Problem 82 - 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 \(k=1.\) $$u(x, t)=e^{-t}(2 \sin x+3 \cos x)$$

Step-by-Step Solution

Answer

Question: Prove that the function $$u(x, t)=e^{-t}(2 \sin x+3 \cos x)$$ satisfies the one-dimensional heat equation $$\frac{\partial u}{\partial t}=k\frac{\partial^{2} u}{\partial x^{2}}$$ with \(k=1\).

Step 1: Find the Partial Derivative of u with respect to t

First, we take the partial derivative of \(u(x, t)\) with respect to \(t\): $$\frac{\partial u}{\partial t} = \frac{\partial}{\partial t}[e^{-t}(2 \sin x+3 \cos x)]$$ By applying the product rule, we get: $$\frac{\partial u}{\partial t} = -e^{-t}(2 \sin x+3 \cos x)$$

Step 2: Find the Partial Derivative of u with respect to x twice

Next, we will find the second partial derivative of \(u(x, t)\) with respect to \(x\). We first find the first partial derivative: $$\frac{\partial u}{\partial x} = \frac{\partial}{\partial x}[e^{-t}(2 \sin x+3 \cos x)]$$ By applying the property of constants and the chain rule, we get: $$\frac{\partial u}{\partial x} = e^{-t}(2 \cos x -3\sin x)$$ Now we find the second partial derivative: $$\frac{\partial^{2} u}{\partial x^{2}}= \frac{\partial}{\partial x}[e^{-t}(2\cos x-3\sin x)]$$ Again, by applying the chain rule and the property of constants, we get: $$\frac{\partial^{2} u}{\partial x^{2}}=e^{-t}(-2\sin x -3\cos x)$$

Step 3: Compare the Partial Derivatives

Now we compare the partial derivative with respect to \(t\) that we found in Step 1 and the second partial derivative with respect to \(x\) that we found in Step 2: - Partial derivative with respect to \(t\): $$\frac{\partial u}{\partial t} = -e^{-t}(2 \sin x+3 \cos x)$$ - Second partial derivative with respect to \(x\): $$\frac{\partial^{2} u}{\partial x^{2}}=e^{-t}(-2\sin x -3\cos x)$$ As we can see, with \(k=1\), the two expressions are equal: $$\frac{\partial u}{\partial t} = k\frac{\partial^{2} u}{\partial x^{2}}$$ This proves that the function $$u(x, t)=e^{-t}(2 \sin x+3 \cos x)$$ satisfies the one-dimensional heat equation with \(k=1\).

Key Concepts

Partial Derivatives

Partial derivatives are essential tools in calculus, especially when we deal with functions of multiple variables. In the context of the heat equation, these derivatives help describe how temperature changes over time and space.
For a given function like \(u(x, t) = e^{-t}(2 \sin x + 3 \cos x)\), the partial derivative with respect to time \(t\) shows the rate of change of temperature as time progresses at any fixed position \(x\).
Similarly, the first and second partial derivatives with respect to space \(x\) indicate how the temperature varies across different points along the bar, including the curvature or bending of the temperature profile.

• First Partial Derivative with respect to \(t\): It uses the product rule to differentiate, treating \(x\) as a constant, giving us insights into the temporal decay of the heat.
• First Partial Derivative with respect to \(x\): It involves differentiating trigonometric functions and applying the chain rule to reveal spatial changes in temperature.
• Second Partial Derivative with respect to \(x\): This further exploration of spatial derivatives tells us about the diffusion of heat, or how heat spreads out over time.

The role of these derivatives is pivotal in proving the validity of solutions to the heat equation, helping to bridge mathematics and physical phenomena.

Conductivity

Conductivity is a measure of how well a material can conduct heat. In the heat equation, the constant \(k\) represents the thermal conductivity of the material along the bar. This constant is directly related to how fast heat can spread through the material over time.
In our heat equation problem, when we have \(k = 1\), it simplifies the mathematical representation but still indicates that the material has a uniform ability to conduct heat. A higher \(k\) would mean the material conducts heat more swiftly, while a lower \(k\) suggests slower heat transfer.

• Understanding \(k\)'s role helps predict heat flow characteristics. In real-world applications, knowing the exact \(k\) is crucial in materials science for designing systems like radiators or insulation.
• The constant \(k\) affects how quickly temperature changes in response to a temperature gradient. It's why materials feel different to the touch; metals with high \(k\) feel cool, while non-metals with low \(k\) feel warm.

In mathematical settings like this exercise, \(k = 1\) is often chosen for simplicity and helps focus on understanding the other variables in play, namely the mathematical behavior of temperature functions and their derivatives.

Mathematical Proof

A mathematical proof is a logical argument that demonstrates the truth of a statement beyond any doubt. In proving that our function \(u(x, t) = e^{-t}(2 \sin x + 3 \cos x)\) satisfies the heat equation, we show that the equation holds for all relevant variables.
The proof involves a sequence of clear steps:

• **Finding the Partial Derivatives**: First, compute the partial derivative with respect to time and the second partial derivative with respect to space.
• **Verification**: Compare these derivatives to see if they satisfy the original heat equation; i.e., whether \(\frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2}\).
• **Conclusion**: Confirming the equivalence of the derivatives supports the assertion that \(u(x, t)\) is a valid solution.

The consistency in this approach ensures that the function behaves precisely as the heat equation predicts. Such proofs are not only academically interesting but essential in applications like engineering, where they help model how heat or other phenomena behave in real-life scenarios. Understanding the logical flow—from deriving equations to verifying them—makes the connection between mathematics and its practical applications more tangible.