#!/usr/bin/env python
# coding: utf-8

# # Stiff ODE

# Consider the stiff ODE
# $$
# y' = -100(y - \sin(t)), \qquad t \ge 0
# $$
# $$
# y(0) = 1
# $$
# The exact solution is
# $$
# y(t) = \frac{10101}{10001}e^{-100t} - \frac{100}{10001}(\cos t - 100 \sin t)
# $$

# In[11]:


import numpy as np
from matplotlib import pyplot as plt


# In[12]:


def f(t,y):
    return -100.0*(y - np.sin(t))

def yexact(t):
    return 10101.0*np.exp(-100*t)/10001.0 - 100.0*(np.cos(t) - 100.0*np.sin(t))/10001.0


# ## Forward Euler

# $$
# y_n = y_{n-1} - 100 h [ y_{n-1} - \sin(t_{n-1})]
# $$

# In[13]:


def ForwardEuler(t0,y0,T,h):
    N = int((T-t0)/h)
    y = np.zeros(N)
    t = np.zeros(N)
    t[0] = t0
    y[0] = y0
    for n in range(1,N):
        y[n] = y[n-1] + h*f(t[n-1],y[n-1])
        t[n] = t[n-1] + h
    return t,y


# In[14]:


t0 = 0
y0 = 1
T  = np.pi

h  = 1.0/52.0
t,y = ForwardEuler(t0,y0,T,h)

plt.plot(t,y,t,yexact(t))
plt.xlim([-0.1,4])
plt.title("Forward Euler, h ="+str(h))
plt.legend(("Euler","Exact"))


# In[15]:


h  = 0.01
t,y = ForwardEuler(t0,y0,T,h)

plt.plot(t,y,t,yexact(t))
plt.xlim([-0.1,4])
plt.title("Forward Euler, h ="+str(h))
plt.legend(("Euler","Exact"))


# ## Backward Euler scheme

# $$
# y_n = y_{n-1} + h [ -100(y_n - \sin(t_n) ]
# $$
# or
# $$
# y_n = \frac{ y_{n-1} + 100 h \sin(t_n)}{1 + 100 h}
# $$

# In[16]:


def BackwardEuler(t0,y0,T,h):
    N = int((T-t0)/h)
    y = np.zeros(N)
    t = np.zeros(N)
    t[0] = t0
    y[0] = y0
    for n in range(1,N):
        t[n] = t[n-1] + h
        y[n] = (y[n-1] + 100.0*h*np.sin(t[n]))/(1.0 + 100.0*h)
    return t,y


# ## Trapezoidal scheme

# $$
# y_n = y_{n-1} + \frac{h}{2}[ -100(y_{n-1}-\sin(t_{n-1})) - 100 (y_n - \sin(t_n))]
# $$
# or
# $$
# y_n = \frac{(1-50h)y_{n-1} + 50h(\sin(t_{n-1}) + \sin(t_n))}{1 + 50 h}
# $$

# In[17]:


def Trapezoidal(t0,y0,T,h):
    N = int((T-t0)/h)
    y = np.zeros(N)
    t = np.zeros(N)
    t[0] = t0
    y[0] = y0
    for n in range(1,N):
        t[n] = t[n-1] + h
        y[n] = ((1.0-50.0*h)*y[n-1] + 50.0*h*(np.sin(t[n-1])+np.sin(t[n])))/(1.0 + 50.0*h)
    return t,y


# In[18]:


h  = 0.1
t,y = BackwardEuler(t0,y0,T,h)
plt.plot(t,y)

t,y = Trapezoidal(t0,y0,T,h)
plt.plot(t,y)

plt.plot(t,yexact(t))
plt.xlim([-0.1,4])
plt.title("Step size h ="+str(h))
plt.legend(("Backward Euler","Trapezoid","Exact"))