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"?
  • You want a bound that is independent of the unknown exact solution.
  • This is obtained by computing the (total) $t$ derivative of the ODE, using the (generalized) chain rule for the composition in $f(t,u(t))$. Then insert again the ODE to replace the derivative.
  • see above, $$u''=\frac{d}{dt}f(t,u(t))=∂_tf(t,u(t))+∂_yf(t,u(t))u'(t)=∂_tf(t,u(t))+∂_yf(t,u(t))f(t,u(t)).$$ The expression is bounded because it is continuous on a compact set.
  • It is not totally clear what you mean with that. The assumptions on $f$ for this proof require a little bit more than the Lipschitz condition and other assumption of the Picard-Lindelöf theorem.

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