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----