lec29

notes.tex

@@ -2676,7 +2676,8 @@ Since $J(t = 0) = 1$, $J \ne 0$ always implies $J(t) > 0$
 \week{}
 \lecture
 
-\begin{ex} Uniform dilation
+
+\begin{ex} Uniform dilatation
 
         Motion according to $A \mapsto X(A,t) = \alpha(t) A$ where
             $\alpha(t) \ge 0, \alpha(0) = 1$.
@@ -2700,7 +2701,7 @@ Since $J(t = 0) = 1$, $J \ne 0$ always implies $J(t) > 0$
             {\alpha(t)} x = (1 - \frac 1 {\alpha(t)}) x$
 
             $v(x,t) = v(\frac 1{\alpha(t)} x, t) = \alpha ' (t) \frac 1
-            {\alphta(t)} x$
+            {\alpha(t)} x$
 
             $\implies \p_t u \ne v$ like in one dimension.
 
@@ -2810,7 +2811,7 @@ so how both the thing moves as a result of its own movement and it being
 transported by the material.
 
 \begin{ex}
-    Uniform dilation
+    Uniform dilatation
 
     $u(x,t) = 1 - \frac 1 {\alpha(t)}x$
 
@@ -2829,6 +2830,139 @@ transported by the material.
 
 \end{ex}
 
+\lecture{}
+
+\subsection{Reynold's Transport theorem in 3D}
+\marginnote{October 8 Lec 29}
+
+The major technical tool to derive it is the fact that for $M(t) \in \R^{3\times
+3}$ (smooth, invertible) it holds that. 
+
+$$\frac d {dt} \det (M(t)) tr(M^{-1} \frac d{dt} M)$$
+
+For $M = F = \nabla_A X$ and $j = \det F$.
+
+$$\frac d {dt} J = J tr (F^{-1} \frac d {dt} F) = \left|\frac d {dt} \nabla_A (X - A)
+= \nabla_A V \right|$$
+
+{In the trace of the matrix, the multiplication commutes $|tr(AB) =
+tr(BA) | \implies J tr(F^{-1} \nabla_A V) = J tr(\nabla_A V F^{-1})$}
+
+
+Chain rule: $\nabla_A V = \nabla_Av(X(A,t),t) = \nabla_{\vec x} \vec v \cdot \nabla_A X = \nabla_{\vec x} \vec v
+\cdot F$
+
+
+    $$J tr(\nabla_A V F^{-1}) = J tr(\nabla_{\vec x } \vec v) = \begin{pmatrix} 
+        v_{x_1}^1 & v_{x_2}^1 &v_{x_3}^1 \\ 
+        v_{x_1}^2 & v_{x_2}^2 &v_{x_3}^2 \\ 
+        v_{x_1}^3 & v_{x_2}^3 &v_{x_3}^3 \\ 
+\end{pmatrix} = J \nabla \cdot \vec v $$
+
+Now consider $R(t = 0) \suybset \R^3$, a piece of volume with a piecewise smooth
+surface and $R(t) = X(R(t = 0), t)$.
+
+\marginnote{Want to create a framework to use the balance laws in 3D.}
+Consider a quantity $f(x,t)$ (smooth).
+
+\marginnote {Change of variables to use material coordinates}
+$$\frac d{dt} \iiint _{R(t)} f(x,t) \; d\vec x = \frac d {dt} \iiint_{R(t=0)} f(X(A,t) ,
+t) \det (F)\; d A_1\: d A_2 \: d A_3 $$
+
+$$= \iiint_{R(0)} \underbrace{\frac d {dt} f(X(A,t),t)}_{D_t f} J + f(X(A,t),t) \p_t J\; d\vec A $$
+
+$$= \iiint_{R(0)} (D_t f + f(\nabla \cdot \vec v)) J \; d\vec A = \iiint_{R(t)}
+(D_t f + f (\nabla \cdot \vec v)) \; d \vec x$$
+
+$$= \iiint_{R(t)} (\p_t f + \underbrace{\nabla \cdot (f \vec v))}_{=\nabla f
+\vec v + f \nabla \cdot \vec v} \; d\vec x = \iiint_{R(t)} \p_t f \; d\vec x +
+\iint_{\p R(t) f \vec v \cdot \vec nn \: dS(\vec x)}$$
+
+Which uses the divergence theorem:
+
+$$\iiint_R \nabla \cdot \vec g \: d \vec x = \iint_{\p R} \vec g \cdot \vec n dS(\vec
+x)$$
+
+
+We get two ways of writing \textbf{Reynold's Transport Theorem} in 3D:
+
+\begin{align*}
+    \frac {d} {dt} \iiint_{R(t)} f (\vec x ,t )\; d\vec x &= \iiint _{R(t)} \p_t f
+d \vec x + \iint_{\p R(t)} f \vec v \cdot \vec n \; dS(\vec x) \\
+                                                          &= \iiint_{R(t)} (D_t
+                                                          f + f(\nabla \cdot
+                                                          \vec v)) \; d\vec x
+\end{align*}
+
+\subsection{Balance Laws in 3D}
+
+Note; $\vec J :=$ flux, $J :=$ jacobi determinant.
+
+$$\frac d {dt} \iint_{R(t)} f(\vec x, t ) \; d\vec x = - \iint_{\p R(t)} \vec J
+\cdot \vec n \; d S(\vec x) + \iiint_{R(t)} Q(x,t) \; dV$$
+
+$Q(\vec x, t) :=$ creation or depletion. 
+
+Using reynold's transport theorem and the divergence theorem: 
+
+$$\iiint_{R(t)} (D_t f + f(\nabla \cdot \vec v) ) \; d \vec x = \iiint_{R(t)}
+(-\nabla \cdot \vec J + Q(\vec x, t)) \; d\vec x$$
+
+Since $R(t) = X(R(0), t) $ is arbitrary, we use the dubois Reymond lemma to
+conclude that the integrand vanishes: 
+
+$$D_t f + f(\nabla \cdot \vec v) = - \nabla \vec J + Q(\vec x, t)$$
+
+$$\p_t f + \nabla \cdot (f \vec v) = - \nabla \cdot \vec J + Q(\vec x, t)$$
+
+Which are the two differential versions of the balance law.
+
+\subsubsection{Continuity Equation}
+
+Consider the mass density $\varrho = \varrho(\vec x, t)$, (mass per volume). 
+
+Mass conservation:  $\vec J = 0, Q = 0$.
+
+$$\frac d {dt} \iiint_{R(t)} \varrho(\vec x ,t) \; d\vec x = 0$$ 
+
+The corresponding differential formulation is:
+
+$$D_t \varrho + \varrho (\nabla \cdot \vec v) = 0$$
+
+$$\p_t \varrho \nabla \cdot (\varrho \vec v) = 0$$
+
+If the material is incompressible $(\varrho = $ a constant $)$, it holds that
+$\p_t \varrho = \p_{x_i} \varrho = 0$ and the continuity equation
+$\p_t \varrho + \nabla_x \varrho \vec v + \varrho(\nabla \cdot \vec v) = 0$
+reduces to
+
+$$\nabla \cdot \vec v = 0$$
+
+\begin{ex} $ $
+
+    \begin{itemize}
+        \item 
+    Shear flow $X(A,t) = \begin{pmatrix} A_1 + \alpha(t) A_2 \\ A_2\\A_3
+        \end{pmatrix}$. In this case we had $\vec v = \begin{pmatrix} \alpha'(t)
+    X_2 \\ 0 \\ 0 \end{pmatrix}   \implies  \nabla  \cdot \vec v = \p_{x_1}
+    (\alpha'(t) X_2) = 0 \implies$ simple sheer is possbile for an
+    incompressible material. 
+        \item 
+            Uniform dilatation: $X(A,t) = \alpha(t) A$
+            We found $\vec v (\vec x, t) = \frac {\alpha '}{\alpha} \vec x$. 
+
+            $$\nabla \cdot v = \nabla \cdot (\frac {\alpha'}{\alpha} \vec x) =
+            \frac {\alpha'}{\alpha}(\p_{x_1}x_1) + \p{x_2} x_2 + \p_{x_3} x_3 = 3
+            \frac {\alpha ' } {\alpha}$$
+
+            So uniform dilatation is not possible with an incompressible
+            material.
+    \end{itemize}
+\end{ex}
+
+\week{}
+\lecture{}
+
 \pagebreak
 \appendix