wk12

notes.tex

@@ -2627,6 +2627,8 @@ if you have $\tau_0 = 0$ (the Kelvin-Voigt model of a viscoelastic solid) $\delt
 \pagebreak
 \chapter{Continuum Mechanics in 3D}
 
+\section{Navier-Stokes Derivation}
+
 \begin{description}
     \item[Material Coordinates]  (lagrangian)
 
@@ -2674,7 +2676,7 @@ 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 $J(t) > 0$ 
 
 \week{}
-\lecture
+\lecture{}
 
 
 \begin{ex} Uniform dilatation
@@ -3050,8 +3052,6 @@ Gives $\vec t = T^T \vec n$  where $T$ is the \emph{Cauchy stress tensor}
     
 
 
-\lecture {}
-
 The left hand side of $( * *  ) $ can be written as 
 
 $$\varrho D_t \vec v + \vec v \cdot D_t \varrho + \vec v \varrho(\nabla \cdot
@@ -3119,6 +3119,7 @@ So the full system we have so far is
 
 What is missing is the constitutive relation for $T$
 
+\lecture{}
 \subsection{Material frame indifference}
 
 Any constitutive law should not depend on translation or rotation: any rigid body motion. 
@@ -3134,9 +3135,520 @@ smooth functions, and $Q$ is a rotation matrix. $Q Q^T = \mathbb I \forall t >
         other coordinate frame. You should be able to compute the stress tensor
         in either and get the same result. 
 
+    \item  Form invariance: any constitutive law $T = G(T) $ stress tensor in
+        terms of another quantity $R$ should be the same in the transformed
+        coordinate frame. 
+\end{enumerate}
+
+These conditions give:
+
+$$G(R^*) = T^* = Q G(R) Q^T \qquad\text{For all } R \text{ and } T$$
+
+
+\begin{ex}
+    $X^* = QX + b(t) \quad |_{\p t}$
+
+    $V^* - Q'(t) X + QV + b'(t) $ this is how the velocity field transforms.
+
+    And a constitutive law such as $T = G(v)$ now in spatial coordinates. (We are
+    using $R = \vec v$ and $R^* = \vec v^*$).
+
+    MFI in this case is
+    
+    $$Q G(v) Q^T = G(Q'(t) x + Q v + b'(t))$$
+
+    For all rotations $Q(t)$ and shifts $b(t)$, but even if $Q = \mathbb I$
+    $C(v) = G(v + b'(t))$ is only true if $b'(t) = 0$ ie $b$ is constant.
+
+
+\end{ex}
+
+\begin{ex}
+    $T = \mu (\nabla v + (\nabla v)^T)  $ ie $R = \nabla v + \nabla v ^T$ is
+    materially frame indifferent. 
+
+\end{ex}
+
+\begin{thm}{}{} Rivilin Ericson Theorem
+
+    Assuming that $T$ and $R$ are symmetric and objective and $T = G(R)$ is form
+    invariant, then it can be written as 
+
+    $$T = \alpha_0 \mathbb I + \alpha_1 R + \alpha_2 R^2$$
+
+    Where $\alpha_0, \alpha_1,\alpha_2 \in \R$ are functions of 
+
+    $I_R = tr(R) =
+    \lambda_1 + \lambda_2 + \lambda_3$
+
+    $II_R = \frac 12(tr(R)^2 - tr(R^2)) = \lambda_1 \lambda_2 + \lambda_1
+    \lambda_3 + \lambda_2 \lambda_3$
+
+    $III_R = \det(R) = \lambda_1 \lambda_2 \lambda_3$
+
+    Which are the principle invariants of $R$, ie, they are conserved under
+    rotations and $\lambda_1 , \lambda_2 \lambda_3$ are eigenvalues of $R$. 
+
+    How to show this: 
+
+    Either constrctive by picking a range of potential rotations, or assuming
+    that $G$ has a (taylor) expansion $G(R) = \sum_{n = 0}^\infty \kappa_n R^n$
+    with constants $\kappa_n$, and using the Cayley-Hamilton theorem: 
+
+    $R \in \R^{3\times 3} $ satisfies its characteristic equation: $\det(R - X
+    \mathbb I) = 0 \implies R^3 I_R R^2 - II_R + III_R \mathbb I$.
+
+$... = \alpha_{2,n} R^2 + \alpha_{1,n} R + \alpha_{0,n} \mathbb I$.
+\end{thm}
+
+\week{}
+\lecture{}
+
+\subsection{Newtonian Fluids}
+
+The stress depends linearly on the strain rate. (In 1D strain rate was $\p_t
+\varepsilon$). In 3D strain rate is pressure. 
+
+\begin{itemize}
+    \item Pressure is modelled by $T = -p \mathbb I$. Acting through an
+        arbitrary plane with unit normal $\vec n$, it is given by $-p \vec n$.
+        Pressure is isotripic. 
+    \item Viscous stress. In 1D this was $\varepsilon_t = \p_t (U_A) = V_A$
+        to model relative velocity differences between neighbouring particles.
+        In 3D spatial coordinates an analog would be $\p_x v, \p_y v, \p_z v$ ie
+        $\nabla \vec v$.
+
+        Decompose this into its symmetric and anti-symmetric component: $\nabla
+        \vec v = D + W $ where:
+
+        $D$ is the rate of deformation tensor defined as:
+$$
+\begin{aligned}
+{D} & \equiv \frac{1}{2}\left(\nabla \vec{v}+\nabla \vec{v}^{T}\right) \\
+&=\left(\begin{array}{ccc}
+\frac{\partial v_{1}}{\partial x} & \frac{1}{2}\left(\frac{\partial v_{1}}{\partial y}+\frac{\partial v_{2}}{\partial x}\right) & \frac{1}{2}\left(\frac{\partial v_{1}}{\partial z}+\frac{\partial v_{3}}{\partial x}\right) \\
+\frac{1}{2}\left(\frac{\partial v_{1}}{\partial y}+\frac{\partial v_{2}}{\partial x}\right) & \frac{\partial v_{2}}{\partial y} & \frac{1}{2}\left(\frac{\partial v_{2}}{\partial z}+\frac{\partial v_{3}}{\partial y}\right) \\
+\frac{1}{2}\left(\frac{\partial v_{1}}{\partial z}+\frac{\partial v_{3}}{\partial x}\right) & \frac{1}{2}\left(\frac{\partial v_{2}}{\partial z}+\frac{\partial v_{3}}{\partial y}\right) & \frac{\partial v_{3}}{\partial z}
+\end{array}\right)
+\end{aligned}
+$$
+
+And $W$ is the vorticity or spin tensor:
+$$
+\begin{aligned}
+{W} & \equiv \frac{1}{2}\left(\nabla \vec{v}-\nabla \vec{v}^{T}\right) \\
+&=\left(\begin{array}{ccc}
+0 & \frac{1}{2}\left(\frac{\partial v_{1}}{\partial y}-\frac{\partial v_{2}}{\partial x}\right) & \frac{1}{2}\left(\frac{\partial v_{1}}{\partial z}-\frac{\partial v_{3}}{\partial x}\right) \\
+\frac{1}{2}\left(\frac{\partial v_{2}}{\partial x}-\frac{\partial v_{1}}{\partial y}\right) & 0 & \frac{1}{2}\left(\frac{\partial v_{2}}{\partial z}-\frac{\partial v_{3}}{\partial y}\right) \\
+\frac{1}{2}\left(\frac{\partial v_{3}}{\partial x}-\frac{\partial v_{1}}{\partial z}\right) & \frac{1}{2}\left(\frac{\partial v_{3}}{\partial y}-\frac{\partial v_{2}}{\partial z}\right) & 0
+\end{array}\right)
+\end{aligned}
+$$
+\end{itemize}
+
+This suggests a constitutive law of the form:
+
+$$T = -p \mathbb I + G(D,w) \qquad(*)$$
+    
+However under rigid body motion $X^* = Q(t) X + b(t)$ and $D$ and $w$ transform
+according to  $D^* = QDQ^T$ $w^* = QwQ^T +
+\underbrace{Q'Q^T}_{\text{an antisimmetric
+matrix}}$
+
+$(*)$ is MFI: $QG(R)Q^T = G(R^*)$ if 
+
+$$QG(D,w) Q^T = G(D^*, w^*) = G(Q D Q^T, QwQ^T, + Q'Q^T)\quad \forall Q(t)$$
+
+But if $Q = \mathbb I$ and $Q' = M$ (anti-symmetric), 
+
+$$G(D,w) = G(D, w + M)$$
+
+Which is only true if $M = \hat 0 \in \R^{3 \times 3}$ is the zero matrix. 
+$\implies T = -p \mathbb I + G(D)$. 
+
+The rivilin-erickson theorem now tells us that 
+
+$$G(D) = \alpha_0 \mathbb I + \alpha_q D + \alpha_2 D^2$$
+
+And $\alpha_0 = 0$ if $D = \mathbf 0$, there is no viscous stress if $D = \mathbf 0$.
+
+Finally, since linearity characterises newtonian fluids, we simplify the model
+by assuming linear dependence of $T$ on $D$.
+
+$\implies \alpha_2 = 0$ because $D^2$ is not linear. 
+
+$\alpha_1 = 2\mu$, a constant, since $D$ is linear. 
+
+$\alpha_0 = \lambda I_D = \lambda tr(D) = \lambda \nabla \cdot \vec v$ since
+$I_d$ is the only principle invariant that is linear.     
+
+
+\begin{itemize}
+    \item  [$\mu$]  Dynamic shear velocity 
+    \item [$p$] pressure
+    \item [$\lambda$] bulk viscosity
+\end{itemize}
+
+$\implies \nabla \cdot T = - \nabla p + \lambda \nabla (\nabla \cdot \vec v )
++ \mu(\Delta \vec v + \nabla (\nabla \cdot \vec v))$
+
+And substitute into the momentum equation to obtain the Navier-Stokes equation. 
+
+$$\varrho D_t \vec v = -\nabla p (\lamba + \mu) \nabla (\nabla \cdot \vec v) +
+\mu \Delta \vec v + \varrho \vec f$$
+
+Typically couple to the continuity equation:
+
+$$D_t \varrho + \varrho \nabla \cdot \vec v = 0$$
+
+Ideal gas law (with known temperature $T$):
+
+$$p = \varrho R T$$
+
+Usually this system is solved numerically, for example using OpenFoam software
+package. 
+
+\lecture{}
+
+A common simplification is to assume incompressibility:  $\p_t \varrho = 0 =
+\p_{\vec x_i } \varrho \implies \p_t \varrho + \Nabla \cdot (\varrho \vec v) = 0$
+
+$\implies \nabvla \cot \vec v = 0$
+
+The stress in this case is $T = -p \I + 2\mu D$ and we obtain the incompressible
+navier-stokes equation:
+
+$$\varrho D_t \vec v = - \nabla p + \mu \Delta \vec v + \varrho \vec f$$
+
+Coupled to : $\nabvla \cdot \vec v = 0$
+
+And treat $p$ as lagrangian multiplier to enforce $\nabla\cdot \vec v = 0 $ 
+
+Then for an incompressible newtonian fluid in cartesian coordinates, letting
+$\vec v = (u,v,w)$ and $\vec f = (f,g,h)$:
+
+\begin{gathered}
+\rho\left(\frac{\partial u}{\partial t}+u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}+w \frac{\partial u}{\partial z}\right)=-\frac{\partial p}{\partial x}+\mu\left(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}+\frac{\partial^{2} u}{\partial z^{2}}\right)+\rho f \\
+\rho\left(\frac{\partial v}{\partial t}+u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}+w \frac{\partial v}{\partial z}\right)=-\frac{\partial p}{\partial y}+\mu\left(\frac{\partial^{2} v}{\partial x^{2}}+\frac{\partial^{2} v}{\partial y^{2}}+\frac{\partial^{2} v}{\partial z^{2}}\right)+\rho g \\
+\rho\left(\frac{\partial w}{\partial t}+u \frac{\partial w}{\partial x}+v \frac{\partial w}{\partial y}+w \frac{\partial w}{\partial z}\right)=-\frac{\partial p}{\partial z}+\mu\left(\frac{\partial^{2} w}{\partial x^{2}}+\frac{\partial^{2} w}{\partial y^{2}}+\frac{\partial^{2} w}{\partial z^{2}}\right)+\rho h \\
+\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0
+\end{gathered}
+
+
+Note: $D_t \vec v = \p_t \vec v + (\vec v \cdot \nabla) \vec v = \p_t
+\vec v + (\nabla\vec v) \vec v $  
+
+$\Delta \vec v = \begin{pmatrix} \Delta u\\ \Delta v \\ \Delta w \end{pmatrix} $
+
+ 
+\subsubsection{Remarks}
+
+\begin{itemize}
     \item 
+Proving global existance in time for the incompressible navier stokes
+equation is one of the millenium problems. 
+\item
+This model is the basis for
+simulations of fluids on all spatial scales: galaxies, earths mantle, oceans,
+hydraulics, in biology; blood flow, tissue, cytoplasm, etc. 
+\item 
+    $\mu$ is the shear viscosity: 
+
+    \begin{center}
+\begin{array}{l|l|l}
+\hline \text { Fluid } & \text { Viscosity (Pa-s) } & \text { Density }\left(\mathrm{kg} / \mathrm{m}^{3}\right) \\
+\hline \text { Air } & 1.8 \times 10^{-5} & 1.18 \\
+\hline \text { Water } & 8.9 \times 10^{-4} & 0.997 \times 10^{3} \\
+\hline \text { Mercury } & 1.5 \times 10^{-3} & 1.3 \times 10^{4} \\
+\hline \text { Olive oil } & 0.08 & 0.92 \times 10^{3} \\
+\hline \text { Honey } & 37 & 1.4 \times 10^{3} \\
+\hline \text { Pitch } & 2.3\times 10^8&                   \\
+\hline
+\end{array}
+\end{center}
+\end{itemize}
+
+\section{Solutions to steady flow problems}
+
+Steady flow is characterised by $\p_t \vec v = 0$, and $\p_t p = 0$ and $\p_t
+\varrho = 0$.
+
+This implies that $D_t \vec v = \p_t \vec v + (\vec v \cdot \nabla) \vec v$ 
+
+So we now the Newtonian Navier--Stokes becomes:
+
+$$\begin{cases} \varrho(\vec v \cdot \nabla) \vec v = - \nabla p + \mu \Delta
+\vec v + \varrho \vec f \\ \nabla \vec v = 0\end{cases}$$
+
+Note that $\vec v, p \varrho $ do not change with time but single particles
+still move, and their trajectories must satisfy in general:
+
+$$\begin{cases}\frac {d X} {dt} = V(A,t) = v(X(A,t), t)\\ X(A,t=0) = A
+\end{cases}$$
+
+So in steady flow:
+
+$$\begin{cases} \frac {dX} {dt} = V(A) = v(X(A,t)) \\ X(A,t=0) = A\end{cases}$$
+
+The trajectory lines do not change with time. 
+
+
+\subsection{Coette plane flow}
+
+Steady flow of fluid between two parallel planes. 
+
+\incfigw{coette}{0.6}
+
+Lower plate is stationary, upper plate has speed in x direction of $u_0$. 
+$dt y = h$.
+
+Use the boundary condition: assume there is a no-slip condition between the fluid
+and the boundary with the plate: the fluid travels at the speed of the plate. 
+
+At the upper plate $\vec v (x, y = h, z) =  (u_0 , 0 , 0)$
+
+At the lower plate $\vec v(x, y= 0, z) = (0,0,0)$
+
+Assume laminar flow: $\vec v = (u,0,0)$: flow is only in the $x$ direction,
+solution is homogeneous in $z$-direction: $\p_z u = 0, \p_z p  = 0$
+
+Now our model becomes:
+
+$$
+(* * *)
+\begin{cases}
+   u_x = 0 \\
+   \varrho u_x u = -p_x + \mu (u_{x x} + u_{y y } + u_{z z}) \\
+   0 = -py + 0  \\
+   0 = 0 + 0 \\
+\end{cases}
+$$
+
+$\p_z u = 0 = \p_x u \implies u = u(y)$
+
+$\p_z p = 0 = \p_y p \implies p = p(x)$
+
+So $(* * * ) $ becomes $0 = -p_x + \mu u_{y y}$, i.e., $p'(x) = \mu u''(y)$.
+
+Use method of separation: $p'(x) = const = \mu u''(y) \implies p' = const
+\implies$ 
+
+$$p(x) = p_0 + x p_1$$
+
+Modelling intuition: $p_1 \ne 0$ would impl;y that the pressure blows up at $x =
+\pm \infty \implies $ assume that $p_1 = 0$, with that $p(x) = p_0$.
+
+\lecture{}
+
+The second equation becomes $\mu u''(y) = 0 \implies u(y) = ay + b$. The
+boundary conditions are $u(y = 0) = 0, u(h = 0) = u_0$, so $a = \frac {u_0} h
+\implies u = \gamma y$ where $\gamma = \frac {u_0} h$ 
+
+$\gamma = \frac {u_0} {h} $ is the \emph{shear rate} which has dimension $\frac
+1t$.
+
+So we have a solution: 
+
+$\vec v = \begin{pmatrix} \gamma y \\ 0 \\0 \end{pmatrix}$  and $p = p_0$
+
+Remark: The stress tensor is given by 
+
+$$T = -p \mathbb I + 2\mu D = -p_0 \mathbb I + \mu(\nabla v + {\nabla v}^T) = -p_0
+         \mathbb I + \mu \begin{pmatrix}0 & \gamma & 0 \\ \gamma & 0 & 0 \\ 0 & 0 & 0   \end{pmatrix} $$
+
+         So the stress through a horizontal plane, for example $ \{ y = 0\} ,
+         \vec n = (0,1,0)$, is given by
+
+
+         $$
+         T \vec n = \left(-p_0 \mathbb I + \mu \begin{pmatrix} 0 & \gamma & 0 \\
+         \gamma & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}  \right) \begin{pmatrix} 0 \\ 1 \\
+         0\end{pmatrix} = \underbrace{-p_0 \begin{pmatrix} 0\\1\\0
+ \end{pmatrix}}_{\vec t_s}  +
+     \underbrace{\mu
+ \begin{pmatrix} \gamma \\ 0 \\ 0 \end{pmatrix} }_{\vec t_p}
+         $$
+
+         $\vec t_s$ --- the shear component.
+
+         $\vec t_p$ --- the normal component.
+
+\subsubsection{Shear Test}
+
+A Shear test can be used to investgate whether a fluid is Newtonian (where
+stress is linear with respect to $\gamma$) and what the shear viscosity is. 
+
+Put the fluid between two plates and move the upper plate with differnet speeds, 
+$\gamma = \frac {u_0}{h}$. Measure the force (stress $\times$ area), to see if
+it is linear with respect to the velocity of the top plate.
+
+Since shear stress $\vec t_s = \mu \gamma \begin{pmatrix} 1 \\ 0 \\ 0
+\end{pmatrix} $, the gradient of your shear stress / shear rate graph is the
+shear viscosity $\mu$.
+
+\subsection{Poiseuille Flow}
+
+Steady flow through a cylindrical pipe of length $L$ with radius $R$. 
+
+Use the incompressible Navier-Stokes equation written in cylindrical
+coordinates. 
+
+Write the problem in polar coordinatnes about the centre of the pipe, with $z$
+running along its length: 
+
+$\vec v = (v_R, v_\theta, v_z) =  v_r \vec e_r + v_\theta \vec e_\theta
++ v_z + \vec e_z$, and 
+
+$\vec f = f_r \vec e_r + f_\theta \vec e_\theta + f_z \vec e_z$, 
+
+$x = r\cos \theta$ and $y = r \sin \theta$ and $z = z$. 
+
+The incompressible Navier--Stokes system for newtonian fluid in
+cyndrical coordinates is:
+
+\begin{aligned}
+&\rho\left(\frac{\partial v_{r}}{\partial t}+v_{r} \frac{\partial v_{r}}{\partial r}+\frac{v_{\theta}}{r} \frac{\partial v_{r}}{\partial \theta}-\frac{v_{\theta}^{2}}{r}+v_{z} \frac{\partial v_{r}}{\partial z}\right)\\
+&\quad=-\frac{\partial p}{\partial r}+\mu\left[\frac{\partial}{\partial r}\left(\frac{1}{r} \frac{\partial}{\partial r}\left(r v_{r}\right)\right)+\frac{1}{r^{2}} \frac{\partial^{2} v_{r}}{\partial \theta^{2}}-\frac{2}{r^{2}} \frac{\partial v_{\theta}}{\partial \theta}+\frac{\partial^{2} v_{r}}{\partial z^{2}}\right]+\rho f_{r}\\\\
+&\rho\left(\frac{\partial v_{\theta}}{\partial t}+v_{r} \frac{\partial v_{\theta}}{\partial r}+\frac{v_{\theta}}{r} \frac{\partial v_{\theta}}{\partial \theta}+\frac{v_{r} v_{\theta}}{r}+v_{z} \frac{\partial v_{\theta}}{\partial z}\right)\\
+&\quad=-\frac{1}{r} \frac{\partial p}{\partial
+\theta}+\mu\left[\frac{\partial}{\partial r}\left(\frac{1}{r}
+\frac{\partial}{\partial r}\left(r v_{\theta}\right)\right)+\frac{1}{r^{2}}
+\frac{\partial^{2} v_{\theta}}{\partial \theta^{2}}+\frac{2}{r^{2}}
+\frac{\partial v_{r}}{\partial \theta}+\frac{\partial^{2} v_{\theta}}{\partial
+z^{2}}\right]+\rho f_{\theta}\\ \\
+&\rho\left(\frac{\partial v_{z}}{\partial t}+v_{r} \frac{\partial v_{z}}{\partial r}+\frac{v_{\theta}}{r} \frac{\partial v_{z}}{\partial \theta}+v_{z} \frac{\partial v_{z}}{\partial z}\right)\\
+&\quad =-\frac{\partial p}{\partial z}+\mu\left[\frac{1}{r}
+\frac{\partial}{\partial r}\left(r \frac{\partial v_{z}}{\partial
+r}\right)+\frac{1}{r^{2}} \frac{\partial^{2} v_{z}}{\partial
+\theta^{2}}+\frac{\partial^{2} v_{z}}{\partial z^{2}}\right]+\rho f_{z}\\  \\
+&\frac{1}{r} \frac{\partial\left(r v_{r}\right)}{\partial r}+\frac{1}{r} \frac{\partial v_{\theta}}{\partial \theta}+\frac{\partial v_{z}}{\partial z}=0
+\end{aligned}
+  
+Unlike before there is pressure difference across the length ($z$) of the pipe 
+and the flow is coming from the pressure difference. 
+
+
+The boundary conditions are $\vec v = \begin{pmatrix} 0\\0\\0 \end{pmatrix} $ at
+$r = R$ (no-slip boundary condition). $p(z = 0) = p_0$ and $p(z = L) = p_1 <
+p_0$.
+
+$\vec v = \begin{pmatrix} v_r \\ v_\theta \\ v_z \end{pmatrix} $
+
+Again assume laminar flow (trajectory not depending on $\theta$ or $r$).
+
+The velocity only has a $z$ component: 
+
+$v_r(z = 0) = 0$ and $v_\theta(z = 0) = 0$
+
+$v_r(z = L) = 0$ and $v_\theta (z = L) = 0$
+
+$v_\tehta = v_r = 0 \implies \vec v = \begin{pmatrix} 0 \\ 0 \\ v_z
+\end{pmatrix} $
+
+And assume the solution satisfies rotational symmetry 
+
+$v_z = v_z(r,z)$ and $p = p(r,z)$.
+
+And assume body forces are $0$.
+
+Now we can get rid of most of the Navier Stokes terms because of these
+assumptions which leaves the governing equations
+
+$$
+\begin{cases}
+    0 = \p_r p \\
+    \varrho v_z \frac {\p v_z} {\p z} =  -p_z + \mu (\frac 1r \p_r( r \p_r
+    v_z ) + \p _{z z} v_z)\\
+    0 = \p_z v_z 
+\end{cases}
+$$
+
+
+$\implies p = p(z) $ and $v_z = (r)$  and $p'(z) = \mu(v_z'' (r) + \frac 1r v_z'
+(r)$
+
+Use method of separation: $p'(z) = constant = \mu (v_z '' + \frac 1r v_z')$
+
+Use $p (z = 0) = p_0$ and $p(z = L) = p_1$ $\implies$
+
+$p = p_0 + \alpha z$ and $l(L) = p_1 = p_0 + \alpha l \implies \alpha = \frac
+{p_1 - p_0 } L$ 
+
+$\implies p(z) = p_0 + \frac {p_1 - p_0}L z$
+
+$\frac {p_1 - p_0} L = \mu (v_z'' + \frac 1r v_z ' ) $
+
+$$\frac r \mu \frac {p_1 - p_0} L = r v_z '' + vz' = (rvz') ' $$
+
+Integrate
+
+$$\frac {r^ 2} 2 \frac {p_1 - p_0} {\mu L} = r v_z ' - C$$
+
+Integrate
+
+$$v_z ' = \frac r2 \frac {p_1 - p_0} {\mu L} + \frac 1r C$$
+
+Integrate using $v_z(R) = 0$
+
+$$v_z = (\frac {r^2}2 - \frac {R^2} 2) \frac {p_1 - p_0} {2\mu L} + C \log r$$
+
+If $C \ne 0$ the velocity would blow up towards the centre of the pipe, so $C =
+0$. 
+
+ $$\implies v_z = 
+ \begin{pmatrix} 0 \\ 0 \\ R^2 - r^2 \end{pmatrix} 
+ \frac {p_0 - p_1} {4 \mu L} $$
+
+ So you get a parabolic velocity profile. 
+
+ The stress tensor in cylindrical coordinates is given by: 
+
+     $$T = \begin{pmatrix}  -p & 0 & \mu v_z' \\ 0 & -p & 0 \\
+     \mu v_z ' & 0 & -p \end{pmatrix} $$
+
+     Eg the stress at the surface of a cylinder with radius $r < R$, with normal
+     to the cylinder of $\vec n = \begin{pmatrix} 1\\0\\0 \end{pmatrix}
+     \implies$ $T \vec n = \begin{pmatrix} -p \\ 0 \\ 0 \end{pmatrix}  + \begin{pmatrix}
+     0 \\ 0 \\ \mu v_z '\end{pmatrix} = \vec t_p + \vec t_s$, shear stress and
+     normal stress.
+
+\section{Closing remarks}
+
+\begin{enumerate}
+    \item Vorticity: the curl of the velocity at a specific point
+
+        $$\vec \omega = \nabla \times \vec v = \begin{pmatrix} \frac {\p
+        w} {\p y} - \frac {\p v} {\p t} \\ \frac {\p v} {\p t} - \frac {\p w }
+    {\p x} \\ \frac {\p v} {\p x} - \frac {\p u} {\p y}  \end{pmatrix} $$
+
+    It is like `how much the particle rotates'. 
+
+    For example in plane Couette flow: $\vec v = \begin{pmatrix} \gamma y \\ 0
+    \\ 0 \end{pmatrix} $, the vorticity is $\vec \omega = \nabla \times \vec v
+    = \begin{pmatrix}  0 \\ 0 \\ - \gamma\end{pmatrix} $
+
+\item 
+    Helmholtz Representation Theorem
+
+    Let $\vec q (\vex x) : \R^3 \to \R^3$ diffble, then 
+
+    $\vec q (\vec x) = \nabla \phi + \nabla\times \vec g $ where 
+
+    $\phi:\R^3 \to
+    R$ and $\vec g: \R^3 \to \R^3, \nabla \cdot \vec g = 0$.
+
+    It can be split into the scalar potential with divergence and no rotational
+    component (curl) and
+    the vector potential that is pure rotation. 
+
+
 \end{enumerate}
 
+
+\week{}
+\lecture{}
+
+
 \pagebreak
 \appendix