Fixing a plot to show a horizontal line instead of a point

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Left Terry
Left Terry 2024-12-31
编辑： Torsten 2024-12-31
Ran in:
Hello. First of all I don't know if my code is correct for the specific task, especially for the error calculations. My problem is that i get a point on the last graph on the errors figure but i need a horizontal line. Why does this happen ?
%--- Exercise 1 - A
%--- Solving the diff. eqn. N'(t) = N(t) - c*N^2(t) with c = 0.5 using
%--- Euler's method, Runge - Kutta method, and predictor-corrector method, and comparing in graph.
%--- I have four initial values for N0.
clc, clear all, close all
tmax = 100;
t = [0:tmax];
c = 0.5;
h = 0.1;
f = @(t,N) N - c*N.^2;
N0 = [0.1 0.5 1.0 2.0];
L = length(t)-1;
N = zeros(1,L);
for i = 1:length(N0)
%--- Exact solution ---
[T,n] = ode45(f,t,N0(i));
figure(1)
subplot(2,2,i), plot(T,n), hold on
title({'N'' = N - cN^2',sprintf('c = %.1f, N_0 = %.1f',c, N0(i))}),xlabel('t'), ylabel('N(t)'), hold on
%--- Euler's method ---
for j = 1:L
NE(1) = N0(i);
NE(1,j+1) = NE(j) + h*f(':',NE(j));
ErrorEuler(j) = abs(n(j) - NE(j))/n(j)*100;
end
plot(t,NE,'r-.'), hold on
%--- Runge - Kutta method ---
for j = 1:tmax
NRK(1) = N0(i);
K1 = f(t(j),NRK(j));
K2 = f(t(j) + h*0.5, NRK(j) + h*K1*0.5);
K3 = f(t(j) + h*0.5, NRK(j) + h*K2*0.5);
K4 = f(t(j) + h, NRK(j) + h*K3);
NRK(j+1) = NRK(j) + h*((K1 + K4)/6 + (K2 + K3)/3);
ErrorRK(j) = abs(n(j) - NRK(j))/n(j)*100;
end
plot(t,NRK,'b.'), legend({'Exact solution','Euler''s method','Runge - Kutta'},'Location','Southeast')
figure(2)
subplot(2,2,i)
plot(NE(1:end-1),ErrorEuler,'.')
hold on
plot(NRK(1:end-1),ErrorRK,'-.')
title({'N(t) vs. Error',sprintf('N_0 = %.1f',N0(i))}), xlabel('N(t)'), ylabel('Error (%)')
legend({'Euler','Runge - Kutta'})
end

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回答（1 个）

Torsten
Torsten 2024-12-31
移动：Torsten 2024-12-31
N(t) = 2 for all t, and the error is 0 %. So plotting a line (vertical or horizontal) would be wrong in the case N_0 = 2 - you only get a single point.
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Torsten
Torsten 2024-12-31
编辑：Torsten 2024-12-31
Ran in:
@Left Terry The problem states tthat we have to use h = 0.1.
But you use h = 1 in the ode45 calculation:
tmax = 100;
t = [0:tmax]
t = 1×101
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
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So either you use h = 0.1 or h = 1 in both calculations.

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