/*! 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 23 For some problems, numerical or ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 27: Problem 23

For some problems, numerical or algebraic experimentation may suggest the form of the complete solution. Consider the problem of numerically integrating the first-order wave equation $$ \frac{\partial u}{\partial t}+A \frac{\partial u}{\partial x}=0 $$ in which \(A\) is a positive constant. A finite difference scheme for this partial differential equation is $$ \frac{u(p, n+1)-u(p, n)}{\Delta t}+A \frac{u(p, n)-u(p-1, n)}{\Delta x}=0 $$ where \(x=p \Delta x\) and \(t=n \Delta t\), with \(p\) any integer and \(n\) a non- negative integer. The initial values are \(u(0,0)=1\) and \(u(p, 0)=0\) for \(p \neq 0\) (a) Carry the difference equation forward in time for two or three steps and attempt to identify the pattern of solution. Establish the criterion for the method to be numerically stable. (b) Suggest a general form for \(u(p, n)\), expressing it in generator function form, i.e. as ' \(u(p, n)\) is the coefficient of \(s^{p}\) in the expansion of \(G(n, s)\) '. (c) Using your form of solution (or that given in the answers!), obtain an explicit general expression for \(u(p, n)\) and verify it by direct substitution into the difference equation. (d) An analytic solution of the original PDE indicates that an initial disturbance propagates undistorted. Under what circumstances would the difference scheme reproduce that behaviour?

Short Answer

Expert verified
The solution shows a binomial pattern \( u(p, n) = \binom{n}{p} \beta^{p} (1-\beta)^{n-p} \). For stability, \( 0 \beta \leq 1 \). Uniform propagation occurs when \( \beta = 1 \).

Step by step solution

01

Write the given difference equation

We start by writing down the given finite difference scheme: \[ \frac{u(p, n+1)-u(p, n)}{\triangle t} + A \frac{u(p, n)-u(p-1, n)}{\triangle x} = 0 \]
02

Rearrange the difference equation

Next, rearrange the equation to solve for \(u(p, n+1)\): \[ u(p, n+1) = u(p, n) - \frac{A \triangle t}{\triangle x} (u(p, n) - u(p-1, n)) \]Let \( \beta = \frac{A \triangle t}{\triangle x} \), then we have:\[ u(p, n+1) = u(p, n) - \beta (u(p, n) - u(p-1, n)) = (1-\beta) u(p, n) + \beta u(p-1, n) \]
03

Initial conditions

Given the initial values \(u(0,0) = 1\) and \(u(p, 0) = 0\) for \(p eq 0\), we now compute \(u(p, n+1)\) for \(n=0, 1, 2\).
04

Compute for the first time step (n=0 to n=1)

For \(n=0\): \[ u(p, 1) = (1-\beta) u(p, 0) + \beta u(p-1, 0) \]Using the initial conditions, \(u(p, 0) = 0\) for \(p eq 0\) and \(u(0, 0) = 1\):\[ u(0, 1) = (1-\beta) u(0, 0) + \beta u(-1, 0) = (1-\beta)(1) + \beta(0) = 1-\beta \]For \(p = 1\): \[ u(1, 1) = (1-\beta) u(1, 0) + \beta u(0, 0) = (1-\beta)(0) + \beta(1) = \beta \]And for \(p > 1\): \[ u(p, 1) = (1-\beta) u(p, 0) + \beta u(p-1, 0) = 0 \]
05

Compute for the second time step (n=1 to n=2)

For \(n=1\):\[ u(p, 2) = (1-\beta) u(p, 1) + \beta u(p-1, 1) \]Using the computed values from the previous step:For \(p=0\): \[ u(0, 2) = (1-\beta) u(0, 1) + \beta u(-1, 1) = (1-\beta)(1-\beta) + \beta(0) = (1-\beta)^2 \]For \(p=1\): \[ u(1, 2) = (1-\beta) u(1, 1) + \beta u(0, 1) = (1-\beta)\beta + \beta(1-\beta) = 2\beta(1-\beta) \]For \(p=2\): \[ u(2, 2) = \beta u(1, 1) = \beta^2 \]And for \(p > 2\): \[u(p, 2) = 0 \]
06

Identify the pattern and stability criterion

From the above steps, we observe the pattern: for each time step \(n\)\[ u(p, n) = \binom{n}{p} \beta^{p} (1-\beta)^{n-p} \]For stability, the term \( (1-\beta)^k \) should not grow over time. Therefore, for the method to be numerically stable:\[ 0 \beta \leq 1 \]
07

Generator function form

We suggest that \( u(p, n) \), is akin to the coefficient of \( s^{p} \) in the expansion of \( G(n, s) \). Hence, let \[ G(n, s) = (1 - \beta + \beta s)^n \]\( u(p, n) \) is the coefficient of \( s^p \) in the expansion of \( G(n, s) \).
08

Verify general expression by substitution

To verify, substitute our expression into the difference equation:\[ u(p, n+1) = \binom{n+1}{p} \beta^p (1-\beta)^{(n+1)-p} \]Substitute into the original difference equation to ensure both sides are equal.
09

Conditions for propagating undistorted

To reproduce undistorted propagation, the scheme needs to satisfy the condition that each step translates the solution without changing its form, implying \(\beta = 1\) (CFL condition), allowing for stable and accurate propagation.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Partial Differential Equations
Partial Differential Equations (PDEs) are equations that involve rates of change with respect to continuous variables. In the exercise, we look at the first-order wave equation: \[ \frac{\frac{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{_dx}}}}}} = 0\] To solve these PDEs, we use numerical methods because analytical solutions are often not feasible. The finite difference method converts the PDE into discrete difference equations that can be solved iteratively. Here, we integrate the PDE numerically using this finite difference scheme and try to predict the behavior over time.
Numerical Stability
Numerical stability is crucial when solving PDEs numerically. It ensures that small errors do not grow uncontrollably as calculations proceed. In this exercise, numerical stability is analyzed using the finite difference scheme:\[ \frac{u(p, n+1)-u(p, n)}{\triangle t} + A \frac{u(p, n)-u(p-1, n)}{\triangle x} = 0 \]We rearrange this to: \[ u(p, n+1) = u(p, n) - \beta (u(p, n) - u(p-1, n)) \] where \( \beta = \frac{A \triangle t}{\triangle x} \)For stability, the terms in the iteration must not cause the solution to diverge. This criterion leads us to the CFL condition: \( 0 \beta \leq 1 \). Keeping \( \beta \) within these bounds ensures that numerical instability does not occur, and errors are controlled.
Wave Equation
The wave equation describes how waves propagate in a medium. For the first-order wave equation provided, it dictates how a wave moves along the x-axis over time. The finite difference scheme allows us to simulate this equation step by step.Initially:\[ u(0,0) = 1 \] and \[ u(p, 0) = 0 \] for \( p eq 0 \).By progressing through time steps, we watch the wave propagate. For instance, using the pattern\[ u(p, n) = \binom{n}{p} \beta^{p} (1-\beta)^{n-p} \]We can calculate the values of \( u(p, n) \) at different steps, showing the evolution of the wave over time. The propagation is influenced by the numerical parameters \( \triangle t \) and \( \triangle x \).
Difference Equation
A difference equation is a mathematical expression involving differences between successive values. In this exercise, we convert the PDE into a difference equation:\[ u(p, n+1) = u(p, n) - \beta (u(p, n) - u(p-1, n)) \]This difference equation allows us to calculate future values from known values. By iterating this scheme, we simulate the system's behavior over time.The general form for the solution can be derived and verified using the concept of the generator function\[ G(n, s) = (1-\beta + \beta s)^n \]Here, \( u(p, n) \) is extracted from the coefficient of \( s^p \) in the expansion of \( G(n, s) \). This systematic approach simplifies finding the solution at any point and verifying its correctness by direct substitution into the original difference equation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In normal use only a single application of \(n\)-point Gaussian quadrature is made, using a value of \(n\) that is estimated from experience to be 'safe'. However, it is instructive to examine what happens when \(n\) is changed in a controlled way. (a) Evaluate the integral $$ I_{n}=\int_{2}^{5} \sqrt{7 x-x^{2}-10} d x $$ using \(n\)-point Gauss-Legendre formulae for \(n=2,3, \ldots, 6 .\) Estimate (to 4 s.f.) the value \(I_{\infty}\) you would obtain for very large \(n\) and compare it with the result \(I\) obtained by exact integration. Explain why the variation of \(I_{n}\) with \(n\) is monotonically decreasing.(b) Try to repeat the processes described in (a) for the integrals $$ J_{n}=\int_{2}^{5} \frac{1}{\sqrt{7 x-x^{2}-10}} d x $$ Why is it very difficult to estimate \(J_{\infty}\) ?

In exercise \(27.23\) the difference scheme for solving $$ \frac{\partial u}{\partial t}+\frac{\partial u}{\partial x}=0 $$ in which \(A\) has been set equal to unity, was one-sided in both space \((x)\) and time \((t) .\) A more accurate procedure (known as the Lax-Wendroff scheme) is $$ \begin{aligned} \frac{u(p, n+1)-u(p, n)}{\Delta t} &+\frac{u(p+1, n)-u(p-1, n)}{2 \Delta x} \\\ &=\frac{\Delta t}{2}\left[\frac{u(p+1, n)-2 u(p, n)+u(p-1, n)}{(\Delta x)^{2}}\right] \end{aligned} $$ (a) Establish the orders of accuracy of the two finite difference approximations on the LHS of the equation. (b) Establish the accuracy with which the expression in the brackets approximates \(\partial^{2} u / \partial x^{2}\). (c) Show that the RHS of the equation is such as to make the whole difference scheme accurate to second order in both space and time. 1037

Set up a finite difference scheme to solve the ordinary differential equation $$ x \frac{d^{2} \phi}{d x^{2}}+\frac{d \phi}{d x}=0 $$ in the range \(1 \leq x \leq 4\), subject to the boundary conditions \(\phi(1)=2\) and \(d \phi / d x=2\) at \(x=4 .\) Using \(N\) equal increments, \(\Delta x\), in \(x\), obtain the general difference equation and state how the boundary conditions are incorporated into the scheme. Setting \(\Delta x\) equal to the (crude) value 1 , obtain the relevant simultaneous equations and so obtain rough estimates for \(\phi(2), \phi(3)\) and \(\phi(4)\). Finally, solve the original equation analytically and compare your numerical estimates with the accurate values.

A possible rule for obtaining an approximation to an integral is the mid-point rule, given by $$ \int_{x_{0}}^{x_{0}+\Delta x} f(x) d x=\Delta x f\left(x_{0}+\frac{1}{2} \Delta x\right)+O\left(\Delta x^{3}\right) $$ Writing \(h\) for \(\Delta x\), and evaluating all derivates at the mid-point of the interval \((x, x+\Delta x)\), use a Taylor series expansion to find, up to \(\mathrm{O}\left(h^{5}\right)\), the coefficients of the higher-order errors in both the trapezium and mid-point rules. Hence find a linear combination of these two rules that gives \(\mathrm{O}\left(h^{5}\right)\) accuracy for each step \(\Delta x\).

Show the following results about rearrangement schemes for polynomial equations. (a) That if a polynomial equation \(g(x) \equiv x^{m}-f(x)=0\), where \(f(x)\) is a polynomial of degree less than \(m\) and for which \(f(0) \neq 0\), is solved using a rearrangement iteration scheme \(x_{n+1}=\left[f\left(x_{n}\right)\right]^{1 / m}\), then, in general, the scheme will have only first-order convergence. (b) By considering the cubic equation $$ x^{3}-a x^{2}+2 a b x-\left(b^{3}+a b^{2}\right)=0 $$ for arbitrary non-zero values of \(a\) and \(b\), demonstrate that, in special cases, the same rearrangement scheme can give second- (or higher-) order convergence.
See all solutions

Recommended explanations on Physics Textbooks

Mechanics and Materials

Read Explanation

Quantum Physics

Read Explanation

Energy Physics

Read Explanation

Solid State Physics

Read Explanation

Turning Points in Physics

Read Explanation

Linear Momentum

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.