Here are two problems in a pset from MIT 18.02 (multivariable calculus).

Problem 2: Let \(f(x,y,z,t)\) be a smooth function, and let \(\nabla f=\left<f_x,f_y,f_z\right>\)​ be the gradient in space variables only. Let \(\mathbf{r}=\mathbf{r}(t)=\left<x(t),y(t),z(t)\right>\) be a smooth curve, and \(\mathbf{v}=\mathbf{r}’(t)\); and suppose we use the notation \(\frac{\textrm{D}f}{\textrm{D}t}=\frac{\textrm{d}}{\textrm{d}t}f(\mathbf{r}(t),t)\).

Use the Chain Rule to show that \(\frac{\textrm{D}f}{\textrm{D}t}=\frac{\partial f}{\partial t}+\mathbf{v}\cdot\nabla f\).

Solution: We have: \[f(\mathbf{r}(t),t)=f(x(t),y(t),z(t),t)\] By the Chain Rule: \[\frac{\textrm{d}}{\textrm{d}t}f(x(t),y(t),z(t),t)=\frac{\partial f}{\partial x}\frac{\textrm{d}x}{\textrm{d}t}+\frac{\partial f}{\partial y}\frac{\textrm{d}y}{\textrm{d}t}+\frac{\partial f}{\partial z}\frac{\textrm{d}z}{\textrm{d}t}+\frac{\partial f}{\partial t}\] Which becomes: \[\frac{\textrm{d}}{\textrm{d}t}f(x(t),y(t),z(t),t)=\frac{\partial f}{\partial t}+\left<\frac{\textrm{d}x}{\textrm{d}t},\frac{\textrm{d}y}{\textrm{d}t},\frac{\textrm{d}z}{\textrm{d}t}\right>\cdot\left<\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right>\] Which is: \[\frac{\partial f}{\partial t}+\mathbf{v}\cdot\nabla f\] As desired. \(\square\)

This function, \(\frac{\textrm{D}f}{\textrm{D}t}\), is called the convective derivative or the material derivative. In fact, there are quite a few names for this. It is important to realize that \(\mathbf{r}\) defines a path or trajectory through space. The function \(f\) then describes something that is changing along a trajectory with time. The next problem makes this clear, letting \(f=\rho\), the density of a fluid.

When \(\rho\) is constant in \(t\), the flow is termed steady. Unsteady flow, as one can imagine by this definition, must be enormously complicated, and it includes phenomena such as turbulence. In the case of steady flow, each trajectory is called a streamline.

A fluid flow is called incompressible if the convective derivative of \(\rho\) is zero. In steady flow, this means that there is no pressure change along a streamline (which makes sense!).

Problem 3a: Suppose that the density function depends only on time \(t\) but is constant in the space variables \((x,y,z)\), that is, \(\rho=\rho(t)\). Then show that the flow is incompressible if and only if the density \(\rho(t)\) is constant in all the variables \((x,y,z,t)\) (in other words, the flow must be steady).

Solution: We want: \[\frac{\partial \rho}{\partial t}+\mathbf{v}\cdot\nabla\rho=0\] But since \(\rho\) does not depend on spatial variables, \(\nabla\rho=0\) and \(\frac{\partial \rho}{\partial t}=\frac{\textrm{d}\rho}{\textrm{d}t}\). Hence: \[\frac{\textrm{d}\rho}{\textrm{d}t}=0\] Integrating both sides WRT \(t\): \[\int{\frac{\textrm{d}\rho}{\textrm{d}t}\textrm{ d}t}=\rho(t)=C\] Hence, the flow is steady. \(\square\)

Problem 3b: Next suppose instead that the density depends only on the space variables \((x,y,z)\) but not (explicitly) on \(t\), so that \(\rho=\rho(x,y,z)\). An incompressible flow in this case is called stratified. Use the result of problem 2 to give the condition on \(\rho\) and \(\mathbf{v}\) for stratified flow.

Solution: In this case: \[\mathbf{v}\cdot\nabla\rho=0\] So the velocity is always orthogonal to direction in which \(\rho\) changes the most. But recall that \(\nabla\rho\) is always orthogonal to the contour surfaces of \(\rho\). It follows that the velocity must always be parallel, and hence tangential to, surfaces of equal density. \(\square\)