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Re: DSolve question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg88751] Re: DSolve question
  • From: Szabolcs Horvát <szhorvat at gmail.com>
  • Date: Fri, 16 May 2008 05:27:41 -0400 (EDT)
  • Organization: University of Bergen
  • References: <g0h82g$mu1$1@smc.vnet.net>

igwood wrote:
> Hi everyone. I'm looking for help using the result of DSolve. 
> In the code below the DSolve statements in In[3] and In[4] find i1[t] and i2[t] respectively in terms of variable t and a constant a1, a2 respectively. Then, numerical values for a1 and a2 will be found by simultaneous equations equating i1[t] and i2[t] in the FindRoot statement in In[7]. 
> However, I havenâ??t been able to make Mathematica relate the i1[t], i2[t] in FindRoot to the i1[t], i2[t] resulting from DSolve. Iâ??ve tried several ways and finally the way below and always get errors. 
> On the other hand if I COPY and PASTE the DSolve output Out[3], Out[4] following the -> into the lhs of i1[t] = lhs, then the FindRoot gives me the a1, and similar for a2.. 
> If anyone can instruct me on how to get the FindRoot to recognize the i1[t], i2[t] found from DSolve I would appreciate it. I guess in other words, how do I get output of DSolve to be a usable function. 
> 
> In my own reply to this post I will paste the example of the code where COPY/PASTE has been used as described above to give numerical values for a1, a2.
> 
> Thanks in advance.
> 
> 
> In[1]:= Duty = .6; SWfreq = 1*10^6; Vin = 3.2; Vout = 1.8; L = 
>  4.7*10^-7; R = 0.008;
> 
> In[2]:= Dt1 = 
>  Duty (1/SWfreq); Dt2 = (1 - Duty) (1/SWfreq); VL2 = - Vout; VL1 = 
>  Vin - Vout; T = 1/SWfreq;
> 
> In[3]:= it1 = 
>  Simplify[DSolve[{L i1'[t] + R i1[t] - VL1 == 0, i1[0] == a1}, i1[t], 
>    t]]

This is unrelated to the question, but it's probably better to work with 
exact numbers when using DSolve.  I don't know if inexact numbers can 
cause trouble with DSolve, but I know that they can be a problem for 
Solve when using certain settings and DSolve does use Solve internally 
(or at least it does emit Solve::* messages sometimes).

So this is a bit better/safer:

it1 = Simplify[
   DSolve[Rationalize[{L i1'[t] + R i1[t] - VL1 == 0, i1[0] == a1}],
    i1[t], t]]


> 
> Out[3]= {{i1[t] -> 175.+ (-175. + 1. a1) E^(-17021.3 t)}}
> 
> In[4]:= it2 = 
>  Simplify[DSolve[{L i2'[t] + R i2[t] - VL2 == 0, i2[0] == a2}, i2[t], 
>    t]]
> 
> Out[4]= {{i2[t] -> -225. + (225.+ 1. a2) E^(-17021.3 t)}}
> 
> In[5]:= i1[t_] = i1[t] /. it1

Try using i1[t_] = i1[t] /. First[it1] here,

> 
> Out[5]= {175.+ (-175. + 1. a1) E^(-17021.3 t)}
> 
> In[6]:= i2[t_] = i2[t] /. it2

and i2[t_] = i2[t] /. First[it2] here, and everything will work fine.

> 
> Out[6]= {-225. + (225.+ 1. a2) E^(-17021.3 t)}
> 
> In[7]:= FindRoot[{i1[0] == i2[T], 
>   i1[Dt1] == i2[Dt1]},
>  {a1, 0}, {a2, 0}]
> 
> During evaluation of In[7]:= Thread::tdlen: Objects of unequal length \
> in {0,{-0.983123}}+{{1.,0}} cannot be combined. >>
> 
> During evaluation of In[7]:= Thread::tdlen: Objects of unequal length \
> in {{1.,0}}+{0,{-0.983123}} cannot be combined. >>
> 
> During evaluation of In[7]:= Thread::tdlen: Objects of unequal length \
> in {0,{-0.989839}}+{{0.989839,0}} cannot be combined. >>
> 
> During evaluation of In[7]:= General::stop: Further output of \
> Thread::tdlen will be suppressed during this calculation. >>
> 
> During evaluation of In[7]:= FindRoot::njnum: The Jacobian is not a \
> matrix of numbers at {a1,a2} = {0.,0.}. >>
> 
> Out[7]= {a1 -> 0., a2 -> 0.}
> 


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