I'm reading about Euler's method to construct approximate solutions to ODEs in Ordinary Differential Equations by Andersson and Böiers. I have questions about properties of the approximate solution. I've tried to be as brief as possible while still being self-contained. Grateful for any help.
Part 1: Approximate solution and error estimate
In what follows $|\cdot|$ denotes either the absolute value of a scalar or the Euclidean norm of a vector (denoted in bold). First a definition of what it means to be an approximate solution:
Definition 1. Let $I$ be an interval on the real axis, and $\Omega$ an open set in $\mathbf R\times\mathbf{R}^n$. Assume that the function $\pmb{f}:\Omega\to\mathbf{R}^n$ is continuous. A continuous function $\pmb{x}(t),\ t\in I$, is called an $\varepsilon$-approximate solution of the system $\pmb{x}'=\pmb{f}(t,\pmb{x})$ if $(t,\pmb{x})\in\Omega$ when $t\in I$ and $$\left|\pmb{x}(t'')-\pmb{x}(t')-\int_{t'}^{t''} \pmb{f}(s,\pmb{x}(s))ds\right|\leq \varepsilon|t''-t'|\quad \text{when } t',t''\in I.$$
Next, a theorem (for the sake of brevity, without proof) on an error estimate on the approximate solution:
Theorem 1. Assume that $\pmb{f}(t,\pmb{x})$ is continuous in $\Omega\subseteq \mathbf{R}\times\mathbf{R}^n$ and satisfies the Lipschitz condition $$|\pmb{f}(t,\pmb{x})-\pmb{f}(t,\pmb{y})|\leq L|\pmb{x}-\pmb{y}|, \quad (t,\pmb{x}),(t,\pmb{y})\in\Omega.$$ Let $\pmb{\tilde{x}}(t)$ be an $\varepsilon$-approximate and $\pmb{x}(t)$ and exact solution of $\pmb{x}'=\pmb{f}(t,\pmb{x})$ in $\Omega$ when $t\in I$. For an arbitrary point $t_0$ in $I$ we then have $$|\pmb{\tilde{x}}(t)-\pmb{x}|\leq |\pmb{\tilde{x}}(t_0)-\pmb{x}(t_0)|e^{L|t-t_0|}+\frac{\varepsilon}{L}(e^{L|t-t_0|}-1),\quad t\in I.$$
Part 2: Euler's method
The Euler's method works as follows. Consider the equation $x'=f(t,x)$ and consider an initial value $(t_0,x(t_0))$. We know the slope of the tangent line through $(t_0,x(t_0))$. Follow this tangent a bit to the right, to $t=t_1=t_0+\delta$ and repeat the procedure for the point $(t_1,x(t_1))$. This way, one gets a broken curve of straight line segments resembling the solution.
To make the definition of the broken curve precise, consider the system $\pmb{x}'=\pmb{f}(t,\pmb{x})$ and an initial value $(t_0,\pmb{x}_0)$. Make a division of the interval $[t_0,t_0+a]$ in equally long subintervals: $t_0<t_1<\ldots<t_m=t_0+a$, and put $\delta=t_j-t_{j-1}$. Then define the function $\pmb{x}_{\delta}$ recursively at the step points $t_j$ by \begin{align} &\pmb{x}_{\delta}(t_0)=\pmb{x}_0, \\ &\pmb{x}_{\delta}(t_{j+1})=\pmb{x}_{\delta}(t_j)+(t_{j+1}-t_j)\pmb{f}(t_j,\pmb{x}_{\delta}(t_j)),\quad j=0,1,\ldots,m-1. \tag1 \end{align} Between the step points the curve of $\pmb{x}_{\delta}$ is supposed to be a straight line, so $$\pmb{x}_{\delta}(t)=\pmb{x}_{\delta}(t_j)+(t-t_j)\pmb{f}(t_j,\pmb{x}_{\delta}(t_j)),\quad t_j\leq t\leq t_{j+1}.\tag2$$ The function $\pmb{x}_{\delta}$ is defined correspondingly in the interval $[t_0-a,t_0]$.
The following theorem shows $\pmb{x}_{\delta}$ is an $\varepsilon$-approximate solution to $\pmb{x}'=\pmb{f}(t,\pmb{x})$ under certain conditions on $\pmb{f}$.
Theorem 2. [Let $B,L$ and $C$ be positive constants.] Assume that
- the function $\pmb{f}(t,\pmb{x})$ is continuous in $\Omega\subseteq\mathbf{R}\times\mathbf{R}^n$, that $|\pmb{f}(t,\pmb{x})|\leq B$ in $\Omega$, and that $a$ is so small that $\Lambda(a)\subseteq\Omega$, where $\Lambda(a)$ denotes the double cone $$\Lambda(a)=\{(t,\pmb{x})\in \mathbf{R}\times\mathbf{R}^n;|t-t_0|\leq a, \ |\pmb{x}-\pmb{x}_0|\leq B|t-t_0|\},$$
- $|\pmb{f}(t,\pmb{x})-\pmb{f}(t,\pmb{y})|\leq L|\pmb{x}-\pmb{y}| \qquad \ \ \ (t,\pmb{x}),(t,\pmb{y})\in\Omega$,
- $|\pmb{f}(t',\pmb{x})-\pmb{f}(t'',\pmb{x})|\leq C|t'-t''| \quad (t',\pmb{x}),(t'',\pmb{x})\in\Omega$.
Then $\pmb{x}_{\delta}$ is an $\varepsilon$-approximate solution to the system $\pmb{x}'=\pmb{f}(t,\pmb{x})$ in the interval $[t_0-a,t_0+a]$, with $\varepsilon=\delta(C+LB)$.
Proof. All line segments in the definition of $\pmb{x}_{\delta}$ have a slope at most $B$. Therefor $(t,\pmb{x}_{\delta}(t))\in\Lambda(a)$ when $t\in[t_0-a,t_0+a]$. Moreover, if $|t'-t''|\le\delta$ then \begin{align} |\pmb{f}(t',\pmb{x}_{\delta}(t'))-\pmb{f}(t'',\pmb{x}_{\delta}(t''))|&\leq |\pmb{f}(t',\pmb{x}_{\delta}(t'))-\pmb{f}(t'',\pmb{x}_{\delta}(t'))| \\ &+|\pmb{f}(t'',\pmb{x}_{\delta}(t'))-\pmb{f}(t'',\pmb{x}_{\delta}(t''))| \\ &\leq C|t'-t''|+L|\pmb{x}_{\delta}(t')-\pmb{x}_{\delta}(t'')| \\ &\leq C|t'-t''|+LB|t'-t''| \\ &\leq \delta(C+LB).\tag3 \end{align} [The first inequality is the triangle inequality, the second follows from 2. and 3. in the assumptions and the third by the fact that each line segments of $\pmb{x}_{\delta}$ have a slope at most $B$, so $|\pmb{x}_{\delta}(t')-\pmb{x}_{\delta}(t'')|\leq B|t'-t''|$.] We must prove that $$\left|\pmb{x}_{\delta}(t'')-\pmb{x}_{\delta}(t')-\int_{t'}^{t''} \pmb{f}(s,\pmb{x}_{\delta}(s))ds\right|\leq \delta(C+LB)|t'-t''|.\tag4$$ If $t'$ and $t''$ belong to the same subinterval $[t_j,t_{j+1}]$ then [from $(2)$] $$\pmb{x}_{\delta}(t'')-\pmb{x}_{\delta}(t')=(t''-t')\pmb{f}(t_j,\pmb{x}_{\delta}(t_j))=\int_{t'}^{t''} \pmb{f}(t_j,\pmb{x}_{\delta}(t_j))ds,$$ and $(4)$ follows from $(3)$. (In particular, $(4)$ is valid when $t'=t_j$ and $t''=t_{j+1}$). If $t'$ and $t''$ belong to different subintervals then use $(4)$ on each one of the intervals $[t',t_{j+1}],[t_{j+1},t_{j+2}],\ldots,[t_{k-1},t_k],[t_k,t'']$ and add the results.
Part 3: Properties of solution and questions
Now comes the part I have some questions about. The authors claim that if we drop assumptions 1. and 2. in theorem 2, it is still possible to show that there is a number $\varepsilon(\delta)$, tending to zero as $\delta\to 0$, such that $\pmb{x}_{\delta}$ is an $\varepsilon(\delta)$-approximate solution in the interval $I(a)=[t_0-a,t_0+a]$. They write:
We know that $(t,\pmb{x}_{\delta}(t))\in\Lambda(a)$ when $t\in I(a)$ and that $$|\pmb{x}_{\delta}(t')-\pmb{x}_{\delta}(t'')|\leq B|t'-t''|\quad t',t''\in I(a).\tag5$$ Put $$\varepsilon(\delta)=\sup\limits_{|t'-t''|\leq\delta}|\pmb{f}(t',\pmb{x}_{\delta}(t'))-\pmb{f}(t'',\pmb{x}_{\delta}(t''))|.$$ The computations in the proof of theorem 2 show that $\pmb{x}_{\delta}$ is an $\varepsilon(\delta)$-approximate solution. Furthermore, $\pmb{f}$ is uniformly continuous on $\Lambda(a)$. From $(5)$ it accordingly follows that $$\lim_{\delta\to0}\varepsilon(\delta)=0.\tag6$$
Question 1: I simply do not understand how $(6)$ follows from $(5)$. Grateful for any help on this.
Question 2: The authors go on to claim that when the IVP $\pmb{x}'=\pmb{f}(t,\pmb{x}), \pmb{x}(t_0)=\pmb{x}_0$ has a solution $\pmb{x}(t)$, then it follows from $(6)$ and theorem 1, that $\pmb{x}_{\delta}$ converges uniformly to $\pmb{x}$ as $\delta\to 0$. The definition of uniform convergence I'm used to is $\lVert f_n-f\rVert\to 0$ as $n\to\infty$, where $\lVert\cdot\rVert$ denotes the sup-norm, but here they are claiming that the sup-norm should tend to $0$ as $\delta\to 0$. This makes me wonder; what is $\pmb{x}_{\delta}$? Is it a sequence? If not, what definition of uniform convergence are the authors using?
Question 1: A function is uniformly continuous if and only if it admits a modulus of continuity. $\varepsilon(\delta)$ is a modulus of continuity for $\pmb f$, which by definition satisfies $\lim_{\delta\to 0^+}\varepsilon(\delta)=0$.
Question 2: $\delta$ depends on the number $m\in\mathbb N$, i.e. how many equally long subintervals the interval $[t_0-a,t_0+a]$ is partitioned into. So $\pmb{x}_{\delta}$ can be viewed as a sequence and converges uniformly as $m\to\infty$ ($\delta\to0$), provided the assumptions of theorem 2 hold.