ODE45 function in matlab takes argument:
(function,[tinitial tfinal],yinitial)
But here, I believe, the span of time is predetermined. How can I assign a vector for it? I mean how do I solve the ODE for the domain 1:0.1:5?
Thank You.
If you need the values at specified points in time, simply go:
tspan = 1:0.1:5
[T Y] = ode45(odefun, tspan, y0)
T should be the same as tspan, and Y will be the corresponding values for each point in time.
You can assign the following vector for time span,and this way, you are saying to ode45 that you want the solution at specific time points (here , every 0.001)
tspan = ti:0.001:tf;
Output vectors ( T, X ) will have as many steps as the tspan vector has.
(But if you mean that you want the solver to take predetermined and constant steps for solving the equation , you can't do that.)
Related
I would like to use numerical approach to calculate the differential with singularities.
For instance, y'=y^2+1 with y(0)=0. The analytical solution is not hard to find, y=tan(x).
The problem arises when I apply the numeric method, see the code in Matlab
tspan = [0 6];
y0 = [0; 0];
ode = #(t, y) y.^2+1;
[t, y] = ode45(ode, tspan, y0);
plot(t, y(:,1))
axis([0 6 -20 20])
which gives the plot below, i.e., it missed the part of solution after the first singularity.
My question: how to find the full numeric solution of a differential equation with singularities?
Thanks in advance!
This is impossible in general. An ODE solution ends where it diverges to infinity (or leaves the domain of the ODE function in any other way).
What you can do is apply domain knowledge. This equation is a Riccati DE. If one knows one solution, one can transform it into a Bernoulli DE that has a solution formula. There is a second transform without knowledge of a particular solution, it works always but is not always helpful. Set y=p/q and select their relationship so that a system of linear equations results. Here p'q-q'p=p^2+q^2 resolves nicely to p'=q, q'=-p, or as second order DE u''+u=0, the harmonic oscillator, for u=q or u=p. This system now has no singularities and the poles of the original solution y are the roots of the denominator q.
My system of equations can be written down as-
y1' = F_1(x,y1,y2)
y2' = F_2(x,y1,y2)
where F1 and F2 are some functions of x, y1, y2. y1 and y2 are functions of x and y1'=dy1_dx and y2'=dy2_dx.
The initial condition is at x=0, y1=y2=0. However I need the solution at y1=1, i.e. as soon as y1=1 is reached, the iteration needs to stop. I want to solve it using a proper solver like ODE45 or ODE15s in MATLAB. However I am not sure how to set the x_span i.e. initial and final values of x. Please guide me.
Note that, setting x at a large value say 1000 will not help, because I don't know where y1=1.
Use an event function (typically equal to y1-1) that you give to the ODE solver:
https://fr.mathworks.com/help/matlab/math/ode-event-location.html
As soon as the solver detect a sign change of your event function, it stops.
I have to solve a second-degree differential equation and I specifically need the value of the first derivative of y at the final time point. My code is the following:
[T Y]=ode45(#(t y)vdp4(t,y,0.3),[0 1],[0.3/4,((3*0.3)^0.5)/2]);
I know the output will contain the values at which ode45 evaluated the function. To get the y values at specific time value at have it has been advised to give more than two time points in the MATLAB documentation. I did that too.
tspan=[0:0.01:1]
[T Y]=ode45(#(t y)vdp4(t,y,0.3),tspan,[0.3/4,((3*0.3)^0.5)/2]);
The T vector still does not have all the values from 0 to 1 (The last value is 0.39). This happens especially after multiple executions of ode45 function. I found something else in the MATLAB documentation: using "sol" structure to deval the values for specific t values. Is that the right way to go?
For reference, my differential equation is in the following function.
function dy = vdp4(t,y,k)
dy = zeros(2,1); % a column vector
dy(1)=y(2);
dy(2)=(y(2)^2-2*t*y(2)+2*y(1))/k+2;
end
Edit: I provided the parameter value. It should now be executable.
Try to plot your solution, you will find the answer
For the given differential equation,
with the initial condition y(0) = 1,
I am trying to find a solution using ODE45 method of MATLAB over the interval [0 5]. However, I noticed that the given equation is actually singular at the given initial point. Given the circumstances, how do you suggest I proceed?
What I have tried is that I can start the interval from 0 + δ for some δ < 1e-5. However, this would be a crude approximation. The equation is not differentiable at the point 0.
Thank you so much,
I'm trying to use ode45 to solve a system of ODE's:
[X,Y]= ode45(#sys,[0, T],y0);
where,
function dy = sys(t,y)
dy(1) = f_1(y)
dy(2) = f_2(y)
dy(3) = f_3(y)
end
The problem is that the function ode45 requires that y0 be initial values [y_1(0), y_2(0), y_3(0)], while in my system, I only have the values [y_2(0), y_3(0), y_3(T)] available.
Mathematically, this set of initial/terminal conditions should be enough to pin down the system, but is there any way I can work with that by ode45 or any other functions in MATLAB?
Thanks!
After digging in the Matlab documentation for a little bit, I think the more elegant way is to use the bvp4c function. bvp4c is a function specifically designed to handle boundary value problems like this, as opposed to ode**, which are really for initial value problems only. In fact, there's a whole set of other functions such as deval and bvpinit in Matlab that really facilitate the use of bvp4c. Here's the link to the Matlab documentation.
I'll post a brief (and perhaps a bit contrived) example here:
function [y1, y2, y3] = test(start,T)
solinit = bvpinit(linspace(0,3,10), [1,1,0]);
sol = bvp4c(#odefun,#bvpbc,solinit);
tspan = linspace(start,T,100);
S = deval(sol, tspan);
y1 = S(1,:);
y2 = S(2,:);
y3 = S(3,:);
plot (tspan,y1)
figure
plot (tspan,y2)
figure
plot (tspan,y3)
%% system definition & BVCs
function dydx = odefun(t,y)
dydx(1) = y(1) + y(2) + t;
dydx(2) = 2*y(1) + y(2);
dydx(3) = 3 * y(1) - y(2);
end
function res = bvpbc(y0,yT)
res= [y0(3) yT(2) yT(3)];
end
end
The test function outputs 3 vectors of solutions points for y1, y2 and y3 respectively, and also plots them.
Here are the variable paths plotted by Matlab:
Also, I found this video by professor Jake Blanchard from WMU to be very helpful.
One approach is to use the shooting method to iteratively solve for the unknown initial state y_1(0) such that the desired final state y_3(T) is satisfied.
The iteration proceeds by solving a nonlinear equation F = 0:
F(y_1(0)) = Y_3(T) - y_3(T)
where the uppercase function Y is the solution obtained by integrating the ODE's forward in time from a set of initial conditions. The task is to guess the value of y_1(0) to obtain F = 0.
Since we're now solving a nonlinear equation, all of the usual approaches apply. Specifically you could use either bisection or Newton's method to update your guess for the unknown initial state y_1(0). Note a couple of things:
The ODE's are integrated on [0,T] at each iteration of the nonlinear solve.
There may be multiple solutions for F = 0, depending on the structure of your ODE's.
Newton's method may converge faster than bisection, but may also be numerically unstable unless you can provide a good starting guess for y_1(0).
Using existing MATLAB functions, the bounded scalar nonlinear solver FMINBND might be a good choice as a nonlinear solver.
Hope this helps.