Verfying partial derivatives for an ODE

Question

Let $\phi(t,x)$ : $[\mathbb{R}$ x $\mathbb{R^N}] \rightarrow \mathbb{R^N}$ have the following properties: for any $t\in\mathbb{R}$ and $x\in\mathbb{R^N}$,

1. $\phi(t,x)$ is differentiable in both arguments and has a differentiable inverse.
2. For any $s\in\mathbb{R}$, $\phi(t+s,x)=\phi(t,\phi(s,x))$.

Define T(t,x) = $\dfrac{\partial\phi(t,x)}{\partial{x}}$ and $f(x)=\dfrac{\partial\phi(t,x)}{\partial{t}}\big|_{t=0}$. Show that $\phi(t,x)$ and $T(t,x)$ satisfy:

a. $\dfrac{\partial\phi(t,x)}{\partial{t}} = f(\phi(t,x))$

b. $\dfrac{\partial{T(t,x)}}{\partial{t}} = Df(\phi(t,x))\hspace{0.5mm}T(t,x)$

where $Df$ is the Jacobian of f.

Here is my proposed solution.

For the first part, I need to show that $\dfrac{\partial\phi(t,x)}{\partial{t}} = f(\phi(t,x))$. Starting from the LHS,

\begin{equation} \begin{split} \dfrac{\partial\phi(t,x)}{\partial{t}} & = \dfrac{\partial\phi(t+0,x)}{\partial{t}} \\ & = \dfrac{\partial\phi(t,\phi(0,x))}{\partial{t}} \\ & = \dfrac{\partial\phi(t,\phi(t,x))}{\partial{t}}\big|_{t=0} \\ & = f(\phi(t,x)) \end{split} \end{equation}

I think that the second to last equality is justified since $\phi(t,x)$ is evaluated at $t=0$. Although, I'm not sure if I have overlooked something and made a mistake.

For the second part, I need to show that $\dfrac{\partial{T(t,x)}}{\partial{t}} = Df(\phi(t,x))\hspace{0.5mm}T(t,x)$. Starting from the LHS,

\begin{equation} \begin{split} \dfrac{\partial{T(t,x)}}{\partial{t}} & = \dfrac{\partial\phi(t,x)}{\partial{x}\hspace{0.5mm}\partial{t}} \\ & = \dfrac{f(\phi(t,x))}{\partial{x}} \\ & = D{f(\phi(t,x))}\hspace{0.5mm}\dfrac{\partial\phi(t,x)}{\partial{x}} \\ & = D{f(\phi(t,x))}\hspace{0.5mm}T(t,x) \end{split} \end{equation}

As $\phi\in{C^1}$, the Jacobian is defined by the third equality.

Is this approach correct? Please let me know if there are any better alternatives.

Answer 1

For the first part, I recommend \begin{align} \frac{\partial \phi}{\partial t}(t, x)=&\ \frac{\partial}{\partial s}\phi(t+s, x)\Big|_{s=0}=\ \frac{\partial}{\partial s}\phi(s, \phi(t, x))\Big|_{s=0}\\ =&\ \frac{\partial \phi}{\partial t}(s, \phi(t, x))\Big|_{s=0} = f(\phi(t, x)) \end{align}

Answer 2

For the first part, we need to show that $\dfrac{\partial\phi(t,x)}{\partial{t}} = f(\phi(t,x))$. Observe that

1. $\phi$ depends on $(t,x)$.
2. $t$ depends on $s$.

So, by the chain rule, we have the following observation (assuming t is fixed)

\begin{equation*} \begin{split} \dfrac{\partial\phi(t+s,x)}{\partial{s}} & = \dfrac{\partial\phi(t+s,x)}{\partial{t}}\cdot\frac{d(t+s)}{ds} \\ & = \dfrac{\partial\phi(t+s,x)}{\partial{t}} \\ \end{split} \end{equation*}

This implies that

$$\dfrac{\partial\phi(t+s,x)}{\partial{s}}\big|_{s=0}=\dfrac{\partial\phi(t,x)}{\partial{t}}$$

Therefore, we can apply the chain rule again to reduce the LHS to

\begin{equation*} \begin{split} \dfrac{\partial\phi(t,x)}{\partial{t}} & = \dfrac{\partial\phi(t+s,x)}{\partial{s}}\big|_{s=0} \\ & = \dfrac{\partial\phi(s,\phi(t,x))}{\partial{s}}\big|_{s=0} \\ & = \dfrac{\partial\phi(s,\phi(t,x))}{\partial{t}}\big|_{s=0}\cdot\frac{d}{ds}(s) \\ & = \dfrac{\partial\phi(s,\phi(t,x))}{\partial{t}}\big|_{s=0} \\ & = f(\phi(t,x)) \end{split} \end{equation*}

For the second part, we need to show that $\dfrac{\partial{T(t,x)}}{\partial{t}} = Df(\phi(t,x))\hspace{0.5mm}T(t,x)$. Starting from the LHS, we can apply the chain rule to form

\begin{equation*} \begin{split} \dfrac{\partial{T(t,x)}}{\partial{t}} & = \dfrac{\partial\phi(t,x)}{\partial{x}\hspace{0.5mm}\partial{t}} \\ & = \dfrac{f(\phi(t,x))}{\partial{x}} \\ & = D{f(\phi(t,x))}\hspace{0.5mm}\dfrac{\partial\phi(t,x)}{\partial{x}} \\ & = D{f(\phi(t,x))}\hspace{0.5mm}T(t,x) \end{split} \end{equation*}

Remark: If $f\in{C^1}$ and is locally Lipschitz continuous, then $\frac{\partial{f(x_0)}}{x}$ is the Jacobian matrix at $x_0$. So, in our example, $Df(\phi(t,x))$ is the Jacobian matrix at $\phi(t,x)$.