def shoot(f, a, b, z1, z2, t, tol):
"""Implements the shooting method to solve second order BVPs
USAGE:
y = shoot(f, a, b, z1, z2, t, tol)
INPUT:
f - function dy/dt = f(y,t). Since we are solving a second-
order boundary-value problem that has been transformed
into a first order system, this function should return a
1x2 array with the first entry equal to y and the second
entry equal to y'.
a - solution value at the left boundary: a = y(t[0]).
b - solution value at the right boundary: b = y(t[n-1]).
z1 - first initial estimate of y'(t[0]).
z1 - second initial estimate of y'(t[0]).
t - array of n time values to determine y at.
tol - allowable tolerance on right boundary: | b - y[n-1] | < tol
OUTPUT:
y - array of solution function values corresponding to the
values in the supplied array t.
NOTE:
This function assumes that the second order BVP has been converted to
a first order system of two equations. The secant method is used to
refine the initial values of y' used for the initial value problems.
"""
from diffeq import rk4
max_iter = 25 # Maximum number of shooting iterations
n = len(t) # Determine the size of the arrays we will generate
# Compute solution to first initial value problem (IVP) with y'(a) = z1.
# Because we are using the secant method to refine our estimates of z =
# y', we don't really need all the solution of the IVP, just the last
# point of it -- this is saved in w1.
y = rk4(f, [a, z1], t)
w1 = y[n - 1, 0]
print "%2d: z = %10.3e, error = %10.3e" % (0, z1, b - w1)
# Begin the main loop. We will compute the solution of a second IVP and
# then use the both solutions to refine our estimate of y'(a). This
# second solution then replaces the first and a new "second" solution is
# generated. This process continues until we either solve the problem to
# within the specified tolerance or we exceed the maximum number of
# allowable iterations.
for i in xrange(max_iter):
# Solve second initial value problem, using y'(a) = z2. We need to
# retain the entire solution vector y since if y(t(n)) is close enough
# to b for us to stop then the first column of y becomes our solution
# vector.
y = rk4(f, [a, z2], t)
w2 = y[n - 1, 0]
print "%2d: z = %10.3e, error = %10.3e" % (i + 1, z2, b - w2)
# Check to see if we are done...
if abs(b - w2) < tol:
break
# Compute the new approximations to the initial value of the first
# derivative. We compute z2 using a linear fit through (z1,w1) and
# (z2,w2) where w1 and w2 are the estimates at t=b of the initial
# value problems solved above with y1'(a) = z1 and y2'(a) = z2. The
# new value for z1 is the old value of z2.
#z1, z2 = ( z2, z1 + ( z2 - z1 ) / ( w2 - w1 ) * ( b - w1 ) )
z1, z2 = (z2, z2 + (z2 - z1) / (w2 - w1) * (b - w2))
w1 = w2
# All done. Check to see if we really solved the problem, and then return
# the solution.
if abs(b - w2) >= tol:
print "\a**** ERROR ****"
print "Maximum number of iterations (%d) exceeded" % max_iter
print "Returned values may not have desired accuracy"
print "Error estimate of returned solution is %e" % (b - w2)
return y[:, 0]
"""
Created on Sat Sep 29 18:38:09 2018
@author: Ben
"""
import numpy as np
from matplotlib import pyplot as pyl
from diffeq import rk4
f = lambda T, t: -r * (T - Ts)
tmax = 60.
T0 = 10.
r = 0.1
Ts = 83.
n = 5
dt = np.zeros(n)
temp = np.zeros(n)
for j in range(0, n):
dt[j] = 1 / (j + 1)
nsteps = int(
tmax /
dt[j]) #the arrays will have different size for different time steps
T = T0
my_temp, my_time = rk4(f, 0, tmax, nsteps)
temp[j] = my_temp[10 * (j + 1)]
pyl.plot(dt, temp, 'o')
pyl.xlabel('dt')
pyl.ylabel('Temp.')
for j in range(0, 20): #arbitrarily increase the size
deltatarr.append(dt)
nsteps = int(
tmax /
dt) #the arrays will have different size for different time steps
my_time = np.linspace(dt, tmax, nsteps)
my_temp = np.zeros(nsteps)
T = T0
for i in range(1, nsteps):
T = euler(T, -r * (T - Ts), dt)
my_temp[i] = T
if dt * i == 10:
t10.append(T)
pyplot.plot(my_time, my_temp, ls='-', lw=0.5)
dt = dt / 2.
print(t10)
print(deltatarr)
pyplot.xlabel('time')
pyplot.ylabel('temperature')
pyplot.xlim(8, 10)
pyplot.ylim(48, 58)
fig2 = pyplot.figure(2)
ax2 = fig2.add_subplot(111)
ax2.scatter(deltatarr, t10)
m = diffeq.rk4(-r * (T - Ts))
def shoot( f, a, b, z1, z2, t, tol ):
"""Implements the shooting method to solve second order BVPs
USAGE:
y = shoot(f, a, b, z1, z2, t, tol)
INPUT:
f - function dy/dt = f(y,t). Since we are solving a second-
order boundary-value problem that has been transformed
into a first order system, this function should return a
1x2 array with the first entry equal to y and the second
entry equal to y'.
a - solution value at the left boundary: a = y(t[0]).
b - solution value at the right boundary: b = y(t[n-1]).
z1 - first initial estimate of y'(t[0]).
z1 - second initial estimate of y'(t[0]).
t - array of n time values to determine y at.
tol - allowable tolerance on right boundary: | b - y[n-1] | < tol
OUTPUT:
y - array of solution function values corresponding to the
values in the supplied array t.
NOTE:
This function assumes that the second order BVP has been converted to
a first order system of two equations. The secant method is used to
refine the initial values of y' used for the initial value problems.
"""
from diffeq import rk4
max_iter = 25 # Maximum number of shooting iterations
n = len( t ) # Determine the size of the arrays we will generate
# Compute solution to first initial value problem (IVP) with y'(a) = z1.
# Because we are using the secant method to refine our estimates of z =
# y', we don't really need all the solution of the IVP, just the last
# point of it -- this is saved in w1.
y = rk4( f, [a,z1], t )
w1 = y[n-1,0]
print "%2d: z = %10.3e, error = %10.3e" % ( 0, z1, b - w1 )
# Begin the main loop. We will compute the solution of a second IVP and
# then use the both solutions to refine our estimate of y'(a). This
# second solution then replaces the first and a new "second" solution is
# generated. This process continues until we either solve the problem to
# within the specified tolerance or we exceed the maximum number of
# allowable iterations.
for i in xrange( max_iter ):
# Solve second initial value problem, using y'(a) = z2. We need to
# retain the entire solution vector y since if y(t(n)) is close enough
# to b for us to stop then the first column of y becomes our solution
# vector.
y = rk4( f, [a,z2], t )
w2 = y[n-1,0]
print "%2d: z = %10.3e, error = %10.3e" % ( i+1, z2, b - w2 )
# Check to see if we are done...
if abs( b - w2 ) < tol:
break
# Compute the new approximations to the initial value of the first
# derivative. We compute z2 using a linear fit through (z1,w1) and
# (z2,w2) where w1 and w2 are the estimates at t=b of the initial
# value problems solved above with y1'(a) = z1 and y2'(a) = z2. The
# new value for z1 is the old value of z2.
#z1, z2 = ( z2, z1 + ( z2 - z1 ) / ( w2 - w1 ) * ( b - w1 ) )
z1, z2 = ( z2, z2 + ( z2 - z1 ) / ( w2 - w1 ) * ( b - w2 ) )
w1 = w2
# All done. Check to see if we really solved the problem, and then return
# the solution.
if abs( b - w2 ) >= tol:
print "\a**** ERROR ****"
print "Maximum number of iterations (%d) exceeded" % max_iter
print "Returned values may not have desired accuracy"
print "Error estimate of returned solution is %e" % ( b - w2 )
return y[:,0]
plt.figure(figsize = (6.75,5))
plt.xlabel('Time', fontsize = 14)
plt.xticks(fontsize = 14)
plt.yticks(fontsize = 14)
plt.title('Huen Method (rk2)')
plt.plot(t, theta2, label = 'Theta')
plt.plot([],[], label = f"R^2 = {format(r2_theta,'.5')}")
plt.plot(t, omega2,'--', label = 'Omega')
plt.plot([],[], label = f"R^2 = {format(r2_omega,'.5')}")
plt.tight_layout()
plt.legend(fontsize = 12)
plt.savefig('huen1.pdf')
# Testing RK4 method, plotting
theta3, omega3, time3 = df.rk4(pend, y0, t)
r3_theta = np.corrcoef(sol1[:,0], theta3)[0,1]
r3_omega = np.corrcoef(sol1[:,1], omega3)[0,1]
plt.figure(figsize = (6.75,5))
plt.xlabel('Time', fontsize = 14)
plt.xticks(fontsize = 14)
plt.yticks(fontsize = 14)
plt.title('RK4 method')
plt.plot(t, theta3, label = 'Theta')
plt.plot([],[], label = f"R^2 = {format(r3_theta,'.5')}")
plt.plot(t, omega3, '--', label = 'Omega')
plt.plot([],[], label = f"R^2 = {format(r3_omega,'.5')}")
plt.tight_layout()
plt.legend(fontsize = 12)
plt.savefig('rk41.pdf')