Exercises — Stream function, velocity potential

Do alag-alag tests. WHY do? Kyunki $\varphi$ ko koi spin nahi chahiye (irrotational) aur $\psi$ ko koi squishing nahi chahiye (incompressible) — dono alag physics hain.

$\psi$ ke liye test (incompressibility, Continuity Equation): $\nabla \cdot \vec{V}=\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=\frac{\partial (-y)}{\partial x}+\frac{\partial (x)}{\partial y}=0+0=0.$ Divergence zero hai ⇒ $\psi$ exist karta hai. ✓

$\varphi$ ke liye test (irrotationality): 2D mein sirf ek vorticity component hai $\omega =\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=\frac{\partial (x)}{\partial x}-\frac{\partial (-y)}{\partial y}=1-(-1)=2\ne 0.$ Non-zero spin ⇒ $\varphi$ exist NAHI karta. ✗

Yeh field solid-body rotation hai (jaise ek spinning turntable): har particle origin ke around circle karta hai, isliye real spin hai, aur koi single "height" scalar isko describe nahi kar sakti.

Parent note se rule: $\psi =\text{const}$ streamlines hain (fluid unke saath behta hai); $\varphi =\text{const}$ equipotentials hain (flow unhe perpendicularly cross karta hai).

• $\psi =3x-2y=\text{const}$ → slope $\frac{dy}{dx}=\frac{3}{2}$ wali straight lines. Yeh streamlines hain.
• $\varphi =2x+3y=\text{const}$ → slope $-\frac{2}{3}$ wali straight lines. Yeh equipotentials hain.

Slopes multiply karte hain $\frac{3}{2}\cdot (-\frac{2}{3})=-1$: yeh perpendicular hain, exactly jaisa ek flow net demand karta hai.

Definitions apply karo ($u={\psi }_y$, $v=-{\psi }_x$). WHY yeh aur $\varphi$ ke wale nahi? Kyunki humein $\psi$ diya gaya hai. $u=\frac{\partial }{\partial y}(x^2-y^2)=-2y,v=-\frac{\partial }{\partial x}(x^2-y^2)=-2x.$ Incompressible? Construction se koi bhi $\psi$ $\nabla \cdot \vec{V}=0$ deta hai. Check karo: $u_x+v_y=0+0=0$ ✓. Irrotational? $\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=\frac{\partial (-2x)}{\partial x}-\frac{\partial (-2y)}{\partial y}=-2-(-2)=0.$ Zero spin ⇒ bhi irrotational. Isliye $\varphi$ bhi exist karta hai — yeh ek ideal flow hai. (Yeh classic "flow toward a corner" / stagnation flow hai.)

Hum integrate karte hain, kyunki $\varphi$ velocity ka anti-derivative hai ($u={\varphi }_x$). $\varphi =\int udx=\int (-2y)dx=-2xy+f(y).$ Yahan $f(y)$ sirf $y$ ka ek unknown function hai — jab hum $x$ ke respect mein differentiate karte hain to yeh zero ho jaata hai, isliye integration abhi isko dekh nahi sakta. Isko doosri equation $v={\varphi }_y$ se pin karo: ${\varphi }_y=-2x+f^{\prime }(y)\stackrel{!}{=}v=-2x\Rightarrow f^{\prime }(y)=0\Rightarrow f=\text{const}.$ Isliye $\varphi =-2xy$ (constant absorbed). Check karo: ${\varphi }_x=-2y=u$ ✓, ${\varphi }_y=-2x=v$ ✓.

Dono scalars ko $u$ aur $v$ par agree karna chahiye. Bridge hai ${\varphi }_x={\psi }_y$ aur ${\varphi }_y=-{\psi }_x$. ${\varphi }_x=2x,{\psi }_y=2x✓$ ${\varphi }_y=2y,-{\psi }_x=-2y.$ Doosri equation: $2y\stackrel{?}{=}-2y$ → sirf $y=0$ par sahi hai. Refuted. Yeh dono ek hi flow describe nahi karte.

Deeper reason: $\varphi =x^2+y^2$ mein ${\nabla }^2\varphi =2+2=4\ne 0$ hai, isliye yeh Laplace Equation ${\nabla }^2\varphi =0$ fail karta hai jo har valid velocity potential ko ek incompressible flow mein satisfy karna chahiye. Yeh pehle se hi ek legal potential tha hi nahi.

$\varphi$ se: $u={\varphi }_x=2x$, $v={\varphi }_y=-2y$. Ab $u={\psi }_y$ se $\psi$ banao: $\psi =\int 2xdy=2xy+g(x).$ $g$ ko $v=-{\psi }_x$ se fix karo: $-{\psi }_x=-(2y+g^{\prime }(x))\stackrel{!}{=}-2y\Rightarrow g^{\prime }(x)=0$. Isliye $\boxed{\psi =2xy}$. Laplace check: ${\nabla }^2\varphi ={\varphi }_{xx}+{\varphi }_{yy}=2+(-2)=0✓,{\nabla }^2\psi ={\psi }_{xx}+{\psi }_{yy}=0+0=0✓.$

WHY superpose? Kyunki Laplace Equation linear hai: agar ${\psi }_1$ aur ${\psi }_2$ dono ${\nabla }^2\psi =0$ solve karte hain, to unka sum bhi karta hai. Streamfunctions add hote hain. $\psi =Uy+\frac{m}{2\pi }\theta ,y=r\sin \theta .$

${\psi }_2$ se Cartesian velocity nikalna — dhyan se karo. Source stream function ${\psi }_2=\frac{m}{2\pi }\theta$ hai. Uski velocity polar coordinates mein stream function ki polar definitions se aati hai, $u_r=\frac{1}{r}\frac{\partial \psi }{\partial \theta },u_{\theta }=-\frac{\partial \psi }{\partial r}.$ Isliye $u_r=\frac{1}{r}\cdot \frac{m}{2\pi }=\frac{m}{2\pi r}$ (pure radial outflow) aur $u_{\theta }=0$. WHY Cartesian mein convert karein? Kyunki stream horizontally flow karti hai ($u$-direction mein), isliye dono flows ko add karne ke liye hum dono ko same $(u,v)$ basis mein express karna padega. Magnitude $u_r$ ki purely radial velocity unit vector $\hat{r}=(x,y)/r$ ke along point karti hai, isliye uske Cartesian components hain $u_r\hat{r}=\frac{m}{2\pi r}\cdot \frac{(x,y)}{r}=\frac{m}{2\pi }\frac{(x,y)}{x^2+y^2},\text{since }{r}^2=x^2+y^2.$ Yahi woh jagah hai jahan se vector form $\frac{m}{2\pi }\frac{(x,y)}{x^2+y^2}$ aata hai — yeh kuch aur nahi bas "radial speed times outward unit vector" hai. Uniform stream $(U,0)$ add karke: $u=U+\frac{m}{2\pi }\frac{x}{x^2+y^2},v=\frac{m}{2\pi }\frac{y}{x^2+y^2}.$ $x$-axis par stagnation: $y=0$ set karo (tab $v=0$ automatically) aur $u=0$: $U+\frac{m}{2\pi }\frac{x}{x^2}=0\Rightarrow U+\frac{m}{2\pi x}=0\Rightarrow x=-\frac{m}{2\pi U}.$ $U=1,m=2\pi$ ke saath: $x=-\frac{2\pi }{2\pi \cdot 1}=-1$. Stagnation point $(-1,0)$ par. Figure mein upar dekho: amber dot $(-1,0)$ par baitha hai, white source dot se upstream (left side mein). Left side ke white arrows incoming uniform stream hain; cyan curves combined $\psi$ ki streamlines hain. Exactly amber dot par source ka outward push (us side mein $-x$ direction mein, left ki taraf) stream ke rightward push ko cancel karta hai — isliye flow ruk jaati hai, split hoti hai, aur "half-body" ke around wrap ho jaati hai. Analytic $x=-1$ exactly wahi amber dot hai.

Parent note se key fact: do streamlines ke beech flux bas $Q={\psi }_2-{\psi }_1$ hai. WHY: flux $=\int (udy-vdx)=\int d\psi =\Delta \psi$.

$\psi =Uy+\frac{m}{2\pi }\theta$ evaluate karo $U=1,\frac{m}{2\pi }=1$ ke saath, isliye $\psi =y+\theta$ (with $\theta \in (-\pi ,\pi ]$).

• $(-1,0)$ par: negative $x$-axis par $\theta =\pi$, $y=0$ ⇒ ${\psi }_1=0+\pi =\pi$.
• $(0,1)$ par: $\theta =\pi /2$, $y=1$ ⇒ ${\psi }_2=1+\pi /2$. $Q={\psi }_2-{\psi }_1=\left(1+\frac{\pi }{2}\right)-\pi =1-\frac{\pi }{2}\approx -\mathrm{0.5708.}$ WHY minus sign — yeh orientation hai, galti nahi. Quantity $Q={\psi }_2-{\psi }_1$ ek curve ko cross karne wala flux measure karta hai streamline $1$ se streamline $2$ tak ek fixed handedness ke saath (positive jab flow left-to-right cross kare jab tum pehli streamline se doosri tak chalo). Yahan $\psi$ actually $(-1,0)$ se $(0,1)$ jaate waqt decrease karta hai, isliye fluid hamare chosen walking order ke opposite sense mein cross karta hai — isliye minus sign. Dono endpoints swap karo ($(0,1)$ se $(-1,0)$ chalo) aur tumhe $+\left(\frac{\pi }{2}-1\right)$ milega. Fluid ki physical amount dono taraf same hai: uska magnitude hai $|Q|=\frac{\pi }{2}-1\approx 0.571\text{units (per unit depth).}$ Sign sirf yeh record karta hai ki flow kis taraf hai; size hi "kitna fluid" ka matlab hai.

Pehle, hume zaroori tool chahiye: polar coordinates mein potential se velocity. Gradient $\nabla \varphi$ "$\varphi$ ki steepest-increase" ka vector hai. Cartesian mein yeh $({\varphi }_x,{\varphi }_y)$ hai. Polar coordinates mein, radial direction mein $dr$ length ka step $\varphi$ ko ${\varphi }_{r}dr$ se change karta hai, lekin angular direction mein ek step arc length $rd\theta$ sweep karta hai (na ki $d\theta$) — kyunki angle $d\theta$ se radius $r$ par ghoomne par tum physically $rd\theta$ distance move karte ho. Kyunki velocity change-in-$\varphi$ per unit distance hai, tangential component ko us arc length se divide karna padta hai: $\nabla \varphi =\left({\varphi }_r,\frac{1}{r}{\varphi }_{\theta }\right)\Rightarrow u_r=\frac{\partial \varphi }{\partial r},u_{\theta }=\frac{1}{r}\frac{\partial \varphi }{\partial \theta }.$ Woh extra $1/r$ hi pura reason hai ki angular formula radial se alag kyun dikhti hai.

(a) Stream function. Polar form mein $u_{\theta }=-\frac{\partial \psi }{\partial r}$. Isliye $-\frac{\partial \psi }{\partial r}=\frac{\Gamma }{2\pi r}\Rightarrow \psi =-\frac{\Gamma }{2\pi }\ln r.$ Streamlines $\psi =\text{const}\Rightarrow r=\text{const}$: concentric circles — sahi hai, fluid chakkar kaata hai.

Velocity potential. Abhi derive ki gayi polar rule use karte hue, $u_{\theta }=\frac{1}{r}\frac{\partial \varphi }{\partial \theta }=\frac{\Gamma }{2\pi r}\Rightarrow \frac{\partial \varphi }{\partial \theta }=\frac{\Gamma }{2\pi }\Rightarrow \varphi =\frac{\Gamma }{2\pi }\theta .$ Note karo yeh $\varphi$ aur $\psi$, parent note mein source ke swapped mirror hain — vortex aur source conjugates hain.

(b) Irrotational check + single-valuedness. Purely azimuthal flow ki vorticity hai ${\omega }_z=\frac{1}{r}\frac{\partial (r{u}_{\theta })}{\partial r}=\frac{1}{r}\frac{\partial }{\partial r}\left(r\cdot \frac{\Gamma }{2\pi r}\right)=\frac{1}{r}\frac{\partial }{\partial r}\left(\frac{\Gamma }{2\pi }\right)=0(r\ne 0).$ Isliye har particle orbit karta hai lekin woh apne axis ke around spin nahi karta — yahi wajah hai ki yeh har jagah irrotational hai, singular core $r=0$ ko chhod kar.

Lekin $\varphi =\frac{\Gamma }{2\pi }\theta$ multivalued hai. Origin ke around ek full loop chalte hain: $\theta$ $2\pi$ se badhta hai, isliye $\varphi$ $\frac{\Gamma }{2\pi }\cdot 2\pi =\Gamma$ jump karta hai chahe tum same physical point par wapas aa gaye ho. Ek genuine single-valued height aisa nahi kar sakti. Yahi L1 caveat exactly action mein hai: flow region (plane minus origin) simply-connected nahi hai — iske paas $r=0$ par ek hole hai, aur us hole ke around loops ko ek point tak shrink nahi kiya ja sakta. Isliye "curl $=0$" hold karta hai phir bhi $\varphi$ single-valued hai sirf tab jab hum plane ko ek ray ke along cut karein (ek branch cut, jaise negative $x$-axis) taki koi path core ko encircle na kar sake. Cut ke across $\Gamma$ ki jump exactly circulation hai — vortex ki defining quantity. (Velocity khud perfectly single-valued rehti hai; sirf potential ambiguity carry karta hai.)

(c) Bernoulli. Irrotational flow ke liye, $p+\frac{1}{2}\rho V^2=\text{const}$ har jagah. Speed $V=u_{\theta }=\frac{\Gamma }{2\pi r}$. $\Gamma =2\pi ,\rho =1$ ke saath: $V=\frac{1}{r}$. $p(r)=C-\frac{1}{2}\rho V^2=C-\frac{1}{2r^2}.$ $p(1)-p(2)=\left(C-\frac{1}{2}\right)-\left(C-\frac{1}{8}\right)=-\frac{1}{2}+\frac{1}{8}=-\frac{3}{8}.$ Isliye centre ke paas pressure $\frac{3}{8}$ units kam hai — exactly wahi reason hai ki ek whirlpool ka middle neeche dip karta hai.

• Streamlines: $\psi =-\frac{\Gamma }{2\pi }\ln r=\text{const}\Rightarrow r=\text{const}$ → concentric circles.
• Equipotentials: $\varphi =\frac{\Gamma }{2\pi }\theta =\text{const}\Rightarrow \theta =\text{const}$ → radial rays.
• Ek circle aur ek radius hamesha $90°$ par milte hain.

Ise upar ke figure se match karo: cyan circles streamlines hain ($r=\text{const}$), amber rays equipotentials hain ($\theta =\text{const}$), aur white dot singular core $r=0$ mark karta hai (jahan $\varphi$ ka branch cut rehta hai). Har jagah jahan ek cyan circle ek amber ray se milta hai, woh right angle par cross karte hain — yahi flow net hai, drawn.

Orthogonality ka algebraic proof (figure se match karta hai). Ek scalar ka gradient uski level curves ke across point karta hai, isliye: $\nabla \varphi =(u,v)\text{(velocity)},\nabla \psi =({\psi }_x,{\psi }_y)=(-v,u).$ Unka dot product: $\nabla \varphi \cdot \nabla \psi =(u)(-v)+(v)(u)=0.$ Zero dot product ⇒ perpendicular gradients ⇒ perpendicular level curves. Isliye algebra ($\nabla \varphi \cdot \nabla \psi =0$) exactly wahi statement hai jo tumhari aankhein figure mein dekhti hain: circles ⟂ rays.

Existence test ::: $\psi$ ko $\nabla \cdot \vec{V}=0$ chahiye; $\varphi$ ko $\nabla \times \vec{V}=0$ aur single-valuedness ke liye simply-connected (hole-free) region chahiye. Recover $\varphi$ from velocity ::: $u$ ko $x$ par integrate karo, phir leftover $f(y)$ ko $v={\varphi }_y$ se fix karo. Superposition rule ::: $\psi$ aur $\varphi$ add hote hain kyunki Laplace's equation linear hai. Polar velocity from a potential ::: $u_r={\varphi }_r,u_{\theta }=\frac{1}{r}{\varphi }_{\theta }$ ($1/r$ arc-length weighting hai). Orbiting vs spinning ::: free vortex ($u_{\theta }\propto 1/r$) irrotational hai (lekin $\varphi$ har loop par $\Gamma$ jump karta hai); solid-body ($u_{\theta }\propto r$) genuinely spin karta hai. Flux meaning of $\psi$ ::: $Q={\psi }_2-{\psi }_1$ do streamlines ke beech; uska sign orientation record karta hai, magnitude amount batata hai.