In my current lecture we derived that the Euler method has consistency of order 1. At one point in the proof it reads:
If $f \in C^1(D)$ on a compact set $D$ around the graph of $u$, we can bound the right hand side.
$$|\tau_k| = \frac{1}{2}\max_{t \in I_k}|u''(t)|h_k = \frac{1}{2}\max_{t \in I_k}|\frac{\partial f}{\partial t}(t, u(t)) + \nabla_yf(t,u(t))u'(t)|h_k$$
$$\leq \underbrace{\frac{1}{2} \max_{(t,y) \in D}|\frac{\partial f}{\partial t}(t,y) + \nabla_yf(t,y)f(t,y)|}_{=:c}\cdot h_k$$Here, we use the assumption that $f$ is sufficiently smooth to conclude that the Euler method is consistent of order $1$ (slightly more than Lipschitz continuous).
I do understand everything up to the second equality I posted above just fine and know that the goal is to find a constant $c$ such that $|\tau_k| \leq c \cdot h_k$. I don't understand:
- Why we can't set $c := \max_{t \in I_k} |u''(t)|$. Is it because we can't assume that $u''(t)$ is bounded on $I_k$?
- How we can split the second derivative into the partial derivative and nabla operator in the second equality.
- What exactly is done in the second equality and why we can then assume that the constant $c$ we are defining is actually bounded.
- and finally why consistency of order 1 means "slighly more than Lipschitz continuous"?
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