Solving two differential equations using ode45

Nagarajan (view profile)

on 3 Jun 2019
Latest activity Commented on by Nagarajan

on 4 Jun 2019

Stephan (view profile)

I have two differential equations in the form
-y"-5y'-y=3x"+4x'^2+2x+6t
-y"-2.3y'-2y=2x"+3x'^2+x
I would like to solve these equations using ODE's
the inital conditions are x(0), y(0), x'(0) and y'(0) are known

Nagarajan

Nagarajan (view profile)

on 3 Jun 2019
I need asssitance in converting the above equations into the ODE solver form
for Ex:
fun=@(t,y)[y(2);......];
Stephan

on 3 Jun 2019
Nagarajan

Nagarajan (view profile)

on 3 Jun 2019
Yeah i do have access
I need assistance while converting the variables
fun=@(t,y)[.....]; % change of variables..
Y0=[x0; x0prime; y0; y0prime]; % initial conditions
tspan=[0 2];
[T,Y]=ode45(fun,tspan,Y0);

Stephan (view profile)

on 3 Jun 2019
Edited by Stephan

Stephan (view profile)

on 3 Jun 2019

This will help:
syms y(t) x(t) t
ode(1) = -(diff(y,t,2))-5*diff(y,t)-y == 3*diff(x,t,2)+4*diff(x,t)^2+2*x+6*t;
ode(2) = -(diff(y,t,2))-2.3*diff(y,t)-2*y == 2*diff(x,t,2)+3*diff(x,t)^2+x;
pretty(ode)
[V,S] = odeToVectorField(ode)
fun = matlabFunction(V,'Vars',{'t','Y'})
x0 = [2 0 3 0.25];
[t, sol] = ode45(fun,[0 2], x0);
subplot(2,2,1)
plot(t,sol(:,1),'-r')
subplot(2,2,2)
plot(t,sol(:,2),'-b')
subplot(2,2,3)
plot(t,sol(:,1),'-g')
subplot(2,2,4)
plot(t,sol(:,2),'-k')

Nagarajan

Nagarajan (view profile)

on 4 Jun 2019
Thank you stephan, a clear explanation.

Steven Lord (view profile)

on 3 Jun 2019

Let v = [x; x'; y; y']. This means v' = [x'; x''; y'; y''].
Can you express x' in terms of some of the elements of v? Sure, it's just v(2). Expressing y' in terms of elements of v is also easy, it's v(4). So we have half the right-hand side of our system of ODEs. We still have two pieces to fill in; I'll represent them by placeholders in the (pseudo)code below.
dvdt = [v(2);
something;
v(4);
somethingElse];
Now things get a bit trickier. You can't isolate y'' or x'' as a function solely of x, x', y, y', and t. So you're going to need to use a Mass matrix. The ODE you're going to solve is not v' = f(t, y) but M*v' = f(t, y). If we rewrite your first ODE to put the second derivatives together on the left side of the equals sign:
-y''-3*x'' = 5*y'+y+4*x'^2+2*x+6*t
The left side is [0 -3 0 -1]*v (to show this, expand it out: 0*x'-3*x''+0*y'-y''). Let's put that in the second row of our mass matrix. The first and third rows are just the corresponding rows of the identity matrix, since [1 0 0 0]*v == x' and v(2) == x' and similarly for [0 0 1 0]*v and v(4) both representing y'.
M = [1 0 0 0;
0 -3 0 -1;
0 0 1 0;
dvdt = [v(2);
something;
v(4);
somethingElse];
Fill in the something part of dvdt using the right side of that rewritten first ODE. Do the same type of manipulation (pull the second derivates to the left side of the equals sign) for the second ODE to fill in aDifferentSomethingElse and somethingElse.
Now write a function to evaluate dvdt as a function of t and v. You will need to use odeset to create an options structure containing your Mass matrix M. Pass the function and the options structure into ode45.
Technically you could have pulled both the first and second derivatives to the left side of the equals sign, and had all the elements of that row of the mass matrix be non-zero. But I wanted to keep this a little simple (too late, I know) so I pulled just the second derivatives into the Mass matrix.

Nagarajan

Nagarajan (view profile)

on 4 Jun 2019
Thank you steven, Your answer gave a clear explanation in solving the ODE's