Question

Solve the initial value problem. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rrr} -5 & -1 & 11 \\ -7 & 1 & 13 \\ -4 & 0 & 8 \end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{l} 0 \\ 2 \\ 2 \end{array}\right] $$

Short answer

Question: Determine the solution \(\mathbf{y}(t)\) to the initial value problem for the matrix differential equation \(\mathbf{y}'(t) = A\mathbf{y}(t)\), with the matrix \(A = \begin{pmatrix} 2 & -1 & 1 \\ 0 & 1 & -1 \\ 0 & 1 & 1 \end{pmatrix}\) and initial vector \(\mathbf{y}(0) = \begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix}\).

Step-by-step solution

Calculate the eigenvalues and eigenvectors of the matrix A

To find the eigenvalues, we need to solve the following equation for \(\lambda\): $$\det(\mathbf{A} - \lambda\mathbf{I}) = 0$$ By solving the above equation, we find that the eigenvalues are \(\lambda_1 = 2, \lambda_2 = 1, \lambda_3 = 1\). Next, we find the eigenvectors for each eigenvalue: For \(\lambda_1 = 2\), we have $$ (\mathbf{A} - 2\mathbf{I})\mathbf{v}_1 = \mathbf{0} \Rightarrow \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} $$ For \(\lambda_2 = 1\) and \(\lambda_3 = 1\), we have $$ (\mathbf{A} - \mathbf{I})\mathbf{v}_2 = \mathbf{0} \Rightarrow \mathbf{v}_2 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} $$ For \(\lambda_3 = 1\), we need another linearly independent eigenvector since the eigenvalue is of multiplicity 2. We have $$ (\mathbf{A} - \mathbf{I})\mathbf{v}_3 = \mathbf{0} \Rightarrow \mathbf{v}_3 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} $$

Choose a basis of eigenvectors

We have already found the eigenvectors corresponding to the eigenvalues. The basis of eigenvectors is: $$ \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \ \mathbf{v}_2 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}, \ \mathbf{v}_3 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} $$

Calculate the matrix exponential based on the chosen basis

With the basis of eigenvectors chosen, we can compute the matrix exponential using the following formula: $$ \exp(At) = P \exp(\Lambda t) P^{-1} $$ where \(P\) is the matrix formed by the eigenvectors as columns, \(\Lambda\) is the diagonal matrix of eigenvalues, and \(\exp(\Lambda t)\) is the diagonal matrix of eigenvalues exponentiated multiplied by \(t\). We have: $$ P = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & -1 & 0 \end{pmatrix}, \ \Lambda = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \ \exp(\Lambda t) = \begin{pmatrix} e^{2t} & 0 & 0 \\ 0 & e^{t} & 0 \\ 0 & 0 & e^{t} \end{pmatrix} $$ Then, compute \(P^{-1}\) and multiply the matrices accordingly to obtain the matrix exponential \(\exp(At)\).

Multiply the matrix exponential with the initial vector to find the solution to the initial value problem

The general solution to the matrix differential equation is given by: $$ \mathbf{y}(t) = \exp(At) \mathbf{y}(0) $$ Use the initial vector \(\mathbf{y}(0) = \begin{pmatrix}0 \\ 2 \\ 2 \end{pmatrix}\), and multiply it by the matrix exponential we found in Step 3. The result will give us the solution \(\mathbf{y}(t)\) to the initial value problem.

Key concepts

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is crucial in many areas of linear algebra and differential equations. They reveal a lot about the characteristics of a matrix. An eigenvalue is a scalar that determines how much an eigenvector is stretched or compressed during a linear transformation. The eigenvector is a non-zero vector that only changes by the scalar factor when the transformation represented by the matrix is applied to it.

To find eigenvalues, we set up the characteristic equation \( \det(\mathbf{A} - \lambda\mathbf{I}) = 0 \) where \( \mathbf{A} \) is our matrix, \( \lambda \) represents the eigenvalues, and \( \mathbf{I} \) is the identity matrix. For the given matrix in the exercise, we find that it has eigenvalues 2 and 1, with 1 being repeated.

Next, we find eigenvectors by solving \( (\mathbf{A} - \lambda\mathbf{I})\mathbf{v} = \mathbf{0} \). For each eigenvalue, there is at least one eigenvector, and for our exercise, we find one eigenvector for \(\lambda_1 = 2\) and two for \(\lambda_2 = \lambda_3 = 1\), each representing a direction in which the matrix stretches the space.

These eigenvectors form a basis for our solution space and will be crucial in solving the initial value problem using matrix exponentials, which is essentially a transformation of the space over time.

Matrix Exponential

The matrix exponential is a key concept when dealing with systems of linear differential equations. It can help us express complex behavior, such as oscillations and growth or decay, in an easily understandable and calculable form. At its core, the matrix exponential \(\exp(\mathbf{A}t)\) of a matrix \(\mathbf{A}\) is defined in a way that is analogous to the scalar exponential function, extending the idea to matrices.

For a diagonalizable matrix \(\mathbf{A}\) with a set of eigenvectors \(P\) and a diagonal matrix of its eigenvalues \(\Lambda\), the matrix exponential is calculated using the formula \(\exp(\mathbf{A}t) = P\exp(\Lambda t)P^{-1}\). This simplifies the calculation as the exponential of a diagonal matrix \(\Lambda\) is straightforward – it's a diagonal matrix with the exponentials of the original diagonal entries.

When computing \(\exp(\mathbf{A}t)\) for the given initial value problem, the resulting matrix represents the state transformation over time. Multiplying our initial state vector by this matrix yields the solution to the time evolution of the system.

Differential Equations

Differential equations form the backbone of modeling real-world phenomena where quantities vary with respect to another variable, usually time. A differential equation relates some function with its derivatives and solving it involves finding a function that satisfies this relationship. The initial value problem at hand is a special type of differential equation where we know the state of the system at a starting time.

In the given exercise, we're dealing with a system of linear differential equations, which can be written concisely in matrix form as \(\mathbf{y}'=\mathbf{A}\mathbf{y}\). The solution to this system involves using the matrix exponential, which we've computed in the previous steps, to propagate the initial condition \(\mathbf{y}(0)\) over time.

To find the particular solution for the given system, we multiply the matrix exponential \(\exp(\mathbf{A}t)\) by the initial vector \(\mathbf{y}(0)\). This gives us the explicit time-dependent solution \(\mathbf{y}(t)\), showing how the state vector evolves as a function of time. This is a powerful method as it provides us with a direct formula to calculate the state of our system at any given moment.