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The texts below are from Ordinary Differential Equations by Arnol'd

Consider the differential equation $\dot{\boldsymbol{x}} = \boldsymbol{v}(t,\boldsymbol{x})$ defined by a vector field $\boldsymbol{v}$ in some domain of the extended phase space $\mathbf{R}^{n+1}$. We define the Picard mapping to be the mapping $A$ that takes the function $\varphi:t \to \boldsymbol{x}$ to the function $A \phi: t \to \boldsymbol{x}$ where $$ (A \varphi)(t) = \boldsymbol{x}_0+ \int_{t_0}^t \boldsymbol{v}(\tau,\varphi(\tau))d\tau $$

And the book says the following Texts from the book

I can't understand this geometrical representation at all. How does the tangent for each $t$ be parallel to $\varphi$? What does it mean "for then $A \varphi$ would be a solution"? A solution of what?

Any help is appreciated!!

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Let's take a simple analogy: The equation $ax=\tan x$ can be seen as a fixed-point equation with iteration $x_{k+1}=\frac1{a}\tan x_k$. However, in general this iteration does not converge or even stay bounded. Now with a simple transformation using the inverse tangent, one gets a family of fixed-point iterations $x_{k+1}=n\pi+\arctan(ax)$ which converges very nicely for every $n\ne 0$.

With differential equations it is the same, any fixed-point form of the equation $x'=v(t,x)$ that contains the derivative contains with it an operation that is unbounded, even more, you would have difficulties to find a space where it could be a fixed-point iteration. Inverting the differentiation by integration, $x(t)=x(0)+\int_0^t v(s,x(s))ds$, results in a fixed-point operator that maps continuous functions to continuous functions, so the space question is trivially solved. The "smoothing" properties of the integration also lead to the contraction property that one needs for convergence.

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  • $\begingroup$ This is a very nice answer. +1 $\endgroup$ Dec 23, 2019 at 4:27
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We are given a vector field $(t,x)\mapsto v(t,x)$ defined on some open $I\times U\subseteq \mathbb R^{n+1}.$ And $\varphi:I\to U$ so we can picture the situation by looking at the graph of $\varphi,$ which is a curve in the $I-U$ "plane." Then, since $(A \varphi)(t) = x_0+ \int_{t_0}^t v(\tau,\varphi(\tau))d\tau,\ (A \varphi)'(t)=v(t,\varphi(t))$ so at the point $(t,\varphi(t)),$ the tangent to the curve $A\varphi$ at $t$ is equal to the value of the given vector field at $(t,\varphi(t)).$ So, geometrically, as $A\varphi(t)$ varies with $t$, of course its tangents also vary with $t$ and are equal to the values of the vector field at $(t,\varphi(t)),$ that is, $\textit{on the graph of}\ \varphi.$

Now, if it happens that these vectors also lie on the graph of $A\varphi,$ then we would have $(t,\varphi(t))=(t,A\varphi(t))$, which implies that $A\varphi=\varphi$, from which we conclude that $\varphi'=v(t,\varphi(t))$ i.e. $A\varphi$ is a solution to the ODE.

Remark: this all seems rather labored to me, probably because I have not read Arnold's book. The point is that the Picard mapping is, under appropriate circumstances, a contraction, from which the existence/uniqueness theorem for ODE's follows easily. I personally think the best way to see the "geometry" is with differential geometry. Lee's book covers this material in great detail, and is very readable!

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  • $\begingroup$ Thanks for the detailed explanations and the recommended book!!! $\endgroup$
    – Tab1e
    Dec 22, 2019 at 3:48
  • $\begingroup$ @UnbelieveTable glad to help! $\endgroup$ Dec 23, 2019 at 4:25

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