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152
notes.tex

@ -2673,6 +2673,158 @@ At $t = 0$, $X(A, t= 0) = A \implies F = \mathbb I \implies J = 1$ @@ -2673,6 +2673,158 @@ At $t = 0$, $X(A, t= 0) = A \implies F = \mathbb I \implies J = 1$
Since $J(t = 0) = 1$, $J \ne 0$ always implies $J(t) > 0$
\begin{ex} Uniform diletion
Motion according to $A \mapsto X(A,t) = \alpha(t) A$ where
$\alpha(t) \ge 0, \alpha(0) = 1$.
Consider a sphere centred at $\hat 0 = \begin{pmatrix} 0\\0\\0
\end{pmatrix} $ with radius $r = 1$ at $t = 0$ and at $t = 1$,
$\alpha(1) = 2$, (the circle is being stretched according to its
radius linearly increasing).
Material Coordinates
$U(A,t) = X(A, t) - A = (\alpha(t) - 1)A $
$V(A,t) = \p_t U = \alpha' (t) A$
Spatial coordinates
Invert $x$: $A = \frac 1 {\alpha (t)} x$
$u(x,t) = U(\frac 1 {\alpha(t)}x, t) = \alpha(t) - 1 \frac 1\
{\alpha(t)} x = (1 - \frac 1 {\alpha(t)}) x$
$v(x,t) = v(\frac 1{\alpha(t)} x, t) = \alpha ' (t) \frac 1
{\alphta(t)} x$
$\implies \p_t u \ne v$ like in one dimension.
$F = \nabla_A X = \nabla _A(\alpha(t) A) = \alpha(t) \nabla_A
\begin{pmatrix} A_1\\A_2\\A_3 \end{pmatrix} = \alpha(t)
\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$
To guarantee $J > 0$, $J = \det F = \det (\alpha(t) \mathbb I) = \alpha(t)^3$
$\implies$ we need $\alpha(t) > 0$.
\end{ex}
\begin{ex}
Simple shear
Motion according to:
\marginnote{There is a shear along the $y$ axis}
$$
X(A,t) = \begin{pmatrix} A_1 + \alpha(t) A_2 \\ A_2 \\ A_3 \end{pmatrix}
\quad\text{Where } \alpha(0) = 0 \text{ so } X(A,0) = A
$$
\incfig{shear}
Material coordinate:
$U(A,t) = \begin{pmatrix} \alpha(t) A_2 \\ 0 \\ 0 \end{pmatrix}
\qquad \qquad
V(A,t) = \begin{pmatrix} \alpha'(t) A_2 \\ 0 \\0 \end{pmatrix}$
Spatial coordinate:
Invert $X = \begin{pmatrix} A_1 + \alpha(t) A_2 \\ A_2 \\ A_3 \end{pmatrix}
= \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} $.
$A_1 = x_1 - \alpha(t) A_2 = x_1 - \alpha(t) x_2$
$A_2 = x_2$ and $A_3 = x_3$ So
$$A = \begin{pmatrix} x_1 - \alpha(t) x_2 \\ x_2 \\ x_3 \end{pmatrix} $$
So
$$u(x,t) = \begin{pmatrix} \alpha(t) x_2 \\ 0 \\0 \end{pmatrix} $$
$$v(x,t) = V\left( \begin{pmatrix} x_1 - \alpha(t) x_2 \\ x_2 \\ x_3 \end{pmatrix}
, t\right) = \begin{pmatrix} \alpha ' (t)x_2 \\ 0\\ 0 \end{pmatrix} $$
Deformation gradient
$F = \nabla_A X = \nabla _a \begin{pmatrix} A_1 + \alpha(t) A_2 \\ A-2 \\ A_3
\end{pmatrix} = \begin{pmatrix} 1&\alpha(t)&0 \\ 0&1&0 \\ 0&0&1
\end{pmatrix} $
$\det F = 1 > 0$ which does not imply any additional constraints.
\end{ex}
\subsection{Material Derivative}
Let $F = (F_1,F_2,F_3)$ respecitvely to $f$ be a quantity in material respective to spatial
coordinates then we ahve
\marginnote{$X = \begin{pmatrix} X_1\\X_2\\X_3 \end{pmatrix} $}
$$F(A,t) = f(X(A,t), t) $$
\begin{align*}
\frac {\p F }{\p t} &= \frac d {dt} f (X(A,t) , t) \\
&= \p_t f + \p _{X_1} f
\frac {\p{X_1}}{\p t} + \p _{X_2} f \frac {\p X_2} {\p t} + \p _{X_3} f \frac
{\p X_3}{\p t} \\
&= \p _t f + \nabla f \cdot \p _ t X \\
&= \p_t f + \nabla f \p _t U \\
&= p_t f + \nabla f \cdot v \\
&= \p_t f + (v \cdot \nabla) f = (\p_t + v \cdot \nabla) f \\
\end{align*}
{$(v \cdot \nabla)$ is notation for a differential operator and does
not make sense on its own, rather it is notation for the previous statement}
\begin{align*}
(v \cdot \nabla) f = \nabla f \cdot v &=
\begin{pmatrix}
f'_{X_1} & f'_{X_2} & f'_{X_3} \\
f''_{X_1} & f''_{X_2} & f''_{X_3} \\
f'''_{X_1} & f'''_{X_2} & f'''_{X_3} \\
\end{pmatrix} \cdot \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}
\\
&= \begin{pmatrix}
f'_{X_1} v_1 + f'_{X_2} v_2 + f'_{X_3} v_3 \\
f''_{X_1}v_1 + f''_{X_2}v_2 + f''_{X_3}v_3 \\
f'''_{X_1}v_1 + f'''_{X_2}v_2 + f'''_{X_3}v_3 \\
\end{pmatrix} \\
&= \begin{pmatrix} \nabla f^1 \cdot v \\ \nabla f^2 \cdot v \\ \nabla
f^3 \cdot v\end{pmatrix}
\end{align*}
$\implies$ the material derivative:
$$D_t f = \p_t f + \nabla f \cdot v$$
The rate of change change from the perspective of fixed world coordinates,
so how both the thing moves as a result of its own movement and it being
transported by the material.
\begin{ex}
Uniform dilation
$u(x,t) = 1 - \frac 1 {\alpha(t)}x$
$v(x,t) = \frac {\alpha'} {\alpha} x$
\begin{align*}
D_t u &= \p _t u + \nabla u \cdot v = \frac {\alpha^2} {\alpha^2} x + (1 -
\frac {\alpha (t)}) \mathbb I \cdot \frac {\alpha ' }{\alpha } x \\
&= \frac {\alpha^2} {\alpha^2}x + (1 - \frac 1 {\alpha(t)}) \cdot
\frac {\alpha '}{\alpha} x \\
&= \frac {\alpha'}{\alpha} x \\
\end{align*}
Which is just the velocity field.
\end{ex}
\pagebreak
\appendix

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