def f(x): ''' Calculates the height of a projectile at a given horizontal distance. theta, v0, and y0 are global variables ''' from scitools.std import tan, cos g = 9.81 # m/s**2 y = x*tan(theta) - 1./(2*v0**2)*(g*x**2)/(cos(theta)**2) + y0 return y
def f(x):
'''
Calculates the height of a projectile at a given
horizontal distance.
theta, v0, and y0 are global variables
'''
from scitools.std import tan, cos
g = 9.81 # m/s**2
y = x * tan(theta) - 1. / (2 * v0**2) * (g * x**2) / (cos(theta)**2) + y0
return y
def T(z, t):
# T0, A, k, and omega are global variables
return T0 + A1 * exp(-a1 * z) * cos(omega1 * t - a1 * z) + \
A2 * exp(-a2 * z) * cos(omega2 * t - a2 * z)
for name in glob.glob('pix/planet*.png'):
os.remove(name)
n = 100 #we want 100 frames
a = 1 #set parameters for the orbit
b = 1
w = 1
delta_t = (2 * pi) / (w * n)
counter = 0
X = []
Y = []
for k in range(n + 1):
tk = k * delta_t #each time through we need to add a new k value
x = a * cos(w * tk)
y = b * sin(w * tk)
X.append(x)
Y.append(y)
XP = array(X)
YP = array(Y)
XF = array([x])
YF = array([y])
velo = w * sqrt(a**2 * sin(w * tk)**2 + b**2 * cos(w * tk)**2)
plot(XP,
YP,
'-r',
XF,
YF,
'bo',
axis=[-1.2, 1.2, -1.2, 1.2],
def T(z, t):
# T0, A, k, and omega are global variables
return T0 + A1 * exp(-a1 * z) * cos(omega1 * t - a1 * z) + \
A2 * exp(-a2 * z) * cos(omega2 * t - a2 * z)
def _dg(x):
return -2*0.1*x*exp(-0.1*x**2)*sin(pi/2*x) + \
pi/2*exp(-0.1*x**2)*cos(pi/2*x)
def _dg(x):
return -2*0.1*x*exp(-0.1*x**2)*sin(pi/2*x) + \
pi/2*exp(-0.1*x**2)*cos(pi/2*x)
def points(N):
x = [0.5 * cos(2 * pi * i / N) for i in range(N + 1)]
y = [0.5 * sin(2 * pi * i / N) for i in range(N + 1)]
return x, y
def points(N):
x = [0.5 * cos(2 * pi * i / N) for i in range(N + 1)]
y = [0.5 * sin(2 * pi * i / N) for i in range(N + 1)]
return x, y
"""
Exercise 5.6: Simulate by hand a vectorized expression
Author: Weiyun Lu
"""
from scitools.std import array, sin, cos, exp, zeros
x = array([0, 2])
t = array([1, 1.5])
y = zeros((2, 2))
f = lambda x, t: cos(sin(x)) + exp(1.0 / t)
for i in range(2):
for j in range(2):
y[i][j] = f(x[i], t[j])
print y
def test_undampedWaves():
#define constants given in exercise
A = 1
mx=7.
my=2.
#define function that give
q=lambda x,y: 1
#define some variables
Lx = 3
Ly = 1.3
T = 1
C = 0.5
dt= 0.1
#define omega so equation holds
w=pi*sqrt((mx/Lx)**2 +(my/Ly)**2)
#help varabeles
kx = pi*mx/Lx
ky = pi*my/Ly
#Exact solution
ue = lambda x,y,t: A*cos(x*kx)*cos(y*ky)*cos(t*w)
#initial condition so we get result we want.
I = lambda x,y: A*cos(x*kx)*cos(y*ky)
#factor dt decreeses per step
step=0.5
#number of steps I want to do
val=5
#array to store errors
E=zeros(val)
for i in range(val):
v='vector'
#solve eqation
u,x,y,t,e=solver(I,None,None,q,0,Lx,Ly,dt*step**(i),T,C,1,mode=v,ue=ue)
E[i]=e
#find convergence rate between diffrent dt values
r =zeros(val-1)
r = log(E[1:]/E[:-1])/log(step)
print "Test convergence for undamped waves:"
for i in range(val):
if i==0:
print "dt: ",dt," Error: ",E[i]
else:
print "dt: ",dt*step**(i)," Error: ",E[i], "rate of con.: ", r[i-1]
#requiere close to 2 in convergence rate for last r.
assert abs(r[-1]-2)<0.01
def test_manufactured():
A=1
B=0
mx=1
my=1
b=1
c=1.1
#define some variables
Lx = 2.
Ly = 2.
T = 1
C = 0.3
dt= 0.1
#help varabeles
kx = pi*mx/Lx
ky = pi*my/Ly
w=1
#Exact solution
ue = lambda x,y,t: A*cos(x*kx)*cos(y*ky)*cos(t*w)*exp(-c*t)
I = lambda x,y: A*cos(x*kx)*cos(y*ky)
V = lambda x,y: -c*A*cos(x*kx)*cos(y*ky)
q = lambda x,y: x**2+y**2
f = lambda x,y,t:A*(-b*(c*cos(t*w) + w*sin(t*w))*cos(kx*x)*cos(ky*y) + c**2*cos(kx*x)*cos(ky*y)*cos(t*w) + 2*c*w*sin(t*w)*cos(kx*x)*cos(ky*y) + kx**2*(x**2 + y**2)*cos(kx*x)*cos(ky*y)*cos(t*w) + 2*kx*x*sin(kx*x)*cos(ky*y)*cos(t*w) + ky**2*(x**2 + y**2)*cos(kx*x)*cos(ky*y)*cos(t*w) + 2*ky*y*sin(ky*y)*cos(kx*x)*cos(t*w) - w**2*cos(kx*x)*cos(ky*y)*cos(t*w))*exp(-c*t)
#factor dt decreeses per step
step=0.5
#number of steps I want to do
val=5
#array to store errors
E=zeros(val)
for i in range(val):
v='vector'
#solve eqation
u,x,y,t,e=solver(I,V,f,q,b,Lx,Ly,dt*step**(i),T,C,8,mode=v,ue=ue)
E[i]=e
#find convergence rate between diffrent dt values
r =zeros(val-1)
r = log(E[1:]/E[:-1])/log(step)
print
print "Convergence rates for manufactured solution:"
for i in range(val):
if i==0:
print "dt: ",dt," Error: ",E[i]
else:
print "dt: ",dt*step**(i)," Error: ",E[i], "rate of con.: ", r[i-1]
#requiere "close" to 2 in convergence rate for last r.
assert abs(r[-1]-2)<0.5
def test_manufactured():
A = 1
B = 0
mx = 1
my = 1
b = 1
c = 1.1
#define some variables
Lx = 2.
Ly = 2.
T = 1
C = 0.3
dt = 0.1
#help varabeles
kx = pi * mx / Lx
ky = pi * my / Ly
w = 1
#Exact solution
ue = lambda x, y, t: A * cos(x * kx) * cos(y * ky) * cos(t * w) * exp(-c *
t)
I = lambda x, y: A * cos(x * kx) * cos(y * ky)
V = lambda x, y: -c * A * cos(x * kx) * cos(y * ky)
q = lambda x, y: x**2 + y**2
f = lambda x, y, t: A * (
-b * (c * cos(t * w) + w * sin(t * w)) * cos(kx * x) * cos(ky * y) + c
**2 * cos(kx * x) * cos(ky * y) * cos(t * w) + 2 * c * w * sin(
t * w) * cos(kx * x) * cos(ky * y) + kx**2 *
(x**2 + y**2) * cos(kx * x) * cos(ky * y) * cos(t * w) + 2 * kx * x *
sin(kx * x) * cos(ky * y) * cos(t * w) + ky**2 *
(x**2 + y**2) * cos(kx * x) * cos(ky * y) * cos(t * w) + 2 * ky * y *
sin(ky * y) * cos(kx * x) * cos(t * w) - w**2 * cos(kx * x) * cos(
ky * y) * cos(t * w)) * exp(-c * t)
#factor dt decreeses per step
step = 0.5
#number of steps I want to do
val = 5
#array to store errors
E = zeros(val)
for i in range(val):
v = 'vector'
#solve eqation
u, x, y, t, e = solver(I,
V,
f,
q,
b,
Lx,
Ly,
dt * step**(i),
T,
C,
8,
mode=v,
ue=ue)
E[i] = e
#find convergence rate between diffrent dt values
r = zeros(val - 1)
r = log(E[1:] / E[:-1]) / log(step)
print
print "Convergence rates for manufactured solution:"
for i in range(val):
if i == 0:
print "dt: ", dt, " Error: ", E[i]
else:
print "dt: ", dt * step**(
i), " Error: ", E[i], "rate of con.: ", r[i - 1]
#requiere "close" to 2 in convergence rate for last r.
assert abs(r[-1] - 2) < 0.5
def test_undampedWaves():
#define constants given in exercise
A = 1
mx = 7.
my = 2.
#define function that give
q = lambda x, y: 1
#define some variables
Lx = 3
Ly = 1.3
T = 1
C = 0.5
dt = 0.1
#define omega so equation holds
w = pi * sqrt((mx / Lx)**2 + (my / Ly)**2)
#help varabeles
kx = pi * mx / Lx
ky = pi * my / Ly
#Exact solution
ue = lambda x, y, t: A * cos(x * kx) * cos(y * ky) * cos(t * w)
#initial condition so we get result we want.
I = lambda x, y: A * cos(x * kx) * cos(y * ky)
#factor dt decreeses per step
step = 0.5
#number of steps I want to do
val = 5
#array to store errors
E = zeros(val)
for i in range(val):
v = 'vector'
#solve eqation
u, x, y, t, e = solver(I,
None,
None,
q,
0,
Lx,
Ly,
dt * step**(i),
T,
C,
1,
mode=v,
ue=ue)
E[i] = e
#find convergence rate between diffrent dt values
r = zeros(val - 1)
r = log(E[1:] / E[:-1]) / log(step)
print "Test convergence for undamped waves:"
for i in range(val):
if i == 0:
print "dt: ", dt, " Error: ", E[i]
else:
print "dt: ", dt * step**(
i), " Error: ", E[i], "rate of con.: ", r[i - 1]
#requiere close to 2 in convergence rate for last r.
assert abs(r[-1] - 2) < 0.01