I want to get the order of accuracy (local truncation error – LTE) of this implicit 2-step method. The first step is Backward Euler to determine an approximation to the value at the midpoint in time, whereas and the second step is the midpoint method using this value. The method is given by
$${U^*} = {U^n} + \frac{k}{2}f\left( {{t_{n + \frac{1}{2}}},{U^*}} \right)$$
$${U^{n + 1}} = {U^n} + kf\left( {{t_{n + \frac{1}{2}}},{U^*}} \right)$$
I know how to start, and how to expand. However, the above relation seems to be recursive? For example, substituting ${U^*}$ (the first approximation) into the the second will give
$${U^{n + 1}} = {U^n} + kf\left( {{t_{n + \frac{1}{2}}},{U^n} + \frac{k}{2}f\left( {{t_{n + \frac{1}{2}}},{U^*}} \right)} \right)$$
and I don't see how would this help! It is apparent that I have to use a 2-variable expansion, but how to get rid of the star (*) term?
Best Answer
This is the midpoint method. With some elementary transformations you find that $$ U^*=\frac{U^n+U^{n+1}}2 $$
Consider for simplicity an autonomous equation $u'=f(u)$. Write the midpoint as increment of the last point, $U^*=U_n+\frac k2 S$. Recursively inserting this in the fixed point equation gives \begin{align} S&=f(U^n+\tfrac k2 S)=f(U^n)+f'(U^n)·\tfrac k2 S+\tfrac12·f''(U^n)·(\tfrac k2 S)^2+O(k^3) \\[.4em] &=f+\tfrac k2·f'· (f+f'·\tfrac k2 S+O(k^2))+\tfrac{k^2}{8}·f''·(f^2+O(k))+O(k^3) \\[.4em] &=f+\tfrac k2·f'f+\tfrac{k^2}{8}(2f'f'f+f''f^2)+O(k^3) \\[.8em] \implies U^{n+1}&=U^n+k·S \\[.4em] &=U^n+k·f+\tfrac{k^2}2·f'f+\tfrac{k^3}{8}(2f'f'f+f''f^2)+O(k^4) \end{align} and compare that with the exact trajectory starting in $(t^n,U^n)$, \begin{align} u(t^{n+1})&=u(t^n+k)=U^n+k·u'(t^n)+\tfrac{k^2}2·u''(t^n)+\tfrac{k^3}6·u'''(t^n)+O(k^4) \\[.4em] &=U^n+k·f(U^n)+\tfrac{k^2}2f'f+\tfrac{k^3}6·(f''f^2+f'f'f)+O(k^4) \end{align}