Question

Solve the given initial value problem. Describe the behavior of the solution as \(t \rightarrow \infty\). $$ \mathbf{x}^{\prime}=\left(\begin{array}{rrr}{1} & {1} & {2} \\ {0} & {2} & {2} \\ {-1} & {1} & {3}\end{array}\right) \mathbf{x}, \quad \mathbf{x}(0)=\left(\begin{array}{l}{2} \\ {0} \\ {1}\end{array}\right) $$

Short answer

Short answer: As t approaches infinity, all the components of the solution vector grow exponentially to infinity, with the second and third components growing at a faster rate compared to the first one.

Step-by-step solution

Eigenvalues and Eigenvectors

Firstly, we need to find the eigenvalues and eigenvectors of matrix: $$ A = \left(\begin{array}{ccc}{1} & {1} & {2} \\\ {0} & {2} & {2} \\\ {-1} & {1} & {3}\end{array}\right) $$ We compute the characteristic polynomial, \(|A - \lambda I|=0\), to find the eigenvalues: $$ \left|\begin{array}{ccc}{1-\lambda} & {1} & {2} \\\ {0} & {2-\lambda} & {2} \\\ {-1} & {1} & {3-\lambda}\end{array}\right| = \left(2-\lambda\right)\left[(1-\lambda)(3-\lambda)+2 \right] = (2-\lambda)^2 (\lambda-4). $$ Eigenvalues are: \(\lambda_1=2\) (with algebraic multiplicity 2) and \(\lambda_2=4\) Next, we find eigenvectors for each eigenvalue: For \(\lambda_1=2\): \((A-2I)v=0\) gives the eigenvectors: $$ v_1=\left(\begin{array}{l}{-1} \\\ {0} \\\ {1}\end{array}\right), \quad v_2=\left(\begin{array}{l}{0} \\\ {1} \\\ {1}\end{array}\right) $$ For \(\lambda_2=4\): \((A-4I)v=0\) gives the eigenvector: $$ v_3=\left(\begin{array}{l}{-2} \\\ {1} \\\ {1}\end{array}\right) $$

Construct Matrix Exponential

Notice that we have a complete set of eigenvectors, meaning that the matrix \(A\) is diagonalizable. We will use diagonalization to compute \(e^{At}\). Let \(P\) be the matrix formed by the eigenvectors: $$ P = \left(\begin{array}{ccc}{-1} & {0} & {-2} \\\ {0} & {1} & {1} \\\ {1} & {1} & {1}\end{array}\right) $$ We compute the inverse of \(P\) and find the diagonal matrix \(D\): $$ P^{-1} = \left(\begin{array}{ccc}{-1} & {-1} & {3} \\\ {1} & {1} & {0} \\\ {1} & {1} & {-2}\end{array}\right), \quad D = \mathrm{diag}(2, 2, 4) $$ Now, we find the matrix exponential: $$ e^{At} = Pe^{Dt}P^{-1} = P \mathrm{diag}(e^{2t}, e^{2t}, e^{4t}) P^{-1}. $$

Multiply with Initial Vector

We multiply \(e^{At}\) with the initial vector: $$ \mathbf{x}(t) = e^{At} \mathbf{x}(0) = P \mathrm{diag}(e^{2t}, e^{2t}, e^{4t}) P^{-1} \left(\begin{array}{l}{2} \\\ {0} \\\ {1}\end{array}\right) $$ After some calculations, we obtain the solution: $$ \mathbf{x}(t) = \left(\begin{array}{l}{e^{2t}(1-2e^{-2t})} \\\ {e^{2t}(e^{2t}-1)} \\\ {e^{2t}(2e^{2t}-1)}\end{array}\right) $$

Analyze the Behavior as t \(\rightarrow \infty\)

As \(t \rightarrow \infty\), we can see that the terms \(e^{-2t}\) will become negligible compared to the other exponential terms. Therefore, the solution, as \(t \rightarrow \infty\), can be rewritten as: $$ \lim_{t \rightarrow \infty} \mathbf{x}(t) = \left(\begin{array}{l}{e^{2t}} \\\ {e^{4t}} \\\ {2e^{4t}}\end{array}\right) $$ This implies that all components of the solution vector grow exponentially to infinity, with the second and third components growing at a faster rate compared to the first one.

Key concepts

Eigenvalues and Eigenvectors

When dealing with differential equations involving matrices, understanding eigenvalues and eigenvectors is crucial. They help you discern critical properties of a matrix that we solve within these systems. An eigenvalue of a matrix is a number that, when subtracted from the diagonal elements and taken through determinant calculations, provides solutions that lead to vectors (eigenvectors) kept in similar directions after the transformation represented by the matrix.
For our problem, the matrix involved is \[ A = \begin{bmatrix} 1 & 1 & 2 \ 0 & 2 & 2 \ -1 & 1 & 3 \end{bmatrix} \] The characteristic polynomial is obtained by solving \(|A - \lambda I| = 0\). After calculations, we find that the eigenvalues are \(\lambda_1 = 2\) with a multiplicity of 2, and \(\lambda_2 = 4\).

• Eigenvalues signify how much the eigenvectors transform in terms of direction and magnitude.
• Having these specific eigenvectors \[ v_1=\begin{bmatrix} -1 \ 0 \ 1 \end{bmatrix}, \quad v_2=\begin{bmatrix} 0 \ 1 \ 1 \end{bmatrix}, \quad v_3=\begin{bmatrix} -2 \ 1 \ 1 \end{bmatrix} \] allows the matrix to be broken down into simpler, more manageable parts.

Exploring eigenvectors offers insight into the directionality of solutions, and when combined with eigenvalues, they tell us how a system will behave over time.

Matrix Exponential

Once we have the eigenvalues and eigenvectors, the next step is to find the matrix exponential. This is crucial for solving systems of linear differential equations. The matrix exponential, denoted as \(e^{At}\), is derived from the diagonalization of matrix \(A\), allowing for easier computation when dealing with the evolution of the system over time.
To compute \(e^{At}\), we use the formula: \[ e^{At} = Pe^{Dt}P^{-1} \]where \(P\) is the matrix formed by the eigenvectors, and \(D\) is the diagonal matrix comprising the eigenvalues. Specifically, we have \[ P = \begin{bmatrix} -1 & 0 & -2 \ 0 & 1 & 1 \ 1 & 1 & 1 \end{bmatrix}\]and \[ D = \text{diag}(2, 2, 4). \]

• The exponential for each eigenvalue simplifies to \(e^{Dt} = \text{diag}(e^{2t}, e^{2t}, e^{4t})\).
• Multiplying by the inverse of \(P\) translates to real-life applications in evolving systems, showcasing how initial conditions change with time.

The matrix exponential is central to exploring how different states of a system evolve, making it easier to predict future behaviors.

Initial Value Problem

The concept of an initial value problem (IVP) forms the heart of many differential equations. An IVP specifies the state of the system at a given point, and from this point, we can find future states using the matrix exponential.
In our exercise, the initial condition is given by \[ \mathbf{x}(0) = \begin{bmatrix} 2 \ 0 \ 1 \end{bmatrix}. \]
Using this initial vector along with the exponential of the matrix \(A\), we find the solution at any future time \(t\).
The process is summarised as:

• Compute \(e^{At}\) and multiply it by \(\mathbf{x}(0)\).
• This multiplication delivers the state \(\mathbf{x}(t)\), predicting future behaviors from known starting conditions.

The beauty of the initial value problem lies in its ability to uniquely determine a system's future evolution using known conditions. It's pivotal for forecasting and controlling systems dynamically.

Asymptotic Behavior

Understanding the asymptotic behavior of a solution to a differential equation tells us how the solution behaves as time progresses towards infinity (\(t \rightarrow \infty\)). For systems like ours, where exponential growth or decay occurs, assessing long-term behavior provides critical insights.
As the time becomes very large, terms such as \(e^{-2t}\) become negligible compared to terms like \(e^{2t}\) or \(e^{4t}\). This results in a simplifying expression:\[\lim_{t \rightarrow \infty} \mathbf{x}(t) = \begin{bmatrix} e^{2t} \ e^{4t} \ 2e^{4t} \end{bmatrix}.\]

• The second and third components grow exponentially faster than the first, indicating their dominant influence as time progresses.
• Exponential terms dictate the speed of growth, showing the stability or instability of the system.

Studying asymptotic behavior is quintessential where understanding the stability and potential unbounded growth of a system matters most, offering foresight into the future dynamics of complex systems.