Exercises — Divergence — definition, physical meaning (flux density)

A scalar (one number at each point). The dot in $\nabla \cdot \vec{F}$ multiplies matching components and adds them: $\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}$. Adding numbers gives a number, not an arrow.

$\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z}=1+1+1=\boxed{3}.$ Each component depends only on its own variable, so each derivative is $1$. Positive and constant → uniform outward spreading everywhere (an expanding gas).

$\frac{\partial (-x)}{\partial x}+\frac{\partial (-y)}{\partial y}+\frac{\partial (-z)}{\partial z}=-1-1-1=\boxed{-3}.$ Negative everywhere → arrows point inward → a sink (a drain).

All three components are constant, so every partial derivative is $0$: $0+0+0=\boxed{0}.$ Big arrows, but nothing changes across the box: whatever enters the left leaves the right. Size $\ne$ spreading.

Take each diagonal partial (treat the other two variables as constants):

• $\frac{\partial }{\partial x}(x^{2}y)=2xy$ (the $y$ rides along as a constant),
• $\frac{\partial }{\partial y}(y^{2}z)=2yz$,
• $\frac{\partial }{\partial z}(z^{2}x)=2zx$. $\mathrm{div}\vec{F}=\boxed{2xy+2yz+2zx}.$

• $\frac{\partial }{\partial x}\sin x=\cos x$,
• $\frac{\partial }{\partial y}{e}^y=e^y$,
• $\frac{\partial }{\partial z}\ln (1+z^2)=\frac{2z}{1+z^2}$ (chain rule: derivative of the inside $1+z^2$ is $2z$, over the inside). $\mathrm{div}\vec{F}=\cos x+e^y+\frac{2z}{1+z^2}.$ At $(0,0,0)$: $\cos 0+e^0+0=1+1+0=\boxed{2}.$

$\frac{\partial (-y)}{\partial x}+\frac{\partial (x)}{\partial y}+0=0+0+0=\boxed{0}.$ $-y$ contains no $x$, so its $x$-derivative is $0$; $x$ contains no $y$, so its $y$-derivative is $0$. The field is a pure rotation: fluid circles the axis but is neither created nor destroyed. (Its curl, not its divergence, is nonzero — see Curl — rotation density.)

Only $F_1=x^2$ is nonzero, so only the two faces perpendicular to $x$ carry flux.

• Right face $x=+h$: outward normal $+\hat{x}$, so flux $=F_1(h)\times \text{area}=h^2\cdot (2h{)}^2=4h^4$.
• Left face $x=-h$: outward normal $-\hat{x}$, so $\vec{F}\cdot \hat{n}=-F_1(-h)=-(-h{)}^2=-h^2$; flux $=-h^2\cdot (2h{)}^2=-4h^4$. The top/bottom/front/back faces have normal $\pm \hat{y},\pm \hat{z}$, and $F_2=F_3=0$, so they contribute $0$. $\text{Net flux}=4h^4-4h^4=0.$ Divide by volume $(2h{)}^3=8h^3$: flux density $=0/8h^3=0$ for every $h$, so the limit is $\boxed{0}$. This matches $2x{|}_{x=0}=0$. ✓ (Because $x^2$ is symmetric, equal outflow leaves both $x$-faces — no net source at the origin.)

Each partial derivative is itself linear: $\frac{\partial }{\partial x}(F_1+G_1)=\frac{\partial F_1}{\partial x}+\frac{\partial G_1}{\partial x}$. Sum the three diagonal terms and regroup: $\mathrm{div}(\vec{F}+\vec{G})=\sum \frac{\partial (F_i+G_i)}{\partial x_i}=\sum \frac{\partial F_i}{\partial x_i}+\sum \frac{\partial G_i}{\partial x_i}=\mathrm{div}\vec{F}+\mathrm{div}\vec{G}.$ Now with $\vec{F}=(x,y,z)$ (div $=3$ from L1.2) and $\vec{G}=(-y,x,0)$ (div $=0$ from L2.3): $\mathrm{div}(\vec{F}+\vec{G})=3+0=\boxed{3}.$ The swirl adds rotation but no spreading, so the total spreading rate is unchanged.

Write $r=\sqrt{x^2+y^2+z^2}$, so $\vec{F}=(x{r}^{-3},y{r}^{-3},z{r}^{-3})$. We need $\partial (x{r}^{-3})/\partial x$. First, $\frac{\partial r}{\partial x}=\frac{x}{r}$ (from $r^2=x^2+y^2+z^2$, differentiate: $2r{r}_x=2x$). Product + chain rule: $\frac{\partial }{\partial x}\left(x{r}^{-3}\right)=1\cdot r^{-3}+x\cdot (-3)r^{-4}\cdot \frac{x}{r}=r^{-3}-\frac{3x^2}{r^5}.$ By symmetry the $y$- and $z$-terms give $r^{-3}-\frac{3y^2}{r^5}$ and $r^{-3}-\frac{3z^2}{r^5}$. Add: $\mathrm{div}\vec{F}=3r^{-3}-\frac{3(x^2+y^2+z^2)}{r^5}=\frac{3}{r^3}-\frac{3r^2}{r^5}=\frac{3}{r^3}-\frac{3}{r^3}=\boxed{0}.$ Meaning: away from the charge the field spreads out geometrically, but its magnitude drops exactly fast enough ($\propto 1/r^2$) that outflow $=$ inflow across any tiny box — zero divergence. All the "source" is concentrated at the origin, where the formula blows up (a job for the Divergence Theorem (Gauss)).

We want ${\nabla }^{2}f=f_{xx}+f_{yy}+f_{zz}=6$. Try $f=x^2+y^2+z^2$:

• $f_x=2x\Rightarrow f_{xx}=2$, and likewise $f_{yy}=2$, $f_{zz}=2$. ${\nabla }^{2}f=2+2+2=6.✓$ So $\boxed{f=x^2+y^2+z^2}$ works. Its gradient is $\vec{F}=(2x,2y,2z)$, whose divergence is $2+2+2=6$ — matching. (Any $f=x^2+y^2+z^2+\text{harmonic}$ also works.)

We need $\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}=0$. Choose $F_1=x$ so $\partial F_1/\partial x=1$; then we need $\partial F_2/\partial y=-1$, e.g. $F_2=-y$. Thus $\boxed{\vec{F}=(x,-y)}.$ Check: $1+(-1)=0$. This is a saddle/stretch flow — it stretches along $x$ and squeezes along $y$, so the fluid is stretched into an ellipse but its area (volume) is preserved. It is not zero and not the circular swirl $(-y,x)$; incompressible yet non-trivial. This "area-preserving stretch" is exactly the kind of flow that keeps the Continuity equation satisfied for an incompressible fluid.

Here $F_1=g(x)$, $F_2=F_3=0$, so $\mathrm{div}\vec{F}=g^{\prime }(x)$. Setting $g^{\prime }(x)=c$ and integrating: $g(x)=cx+b,b\text{any constant}.$ So $\boxed{\vec{F}=(cx+b)\hat{x}}$. The slope of the speed profile is the divergence: a linearly speeding-up flow ($c>0$) creates fluid at a constant rate; the additive constant $b$ is a uniform drift that contributes nothing (as in L1.4).

Proof. The $i$-th diagonal term is $\frac{\partial }{\partial x_i}(\varphi F_i)$. By the ordinary product rule for one derivative: $\frac{\partial }{\partial x_i}(\varphi F_i)=\frac{\partial \varphi }{\partial x_i}{F}_i+\varphi \frac{\partial F_i}{\partial x_i}.$ Sum over $i=1,2,3$: $\mathrm{div}(\varphi \vec{F})=\underset{\nabla \varphi \cdot \vec{F}}{\underbrace{\sum _i\frac{\partial \varphi }{\partial x_i}{F}_i}}+\underset{\varphi \mathrm{div}\vec{F}}{\underbrace{\varphi \sum _i\frac{\partial F_i}{\partial x_i}}}.$ Application. With $\varphi =r^2$, $\nabla \varphi =(2x,2y,2z)=2\vec{F}$, $\mathrm{div}\vec{F}=3$, and $\vec{F}=(x,y,z)$: $\mathrm{div}(r^2\vec{F})=r^2\cdot 3+(2x,2y,2z)\cdot (x,y,z)=3r^2+2(x^2+y^2+z^2)=3r^2+2r^2=\boxed{5r^2}.$ Direct check: $r^2\vec{F}=(r^{2}x,\dots )$, and $\partial (r^{2}x)/\partial x=2x^2+r^2$; summing the three gives $2r^2+3r^2=5r^2$. ✓

Constant density means $\partial \rho /\partial t=0$, so the equation demands $\mathrm{div}({\rho }_0\vec{v})=0$. Since ${\rho }_0$ is constant, $\mathrm{div}({\rho }_0\vec{v})={\rho }_0\mathrm{div}\vec{v}$. Compute $\mathrm{div}\vec{v}=\frac{\partial x}{\partial x}+\frac{\partial (-y)}{\partial y}+0=1-1+0=0.$ So $\mathrm{div}({\rho }_0\vec{v})=0=-\partial \rho /\partial t$. ✓ Yes — the flow is divergence-free (incompressible), so constant density is perfectly consistent: mass is conserved. The stretch-and-squeeze motion preserves volume, so density never changes.

The theorem lets us swap a hard surface integral for an easy volume integral of the divergence. Here $\mathrm{div}\vec{F}=3$ (L1.2), a constant, so $\iint \vec{F}\cdot \hat{n}dS={\iiint }_{V}3dV=3\cdot \text{Vol}(V)=3\cdot \frac{4}{3}\pi R^3=\boxed{4\pi R^3}.$ Sanity check by direct surface reasoning: on the sphere $\hat{n}=(x,y,z)/R$, so $\vec{F}\cdot \hat{n}=(x^2+y^2+z^2)/R=R^2/R=R$, constant; times the sphere's area $4\pi R^2$ gives $R\cdot 4\pi R^2=4\pi R^3$. ✓ Both routes agree — the divergence is the flux density, integrated up.