HW4_Problems_3_4_hints.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Start from the LHS by calculating the volume average of the Cauchy stress tensor $\\left< \\bf T \\right>$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "We know that: \n",
    "${\\bf T} = {J}^{-1}{\\bf S}{\\bf F}^T$, so let us take the volume average (over the deformed volume on both sides)\n",
    "$$\\overline{\\bf T} = \\left< \\bf T \\right> = \\frac{1}{\\left| \\Omega \\right|}\\int_{\\Omega}{\\bf T} d\\Omega  = \\frac{1}{\\left| \\Omega \\right|} \\int_{\\Omega}\n",
    "{J}^{-1}{\\bf S}{\\bf F}^T d\\Omega$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Remember the relation between infinitesimal volume elements: $$ d\\Omega = J d\\Omega_0 \\implies J^{-1} d\\Omega = d\\Omega_0$$ where $d\\Omega$ and $d\\Omega_0$ refer to the infinitesimal volume elements in the deformed and undeformed configurations and $J$ is the determinant of the deformation gradient $\\bf F$. If you substitute this above (notice the change of limits just like you would have in integration) you get the following \n",
    "$$\\left< \\bf T \\right> = \\frac{1}{\\left| \\Omega\\right|}\\int_{\\Omega_0} {\\bf S F}^Td\\Omega_0 $$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Now if you closely look at the RHS (you know that you have to use the divergence theorem and balance of linear momentum; because those are the only things that you know) and rewrite things in indicial notation\n",
    "$$\\frac{1}{\\left| \\Omega\\right|} \\int_{\\Omega_0}\n",
    "S_{ik}F_{jk}d\\Omega_0 =\n",
    "\\frac{1}{\\left| \\Omega\\right|} \\int_{\\Omega_0}\n",
    "S_{ik}\\frac{\\partial x_j}{\\partial X_k} d\\Omega\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Now, for using the divergence theorem, you will need to have \"a derivative\" integrated over the entire volume, so you could write the RHS equivalently (remember integration by parts!!) as \n",
    "$$ \n",
    "\\frac{1}{\\left| \\Omega\\right|} \\int_{\\Omega_0}\n",
    "\\left\\{\n",
    "\\frac{\\partial}{\\partial X_k}\\left( \n",
    "S_{ik}x_j\n",
    "\\right) - \n",
    "x_j \\frac{\\partial S_{ik}}{\\partial X_k}\n",
    "\\right\\} d\\Omega_0\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "You know from \"Balance of linear momentum\" div$({\\bf S}) = {\\bf 0}$, so the second term goes off and once you apply divergence theorem to the first, you are left with \n",
    "$$ \n",
    "\\frac{1}{\\left| \\Omega\\right|} \\int_{\\partial\\Omega_0}\n",
    "S_{ik}N_k x_j d\\Omega_0\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "At this point you have exhausted pretty much everything that you know by now, except: \n",
    "$$\n",
    "\\frac{1}{\\left| \\Omega\\right|} = \\frac{\\left|\\Omega_0 \\right|}{\\left| \\Omega_0\\right|} \\frac{1}{\\left| \\Omega\\right|}\n",
    "$$\n",
    "which, if you carefully notice, gives\n",
    "$$\\frac{1}{\\left| \\Omega\\right|} = \n",
    "\\overline{J}^{-1}\\frac{1}{\\left| \\Omega_0\\right|}\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Substitute this above and you get in the RHS \n",
    "$$ \n",
    "\\overline{J}^{-1}\\frac{1}{\\left| \\Omega_0\\right|} \\int_{\\partial\\Omega_0}\n",
    "S_{ik}N_kx_jd\\Omega_0\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "At this stage there is nothing that remains that you can use to potentially prove the result, which gives you a hint that you need to disprove the result!! Now that you are already on the boundary ($\\partial\\Omega_0$) there could be other types of boundary conditions that you could use. Instead of applying traction/force on the boundary, what other boundary conditions can you prescribe? Apply displacements? So a potential candidate could be \n",
    "$$ {\\bf x} = \\widetilde{\\mathcal{M}}{\\bf X}\\quad \\text{on}\\quad \\partial\\Omega_0$$where, of course, $\\widetilde{M}_{ij}$ is constant on the boundary ($\\partial\\Omega_0$)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "You can substitute this above to get on the RHS\n",
    "$$ \n",
    "\\overline{J}^{-1}\\frac{1}{\\left| \\Omega_0\\right|} \\int_{\\partial\\Omega_0}\n",
    "S_{ik}N_k \\widetilde{M}_{jl} X_ld\\Omega_0\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Or \n",
    "$$ \n",
    "\\overline{J}^{-1}\\widetilde{M}_{jl}\n",
    "\\frac{1}{\\left| \\Omega_0\\right|} \\int_{\\partial\\Omega_0}\n",
    "S_{ik}N_k X_ld\\Omega_0\n",
    "$$apply divergence theorem again to get\n",
    "$$\n",
    "\\frac{1}{\\left| \\Omega_0\\right|} \\int_{\\partial\\Omega_0}\n",
    "S_{ik}X_lN_kd\\Omega_0\n",
    "=\n",
    "\\frac{1}{\\left| \\Omega_0\\right|} \\int_{\\Omega_0}\n",
    "\\frac{\\partial}{\\partial X_k}\n",
    "\\left( \n",
    "S_{ik}X_l\n",
    "\\right)\n",
    "d\\Omega_0 = \n",
    "\\frac{1}{\\left| \\Omega_0\\right|} \\int_{\\Omega_0}\n",
    "\\left( \n",
    "X_l\n",
    "\\frac{\\partial S_{ik}}{X_k}\n",
    "+\n",
    "S_{ik}\\frac{\\partial X_l}{\\partial X_k}\n",
    "\\right)d\\Omega_0\n",
    "$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Again apply balance of linear momentum and proceed from there... You will get something like \n",
    "$$\n",
    "\\overline{J}^{-1}\\overline{\\bf S}\\overline{{\\widetilde{M}}}^T\n",
    "$$on the RHS. What remains to be shown is that $$\n",
    "\\overline{\\widetilde{M}} = \\overline{\\bf F}\n",
    "$$which you should be able to show easily"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [],
   "source": [
    "from sympy import *\n",
    "init_printing()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [],
   "source": [
    "x=Symbol('x')"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [],
   "source": [
    "f = sin(x)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [],
   "source": [
    "g = (cos(x))**2"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": "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\n",
      "text/latex": [
       "$$- \\frac{\\cos^{3}{\\left (x \\right )}}{3}$$"
      ],
      "text/plain": [
       "    3    \n",
       "-cos (x) \n",
       "─────────\n",
       "    3    "
      ]
     },
     "execution_count": 13,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "integrate(f*g)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": "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\n",
      "text/latex": [
       "$$- \\left(\\frac{x}{2} + \\frac{\\sin{\\left (x \\right )} \\cos{\\left (x \\right )}}{2}\\right) \\cos{\\left (x \\right )}$$"
      ],
      "text/plain": [
       " ⎛x   sin(x)⋅cos(x)⎞       \n",
       "-⎜─ + ─────────────⎟⋅cos(x)\n",
       " ⎝2         2      ⎠       "
      ]
     },
     "execution_count": 14,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "integrate(f)*integrate(g)"
   ]
  }
 ],
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