Problem 20 - Differential Equations and Linear Algebra

Question

Solve the initial-value problem $$ \mathbf{x}^{\prime}=A \mathbf{x}, \quad \mathbf{x}(0)=\left[\begin{array}{l} 1 \\ 1 \end{array}\right] $$ when \(A=\left[\begin{array}{rl}0 & 4 \\ -4 & 0\end{array}\right]\). Sketch the solution in the \(x_{1} x_{2}\) plane.

Step-by-Step Solution

Answer

The particular solution to the given initial-value problem is \(\mathbf{x}(t) = \frac{1}{2} e^{4it}\begin{bmatrix}1 \\ -i\end{bmatrix} + \frac{1}{2} e^{-4it}\begin{bmatrix}1 \\ i\end{bmatrix}\). In the \(x_1 x_2\) plane, the solution traces a circle with radius 1 centered at the origin, represented by the parametric equations \(x_1(t) = \cos(4t)\) and \(x_2(t) = \sin(4t)\). The solution starts at \(\mathbf{x}(0) = \left[\begin{array}{l}1 \\ 1\end{array}\right]\) due to the initial condition.

Step 1: Find eigenvalues and eigenvectors of A

First, we need to find the eigenvalues and eigenvectors of the given matrix \(A=\left[\begin{array}{rl}0 & 4 \\\ -4 & 0\end{array}\right]\). To do this, let's solve the following characteristic equation: $$ \text{det}(A - \lambda I) = 0 $$ where \(I\) is the identity matrix and \(\lambda\) represents the eigenvalues. So, we have $$ \text{det}\left(\begin{array}{cc} -\lambda & 4 \\ -4 & -\lambda\end{array}\right) = \lambda^2 + 4^2 = 0 $$ Solving this equation, we get the eigenvalues \(\lambda_1 = 4i\) and \(\lambda_2 = -4i\). Now, let's find the eigenvectors corresponding to these eigenvalues. For \(\lambda_1 = 4i\), we need to solve \((A - 4iI)\mathbf{x} = 0\), where \(\mathbf{x}\) is the eigenvector. This gives us the following system of equations: $$ \left\{\begin{array}{rl} -4i x_1 & + 4x_2 = 0 \\ -4x_1 & - 4i x_2 = 0 \end{array}\right. $$ One possible solution for this system is \(x_1 = 1\) and \(x_2 = -i\). So, eigenvector corresponding to \(\lambda_1 = 4i\) is \(\mathbf{v}_1 = \left[\begin{array}{l}1 \\ -i\end{array}\right]\). Similarly, for \(\lambda_2 = -4i\), we need to solve \((A + 4iI)\mathbf{x} = 0\), which results in the following system of equations: $$ \left\{\begin{array}{rl} 4i x_1 & + 4x_2 = 0 \\ -4x_1 & + 4i x_2 = 0 \end{array}\right. $$ One possible solution for this system is \(x_1 = 1\) and \(x_2 = i\). So, eigenvector corresponding to \(\lambda_2 = -4i\) is \(\mathbf{v}_2 = \left[\begin{array}{l}1 \\ i\end{array}\right]\).

Step 2: Form a general solution

Now, we'll form a general solution to the given differential equation using the eigenvalues and eigenvectors we found. The general solution can be written in the following form: $$ \mathbf{x}(t) = c_1 e^{\lambda_1 t}\mathbf{v}_1 + c_2 e^{\lambda_2 t}\mathbf{v}_2 = c_1 e^{4it}\begin{bmatrix}1 \\ -i\end{bmatrix} + c_2 e^{-4it}\begin{bmatrix}1 \\ i\end{bmatrix} $$

Step 3: Apply initial condition

Now, we need to apply the initial condition, \(\mathbf{x}(0) = \left[\begin{array}{l}1 \\ 1\end{array}\right]\), to find the constants \(c_1\) and \(c_2\). Plugging in \(t=0\) to our general solution, we get: $$ \left[\begin{array}{l}1 \\ 1\end{array}\right] = c_1 e^{4i\cdot 0}\begin{bmatrix}1 \\ -i\end{bmatrix} + c_2 e^{-4i\cdot 0}\begin{bmatrix}1 \\ i\end{bmatrix} = c_1\begin{bmatrix}1 \\ -i\end{bmatrix} + c_2\begin{bmatrix}1 \\ i\end{bmatrix} $$ Solving for \(c_1\) and \(c_2\), we get \(c_1 = \frac{1}{2}\) and \(c_2 = \frac{1}{2}\). Thus, the particular solution is given by: $$ \mathbf{x}(t) = \frac{1}{2} e^{4it}\begin{bmatrix}1 \\ -i\end{bmatrix} + \frac{1}{2} e^{-4it}\begin{bmatrix}1 \\ i\end{bmatrix} $$

Step 4: Sketch the solution in the x1 x2 plane

To sketch the solution, we can write \(\mathbf{x}(t)\) in its real and imaginary parts: $$ x_1(t) = \text{Re}\left(\frac{1}{2} e^{4it}\right)+ \text{Re}\left(\frac{1}{2} e^{-4it}\right) $$ $$ x_2(t) = \text{Im}\left(-\frac{i}{2} e^{4it}\right) + \text{Im}\left(\frac{i}{2} e^{-4it}\right) $$ Now, using Euler's formula, we can write the solution in terms of sine and cosine: $$ x_1(t) = \frac{1}{2}(e^{4it} + e^{-4it}) = cos(4t) $$ $$ x_2(t) = \frac{1}{4i}(-e^{4it} + e^{-4it}) = sin(4t) $$ Thus, $$ \mathbf{x}(t) = \begin{bmatrix}cos(4t) \\ sin(4t)\end{bmatrix} $$ Now, we can sketch the solution in the \(x_1 x_2\) plane, which gives us a circle with radius 1 centered at the origin, since \(x^2_1 + x^2_2=cos^2(4t) + sin^2(4t) =1\). The solution starts at \(\mathbf{x}(0) = \left[\begin{array}{l}1 \\ 1\end{array}\right]\) as given by the initial condition and revolves around the circle.

Key Concepts

Eigenvalues

When we work with systems of differential equations, eigenvalues play a crucial role in determining the behavior of the solutions. Eigenvalues are particular scalars or numbers that, when multiplied by an eigenvector, do not change the direction of that eigenvector. They are solutions to the characteristic equation

• It's formed by taking the determinant of the matrix \(A - \lambda I\), and setting it equal to zero.
• For our exercise with matrix \( A = \begin{bmatrix} 0 & 4 \ -4 & 0 \end{bmatrix} \), the eigenvalues are imaginary numbers: \(\lambda_1 = 4i\) and \(\lambda_2 = -4i\).

Eigenvalues can reveal the oscillatory nature of the solutions when they are complex, as is often the case in systems involving rotations or dynamics governed by differential equations. Imaginary eigenvalues indicate a rotational motion, which in this exercise, translates into a circular path in the solution's sketch.

Initial Value Problem

An initial value problem (IVP) is a differential equation along with a specified value at a given point. This is like setting a starting condition for where your solution curve begins.

• The differential equation tells you how to move from this starting point.
• In our exercise, the initial condition is given as \( \mathbf{x}(0) = \begin{bmatrix} 1 \ 1 \end{bmatrix} \).

This initial condition is crucial because it allows us to find unique values for constants in the solution's general form. Without an initial condition, a differential equation will have an infinite number of solutions, each differing by constant shifts. By using the specified initial condition, we pinpoint the exact path the solution will take in the plane or space, ensuring that it passes through the designated starting point.

Eigenvectors

Once we have eigenvalues, it's time to find the corresponding eigenvectors. These are non-zero vectors that remain almost unchanged except for a scalar factor when a matrix is applied to them. They help form the solutions to differential equations.

• To find eigenvectors, you need to solve the equation \( (A - \lambda I)\mathbf{v} = 0\).
• Each eigenvalue corresponds to its eigenvector. In our example, \(\lambda_1 = 4i\) gives us eigenvector \ \mathbf{v}_1 = \begin{bmatrix} 1 \ -i \end{bmatrix} \.
• Similarly, \(\lambda_2 = -4i\) yields eigenvector \ \mathbf{v}_2 = \begin{bmatrix} 1 \ i \end{bmatrix} \.

Eigenvectors are vital because they help express the general solution of a differential equation. In our case, they determine the orientation and the nature of the paths traced out in the solution's sketch. Eigenvectors represent the direction in which the action of the matrix is simplest, revealing the invariant lines that help us build and understand complex solutions.