Solving heat equation — separation of variables

4.7.9 · Maths › Partial Differential Equations

Step 1 — Separation guess. Maano $u (x, t) = X (x) T (t)$ (space-part times time-part). Yeh step kyun? Agar yeh kaam kare, toh derivatives nicely split ho jaate hain; try karke hum kuch nahi kho rahe — agar is form ka koi solution exist nahi karta, toh hum khud jaanenge.

Tab $u_{t} = X T^{'}$ aur $u_{xx} = X^{''} T$ . Substitute karo: $X T^{'} = α^{2} X^{''} T .$

Step 2 — Variables ko separate karo. $α^{2} X T$ se divide karo: $\frac{T ^{'}}{α ^{2} T} = \frac{X ^{''}}{X} .$ Yeh step kyun? Ab left side sirf $t$ par depend karti hai, right side sirf $x$ par. Ek function of $t$ ka ek function of $x$ ke barabar hone ka ek hi tarika hai sabhi $x, t$ ke liye — dono same constant ke barabar hon. Ise $- λ$ bolte hain. Minus sign kyun? Foresight: hume time mein decaying solutions chahiye, jiske liye $- λ$ with $λ > 0$ chahiye. Hum baaki sign cases verify karenge ki woh fail karte hain.

$\frac{X ^{''}}{X} = - λ, \frac{T ^{'}}{α ^{2} T} = - λ .$

Step 3 — Space ODE solve karo (ek eigenvalue problem). $X^{''} + λ X = 0, X (0) = 0, X (L) = 0.$ Yeh BCs kyun? Kyunki $u (0, t) = X (0) T (t) = 0$ sabhi $t$ ke liye, aur $T \neq \equiv 0$ (warna trivial), toh $X (0) = 0$ chahiye. Usi tarah $L$ par.

$λ$ ka sign test karo (Forecast-then-Verify):

$λ < 0$ , maano $λ = - μ^{2}$ : $X = A e^{μx} + B e^{- μx}$ . BCs force karte hain $A = B = 0$ . Sirf trivial. ✗
$λ = 0$ : $X = A + B x$ . $X (0) = 0 \Rightarrow A = 0$ ; $X (L) = 0 \Rightarrow B = 0$ . Trivial. ✗
$λ > 0$ , maano $λ = k^{2}$ : $X = A cos k x + B sin k x$ . $X (0) = 0 \Rightarrow A = 0$ , toh $X = B sin k x$ . $X (L) = 0 \Rightarrow sin k L = 0 \Rightarrow k L = nπ$ .

Toh eigenvalues aur eigenfunctions hain: $λ_{n} = (\frac{nπ}{L})^{2}, X_{n} (x) = sin \frac{nπ x}{L}, n = 1, 2, 3, \dots$

Step 4 — Time ODE solve karo. $T^{'} = - λ α^{2} T \Rightarrow T_{n} (t) = e^{- λ_{n} α^{2} t} = e^{- α^{2} (nπ / L)^{2} t} .$ Yeh step kyun? Yeh first-order linear hai; iska solution pure exponential decay hai — exactly woh "fading knob."

Step 5 — Product solutions banao aur superpose karo. Har $u_{n} = sin \frac{nπ x}{L} e^{- α^{2} (nπ / L)^{2} t}$ PDE + BCs solve karta hai. Equation linear aur homogeneous hai, toh solutions ke sums bhi solutions hain: $u (x, t) = n = 1 \sum \infty B_{n} sin \frac{nπ x}{L} e^{- α^{2} (nπ / L)^{2} t}$

Step 6 — Initial condition fit karo Fourier coefficients se. $t = 0$ par exponentials $1$ hain: $f (x) = \sum_{n = 1}^{\infty} B_{n} sin \frac{nπ x}{L} .$ Yeh ek Fourier sine series hai. $sin \frac{mπ x}{L}$ se multiply karo, $0$ se $L$ tak integrate karo, orthogonality use karo $\int_{0}^{L} sin \frac{nπ x}{L} sin \frac{mπ x}{L} d x = \frac{L}{2} δ_{nm}$ : $B_{n} = \frac{2}{L} \int_{0}^{L} f (x) sin \frac{nπ x}{L} d x$

Figure — Solving heat equation — separation of variables

Worked examples

Common mistakes

Recall Feynman: 12-saal ke bachche ko explain karo

Socho ek metal ki stick jo beech mein hot hai aur ends par cold hai, jise tum ice mein duba ke rakho taaki ends $0$ ° par rahen. Heat hamesha hot spots se cold spots ki taraf creep karti hai, toh hot middle slowly cool hoti hai aur spread out hoti hai. Hum pretend karte hain ki temperature pattern simple "wave shapes" (humps) se bani hai. Har hump apne timer par fade hota hai — skinny humps (bahut saare wiggles wale) super fast fade hote hain, fat humps slowly fade hote hain. Future predict karne ke liye tum sirf har hump ko uske apne timer par fade hone dete ho aur unhe wapas add kar dete ho. Lamba time baad sirf sabse bada mota hump bachta hai, aur phir woh bhi disappear ho jaata hai. Cold stick. Done.

Flashcards

u_{t} = α^{2} u_{xx}

ke liye separation of variables kaunsa guess se start karta hai?

u (x, t) = X (x) T (t)

, ek space function aur ek time function ka product.

u = X T

substitute karne ke baad, dono sides ko constant set karna kyun justify hota hai?

Ek side sirf

t

par depend karti hai, doosri sirf

x

par; sabhi

x, t

ke liye barabar ⇒ dono constant

- λ

ke barabar hain.

Separation constant

- λ

with

λ > 0

kyun liya jaata hai?

Decaying time solutions

e^{- λ α^{2} t}

aur zero BCs satisfy karne wale nontrivial sine eigenfunctions pane ke liye.

X^{''} + λ X = 0

X (0) = X (L) = 0

ke eigenvalues aur eigenfunctions kya hain?

λ_{n} = (nπ / L)^{2}

X_{n} = sin (nπ x / L)

n = 1, 2, \dots

Mode

n

ka time factor kya hai?

e^{- α^{2} (nπ / L)^{2} t}

Dirichlet heat problem ka full solution kya hai?

u = \sum_{n} B_{n} sin (nπ x / L) e^{- α^{2} (nπ / L)^{2} t}

Coefficients

B_{n}

ka formula kya hai?

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

Kaunsa mode sabse fast decay karta hai aur kyun?

High

n

(wiggly modes); decay rate

\propto n^{2}

Lamba time baad rod ki profile kaisi dikhti hai?

Ek single half-sine bump

sin (π x / L)

(

n = 1

mode) jo zero ki taraf shrink ho raha hota hai.

Yahan sines kyun hain, cosines kyun nahi?

Zero (Dirichlet) end conditions; cosines

x = 0

par nonzero hain,

u (0, t) = 0

violate karte hain.

B_{n}

kaunsa orthogonality relation deta hai?

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

Connections

Fourier Series — coefficients $B_{n}$ exactly Fourier sine coefficients hain.
Sturm-Liouville Theory — $X^{''} + λ X = 0$ ek Sturm–Liouville eigenvalue problem hai; orthogonality guaranteed hai.
Wave Equation — same separation method, lekin time ODE oscillatory hai ( $T^{''} = - λ c^{2} T$ ) decaying nahi.
Laplace Equation — steady-state heat ( $u_{t} = 0$ ) $u_{xx} = 0$ deta hai.
Boundary Conditions — Dirichlet vs Neumann — Neumann (insulated) ends cosine series dete hain.
Superposition Principle — isliye product solutions ka sum legal hai (linear homogeneous PDE).

Solving heat equation — separation of variables

HUM KYA SOLVE KAR RAHE HAIN?

KAISE: scratch se derivation