I wrote this code for my homework on MATLAB about cubic spline interpolation with a tridiagonal matrix. I follow all the steps of the algorithm and I really don't find my error.
The code it's ok with second grade functions but when I put, for example, sin(x) the result is not a spline and don't know why because with the other function I have no problems. Can anyone help me to find the error? Thanks
CUBIC SPLINE SCRIPT:
close all
clear all
clf reset
ff = #(x) sin(x);
x = [-2 0 2 4 6 7 8 9 10 11 12];
for i = 1: length(x),
omega(i) = ff(x(i));
end
n = length(x);
h = zeros(n - 1, 1);
for i = 1: n - 1,
h(i) = x(i + 1) - x(i);
end
a = zeros(n, 1);
b = zeros(n, 1);
d = zeros(n, 1);
f = zeros(n, 1);
for j = 2: n - 1,
a(j) = 2*(h(j) + h(j - 1));
b(j) = h(j - 1);
d(j) = h(j);
f(j) = 6 * (omega(j + 1) - omega(j) / h(j)) - (6 * (omega(j) - omega(j - 1)) / h(j - 1));
end
% Starting conditions
a(1) = -2;
f(1) = 0;
a(n) = 4;
f(n) = 0;
% Coefficents
c = tridiag(a, b, d, f);
t = linspace (x(1), x(n), 301);
for k = 1: length(t),
tk = t(k);
y(k) = spline_aux(x, omega, c, tk);
z(k) = ff(tk);
end
plot(t, z, 'b-', 'linewidth', 2)
hold on;
plot(t, y, 'r.', x, omega, 'go')
grid on
TRIADIAGONAL MATRIX:
function [x] = tridiag(a, b, d, f)
n = length(f);
alfa = zeros(n, 1);
beta = zeros(n, 1);
alfa(1) = a(1);
for i = 2: n,
beta(i) = b(i) / alfa(i - 1);
fprintf (' i: %d beta: %12.8f\n', i, beta(i))
alfa(i) = a(i) - (beta(i)*d(i - 1));
fprintf (' i: %d alfa: %12.8f\n', i, alfa(i))
end
y(1) = f(1);
for i = 2: n,
y(i) = f(i) - beta(i)*y(i - 1);
end
x(n) = y(n) / alfa(n);
for i = n - 1: 1,
x(i) = (y(i) - (d(i)*x(i + 1))) / alfa(i);
end
SPLINE EVALUATION IN tk POINT:
function [s] = spline_aux(x, w, c, tk)
n = length(x);
h = zeros(n - 1, 1);
for i = 1: n - 1,
h(i) = x(i+1) - x(i);
end
for i = 1: n - 1,
if (x(i) <= tk && tk <= x(i+1))
break
end
end
s1 = c(i)*((x(i+1) - tk)^3)/(6*h(i));
s2 = c(i+1)*((tk - x(i))^3)/(6*h(i));
s3 = (w(i)/h(i) - (c(i)*h(i)/6))*(x(i+1) - tk);
s4 = (w(i+1)/h(i) - (c(i+1)*h(i)/6))*(tk - x(i));
s = s1 + s2 + s3 + s4;
Thats because you are not using Matlab's for correctly
in the function function [x] = tridiag(a, b, d, f)
the last for reads for i = n - 1: 1 but that will never execute, you shoudl writte:
for i = n - 1:-1:1
Then works. You should notice that it didt work in ANY previous attempt, not only with sin(x)
Related
I wrote a finite difference algorithm to solve the wave equation which is derived here.
When I ran my code, the plotted graphs of the numerical and analytical solution deviated, which is the problem I am trying to solve. The finite difference algorithm is given in the snippet.
t_min = 0;
t_max = 10;
rps = 400; % Resolution per second
nt = t_max * rps;
t = linspace(t_min, t_max, nt);
dt = t(2) - t(1);
x_min = 0;
x_max = 8;
rpm = 100; % Resolution per menter
nx = x_max * rpm;
x = linspace(x_min, x_max, nx);
dx = x(2) - x(1);
c = 3; % Wave speed
A = pi; % Amplitude
L = x_max; % Rod lenght
w_o = 1; % Circular frequency
Cn = c * (dt / dx); % Courrant number
if Cn > 1 % Stability criteria
error('The stability condition is not satisfied.');
end
U = zeros(nx, nt);
U(:, 1) = zeros(nx, 1);
U(:, 2) = zeros(nx, 1);
U(1, :) = A * sin(w_o * t);
U(end, :) = zeros(1, nt);
for j = 2 : (nt - 1)
for i = 2 : (nx - 1)
U(i, j + 1) = Cn^2 * ( U(i + 1, j) - 2 * U(i, j) + U(i - 1, j) ) + 2 * U(i, j) - U(i, j - 1);
end
end
figure('Name', 'Numeric solution');
surface(t, x, U, 'edgecolor', 'none');
grid on
colormap(jet(256));
colorbar;
xlabel('t ({\its})');
ylabel('x ({\itm})');
title('U(t, x) ({\itm})');
To find the bug, I tryed to change the boundary conditions and see if my graph would look better. It turned out that it did, which means that the code in my double for loop is ok. The boundary conditions are the problem. However, I do not know why the code works with the new boundary conditions and does not with the old ones. I am hoping that somebody will point this out to me. The code I ran is given in the snippet.
t_min = 0;
t_max = 1;
rps = 400; % Resolution per second
nt = t_max * rps;
t = linspace(t_min, t_max, nt);
dt = t(2) - t(1);
x_min = 0;
x_max = 1;
rpm = 100; % Resolution per menter
nx = x_max * rpm;
x = linspace(x_min, x_max, nx);
dx = x(2) - x(1);
c = 3; % Wave speed
A = pi; % Amplitude
L = x_max; % Rod lenght
w_o = 1; % Circular frequency
Cn = c * (dt / dx); % Courrant number
if Cn > 1 % Stability criteria
error('The stability condition is not satisfied.');
end
U = zeros(nx, nt);
U(:, 1) = sin(pi*x);
U(:, 2) = sin(pi*x) * (1 + dt);
U(1, :) = zeros(1, nt);
U(end, :) = zeros(1, nt);
for j = 2 : (nt - 1)
for i = 2 : (nx - 1)
U(i, j + 1) = Cn^2 * ( U(i + 1, j) - 2 * U(i, j) + U(i - 1, j) ) + 2 * U(i, j) - U(i, j - 1);
end
end
figure('Name', 'Numeric solution');
surface(t, x, U, 'edgecolor', 'none');
grid on
colormap(jet(256));
colorbar;
xlabel('t ({\its})');
ylabel('x ({\itm})');
title('U(t, x) ({\itm})');
I am trying to solve this problem. But I keep getting an error.
This is my First Code.
% Program 3.3
function [L, U, P] = lufact(A)
[N, N] = size(A);
X = zeros(N, 1);
Y = zeros(N, 1);
C = zeros(1, N);
R = 1:N;
for p = 1: N-1
[max1, j] = max(abs(A(p:N, p)));
C = A(p,:);
A(p,:) = A(j + p - 1,:);
A(j + p -1, :) = C;
d = R(p);
R(p) = R(j + p -1);
R(j + p - 1) = d;
if A(p,p) == 0
'A is Singular. No unique Solution'
break
end
for k = p + 1:N
mult = A(k,p)/A(p,p);
A(k,p) = mult;
A(k,p + 1:N) = A(k, p + 1:N) - mult *A(p, p + 1:N);
end
I=(1:N)'*ones(1,N,1); J=I';
L = (I>J).*A + eye(N);
U = (J>=I).*A;
P = zeros(N);
for k=1:N
P(k,R(k))=1;
end
end
X(N) = Y(N)/A(N,N);
for k = N-1: -1: 1
X(k) = (Y(k) - A(k, k+1:N)*X(k+1:N))/A(k,k);
end
And This is my 2nd Code which I'm using to solve this problem.
function B = Ques3(A)
% Computes the inverse of a matrix A
[L,U,P] = lufact(A);
N = max(size(A));
I = eye(N);
B = zeros(N);
for j = 1:N
Y = forsub(L,P*I(:,j));
B(:,j) = backsub(U,Y);
end
But I keep getting an error in MATLAB,
>> Ques3(A)
Unrecognized function or variable 'forsub'.
Error in Ques3 (line 12)
Y = forsub(L,P*I(:,j));
I am trying to evaluate two matrixes which I defined outside of the function MetNewtonSist using subs and I get the error Undefined function or variable 'x' whenever I try to run the code.
[edit] I added the code for the GaussPivTot function which determines the solution of a liniear system.
syms x y
f1 = x^2 + y^2 -4;
f2 = (x^2)/8 - y;
J = jacobian( [ f1, f2 ], [x, y]);
F = [f1; f2];
subs(J, {x,y}, {1, 1})
eps = 10^(-6);
[ x_aprox,y_aprox, N ] = MetNewtonSist( F, J, 1, 1, eps )
function [x_aprox, y_aprox, N] = MetNewtonSist(F, J, x0, y0, eps)
k = 1;
x_v(1) = x0;
y_v(1) = y0;
while true
k = k + 1;
z = GaussPivTot(subs(J, {x, y}, {x_v(k-1), y_v(k-1)}),-subs(F,{x, y}, {x_v(k-1), y_v(k-1)}));
x_v(k) = z(1) + x_v(k-1);
y_v(k) = z(1) + y_v(k-1);
if norm(z)/norm([x_v(k-1), y_v(k-1)]) < eps
return
end
end
N = k;
x_aprox = x_v(k);
y_aprox = y_v(k);
end
function [x] = GaussPivTot(A,b)
n = length(b);
A = [A,b];
index = 1:n;
for k = 1:n-1
max = 0;
for i = k:n
for j = k:n
if A(i,j) > max
max = A(i,j);
p = i;
m = j;
end
end
end
if A(p,m) == 0
disp('Sist. incomp. sau comp. nedet.')
return;
end
if p ~= k
aux_line = A(p,:);
A(p,:) = A(k, :);
A(k,:) = aux_line;
end
if m ~= k
aux_col = A(:,m);
A(:,m) = A(:,k);
A(:,k) = aux_col;
aux_index = index(m);
index(m) = index(k);
index(k) = aux_index;
end
for l = k+1:n
M(l,k) = A(l,k)/A(k,k);
aux_line = A(l,:);
A(l,:) = aux_line - M(l,k)*A(k,:);
end
end
if A(n,n) == 0
disp('Sist. incomp. sau comp. nedet.')
return;
end
y = SubsDesc(A, A(:,n+1));
for i = 1:n
x(index(i)) = y(i);
end
end
By default, eps is defined as 2.2204e-16 in MATLAB. So do not overwrite it with your variable and name it any word else.
epsilon = 1e-6;
Coming to your actual issue, pass x and y as input arguments to the MetNewtonSist function. i.e. define MetNewtonSist as:
function [x_aprox, y_aprox, N] = MetNewtonSist(F, J, x0, y0, epsilon, x, y)
%added x and y and renamed eps to epsilon
and then call it with:
[x_aprox, y_aprox, N] = MetNewtonSist(F, J, 1, 1, epsilon, x, y);
I am new to Matlab and trying to find a solution to the error of my code:
Not enough input arguments.
Error in F9>f (line 42)
y = (2 - 2*t*x) / (x^2 + 1) ;
Error in F9 (line 18)
e = euler(f, trange(1), y0_value, h, trange(end));
function [] = F9()
% Euler's Method to solve given functions
% Set initial values
hi = [1/2, 1/4];
trange = [0, 2];
y0_value = 1;
% Set functions' and exact functions' handles
% Calculate and show results
% Loop for functions
for i = 1:2
fprintf('###########\n');
fprintf('Function #%d\n', i)
fprintf('###########\n');
exact_value = f_exact(trange(end));
% Loop for h
for h = hi
% Euler calculations
e = euler(f, trange(1), y0_value, h, trange(end));
fprintf('\nh: %f\n', h);
fprintf('\nEuler: %f \n', e(end));
fprintf('Error: %f\n\n', abs((e(end)-exact_value)/exact_value));
end
fprintf('Exact: %f\n\n', exact_value);
end
end
% Euler's Method
function y = euler(f, t0, y0, h, tn)
n = (tn-t0)/h;
% Initialize t, y
[t, y] = deal(zeros(n, 1));
% Set t0, y0
t(1) = t0;
y(1) = y0;
for i = 1:n
t(i+1) = t(i) + h;
y(i+1) = y(i) + h/2 * (f(t(i), y(i))+ f(t(i+1) , y(i) + h * f(t(i), y(i))));
end
end
% Functions to solve
function y = f(t, x)
y = (2 - 2*t*x) / (x^2 + 1) ;
end
function y = f_exact(x)
y = (2*x + 1) / (x^2 + 1);
end
When you pass f to euler you need to pass it as a handle, i.e. precede it with a #:
e = euler(#f, trange(1), y0_value, h, trange(end));
this is my first time here so I hope that someone can help me.
I'm trying to implementing the Gauss-Seidel method and the power method using a matrix with the storage CSR or called Morse storage. Unfortunately I can't manage to do better then the following codes:
GS-MORSE:
function [y] = gs_morse(aa, diag, col, row, nmax, tol)
[n, n] = size(A);
y = [1, 1, 1, 1];
m = 1;
while m < nmax,
for i = 1: n,
k1 = row(i);
k2 = row(i + 1) - 1;
for k = k1: k2,
y(i) = y(i) + aa(k) * x(col(k));
y(col(k)) = y(col(k)) + aa(k) * diag(i);
end
k2 = k2 + 1;
y(i) = y(i) + aa(k) * diag(i);
end
if (norm(y - x)) < tol
disp(y);
end
m = m + 1;
for i = 1: n,
x(i) = y(i);
end
end
POWER-MORSE:
I was able only to implement the power method but I don't understand how to use the former matrix... so my code for power method is:
function [y, l] = potencia_iterada(A, v)
numiter=100;
eps=1e-10;
x = v(:);
y = x/norm(x);
l = 0;
for k = 1: numiter,
x = A * y;
y = x / norm(x);
l0 = x.' * y;
if abs(l0) < eps
return
end
l = l0;
end
Please anyone can help me for completing these codes or can explain me how can I do that? I really don't understand how to do. Thank you very much