I'm trying to numerically find the solution to A*cos x +B*sin x = C where A and B are two known square matrices of the same size (for example 100x100), and C is a known vector (100x1).
Without the second term (i.e. with a single matrix), I will use Jacobi or Gauss-Seidel to solve this problem and get x but here, I don't see how to proceed to solve this problem in Matlab.
May be, it would be useful to solve the problem as : A*X + B*sqrt(1-X^2) = C.
I would greatly appreciate any help, ideas or advices
Thanks in advance
If I understood you correctly, you could use fsolve like this (c and X are vectors):
A = ones(2,2);
B = ones(2,2);
c = ones(2,1);
% initial point
x0 = ones(length(A), 1);
% call to fsolve
sol = fsolve(#(x) A * cos(x) + B*sin(x) - c, x0);
Here, we solve the nonlinear equation system F(x) = 0 with F: R^N -> R^N and F(x) = A * cos(x) + B*sin(x) - c.
Only for the sake of completeness, here's my previous answer, i.e. how one could do it in case C and X are matrices instead of vectors:
A = ones(2,2);
B = ones(2,2);
C = ones(2,2);
% initial point
x0 = ones(numel(A), 1);
% call to fsolve
fsolve(#(x) fun(x, A, B, C), x0)
function [y] = fun(x, A, B, C)
% Transform the input vector x into a matrix
X = reshape(x, size(A));
% Evaluate the matrix equation
Y = A * cos(X) + B*sin(X) - C;
% flatten the matrix Y to a row vector y
y = reshape(Y, [], 1);
end
Here, the idea is to transform the matrix equation system F: R^(N x N) -> R^(N x N) into a equivalent nonlinear system F: R^(N*N) -> R^(N*N).
Related
I have stumbled upon this matlab code that solves this ODE
y'''(t) + a y(t) = -b y''(t) + u(t)
but I am confused by the ode_system function definition, specifically by the y(2) y(3) part. I would greatly appreciate if someone can shed some light
y(2) y(3) part in the ode_system function confuses me and how it contributes to overaal solution
% Define the parameters a and b
a = 1;
b = 2;
% Define the time horizon [0,1]
time_horizon = [0, 1];
% Define the initial conditions for y, y', and y''
initials = [0; 0; 0];
% Define the function handle for the input function u(t)
%sin(t) is a common example of a time-varying function.
% You can change the definition of u to any other function of time,
% such as a constant, a step function, or a more complex function, depending on your needs
u = #(t) sin(t);
% Define the function handle for the system of ODEs
odefunction = #(t, y) ode_system(t, y, a, b, u);
% Solve the ODEs using ode45
[t, y] = ode45(odefunction, time_horizon, initials);
% Plot the solution
plot(t, y(:,1), '-', 'LineWidth', 2);
xlabel('t');
ylabel('y');
function dydt = ode_system(t, y, a, b, u)
%Define the system of ODEs
dydt = [y(2); y(3); -b*y(3) + u(t) - a*y(1)];
end
This is more of a maths question than a Matlab one.
We would like to rewrite our ODE equation so that there is a single time derivative on the left-hand side and no derivatives on the right.
Currently we have:
y'''(t)+ay(t)=-by''(t)+u(t)
By letting z = y' and x = z' (= y''), we can rewrite this as:
x'(t)+a y(t)=-b x(t)+u(t)
So now we have 3 equations in the form:
y' = z
z' = x
x' = -b * x + u - a *y
We can also think of this as a vector equation where v = (y, z, x).
The right-hand side would then be,
v(1)' = v(2)
v(2)' = v(3)
v(3)' = -b * v(3) + u - a * v(1)
which is what you have in the question.
I have an equation like this:
dy/dx = a(x)*y + b
where a(x) is a non-constant (a=1/x) and b is a vector (10000 rows).
How can I solve this equation?
Let me assume you would like to write a generic numerical solver for dy/dx = a(x)*y + b. Then you can pass the function a(x) as an argument to the right-hand side function of one of the ODE solvers. e.g.
a = #(x) 1/x;
xdomain = [1 10];
b = rand(10000,1);
y0 = ones(10000,1);
[x,y] = ode45(#(x,y,a,b)a(x)*y + b,xdomain,y0,[],a,b);
plot(x,y)
Here, I've specified the domain of x as xdomain, and the value of y at the bottom limit of x as y0.
From my comments, you can solve this without MATLAB. Assuming non-zero x, you can use an integrating factor to get a 10000-by-1 solution y(x)
y_i(x) = b_i*x*ln(x) + c_i*x
with 10000-by-1 vector of constants c, where y_i(x), b_i and c_i are the i-th entries of y(x), b and c respectively. The constant vector c can be determined at some point x0 as
c_i = y_i(x0)/x_0 - b_i*ln(x0)
The GMRES algorithm and its matlab implementation are supposed to solve linear equations system, such as
%Ax = b
A = rand(4);
b = rand(4,1);
x = gmres(A,b);
One can also use a function handle
foo = #(x) A*x + conj(A)*5*x;
y = gmres(foo,b);
What I want is to solve the following
B = rand(4);
H = rand(4);
foo2 = H*B + B*H;
X = gmres(foo2, B) %Will not run!
--Error using gmres (line 94)
--Right hand side must be a column vector of length 30 to match the coefficient matrix.
Mathematically speaking I don't see why gmres couldn't apply to this problem as well.
Note: What I'm really trying to solve is an implicit euler method for a PDE dB/dt = B_xx + B_yy, so H is in fact a second derivative matrix using finite difference.
Thank you
Amir
If I've understood right you want to use GMRES to solve an a sylvester equation
A*X + X*A = C
for n-by-n matrices A, X and C.
(I asked a related question yesterday over at SciComp and got this great answer.)
To use GMRES you can express this matrix-matrix equation as a size n^2 matrix-vector equation. For convenience we can use the Kronecker product, implemented in MATLAB with kron:
A = randn(5);
X = randi(3,[5 5]);
C = A*X + X*A;
% Use the Kronecker product to form an n^2-by-n^2 matrix
% A*X + X*A
bigA = (kron(eye(5),A) + kron(A.',eye(5)));
% Quick check that we're getting the same answer
norm(bigA*X(:) - C(:))
% Use GMRES to calculate X from A and C.
vec_X_gmres = gmres(bigA,C(:));
X_gmres = reshape(vec_X_gmres,5,5);
How do I deduce replacement algorithms forwards and backwards in the solution phase of the Cholesky method?
How do I compare the function choleskiSol?
Here's my code for choleskisol
function x = choleskiSol(L,b)
% Solves [L][L’]{x} = {b}
% USAGE: x = choleskiSol(L,b)
n = length(b);
if size(b,2) > 1
b = b’;
end % {b} must be column vector
for k = 1:n % Solution of [L]{y} = {b}
b(k) = (b(k) - dot(L(k,1:k-1),b(1:k-1)’))/L(k,k);
end
for k = n:-1:1 % Solution of {L}’{x} = {y}
b(k) = (b(k) - dot(L(k+1:n,k),b(k+1:n)))/L(k,k);
end
x = b;
The standard Cholesky decomposition (chol(A)) in matlab decomposes a symmetric (positive-definite) matrix, A, into upper-triangular form. To solve a linear system of equations, you must simply take the upper-triangular form, and solve it via backward substitution. This will yield the variable values for the system.
To complete the solution in matlab w/ parameter matrix A and output vector B:
L = chol(A); % A must be sym and det(A) > 0
x = (L \ (L' \ b)); % L' is lower-triangular
I have a very simple problem.
I have
x=[ 10 25 50];
y=[ 1.2 3 7.5];
I know my curve fitting function
f(x)=(a*x+1)/(bx+c);
How can I get coefficient(a,b,c) solve in matlab and also plot this curve?
Rearrange y = f(x) to make a, b, and c the unknowns:
y = (ax + 1) / (bx + c)
y(bx + c) = ax + 1
ax - bxy - cy = -1;
This describes a system of simultaneous linear equations in a, b, and c when you substitute your three paired values of x and y.
x = [10, 20, 100];
y = [1.2, 0.7, 0.4];
coeffs = [x', (-x.*y)', -y'];
knowns = [-1, -1, -1]';
v = coeffs \ knowns; % v is [a; b; c]
Now you have the coefficients a, b, and c so you can plot the function.
Addendum: plotting
To plot a function, first choose the x-values of the data points
xt = 1:100;
Then calculate the y-values (assuming you've already got a, b, c)
yt = (a*x + 1) ./ (b*x + c)
Then just plot them!
plot(xt, yt);
Read the Matlab help on the plot function for customizing the style of the plot.