Question

Heat equation 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 \(k=1.\) \(u(x, t)=A e^{-a^{2} t} \cos a x,\) for any real numbers \(a\) and \(A\)

Short answer

Question: Verify that the function $$u(x, t)=A e^{-a^{2} t} \cos a x$$ satisfies the heat equation with \(k=1\). Answer: Upon calculating the partial derivatives and plugging them into the heat equation, we have confirmed that the given function $$u(x, t)=A e^{-a^{2} t} \cos a x$$ does satisfy the heat equation with \(k=1\).

Step-by-step solution

Finding partial derivatives

First, find the partial derivatives of \(u(x, t)\) with respect to \(t\) and \(x^2\). Partial derivative with respect to \(t\): $$\frac{\partial u}{\partial t} = \frac{\partial}{\partial t} \left(A e^{-a^{2} t} \cos a x \right)$$ Partial derivative with respect to \(x^2\): $$\frac{\partial^{2} u}{\partial x^{2}} = \frac{\partial^2}{\partial x^{2}} \left( A e^{-a^{2} t} \cos a x \right)$$

Calculate the partial derivatives

Now, calculate the partial derivatives found in step 1. Partial derivative with respect to \(t\): $$\frac{\partial u}{\partial t} = -Aa^{2} e^{-a^{2} t} \cos a x$$ Partial derivative with respect to \(x^2\): First partial derivative with respect to \(x\): $$\frac{\partial u}{\partial x} = -Aa e^{-a^{2} t} \sin a x$$ Second partial derivative with respect to \(x\): $$\frac{\partial^2 u}{\partial x^2} = -Aa^2 e^{-a^{2} t} \cos a x$$

Plug the partial derivatives into the heat equation

Now, plug the partial derivatives into the heat equation formula and check whether it holds true. Heat equation formula: $$\frac{\partial u}{\partial t}=k \frac{\partial^{2} u}{\partial x^{2}}$$ For \(k=1\), the heat equation is: $$\frac{\partial u}{\partial t}= \frac{\partial^{2} u}{\partial x^{2}}$$ Plugging in the values calculated in step 2: $$-Aa^{2} e^{-a^{2} t} \cos a x = -Aa^2 e^{-a^{2} t} \cos a x$$

Check if the equation is true

We can see that both sides of the equation match, thereby confirming that the given function $$u(x, t)=A e^{-a^{2} t} \cos a x$$ satisfies the heat equation with \(k=1\).

Key concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus, especially when dealing with functions of several variables. A partial derivative of a function tells us how the function changes as one of the variables changes, while keeping the others constant. This becomes significant in the heat equation, which is a partial differential equation.
Consider a function of two variables, like the temperature function \(u(x, t)\) in this exercise:

• \(\frac{\partial u}{\partial t}\) is the partial derivative of \(u\) with respect to time \(t\), indicating how the temperature changes at a particular location over time.
• \(\frac{\partial^2 u}{\partial x^2}\) is the second partial derivative of \(u\) with respect to space \(x\), representing how the rate of change of temperature gradient changes over space.

These derivatives help us model the behavior and evolution of phenomena, like heat, that varies with multiple dimensions. In this exercise, these derivatives are essential for verifying whether a proposed function satisfies the heat equation. By calculating and equating both, we ensure that changes in time and space are consistently described by the heat flow model.

Conductivity

Conductivity is a property of materials that describes how well they can conduct heat. It is represented by the constant \(k\) in the heat equation. The value of \(k\) affects how quickly or slowly heat is transferred through a material.
For instance:

• A higher conductivity means the material transfers heat quickly, resulting in faster heat distribution.
• A lower conductivity implies the material is more of an insulator and transfers heat more slowly.

In the given one-dimensional heat equation, it's noted that \(k=1\). This simplifies analysis, allowing us to focus on deriving the relationship between the partial derivatives of temperature concerning time and space.
Understanding conductivity is crucial when analyzing thermal processes as it dictates the speed of heat propagation. Every material has a unique conductivity value, which can substantially affect the outcome of heat transfer scenarios. For example, metals typically have high \(k\) values, explaining why they feel cold to the touch—they easily absorb heat from surroundings.

One-Dimensional Heat Equation

The one-dimensional heat equation is a partial differential equation used to describe how heat diffuses through a medium over time. It is a simplified model that assumes heat conduction only happens in one direction. In the context of this exercise, it's applied on a thin bar:
The equation is given as \[\frac{\partial u}{\partial t}=k \frac{\partial^{2} u}{\partial x^{2}}\]Where:

• \(u(x, t)\): Temperature at position \(x\) and time \(t\).
• \(k\): Thermal conductivity of the material.

Understanding this equation requires recognizing that it equates the rate of change of temperature over time to the spatial derivative of temperature. This representation is crucial in predicting how temperature varies within a material when subjected to heat over time.
In the solved example, verifying a function satisfies this equation involves calculating each side of the equation using partial derivatives and checking equality. This ensures the function adheres to the essence of heat distribution as dictated by the formula. The simplicity of a one-dimensional model helps comprehend fundamental mechanisms before advancing to more complex multi-dimensional cases.