Θ方法的局部误差计算及0-稳定时的方法阶数咨询

Let's work through your two questions using the Taylor expansion progress you've already made—great start on that, by the way!

a) Calculating the Local Truncation Error (LTE) #

First, remember the definition of LTE for a numerical method: we assume $y_n = y(t_n)$ (the exact solution at $t_n$), then compute the difference between the exact solution at $t_{n+1}$ and the approximation $y_{n+1}$ produced by the θ-method. Here's how to formalize your Taylor expansion work:

1. Exact solution Taylor expansion:
   The exact solution at $t_{n+1} = t_n + h$ expands to:
   $$y(t_{n+1}) = y(t_n) + h y'(t_n) + \frac{h^2}{2} y''(t_n) + \frac{h^3}{6} y'''(t_n) + O(h^4)$$

2. θ-method approximation expansion:
   The θ-method is $y_{n+1} = y_n + \theta h f_n + (1-\theta) h f_{n+1}$. Since $y_n = y(t_n)$, we know $f_n = f(t_n, y(t_n)) = y'(t_n)$. For $f_{n+1} = f(t_{n+1}, y(t_{n+1}))$, expand it around $t_n$:
   $$f_{n+1} = y'(t_n) + h y''(t_n) + \frac{h^2}{2} y'''(t_n) + O(h^3)$$
   Substitute this into the θ-method formula:
   $$
   \begin{align*}
   y_{n+1} &= y(t_n) + \theta h y'(t_n) + (1-\theta)h\left(y'(t_n) + h y''(t_n) + \frac{h^2}{2} y'''(t_n) + O(h^3)\right) \
   &= y(t_n) + h y'(t_n) + (1-\theta)h^2 y''(t_n) + \frac{(1-\theta)h^3}{2} y'''(t_n) + O(h^4)
   \end{align*}
   $$

3. Compute the LTE:
   Subtract the approximation from the exact solution:
   $$
   \begin{align*}
   LTE &= y(t_{n+1}) - y_{n+1} \
   &= \left(\frac{1}{2} - (1-\theta)\right)h^2 y''(t_n) + \left(\frac{1}{6} - \frac{(1-\theta)}{2}\right)h^3 y'''(t_n) + O(h^4) \
   &= \left(\theta - \frac{1}{2}\right)h^2 y''(t_n) + \left(\frac{\theta}{2} - \frac{1}{3}\right)h^3 y'''(t_n) + O(h^4)
   \end{align*}
   $$
   This matches the partial expansion you had! The local error is dominated by the highest non-vanishing term here:

   • If $\theta \neq \frac{1}{2}$, the leading term is $O(h^2)$, so the local error is $O(h^2)$.
   • If $\theta = \frac{1}{2}$ (the trapezoidal rule), the $h^2$ term cancels out, leaving the leading term as $O(h^3)$.

b) Order of the Method When 0-Stable #

First, let's clarify 0-stability for the θ-method:
The θ-method is a linear single-step method, and its characteristic polynomial is $\rho(r) = r - 1$. The only root is $r=1$, which is a simple root on the unit circle—this satisfies the 0-stability criteria for linear multistep methods (and single-step methods are inherently 0-stable as long as this condition holds, which it does for all $\theta \in [0,1]$).

Now, the method's order is determined by the highest power of $h$ such that $LTE = O(h^{p+1})$ (where $p$ is the order):

• For $\theta \neq \frac{1}{2}$: The LTE is $O(h^2)$, so the method is 1st-order.
• For $\theta = \frac{1}{2}$: The LTE is $O(h^3)$, so the method is 2nd-order.

In short, when 0-stable (which is always true for θ-methods with $\theta \in [0,1]$), the order is 1 unless $\theta = \frac{1}{2}$, in which case it's 2.

内容的提问来源于stack exchange，提问作者lnbmoco