I am attempting to use some of the functions in MATLAB to numerically solve a pair of coupled second order ODEs of the form
\ddot{x} = f(x,y,\dot{x},\dot{y})
\ddot{y} = f(x,y,\dot{x},\dot{y}).
I am able to get it to work with just one second-order ODE, but the code I am trying to does not work for a pair of ODEs.
The function odeToVectorField effectively takes a second order ODE and writes it as a vector for a pair of coupled first order ODEs. ode45 is the usual Runge-Kutta solution method. xInit and yInit correspond to the initial conditions for x and y and the aim is then to plot both x and y against time over a certain interval of time.
gamma1=0.1;
gamma2=0.1;
a=1;
m=1;
g=9.8;
d=1;
syms x(t) y(t)
eqn1=diff(x,2)== (gamma1*diff(x))/(a + m*d^2 + (m/2)*d^2*cos(y-x)) + (gamma2*diff(y))/(a+ (m/2)*cos(y-x)) - ( (m/2)*d^2*sin(y-x)*(diff(x)^2 - diff(y)^2))/(a + m*d^2 + (m/2)*d^2*cos(y-x)) - ((m/2)*d^2*diff(x)^2*(y-x))/(a+ (m/2)*cos(y-x)) - ((m/2)*d*(3*g*sin(x) + g*sin(y)))/(a + m*d^2 + (m/2)*d^2*cos(y-x)) - ((m/2)*d*g*sin(y))/(a+ (m/2)*cos(y-x))
eqn2=diff(y,2)== (gamma1*diff(x))/((m/2)*d^2*cos(y-x)) + (gamma2*diff(y))/a - ( (m/2)*d^2*sin(y-x)*(diff(x)^2 - diff(y)^2))/((m/2)*d^2*cos(y-x)) - ((m/2)*d^2*diff(x)^2*(y-x))/a - ((m/2)*d*(3*g*sin(x) + g*sin(y)))/((m/2)*d^2*cos(y-x)) - ((m/2)*d*g*sin(y))/a
V = odeToVectorField(eqn1,eqn2)
M = matlabFunction(V,'vars',{'t','Y'})
interval = [0 20];
xInit = [2 0];
yInit = [2 0];
ySol = ode45(M,interval,xInit, yInit);
tValues = linspace(0,20,100);
yValues = deval(ySol,tValues,1);
plot(tValues,yValues)
Just to compare, without using symbolic expressions one would implement this equation as
function dV = M(t,V)
gamma1=0.1;
gamma2=0.1;
a=1;
m=1;
g=9.8;
d=1;
x = V(1); dx = V(2); y = V(3); dy = V(4);
ddx = (gamma1*dx)/(a + m*d^2 + (m/2)*d^2*cos(y-x)) + (gamma2*dy)/(a+ (m/2)*cos(y-x)) - ( (m/2)*d^2*sin(y-x)*(dx^2 - dy^2))/(a + m*d^2 + (m/2)*d^2*cos(y-x)) - ((m/2)*d^2*dx^2*(y-x))/(a+ (m/2)*cos(y-x)) - ((m/2)*d*(3*g*sin(x) + g*sin(y)))/(a + m*d^2 + (m/2)*d^2*cos(y-x)) - ((m/2)*d*g*sin(y))/(a+ (m/2)*cos(y-x));
ddy = (gamma1*dx)/((m/2)*d^2*cos(y-x)) + (gamma2*dy)/a - ( (m/2)*d^2*sin(y-x)*(dx^2 - dy^2))/((m/2)*d^2*cos(y-x)) - ((m/2)*d^2*dx^2*(y-x))/a - ((m/2)*d*(3*g*sin(x) + g*sin(y)))/((m/2)*d^2*cos(y-x)) - ((m/2)*d*g*sin(y))/a;
dV = [dx ddx dy ddy];
end%function
interval = [0 20];
xInit = [2 0];
yInit = [2 0];
vSol = ode45(M,interval,[ xInit yInit]);
tValues = linspace(0,20,100);
xValues = deval(vSol,tValues,1);
plot(tValues,xValues)
This works, but reports a singularity between t=0.244 and t=0.245.
I have created a program in Matlab to try to find the root of f(x) = exp(2x) + 3x - 4 (the function "fopg1" in my code). My code is as follows:
format long
tic;
for dum=1:1000;
x(1) = 0.5;
x(2) = 0.4;
err_tol = 1e-8;
iteration(1) = 1;
n = 3;
while err_estimate > err_tol
iteration(n) = n;
x(n) = x(n-1) - fopg1(x(n-2)) * ((x(n-1) - x(n-2)) / (fopg1(x(n-1)) - fopg1(x(n-2))));
err_estimate(n) = abs(x(n) - x(n-1));
n = n + 1;
end
%end
time = toc;
avgtime = time/1000;
A = [iteration' x' err_estimate' tbd'];
f = '%2i %13.9f %13.9f %7.3f'; compose(f,A)
Unfortunately this does not converge. I feel like it should. Is there a flaw in my program or does it in fact not converge? Thanks in advance.
Maarten
I answered a very similar question here a few days ago. Using the same code, without an iterations limit and with an increased tolerance (1e-8 as per your example), I detect the expected convergence of exp(2x) + 3x - 4 using the secant method:
clear();
clc();
com = Inf;
i = 2;
err_tol = 1e-8;
f = #(x) exp(2*x) + 3*x - 4;
x(1) = 0.5;
x(2) = 0.4;
while (abs(com) > err_tol)
com = f(x(i)) * (x(i) - x(i-1)) / (f(x(i)) - f(x(i-1)));
x(i+1)= x(i) - com;
i = i + 1;
n = n - 1;
end
display(['Root X = ' num2str(x(end))]);
The message I receive is: Root X = 0.47369. It shouldn't be hard for you to implement your additional data within this code.
I am using following objective function for optimization:
function Length_Sum = objective_function( l1,l2,l3 )
Length_Sum = l1 + l2 + l3;
end
With constraints function given below, the constrint function uses another function for calculating values of thetas,
function [c, ceq] = simple_constraint(l1,l2,l3)
c(1) = l3^2 + 200*l3*cos(30) + 10000 - (l1 + l2)^2;
c(2) = (100- l3*cos(30))^2 + (100*sin(30))^2 - (l1-l2)^2;
thetas = inverse_kinematics(l1,l2,l3);
c(3) = thetas(4,1) - 160;
c(4) = thetas(4,2) - 160;
c(5) = thetas(4,3) - 160;
c(6) = 20 - thetas(4,1);
c(7) = 20 - thetas(4,2);
c(8) = 20 - thetas(4,3);
c(9) = thetas(5,1) - 340;
c(10) = thetas(5,2) - 340;
c(11) = thetas(5,3) - 340;
c(12) = 200 - thetas(5,1);
c(13) = 200 - thetas(5,2);
c(14) = 200 - thetas(5,3);
c(15) = thetas(6,1) - 340;
c(16) = thetas(6,2) - 340;
c(17) = thetas(6,3) - 340;
c(18) = 200 - thetas(6,1);
c(19) = 200 - thetas(6,2);
c(20) = 200 - thetas(6,3);
ceq = [];
end
Function called by constraint function is given below,
function thetas = inverse_kinematics(l1,l2,l3)
x = 100;
y = 0;
phi = 210*pi/180:60*pi/180:330*pi/180;
x1 = x - (l3*cos(phi));
y1 = y - (l3*sin(phi));
a = sqrt(x1.^2 + y1.^2);
y2 = -y1./a;
x2 = -x1./a;
gamma = atan2(y2,x2);
c = (- x1.^2 - y1.^2 - l1^2 + l2^2)./(2*l1*a);
d = acos(c);
theta1 = gamma + d;
if theta1 < 0
theta1 = theta1 + 2*pi;
end
theta4 = gamma - d;
if theta4 < 0
theta4 = theta4 + 2*pi;
end
e = (y1 - l1*sin(theta1))/l2;
f = (x1 - l1*cos(theta1))/l2;
theta2 = atan2(e,f) - theta1;
if theta2 < 0
theta2 = theta2 + 2*pi;
end
g = (y1 - l1*sin(theta4))/l2;
h = (x1 - l1*cos(theta4))/l2;
theta5 = atan2(g,h) - theta4;
if theta5 < 0
theta5 = theta5 + 2*pi;
end
theta3 = (phi)- (theta1 + theta2);
if theta3 < 0
theta3 = theta3 + 2*pi;
end
theta6 = (phi)- (theta4 + theta5);
if theta6 < 0
theta6 = theta6 + 2*pi;
end
thetas = [theta1;theta2;theta3;theta4;theta5;theta6].*180/pi;
end
After running this code using ga toolbox, with lower bounds [20 20 20] and upper bounds [100 100 100] and rest parameters set to default, I am getting "Error in running optimization. Not enough input arguments" error. Can someone help?
ga accepts constraint function with input in form of one vector with number of elements corresponding to number of constrainded variables. You should change
function [c, ceq] = simple_constraint(l1,l2,l3)
to
function [c, ceq] = simple_constraint(input)
l1 = input(1);
l2 = input(2);
l3 = input(3);
Next time, I suggest you try File->Generate Code... option from the mentioned toolbox. Then you can debug more easily from Matlab window.
There is also another problem in your program. Try running inverse_kinematics(20,20,20). It fails on line 29, but I won't go into details here, because this is not part of the question.
I am trying to find out the root of an equation constraint using Newton's Method (open to any other method). The equation constraint is dependent on the roots of a quadratic equation which has one unknown term. The equation descriptions (quadratic and constraint) are shown below. The roots of the quadratic equation are assumed to be p1 and p2.
x^2 + x*(c1*unknown/2 + C1*c2)/(c1*c2*c3*unknown/2) + 1/(c1*c2*c3*u/2) = 0
with constraint
(1/(p2-p1))*(exp(-0.3*p2) - 1)*(c1*c2 - p2) - (exp(-0.3*p1) - 1)*(c1*c2 - p1) - 0.04 = 0
I am wondering if there are any other approximation methods to solve this problem if Newton's Method is not going to do so.
Matlab Code
p0=10*10^6;
Cc=0.65*10^-6;Rp=100*10^3;Cp=55*10^-9; z1=1/(Cp*Rp);
N = 100;error = 0.02;
syms 'x'
a = 1;
b = ((Cc*(x/2+Rp))/(Cc*Cp*Rp*x/2));
c = 1/(Cc*Cp*Rp*x/2);
poles1 = (-b + sqrt(b^2 - 4*a*c))/(2*a);
poles2 = (-b - sqrt(b^2 - 4*a*c))/(2*a);
p1_subterm = (exp(-0.3*poles1) - 1)*(z1 - poles1);
p2_subterm = (exp(-0.3*poles2) - 1)*(z1- poles2);
f(x) = (1/(poles2 - poles1))*p2_subterm - p2_subterm - 0.04;
df = diff(f);
while i <= N
p = p0-(f(p0)/df(p0))
if (abs(p - p0)/abs(p)) < error
fprintf('Solution is %f \n', double(p))
return
end
i = i + 1;
p0 = p;
end
fprintf('Solution did not converge within %d iterations \n', N)
I don't think your code for f(x) corresponds to the equation listed above: In the MATLAB code 1/(poles2-poles1) multiplies both the exp((...)poles2) and the exp((...)poles1) terms and you have 1/(c1*c2) - poles1 instead of c1*c2 - poles1.
I have not checked through the poles1 and poles2 terms as I find the 10 sets of parenthesis intimidating. It would suggest making your code more readable by breaking up each of those equations in to a number of shorter, clearer lines.
Noting that we can multiply the quadratic by c1*c2*c3*unknown,
(c1*c2*c3*unknown)*x^2 + x*2*(c1*unknown/2 + C1*c2) + 2 = 0
then you can write the poles and f code as
% Here unknown is x
syms 'x'
a = c1*c2*c3*x;
b = c1*x + 2*c1*c2;
c = 2;
poles1 = (-b + sqrt(b^2 - 4*a*c))/(2*a);
poles2 = (-b - sqrt(b^2 - 4*a*c))/(2*a);
p1_subterm = (exp(-0.3*poles1) - 1)*(c1*c2 - poles1);
p2_subterm = (exp(-0.3*poles2) - 1)*(c1*c2 - poles2);
f(x) = (1/(poles2 - poles1))*p2_subterm - p2_subterm - 0.04;
I have a 1024*1024*51 matrix. I'll do calculations to change some value of the matrix within for loops (change the value of matrix for each iteration). I find that the computing speed gets slower and slower and finally my computer gets into trouble.However the size of the matrix doesn't change. Anyone can shed some light on this problem?
function ActiveContours3D(method,grad,im,mu,nu,lambda1,lambda2,TimeSteps)
epsilon = 10e-10;
tic
fid=fopen('Chr18_z_25of25tiles-C=0_c0_n000.raw','rb','ieee-le');
Xdim = 1024;
Ydim = 1024;
Zdim = 51;
A = fread(fid,[Xdim Ydim*Zdim],'int16');
A = double(A);
size_of_A = size(A)
for(i=1:Zdim)
u0_color(:,:,i) = A(1 : Xdim , (i-1)*Ydim+1 : i*Ydim);
end
fclose(fid)
time = toc
[M,N,P,color] = size(u0_color);
size(u0_color );
u0_color = double(u0_color); % Convert u0_color values to double;
u0 = u0_color(:,:,:,1); % Define the Grayscale volumetric image.
u0_color = uint8(u0_color); % Necessary for color visualization
x = 1:M;
y = 1:N;
z = 1:P;
dx = 1
dy = 1
dz = 1
dim_approx = 2*M*N*P / sqrt(M*N*P);
if(method == 'Explicit')
dt = 0.9 / ((2*mu/(dx^2)) + (2*mu/(dy^2)) + (2*mu/(dz^2))) % 90% CFL
elseif(method == 'Implicit')
dt = (10e7) * 0.9 / ((2*mu/(dx^2)) + (2*mu/(dy^2)) + (2*mu/(dz^2)))
end
[X,Y,Z] = meshgrid(x,y,z);
x0 = (M+1)/2;
y0 = (N+1)/2;
z0 = (P+1)/2;
r0 = min(min(M,N),P)/3;
phi = sqrt((X-x0).^2 + (Y-y0).^2 + (Z-z0).^2) - r0;
phi_visualize = phi; % Use this for visualization in 3D
phi = permute(phi,[2,1,3]); % Use this for computations in 3D
write_to_binary_file(phi_visualize,0); % record initial conditions
tic
for(n=1:TimeSteps)
n
c1 = C1_3d(u0,phi);
c2 = C2_3d(u0,phi);
% x
phi_xp = [phi(2:M,:,:); phi(M,:,:)]; % vertical concatenation
phi_xm = [phi(1,:,:); phi(1:M-1,:,:)]; % (since x values are rows)
% cat(1,A,B) is the same as [A;B]
Dx_m = (phi - phi_xm)/dx; % first derivatives
Dx_p = (phi_xp - phi)/dx;
Dxx = (Dx_p - Dx_m)/dx; % second derivative
% y
phi_yp = [phi(:,2:N,:) phi(:,N,:)]; % horizontal concatenation
phi_ym = [phi(:,1,:) phi(:,1:N-1,:)]; % (since y values are columns)
% cat(2,A,B) is the same as [A,B]
Dy_m = (phi - phi_ym)/dy;
Dy_p = (phi_yp - phi)/dy;
Dyy = (Dy_p - Dy_m)/dy;
% z
phi_zp = cat(3,phi(:,:,2:P),phi(:,:,P));
phi_zm = cat(3,phi(:,:,1) ,phi(:,:,1:P-1));
Dz_m = (phi - phi_zm)/dz;
Dz_p = (phi_zp - phi)/dz;
Dzz = (Dz_p - Dz_m)/dz;
% x,y,z
Dx_0 = (phi_xp - phi_xm) / (2*dx);
Dy_0 = (phi_yp - phi_ym) / (2*dy);
Dz_0 = (phi_zp - phi_zm) / (2*dz);
phi_xp_yp = [phi_xp(:,2:N,:) phi_xp(:,N,:)];
phi_xp_ym = [phi_xp(:,1,:) phi_xp(:,1:N-1,:)];
phi_xm_yp = [phi_xm(:,2:N,:) phi_xm(:,N,:)];
phi_xm_ym = [phi_xm(:,1,:) phi_xm(:,1:N-1,:)];
phi_xp_zp = cat(3,phi_xp(:,:,2:P),phi_xp(:,:,P));
phi_xp_zm = cat(3,phi_xp(:,:,1) ,phi_xp(:,:,1:P-1));
phi_xm_zp = cat(3,phi_xm(:,:,2:P),phi_xm(:,:,P));
phi_xm_zm = cat(3,phi_xm(:,:,1) ,phi_xm(:,:,1:P-1));
phi_yp_zp = cat(3,phi_yp(:,:,2:P),phi_yp(:,:,P));
phi_yp_zm = cat(3,phi_yp(:,:,1) ,phi_yp(:,:,1:P-1));
phi_ym_zp = cat(3,phi_ym(:,:,2:P),phi_ym(:,:,P));
phi_ym_zm = cat(3,phi_ym(:,:,1) ,phi_ym(:,:,1:P-1));
if(grad == 'Dirac')
Grad = DiracDelta(phi); % Dirac delta
%Grad = 1;
elseif(grad == 'Grad ')
Grad = (((Dx_0.^2)+(Dy_0.^2)+(Dz_0.^2)).^(1/2)); % |grad phi|
end
if(method == 'Explicit')
% CURVATURE: *mu*k|grad phi|* (central differences):
K = zeros(M,N,P);
Dxy = (phi_xp_yp - phi_xp_ym - phi_xm_yp + phi_xm_ym) / (4*dx*dy);
Dxz = (phi_xp_zp - phi_xp_zm - phi_xm_zp + phi_xm_zm) / (4*dx*dz);
Dyz = (phi_yp_zp - phi_yp_zm - phi_ym_zp + phi_ym_zm) / (4*dy*dz);
K = ( (Dx_0.^2).*Dyy - 2*Dx_0.*Dy_0.*Dxy + (Dy_0.^2).*Dxx ...
+ (Dx_0.^2).*Dzz - 2*Dx_0.*Dz_0.*Dxz + (Dz_0.^2).*Dxx ...
+ (Dy_0.^2).*Dzz - 2*Dy_0.*Dz_0.*Dyz + (Dz_0.^2).*Dyy) ./ ((Dx_0.^2 + Dy_0.^2 + Dz_0.^2).^(3/2) + epsilon);
phi_temp = phi + dt * Grad .* ( mu.*K + lambda1*(u0 - c1).^2 - lambda2*(u0 - c2).^2 );
elseif(method == 'Implicit')
C1x = 1 ./ sqrt(Dx_p.^2 + Dy_0.^2 + Dz_0.^2 + (10e-7)^2);
C2x = 1 ./ sqrt(Dx_m.^2 + Dy_0.^2 + Dz_0.^2 + (10e-7)^2);
C3y = 1 ./ sqrt(Dx_0.^2 + Dy_p.^2 + Dz_0.^2 + (10e-7)^2);
C4y = 1 ./ sqrt(Dx_0.^2 + Dy_m.^2 + Dz_0.^2 + (10e-7)^2);
C5z = 1 ./ sqrt(Dx_0.^2 + Dy_0.^2 + Dz_p.^2 + (10e-7)^2);
C6z = 1 ./ sqrt(Dx_0.^2 + Dy_0.^2 + Dz_m.^2 + (10e-7)^2);
% m = (dt/(dx*dy)) * Grad .* mu; % 2D
m = (dt/(dx*dy)) * Grad .* mu;
C = 1 + m.*(C1x + C2x + C3y + C4y + C5z + C6z);
C1x_2x = C1x.*phi_xp + C2x.*phi_xm;
C3y_4y = C3y.*phi_yp + C4y.*phi_ym;
C5z_6z = C5z.*phi_zp + C6z.*phi_zm;
phi_temp = (1 ./ C) .* ( phi + m.*(C1x_2x+C3y_4y) + (dt*Grad).*(lambda1*(u0 - c1).^2) - (dt*Grad).*(lambda2*(u0 - c2).^2) );
end
phi = phi_temp;
phi_visualize = permute(phi,[2,1,3]);
write_to_binary_file(phi_visualize,n); % record
end
time = toc
n = n
T = dt*n
clear
clear all
In general Matlab keeps track of all the variables in the form of matrix. When you work with lot of variables with many dimensions the RAM memory will be allocated for storing this variable. Hence on working with lots of variables that is gonna run for multiple iterations it is better to clear the variable from the memory. To do so use the command
clear variable_name_1, variable_name_2,... variable_name_3;
Although keeping all the variables keeps the code to look organised, however when you face such issues try clearing the unwanted variables.
Check this link to use clear command in detail: http://www.mathworks.in/help/matlab/ref/clear.html