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Chapter 26: Problem 1

Interpret and solve the following problem. $$\begin{aligned}\frac{\partial u}{\partial t} &=\frac{\partial^{2} u}{\partial x^{2}}, \quad \text { for } t>0,00 \\\& x \rightarrow 1^{-}, \frac{\partial u}{\partial x} \rightarrow 0, \quad \text { for } t>0 \\\\\text { ANS. } \quad u &=1-\sum_{n=0}^{\infty}(-1)^{n}\left[\operatorname{erfc}\left(\frac{2 n+x}{2 \sqrt{t}}\right)+\operatorname{erfc}\left(\frac{2 n+2-x}{2 \sqrt{t}}\right)\right] \end{aligned}.$$

Short Answer

Expert verified
The proposed solution satisfies both the initial and boundary conditions of the heat equation.

Step by step solution

01

Identify the Equation Type

The given equation \( \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2} \) is a classic form of the heat equation. This type of partial differential equation (PDE) models diffusion processes such as heat conduction.
02

Initial Conditions

Examine the given initial condition \( u \rightarrow 1 \) as \( t \rightarrow 0^+ \) for \( 0 < x < 1 \). This indicates that at time \( t=0 \), the initial temperature distribution across the medium is uniform at 1.
03

Boundary Conditions

Check the boundary conditions given:- As \( x \rightarrow 0^+ \), \( u \rightarrow 0 \) for \( t>0 \). This implies that one end of the rod is kept at zero temperature.- As \( x \rightarrow 1^- \), \( \frac{\partial u}{\partial x} \rightarrow 0 \) for \( t>0 \). This represents a Neumann boundary condition at the other end, meaning there's no heat flux or it is insulated.
04

Analyze the Proposed Solution

The proposed solution is given as: \[ u = 1 - \sum_{n=0}^{\infty}(-1)^n \left[ \operatorname{erfc}\left(\frac{2n + x}{2\sqrt{t}}\right) + \operatorname{erfc}\left(\frac{2n + 2 - x}{2\sqrt{t}}\right) \right] \] This form uses the complementary error function (`erfc`), which is typical for solutions to diffusion problems.
05

Applying Solution Structure

Understand that the use of `erfc` comes from the method of images used in solving PDEs like the heat equation, especially under specific boundary conditions. The infinite sum accounts for the periodic nature imposed by these boundary conditions.
06

Verify Given Conditions with Solution

Ensure the solution meets all initial and boundary conditions. At \( t = 0 \), \( u = 1 \) due to all terms vanishing as error functions approach certain limits, thus meeting the initial condition. For the boundary at \( x = 0 \), as \( t \) increases, \( u \to 0 \). As \( x \to 1 \), the Neumann condition is satisfied because the derivative of \( u \) with respect to \( x \) approaches zero.

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Key Concepts

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

Partial Differential Equations
Partial Differential Equations (PDEs) are a class of mathematical equations used to describe a variety of physical phenomena. These equations involve unknown functions and their partial derivatives with respect to multiple variables. In this exercise, we focus on the Heat Equation, which is a PDE given by: \[\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}\]This particular equation models heat distribution over time in a given medium. It describes how the temperature changes at a point based on the temperature's spatial second derivative. This gives insight into diffusion processes like heat conduction or particle spread. Understanding PDEs is crucial as they form the backbone of many scientific models across physics, engineering, and finance. They can be more complex than ordinary differential equations (ODEs) due to their multidimensional nature. Therefore, solving PDEs usually requires specific methods and techniques like separation of variables, Fourier series, or the method of characteristics.
Boundary Conditions
Boundary Conditions are constraints needed to find a unique solution for a PDE. For the heat equation in this exercise, boundary conditions provide essential information about the system's state at its borders. We are given two boundary conditions here:
  • As \( x \rightarrow 0^+ \), \( u \rightarrow 0 \). This sets the temperature at one boundary of the rod to zero, akin to keeping one end at a fixed cooled temperature.
  • As \( x \rightarrow 1^- \), \( \frac{\partial u}{\partial x} \rightarrow 0 \). This Neumann boundary condition implies no heat is flowing at this boundary, indicating an insulated end.
These conditions are crucial as they allow formulation of solutions like the method of images, which uses symmetrical reflections to satisfy conditions.
Initial Conditions
Initial Conditions provide the system's state at the start of the observation period. Here, at \( t \rightarrow 0^+ \), the condition given is \( u \rightarrow 1 \) for \( 0 < x < 1 \). This means initially, the temperature across the entire rod is uniformly equal to 1.Initial conditions play a pivotal role in determining the evolution of the system over time. Without them, it's impossible to predict how the heat distribution will change. It's like knowing what you have before starting an experiment, to better predict the results at future points.
Diffusion Processes
Diffusion Processes describe how substances like heat or particles spread in a medium over time. The heat equation in this problem models such a process, specifically how heat diffuses along the rod.These diffusion processes are governed by laws like Fick's laws of diffusion, emphasizing how concentration differences drive movement. In the case of the heat equation, it quantifies how heat moves from high-temperature regions to low-temperature ones until equilibrium is reached.Solutions using complementary error functions (\( \operatorname{erfc} \)) are typical here. These functions help solve the PDEs by capturing the diffusion nature in spreading effects, defined by error handling integrals. The infinite series in the solution accounts for periodic boundary conditions, perhaps modeling a repeated or infinite medium.

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Most popular questions from this chapter

Show that erf \(x\) is an odd function of \(x\).

Interpret and solve the problem $$\begin{array}{ll}\frac{\partial^{2} y}{\partial t^{2}}=\frac{\partial^{2} y}{\partial x^{2}}, & \text { for } t>0,00 \\\x \rightarrow 1^{-}, y \rightarrow 0, & \text { for } t>0\end{array}$$ Verify your solution directly. $$\begin{aligned}\text { ANS. } \quad y=x-x^{2}-t^{2} &+\sum_{n=0}^{\infty}(-1)^{n}\left[(t-n-x)^{2} \alpha(t-n-x)\right.\\\&\left.+(t-n-1+x)^{2} \alpha(t-n-1-x)\right]\end{aligned}$$.

Derive the result $$A(s)=L\left\\{t^{-3 / 2} \exp \left(-\frac{k^{2}}{t}\right)\right\\}=\frac{\sqrt{\pi}}{k} \exp (-2 k \sqrt{s}), \quad k>0, s>0$$ directly from the definition of a transform. In the integral $$A(s)=\int_{0}^{\infty} \exp \left(-s t-k^{2} t^{-1}\right) t^{-3 / 2} d t$$ put \(\beta=\sqrt{t}\) to get $$A(s)=2 \int_{0}^{\infty} \beta^{-2} \exp \left(-s \beta^{2}-k^{2} \beta^{-2}\right) d \beta$$ OI $$A(s)=2 \exp (-2 k \sqrt{s}) \int_{0}^{\infty} \beta^{-2} \exp \left[-\left(\beta \sqrt{s}-k \beta^{-1}\right)^{2}\right] d \beta$$ Show that $$\begin{aligned}\frac{d A}{d s} &=-2 \int_{0}^{\infty} \exp \left(-s \beta^{2}-k^{2} \beta^{-2}\right) d \beta \\\&=-2 \exp (-2 k \sqrt{s}) \int_{0}^{\infty} \exp \left[-\left(\beta \sqrt{s}-k \beta^{-1}\right)^{2}\right] d \beta\end{aligned}$$ Thus arrive at the differential equation $$\sqrt{s} \frac{d A}{d s}-k A=-2 \sqrt{\pi} \exp (-2 k \sqrt{s})$$ and from it obtain the desired function \(A(s)\).

Use integration by parts to show that $$\int_{0}^{x} \operatorname{erf} y d y=x \operatorname{erf} x-\frac{1}{\sqrt{\pi}}\left[1-\exp \left(-x^{2}\right)\right].$$

Show that \(\lim _{x \rightarrow 0} \frac{\operatorname{erf} x}{x}=\frac{2}{\sqrt{\pi}}\).
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