Related
My Code right now
% Create some example points x and y
t = pi*[0:.05:1,1.1,1.2:.02:2]; a = 3/2*sqrt(2);
for i=1:size(t,2)
x(i) = a*sqrt(2)*cos(t(i))/(sin(t(i)).^2+1);
y(i) = a*sqrt(2)*cos(t(i))*sin(t(i))/(sin(t(i))^2+1);
end
Please note: The points (x_i|y_i) are not necessarily equidistant, that's why t is created like this. Also t should not be used in further code as for my real problems it is not known, I just get a bunch of x, y and z values in the end. For this example I reduced it to 2D.
Now I'm creating ParametricSplines for the x and y values
% Spline
n=100; [x_t, y_t, tt] = ParametricSpline(x, y, n);
xref = ppval(x_t, tt); yref = ppval(y_t, tt);
with the function
function [ x_t, y_t, t_t ] = ParametricSpline(x,y,n)
m = length(x);
t = zeros(m, 1);
for i=2:m
arc_length = sqrt((x(i)-x(i-1))^2 + (y(i)-y(i-1))^2);
t(i) = t(i-1) + arc_length;
end
t=t./t(length(t));
x_t = spline(t, x);
y_t = spline(t, y);
t_t = linspace(0,1,n);
end
The plot generated by
plot(x,y,'ob',...
xref,yref,'xk',...
xref,yref,'-r'),...
axis equal;
looks like the follows: Plot Spline
The Question:
How do I change the code so I always have one of the resulting points (xref_i|yref_i) (shown as Black X in the plot) directly on the originally given points (x_j|y_j) (shown as Blue O) with additionally n points between (x_j|y_j) and (x_j+1|y_j+1)?
E.g. with n=2 I would like to get the following:
(xref_1|yref_1) = (x_1|y_1)
(xref_2|yref_2)
(xref_3|yref_3)
(xref_4|yref_4) = (x_2|y_2)
(xref_5|yref_5)
[...]
I guess the only thing I need is to change the definition of tt but I just can't figure out how... Thanks for your help!
Use this as your function:
function [ x_t, y_t, tt ] = ParametricSpline(x,y,nt)
arc_length = 0;
n = length(x);
t = zeros(n, 1);
mul_p = linspace(0,1,nt+2)';
mul_p = mul_p(2:end);
tt = t(1);
for i=2:n
arc_length = sqrt((x(i)-x(i-1))^2 + (y(i)-y(i-1))^2);
t(i) = t(i-1) + arc_length;
add_points = mul_p * arc_length + t(i-1);
tt = [tt ; add_points];
end
t=t./t(end);
tt = tt./tt(end);
x_t = spline(t, x);
y_t = spline(t, y);
end
The essence:
You have to construct tt in the same way as your distance vector t plus add additional nt points in between.
Introduction
I am using Matlab to simulate some dynamic systems through numerically solving systems of Second Order Ordinary Differential Equations using ODE45. I found a great tutorial from Mathworks (link for tutorial at end) on how to do this.
In the tutorial the system of equations is explicit in x and y as shown below:
x''=-D(y) * x' * sqrt(x'^2 + y'^2)
y''=-D(y) * y' * sqrt(x'^2 + y'^2) + g(y)
Both equations above have form y'' = f(x, x', y, y')
Question
However, I am coming across systems of equations where the variables can not be solved for explicitly as shown in the example. For example one of the systems has the following set of 3 second order ordinary differential equations:
y double prime equation
y'' - .5*L*(x''*sin(x) + x'^2*cos(x) + (k/m)*y - g = 0
x double prime equation
.33*L^2*x'' - .5*L*y''sin(x) - .33*L^2*C*cos(x) + .5*g*L*sin(x) = 0
A single prime is first derivative
A double prime is second derivative
L, g, m, k, and C are given parameters.
How can Matlab be used to numerically solve a set of second order ordinary differential equations where second order can not be explicitly solved for?
Thanks!
Your second system has the form
a11*x'' + a12*y'' = f1(x,y,x',y')
a21*x'' + a22*y'' = f2(x,y,x',y')
which you can solve as a linear system
[x'', y''] = A\f
or in this case explicitly using Cramer's rule
x'' = ( a22*f1 - a12*f2 ) / (a11*a22 - a12*a21)
y'' accordingly.
I would strongly recommend leaving the intermediate variables in the code to reduce chances for typing errors and avoid multiple computation of the same expressions.
Code could look like this (untested)
function dz = odefunc(t,z)
x=z(1); dx=z(2); y=z(3); dy=z(4);
A = [ [-.5*L*sin(x), 1] ; [.33*L^2, -0.5*L*sin(x)] ]
b = [ [dx^2*cos(x) + (k/m)*y-g]; [-.33*L^2*C*cos(x) + .5*g*L*sin(x)] ]
d2 = A\b
dz = [ dx, d2(1), dy, d2(2) ]
end
Yes your method is correct!
I post the following code below:
%Rotating Pendulum Sym Main
clc
clear all;
%Define parameters
global M K L g C;
M = 1;
K = 25.6;
L = 1;
C = 1;
g = 9.8;
% define initial values for theta, thetad, del, deld
e_0 = 1;
ed_0 = 0;
theta_0 = 0;
thetad_0 = .5;
initialValues = [e_0, ed_0, theta_0, thetad_0];
% Set a timespan
t_initial = 0;
t_final = 36;
dt = .01;
N = (t_final - t_initial)/dt;
timeSpan = linspace(t_final, t_initial, N);
% Run ode45 to get z (theta, thetad, del, deld)
[t, z] = ode45(#RotSpngHndl, timeSpan, initialValues);
%initialize variables
e = zeros(N,1);
ed = zeros(N,1);
theta = zeros(N,1);
thetad = zeros(N,1);
T = zeros(N,1);
V = zeros(N,1);
x = zeros(N,1);
y = zeros(N,1);
for i = 1:N
e(i) = z(i, 1);
ed(i) = z(i, 2);
theta(i) = z(i, 3);
thetad(i) = z(i, 4);
T(i) = .5*M*(ed(i)^2 + (1/3)*L^2*C*sin(theta(i)) + (1/3)*L^2*thetad(i)^2 - L*ed(i)*thetad(i)*sin(theta(i)));
V(i) = -M*g*(e(i) + .5*L*cos(theta(i)));
E(i) = T(i) + V(i);
end
figure(1)
plot(t, T,'r');
hold on;
plot(t, V,'b');
plot(t,E,'y');
title('Energy');
xlabel('time(sec)');
legend('Kinetic Energy', 'Potential Energy', 'Total Energy');
Here is function handle file for ode45:
function dz = RotSpngHndl(~, z)
% Define Global Parameters
global M K L g C
A = [1, -.5*L*sin(z(3));
-.5*L*sin(z(3)), (1/3)*L^2];
b = [.5*L*z(4)^2*cos(z(3)) - (K/M)*z(1) + g;
(1/3)*L^2*C*cos(z(3)) + .5*g*L*sin(z(3))];
X = A\b;
% return column vector [ed; edd; ed; edd]
dz = [z(2);
X(1);
z(4);
X(2)];
I have three vectors x,y,t. For each combination x,y,t there is a (u,v) value associated with it. How to plot this in matlab? Actually I'm trying to plot the solution of 2d hyperbolic equation
vt = A1vx + A2vy where A1 and A2 are 2*2 matrices and v is a 2*1 vector. I was trying scatter3 and quiver3 but being new to matlab I'm not able to represent the solution correctly.
In the below code I have plot at only a particular time-level. How to show the complete solution in just one plot? Any help?
A1 = [5/3 2/3; 1/3 4/3];
A2 = [-1 -2; -1 0];
M = 10;
N = 40;
delta_x = 1/M;
delta_y = delta_x;
delta_t = 1/N;
x_points = 0:delta_x:1;
y_points = 0:delta_y:1;
t_points = 0:delta_t:1;
u = zeros(M+1,M+1,N+1,2);
for i=1:M+1,
for j=1:M+1,
u(i,j,1,1) = (sin(pi*x_points(i)))*sin(2*pi*y_points(j)) ;
u(i,j,1,2) = cos(2*pi*x_points(i));
end
end
for j=1:M+1,
for t=1:N+1,
u(M+1,j,t,1) = sin(2*t);
u(M+1,j,t,2) = cos(2*t);
end
end
for i=1:M+1
for t=1:N+1
u(i,1,t,1) = sin(2*t);
u(i,M+1,t,2) = sin(5*t) ;
end
end
Rx = delta_t/delta_x;
Ry = delta_t/delta_y;
for t=2:N+1
v = zeros(M+1,M+1,2);
for i=2:M,
for j=2:M,
A = [(u(i+1,j,t-1,1) - u(i-1,j,t-1,1)) ; (u(i+1,j,t-1,2) - u(i-1,j,t-1,2))];
B = [(u(i+1,j,t-1,1) -2*u(i,j,t-1,1) +u(i-1,j,t-1,1)) ; (u(i+1,j,t-1,2) -2*u(i,j,t-1,2) +u(i-1,j,t-1,2))];
C = [u(i,j,t-1,1) ; u(i,j,t-1,2)];
v(i,j,:) = C + Rx*A1*A/2 + Rx*Rx*A1*A1*B/2;
end
end
for i=2:M,
for j=2:M,
A = [(v(i,j+1,1) - v(i,j-1,1)) ; (v(i,j+1,2) - v(i,j-1,2)) ];
B = [(v(i,j+1,1) - 2*v(i,j,1) +v(i,j-1,1)) ; (v(i,j+1,2) - 2*v(i,j,2) +v(i,j-1,2))];
C = [v(i,j,1) ; v(i,j,2)];
u(i,j,t,:) = C + Ry*A2*A/2 + Ry*Ry*A2*A2*B/2;
end
end
if j==2
u(i,1,t,2) = u(i,2,t,2);
end
if j==M
u(i,M+1,t,1) = u(i,M,t,1);
end
if i==2
u(1,j,t,:) = u(2,j,t,:) ;
end
end
time_level = 2;
quiver(x_points, y_points, u(:,:,time_level,1), u(:,:,time_level,2))
You can plot it in 3D, but personally I think it would be hard to make sense of.
There's a quiver3 equivalent for your plotting function. z-axis in this case would be time (say, equally spaced), and z components of the vectors would be zero. Unlike 2D version of this function, it does not support passing in coordinate vectors, so you need to create the grid explicitly using meshgrid:
sz = size(u);
[X, Y, Z] = meshgrid(x_points, y_points, 1:sz(3));
quiver3(X, Y, Z, u(:,:,:,1), u(:,:,:,2), zeros(sz(1:3)));
You may also color each timescale differently by plotting them one at a time, but it's still hard to make sense of the results:
figure(); hold('all');
for z = 1:sz(3)
[X, Y, Z] = meshgrid(x_points, y_points, z);
quiver3(X, Y, Z, u(:,:,z,1), u(:,:,z,2), zeros([sz(1:2),1]));
end
I'm trying to solve a system of ordinary differential equations from y_1 to y_2, say
G' = D
D' = f(y,G,D)
with the initial conditions
G(y_1) = 0
D(y_1) = 0
My problem is that I do not know y_1 and y_2, to counter for this I of course need two additional equations which is
F_1(y_1,y_2,G(y_2)) = 0
F_2(y_1,y_2,G(y_2)) = 0
So far I have tried to implement it using fsolve (i guess the new name is fzero) to find the zeros, then from the functions F_1 and F_2 i call the ode45 to solve from y_1 to y_2 in order to calculate the functions. However it does not work and I can not seem to find any mistakes. Therefore i looked for new methods/ideas and I found the method bvp4c, but I'm not sure whether i can use it in my case. Does anyone have experience with bvp4c and know whether and how I should use it, or do you have other ideas?
Any help is appreciated.
Code for reference:
function [Fval,sol,t,G] = EntrepreneurialFinanceNondiversifiableRisk
global sigma rf tau b epsilon gamma theta1 theta2 alpha ya K I taug mu eta...
omega
sigma=0.2236; rf=0.03; tau=0.1129; b=0.85; epsilon=0.2; gamma=2;
theta1 = -0.704; alpha = 0.6; theta2=1.704; ya=0.1438; K=27; I = 10;
taug = 0.1; mu = 0.04; eta = 0.4; omega = 0.1;
tau = 0;
option = optimset('Display','iter');
sol = fsolve(#f,[0.1483,2.8],option);
Fval = f(sol);
[t,G] = solvediff(sol(1),sol(2));
function F = f(x)
yd = x(1);
yu = x(2);
global rf tau b theta1 alpha theta2 K I taug
Vstar =# (y) (1-tau+tau*(1-theta1-(1-alpha)*(1-tau)* theta1/tau)^...
(1/theta1))*y/rf;
VstarPrime = (1-tau+tau*(1-theta1-(1-alpha)*(1-tau)...
*theta1/tau)^(1/theta1))*1/rf;
qbar =#(yd,yu) (yd^theta1-yd^theta2)/(yu^theta2*yd^theta1-yu^theta1*...
yd^theta2);
qunderbar =# (yd,yu) (yu^theta2-yu^theta1)/(yu^theta2*yd^theta1-...
yu^theta1*yd^theta2);
A =# (y) (1-tau)*(y/rf);
V = #(y,yd) A(y) + (tau * b)/rf * (1 - (y/yd)^(theta2)) - (1-alpha)*A(yd)*...
(y/yd)^(theta2);
F0 =# (yd,yu) b/rf-(b/rf-alpha*A(yd))*qunderbar(yd,yu)/(1-qbar(yd,yu));
[t,G] = solvediff(yd,yu);
F = zeros(2,1);
F(1) = Vstar(yu)-F0(yd,yu)-K-taug*(Vstar(yu)-K-I)-G(end,2);
F(2) = (1-taug)*VstarPrime-G(end,2);
function [t,G] = solvediff(yd,yu)
[t,G] = ode45(#diff,[yd,yu],[0,0]);
Assuming the domain of definition is big enough and you have good initial values, a Newton implementation with divided differences for the construction of the Jacobian looks like this:
h=1e-5
for k=1:10 do
Fval = F(yd, yu)
dFdyd = (F(yd+h, yu)-F(yd-h, yu))/(2*h)
dFdyu = (F(yd, yu+h)-F(yd, yu-h))/(2*h)
dF = [ dFdyd dFdyu ]
ynext = [ yd; yu ] - dF^(-1)*Fval;
yd = ynext(1); yu = ynext(2);
end
Got one more problem with matrix multiplication in Matlab. I have to plot Taylor polynomials for the given function. This question is similar to my previous one (but this time, the function is f: R^2 -> R^3) and I can't figure out how to make the matrices in order to make it work...
function example
clf;
M = 40;
N = 20;
% domain of f(x)
x1 = linspace(0,2*pi,M).'*ones(1,N);
x2 = ones(M,1)*linspace(0,2*pi,N);
[y1,y2,y3] = F(x1,x2);
mesh(y1,y2,y3,...
'facecolor','w',...
'edgecolor','k');
axis equal;
axis vis3d;
axis manual;
hold on
% point for our Taylor polynom
xx1 = 3;
xx2 = 0.5;
[yy1,yy2,yy3] = F(xx1,xx2);
% plots one discrete point
plot3(yy1,yy2,yy3,'ro');
[y1,y2,y3] = T1(xx1,xx2,x1,x2);
mesh(y1,y2,y3,...
'facecolor','w',...
'edgecolor','g');
% given function
function [y1,y2,y3] = F(x1,x2)
% constants
R=2; r=1;
y1 = (R+r*cos(x2)).*cos(x1);
y2 = (R+r*cos(x2)).*sin(x1);
y3 = r*sin(x2);
function [y1,y2,y3] = T1(xx1,xx2,x1,x2)
dy = [
-(R + r*cos(xx2))*sin(xx1) -r*cos(xx1)*sin(xx2)
(R + r*cos(xx2))*cos(xx1) -r*sin(xx1)*sin(xx2)
0 r*cos(xx2) ];
y = F(xx1, xx2) + dy.*[x1-xx1; x2-xx2];
function [y1,y2,y3] = T2(xx1,xx2,x1,x2)
% ?
I know that my code is full of mistakes (I just need to fix my T1 function). dy represents Jacobian matrix (total derivation of f(x) - I hope I got it right...). I am not sure how would the Hessian matrix in T2 look, by I hope I will figure it out, I'm just lost in Matlab...
edit: I tried to improve my formatting - here's my Jacobian matrix
[-(R + r*cos(xx2))*sin(xx1), -r*cos(xx1)*sin(xx2)...
(R + r*cos(xx2))*cos(xx1), -r*sin(xx1)*sin(xx2)...
0, r*cos(xx2)];
function [y1,y2,y3]=T1(xx1,xx2,x1,x2)
R=2; r=1;
%derivatives
y1dx1 = -(R + r * cos(xx2)) * sin(xx1);
y1dx2 = -r * cos(xx1) * sin(xx2);
y2dx1 = (R + r * cos(xx2)) * cos(xx1);
y2dx2 = -r * sin(xx1) * sin(xx2);
y3dx1 = 0;
y3dx2 = r * cos(xx2);
%T1
[f1, f2, f3] = F(xx1, xx2);
y1 = f1 + y1dx1*(x1-xx1) + y1dx2*(x2-xx2);
y2 = f2 + y2dx1*(x1-xx1) + y2dx2*(x2-xx2);
y3 = f3 + y3dx1*(x1-xx1) + y3dx2*(x2-xx2);