error in for loop

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hossein
hossein 2024-2-19
编辑： Torsten 2024-2-20
Ran in:
We want to fix the problem of the for loop in the code below without changing the boundary conditions. What is your suggestion? Thank you in advance.
clc;
clear;
close all;
R = linspace(-10, 10, 12);
h = 1e-3;
u_values = zeros(1, length(R)); % Array to store u values
% Calculate u for all R values
for j = 1:length(R)
u = (4 * R(j).^5) - (6 * R(j).^3); % Calculate u for current R
u_values(j) = u; % Store u value
fprintf('u for R = %.2f: %f\n', R(j), u);
if j==1
derivative = (1/(60*h) * (u(j+3) - u(j-3))) - (3/(20*h) * (u(j+2) - u(j-2))) + (3/(4*h) * (u(j+1) - u(j-1)));
end
end
u for R = -10.00: -394000.000000
Index exceeds the number of array elements. Index must not exceed 1.
this error:
Index exceeds the number of array elements (1).
Error in Untitled2 (line 17)
derivative = (1/(60*h) * (u(j+1) - u(j-1))) - (3/(20*h) * (u(j) - u(j-2))) + (3/(4*h) * (u(j) - u(j-1)));

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回答（1 个）

Torsten
Torsten 2024-2-19
移动：Torsten 2024-2-19
u(j-3) is only defined for j>=4, u(j+3) is only defined for j<=length(R)-3. Similar restrictions hold for the other indices. Thus your loop must start with j = 4 and end with length(R)-3.
  2 个评论
Walter Roberson
Walter Roberson 2024-2-19
if j==1
derivative = (1/(60*h) * (u(j+3) - u(j-3))) - (3/(20*h) * (u(j+2) - u(j-2))) + (3/(4*h) * (u(j+1) - u(j-1)));
If you start the loop at j = 4, then j==1 will never be true, so the derivative would never be calculated.
Torsten
Torsten 2024-2-19
编辑：Torsten 2024-2-20
Ran in:
ok, I correct:
Thus your loop must start with j = 4, end with length(R)-3 and you can remove the if-statement.
Further, the vector u must be defined before entering the loop.
R = linspace(-10, 10, 120);
h = R(2)-R(1);
u = 4 * R.^5 - 6 * R.^3;
derivative = nan(size(u));
for j = 4:length(R)-3
derivative(j) = (1/(60*h) * (u(j+3) - u(j-3))) - (3/(20*h) * (u(j+2) - u(j-2))) + (3/(4*h) * (u(j+1) - u(j-1)));
end
hold on
plot(R,derivative)
plot(R,20*R.^4-18*R.^2,'o')
hold off
grid on
If you need derivatives nearer to the boundaries, you will have to use a method of lower order that uses a smaller stencil.

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