1.3.5 Example of Most Accurate Multi-Step Method
Measurable Outcome 1.5 , Measurable Outcome 1.6 , Measurable Outcome 1.8 , Measurable Outcome 1.15
In this example, we will derive the most accurate multi-step method of the following form:
\[v^{n+1} + \alpha _1 v^ n + \alpha _2 v^{n-1} = {\Delta t}\left[ \beta _1 f^ n + \beta _2 f^{n-1} \right]\]
(1.56)
The local truncation error for this method is,
\[\tau = - \alpha _1 u^ n - \alpha _2 u^{n-1} + {\Delta t}\left[ \beta _1 f^ n + \beta _2 f^{n-1} \right] - u^{n+1}\]
(1.57)
Substitution of \(f^ n = u_ t^ n\) and \(f^{n-1} = u_ t^{n-1}\) gives,
\[\tau = - \alpha _1 u^ n - \alpha _2 u^{n-1} + {\Delta t}\left[ \beta _1 u_ t^ n + \beta _2 u_ t^{n-1} \right] - u^{n+1}\]
(1.58)
Then, Taylor series about \(t=t^ n\) are substituted for \(u^{n-1}\), \(u_ t^{n-1}\), and \(u^{n+1}\) to give,
\(\displaystyle \tau\)
\(\displaystyle =\)
\(\displaystyle - \alpha _1 u^ n - \alpha _2\left[ u^ n - {\Delta t}u^ n_ t + \frac{1}{2}{\Delta t}^2 u^ n_{tt} - \frac{1}{6}{\Delta t}^3 u^ n_{ttt} + \frac{1}{24}{\Delta t}^4 u^ n_{tttt} + O({\Delta t}^5) \right]\)
(1.59)
\(\displaystyle + {\Delta t}\beta _1 u_ t^ n + {\Delta t}\beta _2 \left[ u^ n_ t - {\Delta t}u^ n_{tt} + \frac{1}{2}{\Delta t}^2 u^ n_{ttt} - \frac{1}{6}{\Delta t}^3 u^ n_{tttt} + O({\Delta t}^4) \right]\)
(1.60)
\(\displaystyle - \left[ u^ n + {\Delta t}u^ n_ t + \frac{1}{2}{\Delta t}^2 u^ n_{tt} + \frac{1}{6}{\Delta t}^3 u^ n_{ttt} + \frac{1}{24}{\Delta t}^4 u^ n_{tttt} + O({\Delta t}^5) \right]\)
(1.61)
Next, collect the terms in powers of \({\Delta t}\), which gives the following coefficients:
\[\begin{array}{r@{:\: \: }cccccccccc} u^ n & - & \alpha _1 & - & \alpha _2 & & & & & - & 1 \\[0.1in] {\Delta t}u_ t^ n & & & & \alpha _2 & + & \beta _1 & + & \beta _2 & - & 1 \\[0.1in] {\Delta t}^2 u_{tt}^ n & & & - & \frac{\alpha _2}{2} & & & - & \beta _2 & - & \frac{1}{2} \\[0.1in] {\Delta t}^3 u_{ttt}^ n & & & & \frac{\alpha _2}{6} & & & + & \frac{\beta _2}{2} & - & \frac{1}{6} \\[0.1in] {\Delta t}^4 u_{tttt}^ n & & & -& \frac{\alpha _2}{24} & & & - & \frac{\beta _2}{6} & - & \frac{1}{24} \end{array}\]
(1.62)
To find the most accurate multi-step method of the given form, we solve for the values of \(\alpha _1\), \(\alpha _2\), \(\beta _1\), and \(\beta _2\) that result in the coefficients of the first four terms being identically zero.
Exercise 1 Use MATLAB® 's backslash command in the form
x= A\b
Which of the following most closely matches your solution for \([\alpha _1, \alpha _2, \beta _1, \beta _2]^ T\)?
Exercise 1
\([4,-5,2,4]\)
\([1,0,-1,5]\)
\([4,2,-4,1]\)
\([4,-5,4,2]\)
Answer: Note, with these values, the leading error term is \(-\frac{1}{6}{\Delta t}^4 u_{tttt}^ n\). Thus, the scheme is third order accurate (\(p=3\)).
