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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.

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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.

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

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