}_{\nu})\rightarrow({{}^{*}\mathcal{R}},{{}^{*}\mathfrak{E}})$ is measurable, if $f^{-1}:{{}^{*}\mathfrak{E}}\rightarrow\mathcal{D}_{\eta}\times\mathfrak{C}_{\nu}$. Observe that this is equivalent to the definition given in [Loeb]. We will abbreviate this notation to $f:\overline{\mathcal{V}_{\eta}}\rightarrow{{}^{*}\mathcal{R}}$, $f:\overline{\mathcal{V}_{\eta}}\rightarrow{{}^{*}\mathcal{R}}$ or $f:\overline{\mathcal{V}_{\eta}}\times\overline{\mathcal{T}_{\nu}}\rightarrow{{}^{*}\mathcal{R}}$ is measurable, $(*)$. The same applies to $({{}^{*}\mathcal{C}},{{}^{*}\mathfrak{E}})$, the hyper complex numbers, with the transfer of the Borel field $\mathfrak{E}$, generated by the complex topology. Observe that $f:\overline{\mathcal{V}_{\eta}}\rightarrow{{}^{*}\mathcal{C}}$, $f:\overline{\mathcal{T}_{\nu}}\rightarrow{{}^{*}\mathcal{C}}$ $f:\overline{\mathcal{V}_{\eta}}\times\overline{\mathcal{T}_{\nu}}\rightarrow{{}^{*}\mathcal{C}}$ is measurable, in this sense, iff $Re(f)$ and $Im(f)$ are measurable in the sense of $(*)$.

*We let $\overline{\mathcal{S}_{\eta,\nu}}=\overline{\mathcal{V}_{\eta}}\times\overline{\mathcal{T}_{\nu}}$ and;

$V(\overline{\mathcal{V}_{\eta}})=\{f:\overline{\mathcal{V}_{\eta}}\rightarrow{{}^{*}}{\mathcal{C}},\ f\ measurable\ d(\mu_{\eta})\}$

*and, similarly, we define $V(\overline{\mathcal{T}_{\nu}})$. Let;

$V(\overline{\mathcal{S}_{\eta,\nu}})=\{f:\overline{\mathcal{S}_{\eta,\nu}}\rightarrow{{}^{*}}{\mathcal{C}},\ f\ measurable\ d(\mu_{\eta}\times\lambda_{\nu})\}$

$i:(\overline{\mathcal{V}_{\eta}}\times\overline{\mathcal{T}_{\nu}},L(\mathcal{D}_{\eta}\times\mathcal{C}_{\nu}),L(\mu_{\eta}\times\lambda_{\nu}))$

$\rightarrow(\overline{\mathcal{V}_{\eta}}\times\overline{\mathcal{T}_{\nu}},L(\mathcal{D}_{\eta})\times L(\mathcal{C}_{\nu}),L(\mu_{\eta})\times L(\lambda_{\nu}))$

*and the standard part mapping;

$st:(\overline{\mathcal{V}_{\eta}}\times\overline{\mathcal{T}_{\nu}},L(\mathcal{D}_{\eta})\times L(\mathcal{C}_{\nu}),L(\mu_{\eta})\times L(\lambda_{\nu}))\rightarrow[-\pi,\pi]\times\mathcal{R}_{\geq 0}\cup\{+\infty\}$

are measurable and measure preserving.

∎

*Let $f:\overline{\mathcal{V}_{\eta}}\rightarrow{{}^{*}\mathcal{C}}$ be measurable. As in [6], we define the discrete derivative $f^{\prime}$ to be the unique measurable function satisfying;

$f^{\prime}(-\pi+\pi{i\over\eta})={\eta\over 2\pi}(f(-\pi+\pi{i+1\over\eta})-f(-\pi+\pi{i-1\over\eta}))$*;

*for $i\in{{}^{*}\mathcal{N}}_{1\leq i\leq 2\eta-2}$.

$f^{\prime}(\pi-{\pi\over\eta})={\eta\over 2\pi}(f(-\pi)-f(\pi-\pi{2\over\eta}))$

$f^{\prime}(-\pi)={\eta\over 2\pi}(f(-\pi+{\pi\over\eta})-f(\pi-{\pi\over\eta}))$

*Let $f:\overline{\mathcal{T}_{\nu}}\rightarrow{{}^{*}\mathcal{C}}$ be measurable. As in [6], we define the discrete derivative $f^{\prime}$ to be the unique measurable function satisfying;

$f^{\prime}({i\over\nu})=\nu(f({i+1\over\nu})-f({i\over\nu}))$*;

*for $i\in{{}^{*}\mathcal{N}}_{0\leq i\leq\nu^{2}-2}$.

$f^{\prime}({\nu-1\over\nu})=0$*;

*If $f:\overline{\mathcal{V}_{\eta}}\rightarrow{{}^{*}\mathcal{C}}$ is measurable, then we define the shift (left, right);

$f^{lsh}(-\pi+\pi{j\over\eta})=f(-\pi+\pi{j+1\over\eta})$* for $0\leq j\leq 2\eta-2$

$f^{lsh}(\eta-{\pi\over\eta})=f(-\pi)$

$f^{rsh}(-\pi+\pi{j\over\eta})=f(-\pi+\pi{j-1\over\eta})$* for $1\leq j\leq 2\eta-1$

$f^{rsh}(-\pi)=f(\pi-{\pi\over\eta})$

*If $f:\overline{\mathcal{T}_{\nu}}\rightarrow{{}^{*}\mathcal{C}}$ is measurable, then we define the shift (left, right);

$f^{lsh}({j\over\nu})=f({j+1\over\nu})$* for $0\leq j\leq\nu^{2}-2$

$f^{lsh}(\nu-{1\over\nu})=f(0)$

$f^{rsh}({j\over\nu})=f({j-1\over\nu})$* for $1\leq j\leq\nu^{2}-1$

$f^{rsh}(0)=f(\nu-{1\over\nu})$

*If $f:\overline{\mathcal{V}_{\eta}}\times\overline{\mathcal{T}_{\nu}}\rightarrow{{}^{*}\mathcal{C}}$ is measurable. Then we define $\{{\partial f\over\partial x},{\partial f\over\partial t}\}$ to be the unique measurable functions satisfying;

${\partial f\over\partial x}(-\pi+\pi{i\over\eta},t)={\eta\over 2\pi}(f(-\pi+\pi{i+1\over\eta},t)-f(-\pi+\pi{i-1\over\eta},y))$*;

*for $i\in{{}^{*}\mathcal{N}}_{1\leq i\leq 2\eta-2},t\in\overline{\mathcal{T}_{\nu}}$

${\partial f\over\partial x}(\pi-{\pi\over\eta},t)={\eta\over 2\pi}(f(-\pi,t)-f(\pi-\pi{2\over\eta},t))$

${\partial f\over\partial x}(-\pi,t)={\eta\over 2\pi}(f(-\pi+{\pi\over\eta},t)-f(\pi-{\pi\over\eta},t))$

${\partial f\over\partial t}(x,{j\over\nu})=(f(x,{j+1\over\nu})-f(x,{j\over\nu}))$*;

*for $j\in{{}^{*}\mathcal{N}}_{0\leq j\leq\nu^{2}-2},x\in\overline{\mathcal{H}_{\eta}}$

${\partial f\over\partial t}(x,\nu-{1\over\nu})=0$

*We define $\{f^{lsh_{x}},f^{lsh_{t}},f^{rsh_{x}},f^{rsh_{t}}\}$ by;

$f^{lsh_{x}}(x_{0},t_{0})=(f_{t})^{lsh}(x_{0})$

$f^{lsh_{t}}(x_{0},t_{0})=(f_{x})^{lsh}(t_{0})$

$f^{rsh_{x}}(x_{0},t_{0})=(f_{t})^{rsh}(x_{0})$

$f^{rsh_{t}}(x_{0},t_{0})=(f_{x})^{rsh}(t_{0})$

*where, if $(x_{0},t_{0})\in\overline{\mathcal{V}_{\eta}}\times\overline{\mathcal{T}_{\nu}}$;

$f_{t_{0}}(x_{0})=f_{x_{0}}(t_{0})=f(\pi{[{\eta x_{0}\over\pi}]\over\eta},{\nu t_{0}\over\nu})$

$\{{\partial f\over\partial x},{\partial f\over\partial t},{\partial^{2}f\over\partial x^{2}},f_{x},f_{t},f^{lsh_{x}},f^{lsh_{t}},f^{rsh_{x}},f^{rsh_{t}},f^{sh_{x}^{2}},f^{sh_{t}^{2}},f^{rsh_{x}^{2}},f^{rsh_{t}^{2}}\}$

This follows immediately, by transfer, from the corresponding result for the discrete derivatives and shifts of discrete functions $f:{\mathcal{H}_{n}}\times{\mathcal{T}_{m}}\rightarrow\mathcal{C}$, where $n,m\in{\mathcal{N}}$, see Definition 0.15 and Definition 0.18 of [8].

$(i)$*. $\int_{\overline{\mathcal{V}_{\eta}}}g^{\prime}(y)d\mu_{\eta}(y)=0$

$(ii)$*. $(gh)^{\prime}=g^{\prime}h^{lsh}+g^{rsh}h^{\prime}$

$(iii)$*. $\int_{\overline{\mathcal{V}_{\eta}}}(g^{\prime}h)(y)d\mu_{\eta}(y)=-\int_{\overline{\mathcal{V}_{\eta}}}gh^{\prime}d\mu_{\eta}(y)$

$(iv)$*. $\int_{\overline{\mathcal{V}_{\eta}}}g(y)d\mu_{\eta}(y)=\int_{\overline{\mathcal{V}_{\eta}}}g^{lsh}(y)d\mu_{\eta}(y)=\int_{\overline{\mathcal{V}_{\eta}}}g^{rsh}(y)d\mu_{\eta}(y)$

$(v)$*. $(g^{\prime})^{rsh}=(g^{rsh})^{\prime}$, $(g^{\prime})^{lsh}=(g^{lsh})^{\prime}$

$(vi)$*. $\int_{\overline{\mathcal{V}_{\eta}}}(g^{\prime\prime}h)(y)d\mu_{\eta}(y)=\int_{\overline{\mathcal{V}_{\eta}}}(gh^{\prime\prime})(y)d\mu_{\eta}(y)$

*(111Similar results hold for $\{sh_{x},{\partial\over\partial x},{\partial\over\partial t}\}$. Namely, if $g,h:\overline{\mathcal{S}_{\eta,\nu}}\rightarrow{{}^{*}\mathcal{C}}$ are measurable. Then;

$(i)$. $\int_{\overline{\mathcal{S}_{\eta,\nu}}}{\partial g\over\partial x}d(\mu_{\eta}\times\lambda_{\nu})=0$

$(ii)$. ${\partial gh\over\partial x}={\partial g\over\partial x}h^{lsh_{x}}+g^{rsh_{x}}{\partial h\over\partial x}$

$(iii)$. $\int_{\overline{\mathcal{S}_{\eta,\nu}}}{\partial g\over\partial x}hd(\mu_{\eta}\times\lambda_{\nu})=-\int_{\overline{\mathcal{S}_{\eta,\nu}}}g{\partial h\over\partial x}d(\mu_{\eta}\times\lambda_{\nu})$

$(iv)$. $\int_{\overline{\mathcal{S}_{\eta,\nu}}}gd(\mu_{\eta}\times\lambda_{\nu})=\int_{\overline{\mathcal{S}_{\eta,\nu}}}g^{lsh_{x}}d(\mu_{\eta}\times\lambda_{\nu})=\int_{\overline{\mathcal{S}_{\eta,\nu}}}g^{rsh_{x}}d(\mu_{\eta}\times\lambda_{\nu})$

$(v)$. $({\partial g\over\partial x})^{lsh_{x}}={\partial(g^{lsh_{x}})\over\partial x}$, and, similarly, with $rsh_{x}$ replacing $lsh_{x}$.

$(vi)$. $\int_{\overline{\mathcal{S}_{\eta,\nu}}}({\partial^{2}g\over\partial x^{2}}h)d(\mu_{\eta}\times\lambda_{\nu})=\int_{\overline{\mathcal{S}_{\eta,\nu}}}(g{\partial^{2}h\over\partial x^{2}})d(\mu_{\eta}\times\lambda_{\nu})$ $(*)$

For $(i)$, using $(i)$ from the argument in the main proof, we have;

$\int_{\overline{\mathcal{S}_{\eta,\nu}}}{\partial g\over\partial x}d(\mu_{\eta}\times\lambda_{\nu})$

$=\int_{\overline{\mathcal{V}_{\eta}}}(\int_{\overline{\mathcal{T}_{\nu}}}({\partial g\over\partial x})_{t}d\mu_{\eta})d\lambda_{\nu}(t)$

$=\int_{\overline{\mathcal{V}_{\eta}}}(\int_{\overline{\mathcal{T}_{\nu}}}({\partial g_{t}\over\partial x})d\mu_{\eta})d\lambda_{\nu}(t)$

$=\int_{\overline{\mathcal{T}_{\nu}}}0d\lambda_{\nu}(t)=0$

The proofs of $(ii),(iii),(iv)$ are similar to the main proof, relying on the result of $(i)$. $(v)$ follows easily from Definitions 0.17 and $(vi)$ follows, repeating the result of $(iii)$, and applying $(v)$.

)*

$\int_{\overline{\mathcal{V}_{\eta}}}g^{\prime}(y)d\mu_{\eta}(y)$

$={\pi\over\eta}[{{}^{*}}\sum_{1\leq j\leq 2\eta-2}{\eta\over 2\pi}[g(-\pi+\pi({j+1\over\eta}))-g(-\pi+\pi({j-1\over\eta}))]$

$+{\eta\over 2\pi}[g(-\pi+{\pi\over\eta})-g(\pi-{\pi\over\eta})]+{\eta\over 2\pi}[g(-\pi)-g(\pi-2{\pi\over\eta})]]=0$

For $(ii)$, we calculate;

$(gh)^{\prime}(-\pi+\pi{j\over\eta})=$

$={\eta\over 2\pi}(gh(-\pi+\pi{j+1\over\eta})-gh(-\pi+\pi{j-1\over\eta}))$

$={\eta\over 2\pi}(gh(-\pi+\pi{j+1\over\eta})-g(-\pi+\pi{j-1\over\eta})h(-\pi+\pi{j+1\over\eta})$

$+g(-\pi+\pi{j-1\over\eta})h(-\pi+\pi{j+1\over\eta})-gh(-\pi+\pi{j-1\over\eta}))$

$=g^{\prime}(-\pi+\pi{j\over\eta})h(-\pi+\pi{j+1\over\eta})+g(-\pi+\pi{j-1\over\eta})h^{\prime}(-\pi+\pi{j\over\eta})$

$=(g^{\prime}h^{lsh}+g^{rsh}h^{\prime})(-\pi+\pi{j\over\eta})$

Combining $(i),(ii)$, we have;

$0=\int_{\overline{\mathcal{V}_{\eta}}}(gh)^{\prime}(x)d\mu_{\eta}(x)$

$=\int_{\overline{\mathcal{V}_{\eta}}}(g^{\prime}h^{lsh}+g^{rsh}h^{\prime})(x)d\mu_{\eta}(x)$

and, rearranging, that;

$\int_{\overline{\mathcal{V}_{\eta}}}(g^{\prime}h^{lsh})d\mu_{\eta}=-\int_{\overline{\mathcal{H}_{\eta}}}(g^{rsh}h^{\prime})d\mu_{\eta}$

For $(iv)$, we have that;

$\int_{\overline{\mathcal{V}_{\eta}}}g^{rsh}(y)d\mu_{\eta}(y)$

$={\pi\over\eta}({{}^{*}\sum}_{0\leq j\leq 2\eta-1}g^{rsh}(-\pi+\pi{j\over\eta}))$

$={\pi\over\eta}({{}^{*}\sum}_{1\leq j\leq 2\eta-2}g(-\pi+\pi{j-1\over\eta})+g(\pi-{\pi\over\eta}))$

$={\pi\over\eta}({{}^{*}\sum}_{0\leq j\leq 2\eta-1}g(-\pi+\pi{j\over\eta})$

$=\int_{\overline{\mathcal{V}_{\eta}}}g(y)d\mu_{\eta}(y)$

A similar calculation holds with $g^{lsh}$. For $(v)$, we have for $2\leq j\leq 2\eta-2$;

$(g^{\prime})^{rsh}(-\pi+\pi{j\over\eta})$

$=g^{\prime}(-\pi+\pi{j-1\over\eta})$

$={\eta\over 2\pi}(g(-\pi+\pi{j\over\eta})-g(-\pi+\pi{j-2\over\eta}))$

$(g^{rsh})^{\prime}(-\pi+\pi{j\over\eta})$

$={\eta\over 2\pi}(g^{rsh}(-\pi+\pi{j+1\over\eta})-g^{rsh}(-\pi+\pi{j-1\over\eta}))$

$={\eta\over 2\pi}(g(-\pi+\pi{j\over\eta})-g(-\pi+\pi{j-2\over\eta}))$

Similar calculations hold for the remaining $j$ to give that $(g^{\prime})^{rsh}=(g^{rsh})^{\prime}$, and the calculation $(g^{\prime})^{lsh}=(g^{lsh})^{\prime}$ is also similar.

It follows that;

$\int_{\overline{\mathcal{V}_{\eta}}}(g^{\prime}h)d\mu_{\eta}$

$=\int_{\overline{\mathcal{V}_{\eta}}}(g^{\prime}(h^{rsh})^{lsh})d\mu_{\eta}$

$=-\int_{\overline{\mathcal{V}_{\eta}}}(g^{rsh}(h^{rsh})^{\prime})d\mu_{\eta}$

$=-\int_{\overline{\mathcal{V}_{\eta}}}(g^{rsh}(h^{\prime})^{rsh})d\mu_{\eta}$

$=-\int_{\overline{\mathcal{V}_{\eta}}}(gh^{\prime}))d\mu_{\eta}$

which gives $(vi)$, using $(iv),(v)$.

∎

$\overline{f}(-\pi+\pi{2i\over\eta})=f(-\pi+\pi{2i\over\eta})$*;

*for $i\in{{}^{*}\mathcal{N}}_{0\leq i\leq\eta-1}$.

$\overline{f^{\prime}}(-\pi+\pi{2i\over\eta})={\eta\over 2\pi}(f(-\pi+\pi{2i+1\over\eta})-f(-\pi+\pi{2i-1\over\eta}))$*;

*for $i\in{{}^{*}\mathcal{N}}_{1\leq i\leq\eta-1}$.

$\overline{f^{\prime}}(-\pi)={\eta\over 2\pi}(f(-\pi+{\pi\over\eta})-f(\pi-{\pi\over\eta}))$

*and;

$\overline{f^{lsh}}(-\pi+\pi{2j\over\eta})=f(-\pi+\pi{2j+1\over\eta})$* for $0\leq j\leq\eta-1$

$\overline{f^{rsh}}(-\pi+\pi{2j\over\eta})=f(-\pi+\pi{2j-1\over\eta})$* for $1\leq j\leq\eta-1$

$\overline{f^{rsh}}(-\pi)=f(\pi-{\pi\over\eta})$

∎

$(i)$*. $\int_{\overline{\mathcal{V}_{{\eta\over 2}}}}\overline{g^{\prime}}(y)d\mu_{{\eta\over 2}}(y)=0$

$(ii)$*. $\overline{(gh)^{\prime}}=\overline{g^{\prime}}\overline{h^{lsh}}+\overline{g^{rsh}}\overline{h^{\prime}}$

$(iii)$*. $\int_{\overline{\mathcal{V}_{{\eta\over 2}}}}\overline{(g^{\prime}h)}(y)d\mu_{{\eta\over 2}}(y)=-\int_{\overline{\mathcal{V}_{{\eta\over 2}}}}\overline{g^{rsh}(h^{\prime})^{rsh}}d\mu_{\eta}(y)$

$\int_{\overline{\mathcal{V}_{{\eta\over 2}}}}\overline{g^{\prime}}(y)d\mu_{{\eta\over 2}}(y)$

$={2\pi\over\eta}[{{}^{*}}\sum_{1\leq j\leq\eta-1}{\eta\over 2\pi}[g(-\pi+\pi({2j+1\over\eta}))-g(-\pi+\pi({2j-1\over\eta}))]$

$+{\eta\over 2\pi}[g(-\pi+{\pi\over\eta})-g(\pi-{\pi\over\eta})]]=0$

$(ii)$ is clear from the main proof and taking restrictions.

For $(iii)$, integrating both sides of $(ii)$ and using $(i)$, we have that;

$\int_{\overline{\mathcal{V}_{{\eta\over 2}}}}\overline{g^{\prime}h^{lsh}}d\mu_{{\eta\over 2}}(y)=-\int_{\overline{\mathcal{V}_{{\eta\over 2}}}}\overline{g^{rsh}h^{\prime}}d\mu_{{\eta\over 2}}(y)$ $(*)$

Then;

$\int_{\overline{\mathcal{V}_{{\eta\over 2}}}}\overline{g^{\prime}h}d\mu_{{\eta\over 2}}(y)$

$=\int_{\overline{\mathcal{V}_{{\eta\over 2}}}}\overline{g^{\prime}(h^{rsh})^{lsh}}d\mu_{{\eta\over 2}}(y)$

$=-\int_{\overline{\mathcal{V}_{{\eta\over 2}}}}\overline{g^{rsh}(h^{rsh})^{\prime}}d\mu_{{\eta\over 2}}(y)$ by $(*)$

$=-\int_{\overline{\mathcal{V}_{{\eta\over 2}}}}\overline{g^{rsh}(h^{\prime})^{rsh}}d\mu_{{\eta\over 2}}(y)$ by the main proof

∎

${\partial F\over\partial t}={\partial^{2}f\over\partial x^{2}}$

*on $({\overline{\mathcal{T}_{\nu}}\setminus[\nu-{1\over\nu},\nu)})\times\overline{\mathcal{V}_{\eta}}$

*with $F(0,x)=f(x)$, for $x\in\overline{\mathcal{V}_{\eta}}$, $(*)$.

Moreover, if $\eta\leq\sqrt{2\pi}\nu$, and, there exists $M\in\mathcal{R}$, with $max\{f,f^{\prime},f^{\prime\prime}\}\leq M$, then $max\{F,{\partial F\over\partial x},{\partial^{2}F\over\partial x^{2}}\}\leq M$.

${\partial^{2}F\over\partial x^{2}}(-\pi+\pi{i\over\eta},t)={\eta^{2}\over 4\pi^{2}}(F(-\pi+\pi{i+2\over\eta},t)-2F(-\pi+\pi{i\over\eta},t)+F(-\pi+\pi{i-2\over\eta},t))$

$(2\leq i\leq 2\eta-3),t\in\overline{\mathcal{T}_{\nu}}$, with similar results for the remaining $i$.

Therefore, if $F$ satisfies $(*)$, we must have;

$F(0,x)=f(x)$, $(x\in\overline{\mathcal{V}_{\eta}})$

$F({i+1\over\nu},-\pi+\pi{j\over\eta})$

$=F({i\over\nu},-\pi+\pi{j\over\eta})+{\eta^{2}\over 4\pi^{2}\nu}(F({i\over\nu},-\pi+\pi{j+2\over\eta})-2F({i\over\nu},-\pi+\pi{j\over\eta})+F({i\over\nu},-\pi+\pi{j-2\over\eta}))$

$={\eta^{2}\over 4\pi^{2}\nu}F({i\over\nu},-\pi+\pi{j+2\over\eta})+(1-{\eta^{2}\over 2\pi^{2}\nu})(F({i\over\nu},-\pi+\pi{j\over\eta})+{\eta^{2}\over 4\pi^{2}\nu}F({i\over\nu},-\pi+\pi{j-2\over\eta}))$ $(*)$

$(1\leq i\leq\nu^{2}-2,0\leq j\leq 2\eta-1)$

See also the proof of Lemma 0.5 in [7]. The choice of $\eta$ ensures that $1-{\eta^{2}\over 2\pi^{2}\nu}\geq 0$. Hence, inductively, if $|F_{i\over\nu}|\leq M$, then, by $(*)$;

$|F_{i+1\over\nu}|\leq M({\eta^{2}\over 4\pi^{2}\nu}+(1-{\eta^{2}\over 2\pi^{2}\nu})+{\eta^{2}\over 4\pi^{2}\nu})=M$.

We can differentiate $(*)$ and replace $F$ with ${\partial F\over\partial x}$ or ${\partial^{2}F\over\partial x^{2}}$. The same argument, and the assumption on the initial conditions, gives the required bound.

∎

$f_{\eta}(-\pi+\pi{j\over\eta})=f^{*}(-\pi+\pi{j\over\eta})$

$f_{\eta}(x)=f(-\pi+{\pi\over\eta}[{\eta(x+\pi)\over\pi}]$

where $f^{*}$ is the transfer of $f$ to ${{}^{*}[-\pi,\pi)}$, then there exists a constant $M\in\mathcal{R}$, such that $max\{f_{\eta},f_{\eta}^{\prime},f_{\eta}^{\prime\prime}\}\leq M$. In particular, if $F$ solves the nonstandard heat equation, with initial condition $f_{\eta}$, then, $max\{F,{\partial F\over\partial x},{\partial^{2}F\over\partial x^{2}}\}\leq M$ as well.

$|{1\over 2h}(f(x+h)-f(x-h))-f^{\prime}(x)|$

$={1\over 2h}(f(x)+hf^{\prime}(x)+{h^{2}\over 2}f^{\prime\prime}(c)-f(x)+hf^{\prime}(x)-{h^{2}\over 2}f^{\prime\prime}(c^{\prime}))-f^{\prime}(x)|$

$\leq hK$

where $K=max_{[-\pi,\pi)}f^{\prime\prime}$. By transfer, it follows, that, for infinite $\eta$, $(f_{\eta})^{\prime}\simeq(f^{\prime})_{\eta}$. Clearly $(f^{\prime})_{\eta}$ is bounded, as $f^{\prime}$ is, which gives the result for $(f_{\eta})^{\prime}$. The case for $(f_{\eta})^{\prime\prime}$ is similar. The final result is immediate from Lemma 0.24. ∎

$\hat{f}_{\eta}(m)={1\over 2\pi}\int_{\overline{\mathcal{V}_{\eta}}}f(y)exp_{\eta}(-iym)d\mu_{\eta}(y)$

Transposing Lemma 0.9 of [9], (222We have there that the measure on* $\overline{\mathcal{S}_{\eta}}=\lambda_{\eta}$. The result follows using the scalar map $p:\overline{\mathcal{V}_{\eta}}\rightarrow\overline{\mathcal{S}_{\eta}}$, $p(x)={x\over\pi}$, and the fact that $p_{*}(\mu_{\eta})=\lambda_{\eta}$);

$f(x)=\sum_{m\in\mathcal{Z}_{\eta}}\hat{f}_{\eta}(m)exp_{\eta}(ixm)$* $(*)$

*Given a measurable $f:\overline{\mathcal{S}_{\eta,\nu}}\rightarrow{{}^{*}\mathcal{C}}$, we define the nonstandard vertical Fourier transform $\hat{f}:\overline{\mathcal{T}_{\nu}}\times\overline{\mathcal{Z}_{\eta}}\rightarrow{{}^{*}\mathcal{C}}$ by;

$\hat{f}(t,m)={1\over 2\pi}\int_{\overline{\mathcal{V}_{\eta}}}f(t,x)exp_{\eta}(-ixm)d\mu_{\eta}(x)$

*and, given a measurable $g:\overline{\mathcal{T}_{\nu}}\times\overline{\mathcal{Z}_{\eta}}\rightarrow{{}^{*}\mathcal{C}}$, we define the nonstandard inverse vertical Fourier transform by;

$\check{g}(t,x)=\sum_{m\in\mathcal{Z}_{\eta}}g(t,m)exp_{\eta}(ixm)$

*so that, by $(*)$, $f=\check{\hat{f}}$

*Similar to Definition 0.20 of [8], for $f\in\overline{\mathcal{V_{\eta}}}$, we let $\phi_{\eta}:\overline{\mathcal{Z}_{\eta}}\rightarrow{{}^{*}\mathcal{C}}$ be defined by;

$\phi_{\eta}(m)={\eta\over 2\pi}(exp_{\eta}(-im{\pi\over\eta})-exp_{\eta}(im{\pi\over\eta}))$

*We let $\psi_{\eta}:\overline{\mathcal{Z}_{{\eta\over 2}}}\rightarrow{{}^{*}\mathcal{C}}$ be defined by;

$\psi_{\eta}(m)={\eta\over 2\pi}(1-exp_{\eta}(im{2\pi\over\eta}))$

*and, we let $U_{\eta}:\overline{\mathcal{Z}_{{\eta\over 2}}}\rightarrow{{}^{*}\mathcal{C}}$ be defined by;

$U_{\eta}(m)=exp_{\eta}(-im{2\pi\over\eta}))$

The following is the analogue of Lemma 0.14 in [9], using the definition of the discrete derivative in Definition 0.17 and the discrete Fourier coefficients from Definition 0.26;

$\hat{f^{\prime\prime}}(m)=\phi_{\eta}^{2}(m)\hat{f}(m)$

$(\hat{f^{\prime}})(m)={1\over 2\pi}\int_{\overline{\mathcal{V}_{\eta}}}f^{\prime}(x)exp_{\eta}(-ixm)d\mu_{\eta}(y)$

$=-{1\over 2\pi}\int_{\overline{\mathcal{V}_{\eta}}}f(x)(exp_{\eta})^{\prime}(-ixm)d\mu_{\eta}(x)$

A simple calculation shows that;

$(exp_{\eta})^{\prime}(-ixm)=exp_{\eta}(-ixm)\phi_{\eta}(m)$

Therefore;

$(\hat{f^{\prime}})(m)=-\phi_{\eta}(m)\hat{f}(m)$

Then $\hat{f^{\prime\prime}}(m)$

$=-\phi_{\eta}(m)\hat{f^{\prime}}(m)$

$=\phi_{\eta}^{2}(m)\hat{f}(m)$

as required. ∎

$\hat{\overline{f^{\prime\prime}}}(m)=\psi_{\eta}(m)^{2}U_{\eta}(m)(\hat{\overline{f}}(m))$

$\hat{\overline{f^{\prime}}}(m)={1\over 2\pi}\int_{\overline{\mathcal{V}_{{\eta\over 2}}}}\overline{f^{\prime}}(x)exp_{{\eta\over 2}}(-ixm)d\mu_{{\eta\over 2}}(x)$

$={1\over 2\pi}\int_{\overline{\mathcal{V}_{{\eta\over 2}}}}\overline{f^{\prime}}(x)\overline{exp_{\eta}}(-ixm)d\mu_{{\eta\over 2}}(x)$

$={1\over 2\pi}\int_{\overline{\mathcal{V}_{{\eta\over 2}}}}\overline{f^{rsh}}(x)\overline{{exp^{\prime}_{\eta}}^{rsh}}(-ixm)d\mu_{{\eta\over 2}}(x)$

We calculate;

$\overline{{exp^{\prime}_{\eta}}^{rsh}}(-im(-\pi+\pi{2j\over\eta}))$

$={exp^{\prime}_{\eta}}(-im(-\pi+\pi{2j-1\over\eta}))$

$={\eta\over 2\pi}({exp_{\eta}}(-im(-\pi+\pi{2j\over\eta}))-{exp_{\eta}}(-im(-\pi+\pi{2j-2\over\eta})))$

$={exp_{\eta}}(-im(-\pi+\pi{2j\over\eta}))[{\eta\over 2\pi}(1-exp_{\eta}(im({2\pi\over\eta})))]$

$=\psi_{\eta}(m)\overline{exp_{\eta}}(-im(-\pi+\pi{2j\over\eta}))$

$\hat{\overline{f^{\prime}}}(m)={1\over 2\pi}\psi_{\eta}(m)\int_{\overline{\mathcal{V}_{{\eta\over 2}}}}\overline{f^{rsh}}(x)\overline{{exp_{\eta}}}(-ixm)d\mu_{{\eta\over 2}}(x)$

$=\psi_{\eta}(m)(\hat{\overline{f^{rsh}}}(m))$

It follows that;

$\hat{\overline{f^{\prime\prime}}}(m)=\psi_{\eta}(m)(\hat{\overline{(f^{\prime})^{rsh}}}(m))$

$=\psi_{\eta}(m)(\hat{\overline{(f^{rsh})^{\prime}}}(m))$

$=\psi_{\eta}(m)^{2}(\hat{\overline{f^{rsh^{2}}}}(m))$

We calculate;

$\hat{\overline{f^{rsh^{2}}}}(m)$

$={1\over 2\pi}\int_{\overline{\mathcal{V}_{{\eta\over 2}}}}\overline{f^{rsh^{2}}}(x)\overline{exp_{\eta}}(-ixm)d\mu_{{\eta\over 2}}(x)$

$={1\over 2\pi}exp_{\eta}({-2\pi im\over\eta})\int_{\overline{\mathcal{V}_{{\eta\over 2}}}}\overline{f^{rsh^{2}}}(x)\overline{exp_{\eta}^{rsh^{2}}}(-ixm)d\mu_{{\eta\over 2}}(x)$

$={1\over 2\pi}U_{\eta}(m)\int_{\overline{\mathcal{V}_{{\eta\over 2}}}}\overline{f}(x)\overline{exp_{\eta}}(-ixm)d\mu_{{\eta\over 2}}(x)$

$=U_{\eta}(m)\hat{\overline{f}}(m)$

Hence;

$\hat{\overline{f^{\prime\prime}}}(m)=\psi_{\eta}(m)^{2}U_{\eta}(m)(\hat{\overline{f}}(m))$

as required.

∎

$|\hat{\overline{f}}(m)|\leq{F\over m^{2}}$*, for $m\in\mathcal{Z}_{{\eta\over 2}}$.

*Moreover;

$({{}^{\circ}\overline{f}})(x)=\sum_{m\in\mathcal{Z}}{{}^{\circ}((\hat{\overline{f}})(m))}exp(im{{}^{\circ}x})$*, $x\in\overline{\mathcal{V}_{\eta}}$.

$|\hat{\overline{f}}(m)|\leq{F\over m^{2}}$. $(*)$

for $m\in\mathcal{Z}_{\eta\over 2}$, as required. Using the Inversion Theorem from Definition 0.26, we have that;

$\overline{f}(x)={{}^{*}\sum}_{m\in\mathcal{Z}_{\eta\over 2}}\hat{\overline{f}}(m)exp_{{\eta\over 2}}(ixm)=M$, $(x\in\overline{\mathcal{V}_{\eta}})$

Let $L=\sum_{m\in\mathcal{Z}}{{}^{\circ}(\hat{\overline{f}}(m))}exp(im{{}^{\circ}x})$

If $\epsilon>0$, we have, using the result of 0.29, and the fact that $exp_{{\eta\over 2}}$ is $S$-continuous for $m\in\mathcal{Z}$, that;

$|M-L|\leq|M-M_{n}|+|M_{n}-L_{n}|+|L-L_{n}|$

$\leq{{}^{*}\sum}_{n+1\leq|m|\leq{\eta\over 2}}{F\over m^{2}}+\sum_{m=1}^{n}\delta_{i}+\sum_{|m|\geq n+1}{F+1\over m^{2}}$ $(\delta_{i}\simeq 0)$

$\leq{2F({1\over n}-{2\over\eta})}+\delta+{2(F+1)\over n}$ $(\delta\simeq 0)$

$\leq{4(F+1)\over n}<\epsilon$

for $n>{\epsilon\over 4(F+1)}$, $n\in\mathcal{N}$. As $\epsilon$ was arbitrary, we obtain the result.

∎

${{}^{\circ}(\hat{\overline{F}}(m,t))}=e^{-m^{2}{{}^{\circ}t}}(\hat{g})(m)$

*for $m\in\mathcal{Z}$ and finite $t$.

$\overline{{\partial F\over\partial t}}-\overline{{\partial^{2}F\over\partial x^{2}}}=0$

Taking Fourier coefficients, for $m\in\mathcal{Z}$, and, using Lemma 0.28;

${d\hat{\overline{F}}(m,t)\over dt}-\theta_{\eta}(m)\hat{\overline{F}}(m,t)=0$

where $\theta_{\eta}(m)=\psi_{\eta}^{2}(m)U_{\eta}(m)$. Then;

$\nu(\hat{\overline{F}}(m,t+{1\over\nu})-\hat{\overline{F}(m,t)})=\theta_{\eta}(m)\hat{\overline{F}}(m,t)$

Rearranging, we obtain;

$\hat{\overline{F}}(m,t+{1\over\nu})=(1+{\theta_{\eta}(m)\over\nu})\hat{\overline{F}}(m,t)$

and, solving the recurrence;

$\hat{\overline{F}}(m,t)=(1+{\theta_{\eta}(m)\over\nu})^{[\nu t]}\hat{\overline{F}}(m,0)$

Taking standard parts, and using the facts that $lim_{n\rightarrow\infty}(1+{x\over n})^{n}=e^{x}$, and ${{}^{\circ}{\theta_{\eta}(m)}}=-m^{2}$, we obtain;

${{}^{\circ}(\hat{\overline{F}}(m,t))}=(e^{-m^{2}{{}^{\circ}t}}){{}^{\circ}}(\hat{\overline{f}}(m))$

for finite $t$. As $f$ is bounded and $S$-continuous, so is $\overline{f}$, and ${{}^{\circ}f}={{}^{\circ}\overline{f}}$ is integrable. We have that;

${{}^{\circ}\int_{\overline{\mathcal{V}_{\eta\over 2}}}\overline{f}(x)exp_{\eta\over 2}(-imx)}d\mu_{\eta\over 2}=\int_{-\pi}^{\pi}{{}^{\circ}f({{}^{\circ}x})}e^{-ixm}dx=\int_{-\pi}^{\pi}g(x)e^{-ixm}dx$

Hence;

${{}^{\circ}(\hat{\overline{F}}(m,t))}=(e^{-m^{2}{{}^{\circ}t}}\hat{g}(m))$

as required.

∎

${{}^{\circ}\overline{F}(x,t)}=\sum_{m\in\mathcal{Z}}{{}^{\circ}(\hat{\overline{F}}(m,t))}exp(im{{}^{\circ}x})$ $(*)$

By Lemma 0.30;

${{}^{\circ}(\hat{\overline{F}}(m,t))}=e^{-m^{2}{{}^{\circ}t}}\mathcal{F}(g)(m)$ $(**)$

Comparing $(*),(**)$, with the expression;

$G({{}^{\circ}x},{{}^{\circ}t})=\sum_{m\in\mathcal{Z}}e^{-m^{2}{{}^{\circ}t}}\mathcal{F}(g)(m)e^{im{{}^{\circ}x}}$

obtained in Lemma 0.13, gives the result that ${{}^{\circ}\overline{F}(x,t)}=G({{}^{\circ}x},{{}^{\circ}t})$. However, $\overline{F}_{t}$ is $S$-continuous, for finite $t$, by the fact that $({\partial F\over\partial x})_{t}$ is bounded, from Lemma 0.25, hence;

${{}^{\circ}\overline{F}(x,t)}={{}^{\circ}F(x,t)}=G({{}^{\circ}x},{{}^{\circ}t})$

as required.

∎

$\overline{F}(x,t)\simeq\sum_{m\in\mathcal{Z}}exp_{\nu}^{-\theta_{\eta}(m)t}\hat{f}(m)exp_{\eta\over 2}^{imx}$

Taking standard parts, using lemma 0.29, and the fact that $exp_{\nu}^{-\theta_{\eta}(m)t}\simeq 0$, for finite $m$ and infinite $t$, we see that all the coefficients vanish, except when $m=0$, that is;

$\overline{F}(x,t)\simeq\hat{f}(0)=\int_{\overline{\mathcal{V}_{{\eta\over 2}}}}\overline{f}d\mu_{{\eta\over 2}}\simeq\int_{\overline{\mathcal{V}_{\eta}}}fd\mu_{\eta}$

The result follows for $F(x,t)$ by $S$-continuity.

∎