MaplePrimes - answers and comments on Question, phase plots from a differential equation

No sorry I didn't type it in clearly. I am

gdog — Tue, 19 Feb 2013 05:06:24 Z

no sorry that isnt what i mean it was just a differnetial equation

Phase portrait 2nd order DE

Adri van der Meer — Tue, 19 Feb 2013 14:45:55 Z

Your formula is still not unambiguous (unbalanced parentheses). I suppose that you mean:

DE := diff(z(t),t,t) = -((sqrt(diff(z(t)^2,t))+10^10*diff(z(t),t))+10^10*z(t))/diff(z(t),t);

Now rewrite to a system of two first order equations:

 DE2 := diff(z(t),t)=y(t),  subs( {diff(z(t),t)=y(t)}, DE );

A phase portrait with two trajectories:

with(DEtools):
DEplot( {DE2}, [z,y], t=0..1, 
   {[z(0)=5e9,y(0)=2e10],[z(0)=5e9,y(0)=1.5e10]}, 
   z=5e9..2e10, y=5e9..2e10, arrows=smalltwo, linecolor=blue );

By the way, because of the huge constants in the equation, you can omit the sqrt term and simplify to

DE3 := diff(z(t), t) = y(t), diff(y(t), t) = -(10^10*y(t)+10^10*z(t))/y(t);

which gives you the same phase portrait.

Using equation 3.4.82 in the reference

Preben Alsholm — Tue, 19 Feb 2013 21:09:11 Z

Using equation 3.4.82 in the reference you gave and also
diff(phi,t,t) = phidot*diff(phidot,phi) we are back to the second order ode:

ode:=diff(phi(t),t,t)=-(sqrt(3/2)/M^2*sqrt(diff(phi(t),t)^2+m^2*phi(t)^2)*diff(phi(t),t)+m^2*phi(t));
#Following Adri van der Meer we turn the equation into a first order system:
sys:=diff(phi(t),t)=phi1(t),subs(diff(phi(t),t)=phi1(t),ode);
#I pick my own parameters, m = M = 1:
with(DEtools):
DEplot(eval([sys],{M=1,m=1}),[phi(t),phi1(t)],t=0..30,[[phi(0)=1,phi1(0)=0],[phi(0)=-1,phi1(0)=0],[phi(0)=2,phi1(0)=-1],[phi(0)=-2,phi1(0)=1]],color=grey,linecolor=blue,stepsize=.1,thickness=1);

Final attempt

Adri van der Meer — Tue, 19 Feb 2013 21:11:53 Z

I take the constant 1 (instead of 10^10). Then I get:

DE := diff(z(t),t,t) = -((sqrt(diff(z(t),t)^2+z(t)^2)*diff(z(t),t))+z(t));
DE2 := diff(z(t),t)=y(t),  subs( {diff(z(t),t)=y(t)}, DE );
with(DEtools):
DEplot( {DE2}, [z,y], t=0..10, 
   {[z(0)=-2,y(0)=2],[z(0)=-1,y(0)=-2],[z(0)=2,y(0)=-2]}, 
   z=-2..2, y=-2..2, arrows=smalltwo, linecolor=blue );

which looks like figure 3.5

thanks that is awesome, can I ask how you

gdog — Tue, 19 Feb 2013 21:43:43 Z

thanks that is awesome, can I ask how you arose at the intitial conditions that you used because i am always unsure what to do here?