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

Ek scalar (har point par ek number). $\nabla \cdot \vec{F}$ mein dot matching components ko multiply karke unhe add karta hai: $\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}$. Numbers ko add karne se ek number milta hai, koi arrow nahi.

$\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z}=1+1+1=\boxed{3}.$ Har component sirf apne variable par depend karta hai, isliye har derivative $1$ hai. Positive aur constant → har jagah uniform outward spreading (ek expanding gas).

$\frac{\partial (-x)}{\partial x}+\frac{\partial (-y)}{\partial y}+\frac{\partial (-z)}{\partial z}=-1-1-1=\boxed{-3}.$ Har jagah negative → arrows andar ki taraf point kar rahe hain → ek sink (drain).

Teeno components constant hain, isliye har partial derivative $0$ hai: $0+0+0=\boxed{0}.$ Bade arrows hain, lekin box ke across kuch bhi change nahi hota: jo left se enter karta hai woh right se nikalta hai. Size $\ne$ spreading.

Har diagonal partial lo (baaki do variables ko constants maano):

• $\frac{\partial }{\partial x}(x^{2}y)=2xy$ ($y$ ek constant ki tarah saath chalta hai),
• $\frac{\partial }{\partial y}(y^{2}z)=2yz$,
• $\frac{\partial }{\partial z}(z^{2}x)=2zx$. $\mathrm{div}\vec{F}=\boxed{2xy+2yz+2zx}.$

• $\frac{\partial }{\partial x}\sin x=\cos x$,
• $\frac{\partial }{\partial y}{e}^y=e^y$,
• $\frac{\partial }{\partial z}\ln (1+z^2)=\frac{2z}{1+z^2}$ (chain rule: andar ka derivative $1+z^2$ hai jo $2z$ hai, andar se divide). $\mathrm{div}\vec{F}=\cos x+e^y+\frac{2z}{1+z^2}.$ $(0,0,0)$ par: $\cos 0+e^0+0=1+1+0=\boxed{2}.$

$\frac{\partial (-y)}{\partial x}+\frac{\partial (x)}{\partial y}+0=0+0+0=\boxed{0}.$ $-y$ mein koi $x$ nahi hai, isliye uska $x$-derivative $0$ hai; $x$ mein koi $y$ nahi hai, isliye uska $y$-derivative $0$ hai. Yeh field pure rotation hai: fluid axis ke around ghoomta hai lekin na create hota hai, na destroy. (Iska curl, divergence nahi, nonzero hai — dekho Curl — rotation density.)

Sirf $F_1=x^2$ nonzero hai, isliye sirf do faces jo $x$ ke perpendicular hain woh flux carry karti hain.

• Right face $x=+h$: outward normal $+\hat{x}$, toh flux $=F_1(h)\times \text{area}=h^2\cdot (2h{)}^2=4h^4$.
• Left face $x=-h$: outward normal $-\hat{x}$, toh $\vec{F}\cdot \hat{n}=-F_1(-h)=-(-h{)}^2=-h^2$; flux $=-h^2\cdot (2h{)}^2=-4h^4$. Top/bottom/front/back faces ka normal $\pm \hat{y},\pm \hat{z}$ hai, aur $F_2=F_3=0$, isliye unka contribution $0$ hai. $\text{Net flux}=4h^4-4h^4=0.$ Volume $(2h{)}^3=8h^3$ se divide karo: flux density $=0/8h^3=0$ har $h$ ke liye, isliye limit $\boxed{0}$ hai. Yeh $2x{|}_{x=0}=0$ se match karta hai. ✓ (Kyunki $x^2$ symmetric hai, equal outflow dono $x$-faces se nikalta hai — origin par koi net source nahi.)

Har partial derivative khud linear hai: $\frac{\partial }{\partial x}(F_1+G_1)=\frac{\partial F_1}{\partial x}+\frac{\partial G_1}{\partial x}$. Teen diagonal terms ko sum karo aur regroup karo: $\mathrm{div}(\vec{F}+\vec{G})=\sum \frac{\partial (F_i+G_i)}{\partial x_i}=\sum \frac{\partial F_i}{\partial x_i}+\sum \frac{\partial G_i}{\partial x_i}=\mathrm{div}\vec{F}+\mathrm{div}\vec{G}.$ Ab $\vec{F}=(x,y,z)$ ke saath (div $=3$ from L1.2) aur $\vec{G}=(-y,x,0)$ ke saath (div $=0$ from L2.3): $\mathrm{div}(\vec{F}+\vec{G})=3+0=\boxed{3}.$ Swirl rotation add karta hai lekin koi spreading nahi, isliye total spreading rate unchanged rehti hai.

$r=\sqrt{x^2+y^2+z^2}$ likho, toh $\vec{F}=(x{r}^{-3},y{r}^{-3},z{r}^{-3})$. Hume $\partial (x{r}^{-3})/\partial x$ chahiye. Pehle, $\frac{\partial r}{\partial x}=\frac{x}{r}$ (from $r^2=x^2+y^2+z^2$, differentiate karo: $2r{r}_x=2x$). Product + chain rule: $\frac{\partial }{\partial x}\left(x{r}^{-3}\right)=1\cdot r^{-3}+x\cdot (-3)r^{-4}\cdot \frac{x}{r}=r^{-3}-\frac{3x^2}{r^5}.$ Symmetry se $y$- aur $z$-terms $r^{-3}-\frac{3y^2}{r^5}$ aur $r^{-3}-\frac{3z^2}{r^5}$ dete hain. Add karo: $\mathrm{div}\vec{F}=3r^{-3}-\frac{3(x^2+y^2+z^2)}{r^5}=\frac{3}{r^3}-\frac{3r^2}{r^5}=\frac{3}{r^3}-\frac{3}{r^3}=\boxed{0}.$ Matlab: charge se door field geometrically spread hoti hai, lekin uski magnitude exactly utni tezi se drop hoti hai ($\propto 1/r^2$) ki outflow $=$ inflow kisi bhi tiny box ke across — zero divergence. Saara "source" origin par concentrated hai, jahan formula blow up hota hai (yeh kaam Divergence Theorem (Gauss) ka hai).

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

• $f_x=2x\Rightarrow f_{xx}=2$, aur similarly $f_{yy}=2$, $f_{zz}=2$. ${\nabla }^{2}f=2+2+2=6.✓$ Toh $\boxed{f=x^2+y^2+z^2}$ kaam karta hai. Iska gradient $\vec{F}=(2x,2y,2z)$ hai, jiska divergence $2+2+2=6$ hai — match karta hai. (Koi bhi $f=x^2+y^2+z^2+\text{harmonic}$ bhi kaam karega.)

Hume $\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}=0$ chahiye. Choose karo $F_1=x$ toh $\partial F_1/\partial x=1$; phir hume $\partial F_2/\partial y=-1$ chahiye, jaise $F_2=-y$. Isliye $\boxed{\vec{F}=(x,-y)}.$ Check karo: $1+(-1)=0$. Yeh ek saddle/stretch flow hai — yeh $x$ ke along stretch karta hai aur $y$ ke along squeeze karta hai, isliye fluid ek ellipse mein stretch hota hai lekin uska area (volume) preserve hota hai. Yeh zero nahi hai aur circular swirl $(-y,x)$ bhi nahi hai; incompressible phir bhi non-trivial. Yeh "area-preserving stretch" exactly woh flow hai jo incompressible fluid ke liye Continuity equation satisfy rakhta hai.

Yahan $F_1=g(x)$, $F_2=F_3=0$, toh $\mathrm{div}\vec{F}=g^{\prime }(x)$. $g^{\prime }(x)=c$ set karke integrate karo: $g(x)=cx+b,b\text{koi bhi constant}.$ Toh $\boxed{\vec{F}=(cx+b)\hat{x}}$. Speed profile ki slope divergence hai: linearly speeding-up flow ($c>0$) fluid ko constant rate par create karta hai; additive constant $b$ ek uniform drift hai jo kuch contribute nahi karta (jaise L1.4 mein).

Proof. $i$-th diagonal term $\frac{\partial }{\partial x_i}(\varphi F_i)$ hai. Ek derivative ke liye ordinary product rule se: $\frac{\partial }{\partial x_i}(\varphi F_i)=\frac{\partial \varphi }{\partial x_i}{F}_i+\varphi \frac{\partial F_i}{\partial x_i}.$ $i=1,2,3$ par sum karo: $\mathrm{div}(\varphi \vec{F})=\underset{\nabla \varphi \cdot \vec{F}}{\underbrace{\sum _i\frac{\partial \varphi }{\partial x_i}{F}_i}}+\underset{\varphi \mathrm{div}\vec{F}}{\underbrace{\varphi \sum _i\frac{\partial F_i}{\partial x_i}}}.$ Application. $\varphi =r^2$ ke saath, $\nabla \varphi =(2x,2y,2z)=2\vec{F}$, $\mathrm{div}\vec{F}=3$, aur $\vec{F}=(x,y,z)$: $\mathrm{div}(r^2\vec{F})=r^2\cdot 3+(2x,2y,2z)\cdot (x,y,z)=3r^2+2(x^2+y^2+z^2)=3r^2+2r^2=\boxed{5r^2}.$ Direct check: $r^2\vec{F}=(r^{2}x,\dots )$, aur $\partial (r^{2}x)/\partial x=2x^2+r^2$; teeno ko sum karne se $2r^2+3r^2=5r^2$ milta hai. ✓

Constant density matlab $\partial \rho /\partial t=0$, isliye equation demand karta hai $\mathrm{div}({\rho }_0\vec{v})=0$. Kyunki ${\rho }_0$ constant hai, $\mathrm{div}({\rho }_0\vec{v})={\rho }_0\mathrm{div}\vec{v}$. Compute karo $\mathrm{div}\vec{v}=\frac{\partial x}{\partial x}+\frac{\partial (-y)}{\partial y}+0=1-1+0=0.$ Toh $\mathrm{div}({\rho }_0\vec{v})=0=-\partial \rho /\partial t$. ✓ Haan — flow divergence-free (incompressible) hai, isliye constant density perfectly consistent hai: mass conserved hai. Stretch-and-squeeze motion volume preserve karta hai, isliye density kabhi change nahi hoti.

Theorem hume ek mushkil surface integral ko divergence ka ek easy volume integral se swap karne deta hai. Yahan $\mathrm{div}\vec{F}=3$ (L1.2), ek constant, isliye $\iint \vec{F}\cdot \hat{n}dS={\iiint }_{V}3dV=3\cdot \text{Vol}(V)=3\cdot \frac{4}{3}\pi R^3=\boxed{4\pi R^3}.$ Sanity check direct surface reasoning se: sphere par $\hat{n}=(x,y,z)/R$, toh $\vec{F}\cdot \hat{n}=(x^2+y^2+z^2)/R=R^2/R=R$, constant; sphere ki area $4\pi R^2$ se multiply karne par $R\cdot 4\pi R^2=4\pi R^3$ milta hai. ✓ Dono routes agree karte hain — divergence hai flux density, integrate kiya hua.