Derivative of a double integral over a variable circular region

Calculate the following derivative

$$\frac{d}{dt}\iint_{D_t}F(x,y,t) \, \mathrm d x \mathrm d y$$

where

$$D_t = \lbrace (x,y) \mid (x-t)^2+(y-t)^2\leq r^2 \rbrace$$

I've read here$^1$ that the answer to the above question is in the form of: \begin{align}\frac{d}{dt}\iint_{D_t}F(x,y,t)dxdy &=\int_{\partial D_t} F(udy-vdx) + \iint_{D_t}\frac{\partial F}{\partial t}dx dy \\ &=\iint_{D_t}\left[\text{div}(F\mathbf{v})+\frac{\partial F}{\partial t}\right]dx dy\end{align} which is a generalization of Leibniz integral rule.

I don't know how I can calculate $\mathbf{v}$ and $\mathbf{n}$ or $u$ and $v$ in my problem. The paper states that $\mathbf{v}$ with components $u$ and $v$ is the velocity and $\mathbf{n}$ is the normal vector.

Also, I would be grateful if someone could introduce some other references on the derivative of such double integrals to me.

$^1$: Harley Flanders, Differentiation under the integral sign, Am. Math. Mon. 80, 615-627 (1973). ZBL0266.26010.

The following reference was very handy for me:

J. Haslinger and R. Mäkinen, Introduction to shape optimization: theory, approximation, and computation, Society for Industrial and Applied Mathematics (2003).

Solution 1:

Your domain $D_t$ is simply the transport of a disk by the vector $\mathbf v = (1,1)$. Because this is a constant vector, its actually easier to just use a simple change of variables: $$ \frac{d}{d t} \iint_{D_{t}} F(x, y, t) d x d y= \frac{d}{d t}\iint_{D_0} F(x+t,y+t,t)dxdy = \iint_{D_0} \frac{d}{d t}(F(x+t,y+t,t))dxdy= \iint_{D_0} (\partial_xF+\partial_yF+\partial_t F)(x+t,y+t,t)dxdy = \iint_{D_t}\partial_xF+\partial_yF+\partial_t Fdxdy. $$

This formula is a special case of the Reynolds Transport Formula. The great book of Bertozzi and Majda, Vorticity and Incompressible Flow(Google Books) on page 5 has the statement and proof. I'll sketch what $\mathbf v$ is, and then sketch the proof from this book, but perhaps the book A First Course in Continuum Mechanics(Amazon) by Gonzales and Stuart will be more gentle (a first hint of this is that I'd suggest you read from page 137, instead of page 5...)

Suppose $F=(\mathbf x,t)$ is a sufficiently smooth function of $\mathbf x \in \mathbb R^n,t\in\mathbb R$. The velocity acting on particles, $\mathbf v(\mathbf X,t) \in \mathbb R^n$ moves $\mathbf x_0 \in \mathbb R^n$ to the point $\mathbf X=\mathbf X(t,\mathbf x_0) \in \mathbb R^n$ via the particle-trajectory ODE, $$\begin{cases} \frac{d}{dt}{\mathbf X} = \mathbf v(\mathbf X,t),\\ \mathbf X(0,\mathbf x_0) = \mathbf x_0.\end{cases}$$ Applying this to each point in $D_0$, we have $D_t = \mathbf X(t,D_0) = \{ \mathbf X(t,\mathbf x_0) : \mathbf x_0 \in D_0\}$.

In your case, observe that $D_t $ is is the image under translation of $D_0$ by $t\binom{1}{1}$. Hence, $\mathbf v = \binom{1}{1}$ is actually constant.

To prove the formula in the general case, apply a change of variables to write the integral as an integral over the initial positions of the particles $\mathbf x_0 = \mathbf X(t,\cdot)^{-1}(\mathbf x)$. This gives $$ \iint_{D_t} F(\mathbf x,t) d\mathbf x = \iint_{D_0} F(\mathbf X(t,\mathbf x_0),t) J d\mathbf x_0 , $$ where $J=J(t,\mathbf x_0)$ is the Jacobian determinant $J d\mathbf x_0 = d\mathbf x$, $$ J(t,\mathbf x_0) = \det (\nabla \mathbf X(t,\mathbf x_0))=\det (\nabla_{\mathbf x_0} \mathbf X(t,\mathbf x_0)).$$

Taking the gradient in $\mathbf x_0$ of the particle-trajectory ODE gives

$$\begin{cases} \frac{d}{dt}{\nabla \mathbf X} = \nabla \mathbf v(\mathbf X,t)\nabla \mathbf X ,\\ \nabla \mathbf X(0,\mathbf x_0) = I .\end{cases}$$

By Jacobi's formula(Wikipedia page), $$ \frac{d}{d t} \operatorname{det} A(t)=\operatorname{tr}\left(\operatorname{adj}(A(t)) \frac{d A(t)}{d t}\right),$$ we deduce (using $\operatorname{tr}(ABC)= \operatorname{tr}(BCA)$ and $(\operatorname{adj} A) A = \det A$ ) $$\frac{dJ}{dt} = \operatorname{tr}(\nabla \mathbf v(\mathbf X,t))J = \nabla\cdot\mathbf v(\mathbf X,t) J. $$ Pushing the derivative into the integral without worry, then applying chain rule and product rule gives the result, since
$$\frac{d}{dt} \Big( F(\mathbf X,t) J \Big) = \nabla F(\mathbf X,t)\cdot \frac{d}{dt} \mathbf X J + (\partial_t F)(\mathbf X,t)J + F(\mathbf X,t) \frac{d}{dt} J = \Big ( \nabla F(\mathbf X,t) \cdot \mathbf v(X,t) + (\partial_t F)(\mathbf X,t)+ F(\mathbf X,t) \nabla\cdot \mathbf v(\mathbf X,t) \Big) J. $$

Of course in your case this simplifies greatly to the result I first computed, since your $\mathbf v$ is incompressible with $J\equiv 1$.

(In fact, this formula justifies the name "incompressible": in the special case $F\equiv 1$, it shows that $\nabla \cdot \mathbf v = 0$ iff the volume of $D_t$ is constant in $t$.)