Attention

In this section we will assume the reader is familiar with complex numbers and their properties. If preferred, a review of complex numbers can be found in Chapter 10.

9.4. Continuous dynamical systems

9.4.1. Introduction

In this section, we will deal with similar problems as in Section 9.1. There, we were concerned with discrete time. That is, we assumed a certain initial state \(\vect{x}_{0}\) at time \(k=0\), then predicted the next state \(\vect{x}_{1}\) on time \(k=1\), \(\vect{x}_{2}\) on time \(k=2\) and so on. But just as often we want to deal with continuous time. That is, there is no next state but rather a state for every positive real number.

9.4.2. Preliminaries

In order to deal with this new context, we need some preliminaries from calculus. We will consider functions depending on the variable \(t\). The derivative of a function \(f(t)\) will be denoted by \(f'(t)\). To lighten notation, we will often write \(f\) instead of \(f(t)\).

Proposition 9.4.1

Let \(f\) and \(g\) be differentiable functions on \(\R\) and let \(c\) be a real number. Then:

1. \((f+g)'=f'+g'\).

2. \((cf)'=cf'\).

3. if \(f(t)=e^{\lambda t}\), then \(f'(t)=\lambda e^{\lambda t}\).

This last point means that the equation \(x'=\lambda x\) has solution \(x=e^{\lambda t}\). In fact, for every real number \(c\), \(y=ce^{\lambda t}\) is solution of \(x'=\lambda x\). We want to generalise this idea. But first, we need some terminology.

Suppose we have differentiable functions \(x_{1},\ldots,x_{n}\) and real numbers \(a_{ij}\) for \(1\leq i,j\leq n\). A system of equations

\[\begin{split}\begin{cases} x_{1}'&=a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n},\\ x_{2}'&=a_{21}x_{2}+a_{22}x_{2}+\cdots +a_{2n}x_{n},\\ &\,\vdots\\ x_{n}'&=a_{n1}x_{1}+a_{n2}x_{2}+\cdots+a_{nn}x_{n} \end{cases}\end{split}\]

can be conveniently rewritten as \(\vect{x}'=A\vect{x}\) where

\[\begin{split} \vect{x}= \begin{pmatrix} x_{1}\\ x_{2}\\ \vdots\\ x_{n} \end{pmatrix}, \quad \vect{x}'= \begin{pmatrix} x_{1}'\\ x_{2}'\\ \vdots\\ x_{n}' \end{pmatrix} \quad\text{and}\quad A= \begin{pmatrix} a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn} \end{pmatrix}. \end{split}\]

system of linear differential equationssolutionvector-valued functioncomponent functionsderivativesystem of differential equationsdynamical system

Definition 9.4.1

In this context, we call \(\vect{x}'=A\vect{x}\) a system of (linear) differential equations or a dynamical system, \(\vect{x}\) a vector-valued function, \(\vect{x}'\) the derivative of \(\vect{x}\), and the \(x_{i}\)’s the component functions of \(\vect{x}\). Any \(\vect{x}\) for which \(\vect{x}'=A\vect{x}\) holds is called a solution to the system of differential equations.

The following proposition will be quite useful to us. It tells us that, in order to find the full (infinite) solution set, it suffices to find a (finite) basis of solutions.

Proposition 9.4.2

If \(\vect{y}\) and \(\vect{z}\) are solutions of \(\vect{x}'=A\vect{x}\) and \(c\) and \(d\) are arbitrary real numbers, then \(c\vect{y}+d\vect{z}\) is also a solution of \(\vect{x}=A\vect{x}'\).

Proof of Proposition 9.4.2

See Exercise 9.4.1.

Exercise 9.4.1

Prove Proposition 9.4.2

Solution to Exercise 9.4.1

Direct compuations shows the requested:

\[\begin{split} \begin{align*} (c\vect{y}+d\vect{z})' &= (c\vect{y})'+(d\vect{z})' \\ &= c\vect{y}'+d\vect{z}' \\ &= c(A\vect{y})+d(A\vect{z}) \\ &= A(c\vect{y})+A(d\vect{z}) \\ &= A(c\vect{y}+d\vect{z}) \\ \end{align*} \end{split}\]

In view of this result, it makes sense to generalise some concepts which we have seen for vectors to the setting of vectors of functions.

linear combinationlinearly independent

Definition 9.4.2

We say that a vector function \(\vect{x}\) is a linear combination of vector functions \(\vect{x}_{1},\ldots,\vect{x}_{n}\) if there are scalars \(c_{1},\ldots,c_{n}\) in \(\mathbb{R}\) such that:

\[\vect{x}=c_{1}\vect{x}_{1}+\cdots + c_{n}\vect{x}_{n}.\]

In particular, the function \(\vect{0}\) can be written as a linear combination of any set of vector functions \(S=\left\{\vect{x}_{1},\ldots,\vect{x}_{n}\right\}\) by putting \(c_{1}=\cdots=c_{n}=0\). If this is the only way \(\vect{0}\) can be written as a linear combination of vector functions in \(S\), then \(S\) is called linearly independent.

9.4.3. Solving a dynamical system

How do we find the solutions of a dynamical system? Let us see what happens in a simple example.

Example 9.4.1

Consider the system \(\vect{x}'=A\vect{x}\)

\[\begin{split}\vect{x}=\begin{pmatrix} x_{1}\\ x_{2} \end{pmatrix}\quad\text{and}\quad A=\begin{pmatrix} 3&0\\ 0&-1 \end{pmatrix}.\end{split}\]

which can be rewritten as

\[\begin{split}\begin{cases} x_{1}'&=3x_{1}+0x_{2}=3x_{1},\\ x_{2}'&=0x_{1}-1x_{2}=-x_{2}. \end{cases}\end{split}\]

By Proposition 9.4.1, this system has solutions \(x_{1}=c_{1} e^{3t}\), \(x_{2}=c_{2}e^{-t}\). Therefore, the vector functions

\[\begin{split}\vect{x}=c_{1}\begin{pmatrix} 1\\ 0 \end{pmatrix}e^{3t}+c_{2}\begin{pmatrix} 0\\ 1 \end{pmatrix}e^{-t}\end{split}\]

are solutions.

What made this example easy is the fact that \(A\) was diagonal. Indeeed, for any diagonal matrix \(A\), the system \(\vect{x}'=A\vect{x}\) becomes:

\[\begin{split}\begin{cases} x_{1}'&=a_{11}x_{1}+0_{12}x_{2}+\cdots +0_{1n}x_{n}=a_{11}x_{1},\\ x_{2}'&=0_{21}x_{2}+a_{22}x_{2}+\cdots +0_{2n}x_{n}=a_{22}x_{2},\\ &\,\vdots\\ x_{n}'&=0_{n1}x_{1}+0_{n2}x_{2}+\cdots+a_{nn}x_{n}=a_{nn}x_{n}. \end{cases}\end{split}\]

By Proposition 9.4.1, we find the solutions:

\[x_{1}=e^{a_{11}t},\, x_{2}=e^{a_{22}t},\,\ldots,\, x_{n}=e^{a_{nn}t}\]

for the single equations, which allows us to write the solutions as vector functions:

\[\begin{split}\vect{x}=c_{1}\begin{pmatrix} a_{11}\\ 0\\ \vdots\\ 0 \end{pmatrix}e^{\lambda_{1}t}+ c_{2}\begin{pmatrix} 0\\ a_{22}\\ \vdots\\ 0 \end{pmatrix}e^{\lambda_{2}t}+ \cdots+c_{n}\begin{pmatrix} 0\\ 0\\ \vdots\\ a_{nn} \end{pmatrix}e^{\lambda_{n}t}.\end{split}\]

Of course, most matrices are not diagonal. However, most matrices are at least diagonalisable, and we can use this to our advantage. Suppose that the matrix \(A\) can be diagonalised, so \(A=PDP^{-1}\) where \(D\) is the diagonal matrix with the eigenvalues \(\lambda_{1},\ldots,\lambda_{n}\) on the diagonal and \(P\) is the matrix with the corresponding eigenvectors \(\vect{v}_{1},\ldots,\vect{v}_{n}\) as columns. Then we have

\[\begin{split}\begin{align*} \vect{x}'=A\vect{x}&\quad\Longleftrightarrow\quad \vect{x}'=PDP^{-1}\vect{x}\\ &\quad\Longleftrightarrow\quad P^{-1}\vect{x}'=DP^{-1}\vect{x}\\ &\quad\Longleftrightarrow\quad (P^{-1}\vect{x})'=D(P^{-1}\vect{x}) \end{align*}\end{split}\]

and this is a new dynamical system with a diagonal matrix! Hence, we find the solution

\[\begin{split}P^{-1}\vect{x}= c_{1}\begin{pmatrix} a_{11}\\ 0\\ \vdots\\ 0 \end{pmatrix}e^{\lambda_{1}t}+ c_{2}\begin{pmatrix} 0\\ a_{22}\\ \vdots\\ 0 \end{pmatrix}e^{\lambda_{2}t}+ \cdots+c_{n}\begin{pmatrix} 0\\ 0\\ \vdots\\ a_{nn} \end{pmatrix}e^{\lambda_{n}t}.\end{split}\]

The solution of the original system can now be obtained by multiplying from the left with \(P\). This suggests the following proposition:

Proposition 9.4.3

If \(A\) is a diagonalisable matrix with eigenvalues \(\lambda_{1},\ldots,\lambda_{n}\) and a corresponding basis of eigenvectors \(\vect{v}_{1},\ldots,\vect{v}_{n}\), then the system \(\vect{x}'=A\vect{x}\) has general solution:

\[\vect{y}(t)=c_{1}\vect{v}_{1}e^{\lambda{1}t}+c_{2}\vect{v}_{2}e^{\lambda_{2}t}+\cdots+c_{n}\vect{v}_{n}e^{\lambda_{n}t},\]

where \(c_{1},\ldots,c_{n}\) are constants.

eigenfunction

Definition 9.4.3

A function \(\vect{y}=\vect{v}e^{\lambda t}\) is called an eigenfunction of the dynamical system if \(\lambda\) is an eigenvalue with corresponding eigenvector \(\vect{v}\).

Remark 9.4.1

It is easy to check that such an eigenfunction is indeed a solution:

\[ \vect{y}'=(\vect{v}e^{\lambda t})'=\vect{v}\lambda e^{\lambda t}=A\vect{v} e^{\lambda t}=A\vect{y}. \]

Let us now consider an example.

Example 9.4.2

Consider the following system of differential equations:

\[\begin{split} \vect{x}'=A\vect{x}\quad\text{where}\quad A= \begin{pmatrix} 1&4\\ 1&1 \end{pmatrix}. \end{split}\]

A standard computation shows that \(A\) has eigenvalues \(\lambda_{1}=3,\lambda_{2}=-1\) with corresponding eigenvectors

\[\begin{split}\vect{v}_{1}=\begin{pmatrix} 2\\ 1 \end{pmatrix} \quad\text{and}\quad \vect{v}_{2}=\begin{pmatrix} -2\\ 1 \end{pmatrix}.\end{split}\]

Therefore the general solution to the system \(\vect{x}'=A\vect{x}\) is:

\[\begin{split}\vect{y}=c_{1}\begin{pmatrix} 2\\ 1 \end{pmatrix}e^{3t}+c_{2}\begin{pmatrix} -2\\ 1 \end{pmatrix}e^{-t}.\end{split}\]

As long as the matrix \(A\) is diagonalisable, we now know how to solve the system of linear differential equations. But we know more. We also know how a solution \(f(t)\) to such a system will behave as \(t\) goes to infinity. In practical applications, \(t\) usually is time, so this gives us predictions for what happens after a long time.

Example 9.4.3

Suppose some airborn disease is affecting a population. That means that people get sick from the environment, not from other sick people. To keep matters simple, we will assume that the population is constant and that recovery grants full immunity. Let \(S(t)\) be the number of susceptible members and \(I(t)\) the number of infected members of the population at time \(t\). If \(\alpha>0\) is the recovery rate and \(\beta>0\) is the infection rate, then we find:

\[\begin{split} \left\{\begin{array}{cccc} S'(t)&=&-\beta S(t),&\\ I'(t)&=&\beta S(t)&-\alpha I(t). \end{array}\right. \end{split}\]

Define

\[\begin{split} \vect{y}=\begin{pmatrix} S(t)\\ I(t) \end{pmatrix} \quad\text{and}\quad A=\begin{pmatrix} -\beta&0\\ \beta &-\alpha \end{pmatrix}. \end{split}\]

Since \(A\) is an upper diagonal matrix, we can conclude that its eigenvalues are \(-\beta\) and \(-\alpha\), which, for simplicity’s sake, we will assume to be different. Therefore, a solution to the system of linear differential equations \(\vect{y}'=A\vect{y}\) is given by

\[ \vect{y}=c_{1}\vect{v}_{-\beta}e^{-\beta t}+c_{2}\vect{v}_{-\alpha}e^{-\alpha t}, \]

where \(c_{1}\) and \(c_{2}\) are some constants while \(\vect{v}_{-\beta}\) and \(\vect{v}_{-\alpha}\) are the eigenvectors of \(A\) corresponding to \(-\beta \) and \(-\alpha\), respectively. In particular, if \(t\) gets very large, we find very large but negative exponents on the right-hand side. That is, both \(\lim_{t\to\infty}S(t)\) and \(\lim_{t\to\infty} I(t)\) are \(0\). This makes perfect sense intuitively, as we expect all members of the population to get infected and recover. After that, they are neither susceptible nor infected anymore.

Note that, in the long run, we will end up arbitrarily close to \(\vect{0}\) regardless of the starting values of \(S\) and \(I\). That is, if we start in any \(\vect{v}\) and follow the solution \(\vect{y}(t)\) of the system of linear differential equations satisfying the initial condition \(\vect{y}(0)=\vect{v}\), then we will always end up in \(\vect{0}\). In other words, \(\vect{0}\) attracts all points.

For a graphical interpretation of these solutions, we refer to Subsection 9.4.5.

9.4.4. Dealing with complex eigenvalues

Let us once again consider the system \(\vect{y}'=A\vect{y}\). By Proposition 9.4.3, we can find solutions \(\vect{y}=\vect{v}e^{\lambda t}\) where \(\lambda\) is an eigenvalue of \(A\) and \(\vect{v}\) is a corresponding eigenvector. But if \(\lambda\) is not a real number, this does not give a real-valued function. In some applications that’s perfectly fine, but often we’re interested in real solutions to systems of linear differential equations. Can we stil find any of those if some eigenvalues are complex?

Yes, we can! First, we can use the fact that complex eigenvalues come in conjugate pairs \(\lambda=a+bi\), \(\overline{\lambda}=a-bi\) and that the conjugate \(\overline{\vect{v}}\) of an eigenvector \(\vect{v}\) corresponding to \(\lambda\) is an eigenvector corresponding to \(\overline{\lambda}\). Let us write \(\Re{\vect{v}}\) for the vector whose entries are the real parts of the entries of \(\vect{v}\) and \(\Im{\vect{v}}\) for the vector whose entries are the imaginary parts of the entries of \(\vect{v}\). Then, since any linear combination of \(\vect{v}e^{\lambda t}\) and \(\overline{\vect{v}}e^{\overline{\lambda} t}\) is a solution, we have:

\[\begin{split}\begin{align*} \frac{1}{2}\vect{v}e^{\lambda t}+\frac{1}{2}\overline{\vect{v}}e^{\overline{\lambda}t}&=\Re{\vect{v}e^{\lambda t}},\\ \frac{1}{2i}\vect{v}e^{\lambda t}-\frac{1}{2i}\overline{\vect{v}}e^{\overline{\lambda}t}&=\Im{\vect{v}e^{\lambda t}}, \end{align*}\end{split}\]

hence \(\Re{\vect{v}}\) and \(\Im{\vect{v}}\) are solutions, too.

Secondly, we can use the following well-known fact (cf. Definition 10.5.1):

\[e^{(a+bi)t}=e^{at}e^{bit}=e^{at}(\cos(bt)+i\sin(bt))\]

that holds for any real numbers \(a\) and \(b\). In particular, if \(\lambda\) is any complex number, then \(\overline{e^{\lambda}}=e^{\overline{\lambda}}\). So

\[\begin{split}\begin{align*} \vect{v}e^{(a+bi)t}&=(\Re{\vect{v}}+i\Im{\vect{v}})e^{at}(\cos(bt)+i\sin(bt))\\ &=\left[(\Re{\vect{v}}\cos(bt)-\Im{\vect{v}}\sin(bt))+i(\Re{\vect{v}}\sin(bt)+\Im{\vect{v}}\cos(bt))\right]e^{at}. \end{align*}\end{split}\]

From which we conclude that

\[\begin{split}\begin{align*} \Re{\vect{v}e^{\lambda t}}&=(\Re{\vect{v}}\cos(bt)-\Im{\vect{v}}\sin(bt))e^{a t}\quad\text{and}\\ \Im{\vect{v}e^{\lambda t}}&=(\Re{\vect{v}}\sin(bt)+\Im{\vect{v}}\cos(bt))e^{a t} \end{align*}\end{split}\]

If \(A\) is \(2\times 2\), we can summarise this as follows.

Proposition 9.4.4

Let \(A\) be a \(2\times 2\)-matrix with non-real eigenvalue \(\lambda=a+bi\). Let \(\vect{v}\) be an eigenvector associated to \(\lambda.\) Then

\[\begin{split}\begin{align*} \vect{y}_{1}(t)&=\Re{\vect{v}e^{\lambda t}}=(\Re{\vect{v}}\cos(bt)-\Im{\vect{v}}\sin(bt))e^{at}\quad\text{and}\\ \vect{y}_{2}(t)&=\Im{\vect{v}e^{\lambda t}}=(\Re{\vect{v}}\sin(bt)+\Im{\vect{v}}\cos(bt))e^{at} \end{align*}\end{split}\]

are linearly independent solutions to the linear system of differential equations \(\vect{y}'=A\vect{y}\).

For a graphical interpretation of these solutions, we refer to Subsection 9.4.5.

9.4.5. Trajectories

In this section, we will see the geometric interpretation of the several cases we have dealt with. Note that the solution of a dynamical system \(\vect{x}'=A\vect{x}\) contains as many constants as there are rows in \(A\). Therefore, if \(A\) is an \(n\times n\)-matrix and \(\vect{x}_{0}\) is a vector in \(\R^{n}\), there will be one solution of \(\vect{x}'=A\vect{x}\) that satisfies \(\vect{x}(0)=\vect{x}_{0}\).

trajectoryflow map

Definition 9.4.4

Let \(\vect{x}'=A\vect{x}\) be a dynamical system where \(A\) is an \(n\times n\)-matrix. By a trajectory we mean a solution of an initial value problem

\[\vect{x}'=A\vect{x},\quad \vect{x}(0)=\vect{x}_{0}\]

for some \(\vect{x}_{0}\) in \(\R^{n}\). If \(A\) happens to be a \(2\times 2\)-matrix, such a trajectory describes a curve in the plane. By a flow map of a dynamical system, we mean a map in which several such curves have been plotted.

Fig. 9.4.1 On the left a trajectory for the dynamical system associated to a \(2\times2\)-matrix. This trajectory is fully determined by a single initial value, which is indicated by the blue dot. Note that any other initial value which on this trajectory determines the same trajectory. On the right, a flow map for the same dynamical system is plotted. For each trajectory, an initial value is indicated.

It turns out that the eigenvalues and in particular their magnitudes determine what such a flow map will look like. The following definition describes all possible cases.

sourcerepellerunstable spiral pointsinkcentresaddle pointstable spiral pointattractor

Definition 9.4.5

If \(A\) is a \(2\times 2\)-matrix with real eigenvalues \(\lambda_{1}\) and \(\lambda_{2}\), then the origin is called:

• an attractor or a sink if \(\lambda_{1},\lambda_{2}<0\).

• a repeller or a source if \(\lambda_{1},\lambda_{2}>0\).

• a saddle point if \(\lambda_{1}\lambda_{2}<0\), i.e. if \(\lambda_{1}\) and \(\lambda_{2}\) have opposite signs.

The three different behaviours are illustrated in Figure 9.4.2.

Suppose now that \(A\) is a \(2\times 2\)-matrix with complex eigenvalues \(a\pm bi\). Then the origin is called

• a centre if \(a=0\).

• a stable spiral point if \(a<0\).

• an unstable spiral point if \(a>0\).

An example of a spiral point can be seen in Figure 9.4.2.

To see where the names come from, consider the solutions given in Proposition 9.4.4. If \(a<0\) in this proposition, then \(e^{at}\) will become arbitrarily small, so as \(t\) increases, \(\vect{y}(t)\) will approach \(0\). In this case, the trajectory will spiral towards the origin. If \(a>0\), then \(e^{at}\) becomes arbitrarily large and the trajectory will spiral away from the origin. Finally, if \(a=0\) – that is, if we have a purely imaginary eigenvalue – \(\vect{y}(t)\) will move along an elliptic trajectory around the origin.

Fig. 9.4.2 The possible behaviours of the origin illustrated. On the top left, it’s an attractor, on the top right a repeller, on the bottom left a saddle point, and on the bottom right a spiral point. For the spiral point, do you expect the real part of the eigenvalues to be positive or negative, given the figure? That is, do you expect the centre to be a stable spiral point, an unstable spiral point, or a centre?

9.4.6. Systems with higher derivatives

In this section, we will see a trick for dealing with higher derivatives. It works as long as the system under consideration is still linear. Suppose we have a system like this:

\[\begin{split} \begin{cases} x_{1}'''&=ax_{1}+bx_{2},\\ x_{2}'&=cx_{1}+dx_{2}. \end{cases} \end{split}\]

How do we solve this? Our usual method fails because of the higher-order derivative of \(x_{1}\) appearing in the first equation. However, we can remedy this problem by introducing the new dummy variables \(y_{1}=x'_{1}\) and \(y_{2}=x''_{1}\). Now our system becomes:

\[\begin{split} \begin{cases} y_{2}'&=ax_{1}+bx_{2},\\ y_{1}'&=y_{2},\\ x_{1}'&=y_{1},\\ x_{2}'&=cx_{1}+dx_{2} \end{cases} \end{split}\]

and this is of the form we have discussed! It translates to the equation \(\vect{x}'=A\vect{x}\) where

\[\begin{split} A=\begin{pmatrix} 0&0&a&b\\ 1&0&0&0\\ 0&1&0&0\\ 0&0&c&d \end{pmatrix}\quad\text{and}\quad\vect{x}=\begin{pmatrix} y_{2}\\ y_{1}\\ x_{1}\\ x_{2} \end{pmatrix}. \end{split}\]

As an basic application, we can solve second-order linear differential equations:

Proposition 9.4.5

If \(\lambda\) is a root of the quadratic equation \(ax^{2}+bx+c=0\), then the differential equation \(ax''+bx'+cx=0\) has solution \(e^{\lambda t}\).

Proof of Proposition 9.4.5

By putting \(y=x'\), we can rewrite the differential equation \(ax''+bx'+cx=0\) as

\[\begin{split} \begin{cases} y'&=-\frac{b}{a}y-\frac{c}{a}x,\\ x'&=y \end{cases} \end{split}\]

or as \(\vect{x}'=A\vect{x}\) where

\[\begin{split} A=\begin{pmatrix} -\frac{b}{a}&-\frac{c}{a}\\ 1&0 \end{pmatrix} \quad\text{and}\quad \vect{x}=\begin{pmatrix} y\\ x \end{pmatrix}. \end{split}\]

To find solutions, we need the eigenvalues of \(A\). So:

\[\begin{split} \begin{align*} 0&=\begin{vmatrix} -\frac{b}{a}-\lambda&-\frac{c}{a}\\ 1&-\lambda \end{vmatrix}=\lambda^{2}+\frac{b}{a}\lambda+\frac{c}{a}\\ &\Updownarrow\\ 0&=a\lambda^{2}+b\lambda+c \end{align*} \end{split}\]

whence the eigenvalues are precisely the solutions of \(ax^{2}+bx+c=0\). Suppose \(\lambda\) is one of these roots and assume \(\vect{v}\) is a corresponding eigenvector. By Proposition 9.4.3, \(d \vect{v}e^{\lambda x}\) is a solution for any scalar \(d\). As \(x\) is the second component of this solution vector, \(d v_{2}e^{\lambda x}\) is a solution to the original second-order differential equation and so is any scalar multiple of it. Note that \(v_{2}\) is not \(0\), because for any

\[\begin{split}\vect{0}\neq\vect{v}=\begin{pmatrix} v_{1}\\ 0 \end{pmatrix}\quad\text{we have}\quad A\vect{v}=\begin{pmatrix} -\frac{b}{a}v_{1}\\ v_{1} \end{pmatrix}\end{split}\]

which is not a multiple of \(\vect{v}\), so such a \(\vect{v}\) cannot be an eigenvector.

9.4.7. Grasple exercises

Grasple Exercise 9.4.1

Solving an initial value problem step by step.

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Grasple Exercise 9.4.2

A particle moving in a planar force field.

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Grasple Exercise 9.4.3

Another initial value problem.

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Grasple Exercise 9.4.4

General solution of system of first-order differential equations.

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Grasple Exercise 9.4.5

Classify the nature of the origin as an attractor, repeller, saddle point, spiral point or centre point.

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Grasple Exercise 9.4.6

Another classification question.

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Grasple Exercise 9.4.7

One more classification question.

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Grasple Exercise 9.4.8

For which value of a parameter will the origin be a repeller?

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Grasple Exercise 9.4.9

Classify and visualise.

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Grasple Exercise 9.4.10

Classify and visualise once more.

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Grasple Exercise 9.4.11

From a visualisation to eigenvalues and eigenvectors.

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