Fundamental Matrix: Solving A System Of Differential Equations

Let's dive into the world of linear systems and explore how to find the fundamental matrix. This matrix is a cornerstone in understanding the behavior of solutions to systems of differential equations. Specifically, we'll tackle the system:

x' = 16x + 20y

y' = -10x - 14y

with initial conditions x(0) = 0 and y(0) = 0. This looks intimidating at first, but breaking it down step-by-step will make it much clearer. We will explore the concept of the fundamental matrix, and then walk through the process of solving for it in a given system of differential equations. By the end, you'll have a solid understanding of how to find the fundamental matrix and its importance in analyzing linear systems.

Understanding the Fundamental Matrix

The fundamental matrix is like a super key that unlocks all the solutions to a system of linear, homogeneous differential equations. Think of it as a matrix whose columns are linearly independent solutions to your system. These independent solutions form a basis, meaning any other solution can be expressed as a linear combination of them. To fully understand the concept of a fundamental matrix, it's crucial to break down its components and significance.

At its core, the fundamental matrix, often denoted by ${\Phi(t)}$, is a matrix-valued function whose columns consist of linearly independent solutions to a homogeneous system of linear differential equations. Consider a system of first-order linear differential equations expressed in matrix form as ${x' = Ax}$, where ${x}$ is a vector of unknown functions, and ${A}$ is a constant matrix. If we can find ${n}$ linearly independent solutions ${x_1(t), x_2(t), ..., x_n(t)}$ to this system, we can construct the fundamental matrix by arranging these solutions as columns:

${ \Phi(t) = \begin{bmatrix} x_1(t) & x_2(t) & ... & x_n(t) \end{bmatrix} }$

The beauty of the fundamental matrix lies in its ability to represent the general solution of the homogeneous system. Any solution to ${x' = Ax}$ can be written as a linear combination of the columns of ${\Phi(t)}$. In other words, if ${c}$ is a constant vector, then ${x(t) = \Phi(t)c}$ is a solution to the system. This is a powerful result because it provides a concise way to express all possible solutions.

But why is linear independence so crucial? The linear independence of the column vectors ensures that each solution contributes uniquely to the general solution. If the solutions were linearly dependent, it would mean that one or more of them could be expressed as a linear combination of the others, effectively reducing the number of independent solutions and preventing us from forming a complete basis for the solution space. In simpler terms, linearly dependent solutions would provide redundant information, hindering our ability to capture the full spectrum of possible behaviors of the system.

Moreover, the fundamental matrix plays a central role in solving non-homogeneous linear systems. Consider the non-homogeneous system ${x' = Ax + f(t)}$, where ${f(t)}$ is a vector-valued function. Once we have the fundamental matrix ${\Phi(t)}$ for the corresponding homogeneous system ${x' = Ax}$, we can use techniques such as variation of parameters to find a particular solution to the non-homogeneous system. The general solution to the non-homogeneous system is then the sum of the general solution to the homogeneous system and the particular solution to the non-homogeneous system.

The fundamental matrix also connects to the concept of the matrix exponential. If ${A}$ is a constant matrix, the matrix exponential ${e^{At}}$ is defined by the power series:

${ e^{At} = I + At + \frac{(At)^2}{2!} + \frac{(At)^3}{3!} + ... }$

where ${I}$ is the identity matrix. It turns out that ${e^{At}}$ is itself a fundamental matrix for the system ${x' = Ax}$. In fact, it is the unique fundamental matrix that satisfies ${\Phi(0) = I}$. This special fundamental matrix is often called the principal fundamental matrix.

In summary, the fundamental matrix is a crucial tool in the analysis of linear systems of differential equations. Its columns are linearly independent solutions that form a basis for the solution space. It allows us to express the general solution of homogeneous systems, solve non-homogeneous systems using variation of parameters, and connects to the important concept of the matrix exponential. Understanding the fundamental matrix is key to unlocking the behavior of solutions to linear systems and provides a powerful framework for tackling more complex problems in differential equations.

Step-by-Step Solution

Now, let's apply this understanding to our specific problem. We need to find the fundamental matrix for the system:

x' = 16x + 20y

y' = -10x - 14y

with initial conditions x(0) = 0 and y(0) = 0. To solve this, we'll follow a series of steps:

1. Express the System in Matrix Form

First, we rewrite the system in matrix notation. This makes the problem more compact and easier to work with. We can represent the system as x' = Ax, where:

${ A = \begin{bmatrix} 16 & 20 \\ -10 & -14 \end{bmatrix} }$

and

${ x = \begin{bmatrix} x(t) \\ y(t) \end{bmatrix}, \quad x' = \begin{bmatrix} x'(t) \\ y'(t) \end{bmatrix} }$

This matrix form clearly represents the relationships between the derivatives and the original functions.

2. Find the Eigenvalues of the Matrix A

The eigenvalues of matrix A are crucial for determining the form of the solutions. To find them, we solve the characteristic equation, which is given by det(A - λI) = 0, where λ represents the eigenvalues and I is the identity matrix. So, we have:

${ \text{det}(A - \lambda I) = \text{det}\begin{bmatrix} 16 - \lambda & 20 \\ -10 & -14 - \lambda \end{bmatrix} = (16 - \lambda)(-14 - \lambda) - (20)(-10) }$

Expanding this, we get:

${ (16 - \lambda)(-14 - \lambda) + 200 = \lambda^2 - 2\lambda - 224 + 200 = \lambda^2 - 2\lambda - 24 }$

Setting this equal to zero and solving the quadratic equation ${\lambda^2 - 2\lambda - 24 = 0}$, we find the eigenvalues:

${ (\lambda - 6)(\lambda + 4) = 0 \Rightarrow \lambda_1 = 6, \quad \lambda_2 = -4 }$

Thus, we have two distinct real eigenvalues: 6 and -4. These distinct real eigenvalues indicate that the solutions will involve exponential functions.

3. Determine the Eigenvectors for Each Eigenvalue

For each eigenvalue, we find the corresponding eigenvector by solving the equation (A - λI)v = 0, where v is the eigenvector. Let's start with λ₁ = 6:

${ (A - 6I)v_1 = \begin{bmatrix} 16 - 6 & 20 \\ -10 & -14 - 6 \end{bmatrix} \begin{bmatrix} v_{11} \\ v_{12} \end{bmatrix} = \begin{bmatrix} 10 & 20 \\ -10 & -20 \end{bmatrix} \begin{bmatrix} v_{11} \\ v_{12} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} }$

This gives us the equation 10v₁₁ + 20v₁₂ = 0, which simplifies to v₁₁ = -2v₁₂. We can choose v₁₂ = 1, so v₁₁ = -2. Thus, the eigenvector corresponding to λ₁ = 6 is:

${ v_1 = \begin{bmatrix} -2 \\ 1 \end{bmatrix} }$

Now, let's find the eigenvector for λ₂ = -4:

${ (A - (-4)I)v_2 = \begin{bmatrix} 16 + 4 & 20 \\ -10 & -14 + 4 \end{bmatrix} \begin{bmatrix} v_{21} \\ v_{22} \end{bmatrix} = \begin{bmatrix} 20 & 20 \\ -10 & -10 \end{bmatrix} \begin{bmatrix} v_{21} \\ v_{22} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} }$

This gives us the equation 20v₂₁ + 20v₂₂ = 0, which simplifies to v₂₁ = -v₂₂. We can choose v₂₂ = 1, so v₂₁ = -1. Thus, the eigenvector corresponding to λ₂ = -4 is:

${ v_2 = \begin{bmatrix} -1 \\ 1 \end{bmatrix} }$

We now have two linearly independent eigenvectors, which is essential for constructing the fundamental matrix.

4. Construct the General Solution

Using the eigenvalues and eigenvectors, we can write the general solution of the system. Each eigenvalue-eigenvector pair contributes a term to the general solution. The solution corresponding to λ₁ = 6 and v₁ is:

${ x_1(t) = e^{\lambda_1 t}v_1 = e^{6t} \begin{bmatrix} -2 \\ 1 \end{bmatrix} = \begin{bmatrix} -2e^{6t} \\ e^{6t} \end{bmatrix} }$

The solution corresponding to λ₂ = -4 and v₂ is:

${ x_2(t) = e^{\lambda_2 t}v_2 = e^{-4t} \begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} -e^{-4t} \\ e^{-4t} \end{bmatrix} }$

The general solution is a linear combination of these two solutions:

${ x(t) = c_1 x_1(t) + c_2 x_2(t) = c_1 \begin{bmatrix} -2e^{6t} \\ e^{6t} \end{bmatrix} + c_2 \begin{bmatrix} -e^{-4t} \\ e^{-4t} \end{bmatrix} }$

where c₁ and c₂ are arbitrary constants. This general solution represents all possible solutions to the given system of differential equations.

5. Form the Fundamental Matrix

Now we construct the fundamental matrix, ${\Phi(t)}$, by using the linearly independent solutions we found as its columns:

${ \Phi(t) = \begin{bmatrix} -2e^{6t} & -e^{-4t} \\ e^{6t} & e^{-4t} \end{bmatrix} }$

This matrix is fundamental because any solution to the system can be expressed as ${\Phi(t)c}$, where ${c}$ is a constant vector. The fundamental matrix encapsulates all the essential information about the system's solutions.

6. Apply Initial Conditions (if necessary)

If we have initial conditions, we can use them to find the specific solution. In our case, we have x(0) = 0 and y(0) = 0. This means:

${ \begin{bmatrix} x(0) \\ y(0) \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} }$

Using the general solution:

${ x(0) = c_1 \begin{bmatrix} -2e^{0} \\ e^{0} \end{bmatrix} + c_2 \begin{bmatrix} -e^{0} \\ e^{0} \end{bmatrix} = c_1 \begin{bmatrix} -2 \\ 1 \end{bmatrix} + c_2 \begin{bmatrix} -1 \\ 1 \end{bmatrix} }$

Setting this equal to the initial conditions, we get:

${ \begin{bmatrix} -2c_1 - c_2 \\ c_1 + c_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} }$

This gives us a system of linear equations:

-2c₁ - c₂ = 0

c₁ + c₂ = 0

Solving this system, we find c₁ = 0 and c₂ = 0. This means the specific solution that satisfies the initial conditions is the trivial solution:

${ x(t) = 0 \begin{bmatrix} -2e^{6t} \\ e^{6t} \end{bmatrix} + 0 \begin{bmatrix} -e^{-4t} \\ e^{-4t} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} }$

In this particular case, the initial conditions lead to the trivial solution. However, the fundamental matrix we found is still crucial because it provides the framework for understanding all possible solutions to the system, regardless of the initial conditions.

Conclusion

In summary, we've walked through the process of finding the fundamental matrix for the given system of differential equations. We started by converting the system into matrix form, then found the eigenvalues and eigenvectors. Using these, we constructed the general solution and, finally, the fundamental matrix:

${ \Phi(t) = \begin{bmatrix} -2e^{6t} & -e^{-4t} \\ e^{6t} & e^{-4t} \end{bmatrix} }$

While the initial conditions in this specific case led to the trivial solution, the fundamental matrix remains a powerful tool for analyzing the behavior of the system under different conditions. Understanding how to find and interpret the fundamental matrix is a key skill in the study of linear systems and differential equations. If you want to delve deeper into differential equations, consider exploring resources like Khan Academy's Differential Equations section. They offer a wealth of materials to further your understanding.