predictor corrector method error, what should i do?

hey all;

im trying to solve second order ODE using RK4 and predictor-correctoe method without using any built in mtlab function

here is my code :

clear all;
close all;
clc;

h=0.1;          %step size (changable according to the proplem)
x=0:h:1; %the X domain range
yic = [[1;-2],zeros(2, length(x)-1)]; %intial condition in form of matrix
                                      %(changable according to the proplem)
%*********************************************
%             Exact solution
%*********************************************
y_exact=exp(-2*x); %define your equation for exact solution
%**********************************************

%********* Numerical solution *****************
%% RK4th order definition 
for i = 1:6
    if i<=3
        
        K1 = fn(x(i), yic(:, i));
        K2 = fn(x(i) + h/2, yic(:, i) + h*K1/2);
        K3 = fn(x(i) + h/2, yic(:, i) + h*K2/2);
        K4 = fn(x(i) + h,   yic(:, i) + h*K3);
        yic(:, i+1) = yic(:, i) + h/6*(K1 + 2*K2 + 2*K3 + K4);
    else if i>=4
            y_star = yic(:,i)+((h/24).*(55.*fn(x(i),yic(i))-(59.*fn(x(i),yic(:,i-1)))+(37.*fn(x(i),yic(i-2)))-(9.*fn(x(i),yic(i-3)))));
             
            y6 = yic(:,i) + ((h/24).*(9.*fn(x(i+1),y_star(1,:)) +(19.*fn(x(i),yic(:,i)))+(5.*fn(x(i-1),yic(:,i-1))) +(fn(x(i-2) , yic(:,i-2)))));
                 
            
        end
    end
            
end

%% defining the function according to the proplem
function dy = fn(x, y)
    dy = [0,  1
          2, -1] .* y; %change the matrix due to your intital conditins 
                      %and the equation in your proplem
end

here is my error

Unable to perform assignment because the size of the left side is 2-by-1 and the size of the right side is 2-by-2.

Error in HW3_multistep_method (line 32)

yic(:, i+1) = yic(:, i) + h/6*(K1 + 2*K2 + 2*K3 + K4);

what should i do ?

help me pls

Answer 1

[sanidhyak]

Hi @Mohammad Adeeb,

I understand that you are facing some issues while solving a second-order ODE using "RK4" and the "predictor-corrector" method in MATLAB. Upon reviewing the code and the error message I found out that the root cause of this issue lies in the definition of your "fn" function. More specifically, the operation used inside the function is "element-wise multiplication (.*)" between a 2×2 matrix and a 2×1 vector, which leads to a size mismatch. MATLAB expects regular "matrix multiplication (*)" in this case.

To resolve this, please update the "fn" function as shown below:

function dy = fn(x, y)
A = [0, 1;
    2, -1];
dy = A * y;  % Correct matrix multiplication
end

This change ensures that your derivative calculations match the mathematical formulation of your system of ODEs and are compatible with the 2×1 y vector.

Also, kindly make sure all other references to "yic(i)" in your "predictor-corrector" steps are updated to "yic(:, i)" to maintain proper vector indexing. For example,

Before:

fn(x(i), yic(i))

After:

fn(x(i), yic(:, i))

These corrections should completely eliminate the dimension mismatch errors and allow your code to run perfectly.

For further reference, kindly refer to the following official documentation:

• https://www.mathworks.com/help/matlab/matlab_prog/array-vs-matrix-operations.html

I hope this helps!

[Walter Roberson]

More specifically, the operation used inside the function is "element-wise multiplication (.*)" between a 2×2 matrix and a 2×1 vector, which leads to a size mismatch.

For clarity: Since roughly R2015b, MATLAB has had "implicit expansion" that automatically "expands" dimensions of size 1 for binary operators to match the size of the other array. So 2x2 matrix .* 2x1 matrix would be treated similarly to 2x2 matrix .* (repmat, 2x1 matrix, 1, 2) giving a 2 x 2 result. So the 2x2 .* 2x1 giving 2x2 result itself is considered valid.

The challenge is that the assignment is to a 2 x 1 output slot, and a 2 x 2 matrix does not fit into a 2 x 1 output slot.

The real solution in this particular case is as noted, using the * operator instead of the .* operator. But in some cases, the real solution would be to provide a 2 x 2 output area instead of a 2 x 1 output area.

[sanidhyak]

Thank you so much @Walter Roberson for the added clarity!