I have Lotka Volterra Equations where the prey equation is modified. The Equations are


To plot the limit cycles, i used NDSolve to solve the coupled nonlinear differential equations for x[t] and y[t]. I then used ParametricPlot to plot for x[t] and y[t] with respect to t, but I'm not getting appropriate limit cycles.

I did a steady state analysis of the equations and got certain conditions on the value of a, b, c, d and k. The equation has a non-trivial steady state, at $(x,y)=(c/d,(a/b)-(ac/bdk))$. The commands are:

Equations = {x'[t] == a*x[t]*(1 - x[t]/k) - b*x[t]*y[t], x[0] == 2, y[0] == 1, y'[t] == -c*y[t] + d*x[t]*y[t]}; 
s = NDSolve[Equations, {x, y}, {t, 0, time}];
ParametricPlot[Evaluate[{x[t], y[t]} /. s], {t, 0, 30}] 

For the value of a = 2; b = 1; c = 2.5; d = 1.2; k = 2.8, I'm supposed to get a unstable spiral. But instead, the trajectory approaches the steady state of $(x,y) = (2.08333 , 0.511905)$.

Any help will be greatly appreciated.

  • $\begingroup$ Please give the values of the constants you have used, and also the NDSolve and Plotting commands you used. Also, you say you don't get appropriate limit cycles, so what would you have expected instead of what you get? $\endgroup$ Jun 11, 2016 at 20:51
  • $\begingroup$ @MariusLadegårdMeyer - I did a steady state analysis of the equations and got certain conditions on the value of a,b,c,d and k. Now, the equation has a non trivial steady state, at (x,y)=(c/d,(a/b)-(ac/bdk)).S s The commands are - Equations = {x'[t] == ax[t]*(1 - x[t]/k) - bx[t]*y[t], x[0] == 2, y[0] == 1, y'[t] == -cy[t] + dx[t]*y[t]}; s = NDSolve[{Equations}, {x, y}, {t, 0, time}]; ParametricPlot[Evaluate[{x[t], y[t]} /. s], {t, 0, 30}] $\endgroup$ Jun 11, 2016 at 20:53
  • 1
    $\begingroup$ Thanks. Just to have a reasonable starting point, what values did you use for a, b, c, d, k? NDSolve will not do anything without specifying these. $\endgroup$ Jun 11, 2016 at 20:57
  • $\begingroup$ For the value of a = 2; b = 1; c = 2.5; d = 1.2; k = 2.8, I'm supposed to get a unstable spiral. But instead, the Parametric Plot goes to the steady state of (x,y)=(2.08333 , 0.511905) $\endgroup$ Jun 11, 2016 at 20:58
  • $\begingroup$ Please edit your question to include all necessary data; otherwise, people can't help you with your problem. $\endgroup$ Jun 11, 2016 at 21:08

2 Answers 2


Hopf bifurcation analysis

The differential system:

f1[x_,y_]:=a x (1 - x/k) - b x y; 
f2[x_,y_]:=-c y + d x y; 
X={x, y};

$$ \begin{align} &\dot{x}=a x\left(1-\frac{x}{k} \right)- bxy\\ &\dot{y}= dxy - cy \end{align} $$

The Jacobian matrix:


The non-trivial equilibrium point:


$$ \begin{align} P_{0}(x,y)=\left(\frac{c}{d} ,\frac{a}{b}\left(1-\frac{c}{dk} \right)\right) \end{align} $$ The linear approximation at $P_{0}$ (coexistence equilibrium point):


$$ \begin{align} J(P_{0})=\left( \begin{array}{cc} \hspace{-0.25cm}-\displaystyle\frac{a c}{d k} & -\displaystyle\frac{b c}{d}\hspace{0.3cm} \\\\ \hspace{0.2cm}\displaystyle\frac{a (d k-c)}{b k} &\hspace{0.2cm} 0 \\ \end{array} \right) \end{align} $$ Under the Hopf bifurcation conditions, $\text{tr}(J(P_{0},\mu_{0}))=0$ and $\text{det}(J(P_{0},\mu_{0}))>0$, where $\mu_{0}$ is the critical bifurcation value for some parameter of our system. In our case, the parameters are strictly positive and $\text{tr}(J(P_{0}))$ cannot be zero. Therefore, Hopf bifurcation not take place at $P_{0}$. The non-trivial equilibrium $P_{0}$ is always locally stable and the only condition that must be fulfilled is given by the following inequality $$ \frac{c}{d k}<1 $$ Code for time series and phase portrait:

Time series

s = ParametricNDSolve[{x'[t] == a x[t] (1 - x[t]/k) - bx[t]*y[t], 
y'[t] == -c y[t] + d x[t]*y[t], x[0] == 11/5, y[0] == 4/5}, {x, y}, {t, 0, 1000}, {a, b, c, d, k}];
Plot[Evaluate[x[1/4, 1/3, 1/2, 1/4, 10][t] /. s], {t, 0, 300}, PlotRange -> All, PlotPoints -> 500, 
PlotStyle -> {Blue, Thickness[0.003]},AxesStyle -> Directive[Black, Small], Background -> Lighter[Gray, 0.95]]

Time series

Phase portrait:

ParametricPlot[Evaluate[{x[1/4, 1/3, 1/2, 1/4, 10][t], y[1/4, 1/3, 1/2, 1/4, 10][t]} /. s], 
{t, 0, 300}, PlotRange -> All, PlotPoints -> 500, PlotStyle -> {Blue, Thickness[0.003]}, 
AxesStyle -> Directive[Black, Small],Background -> Lighter[Gray, 0.95]]

Phase portrait

Example: Hopf bifurcation in the Brusselator system

Calculation of the first Lyapunov coefficient

The Brusselator system is given by: $$ \begin{align} &\dot{x}=\alpha-(\beta+1)x + x^2 y\\ &\dot{y}= \beta x - x^2 y \end{align} $$

Assuming $\alpha> 0$ fixed and taking $\beta$ as a bifurcation parameter, we show that at $\beta = 1 + \alpha^2$ the system exhibits a supercritical Hopf bifurcation.

The Brusselator system code:

f1[x_, y_] := α - (β + 1) x + x^2 y;
f2[x_, y_] := β x - x^2 y;
F[{x_, y_}, {α_, β_}] := Evaluate@{f1[x, y], f2[x, y]};
X = {x, y};
μ = {α, β};
U = {u, v};
R = {r, s};

The Jacobian matrix and its transpose:

J[{x_, y_}, {α_, β_}] = D[F[X, μ], {X}];
Jt[{x_, y_}, {α_, β_}] = Transpose[J[X, μ]];
MatrixForm[J[X, μ]]
MatrixForm[Jt[X, μ]]

Stability analysis (Routh-Hurwitz criterion):

X0[{α_, β_}] = SolveValues[F[X, μ] == 0, X][[1]]
polJX0 = Collect[CharacteristicPolynomial[J[X0[μ], μ], λ], λ,Simplify];
a0 = CoefficientList[polJX0, λ][[3]];
a1 = CoefficientList[polJX0, λ][[2]];
a2 = CoefficientList[polJX0, λ][[1]];
Reduce[a1 > 0 && a2 > 0 && α > 0 && β > 0, β]
(*α > 0 && 0 < β < 1 + α^2*)

Note that $a_{1}=0$ if and only if $β=β_{0}$, where $β_{0}=1+α^2$. Then, the Brusselator is locally asymptotically stable at $X_{0}(\mu)$ for $β<β_{0}$ and locally asymptotically unstable for $β>β_{0}$ (appears a stable limit cycle surrounded the unstable equilibrium point). We verify the previous conclusion (transversality condition) with the sign of the following derivative:

D[-a1, β]

The analysis at the critical bifurcation value $β_{0}$:

Solve the following system of equations: $$ \left\{\begin{align} F\left((x,y),(\alpha,\beta)\right) &=0, \\ \operatorname{tr}(J((x,y),(\alpha,\beta))) &=0, \end{align}\right. $$ for $(x,y,\beta)$ and we must check that det $\operatorname{det}J((x,y),(\alpha,\beta))>0$ when $\beta = \beta_{0}$ for the solution found, where $\beta_{0}$ is the Hopf critical bifurcation value.

The code for the above system of equations:

X0μ0 = Delete[Part[SolveValues[F[X, μ] == 0 && Tr[J[X, μ]] == 0, {x, y, β}], 1], {3}]
μ0 = Prepend[Delete[Part[SolveValues[F[X, μ] == 0 && Tr[J[X, μ]] == 0, {x, y, β}], 1], {{1}, {2}}], α]
Det[J[X0μ0, μ0]]

Here, the Hopf critical bifurcation value is $\beta_{0}=1+\alpha^2$ and $\operatorname{det}J((x,y),(\alpha,\beta_{0}))=\alpha^2>0$. Thus, the Brusselator at $\beta_{0}=1+\alpha^2$ has the equilibrium $$ \begin{align} X_{0}(\mu_{0})=\left(\alpha, \displaystyle\frac{1+\alpha^2}{\alpha} \right) \end{align} $$ and the linear approximation at $X_{0}(\mu_{0})$ has purely imaginary eigenvalues $\lambda_{1,2}=\pm \omega i$, $\omega=\alpha$.

The code for the linear approximation and its transpose at $X_{0}(\mu_{0})$:

α= ω;
JX0μ0 = Simplify@J[X0μ0, μ0];
JtX0μ0 = Simplify@Transpose@JX0μ0;

The next step is to translate the equilibrium $X_{0}(\mu_{0})$ to the origin of coordinates:

bb = {0, 0};
F0[{x_, y_}, {α_, β_}] = Collect[Expand@F[X + X0μ0, μ0], {x, x^2, y, y^2, x y, x^2 y},Factor]

Now, to obtain the normal form of the Hopf bifurcation, we need the Taylor expansion of the third order for $F_{0}((x,y),(\alpha,\beta))$:

(*Rank 3 tensor*)
D2[{x_, y_}, {α_, β_}] = Simplify@D[F0[X, μ], {X, 2}]
D2X0μ0 = Simplify@D2[bb, μ0]
(*Rank 4 tensor*)
D3[{x_, y_}, {α_, β_}] = Simplify@D[F0[X, μ], {X, 3}]
D3X0μ0 = Simplify@D3[bb, μ0]

Multilinear forms:

(*Bilinear form*)
BB[{x_, y_}, {u_, v_}] = Collect[Expand[D2X0μ0.X.U], {u x, v y, v x, v y}, FullSimplify];
MatrixForm[BB[{x, y}, {u, v}]]
(*Trilinear form*)
CC[{x_, y_}, {u_, v_}, {r_, s_}] = Collect[Expand[D3X0μ0.X.U.R], {r u x, r u y, r v x, r v y, s u x, s v x, s u y, s v y}, FullSimplify]
MatrixForm[CC[{x, y}, {u, v}, {r, s}]]

We verify that the first three terms of the Taylor series expansion of $F_{0}((x,y),(\alpha,\beta))$ are correct:

A=JX0μ0; (*linear approximation*)
MatrixForm@FullSimplify[F0[X, μ] - (A.X + 1/2! BB[X, X] + 1/3! CC[X, X, X]) /. {x -> t x, y -> t y}]

Now, we compute the critical eigenvectors of $J((0,0),\mu_{0})$ and its transpose:

(*Eigenvectors of A=J[X0,μ0]*)
vp = ComplexExpand@Eigenvectors[A]
q = vp[[2]];
qc = vp[[1]];
MatrixForm@Simplify[A.q - I ω q]
(*Eigenvectors of Transpose[A]*)
vpt = ComplexExpand[Eigenvectors[At]]
(*Normalization constant*)
cn = ComplexExpand[Conjugate[vp[[1]] . vpt[[1]]]]
p = Expand@Simplify[vpt[[2]]/cn];
pc = ComplexExpand[Conjugate[p]];
Simplify[At.p-I ω p]

We verify the normalization condition $\langle p,q\rangle=1$


Finally, we compute the first Lyapunov coefficient: $$ \begin{align} l_1(0,\mu_{0})= &\frac{1}{2\omega_0} {\rm Re}\left[\langle p,C(q,q,\bar{q}) \rangle - 2 \langle p, B(q,A_0^{-1}B(q,\bar{q}))\rangle +\\\hspace{0.5cm} \langle p, B(\bar{q},(2i\omega_0 I_n-A_0)^{-1}B(q,q))\rangle \right] \end{align} $$

Before to calculate $l_1(0,\mu_{0})$, we clean $\alpha$


The code for the first Lyapunov coefficient:

Factor@ComplexExpand[Re[1/(2 ω) (p.CC[q, q, qc] - 2 (p.BB[q, Inverse[A].BB[q, qc]]) + p.BB[qc, Inverse[2 I ω*IdentityMatrix[2] - A].BB[q, q]])] /. ω -> ω0]
(*-((2 + α^2)/(2 α (1 + α^2)))*)

$$ \begin{align} l_1(0,\mu_{0})=-\frac{\alpha^2+2}{2 \alpha \left(\alpha^2+1\right)} \end{align} $$ The first Lyapunov coefficient is clearly negative for all positive $\alpha$. Thus, the Hopf bifurcation is nondegenerate and always supercritical.

The above expression is the result that Kuznetsov arrives at on page 105 of his book (see Elements of Applied Bifurcation Theory).

Limit cycle:

Limit cycle

For more details, see: Andronov-Hopf bifurcation.

  • 1
    $\begingroup$ Any recommendations on literature / material about this? $\endgroup$
    – Ruud3.1415
    Nov 21, 2017 at 15:57
  • 2
    $\begingroup$ Elements of Applied Bifurcation Theory - Yuri A. Kuznetsov. Sorry, I hope my recommendation is useful after such a long time. XD $\endgroup$ Feb 6, 2021 at 11:48

I'm afraid that you have calculated the stability incorrectly. Here is the Jacobian of your system:

a = 2; b = 1; c = 2.5; d = 1.2; k = 2.8;
jac[x_, y_] := {{D[a x*(1 - x/k) - b x*y, x], D[a x*(1 - x/k) - b x*y, y]}, 
               {D[-c y + d x*y, x], D[-c y + d x*y, y]}};

At the equilibrium, this is:

jacEq = jac[x, y] //. {x -> 2.08333, y -> 0.511905}

The eigenvalues of this are:

{-0.744047, -0.744047}

so the system is stable about this equilibrium.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.