Added(A)/Deleted(D) following books

 A  Engineering_Thermodynamics_by_Dr._S._S._Khandare/README.txt
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter1.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter10.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter13.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter2.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter21.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter22.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter23.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter24.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter26.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter27.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter28.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter34.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter35.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter4.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter5.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter6.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/chapter9.ipynb
A  Higher_Engineering_Mathematics_by_B._S._Grewal/screenshots/screen1.png
A  Higher_Engineering_Mathematics_by_B._S._Grewal/screenshots/screen2.png
A  Higher_Engineering_Mathematics_by_B._S._Grewal/screenshots/screen3.png
A  sample_notebooks/ManikandanD/Chapter_2_Light_propagation_in_optical_fiber.ipynb

1 files changed, 435 insertions, 0 deletions

diff --git a/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter28.ipynb b/Higher_Engineering_Mathematics_by_B._S._Grewal/chapter28.ipynb
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+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Chapter 28: Numerical Solution Of Partial Differential Equations"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 28.1, page no. 725"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "D=Bˆ2−4AC\n",
+      "if D<0 then elliptic if D=0 then parabolic if D>0 then hyperboic\n",
+      "(i) A=xˆ2, B1−yˆ2 D=4ˆ2−4∗1∗4=0 so the equation is PARABOLIC \n",
+      "(ii) D=4xˆ2(yˆ2−1)\n",
+      "for −inf<x<inf and −1<y<1 D<0 \n",
+      "So the equation is ELLIPTIC \n",
+      "(iii) A=1+xˆ2, B=5+2xˆ2, C=4+xˆ2\n",
+      "D=9>0\n",
+      "So the equation is HYPERBOLIC \n"
+     ]
+    }
+   ],
+   "source": [
+    "print \"D=Bˆ2−4AC\"\n",
+    "print \"if D<0 then elliptic if D=0 then parabolic if D>0 then hyperboic\"\n",
+    "print \"(i) A=xˆ2, B1−yˆ2 D=4ˆ2−4∗1∗4=0 so the equation is PARABOLIC \"\n",
+    "print \"(ii) D=4xˆ2(yˆ2−1)\"\n",
+    "print \"for −inf<x<inf and −1<y<1 D<0 \"\n",
+    "print \"So the equation is ELLIPTIC \"\n",
+    "print \"(iii) A=1+xˆ2, B=5+2xˆ2, C=4+xˆ2\"\n",
+    "print \"D=9>0\"\n",
+    "print \"So the equation is HYPERBOLIC \""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 28.2, page no. 726"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 5,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "See figure in question\n",
+      "From symmetry u7=u1, u8=u2, u9=u3, u3=u1, u6=u4, u9=u7\n",
+      "u5=1/4∗(2000+2000+1000+1000)=1500\n",
+      "u1=1/4(0=1500+1000+2000)=1125\n",
+      "u2=1/4∗(1125+1125+1000+1500)=1188\n",
+      "u4=1/4(2000+1500+1125+1125)=1438\n",
+      "1125 1188 1438 1500\n",
+      "Iterations :\n",
+      "\n",
+      "1301 1414 1164 1031\n",
+      "\n",
+      "1250 1337 1087 1019\n",
+      "\n",
+      "1199 1312 1062 981\n",
+      "\n",
+      "1174 1287 1037 968\n",
+      "\n",
+      "1155 1274 1024 956\n",
+      "\n",
+      "1144 1265 1015 949\n"
+     ]
+    }
+   ],
+   "source": [
+    "print \"See figure in question\"\n",
+    "print \"From symmetry u7=u1, u8=u2, u9=u3, u3=u1, u6=u4, u9=u7\"\n",
+    "print \"u5=1/4∗(2000+2000+1000+1000)=1500\"\n",
+    "u5 = 1500\n",
+    "print \"u1=1/4(0=1500+1000+2000)=1125\"\n",
+    "u1 = 1125\n",
+    "print \"u2=1/4∗(1125+1125+1000+1500)=1188\"\n",
+    "u2 = 1188\n",
+    "print \"u4=1/4(2000+1500+1125+1125)=1438\"\n",
+    "u4 = 1438\n",
+    "print u1,u2,u4,u5 \n",
+    "print \"Iterations :\"\n",
+    "for i in range(1,7):\n",
+    "    u11 = (1000+u2+500+u4)/4\n",
+    "    u22 = (u11+u1+1000+u5)/4 \n",
+    "    u44 = (2000+u5+u11+u1)/4\n",
+    "    u55 = (u44+u4+u22+u2)/4\n",
+    "    print \"\"\n",
+    "    print u55,u44,u22,u11\n",
+    "    u1 = u11 \n",
+    "    u2 = u22 \n",
+    "    u4 = u44 \n",
+    "    u5 = u55"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 28.3, page no. 727"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "See figure in question\n",
+      "To find the initial values of u1 u2 u3 u4 we assume u4=0 \n",
+      "u1=1/4∗(1000+0+1000+2000)=1000\n",
+      "u2=1/4(1000+500+1000+500)=625 \n",
+      "u3=1/4∗(2000+0+1000+500)=875\n",
+      "u4=1/4(875+0+625+0)=375\n",
+      "1000 625 875 375\n",
+      "Iterations:\n",
+      "\n",
+      "437 1000 750 1125\n",
+      "\n",
+      "453 1031 781 1187\n",
+      "\n",
+      "457 1039 789 1203\n",
+      "\n",
+      "458 1041 791 1207\n",
+      "\n",
+      "458 1041 791 1208\n",
+      "\n",
+      "458 1041 791 1208\n"
+     ]
+    }
+   ],
+   "source": [
+    "print \"See figure in question\"\n",
+    "print \"To find the initial values of u1 u2 u3 u4 we assume u4=0 \"\n",
+    "print \"u1=1/4∗(1000+0+1000+2000)=1000\"\n",
+    "u1 = 1000\n",
+    "print \"u2=1/4(1000+500+1000+500)=625 \"\n",
+    "u2 = 625\n",
+    "print \"u3=1/4∗(2000+0+1000+500)=875\"\n",
+    "u3 = 875\n",
+    "print \"u4=1/4(875+0+625+0)=375\"\n",
+    "u4 = 375\n",
+    "print u1,u2,u3,u4 \n",
+    "print \"Iterations:\"\n",
+    "for i in range(1,7):\n",
+    "    u11 = (2000+u2+1000+u3)/4 \n",
+    "    u22 = (u11+500+1000+u4)/4\n",
+    "    u33 = (2000+u4+u11+500)/4\n",
+    "    u44 = (u33+0+u22+0)/4\n",
+    "    print \"\"\n",
+    "    print u44,u33,u22,u11\n",
+    "    u1 = u11 \n",
+    "    u2 = u22 \n",
+    "    u4 = u44 \n",
+    "    u3 = u33"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 28.5, page no. 729"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 10,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Here cˆ2=4, h=1, k=1/8, therefore alpha=(cˆ2)∗k/(hˆ2)\n",
+      "Using bendre−schmidits recurrence relation ie u(i)(j +1)=t∗u(i−1)(j)+t∗u(i+1)(j)+(1−2t)∗u(i,j)\n",
+      "Now since u(0,t)=0=u(8,t) therefore u(0,i)=0 and u(8,j)=0 and u(x,0)=4x−1/2xˆ2 \n",
+      "0.0\n",
+      "-4.0\n",
+      "0.0\n",
+      "4.0\n",
+      "8.0\n",
+      "12.0\n",
+      "16.0\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy\n",
+    "print \"Here cˆ2=4, h=1, k=1/8, therefore alpha=(cˆ2)∗k/(hˆ2)\"\n",
+    "print \"Using bendre−schmidits recurrence relation ie u(i)(j +1)=t∗u(i−1)(j)+t∗u(i+1)(j)+(1−2t)∗u(i,j)\"\n",
+    "print \"Now since u(0,t)=0=u(8,t) therefore u(0,i)=0 and u(8,j)=0 and u(x,0)=4x−1/2xˆ2 \"\n",
+    "c = 2\n",
+    "h = 1\n",
+    "k = 1/8\n",
+    "t = (c**2)*k/(h**2)\n",
+    "A = numpy.ones((9,9))\n",
+    "for i in range(0,9):\n",
+    "    for j in range (0,9):\n",
+    "        A[0,i] = 0\n",
+    "        A[8,i] = 0\n",
+    "        A[i,0] = 4*(i-1)-1/2*(i-1)**2\n",
+    "for i in range(1,8):\n",
+    "    for j in range(1,7):\n",
+    "        A[i,j] = t*A[i-1,j-1]+t*A[i+1,j-1]+(1-2*t)*A[i-1,j-1]\n",
+    "for i in range(1,8):\n",
+    "    j = 2\n",
+    "    print A[i,j]"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 28.6, page no. 730"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Here cˆ2=1, h=1/3, k=1/36, therefore t=(cˆ2)∗k/(hˆ2)=1/4 \n",
+      "So bendre−schmidits recurrence relation ie u(i)(j+1)=1/4(u(i−1)(j)+u(i+1)(j)+2u(i,j)\n",
+      "Now since u(0,t)=0=u(1,t) therefore u(0,i)=0 and u(1,j)=0 and u(x,0)=sin(%pi)x\n",
+      "-0.433012701892\n",
+      "0.216506350946\n",
+      "0.649519052838\n",
+      "0.433012701892\n",
+      "-0.216506350946\n",
+      "-0.649519052838\n",
+      "-0.433012701892\n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy,math\n",
+    "\n",
+    "print \"Here cˆ2=1, h=1/3, k=1/36, therefore t=(cˆ2)∗k/(hˆ2)=1/4 \"\n",
+    "print \"So bendre−schmidits recurrence relation ie u(i)(j+1)=1/4(u(i−1)(j)+u(i+1)(j)+2u(i,j)\"\n",
+    "print \"Now since u(0,t)=0=u(1,t) therefore u(0,i)=0 and u(1,j)=0 and u(x,0)=sin(%pi)x\"\n",
+    "c = 1.\n",
+    "h = 1./3\n",
+    "k = 1./36\n",
+    "t = (c**2)*k/(h**2)\n",
+    "A = numpy.ones((9,9))\n",
+    "for i in range(0,9):\n",
+    "    for j in range(0,9):\n",
+    "        A[0,i] = 0\n",
+    "        A[1,i] = 0\n",
+    "        A[i,0] = math.sin(math.pi/3*(i-1)) \n",
+    "for i in range(1,8):\n",
+    "    for j in range(1,8):\n",
+    "        A[i,j] = t*A[i-1,j-1]+t*A[i+1,j-1]+(1-2*t)*A[i-1,j-1]\n",
+    "for i in range(1,8):\n",
+    "    j = 1\n",
+    "    print A[i,j]"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 28.7, page no. 732"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Here cˆ2=16, taking h=1, finding k such that cˆ2tˆ2=1 \n",
+      "So bendre−schmidits recurrence relation ie u(i)(j+1)=(16tˆ2(u(i−1)(j)+u(i+1)(j))+2(1−16∗tˆ2u(i,j)−u(i)(j−1)\n",
+      "Now since u(0,t)=0=u(5,t) therefore u(0,i)=0 and u(5,j)=0 and u(x,0)=xˆ2(5−x)\n",
+      "Also from 1st derivative (u(i)(j+1)−u(i,j−1))/2k=g(x) and g(x)=0 in this case\n",
+      "So if j=0 this gives u(i)(1)=1/2∗(u(i−1)(0)+u(i+1)(0))\n",
+      "   0.0    0.0    0.0    0.0    0.0 \n",
+      "   0.0    4.0    8.0    12.0    16.0 \n",
+      "   0.0    12.0    24.0    36.0    48.0 \n",
+      "   0.0    18.0    36.0    54.0    72.0 \n",
+      "   0.0    16.0    0.0    0.0    0.0 \n"
+     ]
+    }
+   ],
+   "source": [
+    "import numpy\n",
+    "\n",
+    "print \"Here cˆ2=16, taking h=1, finding k such that cˆ2tˆ2=1 \"\n",
+    "print \"So bendre−schmidits recurrence relation ie u(i)(j+1)=(16tˆ2(u(i−1)(j)+u(i+1)(j))+2(1−16∗tˆ2u(i,j)−u(i)(j−1)\"\n",
+    "print \"Now since u(0,t)=0=u(5,t) therefore u(0,i)=0 and u(5,j)=0 and u(x,0)=xˆ2(5−x)\"\n",
+    "c = 4\n",
+    "h =1 \n",
+    "k = (h/c)\n",
+    "t = k/h\n",
+    "A = numpy.zeros((6,6))\n",
+    "print \"Also from 1st derivative (u(i)(j+1)−u(i,j−1))/2k=g(x) and g(x)=0 in this case\"\n",
+    "print \"So if j=0 this gives u(i)(1)=1/2∗(u(i−1)(0)+u(i+1)(0))\"\n",
+    "for i in range(0,6):\n",
+    "    for j in range(1,9):\n",
+    "        A[0,i] = 0\n",
+    "        A[5,i] = 0;\n",
+    "        A[i,1] = (i)**2*(5-i)\n",
+    "for i in range(0,4):\n",
+    "    A[i+1,2] =1/2*(A[i,1]+A[i+2,1])\n",
+    "for i in range(2,5):\n",
+    "    for j in range(2,5):\n",
+    "        A[i-1,j] = (c*t)**2*(A[i-2,j-1]+A[i,j-1])+2*(1-(c*t)**2)*A[i-1,j-1]-A[i-1,j-2]\n",
+    "for i in range(0,5):\n",
+    "    for j in range(0,5):\n",
+    "        print  \"  \",A[i,j],\n",
+    "    print \"\""
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Example 28.8, page no. 734"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 16,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Here cˆ2=4, taking h=1, finding k such that cˆ2tˆ2=1 \n",
+      "So bendre−schmidits recurrence relation ie u(i)(j+1)=(16tˆ2(u(i−1)(j)+u(i+1)(j))+2(1−16∗tˆ2u(i,j)−u(i)(j−1)\n",
+      "Now since u(0,t)=0=u(4,t) therefore u(0,i)=0 and u(4,j)=0 and u(x,0)=x(4−x) \n",
+      "Also from 1st derivative (u(i)(j+1)−u(i,j−1))/2 k=g(x) and g(x)=0 in this case\n",
+      "So if j=0 this gives u(i)(1)=1/2∗(u(i−1)(0)+u(i+1)(0))\n",
+      "   0.0    0.0    0.0    0.0    0.0 \n",
+      "   3.0    0.0    -3.0    -6.0    -9.0 \n",
+      "   4.0    0.0    -4.0    -8.0    -12.0 \n",
+      "   3.0    0.0    -3.0    -6.0    -9.0 \n",
+      "   0.0    0.0    0.0    0.0    0.0 \n"
+     ]
+    }
+   ],
+   "source": [
+    "print \"Here cˆ2=4, taking h=1, finding k such that cˆ2tˆ2=1 \"\n",
+    "print \"So bendre−schmidits recurrence relation ie u(i)(j+1)=(16tˆ2(u(i−1)(j)+u(i+1)(j))+2(1−16∗tˆ2u(i,j)−u(i)(j−1)\"\n",
+    "print \"Now since u(0,t)=0=u(4,t) therefore u(0,i)=0 and u(4,j)=0 and u(x,0)=x(4−x) \"\n",
+    "c = 2\n",
+    "h = 1\n",
+    "k = (h/c)\n",
+    "t = k/h\n",
+    "A = numpy.zeros((6,6))\n",
+    "print \"Also from 1st derivative (u(i)(j+1)−u(i,j−1))/2 k=g(x) and g(x)=0 in this case\"\n",
+    "print \"So if j=0 this gives u(i)(1)=1/2∗(u(i−1)(0)+u(i+1)(0))\"\n",
+    "for i in range(0,6):\n",
+    "    for j in range(1,9):\n",
+    "        A[0,i] = 0\n",
+    "        A[4,i] = 0\n",
+    "        A[i,0] = (i)*(4-i)\n",
+    "for i in range(0,4):\n",
+    "    A[i+1,2] = 1/2*(A[i,1]+A[i+2,1])\n",
+    "for i in range(2,5):\n",
+    "    for j in range(2,5):\n",
+    "        A[i-1,j] = (c*t)**2*(A[i-2,j-1]+A[i,j-1])+2*(1-(c*t)**2)*A[i-1,j-1]-A[i-1,j-2]\n",
+    "for i in range (0,5):\n",
+    "    for j in range(0,5):\n",
+    "        print \"  \",A[i,j],\n",
+    "    print \"\""
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
+   "version": "2.7.10"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}