Problem 31 - Biocalculus Calculus for the Life Sciences

Question

Our focus has been on systems whose coefficient matrices have distinct eigenvalues. A simple example of a system with repeated eigenvalues is $$\frac{d \mathbf{x}}{d t}=\left[ \begin{array}{rr}{-1} & {0} \\ {0} & {-1}\end{array}\right] \mathbf{x}$$ (a) Show that \(x_{1}(t)=c_{1} e^{-t}\) and \(x_{2}(t)=c_{2} e^{-t}\) is a solution. The origin in this case is called a proper node. (b) Try obtaining this general solution by coefficient matrix. eigenvectors and eigenvalues of the coefficient matrix. Comment on anything unusual that occurs.

Step-by-Step Solution

Answer

(a) Verified: \(x_1(t) = c_1 e^{-t}\), \(x_2(t) = c_2 e^{-t}\) are solutions. (b) Repeated eigenvalues, all vectors eigenvectors; no generalized solution needed.

Step 1: Verify given solution for Part (a)

First, we substitute the given solutions \(x_1(t) = c_1 e^{-t}\) and \(x_2(t) = c_2 e^{-t}\) into the differential equation. The matrix form is \[ \frac{d}{dt} \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix} \begin{bmatrix} x_1(t) \ x_2(t) \end{bmatrix} \]Substitute:\[ \frac{d}{dt} \begin{bmatrix} c_1 e^{-t} \ c_2 e^{-t} \end{bmatrix} = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix} \begin{bmatrix} c_1 e^{-t} \ c_2 e^{-t} \end{bmatrix} \]The derivatives are:\[ \begin{bmatrix} -c_1 e^{-t} \ -c_2 e^{-t} \end{bmatrix} \]The matrix product is:\[ \begin{bmatrix} -c_1 e^{-t} \ -c_2 e^{-t} \end{bmatrix} \]Since both sides are equal, \(x_1(t) = c_1 e^{-t}\) and \(x_2(t) = c_2 e^{-t}\) satisfy the differential equation.

Step 2: Determine eigenvalues for Part (b)

Find the eigenvalues of the coefficient matrix \(A = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}\).The characteristic equation is given by:\[ \det(A - \lambda I) = 0 \]\[ \det\begin{bmatrix} -1 - \lambda & 0 \ 0 & -1 - \lambda \end{bmatrix} = 0 \]This simplifies to \((-1 - \lambda)^2 = 0\).Thus, the eigenvalue is \(\lambda = -1\) with multiplicity 2.

Step 3: Determine eigenvectors for Part (b)

We use \(\lambda = -1\) and find the eigenvectors. Solve:\[ (A + I)\mathbf{v} = \mathbf{0} \]\[ \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix} \]Every vector of the form \(\begin{bmatrix} v_1 \ v_2 \end{bmatrix}\) is an eigenvector.Thus, any vector with same-value components is an eigenvector, indicating the infinite possibilities due to the repeated eigenvalue.

Step 4: Form the general solution

Since \(A\) has a repeated eigenvalue and any vector is an eigenvector, the general solution is a combination of the independent scalar multiples of the same form, \(x(t) = c_1 e^{-t} \begin{bmatrix} 1 \ 0 \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} 0 \ 1 \end{bmatrix}\).This is the same form as given in Part (a):\[ \begin{bmatrix} c_1 e^{-t} \ c_2 e^{-t} \end{bmatrix} \]

Step 5: Comment on anything unusual

The unusual aspect is that despite having repeated eigenvalues, this particular coefficient matrix results in independent solutions \(x_1(t)\) and \(x_2(t)\) because it is a diagonal matrix, which behaves differently from other matrices with repeated eigenvalues. Usually, repeated eigenvalues lead to generalized eigenvectors, which is not needed here.

Key Concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are foundational concepts in linear algebra and are crucial in solving systems of linear differential equations like the one in the exercise. An eigenvalue is a scalar that describes the scaling factor applied to an eigenvector, a non-zero vector that only changes by a scalar factor when a linear transformation is applied using a given matrix.

• To find eigenvalues, solve the characteristic equation \(\det(A - \lambda I) = 0\), where \(A\) is your matrix and \(I\) is the identity matrix of the same size.
• Once eigenvalues are identified, you find eigenvectors by solving \((A - \lambda I)\mathbf{v} = \mathbf{0}\), where \(\mathbf{v}\) is the eigenvector.

In the given system, the matrix \(A\) led to the eigenvalue \(\lambda = -1\). Eigenvectors can be any vector because the matrix was diagonal. This gives an infinite set of possible eigenvectors.

Matrix Diagonalization

Matrix diagonalization is the process of finding a diagonal matrix that represents a given square matrix. Diagonalization helps simplify complex matrix operations, and is particularly useful in solving differential equations. However, not all matrices can be diagonalized.

• A matrix is diagonalizable if it has enough linearly independent eigenvectors to form a basis for its vector space.
• In mathematical terms, a matrix \(A\) is diagonalizable if there exists a diagonal matrix \(D\) and an invertible matrix \(P\) such that \(A = PDP^{-1}\).

For the exercise, the matrix \(A\) is already diagonal. This simplifies solving the system because the matrix form of the solutions is straightforward: each variable can be addressed independently.

Proper Node

The term "proper node" describes the behavior of solutions at a given point, often the origin, in the phase plane of differential equations. It is a type of equilibrium point where trajectories converge directly to the node along linear paths.

• A proper node occurs typically when eigenvalues are real and both have the same sign, leading the solutions to be stable or unstable.
• In the exercise, the repeated eigenvalue \(\lambda = -1\) creates a stable proper node, because the solutions decay exponentially to the origin as \(t\) increases.

Proper nodes provide predictable and simple dynamics, illustrated by straight lines moving into or out of the equilibrium.

Repeated Eigenvalues

Repeated eigenvalues arise when the algebraic multiplicity of an eigenvalue (the number of times it appears as a root of the characteristic equation) is greater than one. This can result in complexities not seen with unique eigenvalues.

• Typically, repeated eigenvalues require finding generalized eigenvectors to ensure a complete solution basis.
• In special cases like the diagonal matrix in this exercise, you don't need generalized eigenvectors, because any vector corresponding to the repeated eigenvalue is a valid eigenvector, maintaining full solutions integrity.

The differentiation with repeated eigenvalues can vary but in this problem, the matrix simplicity allowed straightforward solutions without needing additional calculation.