# Understanding the (Inverse) Flow of an ODE

Background

The problem is taken from section 2.5 of this paper, by Raphael Danchin.

We are concerned with the following Transport Equation:

$$\begin{cases} \partial_t a + v\cdot\nabla a + \lambda a = f, \ \text{in } \mathbb{R^+} \times \mathbb{R^d}, \\ a|_{t=0} = a_0, \ \text{in } \mathbb{R^d}. \end{cases}$$

Here, the initial term $$a_0 = a_0(x)$$, the source term $$f=f(t,x)$$, the coeffictient $$\lambda \geq 0$$, and the transport field $$v=v(t,x)$$ are all given.

We assume $$a\in X$$, for some Banach space $$X$$, $$f \in L^1_{\text{loc}}(\mathbb{R^+} ; X)$$, and $$v$$ is in a nice enough set (depending on choice of $$X$$) for there to be a unique solution.

In particular, $$v$$ is at least integrable-in-time with $$v(t)$$ Lipschitz for all $$t > 0$$, which ensures the existence of a flow, $$\psi$$, with respect to $$v$$.

That is, $$\psi$$ satisfies

\begin{align} \psi_0(x) = x, \quad & \forall x \in \mathbb{R^d} \\ \partial_t \psi_t(x) = v(t, \psi_t(x)), \quad & \forall x \in \mathbb{R^d}, t\geq 0. \end{align}

A flow is meant to satisfy other properties as well, namely being a bijection and forming a semigroup with respect to time. These properties however will not be assumed for this question, as they are part of what I want to prove.

Finally, we claim that the flow, $$\psi$$, allows us to write the following explicit formula for the solution, $$a$$:

$$a(t,x) = e^{-\lambda t} a_0(\psi^{-1}_t (x)) + \int_0^t e^{-\lambda (t-\tau)} f(\tau, \psi_\tau (\psi^{-1}_t (x))) \text{d}\tau.$$

My Problem

I have managed to derive the formula above for $$a$$, by converting from Euclidean to Lagarangian coordinates (i.e. substituting $$x \to \psi_t(x)$$), which changes the material derivative into a simple time derivative:

$$\Big{(}\partial_t + v(t,x)\cdot\nabla \Big{)} a(t,x) = \frac{\text{d}}{\text{d}t} a(t,\psi_t(x)).$$

The problem is thus reduced to an ODE, which after solving gives us the formula above when we undo our substitution to return back to Euclidean coordinates.

My issue is when I try to confirm the result in Euclidean coordinates. That is, I want to explicitly write out

$$\partial_t a(t,x) = \partial_t \Bigg{(}e^{-\lambda t} a_0(\psi^{-1}_t (x)) + \int_0^t e^{-\lambda (t-\tau)} f(\tau, \psi_\tau (\psi^{-1}_t (x))) \text{d}\tau \Bigg{)},$$

etc, and get equality on both sides of the equation that way. To do this, I need all of the terms on the left hand side to cancel out nicely.

In particular, it seems that I need the following identity for the time derivative of $$\psi^{-1}$$:

$$\partial_t \Big{(} \psi^{-1}_t(x) \Big{)} = -v(t,x).$$

This seems natural enough, and I think I recall proving a result like this years ago, but I seem to be running into difficulty with it now.

My Attempt

Our first step is to see if we can write an explicit formula for $$\psi_t(x)$$ when $$t$$ is positive. I believe the following formula should work:

$$\psi_t(x) := x + \int^t_0 v(\tau, \psi_\tau (x)) \text{d}\tau,$$

as it satisfies the two properties laid out in the Background. We then need to try and derive an inverse function from this. We note that any inverse should have the property that $$\psi^{-1}_t(\psi_t(x)) = x$$, for all $$x\in\mathbb{R^d}$$. Then, by our above formula for $$\psi_t$$:

$$x = \psi^{-1}_t(\psi_t(x)) = \psi_t(x) - \int^t_0 v(\tau, \psi_\tau (x)) \text{d}\tau.$$

Now we replace $$\psi_t(x)$$ with a generic Euclidean coordinate $$y$$, to get

$$\psi^{-1}_t(y) = y - \int^t_0 v(\tau, \psi_\tau ( \psi^{-1}_t(y) )) \text{d}\tau.$$

Taking the time derivative and using Leibniz's rule, however, leaves us with a messy unwanted term:

$$\partial_t \psi^{-1}_t(y) = - v(t,y) - \int^t_0 \partial_t v(\tau, \psi_\tau ( \psi^{-1}_t(y) )) \text{d}\tau.$$

My question thus is whether there is a way to show that the messy integral above turns out to be $$0,$$ or if I've messed up somewhere

Edit 01

Just as I was writing this question out, I realised that the property I'm looking for on $$\psi$$ is satisfied if we replace the $$\tau$$ in the integrand with $$t$$. That is, if we set

$$\psi_t(x) := x + \int^t_0 v(\tau , \psi_t(x)) \text{d}\tau.$$

These two possible definitions of $$\psi$$ don't seem to be equivalent, unless perhaps $$v(t,x)$$ is constant wrt $$x$$ along flowlines. That is,

$$v(t,x) = v(t,x_0), \ \forall t\geq 0, \ x \in \{ x \in \mathbb{R^d} \ | \ x = \psi_\tau(x_0) , \ \text{some } \tau \geq 0 \}.$$

So now my question becomes: how do I settle on a definition of $$\psi$$? Are the properties I've listed so far insufficient to find a unique formula?

• I am rather sure that the correct equation for $\psi_t^{-1}$ is $\partial _t \psi_t^{-1} + v \cdot \nabla \psi_t^{-1} = 0$. Jul 16 at 2:28

In fact, the identity you want to prove is wrong. You should prove $$\tag{1} \partial_t\psi^{-1}+v\cdot\partial_x\psi^{-1}=0\,.$$ This can be shown easily by taking $$\partial_t$$ on both sides of $$x=\psi^{-1}_t(\psi_t(x))$$.
Now, I will show how to use (1) to justify that the solution's formula. First, taking $$\partial_t$$, we have $$\partial_t\left(e^{-\lambda t}a_0(\psi^{-1}_t(x))\right)=-\lambda e^{-\lambda t}a_0(\psi^{-1}_t(x))+e^{-\lambda t}\partial_xa_0(\psi^{-1}_t(x))\partial_t(\psi^{-1}_t(x))\,.$$ Since $$\int_0^t e^{-\lambda (t-\tau)} f(\tau, \psi_\tau (\psi^{-1}_t (x))) \text{d}\tau=\int_0^t e^{-\lambda (t-\tau)} f(\tau, \psi^{-1}_{t-\tau} (x)) \text{d}\tau=\int_0^t e^{-\lambda \tau} f(t-\tau, \psi^{-1}_{\tau} (x)) \text{d}\tau\,,$$ we have \begin{align*} \partial_t\left(\int_0^t e^{-\lambda (t-\tau)} f(\tau, \psi_\tau (\psi^{-1}_t (x))) \text{d}\tau\right)=&e^{-\lambda t}f(0,\psi^{-1}_t(x))+\int^t_0e^{-\lambda \tau}\partial_t \left[f(t-\tau,\psi^{-1}_\tau(x))\right]\text{d}\tau\\ \end{align*}
Next, take $$\partial_x$$, we have $$\partial_x\left(e^{-\lambda t}a_0(\psi^{-1}_t(x))\right)=e^{-\lambda t}\partial_xa_0(\psi^{-1}_t(x))\partial_x(\psi^{-1}_t(x))$$ and \begin{align*} \partial_x\left(\int_0^t e^{-\lambda (t-\tau)} f(\tau, \psi_\tau (\psi^{-1}_t (x))) \text{d}\tau\right)=\int^t_0e^{-\lambda \tau}\partial_x f(t-\tau,\psi^{-1}_\tau(x))\partial_x(\psi^{-1}_\tau(x))\text{d}\tau \end{align*}
Using (1), we have $$\partial_t\left(e^{-\lambda t}a_0(\psi^{-1}_t(x))\right)+v(t,x)\cdot\partial_x\left(e^{-\lambda t}a_0(\psi^{-1}_t(x))\right)=-\lambda e^{-\lambda t}a_0(\psi^{-1}_t(x))$$ \begin{aligned} &\partial_t\left(\int_0^t e^{-\lambda (t-\tau)} f(\tau, \psi_\tau (\psi^{-1}_t (x))) \text{d}\tau\right)+v(t,x)\cdot\partial_x\left(\int_0^t e^{-\lambda (t-\tau)} f(\tau, \psi_\tau (\psi^{-1}_t (x))) \text{d}\tau\right)\\ =&e^{-\lambda t}f(0,\psi^{-1}_t(x))\\ &+\int^t_0e^{-\lambda \tau}\left(\partial_t \left[f(t-\tau,\psi^{-1}_\tau(x))\right]+v(t,x)\cdot \partial_x f(t-\tau,\psi^{-1}_\tau(x))\partial_x(\psi^{-1}_\tau(x))\right)\text{d}\tau\\ =&e^{-\lambda t}f(0,\psi^{-1}_t(x))\\ &+\int^t_0e^{-\lambda \tau}\left(\partial_t \left[f(t-\tau,\psi^{-1}_\tau(x))\right]-\partial_x f(t-\tau,\psi^{-1}_\tau(x))\partial_t(\psi^{-1}_\tau(x))\right)\text{d}\tau\\ =&e^{-\lambda t}f(0,\psi^{-1}_t(x))-\int^t_0e^{-\lambda \tau}\frac{\text{d} }{\text{d} \tau}\left[f(t-\tau,\psi^{-1}_\tau(x))\right]\text{d}\tau\\ =&f(t,x)-\lambda \int_0^t e^{-\lambda \tau} f(t-\tau, \psi^{-1}_{\tau} (x)) \text{d}\tau\\ =&f(t,x)-\lambda \int_0^t e^{-\lambda (t-\tau)} f(\tau, \psi_\tau (\psi^{-1}_t (x))) \text{d}\tau\\ \end{aligned} Done.