solving a differential equation coupling together in matrix and vector form with ode45

I have two problem:

1) I used two functions for solving a set of equation not in matrix or vector form but like

[x(2);3+x(4)+x(3);x(4);....................]

then [t,x] = ode45(@(t,x) lili(t,x),tspan,x0);

now I need the vector of dx=[x(2) x(4) x(6)]

but it doesnt return this vector.

2) i want to solve equation like this D2y=d2x+cross(m,n)

its clear cross product is in vector form

how could i solve matrix form of differential equation?

especially initial values is a vector

[Torsten]

1) Unclear what you have done and what your problem is. Please provide the code you are using and possibly the error message you get.

2) ODE45 expects the differential equations to be given in vector format. So convert your equations from matrix form to vector form.

Best wishes

Torsten.

Answer 1

[roya afshar]

clc
clear
x_1=[1 8 9];
x_2=[2 3 4];
x0=[x_1 x_2].';
tspan = [0 20];
one = 1:3;
two = 4:6;
n=[0 1 2]';
f = @(t,x) [x(two);x(one)+n]
[t, xR] = ode45(f, tspan, x0);
x1 = xR(:,one)
 
x2 = xR(:,two)
plot(x1,t)
%plot(x2,t)

I have found a solution could any make me sure?

[Torsten]

You solve

x1'' = x1
x2'' = x2 + 1
x3'' = x3 + 2

with initial conditions

x1(0)=1, x2(0)=8, x3(0)=9, x1'(0)=2, x2'(0)=3, x3'(0)=4

and plot x1, x2 and x3 in plot 1 and x1', x2' and x3' in plot 2.

Is it that what you wanted to achieve ?

[roya afshar]

yes exactly.

thank you so much

Answer 2

[roya afshar]

I have solved 3 second order of differential equations by ode45 and there is no error. as a example:

D2x=x+z+Dx+Dy

D2Y=Dx+z

D2z=Dz+Dx+y

more complicated than above and its nonlinear. and my codes:

function dy=roya(t,y)

dy=[y(2);y(1)+y(5)+y(2)+y(4);y(4);y(2)+y(5);y(6):y(6)+y(2)+y(3)];

end

y0=[0 0 0 0 0 0];

tspan = [0 5];

[t,y] = ode45(@(t,y) roya(t,y),tspan,y0);

now I need vector[y(1) y(3) y(5)] , [y(2) y(4) y(6)]

.........................................................

after solving above I wand to use [y(1) y(3) y(5)] this vector in another differential equation like:

D2m=D2y+cross(a,b)+D1y;

how could I define initial value for this equation

output is a 3by1 vecor?and a,b are 3by3 matrix

[Torsten]

And why don't you solve them all together in one step ?

And what is the outcome you expect from D2y+cross(a,b)+D1y ? D2y and D1y are vectors, cross(a,b) is a 3x3 matrix.

[roya afshar]

the problem is initial value of Dm and m. how could I define initial value for them as a vector? if I solve all of them together I have to define initial value like this?y0=[0 0 0 0 0 0 [1 2 3] [2 3 4]];

a is a 3by1

b 1by3 vector

but if I solve them together and there were no problem still I need to call some of states as an input of kalman filter.

[roya afshar]

clc

clear

y0=[10 5 3];

y1=[1 2 3];

tspan = [0 5];

[t,y] = ode45(@(t,y) bhr(t,y),tspan,[y0 y1]);

function dy = bhr(t,y)

n=[0 1 2];

dy =[y(2);n+y(1)];

end

simple example of what i am looking for is above.

I am trying solve second order deffirential equation in vector form.