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Euler's method for two first order differential equations?
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Yuhong Jin
am 27 Nov. 2019
Beantwortet: Eng
am 23 Apr. 2023
I was trying to solve two first order differential equations like below using the Euler's method and plot two graphs with x and y as a function of t.
The differential equations are:
dxdt = @(x,t) -1.*y-0.1.*x;
dydt = @(y,t) x-0.5.*y;
I tried this script below:
a=0; %initial time
b=2000; %final time
h = 0.01; % time step
N = (b-a)./h;
t=a:h:b;
t(1)=a;
x(1)=0;
y(1)=0;
dxdt = @(x,t) -1.*y-0.1.*x;
dydt = @(y,t) x-0.5.*y;
for n=1:N
x(n+1)=x(n)+h.*dxdt(x(n),t(n));
y(n+1)=y(n)+h.*dydt(y(n),t(n));
t(n+1)=a+n*h;
end
plot(t,x,'-',t,y,'--')
But there was an error saying
Unable to perform assignment because the left and right sides have a different
number of elements.
Error in Q6 (line 15)
x(n+1)=x(n)+h.*dxdt(x(n),t(n));
I did not quite understand why.
Could anyone help me with this please?
Thanks in advance.
2 Kommentare
Akzeptierte Antwort
Jim Riggs
am 27 Nov. 2019
Bearbeitet: Jim Riggs
am 27 Nov. 2019
Try preallocating x and y
nsteps = (b-a)/h + 1; % this is the number of elements in t(a:h:b) (It is = N+1)
x=zeros(1,nsteps);
y=zeros(1,nsteps);
Also, when you define t as
t=a:h:b;
This creates a vector with T(1)=a and t(end)=b, so the following line
t(1)=a
is not necessary.
Inside your for loop, t has already been defined, above, as t=a:h:b, so you don't need to re-define it.
x and y need to be preallocated to size "nsteps", which is 1 greater than N, in order for your loop to work.
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Chris Horne
am 29 Mär. 2022
I figured it out. The function definition delimiters typo. The code produces then 2D plot of the approximating functions.
Chris Horne
am 31 Mär. 2022
Is the term 'forward Euler' the same as 'Euler' in terms of the algorithm? Here is my method for solving 3 equaitons as a vector:
% This code solves u'(t) = F(t,u(t)) where u(t)= t, cos(t), sin(t)
% using the FORWARD EULER METHOD
% Initial conditions and setup
neqn = 3; % set a number of equations variable
h=input('Enter the step size: ') % step size will effect solution size
t=(0:h:4).';%(starting time value 0):h step size
nt = size(t,1); % size of time array
%(the ending value of t ); % the range of t
% define the function vector, F
F = @(t,u)[t,cos(t),sin(t)]; % define the function 'handle', F
% with hard coded vector functions of time
u = zeros(nt,neqn); % initialize the u vector with zeros
v=input('Enter the intial vector values of 3 components using brackets [u1(0),u2(0),u3(0)]: ')
u(1,:)= v; % the initial u value and the first column
%n=numel(u); % the number of u values
% The loop to solve the ODE (Forward Euler Algorithm)
for i = 1:nt-1
u(i+1,:) = u(i,:) + h*F(t(i),u(i,:)); % Euler's formula for a vector function F
end
Weitere Antworten (1)
Eng
am 23 Apr. 2023
% Define the differential equation dydx = @(x,y) x + y;
% Define the initial condition y0 = 1;
% Define the step size and interval h = 0.1; x = 0:h:1;
% Initialize the solution vector y = zeros(size(x)); y(1) = y0;
% Apply the Modified Euler Method for i = 1:length(x)-1 k1 = dydx(x(i), y(i)); k2 = dydx(x(i+1), y(i)+hk1); y(i+1) = y(i) + h/2(k1+k2); end
% Plot the solution plot(x,y) xlabel('x') ylabel('y')
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