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Re: plot variation

  • To: mathgroup at smc.vnet.net
  • Subject: [mg9619] Re: plot variation
  • From: Pasquale Nardone <pnardon at ulb.ac.be>
  • Date: Fri, 14 Nov 1997 21:39:59 -0500
  • Organization: Université Libre de Bruxelles
  • Sender: owner-wri-mathgroup at wolfram.com

Let:
f[u_]:=u^(4/(gamma+1))*(1/2+1/((gamma-1)*u^2))
g[x_]:=x^(4*(gamma-1)/(gamma+1))/(gamma-1)+x^(-(5-3*gamma)/(gamma+1))
gamma=7/5;
equation=Expand[f[u]]-lambda^(-2(gamma-1)/(gamma+1))*Expand[g[x]]

1) Change variable u->v^3,x->y^3
then
f[v] is just (5/2v) + v^5/2
which has a minimum value at v^6=1 (u=v^3=1 or u=-1) so f(1)=3
= or
f(-1)=3
and
g[y] is    ((1/y) +(5 y^2/2))/(lambda^(1/3)) which has a minimum value
at y^3=1/5

2) the general procedure is
a)Plot g[y] to see the minimum and the change in the secoond derivative
b)Plot f[v] then choose a y (a x) and draw a line which corresponds to
the value of
g[y]

3) So If
-3<g[y]<3 then you will not have real solutions for v (for u) and this
depends on lambda
the inequality has 2 or 1 domain in y so in x

You can also just solve numerically

equationNew=PowerExpand[equation/.{u->v^3,x->y^3}]

sol[x_,l_]:=Map[{x,#}&,Select[
Chop[(v/.(NSolve[(equationNew==0)/.{lambda->l,y->x^(1/3)},v]))^3],FreeQ[#,Complex]&]]

this gives for a list of {x,u}
sol[1.0,1.0]
{{1., 0.4006740873368685}, {1., 1.925926790542241}}

Then:

RangeX=Range[-2.,2.,0.1];
results=Flatten[Map[sol[#,1.]&,RangeX],1]

You can then plot this using ListPlot:

ListPlot[res,PlotJoined->False,PlotRange->{{-2,2},{0,5}}]

and that's all folks (just hange lambda and repeat)

--------------------------------------------
 Pasquale Nardone                          *
                                           *
 Universit Libre de Bruxelles             *
 CP 231, Sciences-Physique                 *
 Bld du Triomphe                           *
 1050 Bruxelles, Belgium                   *
 tel: 650,55,15 fax: 650,57,67 (+32,2)     *
http://homepages.ulb.ac.be/~pnardon/       *
        ,,,
       (o o)
----ooO-(_)-Ooo----



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