I want to create a movie in MATLAB which looks like this:
https://www.youtube.com/watch?v=8z_tSVeEFTA
Mathematically, I have the differential equations and the code to solve them numerically:
% lorenz - Program to compute the trajectories of the Lorenz
% equations using the adaptive Runge-Kutta method.
clear; help lorenz;
%* Set initial state x,y,z and parameters r,sigma,b
state = input('Enter the initial position [x y z]: ');
r = input('Enter the parameter r: ');
sigma = 10.; % Parameter sigma
b = 8./3.; % Parameter b
param = [r sigma b]; % Vector of parameters passed to rka
tau = 1; % Initial guess for the timestep
err = 1.e-3; % Error tolerance
%* Loop over the desired number of steps
time = 0;
nstep = input('Enter number of steps: ');
for istep=1:nstep
%* Record values for plotting
x = state(1); y = state(2); z = state(3);
tplot(istep) = time; tauplot(istep) = tau;
xplot(istep) = x; yplot(istep) = y; zplot(istep) = z;
if( rem(istep,50) < 1 )
fprintf('Finished %g steps out of %g\n',istep,nstep);
end
%* Find new state using adaptive Runge-Kutta
[state, time, tau] = rka(state,time,tau,err,'lorzrk',param);
end
%* Print max and min time step returned by rka
fprintf('Adaptive time step: Max = %g, Min = %g \n', ...
max(tauplot(2:nstep)), min(tauplot(2:nstep)));
%* Graph the time series x(t)
figure(1); clf; % Clear figure 1 window and bring forward
plot(tplot,xplot,'-')
xlabel('Time'); ylabel('x(t)')
title('Lorenz model time series')
pause(1) % Pause 1 second
%* Graph the x,y,z phase space trajectory
figure(2); clf; % Clear figure 2 window and bring forward
% Mark the location of the three steady states
x_ss(1) = 0; y_ss(1) = 0; z_ss(1) = 0;
x_ss(2) = sqrt(b*(r-1)); y_ss(2) = x_ss(2); z_ss(2) = r-1;
x_ss(3) = -sqrt(b*(r-1)); y_ss(3) = x_ss(3); z_ss(3) = r-1;
plot3(xplot,yplot,zplot,'-',x_ss,y_ss,z_ss,'*')
view([30 20]); % Rotate to get a better view
grid; % Add a grid to aid perspective
xlabel('x'); ylabel('y'); zlabel('z');
title('Lorenz model phase space');
But I don't know how to plot two different trajectories at once in different colors, and make a clip that looks like this (with the side plot that shows the X coordinate solutions in the same time).
Can anyone help me with this ?
Thank you!
In your example code, you already have a for loop, which goes through all time steps. Now, to generate such an animation, we'll want to plot the figure at each time step. This is done by
for istep=1:nstep
%* Record values for plotting
x = state(1); y = state(2); z = state(3);
tplot(istep) = time; tauplot(istep) = tau;
xplot(istep) = x; yplot(istep) = y; zplot(istep) = z;
%* Find new state using adaptive Runge-Kutta
[state, time, tau] = rka(state,time,tau,err,'lorzrk',param);
%* Create Plot
figure(2);
x_ss(1) = 0; y_ss(1) = 0; z_ss(1) = 0;
x_ss(2) = sqrt(b*(r-1)); y_ss(2) = x_ss(2); z_ss(2) = r-1;
x_ss(3) = -sqrt(b*(r-1)); y_ss(3) = x_ss(3); z_ss(3) = r-1;
plot3(xplot,yplot,zplot,'-',x_ss,y_ss,z_ss,'*')
view([30 20]); % Rotate to get a better view
grid; % Add a grid to aid perspective
xlabel('x'); ylabel('y'); zlabel('z');
title('Lorenz model phase space');
%* Save frame
output_video(istep) = getframe;
end
This will create an array of structs output_video, which is of size 1 x nstep and contains cdata (the image data) and colormap for each frame.
You can view this video in MATLAB with
implay(output_video)
or save it do disk using the VideoWriter class:
v = VideoWriter('my_trajectory_video.avi')
open(v)
writeVideo(v,output_video)
close(v)
Related
I am trying to solve a classification task using logistic regression. Part of my task is to plot the decision boundary. I find that the gradient of the decision boundary seems to be solved correctly by my algorithm but when plotting the boundary is too high and does not separate the points well. I cannot work out why this is and would be grateful for any advice to solve this issue.
data = open('Question5.mat');
x = data.x; y = data.y; % Extract data for ease of use
LR = 0.001; % Set tunable learning rate for gradient descent
w_est = [0; 0; 0]; % Set inital guess for a, b, and c
cost = []; % Initalise array to hold value of cost function
figure;
for i = 1:20000 % Set iteration limit for gradient descent
iteration_cost = 0; grad_a = 0; grad_b = 0; grad_c = 0; % Set innial value of 0 for summed terms
for m = 1:1100 % Iterate through data points
y_hat_est = 1./(1+exp(-w_est'*[x(m,1); x(m,2); 1])); % Calculate value of sigmoid function with estimated coefficients for each datapoint
iteration_cost = iteration_cost + y(m)*log(y_hat_est)+(1-y(m))*log(1-y_hat_est); % Calculate cost function and add it to summed term for each data point
% Calculate each gradient term for each data point and add to
% summed gradient
grad_a = grad_a + (y_hat_est - y(m))*x(m,1);
grad_b = grad_b + (y_hat_est - y(m))*x(m,2);
grad_c = grad_c + (y_hat_est - y(m))*x(m,3);
end
g = [grad_a; grad_b; grad_c]; % Create vector of gradients
w_est = w_est - LR*g; % Update estimate vector with next term
cost(i) = -iteration_cost; % Add the value of the cost function to the array for costs
if mod(i,1000) == 0 % Only plot on some iterations to speed up program
hold off
gscatter(x(:,1),x(:,2),y,'rb'); % Plot scatter plot grouped by class
xlabel('x1'); ylabel('x2'); title(i); % Add title and labels to figure
hold on
x1_plot = -6:4; x2_plot = -3:7; % Create array of values for plotting
plot( -(w_est(1)*x1_plot + w_est(3)) /w_est(2), x2_plot); % Plot decision boundary based on the current coefficient estimates
% pause(1) % Add delay to aid visualisation
end
end
hold off;
figure; plot(cost) % Plot the cost function
title('Cost function'); xlabel('Iteration number'); ylabel('cost');
enter image description here
I've implemented an algorithm for my physics project which does exactly what I want. The problem that I'm stuck which is not the Physics content itself hence I think it might be somewhat trivial to explain what my code does. I'm mainly stuck with the way MATLAB's plotting works if I was to loop over the same algorithm to produce similar graphs with a slight change of a value of my parameter. Here's my code below:
clear; clc; close all;
% Parameters:
z_nn = 4; % Number of nearest-neighbour in lattice (square = 4).
z_nnn = 4; % Number of next-nearest-neighbours in lattice (square = 4).
Lx = 40; % Number of sites along x-axis.
Ly = 40; % Number of sites along y-axis.
sigma = 1; % Size of a site (defines our units of length).
beta = 1.2; % Inverse temperature beta*epsilon.
mu = -2.53; % Chemical potential mu/epsilon.
mu_2 = -2.67; % Chemical potential mu/epsilon for 2nd line.
J = linspace(1, 11, 11);%J points for the line graph plot
potential = zeros(Ly);
attract = 1.6; %wall attraction constant
k = 1; %wall depth
rho_0 = 0.4; % Initial density.
tol = 1e-12; % Convergence tolerance.
count = 30000; % Upper limit for iterations.
alpha = 0.01; % Mixing parameter.
conv = 1; cnt = 1; % Convergence value and counter.
rho = rho_0*ones(Ly); % Initialise rho to the starting guess(i-th rho_old) in Eq(47)
rho_rhs = zeros(Ly); % Initialise rho_new to zeros.
% Solve equations iteratively:
while conv>=tol && cnt<count
cnt = cnt + 1; % Increment counter.
% Loop over all lattice sites:
for j=1:Ly
%Defining the Lennard-Jones potential
if j<k
potential(j) = 1000000000;
else
potential(j) = -attract*(j-k)^(-3);
end
% Handle the periodic boundaries for x and y:
%left = mod((i-1)-1,Lx) + 1; % i-1, maps 0 to Lx.
%right = mod((i+1)-1,Lx) + 1; % i+1, maps Lx+1 to 1.
if j<k+1 %depth of wall
rho_rhs(j) = 0;
rho(j) = 0;
elseif j<(20+k)
rho_rhs(j) = (1 - rho(j))*exp((beta*((3/2)*rho(j-1) + (3/2)*rho(j+1) + 2*rho(j) + mu) - potential(j)));
else
rho_rhs(j) = rho_rhs(j-1);
end
end
conv = sum(sum((rho - rho_rhs).^2)); % Convergence value is the sum of the differences between new and current solution.
rho = alpha*rho_rhs + (1 - alpha)*rho; % Mix the new and current solutions for next iteration.
end
% disp(['conv = ' num2str(conv_2) ' cnt = ' num2str(cnt)]); % Display final answer.
% figure(2);
% pcolor(rho_2);
figure(1);
plot(J, rho(1:11));
hold on;
% plot(J, rho_2(1,1:11));
hold off;
disp(['conv = ' num2str(conv) ' cnt = ' num2str(cnt)]); % Display final answer.
figure(3);
pcolor(rho);
Running this code should give you a graph like this
Now I want to produce a similar graph but with one of the variable's value changed and plotted on the same graph. My approach that I've tried is below:
clear; clc; close all;
% Parameters:
z_nn = 4; % Number of nearest-neighbour in lattice (square = 4).
z_nnn = 4; % Number of next-nearest-neighbours in lattice (square = 4).
Lx = 40; % Number of sites along x-axis.
Ly = 40; % Number of sites along y-axis.
sigma = 1; % Size of a site (defines our units of length).
beta = 1.2; % Inverse temperature beta*epsilon.
mu = [-2.53,-2.67]; % Chemical potential mu/epsilon.
mu_2 = -2.67; % Chemical potential mu/epsilon for 2nd line.
J = linspace(1, 10, 10);%J points for the line graph plot
potential = zeros(Ly, length(mu));
gamma = zeros(Ly, length(mu));
attract = 1.6; %wall attraction constant
k = 1; %wall depth
rho_0 = 0.4; % Initial density.
tol = 1e-12; % Convergence tolerance.
count = 30000; % Upper limit for iterations.
alpha = 0.01; % Mixing parameter.
conv = 1; cnt = 1; % Convergence value and counter.
rho = rho_0*[Ly,length(mu)]; % Initialise rho to the starting guess(i-th rho_old) in Eq(47)
rho_rhs = zeros(Ly,length(mu)); % Initialise rho_new to zeros.
figure(3);
hold on;
% Solve equations iteratively:
while conv>=tol && cnt<count
cnt = cnt + 1; % Increment counter.
% Loop over all lattice sites:
for j=1:Ly
for i=1:length(mu)
y = 1:Ly;
MU = mu(i).*ones(Ly)
%Defining the Lennard-Jones potential
if j<k
potential(j) = 1000000000;
else
potential(j) = -attract*(j-k).^(-3);
end
% Handle the periodic boundaries for x and y:
%left = mod((i-1)-1,Lx) + 1; % i-1, maps 0 to Lx.
%right = mod((i+1)-1,Lx) + 1; % i+1, maps Lx+1 to 1.
if j<k+1 %depth of wall
rho_rhs(j) = 0;
rho(j) = 0;
elseif j<(20+k)
rho_rhs(j) = (1 - rho(j))*exp((beta*((3/2)*rho(j-1) + (3/2)*rho(j+1) + 2*rho(j) + MU - potential(j)));
else
rho_rhs(j) = rho_rhs(j-1);
end
end
end
conv = sum(sum((rho - rho_rhs).^2)); % Convergence value is the sum of the differences between new and current solution.
rho = alpha*rho_rhs + (1 - alpha)*rho; % Mix the new and current solutions for next iteration.
disp(['conv = ' num2str(conv) ' cnt = ' num2str(cnt)]); % Display final answer.
figure(1);
pcolor(rho);
plot(J, rho(1:10));
end
hold off;
The only variable that I'm changing here is mu. I would like to loop my first code so that I can enter an arbitrary amount of different values of mu and plot them on the same graph. Naturally I had to change all of the lists dimension from (1 to size of Ly) to (#of mu(s) to size of Ly), such that when the first code is being looped, the i-th mu value in that loop is being turned into lists with dimension as long as Ly. So I thought I would do the plotting within the loop and use "hold on" encapsulating the whole loop so that every plot that was generated in each loop won't be erased. But I've been spending hours on trying to figure out the semantics of MATLAB but ultimately I can't figure out what to do. So hopefully I can get some help on this!
hold on only applies to the active figure, it is not a generic property shared among all figures. What is does is changing the value of the current figure NextPlot property, which governs the behavior when adding plots to a figure.
hold on is equivalent to set(gcf,'NextPlot','add');
hold off is equivalent to set(gcf,'NextPlot','replace');
In your code you have:
figure(3); % Makes figure 3 the active figure
hold on; % Sets figure 3 'NextPlot' property to 'add'
% Do some things %
while conv>=tol && cnt<count
% Do many things %
figure(1); % Makes figure 1 the active figure; 'hold on' was not applied to that figure
plot(J, rho(1:10)); % plots rho while erasing the previous plot
end
You should try to add another hold on statement after figure(1)
figure(1);
hold on
plot(J, rho(1:10));
I've spent quite some time trying to simulate a simple SISO system using two approaches:
1) Using lsim() in MATLAB
2) By writing down the difference equations myself and iterate over them in a loop.
I was never able to get the same simulation results from both approaches, and I have no idea what I am doing wrong.
I stacked my code in a single m-file so it's easier to follow. Here is the code:
function main()
clear all
clc
simulateUsing_lsim()
simulateUsing_loop()
end
%%%%%% Simulating using lsim %%%%%%%
function simulateUsing_lsim()
% Define the continuous-time closed-loop system
P = getContPlant();
[Kp,Ki,Kd] = get_PIDgains();
C = pid(Kp,Ki,Kd);
clSys_cont = feedback(C*P,1);
% Define the discrete-time closed-loop system
hk = get_sampling_time();
clSys_disc = c2d(clSys_cont,hk);
% Generate the reference signal and the time vector
[r,t] = getReference(hk);
%% Simulate and plot using lsim
figure
lsim(clSys_disc,r,t)
%% Finding and plotting the error
y = lsim(clSys_disc,r);
e = r - y;
figure
p = plot(t,e,'b--');
set(p,'linewidth',2)
legend('error')
xlabel('Time (seconds)')
ylabel('error')
% xlim([-.1 10.1])
end
%%%%%% Simulating using loop iteration (difference equations) %%%%%%%
function simulateUsing_loop()
% Get the cont-time ol-sys
P = getContPlant();
% Get the sampling time
hk = get_sampling_time();
% Get the disc-time ol-sys in SS representation
P_disc = ss(c2d(P,hk));
Ad = P_disc.A;
Bd = P_disc.B;
Cd = P_disc.C;
% Get the PID gains
[Kp,Ki,Kd] = get_PIDgains();
% Generate the reference signal and the time vector
[r,t] = getReference(hk);
%% Perform the system simulation
x = [0 0]'; % Set initial states
e = 0; % Set initial errors
integral_sum = 0; % Set initial integral part value
for i=2:1:length(t)
% Calculate the output signal "y"
y(:,i) = Cd*x;
% Calculate the error "e"
e(:,i) = y(:,i) - r(i);
% Calculate the control signal vector "u"
integral_sum = integral_sum + Ki*hk*e(i);
u(:,i) = Kp*e(i) + integral_sum + (1/hk)*Kd*(e(:,i)-e(:,i-1));
% Saturation. Limit the value of u withing the range [-tol tol]
% tol = 100;
% if abs(u(:,i)) > tol
% u(:,i) = tol * abs(u(:,i))/u(:,i);
% else
% end
% Calculate the state vector "x"
x = Ad*x + Bd*u(:,i); % State transitions to time n
end
%% Subplots
figure
plot(t,y,'b',t,r,'g--')
%% Plotting the error
figure
p = plot(t,e,'r');
set(p,'linewidth',2)
legend('error')
xlabel('Time (seconds)')
ylabel('error')
end
function P = getContPlant()
s = tf('s');
P = 1/(s^2 + 10*s + 20);
end
function [Kp,Ki,Kd] = get_PIDgains()
Kp = 350;
Ki = 300;
Kd = 50;
end
function hk = get_sampling_time()
hk = 0.01;
end
function [r,t] = getReference(hk)
[r,t] = gensig('square',4,10,hk);
end
I got the plant model P and its PID controller from this page (see equation 10), where the system is simulated against a step reference and the result looks pretty much exactly like the lsim() result (just for a single step peak).
However, the result of simulating the system using lsim() is this:
whereas, using the for loop, I got this performance:
I would highly appreciate any help or clarification why I am getting different results.
I have created some MatLab code that plots a plane wave using two different expressions that give the same plane wave. The first expression is in Cartesian coordinates and works fine. However, the second expression is in polar coordinates and when I calculate the plane wave in this case, the plot is distorted. Both plots should look the same. So what am I doing wrong in transforming to/from polar coordinates?
function Plot_Plane_wave()
clc
clear all
close all
%% Step 0. Input paramaters and derived parameters.
alpha = 0*pi/4; % angle of incidence
k = 1; % wavenumber
wavelength = 2*pi/k; % wavelength
%% Step 1. Define various equivalent versions of the incident wave.
f_u_inc_1 = #(alpha,x,y) exp(1i*k*(x*cos(alpha)+y*sin(alpha)));
f_u_inc_2 = #(alpha,r,theta) exp(1i*k*r*cos(theta-alpha));
%% Step 2. Evaluate the incident wave on a grid.
% Grid for field
gridMax = 10;
gridN = 2^3;
g1 = linspace(-gridMax, gridMax, gridN);
g2 = g1;
[x,y] = meshgrid(g1, g2);
[theta,r] = cart2pol(x,y);
u_inc_1 = f_u_inc_1(alpha,x,y);
u_inc_2 = 0*x;
for ir=1:gridN
rVal = r(ir);
for itheta=1:gridN
thetaVal = theta(itheta);
u_inc_2(ir,itheta) = f_u_inc_2(alpha,rVal,thetaVal);
end
end
%% Step 3. Plot the incident wave.
figure(1);
subplot(2,2,1)
imagesc(g1(1,:), g1(1,:), real(u_inc_1));
hGCA = gca; set(hGCA, 'YDir', 'normal');
subplot(2,2,2)
imagesc(g1(1,:), g1(1,:), real(u_inc_2));
hGCA = gca; set(hGCA, 'YDir', 'normal');
end
Your mistake is that your loop is only going through the first gridN values of r and theta. Instead you want to step through the indices of ix and iy and pull out the rVal and thetaVal of the matrices r and theta.
You can change your loop to
for ix=1:gridN
for iy=1:gridN
rVal = r(ix,iy); % Was equivalent to r(ix) outside inner loop
thetaVal = theta(ix,iy); % Was equivalent to theta(iy)
u_inc_2(ix,iy) = f_u_inc_2(alpha,rVal,thetaVal);
end
end
which gives the expected graphs.
Alternatively you can simplify your code by feeding matrices in to your inline functions. To do this you would have to use an elementwise product .* instead of a matrix multiplication * in f_u_inc_2:
alpha = 0*pi/4;
k = 1;
wavelength = 2*pi/k;
f_1 = #(alpha,x,y) exp(1i*k*(x*cos(alpha)+y*sin(alpha)));
f_2 = #(alpha,r,theta) exp(1i*k*r.*cos(theta-alpha));
% Change v
f_old = #(alpha,r,theta) exp(1i*k*r *cos(theta-alpha));
gridMax = 10;
gridN = 2^3;
[x,y] = meshgrid(linspace(-gridMax, gridMax, gridN));
[theta,r] = cart2pol(x,y);
subplot(1,3,1)
contourf(x,y,real(f_1(alpha,x,y)));
title 'Cartesian'
subplot(1,3,2)
contourf(x,y,real(f_2(alpha,r,theta)));
title 'Polar'
subplot(1,3,3)
contourf(x,y,real(f_old(alpha,r,theta)));
title 'Wrong'
I have this Matlab project but for some reason I just cannot stop thinking about it because I could not get it to work.
Objective:
This is a MATLAB script that would calculate the change of pressure, temperature and density of a glider that is being dropped from 10000 feet. As it falls, we want to use those new values calculated and then plugged in a function that has 4 equations that need to be differentiated at every point using ode45 as well as the new values of P T and Rho.
Here is the main code:
% HouseKeeping:
clc
clear all
close all
% Constants:
S = 232; % ft^2
Cd0 = 0.02;
K = 0.07;
W = 11000; % lbf
Cl_max = sqrt(Cd0/K);
Cd_max = 2*K*Cl_max^2;
Rho_10000 = .001756; % slugs/ ft^3
%Initial conditions:
t = 0; % Sec
x = 0; % ft
h = 10000; % ft
v = sqrt((2*W)/(Rho_10000*S*Cl_max)); %ft/s
gamma = -Cd_max/Cl_max;
% Find Endurance:
V_RD= sqrt((2*W)/(S* Rho_10000* sqrt(3*Cd0/K)));
RD= V_RD/((sqrt(3*Cd0/K))/(2*Cd0)) ; % ft/s
Endurcance= h/RD; % 958.3515 sec
% Sea Level values:
TSL = 518.69; % Rankine
PSL = 2116.199414; % lbs/ft^2
RhoSL = 0.0023769; % slugs/ft^3
while h > 0
tspan = [t t+1];
i=1;
X = [x;h;v;gamma;Rho_10000];
Time(i)= t;
% Calling ODE45:
[F] = ode45(# D,tspan, X)
% Hight Varying Parameters:
T = TSL - 0.00356616*h;
P = (1.137193514E-11)*(T)^5.2560613;
Rho = (RhoSL * TSL / PASL)*(P / T);
a = 49.0214 * (T)^.5;
H_Del(i) = (-Cd_max/Cl_max)*(plotted_x(i))+10000;
V_Del(i) = sqrt((2*W)/(Rho*S*Cl_max));
Gamma_Del(i) = -Cd_max/Cl_max;
i= i+1;
X = [ x ; H_Del(i); V_Del(i); Gamma_Del(i); Rho];
end
% Plots:
%1
figure (1)
plot(F(:,1),'-r',F(:,3),'-b')
title('Velocity vs Distance');
xlabel('x (ft)');
ylabel('v (ft/s)');
%2
Figure (2)
plot(F(:,1),'-r',F(:,2),'-b')
title('Altitude vs Distance ');
xlabel('x (ft)');
ylabel('h (ft)');
%3
figure (3)
plot(F(:,1),'-r',F(:,4),'-b')
title('Gamma vs Distance');
xlabel('x (ft)');
ylabel('Gamma (rad)');
%4
figure (4)
plot(t,'-r',F(:,3),'-b')
title('Velocity vs Time');
xlabel(' t (s)');
ylabel('v (ft/s)');
%5
figure (5)
plot(t,'-r',F(:,1),'-b')
title(' Distance vs Time ');
xlabel('t (s)');
ylabel('x (ft)');
%6
figure (6)
plot(t,'-r',F(:,4),'-b')
title('Gamma vs Time ');
xlabel('t (s)');
ylabel('Gamma (rad)');
%7
figure (7)
plot(t,'-r',F(:,3),'-b')
title('Velocity vs Time');
xlabel('t (s)');
ylabel('V (ft/s)');
Here is the Function
function [F] = D(X)
% Constants:
S = 232; % ft^2
Cd0 = 0.02;
K = 0.07;
W = 11000; % lbf
Cl_max = sqrt(Cd0/K);
Cd_max = 2*K*Cl_max^2;
Rho_10000 = .001756; % slugs/ ft^3
% Initial conditions:
t = 0; % Sec
x = 0; % ft
h = 10000; % ft
v = sqrt((2*W)/(Rho_10000*S*Cl_max)); % ft/s
gamma = -Cd_max/Cl_max;
g= 32.2; % ft/s^2
% Equations:
X_Pr = X(3)*cos(X(4));
H_Pr = X(3)*sin(X(4));
V_Pr = (-32.2./W)*(0.5*X(5)*Cd_max*S*X(3)^2 + W*sin(X(4)));
G_Pr = (32.2./(W*X(3)))*(0.5*X(5)*Cl_max*S*X(3)^2 - W*cos(X(4)));
F = [X_Pr;H_Pr;V_Pr;G_Pr];
I am not very good with Matlab but I did my very best! I went to my professors for help but they said they were too busy. I even stalked every senior I knew and they all said they did not know how to do it. My next project will be assigned soon and I think that if I cannot do this then I would not be able to do the next one.
Your code produces the following error:
Error using main>D
Too many input arguments.
This means that ode45 tries to call your provided function D with too many input arguments. You should check the desired odefun format in the ode45 documentation: dydt = odefun(t,y)
So, you should change your function declaration of D to
function [F] = D(t, X)
This should solve your first problem, but a following error pops up:
D returns a vector of length 4, but the length of initial conditions
vector is 5. The vector returned by D and the initial conditions
vector must have the same number of elements.
Again, you should check the ode45 documentation. Your function should return the derivatives of all your input variables X: F = dX/dt. You should also return the derivate of the fifth element Rho_10000.
Next, I got some error about undefined variables such as PASL. Probably because you did not post your full code.
Besides of the errors, you should really check your code again. You have written an infinite while loop while h > 0. You never change h in the loop, nor you use the output of ode45 in your loop. Furthermore you always overwrite your i and X value at the beginning of the loop, which is probably not what you want.
This is not a full answer to your question, but I hope you will be able to continue and post smaller, well defined problems, instead of one big problem which is very difficult to answer completely.