Problem 22 - Calculus for Biology and Medicine

Question

Show that the following system of differential equations has a conserved quantity, and find it: $$ \begin{array}{l} \frac{d x}{d t}=-4 x+2 y \\ \frac{d y}{d t}=-y+2 x \end{array} $$

Step-by-Step Solution

Answer

The conserved quantity is \(C(x, y) = x^2 + 4xy + 4y^2\).

Step 1: Understand the System

We are given a system of linear differential equations with two variables \(x\) and \(y\). The system is written as: \[ \frac{dx}{dt} = -4x + 2y \] and \[ \frac{dy}{dt} = -y + 2x \]. The goal is to find a conserved quantity, which means a function \(C(x, y)\) such that \(\frac{dC}{dt} = 0\).

Step 2: Guess the Form of the Conserved Quantity

Try the form \(C(x, y) = ax^2 + bxy + cy^2\), which is a common form for quadratic conserved quantities. To find \(a\), \(b\), and \(c\), calculate \(\frac{dC}{dt}\) using the chain rule and set it to zero.

Step 3: Calculate the Differential \(\frac{dC}{dt}\)

Use the chain rule: \(\frac{dC}{dt} = \frac{\partial C}{\partial x}\frac{dx}{dt} + \frac{\partial C}{\partial y}\frac{dy}{dt}\). Compute the partial derivatives: \(\frac{\partial C}{\partial x} = 2ax + by\) and \(\frac{\partial C}{\partial y} = 2cy + bx\). Substitute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) into the equation.

Step 4: Substitute and Simplify

Substitute the expressions into \(\frac{dC}{dt}\): \[\frac{dC}{dt} = (2ax + by)(-4x + 2y) + (2cy + bx)(-y + 2x)\]. Expand this expression to obtain: \(\frac{dC}{dt} = (-8ax^2 + 4axy - 4bxy + 2by^2) + (-2cy^2 + 4cxy - bxy + 2bx^2)\).

Step 5: Set Coefficients to Zero

To ensure \(\frac{dC}{dt} = 0\), the coefficients of each term must be zero. Equate coefficients of like terms: For \(x^2\): \(-8a + 2b = 0\), For \(xy\): \(4a - 4b + 4c = 0\), For \(y^2\): \(2b - 2c = 0\).

Step 6: Solve the Equations

Solve the set of equations from Step 5. From \(-8a + 2b = 0\), we get \(b = 4a\). From \(2b - 2c = 0\), we get \(c = b\). Substitute \(b = 4a\) into \(c = b\), giving \(c = 4a\). Use the second equation, \(4a - 4b + 4c = 0\), to verify consistency. This equation simplifies to zero, confirming \(c = b = 4a\).

Step 7: Write the Conserved Quantity

Choose a non-zero value for \(a\), say \(a = 1\), to obtain \(b = 4\) and \(c = 4\). Thus, the conserved quantity is \(C(x, y) = x^2 + 4xy + 4y^2\).

Step 8: Check the Result

Verify by substituting back that \(\frac{dC}{dt} = 0\) with the chosen values and then evaluating the expressions confirm the derivations throughout the solution.

Key Concepts

Conserved Quantity

A conserved quantity in the context of differential equations is essentially a function that remains constant in time as the system evolves. In the exercise, the task is to find a function, say \(C(x, y)\), such that its rate of change with respect to time, \(\frac{dC}{dt}\), is equal to zero. This implies that as time progresses, the value of \(C(x, y)\) does not change, and hence, is conserved.

This concept is important because it provides insight into the system's behavior over time. When a quantity is conserved, it often means there is a symmetry or invariance in the system. For example, in physics, energy or momentum being conserved signifies underlying principles and restrictions about the physical processes involved.

In the given exercise, we hypothesized the form of \(C(x, y)\) to be a quadratic function: \(C(x, y) = ax^2 + bxy + cy^2\). This form was chosen because quadratic forms are common for conserved quantities in linear systems.

• We used the chain rule to differentiate \(C\) with respect to time to identify the coefficients \(a\), \(b\), and \(c\).
• By ensuring the coefficients of the differential expressions vanish, \(C\) becomes conserved.
• Thus, solving these provides constants that ensure \(\frac{dC}{dt} = 0\).

Understanding conserved quantities in differential equations helps highlight integral elements of the system's long-term behavior without solving the system completely.

System of Linear Differential Equations

A system of linear differential equations involves two or more related differential equations with unknown functions. These equations are linear, meaning they involve no products or nonlinear functions of the unknowns or their derivatives.

In the provided exercise, we are given a system of such differential equations:

• \(\frac{dx}{dt} = -4x + 2y\)
• \(\frac{dy}{dt} = -y + 2x\)

These equations describe how two variables, \(x\) and \(y\), change concerning time. Here, the coefficients and terms involve simple multipliers of the variables, making them linear.

The study of these systems involves understanding how the variables influence each other:

• Each equation is dependent not just on one variable but also on the other, creating a coupled system.
• Analyzing these dependencies can unveil significant insights into the dynamic nature of the system.
• Solving these types of systems helps to find expressions for the variables as functions of time, providing clear descriptions of the system's overall dynamic behavior.

Identifying a conserved quantity, as done in the exercise, is often key to simplifying and understanding the solution to these types of systems.

Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate composite functions, allowing the calculation of the derivative of a function based on the derivatives of its inner functions. In the context of differential equations, it's particularly useful for functions of multiple variables.

In the given exercise, we used the chain rule to compute \(\frac{dC}{dt}\) where \(C\) is a function of both \(x\) and \(y\), which are in turn functions of \(t\) or time.

• The expression for \(\frac{dC}{dt}\) was derived by differentiating \(C(x, y)\) using partial derivatives and then substituting in the given equations for \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
• The chain rule formula used is \(\frac{dC}{dt} = \frac{\partial C}{\partial x} \frac{dx}{dt} + \frac{\partial C}{\partial y} \frac{dy}{dt}\).

Utilizing the chain rule allowed for the simplification and manipulation of the function into a form where the conditions for conservation could be readily identified.
Once the partial derivatives of \(C\) were obtained, substitution into this formula yielded a simplified expression, allowing us to adjust the values of \(a\), \(b\), and \(c\) so that \(\frac{dC}{dt} = 0\), confirming conservation.

Understanding and applying the chain rule effectively is critical in navigating complex differential relationships, ensuring precise evolution predictions of systems governed by such equations.