Time derivatives following the motion

Let$H(x,t)$ be some property of a fluid, for example the density, $\rho (x, t)$. There are two different time derivatives that we must distinguish between. The first is the rate of change of $H$ with respect to $t$ at a fixed point in space,

$${\partial H\over \partial t} = \lim_{\delta t \rightarrow 0}{\delta H \over \delta t},$$

where the partial derivative is obtained by keeping the other variables constant (in this case the spatial derivatives). The second is

$${DH\over Dt},$$

which is defined as the rate of change of $H$ with respect to time but attached to a particular fluid elemnet. Thus we consider a particular fluid element and as it moves due to a fluid velocity its position also moves. Thus, moving with the fluid, we calculate how $H$ changes in time but not at a fixed point in space. This `moving with the fluid'is illustrated in Figure 2.3.

Figure 2.3: In response to the velocity component $v_{x}(x,t)$ the fluid element moves from $x$ to $x + v_{x}(x,t)\delta t$.

Thus,

$$\hbox{the rate of change of }H = {H(x + v_{x}\delta t, t + \delta t) - H(x,t)\over \delta t},$$

becomes, in the limit of $\delta t$ tending to zero,

$${DH\over Dt} = {H(x,t) - H(x,t)\over \delta t} + v_{x}{\par... ...ver \partial x} + {\partial H \over \partial t} + O(\delta t).$$

Hence, the final result, as $\delta t$ tends to zero is

$${DH\over Dt} = {\partial H\over \partial t} + v_{x}{\partial H \over \partial x}.$$

In three dimensional problem with $H(x,y,z,t)$ and ${\bf v} = (v_{x}, v_{y}, v_{z})$, the obvious generalisation is simply

$${DH\over Dt} = {\partial H\over \partial t} + v_{x}{\partia... ...al H \over \partial y} + v_{z}{\partial H \over \partial z}.$$

Finally, this may be written in vector form as

$${DH\over Dt} = {\partial H\over \partial t} + {\bf v}\cdot \nabla H.$$ (2.6)

$DH/Dt$ is also called the total time derivative.

Note, you must be very clear about your notation and know the difference between $\partial H/\partial t$ and $DH/Dt$.

Example 2. .1Consider the density profile as $\rho (x,t) = x e^{-2t}$ and the position $x(t)$. If there is a flow so that

$$v_{x} = {dx\over dt} = x,$$

where the form of $v_{x}$ has been selected as simplpy $x$. This equation may be solved to give the position as a function of time as

$$x = x_{0}e^{t},$$

where $x_{0}$ is a constant giving the initial position. Note that $x$ is only a function of time and so we do not need to use partial derivatives in calculating $v_{x}$. Now we calulate the total time derivative of $\rho$ as

$${D\rho \over Dt} = {\partial\rho \over \partial t} + v_{x}{... ...\rho \over \partial x} = - 2xe^{-2t} + xe^{-2t} = - x e^{-2t}.$$

Note that we may eliminate $x$ from the expression for $\rho$ and get $\rho$ purely as a function of $t$, namely,

$$\rho = x_{0}e^{t}e^{-2t} = x_{0}e^{-t}, \qquad {D\rho \over Dt} = -x_{0}e^{-t} = - x e^{-2t},$$

as before.

Example 2. .2If $\rho = x^{2}e^{-2t}$, use the mass continuity equation to determine $v(x)$ satisfying $v(0) = 0$.

$${\partial \rho \over \partial t} + {\partial\over \partial ... ...ho \over \partial x} + \rho {\partial v\over \partial x} = 0.$$

Now the partial derivatives can be evaluated as

$${\partial \rho \over \partial t} = - 2x^{2}e^{-2t}, \qquad {\partial \rho \partial x} = 2 x e^{-2t}.$$

Hence,

$$-2 x^{2} + 2x v + x^{2}{dv\over dx} = 0,$$

and so

$$\begin{eqnarray*} {d\over dx}\left (x^{2} v\right ) & = & 2x^{2} \\ x^{2}v & ... ... \qquad v(0) = 0, \Rightarrow C = 0 \\ v & = & {2\over 3} x. \end{eqnarray*}$$