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Re: an equation containg radicals

  • To: mathgroup at smc.vnet.net
  • Subject: [mg69666] Re: [mg69648] an equation containg radicals
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Wed, 20 Sep 2006 02:44:21 -0400 (EDT)
  • References: <200609190945.FAA28444@smc.vnet.net>

dimmechan at yahoo.com wrote:
> Hello to all.
> 
> In a crack problem appeared the following function.
> 
> K[p_] := 1 - 4*(1 - v)*λ^2*p^2*(1 - Sqrt[e^2 - p^2]/Sqrt[a^2 - p^2])
> a = 1/λ;
> 
> Here are some typical values for the involving constants
> 
> consts = {e -> 1/1000, v -> 3/10, λ -> 10^(-5)};
> 
> Here is the solution obtained with Solve
> 
> sols = FullSimplify[Solve[eq = K[p] == 0, p]]
> {{p -> (-Sqrt[2])*Sqrt[-(1/(λ^2*(-9 + 8*v + Sqrt[-15 + 16*v +
> 64*e^2*(-1 + v)^2*λ^2])))]},
>   {p -> Sqrt[2]*Sqrt[-(1/(λ^2*(-9 + 8*v + Sqrt[-15 + 16*v + 64*e^2*(-1
> + v)^2*λ^2])))]},
>   {p -> (-Sqrt[2])*Sqrt[1/(λ^2*(9 - 8*v + Sqrt[-15 + 16*v + 64*e^2*(-1
> + v)^2*λ^2]))]},
>   {p -> Sqrt[2]*Sqrt[1/(λ^2*(9 - 8*v + Sqrt[-15 + 16*v + 64*e^2*(-1 +
> v)^2*λ^2]))]}}
> 
> What I need now is to see which roots (or if all roots) are extreneous
> (i.e. they do not satisfy the intial equation K[p]=0).
> This is a difficult task for Mathematica.
> 
> TimeConstrained[FullSimplify[eq /. sols], 300, "Failed"]
> "Failed"
> 
> However replacing the values for the constants it is verified that all
> solutions (for this typical values of the constants) are extreneous.
> 
> eq /. sols /. consts
> {False, False, False, False}
> 
> This can be verified also by the following command
> 
> Solve[eq /. consts, p]
> {}
> 
> My question is if it a way to "help" a bit Mathematica in the above
> verification with the symbolic parameters.
> 
> Thanks in advance for any help.

I had some trouble with the fonts so I'm using ii for the last variable 
in your 'consts' replacement list.

K[p_] := 1 - 4*(1 - v)*ii^2*p^2*(1 - Sqrt[e^2 - p^2]/Sqrt[a^2 - p^2])
a = 1/ii;
eq = K[p] == 0;
consts = {e -> 1/1000, v -> 3/10, ii -> 10^(-5)};

If you use Reduce and place some restrictions on teh parameters you will 
see that there are no real-valued solutions in a neighborhood of the 
consts values in parameter space.

In[10]:= sols = Reduce[{eq,0<e<1/10,0<v<1,0<ii<1/100}/.consts, p, Reals]
Out[10]= False

I'm not sure this tells you what you need but it may give some ideas for 
  using Reduce on this equation.

If you want to consider complex valued solutions, another possibility 
might be to use Solve as you have it above, then use FullSimplify as 
above but with assumptions on the parameters. Again, this might just 
hang, but giving assumptions will probably improve the chances that it 
does not, and that it gives a meaningful result.


Daniel Lichtblau
Wolfram Research


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