# coding: utf-8

# # Finite Differences for Boundary Value Problems

# In[1]:


import numpy as np
import matplotlib.pyplot as pt

import scipy.sparse as sps


# We'll solve
# 
# $u''+1000(1+x^2)u=0$ on $(-1,1)$
# 
# with $u(-1)=3$ and $u(1)=-3$.

# In[2]:


#n = 9
n = 200

mesh = np.linspace(-1, 1, n)
h = mesh[1] - mesh[0]


# In[3]:


A = sps.diags(
    [1,-2,1],
    offsets=[-1,0,1], 
    shape=(n, n))

if n < 10:
    print(A.todense())


# Create `second_deriv` as a matrix to apply the second derivative. Can only do that for the interior points!
# 
# * change `shape` and offsets
# * Take `h` into account

# In[4]:


second_deriv = sps.diags(
    [1,-2,1],
    offsets=np.array([-1,0,1])+1,
    shape=(n-2, n))/h**2


if n < 10:
    print(second_deriv.todense())


# In[5]:


factor = sps.diags(
    [1000*(1 + mesh[1:]**2)],
    offsets=[1],
    shape=(n-2, n))

if n < 10:
    print(mesh[1:-1])
    print()
    print(factor.todense())


# In[6]:


A_int = second_deriv+factor

if n < 10:
    print(A_int.todense())


# In[7]:


A = sps.vstack([
    sps.coo_matrix(([1], ([0],[0])), shape=(1, n)),
    A_int,
    sps.coo_matrix(([1], ([0],[n-1])), shape=(1, n)),
    ])
A = sps.csr_matrix(A)

if n < 10:
    print(A.todense())


# Next, assemble the right-hand side as `rhs`:
# 
# Pay special attention to the boundary conditions. What entries of `rhs` do they correspond to?

# In[8]:


rhs = np.zeros(n)
rhs[0] = 3
rhs[-1] = -3


# To wrap up, solve and plot:

# In[9]:


import scipy.sparse.linalg as sla

sol = sla.spsolve(A, rhs)


# In[10]:


pt.plot(mesh, sol)