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Consider the $N$-dimensional autonomous system of ODEs $$\dot{x}= f(x),$$ where a locally unique solution $x(t)$, starting from the initial condition $x$, is denoted as $x(t)=\phi(t,x)$. Assume that

$$\Big(\frac{\partial}{\partial{x}}\phi(t,x)\Big)f(x)=f(\phi(t,x))$$

For the system above, assume that $f(x)$ is analytic (that is, its Taylor series converges to $f$ itself). Let the differential operator $L[\xi]$ be defined as

$$L[\xi]=f(x)\boldsymbol{\cdot}\nabla{\xi}=\sum_{n=1}^{N}f_i(x)\frac{\partial{\xi}}{\partial{x_i}}$$

Show that $\phi(t,x)$ can be expressed as

$$\phi(t,x)=\sum_{n=0}^{\infty}\frac{t^n}{n!}L^n[x]$$

where $L^n[\xi]$ is the shorthand notation for

$$L^n[\xi]=\underbrace{L[L[\cdots{L}[\xi]}_{n\text{-times}}\cdots]]$$

Potentially related questions:

I'm stuck on how to approach this problem. Here is all the information that I have gathered so far -

Through this question, the one dimensional situation states that $e^{a\partial}f(x)=f(a+x)$ (we can think of this as a shift operator).

Inside Ordinary Differential Equations and Dynamical Systems by Teschl, we have the following Lemma (Lemma $6.2$ on page $190$ of the text).

Lemma (Straightening out of vector fields): Suppose $f(x_0)\neq0$. Then, there is a local coordinate transform $y=\varphi(x)$ such that $\dot{x}=f(x)$ is transformed to

$$\dot{y}=(1,0,...,0)$$

Teschl list a similar problem on page $191$ (problem $6.5$ for one-parameter lie groups) in which he states that

Hint: The Taylor coefficients are the derivatives which can be obtained by differentiating the differential equation.

So, I think that I need to apply what was done in this question alongside Lemma 6.2. I will have to consider what a vector field means in this context. I might be able to make the assumption that a vector field is just a linear operator. We are given that

  1. $\dot{x}= f(x)$ is an autonomous system of ODEs
  2. $x(t)=\phi(t,x)$
  3. $\Big(\frac{\partial}{\partial{x}}\phi(t,x)\Big)f(x)=f(\phi(t,x))$
  4. $L[\xi]=f(x)\boldsymbol{\cdot}\nabla{\xi}=\sum_{n=1}^{N}f_i(x)\frac{\partial{\xi}}{\partial{x_i}}$

and we need to show that

$$\phi(t,x)=\sum_{n=0}^{\infty}\frac{t^n}{n!}L^n[x]$$

I also see that Roger Howe wrote a good introduction to lie theory in these notes (he goes through one-parameter lie groups on pages $604-606$).

This appears to be an extremely difficult problem for someone unfamiliar with lie theory. I am going to see if I can figure out a more direct approach.

Bernard
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Axion004
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2 Answers2

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For any differentiable function $B:\Bbb R^n\to\Bbb R^n$ we know from chain rule and differential equation that \begin{align} \frac{∂}{∂t}B(ϕ(t,x))&=\frac{∂B}{∂x}(ϕ(t,x))\cdot \frac{∂}{∂t}ϕ(t,x) \\ &=\frac{∂B}{∂x}(ϕ(t,x))\cdot f(ϕ(t,x)) \\ &=\sum_{i=1}^n \frac{∂B}{∂x_i}(ϕ(t,x)) f_i(ϕ(t,x))=L_{ϕ(t,x)}[B]. \end{align} So along a solution we get $\frac{∂}{∂t}=L_{ϕ(t,x)}$. Now apply this to the translation operator resp. the Taylor expansion $$ ϕ(t,x)=\exp\left(t\frac{∂}{∂s}\right)ϕ(s,x)\Big|_{s=0} =\exp\left(tL_{ϕ(s,x)}\right)[ϕ(s,x)]\Big|_{s=0} =\exp\left(tL_{x}\right)[x]\Big|_{s=0} $$ The same remains true if you replace the exponential by the exponential series.

Lutz Lehmann
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  • I'm having trouble following $\frac{∂B}{∂x}(ϕ(t,x))\cdot f(ϕ(t,x))=\sum_{i=1}^n \frac{∂B}{∂x_i}(ϕ(t,x)) f_i(ϕ(t,x))=L_{ϕ(t,x)}[B]$. For the first equality, $\frac{∂B}{∂x}(ϕ(t,x))\cdot f(ϕ(t,x))=\sum_{i=1}^n \frac{∂B}{∂x_i}(ϕ(t,x)) f_i(ϕ(t,x))$, you must be taking a partial derivative in every direction (since we are working in $\Bbb R^n$). I don't see how one can justify $\sum_{i=1}^n \frac{∂B}{∂x_i}(ϕ(t,x)) f_i(ϕ(t,x))=L_{ϕ(t,x)}[B]$. This must follow from the definition of the differential operator. – Axion004 Jan 04 '19 at 18:02
  • Simple manipulation of linear Taylor polynomials gives $$B(x(t+h))=B(x(t)+f(x(t))h+O(h^2)=B(x(t))+B'(x(t))f(x(t))h+O(h^2).$$ Inside $B'(x)v$ with $v=f(x)h$ is the directional derivative in direction $v$, $B'(x)v=\sum v_i\frac∂{∂x_i}B(x)$. Replacing back $v=f(x)h$ gives exactly the definition of $L$, so that $B(x(t+h))=B(x(t))+LBh+O(h^2)$. There is nothing more to it. – Lutz Lehmann Jan 04 '19 at 18:14
  • Perhaps, to avoid using the Taylor series expansion, one could apply the definition of the directional derivative and conclude that $\dfrac{\partial\xi(x)}{\partial{x}}\cdot{f(x)}=\nabla{\xi(x)}\boldsymbol{\cdot}f(x)=\sum_{n=1}^{N}\frac{\partial{\xi}}{\partial{x_i}}f_i(x)$. – Axion004 Jan 06 '19 at 01:05
  • I'm guessing that there is a logical reason why $\phi(t,x)=\exp\left(t\frac{\partial}{\partial{s}}\right)\phi(s,x)\Big|_{s=0}$. I tried reviewing it tonight and couldn't see how this was formed. I coudn't derive this starting from $\frac{\partial}{\partial{x}}\phi(t,x)f(x)=f(\phi(t,x))$. – Axion004 Jan 06 '19 at 02:40
  • This is just the simple Taylor expansion $x(t)=\sum\frac{x^{(k)}}{k!}t^k=(\exp(tD)x)(0)$. $t$ is here a constant, so it looks bad to have the derivative for $t$, thus changing it to $s$. – Lutz Lehmann Jan 06 '19 at 08:56
  • Unfortunately, one part is still confusing me. We are given that $\phi(t,x)=x(t)=\sum_{n=0}^\infty \frac{x^{(n)}}{n!}{t^n}$. As $exp(tD)=\sum_{n=0}^\infty \frac{(tD)^n}{n!}$, we can compute $exp(tD)x=\sum_{n=0}^\infty \frac{(tD)^n}{n!}x=\sum_{n=0}^\infty \frac{(tD)^n}{n!}\phi(t,x)\Big|{t=0}=\sum{n=0}^\infty \frac{(t\frac{\partial}{\partial{t}})^n}{n!}\phi(t,x)\Big|{t=0}$. As $t$ is a constant, we can replace it with $s$. But, why does $\sum{n=0}^\infty \frac{(t\frac{\partial}{\partial{s}})^n}{n!}\phi(s,x)\Big|{s=0}=\sum{n=0}^\infty \frac{x^{(n)}}{n!}{t^n}$? – Axion004 Jan 06 '19 at 17:04
  • Just use another letter than $t$, as it is too widely used, for instant a constant $a$. Then $x(a)=\exp(a\frac∂{∂t})x(t)|{t=0}=\exp(a\frac∂{∂t})ϕ(t,x)∣|{t=0}$. You then have to replace in the final result $a$ with $t$ to get the form of the claim. The problem is that $t$ and $\frac∂{∂t}$ do not commute, so writing $(t\frac∂{∂t})^n$ is ambiguous and wrong in context in a formally correct interpretation. – Lutz Lehmann Jan 06 '19 at 17:04
  • That makes sense. Although, I don't see how you get the first equality $x(a)=\sum_{n=0}^\infty \frac{x^{(n)}}{n!}a^n=exp(a\frac{\partial}{\partial{t}})x(t)\Big|_{t=0}$. Although, I think that I am asking too many questions and should post this as a new question on Math SE. – Axion004 Jan 06 '19 at 17:34
  • That is the coincidence of the Taylor expansion and the exponential series. I'm not sure where exactly your problem is, as you seemed to have accepted that. There is of course always the convergence of the Taylor series and the coincidence of the limit with the function to consider, as always with Taylor expansions. – Lutz Lehmann Jan 06 '19 at 18:11
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Here is my interpretation of the first answer.

Suppose we have a differentiable function $\xi:\Bbb R^n\to\Bbb R^n$ where $\frac{\partial}{\partial{t}}\phi(t,x)=f(\phi(t,x))$. We know by the chain rule that \begin{align*} \frac{\partial}{\partial{t}}\xi(\phi(t,x))&=\frac{\partial\xi}{\partial{x}}(\phi(t,x))\cdot \frac{\partial}{\partial{t}}\phi(t,x) \\ &=\frac{\partial\xi}{\partial{x}}(\phi(t,x))\cdot f(\phi(t,x)) \\ &=\frac{\partial\xi}{\partial{x}}(x(t))\cdot f(x(t)) \end{align*}

where $\dfrac{\partial\xi(x)}{\partial{x}}\cdot{f(x)}$ is the directional derivative of the function $\xi$ in the direction of $f$. This is defined as

$$\dfrac{\partial\xi(x)}{\partial{x}}\cdot{f(x)}=\nabla{\xi(x)}\boldsymbol{\cdot}f(x)=\sum_{n=1}^{N}\frac{\partial{\xi}}{\partial{x_i}}f_i(x)$$

Therefore,

\begin{align*} \frac{\partial}{\partial{t}}\xi(\phi(t,x))&=\frac{\partial\xi}{\partial{x}}(x(t))\cdot f(x(t))\\ &=\sum_{i=1}^n \frac{\partial\xi}{\partial{x_i}}(x(t)) f_i(x(t)) \\&=\sum_{i=1}^n f_i(\phi(t,x))\frac{\partial\xi}{\partial{x_i}}(\phi(t,x)) =L_{\phi(t,x)}[\xi] \end{align*}

So along a solution we get $\frac{\partial}{\partial{t}}=L_{\phi(t,x)}$. Next, observe that we can write $x(a)$ in series notation as

$$x(a)=\sum_{n=0}^\infty\frac{x^{(n)}}{n!}a^n$$

We also know that

$$\exp(a)=\sum_{n=0}^\infty\frac{a^n}{n!}$$

Applying $\exp(a)$ to the differential operator $a\frac{\partial}{\partial{t}}$ produces

$$\exp\Big(a\frac{\partial}{\partial{t}}\Big)=\sum_{n=0}^\infty\frac{a^n}{n!}\Big(\frac{\partial}{\partial{t}}\Big)^n$$

Hence if we evaluate this at $x(t)$ we have

$$\exp\Big(a\frac{\partial}{\partial{t}}\Big)x(t)=\sum_{n=0}^\infty\frac{a^n}{n!}\Big(\frac{\partial{x(t)}}{\partial{t}}\Big)^n =\sum_{n=0}^\infty\frac{a^n}{n!}x^{n}(t)=\sum_{n=0}^\infty\frac{x^{n}(t)}{n!}a^n$$

Where it is necessary for $t=0$ in order to compute the Maclaurin expansion. Therefore,

$$\exp\Big(a\frac{\partial}{\partial{t}}\Big)x(t)\Big|_{t=0}=\sum_{n=0}^\infty\frac{x^{n}(t)}{n!}a^n=x(a)$$

Noting that $x(t)=\phi(t,x)$ and changing the constants $a\rightarrow{t}$, $t\rightarrow{s}$ so that $x(a)=x(t)$ and $\exp\Big(a\frac{\partial}{\partial{t}}\Big)x(t) = \exp\Big(t\frac{\partial}{\partial{s}}\Big)x(s)$ allows us to write

$$ \phi(t,x)=\exp\left(t\frac{\partial}{\partial{s}}\right)\phi(s,x)\Big|_{s=0} =\exp\left(tL_{\phi(s,x)}\right)[\phi(s,x)]\Big|_{s=0} =\exp\left(tL_{x}\right)[x] $$ If we then replace the exponential with the exponential series, we have $$ \phi(t,x)=\exp\left(tL_{x}\right)[x]=\Big(\sum_{n = 0}^{\infty} \frac{\left(tL_{x}\right)^n}{n!}\Big)[x]=\sum_{n=0}^{\infty}\frac{t^n}{n!}L^n[x]$$

Axion004
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