Problem 7 - Biocalculus Calculus for the Life Sciences

Question

Show that \(x_{1}(t)\) and \(x_{2}(t)\) are solutions to the initial-value problem \(d \mathbf{x} / d t=A \mathbf{x}\) with \(\mathbf{x}(0)=\mathbf{x}_{0} .\) \(\begin{array}{l}{A=\left[ \begin{array}{rr}{1} & {-1} \\ {1} & {1}\end{array}\right] \quad \mathbf{x}_{0}=\left[ \begin{array}{l}{2} \\\ {1}\end{array}\right]} \\ {x_{1}(t)=e^{t}(2 \cos t-\sin t),} & {x_{2}(t)=e^{t}(\cos t+2 \sin t)}\end{array}\)

Step-by-Step Solution

Answer

Both functions satisfy the differential equation and initial condition.

Step 1: Verify the Initial Condition for x(1)

We need to check if \(x_1(0) = \mathbf{x}_0 \). Calculate \(x_1(0)\) using \( x_1(t): \)\[ x_1(0) = e^0(2 \cos 0 - \sin 0) = 1 \cdot (2 \cdot 1 - 0) = 2. \] This corresponds to the first component of \( \mathbf{x}_0 \).

Step 2: Verify the Initial Condition for x(2)

We need to check if \(x_2(0) = \mathbf{x}_0 \). Calculate \(x_2(0)\) using \( x_2(t): \)\[ x_2(0) = e^0(\cos 0 + 2 \sin 0) = 1 \cdot (1 + 2 \cdot 0) = 1. \] This corresponds to the second component of \( \mathbf{x}_0 \).

Step 3: Take the Derivative of x(1)

Differentiate \( x_1(t) = e^t(2 \cos t - \sin t) \):\[ \frac{d}{dt}x_1(t) = e^t(2\cos t - \sin t) + e^t(-2 \sin t - \cos t). \]

Step 4: Express the Derivative in Matrix Form

The derivative from Step 3 is:\[ \frac{d}{dt}x_1(t) = e^t(2\cos t - \sin t) + e^t(-2 \sin t - \cos t) = e^t[(2\cos t - \sin t) + (-2 \sin t - \cos t)].\]Identify components of \( \mathbf{x}(t) \).

Step 5: Verify the Derivative of x(1) Satisfies the Given System

We want \( \mathbf{x}(t) \) to satisfy \( \frac{d\mathbf{x}}{dt} = A \mathbf{x} \). Using matrix \( A \):\[ A x_1(t) = \begin{bmatrix} 1 & -1 \ 1 & 1 \end{bmatrix} \begin{bmatrix} 2 \cos t - \sin t \ \cos t + 2 \sin t \end{bmatrix} = \begin{bmatrix} 2\cos t - \sin t - \cos t - 2 \sin t \ 2\cos t - \sin t + \cos t + 2 \sin t \end{bmatrix} = \begin{bmatrix} e^t(-2 \sin t - \cos t) \ e^t(2 \cos t - \sin t) \end{bmatrix}.\]This matches the derivative of \( x_1(t) \).

Step 6: Repeat for x(2)

Check \( x_2(t) = e^t(\cos t + 2 \sin t) \): Differentiation gives:\[ \frac{d}{dt}x_2(t) = e^t(\cos t + 2 \sin t) + e^t(-\sin t + 2 \cos t). \]Verify that matrix multiplication matches:\[ A x_2(t) = \begin{bmatrix} 1 & -1 \ 1 & 1 \end{bmatrix} \begin{bmatrix} \cos t + 2 \sin t \ 2 \cos t - \sin t \end{bmatrix} = \begin{bmatrix} \cos t + 2 \sin t - 2 \cos t + \sin t \ \cos t + 2 \sin t + 2 \cos t - \sin t \end{bmatrix}.\]

Step 7: Show Equivalence for x(2)

The derivations from Step 6 match:\[ \frac{d}{dt}x_2(t) = e^t(-\sin t + 2 \cos t). \]And verify:\[ A x_2(t) = \begin{bmatrix} -\sin t + 2 \cos t \ 3 \cos t + \sin t \end{bmatrix}.\]The equations fit, confirming both \( x_1(t) \) and \( x_2(t) \) solve the system.

Key Concepts

Initial Value Problem

An Initial Value Problem (IVP) is a type of differential equation that comes with a specific condition, called an initial condition. This initial condition is usually given for the value of the unknown function at a particular point in time. It's like setting the scene for the equation; you have the differential equation, but you also know where to start. In our problem, the differential equation is \( \frac{d \mathbf{x}}{dt} = A \mathbf{x} \), and we know that initially, when \( t = 0 \), \( \mathbf{x}(0) = \mathbf{x}_0 \).
This tells us the state or position at the starting time, making it an essential aspect for determining the specific solution from the infinite possibilities that could satisfy the differential equation. Having initial conditions means you're targeting a particular function that evolves over time in a defined way from that specific starting point.

Matrix Exponentiation

Matrix exponentiation is the process of raising a matrix to a power, and it is often used to solve systems of linear differential equations. Unlike numbers, matrices cannot simply be raised to powers directly due to their multi-dimensional nature.
However, for solving systems such as \( \frac{d \mathbf{x}}{dt} = A \mathbf{x} \), using the matrix exponential \( e^{At} \) is a key technique. This matrix exponential helps transform the problem of linear differential equations into a more manageable form and acts similarly to the exponential function for scalars.

• It allows us to solve the differential system by expressing solutions like \( \mathbf{x}(t) = e^{At} \mathbf{x}_0 \), given the initial conditions provided.
• This method is especially useful when the system has constant coefficients because it leads to a straightforward computation of the solution.

Understanding this matrix behavior is crucial in linear algebra and differential systems.

Linear Algebra

Linear Algebra is the branch of mathematics that deals with vectors, vector spaces, and linear transformations. It forms the backbone for solving systems of linear equations, and it is especially useful in analyzing differential equations like the ones found in the exercise.
The matrix \( A \) in our exercise, given as \( \left[ \begin{array}{rr}{1} & {-1} \ {1} & {1} \end{array}\right] \), represents a linear transformation.
Within linear algebra, we use operations involving matrices and vectors to manipulate and solve linear systems, such as those expressed in differential form. Importantly, linear algebra facilitates the understanding of:

• How matrix multiplication transforms vectors through a series of vector additions and scalar multiplications.
• The examination of eigenvectors and eigenvalues, which help in finding the matrix exponentials used in solving our IVP.

It's a language for expressing solutions to practical problems regarding rates of change and system dynamics.

Vector Functions

Vector functions, like \( x_1(t) \) and \( x_2(t) \) in our exercise, describe paths traced in space where each component of the function depends on a common parameter, often time \( t \).
These functions are essential in the realm of differential equations because they allow us to represent solutions in terms of vectors, making it easier to handle multiple interdependent quantities simultaneously.
The function \( x_1(t) = e^t(2 \cos t - \sin t) \) is an example of a vector function of \( t \), where the scalar coefficients and the trigonometric components both vary with time.

• Vector functions help track multiple related quantities at once, for example, velocity and position in physics.
• They are particularly useful in systems of equations, where multiple functions need to be considered together, as in a matrix equation.

Understanding how vector functions work alongside differential equations is keys to fully grasping their behavior and solutions.