Initialization Cells (Code Needed for Most Plot3D Statements)

ticks[min_,max_]:=Join[
Table[{i,If[i==0, ,Style[i,12]],{0.01,0}},{i,Ceiling[min],Floor[max], 2}],(* Numbered ticks *)
Table[{j, ,{0.01,0}},{j,Round[min],Round[max-1],1}]]         (* Un-numbered ticks *)


(* 2D Options *)
SetOptions[
Plot,
ImageSize->350,
AxesOrigin->{0,0},
AspectRatio->1,
AxesStyle->Directive[Medium, Bold],
Ticks->ticks
];

SetOptions[
ParametricPlot,
ImageSize->350,
AxesOrigin->{0,0},
AspectRatio->1,
AxesStyle->Directive[Medium, Bold],
Ticks->ticks
];

SetOptions[
VectorPlot,
ImageSize->350,
AxesOrigin->{0,0},
AspectRatio->1,
Axes->True,
Frame->False,
AxesStyle->Directive[Medium, Bold],
Ticks->ticks
];

(* 3D Options *)

SetOptions[
Plot3D,
ImageSize->600,
BoxRatios->1,Boxed->False,AxesOrigin->{0,0,0},
AxesStyle->Directive[Medium, Bold],
Ticks->ticks, 
Mesh->None,
ViewPoint->{2,0.9,1},
ViewVertical->{0,0,1}
];

SetOptions[
ContourPlot3D,
ImageSize->600,
BoxRatios->1,Boxed->False,AxesOrigin->{0,0,0},
AxesStyle->Directive[Medium, Bold],
Ticks->ticks, 
Mesh->None,
ViewPoint->{2,0.9,1},
ViewVertical->{0,0,1}
];

SetOptions[
ParametricPlot3D,
ImageSize->600,
BoxRatios->1,Boxed->False,AxesOrigin->{0,0,0},
AxesStyle->Directive[Medium, Bold],
Ticks->ticks, 
Mesh->None,
ViewPoint->{2,0.9,1},
ViewVertical->{0,0,1}
];

SetOptions[
RegionPlot3D,
ImageSize->600,
BoxRatios->1,Boxed->False,AxesOrigin->{0,0,0},
AxesStyle->Directive[Medium, Bold],
Ticks->ticks, 
Mesh->None,
ViewPoint->{2,0.9,1},
ViewVertical->{0,0,1}
];

SetOptions[
VectorPlot3D,
ImageSize->350,
AxesOrigin->{0,0,0},
AspectRatio->1,
BoxRatios->1,
Axes->True,
AxesStyle->Directive[Medium, Bold],
Ticks->ticks,
ViewPoint->{2,0.9,1},
ViewVertical->{0,0,1}
];

SetOptions[
RevolutionPlot3D,
ImageSize->350,
AxesOrigin->{0,0,0},
AspectRatio->1,
BoxRatios->1,
Boxed->False,
Axes->True,
AxesStyle->Directive[Medium, Bold],
Ticks->ticks,
ViewPoint->{2,0.9,1},
ViewVertical->{0,0,1}
];

First Order Differential Equations

Linear Systems  - Finding Solutions

Eigenvalues and Eigenvectors

Defined

Let A be a n ×n matrix. Then a number λ is said to be an eigenvalue if there exists a non-zero vector V such that

({{({{A V=λ V.}})}})

The characteristic equation for a n ×n matrix is defined to be det(A-λ I)=0 where I is the n ×n identity matrix.   But why do we care????

Example

Find the eigenvalues and associated eigenvectors of the matrix A=(({{13, -4}, {-4, 7}})).

A Big Theorem - Real and Distinct Eigenvalues

Suppose the n ×n matrix A has distinct, real eigenvalues (Subscript[λ, 1]), ..., (Subscript[λ, n]) with associated eigenvectors (Subscript[V, 1]),...,(Subscript[V, n]), then the functions

({{({{(Subscript[Y, 1])(t)=(e^((Subscript[λ, 1])t))(Subscript[V, 1]),  (Subscript[Y, 2])(t)=(e^((Subscript[λ, 2])t))(Subscript[V, 2]),⋯,(Subscript[Y, n])(t)=(e^((Subscript[λ, n])t))(Subscript[V, n])}})}})

are all linearly independent and the linear system (dY/dt)=A Y has general solution given by

({{({{Y(t)=(Subscript[C, 1])(e^((Subscript[λ, 1])t))(Subscript[V, 1])+(Subscript[C, 2])(e^((Subscript[λ, 2])t))(Subscript[V, 2])+⋯+(Subscript[C, n])(e^((Subscript[λ, n])t))(Subscript[V, n]).}})}})

Example - The Harmonic Oscillator

Consider a mass attached to a spring.  The mass is m=1 and the spring constant is k=10 and damping coefficient b=7.

Facts:   F=m a=m (((d^2)y)/(dt^2)) (Mass times acceleration)

Spring Constants:  (Subscript[F, s])=-k y (The force is proportional to the position - Hooke’s Law) where k>0.

Damping Constants: (Subscript[F, d])=-b (dy/dt)  (The force is proportional to the velocity) where 	b>0.

Forces must balance:  F=(Subscript[F, s])+(Subscript[F, d]).  (Newton’s Second Law)

Show that a second order DE for this situation is

({{({{(((d^2)y)/(dt^2))+7 (dy/dt)+10 y=0.}})}})

Find the corresponding system where Y(t)=(({{y(t)}, {v(t)}})).

Solve the system.

Create a direction field.  For our sanity, graph y = t(({{(Subscript[v, 1])}, {(Subscript[v, 2])}}))  instead of y = (Subscript[c, 1])(e^((Subscript[λ, 1]) t))(({{(Subscript[v, 1])}, {(Subscript[v, 2])}})) and y = -(Subscript[c, 1])(e^((Subscript[λ, 1]) t))(({{(Subscript[v, 1])}, {(Subscript[v, 2])}})).

ivp={  {-1,-1}, {1,1}, {2,2}, {-2, -2}, {1, -2.5},{-1,2.5}};

Show[
VectorPlot[
{0+v, -10 y-7 v}, {y, -3,3}, {v, -3.5,3.5},
FrameLabel->{y,v},
Axes->True,
VectorScale->{Small, Tiny, None}, VectorStyle->Gray,
StreamScale->Full,
StreamStyle->{Darker[Green], Thick, "Line"},
StreamPoints->{ivp},
PlotLabel->"Direction Field",
Epilog->{PointSize[Large], Point[ivp]}
],
ParametricPlot[
{t, -5 t}, {t, -3,3},
PlotRange->{{-3,3},{-3,3}},
PlotStyle->{Blue, Thick,Dashed}
],
ParametricPlot[
{t, -2 t}, {t, -3,3},
PlotRange->{{-3,3},{-3,3}},
PlotStyle->{Red, Thick,Dashed}
],
Graphics[{Thick,
Arrow[{{0,0},{1/5,-1}}]
}],
Graphics[{Thick,
Arrow[{{0,0},{1,-2}}]
}]
]

For Y(t)=(({{y(t)}, {v(t)}})), plot y(t) and v(t).

Plot[{
Exp[-5 t]+Exp[-2 t], -5 Exp[-5 t]-2 Exp[-2 t]
}, {t, 0, 3},
PlotStyle->{{Blue, Thick}, {Red, Thick}},
AxesLabel->{"t", "y, v"},
PlotLabel->"y(t) Blue, v(t) Red"
]

Sinks, Sources, and Saddles

Given a linear system with two real eigenvalues (Subscript[λ, 1]), (Subscript[λ, 2]). Then,

If (Subscript[λ, 1]), (Subscript[λ, 2])<0, then all solutions tend toward the origin.  In this case, the origin is called a sink.

If (Subscript[λ, 1]), (Subscript[λ, 2])>0,  then the solutions move away from the origin.  In this case, the origin is called a source.

If (Subscript[λ, 1])>0 and (Subscript[λ, 2])<0, then the origin is called a saddle.

ivp={ {2,1}, {-2, -1}, {1, -2.5},{-1,2.5}, {-1,1}, {1,-1}, {-2.5,-1/2},{2.5,0.5}};

GraphicsGrid[
{
{Show[
VectorPlot[
{2 x1+2x2, 1x1+3x2}, {x1, -3,3}, {x2, -3.5,3.5},
FrameLabel->{"(Subscript[x, 1])", "(Subscript[x, 2])"},
Axes->True,
VectorScale->{Small, Tiny, None}, VectorStyle->Gray,
StreamScale->Full,
StreamStyle->{Darker[Green], Thick, "Line"},
StreamPoints->{ivp},
PlotLabel->"Source",
Epilog->{PointSize[Large], Point[ivp]}
],
ParametricPlot[
{t,t}, {t, -3,3},
PlotRange->{{-3,3},{-3,3}},
PlotStyle->{Blue, Thick,Dashed}
],
ParametricPlot[
{-2 t, t}, {t, -3,3},
PlotRange->{{-3,3},{-3,3}},
PlotStyle->{Red, Thick,Dashed}
]
],
Show[
VectorPlot[
{-2 x1-2x2, -1x1-3x2}, {x1, -3,3}, {x2, -3.5,3.5},
FrameLabel->{"(Subscript[x, 1])", "(Subscript[x, 2])"},
Axes->True,
VectorScale->{Small, Tiny, None}, VectorStyle->Gray,
StreamScale->Full,
StreamStyle->{Darker[Green], Thick, "Line"},
StreamPoints->{ivp},
PlotLabel->"Sink",
Epilog->{PointSize[Large], Point[ivp]}
],
ParametricPlot[
{t, t}, {t, -3,3},
PlotRange->{{-3,3},{-3,3}},
PlotStyle->{Blue, Thick,Dashed}
],
ParametricPlot[
{-2 t, t}, {t, -3,3},
PlotRange->{{-3,3},{-3,3}},
PlotStyle->{Red, Thick,Dashed}
]
]
},
{
Show[
VectorPlot[
{8 x1-11x2,6x1-9 x2}, {x1, -3,3}, {x2, -3.5,3.5},
FrameLabel->{"(Subscript[x, 1])", "(Subscript[x, 2])"},
Axes->True,
VectorScale->{Small, Tiny, None}, VectorStyle->Gray,
StreamScale->Full,
StreamStyle->{Darker[Green], Thick, "Line"},
StreamPoints->{ivp},
PlotLabel->"Saddle",
Epilog->{PointSize[Large], Point[ivp]}
],
ParametricPlot[
{t, t}, {t, -3,3},
PlotRange->{{-3,3},{-3,3}},
PlotStyle->{Blue, Thick,Dashed}
],
ParametricPlot[
{11 t,6  t}, {t, -3,3},
PlotRange->{{-3,3},{-3,3}},
PlotStyle->{Red, Thick,Dashed}
]
]
}
},
ImageSize->550]

Homework - For Software, Use Mathematica.  You need to learn some programming, so use the code I have provided!  In particular, stay away from DSolve and NDSolve.

Pages:		277 - 280
Problems:	1, 2, 4, 5, 6, 16, 17, 20

Linear Systems  - Finding (Complex) Solutions

A Big Theorem - Complex and Distinct Eigenvalues

Suppose the n ×n matrix A has distinct, complex eigenvalues (Subscript[λ, 1]), ..., (Subscript[λ, n]) with associated eigenvectors (Subscript[V, 1]),...,(Subscript[V, n]), then the functions

({{({{(Subscript[Y, 1])(t)=(e^((Subscript[λ, 1])t))(Subscript[V, 1]),  (Subscript[Y, 2])(t)=(e^((Subscript[λ, 2])t))(Subscript[V, 2]),⋯,(Subscript[Y, n])(t)=(e^((Subscript[λ, n])t))(Subscript[V, n])}})}})

are all linearly independent and the linear system (dY/dt)=A Y has general solution given by

({{({{Y(t)=(Subscript[C, 1])(e^((Subscript[λ, 1])t))(Subscript[V, 1])+(Subscript[C, 2])(e^((Subscript[λ, 2])t))(Subscript[V, 2])+⋯+(Subscript[C, n])(e^((Subscript[λ, n])t))(Subscript[V, n]).}})}})

Note that n=2 k for this case.

Slight Irritation:

The solution obtained by the “Big Theorem” involves complex numbers.  Here is an interesting way to “fix” this.

Let (dY/dt)=A Y be a differential equation with constant, real matrix A.

Suppose that λ=α+β i is an eigenvalue with associated eigenvector V=a+b i=(({{(Subscript[a, 1])}, {⋱
(Subscript[a, n])}}))+(({{(Subscript[b, 1])}, {⋱
(Subscript[b, n])}})) i of the matrix A.  Then, a solution to the above differential equation is

({{({{(Subscript[Y, 1])(t)=(e^(λ t)) V}}).}})

Also, it should be clear that (Overscript[λ, —])=α-β i is also an eigenvalue with associated eigenvector (Overscript[V, —])=a-b i=(({{(Subscript[a, 1])}, {⋱
(Subscript[a, n])}}))-(({{(Subscript[b, 1])}, {⋱
(Subscript[b, n])}})) i.  Hence, we have another solution (can you verify this?)

({{({{(Subscript[Y, 2])(t)=(e^((Overscript[λ, —]) t)) (Overscript[V, —])}}).}})

Euler’s Formula:

({{({{(e^(α+β i))=(e^α) [cos(β) + i sin(β)] ⟹ }})(e^((α+β i) t))=(e^(α t)) [cos(β t) + i sin(β t)].}})

Note that we now can write

({{({{(Subscript[Y, 1])(t)=(e^(λ t))V}, {=(e^(α t)) [cos(β t) + i sin(β t)](a+b i)}, {=(e^(α t)) [a cos(β t)-b sin(β t)]+i (e^(α t)) [b cos(β t)+a sin(β t)]}, {(Subscript[Y, 2])(t)=(e^((Overscript[λ, —]) t)) (Overscript[V, —])}, {=(e^(α t)) [cos(-β t) + i sin(-β t)](a-b i)}, {=(e^(α t)) [cos(β t) - i sin(β t)](a-b i)}, {=(e^(α t)) [a cos(β t)-b sin(β t)]+i (e^(α t)) [-b cos(β t)-a sin(β t)]}, {=(e^(α t)) [a cos(β t)-b sin(β t)]-i (e^(α t)) [b cos(β t)+a sin(β t)].}})}})

Very Important:

({{({{({{Real((Subscript[Y, 1])(t))=}})(e^(α t)) [a cos(β t)-b sin(β t)]=Real((Subscript[Y, 2])(t))}, {({{Imaginary((Subscript[Y, 1])(t))=}})(e^(α t)) [b cos(β t)+a sin(β t)]=-Imaginary((Subscript[Y, 2])(t)).}})}})

Hence, up to a constant, the complex valued solutions (Subscript[Y, 1]) and (Subscript[Y, 2]) generate the SAME real valued functions, namely

({{({{(e^(α t)) [a cos(β t)-b sin(β t)]   and  }})(e^(α t)) [b cos(β t)+a sin(β t)].}})

Note that these two functions are independent.  Also, each function is a solution to the system differential equations!

So, we only need to do HALF the work...for a given eigenvalue λ, find the corresponding eigenvector V and ignore the complex conjugate of λ.  (Remember, λ and (Overscript[λ, —]) are distinct eigenvalues, so by the theorem we need two solutions...the real and imaginary functions listed above are those solutions!)

The “Real”  Complex Version of the Theorem:

Suppose the n ×n matrix A has 2 k complex eigenvalues (Subscript[λ, j])=(Subscript[α, j])±(Subscript[β, j]) i where j=1,2,...,k.  Further, suppose that for each complex eigenvalue (Subscript[λ, j])=(Subscript[α, j])+(Subscript[β, j]) i there is an associated eigenvector (Subscript[V, j])=(Subscript[a, j])+(Subscript[b, j]) i, for j=1,2,...,k.  Then,

({{({{(e^((Subscript[α, j]) t)) [(Subscript[a, j]) cos((Subscript[β, j]) t)-(Subscript[b, j]) sin((Subscript[β, j]) t)]   and  }})(e^((Subscript[α, j]) t)) [(Subscript[b, j]) cos((Subscript[β, j]) t)+(Subscript[a, j]) sin((Subscript[β, j]) t)].}})

are all linearly independent.

Furthermore, the function

({{({{Y(t)=(∑_(j=1)^k)(Subscript[c, j])((e^((Subscript[α, j]) t)) [(Subscript[a, j]) cos((Subscript[β, j]) t)-(Subscript[b, j]) sin((Subscript[β, j]) t)] )+(Subscript[d, j])((e^((Subscript[α, j]) t)) [(Subscript[b, j]) cos((Subscript[β, j]) t)+(Subscript[a, j]) sin((Subscript[β, j]) t)]).}})}})

is a solution to the linear system (dY/dt)=A Y.

Example

Consider the system

({{({{((Subscript[dx, 1])/dt)=4 (Subscript[x, 1])-3 (Subscript[x, 2])}, {((Subscript[dx, 2])/dt)=3 (Subscript[x, 1])+4 (Subscript[x, 2.])}})}})

Solve the system.

Create a direction field.

ivp={  {-1,-1}, {1,1}, {2,2}, {-2, -2}, {1, -2.5},{-1,2.5}};

VectorPlot[
{4 x1-3x2, 3x1+4x2}, {x1, -3,3}, {x2, -3.5,3.5},
FrameLabel->{"(Subscript[x, 1])", "(Subscript[x, 2])"},
Axes->True,
VectorScale->{Small, Tiny, None}, VectorStyle->Gray,
StreamScale->Full,
StreamStyle->{Darker[Green], Thick, "Line"},
StreamPoints->{ivp},
PlotLabel->"Direction Field",
Epilog->{PointSize[Large], Point[ivp]}
]

For Y(t)=(({{(Subscript[x, 1])(t)}, {(Subscript[x, 2])(t)}})), plot (Subscript[x, 1])(t) and (Subscript[x, 2])(t) if Y(0)=(({{1}, {1}})).

DSolve[{x1'[t]==4 x1[t]-3x2[t], x2'[t]==3x1[t]+4x2[t], x1[0]==1, x2[0]==1}, {x1[t],x2[t]},t]

Plot[
{
Exp[4 t] (Cos[3 t]-Sin[3 t]), Exp[4 t] (Cos[3 t]+Sin[3 t])
}, {t, 0,1.75},Ticks->Automatic,PlotRange->{-560,500},
PlotStyle->{{Blue, Thick}, {Red, Thick}},
AxesLabel->{"t", "(Subscript[x, 1]), (Subscript[x, 2])"},
PlotLabel->"(Subscript[x, 1])(t) Blue, (Subscript[x, 2])(t) Red"
]

Sinks, Sources, and Centers

Given a linear system with complex eigenvalues λ=α±β i, where β>0, the soluton curves spiral around the origin in the phase plane with a natural period of 2 π/β and natural frequency of β/(2 π).  Moreover,

If α<0, then the solutions spiral toward the origin.  In this case, the origin is called a (spiral) sink.

If α>0, then the solutions spiral away the origin.  In this case, the origin is called a (spiral) source.

If α=0, then the solutions are periodic.  They repeat the same closed curve.  In this case, the origin is called a center.

ivp={  {-1,-1}, {1,1}, {2,2}, {-2, -2}, {1, -2.5},{-1,2.5}};

GraphicsGrid[
{
{VectorPlot[
{4 x1-3x2, 3x1+4x2}, {x1, -3,3}, {x2, -3.5,3.5},
FrameLabel->{"(Subscript[x, 1])", "(Subscript[x, 2])"},
Axes->True,
VectorScale->{Small, Tiny, None}, VectorStyle->Gray,
StreamScale->Full,
StreamStyle->{Darker[Green], Thick, "Line"},
StreamPoints->{ivp},
PlotLabel->"Source",
Epilog->{PointSize[Large], Point[ivp]}
],
VectorPlot[
{-2 x1-3x2, 3x1-2x2}, {x1, -3,3}, {x2, -3.5,3.5},
FrameLabel->{"(Subscript[x, 1])", "(Subscript[x, 2])"},
Axes->True,
VectorScale->{Small, Tiny, None}, VectorStyle->Gray,
StreamScale->Full,
StreamStyle->{Darker[Green], Thick, "Line"},
StreamPoints->{ivp},
PlotLabel->"Sink",
Epilog->{PointSize[Large], Point[ivp]}
]
},
{
VectorPlot[
{1x2,-2x1}, {x1, -3,3}, {x2, -3.5,3.5},
FrameLabel->{"(Subscript[x, 1])", "(Subscript[x, 2])"},
Axes->True,
VectorScale->{Small, Tiny, None}, VectorStyle->Gray,
StreamScale->Full,
StreamStyle->{Darker[Green], Thick, "Line"},
StreamPoints->{ivp},
PlotLabel->"Center",
Epilog->{PointSize[Large], Point[ivp]}
],
}
},
ImageSize->600]

Homework - For Software, Use Mathematica.  You need to learn some programming, so use the code I have provided!  In particular, stay away from DSolve and NDSolve.

Pages:		310 - 311
Problems:	1, 3, 5, 7, 9, 11, 13, 16, 18, 21, 23

Linear Systems  - Repeated Eigenvalues; Generalized Eigenvectors

Two Little Examples

Find the eigenvectors

A=(({{2, 0}, {0, 2}}))

DSolve[{x1'[t]==2 x1[t], x2'[t]==2 x2[t]}, {x1[t],x2[t]},t]

B=(({{2, -1}, {0, 2}}))

DSolve[{x1'[t]==2 x1[t]-x2[t], x2'[t]==2 x2[t]}, {x1[t],x2[t]},t]

Some Definitions and Notes

The algebraic multiplicity (A.M.) of an eigenvalue λ is the number of times it appears as a root of the characteristic equation det(A=λ I)=0.

The geometric multiplicity (G.M.) of an eigenvalue λ is the number of linearly independent eigenvetors corresponding to λ (i.e. the dimension of the eigenspace, for the Linear Algebra students).

Note that 1 ≤ G.M. ≤ A.M.  It is most interesting when G.M. < A. M.

If λ is an eigenvalue for a matrix A, then v≠ 0 is a generalized eigenvector if (((A-λ I))^d) v=0 for some positive integer d.

Little odd note:  If (A-λI)(Subscript[V, 1])=({{0}}) (that is, we have an eigenvector), then

({{({{(A-λI)(Subscript[V, 2])=({{(Subscript[V, 1])}})⟹(((A-λI))^2)(Subscript[V, 2])=(A-λI)(Subscript[V, 1])=0}})}})

which means we just found a generalize eigenvector! (assuming, of course, that (Subscript[V, 2]) is non-zero.)

A Tiny Theorem - Repeated Eigenvalues, not Enough Eigenvectors

Suppose the 2 ×2 matrix A has a repeated eigenvalue, λ, of multiplicity 2 with a single eigenvector (Subscript[V, 1]).  Then a solution to the linear system (dY/dt)=A Y is

({{({{Y(t)=(e^(λ t))(Subscript[V, 2])}})+t (e^(λ t))(Subscript[V, 1]),}})

where

({{(A-λI)(Subscript[V, 2])=({{(Subscript[V, 1])}}).}})

Moreover, the general solution is given by

({{({{({{({{Y(t)=(Subscript[c, 1 ])(e^(λ t))(Subscript[V, 1])+(Subscript[c, 2])(e^(λ t))((Subscript[V, 2])}})+t (Subscript[V, 1])).}})}})}})

Example

Solve the system (dY/dt)=B Y where B=(({{2, -1}, {0, 2}})).

The Big Theorem - Repeated Eigenvalus, not Enough Eigenvectors

Suppose the n ×n matrix A has a repeated eigenvalue, λ, of multiplicity k with an eigenvector (Subscript[V, 1]).  Then a solution to the linear system (dY/dt)=A Y is found by first computing generalized eigenvectors (Subscript[V, 2]), (Subscript[V, 3]), ..., (Subscript[V, k]) where

({{({{(A-λI)(Subscript[V, 1])=0  (OK, this is the eigenvector we already know)}, {(A-λI)(Subscript[V, 2])=(Subscript[V, 1]) }, {⋮}, {(A-λI)(Subscript[V, k])=({{(Subscript[V, k-1])}}).}})}})

Next, k linearly independent solutions (possibly part of a larger one) are found by

({{({{({{(Subscript[Y, 1])(t)= (Subscript[V, 1]) (e^(λ t))}, {({{(Subscript[Y, 2])(t)= ((Subscript[V, 2])+(Subscript[V, 1]) t) (e^(λ t))}, {⋮}, {({{(Subscript[Y, k])(t)= ((Subscript[V, k])+(Subscript[V, k-1]) t+(Subscript[V, k-2]) ((t^2)/2)+⋯+(Subscript[V, 1]) ((t^(k-1))/((k-1)!))) (e^(λ t)).}})}})}})}})}})

Thus, a solution (possibly part of a larger solution), based upon λ, is given by

({{({{(Subscript[Y, λ])(t)=(Subscript[C, 1 ]) (Subscript[Y, 1])(t)+(Subscript[C, 2])(Subscript[Y, 2])(t)+⋯+(Subscript[C, k])(Subscript[Y, k])(t).}})}})

Example

Consider the system (dY/dt)=A Y where A=(({{-1, 2, 1}, {0, -1, 0}, {-1, -3, -3}})).

Note that the characteristic equation is -(λ+1)(((λ+2))^2)=0.

An eigenvector for λ=-1 is (({{1}, {1}, {-2}})).

Find the general solution.

A Zero Eigenvalue

Consider the system (dY/dt)=(({{-2, 1}, {2, -1}}))Y.  The eigenvalues and eigenvectors are

Eigensystem[
{
{-2,1},
{2,-1}
}
]

Hence, the general solution is given by

({{({{Y(t)=(Subscript[c, 1]) (e^(-3 t)) (({{-1}, {1}}))}})+(Subscript[c, 2]) (({{1}, {2}})).}})

Since one eigenvalue is less than zero, then Y(t)⟶(Subscript[c, 2]) (({{1}, {2}})) as t->∞.  So all solutions move towards the constant vector (Subscript[c, 2]) (({{1}, {2}})).  (Of course, if the eigenvalue had been positive, then all solutions would move away from the constant vector).

ivp={ {2,1}, {-2, -1}, {1, -2.5},{-1,2.5}};

GraphicsGrid[
{
{Show[
VectorPlot[
{-2 x1+x2, 2x1-1x2}, {x1, -3,3}, {x2, -3.5,3.5},
FrameLabel->{"(Subscript[x, 1])", "(Subscript[x, 2])"},
Axes->True,
VectorScale->{Small, Tiny, None}, VectorStyle->Gray,
StreamScale->Full,
StreamStyle->{Darker[Green], Thick, "Line"},
StreamPoints->{ivp},
PlotLabel->"Zero Eigenvalue - Towards",
Epilog->{PointSize[Large], Point[ivp]}
],
ParametricPlot[
{1 t, 2 t}, {t, -3,3},
PlotRange->{{-3,3},{-3,3}},
PlotStyle->{Blue, Thick,Dashed}
],
ParametricPlot[
{- Exp[-3 t], Exp[-3 t]}, {t, -3,3},
PlotRange->{{-3,3},{-3,3}},
PlotStyle->{Red, Thick,Dashed}
],
ParametricPlot[
{ Exp[-3 t],- Exp[-3 t]}, {t, -3,3},
PlotRange->{{-3,3},{-3,3}},
PlotStyle->{Red, Thick,Dashed}
]
],
Show[
VectorPlot[
{2 x1-x2,- 2x1+1x2}, {x1, -3,3}, {x2, -3.5,3.5},
FrameLabel->{"(Subscript[x, 1])", "(Subscript[x, 2])"},
Axes->True,
VectorScale->{Small, Tiny, None}, VectorStyle->Gray,
StreamScale->Full,
StreamStyle->{Darker[Green], Thick, "Line"},
StreamPoints->{ivp},
PlotLabel->"Zero Eigenvalue - Away",
Epilog->{PointSize[Large], Point[ivp]}
],
ParametricPlot[
{1 t, 2 t}, {t, -3,3},
PlotRange->{{-3,3},{-3,3}},
PlotStyle->{Blue, Thick,Dashed}
],
ParametricPlot[
{- Exp[3 t], Exp[3 t]}, {t, -3,3},
PlotRange->{{-3,3},{-3,3}},
PlotStyle->{Red, Thick,Dashed}
],
ParametricPlot[
{ Exp[3 t],- Exp[3 t]}, {t, -3,3},
PlotRange->{{-3,3},{-3,3}},
PlotStyle->{Red, Thick,Dashed}
]
]
}
},
ImageSize->600]

Homework - For Software, Use Mathematica.  You need to learn some programming, so use the code I have provided!  In particular, stay away from DSolve and NDSolve.

Pages:		327 - 330
Problems:	5, 6, 7, 8, 9, 10, 11, 12, 18