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


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


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


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


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


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


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 $B:\Bbb R^n\to\Bbb R^n$ 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.

  • $\begingroup$ 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. $\endgroup$ – Axion004 Jan 4 at 18:02
  • $\begingroup$ 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 $B(x(t+h))=B(x(t))+L[B](x(t))h+O(h^2)$. There is nothing more to it. $\endgroup$ – LutzL Jan 4 at 18:14
  • $\begingroup$ Perhaps, to avoid using the Taylor series expansion, one could apply the definition of the directional derivative and conclude that $\dfrac{\partial\xi(x)}{\partial{x}}\cdot{f(x)}=\nabla{\xi(x)}\boldsymbol{\cdot}f(x)=\sum_{n=1}^{N}\frac{\partial{\xi}}{\partial{x_i}}f_i(x)$. $\endgroup$ – Axion004 Jan 6 at 1:05
  • $\begingroup$ 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 $\frac{\partial}{\partial{x}}\phi(t,x)f(x)=f(\phi(t,x))$. $\endgroup$ – Axion004 Jan 6 at 2:40
  • $\begingroup$ 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 $s$. $\endgroup$ – LutzL Jan 6 at 8:56

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



\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


We also know that


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


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,


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.