Exercises — Streamlines, pathlines, streaklines

(a) Streakline — bahut saare puffs, ek source (chimney), sab abhi dekhe ja rahe hain. (b) Pathline — ek object (balloon), uska poora trajectory time ke saath. (c) Streamline — ek single instant ka direction field; har jagah $\vec{v}$ ke tangent. Mnemonic check: SNAP, TRAIL, SMOKE. (c)=SNAP, (b)=TRAIL, (a)=SMOKE.

False. Time frozen hota hai — yeh ek constant parameter hai, integration ka variable nahi. Hum sirf ek instant par spatial relationship integrate karte hain. (Time ke upar integrate karna woh kaam hai jo pathline karti hai.)

Yeh sab identical hain. Ek steady field ($\partial \vec{v}/\partial t=0$) kabhi nahi badalti, isliye ek snapshot, ek particle ki trail, aur ek smoke plume ke beech koi "history" ka fark nahi hota. Yeh parent note ka golden rule hai. Dekho Steady vs unsteady flow.

Step 1 — ODE likho. $\frac{dx}{2}=\frac{dy}{3}$. Kyun: $u=2,v=3$ ko $\frac{dx}{u}=\frac{dy}{v}$ mein plug karo. Yahan dono components nonzero constants hain ($u=2\ne 0$, $v=3\ne 0$), isliye is instant mein koi degenerate points nahi hain aur $u$ aur $v$ se divide karna har jagah legal hai — isliye compact form is problem ke liye safe hai. Step 2 — cross-multiply. $3dx=2dy$. Kyun: dono sides ko $2\cdot 3=6$ se multiply karo. Kyunki $u,v$ constants hain, hum unhe differentials ke past freely move kar sakte hain; yeh step exactly safe cross-multiplied form $vdx=udy$ hai jahan $v=3,u=2$. Step 3 — integrate karo. $3x=2y+C\Rightarrow y=\frac{3}{2}x+C^{\prime }$. Result: slope $\frac{3}{2}$ wali parallel straight lines — exactly $\vec{v}=(2,3)$ ki direction (rise $3$ over run $2$).

Figure kya dikhata hai. Neeche diye figure mein, blue arrows constant velocity $\vec{v}=(2,3)$ hain jo ek grid of points par draw kiye gaye hain — notice karo ki har arrow identical hai (uniform flow). Orange lines streamlines $y=\frac{3}{2}x+C$ hain; kisi bhi ek ko follow karo aur tum dekhoge ki blue arrows exactly uske along hain (tangent = velocity). Red annotation slope $v/u=3/2$ mark karta hai: har run $2$ ke liye rise $3$. Yeh "tangent parallel to $\vec{v}$" ka visual meaning hai.

Step 1 — ODE. $\frac{dx}{x}=\frac{dy}{y}$, valid jahan $x\ne 0$ aur $y\ne 0$. Step 2 — dono sides integrate karo. $\ln |x|=\ln |y|+C$. Kyun: $\int \frac{dx}{x}=\ln |x|$; yeh natural log hai kyunki integrand $1/x$ hai. Step 3 — exponentiate karo. $|x|=e^C|y|\Rightarrow y=kx$ constant $k$ ke liye. Edge cases. $y$-axis par $x=0$ isliye $u=0$: $vdx=udy\Rightarrow ydx=0\Rightarrow dx=0$ par fall back karo, ek vertical streamline — $y$-axis khud. $x$-axis par $y=0$ isliye $v=0$: $ydx=xdy\Rightarrow xdy=0\Rightarrow dy=0$, ek horizontal streamline — $x$-axis. Origin par dono $u=v=0$: ek stagnation point jahan saari radial lines milti hain. Result: origin se guzarne wali straight lines ($y=kx$ family aur dono axes). Flow radially outward push karta hai, isliye streamlines radial spokes hain. ✔ (Continuity equation intuition se match karta hai: fluid rays ke along bahar stream kar raha hai.)

Step 1. $\frac{dx}{dt}=1\Rightarrow x=t$. Step 2. $\frac{dy}{dt}=2\Rightarrow y=2t$. Step 3 — $t$ eliminate karo. $t=x$, isliye $y=2x$. Result: line $y=2x$. Kyunki flow steady hai, yeh pathline ek streamline bhi hai (slope $v/u=2$). ✔

Step 1 — $t=2$ freeze karo, toh $v=2$ ek constant hai. ODE: $\frac{dx}{1}=\frac{dy}{2}$. (Dono $u=1$ aur $v=2$ nonzero hain, isliye is instant mein koi degenerate points nahi hain.) Kyun: streamlines ka golden rule — $t$ ko ek fixed number treat karo, yahan $2$. Step 2 — integrate karo. $2x=y+C\Rightarrow y=2x+C$. Result: $t=2$ par streamlines slope $2$ wali straight lines hain. Isse pathline (neeche) se compare karo — yeh alag hain!

Step 1. $\frac{dx}{dt}=1\Rightarrow x=t$ (kyunki $x_0=0,t_0=0$). Step 2. $\frac{dy}{dt}=t\Rightarrow y=\frac{1}{2}{t}^2$. Step 3 — $t$ eliminate karo. $t=x$, isliye $y=\frac{1}{2}{x}^2$. Result: ek parabola, straight line nahi. Toh pathline (curve) ≠ $t=2$ streamline (line). Yahi unsteady flow ka poora point hai.

Figure kya dikhata hai. Neeche diya figure is field ke liye teeno curves ko ek axis par overlay karta hai taaki split ek nazar mein visible ho: blue straight line $t=2$ streamline $y=2x$ hai (ek snapshot), orange parabola yeh pathline $y=\frac{1}{2}{x}^2$ hai (ek particle ki poori trail), aur green parabola L3.3 ki streakline hai. Red dot at $(2,2)$ woh single particle hai jo $\tau =0$ par release hua tha — notice karo ki pathline aur streakline dono isse pass karti hain, kyunki yeh dono constructions mein common real particle hai. Gray origin se curves ko bahar follow karo aur tum dekh sakte ho ki ek snapshot (line) genuinely unsteady flow mein ek history (parabolas) se alag hoti hai.

Step 1 — time $\tau$ par release hue particle ki pathline (origin par $t=\tau$ par start): $x={\int }_{\tau }^{t}1d{t}^{\prime }=t-\tau ,y={\int }_{\tau }^t{t}^{\prime }d{t}^{\prime }=\frac{1}{2}(t^2-{\tau }^2).$ Step 2 — $t=2$ fix karo, $\tau \le 2$ sweep karo. $x=2-\tau ,y=\frac{1}{2}(4-{\tau }^2).$ Step 3 — $\tau$ eliminate karo. $x=2-\tau$ se: $\tau =2-x$. Phir $y=\frac{1}{2}(4-(2-x{)}^2)=\frac{1}{2}(4-4+4x-x^2)=2x-\frac{1}{2}{x}^2.$ Result: streakline $y=2x-\frac{1}{2}{x}^2$ hai, pathline $y=\frac{1}{2}{x}^2$ se alag parabola. Teeno curves origin se ab alag hain — prove karta hai ki flow unsteady hai. ✔ (Yeh L3.2 figure mein green curve hai.)

Quick sanity checks: $x=0$ par (yaani $\tau =2$, abhi release hua particle), $y=0$ ✔. $\tau =0$ par release hua pathline wala particle $t=2$ par $(x,y)=(2,2)$ par hai; streakline $x=2$ par $y=2(2)-\frac{1}{2}(4)=2$ deta hai ✔ — dono curves us shared particle par milni hi chahiye.

Steadiness: $u=-y$ aur $v=x$ mein koi explicit $t$ nahi hai, isliye $\partial \vec{v}/\partial t=0$. Flow steady hai. Golden rule: steady ⇒ pathlines = streamlines. Isliye pathlines same concentric circles $x^2+y^2=R^2$ hain. Radius $R$ par ek particle us circle par forever orbit karta hai. Edge note: origin par $u=-0=0$ aur $v=0$, isliye origin ek stagnation point hai — swirl ka centre, jahan fluid move nahi karta aur koi unique streamline direction exist nahi karta. (Yeh Lagrangian vs Eulerian description se connect hota hai: Eulerian field aur Lagrangian trajectory yahan agree karte hain.)

Step 1 — "constant $\psi$" ka matlab. Ek curve ke along jahan $\psi$ nahi badalta, uska total differential zero hai: $d\psi =\frac{\partial \psi }{\partial x}dx+\frac{\partial \psi }{\partial y}dy=0.$ Step 2 — definitions substitute karo. $\frac{\partial \psi }{\partial x}=-v$ aur $\frac{\partial \psi }{\partial y}=u$, isliye $-vdx+udy=0\Rightarrow vdx-udy=0\Rightarrow \frac{dx}{u}=\frac{dy}{v}.$ Step 3 — pehchano. Yeh exactly streamline ODE hai — aur notice karo yeh naturally safe cross-multiplied form $vdx-udy=0$ mein aata hai, jo tab bhi valid rehta hai jahan $u=0$ ya $v=0$ ho. Isliye constant $\psi$ ki lines streamlines hain. Yahi wajah hai ki Stream function ψ itna powerful shortcut hai — tum streamlines sirf ek scalar contour karke draw kar sakte ho.

Step 1. Chahiye $u=\frac{\partial \psi }{\partial y}=-y\Rightarrow \psi =-\frac{1}{2}{y}^2+f(x)$. Step 2. Chahiye $v=-\frac{\partial \psi }{\partial x}=x\Rightarrow \frac{\partial \psi }{\partial x}=-x\Rightarrow f(x)=-\frac{1}{2}{x}^2+\text{const}$. Step 3. Toh $\psi =-\frac{1}{2}(x^2+y^2)$. Contours $\psi =\text{const}$ dete hain $x^2+y^2=\text{const}$ — concentric circles, parent note se match karta hai. ✔

Step 1 — curves ki slope nikalo. $xy=C$ differentiate karo: $y+x\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{y}{x}$. Step 2 — $\frac{dy}{dx}=\frac{v}{u}$ se match karo. Chahiye $\frac{v}{u}=-\frac{y}{x}$. Sabse simple choice: $u=x,v=-y$. Step 3 — verify karo. Streamline ODE $\frac{dx}{x}=\frac{dy}{-y}\Rightarrow \ln |x|=-\ln |y|+C\Rightarrow xy=\text{const}$. ✔ Edge cases: axes par ek component vanish ho jaata hai — origin phir se ek stagnation point hai ($u=v=0$), aur axes khud degenerate streamlines $C=0$ hain. Result: $\vec{v}=(x,-y)$ (ek "saddle" / straining flow) hyperbolic streamlines deta hai. Koi bhi scalar multiple $\vec{v}=(kx,-ky)$ bhi kaam karta hai — direction, magnitude nahi, streamline shape set karta hai.

(a) Streamline at $t=2$ ($t$ freeze karo, toh $u=2$): $\frac{dx}{2}=\frac{dy}{1}\Rightarrow x=2y+C$. Origin se: $x=2y$, yaani $y=\frac{1}{2}x$ — ek straight line.

(b) Pathline ($\tau =0$ par release): $\frac{dx}{dt}=t\Rightarrow x=\frac{1}{2}{t}^2$; $\frac{dy}{dt}=1\Rightarrow y=t$. $t=y$ eliminate karo: $x=\frac{1}{2}{y}^2$ — ek sideways parabola. $t=2$ par particle $(2,2)$ par hai.

(c) Streakline at $t=2$ origin se. $\tau$ par release hua particle: $x={\int }_{\tau }^2{t}^{\prime }d{t}^{\prime }=\frac{1}{2}(4-{\tau }^2)$, $y={\int }_{\tau }^{2}d{t}^{\prime }=2-\tau$. $y=2-\tau$ se: $\tau =2-y$. Phir $x=\frac{1}{2}(4-(2-y{)}^2)=\frac{1}{2}(4y-y^2)=2y-\frac{1}{2}{y}^2$ — ek alag parabola.

Count: teen distinct curves (line, parabola $x=\frac{1}{2}{y}^2$, parabola $x=2y-\frac{1}{2}{y}^2$) — flow unsteady hai ($u=t$, $t$ par depend karta hai), isliye teeno alag hain. Shared-particle check: pathline aur streakline $\tau =0$ par release hue particle par milni chahiye: pathline $t=2$ par $(2,2)$ deta hai; streakline $y=2$ par $x=2(2)-\frac{1}{2}(4)=2$ deta hai ✔.

Unsteady flow mein reliable nahi. Ek streakline bahut saare past times par release hue particles se bani hoti hai, har ek us samay ke field ka imprint carry karta hai jab woh tha. Curve ki shape history ka blend hai, present velocity ka snapshot nahi. Sirf source point par tangent current velocity direction ke barabar hota hai (abhi release hua particle, $\tau \to t$). Streak ke baaki har jagah purani flow states reflect hoti hain. Steady flow mein claim sach ho jaata hai, kyunki past aur present fields identical hain. Yahi deep reason hai ki experiments (Flow visualization techniques (dye, smoke, PIV)) ko carefully interpret karna padta hai.

Unsteady mein distinct, steady mein identical ::: teenon curves (streamline, pathline, streakline) Streamline ke liye kya frozen hota hai ::: time $t$ Streakline ke liye kya sweep hota hai ::: release time $\tau$, fixed observation time par Constant-$\psi$ contours hain ::: streamlines Streamline ODE ki safe form jahan ek component zero ho ::: $vdx=udy$ Point jahan $u=v=0$ ::: ek stagnation point