/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 The edges \(x=0, a\) and \(y=b\)... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 3

The edges \(x=0, a\) and \(y=b\) of the rectangle \(0 \leqslant x \leqslant a, 0 \leqslant y \leqslant b\) are maintained at zero temperature while the temperature along the edge \(y=0\) is made to vary according to the rule \(\theta(x, 0, t)=f(x), 0 \leqslant x \leqslant a, t>0\) If the initial temperature in the rectangle is zero, find the temperature at any subsequent time \(t\), and deduce that the steady-state temperature is $$ \frac{2}{a} \sum_{m=1}^{\infty} \frac{\sinh [m \pi(b-y) / a]}{\sinh (m \pi b / a)} \sin \left(\frac{m \pi x}{a}\right) \int_{0}^{a} f(u) \sin \left(\frac{m \pi u}{a}\right) d u $$

Short Answer

Expert verified
The steady-state temperature is given by the series involving a hyperbolic sine function and sine integrals.

Step by step solution

01

Set up the heat equation

The problem is governed by the heat equation for a two-dimensional region, given by \( \frac{\partial \theta}{\partial t} = \alpha \left( \frac{\partial^2 \theta}{\partial x^2} + \frac{\partial^2 \theta}{\partial y^2} \right) \), where \( \alpha \) is the thermal diffusivity. Since we are looking for steady-state solutions, we set \( \frac{\partial \theta}{\partial t} = 0 \) which simplifies the equation to the Laplace equation \( \frac{\partial^2 \theta}{\partial x^2} + \frac{\partial^2 \theta}{\partial y^2} = 0 \).
02

Boundary conditions

The boundary conditions are given as: \( \theta(x,0,t) = f(x) \) for \( 0 \leq x \leq a \) and \( t > 0 \), while \( \theta(x,b,t) = 0 \), \( \theta(0,y,t) = 0 \), and \( \theta(a,y,t) = 0 \). This suggests using the method of separation of variables to find \( \theta(x,y,t) = X(x)Y(y)T(t) \).
03

Solve for spatial components

Applying separation of variables, we assume \( \theta(x,y,t) = X(x)Y(y) \). Plugging into the Laplace equation, we obtain two ordinary differential equations (ODEs): \( X''(x) + \lambda X(x) = 0 \) and \( Y''(y) - \lambda Y(y) = 0 \). The boundary conditions require \( X(0) = 0 \) and \( X(a) = 0 \), leading to non-trivial solutions where \( \lambda = \left(\frac{m\pi}{a}\right)^2 \). Thus \( X(x) = \sin \left(\frac{m\pi x}{a}\right) \).
04

Solve for temperature distribution in y-direction

For \( Y(y) \), the solutions are of the form \( Y(y) = A\sinh\left(\frac{m\pi y}{a}\right) + B\cosh\left(\frac{m\pi y}{a}\right) \). Applying boundary conditions \( Y(b) = 0 \) gives \( B = 0 \), hence \( Y(y) = A\sinh\left(\frac{m\pi y}{a}\right) \).
05

Form the complete solution

Combine solutions to get \( \theta_m(x, y) = \sin\left(\frac{m\pi x}{a}\right) \sinh\left(\frac{m\pi y}{a}\right) \). The general solution, sum of such solutions, is \( \theta(x,y) = \sum_{m=1}^{\infty} C_m \sin\left(\frac{m\pi x}{a}\right)\sinh\left(\frac{m\pi(b-y)}{a}\right) \).
06

Determine coefficients

The coefficients \( C_m \) are determined using the initial condition \( \theta(x,0,t) = f(x) \). Using the orthogonality of the sine function, we have: \[ C_m \sinh\left(\frac{m\pi b}{a}\right) = \frac{2}{a}\int_0^a f(u) \sin\left(\frac{m\pi u}{a}\right) du \]. Solving for \( C_m \), we get \( C_m = \frac{2}{a\sinh\left(\frac{m\pi b}{a}\right)}\int_0^a f(u) \sin\left(\frac{m\pi u}{a}\right) du \).
07

Rewrite the steady-state temperature formula

The steady-state solution is given by assuming the absence of the time-dependent equation, thus our solution is: \[ \theta(x,y) = \frac{2}{a} \sum_{m=1}^{\infty} \frac{\sinh\left(\frac{m\pi(b-y)}{a}\right)}{\sinh\left(\frac{m\pi b}{a}\right)} \sin\left(\frac{m\pi x}{a}\right) \int_0^a f(u) \sin\left(\frac{m\pi u}{a}\right) du \].

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Key Concepts

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

Heat Equation
The heat equation is a fundamental principle describing the process of heat distribution over time. In mathematics, it is formulated as a partial differential equation, commonly expressed in two dimensions as: \[ \frac{\partial \theta}{\partial t} = \alpha \left( \frac{\partial^2 \theta}{\partial x^2} + \frac{\partial^2 \theta}{\partial y^2} \right) \] where \( \theta \) represents the temperature, \( \alpha \) is the thermal diffusivity, and \( x \), \( y \), and \( t \) correspond to spatial coordinates and time. This equation serves as a model to predict how heat diffuses through a physical medium.
  • **Steady-state assumption**: In steady-state situations, the temperature distribution does not evolve over time. This means \( \frac{\partial \theta}{\partial t} = 0 \), leading to simplifications.
  • **Resulting Laplace equation**: Setting \( \frac{\partial \theta}{\partial t} = 0 \) simplifies the heat equation to the Laplace equation.
Separation of Variables
Separation of Variables is a powerful mathematical method used to solve partial differential equations (PDEs), such as the heat equation. It transforms a complex PDE into simpler, separate ordinary differential equations (ODEs) by assuming that the solution can be represented as a product of functions. In this context, we express the solution as \( \theta(x,y,t) = X(x)Y(y)T(t) \). By substituting this form into the heat equation, the multi-variable equation becomes easier to manage as it breaks down into distinct components, each dependent on a single variable:
  • **Spatial Component for \( x \)**: \( X''(x) + \lambda X(x) = 0 \)
  • **Spatial Component for \( y \)**: \( Y''(y) - \lambda Y(y) = 0 \)
This strategy is fundamental in facilitating the approach to solve heat distribution problems by leveraging the structure provided by boundary conditions and eigenvalues \( \lambda \).
Laplace Equation
Derived from the heat equation under steady-state conditions, the Laplace equation is a cornerstone in thermal analyses. It takes the form: \[ \frac{\partial^2 \theta}{\partial x^2} + \frac{\partial^2 \theta}{\partial y^2} = 0 \] This equation focuses on spatial variables, providing solutions for temperature excluding time dependency. The simplicity of the Laplace equation lies in its linearity, which allows us to apply superposition and construct solutions satisfying specific boundary conditions.
  • **Importance in steady states**: Without time as a factor, the equation describes how temperature spreads uniformly within a given region.
  • **Boundary conditions role**: The boundary conditions play a crucial role in determining unique solutions within the relevant domain.
Boundary Conditions
Boundary conditions are essential in solving differential equations like the heat equation. They prescribe the behavior of a function along the domain's edges, ensuring the solution's physical appropriateness. In our scenario, we have specific conditions: - \( \theta(x,0,t) = f(x) \) for \( 0 \leq x \leq a \) and \( t > 0 \) - \( \theta(x,b,t) = 0 \), \( \theta(0,y,t) = 0 \), \( \theta(a,y,t) = 0 \) The boundary conditions dictate how heat extends across and out of the rectangle's boundary, including:
  • **Zero condition at edges**: Ensures the temperature remains zero along certain borders, simplifying the equation in certain areas.
  • **Variable condition**: At \( y = 0 \), \( \theta \) varies with \( x \), modeling the influence of initial heat disturbance over time.
These conditions are pivotal in shaping the final temperature distribution within the rectangle.

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Most popular questions from this chapter

A circular cylinder of radius \(a\) has its surface kept at a constant temperature \(\theta_{0}\). If the initial temperature is zero throughout the cylinder, prove that for \(t>0\) $$ \theta(r, t)=\theta_{0}\left\\{1-\frac{2}{a} \sum_{n=1}^{\infty} \frac{J_{0}\left(\xi_{n} a\right)}{\xi_{n} J_{1}\left(\xi_{n} a\right)} e^{-\xi_{n}^{2} \kappa t}\right\\} $$ where \(\pm \xi_{1}, \pm \xi_{0}, \ldots, \pm \xi_{n}, \ldots\) are the roots of \(J_{0}(\xi a)=0\).

If \(\theta_{r}\left(x_{r}, t\right) r=1,2,3\) is the solution of the one- dimensional diffusion equation $$ \frac{\partial^{2} \theta_{r}}{\partial x_{r}^{2}}=\frac{1}{\kappa} \frac{\partial \theta_{r}}{\partial t} \quad a_{r}0 $$ satisfying the initial condition \(\theta_{r}\left(x_{r}, \theta\right)=f_{r}\left(x_{r}\right)\) and the boundary conditions $$ \begin{array}{lll} \alpha_{r} \frac{\partial \theta_{r}}{\partial x_{r}}-\alpha_{r}^{\prime} \theta_{r}=0, & x_{r}=a_{r} & t>0 \\ \beta_{r} \frac{\partial \theta_{r}}{\partial x_{r}}+\beta_{r}^{\prime} \theta_{r}=0, & x_{r}=b_{r} & t>0 \end{array} $$ then the solution of $$ \frac{\partial^{2} \theta}{\partial x_{1}^{2}}+\frac{\partial^{2} \theta}{\partial x_{2}^{2}}+\frac{\partial^{2} \theta}{\partial x_{3}^{2}}=\frac{1}{\kappa} \frac{\partial \theta}{\partial t} $$ in the rectangular parallelepiped \(a_{1}0, r=1,2,3, \\ \beta_{r} \frac{\partial \theta}{\partial x_{r}}+\beta_{r}^{\prime} \theta=0, & x_{r}=b_{r} \quad t>0, r=1,2,3 \end{array} $$ and the initial condition \(\theta=f_{1}\left(x_{1}\right) f_{2}\left(x_{2}\right) f_{3}\left(x_{3}\right)\) is $$ \theta\left(x_{1}, x_{2}, x_{3}, t\right)=\theta_{1}\left(x_{1}, t\right) \theta_{2}\left(x_{2}, t\right) \theta_{3}\left(x_{3}, t\right) $$

Elementary Solutions of the Diffusion Equation In this section we shall consider elementary solutions of the onedimensional diffusion equation $$ \frac{\partial^{2} \theta}{\partial x^{2}}=\frac{1}{\kappa} \frac{\partial \theta}{\partial t} $$ We begin by considering the expression $$ \theta=\frac{1}{\sqrt{t}} \exp \left(-\frac{x^{2}}{4 \kappa t}\right) $$ For this function it is readily seen that and $$ \begin{aligned} \frac{\partial^{2} \theta}{\partial x^{2}} &=\frac{x^{2}}{4 \kappa^{2} t^{5.2}} e^{-x^{2} / 4 x t}-\frac{1}{2 \kappa t^{3 / 2}} e^{-x^{2} / 4 \kappa t} \\ \frac{\partial \theta}{\partial t} &=\frac{x^{2}}{4 \kappa t^{5 / 2}} e^{-x^{2} / 4 \kappa t}-\frac{1}{2 t^{3 / 2}} e^{-x^{2} / 4 \kappa t} \end{aligned} $$ showing that the function (2) is a solution of the equation (1). It follows immediately that $$ \frac{1}{2 \sqrt{\pi \kappa t}} e^{-(x-\xi)^{2} / 4 \kappa t} $$ where \(\xi\) is an arbitrary real constant, is also a solution. Furthermore, if the function \(\phi(x)\) is bounded for all real values of \(x\), then it is possible that the integral $$ \frac{1}{2 \sqrt{\pi \kappa t}} \int_{-\infty}^{\infty} \phi(\xi) \exp \left\\{-\frac{(x-\xi)^{2}}{4 \kappa t}\right\\} d \xi $$ is also, in some sense, a solution of the equation (1). It may readily be proved that the integral (4) is convergent if \(t>0\) and that the integrals obtained from it by differentiating under the integral sign with respect to \(x\) and \(t\) are uniformly convergent in the neighborhood of the point \((x, t) .\) The function \(\theta(x, t)\) and its derivatives of all orders therefore exist for \(t>0\), and since the integrand satisfies the one-dimensional diffusion equation, it follows that \(\theta(x, t)\) itself satisfies that equation for \(t>0\). Now where $$ \begin{array}{c} \left|\frac{1}{2(\pi \kappa t)^{\frac{1}{2}}} \int_{-\infty}^{\infty} \phi(\xi) \exp \left\\{-\frac{(x-\xi)^{2}}{4 \kappa t}\right\\} d \xi-\phi(x)\right| \\ \quad=\left|I_{1}+I_{2}+I_{3}-I_{4}\right| \\ I_{1}=\frac{1}{\sqrt{\pi}} \int_{-N}^{N}\\{\phi(x+2 u \sqrt{\kappa t})-\phi(x)\\} e^{-u^{2}} d u \end{array} $$ $$ \begin{array}{l} I_{2}=\frac{1}{\sqrt{\pi}} \int_{N}^{\infty} \phi(x+2 u \sqrt{\kappa t}) e^{-u^{2}} d u \\ I_{3}=\frac{1}{\sqrt{\pi}} \int_{-\infty}^{-N} \phi(x+2 u \sqrt{\kappa t}) e^{-u^{2}} d u \\ I_{4}=\frac{2 \phi(x)}{\sqrt{\pi}} \int_{N}^{\infty} e^{-u^{2}} d u \end{array} $$ If the function \(\phi(x)\) is bounded, we can make each of the integrals \(I_{2}, I_{3}\), \(I_{4}\) as small as we please by taking \(N\) to be sufficiently large, and by the continuity of the function \(\phi\) we can make the integral \(I_{1}\) as small as we please by taking \(t\) sufficiently small. Thus as \(t \rightarrow 0, \theta(x, t) \rightarrow \phi(x)\). Thus the Poisson integral $$ \theta(x, t)=\frac{1}{2(\pi \kappa t)^{\frac{1}{2}}} \int_{-\infty}^{\infty} \phi(\xi) \exp \left\\{-\frac{(x-\xi)^{2}}{4 \kappa t}\right\\} d \xi $$ is the solution of the initial value problem $$ \begin{aligned} \frac{\partial^{2} \theta}{\partial x^{2}} &=\frac{1}{\kappa} \frac{\partial \theta}{\partial t} \quad-\infty0 \\ \theta(0, t)=0 & t>0\end{array}\)

The slowing-down density \(\theta\) of neutrons in the infinite pile \(0 \leqslant x \leqslant a\), \(0 \leqslant y \leqslant b,-\infty

Solve the one-dimensional diffusion equation in the region \(0 \leqslant x \leqslant \pi\), \(t \geqslant 0\), when (i) \(\theta\) remains finite as \(t \rightarrow \infty\); (ii) \(\quad \theta=0\) if \(x=0\) or \(\pi\), for all values of \(t\); (iii) At \(t=0,\left\\{\begin{array}{ll}\theta=x & 0 \leqslant x \leqslant \frac{1}{2} \pi \\ \theta=\pi-x & \frac{1}{2} \pi \leqslant x \leqslant \pi .\end{array}\right.\)
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