{
  "index": "1973-A-5",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "A-5. A particle moves in 3-space according to the equations:\n\\[\n\\frac{d x}{d t}=y z, \\quad \\frac{d y}{d t}=z x, \\quad \\frac{d z}{d t}=x y .\n\\]\n[Here \\( x(t), y(t), z(t) \\) are real-valued functions of the real variable \\( t \\).] Show that:\n(a) If two of \\( x(0), y(0), z(0) \\) equal zero, then the particle never moves.\n(b) If \\( x(0)=y(0)=1, z(0)=0 \\), then the solution is:\n\\[\nx=\\sec t, y=\\sec t, z=\\tan t\n\\]\nwhereas if \\( x(0)=y(0)=1, z(0)=-1 \\), then\n\\[\nx=1 /(t+1), y=1 /(t+1), z=-1 /(t+1)\n\\]\n(c) If at least two of the values \\( x(0), y(0), z(0) \\) are different from zero, then either the particle moves to infinity at some finite time in the future, or it came from infinity at some finite time in the past. [A point \\( (x, y, z) \\) in 3-space \"moves to infinity\" if its distance from the origin approaches infinity.]",
  "solution": "A-5. (a) If two of \\( x(0), y(0), z(0) \\) vanish, then \\( x^{\\prime}(0)=y^{\\prime}(0)=z^{\\prime}(0)=0 \\), and the uniqueness theorem applies (the equations are clearly \"Lipschitz').\n(b) Clear (this was intended as a hint for part (c)).\n(c) Now write the equations in the symmetric form:\n\\[\nx x^{\\prime}=y y^{\\prime}=z z^{\\prime}=x y z\n\\]\n\nThus \\( x^{2}-c_{1}=y^{2}-c_{2}=z^{2}-c_{3} \\) with constant \\( c_{i} \\). Say without loss of generality that \\( c_{1} \\geqq c_{2} \\geqq c_{3} \\), and then set \\( c_{3}=0 \\). Thus \\( z^{2} \\leqq y^{2} \\leqq x^{2} \\), and:\n\\[\n\\begin{array}{c}\nz^{2}=x^{2}-c_{1}=y^{2}-c_{2}, c_{i} \\geqq 0 \\\\\n\\frac{d z}{d t}= \\pm \\sqrt{\\left(z^{2}+c_{1}\\right)\\left(z^{2}+c_{2}\\right)}\n\\end{array}\n\\]\n\nNow let time \\( t \\) move in the direction which makes \\( |z| \\) increase (this depends on the sign of \\( z \\) and on the \\( \\pm \\) sign in the square root).\n\nFor simplicity assume that \\( z(0) \\geqq 0 \\), and that the sign on the square root is + ; then let time move positively. Since\n\\[\nt=\\int d z / \\sqrt{\\left(z^{2}+c_{1}\\right)\\left(z^{2}+c_{2}\\right)}\n\\]\nand the \\( z \\)-integral converges, a finite amount of time suffices to push \\( z \\) out to infinity.",
  "vars": [
    "t",
    "x",
    "y",
    "z"
  ],
  "params": [
    "c_1",
    "c_2",
    "c_3",
    "c_i"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "t": "timevar",
        "x": "abscissa",
        "y": "ordinate",
        "z": "altitude",
        "c_1": "constone",
        "c_2": "consttwo",
        "c_3": "constthr",
        "c_i": "constgen"
      },
      "question": "A-5. A particle moves in 3-space according to the equations:\n\\[\n\\frac{d abscissa}{d timevar}=ordinate altitude, \\quad \\frac{d ordinate}{d timevar}=altitude abscissa, \\quad \\frac{d altitude}{d timevar}=abscissa ordinate .\n\\]\n[Here \\( abscissa(timevar), ordinate(timevar), altitude(timevar) \\) are real-valued functions of the real variable \\( timevar \\).] Show that:\n(a) If two of \\( abscissa(0), ordinate(0), altitude(0) \\) equal zero, then the particle never moves.\n(b) If \\( abscissa(0)=ordinate(0)=1, altitude(0)=0 \\), then the solution is:\n\\[\nabscissa=\\sec timevar, ordinate=\\sec timevar, altitude=\\tan timevar\n\\]\nwhereas if \\( abscissa(0)=ordinate(0)=1, altitude(0)=-1 \\), then\n\\[\nabscissa=1 /(timevar+1), ordinate=1 /(timevar+1), altitude=-1 /(timevar+1)\n\\]\n(c) If at least two of the values \\( abscissa(0), ordinate(0), altitude(0) \\) are different from zero, then either the particle moves to infinity at some finite time in the future, or it came from infinity at some finite time in the past. [A point \\( (abscissa, ordinate, altitude) \\) in 3-space \"moves to infinity\" if its distance from the origin approaches infinity.]",
      "solution": "A-5. (a) If two of \\( abscissa(0), ordinate(0), altitude(0) \\) vanish, then \\( abscissa^{\\prime}(0)=ordinate^{\\prime}(0)=altitude^{\\prime}(0)=0 \\), and the uniqueness theorem applies (the equations are clearly \"Lipschitz\").\n(b) Clear (this was intended as a hint for part (c)).\n(c) Now write the equations in the symmetric form:\n\\[\nabscissa\\,abscissa^{\\prime}=ordinate\\,ordinate^{\\prime}=altitude\\,altitude^{\\prime}=abscissa\\;ordinate\\;altitude\n\\]\nThus \\( abscissa^{2}-constone=ordinate^{2}-consttwo=altitude^{2}-constthr \\) with constant \\( constgen \\). Say without loss of generality that \\( constone \\geqq consttwo \\geqq constthr \\), and then set \\( constthr=0 \\). Thus \\( altitude^{2} \\leqq ordinate^{2} \\leqq abscissa^{2} \\), and:\n\\[\n\\begin{array}{c}\naltitude^{2}=abscissa^{2}-constone=ordinate^{2}-consttwo,\\; constgen \\geqq 0 \\\\\n\\frac{d altitude}{d timevar}= \\pm \\sqrt{\\left(altitude^{2}+constone\\right)\\left(altitude^{2}+consttwo\\right)}\n\\end{array}\n\\]\nNow let time \\( timevar \\) move in the direction which makes \\( |altitude| \\) increase (this depends on the sign of \\( altitude \\) and on the \\( \\pm \\) sign in the square root).\n\nFor simplicity assume that \\( altitude(0) \\geqq 0 \\), and that the sign on the square root is + ; then let time move positively. Since\n\\[\ntimevar=\\int d altitude / \\sqrt{\\left(altitude^{2}+constone\\right)\\left(altitude^{2}+consttwo\\right)}\n\\]\nand the \\( altitude \\)-integral converges, a finite amount of time suffices to push \\( altitude \\) out to infinity."
    },
    "descriptive_long_confusing": {
      "map": {
        "t": "momentum",
        "x": "latitude",
        "y": "longitude",
        "z": "altitude",
        "c_1": "compassone",
        "c_2": "compasstwo",
        "c_3": "compassthree",
        "c_i": "compassindex"
      },
      "question": "A-5. A particle moves in 3-space according to the equations:\n\\[\n\\frac{d latitude}{d momentum}=longitude\\, altitude, \\quad \\frac{d longitude}{d momentum}=altitude\\, latitude, \\quad \\frac{d altitude}{d momentum}=latitude\\, longitude .\n\\]\n[Here \\( latitude(momentum), longitude(momentum), altitude(momentum) \\) are real-valued functions of the real variable \\( momentum \\).] Show that:\n(a) If two of \\( latitude(0), longitude(0), altitude(0) \\) equal zero, then the particle never moves.\n(b) If \\( latitude(0)=longitude(0)=1, altitude(0)=0 \\), then the solution is:\n\\[\nlatitude=\\sec momentum, longitude=\\sec momentum, altitude=\\tan momentum\n\\]\nwhereas if \\( latitude(0)=longitude(0)=1, altitude(0)=-1 \\), then\n\\[\nlatitude=1 /(momentum+1), longitude=1 /(momentum+1), altitude=-1 /(momentum+1)\n\\]\n(c) If at least two of the values \\( latitude(0), longitude(0), altitude(0) \\) are different from zero, then either the particle moves to infinity at some finite time in the future, or it came from infinity at some finite time in the past. [A point \\( (latitude, longitude, altitude) \\) in 3-space \"moves to infinity\" if its distance from the origin approaches infinity.]",
      "solution": "A-5. (a) If two of \\( latitude(0), longitude(0), altitude(0) \\) vanish, then \\( latitude^{\\prime}(0)=longitude^{\\prime}(0)=altitude^{\\prime}(0)=0 \\), and the uniqueness theorem applies (the equations are clearly \"Lipschitz').\n(b) Clear (this was intended as a hint for part (c)).\n(c) Now write the equations in the symmetric form:\n\\[\nlatitude\\, latitude^{\\prime}=longitude\\, longitude^{\\prime}=altitude\\, altitude^{\\prime}=latitude\\, longitude\\, altitude\n\\]\n\nThus \\( latitude^{2}-compassone=longitude^{2}-compasstwo=altitude^{2}-compassthree \\) with constant \\( compassindex \\). Say without loss of generality that \\( compassone \\geqq compasstwo \\geqq compassthree \\), and then set \\( compassthree=0 \\). Thus \\( altitude^{2} \\leqq longitude^{2} \\leqq latitude^{2} \\), and:\n\\[\n\\begin{array}{c}\naltitude^{2}=latitude^{2}-compassone=longitude^{2}-compasstwo, compassindex \\geqq 0 \\\\\n\\frac{d altitude}{d momentum}= \\pm \\sqrt{\\left(altitude^{2}+compassone\\right)\\left(altitude^{2}+compasstwo\\right)}\n\\end{array}\n\\]\n\nNow let time \\( momentum \\) move in the direction which makes \\( |altitude| \\) increase (this depends on the sign of \\( altitude \\) and on the \\( \\pm \\) sign in the square root).\n\nFor simplicity assume that \\( altitude(0) \\geqq 0 \\), and that the sign on the square root is + ; then let time move positively. Since\n\\[\nmomentum=\\int d altitude / \\sqrt{\\left(altitude^{2}+compassone\\right)\\left(altitude^{2}+compasstwo\\right)}\n\\]\nand the \\( altitude \\)-integral converges, a finite amount of time suffices to push \\( altitude \\) out to infinity."
    },
    "descriptive_long_misleading": {
      "map": {
        "t": "timelessness",
        "x": "emptiness",
        "y": "nothingy",
        "z": "voidness",
        "c_1": "fluidone",
        "c_2": "fluidtwo",
        "c_3": "fluidthree",
        "c_i": "fluidany"
      },
      "question": "A-5. A particle moves in 3-space according to the equations:\n\\[\n\\frac{d emptiness}{d timelessness}=nothingy\\,voidness, \\quad \\frac{d nothingy}{d timelessness}=voidness\\,emptiness, \\quad \\frac{d voidness}{d timelessness}=emptiness\\,nothingy .\n\\]\n[Here \\( emptiness(timelessness), nothingy(timelessness), voidness(timelessness) \\) are real-valued functions of the real variable \\( timelessness \\).] Show that:\n(a) If two of \\( emptiness(0), nothingy(0), voidness(0) \\) equal zero, then the particle never moves.\n(b) If \\( emptiness(0)=nothingy(0)=1, voidness(0)=0 \\), then the solution is:\n\\[\nemptiness=\\sec timelessness, \\; nothingy=\\sec timelessness, \\; voidness=\\tan timelessness\n\\]\nwhereas if \\( emptiness(0)=nothingy(0)=1, voidness(0)=-1 \\), then\n\\[\nemptiness=1 /(timelessness+1), \\; nothingy=1 /(timelessness+1), \\; voidness=-1 /(timelessness+1)\n\\]\n(c) If at least two of the values \\( emptiness(0), nothingy(0), voidness(0) \\) are different from zero, then either the particle moves to infinity at some finite time in the future, or it came from infinity at some finite time in the past. [A point \\( (emptiness, nothingy, voidness) \\) in 3-space \"moves to infinity\" if its distance from the origin approaches infinity.]",
      "solution": "A-5. (a) If two of \\( emptiness(0), nothingy(0), voidness(0) \\) vanish, then \\( emptiness^{\\prime}(0)=nothingy^{\\prime}(0)=voidness^{\\prime}(0)=0 \\), and the uniqueness theorem applies (the equations are clearly \"Lipschitz\").\n\n(b) Clear (this was intended as a hint for part (c)).\n\n(c) Now write the equations in the symmetric form:\n\\[\nemptiness\\,emptiness^{\\prime}=nothingy\\,nothingy^{\\prime}=voidness\\,voidness^{\\prime}=emptiness\\,nothingy\\,voidness\n\\]\n\nThus \\( emptiness^{2}-fluidone=nothingy^{2}-fluidtwo=voidness^{2}-fluidthree \\) with constant \\( fluidany \\). Say without loss of generality that \\( fluidone \\geqq fluidtwo \\geqq fluidthree \\), and then set \\( fluidthree=0 \\). Thus \\( voidness^{2} \\leqq nothingy^{2} \\leqq emptiness^{2} \\), and:\n\\[\n\\begin{array}{c}\nvoidness^{2}=emptiness^{2}-fluidone=nothingy^{2}-fluidtwo, \\; fluidany \\geqq 0 \\\\\n\\frac{d voidness}{d timelessness}= \\pm \\sqrt{\\left(voidness^{2}+fluidone\\right)\\left(voidness^{2}+fluidtwo\\right)}\n\\end{array}\n\\]\n\nNow let time \\( timelessness \\) move in the direction which makes \\( |voidness| \\) increase (this depends on the sign of \\( voidness \\) and on the \\( \\pm \\) sign in the square root).\n\nFor simplicity assume that \\( voidness(0) \\geqq 0 \\), and that the sign on the square root is + ; then let time move positively. Since\n\\[\ntimelessness=\\int d voidness / \\sqrt{\\left(voidness^{2}+fluidone\\right)\\left(voidness^{2}+fluidtwo\\right)}\n\\]\nand the \\( voidness \\)-integral converges, a finite amount of time suffices to push \\( voidness \\) out to infinity."
    },
    "garbled_string": {
      "map": {
        "t": "qzxwvtnp",
        "x": "hjgrksla",
        "y": "mnpqrzso",
        "z": "cvbnduke",
        "c_1": "plmqjast",
        "c_2": "ghtresaz",
        "c_3": "kilderma",
        "c_i": "swemprat"
      },
      "question": "A particle moves in 3-space according to the equations:\n\\[\n\\frac{d hjgrksla}{d qzxwvtnp}= mnpqrzso\\, cvbnduke, \\quad \\frac{d mnpqrzso}{d qzxwvtnp}= cvbnduke\\, hjgrksla, \\quad \\frac{d cvbnduke}{d qzxwvtnp}= hjgrksla\\, mnpqrzso .\n\\]\n[Here \\( hjgrksla(qzxwvtnp), mnpqrzso(qzxwvtnp), cvbnduke(qzxwvtnp) \\) are real-valued functions of the real variable \\( qzxwvtnp \\).] Show that:\n(a) If two of \\( hjgrksla(0), mnpqrzso(0), cvbnduke(0) \\) equal zero, then the particle never moves.\n(b) If \\( hjgrksla(0)=mnpqrzso(0)=1, cvbnduke(0)=0 \\), then the solution is:\n\\[\nhjgrksla=\\sec qzxwvtnp, \\, mnpqrzso=\\sec qzxwvtnp, \\, cvbnduke=\\tan qzxwvtnp\n\\]\nwhereas if \\( hjgrksla(0)=mnpqrzso(0)=1, cvbnduke(0)=-1 \\), then\n\\[\nhjgrksla=1 /(qzxwvtnp+1), \\, mnpqrzso=1 /(qzxwvtnp+1), \\, cvbnduke=-1 /(qzxwvtnp+1)\n\\]\n(c) If at least two of the values \\( hjgrksla(0), mnpqrzso(0), cvbnduke(0) \\) are different from zero, then either the particle moves to infinity at some finite time in the future, or it came from infinity at some finite time in the past. [A point \\( (hjgrksla, mnpqrzso, cvbnduke) \\) in 3-space \"moves to infinity\" if its distance from the origin approaches infinity.]",
      "solution": "A-5. (a) If two of \\( hjgrksla(0), mnpqrzso(0), cvbnduke(0) \\) vanish, then \\( hjgrksla^{\\prime}(0)=mnpqrzso^{\\prime}(0)=cvbnduke^{\\prime}(0)=0 \\), and the uniqueness theorem applies (the equations are clearly \"Lipschitz').\n(b) Clear (this was intended as a hint for part (c)).\n(c) Now write the equations in the symmetric form:\n\\[\nhjgrksla\\, hjgrksla^{\\prime}= mnpqrzso\\, mnpqrzso^{\\prime}= cvbnduke\\, cvbnduke^{\\prime}= hjgrksla\\, mnpqrzso\\, cvbnduke\n\\]\n\nThus \\( hjgrksla^{2}-plmqjast= mnpqrzso^{2}- ghtresaz= cvbnduke^{2}- kilderma \\) with constant \\( swemprat \\). Say without loss of generality that \\( plmqjast \\geqq ghtresaz \\geqq kilderma \\), and then set \\( kilderma=0 \\). Thus \\( cvbnduke^{2} \\leqq mnpqrzso^{2} \\leqq hjgrksla^{2} \\), and:\n\\[\n\\begin{array}{c}\ncvbnduke^{2}= hjgrksla^{2}- plmqjast= mnpqrzso^{2}- ghtresaz, \\quad swemprat \\geqq 0 \\\\\n\\frac{d cvbnduke}{d qzxwvtnp}= \\pm \\sqrt{\\left(cvbnduke^{2}+ plmqjast\\right)\\left(cvbnduke^{2}+ ghtresaz\\right)}\n\\end{array}\n\\]\n\nNow let time \\( qzxwvtnp \\) move in the direction which makes \\( |cvbnduke| \\) increase (this depends on the sign of \\( cvbnduke \\) and on the \\( \\pm \\) sign in the square root).\n\nFor simplicity assume that \\( cvbnduke(0) \\geqq 0 \\), and that the sign on the square root is + ; then let time move positively. Since\n\\[\nqzxwvtnp=\\int d cvbnduke / \\sqrt{\\left(cvbnduke^{2}+ plmqjast\\right)\\left(cvbnduke^{2}+ ghtresaz\\right)}\n\\]\nand the \\( cvbnduke \\)-integral converges, a finite amount of time suffices to push \\( cvbnduke \\) out to infinity."
    },
    "kernel_variant": {
      "question": "Let $X(\\tau),\\,Y(\\tau),\\,Z(\\tau),\\,W(\\tau)$ be real–valued functions that solve the autonomous system  \n\\[\n(S_{4})\\qquad \n\\begin{cases}\n\\dfrac{dX}{d\\tau}=YZW,\\\\[6pt]\n\\dfrac{dY}{d\\tau}=ZWX,\\\\[6pt]\n\\dfrac{dZ}{d\\tau}=WXY,\\\\[6pt]\n\\dfrac{dW}{d\\tau}=XYZ,\n\\end{cases}\\qquad \\tau\\in I\\subset\\mathbf R,\n\\]\nand set $(X,Y,Z,W)(0)=(X_{0},Y_{0},Z_{0},W_{0})$.\n\n(a)\\; (Zero–pattern test)  \nProve that the vector field in $(S_{4})$ vanishes at a point $(X,Y,Z,W)$ if and only if at least two of the four coordinates are $0$.  Deduce that the trajectory through $(X_{0},Y_{0},Z_{0},W_{0})$ is constant in $\\tau$ precisely when at least two coordinates of the initial point are zero.\n\n(b)\\; (First integrals)  \nIntroduce the quadratic quantities\n\\[\n\\Delta_{1}:=X^{2}-Y^{2},\\qquad\n\\Delta_{2}:=X^{2}-Z^{2},\\qquad\n\\Delta_{3}:=X^{2}-W^{2}.\n\\]\n(i)\\; Show that $\\Delta_{1},\\Delta_{2},\\Delta_{3}$ are conserved along every solution of $(S_{4})$.  \n(ii)\\; Prove that on the open set\n\\[\n\\mathcal U:=\\{(X,Y,Z,W)\\in\\mathbf R^{4}\\mid X Y Z W\\neq0\\}\n\\]\nthe three integrals are functionally independent.  \n(iii)\\; Deduce from the regular–value theorem that every connected component of a common level set $\\{\\Delta_{1}=c_{1},\\,\\Delta_{2}=c_{2},\\,\\Delta_{3}=c_{3}\\}\\subset\\mathcal U$ is a one–dimensional submanifold; hence $\\{\\Delta_{1},\\Delta_{2},\\Delta_{3}\\}$ form a complete set of independent first integrals on $\\mathcal U$.\n\n(c)\\; (Full–symmetry line)  \nAssume $X_{0}=Y_{0}=Z_{0}=W_{0}=a\\neq0$.  \n(i)\\; Find an explicit formula for the solution and determine its maximal interval of existence.  \n(ii)\\; Show that the trajectory reaches infinite distance from the origin at the finite blow–up time $\\tau^{\\ast}>0$.\n\n(d)\\; (Exactly two non–zero coordinates: saddle–centre equilibria)  \nSuppose exactly two of $X_{0},Y_{0},Z_{0},W_{0}$ are non–zero; without loss of generality\n\\[\n(X_{0},Y_{0},Z_{0},W_{0})=(a,b,0,0)\\qquad(a b\\neq0).\n\\]\n(i)\\; Show that the corresponding trajectory is the constant point $(a,b,0,0)$.  \n(ii)\\; Compute the Jacobian of the vector field at $(a,b,0,0)$ and show that its eigenvalues are $\\{0,0,a b,-a b\\}$.  Conclude that the equilibrium possesses a two–dimensional centre subspace and one–dimensional stable/unstable subspaces (the signs depend on $\\operatorname{sign}(a b)$); it is therefore a partially hyperbolic saddle–centre.  \n(iii)\\; Prove that every neighbourhood of $(a,b,0,0)$ contains non–stationary orbits that blow up in finite time.  Conclude that the equilibrium is (Lyapunov) unstable.\n\n(e)\\; (Generic data: reduction to one elliptic quadrature)  \nAssume at least three of $X_{0},Y_{0},Z_{0},W_{0}$ are non–zero.\n\n(i)\\; Show that on every connected time interval on which none of the coordinates vanishes one can write  \n\\[\nX(\\tau)=\\sigma_{1}\\sqrt{W(\\tau)^{2}+\\kappa_{1}},\\qquad\nZ(\\tau)=\\sigma_{2}\\sqrt{W(\\tau)^{2}+\\kappa_{2}},\\qquad\nY(\\tau)=\\sigma_{3}\\sqrt{W(\\tau)^{2}+\\kappa_{3}},\n\\]\nwhere the constants $\\kappa_{1},\\kappa_{2},\\kappa_{3}$ are  \n\\[\n\\kappa_{1}:=\\Delta_{3},\\qquad\n\\kappa_{2}:=\\Delta_{3}-\\Delta_{2},\\qquad\n\\kappa_{3}:=\\Delta_{3}-\\Delta_{1},\n\\]\nand $\\sigma_{1},\\sigma_{2},\\sigma_{3}\\in\\{+1,-1\\}$ are fixed signs on that interval.\n\n(ii)\\; Deduce the single scalar differential equation  \n\\[\n\\Bigl(\\frac{dW}{d\\tau}\\Bigr)^{2}\n      =(W^{2}+\\kappa_{1})(W^{2}+\\kappa_{2})(W^{2}+\\kappa_{3})\n      =:P(W^{2}).\n\\]\n\n(iii)\\; Prove that for every non–stationary solution there exists a time shift so that  \n$P(W(0)^{2})>0$.  Show that $P(W(\\tau)^{2})>0$ and $\\dfrac{dW}{d\\tau}\\neq0$ on an open interval that contains $\\tau=0$, deduce that $W$ is strictly monotone there, and conclude that the maximal interval of existence of every non–stationary solution is finite: the orbit either emanates from \\emph{infinity} in finite past time or escapes to \\emph{infinity} in finite future time (or both).\n\n(iv)\\; With $t=W^{2}$ write the quadrature $\\displaystyle\\int\\dfrac{dt}{\\sqrt{P(t)}}$.  Show that it can be carried out in elementary functions if and only if $P$ has a repeated root, i.e.\\ iff at least two of $\\kappa_{1},\\kappa_{2},\\kappa_{3}$ coincide; otherwise one obtains an elliptic integral of the first kind.\n\n(f)\\; (Higher–dimensional outlook — brief)  \nFor $n\\ge3$ consider\n\\[\n\\frac{dx_{i}}{d\\tau}= \\prod_{j\\neq i} x_{j}\\qquad(i=1,\\dots ,n).\n\\]\nState (no proofs required) the analogue of parts (a)–(e) for this $n$–dimensional system and indicate which steps demand genuinely new ideas when $n>4$.",
      "solution": "Throughout $\\dot{\\phantom X}$ denotes $d/d\\tau$.\n\n(a)\\; \\textbf{Zero–pattern test.}  \nEach component of the right–hand side in $(S_{4})$ is a product of three coordinates.  Hence the vector field vanishes at $(X,Y,Z,W)$ exactly when every such triple product contains a zero factor, i.e.\\ when at least two of $X,Y,Z,W$ are $0$.  Conversely, if at least two coordinates are zero, every component of the right–hand side equals $0$ so the point is an equilibrium.  \nUniqueness of solutions of an autonomous Lipschitz system then implies that the trajectory through $(X_{0},Y_{0},Z_{0},W_{0})$ is constant precisely when at least two coordinates of the initial point are $0$.\n\n(b)\\; \\textbf{First integrals.}\n\n(i)\\; \\emph{Conservation.}  A direct computation gives  \n\\[\n\\dot\\Delta_{1}=2(X\\dot X-Y\\dot Y)=2(X\\,YZW-Y\\,ZWX)=0,\n\\]\nand cyclic permutations give $\\dot\\Delta_{2}=0=\\dot\\Delta_{3}$.\n\n(ii)\\; \\emph{Functional independence on $\\mathcal U$.}  \nOn $\\mathcal U$ the gradients\n\\[\n\\nabla\\Delta_{1}=(2X,-2Y,0,0),\\quad\n\\nabla\\Delta_{2}=(2X,0,-2Z,0),\\quad\n\\nabla\\Delta_{3}=(2X,0,0,-2W)\n\\]\nsatisfy, for any $(\\alpha_{1},\\alpha_{2},\\alpha_{3})\\in\\mathbf R^{3}$,\n\\[\n\\alpha_{1}\\nabla\\Delta_{1}+\\alpha_{2}\\nabla\\Delta_{2}+\\alpha_{3}\\nabla\\Delta_{3}=0\n\\Longrightarrow\n\\begin{cases}\n2X(\\alpha_{1}+\\alpha_{2}+\\alpha_{3})=0,\\\\\n-2Y\\alpha_{1}=0,\\\\\n-2Z\\alpha_{2}=0,\\\\\n-2W\\alpha_{3}=0.\n\\end{cases}\n\\]\nBecause $X,Y,Z,W\\neq0$ on $\\mathcal U$, we obtain $\\alpha_{1}=\\alpha_{2}=\\alpha_{3}=0$; hence the gradients are linearly independent and the three $\\Delta_{j}$ are functionally independent.\n\n(iii)\\; \\emph{Regular–value theorem.}  \nSince $\\operatorname{rank}(d\\Delta)=3$ on $\\mathcal U$, each common level  \n\\[\nM_{c}:=\\{\\Delta_{1}=c_{1},\\Delta_{2}=c_{2},\\Delta_{3}=c_{3}\\}\\cap\\mathcal U\n\\]\nis a $1$–dimensional embedded submanifold.  Because the $\\Delta_{j}$ are conserved, the vector field of $(S_{4})$ is tangent to $M_{c}$; the set $\\{\\Delta_{1},\\Delta_{2},\\Delta_{3}\\}$ therefore forms a complete set of independent first integrals on $\\mathcal U$.\n\n(c)\\; \\textbf{Full–symmetry line.}  \nPutting $X=Y=Z=W=:U$ gives $U'=U^{3}$, which separates to\n\\[\n\\int\\frac{dU}{U^{3}}=-\\tfrac12\\tau+C\n\\Longrightarrow\nU(\\tau)=\\frac{a}{\\sqrt{1-2a^{2}\\tau}},\\qquad\n\\tau<\\tau^{\\ast}:=\\frac1{2a^{2}}.\n\\]\nThus\n\\[\n(X,Y,Z,W)(\\tau)=\\frac{a}{\\sqrt{1-2a^{2}\\tau}}\\,(1,1,1,1),\\qquad\n\\tau\\in\\bigl(-\\infty,\\tau^{\\ast}\\bigr),\n\\]\nand $\\lVert(X,Y,Z,W)(\\tau)\\rVert\\to\\infty$ as $\\tau\\uparrow\\tau^{\\ast}$.\n\n(d)\\; \\textbf{Two–zero equilibria.}\n\n(i)\\;  If $(X_{0},Y_{0},Z_{0},W_{0})=(a,b,0,0)$ each right–hand side in $(S_{4})$ vanishes at the initial point; the solution is the constant $(a,b,0,0)$.\n\n(ii)\\;  Denote by $F$ the vector field in $(S_{4})$.  At $(a,b,0,0)$\n\\[\nDF(a,b,0,0)=\n\\begin{pmatrix}\n0&0&0&0\\\\\n0&0&0&0\\\\\n0&0&0& a b\\\\\n0&0& a b&0\n\\end{pmatrix},\n\\qquad\\operatorname{spec}DF=\\{0,0,a b,-a b\\}.\n\\]\nHence the equilibrium possesses a $2$–dimensional centre subspace and one–dimensional stable/unstable subspaces corresponding to $\\mp\\operatorname{sign}(a b)$; it is a partially hyperbolic saddle–centre.\n\n(iii)\\;  \\emph{Lyapunov instability.}  \nChoose $\\varepsilon>0$ small and perturb the initial data to  \n\\[\n(X,Y,Z,W)(0)=(a,b,\\varepsilon,\\varepsilon).\n\\]\nNow at least three coordinates are non–zero, so part (e) applies.  In particular\n\\[\n\\dot W^{2}=P(W^{2}),\\qquad\nP(t):=(t+\\kappa_{1})(t+\\kappa_{2})(t+\\kappa_{3}),\n\\]\nwith\n\\[\n\\kappa_{1}=a^{2}-\\varepsilon^{2},\\qquad\n\\kappa_{2}=0,\\qquad\n\\kappa_{3}=b^{2}-\\varepsilon^{2}.\n\\]\nThen\n\\[\nP\\bigl(W(0)^{2}\\bigr)=P(\\varepsilon^{2})\n                   =\\varepsilon^{2}a^{2}b^{2}>0,\n\\qquad\n\\dot W(0)=a b\\varepsilon\\neq0.\n\\]\nPart (e)(iii) (proved below) implies that this solution blows up in finite (forward or backward) time.  Since $\\varepsilon$ can be arbitrarily small, every neighbourhood of $(a,b,0,0)$ contains orbits that escape to infinity in finite time; the equilibrium is therefore Lyapunov–unstable.\n\n(e)\\; \\textbf{Generic data.}\n\n(i)\\;  From the conserved quantities,\n\\[\nX^{2}=W^{2}+\\kappa_{1},\\quad\nZ^{2}=W^{2}+\\kappa_{2},\\quad\nY^{2}=W^{2}+\\kappa_{3},\n\\]\nand on any interval where the signs of $X,Y,Z$ are fixed we can write\n\\[\nX=\\sigma_{1}\\sqrt{W^{2}+\\kappa_{1}},\\quad\nZ=\\sigma_{2}\\sqrt{W^{2}+\\kappa_{2}},\\quad\nY=\\sigma_{3}\\sqrt{W^{2}+\\kappa_{3}},\n\\]\nas required.\n\n(ii)\\;  Substituting in $\\dot W=X Y Z$ and squaring yields\n\\[\n\\dot W^{2}=(W^{2}+\\kappa_{1})(W^{2}+\\kappa_{2})(W^{2}+\\kappa_{3})\n           =P(W^{2}).\n\\]\n\n(iii)\\;  \\emph{Existence of a time where $P>0$ and finite life–time of solutions.}\n\n\\textbf{Step 1:  $P(W^{2})=\\dot W^{2}$.}  \nBecause $W^{2}+\\kappa_{1}=X^{2}$, $W^{2}+\\kappa_{2}=Z^{2}$, $W^{2}+\\kappa_{3}=Y^{2}$, we have\n\\[\nP(W^{2})=X^{2}Y^{2}Z^{2}=(XYZ)^{2}=\\dot W^{2}.\n\\]\n\n\\textbf{Step 2:  Zeros of $P$ are isolated.}  \nIf $P(W(\\tau_{0})^{2})=0$, then exactly one of $X,Y,Z$ vanishes at $\\tau_{0}$ (because at least three coordinates are non–zero).  Suppose $X(\\tau_{0})=0$.  Since $Y(\\tau_{0})Z(\\tau_{0})W(\\tau_{0})\\neq0$,\n\\[\n\\dot X(\\tau_{0})=YZW\\bigl|_{\\tau=\\tau_{0}}\\neq0,\n\\]\nso $X$ crosses the plane $X=0$ linearly and is non–zero for $\\tau\\neq\\tau_{0}$ close to $\\tau_{0}$.  Consequently $XYZ$ and hence $P(W^{2})$ become strictly positive (in one time direction).  Therefore the zeros of $P\\circ W^{2}$ are isolated points, and every non–stationary solution enters an open interval on which $P(W^{2})>0$ and $\\dot W\\neq0$.\n\n\\textbf{Step 3:  Time shift.}  \nTranslating the time variable so that $\\tau=0$ lies in such an interval yields the claimed normalisation $P(W(0)^{2})>0$ and $\\dot W(0)\\neq0$.  On that interval $W$ is strictly monotone.\n\n\\textbf{Step 4:  $W$ escapes to infinity in finite time.}  \nWithout loss of generality assume $\\dot W>0$ (the case $\\dot W<0$ is symmetric).  Pick\n\\[\nM:=2\\max_{j}\\sqrt{|\\kappa_{j}|}+1>0 .\n\\]\nFor $|W|\\ge M$ we have $|\\;W^{2}+\\kappa_{j}\\;|\\ge\\tfrac12 W^{2}$, hence\n\\[\n|\\dot W|=\\sqrt{P(W^{2})}\n        \\ge\\sqrt{\\bigl(\\tfrac12 W^{2}\\bigr)^{3}}\n        =\\frac{|W|^{3}}{2\\sqrt2}\\qquad(|W|\\ge M).\n\\]\nBecause $W$ is increasing and $\\dot W(0)\\neq0$, it reaches the value $|W|=M$ in some finite time $\\tau_{1}>0$.  Integrating the above inequality from $\\tau_{1}$ onwards gives\n\\[\n\\frac1{W^{2}(\\tau)}\\le\\frac1{M^{2}}-\\frac{\\tau-\\tau_{1}}{\\sqrt2}\\qquad\n(\\tau\\ge\\tau_{1}),\n\\]\nso $\\tfrac1{W^{2}}$ becomes $0$ at most $\\sqrt2\\,M^{-2}$ units of time after $\\tau_{1}$.  Thus $|W|\\to\\infty$ in finite forward time.  Reversing time gives the corresponding statement for the past.  Hence every non–stationary solution blows up in finite time in at least one time direction.\n\n(iv)\\;  \\emph{Quadrature.}  With $t=W^{2}$ we obtain\n\\[\n\\tau-\\tau_{0}= \\int_{t_{0}}^{t} \\frac{ds}{\\sqrt{P(s)}}.\n\\]\nIf, say, $\\kappa_{1}=\\kappa_{2}$, then $P(s)=(s+\\kappa_{1})^{2}(s+\\kappa_{3})$, partial fractions reduce the integral to elementary functions, and the solution can be written in closed form.  Otherwise $P$ has three simple (possibly complex) roots, so the integral is an elliptic integral of the first kind.\n\n(f)\\; \\textbf{Higher–dimensional outlook ($n\\ge3$).}\n\nFor  \n\\[\n\\dot x_{i}= \\prod_{j\\neq i}x_{j}\\qquad (i=1,\\dots ,n)\n\\]\none has:\n\n\\begin{itemize}\n\\item \\emph{Stationary points:} the vector field vanishes iff at least two coordinates are $0$.\n\n\\item \\emph{Quadratic first integrals:}  \n      $x_{1}^{2}-x_{k}^{2}$, $k=2,\\dots ,n$, are conserved and constitute $n-1$ independent first integrals on $\\{\\prod_{j}x_{j}\\neq0\\}$.\n\n\\item \\emph{Full–symmetry line:}  \n      If all coordinates equal $u$, then $u'=u^{\\,n-1}$, so $|u|$ blows up in finite time.\n\n\\item \\emph{Equilibria with exactly two non–zero coordinates:}  \n      Such points are equilibria.  \n      For $n=4$ the Jacobian has eigenvalues $\\{0,0,\\lambda,-\\lambda\\}$ (saddle–centre);  \n      for $n\\ge5$ the Jacobian is the zero matrix, so centre–manifold or higher–order analysis is required.\n\n\\item \\emph{Generic orbits:}  \n      Eliminating $n-2$ variables yields  \n      \\[\n      \\Bigl(\\frac{dx_{n}}{d\\tau}\\Bigr)^{2}\n          =\\prod_{k=1}^{\\,n-1}\\bigl(x_{n}^{2}+c_{k}\\bigr),\n      \\]\n      a polynomial of degree $n-1$ in $x_{n}^{2}$.  \n      For $n\\le5$ the quadrature is elliptic (genus $1$); for $n\\ge6$ it becomes hyper–elliptic of genus $\\lfloor\\tfrac{n-2}{2}\\rfloor$, so integrating the flow requires the theory of Riemann surfaces of higher genus.\n\\end{itemize}",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.608794",
        "was_fixed": false,
        "difficulty_analysis": "• Dimensional uplift:  The original 3-variable system is replaced by a 4-variable\n  system in which each derivative involves a triple product.  The phase space jumps\n  from ℝ³ to ℝ⁴; the dynamics no longer reduce to a single quadrature of a quartic\n  but to one of a sextic.\n\n• More invariants & algebra:  Three independent quadratic first integrals must be\n  detected, their inter-relations analysed, and completeness proved.\n\n• Richer taxonomy:  Separate treatments are required for 0, 1, 2, 3, 4 non-zero\n  coordinates, each giving different analytic behaviour (stationary, periodic,\n  exponential, elliptic).\n\n• Higher transcendental functions:  Generic solutions cannot be expressed by\n  elementary or even trigonometric/hyperbolic functions; elliptic integrals appear\n  naturally, demanding familiarity with the theory of elliptic functions.\n\n• Finite-time escape:  Establishing blow-up uses comparison theorems and precise\n  asymptotic estimates beyond the elementary argument in the original problem.\n\n• Open-ended extension:  Part (f) indicates how complexity grows further in\n  n > 4, touching on hyper-elliptic integrals and the energy–momentum map, topics\n  absent from the original statement.\n\nThese additions force competitors to juggle invariant theory, asymptotic analysis,\nexplicit quadratures, and special-function theory, substantially raising the level\nof technical sophistication needed to solve the problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let $X(\\tau),\\,Y(\\tau),\\,Z(\\tau),\\,W(\\tau)$ be real–valued functions that solve the autonomous system  \n\\[\n(S_{4})\\qquad \n\\begin{cases}\n\\dfrac{dX}{d\\tau}=YZW,\\\\[6pt]\n\\dfrac{dY}{d\\tau}=ZWX,\\\\[6pt]\n\\dfrac{dZ}{d\\tau}=WXY,\\\\[6pt]\n\\dfrac{dW}{d\\tau}=XYZ,\n\\end{cases}\\qquad \\tau\\in I\\subset\\mathbf R,\n\\]\nand set $(X,Y,Z,W)(0)=(X_{0},Y_{0},Z_{0},W_{0})$.\n\n(a)\\; (Zero–pattern test)  \nProve that the vector field in $(S_{4})$ vanishes at a point $(X,Y,Z,W)$ if and only if at least two of the four coordinates are $0$.  Deduce that the trajectory through $(X_{0},Y_{0},Z_{0},W_{0})$ is constant in $\\tau$ precisely when at least two coordinates of the initial point are zero.\n\n(b)\\; (First integrals)  \nIntroduce the quadratic quantities\n\\[\n\\Delta_{1}:=X^{2}-Y^{2},\\qquad\n\\Delta_{2}:=X^{2}-Z^{2},\\qquad\n\\Delta_{3}:=X^{2}-W^{2}.\n\\]\n(i)\\; Show that $\\Delta_{1},\\Delta_{2},\\Delta_{3}$ are conserved along every solution of $(S_{4})$.  \n(ii)\\; Prove that on the open set\n\\[\n\\mathcal U:=\\{(X,Y,Z,W)\\in\\mathbf R^{4}\\mid X Y Z W\\neq0\\}\n\\]\nthe three integrals are functionally independent.  \n(iii)\\; Deduce from the regular–value theorem that every connected component of a common level set $\\{\\Delta_{1}=c_{1},\\,\\Delta_{2}=c_{2},\\,\\Delta_{3}=c_{3}\\}\\subset\\mathcal U$ is a one–dimensional submanifold; hence $\\{\\Delta_{1},\\Delta_{2},\\Delta_{3}\\}$ form a complete set of independent first integrals on $\\mathcal U$.\n\n(c)\\; (Full–symmetry line)  \nAssume $X_{0}=Y_{0}=Z_{0}=W_{0}=a\\neq0$.  \n(i)\\; Find an explicit formula for the solution and determine its maximal interval of existence.  \n(ii)\\; Show that the trajectory reaches infinite distance from the origin at the finite blow–up time $\\tau^{\\ast}>0$.\n\n(d)\\; (Exactly two non–zero coordinates: saddle–centre equilibria)  \nSuppose exactly two of $X_{0},Y_{0},Z_{0},W_{0}$ are non–zero; without loss of generality\n\\[\n(X_{0},Y_{0},Z_{0},W_{0})=(a,b,0,0)\\qquad(a b\\neq0).\n\\]\n(i)\\; Show that the corresponding trajectory is the constant point $(a,b,0,0)$.  \n(ii)\\; Compute the Jacobian of the vector field at $(a,b,0,0)$ and show that its eigenvalues are $\\{0,0,a b,-a b\\}$.  Conclude that the equilibrium possesses a two–dimensional centre subspace and one–dimensional stable/unstable subspaces (the signs depend on $\\operatorname{sign}(a b)$); it is therefore a partially hyperbolic saddle–centre.  \n(iii)\\; Prove that every neighbourhood of $(a,b,0,0)$ contains non–stationary orbits that blow up in finite time.  Conclude that the equilibrium is (Lyapunov) unstable.\n\n(e)\\; (Generic data: reduction to one elliptic quadrature)  \nAssume at least three of $X_{0},Y_{0},Z_{0},W_{0}$ are non–zero.\n\n(i)\\; Show that on every connected time interval on which none of the coordinates vanishes one can write  \n\\[\nX(\\tau)=\\sigma_{1}\\sqrt{W(\\tau)^{2}+\\kappa_{1}},\\qquad\nZ(\\tau)=\\sigma_{2}\\sqrt{W(\\tau)^{2}+\\kappa_{2}},\\qquad\nY(\\tau)=\\sigma_{3}\\sqrt{W(\\tau)^{2}+\\kappa_{3}},\n\\]\nwhere the constants $\\kappa_{1},\\kappa_{2},\\kappa_{3}$ are  \n\\[\n\\kappa_{1}:=\\Delta_{3},\\qquad\n\\kappa_{2}:=\\Delta_{3}-\\Delta_{2},\\qquad\n\\kappa_{3}:=\\Delta_{3}-\\Delta_{1},\n\\]\nand $\\sigma_{1},\\sigma_{2},\\sigma_{3}\\in\\{+1,-1\\}$ are fixed signs on that interval.\n\n(ii)\\; Deduce the single scalar differential equation  \n\\[\n\\Bigl(\\frac{dW}{d\\tau}\\Bigr)^{2}\n      =(W^{2}+\\kappa_{1})(W^{2}+\\kappa_{2})(W^{2}+\\kappa_{3})\n      =:P(W^{2}).\n\\]\n\n(iii)\\; Show that for every non–stationary solution there exists a shift of the time variable after which  \n$P(W(0)^{2})>0$.  Deduce that $W$ is strictly monotone on every connected component of  \n$\\{\\tau\\mid P(W(\\tau)^{2})>0\\}$.  Conclude that the maximal interval of existence of any non–stationary solution is finite: the orbit either emanates from \\emph{infinity} in a finite past time or escapes to \\emph{infinity} in a finite future time (or both).\n\n(iv)\\; With $t=W^{2}$ write the quadrature $\\displaystyle\\int\\dfrac{dt}{\\sqrt{P(t)}}$.  Show that it can be carried out in elementary functions if and only if $P$ has a repeated root, i.e.\\ iff at least two of $\\kappa_{1},\\kappa_{2},\\kappa_{3}$ coincide; otherwise one obtains an elliptic integral of the first kind.\n\n(f)\\; (Higher–dimensional outlook — brief)  \nFor $n\\ge3$ consider\n\\[\n\\frac{dx_{i}}{d\\tau}= \\prod_{j\\neq i} x_{j}\\qquad(i=1,\\dots ,n).\n\\]\nState (no proofs required) the analogue of parts (a)–(e) for this $n$–dimensional system and indicate which steps demand genuinely new ideas when $n>4$.",
      "solution": "Throughout a dot denotes $d/d\\tau$.\n\n(a)\\; \\textbf{Zero–pattern test.}  \nEvery component of the right–hand side in $(S_{4})$ is a product of three coordinates.  The vector field therefore vanishes at $(X,Y,Z,W)$ exactly when all such products contain a zero factor, that is, when at least two of $X,Y,Z,W$ are $0$.  Conversely, if at least two coordinates are $0$, each component of the right–hand side is $0$, so the point is an equilibrium.  By uniqueness of solutions the trajectory through $(X_{0},Y_{0},Z_{0},W_{0})$ is constant precisely when at least two coordinates of the initial point are $0$.\n\n(b)\\; \\textbf{First integrals.}\n\n(i)\\; \\emph{Conservation.}\n\\[\n\\dot\\Delta_{1}=2\\bigl(X\\dot X-Y\\dot Y\\bigr)\n              =2\\bigl(X\\cdot YZW-Y\\cdot ZWX\\bigr)=0,\n\\]\nand the same computation gives $\\dot\\Delta_{2}=0=\\dot\\Delta_{3}$.\n\n(ii)\\; \\emph{Functional independence on $\\mathcal U$.}  \nOn $\\mathcal U$ none of the four coordinates is $0$, and  \n\\[\n\\nabla\\Delta_{1}=(2X,-2Y,0,0),\\quad\n\\nabla\\Delta_{2}=(2X,0,-2Z,0),\\quad\n\\nabla\\Delta_{3}=(2X,0,0,-2W)\n\\]\nare clearly linearly independent; hence $\\Delta_{1},\\Delta_{2},\\Delta_{3}$ are functionally independent.\n\n(iii)\\; \\emph{Regular–value theorem.}  \nBecause the three gradients above are independent on $\\mathcal U$, every common regular level surface  \n\\[\nM_{c}:=\\{\\Delta_{1}=c_{1},\\Delta_{2}=c_{2},\\Delta_{3}=c_{3}\\}\\cap\\mathcal U\n\\]\nis a one–dimensional embedded submanifold.  The vector field of $(S_{4})$ is tangent to $M_{c}$ since the $\\Delta_{j}$ are first integrals, so the triple $(\\Delta_{1},\\Delta_{2},\\Delta_{3})$ is complete on $\\mathcal U$.\n\n(c)\\; \\textbf{Full–symmetry line.}  \nOn the diagonal $X=Y=Z=W=:U$ system $(S_{4})$ reduces to $U'=U^{3}$.  Separation of variables gives\n\\[\n\\int\\frac{dU}{U^{3}}=-\\frac12\\,\\tau+C\n\\quad\\Longrightarrow\\quad\nU(\\tau)=\\frac{a}{\\sqrt{1-2a^{2}\\tau}},\n\\]\nvalid as long as $1-2a^{2}\\tau>0$.  Hence  \n\\[\n(X,Y,Z,W)(\\tau)=U(\\tau)(1,1,1,1),\\qquad\n\\tau\\in\\bigl(-\\infty,\\tau^{\\ast}\\bigr),\\qquad\n\\tau^{\\ast}:=\\frac1{2a^{2}}>0,\n\\]\nand $\\lVert(X,Y,Z,W)(\\tau)\\rVert\\to\\infty$ as $\\tau\\uparrow\\tau^{\\ast}$.\n\n(d)\\; \\textbf{Two–zero equilibria.}\n\n(i)\\;  If $(X_{0},Y_{0},Z_{0},W_{0})=(a,b,0,0)$ each right–hand side in $(S_{4})$ vanishes at the initial point; the solution is therefore the constant $(a,b,0,0)$.\n\n(ii)\\;  Let $F$ be the vector field of $(S_{4})$.  At $(a,b,0,0)$\n\\[\nDF(a,b,0,0)=\n\\begin{pmatrix}\n0&0&0&0\\\\\n0&0&0&0\\\\\n0&0&0& a b\\\\\n0&0& a b&0\n\\end{pmatrix},\n\\qquad\\operatorname{spec}DF=\\{0,0,a b,-a b\\}.\n\\]\nThus the equilibrium has a two–dimensional centre subspace (eigenvalue $0$) and one–dimensional subspaces associated with $\\pm a b$.  The subspace corresponding to the positive (negative) eigenvalue is unstable (stable) when $a b>0$ and vice versa when $a b<0$.  In all cases the point is a partially hyperbolic saddle–centre.\n\n(iii)\\;  \\emph{Lyapunov instability.}  \nFix $\\varepsilon>0$ and consider the perturbed initial condition  \n\\[\n(X,Y,Z,W)(0)=(a,b,\\varepsilon,\\varepsilon),\n\\]\nwith $0<\\varepsilon\\ll1$.  Because $\\varepsilon\\neq0$ at least three coordinates are non–zero, part (e) applies and yields\n\\[\n\\dot W^{2}=P(W^{2}),\\qquad\nP(t):=(t+\\kappa_{1})(t+\\kappa_{2})(t+\\kappa_{3}),\n\\]\nwith\n\\[\n\\kappa_{1}=a^{2}-\\varepsilon^{2},\\qquad\n\\kappa_{2}=0,\\qquad\n\\kappa_{3}=b^{2}-\\varepsilon^{2}.\n\\]\nConsequently\n\\[\nP\\bigl(W(0)^{2}\\bigr)=P(\\varepsilon^{2})=\\varepsilon^{2}a^{2}b^{2}>0,\n\\qquad\n\\dot W(0)=a b\\varepsilon\\neq0.\n\\]\nBy continuity of $\\dot W$ there exists $\\delta>0$ such that $P(W(\\tau)^{2})>0$ and therefore $\\dot W\\neq0$ for $|\\tau|<\\delta$.  Shifting $\\tau$ if necessary we may assume $P(W(0)^{2})>0$.  Part (e)(iii) then shows that the corresponding solution blows up in finite forward or backward time.  Since $\\varepsilon$ can be chosen arbitrarily small, every neighbourhood of $(a,b,0,0)$ contains non–stationary solutions that become unbounded in finite time; the equilibrium is therefore Lyapunov–unstable.\n\n(e)\\; \\textbf{Generic data.}\n\n(i)\\;  The conserved quantities from (b) give\n\\[\nX^{2}=W^{2}+\\kappa_{1},\\quad\nZ^{2}=W^{2}+\\kappa_{2},\\quad\nY^{2}=W^{2}+\\kappa_{3},\n\\]\nwith $\\kappa_{j}$ as stated.  On an interval where the signs of $X,Y,Z$ do not change,\n\\[\nX=\\sigma_{1}\\sqrt{W^{2}+\\kappa_{1}},\\quad\nZ=\\sigma_{2}\\sqrt{W^{2}+\\kappa_{2}},\\quad\nY=\\sigma_{3}\\sqrt{W^{2}+\\kappa_{3}},\n\\]\nwhere $\\sigma_{j}\\in\\{\\pm1\\}$ are fixed.\n\n(ii)\\;  Substituting into $\\dot W=X Y Z$ and squaring gives the scalar equation $\\dot W^{2}=P(W^{2})$ with \n\\[\nP(t):=(t+\\kappa_{1})(t+\\kappa_{2})(t+\\kappa_{3}).\n\\]\n\n(iii)\\;  \\emph{Finite life–time.}  \nBecause the solution is non–stationary, at least one of $X,Y,Z,W$ is non–zero.  \nIf $P(W(0)^{2})=0$ then $W(0)=0$ and (since three coordinates are non–zero) $\\dot W(0)=X(0)Y(0)Z(0)\\neq0$, so $P(W(\\tau)^{2})>0$ for sufficiently small $|\\tau|$.  Shifting $\\tau$ into that interval we may \\emph{without loss of generality} assume\n\\[\nP\\bigl(W(0)^{2}\\bigr)>0.\n\\]\nContinuity of $\\dot W$ implies that $\\dot W$ keeps a constant non–zero sign on some open interval around $\\tau=0$; hence $W$ is strictly monotone there.\n\nChoose\n\\[\nM:=2\\max_{j}\\sqrt{|\\kappa_{j}|}+1>0 .\n\\]\nFor $|W|\\ge M$ we have\n\\[\n|W^{2}+\\kappa_{j}|\\ge W^{2}-|\\kappa_{j}|\n                \\ge W^{2}-\\tfrac12 W^{2}\n                =\\tfrac12 W^{2},\n\\]\nbecause $|W|^{2}\\ge4|\\kappa_{j}|$ by construction of $M$.  Consequently\n\\[\n|\\dot W|=\\sqrt{P(W^{2})}\n        \\ge\\sqrt{\\bigl(\\tfrac12 W^{2}\\bigr)^{3}}\n        =\\frac{|W|^{3}}{2\\sqrt2}\\qquad(|W|\\ge M).\n\\]\nTherefore\n\\[\n\\Bigl|\\frac{d}{d\\tau}\\frac1{W^{2}}\\Bigr|\n      =\\frac{2|\\dot W|}{W^{3}}\n      \\ge\\frac1{\\sqrt2}\\qquad(|W|\\ge M).\n\\]\nStarting from the first time $\\tau_{1}$ at which $|W|=M$, integration shows that $\\tfrac1{W^{2}}$ reaches $0$ in at most $\\sqrt2\\,M^{-2}$ time; equivalently $|W|\\to\\infty$ within this finite additional time.  Adding the finite time $\\tau_{1}$ needed to reach $|W|=M$ we conclude that every non–stationary solution blows up in finite forward or backward time (or both).\n\n(iv)\\;  \\emph{Quadrature.}  \nSetting $t=W^{2}$ yields\n\\[\n\\tau-\\tau_{0}= \\int_{t_{0}}^{t} \\frac{ds}{\\sqrt{P(s)}}.\n\\]\nIf at least two of the $\\kappa_{j}$ coincide, say $\\kappa_{1}=\\kappa_{2}$, then $P(s)=(s+\\kappa_{1})^{2}(s+\\kappa_{3})$; a partial–fraction decomposition reduces the integral to elementary functions.  Otherwise the roots are simple and the integral is an elliptic integral of the first kind.\n\n(f)\\; \\textbf{Higher–dimensional outlook ($n\\ge3$).}  \n\nFor  \n\\[\n\\dot x_{i}= \\prod_{j\\neq i}x_{j}\\qquad (i=1,\\dots ,n)\n\\]\none has:\n\n\\begin{itemize}\n\\item \\emph{Stationary points:} the vector field vanishes iff at least two coordinates are $0$.\n\n\\item \\emph{Quadratic first integrals:}  \n      $x_{1}^{2}-x_{k}^{2}$, $k=2,\\dots ,n$, are conserved and constitute $n-1$ independent first integrals on $\\{\\prod_{j}x_{j}\\neq0\\}$.\n\n\\item \\emph{Full–symmetry line:}  \n      If all coordinates equal $u$, then $u'=u^{\\,n-1}$, so $|u|$ blows up in finite time.\n\n\\item \\emph{Equilibria with exactly two non–zero coordinates:}  \n      When precisely two coordinates are non–zero the point is an equilibrium.  \n      For $n=4$ the Jacobian has eigenvalues $\\{0,0,\\lambda,-\\lambda\\}$ (saddle–centre);  \n      for $n\\ge5$ the Jacobian is the zero matrix, so higher–order analysis (e.g.\\ centre–manifold theory) is required.\n\n\\item \\emph{Generic orbits:}  \n      Eliminating $n-2$ variables gives  \n      \\[\n      \\Bigl(\\frac{dx_{n}}{d\\tau}\\Bigr)^{2}\n          =\\prod_{k=1}^{\\,n-1}\\bigl(x_{n}^{2}+c_{k}\\bigr),\n      \\]\n      a polynomial of degree $n-1$ in $x_{n}^{2}$.  \n      For $n\\le5$ the quadrature is elliptic (genus $1$); for $n\\ge6$ it becomes hyper–elliptic of genus $\\lfloor\\tfrac{n-2}{2}\\rfloor$, so integrating the flow requires methods from the theory of Riemann surfaces of higher genus.\n\\end{itemize}",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.487062",
        "was_fixed": false,
        "difficulty_analysis": "• Dimensional uplift:  The original 3-variable system is replaced by a 4-variable\n  system in which each derivative involves a triple product.  The phase space jumps\n  from ℝ³ to ℝ⁴; the dynamics no longer reduce to a single quadrature of a quartic\n  but to one of a sextic.\n\n• More invariants & algebra:  Three independent quadratic first integrals must be\n  detected, their inter-relations analysed, and completeness proved.\n\n• Richer taxonomy:  Separate treatments are required for 0, 1, 2, 3, 4 non-zero\n  coordinates, each giving different analytic behaviour (stationary, periodic,\n  exponential, elliptic).\n\n• Higher transcendental functions:  Generic solutions cannot be expressed by\n  elementary or even trigonometric/hyperbolic functions; elliptic integrals appear\n  naturally, demanding familiarity with the theory of elliptic functions.\n\n• Finite-time escape:  Establishing blow-up uses comparison theorems and precise\n  asymptotic estimates beyond the elementary argument in the original problem.\n\n• Open-ended extension:  Part (f) indicates how complexity grows further in\n  n > 4, touching on hyper-elliptic integrals and the energy–momentum map, topics\n  absent from the original statement.\n\nThese additions force competitors to juggle invariant theory, asymptotic analysis,\nexplicit quadratures, and special-function theory, substantially raising the level\nof technical sophistication needed to solve the problem."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}