Problem Sheet 3 Question 1

The general form of the population growth differential equation

(161)\[\begin{equation} y^{'}=t-y, \ \ (0 \leq t \leq 4) \end{equation}\]

with the initial condition

(162)\[\begin{equation}y(0)=1,\end{equation}\]

Has the exact soulation. \begin{equation} y= 2e^{-t}+t-1\end{equation}

Setting up Libraries

import numpy as np
import math 

%matplotlib inline
import matplotlib.pyplot as plt # side-stepping mpl backend
import matplotlib.gridspec as gridspec # subplots
import warnings

warnings.filterwarnings("ignore")

Defining the function

\begin{equation}f(t,y)=t-y$$

def myfun_ty(t,y):
    return t-y

Initial Setup

Defining the step size \(h\) from the interval range \(a\leq t \leq b\) and number of steps \(N\)

(163)\[\begin{equation}h=\frac{b-a}{h}.\end{equation}\]

This gives the discrete time steps,

(164)\[\begin{equation}t_{i}=t_0+ih,\end{equation}\]

where \(t_0=a\).

# Start and end of interval
b=4
a=0
# Step size
N=8
h=(b-a)/(N)
t=np.arange(a,b+h,h)

Setting up the initial conditions of the equation

(165)\[\begin{equation}w_0=IC\end{equation}\]

# Initial Condition
IC=1
w=np.zeros(N+1)
y=(IC+1)*np.exp(-t)+t-1#np.zeros(N+1)
w[0]=IC

2nd Order Runge Kutta (Mid-method)

(166)\[\begin{equation}k_1=f(t,y),\end{equation}\]

(167)\[\begin{equation}k_2=f(t+\frac{h}{2},y+\frac{h}{2}k_2),\end{equation}\]

(168)\[\begin{equation}w_{i+1}=w_{i}+h(k_2).\end{equation}\]

for k in range (0,N):
    k1=myfun_ty(t[k],w[k])
    k2=myfun_ty(t[k]+h/2,w[k]+h/2*k1)
    w[k+1]=w[k]+h*(k2)

Plotting Results

fig = plt.figure(figsize=(10,4))
# --- left hand plot
ax = fig.add_subplot(1,3,1)
plt.plot(t,w, '--',color='blue')
#ax.legend(loc='best')
plt.title('Numerical Solution h=%s'%(h))

ax = fig.add_subplot(1,3,2)
plt.plot(t,y,color='black')
plt.title('Exact Solution ')

ax = fig.add_subplot(1,3,3)
plt.plot(t,y-w, 'o',color='red')
plt.title('Error')
# --- title, explanatory text and save
fig.suptitle(r"$y'=t-y,   y(0)=%s$"%(IC), fontsize=20)
plt.tight_layout()
plt.subplots_adjust(top=0.85)