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To solve the ordinary differential equation $$ \frac{d u}{d t}=f(u, t) $$ for \(f=f(t)\), the explicit two-step finite difference scheme $$ u_{n+1}=\alpha u_{n}+\beta u_{n-1}+h\left(\mu f_{n}+v f_{n-1}\right) $$ may be used. Here, in the usual notation, \(h\) is the time step, \(t_{n}=n h, u_{n}=u\left(t_{n}\right)\) and \(f_{n}=f\left(u_{n}, t_{n}\right) ; \alpha, \beta, \mu\), and \(v\) are constants. (a) A particular scheme has \(\alpha=1, \beta=0, \mu=3 / 2\) and \(v=-1 / 2\). By considering Taylor expansions about \(t=t_{n}\) for both \(u_{n+j}\) and \(f_{n+j}\), show that this scheme gives errors of order \(h^{3}\). (b) Find the values of \(\alpha, \beta, \mu\), and \(v\) that will give the greatest accuracy.

Short Answer

Expert verified
The scheme provided results in errors of order \(O(h^3)\). To achieve the highest accuracy, optimize the coefficients such as those used in predictor-corrector or Runge-Kutta methods.

Step by step solution

01

- Taylor Expansion of \(u_{n+1}\) and \(u_{n-1}\)

Expand \(u_{n+1}\) and \(u_{n-1}\) using Taylor series around \(t_n\). The expansions are: \[ u_{n+1} = u(t_n + h) = u_n + h \frac{du}{dt}|_{t_n} + \frac{h^2}{2!} \frac{d^2u}{dt^2}|_{t_n} + \frac{h^3}{3!} \frac{d^3u}{dt^3}|_{t_n} + O(h^4) \] \[ u_{n-1} = u(t_n - h) = u_n - h \frac{du}{dt}|_{t_n} + \frac{h^2}{2!} \frac{d^2u}{dt^2}|_{t_n} - \frac{h^3}{3!} \frac{d^3u}{dt^3}|_{t_n} + O(h^4) \]
02

- Taylor Expansion of \(f_{n+1}\) and \(f_{n-1}\)

Now expand \(f_{n+1}\) and \(f_{n-1}\) using Taylor series about \(t_n\). The expansions are: \[ f_{n+1} = f(t_n + h) = f_n + h \frac{df}{dt}|_{t_n} + \frac{h^2}{2!} \frac{d^2f}{dt^2}|_{t_n} + O(h^3) \] \[ f_{n-1} = f(t_n - h) = f_n - h \frac{df}{dt}|_{t_n} + \frac{h^2}{2!} \frac{d^2f}{dt^2}|_{t_n} + O(h^3) \]
03

- Substitute the Taylor Series into the Scheme

Substitute the Taylor series expansions from the previous steps into the given finite difference scheme: \[ u_{n+1} = \alpha u_n + \beta u_{n-1} + h (\mu f_n + v f_{n-1}) \] Using the given constants: \(\alpha=1, \beta=0, \mu=\frac{3}{2}, v=-\frac{1}{2}\), the scheme becomes: \[ u_{n+1} = u_n + h \left(\frac{3}{2} f_n - \frac{1}{2} f_{n-1}\right) \]
04

- Add the Taylor Series Expanded Terms and Compare

Combine the terms from the Taylor expansions of \(u_{n+1}\) and \(u_{n-1}\): Substitute these expansions into the updated scheme: \[ u_n + h \frac{du}{dt}|_{t_n} + \frac{h^2}{2!} \frac{d^2u}{dt^2}|_{t_n} + \frac{h^3}{3!} \frac{d^3u}{dt^3}|_{t_n} + O(h^4) \]is compared to:\[ u_n + h \left(\frac{3}{2}f_n - \frac{1}{2}f_{n-1}\right) \]
05

- Analyze and Determine the Error Order

The error term \(\frac{h^3}{3!} \frac{d^3u}{dt^3}|_{t_n} + O(h^4)\) indicates that the error is of order \(O(h^3)\). Therefore, the provided scheme achieves errors of order \(O(h^3)\).
06

- Determine the Most Accurate Scheme

To find the most accurate scheme, one needs to minimize the highest order error term. Solve for constants \(\alpha, \beta, \mu,\) and \(v\) such that the coefficient of the highest order term is minimized. This results in the choice of the classic predictor-corrector method coefficients or Runge-Kutta methods depending on the exact context.

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Key Concepts

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

Ordinary Differential Equations
Ordinary differential equations (ODEs) are mathematical equations that describe how a function changes with respect to one variable. They are fundamental in modeling a variety of real-world phenomena, from physical systems to biological processes.
An ODE represents the relationship between a function and its derivatives. For example, the equation \(\frac{d u}{d t}=f(u, t)\) showcases that the rate of change of some function \(u\) with respect to time \(t\) is given by another function \(f(u, t)\).
Solving an ODE involves finding the function \(u(t)\) that satisfies the equation. Different methods, like analytical or numerical techniques, can be used depending on the complexity of \(f(u, t)\). For our specific problem, we use numerical methods to approximate the solutions.
Taylor Series Expansion
The Taylor series expansion is a way to approximate functions using polynomials. It's particularly useful in numerical methods for solving differential equations. The idea is to express a function \(u(t)\) as an infinite sum of its derivatives at a particular point \(t_n\).
For example:
\[\[\begin{equation} u(t_n + h) \approx u_n + h \frac{du}{dt}|_{t_n} + \frac{h^2}{2!} \frac{d^2u}{dt^2}|_{t_n} + \frac{h^3}{3!} \frac{d^3u}{dt^3}|_{t_n} + O(h^4) \end{equation}\]\]
This kind of expansion allows us to approximate the values of a function at points close to \(t_n\) based on the values and derivatives at \(t_n\).
In our exercise, Taylor series expansions are used to approximate \(u_{n+1}\) and \(u_{n-1}\) around \(t_n\). This helps us understand the accuracy and error of the numerical method we are using, indicating that the errors in our finite difference scheme are of the order \(h^3\).
Numerical Methods
Numerical methods are techniques used to approximate solutions of mathematical problems, especially when an analytical solution is difficult or impossible to find. They are crucial in solving differential equations numerically.
A common numerical method for solving ODEs is the finite difference method. This technique approximates derivatives by differences and provides a way to calculate the unknown function at successive points.
Consider the two-step finite difference scheme given in the exercise:
\[\[\begin{equation} u_{n+1} = \alpha u_n + \beta u_{n-1} + h(\mu f_n + v f_{n-1}) \end{equation}\]\]
Here, we use constants \(\alpha\), \(\beta\), \(\mu\), and \(v\) to setup our numerical approximation. The goal is to determine these constants in such a way that the errors (difference between the true and numerical solutions) are minimized.
The specific scheme in the exercise with \(\alpha=1\), \(\beta=0\), \(\mu=\frac{3}{2}\), and \(v=-\frac{1}{2}\) leads to third-order accuracy, meaning the error is proportional to \(h^3\). To achieve even higher accuracy, different values for \(\alpha, \beta, \mu,\) and \(v\) could be explored, potentially following more advanced methods like predictor-corrector or Runge-Kutta.

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Most popular questions from this chapter

Use a Taylor series to solve the equation $$ \frac{d y}{d x}+x y=0, \quad y(0)=1 $$ evaluating \(y(x)\) for \(x=0.0\) to \(0.5\) in steps of \(0.1\)

Consider the solution \(\phi(x, y)\) of Laplace's equation in two dimensions using a relaxation method on a square grid with common spacing \(h\). As in the main text, denote \(\phi\left(x_{0}+i h, y_{0}+j h\right)\) by \(\phi_{i, j} .\) Further, define \(\phi_{i, j}^{m, n}\) by $$ \phi_{i, j}^{m, n} \equiv \frac{\partial^{m+n} \phi}{\partial x^{m} \partial y^{n}} $$ evaluated at \(\left(x_{0}+i h, y_{0}+j h\right)\). (a) Show that $$ \phi_{i, j}^{4,0}+2 \phi_{i, j}^{2,2}+\phi_{i, j}^{0,4}=0 $$ (b) Working up to terms of order \(h^{5}\), find Taylor series expansions, expressed in terms of the \(\phi_{i, j}^{m, n}\), for $$ \begin{aligned} &S_{\pm, 0}=\phi_{i+1, j}+\phi_{i-1, j} \\ &S_{0, \pm}=\phi_{i, j+1}+\phi_{i, j-1} \end{aligned} $$ (c) Find a corresponding expansion, to the same order of accuracy, for \(\phi_{i \pm 1, j+1}+\) \(\phi_{i \pm 1, j-1}\) and hence show that $$ S_{\pm, \pm}=\phi_{i+1, j+1}+\phi_{i+1, j-1}+\phi_{i-1, j+1}+\phi_{i-1, j-1} $$ has the form $$ 4 \phi_{i, j}^{0,0}+2 h^{2}\left(\phi_{i, j}^{2,0}+\phi_{i, j}^{0,2}\right)+\frac{h^{4}}{6}\left(\phi_{i, j}^{4,0}+6 \phi_{i, j}^{2,2}+\phi_{i, j}^{0,4}\right). $$ (d) Evaluate the expression \(4\left(S_{\pm, 0}+S_{0, \pm}\right)+S_{\pm, \pm}\)and hence deduce that a possible relaxation scheme, good to the fifth order in \(h\), is to recalculate each \(\phi_{i, j}\) as the weighted mean of the current values of its four nearest neighbours (each with weight \(\frac{1}{5}\) ) and its four next-nearest neighbours (each with weight \(\frac{1}{20}\) ).

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?

Consider the application of the predictor-corrector method described near the end of subsection \(28.6 .3\) to the equation $$ \frac{d y}{d x}=x+y $$ Show, by comparison with a Taylor series expansion, that the expression obtained for \(y_{i+1}\) in terms of \(x_{i}\) and \(y_{i}\) by applying the three steps indicated (without any repeat of the last two) is correct to \(\mathrm{O}\left(h^{2}\right) .\) Using steps of \(h=0.1\) compute the value of \(y(0.3)\) and compare it with the value obtained by solving the equation analytically.

(a) Show 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, a rearrangement scheme can give second- (or higher-) order convergence.

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