Re: FindRoot with NDSolve inside of it doesn't work

*To*: mathgroup at smc.vnet.net*Subject*: [mg111367] Re: FindRoot with NDSolve inside of it doesn't work*From*: Sam Takoy <sam.takoy at yahoo.com>*Date*: Thu, 29 Jul 2010 06:42:57 -0400 (EDT)

One more question. I noticed that I do have to "Clear[f]", otherwise the change is not made. How come? I don't have that problem here: f[x_] := 5 f[x_] := 6 f[x] Thanks again, Sam ________________________________ From: Leonid Shifrin <lshifr at gmail.com> To: Sam Takoy <sam.takoy at yahoo.com>; mathgroup at smc.vnet.net Sent: Wed, July 28, 2010 2:52:56 PM Subject: [mg111367] Re: [mg111343] FindRoot with NDSolve inside of it doesn't work Sam, The difference is that in your last example with exponents the r.h.s of the function does not necessarily involve numerical steps, while when you invoke NDSolve, it does. So, for functions of the former type, you can get away with generic patterns, while for functions of the latter type you can not. This is related to the order in which expressions are evaluated. Inside FindRoot there is f[x0, y0] == {2.7, 5.4}. When this is evaluated, first f[x0, y0] is evaluated. If you have generic patterns x_,y_ in the definition of <f>, then the definition for <f> happily applies at this moment, while x0 and y0 are still symbolic. This is wrong since then NDSolve can not do its job and returns back the set of equations. To postpone the evaluation of <f> until numeric values are supplied by FindRoot in place of x0, y0, we restrict the pattern. Then, evaluation of f[x0, y0] == {2.7, 5.4} just yields back f[x0, y0] == {2.7, 5.4}, since the pattern is not matched by symbolic x0,y0, and then later x0 and y0 are bound to (replaced by) the numeric values generated by FindRoot, which results in correct execution. Cases where one needs to use this trick are interesting since the interaction between built-in functions (FindRoot here) and user - defined ones (f here) mediated by this trick are rather non-trivial. In some sense, we go a little under the hood of the built-in commands here. However they work, we know for sure that they are bound to call Mathematica evaluator on <f> at some point since <f> was defined at top level. And that means that we have all the usual possibilities to alter that part of evaluation that the evaluator normally gives us. Now, since FindRoot holds all its arguments, one may ask why does it not bind equations to numerical arguments right away. My guess is that sometimes it attempts to do some symbolic preprocessing with the equations, a possibility which would be ruled out in the latter scenario. FindRoot does not really care which type the equation is, symbolic or numeric or mixed, but the real problem was in the way you defined <f> - on some (symbolic say) arguments it can produce nonsense, and that means a flawed design of your function. If it makes sense to call it only on numeric arguments (like in your original case), this should be reflected in its definition. In all other cases, it should return unevaluated (this is a general rule for functions in Mathematica), and this is what the pattern-based definition semantics gives you. Hope this helps. Regards, Leonid On Wed, Jul 28, 2010 at 9:17 PM, Sam Takoy <sam.takoy at yahoo.com> wrote: Hi, > > >Thanks for the response. > > >I have determined that introducing ?NumericQ alone works and I am wondering >why? > > >How is the "f" that involves NDSolve fundamentally different from > > >f[x0_,y0_]:={x0 Exp[1],y0 Exp[1]} > > >? > > >Thanks again, > >Sam > > > ________________________________ From: Leonid Shifrin <lshifr at gmail.com> >To: Sam Takoy <sam.takoy at yahoo.com>; mathgroup at smc.vnet.net >Sent: Wed, July 28, 2010 6:03:47 AM >Subject: Re: [mg111343] FindRoot with NDSolve inside of it doesn't work > > > >Sam, > >This modification will work: > > >In[31]:= >Clear[f]; >f[x0_?NumericQ, y0_?NumericQ] := > Module[{sol, x, y}, > (sol = NDSolve[{x'[t] == x[t], y'[t] == y[t], x[0] == x0, > y[0] == y0}, {x, y}, {t, 0, 1}]; > {x[1], y[1]} /. sol[[1]])] > > >In[33]:= > (*f[x0_,y0_]:={x0 Exp[1],y0 Exp[1]};(*Works for this f*)*) >FindRoot[f[x0, y0] == {2.7, 5.4}, {{x0, 1}, {y0, 2}}] > > > > >Out[33]= {x0 -> 0.993274, y0 -> 1.98655} > > >Apart from localizing your variables, the main thing here was to restrict x0 and >y0 >as input parameters to <f>, to only numeric values, by appropriate patterns. > > >Regards, >Leonid > > > >On Wed, Jul 28, 2010 at 10:54 AM, Sam Takoy <sam.takoy at yahoo.com> wrote: > >Hi, >> >>I think it's completely self-explanatory what I'm trying to do in this >>model example: >> >>f[x0_, y0_] := ( >> sol = NDSolve[{x'[t] == x[t], y'[t] == y[t], x[0] == x0, >> y[0] == y0}, {x, y}, {t, 0, 1}]; >> {x[1], y[1]} /. sol[[1]] >> ) >> >>(*f[x0_, y0_]:={x0 Exp[1], y0 Exp[1]}; (* Works for this f *) *) >>FindRoot[f[x0, y0] == {2.7, 5.4}, {{x0, 1}, {y0, 2}}] >> >>Can someone please help me fix this and explain why it's not currently >>working? >> >>Many thanks in advance! >> >>Sam >> >> > >

**Re: FindRoot with NDSolve inside of it doesn't work**

**Re: FindRoot with NDSolve inside of it doesn't work**

**Re: FindRoot with NDSolve inside of it doesn't work**

**Re: FindRoot with NDSolve inside of it doesn't work**