predictor corrector method error, what should i do?

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Mohammad Adeeb
Mohammad Adeeb 2021-3-25
评论： sanidhyak 2025-6-26
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

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回答（1 个）

sanidhyak
sanidhyak 2025-6-25
编辑：sanidhyak 2025-6-25
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:
I hope this helps!
  2 个评论
Walter Roberson
Walter Roberson 2025-6-25
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.

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