Jtdylktuy

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

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

Potentially related questions:

• How to properly apply the Lie Series


• Exponential of a function times derivative


• How to derive these Lie Series formulas

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 $e^{apartial}f(x)=f(a+x)$ (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.

For any differentiable function $B:Bbb R^ntoBbb R^n$ 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
$$
ϕ(t,x)=expleft(tfrac{∂}{∂s}right)ϕ(s,x)Big|_{s=0}
=expleft(tL_{ϕ(s,x)}right)[ϕ(s,x)]Big|_{s=0}
=expleft(tL_{x}right)[x]Big|_{s=0}
$$
The same remains true if you replace the exponential by the exponential series.

Here is my interpretation of the first answer.

Suppose we have a differentiable function $xi:Bbb R^ntoBbb R^n$ 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{partialxi}{partial{x}}(phi(t,x))cdot frac{partial}{partial{t}}phi(t,x)
\
&=frac{partialxi}{partial{x}}(phi(t,x))cdot f(phi(t,x))
\
&=frac{partialxi}{partial{x}}(x(t))cdot f(x(t))
end{align*}

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

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

Therefore,

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

$$x(a)=sum_{n=0}^inftyfrac{x^{(n)}}{n!}a^n$$

We also know that

$$exp(a)=sum_{n=0}^inftyfrac{a^n}{n!}$$

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

$$expBig(afrac{partial}{partial{t}}Big)=sum_{n=0}^inftyfrac{a^n}{n!}Big(frac{partial}{partial{t}}Big)^n$$

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

$$expBig(afrac{partial}{partial{t}}Big)x(t)=sum_{n=0}^inftyfrac{a^n}{n!}Big(frac{partial{x(t)}}{partial{t}}Big)^n =sum_{n=0}^inftyfrac{a^n}{n!}x^{n}(t)=sum_{n=0}^inftyfrac{x^{n}(t)}{n!}a^n$$

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

$$expBig(afrac{partial}{partial{t}}Big)x(t)Big|_{t=0}=sum_{n=0}^inftyfrac{x^{n}(t)}{n!}a^n=x(a)$$

Noting that $x(t)=phi(t,x)$ and changing the constants $arightarrow{t}$, $trightarrow{s}$ so that $x(a)=x(t)$ and $expBig(afrac{partial}{partial{t}}Big)x(t) = expBig(tfrac{partial}{partial{s}}Big)x(s)$ allows us to write

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