Problem 27 - Biocalculus Calculus for the Life Sciences

Question

Solve the initial value problem \(d \mathbf{x} / d t=A \mathbf{x}\) with \(\mathbf{x}(0)=\mathbf{x}_{0} .\) \(A=\left[ \begin{array}{rr}{2} & {-5} \\ {2} & {1}\end{array}\right] \quad \mathbf{x}_{0}=\left[ \begin{array}{l}{1} \\ {1}\end{array}\right]\)

Step-by-Step Solution

Answer

The solution is the combination of eigenvectors scaled by constants determined from the initial condition.

Step 1: Eigenvalue Calculation

First, we need to find the eigenvalues of the matrix \( A \). The eigenvalues \( \lambda \) are solutions to the characteristic equation: \( \det(A - \lambda I) = 0 \). For matrix \( A = \begin{bmatrix} 2 & -5 \ 2 & 1 \end{bmatrix} \), we form the determinant as follows:\[ \det \left( \begin{bmatrix} 2 - \lambda & -5 \ 2 & 1 - \lambda \end{bmatrix} \right) = (2 - \lambda)(1 - \lambda) - (-5)(2) \]Simplifying gives the characteristic equation:\[ (2 - \lambda)(1 - \lambda) + 10 = \lambda^2 - 3\lambda + 12 = 0 \]

Step 2: Solve the Characteristic Equation

We solve the characteristic equation \( \lambda^2 - 3\lambda + 12 = 0 \). The roots of this equation give us the eigenvalues, which can be found using the quadratic formula:\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substituting \( a = 1, b = -3, c = 12 \):\[ \lambda = \frac{3 \pm \sqrt{(-3)^2 - 4 \times 1 \times 12}}{2} = \frac{3 \pm \sqrt{9 - 48}}{2} = \frac{3 \pm i\sqrt{39}}{2} \]The eigenvalues are \( \lambda_1 = \frac{3 + i \sqrt{39}}{2} \) and \( \lambda_2 = \frac{3 - i \sqrt{39}}{2} \).

Step 3: Eigenvector Calculation

For each eigenvalue, find the corresponding eigenvector \( \mathbf{v} \) by solving \( (A - \lambda I)\mathbf{v} = 0 \). Using \( \lambda_1 = \frac{3 + i \sqrt{39}}{2} \) as an example:\[ \left( \begin{bmatrix} 2 & -5 \ 2 & 1 \end{bmatrix} - \frac{3 + i \sqrt{39}}{2} \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \right) \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = 0 \]Substitute \( \lambda_1 \) and solve for \( v_1 \) and \( v_2 \). Repeat for \( \lambda_2 \).

Step 4: Construct General Solution

The solution to \( \frac{d \mathbf{x}}{dt} = A \mathbf{x} \) is given by \( \mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 \). Substitute \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) obtained from earlier to construct this solution.

Step 5: Apply Initial Conditions

Use the given initial condition \( \mathbf{x}(0) = \mathbf{x}_0 = \begin{bmatrix} 1 \ 1 \end{bmatrix} \) to find constants \( c_1 \) and \( c_2 \). Substitute \( t = 0 \) into the general solution to form the equation system:\[ \begin{bmatrix} c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 \end{bmatrix} = \begin{bmatrix} 1 \ 1 \end{bmatrix} \]Solve the system to find \( c_1 \) and \( c_2 \).

Step 6: Write the Particular Solution

With \( c_1 \) and \( c_2 \) determined, substitute them back into the general solution to write the particular solution for the initial value problem. This represents the time evolution of \( \mathbf{x}(t) \).

Key Concepts

Eigenvalue Calculation

To solve an initial value problem of the form \(\frac{d \mathbf{x}}{dt} = A \mathbf{x}\), the first important step is determining the eigenvalues of matrix \( A \).
This is crucial because eigenvalues help us understand the system's behavior over time.
The eigenvalues are found by solving the equation \(\det(A - \lambda I) = 0\), where \(\lambda\) represents the eigenvalues and \(I\) is the identity matrix of the same dimensions as \(A\).

For our example, with matrix \(A = \begin{bmatrix} 2 & -5 \ 2 & 1 \end{bmatrix}\), the determinant of \(A - \lambda I\) is set to zero:

• Form the new matrix: \( \begin{bmatrix} 2 - \lambda & -5 \ 2 & 1 - \lambda \end{bmatrix} \)
• Calculate the determinant: \((2 - \lambda)(1 - \lambda) + 10\)
• Simlify to get the characteristic equation: \(\lambda^2 - 3\lambda + 12 = 0\)

Finding these eigenvalues is the first step towards constructing a solution to the differential equation.

Characteristic Equation

The characteristic equation arises from the determinant calculation \(\det(A-\lambda I) = 0\).
It is a polynomial equation in terms of \(\lambda\).
In our example, the characteristic equation we obtain is \(\lambda^2 - 3\lambda + 12 = 0\).

To solve this equation, the quadratic formula is applied:

• The general quadratic formula is \(\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
• Substitute \(a = 1\), \(b = -3\), and \(c = 12\) into the formula.
• This results in \(\lambda = \frac{3 \pm i \sqrt{39}}{2}\).

These complex roots indicate oscillatory solutions, which will be reflected in the solved system's behavior.
This leads to solutions involving complex exponentials, indicating a system that oscillates over time.

General Solution

Once the eigenvalues and their corresponding eigenvectors are found, the general solution to the differential equation \(\frac{d \mathbf{x}}{dt} = A \mathbf{x}\) can be constructed.
For our system, the general solution takes the form:

• \(\mathbf{x}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2\)

Where \(c_1\) and \(c_2\) are constants determined by initial conditions, and \(\mathbf{v}_1\) and \(\mathbf{v}_2\) are eigenvectors associated with \(\lambda_1\) and \(\lambda_2\), respectively.
This general solution represents the most comprehensive expression for the system's time evolution, capturing the impact of both modes of the system.
The presence of complex exponentials ensures the inclusion of oscillations rooted in the system's dynamics.

Initial Conditions

Initial conditions are given values that help determine the unique solution acknowledging the problem's constraints.
In our initial value problem, the initial condition is \(\mathbf{x}(0) = \mathbf{x}_0 = \begin{bmatrix} 1 \ 1 \end{bmatrix}\).
To incorporate this into our general solution:

• Substitute \(t = 0\) in the general solution.
• Equate it to the given initial condition: \(\begin{bmatrix} c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 \end{bmatrix} = \begin{bmatrix} 1 \ 1 \end{bmatrix}\)
• Solve this system of equations to find the specific values for \(c_1\) and \(c_2\).

Applying the initial conditions allows us to precise the initial setup and obtain a particular solution.
This defines the system's exact response over time, with all constants in place.