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Chapter 12: Problem 46

The wave equation \(c^{2} \partial^{2} u / \partial x^{2}=\partial^{2} u / \partial t^{2}\) and the heat equation \(c \partial^{2} u / \partial x^{2}=\partial u / \partial t\) are two of the most important equations in physics ( \(c\) is a constant). These are called partial differential equations. Show each of the following: (a) \(u=\cos x \cos c t\) and \(u=e^{x}\) cosh \(c t\) satisfy the wave equation. (b) \(u=e^{-c t} \sin x\) and \(u=t^{-1 / 2} e^{-x^{2} /(4 c t)}\) satisfy the heat equation.

Short Answer

Expert verified
Both functions in each part satisfy their respective equations.

Step by step solution

01

Verify wave equation for \(u = \cos x \cos ct\)

First, compute the partial derivatives. Find \(\frac{\partial u}{\partial x}\):\[ \frac{\partial}{\partial x}(\cos x \cos ct) = -\sin x \cos ct \]Next, compute \(\frac{\partial^2 u}{\partial x^2}\):\[ \frac{\partial^2}{\partial x^2}(\cos x \cos ct) = -\frac{\partial}{\partial x}(\sin x \cos ct) = -\cos x \cos ct \]Now, compute the time derivatives. Find \(\frac{\partial u}{\partial t}\):\[ \frac{\partial}{\partial t}(\cos x \cos ct) = -c \cos x \sin ct \]And then \(\frac{\partial^2 u}{\partial t^2}\):\[ \frac{\partial^2}{\partial t^2}(\cos x \cos ct) = -c^2 \cos x \cos ct \]Substitute into the wave equation:\[ c^2(-\cos x \cos ct) = -c^2 \cos x \cos ct \]Both sides are equal, so \(u = \cos x \cos ct\) satisfies the wave equation.
02

Verify wave equation for \(u = e^x \cosh ct\)

Begin with the spatial derivatives. First, compute \(\frac{\partial u}{\partial x}\):\[ \frac{\partial}{\partial x}(e^x \cosh ct) = e^x \cosh ct \]Then, find \(\frac{\partial^2 u}{\partial x^2}\):\[ \frac{\partial^2}{\partial x^2}(e^x \cosh ct) = e^x \cosh ct \]Now, compute the time derivatives. First, \(\frac{\partial u}{\partial t}\):\[ \frac{\partial}{\partial t}(e^x \cosh ct) = c e^x \sinh ct \]Then, \(\frac{\partial^2 u}{\partial t^2}\):\[ \frac{\partial^2}{\partial t^2}(e^x \cosh ct) = c^2 e^x \cosh ct \]Substitute into the wave equation:\[ c^2(e^x \cosh ct) = c^2 e^x \cosh ct \]Both sides are equal, verifying that \(u = e^x \cosh ct\) satisfies the wave equation.
03

Verify heat equation for \(u = e^{-ct} \sin x\)

First, compute the spatial and temporal derivatives of \(u\). For the spatial derivative:\[ \frac{\partial u}{\partial x} = e^{-ct} \cos x \]And the second spatial derivative:\[ \frac{\partial^2 u}{\partial x^2} = -e^{-ct} \sin x \]For the temporal derivative:\[ \frac{\partial u}{\partial t} = -ce^{-ct} \sin x \]Substitute into the heat equation:\[ c(-e^{-ct} \sin x) = -ce^{-ct} \sin x \]Both sides match, showing that \(u = e^{-ct} \sin x\) satisfies the heat equation.
04

Verify heat equation for \(u = t^{-1/2} e^{-x^2/(4ct)}\)

Start by finding the spatial derivatives. First, calculate \(\frac{\partial u}{\partial x}\):\[ \frac{\partial}{\partial x} \left(t^{-1/2} e^{-x^2/(4ct)}\right) = t^{-1/2} \cdot \left(-\frac{x}{2ct} \right) \cdot e^{-x^2/(4ct)} = -\frac{x}{2ct^{3/2}} e^{-x^2/(4ct)} \]Next, the second spatial derivative:\[ \frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x} \left(-\frac{x}{2ct^{3/2}} e^{-x^2/(4ct)}\right) = \frac{1}{2ct^{3/2}} e^{-x^2/(4ct)} - \frac{x^2}{4c^2t^{5/2}} e^{-x^2/(4ct)} \]For the temporal derivative:\[ \frac{\partial u}{\partial t} = \frac{1}{2t^{3/2}} e^{-x^2/(4ct)} + \frac{x^2}{4ct^{5/2}} e^{-x^2/(4ct)} \]Substitute into the heat equation:\[ c \left( \frac{1}{2ct^{3/2}} e^{-x^2/(4ct)} - \frac{x^2}{4c^2t^{5/2}} e^{-x^2/(4ct)} \right) = \frac{1}{2t^{3/2}} e^{-x^2/(4ct)} + \frac{x^2}{4ct^{5/2}} e^{-x^2/(4ct)} \]Both sides of the equation are identically equal, hence \(u = t^{-1/2} e^{-x^2/(4ct)}\) satisfies the heat equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wave Equation
The wave equation is a fundamental concept in physics and mathematical modeling that describes how waves propagate in various mediums. It is a type of partial differential equation (PDE) that embodies the relationship between space and time variations of a wave. The general form of the wave equation is: \[ c^2 \frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 u}{\partial t^2} \] Where \(u\) represents the wave function, \(c\) is the wave speed, \(x\) is the spatial coordinate, and \(t\) is time. Key insights about the wave equation include:
  • It models waves like sound and light that travel through different media.
  • The equation indicates that the wave's acceleration in time is proportionally related to its curvature in space.
  • This equation can be used to study how a wave changes shape and speed as it moves.
When verifying solutions, as shown in the exercise, we focus on calculating partial derivatives with respect to \(x\) and \(t\) to show the equation holds true. This often involves trigonometric or exponential functions, reflecting oscillating or growing/decaying wave behaviors.
Heat Equation
The heat equation is another critical partial differential equation often used to describe the diffusion of heat (or other substances) in various materials over time. It relates the change in temperature of a medium to the flow of heat through the medium. The canonical form of the heat equation is: \[ c \frac{\partial^2 u}{\partial x^2} = \frac{\partial u}{\partial t} \] Where \(u\) is the temperature distribution, \(c\) is a constant related to thermal conductivity, \(x\) is the spatial variable, and \(t\) represents time. Essential aspects of the heat equation include:
  • It is used for modeling how heat diffuses through a solid over time.
  • The equation describes how temperature gradients drive heat flow, which aligns with Fourier's law of heat conduction.
  • Solving the heat equation helps in predicting temperature changes and is vital in engineering and environmental sciences.
In solving the exercise, we identify solutions by calculating derivatives and substituting back into the heat equation to confirm the equation holds true. This confirms the appropriateness of the temperature distribution function.
Partial Derivatives
Partial derivatives are at the heart of solving partial differential equations like the wave and heat equations. They measure the rate at which a function changes as one of its variables changes, while keeping other variables constant. In mathematical notation, a partial derivative of a function \(u(x, t)\) with respect to \(x\) is denoted as \(\frac{\partial u}{\partial x}\), and with respect to \(t\), it's \(\frac{\partial u}{\partial t}\). Why partial derivatives matter:
  • They allow us to analyze the behavior of functions of multiple variables, revealing insights into how each variable influences the function.
  • In wave and heat equations, nested derivatives (\(\frac{\partial^2 u}{\partial x^2}\) and \(\frac{\partial^2 u}{\partial t^2}\)) relate spatial and temporal changes, crucial for finding solutions.
  • These derivatives underpin vector calculus and are instrumental in fields like fluid dynamics, electromagnetism, and economic modeling.
To solve PDEs, one calculates these partial derivatives and substitutes them back into the given equation to check if assumed solutions satisfy the original PDE conditions.

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Most popular questions from this chapter

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