Question

Suppose the \(n \times n\) matrix function \(A\) and the \(n-\) vector function \(\mathbf{f}\) are continuous on \((a, b)\). Let \(t_{0}\) be in \((a, b),\) let \(\mathbf{k}\) be an arbitrary constant vector, and let \(Y\) be a fundamental matrix for the homogeneous system \(\mathbf{y}^{\prime}=A(t) \mathbf{y}\). Use variation of parameters to show that the solution of the initial value problem $$ \mathbf{y}^{\prime}=A(t) \mathbf{y}+\mathbf{f}(t), \quad \mathbf{y}\left(t_{0}\right)=\mathbf{k} $$ is $$ \mathbf{y}(t)=Y(t)\left(Y^{-1}\left(t_{0}\right) \mathbf{k}+\int_{t_{0}}^{t} Y^{-1}(s) \mathbf{f}(s) d s\right) . $$

Short answer

Short Answer: Given an initial value problem of the form $\mathbf{y}^{\prime}=A(t) \mathbf{y}+\mathbf{f}(t)$, with $\mathbf{y}\left(t_{0}\right)=\mathbf{k}$ and $Y$ being a fundamental matrix for the homogeneous system, the solution using the variation of parameters method is given by: $$\mathbf{y}(t) = Y(t)\left[Y^{-1}(t_0)\mathbf{k} + \int_{t_0}^t \left[ - Y^{-1}(s)Y^{\prime}(s)u(s) + Y^{-1}(s)A(s)Y(s)u(s) + Y^{-1}(s)\mathbf{f}(s)\right]ds\right].$$

Step-by-step solution

Introduce a new function, \(u(t)\), and express the desired solution \(\mathbf{y}(t)\) as the product of the fundamental matrix \(Y(t)\) and this new function: $$\mathbf{y}(t) = Y(t) u(t).$$ #Step 2: Differentiate the solution function#

Differentiate \(\mathbf{y(t)}\) with respect to \(t\): $$\mathbf{y}^{\prime}(t) = Y^{\prime}(t) u(t) + Y(t) u^{\prime}(t).$$ #Step 3: Substitute the original equation and apply initial conditions#

Substitute the original equation \(\mathbf{y}^{\prime}(t) = A(t)\mathbf{y}(t) + \mathbf{f}(t)\) and the definition of the homogeneous solution \(\mathbf{y}(t) = Y(t)u(t)\) into the differentiated equation from Step 2: $$A(t)Y(t)u(t) + \mathbf{f}(t) = Y^{\prime}(t)u(t) + Y(t)u^{\prime}(t).$$ As well, consider the initial condition \(\mathbf{y}(t_0) = \mathbf{k}\), which implies \(Y(t_0)u(t_0) = \mathbf{k}\). #Step 4: Solve for u'(t)#

Notice that we can write the equation from Step 3 as $$Y(t)u^{\prime}(t) = - Y^{\prime}(t)u(t) + A(t)Y(t)u(t) + \mathbf{f}(t).$$ By multiplying both sides by \(Y^{-1}(t)\), we find $$u^{\prime}(t) = - Y^{-1}(t)Y^{\prime}(t)u(t) + Y^{-1}(t)A(t)Y(t)u(t) + Y^{-1}(t)\mathbf{f}(t).$$ #Step 5: Integrate u'(t) to find u(t)#

Integrate \(u^{\prime}(t)\) over the interval \((t_0, t)\) to find \(u(t)\): $$u(t) = u(t_0) + \int_{t_0}^t \left[- Y^{-1}(s)Y^{\prime}(s)u(s) + Y^{-1}(s)A(s)Y(s)u(s) + Y^{-1}(s)\mathbf{f}(s)\right]ds.$$ Using the initial condition, \(Y(t_0)u(t_0) = \mathbf{k}\), we find \(u(t_0) = Y^{-1}(t_0)\mathbf{k}\): $$u(t) = Y^{-1}(t_0)\mathbf{k} + \int_{t_0}^t \left[ - Y^{-1}(s)Y^{\prime}(s)u(s) + Y^{-1}(s)A(s)Y(s)u(s) + Y^{-1}(s)\mathbf{f}(s)\right]ds.$$ #Step 6: Write the final answer using the form of the solution function#

Now that we have an expression for \(u(t)\), substitute it back into the solution function \(\mathbf{y}(t) = Y(t)u(t)\): $$\mathbf{y}(t) = Y(t)\left[Y^{-1}(t_0)\mathbf{k} + \int_{t_0}^t \left[ - Y^{-1}(s)Y^{\prime}(s)u(s) + Y^{-1}(s)A(s)Y(s)u(s) + Y^{-1}(s)\mathbf{f}(s)\right]ds\right].$$ This is the desired solution for the given initial value problem.

Key concepts

Fundamental Matrix

To understand the fundamental matrix, think of it as a key piece in solving differential equations, especially when matrices are involved. Imagine a system of differential equations in the form \( \mathbf{y}' = A(t) \mathbf{y} \). A fundamental matrix \( Y(t) \) is composed of solutions to this homogeneous system, arranged in columns, and it aids in expressing the general solution for the system.

A fundamental matrix \( Y(t) \) has a unique role:

• It consists of linearly independent solutions.
• The determinant of \( Y(t) \) is non-zero, indicating that the matrix is invertible.

Knowing \( Y(t) \) allows us to solve the differential equation as it provides a structure to express any solution through linear combinations. For initial value and non-homogeneous problems, \( Y(t) \) becomes more crucial as it interacts with other functions to determine the behavior of solutions.

Initial Value Problem

An initial value problem (IVP) is one where we seek a particular solution to a differential equation that passes through a specific point, known as the initial condition. In our context, the IVP is given by \( \mathbf{y}^{\prime}=A(t) \mathbf{y}+\mathbf{f}(t), \mathbf{y}(t_{0})=\mathbf{k} \). Here:

• \( \mathbf{y}(t) \) is our unknown function that we're trying to find.
• \( t_0 \) is the initial time where the condition is applied.
• \( \mathbf{k} \) is the value of the solution at \( t_0 \).

This setup ensures that not only do we solve the equation but we also match the initial conditions. The initial condition \( \mathbf{y}(t_0) = \mathbf{k} \) acts as an anchor from which the solution develops based on the differential equation. By specifying \( \mathbf{k} \), we remove the ambiguity of infinite solutions, focusing only on the one that meets our specific conditions.

Matrix Differential Equation

A matrix differential equation is similar to regular differential equations but involves matrices, which can handle multiple dimensions and complexity more efficiently. The differential equation \( \mathbf{y}^{\prime} = A(t) \mathbf{y} + \mathbf{f}(t) \) is an example, where:

• \( A(t) \) is a matrix function, defining how the system changes over time.
• \( \mathbf{y}(t) \) is a vector function representing the solution's state.
• \( \mathbf{f}(t) \) is an inhomogeneous term adding external inputs.

Understanding and solving such equations involves recognizing the roles these components play. The matrix \( A(t) \) describes the internal dynamics, while \( \mathbf{f}(t) \) introduces influences that impact \( \mathbf{y}(t) \).'s evolution. By using methods like variation of parameters, solving these equations necessitates intricate manipulation of matrices and their inverses, demonstrating the sophisticated level of mathematics involved in describing systems with time-dependent changes. This approach allows us to predict outcomes under varying conditions, crucial for fields that require precise modeling, such as physics, engineering, and more.