# Understanding proof of Euler method having consistency of order 1

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"?

• 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.