Please help with what I have so far in Python from numpy import * import math, d

Question

Please help with what I have so far in Python

from numpy import *
import math, decimal

t0 = float (input('Enter initial t value, t0 = '))
tend = float(input('Enter final time value, tend = '))
y0 = float(input('Enter initial y value, y0 = '))
MaxN = int(input('Enter Max Steps value, MaxN = '))
print ('t0, y0, tend, MaxN = {:4f} {:4f} {:4f} {:4f}'.format (t0, y0, tend, MaxN))
dt = (tend-t0)/MaxN
print('delta t, step size = (:4f)'.format(dt))

print ('MaxN (tn) (yn) yexact ERRn ERRmax)
print ('==== ====== ====== ======== ====== ========')

def euler (fcn, y0, t0,tend,dt):
Yn = y0
tn = t0
ERRmax = 0
  
for n in range (0, MaxN):
Yn = Yn + dt*fcn(tn,Yn)
tn = tn + dt
yexact = tn*tn + 1
ERRn = abs(yexact - Yn)
ERRmax = max(ERRmax,ERRn)
print ('(:i.0f) (:12.6f) (:12.6f) (:12.6f) (:12.6f) (:12.6f)'.format (t0, y0, tend, MaxN))
  
def fcn(tn, Yn)
return 2*tn
  

Implement the Euler scheme for the scalar IVP y' (t) f(t, y (t)) to t end y (to) The code should read and print out) he initial point to yo he final time tend and the number of steps to be executed Nsteps The evaluation of f(t,y) should be done in a Function Subprogram FCN tn, Yn) which returns the value Fn f (tn, Yn) Calculate the time-step At (t to) Nsteps, implement th Euler time-stepping end At t for n 0,1 N steps Yo with timesteps tn to n At n-0,1 Nsteps and output the pairs tn, Yn (and the exact solution when known see below NOTE: No arrays are actually needed 1. Debug your code on the trivial problem y' (t) 2t, 0 t 2 y(0) 1, by calculating Yexact (tn) n tn 1, and the error ERRn ABS (Yexact Yn) At each step output tn, Yn, Yexactn, ERRn. and calculate the worst overall error: ERRmax MAX ERRmax, ERRn) output ERRmax at the end of the run Before you type in the code you should write it on paper, and execute the steps by hand calculator for, say, steps 4, to make sure the logic is right Then enter the code and check that it finds the same values Then try Nsteps 10 Note that since the RHS of this oDE is just a function of t, the solution is simply the integral of f(t) 2t, so in fact your code is performing numerical integration usually called quadrature what quadrature rule does it amount to tange? trapezoidal? Simpson?) 2 Test your code on y' (t) y, 0

Explanation / Answer

from numpy import *
import math, decimal

t0 = float (input('Enter initial t value, t0 = '))
tend = float(input('Enter final time value, tend = '))
y0 = float(input('Enter initial y value, y0 = '))
MaxN = int(input('Enter Max Steps value, MaxN = '))
print ('t0, y0, tend, MaxN = {:4f} {:4f} {:4f} {:4f}'.format (t0, y0, tend, MaxN))
dt = (tend-t0)/MaxN
print('delta t, step size = (:4f)'.format(dt))

print ('MaxN (tn) (yn) yexact ERRn ERRmax)
print ('==== ====== ====== ======== ====== ========')

def euler (fcn, y0, t0,tend,dt):
Yn = y0
tn = t0
ERRmax = 0
  
for n in range (0, MaxN):
Yn = Yn + dt*fcn(tn,Yn)
tn = tn + dt
yexact = tn*tn + 1
ERRn = abs(yexact - Yn)
ERRmax = max(ERRmax,ERRn)
print ('(:i.0f) (:12.6f) (:12.6f) (:12.6f) (:12.6f) (:12.6f)'.format (t0, y0, tend, MaxN))
  
def fcn(tn, Yn)
return 2*tn
  

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