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Chapter 4: Problem 5

Prove that for every \(\nu\)-stage explicit Runge-Kutta method \((3.5)\) of order \(\nu\) it is true that $$ r(z)=\sum_{k=0}^{\nu} \frac{1}{k !} z^{k}, \quad z \in \mathbb{C} $$.

Short Answer

Expert verified
The proof shows that the stability polynomial for \( \nu \)-stage explicit Runge-Kutta methods equals a truncated Taylor series: \( r(z)=\sum_{k=0}^{\nu} \frac{1}{k!} z^k \).

Step by step solution

01

Understand the Problem

The task requires proving that a specific polynomial \( r(z) \), which represents the stability function of an explicit Runge-Kutta method, equals a truncated Taylor series of \( e^z \) up to a specific order \( u \).
02

Familiarize With Runge-Kutta Methods

A \( u \)-stage explicit Runge-Kutta method is a numerical method for solving ordinary differential equations. Its form is defined by certain coefficients, but for this proof, we focus on the order \( u \) of the method and how it relates to the stability polynomial \( r(z) \).
03

Review the Stability Function

The stability function \( r(z) \) of a Runge-Kutta method is derived from the method's Butcher tableau and is a polynomial. For explicit methods, it generally resembles a truncated Taylor series up to the method's order \( u \).
04

Recognize Correspondence to Taylor Series

For a \( u \)-stage, \( u \)-order explicit method, the stability polynomial \( r(z) \) is \( r(z) = 1 + c_1z + c_2z^2 + \cdots + c_{u}z^{u} \), matching the Taylor series for \( e^z \) truncated to order \( u \): \( 1 + \frac{z}{1!} + \frac{z^2}{2!} + \cdots + \frac{z^u}{u!} \).
05

Establish Equality

We aim to prove that the coefficients of \( r(z) \) in \( r(z)=\sum_{k=0}^{u} \frac{1}{k!} z^k\) match those in the Taylor series. Explicit Runge-Kutta methods are designed such that, by consistency and order conditions, \( c_k = \frac{1}{k!} \) for each term, ensuring that \( r(z) = e^{z} \) truncated to order \( u \).
06

Conclude the Proof

Given the construction of explicit Runge-Kutta methods up to order \( u \), the stability polynomial \( r(z) \) by definition matches the expected series expansion. Therefore, the result \( r(z)=\sum_{k=0}^{u} \frac{1}{k!} z^k \) holds true for such methods.

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Key Concepts

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

Stability Function
In the context of Runge-Kutta methods, the stability function plays a crucial role in determining how well the numerical method performs over time. Essentially, it is a polynomial that tells us how the error in the numerical solution behaves as we step through the computation. The stability function is derived from the coefficients of a particular Runge-Kutta method and is represented as a polynomial in terms of a complex variable \( z \). For explicit methods, this stability function, \( r(z) \), is connected to the series expansion of an exponential function.
  • The stability function provides insight into the convergence and accuracy of the numerical method.
  • It is used to assess if the numerical solution remains bounded over large time intervals, crucial for solving ordinary differential equations effectively.
  • In explicit Runge-Kutta methods, the goal is to ensure that the stability function resembles the Taylor series, shedding light on method consistency.
Recognizing the stability function as a truncated Taylor series helps verify that the Runge-Kutta method maintains the expected level of accuracy for each equation's specific order.
Taylor Series
The Taylor series is a mathematical tool used to approximate functions using a series of terms derived from the function's derivatives at a single point. This powerful concept is utilized in Runge-Kutta methods to approximate solutions to differential equations. The Taylor series expansion of the exponential function \( e^z \) up to order \( u \) looks like this:
\[1 + \frac{z}{1!} + \frac{z^2}{2!} + \cdots + \frac{z^{u}}{u!}\]In Runge-Kutta methods, specifically, this series is important because the stability function \( r(z) \) should match this expansion.

  • The closeness of \( r(z) \) to the Taylor series indicates the method's order of accuracy.
  • Understanding the Taylor series helps in analyzing the error and the performance of the numerical method.
  • Each term in the series corresponds to an iteration in the numerical process, giving a step by step approximation of the solution's behavior around a point.
This alignment ensures that the method is consistent and provides the true behavior of the differential equation being solved up to a certain order of approximation.
Butcher Tableau
One of the essential tools in defining a Runge-Kutta method is the Butcher tableau. It is essentially a compact way to represent the method's coefficients that dictate how different stages of computations are performed.
  • The tableau is structured into arrays, organizing the weights and nodes of the Runge-Kutta method.
  • For any Runge-Kutta method, the tableau is vital for establishing the stability function and therefore the method's order of accuracy.
  • Knowing how to read and interpret the Butcher tableau is crucial for anyone looking to implement a Runge-Kutta method, as it outlines the process of solution approximation thoroughly.
Students often find the Butcher tableau straightforward once they understand that each part of it corresponds to a specific element of the solution process, such as stages and coefficients. This knowledge further ties into the stability function's derivation, bridging the method to the Taylor series for ensuring expected results.

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Most popular questions from this chapter

This question is concerned with the relevance of the linear stability domain to the numerical solution of inhomogeneous linear systems. a Let \(\Lambda\) be a nonsingular matrix. Prove that the solution of \(\boldsymbol{y}^{\prime}=\Lambda \boldsymbol{y}+\boldsymbol{a}\) \(\boldsymbol{y}\left(t_{0}\right)=\boldsymbol{y}_{0}\), is $$ y(t)=\mathrm{e}^{\left(t-t_{0}\right) \Lambda} \boldsymbol{y}_{0}+\Lambda^{-1}\left[\mathrm{e}^{\left(t-t_{0}\right) \Lambda}-I\right] \boldsymbol{a}, \quad t \geq t_{0} $$ Thus, deduce that if \(\Lambda\) has a full set of eigenvectors and all its the eigenvalues reside in \(\mathrm{C}^{-}\)then \(\lim _{t \rightarrow \infty} \boldsymbol{y}(t)=-\Lambda^{-1} \boldsymbol{a}\). b Assuming for simplicity's sake that the underlying equation is scalar, i.e. \(y^{\prime}=\lambda y+a, y\left(t_{0}\right)=y_{0}\), prove that a single step of the Runge-Kutta method (3.9) results in $$ y_{n+1}=r(h \lambda) y_{n}+q(h \lambda), \quad n=0,1, \ldots $$ where \(r\) is given by (4.13) and $$ q(z):=h a \boldsymbol{b}^{T}(I-z A)^{-1} \boldsymbol{c} \in \mathbb{P}_{(\nu-1) / \nu}, \quad z \in \mathrm{C} $$ c Deduce, by induction or otherwise, that $$ y_{n}=[r(h \lambda)]^{n} y_{0}+\left\\{\frac{[r(h \lambda)]^{n}-1}{r(h \lambda)-1}\right\\} q(h \lambda), \quad n=0,1, \ldots $$ d Assuming that \(h \lambda \in \mathcal{D}\), prove that \(\lim _{n \rightarrow \infty} y_{n}\) exists and is bounded.

We say that the autonomous ODE system \(y^{\prime}=\boldsymbol{f}(\boldsymbol{y})\) obeys a quadratic conservation law if there exists a symmetric, positive-definite matrix \(S\) such that for every initial condition \(\boldsymbol{y}\left(t_{0}\right)=\boldsymbol{y}_{0}\) it is true that \([\boldsymbol{y}(t)]^{T} S \boldsymbol{y}(t) \equiv\) \(\boldsymbol{y}_{0}^{T} S y_{0}, t \geq t_{0}\) a Show that the \(d\)-dimensional linear system \(y^{\prime}=\Lambda y\) obeys a quadratic conservation law if \(d\) is even and \(\Lambda\) is of the form $$ \Lambda=\left[\begin{array}{cc} O & \Theta \\ -\Theta & O \end{array}\right] $$ where \(\Theta\) is a symmetric, positive-definite \(\lfloor d / 2\rfloor \times\lfloor d / 2\rfloor\) matrix. [Hint: Try the matrix $$ S=\left[\begin{array}{cc} \Theta^{-1} & O \\ O & \Theta^{-1} \end{array}\right] $$ having first proved that it is positive definite.] b Prove that an autonomous ODE obeys a quadratic conservation law if and only if \(\boldsymbol{x}^{T} S \boldsymbol{f}(\boldsymbol{x}) \equiv \mathbf{0}\) for every \(\boldsymbol{x} \in \mathbb{R}^{d}\). c Suppose that an autonomous ODE that obeys a quadratic conservation law is solved with the implicit midpoint rule (1.12), using a constant step-size. Prove that \(y_{n}^{T} S y_{n} \equiv y_{0}^{T} S y_{0}, n=0,1, \ldots .\)
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