---
# System prepended metadata

title: Setting up

---

___
# Setting up
$$x^2$$

$$
\def \mb#1{\mathbf{#1}}
$$

$\mb e,\u{e}$

$$
\renewcommand{\u}[1]{\boldsymbol{#1}}
\renewcommand{\t}[1]{\tilde{#1}}
\newcommand{\m}[1]{\mathbb{#1}}
\renewcommand{\c}[1]{\mathcal{#1}}
\newcommand{\lsc}[2][\mathscr{l}]{{}^{ #1 }\! #2}
\newcommand{\lscH}[1]{\lsc[H]{#1}}
\newcommand{\dsf}[1]{\Delta\boldsymbol{\sf #1}}
\newcommand{\tpsb}[1]{\left. #1 \right.^{\sf T}}
\newcommand{\tps}[1]{\left( #1 \right)^{\sf T}}
\newcommand{\usf}[1]{\boldsymbol{\sf #1}}
\newcommand{\busf}[1]{\bar{\usf{ #1}}}
\newcommand{\tu}[1]{\tilde{\u{ #1}}}
\newcommand{\tusf}[1]{\tilde{\usf{ #1}}}
\newcommand{\btusf}[1]{\bar{\tusf{ #1}}}
\newcommand{\pr}[1]{\left( #1 \right)}
$$
___




# Wavy peeling, supplementary writing

Overbar for re-scaled ("non-dimensionalized").
Hat for unit vectors.


Point $X$
co-ordinates in $(\boldsymbol{e}_i)$ basis of $X$ is $x_i$
co-ordinates in $(\boldsymbol{f}_i)$ basis of $X$ is $\mathscr{x}_i$



Point $P$
co-ordinates in $(\boldsymbol{e}_i)$ basis of $P$ is $p_i$
co-ordinates in $(\boldsymbol{f}_i)$ basis of $P$ is $\mathscr{p}_i$


Point $\tilde{P}$
co-ordinates in $(\boldsymbol{e}_i)$ basis of $\tilde{P}$ is $\tilde{p}_i$
co-ordinates in $(\boldsymbol{f}_i)$ basis of $\tilde{P}$ is $\tilde{\mathscr{p}}_i$

$\mathfrak{p}_i$
$\mathfrak{p}_i$


### Notation and variables


| Column 1 | Column 2 | Column 3 |
| -------- | -------- | -------- |
| $\mb e_1$     | Length     | a basis vector of $\mathbb{E}$     |
| $\mb e_2$     | Length     | a basis vector of $\mathbb{E}$     |
 |  $\lambda$  | Non-dimensional | non-dimensioanal periodicity |
 | $A$     | Non-dimensional     | non-dimensioanl amplitude |
| $l$     | Non-dimensional     | non-dimensioanl peeled length |
| $\bar{x}_1$     | Non-dimensional     | res-scaled $x_1$ co-ordinate |
| $a$     | Non-dimensional     | de-adhered length |
| $\bar{a}:=a/\lambda$     | Non-dimensional     | scaled de-adhered length |
| $\bar{x}_1:=x_1/\lambda$     | Non-dimensional     | scaled $x_1$ co-ordinate    |
| $\mathrm{B}$     | non-dimensional     | Width of the thin film   |
| $B:=\mathrm{B}/\mathrm{m}$     | $\mathrm{m}$     | Width of the thin film   |
| $\mathscr{h}$     | $\mathrm{m}$     | Thickness of the thin film   |
| $\mathscr{E}$     | $\mathrm{N}~\mathrm{m}^{-2}$     | Young's modulus   |
| $\mathscr{w}$     | $\mathrm{N}~\mathrm{m}^{-1}$     | Work of adhesion   |
| $\bar{\mathscr{w}}:=\frac{\mathrm{w}}{\mathrm{E}\,\mathrm{h}}$     | $\mathrm{N}~\mathrm{L}^{-2}$     | non-dimensional work of adhesion   |
| $\boldsymbol{\mathscr{F}}$     | $\mathrm{N}$     | dimensional  force   |
| $\bar{\mb{F}}:=F $     | $\mathrm{N}$     | dimensional  force   |



#### Peeled length

The units of $\mb{e}_1$, and $\mb{e}_2$ is $\lambda\, \mathrm{m}$. The units of $\mb{e}_3$ is $\mathrm{m}$. The metric co-efficients corresponding to these basis vectors are $g_{11}=g_{22}=(\lambda\, \mathrm{m})^2$, and $g_{12}=g_{21}=0$.

The length of the peeled film
$$
\begin{aligned}
d\mb{r}&=dx_1\mb{e}_1+dx_2\mb{e}_2 \\
d\mathrm{l}^2 &=d\mb{r}\cdot d\mb{r}\\
d\mathrm{l}^2&= g_{ij}dx_i dx_j\\
d\mathrm{l}&=\lambda\, \mathrm{m} \left(dx_1^2+dx_2^2\right)^{1/2}\\
\mathrm{l}&=\lambda\, \mathrm{m}\int_{0}^{a} \left(dx_1^2+dx_2^2\right)^{1/2}\\
&= \lambda\, \mathrm{m}\int_{0}^{a}dx_1 \left(1+\alpha^2 \varrho'(x_2)^2\right)^{1/2}
\end{aligned}
$$


Setting the peeled length of the film to be is $\lambda l\,\mathrm{m}$. So we get that

$$
\begin{aligned}
l(a)&=\int_{0}^{a}dx_1 \left(1+\alpha^2 \varrho'(x_2)^2\right)^{1/2}.
\end{aligned}
$$


#### Change in the force potential energy
We are going to think of forces as a linear maps from $\mathbb{E}$ into $\mathbb{U}$ (The energy vector space).  Let the thin film's Young's modulus be $\mathrm{E}:=E\,\mathrm{N m^{-2}}$, its thickness be $\mathrm{h}:=h\, \mathrm{m}$, and its width be $\mathrm{B}:=B\, \mathrm{m}$. The quantity $\mathrm{E}\,\mathrm{B}\,\mathrm{h}$ has units of $\mathrm{N}$.

It can be shown that the set of all forces from $\mathbb{E}$ into $\mathbb{U}$ form a vectors space. We denote that force-vector spaces as $\mathbb{F}$. Take $(\mb{f}_i)$ be set of basis vectors for $\mathbb{F}$. The vectors $\mb{f}_i$ are defined such that $\mb{f}_i(\hat{\mb{e}}_j)=E B h \lambda \delta_{ij} \, \mathrm{Joules}$ for all $i,~j\in \mathcal{I}$ except for $i=j=3$ and $\mb{f}_3(\mb{e}_3)= E B h \, \mathrm{Joules}$. Thus, the units of each of $\mb{f}_i$ is $\mathrm{Newtons}$ and each of them are of magnitude $E B h$.

The force acting on the thin film can be written as  
$$
\begin{aligned}
\mb{F}=F\left(-\cos(\theta_0)\mb{f}_{1}+\sin(\theta_0)\mb{f}_{2}\right),
\end{aligned}
 $$
where $F\ge0$. Let us compute the work done by $\mb{F}$ on a displacement of $\delta \mb{u}=\delta x_i \mb{e}_i$.


$$
\begin{aligned}
\delta\Pi_{\rm force}&=-\mb{F}\delta \mb{u}\\
&=F\left(-\cos(\theta_0)\mb{f}_{1}+\sin(\theta_0)\mb{f}_{2}\right)(\delta x_1 \mb{e}_1+\delta x_2 \mb{e}_2)\\
&=F E B h \lambda \left(-\cos(\theta_0)\delta x_1+\sin(\theta_0)\delta x_2\right) \mathrm{Joules}
\end{aligned}
$$

W denite the non-dimensional increment in the force-potential energy as
$$
\delta \bar{\Pi}_{\mathrm{force}}=\frac{\delta {\Pi}_{\mathrm{force}}}{E B h \lambda\, \mathrm{Joules}}
$$
Thus we get,
$$
\delta \bar{\Pi}_{\mathrm{force}}=F  \left(-\cos(\theta_0)\delta x_1+\sin(\theta_0)\delta x_2\right)
$$


<!-- The force potential energy is
$$
\begin{aligned}
\Pi_{\rm force}=-\mb{F}\cdot \mb{u}
\end{aligned}
$$ -->




Let's define the non-dimensional,  change in the potential energy that is associated with $\mathbf{F}$ as


#### Change in strain energy
The stress the thin film is $\sigma=\lVert \mb{F}\rVert/(\mathrm{B} \mathrm{h})$, which $\sigma=F E B h\,\mathrm{N}/(\mathrm{B} \mathrm{h})$. The strain is
$$
\begin{aligned}
\epsilon&=\frac{F E B h\,\mathrm{N}}{(\mathrm{E}\,\mathrm{B} \mathrm{h})}\\
&=\frac{F E B h\,\mathrm{N}}{E B h \, \mathrm{N}}\\
&=F \\
\end{aligned}
$$

The elastic strain energy is
$$
\begin{aligned}
 \Pi_{\rm strain}&=
\frac{1}{2}\sigma \epsilon \mathrm{B} \mathrm{h} \lambda l \mathrm{m}\\
&=\frac{1}{2} \frac{F E B h\,\mathrm{N}}{\mathrm{B} \mathrm{h}}F\, \mathrm{B} \mathrm{h} \lambda l \mathrm{m}\\
&=\frac{1}{2} F^2 (E B h \lambda) l\, \mathrm{J}
\end{aligned}
$$

The non-dimensional energy is
$$
\begin{aligned}
 \bar{\Pi}_{\rm strain}&=\frac{\delta \Pi_{\rm strain}}{E B h \lambda \, \mathrm{J}}\\
&=\frac{1}{2} \frac{F^2 (E B h \lambda) l\, \mathrm{J}}{E B h \lambda \, \mathrm{J}}\\
&=\frac{1}{2} F^2  l\\
\end{aligned}
$$


The change in non-dimensional strain energy is
$$
\begin{aligned}
 \delta \bar{\Pi}_{\rm strain}
&=\frac{1}{2} F  \delta l\\
\end{aligned}
$$


#### Change in surface energy

The interfacial potential energy of the system is
$$
\begin{aligned}
\mathrm{w}\, (B\, \mathrm{m}) \left(\lambda \, l\, \mathrm{m}\right)
\end{aligned}
$$

The change in interfacial potential energy when $l$ changes to $l+\delta l$ is
$$
\begin{aligned}
\delta \Pi_{\rm surf}&= \mathrm{w}\, (B\,\mathrm{m}) \left(\lambda\, \delta l\, \mathrm{m}\right)
\end{aligned}
$$

Let's define the non-dimensional,  change in interfacial potential energy as
$$
\delta \bar{\Pi}_{\rm surf}
:=
\frac{\delta \Pi_{\rm surf}}{\mathrm{E}\:\mathrm{B}\:\mathrm{h}\: \lambda\, \mathrm{m}},
$$
which simplifies to
$$
\delta \bar{\Pi}_{\rm surf}:=
w\, \delta l
$$

### Evoluation law for de-adherence

<details> <summary> Evoluation law for de-adherence (first Outline) </summary>


Consider a configuration $\boldsymbol{\kappa}$ in which the delamination length is $a$ and the peeled length is $l_p$. The configuraton we consider can be quite complicated. For example, it may involve multiple contacts between the peeled part of the film and the  unpeeled part of the thin film  and/or the section of the the substrate's surface from which the thin film has been peeled off. We assume that when these type of contacts occurs they are frictionless and non-adhesive.  


Consider a configulation $\tilde{\boldsymbol{\kappa}}$, which is close to $\boldsymbol{\kappa}$, in which delamination length is $a+\Delta a$. By "close" we mean that the number of non-adhesive contact patches in $\boldsymbol{\kappa}$ and $\tilde{\boldsymbol{\kappa}}$ are the same.
The change in the systems pontential energy  between $\boldsymbol{\kappa}$ and $\tilde{\boldsymbol{\kappa}}$ is
$$
\Delta \Pi\left(\Delta a; F\right)=\frac{F^2}{B h E}\Delta l+w B \Delta l+\mathbf{F}\cdot \Delta \mathbf{u}.
$$

The first term in the above equation is blag, the second is something, the thirtd is something.

The configurations can be placed into the following categories.  


* Case (i) There is no contact between the peeled part of the film with either .... or .....
*  Case (ii) non-adhesive contact patches
    * Case (ii.A) non-adhesive contact patches, forward peeling
    * Case (ii.B) non-adhesive contact, reverse peeling.
        * Case (ii.B.1) non-adhesive contact, reverse peeling.
        * Case (ii.B.2) non-adhesive contact, reverse peeling.


Case (i) occurs when

$$
\thet_0....
$$

Case (ii) occurs .....

We analyze the above cases in detail in the following section.

###

We have analysed how the $\Delta u$, which is the difference in the position vector of theleft end of the thin film varies in each of ther above case. The details can be found in \S. We found that in each of the above case $\Delta u$ varies as


</details>
