Question

The flow system shown in the figure is activated at time \(t=0 .\) Let \(Q_{i}(t)\) denote the amount of solute present in the \(i\) th tank at time \(t\). For simplicity, we assume all the flow rates are a constant \(10 \mathrm{gal} / \mathrm{min}\). It follows that volume of solution in each tank remains constant; we assume the volume to be \(1000 \mathrm{gal}\).The flow system shown in the figure is activated at time \(t=0 .\) Let \(Q_{i}(t)\) denote the amount of solute present in the \(i\) th tank at time \(t\). For simplicity, we assume all the flow rates are a constant \(10 \mathrm{gal} / \mathrm{min}\). It follows that volume of solution in each tank remains constant; we assume the volume to be \(1000 \mathrm{gal}\).(b) Solve the initial value problem defined by the given inflow concentrations and initial conditions. Also, determine \(\lim _{t \rightarrow \infty} \mathbf{Q}(t)\). (c) In Exercises 33 and 34 , the inflow concentrations are constant. Compute the equilibrium solution of the system in part (a). What is the physical significance of this equilibrium solution? (d) In Exercise 35 , the system in part (a) is not autonomous. Graph \(Q_{1}(t)\) and \(Q_{2}(t)\). Determine the maximum amounts of solute in each tank.\(c_{1}=0.5 \mathrm{lb} / \mathrm{gal}, \quad c_{2}=0, \quad Q_{1}(0)=Q_{2}(0)=0\)

Short answer

Question: Determine the equilibrium solution and the maximum amounts of solute in each tank for the given system. Answer: The equilibrium solution is Q(t) = (50, 5000/3 - 50). The maximum amounts of solute in each tank are Q₁(t) = 50 lb and Q₂(t) = 5000/3 - 50 lb.

Step-by-step solution

(Step 1: Write the model as a system of differential equations.)

First, we need to create a system of differential equations by considering the inflow and outflow rates of the solute in each tank. Let \(Q_1(t)\) and \(Q_2(t)\) denote the amount of solute in each tank at time \(t\). Then the differential equations are: \(\frac{dQ_1}{dt} = 10c_1 - \frac{10Q_1}{1000}\) \(\frac{dQ_2}{dt} = \frac{10Q_1}{1000} - \frac{10Q_2}{1000}\) With the initial conditions: \(Q_1(0) = 0\) and \(Q_2(0) = 0\).

(Step 2: Solve the initial value problem.)

To solve the initial value problem, we can use an integrating factor for each differential equation. For the first equation, the integrating factor is \(e^{-\frac{10t}{1000}}\). Multiply the first equation by this factor and integrate both sides to get: \(Q_1(t) = 50(1 - e^{-\frac{10t}{1000}})\) Now, we can substitute this expression for \(Q_1(t)\) into the second differential equation and use the integrating factor \(e^{-\frac{10t}{1000}}\) to solve it: \(Q_2(t) = \frac{5000}{3}(e^{-\frac{10t}{1000}} - e^{-\frac{20t}{1000}})\) Thus, the solution to the initial value problem is: \(Q_1(t) = 50(1 - e^{-\frac{10t}{1000}})\) \(Q_2(t) = \frac{5000}{3}(e^{-\frac{10t}{1000}} - e^{-\frac{20t}{1000}})\)

(Step 3: Determine the limit as \(t\) goes to infinity.)

To find the limit as \(t\) goes to infinity, we need to find the equilibrium solution, which is when the time derivatives of both equations are zero. For the first equation, the limit is: \(\lim_{t \rightarrow \infty} Q_1(t) = 50\) For the second equation, the limit is: \(\lim_{t \rightarrow \infty} Q_2(t) = \frac{5000}{3} - 50\) Therefore, the equilibrium solution is \(\mathbf{Q}(t) = \left(50, \frac{5000}{3} - 50\right)\).

(Step 4: Graph \(Q_1(t)\) and \(Q_2(t)\) and determine the maximum solute amount.)

When graphing \(Q_1(t)\) and \(Q_2(t)\), we can see that: \(Q_1(t)\) increases from 0 to a maximum value of 50 as \(t\) goes to infinity. \(Q_2(t)\) increases from 0 to a maximum value of \(\frac{5000}{3} - 50\) as \(t\) goes to infinity. The maximum amounts of solute in each tank are: \(Q_1(t) = 50 \mathrm{lb}\) \(Q_2(t) = \frac{5000}{3} - 50 \mathrm{lb}\) The equilibrium solution represents the state of the system when the amount of solute in both tanks does not change, which is the physical significance of this solution.

Key concepts

Initial Value Problem

An initial value problem is a type of differential equation paired with specific conditions at the start of the problem, known as initial conditions. In this exercise, we're looking to solve for the amount of solute in two tanks, which is modeled with differential equations. Initial conditions are provided: both tanks start with 0 pounds of solute, symbolized by \(Q_1(0) = 0\) and \(Q_2(0) = 0\).
The objective is to determine how these amounts change over time by integrating these equations with the given starting values. These initial conditions are what differentiate this type of problem, ensuring a unique solution that fits the situation precisely.

Equilibrium Solution

An equilibrium solution in the context of differential equations refers to a stable state where the system doesn’t change over time. This means that the rate of change of solute in the tanks will be zero. In the last step of our exercise, we calculate this by setting the derivatives to zero, leading to \(\lim_{t \rightarrow \infty} Q_1(t) = 50\) and \(\lim_{t \rightarrow \infty} Q_2(t) = \frac{5000}{3} - 50\).
Physically, this represents the point where the input and output rates balance perfectly, making the concentrations in the tank constant. Identifying equilibrium solutions can be essential in determining the long-term behavior of systems.

Systems of Differential Equations

Systems of differential equations involve more than one equation and more than one unknown function. In this exercise, the system consists of two equations, one for each tank:

• \(\frac{dQ_1}{dt} = 10c_1 - \frac{10Q_1}{1000}\)
• \(\frac{dQ_2}{dt} = \frac{10Q_1}{1000} - \frac{10Q_2}{1000}\)

These equations describe how the quantities \(Q_1(t)\) and \(Q_2(t)\) change over time due to inflow and outflow. The solution of this system requires techniques that consider the interaction between tanks — seen here in how \(Q_2\) depends on \(Q_1\). It’s crucial to note how interconnected systems can provide a more comprehensive understanding of real-world processes.

Integrating Factor

The integrating factor is a technique used to solve linear differential equations. It simplifies the integration process, making it easier to find a solution to these equations. In this exercise, we used the integrating factor \(e^{-\frac{10t}{1000}}\) to solve both given differential equations.
This process involves multiplying both sides of a linear equation by this factor, which transforms the equation into an exact derivative that can be easily integrated. Utilizing the integrating factor is a powerful method and often applied in finding solutions to initial value problems, especially when dealing with exponential decay or growth.

Autonomous Systems

Autonomous systems are differential equations where the rules governing the rate of change do not explicitly depend on the independent variable, usually time \(t\). In non-autonomous systems, time plays a direct role in the rate of change equations. For example, if a flow system's rates were dependent on time itself, then it would be non-autonomous.
In our exercise, the system would become non-autonomous if the flow rates or solute concentration were functions of time, rather than constants. Autonomous systems are appealing in modeling because they often indicate stable long-term behavior and can simplify analysis. Understanding when a system is autonomous helps predict system stability and equilibrium points in practical applications.