Visual walkthrough — Solving heat equation — separation of variables

4.7.9 · D2 · Maths › Partial Differential Equations › Heat Equation ko Separation of Variables se Solve karna

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Step 0 — Pictures kis cheez ki hain

Kisi bhi algebra se pehle, stage fix karte hain.

• Rod ek line hai $x=0$ se $x=L$ tak. Letter $x$ ka matlab hai "rod par kitni door ho."
• Temperature $u$ ek height hai: har jagah $x$ aur har time $t$ par, $u(x,t)$ batata hai ki woh jagah abhi kitni garm hai. Toh temperature profile ek rod ke upar khinchi hui curve hoti hai — jahan garam ho wahan oonchi, jahan thandi ho wahan neech.
• Dono ends hamesha $0$ par rakhe jaate hain: $u(0,t)=0$ aur $u(L,t)=0$. Picture mein, curve dono ends par floor se chipki rehti hai.

Game ka rule — heat equation khud — yeh hai:

$\underset{\text{garm hone ki rate}}{\underbrace{u_t}}=\underset{\begin{array}{c}\text{heat kitni tezi se}\\ \text{failti hai}\end{array}}{\underbrace{{\alpha }^2}}\underset{\text{curvature}}{\underbrace{u_{xx}}}$

Ise words mein padho: ek jagah wahan tezi se garm hoti hai jahan curve upar ko modhti hai (ek dip), aur wahan thandi hoti hai jahan neech ko modhti hai (ek bump). Yeh akela sentence poori physics hai; neeche sab kuch ise solve karna hai.

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Step 1 — Guess: shape × dimmer-knob

KYA. Hum guess karte hain ki answer ki ek khaas shape hai: $u(x,t)=X(x)T(t).$ $X(x)$ rod ke saath ek fixed shape hai (sirf jagah par depend karta hai). $T(t)$ ek single dimmer knob hai (sirf time par depend karta hai) jo us poori shape ko upar ya neeche scale karta hai.

KYUN. Agar shape kabhi nahi badalti aur sirf uski overall brightness fade hoti hai, toh mushkil PDE — jo space aur time ko mix karta hai — do alag aasaan problems mein split ho sakti hai. Hum ise try karte hain kyunki agar yeh fail ho, toh algebra hame turant bata deta hai, aur hum ne kuch nahi khoya.

PICTURE. Frozen shape $X(x)$ har time par wahi curve hai; knob $T(t)$ use waqt ke saath floor ki taraf dabata jaata hai.

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Step 2 — Equation ko split hone par majboor karo

KYA. Guess ko $u_t={\alpha }^2{u}_{xx}$ mein daalo.

Kyunki $X$ time ko ignore karta hai aur $T$ space ko: $u_t=X(x)T^{\prime }(t),u_{xx}=X^{\prime \prime }(x)T(t).$ Yahan $T^{\prime }$ ka matlab hai "knob kitni tezi se badalti hai," aur $X^{\prime \prime }$ ka matlab hai "shape ki curvature." Substitute karne par: $X{T}^{\prime }={\alpha }^2{X}^{\prime \prime }T.$

Ab dono sides ko ${\alpha }^{2}XT$ se divide karo: $\underset{\text{sirf }t\text{ yahan hai}}{\underbrace{\frac{T^{\prime }}{{\alpha }^{2}T}}}=\underset{\text{sirf }x\text{ yahan hai}}{\underbrace{\frac{X^{\prime \prime }}{X}}}.$

KYUN. Left side tab hi badalti hai jab time badalta hai; right side tab hi badalti hai jab position badalta hai. Agar main ruka rahun ($x$ fixed) aur time chalne dun, toh right side freeze hai — toh left side bhi nahi hilegi. Agar main time freeze karun aur rod par chalun, toh left freeze hai — toh right bhi nahi hil sakti. Sirf ek hi tarika hai jis se aisi do cheezein hamesha equal rahen — agar dono ek hi constant hon.

PICTURE. Do dials hamesha ek hi number dikhate hain: "time" dial ghoomana reading nahi badal sakta, aur "space" dial ghoomana bhi nahi — toh reading lock hai.

Us locked constant ka naam $-\lambda$ rakhte hain: $\frac{X^{\prime \prime }}{X}=-\lambda ,\frac{T^{\prime }}{{\alpha }^{2}T}=-\lambda .$

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Step 3 — Space problem, aur kyun $\lambda$ ka sign sab kuch decide karta hai

KYA. $\frac{X^{\prime \prime }}{X}=-\lambda$ se hame space ODE milti hai: $X^{\prime \prime }+\lambda X=0,X(0)=0,X(L)=0.$ Boundary conditions pinned ends se aate hain: $u(0,t)=X(0)T(t)=0$ sab $t$ ke liye, aur $T$ zero nahi hai (woh toh ek dead rod hogi), toh $X(0)=0$. Same $L$ par.

Teen signs kyun test karein? $X^{\prime \prime }+\lambda X=0$ ke solution ki shape is par bilkul alag hoti hai ki $\lambda$ negative hai, zero hai, ya positive. Hame teeno check karne chahiye — reader ko kabhi nahi sochna chahiye "lekin agar $\lambda <0$ ho toh?"

PICTURE. Teen candidate shapes, har ek dono ends par floor ko touch karne ki koshish kar raha hai:

• Case $\lambda <0$ ($\lambda =-{\mu }^2$ likho): solutions hain $X=A{e}^{\mu x}+B{e}^{-\mu x}$ — pure growing/shrinking exponentials, koi wiggle nahi. Sirf exponentials se bani curve do baar floor par nahi dip kar sakti aur wapas aa sakti. Dono ends par pin karne se $A=B=0$ force hota hai: flat-zero line. Trivial — reject.
• Case $\lambda =0$: solutions seedhi lines hain $X=A+Bx$. $x=0$ aur $x=L$ dono par floor se guzarne wali line floor hi hai. $A=B=0$. Trivial — reject.
• Case $\lambda >0$ ($\lambda =k^2$ likho): solutions hain $X=A\cos kx+B\sin kx$ — asli waves. Waves wapas floor par aa sakti hain. Yeh eklauta case hai jis mein zindagi hai. Rakho.

Achhe case ko finish karo. $X(0)=0$ cosine ko khatam kar deta hai (kyunki $\cos 0=1\ne 0$), $X=B\sin kx$ bachta hai. Phir $X(L)=0$ ke liye chahiye: $\sin kL=0⟹kL=n\pi ,n=1,2,3,\dots$ Sirf yeh khaas $k$ wave ko $x=L$ par exactly floor par land karne dete hain.

Toh surviving shapes — eigenfunctions — aur unke eigenvalues hain: $\boxed{X_n(x)=\sin \frac{n\pi x}{L},{\lambda }_n={\left(\frac{n\pi }{L}\right)}^2}$ $n=1$ ek hump hai, $n=2$ do humps hain, wagera — har ek rod mein poori number of half-waves fit karta hai. ("Boundary" ko kaun si shapes survive karti hain, yeh sawaal Sturm-Liouville Theory ka dil hai.)

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Step 4 — Time problem: dimmer fade hota hai

KYA. Ab ${\lambda }_n$ ko time side $\frac{T^{\prime }}{{\alpha }^{2}T}=-{\lambda }_n$ mein daalo, yaani $T^{\prime }=-{\lambda }_n{\alpha }^{2}T.$ Yeh kehta hai "knob ki change ki rate khud ka negative multiple hai" — exponential decay ki pehchaan. Iska solution: $\boxed{T_n(t)=e^{-{\alpha }^2{\lambda }_{n}t}=e^{-{\alpha }^2(n\pi /L{)}^{2}t}}$

KYUN. $T^{\prime }=(\text{negative})\times T$ woh ek equation hai jiska answer ek shrinking exponential hai. Har mode ko apna decay speed milta hai: exponent mein ${\lambda }_n=(n\pi /L{)}^2$ hai, toh mode $n$, $n^2$ ke proportional rate se fade karta hai. Ab Step 2 ke minus sign ka faida dikha — $+\lambda$ ke saath yeh exponential grow karta aur rod jal jaati, jo frozen ends ke saath impossible hai.

PICTURE. Mode number jitna zyada, zero tak plunge utna steep.

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Step 5 — Superpose: kai fading humps add karo

KYA. Har ek single mode $u_n(x,t)=\sin \frac{n\pi x}{L}{e}^{-{\alpha }^2(n\pi /L{)}^{2}t}$ PDE aur dono pinned-end conditions solve karta hai. Kyunki heat equation linear aur homogeneous hai (koi $u^2$ nahi, koi stray constant nahi), solutions ka koi bhi sum phir se ek solution hai — yeh Superposition Principle hai. Toh hum har mode ko ek saath allow karte hain, har ek ki apni strength $B_n$ ke saath: $\boxed{u(x,t)=\sum _{n=1}^{\infty }{B}_n\sin \frac{n\pi x}{L}{e}^{-{\alpha }^2(n\pi /L{)}^{2}t}}$

KYUN. Ek pure sine almost kabhi bhi real starting profile se match nahi karti. Lekin sines ko stack karna koi bhi pinned-end shape bana sakta hai — exactly yahi ek Fourier Series karta hai. $B_n$ har hump ki volume settings hain.

PICTURE. Ek tall one-hump, ek shorter two-hump, ek smaller three-hump stack karo… aur unka sum ek lumpy profile hai — initial temperature.

Har symbol, jahan bhi hai: $u(x,t)={\sum }_{n\ge 1}\underset{\begin{array}{c}\text{mode }n\text{ kitna}\\ \text{loud hai}\end{array}}{\underbrace{B_n}}\underset{\begin{array}{c}\text{uski frozen}\\ \text{shape}\end{array}}{\underbrace{\sin \frac{n\pi x}{L}}}\underset{\begin{array}{c}\text{uska personal}\\ \text{fade timer}\end{array}}{\underbrace{e^{-{\alpha }^2(n\pi /L{)}^{2}t}}}$

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Step 6 — Start fit karo: har $B_n$ dhundho

KYA. $t=0$ par har fade timer $1$ ke barabar hota hai ($e^0=1$), toh solution ko initial profile $f(x)$ ban-na chahiye: $f(x)={\sum }_{n=1}^{\infty }{B}_n\sin \frac{n\pi x}{L}.$ Ek coefficient $B_m$ nikalne ke liye, dono sides ko $\sin \frac{m\pi x}{L}$ se multiply karo aur rod par integrate karo. Jaadu hai orthogonality: ${\int }_0^L\sin \frac{n\pi x}{L}\sin \frac{m\pi x}{L}dx=\{\begin{array}{ll}L/2,& n=m\\ 0,& n\ne m\end{array}$ Har mismatched pair integrate hokar zero ho jaata hai, toh sirf $n=m$ term bachta hai: ${\int }_0^{L}f(x)\sin \frac{m\pi x}{L}dx=B_m\cdot \frac{L}{2}⟹\boxed{B_m=\frac{2}{L}{\int }_0^{L}f(x)\sin \frac{m\pi x}{L}dx}$

KYUN. Orthogonality ek sieve hai: ek mode se multiply karke integrate karna us mode ki strength filter kar leta hai jabki baaki sab ko mita deta hai. $\frac{2}{L}$ sirf mode ke apne "squared length" $L/2$ se divide karna hai.

PICTURE. Matching-frequency sines ek positive area tak overlap karti hain (woh sieve se nikal jaati hain); alag-frequency sines zero area tak cancel ho jaati hain.

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Ek-picture summary

Pipeline ko upar se neeche padho: guess → split → sine shapes → decay timers → sum → fit. Neecha panel asli payoff dikhata hai: ek lumpy hot start, time ke saath dekha gaya, hamesha ek single half-sine hump mein relax hota hai (mode $n=1$ sabse zyada time tak rehta hai kyunki yeh sabse dhire decay karta hai), phir woh hump bhi ek cold rod mein doob jaata hai.

Recall Feynman: poora walkthrough plain words mein

Hamare paas ek hot rod thi jiske ends frozen the aur hum uska future jaanna chahte the. Pehle humne guess kiya ki temperature ek fixed shape hai jise ek fading dimmer-knob se multiply kiya gaya hai. Woh daalne par equation space part aur time part mein split ho gayi jo ek hi locked number ke barabar rehne par majboor thi — kyunki ek part sirf position ke saath badal sakta tha aur doosra sirf time ke saath, toh koi bhi hil nahi sakta tha. Us number ka sign test karne par, sirf ek positive choice ne asli wave shapes di jo frozen floor ko dono ends par touch kar sakti thi; woh shapes sine humps nikle jo rod mein poori number of half-waves fit karti hain. Time part ek fading exponential ban gaya, aur pateeli humps (zyada wiggles) bahut tezi se fade hoti hain. Kyunki equation linear thi, humne sab humps add kar diye, har ek apne timer par fade hota hua. Aakhir mein, asli starting warmth match karne ke liye, humne har hump ki loudness ek Fourier sieve se nikali — ek hump se multiply karo, integrate karo, aur har doosra hump cancel ho jaata hai. Time chalne do, aur rod hamesha cold hone se pehle ek single fat half-sine bump mein pighal jaati hai. Yahi poori kahaani hai.

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Related builds: Wave Equation (same separation, oscillating time part), Laplace Equation (steady state, bilkul time nahi), aur Fourier Series coefficient machinery ke liye.