Re: Finding a continuous solution of a cubic

• To: mathgroup at smc.vnet.net
• Subject: [mg86006] Re: Finding a continuous solution of a cubic
• From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
• Date: Sat, 1 Mar 2008 04:44:59 -0500 (EST)
• References: <fq3agu\$fq0\$1@smc.vnet.net>

```Hugh Goyder <h.g.d.goyder at cranfield.ac.uk> wrote:
> I am investigating the maxima and minima of the expression, e, below.
> The expression is a quartic in x and has two parameters a and b which,
> for my problem, are restricted to the range -1 < a < 1 and 0 < b (the
> interesting part is 0<b<1).
>
> I start with a simple approach and take the derivative with respect to
> x and solve to find the three turning points. The solution is
> complicated and not easy to examine. I work out the discriminant to
> see when I will have three real roots or one and plot the region. I
> work out three functions for the three roots r1, r2 and r3, and three
> functions for the value of e at the roots, f1, f2, and f3. I am
> expecting three real functions when within the region defined by the
> discriminant and one outside this region. I plot the three roots and
> the three functions (there are some artifacts where the functions
> turn from real to complex, this is not a problem).
>
> Now my problem. The function f3 is continuous in the region where
> there are three solutions and this I like. Functions f1 and f2 are not
> continuous but contain holes. If functions f1 and f2 are plotted
> together then they each fill in the others holes. The holes makes my
> life difficult because I do not have continuous functions.
>
> Is it possible to make Solve find roots that will give rise to three
> functions one corresponding to the maxima and two corresponding to
> each minima without the functions crossing over and jumping between
> the turning points?

Andrzej and Daniel have already responded, but I'm not sure they gave you
quite what you want.

I believe that the three functions you want, the solutions of d == 0,
can be expressed as

x -> Root[-2 a b + (-1 + b + a^2 b) #1 + #1^3 &, 1]
applicable when  27 a^2 b^2 <= (1 - b - a^2 b)^3 Or a <= 0

x -> Root[-2 a b + (-1 + b + a^2 b) #1 + #1^3 &, 2]
applicable when  27 a^2 b^2 <= (1 - b - a^2 b)^3

x -> -Root[2 a b + (-1 + b + a^2 b) #1 + #1^3 &, 1]
applicable when  27 a^2 b^2 <= (1 - b - a^2 b)^3 Or a >= 0

David W. Cantrell

> I also notice, from the plots when shown together,  that my desired
> functions are even with respect to the parameter a. Is this correct
> and how would I show this if I can't assemble them as continuous
> functions?
>
> Thanks
>
> Hugh Goyder
>
> e = ((1 - x)^2 + b*(-1 + a)^2)*((1 + x)^2 +
>           b*(1 + a)^2);
>
> d = Simplify[D[e, x]];
>
> dis = Discriminant[d, x];
>
> ContourPlot[dis, {a, -1, 1}, {b, 0, 1},
>    ContourShading -> False, Contours -> {0},
>    FrameLabel -> {"a", "b"}]
>
> sol = Solve[d == 0, x];
>
> ClearAll[x1, x2, x3, f1, f2, f3];
> x1[a_, b_] := Evaluate[x /. sol[[1]]];
> x2[a_, b_] := Evaluate[x /. sol[[2]]];
> x3[a_, b_] := Evaluate[x /. sol[[3]]];
> f1[a_, b_] := Evaluate[e /. sol[[1]]];
> f2[a_, b_] := Evaluate[e /. sol[[2]]];
> f3[a_, b_] := Evaluate[e /. sol[[3]]];
>
> r1 = Plot3D[x1[a, b], {a, -1, 1}, {b, 0, 1}]
>
> r2 = Plot3D[x2[a, b], {a, -1, 1}, {b, 0, 1}]
>
> r3 = Plot3D[x3[a, b], {a, -1, 1}, {b, 0, 1}]
>
> Show[r1, r2, r3, PlotRange -> All]
>
> s1 = Plot3D[f1[a, b], {a, -1, 1}, {b, 0, 1}]
>
> s2 = Plot3D[f2[a, b], {a, -1, 1}, {b, 0, 1}]
>
> s3 = Plot3D[f3[a, b], {a, -1, 1}, {b, 0, 2}]
>
> Show[s1, s2]

```

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