Question

Let \(A\) be a \((2 \times 2)\) constant matrix, and let \((\lambda, \mathbf{u})\) be an eigenpair for \(A\). Assume that \(\lambda\) is real, \(\lambda \neq 0\), and $$ \mathbf{u}=\left[\begin{array}{l} u_{1} \\ u_{2} \end{array}\right] $$ Consider the phase plane for the autonomous linear system \(\mathbf{y}^{\prime}=A \mathbf{y}\). We can define a phase-plane line through the origin by the parametric equations \(x=\tau u_{1}, y=\tau u_{2}\), \(-\infty<\tau<\infty\). Let \(P\) be any point on this line, say \(P=\left(\tau_{0} u_{1}, \tau_{0} u_{2}\right)\) for some \(\tau_{0} \neq 0\). (a) Show that at the point \(P, x^{\prime}=\tau_{0} \lambda u_{1}\) and \(y^{\prime}=\tau_{0} \lambda u_{2}\). (b) How is the velocity vector \(\mathbf{v}(t)=x^{\prime}(t) \mathbf{i}+y^{\prime}(t) \mathbf{j}\) at point \(P\) oriented relative to the line?

Short answer

Short Answer: At point P, the derivatives are \(x^{\prime} = \tau_{0}\lambda u_{1}\) and \(y^{\prime} = \tau_{0}\lambda u_{2}\). The orientation of the velocity vector \(\mathbf{v}(t)\) at point P is parallel to the line at the point P.

Step-by-step solution

Given the eigenpair \((\lambda, \mathbf{u})\), we know that \(A\mathbf{u} = \lambda\mathbf{u}\). This means that $$ A\left[\begin{array}{l} u_{1} \\\ u_{2} \end{array}\right] = \lambda\left[\begin{array}{l} u_{1} \\\ u_{2} \end{array}\right]. $$ #Step 2: Parametric Equations for the Phase-plane Line#

We are given the parametric equations for the phase-plane line through the origin: \(x=\tau u_{1}, y=\tau u_{2}\). We can write this as a vector equation: $$ \mathbf{y} = \tau\mathbf{u} = \tau\left[\begin{array}{l} u_{1} \\\ u_{2} \end{array}\right]. $$ #Step 3: Calculate \(x^{\prime}\) and \(y^{\prime}\) at Point \(P\)#

To find the derivatives \(x^{\prime}\) and \(y^{\prime}\) at point \(P\), we need to differentiate the parametric equations with respect to time and then evaluate them at \(\tau = \tau_{0}\). So, if we take the derivative of \(\mathbf{y}\) with respect to time, we get \(\mathbf{y}^{\prime} = \frac{d\mathbf{u}}{dt} = A\mathbf{u}\). From step 1, we know that \(A\mathbf{u} = \lambda\mathbf{u}\), so $$ \mathbf{y}^{\prime} = \lambda\mathbf{u} = \lambda\left[\begin{array}{l} u_{1} \\\ u_{2} \end{array}\right]. $$ Now, we need to multiply this by \(\tau_0\) to get the derivatives at point \(P\). So, $$ \tau_{0}\mathbf{y}^{\prime} = \tau_{0}\lambda\left[\begin{array}{l} u_{1} \\\ u_{2} \end{array}\right] = \left[\begin{array}{l} \tau_{0}\lambda u_{1} \\\ \tau_{0}\lambda u_{2} \end{array}\right]. $$ This means that at point \(P\), we have \(x^{\prime} = \tau_{0}\lambda u_{1}\) and \(y^{\prime} = \tau_{0}\lambda u_{2}\). #Step 4: Orientation of the Velocity Vector \(\mathbf{v}(t)\)#

Now, we need to find the orientation of the velocity vector \(\mathbf{v}(t)\) at point \(P\) relative to the line. We have $$ \mathbf{v}(t)=x^{\prime}(t) \mathbf{i}+y^{\prime}(t) \mathbf{j} = \tau_{0}\lambda u_{1}\mathbf{i} + \tau_{0}\lambda u_{2}\mathbf{j}. $$ Since the vector \(\mathbf{v}(t)\) is a scalar multiple of \(\mathbf{u}\) (i.e., \(\tau_{0}\lambda\)), it is parallel to the line defined by the parametric equations. Therefore, the velocity vector \(\mathbf{v}(t)\) at point \(P\) is oriented parallel to the line at the point \(P\).

Key concepts

Eigenpair

An eigenpair of a matrix consists of an eigenvalue and an eigenvector. Specifically, for a matrix A, if multiplying the matrix by a non-zero vector u is the same as multiplying that vector by a scalar λ, then (λ, u) is an eigenpair of A. In mathematical terms, A · u = λ u.

This concept is critical in understanding the behavior of systems of differential equations, particularly in how solutions evolve over time. In our exercise, we're told λ is real and non-zero, which suggests the system has a real and distinct direction of stretching or compressing along the direction of the eigenvector u.

Autonomous Linear System

An autonomous linear system can be described by a set of linear differential equations that do not explicitly depend on the independent variable, often time (t). The general form of such a system is y′ = A y, where A is a constant matrix and y is a vector of dependent variables.

These systems are noteworthy for their predictability and the ease with which their long-term behavior can be analyzed, especially through phase-plane analysis. Such a system's behavior is entirely determined by the matrix A and can involve equilibrium points, and characteristic directions defined by the eigenvectors of A.

Velocity Vector

In the context of our system, the velocity vector represents the instantaneous rate of change in the system's state. It is denoted by v(t) = x′(t) i + y′(t) j, where i and j are the unit vectors in the Cartesian coordinate system.

The velocity vector's direction and magnitude at a given point provide critical information about the system's trajectory around that point. In our exercise, determining the velocity vector at point P helps us understand the system's local behavior and how the state is evolving precisely at P. Notably, the velocity vector at any point on the phase-plane line through the origin will be parallel to the eigenvector associated with that line.

Parametric Equations

Parametric equations are a way of expressing a mathematical relationship by defining each variable in terms of one or more independent parameters. For the phase-plane line in our exercise, the parametric equations x = τ u1 and y = τ u2 define every point on the line using the parameter τ.

Using parametric equations allows us to describe complex curves and lines in space more simply and can be particularly useful in systems of differential equations. Here, they define a straight line in the phase plane, highlighting the system's behavior along the direction of the eigenvector u. The parameter τ scales the eigenvector, and by changing its value, we can find every point along the phase-plane line.