How do I get I pass out the input parameter to ode45 integration

I want to integrate velocity and obtain posiiton. Is there a way to do this with ode45 and obtain the set of velocities as additonal output to the ode45. Here is a code I have written but I am not sure it works correctly.

close all; clear all;clf
x0=[0.5;0.5];
tf=5; N=10000
tspan = linspace(0, tf,N);
dt =tf/N;
[t,x]= ode45(@(t,x)position_calc(x), tspan, x0);
figure
plot(t,x(:,1),'b--','linewidth',2)
hold on
% figure
plot(t,x(:,2),'g--');

grid
legend(["x","vel","pos"])


function [dstate]= position_calc(state)
x=state(1);
dx= state(2);
vel=x^2;
dstate= [x;vel];
end

I do not want to run the function inside ode again to obtain the velocity

[Torsten]

x(:,2) is the velocity, isn't it ?

You integrate acceleration in your code. Thus what you call "vel" is acceleration, not velocity.

So dstate should be [state(2),acceleration].

If you want to integrate velocity, you have one differential equation, not two.

[Hafeez Jimoh]

vel= x^2. But I also need the values of x itself. I want to integrate velocity to obtain position. So x(:,2) is intended to be velocity and x(:,1) is intended to be position. Kindly let me know if I am doing something wrong. Thanks

[Torsten]

If x(:,1) is intended to be position and x(:,2) is intended to be velocity, your odes read

y''(t) = a

where "a" is acceleration.

Written as a system

y1' = y2
y2' = a

where y1 is position and y2 is velocity.

So I don't understand your function "position_calc".

Answer 1

[William Rose]

@Hafeez Jimoh,

I think @Torsten, is right, as usual. I think you are simulating a one-dimensional system in which the velocity equals the square of the position:

v = x^2.

Therefore I recommend that x represent position only. In other words, x should be a vector, rather than an array.

After you do the integration, you can compute the velocity.

Here is a code fragment that integrates the system for 1.5 seconds. I find that by t=2, it blows up.

x0=0.5;                     %initial position
tf=1.5;                     %final time
[t,x]= ode45(@(t,x) x*x, [0 tf], x0);
v=x.*x;                     %velocity
plot(t,x,'b',t,v,'-r')      %plot results
grid on; legend(["Position","Velocity"]); xlabel('Time')

Good luck.

[Hafeez Jimoh]

Is there anyway to avoid doing this :

v=x.*x;                     %velocity

and getting the velocity directly from the ode45 line?

[Sam Chak]

Hafeez, I suggest you show the Differential equation that you want to solve. Only math, no MATLAB code. This will allow us to clearly look into the math problem and then check if it can be solved using ode45 solver.

What you described sounds like a kinematic differential equation. But what you want to obtain the velocity state directly from the ode45 solution indicates a dynamical differential equation.

[Torsten]

Is there anyway to avoid doing this :

v=x.*x; %velocity

and getting the velocity directly from the ode45 line?

Not really. The usual way is to call the function in which you supply the derivatives with the result vectors from ode45 after integration has finished:

x0 = 0.5;                   %initial position
tf = 1.5;                   %final time
[t,x] = ode45(@fun, [0 tf], x0);
v = fun(t,x);               %velocity
plot(t,x,'b',t,v,'-r')      %plot results
grid on; legend(["Position","Velocity"]); xlabel('Time')

function dx = fun(t,x)
  dx = x.^2;
end

Answer 2

[Sulaymon Eshkabilov]

There are a couple of overlooked points as Torsten pinpointed out. Here is one simple explae of dx = 2x example solution very similar to your exercise:

% Let's say: dx = 2*x;  which is the velocity formulation: f(t,x) = 2*x;
% then  solution x(t) can be found using dsolve() that gives an analytical
% solution or ode45 gives numerical solution

close all; clearvars
x0=0.5;         % Note: dx has one Initial Condition, i.e., x(0) = 0.5, for example
tf=5; N=10000;
tspan = linspace(0, tf,N);
[t,x]= ode45(@(t,x)position_calc(x), tspan, x0);
figure
yyaxis left
plot(t,x,'bo-','linewidth',2, 'DisplayName','Displacement')
ylabel('$x(t), \ [m]$', 'Interpreter','latex')

% Now, the velocity values can be obtained from the integrated displacement values,
v =diff(x);
yyaxis right
plot(t(1:end-1),v,'r-.', 'LineWidth',1.75, 'DisplayName','Velocity');
ylabel('$v(t), \ [m/s]$', 'Interpreter','latex')
xlabel('$time, \ [s]$', 'Interpreter','latex')
grid on
legend('Show', 'Location', 'Best')

figure
D = array2table(t(1:end-1));
D = renamevars(D, 'Var1', 'time');
D.x = x(1:end-1);
D.v = v;
H=stackedplot(D, 'XVariable', 'time');
H.LineWidth=2;
H.LineProperties(1).Color = 'b';
H.LineProperties(1).Marker = 'o';
H.LineProperties(2).Color = 'r';
H.LineProperties(2).LineStyle = '-.';
grid on

function dx= position_calc(state)
dx= 2*state;
end