{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter 5 - Boundary Value Problems"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exa 5.1 Page 87"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the value of y at x=.5 from finite difference method is -0.25\n",
      "the value of y from exact soln at x=.5 is -0.25\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "\n",
    "#finite difference method\n",
    "#given d2y_by_dx2-2=0 hence it is dirichlet's problem\n",
    "\n",
    "x_1=0 ; y_1=0               #initial boundary conditions\n",
    "x_3=1 ; y_3=0\n",
    "\n",
    "delta_x=.5                #since we have to find y_2 at x=.5 so x_2=.5 \n",
    "#in the central difference method substituting i=2 we have\n",
    "#(y_3-2*y_2+y_1)/(delta_x)**2=2\n",
    "#since y_1 & y_3=0 as per B.C.\n",
    "y_2=(y_3+y_1-2*delta_x**2)/2\n",
    "print \"the value of y at x=.5 from finite difference method is\",y_2\n",
    "x=.5\n",
    "y_exact=x**2-x\n",
    "print \"the value of y from exact soln at x=.5 is\",y_exact"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exa 5.2 Page 89"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "to solve this problem we will take delta x=.5 since we have to find the value at x=.5\n",
      "the value of y at x=.5 is -0.25\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "#boundary conditions are: x=0 at y=0# dy/dx=1 at x=1\n",
    "print \"to solve this problem we will take delta x=.5 since we have to find the value at x=.5\"\n",
    "delta_x=.5\n",
    "y_1=0\n",
    "#using central difference eqn\n",
    "dy_by_dx=1                 #at x=1, i=3\n",
    "#y_4=dy/dx*2*delta_x+y_2             sincefrom B.C. at node 3\n",
    "\n",
    "#y_2=delta_x**2+y_3-delta_x           on substituting the value of y_4\n",
    "y_3=-(2*delta_x**2+2*(delta_x**2-delta_x)-y_1) #on substituting the value of y_2\n",
    "y_2=delta_x**2+y_3-delta_x\n",
    "print \"the value of y at x=.5 is\",y_2"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exa 5.3 Page 89"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " the solution of eg 5.3 -->Discretization in 1-D space\n",
      "the value of y at x= 0.333333333333 0.666666666667 is respectively 0.111111111111 -0.222222222222\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "from numpy import arange\n",
    "print \"the solution of eg 5.3 -->Discretization in 1-D space\"\n",
    "#given the source term f(x)=4x**2-2x-4\n",
    "#given eqn d2y/dx2-2y=f(x)\n",
    "y_1=0\n",
    "y_4=-1\n",
    "delta_x=1/3           #since given 3 parts and length=1\n",
    "for i in range(0,4):\n",
    "    j=arange(0,1+delta_x,delta_x)\n",
    "\n",
    "#given to divide the line in 3 parts\n",
    "#at node 2\n",
    "#y_3-2*y_2\n",
    "def fx3(x):\n",
    "    d=(4*x**2-2*x-4)\n",
    "    return d\n",
    "f2=fx3(j[1])\n",
    "f3=fx3(j[2])\n",
    "y_3=((f2)*delta_x**2+(2+2*delta_x**2)*((f3)*delta_x**2-y_4)-y_1)/(1-(2+2*delta_x**2)**2)\n",
    "y_2=(f3+2*y_3)*delta_x**2+2*y_3-y_4\n",
    "print \"the value of y at x=\",j[1],j[2],\"is respectively\",y_2,y_3"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    " ## Exa 5.4 Page 90"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the exact analytical soln are\n",
      "0.111111111111\n",
      "-0.222222222222\n",
      "-1.0\n",
      "the value of f(x) at node 2 3 and 4 are -4.22222222222 -3.55555555556 -2.0\n",
      "the values of y2, y3, y4 by TDMA method are\n",
      "0.111111111111\n",
      "-0.222222222222\n",
      "-1.0\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "from numpy import arange,zeros\n",
    "y_1=0\n",
    "dy_by_dx=-3            #at x=1\n",
    "delta_x=1/3           #since given 3 parts and length=1\n",
    "for i in range(0,4):\n",
    "    j=arange(0,1+delta_x,delta_x)\n",
    "\n",
    "#given to divide the line in 3 parts\n",
    "def fx3(x):\n",
    "    d=(4*x**2-2*x-4)\n",
    "    return d\n",
    "\n",
    "f2=fx3(j[1])\n",
    "f3=fx3(j[2])\n",
    "f4=fx3(j[3])\n",
    "print \"the exact analytical soln are\"\n",
    "y=[]\n",
    "for i in range(1,4):\n",
    "    y.append(-2*j[i]**2+j[i])\n",
    "    print y[i-1]\n",
    "\n",
    "#using B.C.-2 at node 4 we get\n",
    "#(y_5-y_3)/(2*delta_x)=-3\n",
    "#eqn at node 2\n",
    "#-20*y_2+9*y_3=f2\n",
    "#at node 3\n",
    "#9*y_2-20*y_3+9*y_4=f3\n",
    "#at node 4\n",
    "#18*y_3-20*y_4=16\n",
    "print \"the value of f(x) at node 2 3 and 4 are\",f2,f3,f4\n",
    "#now solving the equations using TDMA method \n",
    "\n",
    "a=[0,9,18]                   #sub diagonal assignment\n",
    "\n",
    "b=[]\n",
    "for j in range(1,4):\n",
    "    b.append(-20)            #main diagonal assignment\n",
    "c=[]\n",
    "for k in range(1,3):\n",
    "    c.append(9)            #super diagonal assignment\n",
    "\n",
    "d=[f2,f3,16]                     #given values assignment\n",
    "\n",
    "i=1#\n",
    "n=3#\n",
    "beta1=[b[i-1]]#                #initial b is equal to beta since a1=0\n",
    "gamma1=[d[i-1]/beta1[i-1]]       #since c7=0\n",
    "m=i+1#\n",
    "for j in range(m,n+1):\n",
    "    beta1.append(b[j-1]-a[j-1]*c[j-2]/beta1[j-2])\n",
    "    gamma1.append((d[j-1]-a[j-1]*gamma1[j-2])/beta1[j-1])\n",
    "x=zeros(n)\n",
    "x[-1]=gamma1[n-1]               #since c7=0\n",
    "n1=n-i#\n",
    "for k in range(1,n1+1):\n",
    "    j=n-k# \n",
    "    x[j-1]=gamma1[j-1]-c[j-1]*x[j]/beta1[j-1]\n",
    "\n",
    "print \"the values of y2, y3, y4 by TDMA method are\"\n",
    "for i in range(1,4):\n",
    "    print x[i-1]\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exa 5.5 Page 91"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the analytical soln are\n",
      "0.648054273664\n",
      "0.651297246159\n",
      "0.661058620401\n",
      "0.677436091507\n",
      "0.70059357071\n",
      "0.730762825846\n",
      "0.768245800962\n",
      "0.81341763827\n",
      "0.8667304327\n",
      "0.92871775662\n",
      "the values of y from y1 to y10 by TDMA method are\n",
      "0.648259699273\n",
      "0.65150099777\n",
      "0.661257306244\n",
      "0.67762618778\n",
      "0.700771331194\n",
      "0.730924187921\n",
      "0.768386286526\n",
      "0.813532247997\n",
      "0.866813531948\n",
      "0.928762951218\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "from math import cosh\n",
    "from numpy import zeros\n",
    "delta_x=.1\n",
    "y_11=1\n",
    "dy_by_dx_1=0            #dy/dx at x=0\n",
    "# given eqn d2y/dx2=y\n",
    "print \"the analytical soln are\"\n",
    "y_an=[]\n",
    "for i in range(1,11):\n",
    "    y_an.append(cosh((i-1)/10)/cosh(1))\n",
    "    print y_an[i-1]\n",
    "\n",
    "#using central difference method we have \n",
    "#y(i-1) - (2+delat_x**2)y(i) + y(i+1)=0\n",
    "#therefore the eqns can be solved using TDMA method\n",
    "a=[None]\n",
    "for i in range(2,11):\n",
    "    a.append(1)            #sub diagonal assignment\n",
    "b=[]\n",
    "for j in range(1,11):\n",
    "    b.append(-2.01)           #main diagonal assignment\n",
    "\n",
    "c=[2]\n",
    "for k in range(2,10):\n",
    "    c.append(1)            #super diagonal assignment\n",
    "d=[]\n",
    "for l in range(1,10):\n",
    "    d.append(0)\n",
    "                 #given values assignment\n",
    "d.append(-1)\n",
    "i=1#\n",
    "n=10#\n",
    "beta1=[b[i-1]]                #initial b is equal to beta since a1=0\n",
    "gamma1=[d[i-1]/beta1[i-1]]      #since c7=0\n",
    "m=i+1#\n",
    "for j in range(m,n+1):\n",
    "    beta1.append(b[j-1]-a[j-1]*c[j-2]/beta1[j-2])\n",
    "    gamma1.append((d[j-1]-a[j-1]*gamma1[j-2])/beta1[j-1])\n",
    "\n",
    "x=zeros(10)\n",
    "x[n-1]=gamma1[n-1]                #since c7=0\n",
    "n1=n-i#\n",
    "for k in range(1,n1+1):\n",
    "    j=n-k\n",
    "    x[j-1]=gamma1[j-1]-c[j-1]*x[j]/beta1[j-1]\n",
    "\n",
    "print \"the values of y from y1 to y10 by TDMA method are\"\n",
    "for i in range(1,11):\n",
    "    print x[i-1]"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exa 5.6 Page 92"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the values of T from T2 to T9 by TDMA method are\n",
      "111.111111111\n",
      "122.222222222\n",
      "133.333333333\n",
      "144.444444444\n",
      "155.555555556\n",
      "166.666666667\n",
      "177.777777778\n",
      "188.888888889\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "from numpy import zeros\n",
    "T_1=100 ;  T_10=200\n",
    "delta_x=(T_10-T_1)/9               #since 9 divisions are to be made\n",
    "#using central difference method we get\n",
    "#for node 2--> 2*T_2-T_3=100\n",
    "a=[0]\n",
    "for i in range(2,9):\n",
    "    a.append(-1)            #sub diagonal assignment\n",
    "b=[]\n",
    "for j in range(1,9):\n",
    "    b.append(2)           #main diagonal assignment\n",
    "c=[]\n",
    "for k in range(1,8):\n",
    "    c.append(-1)            #super diagonal assignment\n",
    "\n",
    "d=[100]\n",
    "for l in range(2,8):\n",
    "    d.append(0);         #given values assignment\n",
    "\n",
    "d.append(200)\n",
    "\n",
    "\n",
    "i=1;\n",
    "n=8;\n",
    "beta1=[b[i-1]]                #initial b is equal to beta since a1=0\n",
    "gamma1=[d[i-1]/beta1[i-1]];      #since c7=0\n",
    "m=i+1;\n",
    "for j in range(m,n+1):\n",
    "    beta1.append(b[j-1]-a[j-1]*c[j-2]/beta1[j-2])\n",
    "    gamma1.append((d[j-1]-a[j-1]*gamma1[j-2])/beta1[j-1])\n",
    "\n",
    "x=zeros(n)    \n",
    "x[-1]=gamma1[n-1]               #since c7=0\n",
    "n1=n-i;\n",
    "for k in range(1,n1+1):\n",
    "    j=n-k\n",
    "    x[j-1]=gamma1[j-1]-c[j-1]*x[j]/beta1[j-1];\n",
    "\n",
    "print \"the values of T from T2 to T9 by TDMA method are\"\n",
    "for i in range(1,9):\n",
    "    print x[i-1]\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exa 5.7 Page 93"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the values of T from T1 to T4 in Celsius by TDMA method are\n",
      "231.893171899\n",
      "199.529667042\n",
      "180.886766374\n",
      "174.799288605\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "from math import pi,sqrt\n",
    "from numpy import zeros\n",
    "dia=.02\n",
    "l=.05\n",
    "T_0=320\n",
    "delta_x=l/4\n",
    "k=50\n",
    "h=100\n",
    "T_surr=20\n",
    "#B.C--> d(theta)/dx+h(theta)/k=0 at x=0.05\n",
    "#let m=sqrt(hP/kA)\n",
    "P=pi*dia\n",
    "A=pi*dia**2/4\n",
    "m=sqrt(h*P/(k*A))#\n",
    "#using central difference method we get \n",
    "#-theta(i-1)+(2+(m*delta_x**2)*theta(i)+theta(i+1))=0\n",
    "theta_0=T_0-T_surr\n",
    "#using B.C. at node 4 we get--> theta(5)=theta(3)-2h*theta(4)*delta_x/k\n",
    "#now the eqns can be solved using TDMA method\n",
    "a=[None]\n",
    "for i in range(2,4):\n",
    "    a.append(-1 )           #sub diagonal assignment\n",
    "\n",
    "a.append(-2)\n",
    "b=[]\n",
    "for j in range(1,4):\n",
    "    b.append(2.0625)#           #main diagonal assignment\n",
    "\n",
    "b.append(2.1125)\n",
    "for k in range(1,4):\n",
    "    c.append(-1)            #super diagonal assignment\n",
    "d=[300]\n",
    "for l in range(2,5):\n",
    "    d.append(0)\n",
    "                  #given values assignment\n",
    "\n",
    "i=1#\n",
    "n=4#\n",
    "beta1=[b[i-1]]                #initial b is equal to beta since a1=0\n",
    "gamma1=[d[i-1]/beta1[i-1]]      #since c7=0\n",
    "m=i+1\n",
    "for j in range(m,n+1):\n",
    "    beta1.append(b[j-1-1]-a[j-1]*c[j-2-1]/beta1[j-1-1])\n",
    "    gamma1.append((d[j-1]-a[j-1]*gamma1[j-2])/beta1[j-1])\n",
    "\n",
    "x=zeros(n)    \n",
    "x[n-1]=gamma1[n-1]#               #since c7=0\n",
    "n1=n-i#\n",
    "for k in range(1,n1+1):\n",
    "    j=n-k#\n",
    "    x[j-1]=gamma1[j-1]-c[j-1]*x[j]/beta1[j-1]\n",
    "\n",
    "print \"the values of T from T1 to T4 in Celsius by TDMA method are\"\n",
    "for i in range(1,5):\n",
    "    print x[i-1]-T_surr\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exa 5.8 Page 94"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "the values of conc. at x=.5mm or at the 50th node is 0.730764441227\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "from math import pi,sqrt\n",
    "from numpy import zeros\n",
    "\n",
    "lnght=.001\n",
    "k_const=.001\n",
    "D=10**-9\n",
    "delta_x=l/100\n",
    "C=1                 #in mol/m3\n",
    "#B.C. are C=1 at x=0\n",
    "#         dC/dx=0 at x=10**-3    since at the end point conc. is const.\n",
    "#using central difference method we get the following eqns which can be solved using TDMA method\n",
    "a=[0]\n",
    "for i in range(2,100):\n",
    "    a.append(1)            #sub diagonal assignment\n",
    "\n",
    "a.append(2)                  #since C99=C100 using B.C. \n",
    "\n",
    "b=[]\n",
    "for j in range(1,101):\n",
    "    b.append(-2.0001)           #main diagonal assignment\n",
    "\n",
    "c=[]\n",
    "for k in range(1,100):\n",
    "    c.append(1)            #super diagonal assignment\n",
    "\n",
    "d=[-1]\n",
    "for l in range(2,101):\n",
    "    d.append(0)\n",
    "          #given values assignment\n",
    "i=1#\n",
    "n=100#\n",
    "beta1 =[b[i-1]]                #initial b is equal to beta since a1=0\n",
    "gamma1=[d[i-1]/beta1[i-1]]      #since c7=0\n",
    "m=i+1#\n",
    "for j in range(m,n+1):\n",
    "    beta1.append(b[j-1]-a[j-1]*c[j-2]/beta1[j-2])\n",
    "    gamma1.append((d[j-1]-a[j-1]*gamma1[j-2])/beta1[j-1])\n",
    "\n",
    "x=zeros(n)    \n",
    "x[n-1]=gamma1[n-1]               #since c7=0\n",
    "n1=n-i#\n",
    "for k in range(1,n1+1):\n",
    "    j=n-k# \n",
    "    x[j-1]=gamma1[j-1]-c[j-1]*x[j]/beta1[j-1]\n",
    "\n",
    "print \"the values of conc. at x=.5mm or at the 50th node is\",x[49]"
   ]
  }
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