## Transcribed Text

Exercise 100%
We will apply backward and forward Euler to approximate the solution
system
sapplemented by the initial condition
x(0)
Here =.==(j-1)6 for j = 1,,2N and the matrix A isgiven by
-1
-1
A
-1
In short
Aij=
ifi=j£1 mod 2N.
otherwise.
This corresponds tothe PDE Example 27.1 (Lecture 27)but with (0. replaced by (0. 2x) and
the boundary condition replaced by the periodic boundary conditions
The reason for these modifications that the exact solution the system can be computed using
the code obtained in Homework This because the eigenvectors for are given by
with eigenvalue
&
1.
Run backward Euler Take 10 until E using NT time of size
To
assess the efficiency of the method. you will need to compute the error
ENT
where is the exact solution at point x=x; and time t = and cistheresult from
the
backward Euler at L=1. The exact solution can be obtained by executing
EXACTODE(N.1)
Run for NT 10. 100. 1000. 10000 ablt to see first order
convergence *= **
2. Modify your routinesoit computes the forward Nom for N 10. 20. 40.80
try get rough idem where the scheme becomes unstable by running different values
NT. Note that you may need to tun more steps with the same N and to see the blow up.
For example:
NT
max
clusiom
19
clear
probably blowing up
definitely
blowing
up.
For
the
last
note
that
(==
1/19.
be
able
see
that
N
10
and 1/20still blows /22 Thus the blow up value
for
N= 10is between &= 1/22 Do
3.
The
Crank-
Nicholson
or
This scheme is A-stable and secoad order. Modify your code so it runs Crank -Nicholsor
and report the error as a function of NT 4.8 16.32.64 Yom should be able to see the
second order convergence.
EXACTODE.r
% computes the exact solution of the ODE at
function [ex]=EXACTODE(N,t)
forj=1:2*N
xj=pi*(j-1)/N;
f(j)=xj*(2*pi-xj);
end
x=TRIGCOEF(N,f);
hs=(pi/N)^2;
% multiply by the symbol
or ==1:2*N
js=-N+j;
cx(j)=cx(j)*exp(-(2-2°cos(js*pi/N))/hs*t);
end
ex=TRIGEVAL(N,cx):
% From exercise 5
function [tf]=TRIGEVAL(n,tc)
for i=1:n+1
tcl(j)=tc(j+n- 1);
end
for j=2:n
tcl(j+n)=tc(j-1);
end
tf=fft(tc1);
function [cx]=TRIGCOEF(n.f)
cx1=ifft(f);
for j=1:n+1
cx(j+n-1)=cxl(j);
end
forj=2:n
cx(j-1)=cx1(j+n);
end

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Q1

clear; clc; close all;

N = 10; t = 1;

yEx = real(EXACTODE(N,t));

NT = power(10,1:4);

eNT = [];

for i=1:length(NT)

tv = linspace(0,t,NT(i));

h = pi/N;

j =1:2*N;

xv = (j-1)*h;

v0 = xv.*(2*pi-xv);

yNum = eulerBackward(tv,@myODE,v0,N);

eNT = cat(2,eNT, max(abs(yEx-yNum(end,:))));

end

str1 = 'Error plot for N=10:Backward Euler';

figure('Name',str1);

loglog(NT,eNT); hold on; xlabel('NT'); ylabel('e_N_T');

title(str1);...