Qualitative Behavior: Phase Portraits

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[2020-02-13 Thu 17:57]

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Qualitative Behavior: Phase Portraits | Unit IV: First-order Systems | Differential Equations | Mathematics | MIT OpenCourseWare

1 What is the phase plane for this system?

1.1 Front

What is the phase plane for this system?

\({\displaystyle \dot{\vb{x}} = A \vb{x}}\), where \({\displaystyle A = \begin{pmatrix}a & b \\ c & d\end{pmatrix}}\)

1.2 Back

It’s the \(xy\text{-plane}\) itself, where you can draw the trajectory of a solution with an arrow to indicate the direction of increasing time.

2 What is a phase portrait?

2.1 Front

What is a phase portrait?

2.2 Back

It is the graphs of enough trajectories in the phase plane to give a good sense of all solutions to the system

3 What is a critical point for this system?

3.1 Front

What is a critical point for this system?

\({\displaystyle \dot{\vb{u}} = A \vb{u}}\), where \({\displaystyle A = \begin{pmatrix}a & b \\ c & d\end{pmatrix}}\)

3.2 Back

The point \((x_0, y_0)\) is a critical point of the system if

\({\displaystyle \begin{pmatrix}\dot{x} \\ \dot{y}\end{pmatrix} = A \begin{pmatrix}x_0 \\ y_0\end{pmatrix} = \begin{pmatrix}0 \\ 0\end{pmatrix}}\)

The equations of the system show this is equivalent to

• \(x = x_0\)
• \(y = y_0\)

is a constant solution to the system

4 Why are important the critical points for this system?

4.1 Front

Why are important the critical points for this system?

\({\displaystyle \dot{\vb{u}} = A \vb{u}}\), where \({\displaystyle A = \begin{pmatrix}a & b \\ c & d\end{pmatrix}}\)

4.2 Back

Critical points are the key to our qualitative view of systems. We classify the linear systems by their behavior near critical points.

5 Which is the critical point that have all systems?

5.1 Front

Which is the critical point that have all systems?

\({\displaystyle \dot{\vb{u}} = A \vb{u}}\), where \({\displaystyle A = \begin{pmatrix}a & b \\ c & d\end{pmatrix}}\)

5.2 Back

The point \((0,0)\)

6 When can we say that the system only has the point 0 as critical point?

6.1 Front

When can we say that the system only has the point 0 as critical point?

\({\displaystyle \dot{\vb{u}} = A \vb{u}}\), where \({\displaystyle A = \begin{pmatrix}a & b \\ c & d\end{pmatrix}}\)

6.2 Back

If the matrix \(A\) is invertible, then \((0,0)\) is the only critical point

7 Can 2 solution to the system intersect in the phase portrait?

7.1 Front

Can 2 solution to the system intersect in the phase portrait?

\({\displaystyle \dot{\vb{u}} = A \vb{u}}\), where \({\displaystyle A = \begin{pmatrix}a & b \\ c & d\end{pmatrix}}\)

7.2 Back

No, two trajectories of the solution for the system cannot intersect.

It’s similar to integral curves for direction fields, where the integral curves did not cross

8 How can we draw the easiest trajectories of this system in a phase portrait?

8.1 Front

How can we draw the easiest trajectories of this system in a phase portrait?

• \({\displaystyle \dot{x} = y}\)
• \({\displaystyle \dot{y} = x}\)

Which solution is \({\displaystyle \vb{u}(t) = c_1 \begin{pmatrix}1 \\ 1\end{pmatrix} e^t + c_2 \begin{pmatrix}1 \\ -1\end{pmatrix} e^{-t}}\)

8.2 Back

Setting up \(c_{1} = c_2 = \pm 1\), we get 4 solutions

• \({\displaystyle \begin{pmatrix}1 \\ 1\end{pmatrix}e^t}\)
• \({\displaystyle - \begin{pmatrix}1 \\ 1\end{pmatrix} e^t}\)
• \({\displaystyle \begin{pmatrix}1 \\ -1\end{pmatrix}e^{-t}}\)
• \({\displaystyle - \begin{pmatrix}1 \\ -1\end{pmatrix}e^{-t}}\)

The first solution, when \(t = 0\), the point is at \((1,1)\). As \(t\) increases, the point moves outward along the line \(y = x\). As \(t\) decreases through negative values, the point moves inwards along the line, toward \((0,0)\)

Since \(t\) is always understood to be increasing on the trajectory, the whole trajectory consists of the ray \(y=x\) in the first quadrant, excluding the origin (which is not reached in finite negative time), with an outward direction of motion

Similar analysis can be made for the other 3 solutions,

Each of the four solutions has as its trajectory one of the four rays. The indicated direction of motion is outward or inward according to whether the exponential factor increases or decreases as \(t\) increases

The origin is a fifth trajectory itself, which is a stationary point

So the intersecting diagonal lines represent five trajectories, no two of which intersect

9 Are the trajectories intersecting at point 0 for this system?

9.1 Front

Are the trajectories intersecting at point 0 for this system?

\({\displaystyle \dot{\vb{u}} = A \vb{u}}\)

9.2 Back

No, at point \((0,0)\), in this point its a solution itself

10 What can we say about this system from its phase portrait?

10.1 Front

What can we say about this system from its phase portrait?

10.2 Back

It’s a unstable saddle. It is called unstable because the trajectories go off to infinity as \(t\) increases, except at 0 and line \(y = -x\)

It’s called a saddle because of its general resemblance to the level curves of a saddle-shaped surface in 3-space

11 What can we say about this system from its phase portrait?

11.1 Front

What can we say about this system from its phase portrait?

11.2 Back

It’s called an asymptotically stable node or sink node

Means that all the trajectories approach the critical point as \(t\) increases

12 What can we say about this system from its phase portrait?

12.1 Front

What can we say about this system from its phase portrait?

12.2 Back

It’s called unstable node or source node

13 What can we say about this system from its phase portrait?

13.1 Front

What can we say about this system from its phase portrait?

13.2 Back

It’s a stable center linear system. The word stable means that any trajectory stays within a bounded region of the phase plane as \(t\) increases or decreases indefinitely

Note: We cannot use “asymptotically stable”, since the trajectories do not approach the critical point \((0,0)\) as \(t\) increases

In this case the trajectories are circles, but you can use the word center if the curves were ellipsses having the origin as cneter

14 What can we say about this system from its phase portrait?

14.1 Front

What can we say about this system from its phase portrait?

14.2 Back

Their trajectories are therefore traced out by the tip of an origin vector that rotates clockwise at a constant rate, while its magnitude shrinks exponentially to \(0\)

In other words, the trajectories spiral in toward the origin as \(t\) increases

We call this pattern as asymptotically stable spiral

15 What can we say about this system from its phase portrait?

15.1 Front

What can we say about this system from its phase portrait?

15.2 Back

The magnitude of the rotating vector increases as \(t\) increases, giving as pattern an unstable spiral

16 Draw a sketch of this linear system

16.1 Front

Draw a sketch of this linear system

• \({\displaystyle \dot{x} = y}\)
• \({\displaystyle \dot{y} = x}\)

Which solution is \({\displaystyle \vb{u}(t) = c_1 \begin{pmatrix}1 \\ 1\end{pmatrix} e^t + c_2 \begin{pmatrix}1 \\ -1\end{pmatrix} e^{-t}}\)

16.2 Back

For these trajectories we can do a little algebra

• \(x = c_1 e^t + c_2 e^{-t}\)
• \(y = c_1 e^t - c_2 e^{-t}\)

This easily gives \(x^2 - y^2 = 4c_1 c_2 = k\), where \(k\) is a constant, which is the equation of a hyperbola oriented with the axes - we sketch in some of the hyperbolas.

We know which direction to point the arrows indicating the direction of motion as \(t\) increases since they must be compatible with the motion along the rays - for by continuity, nearby trajectories must have arrowheads pointing in similar directions

It’s unstable saddle

17 Draw a sketch of this linear system

17.1 Front

Draw a sketch of this linear system

• \({\displaystyle \dot{x} = -x}\)
• \({\displaystyle \dot{y} = -2y}\)

Which solution is \({\displaystyle \vb{u}(t) = c_1 \begin{pmatrix}1 \\ 0\end{pmatrix} e^{-t} + c_2 \begin{pmatrix}0 \\ 1\end{pmatrix} e^{-2t}}\)

17.2 Back

• \(x = c_1 e^{-t}\)
• \(y = c_2 e^{-2t}\)

Trajectories are a family of parabolas \(y = cx^2\)

We can see that the term \(e^{-t}\) is dominant when \(t \to \infty\), so as \(t\) increasing the solution will be parallel to the ray \({\displaystyle \begin{pmatrix}1 \\ 0\end{pmatrix}}\)

When \(t \to -\infty\), the dominant term is \(e^{-2t}\), so as \(t\) decreases the trajectories will be parallel to ray \({\displaystyle \begin{pmatrix}0 \\ 1\end{pmatrix}}\)

It’s a asymptotically stable node or sink node

18 Draw a sketch of this linear system

18.1 Front

Draw a sketch of this linear system

• \({\displaystyle \dot{x} = x}\)
• \({\displaystyle \dot{y} = 2y}\)

Which solution is \({\displaystyle \vb{u}(t) = c_1 \begin{pmatrix}1 \\ 0\end{pmatrix} e^{t} + c_2 \begin{pmatrix}0 \\ 1\end{pmatrix} e^{2t}}\)

18.2 Back

It’s an unstable node or source node

19 Draw a sketch of this linear system

19.1 Front

Draw a sketch of this linear system

• \({\displaystyle \dot{x} = y}\)
• \({\displaystyle \dot{y} = -x}\)

Which solution is \({\displaystyle \vb{u}(t) = c_1 \begin{pmatrix} \sin(t) \\ \cos(t)\end{pmatrix} + c_2 \begin{pmatrix} \cos(t) \\ - \sin(t)\end{pmatrix}}\)

19.2 Back

Dividing \(y’/x’\) converts to a separable first order ODE

\({\displaystyle \frac{\dv{y}{t}}{\dv{x}{t}} = \frac{\dd{y}}{\dd{x}} = - \frac{x}{y}}\)

\(x^2 + y^2 = c\)

The trajectories are the family of circles centered at the origin. The direction of motion is determined by its velocity field. You can get a velocity vector at \((1,0\) is \(\ev{0, -1} = -\vu{j}\)

So the motion is clockwise.

It’s a stable center linear system

20 Draw a sketch of this linear system

20.1 Front

Draw a sketch of this linear system

• \({\displaystyle \dot{x} = -x + y}\)
• \({\displaystyle \dot{y} = -x - y}\)

Which solution is \({\displaystyle \vb{u}(t) = c_1 \begin{pmatrix}\sin(t) \\ \cos(t)\end{pmatrix} e^{-t} + c_2 \begin{pmatrix}\cos(t) \\ - \sin(t)\end{pmatrix} e^{-t}}\)

20.2 Back

Their trajectories are therefore traced out by the tip of an origin vector that rotates clockwise at a constant rate, while its magnitude shrinks exponentially to \(0\)

In other words, the trajectories spiral in toward the origin as \(t\) increases

We call this pattern as asymptotically stable spiral

21 Draw a sketch of this linear system

21.1 Front

Draw a sketch of this linear system

• \({\displaystyle \dot{x} = x + y}\)
• \({\displaystyle \dot{y} = -x + y}\)

Which solution is \({\displaystyle \vb{u}(t) = c_1 \begin{pmatrix} \sin(t) \\ \cos(t)\end{pmatrix} e^{t} + c_2 \begin{pmatrix}\cos(t) \\ - \sin(t)\end{pmatrix} e^{t}}\)

21.2 Back

The magnitude of the rotating vector increases as \(t\) increases, giving as pattern an unstable spiral

22 Describe a general sketch for this linear system

22.1 Front

Describe a general sketch for this linear system

\({\displaystyle \dot{\vb{x}} = A \vb{x}}\), where \(A = 2 \cross 2\) constant matrix

Which solution is \({\displaystyle \vb{u} = c_1 \vb{\alpha_1} e^{\lambda_1 t} + c_2 \vb{\alpha_2} e^{\lambda_2 t}}\)

Suppose that its eigenvalues have opposite signs: \(\lambda_1 \gt 0\) and \(\lambda_2 \lt 0\)

22.2 Back

This is unstable saddle

• Draw 4 simple solution as rays, setting \(c_1 = \pm 1\) and \(c_2 = \pm 1\) and the other to \(0\)

  • \({\displaystyle \vb{x} = \pm \vb{\alpha_1} e^{\lambda_1 t}}\)
    • \(\lambda_1 \gt 0\), then as \(t\) goes from \(-\infty\) to \(\infty\), this scalar factor increases from \(0\) to \(\infty\)
    • A ray going out from the origin in the direction of the vector \(\vb{\alpha_1}\)
    • Similar way for \(\vb{x} = - \vb{\alpha_1} e^{\lambda_1 t}\)
  • \({\displaystyle \vb{x} = \pm \vb{\alpha_2} e^{\lambda_2 t}}\)
    • \(\lambda_2 \gt 0\), then as \(t\) goes from \(-\infty\) to \(\infty\), this scalar factor increases from \(\infty\) to \(0\)
    • A ray going out towards the origin in the direction of the vector \(\vb{\alpha_2}\)
    • Similar way for \(\vb{x} = - \vb{\alpha_2} e^{\lambda_2 t}\)

• To complete the picture, we sketch some nearby trajectories

  • Smooth curves generally following the directions of the four rays described above
  • They look something like hyperbolas, and they do have the rays as asymptotes
  • \({\displaystyle \vb{u} = c_1 \vb{\alpha_1} e^{\lambda_1 t} + C_2 \vb{\alpha_2} e^{\lambda_2 t}}\)

  

23 Describe a general sketch for this linear system

23.1 Front

Describe a general sketch for this linear system

\({\displaystyle \dot{\vb{x}} = A \vb{x}}\), where \(A = 2 \cross 2\) constant matrix

Which solution is \({\displaystyle \vb{u} = c_1 \vb{\alpha_1} e^{\lambda_1 t} + c_2 \vb{\alpha_2} e^{\lambda_2 t}}\)

Suppose that its eigenvalues have same signs: \(\lambda_1 \lt \lambda_2 \lt 0\)

23.2 Back

It’s a asymptotically stable / sink node

• The rays coming in towards the origin as \(t\) increases, since both of the \(\lambda_i\) are negative

• The parabolic curves will certainly come in to the origin as \(t\) increases

  • Near the origin:
    • Corresponds to large positive \(t\) (\(t \gg 1\))
    • The dominant term is \({\displaystyle \vb{u} \approx c_2 \vb{\alpha_2} e^{-\lambda_2 t}}\), so solution near origin are parallel to \(\alpha_2\) eigenvector
  • Far from the origin:
    • Corresponds to large negative \(t\) (\(t \ll -1\))
    • The dominant term is \({\displaystyle \vb{u} \approx c_1 \vb{\alpha_1} e^{\lambda_1 t}}\), so solution far from the origin are parallel to \(\alpha_1\) eigenvector

  

24 Describe a general sketch for this linear system

24.1 Front

Describe a general sketch for this linear system

\({\displaystyle \dot{\vb{x}} = A \vb{x}}\), where \(A = 2 \cross 2\) constant matrix

Which solution is \({\displaystyle \vb{u} = c_1 \vb{\alpha_1} e^{\lambda_1 t} + c_2 \vb{\alpha_2} e^{\lambda_2 t}}\)

Suppose that its eigenvalues have same signs: \(\lambda_1 \gt \lambda_2 \gt 0\)

24.2 Back

It’s an unstable source node

• The rays coming from the origin is outwards, since the \(\lambda_i \gt 0\)

• The parabolic curves will certainly come from the origin as \(t\) increases

  • Near the origin:
    • Corresponds to large negative \(t\) (\(t \ll -1\))
    • The dominant term is \({\displaystyle \vb{u} \approx c_2 \vb{\alpha_2} e^{\lambda_2 t}}\), so solution near origin are parallel to \(\alpha_2\) eigenvector
  • Far from the origin:
    • Corresponds to large positive \(t\) (\(t \gg 1\))
    • The dominant term is \({\displaystyle \vb{u} \approx c_1 \vb{\alpha_1} e^{\lambda_1 t}}\), so solution far from the origin are parallel to \(\alpha_1\) eigenvector

  

25 Describe a general sketch for this linear system

25.1 Front

Describe a general sketch for this linear system

\({\displaystyle \dot{\vb{x}} = A \vb{x}}\), where \(A = 2 \cross 2\) constant matrix

Which solution is \({\displaystyle \vb{u} = c_1 \vb{\alpha_1} e^{\lambda_1 t} + c_2 \vb{\alpha_2} e^{\lambda_2 t}}\)

Suppose that its eigenvalues are pure imaginary: \(\lambda = \pm bi\), \(b \gt 0\)

25.2 Back

It’s a stable center linear system

Its solution has the form \({\displaystyle \vb{u} = \vb{c_1} \cos(bt) + \vb{c_2} \sin(bt)}\), where \(\vb{c_1}, \vb{c_2}\) constant vectors

Every solution is periodic, with period \({\displaystyle \frac{2\pi}{b}}\), the moving point representing it retraces its path at this interval. The trajectories are ellipses

It’s enough if you determine whether the motion is clockwise or counterclockwise. So you can get a single velocity vector \(\dot{\vb{u}}\) of the velocity field

26 Describe a general sketch for this linear system

26.1 Front

Describe a general sketch for this linear system

\({\displaystyle \dot{\vb{x}} = A \vb{x}}\), where \(A = 2 \cross 2\) constant matrix

Which solution is \({\displaystyle \vb{u} = c_1 \vb{\alpha_1} e^{\lambda_1 t} + c_2 \vb{\alpha_2} e^{\lambda_2 t}}\)

Suppose that its eigenvalues are complex: \(\lambda = a \pm bi\), \(a \lt 0\) and \(b \gt 0\)

26.2 Back

It’s asymptotically stable (sink) spiral

The solutions has the form \({\displaystyle \vb{u} = e^{at} (\vb{c_1} \cos(bt) + c_2 \sin(bt))}\), where \(\vb{c_1}, \vb{c_2}\) constant vectors

Decreases towards \(0\) as \(t \to \infty\)

Thus the point \(\vb{u}\) travels in a trajectory which is like an ellipse, except that the distance from the origin is steadily shrinking towards the origin.

Te direction of motion is calculated from a single velocity vector \(\dot{\vb{u}}\)

27 Describe a general sketch for this linear system

27.1 Front

Describe a general sketch for this linear system

\({\displaystyle \dot{\vb{x}} = A \vb{x}}\), where \(A = 2 \cross 2\) constant matrix

Which solution is \({\displaystyle \vb{u} = c_1 \vb{\alpha_1} e^{\lambda_1 t} + c_2 \vb{\alpha_2} e^{\lambda_2 t}}\)

Suppose that its eigenvalues are complex: \(\lambda = a \pm bi\), \(a \gt 0\) and \(b \gt 0\)

27.2 Back

It’s unstable (source) spiral

The solutions has the form \({\displaystyle \vb{u} = e^{at} (\vb{c_1} \cos(bt) + c_2 \sin(bt))}\), where \(\vb{c_1}, \vb{c_2}\) constant vectors

Increases towards \(\infty\) as \(t \to \infty\)

Thus the point \(\vb{u}\) travels in a trajectory which is like an ellipse, except that the distance from the origin is steadily expanding away from the origin.

Te direction of motion is calculated from a single velocity vector \(\dot{\vb{u}}\)

28 Describe a general sketch for this linear system

28.1 Front

Describe a general sketch for this linear system

\({\displaystyle \dot{\vb{x}} = A \vb{x}}\), where \(A = 2 \cross 2\) constant matrix

Which solution is \({\displaystyle \vb{u} = c_1 \vb{\alpha_1} e^{\lambda_1 t} + c_2 \vb{\alpha_2} e^{\lambda_2 t}}\)

Suppose that its eigenvalues are repeated: \(\lambda \neq 0\), defective

28.2 Back

It’s incomplete, one independent eigenvector

It’s called defective node

If \(\lambda \gt 0\) is unstable, if \(\lambda \lt 0\) is asymptotically stable

29 Describe a general sketch for this linear system

29.1 Front

Describe a general sketch for this linear system

\({\displaystyle \dot{\vb{x}} = A \vb{x}}\), where \(A = 2 \cross 2\) constant matrix

Which solution is \({\displaystyle \vb{u} = c_1 \vb{\alpha_1} e^{\lambda_1 t} + c_2 \vb{\alpha_2} e^{\lambda_2 t}}\)

Suppose that its eigenvalues are repeated: \(\lambda \neq 0\), complete

29.2 Back

You have 2 independent eigenvector corresponding to this unique \(\lambda\)

It’s called star node

If \(\lambda \gt 0\) is unstable, and if \(\lambda \lt 0\) is asymptotically stable

30 Describe a general sketch for this linear system

30.1 Front

Describe a general sketch for this linear system

\({\displaystyle \dot{\vb{x}} = A \vb{x}}\), where \(A = 2 \cross 2\) constant matrix

Which solution is \({\displaystyle \vb{u} = c_1 \vb{\alpha_1} e^{\lambda_1 t} + c_2 \vb{\alpha_2} e^{\lambda_2 t}}\)

Suppose that its eigenvalues \(\lambda_1 = 0 \gt \lambda_2\)

30.2 Back

The critical points are not isolated - they lie on the line through \(0\) with direction \(\alpha_1\)

\({\displaystyle \vb{u} = c_1 \alpha_1 +c_2 \alpha_2 e^{\lambda_2 t}}\)

As \(t \to \infty\), \(\vb{u} \to c_1 \alpha_1\) along a line parallel to \(\alpha_2\)

31 Describe a general sketch for this linear system

31.1 Front

Describe a general sketch for this linear system

\({\displaystyle \dot{\vb{x}} = A \vb{x}}\), where \(A = 2 \cross 2\) constant matrix

Which solution is \({\displaystyle \vb{u} = c_1 \vb{\alpha_1} e^{\lambda_1 t} + c_2 \vb{\alpha_2} e^{\lambda_2 t}}\)

Suppose that its eigenvalues \(\lambda_1 = 0 \lt \lambda_2\)

31.2 Back

The critical points are not isolated - they lie on the line through \(0\) with direction \(\alpha_1\)

\({\displaystyle \vb{u} = c_1 \alpha_1 +c_2 \alpha_2 e^{\lambda_2 t}}\)

As \(t \to \infty\), \(\vb{u} \to \infty\) along a line parallel to \(\alpha_2\)

32 Describe a general sketch for this linear system

32.1 Front

Describe a general sketch for this linear system

\({\displaystyle \dot{\vb{x}} = A \vb{x}}\), where \(A = 2 \cross 2\) constant matrix

Which solution is \({\displaystyle \vb{u} = c_1 \vb{\alpha_1} e^{\lambda_1 t} + c_2 \vb{\alpha_2} e^{\lambda_2 t}}\)

Suppose that \(\lambda_1 = \lambda_2 = 0\)

32.2 Back

It’s a line of critical points

\({\displaystyle \vb{u} = c_1 \alpha_1 + c_2 (t \alpha_1 + \alpha_2)}\)

Trajectories are parallel to \(\alpha_1\)

33 What is the Trace Determinant graph (T-D graph)?

33.1 Front

What is the Trace Determinant graph (T-D graph)?

33.2 Back

It’s a graph where you can show the behaviour of the linear system through it’s characteristic equation \(p_A(\lambda)\)

Its abbreviations stands for \(T = \operatorname{tr}(A)\) and \(D = \operatorname{det}(A)\)

34 What is the curve which separate the real and complex eigenvalues in T-D graph?

34.1 Front

What is the curve which separate the real and complex eigenvalues in T-D graph?

34.2 Back

It’s the curve \({\displaystyle D = \frac{T^2}{4}}\)

If the discriminant of quadratic formula of the characteristic polynomial \(p_A(\lambda) = \lambda^2 - T\lambda + D\)

So, it’s root come from \({\displaystyle \lambda = \frac{T \pm \sqrt{T^2 - 4D}}{2}}\)

• \(\lambda\) is real if \(T^2 - 4D \gt 0\)
• \(\lambda\) is complex if \(T^2 - 4D \lt 0\)

35 What does mean this condition in the T-D graph?

35.1 Front

What does mean this condition in the T-D graph?

\(D \lt 0\)

Explain how is the response

35.2 Back

\({\displaystyle \lambda = \frac{T \pm \sqrt{T^2 - 4D}}{2}}\)

The eigenvalues are real and of opposite sign, hence the phase portrait is a unstable saddle

36 What does mean this condition in the T-D graph?

36.1 Front

What does mean this condition in the T-D graph?

\(0 \lt D \lt T^2/4\)

Explain how is the response

36.2 Back

\({\displaystyle \lambda = \frac{T \pm \sqrt{T^2 - 4D}}{2}}\)

The eigenvalues are real, distinct, and of the same sign, hence the phase portrait is a node

If \(T \lt 0\) is stable

If \(T \gt 0\) is unstable

37 What does mean this condition in the T-D graph?

37.1 Front

What does mean this condition in the T-D graph?

\(0 \lt T^2/4 \lt D\)

Explain how is the response

37.2 Back

\({\displaystyle \lambda = \frac{T \pm \sqrt{T^2 - 4D}}{2}}\)

The eigenvalues are neither real nor purely imaginary (are complex), hence the phase portrait is a spiral

If \(T \lt 0\) is stable

If \(T \gt 0\) is unstable

38 Describe all possible responses from the T-D graph?

38.1 Front

Describe all possible responses from the T-D graph?

38.2 Back

39 How can we determine the direction of the trajectory from this matrix

39.1 Front

How can we determine the direction of the trajectory from this matrix

\({\displaystyle \begin{pmatrix}-1 & 1 \\ 1 & -1\end{pmatrix}}\)

39.2 Back

The linear system is \({\displaystyle \dot{\vb{u}} = A \vb{u}}\), and the direction field is given by \(\dot{\vb{u}}\)

\({\displaystyle \dot{\vb{u}} = \begin{pmatrix}-1 & 1 \\ 1 & -1\end{pmatrix}}\)

• \(\dot{x} = - x + y\)
• \(\dot{y} = x - y\)

At point \((1,0)\), the velocity vector is \(\ev{-1, 1}\), which means that the trajectory is upwards

Another check is using T-D diagram for view which kind of solution is, and check the direction of velocity field with the entry \(A_{21}\) of the matrix. In this case is positive, so that means that it’s upwards.

40 On the T-D plane, where can you guarantee that any matrix with that value of trace and determinant is stable?

40.1 Front

On the T-D plane, where can you guarantee that any matrix with that value of trace and determinant is stable?

40.2 Back

Stable solutions occur if and only if both eigenvalues have negative real part. This happens exactly when the trace is negative an the determinant is positive - so, at all points inside the second quadrant in the T-D plane

41 On the T-D plane, where can you guarantee that any matrix with that value of trace and determinant is unstable?

41.1 Front

On the T-D plane, where can you guarantee that any matrix with that value of trace and determinant is unstable?

41.2 Back

Unstable solutions occur when there is at least one eigenvalue with positive real part. This happens either when the determinant is negative, or when the trace is positive - so, at all points inside the first, third, and fourth quadrants on the T-D plane

42 On the T-D plane, where can you guarantee that any matrix with that value of trace and determinant is neutrally stable?

42.1 Front

On the T-D plane, where can you guarantee that any matrix with that value of trace and determinant is neutrally stable?

42.2 Back

Neutrally stable solutions occur either when the trace is zero and the determinant is positive (giving periodic solutions), or when the determinant is zero and the trace is negative (giving combs)

43 On the T-D plane, are there any values of the trace and determinant for which there are matrices exhibiting more than one type of limiting behaviour?

43.1 Front

On the T-D plane, are there any values of the trace and determinant for which there are matrices exhibiting more than one type of limiting behaviour?

43.2 Back

When the trace and determinant are both zero, the zero matrix gives neutrally stable behaviour, while the matrix that has exactly one nonzero entry, in the upper right, gives unstable behaviour.

44 What is the normal mode and general solution for this linear system?

44.1 Front

What is the normal mode and general solution for this linear system?

\({\displaystyle \dot{\vb{u}} = A \vb{u}}\), where \({\displaystyle A = \begin{pmatrix}0.5 & 1 \\ -2.25 & 0.5\end{pmatrix}}\)

44.2 Back

Characteristic polynomial: \({\displaystyle p_A(\lambda) = \lambda^2 - \lambda + 2.5}\)

Eigenvalues: \({\displaystyle \lambda = \frac{1 \pm 3i}{2}}\)

Eigenvector for \({\displaystyle \lambda_1 = \frac{1 + 3i}{2}}\) satisfies \((A - \lambda_1 I) \vb{v_1}\)

Hence, \({\displaystyle \begin{pmatrix}-3i/2 & 1 \\ -9/4 & -3i/2\end{pmatrix} \vb{v_1} = \vb{0}}\)

Eigenvector: \({\displaystyle \vb{v_1} = \begin{pmatrix}1 \\ 3i/2\end{pmatrix}}\)

The normal modes: \({\displaystyle e^{(1 + 3i)t} \begin{pmatrix}1 \\ 3i/2\end{pmatrix}}\), which has

Real part of normal mode: \({\displaystyle \vb{u_1} = e^{t/2} \begin{pmatrix}\cos(3t/2) \\ - (3/2) \sin(3t/2)\end{pmatrix}}\)

Imaginary part of normal mode: \({\displaystyle \vb{u_2} = e^{t/2} \begin{pmatrix}\sin(3t/2) \\ (3/2) \cos(3t/2)\end{pmatrix}}\)

45 Suppose this IVP, is x and y increasing at t = 0?

45.1 Front

Suppose this IVP, is x and y increasing at t = 0?

\({\displaystyle A = \begin{pmatrix}0.5 & 1 \\ -2.25 & 0.5\end{pmatrix}}\)

At point \((1,0)\) of the phase portrait

45.2 Back

\({\displaystyle \dot{\vb{u}} = A \vb{u}}\), at \(t=0\) we have \(\vb{u} = \begin{pmatrix}1 \\ 0\end{pmatrix}\)

So, \({\displaystyle \dot{\vb{u}} = \begin{pmatrix}0.5 & 1 \\ -2.25 & 0.5\end{pmatrix} \begin{pmatrix}1 \\ 0\end{pmatrix} = \begin{pmatrix}0.5 \\ -2.25\end{pmatrix}}\)

So the values of \(x\) increases at \(t = 0\) with a ratio of \(0.5\), but the values of \(y\) decreases at \(t=0\) with a ratio of \(-2.25\)