Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.



Re: InterpolatingFunction
Posted:
Oct 14, 1996 2:19 AM


Andr=E9 Hautot wrote: > =
> Who can help me? I am studying mathematica and I ask the following > question. Suppose I wish to study the function y[t] defined by the > oversimplified differential equation > =
> sol:=3DNDSolve[{y''[t]+y[t]=3D=3D0,y[0]=3D=3D1,y'[0]=3D=3D1},y,{t,0,20}= , > MaxSteps>2000] > =
> How can I find ALL the roots of y[t]=3D0 in the range {t,0,20}? > I have tried a lot of things but they don't work. Mathematica user's > guide does not help me very much. > This kind of question arises when trying to draw Poincare sections of > chaotic motions in dynamics. > Since I am not yet registered in a mathematica group, please reply to > my personal address: ahautot@ulg.ac.be > Thank you in advance, AH.
Well, it would be nice if there was an "automatic" way. I know of none. Using intuition and observation, though, one can arrive at a complete solution. I'd like to seee other suggestions, but for now, here's mine:
(1) Why use NDSolve in this case? It's so much slower than simply
In[11]:=3D DSolve[{y''[t]+y[t]=3D=3D0,y[0]=3D=3D1,y'[0]=3D=3D1}, y[t], t] Out[11]=3D {{y[t] > Cos[t] + Sin[t]}} For referential simplicity, set =
In[12]:=3D f2[t_] :=3D Evaluate[ y[t]/.First[Flatten[%]] ]
(2) Look at the picture of the graph. There's only a small number of zeroes in the interval of interest, so one can get a list of estimates of their location by roughly sampling the interval and looking for signchanges:
In[29]:=3D t1 =3D Table[ {i, f2[i]}, {i, 0, 20, 0.5} ]; In[37]:=3D zcL =3D (Sign[First[#]] Sign[Last[#]])& /@ Partition[ Last[Transpose[t1]], 2, 1]; In[41]:=3D pL =3D Flatten[Position[zcL, 1]] Out[41]=3D {5, 11, 18, 24, 30, 37}
pL is the list of the positions of sign changes in the sample list t1. Now find the tcoordinates of the points _just before_ the sign changes. Those will be the starting estimates for FindRoot[].
In[42]:=3D svL =3D First[Transpose[t1]][[#]]& /@ pL Out[42]=3D {2., 5., 8.5, 11.5, 14.5, 18.}
(3) Apply FindRoot[] to all initial guesses:
In[45]:=3D FindRoot[ f2[t] =3D=3D 0, {t, #} ]& /@ svL Out[45]=3D {{t > 2.35619}, {t > 5.49779}, {t > 8.63938}, {t > 11.781}, =
{t > 14.9226}, {t > 18.0642}} In[46]:=3D zeroes =3D (t/.#& /@ Flatten[%]) Out[46]=3D {2.35619, 5.49779, 8.63938, 11.781, 14.9226, 18.0642}
If one is concerned about duplicate roots, one can run Union[zeroes].
Check the results with f2[#]& /@ zeroes//Chop (nned to chop!) or
Plot[ f2[t], {t, 0, 20}, Epilog > { GrayLevel[0.5], PointSize[0.02], Point[{#,0}]& /@ zeroes } ];
This is a bit of a roundabout approach, so let's see who can do it more elegantly ...
 =
______________________________________________________________________ Harald Berndt, Ph.D. Research Specialist, =
Consultant =
Phone: 5106525974 FAX: 5102154299 ______________________________________________________________________ "I am what I am".................................. .Popeye the Sailor
(KRYTEN:.............Are you sure? I always thought it was Descartes!) (LISTER:.So did I, man! It's so easy to get those two dudes mixed up!)



