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"source": [
"# Chapter 14 - Boundary value and eigen value problems"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. 14_01 Pg No. 467"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": false
},
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{
"name": "stdout",
"output_type": "stream",
"text": [
"B1 = 7.75\n",
"Since B1 is less than B , let z(1) = y(1) = 4*(M2)\n",
"B2 = 9.75\n",
"Since B2 is larger than B ,let us have third estimate of z(1) = M3 \n",
"B3 = 9.0\n",
"The solution is [2, 4.375, 9.0]\n"
]
}
],
"source": [
"#Shooting Method\n",
"\n",
"def heun(f,x0,y0,z0,h,xf):\n",
" x = [x0] #\n",
" global y\n",
" y = [y0] #\n",
" z = [z0] #\n",
" n = (xf - x0)/h\n",
" m1=[0,0];m2=[0,0];m=[0,0]\n",
" for i in range(0,int(n)):\n",
" m1[0] = z[i] \n",
" m1[1] = f(x[i],y[i])\n",
" m2[0] = z[i] + h*m1[1]\n",
" m2[1] = f(x[i]+h,y[i]+h*m1[0])\n",
" m[0] = (m1[0] + m2[0])/2 \n",
" m[1] = ( m1[1] + m2[1] )/2\n",
" x.append(x[(i)] + h)\n",
" y.append(y[(i)] + h*m[0])\n",
" z.append(z[(i)] + h*m[1])\n",
" \n",
" B = y[int(n)]\n",
" return B\n",
"\n",
"\n",
"def f(x,y):\n",
" F = 6*x\n",
" return F\n",
"\n",
"x0 = 1 #\n",
"y0 = 2 #\n",
"h = 0.5 #\n",
"z0 = 2\n",
"M1 = z0 \n",
"xf = 2\n",
"B = 9\n",
"B1 = heun(f,x0,y0,z0,h,xf)\n",
"print 'B1 = ',B1\n",
"\n",
"if B1 != B:\n",
" print 'Since B1 is less than B , let z(1) = y(1) = 4*(M2)'\n",
" z0 = 4\n",
" M2 = z0\n",
" B2 = heun(f,x0,y0,z0,h,xf)\n",
" print 'B2 = ',B2\n",
" if B2 != B:\n",
" print 'Since B2 is larger than B ,let us have third estimate of z(1) = M3 '\n",
" M3 = M2 - (B2 - B)*(M2 - M1)/(B2 - B1)\n",
" z0 = M3 \n",
" B3= heun(f,x0,y0,z0,h,xf)\n",
" print 'B3 = ',B3\n",
" print 'The solution is ',y"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. 14_02 Pg No. 470"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
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"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" \n",
" The solution is \n",
" y1 = y(0.25) = -0.100203 \n",
" y2 = y(0.5) = -0.137907 \n",
" y3 = y(0.75) = -0.109079 \n",
" \n"
]
}
],
"source": [
"from numpy import array,exp,zeros,vstack,hstack,linalg\n",
"#Finite Difference Method\n",
"\n",
"def d2y(x):\n",
" D2Y = exp(x**2)\n",
" return D2Y\n",
"x_1 = 0#\n",
"y_0 = 0 #\n",
"y_1 = 0 #\n",
"h = 0.25\n",
"xf = 1\n",
"n = (xf-x_1)/h\n",
"A=zeros([int(n-1),int(n-1)])\n",
"B=zeros([int(n-1),1])\n",
"for i in range(0,int(n-1)):\n",
" A[i,:] = [1, -2, 1]\n",
" B[i,0] = exp((x_1 + i*h)**2)*h**2\n",
"\n",
"A[0,0] = 0 # #since we know y0 and y4\n",
"A[2,2] = 0 #\n",
"A[0,:] = hstack([ A[0,1:3], [0]]) #rearranging terms\n",
"A[2,0:3] = hstack([[ 0], A[2,0:2]]) \n",
"C=linalg.solve(A,B)\n",
"print ' \\n The solution is \\n y1 = y(0.25) = %f \\n y2 = y(0.5) = %f \\n y3 = y(0.75) = %f \\n '%(C[0],C[1],C[2]) "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. 14_03 Pg No. 473"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" The roots are \n",
" lamda1 = 4.000000 \n",
" lamda2 = 6.000000 \n",
" \n",
"X1 = Matrix([[1.00000000000000], [1]])\n",
"X2 = Matrix([[2.00000000000000], [1]])\n"
]
}
],
"source": [
"#Eigen Vectors\n",
"from numpy import eye\n",
"from sympy import symbols,Matrix,det,solve\n",
"A = [[8 ,-4],[ 2, 2 ] ]\n",
"A=Matrix(A)\n",
"lamd = symbols('lamd')\n",
"p = det(A - lamd*eye(2))\n",
"root = solve(p,lamd)\n",
"print '\\n The roots are \\n lamda1 = %f \\n lamda2 = %f \\n '%(root[0],root[1])\n",
"A1 = A - root[0]*eye(2)\n",
"X1 = Matrix([[-1*A1[0,1]/A1[0,0]],[1]])\n",
"print 'X1 = ',X1\n",
"A2 = A - root[1]*eye(2)\n",
"X2 = Matrix([[-1*A2[0,1]/A2[0,0]],[ 1]])\n",
"print 'X2 = ',X2"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. 14_04 Pg No. 474"
]
},
{
"cell_type": "code",
"execution_count": 28,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
"A2 = \n",
"[[-5. 0. 0.]\n",
" [ 1. -8. 9.]\n",
" [ 0. 4. -9.]]\n",
"\n",
"p2 = -11.000000\n",
"\n",
"\n",
"A3 = \n",
"[[ -6. 0. 0.]\n",
" [ 1. -6. 27.]\n",
" [ 0. 8. -6.]]\n",
"\n",
"p3 = -6.000000\n",
"\n"
]
}
],
"source": [
"from numpy import array,shape,trace,poly\n",
"#Fadeev - Leverrier method\n",
"\n",
"A = [[ -1, 0, 0],[ 1, -2, 3],[ 0, 2, -3 ]]\n",
"A=array(A)\n",
"r,c= shape(A)[0],shape(A)[1]\n",
"A1 = A\n",
"p= [trace(A1)]\n",
"for i in range(1,r):\n",
" A1 = A*( A1 - p[(i-1)]*eye(3))\n",
" p.append(trace(A1)/(i+1))\n",
" print '\\nA%d = '%(i+1)\n",
" print A1\n",
" print '\\np%d = %f\\n'%((i+1),p[i])"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. 14_05 Pg No. 476"
]
},
{
"cell_type": "code",
"execution_count": 46,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"\n",
" Eigen vector - 1 \n",
" for lamda1 = 0.000000 \n",
" X1 = \n",
"[-inf nan nan]\n",
"\n",
" Eigen vector - 2 \n",
" for lamda2 = 0.000000 \n",
" X2 = \n",
"[ 0. 0. 0.]\n",
"\n",
" Eigen vector - 3 \n",
" for lamda3 = 0.000000 \n",
" X3 = \n",
"[ 0. 0. -1.]\n"
]
}
],
"source": [
"import numpy as np\n",
"np.seterr(divide='ignore', invalid='ignore')\n",
"from __future__ import division\n",
"from numpy.linalg import eig\n",
"from numpy import array,diag\n",
"#Eigen Vectors\n",
"\n",
"A = [[ -1, 0, 0],[1, -2, 3],[0, 2, -3]]\n",
"A=array(A)\n",
"evectors,evalues = eig(A)\n",
"evectors=diag(evectors)\n",
"for i in range(0,3):\n",
" print '\\n Eigen vector - %d \\n for lamda%d = %f \\n X%d = '%(i+1,i+1,evalues[0,0],i+1)\n",
" evectors[:,0] = evectors[:,0]/evectors[1,i]\n",
" print evectors[:,i]"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Example No. 14_06 Pg No. 478"
]
},
{
"cell_type": "code",
"execution_count": 48,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Iterations:\n",
" 0 1 2 3 4 5 6 7 \n",
"X = [[ 0. 1. 0.8 1. 0.97560976 1.\n",
" 0.99726027 1. ]\n",
" [ 1. 0.5 1. 0.92857143 1. 0.99180328\n",
" 1. 0.99908592]\n",
" [ 0. 0. 0. 0. 0. 0. 0.\n",
" 0. ]]\n",
"\n",
"Y = [[None 2.0 2.0 2.8 2.8571428571428577 2.975609756097561 2.9836065573770494\n",
" 2.9972602739726026]\n",
" [None 1.0 2.5 2.6 2.928571428571429 2.951219512195122 2.9918032786885247\n",
" 2.9945205479452053]\n",
" [None 0.0 0.0 0.0 0.0 0.0 0.0 0.0]]\n"
]
}
],
"source": [
"from numpy import array, zeros,expand_dims, dot, hstack\n",
"A = array([[ 1, 2, 0],[2, 1 ,0],[0, 0, -1 ]])\n",
"X=zeros([3,8])\n",
"Y=zeros([3,7])\n",
"X[:,0] =[0,1,0]\n",
"\n",
"for i in range(1,8):\n",
" Y[:,i-1] = [xx for xx in A.dot(expand_dims(X[:,i-1], axis=1))]\n",
" \n",
" X[:,i] = Y[:,i-1]/max(Y[:,i-1])\n",
"\n",
"print 'Iterations:\\n 0 1 2 3 4 5 6 7 '\n",
"print 'X = ',X\n",
"Y=hstack([[[None],[None],[None]],Y])\n",
"print '\\nY = ',Y\n"
]
}
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