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Chapter 12: Problem 5

Solve the boundary-value problem $$ \begin{aligned} &k \frac{\partial^{2} u}{\partial x^{2}}+A e^{-\beta z}=\frac{\partial u}{\partial t}, \quad \beta>0, \quad 00 \\ &u(0, t)=0, \quad u(1, t)=0, \quad t>0 \\ &u(x, 0)=f(x), \quad 0

Short Answer

Expert verified
The solution is expressed as a Fourier series: \( u(x, t) = \sum_{n=1}^{\infty} \left[ C_n e^{-(k(n\pi)^2)t} + \frac{Ae^{-\beta z}}{k(n\pi)^2} \right] \text{sin}(n\pi x) \).

Step by step solution

01

Analyze the given problem

The given problem is a partial differential equation (PDE) resembling the heat equation with a source term. The equation is \( k \frac{\partial^{2} u}{\partial x^{2}} + A e^{-\beta z} = \frac{\partial u}{\partial t} \). The domain is \( 0 < x < 1 \) and \( t > 0 \), with boundary conditions \( u(0, t) = 0 \), \( u(1, t) = 0 \), and initial condition \( u(x, 0) = f(x) \).
02

Apply Separation of Variables

Assume a solution of the form \( u(x, t) = X(x)T(t) \). Substituting it into the PDE gives: \( k X''(x) T(t) + A e^{-\beta z} = X(x) T'(t) \). Divide the equation by \( X(x)T(t) \), yielding \( \frac{T'(t)}{kT(t)} = \frac{X''(x)}{X(x)} - \frac{A e^{-\beta z}}{kT(t)} \).
03

Solve the Spatial Part

The spatial part \( \frac{X''(x)}{X(x)} = -\lambda \) leads to the equation \( X''(x) + \lambda X(x) = 0 \) with boundary conditions \( X(0) = 0 \) and \( X(1) = 0 \). A suitable solution is \( X_n(x) = \,\text{sin}(n\pi x) \) with \( \lambda = (n\pi)^2 \), \( n = 1, 2, 3, \ldots \).
04

Solve the Time Part

For the time part, \( \frac{T'(t)}{k} + \lambda T(t) = A e^{-\beta z} \). Solving this ordinary differential equation gives \( T_n(t) = C_n e^{-(k(n\pi)^2)t} + \frac{Ae^{-\beta z}}{k(n\pi)^2} \).
05

Construct General Solution

The general solution is \( u(x, t) = \sum_{n=1}^{\infty} \left[ C_n e^{-(k(n\pi)^2)t} + \frac{Ae^{-\beta z}}{k(n\pi)^2} \right] \text{sin}(n\pi x) \).
06

Apply Initial Condition

Apply the initial condition \( u(x, 0) = f(x) \) to find \( C_n \): \( f(x) = \sum_{n=1}^{\infty} \left[ C_n + \frac{Ae^{-\beta z}}{k(n\pi)^2} \right] \text{sin}(n\pi x) \). Use Fourier coefficients to solve: \( C_n = \int_0^1 f(x) \text{sin}(n\pi x) \; dx \) minus the contribution from the source term.

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Key Concepts

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

Boundary-value problems
Boundary-value problems are core aspects of solving partial differential equations (PDEs), where we determine a solution satisfying certain conditions on the boundaries of the domain. For the provided exercise, we're dealing with a PDE within the region
  • The spatial domain is defined as \(0 < x < 1\)
  • The boundary conditions specify that \(u(0, t) = 0\) and \(u(1, t) = 0\) for all times \(t > 0\).
  • These boundary conditions ensure that the solution remains zero at both ends of the interval, resembling the scenario of a rod maintained at zero temperature at both ends.
In addition to boundary conditions, boundary-value problems also include initial conditions. Here, the initial condition is \(u(x, 0) = f(x)\). The task is to find the solution \(u(x, t)\) for all \((x, t)\) within our specified domains, fulfilling both boundary and initial conditions. This ensures our solution is correctly defined at the problems' boundaries as well as its starting configuration.
Heat equation
The heat equation is a classical partial differential equation describing the distribution of heat (or temperature variations) in a given region over time. In our problem, the equation is slightly modified with a source term due to radioactive decay
  • The equation: \(k\frac{\partial^{2} u}{\partial x^{2}} + A e^{-\beta z} = \frac{\partial u}{\partial t}\), helps us understand how heat in a thin rod evolves as time progresses.
  • Here, \(k\) represents thermal conductivity, which affects how quickly heat spreads through the medium.
  • The term \(A e^{-\beta z}\) signifies the heat generated by some source - in this instance, the radioactive decay impacting heat distribution.
Our task involves finding \(u(x, t)\), the temperature at any position \(x\) at time \(t\), given both the boundary and initial conditions. The presence of the source term implies additional heat input, making the solution more intricate than the standard heat equation.
Separation of variables
Separation of variables is a powerful technique to solve linear partial differential equations with boundary conditions. The objective is generally to express the solution as a product of functions, each dependent on a single coordinate
  • For example, assume \(u(x, t) = X(x)T(t)\), separating spatial and temporal components.
  • Substituting this form into the PDE transforms it, allowing us to split the PDE into two ODEs: one for space \(X(x)\) and another for time \(T(t)\).
This turns the complex PDE into two simpler problems, each solvable independently. Here:
  • The spatial part: \(X''(x) + \lambda X(x) = 0\), with boundary conditions \(X(0) = 0\) and \(X(1) = 0\).
  • The time-dependent equation: \(T'(t) + k\lambda T(t) = A e^{-\beta z}\).
Decomposing the PDE via separation of variables eases the process of finding a precise solution that adheres to both input conditions and natural characteristics of the problem.
Fourier series
Fourier series are key tools when solving boundary-value problems, chiefly for separating variables. They help express functions in terms of sinusoidal functions, making it easier to tackle the resulting equations
  • They express a function \(f(x)\) as an infinite sum: \(f(x) = \sum_{n=1}^{\infty} a_n \sin(n\pi x)\).
  • In our example, solutions for \(X(x)\) involve \(\sin(n\pi x)\), natural harmonics satisfying our boundary conditions.
  • The coefficients \(C_n\) are determined using Fourier analysis, calculated by integrating against the sine basis.
Thus, the solution to our problem is constructed as a Fourier series, combining both space and time components. This technique transforms complex expressions into manageable forms and ensures the adherence to both initial and boundary conditions, especially when internal heat sources complicate the solution.

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Most popular questions from this chapter

Solve the boundary-value problem $$ \begin{aligned} &k \frac{\partial^{2} u}{\partial x^{2}}+\sin 2 \pi x=\frac{\partial u}{\partial t}, \quad 00 \\ &u(0, t)=0, \quad u(1, t)=0, \quad t>0 \\ &u(x, 0)=\sin \pi x, \quad 0

Find a steady-state solution \(\psi(x)\) of the boundary-value problem $$ \begin{array}{ll} k \frac{\partial^{2} u}{\partial x^{2}}-h\left(u-u_{0}\right)=\frac{\partial u}{\partial t}, \quad 00 \\ u(0, t)=u_{0}, \quad u(1, t)=0, \quad t>0 \\ u(x, 0)=f(x), \quad 0

Solve the boundary-value problem $$ \begin{array}{ll} a^{2} \frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial^{2} u}{\partial t^{2}}, & 00 \\ u(0, t)=0, & \left.E \frac{\partial u}{\partial x}\right|_{x=L}=F_{0}, \quad t>0 \\ u(x, 0)=0, & \left.\frac{\partial u}{\partial t}\right|_{t=0}=0, \quad 0

Find the steady-state temperature for a rectangular plate with beundary eonditions as given $$ \begin{array}{ll} u(0, y)=0, & u(a, y)=0 \\ u(x, 0)=f(x), & u(x, b)=0 \end{array} $$

Solve the boundary-value problem $$ \begin{aligned} &k \frac{\partial^{2} u}{\partial x^{2}}-h u=\frac{\partial u}{\partial t}, \quad 00 \\ &u(0, t)=0, \quad u(\pi, t)=u_{0}, \quad t>0 \\ &u(x, 0)=0, \quad 0
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