4.7.13 · Maths › Partial Differential Equations
Intuition Ek sentence mein badi idea
Laplace's equation uss system ko describe karta hai jo time ke saath change karna band kar chuka hai — ek steady state — jahan har point ki value apne neighbours ka average hoti hai, isliye kuch bhi kisi ko kahi push nahi karta.
Definition Laplace's equation
n dimensions mein, ek scalar field u ( x ) ke liye:
∇ 2 u = 0
2D Cartesian coordinates mein:
∂ x 2 ∂ 2 u + ∂ y 2 ∂ 2 u = 0
Jo function u , ∇ 2 u = 0 satisfy kare use harmonic kehte hain. Operator ∇ 2 ko Laplacian kehte hain.
Yeh prototype elliptic PDE hai. "Elliptic" naam aata hai A u xx + B u x y + C u y y + ⋯ = 0 ko discriminant B 2 − 4 A C se classify karne se. Laplace ke liye, A = C = 1 , B = 0 , toh B 2 − 4 A C = − 4 < 0 → elliptic (jaise ellipse ki equation x 2 + y 2 = 1 ).
Sabse saaf derivation heat (diffusion) equation se shuru hoti hai aur poochti hai "jab time matter karna band kar de toh kya hoga?"
∇ 2 u ko padhna
Kisi point par ∇ 2 u measure karta hai ki wahan u apne nearby values ke average se kitna alag hai. ∇ 2 u > 0 ⇒ point ek dip hai (apne surroundings se neeche hai, toh heat andar aati hai, yeh warm hota hai). ∇ 2 u = 0 ⇒ point pehle se apne surroundings ke average ke barabar hai ⇒ koi net flow nahi ⇒ steady.
Yeh sach kyun hai (sketch). u ˉ ( r ) = radius r ke circle par u ka average define karo. Differentiate karo aur divergence theorem use karo:
d r d u ˉ = 2 π r 1 ∮ ∂ n ∂ u d s = 2 π r 1 ∬ d i s k ∇ 2 u d A = 0.
Toh u ˉ ( r ) r mein constant hai; r → 0 shrink karne par u ˉ = u ( x 0 ) milta hai.
Intuition Consequence: Maximum Principle
Ek harmonic function ke koi interior maxima ya minima nahi hote — extremes sirf boundary par rehte hain. Kyun? Agar koi point strict maximum hota toh woh apne neighbours ke average se zyada hota, jo mean-value property ko contradict karta. Physically: steady temperature mein beech mein koi hot spot nahi ho sakta jo thande surroundings se ghira ho — woh hot spot heat leak kar raha hota, toh woh steady hi nahi tha. Sabse garm aur thande jagah edge par hote hain.
Elliptic problems ko poori closed boundary par conditions chahiye (time mein initial conditions nahi, kyunki time hai hi nahi!).
Dirichlet : boundary par u specified (edge temperature fix karo).
Neumann : ∂ u / ∂ n specified (edge ke through heat flux fix karo; insulated ⇒ = 0 ).
Intuition "Initial condition" kyun nahi?
Parabolic/hyperbolic PDEs time mein aage evolve karte hain, isliye aap ek starting state dete ho. Laplace ne time hataa diya hai — answer purely wahi hota hai jo spatial boundary par hota hai. Interior ko averaging se "smoothly fill in" kiya jaata hai.
Worked example 1D Laplace — sabse simple case
1D mein, ∇ 2 u = u ′′ ( x ) = 0 .
Solve: do baar integrate karo ⇒ u ( x ) = a x + b . Yeh step kyun? u ′′ = 0 ka matlab slope constant hai ⇒ ek straight line.
u ( 0 ) = 10 , u ( L ) = 30 ke saath: b = 10 , a = 20/ L , toh u ( x ) = 10 + L 20 x .
Physical meaning: dono ends par heated rod mein steady temperature ek straight-line profile hoti hai — exactly average-of-neighbours rule (har interior point apne dono sides ka midpoint average hai).
Worked example 2D harmonic function verify karna
Kya u ( x , y ) = x 2 − y 2 harmonic hai?
u xx = 2 , u y y = − 2 , toh u xx + u y y = 2 − 2 = 0 . ✔ Harmonic.
Yeh step kyun? Bas check karo ki Laplacian zero hai. (Yeh z 2 = ( x + i y ) 2 ka real part hai; kisi bhi analytic function ke real/imaginary parts harmonic hote hain — complex analysis se ek gehri link.)
Worked example Mean-value check
Harmonic u = x lo. Ise ( 2 , 0 ) par centred unit circle par average karo: points hain ( 2 + cos θ , sin θ ) , toh u = 2 + cos θ . θ ∈ [ 0 , 2 π ) par 2 + cos θ = 2 ka average.
Centre value hai u ( 2 , 0 ) = 2 . ✔ Yeh match karte hain, mean-value property confirm karti hai.
Yeh step kyun? Abstract "value = neighbours ka average" ko ek concrete arithmetic check mein badal deta hai.
Common mistake "Laplace's equation describe karta hai ki cheezein time ke saath kaise evolve karti hain."
Kyun sahi lagta hai: Yeh heat equation jaisi dikhti hai, jo time evolution ke baare mein hai. Fix: Laplace, heat equation hai jisme u t = 0 — yeh final, settled state describe karta hai, jisme koi time variable bilkul nahi . Yeh ek snapshot hai, movie nahi.
Common mistake "Main isme wave/heat equation ki tarah initial condition dunga."
Kyun sahi lagta hai: Pehle milne wali zyaadatar PDEs evolution equations hoti hain jinhe initial data chahiye. Fix: Elliptic equations boundary-value problems hain — data poori spatial boundary par specify karo. Koi "initial time" nahi hai shuru karne ke liye.
Common mistake "Ek harmonic function region ke beech mein peak ho sakta hai."
Kyun sahi lagta hai: Bahut saare smooth bumpy functions exist karte hain. Fix: Maximum principle harmonic functions ke liye interior extrema forbid karta hai. Ek "peak" value = average of neighbours ko violate karta. Extremes boundary par hote hain.
∇ 2 u = 0 ka matlab u constant hai."
Kyun sahi lagta hai: Zero aksar "kuch nahi ho raha" suggest karta hai. Fix: ∇ 2 u ek second-order combination hai; u = x 2 − y 2 ya u = x y kaafi vary karte hain phir bhi harmonic hain. Zero Laplacian ka matlab balanced curvature hai, flat nahi.
Recall Feynman: 12-saal ke bachche ko explain karo
Ek metal plate imagine karo. Tum edges ko fixed temperatures par pakdte ho — ek edge garam, ek thandi — aur bahut, bahut lambe time tak wait karte ho. Aakhir mein heat idhar-udhar hona band ho jaati hai; har spot ek aisi temperature par settle ho jaata hai jo bas apne aas-paas ki temperatures ka average hai. Woh settled picture hi hai jo Laplace's equation describe karta hai. Kyunki har point apne neighbours ka average hai, tum kabhi beech mein ek akela super-garam dot nahi rakh sakte jo thande surroundings se ghira ho — woh instantly apni heat share karna shuru kar deta aur "settled" rehna band kar deta. Toh sabse garam aur thande jagah hamesha un edges par hote hain jo tum pakde ho.
"L.A.P." — Laplace = Average = Plateau (koi inner peaks nahi).
Yeh bhi: Elliptic = "ek band bowl mein still water ki tarah settled" — boundary poora interior fix karti hai.
Symbols mein Laplace's equation ∇ 2 u = 0 (the Laplacian of u is zero)
∇ 2 u = 0 satisfy karne wale function ko kya kehte hain?harmonic
Laplace's equation kaun si physical situation describe karta hai? Ek steady-state (time-independent) field, jaise equilibrium temperature, electrostatic potential, ya incompressible irrotational flow
Steady-state heat flow se ∇ 2 u = 0 kyun milta hai? u t = 0 set karne par heat equation u t = α ∇ 2 u mein ∇ 2 u = 0 bachta hai
Laplace's equation ka PDE-classification type kya hai? Elliptic (B 2 − 4 A C < 0 jisme A = C = 1 , B = 0 )
Mean-value property state karo Ek harmonic function ki kisi point par value us point ke surrounding circle/sphere par uski values ke average ke barabar hoti hai
Maximum principle kya kehta hai? Ek harmonic function ke koi interior maxima ya minima nahi hote; extremes sirf boundary par aate hain
Ek elliptic PDE ko kis tarah ki conditions chahiye? Poori closed boundary par boundary conditions (Dirichlet ya Neumann), initial conditions nahi
Dirichlet vs Neumann condition Dirichlet, boundary par u fix karta hai; Neumann, normal derivative ∂ u / ∂ n (flux) fix karta hai
Kya u = x 2 − y 2 harmonic hai? Haan: u xx + u y y = 2 − 2 = 0
u ′′ ( x ) = 0 ko u ( 0 ) = 10 , u ( L ) = 30 ke saath solve karou ( x ) = 10 + L 20 x (ek straight line)
Kisi point par ∇ 2 u > 0 ka sign interpretation Point apne neighbours ke average se neeche hai; heat andar aati hai, yeh warm hone ki tendency hai
Heat equation ut = alpha lap u
Fourier law q = -k grad u
Laplace equation lap u = 0
Value = average of neighbours