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Bifurcation Points - Differential Equations

  1. Jul 13, 2005 #1
    "Bifurcation Points. In many physical problems some observable quantity, such as a velocity, waveform, or chemical reaction, depends on a parameter describing the physical state. As this parameter is increased, a critical value is reached at which the velocity, or waveform, or reaction suddenly changes its character. For example, as the amount of one of the chemicals in a certain mixture is increased, spiral wave patterns of varying color suddenly emerge in an originally quiescent fluid. In many such cases the mathematical analysis ultimately leads to an equation of the form

    [tex]\frac{dx}{dt} = \left( R - R_c \right) x - ax^3 \mbox{.}[/tex]

    Here [tex]a[/tex] and [tex]R_c[/tex] are positive constants, and [tex]R[/tex] is a parameter that may take on various values. For example, [tex]R[/tex] may measure the amount of a certain chemical and [tex]x[/tex] may measure a chemical reaction.

    (a) If [tex]R < R_c \mbox{,}[/tex] show that there is only one equilibrium solution [tex]x = 0[/tex] and that it is asymptotically stable.

    (b) If [tex]R > R_c \mbox{,}[/tex] show that there are three equilibrium solutions, [tex]x=0[/tex] and [tex]x = \pm \sqrt{ \left( R - R_c \right) /a} \mbox{,}[/tex] and that the first solution is unstable while the other two are asymptotically stable.

    (c) Draw a graph in the [tex]Rx[/tex]-plane showing all equilibrium solutions and label each one as asymptotically stable or unstable.

    The point [tex]R = R_c[/tex] is called a bifurcation point. For R < Rc one observes the asymptotically stable equilibrium solution [tex]x = 0\mbox{.}[/tex] However, this solution loses its stability as [tex]R[/tex] passes through the value Rc, and for R > Rc the asymptotically stable (and hence the observable) solutions are [tex]x = \sqrt{\left( R - R_c \right)/a}[/tex] and [tex]x = -\sqrt{\left( R - R_c \right)/a} \mbox{.}[/tex] Because of the way in which the solutions branch at [tex]R_c[/tex], this type of bifurcation is called a pitchfork bifurcation; your sketch should suggest that this name is appropriate."

    My work

    We are given the following:

    [tex]\frac{dx}{dt} = \left( R - R_c \right) x - ax^3 \mbox{,} \qquad R_c > 0 \mbox{,} \qquad a > 0 \mbox{.}[/tex]

    (a)

    If [tex]R<R_c\mbox{,}[/tex] then we obtain:

    [tex]f\left( x \right) = \left( R - R_c \right) x - ax^3 = 0 \Rightarrow x = 0 = \phi \left( t \right) \mbox{.}[/tex]

    Note that [tex]R - R_c \neq ax^2[/tex]. Thus, [tex]\phi \left( t \right)[/tex] is the only critical point. Also, consider the following:

    [tex]f ^{\prime} \left( x \right) = R - R_c - 3ax^2 \Rightarrow f ^{\prime} \left( \phi \right) = R - R_c < 0[/tex]

    which implies that [tex]\phi \left( t \right)[/tex] is asymptotically stable.

    (b)

    If [tex]R>R_c\mbox{,}[/tex] then we obtain:

    [tex] f\left( x \right) = \left( R - R_c \right) x - ax^3 = 0 \Rightarrow \left\{ \begin{array}{l} \displaystyle x = 0 = \zeta _1 (t) \mbox{,} \vspace*{6pt} \\ \displaystyle x = \sqrt{\frac{R - R_c}{a}} = \zeta _2 (t) \mbox{,} \vspace*{6pt} \\ \displaystyle x = -\sqrt{\frac{R - R_c}{a}} = \zeta _3 (t) \mbox{.} \end{array} \right.[/tex]

    Also, consider the following:

    [tex]f ^{\prime} \left( x \right) = R - R_c - 3ax^2 \Rightarrow \left\{ \begin{array}{l} f ^{\prime} \left( \zeta _1 \right) = R - R_c > 0 \mbox{,} \\ f ^{\prime} \left( \zeta _2 \right) = -2\left( R - R_c \right) < 0 \mbox{,} \\ f ^{\prime} \left( \zeta _3 \right) = 4\left( R - R_c \right) > 0 \mbox{,} \end{array} \right.[/tex]

    which implies that [tex]\zeta _1 \left( t \right)[/tex] is an unstable critical point, and [tex]\zeta _2 \left( t \right)[/tex] is an asymptotically stable critical point, and [tex]\zeta _3 \left( t \right)[/tex] is an asymptotically stable critical point.

    (c)

    First, let's find any inflection points there may be.

    [tex]f ^{\prime} \left( x \right) = R - R_c - 3ax^2 = 0 \Rightarrow x = \pm \sqrt{\frac{R-R_c}{3a}}[/tex]

    I know how a pitchfork looks like, but I'm not sure how those inflection points are going to affect this particular picture. It'd be easier if I knew how to plot this using mathematica. Any tips?

    Any help is highly appreciated.
     
    Last edited: Jul 14, 2005
  2. jcsd
  3. Jul 14, 2005 #2

    CarlB

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    In part (b), there is a mistake in the calculation of one of the f' that is inconsistent with the conclusion you've given.

    For part (c), you will need to plot the stable/unstable points (i.e. x values) against R. This means that you need to derive an equation that relates R and x. Of course that equation is just going to be dx/dt = 0. The equation will be a third order polynomial in x and a first order polynomial in R.

    You will need to recall your calculus (or was it algebra? for me it was a very long time ago) class that covered the algebraic equations that correspond to various conic sections. Figure out which conic sections correspond to the solutions and you will find that two of these form the pitch fork. That is, one will give the (straight) center tine and the handle, while the other will give the other two (curved) tines.

    Carl
     
    Last edited: Jul 14, 2005
  4. Jul 14, 2005 #3
    Thanks for your input, Carl.

    I found a typo in part (b) - it's been corrected now. Other than that, I suppose it is relevant to point out that I found in an earlier exercise that when: [tex]f^{\prime} (\xi) < 0[/tex] it follows that [tex]\xi[/tex] is an asymptotically stable equilibrium solution; [tex]f^{\prime} (\xi) > 0[/tex] it follows that [tex]\xi[/tex] is an unstable equilibrium solution; [tex]f^{\prime} (\xi) = 0[/tex] it follows that [tex]\xi[/tex] is a semistable equilibrium solution. That's how I worked it out.

    Regarding part (c), I've got a sketch of what I have in mind at

    http://mygraph.cjb.net/

    It's not done, but it seems to join the solutions found in parts (a) and (b) without a problem. Again, I'm stuck with the asymptotes in this graph. There may be a mistake there.

    Thanks!
     
  5. Jul 14, 2005 #4

    saltydog

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    Here's (a) in Mathematica:

    Code (Text):
    a = 1;
    R = 1;
    rc = 2;
    sol1 = NDSolve[{x'[t] == (R - rc) x[t] - a x[t]^3, x[0] == 5}, x, {t, 0, 20}]
    f1[t_] := Evaluate[x[t] /. Flatten[sol1]];
    pt1 = Plot[f1[t], {t, 0, 20}]
    sol2 = NDSolve[{x'[t] == (R - rc) x[t] - a x[t]^3, x[0] == -5},
        x, {t, 0, 20}]
    f2[t_] := Evaluate[x[t] /. Flatten[sol2]];
    pt2 = Plot[f[t], {t, 0, 20}]
    Show[{pt1, pt2}, PlotRange -> {{0, 10}, {-5, 5}}]
    A plot of two solutions exhibiting the asymptotic tend to the equilibrium point is attached.
     

    Attached Files:

  6. Jul 14, 2005 #5

    CarlB

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    The graph is close, but you may get docked some points for some subtle inaccuracies. First, there are no assymptotes, second, the point where the curves meet is going to make four right angles.

    When I taught college algebra, trig and calculus, graphing was one of the most difficult subjects. For this problem, it's useful to have a good understanding of what the equations for the various conic sections look like.

    Carl
     
  7. Jul 14, 2005 #6

    saltydog

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    Here's part (b) in Mathematica:

    Code (Text):

    a=1;
    R = 2;
    rc = 1;
    r1 = 0;
    [tex]r2=\sqrt{\frac{R-rc}{a}};\quad r3=-\sqrt{\frac{R-rc}{a}}[/tex]


    line1 = Graphics[Line[{{0, r1}, {10, r1}}]];
    line2 = Graphics[{Dashing[{0.05, 0.05}], Line[{{-10, r2}, {10, r2}}]}];
    line3 = Graphics[{Dashing[{0.05, 0.05}], Line[{{-10, r3}, {10, r3}}]}];

    sol3 = NDSolve[{x'[t] == (R - rc) x[t] - a x[t]^3, x[0] == 5}, x, {t, 0, 20}]
    f3[t_] := Evaluate[x[t] /. Flatten[sol3]];
    pt3 = Plot[f3[t], {t, 0, 10}, PlotRange -> {{-5, 5}, {-5, 5}},
        AspectRatio -> 1]

    sol4 = NDSolve[{x'[t] == (R - rc) x[t] - a x[t]^3, x[0] == 0.1},
        x, {t, 0, 20}]
    f4[t_] := Evaluate[x[t] /. Flatten[sol4]];
    pt4 = Plot[f4[t], {t, 0, 20}, PlotRange -> {{-5, 5}, {-5, 5}},
        AspectRatio -> 1]

    sol5 = NDSolve[{x'[t] == (R - rc) x[t] - a x[t]^3, x[0] == -0.1},
        x, {t, 0, 5}]
    f5[t_] := Evaluate[x[t] /. Flatten[sol5]];
    pt5 = Plot[f5[t], {t, 0, 5}, PlotRange -> {{-5, 5}, {-5, 5}},
        AspectRatio -> 1]

    sol6 = NDSolve[{x'[t] == (R - rc) x[t] - a x[t]^3, x[0] == -5}, x, {t, 0, 5}]
    f6[x_] := Evaluate[x[t] /. Flatten[sol6]];
    pt6 = Plot[f6[t], {t, 0, 5}, PlotRange -> {{-5, 5}, {-5, 5}},
        AspectRatio -> 1]
    Show[{pt3, pt4, pt5, pt6, line2, line3}]
    A plot showing how solutions near the unstable equilibrium point move away from it and how others move to the stable equilibrium points is attached.
     

    Attached Files:

  8. Jul 14, 2005 #7

    saltydog

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    Here's part (c) in Mathematica for the case of Rc=1 and R varies from 0 to 2:

    Code (Text):
    plotarray = Table[{0, 0}, {5000}];
    index = 0;
    For[R = 0, R <= 2, R += 0.001,
        rootarray = x /. Solve[R x - x - x^3 == 0, x];
        For[rnum = 1, rnum <= Length[rootarray], rnum++,
          If[Element[rootarray[[rnum]], Reals],
              index++;
              plotarray[[index]] = {R, rootarray[[rnum]]};
              ];
          ];
        ];
    plotarray = Take[plotarray, index];
    ListPlot[plotarray, Axes -> None]
    ListPlot[plotarray]
    The plots below exhibit this pitch-fork bifurcation (with and without axes to illustrate middle line). There are no asymptotes.

    Yea Thiago, these are my favorites. Now, how about this one:

    [tex]\frac{dx}{dt}=c+kx-x^3[/tex]

    This is a two-parameter differential equation which exhibits the "cusp catastrophe".
     

    Attached Files:

    Last edited: Jul 14, 2005
  9. Jul 14, 2005 #8
    Thank you so much, guys. Now it's crystal clear.

    Hey, saltydog, how did you learn to plot that way? Is there any book or other resource you would recommend me (other than the documentation that ships with Mathematica)?

    Thiago
     
  10. Jul 15, 2005 #9

    saltydog

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    Hello Thiago. I just learned it by doing. Probably have books though. What about:

    [tex]\frac{dx}{dt}=c+dx-x^3?[/tex]

    Instead of a bifurcation point, this one has a bifurcation curve and a cusp catastrophe surface above it. See attached.

    I imagine life 600 million years ago on the top fold of such a cusp. Something happen. The system bifurcated. Disparate animal clans emerged on the bottom surface.

    Interesting how such a simple equation can give insight into such a profound and global phenomenon as Punctuated Equilibrium. :smile:
     

    Attached Files:

  11. Jul 15, 2005 #10
    Wow! That goes against the traditional Darwinian theory of evolution, doesn't it? Does it illustrate a mutation at the highest level (catastrophe surface) over a short period of time, being responsible for new species (bifurcation curve)?
     
  12. Jul 15, 2005 #11

    saltydog

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    I believe the fossil record supports Punctuated Equilibrium: periods of relative stasis punctuated by periods of rapid change. Pretty sure anyway.

    Our world is rife with catastrophe: extinctions, geological events, stock-market crashes, human conflict, astronomical events. Many processes have the potential for catastrophe: a sudden and qualitative change in a system as is the case when it passes through it's bifurcation point.

    I can't speculate on what specific event caused the Cambrain explosion but it appears to me it represented a sudden and qualitative change that looks to me like some sort of bifurcation. Perhaps many pecularities in nature can be due to this: why do men and chimps share 98% of genes but are so different? A bifurcation perhaps?

    And if ODEs exhibit this sort of phenomenon, what then do PDEs exhibit? :smile:
     
  13. Jul 15, 2005 #12
    You've got a point there. It sounds reasonable to me.

    I haven't studied PDE's so far. Anyhow, I do have guess... chaos? :smile:
     
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