**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; F[{x_,y_},{a_,b_,c_,d_,k_}]:=Evaluate@{f1[x,y],f2[x,y]}; X={x, y}; μ={a,b,c,d,k}; $$ \begin{align} &\dot{x}=a x\left(1-\frac{x}{k} \right)- bxy\\ &\dot{y}= dxy - cy \end{align} $$ The Jacobian matrix: J[{x_,y_},{a_,b_,c_,d_,k_}]:=Evaluate@D[F[X,μ],{X}] The non-trivial equilibrium point: X0=Normal[Simplify[SolveValues[F[X,μ]==0&&Variables[F[X,μ]]>0,X]]][[1]]; MatrixForm@X0 $$ \begin{align} P_{0}(x,y)=\left(\frac{c}{d} ,\frac{a}{b}\left(1-\frac{c}{dk} \right)\right) \end{align} $$ The linear approximation at $P_{0}$ (coexistence equilibrium point): J0=Simplify@J[X0,μ]; MatrixForm@J0 $$ \begin{align} J(P_{0})=\left( \begin{array}{cc} \hspace{-0.25cm}-\displaystyle\frac{a c}{d k} & -\displaystyle\frac{b c}{d}\hspace{0.3cm} \\\\ \hspace{0.2cm}\displaystyle\frac{a (d k-c)}{b k} &\hspace{0.2cm} 0 \\ \end{array} \right) \end{align} $$ 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][1]][1] 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][2]][2] **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}; 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] (*1*) 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; MatrixForm@JX0μ0 MatrixForm@JtX0μ0 Eigenvalues[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] MatrixForm@F0[bb,μ] 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: MatrixForm@FullSimplify[F0[X, μ] - (JX0μ0.X + 1/2! BB[X, X] + 1/3! CC[X, X, X]) /. {x -> t x, y -> t y}] (*{0,0}*) Now, we compute the critical eigenvectors of $J((0,0),\mu_{0})$ and its transpose: (*Eigenvectors for J[X0,μ0]*) vp = ComplexExpand@Eigenvectors[JX0μ0] q = vp[[2]]; qc = vp[[1]]; MatrixForm@q MatrixForm@qc MatrixForm@Simplify[JX0μ0.q - I ω q] (*Eigenvectors for Transpose[J[X0,μ0]]*) vpt = ComplexExpand[Eigenvectors[JtX0μ0]] (*Normalization constant*) cn = ComplexExpand[Conjugate[vp[[1]] . vpt[[1]]]] p = Expand@Simplify[vpt[[2]]/cn]; pc = ComplexExpand[Conjugate[p]]; MatrixForm@p MatrixForm@pc Simplify[JtX0μ0.p-I ω p] We verify the normalization condition $\langle p,q\rangle=1$ Simplify@(p.q) (*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 $\alpha$ Clear[α] ω0=α; The code for the first Lyapunov coefficient: Factor@ComplexExpand[Re[1/(2 ω) (p.D3X0μ0.q.q.qc- 2(p.D2X0μ0.q.Inverse[JX0μ0].D2X0μ0.q. qc) + p.D2X0μ0.qc. Inverse[2 I ω*IdentityMatrix[2]-JX0μ0].D2X0μ0.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][3]). Limit cycle: [![Limit cycle][4]][4] For more details see: [Andronov-Hopf bifurcation][5]. ###### In case anyone wishes to collaborate on these topics, please, communicate. [1]: https://i.sstatic.net/pRAqT.png [2]: https://i.sstatic.net/rQ6Nk.png [3]: https://doi.org/10.1007/978-1-4757-3978-7 [4]: https://i.sstatic.net/SCEwC.png [5]: http://www.scholarpedia.org/article/Andronov-Hopf_bifurcation