# Derivative of line integral with respect to initial conditions

In a fluid mechanics application I compute the following integral to find position given a known velocity field that varies in both time and space: $$\mathbf{x}(t) = \mathbf{x}_0 + \int_0^t \mathbf{v}(\mathbf{x}(t'),t') dt'$$ Let's assume that I know explicitly the form of $\mathbf{v}(\mathbf{x}(t'),t')$.

Several important properties of the flow depend on the gradient of the map defined by this integral, taken with respect to the initial conditions (in my case, it is the Cauchy-Green tensor). Extending the notion to make the dependence on the starting point explicit, $$\mathbb{F} \equiv \nabla_{\mathbf{x}_0} \,\mathbf{x}(t; \mathbf{x}_0)$$

Is there any way at all to put the gradient operator underneath the integral sign? In other words, can $\mathbb{F}$ be written in such a way that time integration is the outermost operation? I have a feeling that this is related to integration by parts, but I am otherwise stuck.

Note: I am particularly interested in the case where the velocity field is two-dimensional and separable, $$\mathbf{v}(\mathbf{x}(t'),t') = A(t) \mathbf{u}(\mathbf{x})$$

• Could you precise if/how $x_0$ is involved in the integral? And just to make sure, the same $x(t)$ appears on both sides of your equation, right? So $x(t)$ is implicitly defined? – user130529 Dec 30 '16 at 20:08
• @claudechuber The velocity field depends explicitly on space and time. So if I start out a massless tracer particle at $\mathbf{x}_0$ at $t=0$, it always moves with local velocity given by the field. In that sense, its final position depends on where it's been in the past, as well as when it was there. I believe that would make this integral implicit---I will edit the notation to make this clear – wil3 Dec 30 '16 at 21:06
• I would direct you to Leibniz's integral rule, which has a generalized form for higher dimensions. – MSha Dec 31 '16 at 2:18
• @MSha thanks for your comment, but I'm not sure I follow exactly how Liebniz's rule can be used for a line integral? – wil3 Dec 31 '16 at 22:41

Maybe I'm overlooking something, but doesn't the chain rule work here? For using the chain rule, I find it more convenient to use derivatives instead of gradients (for example, using your notation, $\mathbb{F}\mathbf{h} = D_{x_0}(\mathbf{x}(t))\mathbf{h}$). From $$\mathbf{x}(t) = \mathbf{x}_0 + \int_0^t \mathbf{v}(\mathbf{x}(t'),t') dt',$$ taking the derivative on both sides, it follows that $$D_{x_0}(\mathbf{x}(t))\mathbf{h} = \mathbf{h} + \int_0^t D_{\mathbf{x}}(\mathbf{v}(\mathbf{x}(t'),t'))D_{x_0}(\mathbf{x}(t'))\mathbf{h} \ dt'$$ where $\mathbf{h}$ is a variation of the initial position $\mathbf{x}_0$.
In the case where $$\mathbf{v}(\mathbf{x}(t'),t') = A(t') \mathbf{u}(\mathbf{x}(t')),$$ you have $$D_{\mathbf{x}}(\mathbf{v}(\mathbf{x}(t'),t'))D_{x_0}(\mathbf{x}(t'))\mathbf{h} = A(t') D\mathbf{u}(\mathbf{x}(t'))D_{x_0}(\mathbf{x}(t'))\mathbf{h}.$$