{
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{
"cell_type": "markdown",
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"source": [
"# Chapter 15 - Solution of partial differential equations"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. 15_01 Pg No. 488"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" The solution of the system is \n",
" f1 = 75 \n",
" f2 = 50 \n",
" f3 = 50 \n",
" f4 = 24 \n",
" \n"
]
}
],
"source": [
"from numpy import zeros,mat\n",
"from numpy.linalg import solve\n",
"#Elliptic Equations\n",
"\n",
"l = 15\n",
"h = 5\n",
"n = 1 + 15/5\n",
"f=zeros([4,4])\n",
"f[0,0:3]=100\n",
"f[0:3,0]=100\n",
"\n",
"#At point 1 : f2 + f3 - 4f1 + 100 + 100 = 0\n",
"#At point 2 : f1 + f4 - 4f2 + 100 + 0 = 0\n",
"#At point 3 : f1 + f4 - 4f3 + 100 + 0 = 0\n",
"#At point 4 : f2 + f3 - 4f4 + 0 + 0 = 0\n",
"#\n",
"#Final Equations are\n",
"# -4f1 + f2 + f3 + 0 = -200\n",
"# f1 - 4f2 + 0 + f4 = -100\n",
"# f1 + 0 - 4f3 + f4 = -100\n",
"# 0 + f2 + f3 - 4f4 = 0\n",
"A = mat([[ -4, 1 ,1 ,0],[1, -4, 0 ,1],[1, 0 ,-4 ,1],[0, 1, 1, -4 ]])\n",
"B = mat([[-200],[-100],[-100],[0]])\n",
"C = solve(A,B)\n",
"print '\\n The solution of the system is \\n f1 = %d \\n f2 = %d \\n f3 = %d \\n f4 = %d \\n '%(C[0],C[1],C[2],C[3])"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. 15_02 Pg No. 489"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"First row of below matrix represents iteration number:\n",
"[[ 0. 1. 2. 3. 4.]\n",
" [ 75. 75. 75. 75. 75.]\n",
" [ 0. 0. 0. 0. 0.]\n",
" [ 0. 0. 0. 0. 0.]\n",
" [ 0. 50. 50. 50. 50.]]\n"
]
}
],
"source": [
"from numpy import zeros,mat\n",
"from numpy.linalg import solve\n",
"\n",
"#Liebmann's Iterative method\n",
"\n",
"f=zeros([4,4])\n",
"f[0,0:3]=100\n",
"f[0:3,0]=100\n",
"A=zeros([5,5])\n",
"A[0,0:5] = range(0,5)\n",
"for n in range(1,6):\n",
" for i in range(2,4):\n",
" for j in range(2,4):\n",
" if n == 1 and i == 2 and j == 2 :\n",
" f[i-1,j-1] = ( f[i,j] + f[i-2,j-2] + f[i-2,j] + f[i,j-2] )/4\n",
" else:\n",
" f[i,j] = ( f[i,j-1] + f[i-2,j-1] + f[i-1,j]+ f[i-1,j-2] )/4\n",
" A[1:5,n-1] = mat([f[1,1],f[1,2],f[2,1],f[2,2] ])\n",
"\n",
"\n",
"print 'First row of below matrix represents iteration number:\\n',A"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. 15_03 Pg No. 490"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The solution is \n",
" f1 = -5.500000 \n",
" f2 = -10.750000 \n",
" f3 = -3.250000 \n",
" f4 = -5.500000 \n",
" \n"
]
}
],
"source": [
"from numpy import zeros,mat\n",
"from numpy.linalg import solve\n",
"\n",
"#Poisson's Equation\n",
"\n",
"#D2f = 2*x**2 * y**2\n",
"# f = 0\n",
"# h = 1 \n",
"#Point 1 : 0 + 0 + f2 + f3 - 4f1 = 2(1)**2 * 2**2\n",
"# f2 + f3 - 4f1 = 8\n",
"#Point 2 : 0 + 0 + f1 + f4 -4f2 = 2*(2)**2*2**2\n",
"# f1 - 4f2 = f4 = 32\n",
"#Point 3 : 0 + 0 + f1 + f4 - 4f4 = 2*(1**2)*1**2\n",
"# f1 -4f3 + f4 = 2\n",
"#Point 4 : 0 + 0 + f2 + f3 - 4f4 = 2* 2**2 * 1**2\n",
"# f2 + f3 - 4f4 = 8\n",
"#Rearranging the equations\n",
"# -4f1 + f2 + f3 = 8\n",
"# f1 - 4f2 + f4 = 32\n",
"# f1 - 4f3 + f4 = 2\n",
"# f2 + f3 - 4f4 = 8\n",
"A = mat([[ -4, 1, 1, 0],[1, -4 ,0 ,1],[1, 0, -4, 1],[0, 1, 1, -4]])\n",
"B = mat([[ 8],[32],[2],[8 ]])\n",
"C = solve(A,B)\n",
"print 'The solution is \\n f1 = %f \\n f2 = %f \\n f3 = %f \\n f4 = %f \\n '%( C[0],C[1],C[2],C[3])"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. 15_04 Pg No. 491"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Iteration 1\n",
" f1 = -2.000000, f2 = -9.000000, f3 = -2.000000, f4 = -2.000000\n",
"\n",
"\n",
" Iteration 2\n",
" f1 = -5.000000, f2 = -11.000000, f3 = -3.000000, f4 = -5.000000\n",
"\n",
"\n",
" Iteration 3\n",
" f1 = -6.000000, f2 = -11.000000, f3 = -4.000000, f4 = -6.000000\n",
"\n",
"\n",
" Iteration 4\n",
" f1 = -6.000000, f2 = -11.000000, f3 = -4.000000, f4 = -6.000000\n",
"\n"
]
}
],
"source": [
"#Gauss-Seidel Iteration\n",
"\n",
"f2 = 0\n",
"f3 = 0\n",
"for i in range(1,5):\n",
" f1 = (f2 + f3 - 8)/4\n",
" f4 = f1 \n",
" f2 = (f1 + f4 -32)/4\n",
" f3 = (f1 + f4 - 2)/4\n",
" print '\\n Iteration %d\\n f1 = %f, f2 = %f, f3 = %f, f4 = %f\\n'%(i,f1,f2,f3,f4)\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. 15_05 Pg No. 494"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The final results are \n",
"[[ 0. 150. 100. 50. 0. ]\n",
" [ 0. 50. 100. 50. 0. ]\n",
" [ 0. 50. 50. 50. 0. ]\n",
" [ 0. 25. 50. 25. 0. ]\n",
" [ 0. 25. 25. 25. 0. ]\n",
" [ 0. 12.5 25. 12.5 0. ]\n",
" [ 0. 12.5 12.5 12.5 0. ]]\n"
]
}
],
"source": [
"from numpy import zeros,mat\n",
"\n",
"#Initial Value Problems\n",
"\n",
"h = 1 #\n",
"k = 2 #\n",
"tau = h**2/(2*k)\n",
"f=zeros([7,5])\n",
"for i in range(1,5):\n",
" f[0,i] = 50*( 4 - (i) )\n",
"\n",
"f[0:7,0] = 0\n",
"f[0:7,4] = 0 \n",
"for j in range(0,6):\n",
" for i in range(1,4):\n",
" f[j+1,i] = ( f[j,i-1] + f[j,i+1] )/2 \n",
"print 'The final results are \\n',f"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. 15_06 Pg No. 497"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The final solution using crank nicholson implicit method is :\n",
"[[ 0. 150. 100. 50. 0. ]\n",
" [ 0. 42.85714286 71.42857143 42.85714286 0. ]\n",
" [ 0. 26.53061224 34.69387755 26.53061224 0. ]\n",
" [ 0. 13.70262391 20.11661808 13.70262391 0. ]\n",
" [ 0. 7.70512287 10.70387339 7.70512287 0. ]]\n"
]
}
],
"source": [
"from numpy import zeros,mat,linalg\n",
"#Crank-Nicholson Implicit Method\n",
"\n",
"h = 1 #\n",
"k = 2 #\n",
"tau = h**2/(2*k)\n",
"f=zeros([5,5])\n",
"for i in range(1,4):\n",
" f[0,i] = 50*( 4 - (i) )\n",
"f[0:5,0] = 0 \n",
"f[0:5,4] = 0 \n",
"A = mat([[4, -1, 0 ],[-1 , 4 , -1 ],[0 , -1 , 4]])\n",
"B=zeros([3,1])\n",
"for j in range(0,4):\n",
" for i in range(1,4):\n",
" B[i-1,0] = f[j,i-1] + f[j,i+1]\n",
" \n",
" C = linalg.solve(A,B)\n",
" f[j+1,1] = C[0]\n",
" f[j+1,2] = C[1]\n",
" f[j+1,3] = C[2]\n",
"\n",
"print 'The final solution using crank nicholson implicit method is :\\n',f"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. 15_07 Pg No. 500"
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"The values estimated are :\n",
"[[ 0. 4. 6. 6. 4. 0.]\n",
" [ 0. 3. 5. 5. 3. 0.]\n",
" [ 0. 1. 2. 2. 1. 0.]\n",
" [ 0. -1. -2. -2. -1. 0.]\n",
" [ 0. -3. -5. -5. -3. 0.]\n",
" [ 0. -4. -6. -6. -4. 0.]]\n"
]
}
],
"source": [
"from __future__ import division\n",
"from numpy import sqrt\n",
"#Hyperbolic Equations\n",
"\n",
"h = 1\n",
"Tbyp = 4\n",
"tau = sqrt(h**2 /4)\n",
"r = int(1+(2.5 - 0)/tau)\n",
"c = int(1+(5 - 0)/h)\n",
"\n",
"f=zeros([6,6])\n",
"for i in range(1,int(c)-1):\n",
" f[0,i] = (i)*(5 - (i) )\n",
"\n",
"f[0:r-1,0] = 0\n",
"f[0:r-1,c-1] = 0\n",
"g=[]\n",
"for i in range(1,c-1):\n",
" g.append(0)\n",
" f[1,i] = (f[0,i+1] + f[0,i-1])/2 + tau*g[i-1] \n",
"\n",
"for j in range(1,r-1):\n",
" for i in range(1,c-1):\n",
" f[j+1,i] = -f[j-1,i] + f[j,i+1] + f[j,i-1]\n",
" \n",
"\n",
"print 'The values estimated are :\\n',f"
]
}
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