Problem 9 - Differential Equations and Linear Algebra

Question

Solve \(\mathbf{x}^{\prime}=A \mathbf{x}\) by determining \(n\) linearly independent solutions of the form \(\mathbf{x}(t)=e^{A t} \mathbf{v}\). \(A=\left[\begin{array}{rrr}-8 & 6 & -3 \\ -12 & 10 & -3 \\ -12 & 12 & -2\end{array}\right] .\) You may assume that \(p(\lambda)=\)\ \(-(\lambda+2)^{2}(\lambda-4)\).

Step-by-Step Solution

Answer

We found the eigenvalues \(\lambda_1 = -2\) (with multiplicity 2) and \(\lambda_2 = 4\), and their corresponding eigenvectors \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}\), \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}\), and \(\mathbf{v}_3 = \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix}\). Using the given formula, we found three linearly independent solutions: \(\mathbf{x}_1(t) = e^{-2t}\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}\), \(\mathbf{x}_2(t) = e^{-2t}\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}\), and \(\mathbf{x}_3(t) = e^{4t}\begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix}\).

Step 1: Find the eigenvalues

We are given the characteristic polynomial \(p(\lambda) = -(\lambda + 2)^2(\lambda - 4)\). To find the eigenvalues, we set \(p(\lambda) = 0\): \(-(\lambda + 2)^2(\lambda - 4) = 0\) \( (\lambda + 2)^2(\lambda - 4) = 0\) The eigenvalues are thus \(\lambda_1 = -2\) (with multiplicity 2) and \(\lambda_2 = 4\).

Step 2: Find the eigenvectors

Next, we need to find eigenvectors associated with each eigenvalue. For \(\lambda_1 = -2\), we solve the system \((A - (-2)I)\mathbf{v} = \mathbf{0}\): \(\begin{pmatrix} -8+2 & 6 & -3 \\ -12 & 10+2 & -3 \\ -12 & 12 & -2+2 \end{pmatrix}\mathbf{v} = \begin{pmatrix} -6 & 6 & -3 \\ -12 & 12 & -3 \\ -12 & 12 & 0\end{pmatrix}\mathbf{v} = \mathbf{0}\) We find that there are 2 linearly independent eigenvectors associated with this eigenvalue: \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}\) and \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}\). For \(\lambda_2 = 4\), we solve the system \((A - 4I)\mathbf{v} = \mathbf{0}\): \(\begin{pmatrix} -8-4 & 6 & -3 \\ -12 & 10-4 & -3 \\ -12 & 12 & -2-4 \end{pmatrix}\mathbf{v} = \begin{pmatrix} -12 & 6 & -3 \\ -12 & 6 & -3 \\ -12 & 12 & -6\end{pmatrix}\mathbf{v} = \mathbf{0}\) We find that there is 1 linearly independent eigenvector associated with this eigenvalue: \(\mathbf{v}_3 = \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix}\).

Step 3: Find the linearly independent solutions

We are given that the linearly independent solutions have the form \(\mathbf{x}(t) = e^{A t}\mathbf{v}\). Since we have found the eigenvectors, we can substitute them into the given formula. For \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}\) and \(\lambda_1 = -2\): \(\mathbf{x}_1(t) = e^{-2t}\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}\) For \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}\) and \(\lambda_1 = -2\): \(\mathbf{x}_2(t) = e^{-2t}\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}\) For \(\mathbf{v}_3 = \begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix}\) and \(\lambda_2 = 4\): \(\mathbf{x}_3(t) = e^{4t}\begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix}\) We have found the following three linearly independent solutions: \(\mathbf{x}_1(t) = e^{-2t}\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}, \mathbf{x}_2(t) = e^{-2t}\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}, \mathbf{x}_3(t) = e^{4t}\begin{pmatrix} 1 \\ 2 \\ 4 \end{pmatrix}\)

Key Concepts

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is key to solving differential equations. Eigenvalues are the special values of \( \lambda \) that satisfy the characteristic equation. These scalar values tell us how much an eigenvector is stretched in a given transformation.
An eigenvector is a non-zero vector that only changes in magnitude, not direction, when a linear transformation is applied. This vector remains "in line" with the transformation. In other words, if we apply a matrix transformation to an eigenvector, we end up with a scalar multiple of that eigenvector.To find these, solve the equation \((A - \lambda I)\mathbf{v} = \mathbf{0}\).

• Important result: Eigenvectors corresponding to distinct eigenvalues are linearly independent.
• Applications: Eigenvalues and eigenvectors are crucial in stability analysis and systems of differential equations.

Linear Independence

Linear independence refers to a set of vectors that do not depend linearly on one another. This means, no vector in the set can be formed by a linear combination of other vectors in the set.
In the context of differential equations, linear independence of solutions ensures we have enough unique solutions to construct the general solution. Linearly independent solutions can span the solution space of a differential equation.

• Check for Linear Independence: Use the Wronskian determinant or simple observation of vectors.
  • For eigenvectors, simply ensure they correspond to distinct eigenvalues or have different form.
• Significance: Ensures all possible directions in the vector space are covered by solutions.

Characteristic Polynomial

The characteristic polynomial is closely linked to the matrix of a system of equations. It is usually represented by \( \text{det}(A - \lambda I) = p(\lambda) \).It is derived from the determinant of the matrix \( (A - \lambda I) \), where \( I \) is the identity matrix.
Finding the roots of this polynomial gives us the eigenvalues, which are critical in solving systems of differential equations.

• Calculation: The polynomial results from expanding the determinant \( \text{det}(A - \lambda I) \).
  • For each eigenvalue \( \lambda \), you have \( (\lambda_i + a)^{b} \), where \( a \) and \( b \) are constants.
• Importance: Provides a method to obtain eigenvalues directly.