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

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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
tf=5; N=10000
tspan = linspace(0, tf,N);
dt =tf/N;
[t,x]= ode45(@(t,x)position_calc(x), tspan, x0);
hold on
% figure
function [dstate]= position_calc(state)
dx= state(2);
dstate= [x;vel];
I do not want to run the function inside ode again to obtain the velocity
Torsten on 31 Jan 2023
Edited: Torsten on 31 Jan 2023
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".

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Accepted Answer

William Rose
William Rose on 31 Jan 2023
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.
Torsten on 31 Jan 2023
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;

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More Answers (1)

Sulaymon Eshkabilov
Sulaymon Eshkabilov on 31 Jan 2023
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);
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')
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.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;


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