Consider the Runge-Kutta method associated with the Butcher Array: 0 | 0 0 0 1/2 | 1/2 0 0 1 | -1 2 0 --------------- | 1/6 2/3 1/6 1. Formulate the method in a comprehe... Consider the Runge-Kutta method associated with the Butcher Array: 0 | 0 0 0 1/2 | 1/2 0 0 1 | -1 2 0 --------------- | 1/6 2/3 1/6 1. Formulate the method in a comprehensive explicit manner. 2. Prove that the scheme is at least third order accurate. Read more

Steps to Solve

1. Formulate the Runge-Kutta method

The Butcher array provides all the coefficients needed to formulate the Runge - Kutta method. Given the Butcher array:

$$ \begin{array}{c|ccc} 0 & 0 & 0 & 0 \ 1/2 & 1/2 & 0 & 0 \ 1 & -1 & 2 & 0 \ \hline & 1/6 & 2/3 & 1/6 \end{array} $$

Let's denote the entries as follows:

$c = [0, 1/2, 1]^T$, $b = [1/6, 2/3, 1/6]$, and $A = \begin{bmatrix} 0 & 0 & 0 \ 1/2 & 0 & 0 \ -1 & 2 & 0 \end{bmatrix}$.

The Runge-Kutta method is given by:

$$ \begin{aligned} k_1 &= f(t_n, y_n) \ k_2 &= f(t_n + \frac{1}{2}h, y_n + \frac{1}{2}hk_1) \ k_3 &= f(t_n + h, y_n + h(-k_1 + 2k_2)) \ y_{n+1} &= y_n + h(\frac{1}{6}k_1 + \frac{2}{3}k_2 + \frac{1}{6}k_3) \end{aligned} $$

2. Prove Third-Order Accuracy

To prove that the method is at least third-order accurate, we need to show that its Taylor series expansion matches the Taylor series expansion of the exact solution up to the $h^3$ term. Let's consider the Taylor series expansion of the exact solution $y(t_{n+1})$ around $t_n$:

$y(t_{n+1}) = y(t_n + h) = y(t_n) + hy'(t_n) + \frac{h^2}{2}y''(t_n) + \frac{h^3}{6}y'''(t_n) + O(h^4)$.

Since $y'(t) = f(t, y(t))$, we have:

$y(t_{n+1}) = y_n + hf_n + \frac{h^2}{2}(f_t + f_y f)n + \frac{h^3}{6}(f{tt} + 2f_{ty}f + f_{yy}f^2 + f_y(f_t + f_y f))_n + O(h^4)$.

Now, let's expand the Runge-Kutta method:

$y_{n+1} = y_n + h(\frac{1}{6}k_1 + \frac{2}{3}k_2 + \frac{1}{6}k_3)$.

$k_1 = f_n$

$k_2 = f(t_n + \frac{1}{2}h, y_n + \frac{1}{2}hf_n) = f_n + \frac{1}{2}h(f_t + f_y f)n + \frac{1}{8}h^2(f{tt} + 2f_{ty}f + f_{yy}f^2)_n + O(h^3)$

$k_3 = f(t_n + h, y_n + h(-f_n + 2k_2)) = f(t_n + h, y_n + h(-f_n + 2(f_n + \frac{1}{2}h(f_t + f_y f)_n + O(h^2)))) = f(t_n + h, y_n + h(f_n + h(f_t + f_y f)_n + O(h^2)))$

$k_3 = f_n + h(f_t + f_y f)n + \frac{h^2}{2}(f{tt} + 2f_{ty}f + f_{yy}f^2)_n + h^2f_y(f_t + f_y f)_n + O(h^3)$

Substitute these expansions into the Runge-Kutta formula:

$y_{n+1} = y_n + h[\frac{1}{6}f_n + \frac{2}{3}(f_n + \frac{1}{2}h(f_t + f_y f)n + \frac{1}{8}h^2(f{tt} + 2f_{ty}f + f_{yy}f^2)n) + \frac{1}{6}(f_n + h(f_t + f_y f)n + \frac{h^2}{2}(f{tt} + 2f{ty}f + f_{yy}f^2)_n + h^2f_y(f_t + f_y f))] + O(h^4)$

$y_{n+1} = y_n + h[f_n + h(\frac{1}{3} + \frac{1}{6})(f_t + f_y f)n + h^2(\frac{1}{12} + \frac{1}{12})(f{tt} + 2f_{ty}f + f_{yy}f^2)_n + \frac{1}{6}f_y(f_t + f_y f)_n] + O(h^4)$

$y_{n+1} = y_n + hf_n + \frac{h^2}{2}(f_t + f_y f)n + h^3[\frac{1}{12}(f{tt} + 2f_{ty}f + f_{yy}f^2)_n + \frac{1}{6}f_y(f_t + f_y f)_n] + O(h^4)$.

$y_{n+1} = y_n + hf_n + \frac{h^2}{2}(f_t + f_y f)n + \frac{h^3}{6}(f{tt} + 2f_{ty}f + f_{yy}f^2 + 2f_y(f_t + f_y f))_n + O(h^4)$.

For the method to be third-order accurate the following conditions must be satisfied:

• $\sum_{i=1}^{s} b_i = 1$
• $\sum_{i=1}^{s} b_i c_i = \frac{1}{2}$
• $\sum_{i=1}^{s} b_i c_i^2 = \frac{1}{3}$
• $\sum_{i,j=1}^{s} b_i a_{ij} c_j = \frac{1}{6}$
• $\sum_{i=1}^{s} b_i c_i^3 = \frac{1}{4}$
• $\sum_{i,j=1}^{s} b_i c_i a_{ij} = \frac{1}{8}$
• $\sum_{i, j=1}^{s} b_i a_{ij} c_j^2 = \frac{1}{12}$
• $\sum_{i,j,k=1}^{s} b_i a_{ij} a_{jk} = \frac{1}{24}$

For our Butcher Tableau we have:

• $\frac{1}{6} + \frac{2}{3} + \frac{1}{6} = 1$
• $\frac{1}{6}(0) + \frac{2}{3}(\frac{1}{2}) + \frac{1}{6}(1) = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$
• $\frac{1}{6}(0)^2 + \frac{2}{3}(\frac{1}{2})^2 + \frac{1}{6}(1)^2 = \frac{2}{3} \cdot \frac{1}{4} + \frac{1}{6} = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}$
• $\frac{1}{6}(\frac{1}{2} \cdot 0 + 2 \cdot \frac{1}{2} ) + \frac{2}{3}(0) + \frac{1}{6}(0) = \frac{1}{6}$
• $\frac{1}{6} \cdot 0^3 + \frac{2}{3} (\frac{1}{2})^3 + \frac{1}{6} \cdot 1^3 = \frac{2}{3} \cdot \frac{1}{8} + \frac{1}{6} = \frac{1}{12} + \frac{1}{6} = \frac{1}{4}$
• $\frac{1}{6} \cdot 0 \cdot 0 + \frac{2}{3} \cdot \frac{1}{2} \cdot 0 + \frac{1}{6} \cdot 1 (\frac{1}{2} \cdot 0 + 2 \cdot \frac{1}{2}) = \frac{1}{6}$
• $\frac{1}{6} \cdot 0 \cdot 0 + \frac{2}{3} \cdot \frac{1}{2} \cdot (\frac{1}{2})^2 + \frac{1}{6} \cdot 0 = \frac{1}{12}$
• $\frac{1}{6} \cdot 0 \cdot 0 + \frac{2}{3} (\frac{1}{2} \cdot 0) + \frac{1}{6} [(-1) \cdot 0 + 2 \cdot \frac{1}{2}] = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$

Since not all order conditions are satisfied, the method is second order accurate only. The error term is: $\frac{h^3}{6}(f_{tt} + 2f_{ty}f + f_{yy}f^2 + f_y(f_t + f_y f))n - h^3[\frac{1}{12}(f{tt} + 2f_{ty}f + f_{yy}f^2)_n + \frac{1}{6}f_y(f_t + f_y f)_n]$