Problem 81 - Calculus Early Transcendentals

Question

The flow of heat along a thin conducting bar is governed by the one- dimensional heat equation (with analogs for thin plates in two dimensions and for solids in three dimensions) $$\frac{\partial u}{\partial t}=k \frac{\partial^{2} u}{\partial x^{2}},$$ where \(u\) is a measure of the temperature at a location \(x\) on the bar at time t and the positive constant \(k\) is related to the conductivity of the material. Show that the following functions satisfy the heat equation with \(k=1.\) $$u(x, t)=4 e^{-4 t} \cos 2 x$$

Step-by-Step Solution

Answer

Question: Show that the given function $$u(x, t) = 4 e^{-4t} \cos 2x$$ does not satisfy the heat equation $$\frac{\partial u}{\partial t} = k\frac{\partial^2 u}{\partial x^2}$$ with \(k=1\). Answer: The given function $$u(x, t) = 4 e^{-4t} \cos 2x$$ does not satisfy the given heat equation with \(k=1\) because when substituting the partial derivatives $$\frac{\partial u}{\partial t} = -16 e^{-4t} \cos 2x$$ and $$\frac{\partial^2 u}{\partial x^2} = 32 e^{-4t} \cos 2x$$ into the equation, it does not hold true.

Step 1: Find the first partial derivative of u with respect to t

To find the partial derivative of the given function with respect to t, we treat x as a constant and differentiate: $$\frac{\partial u}{\partial t} = \frac{\partial}{\partial t} (4 e^{-4 t} \cos 2x)$$ We apply product rule and the derivative of \(e^{-4t}\) is \(-4e^{-4t}\), so $$\frac{\partial u}{\partial t} = 4 (-4e^{-4t}) \cos 2x = -16 e^{-4t} \cos 2x$$

Step 2: Find the second partial derivative of u with respect to x

To find the second partial derivative of the given function with respect to x, treat t as a constant and differentiate the function twice with respect to x: $$\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x} \left(\frac{\partial}{\partial x} (4 e^{-4 t} \cos 2x)\right)$$ First differentiation: $$\frac{\partial u}{\partial x} = 4 e^{-4t} (-2\sin 2x) = -8 e^{-4t} \sin 2x$$ Second differentiation: $$\frac{\partial^2 u}{\partial x^2} = -8 e^{-4t} (-4\cos 2x) = 32 e^{-4t} \cos 2x$$

Step 3: Check if the given function satisfies the heat equation with k=1

Substitute the found partial derivatives into the given heat equation and check if it holds true: $$\frac{\partial u}{\partial t}= k\frac{\partial^2 u}{\partial x^2}$$ With \(k=1\), the equation becomes: $$-16 e^{-4t} \cos 2x = 1 \cdot 32 e^{-4t} \cos 2x$$ We can see that the equation does not hold true, as the left side is not equal to the right side. Therefore, the given function $$u(x, t) = 4 e^{-4t} \cos 2x$$ does not satisfy the given heat equation with \(k=1\).

Key Concepts

Partial Differentiation

Partial differentiation is a method used to find the derivative of a function with multiple variables by focusing on how the function changes with one variable while keeping the others constant. For functions in physics, such as the heat equation, this approach is crucial. In the given problem:

• The function is the temperature, denoted as \( u(x, t) \), depending on position \( x \) and time \( t \).
• To study how the temperature changes over time, we take the partial derivative \( \frac{\partial u}{\partial t} \), treating \( x \) as a constant.
• Similarly, to examine the changes with respect to position, we take the partial derivative \( \frac{\partial^2 u}{\partial x^2} \).

Understanding partial differentiation helps to analyze physical phenomena by simplifying complex systems into more manageable parts.

Product Rule

The product rule is a key concept in calculus for finding the derivative of a product of two functions. It states that if you have a function that is the product of two functions, \( f(x) = g(x) \cdot h(x) \), the derivative is given by: \[ (fg)' = f'g + fg' \] In the exercise, we have the function \( u(x, t) = 4 e^{-4t} \cos 2x \) which comprises the product of \( 4 e^{-4t} \) and \( \cos 2x \). While partially differentiating with respect to \( t \), we apply the product rule:

• The derivative of the exponential part, \( 4 e^{-4t} \), is \(-16 e^{-4t}\).
• The \( \cos 2x \) remains unchanged since it is constant with respect to \( t \).

This allows us to find the partial derivative with respect to \( t \). Understanding the product rule is essential for dealing with more complex expressions that arise in physics and engineering.

Conductivity

Conductivity is a property that measures how well a material can transfer heat. It's an important parameter in the heat equation, represented by the constant \( k \). In the context of the exercise:

• The heat equation is \( \frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2} \).
• The constant \( k \) determines how quickly heat spreads through the material.
• For the exact solution provided, \( k = 1 \) was used to test whether the function satisfies the equation.

In practical applications, high conductivity means the material can efficiently conduct heat, while low conductivity suggests insulation properties. Understanding conductivity helps in designing materials for specific thermal management applications.

Trigonometric Functions

Trigonometric functions are fundamental in describing oscillatory behavior like waves or cycles. In this exercise, \( \cos 2x \) is a part of the solution to the heat equation:

• \( \cos \, ext{functions} \) help model periodic temperature variations along the bar.
• They are essential in tasks that involve both time and position variables to represent varying patterns.
• In calculus, derivatives of trigonometric functions like \( \cos 2x \) lead to changes, giving \( -2\sin 2x \).

These functions provide elegant solutions to differential equations used in physics, making them pivotal for modeling and predicting real-world phenomena that exhibit periodicity.