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== 8.6#h(0.6em)Exercises


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/8%3A_Nonlinear_Systems/8.1%3A_Linearization_critical_points_and_equilibria")[8.1: Linearization, critical points, and equilibria]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Sketch the phase plane vector field for:


      + #math.equation(block: false, alt: "x prime equals x squared , y prime equals y squared")[$x^(′) = x^(2) , " " " " y^(′) = y^(2)$],
+ #math.equation(block: false, alt: "x prime equals open parenthesis x minus y close parenthesis squared , y prime equals minus x")[$x^(′) = ( x − y )^(2) , " " " " y^(′) = − x$],
+ #math.equation(block: false, alt: "x prime equals e to the power y , y prime equals e to the power x")[$x^(′) = e^(y) , " " " " y^(′) = e^(x)$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Match systems


      + #math.equation(block: false, alt: "x prime equals x squared")[$x^(′) = x^(2)$], #math.equation(block: false, alt: "y prime equals y squared")[$y^(′) = y^(2)$],
+ #math.equation(block: false, alt: "x prime equals x y")[$x^(′) = x y$], #math.equation(block: false, alt: "y prime equals 1 plus y squared")[$y^(′) = 1 + y^(2)$],
+ #math.equation(block: false, alt: "x prime equals sin open parenthesis π y close parenthesis")[$x^(′) = sin ( π y )$], #math.equation(block: false, alt: "y prime equals x")[$y^(′) = x$],


      to the vector fields below. Justify.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the critical points and linearizations of the following systems.


      + #math.equation(block: false, alt: "x prime equals x squared minus y squared")[$x^(′) = x^(2) − y^(2)$], #math.equation(block: false, alt: "y prime equals x squared plus y squared minus 1")[$y^(′) = x^(2) + y^(2) − 1$],
+ #math.equation(block: false, alt: "x prime equals minus y")[$x^(′) = − y$], #math.equation(block: false, alt: "y prime equals 3 x plus y x squared")[$y^(′) = 3 x + y x^(2)$],
+ #math.equation(block: false, alt: "x prime equals x squared plus y")[$x^(′) = x^(2) + y$], #math.equation(block: false, alt: "y prime equals y squared plus x")[$y^(′) = y^(2) + x$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
For the following systems, verify they have critical point at #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$], and find the linearization at #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$].


      + #math.equation(block: false, alt: "x prime equals x plus 2 y plus x squared minus y squared")[$x^(′) = x + 2 y + x^(2) − y^(2)$], #math.equation(block: false, alt: "y prime equals 2 y minus x squared")[$y^(′) = 2 y − x^(2)$]
+ #math.equation(block: false, alt: "x prime equals minus y")[$x^(′) = − y$], #math.equation(block: false, alt: "y prime equals x minus y cubed")[$y^(′) = x − y^(3)$]
+ #math.equation(block: false, alt: "x prime equals a x plus b y plus f open parenthesis x , y close parenthesis")[$x^(′) = a x + b y + f ( x , y )$], #math.equation(block: false, alt: "y prime equals c x plus d y plus g open parenthesis x , y close parenthesis")[$y^(′) = c x + d y + g ( x , y )$], where #math.equation(block: false, alt: "f open parenthesis 0 , 0 close parenthesis equals 0")[$f ( 0 , 0 ) = 0$], #math.equation(block: false, alt: "g open parenthesis 0 , 0 close parenthesis equals 0")[$g ( 0 , 0 ) = 0$], and all first partial derivatives of #math.equation(block: false, alt: "f")[$f$] and #math.equation(block: false, alt: "g")[$g$] are also zero at #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$], that is, #math.equation(block: false, alt: "the fraction ∂ f over ∂ x open parenthesis 0 , 0 close parenthesis equals the fraction ∂ f over ∂ y open parenthesis 0 , 0 close parenthesis equals the fraction ∂ g over ∂ x open parenthesis 0 , 0 close parenthesis equals the fraction ∂ g over ∂ y open parenthesis 0 , 0 close parenthesis equals 0")[$frac(∂ f, ∂ x) ( 0 , 0 ) = frac(∂ f, ∂ y) ( 0 , 0 ) = frac(∂ g, ∂ x) ( 0 , 0 ) = frac(∂ g, ∂ y) ( 0 , 0 ) = 0$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Take #math.equation(block: false, alt: "x prime equals open parenthesis x minus y close parenthesis squared")[$x^(′) = ( x − y )^(2)$], #math.equation(block: false, alt: "y prime equals open parenthesis x plus y close parenthesis squared")[$y^(′) = ( x + y )^(2)$].


      + Find the set of critical points.
+ Sketch a phase diagram and describe the behavior near the critical point(s).
+ Find the linearization. Is it helpful in understanding the system?
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Take #math.equation(block: false, alt: "x prime equals x squared")[$x^(′) = x^(2)$], #math.equation(block: false, alt: "y prime equals x cubed")[$y^(′) = x^(3)$].


      + Find the set of critical points.
+ Sketch a phase diagram and describe the behavior near the critical point(s).
+ Find the linearization. Is it helpful in understanding the system?
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the critical points and linearizations of the following systems.


      + #math.equation(block: false, alt: "x prime equals sin open parenthesis π y close parenthesis plus open parenthesis x minus 1 close parenthesis squared")[$x^(′) = sin ( π y ) + ( x − 1 )^(2)$], #math.equation(block: false, alt: "y prime equals y squared minus y")[$y^(′) = y^(2) − y$],
+ #math.equation(block: false, alt: "x prime equals x plus y plus y squared")[$x^(′) = x + y + y^(2)$], #math.equation(block: false, alt: "y prime equals x")[$y^(′) = x$],
+ #math.equation(block: false, alt: "x prime equals open parenthesis x minus 1 close parenthesis squared plus y")[$x^(′) = ( x − 1 )^(2) + y$], #math.equation(block: false, alt: "y prime equals x squared plus y")[$y^(′) = x^(2) + y$].


      Answer


      + Critical points #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] and #math.equation(block: false, alt: "open parenthesis 0 , 1 close parenthesis")[$( 0 , 1 )$]. At #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] using #math.equation(block: false, alt: "u equals x")[$u = x$], #math.equation(block: false, alt: "v equals y")[$v = y$] the linearization is #math.equation(block: false, alt: "u prime equals minus 2 u minus open parenthesis the fraction 1 over π close parenthesis v")[$u^(′) = − 2 u − ( frac(1, π) ) v$], #math.equation(block: false, alt: "v prime equals minus v")[$v^(′) = − v$]. At #math.equation(block: false, alt: "open parenthesis 0 , 1 close parenthesis")[$( 0 , 1 )$] using #math.equation(block: false, alt: "u equals x")[$u = x$], #math.equation(block: false, alt: "v equals y minus 1")[$v = y − 1$] the linearization is #math.equation(block: false, alt: "u prime equals minus 2 u plus open parenthesis the fraction 1 over π close parenthesis v")[$u^(′) = − 2 u + ( frac(1, π) ) v$] #math.equation(block: false, alt: "v prime equals v")[$v^(′) = v$].
+ Critical point #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$]. Using #math.equation(block: false, alt: "u equals x")[$u = x$], #math.equation(block: false, alt: "v equals y")[$v = y$] the linearization is #math.equation(block: false, alt: "u prime equals u plus v")[$u^(′) = u + v$], #math.equation(block: false, alt: "v prime equals u")[$v^(′) = u$].
+ Critical point #math.equation(block: false, alt: "open parenthesis the fraction 1 over 2 , minus the fraction 1 over 4 close parenthesis")[$( frac(1, 2) , − frac(1, 4) )$]. Using #math.equation(block: false, alt: "u equals x minus the fraction 1 over 2")[$u = x − frac(1, 2)$], #math.equation(block: false, alt: "v equals y plus the fraction 1 over 4")[$v = y + frac(1, 4)$] the linearization is #math.equation(block: false, alt: "u prime equals minus u plus v")[$u^(′) = − u + v$], #math.equation(block: false, alt: "v prime equals u plus v")[$v^(′) = u + v$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Match systems


      + #math.equation(block: false, alt: "x prime equals y squared")[$x^(′) = y^(2)$], #math.equation(block: false, alt: "y prime equals minus x squared")[$y^(′) = − x^(2)$],
+ #math.equation(block: false, alt: "x prime equals y")[$x^(′) = y$], #math.equation(block: false, alt: "y prime equals open parenthesis x minus 1 close parenthesis open parenthesis x plus 1 close parenthesis")[$y^(′) = ( x − 1 ) ( x + 1 )$],
+ #math.equation(block: false, alt: "x prime equals y plus x squared")[$x^(′) = y + x^(2)$], #math.equation(block: false, alt: "y prime equals minus x")[$y^(′) = − x$],


      to the vector fields below. Justify.


      
      Answer


      + is c),
+ is a),
+ is b)
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
The idea of critical points and linearization works in higher dimensions as well. You simply make the Jacobian matrix bigger by adding more functions and more variables. For the following system of 3 equations find the critical points and their linearizations:


      #math.equation(block: true, alt: "x prime equals x plus z squared , y prime equals z squared minus y , z prime equals z plus x squared .")[$x^(′) = x + z^(2) , #h(1em) y^(′) = z^(2) − y , #h(1em) z^(′) = z + x^(2) .$]


      Answer


      Critical points are #math.equation(block: false, alt: "open parenthesis 0 , 0 , 0 close parenthesis")[$( 0 , 0 , 0 )$], and #math.equation(block: false, alt: "open parenthesis − 1 , 1 , − 1 close parenthesis")[$( − 1 , 1 , − 1 )$]. The linearization at the origin using variables #math.equation(block: false, alt: "u equals x")[$u = x$], #math.equation(block: false, alt: "v equals y")[$v = y$], #math.equation(block: false, alt: "w equals z")[$w = z$] is #math.equation(block: false, alt: "u prime equals u")[$u^(′) = u$], #math.equation(block: false, alt: "v prime equals minus v")[$v^(′) = − v$], #math.equation(block: false, alt: "z prime equals w")[$z^(′) = w$]. The linearization at the point #math.equation(block: false, alt: "open parenthesis − 1 , 1 , − 1 close parenthesis")[$( − 1 , 1 , − 1 )$] using variables #math.equation(block: false, alt: "u equals x plus 1")[$u = x + 1$], #math.equation(block: false, alt: "v equals y minus 1")[$v = y − 1$], #math.equation(block: false, alt: "w equals z plus 1")[$w = z + 1$] is #math.equation(block: false, alt: "u prime equals u minus 2")[$u^(′) = u − 2$], #math.equation(block: false, alt: "v prime equals minus v minus 2 w")[$v^(′) = − v − 2 w$], #math.equation(block: false, alt: "w prime equals w minus 2 u")[$w^(′) = w − 2 u$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Any two-dimensional non-autonomous system #math.equation(block: false, alt: "x prime equals f open parenthesis x , y , t close parenthesis")[$x^(′) = f ( x , y , t )$], #math.equation(block: false, alt: "y prime equals g open parenthesis x , y , t close parenthesis")[$y^(′) = g ( x , y , t )$] can be written as a three-dimensional autonomous system (three equations). Write down this autonomous system using the variables #math.equation(block: false, alt: "u")[$u$], #math.equation(block: false, alt: "v")[$v$], #math.equation(block: false, alt: "w")[$w$].


      Answer


      #math.equation(block: true, alt: "u prime equals f open parenthesis u , v , w close parenthesis")[$u^(′) = f ( u , v , w )$], #math.equation(block: true, alt: "v prime equals g open parenthesis u , v , w close parenthesis")[$v^(′) = g ( u , v , w )$], #math.equation(block: true, alt: "w prime equals 1")[$w^(′) = 1$].
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/8%3A_Nonlinear_Systems/8.2%3A_Stability_and_classication_of_isolated_critical_points")[8.2: Stability and classification of isolated critical points]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
For the systems below, find and classify the critical points, also indicate if the equilibria are stable, asymptotically stable, or unstable.


      + #math.equation(block: false, alt: "x prime equals minus x plus 3 x squared , y prime equals minus y")[$x^(′) = − x + 3 x^(2) , y^(′) = − y$]
+ #math.equation(block: false, alt: "x prime equals x squared plus y squared minus 1")[$x^(′) = x^(2) + y^(2) − 1$],#math.equation(block: false, alt: "y prime equals x")[$y^(′) = x$]
+ #math.equation(block: false, alt: "x prime equals y e to the power x")[$x^(′) = y e^(x)$],#math.equation(block: false, alt: "y prime equals y minus x plus y squared")[$y^(′) = y − x + y^(2)$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the implicit equations of the trajectories of the following conservative systems. Next find their critical points (if any) and classify them.


      + #math.equation(block: false, alt: "x double prime plus x plus x cubed equals 0")[$x^(″) + x + x^(3) = 0$]
+ #math.equation(block: false, alt: "θ double prime plus sin θ equals 0")[$θ^(″) + sin θ = 0$]
+ #math.equation(block: false, alt: "z double prime plus open parenthesis z minus 1 close parenthesis open parenthesis z plus 1 close parenthesis equals 0")[$z^(″) + ( z − 1 ) ( z + 1 ) = 0$]
+ #math.equation(block: false, alt: "x double prime plus x squared plus 1 equals 0")[$x^(″) + x^(2) + 1 = 0$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find and classify the critical point(s) of #math.equation(block: false, alt: "x prime equals minus x squared")[$x^(′) = − x^(2)$],#math.equation(block: false, alt: "y prime equals minus y squared")[$y^(′) = − y^(2)$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose #math.equation(block: false, alt: "x prime equals minus x y")[$x^(′) = − x y$],#math.equation(block: false, alt: "y prime equals x squared minus 1 minus y")[$y^(′) = x^(2) − 1 − y$].


      + Show there are two spiral sinks at #math.equation(block: false, alt: "open parenthesis minus 1 , 0 close parenthesis")[$( − 1 , 0 )$] and #math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$].
+ For any initial point of the form #math.equation(block: false, alt: "open parenthesis 0 , y sub 0 close parenthesis")[$( 0 , y_(0) )$],find what is the trajectory.
+ Can a trajectory starting at #math.equation(block: false, alt: "open parenthesis x sub 0 , y sub 0 close parenthesis")[$( x_(0) , y_(0) )$] where #math.equation(block: false, alt: "x sub 0 greater than 0")[$x_(0) > 0$] spiral into the critical point at #math.equation(block: false, alt: "open parenthesis minus 1 , 0 close parenthesis")[$( − 1 , 0 )$]? Why or why not?
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
In the example #math.equation(block: false, alt: "x prime equals y")[$x^(′) = y$],#math.equation(block: false, alt: "y prime equals y cubed minus x")[$y^(′) = y^(3) − x$] show that for any trajectory, the distance from the origin is an increasing function. Conclude that the origin behaves like is a spiral source. Hint: Consider #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals open parenthesis x open parenthesis t close parenthesis close parenthesis squared plus open parenthesis y open parenthesis t close parenthesis close parenthesis squared")[$f ( t ) = attach(( x ( t ) ), t: 2) + attach(( y ( t ) ), t: 2)$] and show it has positive derivative.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose #math.equation(block: false, alt: "f")[$f$] is always positive. Find the trajectories of #math.equation(block: false, alt: "x double prime plus f open parenthesis x prime close parenthesis equals 0")[$x^(″) + f ( x^(′) ) = 0$]. Are there any critical points?
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose that #math.equation(block: false, alt: "x prime equals f open parenthesis x , y close parenthesis")[$x^(′) = f ( x , y )$],#math.equation(block: false, alt: "y prime equals g open parenthesis x , y close parenthesis")[$y^(′) = g ( x , y )$]. Suppose that #math.equation(block: false, alt: "g open parenthesis x , y close parenthesis greater than 1")[$g ( x , y ) > 1$] for all #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$]. Are there any critical points? What can we say about the trajectories at #math.equation(block: false, alt: "t")[$t$] goes to infinity?
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
For the systems below, find and classify the critical points.


      + #math.equation(block: false, alt: "x prime equals minus x plus x squared")[$x^(′) = − x + x^(2)$],#math.equation(block: false, alt: "y prime equals y")[$y^(′) = y$]
+ #math.equation(block: false, alt: "x prime equals y minus y squared minus x")[$x^(′) = y − y^(2) − x$],#math.equation(block: false, alt: "y prime equals minus x")[$y^(′) = − x$]
+ #math.equation(block: false, alt: "x prime equals x y")[$x^(′) = x y$],#math.equation(block: false, alt: "y prime equals x plus y minus 1")[$y^(′) = x + y − 1$]


      Answer


      + #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$]: saddle (unstable), #math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$]: source (unstable),
+ #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$]: spiral sink (asymptotically stable), #math.equation(block: false, alt: "open parenthesis 0 , 1 close parenthesis")[$( 0 , 1 )$]: saddle (unstable),
+ #math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$]: saddle (unstable), #math.equation(block: false, alt: "open parenthesis 0 , 1 close parenthesis")[$( 0 , 1 )$]: source (unstable)
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the implicit equations of the trajectories of the following conservative systems. Next find their critical points (if any) and classify them.


      + #math.equation(block: false, alt: "x double prime plus x squared equals 4")[$x^(″) + x^(2) = 4$]
+ #math.equation(block: false, alt: "x double prime plus e to the power x equals 0")[$x^(″) + e^(x) = 0$]
+ #math.equation(block: false, alt: "x double prime plus open parenthesis x plus 1 close parenthesis e to the power x equals 0")[$x^(″) + ( x + 1 ) e^(x) = 0$]


      Answer


      + #math.equation(block: false, alt: "the fraction 1 over 2 y squared plus the fraction 1 over 3 x cubed minus 4 x equals C")[$frac(1, 2) y^(2) + frac(1, 3) x^(3) − 4 x = C$], critical points: #math.equation(block: false, alt: "open parenthesis minus 2 , 0 close parenthesis")[$( − 2 , 0 )$], an unstable saddle, and #math.equation(block: false, alt: "open parenthesis 2 , 0 close parenthesis")[$( 2 , 0 )$], a stable center.
+ #math.equation(block: false, alt: "the fraction 1 over 2 y squared plus e to the power x equals C")[$frac(1, 2) y^(2) + e^(x) = C$], no critical points.
+ #math.equation(block: false, alt: "the fraction 1 over 2 y squared plus x e to the power x equals C")[$frac(1, 2) y^(2) + x e^(x) = C$], critical point at #math.equation(block: false, alt: "open parenthesis minus 1 , 0 close parenthesis")[$( − 1 , 0 )$] is a stable center.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
The conservative system #math.equation(block: false, alt: "x double prime plus x cubed equals 0")[$x^(″) + x^(3) = 0$] is not almost linear. Classify its critical point(s) nonetheless.


      Answer


      Critical point at #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$]. Trajectories are #math.equation(block: false, alt: "y equals ± the square root of 2 C minus open parenthesis the fraction 1 over 2 close parenthesis x to the power 4")[$y = ± sqrt(2 C − ( frac(1, 2) ) x^(4))$], for #math.equation(block: false, alt: "C greater than 0")[$C > 0$], these give closed curves around the origin, so the critical point is a stable center.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Derive an analogous classification of critical points for equations in one dimension, such as #math.equation(block: false, alt: "x prime equals f open parenthesis x close parenthesis")[$x^(′) = f ( x )$] based on the derivative. A point #math.equation(block: false, alt: "x sub 0")[$x_(0)$] is critical when #math.equation(block: false, alt: "f open parenthesis x sub 0 close parenthesis equals 0")[$f ( x_(0) ) = 0$] and almost linear if in addition #math.equation(block: false, alt: "f prime open parenthesis x sub 0 close parenthesis ⧸ equals 0")[$f^(′) ( x_(0) ) "⧸" = 0$]. Figure out if the critical point is stable or unstable depending on the sign of #math.equation(block: false, alt: "f prime open parenthesis x sub 0 close parenthesis")[$f^(′) ( x_(0) )$]. Explain. Hint: see #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/1%3A_First_order_ODEs/1.6%3A_Autonomous_equations")[Section 1.6].


      Answer


      A critical point #math.equation(block: false, alt: "x sub 0")[$x_(0)$] is stable if #math.equation(block: false, alt: "f prime open parenthesis x sub 0 close parenthesis less than 0")[$f^(′) ( x_(0) ) < 0$] and unstable when #math.equation(block: false, alt: "f prime open parenthesis x sub 0 close parenthesis less than 0")[$f^(′) ( x_(0) ) < 0$].
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/8%3A_Nonlinear_Systems/8.3%3A_Applications_of_Nonlinear_Systems")[8.3: Applications of nonlinear systems]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Take the #emph[damped nonlinear pendulum equation] #math.equation(block: false, alt: "θ double prime plus μ θ prime plus open parenthesis the fraction g over L close parenthesis sin θ equals 0")[$θ^(″) + μ θ^(′) + ( frac(g, L) ) sin θ = 0$] for some #math.equation(block: false, alt: "μ greater than 0")[$μ > 0$] (that is, there is some friction).


      + Suppose #math.equation(block: false, alt: "μ equals 1")[$μ = 1$] and #math.equation(block: false, alt: "the fraction g over L equals 1")[$frac(g, L) = 1$] for simplicity, find and classify the critical points.
+ Do the same for any #math.equation(block: false, alt: "μ greater than 0")[$μ > 0$] and any #math.equation(block: false, alt: "g")[$g$] and #math.equation(block: false, alt: "L")[$L$], but such that the damping is small, in particular, #math.equation(block: false, alt: "μ squared less than 4 open parenthesis the fraction g over L close parenthesis")[$μ^(2) < 4 ( frac(g, L) )$].
+ Explain what your findings mean, and if it agrees with what you expect in reality.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose the hares do not grow exponentially, but logistically. In particular consider


      #math.equation(block: true, alt: "x prime equals open parenthesis 0.4 minus 0.01 y close parenthesis x minus γ x squared , y prime equals open parenthesis 0.003 x minus 0.3 close parenthesis y .")[$x^(′) = ( 0.4 − 0.01 y ) x − γ x^(2) , " " " " " " " " " " y^(′) = ( 0.003 x − 0.3 ) y .$]


      For the following two values of #math.equation(block: false, alt: "γ")[$γ$], find and classify all the critical points in the positive quadrant, that is, for #math.equation(block: false, alt: "x greater than or equal to 0")[$x ≥ 0$] and #math.equation(block: false, alt: "y greater than or equal to 0")[$y ≥ 0$]. Then sketch the phase diagram. Discuss the implication for the long term behavior of the population.


      + #math.equation(block: false, alt: "γ equals 0.001")[$γ = 0.001$],
+ #math.equation(block: false, alt: "γ equals 0.01")[$γ = 0.01$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
+ Suppose #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] are positive variables. Show #math.equation(block: false, alt: "the fraction y x over e to the power x plus y")[$frac(y x, e^(x + y))$] attains a maximum at #math.equation(block: false, alt: "open parenthesis 1 , 1 close parenthesis")[$( 1 , 1 )$].
+ Suppose #math.equation(block: false, alt: "a , b , c , d")[$a , b , c , d$] are positive constants, and also suppose #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] are positive variables. Show #math.equation(block: false, alt: "the fraction y to the power a x to the power d over e to the power c x plus b y")[$frac(y^(a) x^(d), e^(c x + b y))$] attains a maximum at #math.equation(block: false, alt: "open parenthesis the fraction d over c , the fraction a over b close parenthesis")[$( frac(d, c) , frac(a, b) )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose that for the pendulum equation we take a trajectory giving the spinning-around motion, for example #math.equation(block: false, alt: "ω equals the square root of the fraction 2 g over L cos θ plus the fraction 2 g over L plus ω 0 2")[$ω = sqrt(frac(2 g, L) cos θ + frac(2 g, L) + ω_(0)^(2))$]. This is the trajectory where the lowest angular velocity is #math.equation(block: false, alt: "ω 0 2")[$ω_(0)^(2)$]. Find an integral expression for how long it takes the pendulum to go all the way around.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[(challenging)]


      Take the pendulum, suppose the initial position is #math.equation(block: false, alt: "θ equals 0")[$θ = 0$].


      + Find the expression for #math.equation(block: false, alt: "ω")[$ω$] giving the trajectory with initial condition #math.equation(block: false, alt: "open parenthesis 0 , ω sub 0 close parenthesis")[$( 0 , ω_(0) )$]. Hint: Figure out what #math.equation(block: false, alt: "C")[$C$] should be in terms of #math.equation(block: false, alt: "ω sub 0")[$ω_(0)$].
+ Find the crucial angular velocity #math.equation(block: false, alt: "ω sub 1")[$ω_(1)$], such that for any higher initial angular velocity, the pendulum will keep going around its axis, and for any lower initial angular velocity, the pendulum will simply swing back and forth. Hint: When the pendulum doesn't go over the top the expression for #math.equation(block: false, alt: "ω")[$ω$] will be undefined for some #math.equation(block: false, alt: "θ")[$θ$]s.
+ What do you think happens if the initial condition is #math.equation(block: false, alt: "open parenthesis 0 , ω sub 1 close parenthesis")[$( 0 , ω_(1) )$], that is, the initial angle is 0, and the initial angular velocity is exactly #math.equation(block: false, alt: "ω sub 1")[$ω_(1)$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Take the damped nonlinear pendulum equation #math.equation(block: false, alt: "θ double prime plus μ θ prime plus open parenthesis the fraction g over L close parenthesis sin θ equals 0")[$θ^(″) + μ θ^(′) + ( frac(g, L) ) sin θ = 0$] for some #math.equation(block: false, alt: "μ greater than 0")[$μ > 0$] (that is, there is friction). Suppose the friction is large, in particular #math.equation(block: false, alt: "μ squared greater than 4 open parenthesis the fraction g over L close parenthesis")[$μ^(2) > 4 ( frac(g, L) )$].


      + Find and classify the critical points.
+ Explain what your findings mean, and if it agrees with what you expect in reality.


      Answer


      + Critical points are #math.equation(block: false, alt: "ω equals 0")[$ω = 0$], #math.equation(block: false, alt: "θ equals k π")[$θ = k π$] for any integer #math.equation(block: false, alt: "k")[$k$]. When #math.equation(block: false, alt: "k")[$k$] is odd, we have a saddle point. When #math.equation(block: false, alt: "k")[$k$] is even we get a sink.
+ The findings mean the pendulum will simply go to one of the sinks, for example #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] and it will not swing back and forth. The friction is too high for it to oscillate, just like an overdamped mass-spring system.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose we have the system predator-prey system where the foxes are also killed at a constant rate #math.equation(block: false, alt: "h")[$h$] (#math.equation(block: false, alt: "h")[$h$] foxes killed per unit time): #math.equation(block: false, alt: "x prime equals open parenthesis a minus b y close parenthesis x ,")[$x^(′) = ( a − b y ) x ,$] #math.equation(block: false, alt: "y prime equals open parenthesis c x minus d close parenthesis y minus h")[$y^(′) = ( c x − d ) y − h$].


      + Find the critical points and the Jacobin matrices of the system.
+ Put in the constants #math.equation(block: false, alt: "a equals 0.4")[$a = 0.4$], #math.equation(block: false, alt: "b equals 0.01")[$b = 0.01$], #math.equation(block: false, alt: "c equals 0.003")[$c = 0.003$], #math.equation(block: false, alt: "d equals 0.3")[$d = 0.3$], #math.equation(block: false, alt: "h equals 10")[$h = 10$]. Analyze the critical points. What do you think it says about the forest?


      Answer


      + Solving for the critical points we get #math.equation(block: false, alt: "open parenthesis 0 , minus the fraction h over d close parenthesis")[$( 0 , − frac(h, d) )$] and #math.equation(block: false, alt: "open parenthesis the fraction b h plus a d over a c , the fraction a over b close parenthesis")[$( frac(b h + a d, a c) , frac(a, b) )$]. The Jacobian matrix at #math.equation(block: false, alt: "open parenthesis 0 , minus the fraction h over d close parenthesis")[$( 0 , − frac(h, d) )$] is #math.equation(block: false, alt: "a plus the fraction b h over d 0; minus the fraction c d over d minus d")[$a + frac(b h, d) 0 \ − frac(c d, d) − d$] whose eigenvalues are #math.equation(block: false, alt: "a plus the fraction b h over d")[$a + frac(b h, d)$] and #math.equation(block: false, alt: "minus d")[$− d$]. The eigenvalues are real of opposite signs and we get a saddle. (In the application, however, we are only looking at the positive quadrant so this critical point is irrelevant.) At #math.equation(block: false, alt: "open parenthesis the fraction b h plus a d over a c , the fraction a over b close parenthesis")[$( frac(b h + a d, a c) , frac(a, b) )$] we get Jacobian matrix #math.equation(block: false, alt: "0 minus the fraction b open parenthesis b h plus a d close parenthesis over a c; the fraction a c over b the fraction b h plus a d over a minus d")[$0 − frac(b ( b h + a d ), a c) \ frac(a c, b) frac(b h + a d, a) − d$].
+ For the specific numbers given, the second critical point is #math.equation(block: false, alt: "open parenthesis the fraction 550 over 3 , 40 close parenthesis")[$( frac(550, 3) , 40 )$] the matrix is #math.equation(block: false, alt: "0 minus the fraction 11 over 6; the fraction 3 over 25 the fraction 1 over 4")[$0 − frac(11, 6) \ frac(3, 25) frac(1, 4)$], which has eigenvalues #math.equation(block: false, alt: "the fraction 5 ± i the square root of 327 over 40")[$frac(5 ± i sqrt(327), 40)$]. Therefore there is a spiral source; the solution spirals outwards. The solution eventually hits one of the axes, #math.equation(block: false, alt: "x equals 0")[$x = 0$] or #math.equation(block: false, alt: "y equals 0")[$y = 0$], so something will die out in the forest.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[(challenging)]


      Suppose the foxes never die. That is, we have the system #math.equation(block: false, alt: "x prime equals open parenthesis a minus b y close parenthesis x ,")[$x^(′) = ( a − b y ) x ,$] #math.equation(block: false, alt: "y prime equals c x y")[$y^(′) = c x y$]. Find the critical points and notice they are not isolated. What will happen to the population in the forest if it starts at some positive numbers. Hint: Think of the constant of motion.


      Answer


      The critical points are on the line #math.equation(block: false, alt: "x equals 0")[$x = 0$]. In the positive quadrant the #math.equation(block: false, alt: "y prime")[$y^(′)$] is always positive and so the fox population always grows. The constant of motion is #math.equation(block: false, alt: "C equals y to the power a e to the power minus c x minus b y")[$C = y^(a) e^(− c x − b y)$], for any #math.equation(block: false, alt: "C")[$C$] this curve must hit the #math.equation(block: false, alt: "y")[$y$]-axis (why?), so the trajectory will simply approach a point on the #math.equation(block: false, alt: "y")[$y$] axis somewhere and the number of hares will go to zero.
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/8%3A_Nonlinear_Systems/8.4%3A_Limit_cycles")[8.4: Limit cycles]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Show that the following systems have no closed trajectories.


      + #math.equation(block: false, alt: "x prime equals x cubed plus y , y prime equals y cubed plus x squared")[$x^(′) = x^(3) + y , y^(′) = y^(3) + x^(2)$],
+ #math.equation(block: false, alt: "x prime equals e to the power x minus y , y prime equals e to the power x plus y")[$x^(′) = e^(x − y) , y^(′) = e^(x + y)$],
+ #math.equation(block: false, alt: "x prime equals x plus 3 y squared minus y cubed , y prime equals y cubed plus x squared")[$x^(′) = x + 3 y^(2) − y^(3) , y^(′) = y^(3) + x^(2)$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Formulate a condition for a 2-by-2 linear system #math.equation(block: false, alt: "x → prime equals A x →")[$arrow(x)^(′) = A arrow(x)$] to not be a center using the Bendixson-Dulac theorem. That is, the theorem says something about certain elements of #math.equation(block: false, alt: "A")[$A$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Explain why the Bendixson-Dulac Theorem does not apply for any conservative system #math.equation(block: false, alt: "x double prime plus h open parenthesis x close parenthesis equals 0")[$x^(″) + h ( x ) = 0$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
A system such as #math.equation(block: false, alt: "x prime equals x , y prime equals y")[$x^(′) = x , y^(′) = y$] has solutions that exist for all time #math.equation(block: false, alt: "t")[$t$], yet there are no closed trajectories or other limit cycles. Explain why the Poincare-Bendixson Theorem does not apply.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Differential equations can also be given in different coordinate systems. Suppose we have the system #math.equation(block: false, alt: "r prime equals 1 minus r squared")[$r^(′) = 1 − r^(2)$], #math.equation(block: false, alt: "θ prime equals 1")[$θ^(′) = 1$] given in polar coordinates. Find all the closed trajectories and check if they are limit cycles and if so, if they are asymptotically stable or not.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Show that the following systems have no closed trajectories.


      + #math.equation(block: false, alt: "x prime equals x plus y squared")[$x^(′) = x + y^(2)$], #math.equation(block: false, alt: "y prime equals y plus x squared")[$y^(′) = y + x^(2)$],
+ #math.equation(block: false, alt: "x prime equals minus x sin squared open parenthesis y close parenthesis")[$x^(′) = − x sin^(2) ( y )$], #math.equation(block: false, alt: "y prime equals e to the power x")[$y^(′) = e^(x)$],
+ #math.equation(block: false, alt: "x prime equals x y")[$x^(′) = x y$], #math.equation(block: false, alt: "y prime equals x plus x squared")[$y^(′) = x + x^(2)$].


      Answer


      Use Bendixson–Dulac Theorem.


      + #math.equation(block: false, alt: "f sub x plus g sub y equals 1 plus 1 greater than 0")[$f_(x) + g_(y) = 1 + 1 > 0$], so no closed trajectories.
+ #math.equation(block: false, alt: "f sub x plus g sub y equals minus sin squared open parenthesis y close parenthesis plus 0 less than 0")[$f_(x) + g_(y) = − sin^(2) ( y ) + 0 < 0$] for all #math.equation(block: false, alt: "x , y")[$x , y$] except the lines given by #math.equation(block: false, alt: "y equals k π")[$y = k π$] (where we get zero), so no closed trajectories.
+ #math.equation(block: false, alt: "f sub x plus g sub y equals y plus 0 greater than 0")[$f_(x) + g_(y) = y + 0 > 0$] for all #math.equation(block: false, alt: "x , y")[$x , y$] except the line given by #math.equation(block: false, alt: "y equals 0")[$y = 0$] (where we get zero), so no closed trajectories.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose an autonomous system in the plane has a solution #math.equation(block: false, alt: "x equals cos open parenthesis t close parenthesis plus e to the power minus t")[$x = cos ( t ) + e^(− t)$], #math.equation(block: false, alt: "y equals sin open parenthesis t close parenthesis plus e to the power minus t")[$y = sin ( t ) + e^(− t)$]. What can you say about the system (in particular about limit cycles and periodic solutions)?


      Answer


      Using Poincaré–Bendixson Theorem, the system has a limit cycle, which is the unit circle centered at the origin, as #math.equation(block: false, alt: "x equals cos open parenthesis t close parenthesis plus e to the power minus t")[$x = cos ( t ) + e^(− t)$], #math.equation(block: false, alt: "y equals sin open parenthesis t close parenthesis plus e to the power minus t")[$y = sin ( t ) + e^(− t)$] gets closer and closer to the unit circle. Thus #math.equation(block: false, alt: "x equals cos open parenthesis t close parenthesis")[$x = cos ( t )$], #math.equation(block: false, alt: "y equals sin open parenthesis t close parenthesis")[$y = sin ( t )$] is the periodic solution.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Show that the limit cycle of the Van der Pol oscillator (for #math.equation(block: false, alt: "μ greater than 0")[$μ > 0$]) must not lie completely in the set where #math.equation(block: false, alt: "minus the square root of the fraction 1 plus μ over μ less than x less than the square root of the fraction 1 plus μ over μ")[$− sqrt(frac(1 + μ, μ)) < x < sqrt(frac(1 + μ, μ))$]. Compare with Figure 8.4.1.


      Answer


      #math.equation(block: false, alt: "f open parenthesis x , y close parenthesis equals y")[$f ( x , y ) = y$], #math.equation(block: false, alt: "g open parenthesis x , y close parenthesis equals μ open parenthesis 1 minus x squared close parenthesis y minus x")[$g ( x , y ) = μ ( 1 − x^(2) ) y − x$]. So #math.equation(block: false, alt: "f sub x plus g sub y equals μ open parenthesis 1 minus x squared close parenthesis")[$f_(x) + g_(y) = μ ( 1 − x^(2) )$]. The Bendixson–Dulac Theorem says there is no closed trajectory lying entirely in the set #math.equation(block: false, alt: "x squared less than 1")[$x^(2) < 1$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose we have the system #math.equation(block: false, alt: "r prime equals sin open parenthesis r close parenthesis")[$r^(′) = sin ( r )$], #math.equation(block: false, alt: "θ prime equals 1")[$θ^(′) = 1$] given in polar coordinates. Find all the closed trajectories.


      Answer


      The closed trajectories are those where #math.equation(block: false, alt: "sin open parenthesis r close parenthesis equals 0")[$sin ( r ) = 0$], therefore, all the circles centered at the origin with radius that is a multiple of #math.equation(block: false, alt: "π")[$π$] are closed trajectories.
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/8%3A_Nonlinear_Systems/8.5%3A_Chaos")[8.5: Chaos]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
For the non-chaotic equation #math.equation(block: false, alt: "x double prime plus 2 p x prime plus ω 0 2 x equals the fraction F sub 0 over m cos open parenthesis ω t close parenthesis")[$x^(″) + 2 p x^(′) + ω_(0)^(2) x = frac(F_(0), m) cos ( ω t )$], suppose we strobe with frequency #math.equation(block: false, alt: "ω")[$ω$] as we mentioned above. Use the known steady periodic solution to find precisely the point which is the attractor for the Poincar#math.equation(block: false, alt: "e `")[$limits(e)^(`)$] section.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[(project)]


      A simple fractal attractor can be drawn via the following chaos game. Draw three points of a triangle (just the vertices) and number them, say #math.equation(block: false, alt: "p sub 1")[$p_(1)$], #math.equation(block: false, alt: "p sub 2")[$p_(2)$] and #math.equation(block: false, alt: "p sub 3")[$p_(3)$]. Start with some random point #math.equation(block: false, alt: "p")[$p$] (does not have to be one of the three points above) and draw it. Roll a die, and use it to pick of the #math.equation(block: false, alt: "p sub 1")[$p_(1)$], #math.equation(block: false, alt: "p sub 2")[$p_(2)$], or #math.equation(block: false, alt: "p sub 3")[$p_(3)$] randomly (for example 1 and 4 mean #math.equation(block: false, alt: "p sub 1")[$p_(1)$], 2 and 5 mean #math.equation(block: false, alt: "p sub 2")[$p_(2)$], and 3 and 6 mean #math.equation(block: false, alt: "p sub 3")[$p_(3)$]). Suppose we picked #math.equation(block: false, alt: "p sub 2")[$p_(2)$], then let #math.equation(block: false, alt: "p sub new")[$p_("new")$] be the point exactly halfway between #math.equation(block: false, alt: "p")[$p$] and #math.equation(block: false, alt: "p sub 2")[$p_(2)$]. Draw this point and let #math.equation(block: false, alt: "p")[$p$] now refer to this new point #math.equation(block: false, alt: "p sub new")[$p_("new")$]. Rinse, repeat. Try to be precise and draw as many iterations as possible. Your points should be attracted to the so-called #emph[Sierpinski triangle]. A computer was used to run the game for 10,000 iterations to obtain the picture in Figure #math.equation(block: false, alt: "1")[$1$].


      #figure(figph[10,000 iterations of the chaos game producing the Sierpinski triangle.  Fractal graph], alt: "10,000 iterations of the chaos game producing the Sierpinski triangle. Fractal graph", caption: none)


      Figure #math.equation(block: false, alt: "1")[$1$]: #math.equation(block: false, alt: "10 , 000")[$10 , 000$] iterations of the chaos game producing the Sierpinski triangle.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[(project)]


      Construct the double pendulum described in the text with a string and two nuts (or heavy beads). Play around with the position of the middle nut, and perhaps use different weight nuts. Describe what you find.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[(computer project)]


      Use a computer software (such as Matlab, Octave, or perhaps even a spreadsheet), plot the solution of the given forced Duffing equation with Euler's method. Plotting the solution for #math.equation(block: false, alt: "t")[$t$] from #math.equation(block: false, alt: "0")[$0$] to #math.equation(block: false, alt: "100")[$100$] with several different (small) step sizes. Discuss.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find critical points of the Lorenz system and the associated linearizations.


      Answer


      Critical points: #math.equation(block: false, alt: "open parenthesis 0 , 0 , 0 close parenthesis")[$( 0 , 0 , 0 )$], #math.equation(block: false, alt: "open parenthesis 3 the square root of 8 , 3 the square root of 8 , 27 close parenthesis")[$( 3 sqrt(8) , 3 sqrt(8) , 27 )$], #math.equation(block: false, alt: "open parenthesis minus 3 the square root of 8 , minus 3 the square root of 8 , 27 close parenthesis")[$( − 3 sqrt(8) , − 3 sqrt(8) , 27 )$]. Linearization at #math.equation(block: false, alt: "open parenthesis 0 , 0 , 0 close parenthesis")[$( 0 , 0 , 0 )$] using #math.equation(block: false, alt: "u equals x")[$u = x$], #math.equation(block: false, alt: "v equals y")[$v = y$], #math.equation(block: false, alt: "w equals z")[$w = z$] is #math.equation(block: false, alt: "u prime equals minus 10 u plus 10 v")[$u^(′) = − 10 u + 10 v$], #math.equation(block: false, alt: "v prime equals 28 u minus v")[$v^(′) = 28 u − v$], #math.equation(block: false, alt: "w prime equals minus open parenthesis the fraction 8 over 3 close parenthesis w")[$w^(′) = − ( frac(8, 3) ) w$]. Linearization at #math.equation(block: false, alt: "open parenthesis 3 the square root of 8 , 3 the square root of 8 , 27 close parenthesis")[$( 3 sqrt(8) , 3 sqrt(8) , 27 )$] using #math.equation(block: false, alt: "u equals x minus 3 the square root of 8")[$u = x − 3 sqrt(8)$], #math.equation(block: false, alt: "v equals y minus 3 the square root of 8")[$v = y − 3 sqrt(8)$], #math.equation(block: false, alt: "w equals z minus 27")[$w = z − 27$] is #math.equation(block: false, alt: "u prime equals minus 10 u plus 10 v")[$u^(′) = − 10 u + 10 v$], #math.equation(block: false, alt: "v prime equals u minus v minus 3 the square root of 8 w")[$v^(′) = u − v − 3 sqrt(8) w$], #math.equation(block: false, alt: "w prime equals 3 the square root of 8 u plus 3 the square root of 8 v minus open parenthesis the fraction 8 over 3 close parenthesis w")[$w^(′) = 3 sqrt(8) u + 3 sqrt(8) v − ( frac(8, 3) ) w$]. Linearization at #math.equation(block: false, alt: "open parenthesis minus 3 the square root of 8 , minus 3 the square root of 8 , 27 close parenthesis")[$( − 3 sqrt(8) , − 3 sqrt(8) , 27 )$] using #math.equation(block: false, alt: "u equals x plus 3 the square root of 8")[$u = x + 3 sqrt(8)$], #math.equation(block: false, alt: "v equals y plus 3 the square root of 8")[$v = y + 3 sqrt(8)$], #math.equation(block: false, alt: "w equals z minus 27")[$w = z − 27$] is #math.equation(block: false, alt: "u prime equals minus 10 u plus 10 v")[$u^(′) = − 10 u + 10 v$], #math.equation(block: false, alt: "v prime equals u minus v plus 3 the square root of 8 w")[$v^(′) = u − v + 3 sqrt(8) w$], #math.equation(block: false, alt: "w prime equals minus 3 the square root of 8 u minus 3 the square root of 8 v minus open parenthesis the fraction 8 over 3 close parenthesis w")[$w^(′) = − 3 sqrt(8) u − 3 sqrt(8) v − ( frac(8, 3) ) w$].
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