{
  "index": "1969-A-5",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "A-5. Let \\( u(t) \\) be a continuous function in the system of differential equations\n\\[\n\\frac{d x}{d t}=-2 y+u(t), \\quad \\frac{d y}{d t}=-2 x+u(t)\n\\]\n\nShow that, regardless of the choice of \\( u(t) \\), the solution of the system which satisfies \\( x=x_{0}, y=y_{0} \\) at \\( t=0 \\) will never pass through \\( (0,0) \\) unless \\( x_{0}=y_{0} \\). When \\( x_{0}=y_{0} \\), show that, for any positive value \\( t_{0} \\) of \\( t \\), it is possible to choose \\( u(t) \\) so the solution is at \\( (0,0) \\) when \\( t=t_{0} \\).",
  "solution": "A-5 Subtracting the two equations eliminates \\( u(t) \\) and provides the simpler equation \\( d(x-y) / d t=2(x-y) \\), which has the solution \\( x-y=\\left(x_{0}-y_{0}\\right) e^{2 t} \\). If \\( x_{0} \\neq y_{0} \\), the right hand side is never zero and so \\( x=y=0 \\) can never occur.\n\nFor the second part, \\( x_{0}=y_{0} \\). In this case, \\( x(t)=y(t) \\) and so every solution is a parametrization of the line \\( x=y \\). We can attempt to get a solution of the form \\( x=x_{0}-a t, y=y_{0}-a t \\). This will be a solution if \\( u(t)=2\\left(x_{0}-a t\\right)-a \\). By taking \\( a=x_{0} / t_{0}, x=y=0 \\) at \\( t=t_{0} \\).",
  "vars": [
    "x",
    "y",
    "t"
  ],
  "params": [
    "u",
    "a",
    "x_0",
    "y_0",
    "t_0"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "positionx",
        "y": "positiony",
        "t": "timevar",
        "u": "controlf",
        "a": "slopepar",
        "x_0": "initialx",
        "y_0": "initialy",
        "t_0": "targett"
      },
      "question": "A-5. Let \\( controlf(timevar) \\) be a continuous function in the system of differential equations\n\\[\n\\frac{d positionx}{d timevar}=-2 positiony+controlf(timevar), \\quad \\frac{d positiony}{d timevar}=-2 positionx+controlf(timevar)\n\\]\n\nShow that, regardless of the choice of \\( controlf(timevar) \\), the solution of the system which satisfies \\( positionx=initialx, positiony=initialy \\) at \\( timevar=0 \\) will never pass through \\( (0,0) \\) unless \\( initialx=initialy \\). When \\( initialx=initialy \\), show that, for any positive value \\( targett \\) of \\( timevar \\), it is possible to choose \\( controlf(timevar) \\) so the solution is at \\( (0,0) \\) when \\( timevar=targett \\).",
      "solution": "A-5 Subtracting the two equations eliminates \\( controlf(timevar) \\) and provides the simpler equation \\( d(positionx-positiony) / d timevar=2(positionx-positiony) \\), which has the solution \\( positionx-positiony=\\left(initialx-initialy\\right) e^{2 timevar} \\). If \\( initialx \\neq initialy \\), the right hand side is never zero and so \\( positionx=positiony=0 \\) can never occur.\n\nFor the second part, \\( initialx=initialy \\). In this case, \\( positionx(timevar)=positiony(timevar) \\) and so every solution is a parametrization of the line \\( positionx=positiony \\). We can attempt to get a solution of the form \\( positionx=initialx-slopepar timevar, positiony=initialy-slopepar timevar \\). This will be a solution if \\( controlf(timevar)=2\\left(initialx-slopepar timevar\\right)-slopepar \\). By taking \\( slopepar=initialx / targett, positionx=positiony=0 \\) at \\( timevar=targett \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "orchidpetal",
        "y": "marblegrain",
        "t": "canyontrail",
        "u": "velvetquill",
        "a": "lanternbeam",
        "x_0": "orchidseed",
        "y_0": "marbleseed",
        "t_0": "canyoncrest"
      },
      "question": "A-5. Let \\( velvetquill(canyontrail) \\) be a continuous function in the system of differential equations\n\\[\n\\frac{d orchidpetal}{d canyontrail}=-2 marblegrain+velvetquill(canyontrail), \\quad \\frac{d marblegrain}{d canyontrail}=-2 orchidpetal+velvetquill(canyontrail)\n\\]\n\nShow that, regardless of the choice of \\( velvetquill(canyontrail) \\), the solution of the system which satisfies \\( orchidpetal=orchidseed, marblegrain=marbleseed \\) at \\( canyontrail=0 \\) will never pass through \\( (0,0) \\) unless \\( orchidseed=marbleseed \\). When \\( orchidseed=marbleseed \\), show that, for any positive value \\( canyoncrest \\) of \\( canyontrail \\), it is possible to choose \\( velvetquill(canyontrail) \\) so the solution is at \\( (0,0) \\) when \\( canyontrail=canyoncrest \\).",
      "solution": "A-5 Subtracting the two equations eliminates \\( velvetquill(canyontrail) \\) and provides the simpler equation \\( d(orchidpetal-marblegrain) / d canyontrail=2(orchidpetal-marblegrain) \\), which has the solution \\( orchidpetal-marblegrain=\\left(orchidseed-marbleseed\\right) e^{2 canyontrail} \\). If \\( orchidseed \\neq marbleseed \\), the right hand side is never zero and so \\( orchidpetal=marblegrain=0 \\) can never occur.\n\nFor the second part, \\( orchidseed=marbleseed \\). In this case, \\( orchidpetal(canyontrail)=marblegrain(canyontrail) \\) and so every solution is a parametrization of the line \\( orchidpetal=marblegrain \\). We can attempt to get a solution of the form \\( orchidpetal=orchidseed-lanternbeam \\\\, canyontrail, marblegrain=marbleseed-lanternbeam \\\\, canyontrail \\). This will be a solution if \\( velvetquill(canyontrail)=2\\left(orchidseed-lanternbeam \\\\, canyontrail\\right)-lanternbeam \\). By taking \\( lanternbeam=orchidseed / canyoncrest, orchidpetal=marblegrain=0 \\) at \\( canyontrail=canyoncrest \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "t": "spacevalue",
        "u": "staticinput",
        "a": "curvedness",
        "x_0": "finalvertical",
        "y_0": "terminalhorizontal",
        "t_0": "spacestart"
      },
      "question": "A-5. Let \\( staticinput(spacevalue) \\) be a continuous function in the system of differential equations\n\\[\n\\frac{d verticalaxis}{d spacevalue}=-2 horizontalaxis+staticinput(spacevalue), \\quad \\frac{d horizontalaxis}{d spacevalue}=-2 verticalaxis+staticinput(spacevalue)\n\\]\n\nShow that, regardless of the choice of \\( staticinput(spacevalue) \\), the solution of the system which satisfies \\( verticalaxis=finalvertical, horizontalaxis=terminalhorizontal \\) at \\( spacevalue=0 \\) will never pass through \\((0,0)\\) unless \\( finalvertical=terminalhorizontal \\). When \\( finalvertical=terminalhorizontal \\), show that, for any positive value \\( spacestart \\) of \\( spacevalue \\), it is possible to choose \\( staticinput(spacevalue) \\) so the solution is at \\((0,0)\\) when \\( spacevalue=spacestart \\).",
      "solution": "A-5 Subtracting the two equations eliminates \\( staticinput(spacevalue) \\) and provides the simpler equation \\( d(verticalaxis-horizontalaxis) / d spacevalue = 2(verticalaxis-horizontalaxis) \\), which has the solution \\( verticalaxis-horizontalaxis=(finalvertical-terminalhorizontal)e^{2 spacevalue} \\). If \\( finalvertical \\neq terminalhorizontal \\), the right-hand side is never zero and so \\( verticalaxis=horizontalaxis=0 \\) can never occur.\n\nFor the second part, \\( finalvertical=terminalhorizontal \\). In this case, \\( verticalaxis(spacevalue)=horizontalaxis(spacevalue) \\) and so every solution is a parametrization of the line \\( verticalaxis=horizontalaxis \\). We can attempt to get a solution of the form \\( verticalaxis=finalvertical-curvedness\\,spacevalue, \\; horizontalaxis=terminalhorizontal-curvedness\\,spacevalue \\). This will be a solution if \\( staticinput(spacevalue)=2(finalvertical-curvedness\\,spacevalue)-curvedness \\). By taking \\( curvedness=finalvertical / spacestart, \\; verticalaxis=horizontalaxis=0 \\) at \\( spacevalue=spacestart \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "t": "vmbnclke",
        "u": "fkdajshg",
        "a": "rplxjgmu",
        "x_0": "lqnsdyeo",
        "y_0": "ztpfrkhd",
        "t_0": "bsmnqwer"
      },
      "question": "A-5. Let \\( fkdajshg(vmbnclke) \\) be a continuous function in the system of differential equations\n\\[\n\\frac{d qzxwvtnp}{d vmbnclke}=-2 hjgrksla+fkdajshg(vmbnclke), \\quad \\frac{d hjgrksla}{d vmbnclke}=-2 qzxwvtnp+fkdajshg(vmbnclke)\n\\]\n\nShow that, regardless of the choice of \\( fkdajshg(vmbnclke) \\), the solution of the system which satisfies \\( qzxwvtnp=lqnsdyeo, hjgrksla=ztpfrkhd \\) at \\( vmbnclke=0 \\) will never pass through \\( (0,0) \\) unless \\( lqnsdyeo=ztpfrkhd \\). When \\( lqnsdyeo=ztpfrkhd \\), show that, for any positive value \\( bsmnqwer \\) of \\( vmbnclke \\), it is possible to choose \\( fkdajshg(vmbnclke) \\) so the solution is at \\( (0,0) \\) when \\( vmbnclke=bsmnqwer \\).",
      "solution": "A-5 Subtracting the two equations eliminates \\( fkdajshg(vmbnclke) \\) and provides the simpler equation \\( d(qzxwvtnp-hjgrksla) / d vmbnclke=2(qzxwvtnp-hjgrksla) \\), which has the solution \\( qzxwvtnp-hjgrksla=(lqnsdyeo-ztpfrkhd) e^{2 vmbnclke} \\). If \\( lqnsdyeo \\neq ztpfrkhd \\), the right hand side is never zero and so \\( qzxwvtnp=hjgrksla=0 \\) can never occur.\n\nFor the second part, \\( lqnsdyeo=ztpfrkhd \\). In this case, \\( qzxwvtnp(vmbnclke)=hjgrksla(vmbnclke) \\) and so every solution is a parametrization of the line \\( qzxwvtnp=hjgrksla \\). We can attempt to get a solution of the form \\( qzxwvtnp=lqnsdyeo-rplxjgmu vmbnclke, hjgrksla=ztpfrkhd-rplxjgmu vmbnclke \\). This will be a solution if \\( fkdajshg(vmbnclke)=2(lqnsdyeo-rplxjgmu vmbnclke)-rplxjgmu \\). By taking \\( rplxjgmu=lqnsdyeo / bsmnqwer \\), \\( qzxwvtnp=hjgrksla=0 \\) at \\( vmbnclke=bsmnqwer \\)."
    },
    "kernel_variant": {
      "question": "Fix an integer n ≥ 3, a transfer time t₀ > 0, an energy budget E > 0 and an amplitude budget M > 0.  \n\nThroughout let u:[0,t₀] → ℝ be any Lebesgue–measurable control that satisfies  \n\n (Amplitude) |u(t)| ≤ M for a.e. t∈[0,t₀],  \n (Energy)  ∫₀^{t₀} u(t)² dt ≤ E.  \n\nOn ℝ^{2n} consider the block–diagonal controlled system  \n\n ẋᵢ = 4 xᵢ + u(t),  i = 1,…,n     (1a)  \n ẏᵢ = −3 yᵢ + u(t),  i = 1,…,n     (1b)\n\nwith initial state X₀ = (x_{10},…,x_{n0},y_{10},…,y_{n0}).  \nThe target set is the two–dimensional affine subspace  \n\n S = { (α,…,α︸ n , β,…,β︸ n ) : α,β∈ℝ }.  \n\n(A) Show that for every control u the (2n–2) functions  \n\n Δ_{ij}(t)=e^{−4t}(x_i(t)−x_j(t)), Γ_{ij}(t)=e^{3t}(y_i(t)−y_j(t))\n\nare invariants of motion.\n\n(B) Deduce that the trajectory can reach S at some time τ > 0 only if  \n\n x_{10}=⋯=x_{n0}, y_{10}=⋯=y_{n0}.       (2)\n\n(C) Assume (2) and denote α₀:=x_{10}=⋯=x_{n0}, β₀:=y_{10}=⋯=y_{n0}.  \nShow that the 2n–dimensional problem reduces to the two–dimensional system  \n\n α̇ = 4α+u(t), β̇ = −3β+u(t), (α(0),β(0))=(α₀,β₀).  (3)\n\nGiven an arbitrary target (α₁,β₁)∈ℝ², prove that all controls steering (3) to (α₁,β₁) in the fixed time t₀ form the affine subspace  \n\n 𝕌 = { u∈L²[0,t₀] : T u = ĥ },         (4)\n\nwhere  \n\n T:L²[0,t₀]→ℝ², Tu=(∫₀^{t₀}e^{−4s}u(s)ds, ∫₀^{t₀}e^{3s}u(s)ds),  \n ĥ=(e^{−4t₀}α₁−α₀, e^{3t₀}β₁−β₀).\n\n(D) Introduce the controllability Gramian W(t₀):=T T*.  \n\n(i) Prove that W(t₀) is positive definite for every t₀ > 0.  \n\n(ii) Show that 𝕌 contains a unique element u* of minimum L²–energy, namely  \n\n u* = T* W(t₀)^{−1} ĥ,  E_min(t₀):=∫₀^{t₀}u*(t)²dt=⟨ĥ, W(t₀)^{−1}ĥ⟩.\n\n(iii) Express u* in the closed form  \n\n u*(t)=a e^{−4t}+b e^{3t}, 0≤t≤t₀,\n\nand give a,b explicitly in terms of (t₀,α₀,α₁,β₀,β₁).\n\n(E) Amplitude analysis.\n\n(i) (A universal necessary bound)  \nProve that every admissible steering control must satisfy  \n\n M ≥ M_low:=max{4|ĥ₁|/(1−e^{−4t₀}), 3|ĥ₂|/(e^{3t₀}−1)}, ĥ=(ĥ₁,ĥ₂).\n\n(ii) (Amplitude of the minimum–energy control)  \nFor u*(t)=a e^{−4t}+b e^{3t} show that\n  (a) u* is strictly monotone when ab ≤ 0;  \n  (b) when ab > 0 it has a unique stationary point  \n\n    t_c = (1/7) ln(4a/3b)  \n\n   which is a strict local minimum of u* if a,b > 0 and a strict local maximum of u* if a,b < 0.  \n\nDeduce that in every case  \n\n M_* := ess sup_{t∈[0,t₀]}|u*(t)| = max{|a+b|, |a e^{−4t₀}+b e^{3t₀}|}.\n\n(iii) (Simultaneous budgets)\n\n (a) Show that if E ≥ E_min(t₀) and M ≥ M_* then the 2n–system (1) is reachable in time t₀ (sufficiency).  \n\n (b) Show that if either E < E_min(t₀) or M < M_low the transfer is impossible (necessity).  \n\n (c) Explain why in general M_low ≤ M_*, with strict inequality possible.  \nGive an explicit numerical example (specify n,t₀,α₀,β₀,α₁,β₁) for which M_low < M_*, and construct a steering control that respects the smaller bound M_low (necessarily of higher energy than u*).\n\n(F) Give an explicit initial condition that violates (2) (e.g. x_{10}=1,x_{20}=0) and use the invariants from (A) to prove that no admissible control can make the trajectory reach S in finite time.",
      "solution": "(A) Invariance.  \nDifferentiate Δ_{ij}(t)=e^{−4t}(x_i−x_j):\n\n dΔ_{ij}/dt = e^{−4t}(4x_i+u−4x_j−u)−4e^{−4t}(x_i−x_j)=0.\n\nThe same computation with Γ_{ij}(t)=e^{3t}(y_i−y_j) gives dΓ_{ij}/dt=0.  \nHence the (2n−2) quantities are conserved for every admissible control u.\n\n(B) Condition (2).  \nIf X(τ)∈S then x_i(τ)=x_j(τ) and y_i(τ)=y_j(τ); consequently Δ_{ij}(τ)=Γ_{ij}(τ)=0.  \nBecause the invariants are constant, they must already vanish at t=0, which is exactly (2).\n\n(C) Reduction to two dimensions.  \nUnder (2) set α:=x_1=⋯=x_n and β:=y_1=⋯=y_n; substituting into (1) yields (3).  \nSolving (3) on [0,t₀] and imposing (α(t₀),β(t₀))=(α₁,β₁) produces\n\n Tu = (∫₀^{t₀}e^{−4s}u(s)ds, ∫₀^{t₀}e^{3s}u(s)ds)=ĥ,\n\ni.e. the affine subspace (4).\n\n(D) Minimum–energy synthesis.\n\n(i) Let c=(c₁,c₂)≠0.  If T* c=0, then c₁e^{−4t}+c₂e^{3t}=0 for all t∈[0,t₀], impossible because the two exponentials are linearly independent. Hence T* is injective and W=T T* is positive definite.\n\n(ii) Because Range T is closed, the orthogonal projection of 0 onto 𝕌 is u*=T*W^{−1}ĥ; its energy is ⟨ĥ,W^{−1}ĥ⟩.\n\n(iii)  Computations:\n\n w_{11}=∫₀^{t₀}e^{−8s}ds=(1−e^{−8t₀})/8,  \n w_{22}=∫₀^{t₀}e^{6s}ds=(e^{6t₀}−1)/6,  \n w_{12}=w_{21}=∫₀^{t₀}e^{−s}ds=(1−e^{−t₀}).\n\nLet Δ := w_{11}w_{22}−w_{12}² > 0.  Then\n\n a=(w_{22}ĥ₁−w_{12}ĥ₂)/Δ, b=(−w_{12}ĥ₁+w_{11}ĥ₂)/Δ,\n\nso u*(t)=a e^{−4t}+b e^{3t}.\n\n(E) Amplitude questions.\n\n(i)  By Hölder\n\n |ĥ₁| ≤ M∫₀^{t₀}e^{−4s}ds = M(1−e^{−4t₀})/4,  \n |ĥ₂| ≤ M∫₀^{t₀}e^{3s}ds  = M(e^{3t₀}−1)/3,\n\nwhence M ≥ M_low.\n\n(ii)  Put f(t)=u*(t)=a e^{−4t}+b e^{3t}.  \n • If ab≤0, f′(t)=−4a e^{−4t}+3b e^{3t} keeps a constant sign, so |f| is monotone and attains its maximum at one of the endpoints.  \n • If ab>0, f′ vanishes only at  \n\n  t_c=(1/7)ln(4a/3b).  \n\n  Because f″(t)=16a e^{−4t}+9b e^{3t}, we obtain  \n\n   f″(t_c)=28a e^{−4t_c}.  \n\n  Hence t_c is a strict local minimum of f when a,b>0 and a strict local maximum of f when a,b<0.  In either case it is a strict local minimum of |f|, so the essential supremum of |u*| is still attained at an endpoint.  \n\nConsequently  \n\n M_* = max{|f(0)|,|f(t₀)|} = max{|a+b|, |a e^{−4t₀}+b e^{3t₀}|}.\n\n(iii) Simultaneous budgets.\n\n(a)  If E ≥ E_min and M ≥ M_*, the control u* satisfies both constraints, hence (1) is reachable.\n\n(b)  If E<E_min no element of 𝕌 has enough energy; if M<M_low no element of 𝕌 can obey the amplitude bound.  Reachability is impossible.\n\n(c)  The inequality M_low ≤ M_* is immediate because u* itself belongs to 𝕌.  Strict inequality can occur.\n\nExample with strict inequality and explicit admissible control.  \nTake  \n\n n = 3, t₀ = 1, α₀ = β₀ = 0, α₁ = 1, β₁ = β̂,\n\nwhere β̂ is chosen so that the two terms in M_low coincide:\n\n 4 e^{−4}α₁ /(1−e^{−4}) = 3 e^{3}β̂ /(e^{3}−1).\n\nSolving gives  \n\n β̂ = α₁·[4 e^{−4}(e^{3}−1)]/[3 e^{3}(1−e^{−4})] ≈ 2.364 × 10^{−2}.\n\nFor these data  \n\n ĥ = (e^{−4}α₁, e^{3}β̂) ≈ (1.8315·10^{−2}, 4.740·10^{−1}),  \n\n M_low = 4|ĥ₁|/(1−e^{−4}) ≈ 7.45 × 10^{−2}.  \n\nThe minimum–energy control computed above yields  \n\n M_* ≈ 1.23 × 10^{−1} > M_low.\n\nA steering control that respects the smaller bound is the constant control  \n\n u_const(t) = M_low·sign(α₁) (here positive), 0 ≤ t ≤ t₀.\n\nIndeed\n\n Tu_const = ( M_low(1−e^{−4})/4 , M_low(e^{3}−1)/3 ) = ĥ,\n\nso the state reaches the target in time t₀ while |u_const(t)| = M_low for all t.  \nIts energy E_const = M_low² t₀ is strictly larger than E_min, thereby illustrating the requested phenomenon.\n\n(F)  Take X₀=(1,0,0,…,0,0,…,0).  Then Δ_{12}(0)=1, hence Δ_{12}(t)≡1 and x₁(t)≠x₂(t) for all t≥0.  Therefore the trajectory can never lie in S, regardless of the control.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.584124",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: The problem is posed in 2n dimensions (n≥3), instead of 2 or 3.  \n2. Multiple constraints: Both an L∞ (amplitude) and an L² (energy) constraint are enforced simultaneously.  \n3. Reduction & functional analysis: Reachability is reduced to an infinite–dimensional moment problem; solving it calls for Hilbert–space projection arguments and computation of a controllability Gramian, techniques absent from the original problem.  \n4. Optimisation: The solver must identify and compute the minimum–energy control (quadratic optimisation in function space) and then sharpen it to an amplitude–optimal control, mixing optimal–control theory with real analysis.  \n5. Necessity & sufficiency: One has to prove precise reachability conditions (parts (B), (E), (F)), not merely supply a single steering control.  \n6. Interaction of concepts: Linear invariants, controllability, Gramian inversion, and optimisation under mixed L²/L∞ constraints all interact, greatly deepening the technical load relative to the original question, which used only a single elementary invariant and elementary integration."
      }
    },
    "original_kernel_variant": {
      "question": "Fix an integer n ≥ 3, a transfer time t₀ > 0, an energy budget E > 0 and an amplitude budget M > 0.  \n\nThroughout let u:[0,t₀] → ℝ be any Lebesgue–measurable control that satisfies  \n\n (Amplitude) |u(t)| ≤ M for a.e. t∈[0,t₀],  \n (Energy)  ∫₀^{t₀} u(t)² dt ≤ E.  \n\nOn ℝ^{2n} consider the block–diagonal controlled system  \n\n ẋᵢ = 4 xᵢ + u(t),  i = 1,…,n     (1a)  \n ẏᵢ = −3 yᵢ + u(t),  i = 1,…,n     (1b)\n\nwith initial state X₀ = (x_{10},…,x_{n0},y_{10},…,y_{n0}).  \nThe target set is the two–dimensional affine subspace  \n\n S = { (α,…,α︸ n , β,…,β︸ n ) : α,β∈ℝ }.  \n\n(A) Show that for every control u the (2n–2) functions  \n\n Δ_{ij}(t)=e^{−4t}(x_i(t)−x_j(t)), Γ_{ij}(t)=e^{3t}(y_i(t)−y_j(t))\n\nare invariants of motion.\n\n(B) Deduce that the trajectory can reach S at some time τ > 0 only if  \n\n x_{10}=⋯=x_{n0}, y_{10}=⋯=y_{n0}.       (2)\n\n(C) Assume (2) and denote α₀:=x_{10}=⋯=x_{n0}, β₀:=y_{10}=⋯=y_{n0}.  \nShow that the 2n–dimensional problem reduces to the two–dimensional system  \n\n α̇ = 4α+u(t), β̇ = −3β+u(t), (α(0),β(0))=(α₀,β₀).  (3)\n\nGiven an arbitrary target (α₁,β₁)∈ℝ², prove that all controls steering (3) to (α₁,β₁) in the fixed time t₀ form the affine subspace  \n\n 𝕌 = { u∈L²[0,t₀] : T u = ĥ },         (4)\n\nwhere  \n\n T:L²[0,t₀]→ℝ², Tu=(∫₀^{t₀}e^{−4s}u(s)ds, ∫₀^{t₀}e^{3s}u(s)ds),  \n ĥ=(e^{−4t₀}α₁−α₀, e^{3t₀}β₁−β₀).\n\n(D) Introduce the controllability Gramian W(t₀):=T T*.  \n\n(i) Prove that W(t₀) is positive definite for every t₀ > 0.  \n\n(ii) Show that 𝕌 contains a unique element u* of minimum L²–energy, namely  \n\n u* = T* W(t₀)^{−1} ĥ,  E_min(t₀):=∫₀^{t₀}u*(t)²dt=⟨ĥ, W(t₀)^{−1}ĥ⟩.\n\n(iii) Express u* in the closed form  \n\n u*(t)=a e^{−4t}+b e^{3t}, 0≤t≤t₀,\n\nand give a,b explicitly in terms of (t₀,α₀,α₁,β₀,β₁).\n\n(E) Amplitude analysis.\n\n(i) (A universal necessary bound)  \nProve that every admissible steering control must satisfy  \n\n M ≥ M_low:=max{4|ĥ₁|/(1−e^{−4t₀}), 3|ĥ₂|/(e^{3t₀}−1)}, ĥ=(ĥ₁,ĥ₂).\n\n(ii) (Amplitude of the minimum–energy control)  \nFor u*(t)=a e^{−4t}+b e^{3t} show that\n  (a) u* is strictly monotone when ab ≤ 0;  \n  (b) when ab > 0 it has a unique stationary point  \n\n    t_c = (1/7) ln(4a/3b)  \n\n   which is a strict local minimum of u* if a,b > 0 and a strict local maximum of u* if a,b < 0.  \n\nDeduce that in every case  \n\n M_* := ess sup_{t∈[0,t₀]}|u*(t)| = max{|a+b|, |a e^{−4t₀}+b e^{3t₀}|}.\n\n(iii) (Simultaneous budgets)\n\n (a) Show that if E ≥ E_min(t₀) and M ≥ M_* then the 2n–system (1) is reachable in time t₀ (sufficiency).  \n\n (b) Show that if either E < E_min(t₀) or M < M_low the transfer is impossible (necessity).  \n\n (c) Explain why in general M_low ≤ M_*, with strict inequality possible.  \nGive an explicit numerical example (specify n,t₀,α₀,β₀,α₁,β₁) for which M_low < M_*, and construct a steering control that respects the smaller bound M_low (necessarily of higher energy than u*).\n\n(F) Give an explicit initial condition that violates (2) (e.g. x_{10}=1,x_{20}=0) and use the invariants from (A) to prove that no admissible control can make the trajectory reach S in finite time.",
      "solution": "(A) Invariance.  \nDifferentiate Δ_{ij}(t)=e^{−4t}(x_i−x_j):\n\n dΔ_{ij}/dt = e^{−4t}(4x_i+u−4x_j−u)−4e^{−4t}(x_i−x_j)=0.\n\nThe same computation with Γ_{ij}(t)=e^{3t}(y_i−y_j) gives dΓ_{ij}/dt=0.  \nHence the (2n−2) quantities are conserved for every admissible control u.\n\n(B) Condition (2).  \nIf X(τ)∈S then x_i(τ)=x_j(τ) and y_i(τ)=y_j(τ); consequently Δ_{ij}(τ)=Γ_{ij}(τ)=0.  \nBecause the invariants are constant, they must already vanish at t=0, which is exactly (2).\n\n(C) Reduction to two dimensions.  \nUnder (2) set α:=x_1=⋯=x_n and β:=y_1=⋯=y_n; substituting into (1) yields (3).  \nSolving (3) on [0,t₀] and imposing (α(t₀),β(t₀))=(α₁,β₁) produces\n\n Tu = (∫₀^{t₀}e^{−4s}u(s)ds, ∫₀^{t₀}e^{3s}u(s)ds)=ĥ,\n\ni.e. the affine subspace (4).\n\n(D) Minimum–energy synthesis.\n\n(i) Let c=(c₁,c₂)≠0.  If T* c=0, then c₁e^{−4t}+c₂e^{3t}=0 for all t∈[0,t₀], impossible because the two exponentials are linearly independent. Hence T* is injective and W=T T* is positive definite.\n\n(ii) Because Range T is closed, the orthogonal projection of 0 onto 𝕌 is u*=T*W^{−1}ĥ; its energy is ⟨ĥ,W^{−1}ĥ⟩.\n\n(iii)  Computations:\n\n w_{11}=∫₀^{t₀}e^{−8s}ds=(1−e^{−8t₀})/8,  \n w_{22}=∫₀^{t₀}e^{6s}ds=(e^{6t₀}−1)/6,  \n w_{12}=w_{21}=∫₀^{t₀}e^{−s}ds=(1−e^{−t₀}).\n\nLet Δ := w_{11}w_{22}−w_{12}² > 0.  Then\n\n a=(w_{22}ĥ₁−w_{12}ĥ₂)/Δ, b=(−w_{12}ĥ₁+w_{11}ĥ₂)/Δ,\n\nso u*(t)=a e^{−4t}+b e^{3t}.\n\n(E) Amplitude questions.\n\n(i)  By Hölder\n\n |ĥ₁| ≤ M∫₀^{t₀}e^{−4s}ds = M(1−e^{−4t₀})/4,  \n |ĥ₂| ≤ M∫₀^{t₀}e^{3s}ds  = M(e^{3t₀}−1)/3,\n\nwhence M ≥ M_low.\n\n(ii)  Put f(t)=u*(t)=a e^{−4t}+b e^{3t}.  \n • If ab≤0, f′(t)=−4a e^{−4t}+3b e^{3t} keeps a constant sign, so |f| is monotone and attains its maximum at one of the endpoints.  \n • If ab>0, f′ vanishes only at  \n\n  t_c=(1/7)ln(4a/3b).  \n\n  Because f″(t)=16a e^{−4t}+9b e^{3t}, we obtain  \n\n   f″(t_c)=28a e^{−4t_c}.  \n\n  Hence t_c is a strict local minimum of f when a,b>0 and a strict local maximum of f when a,b<0.  In either case it is a strict local minimum of |f|, so the essential supremum of |u*| is still attained at an endpoint.  \n\nConsequently  \n\n M_* = max{|f(0)|,|f(t₀)|} = max{|a+b|, |a e^{−4t₀}+b e^{3t₀}|}.\n\n(iii) Simultaneous budgets.\n\n(a)  If E ≥ E_min and M ≥ M_*, the control u* satisfies both constraints, hence (1) is reachable.\n\n(b)  If E<E_min no element of 𝕌 has enough energy; if M<M_low no element of 𝕌 can obey the amplitude bound.  Reachability is impossible.\n\n(c)  The inequality M_low ≤ M_* is immediate because u* itself belongs to 𝕌.  Strict inequality can occur.\n\nExample with strict inequality and explicit admissible control.  \nTake  \n\n n = 3, t₀ = 1, α₀ = β₀ = 0, α₁ = 1, β₁ = β̂,\n\nwhere β̂ is chosen so that the two terms in M_low coincide:\n\n 4 e^{−4}α₁ /(1−e^{−4}) = 3 e^{3}β̂ /(e^{3}−1).\n\nSolving gives  \n\n β̂ = α₁·[4 e^{−4}(e^{3}−1)]/[3 e^{3}(1−e^{−4})] ≈ 2.364 × 10^{−2}.\n\nFor these data  \n\n ĥ = (e^{−4}α₁, e^{3}β̂) ≈ (1.8315·10^{−2}, 4.740·10^{−1}),  \n\n M_low = 4|ĥ₁|/(1−e^{−4}) ≈ 7.45 × 10^{−2}.  \n\nThe minimum–energy control computed above yields  \n\n M_* ≈ 1.23 × 10^{−1} > M_low.\n\nA steering control that respects the smaller bound is the constant control  \n\n u_const(t) = M_low·sign(α₁) (here positive), 0 ≤ t ≤ t₀.\n\nIndeed\n\n Tu_const = ( M_low(1−e^{−4})/4 , M_low(e^{3}−1)/3 ) = ĥ,\n\nso the state reaches the target in time t₀ while |u_const(t)| = M_low for all t.  \nIts energy E_const = M_low² t₀ is strictly larger than E_min, thereby illustrating the requested phenomenon.\n\n(F)  Take X₀=(1,0,0,…,0,0,…,0).  Then Δ_{12}(0)=1, hence Δ_{12}(t)≡1 and x₁(t)≠x₂(t) for all t≥0.  Therefore the trajectory can never lie in S, regardless of the control.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.470423",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension: The problem is posed in 2n dimensions (n≥3), instead of 2 or 3.  \n2. Multiple constraints: Both an L∞ (amplitude) and an L² (energy) constraint are enforced simultaneously.  \n3. Reduction & functional analysis: Reachability is reduced to an infinite–dimensional moment problem; solving it calls for Hilbert–space projection arguments and computation of a controllability Gramian, techniques absent from the original problem.  \n4. Optimisation: The solver must identify and compute the minimum–energy control (quadratic optimisation in function space) and then sharpen it to an amplitude–optimal control, mixing optimal–control theory with real analysis.  \n5. Necessity & sufficiency: One has to prove precise reachability conditions (parts (B), (E), (F)), not merely supply a single steering control.  \n6. Interaction of concepts: Linear invariants, controllability, Gramian inversion, and optimisation under mixed L²/L∞ constraints all interact, greatly deepening the technical load relative to the original question, which used only a single elementary invariant and elementary integration."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}