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

# # Stationary Iterative Methods for Linear Systems

# In[2]:


import numpy as np
import scipy.linalg as la

import matplotlib.pyplot as pt

# Let's solve $u''=-30x^2$ with $u(0)=1$ and $u(1)=-1$.

# In[3]:


n = 50

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

# Set up the system matrix `A` to carry out centered finite differences
# 
# $$
# u''(x)\approx \frac{u(x+h) - 2u(x) + u(x-h)}{h^2}.
# $$
# 
# Use `np.eye(n, k=...)`. What needs to be in the first and last row?

# In[5]:


A = (np.eye(n, k=1) + -2*np.eye(n) + np.eye(n, k=-1))/h**2
A[0] = 0
A[-1] = 0
A[0,0] = 1
A[-1,-1] = 1

# Next, fix the right hand side:

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b = -30*mesh**2
b[0] = 1
b[-1] = -1

# Compute a reference solution `x_true` to the linear system:

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x_true = la.solve(A, b)
pt.plot(mesh, x_true)

# Next, we'll try all the stationary iterative methods we have seen.

# ## Jacobi

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x = np.zeros(n)

# Next, apply a Jacobi step:

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x_new = np.empty(n)

for i in range(n):
    x_new[i] = b[i]
    for j in range(n):
        if i != j:
            x_new[i] -= A[i,j]*x[j]
        
    x_new[i] = x_new[i] / A[i,i]

x = x_new

# In[43]:


pt.plot(mesh, x)
pt.plot(mesh, x_true, label="true")
pt.legend()

# * Ideas to accelerate this?
# * Multigrid

# ## Gauss-Seidel

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x = np.zeros(n)

# In[64]:


x_new = np.empty(n)

for i in range(n):
    x_new[i] = b[i]
    for j in range(i):
        x_new[i] -= A[i,j]*x_new[j]
    for j in range(i+1, n):
        x_new[i] -= A[i,j]*x[j]
        
    x_new[i] = x_new[i] / A[i,i]

x = x_new
pt.plot(mesh, x)
pt.plot(mesh, x_true, label="true")
pt.legend()

# ### And now Successive Over-Relaxation ("SOR")

# In[92]:


x = np.zeros(n)

# In[106]:


x_new = np.empty(n)

for i in range(n):
    x_new[i] = b[i]
    for j in range(i):
        x_new[i] -= A[i,j]*x_new[j]
    for j in range(i+1, n):
        x_new[i] -= A[i,j]*x[j]
        
    x_new[i] = x_new[i] / A[i,i]

direction = x_new - x
omega = 1.5
x = x + omega*direction

pt.plot(mesh, x)
pt.plot(mesh, x_true, label="true")
pt.legend()
pt.ylim([-1.3, 1.3])

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