龙格-库塔方法技术问题：推导对应公式并确定方法阶数

Part a: Corresponding Calculation Formulas #

First, let's map the given RK matrix to standard 4-stage Runge-Kutta notation. The matrix gives us these key coefficients:

• Stage time offsets: $c = \left[0, \frac{1}{2}, 1, 1\right]$
• Stage weight matrix:
  $$A = \begin{pmatrix}
  0 & 0 & 0 & 0 \
  \frac{1}{2} & 0 & 0 & 0 \
  0 & 1 & 0 & 0 \
  0 & 0 & 1 & 0
  \end{pmatrix}$$
• Final update weights: $b = \left[\frac{1}{6}, \frac{2}{3}, 0, \frac{1}{6}\right]$

Using these, here's the full step-by-step formula for solving $y' = f(t,y)$:

1. Calculate the four stage values:

   • $k_1 = f(t_n, y_n)$
   • $k_2 = f\left(t_n + \frac{1}{2}h, y_n + \frac{1}{2}h k_1\right)$
   • $k_3 = f\left(t_n + h, y_n + h k_2\right)$
   • $k_4 = f\left(t_n + h, y_n + h k_3\right)$

2. Update the solution to the next time step:
   $$y_{n+1} = y_n + h\left( \frac{1}{6}k_1 + \frac{2}{3}k_2 + 0 \cdot k_3 + \frac{1}{6}k_4 \right)$$

Part b: Determining the Method's Order #

We'll verify standard RK order conditions to find the highest satisfied order:

Order 1 Condition #

Sum of update weights equals 1:
$$\sum_{i=1}^4 b_i = \frac{1}{6} + \frac{2}{3} + 0 + \frac{1}{6} = 1$$
✅ Satisfied.

Order 2 Condition #

$\sum_{i=1}^4 b_i c_i = \frac{1}{2}$:
$$\frac{1}{6} \cdot 0 + \frac{2}{3} \cdot \frac{1}{2} + 0 \cdot 1 + \frac{1}{6} \cdot 1 = 0 + \frac{1}{3} + 0 + \frac{1}{6} = \frac{1}{2}$$
✅ Satisfied.

Order 3 Conditions #

Two core conditions for order 3:

1. $\sum_{i=1}^4 b_i c_i^2 = \frac{1}{3}$:
   $$\frac{1}{6} \cdot 0^2 + \frac{2}{3} \cdot \left(\frac{1}{2}\right)^2 + 0 \cdot 1^2 + \frac{1}{6} \cdot 1^2 = 0 + \frac{1}{6} + 0 + \frac{1}{6} = \frac{1}{3}$$
   ✅ Satisfied.
2. $\sum_{i=1}^4 b_i \sum_{j=1}^4 A_{ij} c_j = \frac{1}{6}$:
   Calculating each term:
   • $i=1$: $0$; $i=2$: $0$; $i=3$: $0 \cdot \frac{1}{2} = 0$; $i=4$: $\frac{1}{6} \cdot 1 = \frac{1}{6}$
     Sum is $\frac{1}{6}$ ✅ Satisfied.

Order 4 Conditions #

We check a critical order 4 condition: $\sum_{i=1}^4 b_i \sum_{j=1}^4 A_{ij} c_j^2 = \frac{1}{8}$:
$$\frac{1}{6} \cdot 0 + \frac{2}{3} \cdot 0 + 0 \cdot \frac{1}{4} + \frac{1}{6} \cdot 1 = \frac{1}{6} \neq \frac{1}{8}$$
❌ Not satisfied.

Since all order 3 conditions hold but order 4 conditions fail, this method is 3rd order.

内容的提问来源于stack exchange，提问作者lnbmoco