Heat equation (parabolic) 1D — derivation from Fourier's law: Question bank

4.7.8 · D5 · Maths › Partial Differential Equations › Heat equation (parabolic) 1D — derivation from Fourier's law

True or false — justify karo

Woh point jahan temperature profile concave up ho (

u_{xx} > 0

), abhi cool ho rahi hai.

False. Concave up ek valley hai; uske neighbours zyada garam hain, toh heat andar flow karti hai aur

u_{t} = α u_{xx} > 0

— woh warm hota hai. Hills (

u_{xx} < 0

) cool hoti hain.

Heat equation time reverse karne par symmetric hai (

t \to - t

), jaise Newton ke laws hote hain.

False.

t \to - t

replace karne par

u_{t} = α u_{xx}

ban jaati hai

- u_{t} = α u_{xx}

, jo ek alag (backward, unstable) equation hai. Heat spread hoti hai lekin kabhi spontaneously un-spread nahi hoti — woh broken symmetry hi time ka arrow hai.

Agar temperature profile har jagah ek straight line hai, toh koi bhi point apna temperature change nahi karta.

True. Ek straight line mein har point par

u_{xx} = 0

hoti hai, toh

u_{t} = α \cdot 0 = 0

. Linear profile 1D steady state hai.

Thermal conductivity

k

ko double karne par rod hamesha exactly double speed se equilibrate hoti hai.

Generally nahi. Time-scale

α = k / (ρ c)

se set hota hai;

k

double karne par

α

double hoti hai sirf tabhi jab

ρ c

fixed rahe. Bade

ρ c

wali material mein bade

k

ke bawajood modest

α

ho sakti hai.

Flux

q = - k u_{x}

nonzero ho sakta hai even jahan temperature

u

khud zero ho.

True. Flux slope

u_{x}

par depend karta hai,

u

ki value par nahi.

0 K

par baitha ek point steep gradient par abhi bhi heat pass karta hai.

Rod ke ek end par disturbance ko far end tak feel hone mein kuch finite time lagta hai.

False (mathematically). Heat equation parabolic hai jisme infinite signal speed hoti hai: koi bhi change instantly har jagah feel hoti hai, halaanki door ka magnitude exponentially tiny hota hai. (Wave equation finite-speed wali hoti hai.)

Temperature ke local maximum par, heat equation force karti hai ki wahan temperature non-increasing rahe.

True. Local max par

u_{xx} \leq 0

hoti hai, toh

u_{t} = α u_{xx} \leq 0

. Yahi maximum principle ki neenv hai: interior hot spots sirf fade ho sakte hain.

Equation

ρ c u_{t} = k u_{xx} + f

mein

f > 0

hone par ek straight-line profile bhi heat up ho sakti hai.

True.

u_{xx} = 0

ke saath akela source deta hai

u_{t} = f / (ρ c) > 0

. Ek heat source flat profile ko garam karta hai jo otherwise steady rehti.

Galti dhundho

"Fourier's law hai

q = + k u_{x}

kyunki zyada gradient matlab zyada flux."

Sign galat hai. Heat hot se cold ki taraf flow karti hai, yaani gradient ke neeche, toh

q = - k u_{x}

. Magnitude gradient ke saath badhti zaroor hai — lekin direction

u_{x}

ke opposite hoti hai.

"Kyunki energy conserved aur reversible hai, heat equation

u_{tt} = α u_{xx}

honi chahiye, time mein second order."

Energy conservation deta hai

ρ c u_{t} = - q_{x}

, jo

t

mein first order hai — storage

ρ c u

mein ek time derivative hai, Fourier's flux mein koi nahi. Sirf

u_{tt}

ke liye ek second physical law chahiye hota (jaise waves ke liye Newton ka). Diffusion reversible nahi hai.

"Diffusivity bas conductivity hai,

α = k

Storage term missing hai.

α = k / (ρ c)

ke units

m^{2} / s

hain, jabki

k

ke units

W/m\cdotK

hain — yeh dimensionally bhi same nahi hain, toh

α = k

impossible hai.

"Step 5 mein right-hand side

+ \partial q / \partial x

ban jaata hai."

Yeh $- \partial q / \partial x$ banta hai. Net-in

= q (x) - q (x + Δ x)

, toh

Δ x

se divide karne par difference quotient ka negative milta hai, yaani

- q_{x}

. Woh minus drop karne par poori PDE ka sign palat jaata hai.

u_{xx}

temperature ka slope measure karta hai, toh

u_{xx} = 0

matlab temperature constant hai."

u_{xx}

curvature hai, slope nahi;

u_{x}

slope hai.

u_{xx} = 0

matlab profile straight hai (constant slope), jo rod ke across abhi bhi rise kar sakta hai — zaroori nahi ki constant ho.

"Kyunki

q (x, t) - q (x + Δ x, t)

net heat in hai, aur heat leave hona bura hai, toh hamen net-in

= q (x + Δ x) - q (x)

likhna chahiye."

Nahi. Dono fluxes

+ x

direction mein point karti hain. Energy left face se rate

q (x)

par enter karti hai aur right face se rate

q (x + Δ x)

par leave karti hai, toh net gain

= q (x) - q (x + Δ x)

. Proposed version in aur out ko reverse kar deta hai.

"Steady state matlab

u = 0

har jagah."

Steady state matlab

u_{t} = 0

, yaani

u_{xx} = 0

, jo deta hai ek linear profile

u = a + b x

. Temperature large aur space mein varying ho sakti hai; bas time mein change nahi hoti.

Why questions

Ek single

α

kyun

u_{xx}

ko multiply karta hai separate constants

k

ρ

c

ki jagah?

Balance

ρ c u_{t} = k u_{xx}

ρ c

se divide karne par teeno material constants collapse ho jaate hain ek single ratio

α = k / (ρ c)

mein, jo akela combination hai jo diffusion time-scale set karta hai.

Fourier's law mein minus sign kyun zaroori hai taaki final PDE physically stable rahe?

Minus

- \partial_{x} (- k u_{x})

+ k u_{xx}

mein badal deta hai, jisse milta hai

u_{t} = + α u_{xx}

— smoothing aur stable. Missing minus deta hai

u_{t} = - α u_{xx}

, backward heat equation, jo tiny wiggles ko explosively amplify karta hai.

Parent derivation ke Step 2 mein hum integral ke andar differentiate kyun kar sakte hain?

Integration limits

x

aur

x + Δ x

space mein fixed hain; sirf integrand

t

par depend karta hai, toh

d / d t

integral ke through pass hoti hai aur sirf

u

ko hit karti hai, deta hai

\int ρ c u_{t} A d s

Curvature (slope nahi) kyun decide karti hai ki ek point warm hoga ya cool?

Jo matter karta hai woh yeh hai ki ek point apne neighbours ke average se hotter hai ya colder. Curvature

u_{xx}

exactly woh gap measure karti hai; slope

u_{x}

sirf yeh batata hai ki temperature kis taraf tilt hai, jo akele point mein koi net energy nahi laata.

Heat equation ko parabolic classify kyun kiya jaata hai, hyperbolic nahi?

0 \cdot u_{tt} + 0 \cdot u_{x t} - α u_{xx} + u_{t} = 0

likhne par, second-order part mein

B^{2} - A C = 0

hota hai (kyunki

A = 0

B = 0

). Zero discriminant parabolic case hai — diffusive, ek real characteristic direction.

e^{- α π^{2} t} sin (π x)

jaisa solution grow kyun nahi karta, decay kyun karta hai?

Mode

sin (π x)

ki curvature

- π^{2} sin (π x)

hai, toh uski peaks hills hain jo cool hoti hain aur troughs valleys hain jo warm hoti hain — diffusion use flatten karta hai, aur math ise shrinking factor

e^{- α π^{2} t}

mein pack kar deta hai.

Edge cases

Agar initial temperature poori rod par uniform (constant) ho, toh

u_{t}

har jagah kya hogi?

Dono

u_{x} = 0

aur

u_{xx} = 0

hain, toh

u_{t} = 0

: ek uniform rod already equilibrium mein hai aur wahin rehti hai (bina kisi source ke).

Temperature profile ke ek smooth interior peak par flux

q

ka kya hota hai?

Peak par

u_{x} = 0

, toh

q = - k u_{x} = 0

us single point par — lekin

q

dono taraf nonzero hai, peak se door flow kar raha hai, isliye peak abhi bhi cool hoti hai.

Agar

α = 0

ho (perfect insulator,

k = 0

), toh equation kya kehti hai?

u_{t} = 0

har jagah: conductivity zero hone par koi heat move nahi hoti, toh temperature apni initial profile par forever frozen rehti hai, chahe woh kitni bhi lumpy ho.

Sharp square-wave initial temperature ke bilkul pehle instant mein, corner point ka behaviour well-defined hai?

Nahi — ek corner mein undefined (infinite) curvature

u_{xx}

hoti hai, toh

u_{t}

wahan

t = 0

par finite nahi hoti. Equation corner ko instantly smooth kar deti hai; kisi bhi

t > 0

ke liye profile smooth hoti hai aur sab kuch well-defined hota hai.

Insulated boundary par (koi heat cross nahi hoti), Fourier's law profile par kya condition force karti hai?

Zero flux matlab

q = - k u_{x} = 0

, toh us end par

u_{x} = 0

— temperature profile insulated wall se zero slope ke saath milti hai (wall ke against flat).

Agar source

f

flat profile par negative ho (heat sink), toh temperature kis taraf jaati hai?

ρ c u_{t} = k \cdot 0 + f = f < 0

, toh

u_{t} < 0

: ek perfectly flat, otherwise-steady profile bhi wahan cool hoti hai jahan energy drain ho rahi hai.

Connections

Heat equation (parabolic) 1D — derivation from Fourier's law
Fourier's Law of Conduction
Conservation Laws and Continuity Equation
Classification of PDEs (elliptic, parabolic, hyperbolic)
Wave Equation (hyperbolic) 1D
Laplace Equation (steady-state heat)
Thermal Diffusivity and Material Properties
Separation of Variables for the Heat Equation
Fourier Series Solutions