Calculate the following derivative

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


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

I've read here 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.


Your domain $D_t$ is quite simple, you should be able to compute whatever you need. But also because of how simple it is you can 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 might be called 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 $t$.)

  • $\begingroup$ Thank you for the wonderful idea of change of variables. I didn't think of the problem in this way! I deduce from your answer that $\text{div}(F\mathbf{v})=\partial_xF+\partial_yF$ so $\mathbf{v}=(1,1)$. But I don't understand why this is true. $\endgroup$ – SMA.D Jun 9 '19 at 6:32
  • $\begingroup$ @SMA.D actually the change of variables is the first step in the proof of the general case. It's $(1,1)$ because this is the velocity of the set $D_t$. I suggest drawing a few of the sets if this is not clear. $\endgroup$ – Calvin Khor Jun 9 '19 at 6:44
  • $\begingroup$ Intuitively it's correct. But to drive it analytically should I calculate $\frac{dx}{dt}$ and $\frac{dy}{dt}$ over $(x-t)^2+(y-t)^2=r^2$? $\endgroup$ – SMA.D Jun 9 '19 at 6:49
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    $\begingroup$ I couldn't find this generalization of Leibniz integral rule in intermediate calculus books. Do you know any reference book for this generalization? $\endgroup$ – SMA.D Jun 9 '19 at 7:00
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    $\begingroup$ @SMA.D I've added a reference and a sketch of the proof in the general case. $\endgroup$ – Calvin Khor Jun 9 '19 at 8:06

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