Solving heat equation — separation of variables: Question bank

4.7.9 · D5 · Maths › Partial Differential Equations › Solving heat equation — separation of variables

Jo machinery probe ho rahi hai: guess $u = X (x) T (t)$ , separation constant $- λ$ , Sturm-Liouville Theory se aane wale sine eigenfunctions, Fourier Series fit, aur Superposition Principle jo modes ko jodhta hai.

True or false — justify karo

"Separation of variables kisi bhi PDE mein, kisi bhi

f (x)

ke saath hamesha kaam karta hai" — yeh claim true hai.

False — iske liye ek linear, homogeneous PDE chahiye with homogeneous boundary conditions, taaki superposition se koi bhi

f

rebuild ho sake; ek nonlinear equation ya nonzero (

u (0, t) = 5

) BC is method ko tod deta hai.

u_{t} = α^{2} u_{xx}

ka har solution product form

u = X (x) T (t)

mein hona chahiye.

False — building blocks

u_{n} = X_{n} T_{n}

is form mein hote hain, lekin actual solution ek sum

\sum B_{n} X_{n} T_{n}

hota hai, jo generally ek single product nahi hota. Separation ek basis dhundh ta hai, poora answer nahi.

Separation constant ko

+ λ

with

λ > 0

likh sakte hain aur same physics milegi.

False —

+ λ

space mein

cosh, sinh

force karta hai (do zero ends satisfy nahi kar sakta) aur time mein

e^{+ λ α^{2} t}

deta hai, jo unbounded grow karta hai. Zero BCs plus physical decay

- λ

λ > 0

force karte hain.

λ = 0

Dirichlet rod ke liye ek valid eigenvalue hai.

False —

λ = 0

X = A + B x

milta hai; dono zero BCs

A = B = 0

force karte hain, sirf trivial solution. Yeh kuch contribute nahi karta, isliye exclude hai.

Higher-frequency modes (

n

large) lower ones se zyada fast decay karte hain.

True — decay rate

α^{2} (nπ / L)^{2} \propto n^{2}

hai, toh wiggly modes exponentially zyada jaldi crush hote hain; yahi reason hai ki heat pehle sharp features smooth karti hai.

Agar ends frozen ki jagah insulated (

u_{x} (0, t) = u_{x} (L, t) = 0

) hote, toh eigenfunctions phir bhi sines hote.

False — Neumann (insulated) BCs cosines

cos (nπ x / L)

select karte hain aur ek

n = 0

constant term add hota hai; Boundary Conditions — Dirichlet vs Neumann dekho. Hamesha eigenfunction ko BC se match karo.

Diffusivity

α^{2}

double karne se eigenfunctions ki shape badal jaati hai.

False —

α^{2}

sirf time factor

e^{- α^{2} (nπ / L)^{2} t}

mein aata hai, toh yeh modes kitni fast fade hoti hain yeh change karta hai, spatial sine shapes nahi.

Is problem ka steady-state (

t \to \infty

) ek non-trivial temperature profile hai.

False — dono ends

0

par hain aur koi source nahi, isliye har mode

\to 0

hoti hai, toh

u \to 0

everywhere. Rod uniformly cold ho jaati hai.

Spot the error

u = X T

, toh

u_{t} = X^{'} T^{'}

." — galti dhundho.

u_{t}

sirf time mein differentiate karta hai:

u_{t} = X T^{'}

(space factor

X

t

ke respect mein constant hai). Similarly

u_{xx} = X^{''} T

. Dono factors ko ek saath differentiate karna galti hai.

"Divide karne ke baad,

\frac{T ^{'}}{α ^{2} T} = \frac{X ^{''}}{X}

— kyunki dono sides equal hain, toh dono

x

aur

t

ki koi function hain." — flaw kahan hai?

Galat conclusion: left side sirf

t

par depend karti hai, right side sirf

x

par; sab

x, t

ke liye equality dono ko same constant hone par force karti hai, koi function nahi. Wahi constant poori baat ka point hai.

X (0) = 0

kyunki hamein bataya gaya hai

u (0, t) = 0

, toh

X (0) T (t) = 0

, toh

T (t) = 0

." — kya galat hai?

Agar

T \equiv 0

toh solution trivially zero everywhere hai. Nontrivial solution rakhne ke liye hum doosre factor ko zero demand karte hain:

X (0) = 0

sin k L = 0

toh

k L = nπ

, se

n = 0, 1, 2, \dots

" — kya

n = 0

allowed hai?

Nahi —

n = 0

X = sin 0 = 0

milta hai, trivial eigenfunction. Hum

n = 1

se start karte hain. (Negative

n

wahi sines sign tak repeat karte hain, kuch naya nahi dete.)

f (x)

ko full Fourier series

\frac{a _{0}}{2} + \sum (a_{n} cos + b_{n} sin)

se fit karo." — yahan yeh galat kyun hai?

Cosines (aur constant)

x = 0

par nonzero hain,

u (0, t) = 0

violate karta hai. Dirichlet zero BCs sirf pure sine series select karte hain.

B_{n} = \int_{0}^{L} f (x) sin \frac{nπ x}{L} d x

." — kya missing hai?

Normalizing factor

\frac{2}{L}

. Kyunki

\int_{0}^{L} sin^{2} \frac{nπ x}{L} d x = \frac{L}{2}

hai, orthogonality deta hai

B_{n} = \frac{2}{L} \int_{0}^{L} f sin d x

. Isko bhool jaane se har coefficient galat scale hoti hai.

"Kyunki har

u_{n}

PDE solve karta hai, unka sum isse sirf tabhi solve karta hai jab hum lucky hote hain." — reasoning correct karo.

Lucky nahi — equation linear aur homogeneous hai, toh Superposition Principle ke according solutions ka koi bhi (yahan tak ki infinite) sum automatically ek solution hai. Isliye hum superpose karne ke liye allowed hain.

Why questions

Hum

u = X T

guess kyun karte hain, derive kyun nahi karte?

Kyunki yeh ek ansatz (trial form) hai: agar yeh solutions produce karta hai toh kuch nahi khoote, aur agar koi product solution exist nahi karta toh hum jaante jab algebra fail hoti. Yeh ek licensed gamble hai jo linear PDEs ke liye pay off karta hai.

Ek hard PDE ko do ODEs mein split karna progress kyun maana jaata hai?

ODEs in a single variable standard techniques (characteristic roots, integrating factors) se solvable hain; PDE ne

x

aur

t

derivatives ko inseparably mix kiya tha. Separation ek hard problem ko do easy problems se trade karta hai.

Space problem ek plain ODE ki jagah eigenvalue problem kyun hona chahiye?

Dono ends par zero BCs

X^{''} + λ X = 0

ko over-constrain karte hain; sirf special

λ

values (eigenvalues) nonzero solutions admit karte hain. Yeh exactly Sturm-Liouville Theory ka setup hai.

Initial condition ko last mein kyun handle kiya jaata hai?

Kyunki yeh akela non-homogeneous condition hai. Pehle hum PDE aur zero BCs satisfy karne wale saare solutions build karte hain, phir coefficients

B_{n}

choose karte hain

f (x)

match karne ke liye — last free knobs.

Long-time profile hamesha ek half-sine bump ki taraf kyun approach karta hai?

Har mode

n \geq 2

n = 1

n^{2}

times faster decay karta hai, toh kaafi time baad sirf

n = 1

term

sin \frac{π x}{L}

meaningfully survive karta hai — ek universal shape starting

f

se independent.

Flat initial profile (

f =

const) sirf odd modes kyun excite karta hai?

Ek constant rod ke midpoint ke baare mein symmetric hai; even sines

sin \frac{2 π x}{L}, sin \frac{4 π x}{L}, \dots

wahan antisymmetric hain, toh unke projection integrals vanish ho jaate hain. Sirf odd sines symmetry share karte hain.

Same

sin \frac{nπ x}{L}

eigenfunction aur Fourier Series basis dono kyun use hota hai?

Kyunki self-adjoint boundary problem ke eigenfunctions orthogonal aur complete hote hain — yeh us basis ki tarah bhi kaam karte hain jisme hum kisi bhi

f

ko expand karte hain. Sturm-Liouville yeh guarantee karta hai.

Edge cases

Kya hota hai agar

f (x)

already ek single sine hai, jaise

sin \frac{3 π x}{L}

Koi integral nahi chahiye — seedha

B_{3} = 1

padh lo, baaki sab

0

, aur uska decay attach karo:

u = sin \frac{3 π x}{L} e^{- 9 α^{2} π^{2} t / L^{2}}

. Ek single mode akele evolve karta hai.

Agar

f (x)

discontinuous ho ya corner ho (jaise step)?

Sine series phir bhi (mean mein) converge karti hai aur solution

t > 0

ke liye valid hai; diffusion discontinuity ko instantly smooth kar deta hai kyunki high-

n

modes damped hain, toh

u

t = 0

ke baad turant smooth ban jaata hai.

t = 0

par endpoints par series se kya

u (x, 0)

predict hota hai agar

f (0) \neq = 0

ho?

Sine series

x = 0, L

par

0

par pinned hai, toh wahan sirf

0

par converge kar sakti hai chahe

f

kuch bhi ho; agar

f (0) \neq = 0

hai toh ends par jump (Gibbs) milta hai — yeh sign hai ki IC BCs ke saath incompatible hai.

Agar initial condition

f (x) = 0

ho toh kya hoga?

Har

B_{n} = 0

, toh

u \equiv 0

sab time ke liye — trivial solution, aur correctly so: ek rod jo cold start karti hai cold ends ke saath kabhi heat nahi hoti.

Jab

α^{2} \to 0

(perfect insulator), solution kya karta hai?

Time factors

e^{- α^{2} (nπ / L)^{2} t} \to 1

, toh

u \to f (x)

time mein frozen — koi heat conduct nahi hoti, profile kabhi change nahi hoti. Math physics se match karta hai.

Agar

L \to \infty

(bahut lamba rod) ho toh kya?

Eigenvalue spacing

\frac{nπ}{L} \to 0

ho jaati hai, discrete sum ek integral ban jaata hai, aur sine series ek sine transform ban jaati hai — wahi idea, continuum version, unbounded rod ke liye.

Recall Ek-line self-test

Ends $0$ par frozen, koi source nahi, wiggly start — kaafi time baad rod kaisi dikhti hai... ::: ek single half-sine hump $sin \frac{π x}{L}$ , cold hoti hui.

Related equations jo yahi machinery reuse karti hain: Wave Equation (second time derivative → decaying ki jagah oscillating time factor) aur Laplace Equation (steady state, bilkul time nahi).