Unable to perform assignment because the left and right sides have a different number of elements.

I am getting the error code in the title for line 40.

%% Clear MATLAB

clear;

clc;

%% Known Values

h=0.01;

tlim=40;

t=0:h:tlim;

F0=20;

wf=3;

k=5.15;

c=12.2;

P=F0*sin(wf*t);

K=zeros(1,4001);

L=zeros(1,4001);

%% Initial Conditions

x(1)=-0.02;

xdot(1)=0;

y(1)=xdot(1);

%% Define Equations

f1=@(x,y,t) y;

f2=@(x,y,t) (P-k*x-c*xdot);

%% RK4 Loop

for i=1:tlim

t(i+1)=t(i)+h;

K1=h*f1(x(i),y(i),t(i));

L1=h*f2(x(i),y(i),t(i));

K2=h*f1(x(i)+h*K1/2,y(i)+h*K1/2,t(i)+h/2);

L2=h*f2(x(i)+h*L1/2,y(i)+h*L1/2,t(i)+h/2);

K3=h*f1(x(i)+h*K2/2,y(i)+h*K2/2,t(i)+h/2);

L3=h*f2(x(i)+h*L2/2,y(i)+h*L2/2,t(i)+h/2);

K4=h*f1(x(i)+h*K3,y(i)+h*K3,t(i)+h);

L4=h*f1(x(i)+h*L3,y(i)+h*L3,t(i)+h);

K(i+1)=K(i)+(1/6)*(K1+2*K2+2*K3+K4);

L(i+1)=L(i)+(1/6)*(L1+2*L2+2*L3+L4);

end

Answer 1

[VBBV]

K=zeros(1,4100);
L=zeros(1,4100);

Check the size of vectors K and L

[joe brady]

Hi Thanks for the reply,

There are 4001 time intervals, shouldnt this mean that there is also 4001 K and L values?

[VBBV]

L4=h*f2(x(i)+h*L3,y(i)+h*L3,t(i)+h);
  L(i+1,:)=L(i,:)+(1/6)*(L1+2*L2+2*L3+L4);

Also the above changes as well

[joe brady]

Thanks for your response. Making the changes you suggested i am now getting the error: "Arrays have incompatible sizes for this operation" at line 40:

%% Clear MATLAB
clear;
clc;
%% Known Values
h=0.01;
tlim=40;
t=0:h:tlim;
F0=20;
wf=3;
k=5.15;
c=12.2;
P=F0*sin(wf*t);
K=zeros(1,4100);
L=zeros(1,4100);
%% Initial Conditions
x(1)=-0.02;
xdot(1)=0;
y(1)=xdot(1);
%% Define Equations
f1=@(x,y,t) y;
f2=@(x,y,t) (P-k*x-c*xdot);
%% RK4 Loop
for i=1:tlim
    t(i+1)=t(i)+h;
    K1=h*f1(x(i),y(i),t(i));
    L1=h*f2(x(i),y(i),t(i));
    K2=h*f1(x(i)+h*K1/2,y(i)+h*K1/2,t(i)+h/2);
    L2=h*f2(x(i)+h*L1/2,y(i)+h*L1/2,t(i)+h/2);
    K3=h*f1(x(i)+h*K2/2,y(i)+h*K2/2,t(i)+h/2);
    L3=h*f2(x(i)+h*L2/2,y(i)+h*L2/2,t(i)+h/2);
    K4=h*f1(x(i)+h*K3,y(i)+h*K3,t(i)+h);
    L4=h*f2(x(i)+h*L3,y(i)+h*L3,t(i)+h);
    K(i+1,:)=K(i,:)+(1/6)*(K1+2*K2+2*K3+K4);
    L(i+1,:)=L(i,:)+(1/6)*(L1+2*L2+2*L3+L4);
    
end

[VBBV]

K=zeros(tlim,4100);
L=zeros(tlim,4100);

Sorry, i actually meant above

[joe brady]

I'm still getting an error message saying "Arrays have incompatible sizes for this operation.", i notice when i make the suggested K and L changes the matricies is a 2x4100, whereas my values for L1,2,3,4 K1,2,3,4 are in the form 1x4001?

[VBBV]

As @imageanalyst told, you need to assign to check with x and y vectors inside the for loop

x(i+1) = x(i) + something;
y(i+1) = y(i) + something;
for i = 1:tlim-1 % this too

Answer 2

[Image Analyst]

4001 is different than 4100. you might need to do this:

h=0.01;
tlim=40;
t=0:h:tlim;
numElements = numel(t)
F0=20;
wf=3;
k=5.15;
c=12.2;
P=F0*sin(wf*t);
K=zeros(1, numElements);
L=zeros(1, numElements);

but then you just run into trouble later because you're trying to use x(2), etc. in

for i=1: numElements - 1
    t(i+1)=t(i)+h;
    K1=h*f1(x(i),y(i),t(i));

when there is no second element to x or y. They're scalars.

[joe brady]

Thanks for the response, for context x is the displacement of an object at time intervals of h up to tlim, and y is the velocity, how would i transform my code?

[Image Analyst]

You need to define x and y in advance of entering your loop. What do you expect or want them to be? I have no idea. If you leave it up to me I'd just assign them as all zeros

x = zeros(1, numElements);
y = zeros(1, numElements);

or maybe random numbers

x = rand(1, numElements);
y = rand(1, numElements);

but you probably woudn't want that