4.4.24 · D3 · Maths › Multivariable Calculus › Divergence — definition, physical meaning (flux density)
Yeh page ek drill hai. Divergence — definition, physical meaning (flux density) parent note ne idea (flux per tiny volume) aur formula build kiya tha. Yahaan hum har tarah ki problem cover karte hain jo aa sakti hai, taaki koi bhi case aisa na ho jo tumne pehle dekha na ho.
Kuch bhi compute karne se pehle, ek reminder plain words mein. Vector field ka matlab sirf itna hai: space ke har point par ek arrow. Arrow F = ( F 1 , F 2 , F 3 ) mein teen numbers hote hain — kitna east (F 1 ), kitna north (F 2 ), kitna upar (F 3 ) point karta hai. Divergence poochta hai: is point ke aas-paas, kya arrow-stuff net bahar ja raha hai (ek tap) ya net aa raha hai (ek drain)? Hum ise measure karte hain
div F = ∂ x ∂ F 1 + ∂ y ∂ F 2 + ∂ z ∂ F 3 .
Symbol ∂ x ∂ F 1 (padho "partial of F 1 with respect to x ") ka matlab hai: east-arrow F 1 kitni tezi se badhta hai jab tum east ki taraf step karo , y , z ko frozen rakhte hue. Hum sirf matching partial lete hain — F 1 ke saath x , F 2 ke saath y , F 3 ke saath z . Yeh "diagonal" rule hi poora khel hai.
Yeh un sabhi case-classes ki poori list hai jo yeh topic tumpar throw kar sakta hai. Neeche har cell ko aage ek worked example se cover kiya gaya hai.
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Case class
Kya tricky hai
Example
A
Positive constant divergence (source)
ek uniform "tap" padhna
Ex 1
B
Negative constant divergence (sink)
ek "drain" ka sign
Ex 2
C
Zero divergence, bade arrows (rotation)
size ≠ spreading
Ex 3
D
Sign-varying divergence (yahaan positive, wahaan negative)
ek field, kai behaviours
Ex 4
E
Degenerate: uniform field
sabse bade arrows, div = 0
Ex 5
F
Cross-term trap (∂ F 2 / ∂ x )
kaun se partials count karte hain
Ex 6
G
Limiting check: ek real box shrink karo, V → 0 lo
definition = formula
Ex 7
H
Real-world word problem (fluid tap)
units, interpretation
Ex 8
I
Exam twist: woh jagah dhundho jahan div = 0
solve karo, sirf compute nahi
Ex 9
J
2D radial with a singularity at the origin
zero everywhere phir bhi ek source
Ex 10
Mnemonic Har cell ke peeche ek rule
Diagonal Out = Tap. Sirf diagonal partials add karo; positive total = tap (source), negative = drain (sink), zero = jo bhi aata hai woh bahar bhi jaata hai.
Worked example Example 1: pure outward field
F = ( 2 x , 2 y , 2 z ) . div F compute karo aur interpret karo.
Forecast: arrows seedhe bahar point karte hain aur distance ke saath badhte hain — guess: positive, aur wahi number har jagah. Kaunsa number?
∂ x ∂ ( 2 x ) = 2 .
Yeh step kyun? Diagonal rule: F 1 = 2 x ko x ke saath pair karo. Kyunki 2 x har east step par rate 2 se badhta hai, uska partial 2 hai.
∂ y ∂ ( 2 y ) = 2 aur ∂ z ∂ ( 2 z ) = 2 bhi usi reasoning se.
Diagonal add karo: div F = 2 + 2 + 2 = 6 .
Matlab: + 6 har jagah → ek uniform tap (source), constant rate par spreading.
Verify: har component sirf apne variable par depend karta hai, toh koi cross-terms nahi, aur 2 + 2 + 2 = 6 . Positive "arrows point outward" se match karta hai. ✓
Worked example Example 2: pure inward field
F = ( − 3 x , − 3 y , − 3 z ) . Compute karo aur interpret karo.
Forecast: arrows har jagah andar point karte hain. Sign hona chahiye...?
∂ x ∂ ( − 3 x ) = − 3 .
Yeh step kyun? Jab tum east step karte ho, east-arrow rate − 3 par aur zyada negative hota jaata hai.
y aur z ke liye bhi same: har ek − 3 deta hai.
Sum: div F = − 3 − 3 − 3 = − 9 .
Matlab: − 9 < 0 → har jagah ek drain (sink); cheezein khatam ho rahi hain.
Verify: negative, inward arrows se match karta hai; − 3 ⋅ 3 = − 9 . ✓
Worked example Example 3 (Cell C): rotation — bade arrows, zero divergence
F = ( − y , x , 0 ) , z -axis ke aas-paas ek swirl (figure dekho).
Forecast: axis se door arrows bahut bade ho sakte hain. Bade arrows ⇒ bada divergence? Apna gut test karo.
∂ x ∂ ( − y ) = 0 .
Yeh step kyun? − y mein koi x nahi hai, toh east step karne par yeh bilkul nahi badalta → rate 0 .
∂ y ∂ ( x ) = 0 (x ke andar koi y nahi), aur ∂ 0/ ∂ z = 0 .
Sum: div F = 0 + 0 + 0 = 0 .
Matlab: fluid sirf ghoomta hai — kuch create ya destroy nahi hota. Is field mein rotation hai (Curl — rotation density ), spreading nahi.
Verify: figure mein, chhote dotted balloon ko dekho: exactly utne hi arrows andar jaate hain jitne bahar. Net = 0 . ✓
Worked example Example 5 (Cell E): biggest-arrow trap — uniform field
F = ( 100 , 0 , 0 ) — ek strong steady wind har jagah east mein beh raha hai.
Forecast: magnitude 100 ! Zaroor bada divergence hoga?
∂ ( 100 ) / ∂ x = 0 — ek constant move karne par nahi badalta.
Baaki diagonal partials bhi 0 hain.
div F = 0 .
Matlab: jo kuch balloon ke west side se andar aata hai woh east side se bahar jaata hai — perfectly balanced. Size spreading nahi hai.
Verify: sabhi components constant ⇒ har partial 0 ⇒ sum 0 . ✓
Common mistake "Bade arrows ka matlab bada divergence hai."
Kyun sahi lagta hai: strong flow dramatic lagta hai . Fix: Examples 3 aur 5 mein bade arrows hain aur divergence exactly 0 hai. Divergence change measure karta hai (net in vs out), kabhi length nahi.
Worked example Example 4: kahin positive, kahin negative
F = ( x 2 − y 2 , 0 , 0 ) ek banaya hua 1D-ish flow hai. Yeh kahaan source hai, kahaan sink?
Forecast: sirf F 1 vary karta hai, aur sirf x ke saath. Toh div ka sign kisi expression in x ke sign ko follow karna chahiye.
div F = ∂ x ∂ ( x 2 − y 2 ) + 0 + 0 = 2 x .
Yeh step kyun? Diagonal partial: x 2 − y 2 ko x ke respect mein differentiate karo, y ko frozen treat karo, toh − y 2 khatam ho jaata hai aur x 2 → 2 x .
Sign table set up karo: 2 x > 0 jab x > 0 ; 2 x < 0 jab x < 0 ; 2 x = 0 plane x = 0 par.
Matlab: wahi field region x > 0 mein ek tap hai, x < 0 mein ek drain hai, aur x = 0 par dono nahi. Ek field apni position ke hisaab se har hat pehen sakta hai.
Verify: x = 3 par: div = 6 > 0 (source). x = − 3 par: div = − 6 < 0 (sink). x = 0 par: div = 0 . ✓
Worked example Example 6: partials jo relevant lagte hain par hain nahi
F = ( 3 y , 5 x , 7 z ) . div F compute karo.
Forecast: pehle slot mein y hai aur doosre mein x — unhe differentiate karne ka man karta hai. Answer guess karo, phir check karo ki woh andar jaate hain ya nahi.
F 1 = 3 y : matching partial hai ∂ ( 3 y ) / ∂ x = 0 (3 y ke andar koi x nahi).
Yeh step kyun? Divergence sirf ∂ F 1 / ∂ x use karta hai — ∂ F 1 / ∂ y nahi . Cross-partial ∂ ( 3 y ) / ∂ y = 3 curl ka kaam hai, yahaan nahi.
F 2 = 5 x : matching partial ∂ ( 5 x ) / ∂ y = 0 .
F 3 = 7 z : matching partial ∂ ( 7 z ) / ∂ z = 7 .
Diagonal sum karo: 0 + 0 + 7 = 7 .
Matlab: bahut saari cross-coupling ke bawajood, net spreading poori tarah z -part se aati hai.
Verify: sirf F 3 apne axis ke saath badhta hai; div = 7 . (Agar tumne galti se cross-terms 3 aur 5 use kiye hote, toh 15 aata — woh ek curl-flavoured mistake hai.) ✓
Worked example Example 7: ek real box shrink karo aur formula recover karo
F = ( x , 0 , 0 ) . Formula kehta hai div = 1 . Ise geometric definition (flux per volume) se prove karo, origin par centred side 2 h ke cube par.
Forecast: kya actual flux ÷ volume exactly 1 deta hai, h se independent?
Sirf do x -faces matter karte hain. Top/bottom/front/back faces par outward normal ± y ^ ya ± z ^ hai, aur F = ( x , 0 , 0 ) mein zero y , z parts hain, toh F ⋅ n ^ = 0 wahaan.
Yeh step kyun? Flux hai F ⋅ n ^ ; ek field jisme koi y , z components nahi hain woh y , z ki taraf face karne wale faces se koi flux pass nahi karta.
Right face x = h : F 1 = h , outward normal + x ^ , area ( 2 h ) 2 = 4 h 2 . Flux out = h ⋅ 4 h 2 = 4 h 3 .
Left face x = − h : F 1 = − h , outward normal − x ^ , toh F ⋅ n ^ = − ( − h ) = + h . Flux out = h ⋅ 4 h 2 = 4 h 3 .
Yeh step kyun? Left par, field value aur normal dono negative hain — do minus signs outflow ko positive banate hain. Fluid sachchi mein dono faces se bahar jaata hai.
Total flux = 4 h 3 + 4 h 3 = 8 h 3 . Volume = ( 2 h ) 3 = 8 h 3 .
Flux per volume = 8 h 3 8 h 3 = 1 , har h ke liye. h → 0 lo: phir bhi 1 .
Matlab: geometric definition aur formula exactly agree karte hain — figure mein balloon ko shrink hote dekho.
Verify: 8 h 3 /8 h 3 = 1 = ∂ x / ∂ x . h se independent, jaise limit demand karti hai. ✓
Worked example Example 8: a leaky pipe field
Ek tank mein water velocity v = ( 0.4 x , 0.4 y , − 0.8 z ) hai, metres per second mein measure ki gayi, position metres mein. Kya origin par paani create ho raha hai ya destroy, aur per unit volume kitni rate par?
Forecast: x , y mein do positive spreads aur z mein ek squeeze — kya woh cancel karenge?
∂ x ∂ ( 0.4 x ) = 0.4 s − 1 .
Yeh step kyun? Units: velocity (m/s) ko length (m) se differentiate karne par per-second (s − 1 ) milta hai — ek rate , exactly wahi jo "per unit volume per second" ko chahiye.
∂ y ∂ ( 0.4 y ) = 0.4 s − 1 ; ∂ z ∂ ( − 0.8 z ) = − 0.8 s − 1 .
Sum: div v = 0.4 + 0.4 − 0.8 = 0 s − 1 .
Matlab: divergence 0 har jagah → flow incompressible hai; kuch create ya destroy nahi hota. Sideways spread karna downward squeeze se exactly compensate hota hai. Yeh constant density ke saath Continuity equation hai: ∇ ⋅ v = 0 .
Verify: 0.4 + 0.4 − 0.8 = 0 ; units sabhi s − 1 , consistent. ✓
Worked example Example 9: source-free surface dhundho
F = ( x 2 , y 2 , z 2 ) . Kis surface par flow na source hai na sink?
Forecast: har partial 2 x , 2 y , 2 z hai; unka sum zero set karne par ek flat surface (ek plane) define hona chahiye.
div F = ∂ x ∂ x 2 + ∂ y ∂ y 2 + ∂ z ∂ z 2 = 2 x + 2 y + 2 z .
Yeh step kyun? x 2 , y 2 , z 2 ke diagonal partials 2 x , 2 y , 2 z hain.
div = 0 set karo: 2 x + 2 y + 2 z = 0 ⟺ x + y + z = 0 .
Yeh step kyun? "Na source na sink" precisely div = 0 hai; 2 se divide karo.
Matlab: flow exactly plane x + y + z = 0 par source-free hai; ek taraf tap, doosri taraf drain.
Verify: point ( 1 , − 1 , 0 ) par: 2 ( 1 ) + 2 ( − 1 ) + 0 = 0 ✓ (plane par). ( 1 , 1 , 1 ) par: 2 + 2 + 2 = 6 > 0 (source, plane se bahar). ✓
Worked example Example 10: woh field jo har jagah source-free hai phir bhi tap ki tarah behave karta hai
Plane mein, F = ( x 2 + y 2 x , x 2 + y 2 y ) lo (arrows bahar point karte hain, par distance ke saath shrink hote hain — figure dekho). Iska divergence origin se door compute karo, aur origin explain karo.
Forecast: arrows bahar point karte hain, toh tum positive source guess karte. Par woh baahri taraf kamzor hote hain — shayad woh kamzori spreading ko exactly cancel kar deti hai?
r 2 = x 2 + y 2 likho, toh F 1 = x / r 2 . Quotient rule use karo ∂ r 2 / ∂ x = 2 x ke saath:
∂ x ∂ ( r 2 x ) = r 4 1 ⋅ r 2 − x ⋅ 2 x = r 4 r 2 − 2 x 2 = r 4 y 2 − x 2 .
Yeh step kyun? F 1 ek fraction hai jiska top aur bottom dono x rakhte hain; quotient rule handle karta hai "ek ratio kitni tezi se change hota hai."
Symmetry se (x ↔ y ): ∂ y ∂ ( r 2 y ) = r 4 x 2 − y 2 .
Add karo: div F = r 4 y 2 − x 2 + r 4 x 2 − y 2 = 0 (r = 0 ke liye).
Matlab: origin se door outward spreading exactly arrows ke shrink hone se undo ho jaati hai — divergence 0 har jagah except ek point. Par origin ke aas-paas kisi bhi circle se total flux ek fixed positive number hai: saara "source" singular origin par concentrated hai , jahan formula break down karta hai (zero se division). Yeh point sources aur Divergence Theorem (Gauss) ke peeche ka flavor hai.
Verify: do fractions exact negatives hain, r = 0 ke liye 0 tak sum hote hain; ( 1 , 0 ) par check karo: ( 0 − 1 ) /1 + ( 1 − 0 ) /1 = 0 ✓. ( 1 , 2 ) par: ( 4 − 1 ) /25 + ( 1 − 4 ) /25 = 3/25 − 3/25 = 0 ✓.
Recall Kaun sa cell tumhare liye sabse mushkil tha?
Zero-divergence-with-big-arrows (C/E) aur singular source (J) sabse zyada logon ko trip karte hain. Dono ek hi baat kehte hain: divergence net leaving measure karta hai, arrow size nahi, aur appearance nahi.
Big arrows wala field phir bhi zero divergence rakh sakta hai — true ya false? True; e.g. ( 100 , 0 , 0 ) ya swirl ( − y , x , 0 ) , dono div = 0 .
div ( 2 x , 2 y , 2 z ) = ? 2 + 2 + 2 = 6 .
Kis surface par ( x 2 , y 2 , z 2 ) ka zero divergence hai? Plane x + y + z = 0 .
( x / r 2 , y / r 2 ) ka origin se door zero divergence kyun hai?Outward spreading exactly arrows ke 1/ r se shrink hone se cancel ho jaati hai; source singular origin par concentrated hai.
F = ( 3 y , 5 x , 7 z ) ke liye divergence mein kaun se partials jaate hain?Sirf diagonal wale ∂ ( 3 y ) / ∂ x , ∂ ( 5 x ) / ∂ y , ∂ ( 7 z ) / ∂ z = 0 + 0 + 7 = 7 .