4.7.14 · Maths › Partial Differential Equations
Laplace's equation ∇ 2 u = u xx + u y y = 0 ek steady state describe karti hai — ek metal plate mein temperature jo settle ho gayi ho, ek electrostatic potential, ya ek stretched soap film. Time mein kuch bhi nahi badalta; har interior point ki value uske neighbours ki average hoti hai. Ek rectangle par hume charon edges ki values di jaati hain (boundary conditions) aur andar fill karna hota hai.
Trick yeh hai: guess karo ki solution ek product hai u ( x , y ) = X ( x ) Y ( y ) . Isse do variables wali ek mushkil PDE, do aasaan ODEs mein badal jaati hai. Phir hum aisi infinitely many product solutions ko stack karte hain (ek Fourier series) taaki jo bhi boundary data ho usse match kar sakein.
Definition Rectangle par Dirichlet problem
0 < x < a , 0 < y < b par u ( x , y ) dhundho jo satisfy kare
u xx + u y y = 0
charon edges par prescribed values ke saath. Standard "ek nonzero side" case:
u ( 0 , y ) = 0 , u ( a , y ) = 0 , u ( x , 0 ) = 0 , u ( x , b ) = f ( x ) .
Teen homogeneous (zero) sides + ek given side. Koi bhi rectangle problem chaar aise cases se banta hai superposition se (har nonzero side ko alag solve karo, phir jod do).
Ek waqt mein sirf ek nonzero side KYU? Laplace's equation linear aur homogeneous hai, isliye solutions ke sums bhi solutions hote hain. Agar chaaron sides nonzero hain, toh chaar sub-problems mein toddo jisme har ek mein teen zero edges hain, har ek solve karo, aur jod do. Toh ek-side case master karna matlab sab kuch master karna.
Maano u ( x , y ) = X ( x ) Y ( y ) . Plug in karo:
X ′′ ( x ) Y ( y ) + X ( x ) Y ′′ ( y ) = 0.
X ( x ) Y ( y ) se divide karo (valid hai jab dono nonzero ho):
X X ′′ + Y Y ′′ = 0 ⟹ X X ′′ = − Y Y ′′ .
Left side sirf x par depend karti hai, right side sirf y par . Ek pure-x cheez ek pure-y cheez ke barabar sabhi x , y ke liye tabhi ho sakti hai jab dono ek hi constant ke barabar hon. Ise − λ kaho.
X X ′′ = − λ , Y Y ′′ = λ .
Hamare paas u ( 0 , y ) = u ( a , y ) = 0 hai, yaani X ( 0 ) = X ( a ) = 0 . x mein do zero boundary conditions ⇒ hume ek oscillating X chahiye (ek sine), jo λ > 0 force karta hai. λ = k 2 likho.
Common mistake Steel-man: "
X ke liye λ < 0 kyun na choose karein?"
Yeh theek lagta hai kyunki X ′′ = λ X with λ < 0 ke bhi solutions hote hain (e ± k x ). Kyun tempting lagta hai: ODE ki algebra dono taraf kaam karti hai. Kyun galat hai: λ ≤ 0 ke saath X ′′ = λ X ka, X ( 0 ) = X ( a ) = 0 ke saath, sirf trivial solution X ≡ 0 hai. Exponentials/lines do points par bina identically zero hue vanish nahi kar sakते. Fix: boundary conditions sign choose karti hain . Do homogeneous endpoints ⇒ sines ⇒ λ = k 2 > 0 .
X ′′ + k 2 X = 0 ⇒ X = A cos k x + B sin k x .
X ( 0 ) = 0 ⇒ A = 0 . Toh X = B sin k x .
X ( a ) = 0 ⇒ sin k a = 0 ⇒ k a = nπ , n = 1 , 2 , 3 , …
k n = a nπ , X n ( x ) = sin a nπ x
Yeh eigenfunctions hain; λ n = ( nπ / a ) 2 eigenvalues hain.
Y ′′ − k n 2 Y = 0 ⇒ Y = C cosh a nπ y + D sinh a nπ y .
u ( x , 0 ) = 0 ⇒ Y ( 0 ) = 0 ⇒ C = 0 apply karo. Toh
Y n ( y ) = sinh a nπ y .
y mein hyperbolic kyun, sine kyun nahi?
y -direction mein sirf ek homogeneous side hai (y = 0 ) aur ek given side (y = b ). Ek zero BC oscillation force nahi karta. λ ka sign pehle se fix ho chuka hai (positive) x ki wajah se, toh Y ′′ = + k 2 Y growing/decaying solutions cosh , sinh deta hai. Physically: top par inject ki gayi heat plate mein neeche jaate waqt exponentially decay karti hai — yeh wiggle nahi karti.
Har u n = sin a nπ x sinh a nπ y teen zero edges ke saath Laplace solve karta hai. Inhe sum karo:
u ( x , y ) = ∑ n = 1 ∞ B n sin a nπ x sinh a nπ y .
Final condition u ( x , b ) = f ( x ) impose karo:
f ( x ) = ∑ n = 1 ∞ ( B n sinh a nπ b ) sin a nπ x .
Yeh sirf f ki Fourier sine series hai. Bracket coefficient hai:
B n sinh a nπ b = a 2 ∫ 0 a f ( x ) sin a nπ x d x .
Worked example Example 1 — constant top edge
f ( x ) = u 0
Square plate a = b = 1 , u ( x , 1 ) = u 0 , baaki teen sides 0 .
Coefficient integral. ∫ 0 1 u 0 sin ( nπ x ) d x = u 0 ⋅ nπ 1 − c o s nπ = nπ u 0 ( 1 − ( − 1 ) n ) .
Yeh step kyun? Top edge ko match karna constant u 0 ki Fourier sine expansion hai.
Even n ke liye yeh 0 hai, aur odd n ke liye nπ 2 u 0 hai.
sinh nπ se divide karo (y = b = 1 par Y normalization):
B n = s i n h nπ 2 ⋅ nπ u 0 ( 1 − ( − 1 ) n ) .
Kyun? Kyunki u ( x , b ) mein sinh a nπ b ka factor hota hai jise utaarna padta hai.
Answer:
u ( x , y ) = π 4 u 0 ∑ n odd n 1 s i n h ( nπ ) s i n h ( nπ y ) sin ( nπ x ) .
Sanity check: y = 1 par sinh ratio = 1 hai, jo u 0 ke liye classic square-wave Fourier series deta hai. ✓ y = 0 par, sinh 0 = 0 ⇒ u = 0 . ✓
Worked example Example 2 — single-mode top
f ( x ) = 5 sin a 3 π x
Pehle se hi ek single eigenfunction hai! Koi integral nahi chahiye.
Yeh step kyun? f ki sine series mein sirf n = 3 term hai, B 3 sinh a 3 π b = 5 .
u ( x , y ) = 5 s i n h a 3 π b s i n h a 3 π y sin a 3 π x .
Kyun? Har mode independent hai (sines ki orthogonality), toh andar pure mode andar jaaye toh bahar pure mode aata hai — lekin vertically sinh ratio se reshape hoke. Yeh 80/20 insight hai: zyaadatar exam problems mein f pehle se sines mein diya hota hai toh integral skip hota hai.
Worked example Example 3 — LEFT edge nonzero
u ( 0 , y ) = g ( y ) , baaki teen sides zero, 0 < x < a , 0 < y < b par.
Yeh step kyun? Ab homogeneous direction y hai (do zero edges y = 0 , b par), toh roles swap karo: eigenfunctions sin b nπ y , aur x -solution hyperbolic hai.
Kyunki nonzero edge x = 0 par hai (na ki x = a par), sinh b nπ ( a − x ) use karo taaki x = a par vanish ho:
u ( x , y ) = ∑ n B n s i n h b nπ a s i n h b nπ ( a − x ) sin b nπ y , B n = b 2 ∫ 0 b g ( y ) sin b nπ y d y .
Shifted argument kyun? Hume hyperbolic factor chahiye jo opposite (zero) edge x = a par = 0 ho aur x = 0 par matching ke saath = 1 ·ho. sinh b nπ ( a − x ) choose karna exactly yahi karta hai.
Common mistake Aur steel-manned errors
"y mein bhi sin use karunga — symmetry!" Sahi lagta hai kyunki x aur y ∇ 2 u = 0 mein symmetric dikhte hain. Lekin boundary conditions symmetry tod deti hain: jis direction mein do zero edges hain use sines milti hain, doosre ko sinh milta hai. Fix: pehle har direction mein homogeneous edges count karo.
"sinh ( nπ b / a ) se divide karna bhool gaya." Tempting lagta hai kyunki yeh plain Fourier series jaisa dikhta hai. Fix: Fourier coefficient B n nahi balki B n sinh a nπ b ke barabar hai — B n isolate karo.
"Y ke liye cosh use kiya." cosh 0 = 1 = 0 , toh u ( x , 0 ) = 0 fail hota hai. Sirf sinh zero edge par vanish karta hai.
Recall Feynman: 12-saal ke bacche ko samjhao
Ek metal plate imagine karo jo teen sides par thandi (0°) hai aur upar se garam hai. Thodi der baad heat failti hai aur settle ho jaati hai. Hum jaanna chahte hain ki andar har point kitna garam hai. Smart trick: maano temperature "left-right pattern" times "upar-neeche pattern" hai. Left-right pattern dono side-walls par 0 hona chahiye, toh yeh ek wave hai jo exactly fit hoti hai (jaise guitar string: 1 hump, 2 humps, 3 humps...). Upar-neeche pattern ko neeche thande bottom ki taraf jaate waqt khatam hona chahiye — yahi sinh hai, jo bottom par 0 hoti hai aur garam top ki taraf badhti hai. Sabhi guitar-string waves ko sahi "loudness" ke saath joddo taaki top edge match ho — ho gaya!
Mnemonic Split yaad rakho
"Do zeros gaate hain, ek zero jhuulta hai."
Kisi direction mein do homogeneous edges ⇒ sin e (gaana/oscillate karna). Ek homogeneous edge ⇒ sinh (jhuulna/badhna). Aur: "sinh thandi side par marta hai" (sinh 0 = 0 ).
#flashcards/maths
Separation constant ko constant kyun hona chahiye? Ek side sirf x par depend karti hai, doosri sirf y par; sabhi x , y ke liye equal ⇒ dono ek common constant ke barabar hain.
Kaunsi direction ko sine eigenfunctions milti hain aur kyun? Jis direction mein DO homogeneous (zero) boundaries hain; do zero endpoints oscillation force karte hain, sirf sines dono par vanish karti hain.
X ( 0 ) = X ( a ) = 0 ke liye eigenvalues/eigenfunctions kya hain?λ n = ( nπ / a ) 2 , X n = sin ( nπ x / a ) , n = 1 , 2 , …
Yahan y -direction mein sinh kyun (na cosh ya sine)? λ ka sign x se positive fix hota hai, Y ′′ = k 2 Y deta hai ⇒ hyperbolic; Y ( 0 ) = 0 chahiye toh sinh choose karo kyunki sinh 0 = 0 .
Top edge f ( x ) ke liye u ( x , y ) ka final form? ∑ n B n sin a nπ x sinh a nπ y .
B n ka formula?B n = a sinh ( nπ b / a ) 2 ∫ 0 a f ( x ) sin a nπ x d x .
Charon edges nonzero hone par kya karte ho? Chaar problems mein toddo (har ek mein ek nonzero edge), alag-alag solve karo, superpose karo (linearity).
Agar f ( x ) = 5 sin ( 3 π x / a ) ho, toh u kya hoga? 5 sinh ( 3 π b / a ) sinh ( 3 π y / a ) sin a 3 π x (single mode, koi integral nahi).
Y ke liye cosh use karne par kya galat hoga?cosh 0 = 1 = 0 , u ( x , 0 ) = 0 violate hota hai.
Laplace's equation ka physical meaning? Steady state; har point ki value uske neighbours ki average hoti hai (koi source nahi, koi time change nahi).
Fourier sine series — woh engine jo B n compute karta hai.
Separation of variables — general method (bhi use hota hai Heat equation 1D , Wave equation 1D ke liye).
Sturm–Liouville eigenvalue problems — kyun sin ( nπ x / a ) orthogonal eigenfunctions hain.
Hyperbolic functions sinh cosh — decaying/growing y -profiles.
Superposition principle — chaar-edge problem ko split karna.
Harmonic functions and mean value property — "neighbours ki average" wali picture.
Heat equation 1D — same eigenfunctions, lekin sinh ki jagah e − λ t .
Dirichlet problem on rectangle
lambda=k^2 greater than 0
Eigenfunctions sin nPix/a
Fourier series superposition