Solution

Problem 2.1: We are asked to show that the derivative of an integral is given by the following:

$$\frac{d}{dt} \int_{x_1(t)}^{x_2(t)} \phi(x,t) dV = \int_{x_1(t)}^... ...phi}{\partial t} dV + \phi(x_2,t) \frac{dx_2}{dt} - \phi(x_1,t) \frac{dx_1}{dt}$$

We start with the fundemental theorem of calculus and define $\Phi(x,t)$ as the antiderivative of $\phi(x,t)$ with respect to $dV$ (i.e. $dx$ for the 1D case). Then we continue by computing the total derivative with respect to time using the chain rule. Next, we can invoke the fundemental theorem of calculus again to rewrite the equation in terms of $\phi$. Finally, we note that the definition of the partial derivative holds all other variables constant, i.e. $x$, and thus we can move the partial derivative with respect to time inside the integral as both are linear operators with standard continuity assuptions.

$$\frac{d}{dt} \int_{x_1(t)}^{x_2(t)} \phi(x,t) dV = \frac{d}{dt} \left[ \Phi(x_2(t),t) - \Phi(x_1(t),t) \right]$$

$$= \left( \frac{\partial}{\partial t} \left[ \Phi(x_2,t) \right] +... ...ft[ \Phi(x_1, t) \right] \frac{\partial}{\partial t} \left[ x_1 \right] \right)$$

$$= \left( \frac{\partial}{\partial t} \left[ \Phi(x_2,t) \right] +... ...left[ \Phi(x_1,t) \right] + \phi(x_1,t) \frac{\partial x_1}{\partial t} \right)$$

$$= \left( \frac{\partial}{\partial t} \left[ \Phi(x_2,t) \right] -... ...) \frac{\partial x_2}{\partial t} - \phi(x_1,t) \frac{\partial x_1}{\partial t}$$

$$= \left(\left.\frac{\partial}{\partial t} \int_{x_1}^{x_2} \phi(x... ...ight\vert _t + \phi(x_2,t) \frac{dx_2}{dt} - \phi(x_1,t) \frac{dx_1}{dt}\right)$$

$$= \int_{x_1(t)}^{x_2(t)} \frac{\partial \phi}{\partial t} dV + \phi(x_2,t) \frac{dx_2}{dt} - \phi(x_1,t) \frac{dx_1}{dt}$$

Thus, we have derived what is known as Leibniz's integral rule in one dimension.

Problem 2.2: We are now asked to generalize this result to 3D. Thus, we want to show that:

$$\frac{D}{Dt} \int_{V(t)} \phi(\vec{x},t) dV = \int_{V(t)} \frac{\partial \phi} {\partial t} dV + \int_{\partial V(t)} \phi \vec{v}\cdot d\vec{S}$$

We start by writing the integral three equivalent ways in terms of the 3 relevant planes $A_x$, $A_y$, and $A_z$.

$$\frac{D}{Dt} \int_{V(t)} \phi(\vec{x},t) dV = \frac{D}{Dt} \int_{x_1(t)}^{x_2(t)} \int_{S_x(x,t)} \phi(x,y,z,t) dS_x dx$$

$$= \frac{D}{Dt} \int_{y_1(t)}^{y_2(t)} \int_{S_y(y,t)} \phi(x,y,z,t) dS_y dy$$

$$= \frac{D}{Dt} \int_{z_1(t)}^{z_2(t)} \int_{S_z(z,t)} \phi(x,y,z,t) dS_z dz$$

We now evaluate the each of these integrals using the 1D Leibniz integral rule from Problem 2.1. We then apply the fundemental theorem of calculus to rewrite the results in terms of $\phi$. Finally, we combine the results into a single equation using vector notation and the definition of the velocity vector, $\vec{v} = (dx/dt, dy/dt, dz/dt)$ and the del operator, $\nabla = (\partial/\partial x, \partial/\partial y, \partial/\partial z)$.

$$\frac{D}{Dt} \int_{V(t)} \phi(\vec{x},t) dV = \frac{D}{Dt} \int_{x_1(t)}^{x_2(t)} \int_{S_x(x,t)} \phi(x,y,z,t) dS_x dx$$

$$= \int_{V(t)} \frac{\partial \phi}{\partial t} dV + \int_{S_x} \p... ...2,y,z,t) dS_x \frac{dx_2}{dt} - \int_{S_x} \phi(x_1,y,z,t) dS_x \frac{dx_1}{dt}$$

$$= \int_{V(t)} \frac{\partial \phi}{\partial t} dV + \int_{S_y} \p... ...y_2,z,t) dS_y \frac{dy_2}{dt} - \int_{S_y} \phi(x,y_1,z,t) dS_y \frac{dy_1}{dt}$$

$$= \int_{V(t)} \frac{\partial \phi}{\partial t} dV + \int_{S_z} \p... ...y,z_2,t) dS_z \frac{dz_2}{dt} - \int_{S_z} \phi(x,y,z_1,t) dS_z \frac{dz_1}{dt}$$

This set of equations implies that:

$$\int_{S_y} \phi(x,y_2,z,t) dS_y \frac{dy_2}{dt} - \int_{S_y} \phi(x,y_1,z,t) dS_y \frac{dy_1}{dt} = \int_{S_z} \phi(x,y,z_2,t) dS_z \frac{dz_2}{dt} - \int_{S_z} \phi(x,y,z_1,t) dS_z \frac{dz_1}{dt}$$

This means that we can add zero to the first equation without changing its value. We then choose a paramaterization that allows us to combine all six surface terms into a integral (i.e. resulting in the normal vector for each surface). Thus, we have:

$$\frac{D}{Dt} \int_{V(t)} \phi(\vec{x},t) dV = \frac{D}{Dt} \int_{x_1(t)}^{x_2(t)} \int_{S_x(x,t)} \phi(x,y,z,t) dS_x dx$$

$$= \int_{V(t)} \frac{\partial \phi}{\partial t} dV + \int_{S_x} \p... ...2,y,z,t) dS_x \frac{dx_2}{dt} - \int_{S_x} \phi(x_1,y,z,t) dS_x \frac{dx_1}{dt}$$

$$\quad \quad - \left(\int_{S_y} \phi(x,y_2,z,t) dS_y \frac{dy_2}{dt} - \int_{S_y} \phi(x,y_1,z,t) dS_y \frac{dy_1}{dt} \right)$$

$$\quad \quad + \int_{S_z} \phi(x,y,z_2,t) dS_z \frac{dz_2}{dt} - \int_{S_z} \phi(x,y,z_1,t) dS_z \frac{dz_1}{dt}$$

$$=\int_{V(t)} \frac{\partial \phi} {\partial t} dV + \int_{\partial V(t)} \phi \vec{v}\cdot d\vec{S}$$

Problem 2.3: We are now asked to generalize this result to 3D. Thus, we want to show that:

$$\frac{D}{Dt} \int_{V(t)} \phi(\vec{x},t) dV = \int_{V(t)} \frac{\partial \phi}{\partial t} + \nabla \cdot (\vec{v} \phi) dV$$

We start by writing out the integral and expainding it to three explicit integrals over the $x$, $y$, and $z$ dimensions using Fubini's theorem. We define the space of integration as $V = [x_1(t), x_2(t)] \times [y_1(x,t), y_2(x,t)] \times [z_1(x,y,t), z_2(x,y,t)]$. We then apply the 1D Leibniz integral rule to the outermost integral, apply the fundemental theorem of calculus, and then expand using the product rule.

$$\frac{D}{Dt} \int_{V(t)} \phi(\vec{x},t) dV = \frac{D}{Dt} \int_{... ... \int_{y_1(x,t)}^{y_2(x,t)} \int_{z_1(x,y,t)}^{z_2(x,y,t)} \phi(x,y,z,t) dzdydx$$

$$= \int_{V(t)} \frac{\partial \phi}{\partial t} dV + \int_{y_1}^{y... ...2}{dt} - \int_{y_1}^{y_2} \int_{z_1}^{z_2} \phi(x_1,y,z,t) dzdy \frac{dx_1}{dt}$$

$$= \int_{V(t)} \frac{\partial \phi}{\partial t} dV + \int_{x_1}^{x... ...t[\int_{y_1}^{y_2} \int_{z_1}^{z_2} \phi(x,y,z,t) dzdy \frac{dx}{dt} \right] dx$$

$$= \int_{V(t)} \frac{\partial \phi}{\partial t} dV + \int_{x_1}^{x... ... \int_{z_1}^{z_2} \phi(x,y,z,t) dzdy \frac{d}{dx}\left[ \frac{dx}{dt}\right] dx$$

Now, we need to apply Leibniz integral rule but now with respect to $x$. We then use the fundemental theorem of calculus to rewrite the result in terms of an integral and apply the product rule to expand the expression.

$$\frac{d}{dx} \int_{y_1}^{y_2} \int_{z_1}^{z_2} \phi(x,y,z,t) dzdy... ...2}\frac{d}{dy} \left[ \int_{z_1}^{z_2} \phi(x,y,z,t) dz \frac{dy}{dx}\right] dy$$

$$= \int_{y_1}^{y_2}\int_{z_1}^{z_2} \frac{\partial \phi}{\partial ... ..._2} \int_{z_1}^{z_2} \phi(x,y,z,t) dz \frac{d}{dy}\left[\frac{dy}{dx}\right] dy$$

We use Leibniz integral rule again, this time with respect to $y$. We then use the fundemental theorem of calculus to rewrite the result in terms of an integral and apply the product rule.

$$\frac{d}{dy} \int_{z_1}^{z_2} \phi(x,y,z,t) dz = \int_{z_1}^{z_2} \frac{\partial}{\partial y} \phi(x,y,z,t) dz + \int_{z_1}^{z_2} \frac{d}{dz} \left[ \phi(x,y,z,t) \frac{dz}{dy} \right] dz$$

$$= \int_{z_1}^{z_2} \frac{\partial \phi}{\partial y} dz + \int_{z_... ...} dz + \int_{z_1}^{z_2} \phi(x,y,z,t) \frac{d}{dz}\left[\frac{dz}{dy}\right] dz$$

Now, we need to substitute these results back into the original equations and combine the terms into vector operations. We then undo the product rule using the chain rule and then apply the definition of the velocity vector, $\vec{v} = (dx/dt, dy/dt, dz/dt)$ and the del operator, $\nabla = (\partial/\partial x, \partial/\partial y, \partial/\partial z)$. Lastly, we use the product rule to combine the terms into the final result.

$$\frac{D}{Dt} \int_{V(t)} \phi(\vec{x},t) dV$$

$$= \int_{V(t)} \frac{\partial \phi}{\partial t} + \frac{\partial \... ...dt} + \phi \frac{d}{dz}\left[ \frac{dz}{dy}\frac{dy}{dx}\frac{dx}{dt}\right] dV$$

$$= \int_{V(t)} \frac{\partial \phi}{\partial t} + \frac{\partial \... ...hi}{\partial z} \frac{dz}{dt} + \phi \frac{d}{dz}\left[ \frac{dz}{dt}\right] dV$$

$$= \int_{V(t)} \frac{\partial \phi}{\partial t} + \nabla \phi \cdot \vec{v} + \phi (\nabla \cdot \vec{v}) dV$$

$$= \int_{V(t)} \frac{\partial \phi}{\partial t} + \nabla \cdot (\vec{v} \phi) dV$$