% solving the equation analyticaly
clear all
clc
s = dsolve('Dx = -2.3*x','x(0) = 1', 't');
sol = simplify(s)
% plot the analytical solution in the time interval
t0 = 0;
h = 0.01;
tend = 5;
t = [t0:h:tend];
sol = exp(-(2.3*t));
plot (t, sol, 'k-')
hold on;
% solving the equation numerically using explicit Euler's method
x0 = 1;
h1 = 1;
N = (tend - t0)/h1;
t = [t0:h1:tend];
x(1) = x0;
for i = 1:N
x(i+1) = x(i)+0.5*(h1*-2.3*x(i)+h1*-2.3*(h1*-2.3*x(i)));
end
plot(t,x,'o--')
title('Figure 3.5: Stiff ODE with Euler method when h = 1')
legend({'exact', 'Euler'},'location','southeast')
xlabel('t')
ylabel('x')
ax = gca;
ax.XTick = 0:1:5;
ax.XAxisLocation = 'origin';
ax.YAxisLocation = 'origin';
box off;
hold off;
Is there something wrong with my code??
As the correct graph is not showing.

3 commentaires

John D'Errico
John D'Errico le 11 Juin 2021
What did result? Do you recognize that since the problem is stiff, and that an explicit Euler method is often NOT expected to produce a viable result on stiff problems. That is why you are learning about stiff problems.
Jan
Jan le 11 Juin 2021
As far as I see, this comment is missleading:
% solving the equation numerically using explicit Euler's method
Paul
Paul le 11 Juin 2021
Is there a reference for the equation
x(i+1) = ...
Also, what makes this first order differential equation with exponential solution stiff?

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 Réponse acceptée

Paul
Paul le 15 Juin 2021

2 votes

Correct implementation of Heun's method, as I understand it, is given below and compare to the analytic solution and other equations given in this thread:
xdot = @(t,x)(-2.3*x);
c1 = 0; c2 = 1;
a11 = 0; a12 = 0; a21 = 1; a22 = 0;
b1 = 1/2; b2 = 1/2;
h1 = 0.25;
t = 0:h1:5;
xc = 0*t;
x = 0*t;
y = 0*t;
xc(1) = 1;
x(1) = 1;
y(1) = 1;
for ii = 1:(numel(t)-1)
k1 = xdot(t(ii),xc(ii));
k2 = xdot(t(ii)+c2*h1,xc(ii)+h1*a21*k1);
xc(ii+1) = xc(ii) + h1*(b1*k1 + b2*k2);
x(ii+1) = x(ii)+0.5*(h1*-2.3*x(ii)+h1*-2.3*(h1*-2.3*x(ii))); %Original question
y(ii+1) = y(ii)+0.5*h1*(-2.3*y(ii)+y(ii)+-2.3*h1*y(ii)); %Answer1
end
plot(t,exp(-2.3*t),'-x',t,x,'-o',t,y,'-o',t,xc,'-o'),grid
legend('analytic','xc','x','y');

Plus de réponses (1)

Sulaymon Eshkabilov
Sulaymon Eshkabilov le 12 Juin 2021

1 vote

There are a few crucial errs in your code. Here is the completely corrected code. Note that the step size of h1 = 1 gives very crude numerical soutions and thus, h1 has to be chosen carefully.
clearvars; clc % It is better start with clearvars instead of clear all that takes some time
syms x(t) % This one is recommended syntax of Symbolic Math toolbox
Dx = diff(x, t); % This one is recommended syntax
s = dsolve(Dx == -2.3*x,x(0) == 1); % This one is recommended syntax
sol = simplify(s)
% plot the analytical solution in the time interval
t0 = 0;
h = 0.01; % This is a reasonably small step size
tend = 5;
t = t0:h:tend;
sol = subs(sol, t); % Use the above computed symbolic solution
plot(t, sol, 'k-')
hold on
% solving the equation numerically using explicit Euler's method
x0 = 1;
h1 = 0.25; % Step size needs to be selected carefully
t = t0:h1:tend;
y = [x0, zeros(1, numel(t)-1)]; % Memory allocation
for i = 1:numel(t)-1
y(i+1) = y(i)+0.5*h1*(-2.3*y(i)+y(i)+-2.3*h1*y(i)); % See the corrected loop
end
plot(t,y,'o--')
title('Figure 3.5: Stiff ODE with Euler method when h = 1')
legend({'exact', 'Euler'},'location','southeast')
xlabel('t')
ylabel('x')
ax = gca;
ax.XTick = 0:1:5;
ax.XAxisLocation = 'origin';
ax.YAxisLocation = 'origin';
box off;
hold off;

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