2nd order numerical differential equation system solving

3 visualizzazioni (ultimi 30 giorni)
Hi!
Could you guys please help me with the following 2nd order equation system?
  1. G=6.673*10^-11;
  2. m1=1; m2=2; m3=3;
  3. syms x1(t) x2(t) x3(t);
  4. syms y1(t) y2(t) y3(t);
  5. syms u1(t) u2(t) u3(t);
  6. syms v1(t) v2(t) v3(t);
  7. %Körper 1/Mass1
  8. ode1 = u1==diff(x1,t);
  9. ode2 = v1==diff(y1,t);
  10. ode3 = diff(u1,t)*m1==((G*m1*m2)/((x2-x1)^2+(y2-y1)^2)^(3/2))*(x2-x1)+((G*m1*m3)/((x3-x1)^2+(y3-y1)^2)^(3/2))*(x3-x1);
  11. ode4 = diff(v1,t)*m1==((G*m1*m2)/((x2-x1)^2+(y2-y1)^2)^(3/2))*(y2-y1)+((G*m1*m3)/((x3-x1)^2+(y3-y1)^2)^(3/2))*(y3-y1);
  12. %Körper 2/Mass2
  13. ode5 = u2==diff(x2,t);
  14. ode6 = v2==diff(y2,t);
  15. ode7 = diff(u2,t)*m2==((G*m2*m3)/((x3-x2)^2+(y3-y2)^2)^(3/2))*(x3-x2)+((G*m1*m2)/((x1-x2)^2+(y1-y2)^2)^(3/2))*(x1-x2);
  16. ode8 = diff(v2,t)*m2==((G*m2*m3)/((x3-x2)^2+(y3-y2)^2)^(3/2))*(y3-y2)+((G*m1*m2)/((x1-x2)^2+(y1-y2)^2)^(3/2))*(y1-y2);
  17. %Körper 3/Mass3
  18. ode9 = u3==diff(x3,t);
  19. ode10 = v3==diff(y3,t);
  20. ode11 = diff(u3,t)*m3==((G*m3*m1)/((x1-x3)^2+(y1-y3)^2)^(3/2))*(x1-x3)+((G*m3*m2)/((x2-x3)^2+(y2-y3)^2)^(3/2))*(x2-x3);
  21. ode12 = diff(v3,t)*m3==((G*m3*m1)/((x1-x3)^2+(y1-y3)^2)^(3/2))*(y1-y3)+((G*m3*m2)/((x2-x3)^2+(y2-y3)^2)^(3/2))*(y2-y3);
  22. cond1 = x1(0) == 0;
  23. cond2 = x2(0) == 1;
  24. cond3 = x3(0) == 2;
  25. cond4 = y1(0) == 5;
  26. cond5 = y2(0) == 4;
  27. cond6 = y3(0) == 3;
  28. cond7 = u1(0) == 1;
  29. cond8 = u2(0) == 1;
  30. cond9 = u3(0) == 1;
  31. cond10 = v1(0) == 1;
  32. cond11 = v2(0) == 1;
  33. cond12 = v3(0) == 1;
  34. conds = [cond1; cond2; cond3; cond4; cond5; cond6; cond7; cond8; cond9; cond10; cond11; cond12];
  35. odes = [ode1; ode2; ode3; ode4; ode5; ode6; ode7; ode8; ode9; ode10; ode11; ode12];
I tried to solve it with dsolve. How could it be solved with ode45? Thanks in advance!

Risposta accettata

Torsten
Torsten il 6 Ott 2017
function main
y0=[0; 5; 1; 1; 1; 4; 1; 1; 2; 3; 1; 1];
t0=0;
tfinal=10;
[T Y] = ode45(@odesNew,[t0 tfinal],y0)
function dy = odesNew(t,y)
G=6.673*10^-11;
m1=1; m2=2; m3=3;
dy=zeros(12,1);
x1=y(1);
x2=y(2);
x3=y(3);
y1=y(4);
y2=y(5);
y3=y(6);
u1=y(7);
u2=y(8);
u3=y(9);
v1=y(10);
v2=y(11);
v3=y(12);
%Körper 1/Mass1
dy(1)=u1;
dy(4)=v1;
dy(7)=(((G*m1*m2)/((x2-x1)^2+(y2-y1)^2)^(3/2))*(x2-x1)+((G*m1*m3)/((x3-x1)^2+(y3-y1)^2)^(3/2))*(x3-x1))/m1;
dy(10)=(((G*m1*m2)/((x2-x1)^2+(y2-y1)^2)^(3/2))*(y2-y1)+((G*m1*m3)/((x3-x1)^2+(y3-y1)^2)^(3/2))*(y3-y1))/m1;
%Körper 2/Mass2
dy(2)=u2;
dy(5)=v2;
dy(8)=(((G*m2*m3)/((x3-x2)^2+(y3-y2)^2)^(3/2))*(x3-x2)+((G*m1*m2)/((x1-x2)^2+(y1-y2)^2)^(3/2))*(x1-x2))/m2;
dy(11)=(((G*m2*m3)/((x3-x2)^2+(y3-y2)^2)^(3/2))*(y3-y2)+((G*m1*m2)/((x1-x2)^2+(y1-y2)^2)^(3/2))*(y1-y2))/m2;
%Körper 3/Mass3
dy(3)=u3;
dy(6)=v3;
dy(9)=(((G*m3*m1)/((x1-x3)^2+(y1-y3)^2)^(3/2))*(x1-x3)+((G*m3*m2)/((x2-x3)^2+(y2-y3)^2)^(3/2))*(x2-x3))/m3;
dy(12)=(((G*m3*m1)/((x1-x3)^2+(y1-y3)^2)^(3/2))*(y1-y3)+((G*m3*m2)/((x2-x3)^2+(y2-y3)^2)^(3/2))*(y2-y3))/m3;
Best wishes
Torsten.
  2 Commenti
ahkrit
ahkrit il 10 Ott 2017
Thanks a lot! I would like to just know, is there another way of solving this with the original form of the equations? I mean with more functions as i wrote it first.
Thanks in advance!
Torsten
Torsten il 11 Ott 2017
https://de.mathworks.com/help/symbolic/solve-differential-algebraic-equations.html#bvh12tx-2
might help.
Best wishes
Torsten.

Accedi per commentare.

Più risposte (1)

Josh Meyer
Josh Meyer il 5 Ott 2017
Use the Symbolic Math Toolbox function odeFunction to convert the odes variable into a function handle. Once you have that, you just need to construct a numeric vector of initial conditions y0 (similar to conds) and decide what time span to solve over. The syntax will be
[t,y] = ode45(@odesNew,[t0 tfinal],y0)
  1 Commento
ahkrit
ahkrit il 5 Ott 2017
Hi! Could you write down exactly how it should look like for the first 4 equations, i just cannot figure it out how it should work.
Thanks in advance!

Accedi per commentare.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by