We give the multiplicative Taylor theorem for two variables given in [6] to derive the multiplicative finite differences scheme.

#### Theorem 2.1

(Multiplicative Taylor theorem). *Let* (*a*, *b*) × (*c*, *d*)⊂ ℝ^{2}. *Assume that* *u* : (*a*, *b*) × (*c*, *d*)→ ℝ *has all partial*^{*} *derivatives of order n* + 1 *times on* (*a*, *b*)×(*c*, *d*). *If x*_{0}∈[*a*, *b*], *and t*_{0} ∈ [*c*, *d*] *then for every x*∈[*a*, *b*] *and t*∈[*c*, *d*] *with x* ≢ *x*_{0}, *and t* ≢ *t*_{0}, *there exist x*_{1}∈(*x*, *x*_{0}) *and t*_{1}∈(*t*, *t*_{0}) *such that*
$$\begin{array}{}{\displaystyle u(x,t)=\prod _{m=0}^{n}\prod _{i=0}^{m}({u}_{{x}^{i}{t}^{m-i}}^{\ast (m)}({x}_{0},{t}_{0}){)}^{\frac{{h}^{i}{k}^{m-i}}{i!(m-i)!}}\prod _{i=0}^{n+1}({u}_{{x}^{i}{t}^{n+1-i}}^{\ast (n+1)}({x}_{1},{t}_{1}){)}^{\frac{{h}^{i}{k}^{n+1-i}}{i!(n+1-i)!}},}\end{array}$$(7)

*where h* = *x* − *x*_{0} *and k* = *t* − *t*_{0}.

From (7), the multiplicative Taylor expansion of *u*(*x*+*ϵ h*, *t* + *k*) and *u*(*x* + *ϵh*, *t*) about the virtual node $(x,t+\frac{k}{2})$ for *ϵ* ∈ {−1, 0, 1} are written as follows:
$$\begin{array}{}{\displaystyle u(x+\u03f5h,\phantom{\rule{thinmathspace}{0ex}}t+k)=u(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2})[{u}_{x}^{\ast}(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2}){]}^{\u03f5h}[{u}_{t}^{\ast}(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2}){]}^{\frac{k}{2}}}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle \times [{u}_{{x}^{2}}^{\ast \ast}(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2}){]}^{\frac{(\u03f5h{)}^{2}}{2!}}[{u}_{{t}^{2}}^{\ast \ast}(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2}){]}^{\frac{(k/2{)}^{2}}{2!}}}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle \times [{u}_{xt}^{\ast \ast}(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2}){]}^{\frac{(2\u03f5h(k/2))}{2!}}[{u}_{{x}^{3}}^{\ast \ast \ast}(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2}){]}^{\frac{(\u03f5h{)}^{3}}{{3}^{1}}}}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle \times [{u}_{{x}^{2}t}^{\ast \ast \ast}(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2}){]}^{\frac{3(k/2)(\u03f5h{)}^{2}}{3!}}[{u}_{x{t}^{2}}^{\ast \ast \ast}(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2}){]}^{\frac{3(k/2{)}^{2}\u03f5h}{3!}}}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle \times [{u}_{{t}^{3}}^{\ast \ast \ast}(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2}){]}^{\frac{(k/2{)}^{3}}{3!}}\dots ,}\end{array}$$(8)
$$\begin{array}{}{\displaystyle u(x+\u03f5h,\phantom{\rule{thinmathspace}{0ex}}t)=u(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2})[{u}_{x}^{\ast}(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2}){]}^{\u03f5h}[{u}_{t}^{\ast}(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2}){]}^{\frac{-k}{2}}}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle \times [{u}_{{x}^{2}}^{\ast \ast}(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2}){]}^{\frac{(\u03f5h{)}^{2}}{2!}}[{u}_{{t}^{2}}^{\ast \ast}(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2}){]}^{\frac{(-k/2{)}^{2}}{2!}}}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle \times [{u}_{xt}^{\ast \ast}(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2}){]}^{\frac{(2\u03f5h(-k/2))}{2!}}[{u}_{{x}^{3}}^{\ast \ast \ast}(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2}){]}^{\frac{(\u03f5h{)}^{3}}{3!}}}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle \times [{u}_{{x}^{2}t}^{\ast \ast \ast}(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2}){]}^{\frac{3(-k/2)(\u03f5h{)}^{2}}{3!}}[{u}_{x{t}^{2}}^{\ast \ast \ast}(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2}){]}^{\frac{3(-k/2{)}^{2}\u03f5h}{3!}}}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle \times [{u}_{{t}^{3}}^{\ast \ast \ast}(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2}){]}^{\frac{(-k/2{)}^{3}}{3!}}\dots .}\end{array}$$(9)

The approximation of the first order multiplicative partial derivative with respect to *t* is obtained by dividing (8) to (9) with *ϵ* = 0
$$\begin{array}{}{\displaystyle {u}_{t}^{\ast}(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2})\approx [\frac{u(x,t+k)}{u(x,t)}{]}^{\frac{1}{k}}.}\end{array}$$(10)

Analogously, to approximate the second order multiplicative partial derivative ${u}_{{x}^{2}}^{\ast \ast}(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2}),$ we use the geometric mean of the second multiplicative centered differences for ${u}_{{x}^{2}}^{\ast \ast}(x,\phantom{\rule{thinmathspace}{0ex}}t)\text{\hspace{0.17em}and\hspace{0.17em}}{u}_{{x}^{2}}^{\ast \ast}(x,\phantom{\rule{thinmathspace}{0ex}}t+k)$ in (8),
$$\begin{array}{}{\displaystyle {u}_{{x}^{2}}^{\ast \ast}(x,\phantom{\rule{thinmathspace}{0ex}}t+\frac{k}{2})\approx [\frac{u(x+h,t+k)u(x-h,t+k)}{u(x,t+k{)}^{2}}\frac{u(x+h,t)u(x-h,t)}{u(x,t{)}^{2}}{]}^{\frac{1}{2{h}^{2}}}.}\end{array}$$(11)

We consider the initial value problem (3) with some boundary conditions in one space dimension,
$$\begin{array}{}{u}_{t}^{\ast}={u}_{{x}^{2}}^{\ast \ast},\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\text{for\hspace{0.17em}}\phantom{\rule{1em}{0ex}}0\le x\le L,\phantom{\rule{1em}{0ex}}t\ge 0,\\ u(x,0)=f(x),\phantom{\rule{2em}{0ex}}\text{for\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{1em}{0ex}}0\le x\le L,\\ u(0,t)={g}_{0}(t),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}u(L,t)={g}_{1}(t)\phantom{\rule{2em}{0ex}}\text{\hspace{0.17em}for}\phantom{\rule{2em}{0ex}}t\ge 0\end{array}$$(12)

We remark that the length between two consecutive points need not be equal but for simplicity we denote *x*_{i} = *ih*, for *i* = 0, 1, 2, *n* with the space step size $h=\frac{L}{n}$ and *t*_{j} = *jk*, for *j* = 0, 1, *m* with the time step size $k=\frac{{t}_{max}}{m}$ where *t*_{max} is the maximum time for the desired solution.

The multiplicative methods for (12) is proposed in the form:
$$\begin{array}{}{\displaystyle (\frac{{u}_{i,j+1}}{{u}_{i,j}}{)}^{\frac{1}{k}}=\left[(\frac{{u}_{i+1,j}{u}_{i-1,j}}{({u}_{i,j}{)}^{2}}{)}^{\frac{1-\theta}{{h}^{2}}}(\frac{{u}_{i+1,j+1}{u}_{i-1,j+1}}{({u}_{i,j+1}{)}^{2}}{)}^{\frac{\theta}{{h}^{2}}}\right],}\end{array}$$(13)

where *u*_{i,j} = *u*(*x*_{i}, *t*_{j}), and *θ*∈[0, 1]. If we set $r=\frac{k}{{h}^{2}}$
which is called parabolic mesh ratio, then the finite difference equation (13) is rewritten as
$$\begin{array}{}({u}_{i+1,j+1}{)}^{-r\theta}({u}_{i,j+1}{)}^{1+2r\theta}({u}_{i-1,j+1}{)}^{-r\theta}=({u}_{i+1,j}{)}^{r(1-\theta )}({u}_{i,j}{)}^{1-2r(1-\theta )}({u}_{i-1,j}{)}^{r(1-\theta )}.\end{array}$$(14)

It is noted that the numerical scheme (14) for *θ* = 0 and *θ* = 1 yields the multiplicative explicit Euler method and multiplicative implicit Euler method respectively. When *θ* = 1/2, (14) yields the multiplicative Crank-Nicolson method. If we set $\overrightarrow{U}{\phantom{\rule{thinmathspace}{0ex}}}^{j}=[\mathrm{ln}{u}_{1,j},\mathrm{ln}{u}_{2,j},\dots ,\mathrm{ln}{u}_{n-1,j}{]}^{T}\phantom{\rule{thinmathspace}{0ex}},$ we rewrite (14) in the matrix form
$$\begin{array}{}(I-r\theta A)\overrightarrow{U}{\phantom{\rule{thinmathspace}{0ex}}}^{j+1}=(I+r(1-\theta )A)\overrightarrow{U}{\phantom{\rule{thinmathspace}{0ex}}}^{j}+\overrightarrow{b},\end{array}$$(15)

where *I* is the unit matrix,
$$\begin{array}{}A=\left[\begin{array}{ccccc}-2& 1& 0& \cdots & 0\\ 1& -2& 1& 0& 0\\ 0& \ddots & \ddots & \ddots & \vdots \\ 0& \ddots & 1& -2& 1\\ 0& \cdots & 0& 1& -2\end{array}\right]\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{\hspace{0.17em}and\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\overrightarrow{b}=r\left[\begin{array}{c}\mathrm{ln}({g}_{0}({t}_{m}))+\mathrm{ln}({g}_{0}({t}_{m+1}))\\ 0\\ \vdots \\ 0\\ \mathrm{ln}({g}_{1}({t}_{m}))+\mathrm{ln}({g}_{1}({t}_{m+1}))\end{array}\right].\end{array}$$

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