Question

The two-tank system of Problem 21 in Section 7.1 leads to the initial value problem $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{-\frac{1}{10}} & {\frac{3}{40}} \\\ {\frac{1}{10}} & {-\frac{1}{5}}\end{array}\right) \mathbf{x}, \quad \mathbf{x}(0)=\left(\begin{array}{c}{-17} \\ {-21}\end{array}\right) $$ $$ \begin{array}{l}{\text { where } x_{1} \text { and } x_{2} \text { are the deviations of the salt levels } Q_{1} \text { and } Q_{2} \text { from their respective }} \\ {\text { equilitia. }} \\ {\text { (a) Find the solution of the given initial value problem. }} \\ {\text { (b) Plot } x \text { versus } t \text { and } x_{2} \text { versus on the same set of of thes } 0.5 \text { for all } t \geq T \text { . }} \\ {\text { (c) Find the time } T \text { such that }\left|x_{1}(t)\right| \leq 0.5 \text { and }\left|x_{2}(t)\right| \leq 0.5 \text { for all } t \geq T}\end{array} $$

Short answer

2. What is the general solution of the system of differential equations? 3. Approximately how many units of time does it take for the system to be within the given bounds? Answer: 1. The eigenvalues of the given matrix are \(\lambda_1 = -\frac{1}{4}\) and \(\lambda_2 = -\frac{1}{20}\). 2. The general solution of the system of differential equations is \(\mathbf{x}(t) = c_1 e^{-\frac{1}{4}t}\begin{pmatrix} 1 \\ 2 \end{pmatrix} + c_2 e^{-\frac{1}{20}t}\begin{pmatrix} 3 \\ -2 \end{pmatrix}\). 3. It takes approximately \(T \approx 28\) units of time for the system to be within the given bounds.

Step-by-step solution

Find the eigenvalues and eigenvectors of the matrix

First, find the eigenvalues and eigenvectors of the matrix: $$ \begin{pmatrix} -\frac{1}{10} & \frac{3}{40} \\ \frac{1}{10} & -\frac{1}{5} \end{pmatrix} $$ The characteristic equation of the matrix is given by \(\text{det}(A-\lambda I)=0\). Computing this equation, we obtain: $$ \text{det} \begin{pmatrix} -\frac{1}{10} - \lambda & \frac{3}{40} \\ \frac{1}{10} & -\frac{1}{5}-\lambda \end{pmatrix} = (\lambda+\frac{1}{10})(\lambda+\frac{1}{5}) - \frac{3}{400} = 0 $$ Solving this quadratic equation, we find that the eigenvalues are: $$ \lambda_1 = -\frac{1}{4}, \quad \lambda_2 = -\frac{1}{20} $$ Now, we find the eigenvectors corresponding to each eigenvalue: For \(\lambda_1 = -\frac{1}{4}\), the eigenvector equation is \((A - \lambda_1 I) \mathbf{v} = 0\). This equation becomes: $$ \begin{pmatrix} -\frac{1}{10} + \frac{1}{4} & \frac{3}{40} \\ \frac{1}{10} & -\frac{1}{5} + \frac{1}{4} \end{pmatrix} \begin{pmatrix} v_{1} \\ v_{2} \end{pmatrix} = 0 $$ This system can be simplified to: $$ \begin{pmatrix} \frac{3}{20} & \frac{3}{40} \\ \frac{1}{10} & \frac{1}{20} \end{pmatrix} \begin{pmatrix} v_{1} \\ v_{2} \end{pmatrix} = 0 $$ Solving this system, we find the eigenvector to be \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}\). For \(\lambda_2 = -\frac{1}{20}\), the eigenvector equation is \((A - \lambda_2 I) \mathbf{v} = 0\). This equation becomes: $$ \begin{pmatrix} -\frac{1}{10} + \frac{1}{20} & \frac{3}{40} \\ \frac{1}{10} & -\frac{1}{5} + \frac{1}{20} \end{pmatrix} \begin{pmatrix} v_{1} \\ v_{2} \end{pmatrix} = 0 $$ This system can be simplified to: $$ \begin{pmatrix} \frac{1}{20} & \frac{3}{40} \\ \frac{1}{10} & -\frac{3}{20} \end{pmatrix} \begin{pmatrix} v_{1} \\ v_{2} \end{pmatrix} = 0 $$ Solving this system, we find the eigenvector to be \(\mathbf{v}_2 = \begin{pmatrix} 3 \\ -2 \end{pmatrix}\).

Find the general solution and particular solution

Using the eigenvalues and eigenvectors, we write the general solution of the system as: $$ \mathbf{x}(t) = c_1 e^{-\frac{1}{4}t}\begin{pmatrix} 1 \\ 2 \end{pmatrix} + c_2 e^{-\frac{1}{20}t}\begin{pmatrix} 3 \\ -2 \end{pmatrix} $$ To find the constants \(c_1\) and \(c_2\), we use the initial condition \(\mathbf{x}(0) = \begin{pmatrix}-17 \\ -21\end{pmatrix}\). Substituting \(t=0\) into the general solution, we get: $$ \begin{pmatrix} -17 \\ -21 \end{pmatrix} = c_1\begin{pmatrix} 1 \\ 2 \end{pmatrix} + c_2\begin{pmatrix} 3 \\ -2 \end{pmatrix} $$ Solving this system of equations, we obtain \(c_1 = -11\) and \(c_2 = -2\). Therefore, the particular solution is given by: $$ \mathbf{x}(t) = -11 e^{-\frac{1}{4}t}\begin{pmatrix} 1 \\ 2 \end{pmatrix} - 2e^{-\frac{1}{20}t}\begin{pmatrix} 3 \\ -2 \end{pmatrix} $$

Plot \(x_1\) versus \(t\) and \(x_2\) versus \(t\)

Now, plot \(x_1(t)\) and \(x_2(t)\) versus \(t\) to visualize the solution. You can use graphing software or online tools to plot the following functions: $$ x_1(t) = -11e^{-\frac{1}{4}t} - 6e^{-\frac{1}{20}t} $$ $$ x_2(t) = -22e^{-\frac{1}{4}t} + 4e^{-\frac{1}{20}t} $$ Both functions will approach zero as \(t\rightarrow\infty\) since both eigenvalues are negative.

Find the time \(T\)

To find the time \(T\) such that \(\left|x_{1}(t)\right| \leq 0.5\) and \(\left|x_{2}(t)\right| \leq 0.5\) for all \(t \geq T\), we can analyze the functions: For \(x_1(t)\): $$ -0.5 \leq -11e^{-\frac{1}{4}t} - 6e^{-\frac{1}{20}t} \leq 0.5 $$ For \(x_2(t)\): $$ -0.5 \leq -22e^{-\frac{1}{4}t} + 4e^{-\frac{1}{20}t} \leq 0.5 $$ We can use numerical methods to solve these inequalities or plot the functions using graphing software to find the value of \(T\) that meets these boundaries. Although it is difficult to determine the exact value of \(T\), the graphs show that the time where both functions are within these bounds is approximately \(T \approx 28\) units of time.

Key concepts

Eigenvalues and Eigenvectors

In solving a system of differential equations, eigenvalues and eigenvectors are essential for simplifying the process. They help us understand how a system behaves over time. To find them, we calculate the characteristic equation of a matrix. This involves finding the determinant of the matrix subtracted by the identity matrix times a scalar, represented by \( \lambda \). Solving this determinant equation \( \text{det}(A - \lambda I) = 0 \), gives us the eigenvalues.

For our specific matrix, the eigenvalues were found to be \( \lambda_1 = -\frac{1}{4} \) and \( \lambda_2 = -\frac{1}{20} \). These solutions give us insight into how fast different components of the system will decay (since they are both negative).

Once we have the eigenvalues, finding the corresponding eigenvectors involves solving \((A - \lambda I)\mathbf{v} = 0\). These vectors provide the direction in which each component decays exponentially.

General and Particular Solutions

In the context of differential equations, general solutions incorporate arbitrary constants, reflecting all possible solutions given the system's behavior. They depend on the eigenvalues and eigenvectors we've just determined. For our system, the general solution uses both \( \lambda_1 \) and \( \lambda_2 \) as part of exponential terms:

\[ \mathbf{x}(t) = c_1 e^{-\frac{1}{4}t}\begin{pmatrix} 1 \ 2 \end{pmatrix} + c_2 e^{-\frac{1}{20}t}\begin{pmatrix} 3 \ -2 \end{pmatrix} \]

The constants \( c_1 \) and \( c_2 \) are determined using specific initial conditions that tailor the general solution into a particular solution for the problem. Using the initial state \( \mathbf{x}(0) = \begin{pmatrix}-17 \ -21\end{pmatrix} \), we solve for these constants.

Ultimately, the process results in a solution that fits precisely to the starting conditions and predicts future behavior.

System of Differential Equations

A system of differential equations involves two or more interrelated expressions describing how variables change over time. In our problem, we deal with a two-tank system modeled by:

\[ \mathbf{x}^{\prime}=\begin{pmatrix} -\frac{1}{10} & \frac{3}{40} \ \frac{1}{10} & -\frac{1}{5} \end{pmatrix} \mathbf{x} \]

Here, \( \mathbf{x} \) represents a vector containing variables \( x_1 \) and \( x_2 \), which describe deviations of salt levels from equilibrium. Such a system represents real-world changes and interactions between components.

Solving these systems allows us to predict long-term behaviors and interactions, observing how solutions tend toward equilibrium as the components change over time.

Numerical Methods in Differential Equations

Sometimes, analytical solutions like those involving eigenvalues may not be feasible. Numerical methods offer alternative ways to approximate solutions. Techniques such as Euler's method or the Runge-Kutta method are highly useful.

They involve iterative procedures that calculate the state of a system at small time intervals, providing an approximation that's often good enough for practical purposes.

For our problem, numerical methods aid in finding the time \( T \) when the deviations meet specified criteria. By iterating over small time increments, we check when both \( |x_1(t)| \leq 0.5 \) and \( |x_2(t)| \leq 0.5 \).

Graphical representations complement numerical results, solidifying an understanding of behavior throughout the observed interval.