My ellipse wont close after running the code .

Hy guys . In this task , i have to implement periodic boundary conditons to my ellipse . However , the boundary conditions are not being implemented and i dont know what mistake i'm making if you comment them, the whole code still runs and you can see that the ellipse wont close at the end because the boundary conditions are not working at all. Any suggestions and solutions on how to fix this are highly appreciated.

Below is the code and it has to be saved in 3 files to run it since i use 2 functions and the 3rd file i call the 2 functions.

%---------------------------------Example (i use to call both functions)--------------------------------------
% Example with an ellipse
theta = linspace(0, 2 * pi, 8);
t_x = cos(theta);
t_y = sin(theta);

% Interpolate x-coordinate
[b_x, c_x, d_x] = periodic_cubic_spline_interpolation(theta, t_x);

% Interpolate y-coordinate
[b_y, c_y, d_y] = periodic_cubic_spline_interpolation(theta, t_y);

% Evaluate the spline at points
t_eval = linspace(0, 2 * pi, 500);
x_eval = arrayfun(@(t) evaluate_cubic_spline(theta, t_x, b_x, c_x, d_x, t), t_eval);
y_eval = arrayfun(@(t) evaluate_cubic_spline(theta, t_y, b_y, c_y, d_y, t), t_eval);

% Plot the results
figure;
plot(t_x, t_y, 'ko', 'LineWidth', 1.5);
hold on;
grid on;
plot(x_eval, y_eval, 'r-', 'LineWidth', 2);
title('Cubic Spline Interpolation of Ellipse');
legend('True Ellipse', 'Cubic Spline');
xlabel('X-axis');
ylabel('Y-axis');

%---------------------------------------function 1--------------------------------------------------------
function [b, c, d] = periodic_cubic_spline_interpolation(t, x)
n = length(t);
h = diff(t);

alpha = zeros(n, 1);
beta = zeros(n, 1);
for i = 2:n - 1
    alpha(i) = 3 * (x(i + 1) - x(i)) / h(i) - 3 * (x(i) - x(i - 1)) / h(i - 1);
    beta(i) = 2 * (h(i - 1) + h(i));
end

% Periodic boundary conditions------%(these are not being implemented!!!)%--------------------------------
alpha(1) = 3 * (x(2) - x(1)) / h(1) - 3 * (x(1) - x(n)) / h(n - 1);
alpha(n) = 3 * (x(1) - x(n)) / h(n - 1) - 3 * (x(n) - x(n - 1)) / h(n - 2);
beta(1) = 2 * (h(n - 1) + h(1));
beta(n) = 2 * (h(n - 1) + h(n - 2));




l = zeros(n, 1);
mu = zeros(n, 1);
z = zeros(n, 1);
l(1) = 1;
mu(1) = 0;
z(1) = 0;

for i = 2:n - 1
    l(i) = 2 * (t(i + 1) - t(i - 1)) - h(i - 1) * mu(i - 1);
    mu(i) = h(i) / l(i);
    z(i) = (alpha(i) - h(i - 1) * z(i - 1)) / l(i);
end

l(n) = 1;
z(n) = 0;
c = zeros(n, 1);
b = zeros(n, 1);
d = zeros(n, 1);

for j = n - 1:-1:1
    c(j) = z(j) - mu(j) * c(j + 1);
    b(j) = (x(j + 1) - x(j)) / h(j) - h(j) * (c(j + 1) + 2 * c(j)) / 3;
    d(j) = (c(j + 1) - c(j)) / (3 * h(j));
end
end



%------------------------------------function 2---------------------------------------------------------

function spline_eval = evaluate_cubic_spline(t, x, b, c, d, t_eval)
n = length(t);
j = max(min(find(t >= t_eval, 1) - 1, n - 2), 1);
tj = t(j);
spline_eval = x(j) + b(j) * (t_eval - tj) + c(j) * (t_eval - tj) ^ 2 + d(j) * (t_eval - tj) ^ 3;
end

Answer 1

[Matt J]

Can't you just use the native spline() command?

theta = linspace(0, 2 * pi, 8);
t_x = 2*cos(theta);
t_y = sin(theta);

dt_x=0;  %end slope x
dt_y=(t_y(2)-t_y(1))./(theta(2)-theta(1)); %end slope y
theta_eval = linspace(0, 2 * pi, 500);

t_eval=periodicSpline(theta,[t_x;t_y], [dt_x; dt_y], theta_eval);


plot(t_x, t_y, 'ko', 'LineWidth', 1.5);
hold on;
grid on;
plot(t_eval(1,:), t_eval(2,:), 'r-', 'LineWidth', 2); axis equal; axis padded
title('Cubic Spline Interpolation of Ellipse');
legend('True Ellipse', 'Cubic Spline');
xlabel('X-axis');
ylabel('Y-axis');
hold off

function out = periodicSpline(tdata,xydata,xyslope, tquery)

  xydata(:,end)=xydata(:,1);
  
  xydata=[xyslope, xydata,xyslope];
  
  out=spline(tdata,xydata,tquery);

end

[Panda05]

not really , because the goal of this work is to make sure that i create my own spline function that has a periodic boundary condition that i specified in the first function . However , i dont seem to find a way of making it get implemented in the function itsself.the intention is that if you replace the 8 in theta = linspace(0, 2 * pi, 8); to lets say 20, that ellipse gets better , hence even closes better at the end

[Torsten]

It seems you didn't implement conditions that connect p_1(t_1) and p_(n-1)(t_n) (and the derivatives) because the curve doesn't close (smoothly).

http://www.mosismath.com/PeriodicSplines/PeriodicSplines.html

I think this will destroy the tridiagonal structure of the linear system to be solved and the direct application of Thomas' Algorithm.

[Matt J]

not really , because the goal of this work is to make sure that i create my own spline function that has a periodic boundary condition that i specified in the first function

But you can do this just by making the first and last data point the same before giving it to the spline() command. The native spline command also let's you set the endslopes the same, which I've now modified the answer to do.

[Torsten]

It seems to be a homework problem, and the OP is not allowed to use MATLAB's "spline" function.

[Panda05]

yeah i can't use spline function in this problem . because i have to create one. Yeah it is a homework problem

[Image Analyst]

I've tagged it as homework for you. You might want to read the link below to get started:

How do I get help on homework questions on MATLAB Answers? - MATLAB Answers - MATLAB Central

Obviously we can't give you the full solution because you're not allowed to turn in our code as your own but we can give you suggestions about your own code.

[Torsten]

@MAKEERA

Make the ansatz

p_j(x) = a_j + b_j*(x-x_j) + c_j*(x-x_j)^2 + d_j*(x-x_j)^3  (j=1,...,N-1)

where N is the number of x-data points.

So in total you have 4*(N-1) unknowns a_j, b_j, c_j and d_j that are to be determined from the 4*(N-1) conditions

p_j(x_j) = y_j (j=1,...,N-1)
p_(N-1)(x_N) = p_1(x_1)
p_j(x_j) = p_(j-1)(x_j) (j=2,...N-1)
p_(N-1)'(x_N) = p_1'(x_1)
p_j'(x_j) = p_(j-1)'(x_j)  (j=2,...,N-1)
p_(N-1)''(x_N) = p_1''(x_1)
p_j''(x_j) = p_(j-1)''(x_j) (j=2,...,N-1)

This gives a linear system of equations in the unknowns a_j, b_j, c_j and d_j. Solve it.

[Panda05]

Thank you so much everyone ! I tried to implement that and it worked . May God bless you