Solving heat equation — separation of variables: Worked examples

4.7.9 · D3 · Maths › Partial Differential Equations › Solving heat equation — separation of variables

Yeh page ek shooting gallery hai. Hum har tarah ki situation list karte hain jo linear-homogeneous heat problem aapke saamne rakh sakta hai, phir har ek ko ek poore worked example se solve karte hain. Parent recipe paas mein rakho: Solving heat equation — separation of variables.

Recall "Har scenario" ke baare mein ek baat (scope, honestly boli)

Yeh page linear homogeneous rod ko poora cover karta hai: sabhi homogeneous boundary types — Dirichlet ( $u = 0$ ), Neumann ( $u_{x} = 0$ ), aur mixed / Robin ends ka ek note — plus har tarah ki input shape (single/multiple/constant/piecewise/zero) aur dono time limits. Jo hum yahan nahi cover karte (unhe extra machinery chahiye aur woh doosre pages par hain): non-homogeneous BCs (ends jo nonzero ya time-dependent temperature par rakhe hain — steady state subtract karke handle hota hai), source terms $u_{t} = α^{2} u_{xx} + Q$ , aur full Robin eigenvalue transcendental equations. Toh "har scenario" ka matlab hai homogeneous separable problem ke har scenario, jo is chapter mein build kiya gaya hai. Sabhi BC types ke peeche eigenfunction/eigenvalue machinery yahan hai: Sturm-Liouville Theory.

The scenario matrix

Neeche har worked example us cell ke saath tagged hai jise woh cover karta hai.

Cell	Kya special hai	Example
A. Single mode	$f (x)$ already ek sine hai — read off karo, koi integral nahi	Ex 1
B. Sum of modes	$f (x)$ kuch sines ka sum hai — kai read off karo	Ex 2
C. Constant profile	$f (x) =$ const — integrate karna padega, sirf odd $n$ bachte hain	Ex 3
D. Triangle / tent	piecewise-linear peak — integration by parts	Ex 4
E. Degenerate input	$f (x) \equiv 0$ — trivial (dead) rod, uniqueness	Ex 5
F. Limiting behaviour	$t \to \infty$ aur $t \to 0^{+}$ — convergence ki subtleties	Ex 6
G. Neumann + mixed twist	insulated $u_{x} = 0$ (cosines) aur ek mixed end	Ex 7
H. Word problem	real rod, real numbers, "cool hone mein kitna time?"	Ex 8

Cases A–H cover karte hain: single/multiple/constant/piecewise inputs, zero-degenerate input, dono time limits (convergence ke saath), Neumann aur mixed boundary families, aur ek physical estimate.

Ex 1 — Cell A: ek single sine mode

Worked example Statement

$0 < x < L$ par rod problem solve karo jahan $u (0, t) = u (L, t) = 0$ aur initial shape $f (x) = 5 sin \frac{2 π x}{L} .$

Forecast: Aage padhne se pehle — kya tumhe lagta hai yahan integral compute karna padega? Guess karo kaun se $B_{n}$ nonzero hain.

Step 1 — Template se match karo. $t = 0$ par general series hai $f (x) = \sum_{n} B_{n} sin \frac{nπ x}{L}$ . Yeh step kyun? Kyunki $sin \frac{nπ x}{L}$ orthogonal hain (har ek independent hai), left side ka sine sirf same sine se right side par match kar sakta hai. Agar $f$ pehle se unme se ek hai, toh koi integral nahi chahiye — bas coefficients read karo.

Step 2 — Read off karo. $5 sin \frac{2 π x}{L}$ ko $n = 2$ term se compare karo: $B_{2} = 5$ , aur baaki har $B_{n} = 0$ .

Step 3 — Decay attach karo. Mode $n = 2$ decay karta hai $e^{- α^{2} (2 π / L)^{2} t} = e^{- 4 α^{2} π^{2} t / L^{2}}$ ki tarah. 4 kyun? Decay rate $\propto n^{2} = 2^{2} = 4$ hai. $u (x, t) = 5 sin \frac{2 π x}{L} e^{- 4 α^{2} π^{2} t / L^{2}} .$

Verify karo: $t = 0$ par, $e^{0} = 1$ , toh $u (x, 0) = 5 sin \frac{2 π x}{L} = f (x)$ . ✓ Ends par: $sin 0 = 0$ aur $sin 2 π = 0$ , toh $u (0, t) = u (L, t) = 0$ . ✓

Ex 2 — Cell B: modes ka sum

Worked example Statement

Same rod, initial shape ek combination: $f (x) = 3 sin \frac{π x}{L} - 2 sin \frac{4 π x}{L} .$

Forecast: Final sum mein kitne terms honge? $3$ -term ya $2$ -term, kaun zyada jaldi fade hoga?

Step 1 — Har piece read off karo. Yeh step kyun? Same orthogonality reasoning jaise Ex 1 — sines ka linear combination term-by-term map hota hai. $B_{1} = 3$ , $B_{4} = - 2$ , baaki sab $0$ .

Step 2 — Har ek ko apna timer do. $n = 1$ decay karta hai rate $α^{2} (π / L)^{2}$ par; $n = 4$ decay karta hai rate $16 α^{2} (π / L)^{2}$ par (kyunki $4^{2} = 16$ ). $u (x, t) = 3 sin \frac{π x}{L} e^{- α^{2} π^{2} t / L^{2}} - 2 sin \frac{4 π x}{L} e^{- 16 α^{2} π^{2} t / L^{2}} .$

Step 3 — Alag-alag fates observe karo. $n = 4$ hump, $n = 1$ hump se $16/1 = 16$ guna zyada tezi se decay karta hai. Superposition Principle ki wajah se hum independently-fading pieces ko simply add karte hain.

Verify karo: $t = 0$ par $3 sin \frac{π x}{L} - 2 sin \frac{4 π x}{L}$ wapas milta hai. ✓ Decay rates ka ratio $16$ hai. ✓

Ex 3 — Cell C: ek constant (flat hot) rod

Worked example Statement

$L = π$ , $α^{2} = 1 m^{2} / s$ , aur poori rod $u (x, 0) = 100$ ° par start hoti hai, ends $0$ ° par dunk kiye hain.

Forecast: Ek flat line bilkul sine jaisi nahi hoti — kya tumhe lagta hai sabhi $B_{n}$ , sirf odd, ya sirf even honge?

Step 1 — Projection integral set up karo. Yeh step kyun? Ek constant ek single eigenfunction nahi hai, toh humein ise coefficient formula use karke har sine par "project" karna padega. $B_{n} = \frac{2}{π} \int_{0}^{π} 100 sin (n x) d x .$

Step 2 — Integral karo. $\int_{0}^{π} sin (n x) d x = [- \frac{c o s n x}{n}]_{0}^{π} = \frac{1 - c o s nπ}{n} = \frac{1 - ( - 1 ) ^{n}}{n} .$ $(- 1)^{n}$ kyun? Kyunki $cos nπ$ alternate karta hai: $- 1, + 1, - 1, \dots$ Toh bracket $0$ hota hai even $n$ ke liye aur $2$ odd $n$ ke liye.

Step 3 — Simplify karo. $B_{n} = \frac{200}{π} \cdot \frac{1 - ( - 1 ) ^{n}}{n} = ⎩ ⎨ ⎧ \frac{400}{π n} 0 n odd n even .$ Sirf odd kyun bachte hain? Ek flat, left–right symmetric hump mein odd sines jaisi symmetry hoti hai; even sines midpoint ke baare mein antisymmetric hote hain aur cancel ho jaate hain.

Step 4 — Assemble karo. $u (x, t) = \frac{400}{π} n odd \sum \frac{1}{n} sin (n x) e^{- n^{2} t} .$

Verify karo: $t = 0$ , $x = π /2$ par: terms hain $\frac{400}{π} \cdot \frac{s i n ( nπ /2 )}{n}$ . Ab $sin (nπ /2) = + 1, - 1, + 1, \dots$ $n = 1, 3, 5, \dots$ ke liye, yaani $(- 1)^{(n - 1) /2}$ , toh sum hai $\frac{400}{π} (1 - \frac{1}{3} + \frac{1}{5} - \dots)$ . Yeh alternating sum Leibniz / Gregory series hai $arctan 1 = π /4$ ke liye (ek standard Fourier/Taylor result: $arctan x = x - \frac{x ^{3}}{3} + \frac{x ^{5}}{5} - \dots$ mein $x = 1$ rakh do). Isliye total hai $\frac{400}{π} \cdot \frac{π}{4} = 100.$ ✓ Yeh constant $100$ ki Fourier series hai midpoint par evaluate ki — jahan woh function value par converge karti hai.

Ex 4 — Cell D: ek triangular (tent) initial profile

Worked example Statement

$L = 1$ . Rod ek symmetric tent se start hoti hai jo middle mein peak karti hai: $f (x) = {x 1 - x 0 \leq x \leq \frac{1}{2} \frac{1}{2} \leq x \leq 1.$ $B_{n}$ nikalo (jahan $L = 1$ hai, toh $sin nπ x$ ).

Figure s01 (alt-text/caption): orange curve tent

f

hai; plum dashed vertical line

x = \frac{1}{2}

par symmetry axis hai; teal segment neeche dono ends ko

0

par pinned dikhata hai. Picture ka message:

f

x = \frac{1}{2}

ke baare mein symmetric hai, toh sirf symmetric (odd-

n

) sine humps ise bana sakte hain — even humps, jo midpoint ke baare mein antisymmetric hote hain, ka weight zero hona chahiye.

Forecast: Plum dashed line dekho. Kya even modes aayenge?

Step 1 — Peak par coefficient integral ko split karo. Yeh step kyun? $f$ do alag formulas se define hoti hai $x = \frac{1}{2}$ ke dono taraf, toh single integral do mein split ho jaata hai. $B_{n} = 2 \int_{0}^{1} f (x) sin (nπ x) d x = 2 [I_{1} \int_{0}^{1/2} x sin (nπ x) d x + I_{2} \int_{1/2}^{1} (1 - x) sin (nπ x) d x] .$ (Yahan $\frac{2}{L} = 2$ kyunki $L = 1$ hai.)

Step 2 — $I_{1}$ par integration by parts explicitly karo. Integration by parts kyun? Integrand hai (polynomial) $\times$ (sine); parts polynomial degree ko tab tak kam karta hai jab tak sirf sines/cosines na bachein. Maano $u = x, d v = sin (nπ x) d x$ , toh $d u = d x, v = - \frac{cos ( nπ x )}{nπ}$ . Phir $I_{1} = [- \frac{x c o s ( nπ x )}{nπ}]_{0}^{1/2} + \frac{1}{nπ} \int_{0}^{1/2} cos (nπ x) d x .$ Bacha hua integral hai $\frac{1}{nπ} [\frac{sin ( nπ x )}{nπ}]_{0}^{1/2} = \frac{sin ( nπ /2 )}{( nπ ) ^{2}}$ . Bracket ko $x = \frac{1}{2}$ par evaluate karo (lower limit $x = 0$ se $0$ milta hai): $I_{1} = - \frac{\frac{1}{2} c o s ( nπ /2 )}{nπ} + \frac{s i n ( nπ /2 )}{( nπ ) ^{2}} .$

Step 3 — $I_{2}$ karo aur boundary terms cancel hote dekhो. Likho $I_{2} = \int_{1/2}^{1} sin (nπ x) d x - \int_{1/2}^{1} x sin (nπ x) d x$ . Pehla piece hai $[- \frac{c o s ( nπ x )}{nπ}]_{1/2}^{1} = \frac{cos ( nπ /2 ) - cos ( nπ )}{nπ}$ . Doosra, Step 2 jaisi same parts se, $\int_{1/2}^{1} x sin (nπ x) d x = [- \frac{x c o s ( nπ x )}{nπ}]_{1/2}^{1} + \frac{s i n ( nπ x )}{( nπ ) ^{2}}_{1/2}^{1} = - \frac{c o s ( nπ )}{nπ} + \frac{\frac{1}{2} c o s ( nπ /2 )}{nπ} - \frac{s i n ( nπ /2 )}{( nπ ) ^{2}} .$ Ab $I_{1} + I_{2}$ add karo. $\frac{\frac{1}{2} cos ( nπ /2 )}{nπ}$ terms collect karo: $I_{1}$ deta hai $- \frac{\frac{1}{2} cos ( nπ /2 )}{nπ}$ ; $I_{2}$ se pehle piece ka $cos (nπ /2) / nπ$ aur subtracted doosre piece ka $- \frac{1}{2} cos (nπ /2) / nπ$ milke $+ \frac{\frac{1}{2} cos ( nπ /2 )}{nπ}$ deta hai. Yeh dono exactly cancel hote hain — yahi hai "symmetry se $x = \frac{1}{2}$ par boundary terms cancel hona." $cos (nπ) / nπ$ terms bhi cancel ho jaate hain. Sirf sine terms bachte hain: $I_{1} + I_{2} = \frac{s i n ( nπ /2 )}{( nπ ) ^{2}} + \frac{s i n ( nπ /2 )}{( nπ ) ^{2}} = \frac{2 s i n ( nπ /2 )}{( nπ ) ^{2}} .$ Isliye $B_{n} = 2 (I_{1} + I_{2}) = \frac{4}{( nπ ) ^{2}} sin \frac{nπ}{2} .$ $sin \frac{nπ}{2}$ kyun? Yeh even $n$ ke liye $0$ aur odd $n$ ke liye $\pm 1$ hota hai — tent, symmetric hone ki wajah se, sirf odd modes rakhta hai, exactly jaise figure ne predict kiya tha.

Step 4 — Assemble karo. $u (x, t) = \sum_{n odd} \frac{4}{( nπ ) ^{2}} sin \frac{nπ}{2} sin (nπ x) e^{- α^{2} n^{2} π^{2} t} .$

Verify karo: Pehla mode $B_{1} = \frac{4}{π ^{2}} sin \frac{π}{2} = \frac{4}{π ^{2}} \approx 0.4053$ . Check karo ki series peak recover karta hai: $x = \frac{1}{2}$ par, $f = \frac{1}{2}$ , aur $\sum_{n odd} \frac{4}{( nπ ) ^{2}} sin^{2} \frac{nπ}{2} = \frac{4}{π ^{2}} \sum_{n odd} \frac{1}{n ^{2}} = \frac{4}{π ^{2}} \cdot \frac{π ^{2}}{8} = \frac{1}{2}$ . ✓ (Identity $\sum_{n odd} \frac{1}{n ^{2}} = \frac{π ^{2}}{8}$ ek standard Fourier/Basel result hai.)

Ex 5 — Cell E: degenerate / zero input

Worked example Statement

$f (x) \equiv 0$ : rod har jagah $0$ ° par start hoti hai, ends $0$ ° par rakhe hain.

Forecast: $t = 5$ seconds par, ya kisi bhi time par, temperature guess karo.

Step 1 — Saare coefficients compute karo. $B_{n} = \frac{2}{L} \int_{0}^{L} 0 \cdot sin \frac{nπ x}{L} d x = 0 for every n .$ Kyun bother karo? Yeh dikhane ke liye ki machinery boring case mein bhi consistent hai — koi hidden nonzero mode andar nahi ghusta.

Step 2 — Assemble karo. $u (x, t) = 0 for all x, t .$ Yeh trivial solution hai.

Step 3 — Yeh sirf solution kyun hai (uniqueness, properly stated). Koi aur solution exist kyun nahi kar sakta? Do independent guarantees: (i) Maximum principle. Heat equation mein koi source nahi hone par, $u$ ka maximum aur minimum $[0, L] \times [0, T]$ par "parabolic boundary" par hona chahiye — ya toh $t = 0$ par ya dono ends par. Yahan in sabhi par $u = 0$ hai, toh $0 \leq u \leq 0$ har jagah: $u \equiv 0$ . Isko same data wale kisi bhi do solutions ke difference par apply karo, yeh unhe equal force karta hai — isliye poore Dirichlet problem ke liye uniqueness milti hai, sirf is case ke liye nahi. (ii) Sturm–Liouville / energy view. Eigenfunctions $sin \frac{nπ x}{L}$ ek complete orthogonal basis banate hain (Sturm-Liouville Theory); jis function ka har coefficient $0$ ho woh zero function hai. General uniqueness bhi dekho Boundary Conditions — Dirichlet vs Neumann par.

Verify karo: $u_{t} = 0$ , $u_{xx} = 0$ , toh $u_{t} = α^{2} u_{xx}$ hold karta hai ( $0 = 0$ ). BCs aur IC sab satisfied hain. ✓

Ex 6 — Cell F: limiting behaviour aur convergence

Worked example Statement

Ex 4 ka tent solution lo ( $L = 1$ , $α^{2} = 1$ ). Profile describe karo (a) bahut bade $t$ ke liye, aur (b) jab $t \to 0^{+}$ — aur honestly bolo ki series har case mein kaise converge karti hai.

Figure s02 (alt-text/caption): orange = sharp tent jab

t \to 0^{+}

recover hoti hai; plum dash-dot = ek intermediate time jahan corner already round ho chuka hai; teal = large

t

par akela surviving half-sine bump. Message: high modes sharp corner uthate hain aur pehle marte hain, toh profile upar se neeche smooth hoti hai phir ek single hump mein collapse hoti hai.

Forecast: Kaun sa single shape har Dirichlet rod marne se pehle collapse hota hai?

Step 1 — Large- $t$ limit (aur kyun yeh limit legal hai). Mode $n$ mein $e^{- n^{2} π^{2} t}$ hota hai. $n = 3$ term ka $n = 1$ term se ratio hai $e^{- (9 - 1) π^{2} t} = e^{- 8 π^{2} t} \to 0$ . Convergence note: kisi bhi fixed $t_{0} > 0$ ke liye coefficients times $e^{- n^{2} π^{2} t}$ , $n$ ki kisi bhi power se zyada tezi se decay karte hain, toh series aur uske sabhi $x$ - aur $t$ -derivatives uniformly converge karte hain $[0, 1] \times [t_{0}, \infty)$ par (Weierstrass $M$ -test with $M_{n} = ∣ B_{n} ∣ e^{- n^{2} π^{2} t_{0}}$ ). Uniform convergence exactly wahi hai jo term-by-term limiting aur differentiation ki permission deta hai — toh sirf $n = 1$ rakhna rigorous hai: $u (x, t) \approx B_{1} sin (π x) e^{- π^{2} t} = \frac{4}{π ^{2}} sin (π x) e^{- π^{2} t} .$ Universal shape: ek single half-sine bump (teal curve), chahe start kaisa bhi tha.

Step 2 — Small- $t$ limit (ek delicate end). Jab $t \to 0^{+}$ har $e^{- n^{2} π^{2} t} \to 1$ , toh formally $u (x, 0^{+}) = \sum B_{n} sin (nπ x) = f (x)$ . Convergence note: $t = 0$ par smoothing exponentials gone hain. Yahan $B_{n} \sim 1/ n^{2}$ hai, jo absolutely summable hai, toh tent ki series continuous tent par uniformly converge karti hai (tent mein koi jumps nahi hain). Ex 3 ke constant- $100$ profile se contrast karo: wahan $B_{n} \sim 1/ n$ hai, series sirf pointwise converge karti hai (uniformly nahi) $(0, π)$ ke andar aur ends par sum $0 \neq = 100$ hai — endpoints aur koi bhi jump midpoint value par converge karte hain (Gibbs behaviour). Toh " $u (x, 0) = f (x)$ " interior mein hold karta hai lekin discontinuities aur boundaries par carefully padhna chahiye.

Step 3 — Smoothing ki kahani. Kisi bhi $t > 0$ ke liye solution infinitely smooth hoti hai (exponentials saari roughness maar dete hain): high- $n$ wiggles pehle marte hain, toh tent ka corner instantly round ho jaata hai, phir poora bump fade hota hai. Yeh instant smoothing wajah hai ki heat flow sharp features mita deta hai.

Verify karo: Amplitude ratio $n = 3$ vs $n = 1$ at $t = 0.1$ hai $\frac{B _{3}}{B _{1}} e^{- 8 π^{2} (0.1)} = - \frac{1}{9} e^{- 0.8 π^{2}} \approx - 4.15 \times 1 0^{- 5}$ , negligible — single-mode approximation excellent hai. ✓

Ex 7 — Cell G: Neumann aur mixed twists (cosines, aur ek ek taraf ka end)

Worked example Statement

(a) Neumann: same rod, lekin ends insulated: $u_{x} (0, t) = u_{x} (L, t) = 0$ , initial $f (x)$ . Building blocks nikalo. (b) Mixed: ends $u (0, t) = 0$ (fixed) aur $u_{x} (L, t) = 0$ (insulated). Ab kaun se eigenfunctions hain?

Forecast: (a) mein sines forbidden hain, toh kaun se functions dono ends par zero slope rakhte hain? (b) mein kaun se quarter-waves fit hote hain?

Step 1 — Space ODE ko Neumann BCs ke saath redo karo. Kyun redo karo? Eigenfunctions BCs satisfy karne ke liye choose ki jaati hain; alag BCs ⇒ alag eigenfunctions. Dekho Boundary Conditions — Dirichlet vs Neumann. $X^{''} + λ X = 0, X^{'} (0) = 0, X^{'} (L) = 0.$ $λ = k^{2} > 0$ ke liye: $X = A cos k x + B sin k x$ , $X^{'} = - A k sin k x + B k cos k x$ . $X^{'} (0) = B k = 0 \Rightarrow B = 0$ ; $X^{'} (L) = - A k sin k L = 0 \Rightarrow sin k L = 0 \Rightarrow k L = nπ$ . Ab cosines kyun? Zero-slope ends ek aisi curve chahte hain jo dono ends par flat ho — $cos \frac{nπ x}{L}$ wahan horizontal tangents rakhta hai. Saath hi $λ = 0$ ab kaam karta hai ( $X_{0} = 1$ , slope zero har jagah) — ek aisa mode jo Dirichlet ne reject kar diya tha! $X_{n} (x) = cos \frac{nπ x}{L}, n = 0, 1, 2, \dots$

Step 2 — Neumann solution assemble karo. $u (x, t) = \frac{A _{0}}{2} + n = 1 \sum \infty A_{n} cos \frac{nπ x}{L} e^{- α^{2} (nπ / L)^{2} t}, A_{n} = \frac{2}{L} \int_{0}^{L} f (x) cos \frac{nπ x}{L} d x .$ Yeh Fourier cosine series hai (Fourier Series). $n = 0$ term $\frac{A _{0}}{2} = \frac{1}{L} \int_{0}^{L} f d x$ ka decay rate $0$ hai — yeh kabhi fade nahi hota. Ek mode forever kyun bachta hai? Insulated ends se koi heat escape nahi hoti, toh total heat conserved hoti hai; rod apne average temperature par settle hoti hai, zero par nahi. Ex 6 se contrast karo, jahan Dirichlet ends ne saari heat $0$ par drain kar di thi.

Step 3 — Mixed case (ek fixed, ek insulated). $X (0) = 0 \Rightarrow A = 0$ , toh $X = B sin k x$ ; phir $X^{'} (L) = B k cos k L = 0 \Rightarrow cos k L = 0 \Rightarrow k L = (n - \frac{1}{2}) π$ . $X_{n} (x) = sin \frac{( 2 n - 1 ) π x}{2 L}, n = 1, 2, \dots$ "Quarter-waves" kyun? $sin$ $0$ se start hota hai (fixed end) aur $x = L$ par flat top (zero slope) par pahunchna chahiye — yeh ek full sine wave ka quarter, three-quarters, … hota hai. Eigenvalues $λ_{n} = (\frac{( 2 n - 1 ) π}{2 L})^{2}$ . Yeh Sturm-Liouville Theory framework mein fit karta hai; sirf boundary equation change hui.

Verify karo: Neumann with constant $f = 100$ , $L = π$ : $\frac{A _{0}}{2} = \frac{1}{π} \int_{0}^{π} 100 d x = 100$ , aur $A_{n} = \frac{2}{π} \int_{0}^{π} 100 cos (n x) d x = 0$ for $n \geq 1$ . Steady state $= 100$ (start average). ✓ Mixed check: smallest eigenvalue $k_{1} = \frac{π}{2 L}$ deta hai $cos (k_{1} L) = cos \frac{π}{2} = 0$ . ✓

Ex 8 — Cell H: ek real-world word problem

Worked example Statement

Ek metal rod $L = π$ metres, thermal diffusivity $α^{2} = 1 m^{2} / s$ , single mode $u (x, 0) = 80 sin x$ °C se start hoti hai, ends $0$ °C par rakhe hain. Midpoint temperature $10$ °C tak drop hone mein kitna time lagega?

Forecast: Estimate karo — kuch tenths of a second, ya kai seconds?

Step 1 — Solution likho. Yeh step kyun? $f = 80 sin x$ ek pure $n = 1$ mode hai ( $L = π \Rightarrow sin \frac{nπ x}{π} = sin n x$ ), toh $B_{1} = 80$ , decay rate $α^{2} (1)^{2} = 1$ . $u (x, t) = 80 sin x e^{- t} .$

Step 2 — Midpoint $x = π /2$ par evaluate karo. $sin \frac{π}{2} = 1$ , toh $u (π /2, t) = 80 e^{- t}$ . Midpoint kyun? $sin x$ wahan peak karta hai, toh yeh sabse hot, sabse zyada time tak rehne wala point hai — cooling ke liye "worst case."

Step 3 — Time ke liye solve karo. Set karo $80 e^{- t} = 10 \Rightarrow e^{- t} = \frac{1}{8} \Rightarrow t = ln 8 \approx 2.079 s .$ $ln$ kyun? Exponential decay ko logarithm undo karta hai — "value $V$ tak pahunchne mein kitna time?" ke liye natural tool.

$t = ln 8 \approx 2.08 seconds .$

Verify karo (units + value): rate $α^{2} (nπ / L)^{2} = 1 \cdot (1/1)^{2} = 1 s^{- 1}$ , toh exponent $- t$ dimensionless hai. ✓ Wapas plug karo: $80 e^{- l n 8} = 80/8 = 10$ °C. ✓

Recall Matrix par quick self-test

Single-sine $f$ ke liye $B_{n}$ nikalne mein integral chahiye? ::: Nahi — read off karo (Cell A). Constant $f$ kaun se modes rakhta hai? ::: Sirf odd $n$ (Cell C). $t \to \infty$ par ek Dirichlet rod kaun se shape mein jaati hai? ::: Ek half-sine bump, phir $0$ (Cell F). Large $t$ par sirf $n = 1$ rakhna rigorous kyun hai? ::: Fixed $t_{0} > 0$ ke liye series uniformly converge karti hai (Weierstrass $M$ -test), jo term-by-term limits ki permission deta hai. Insulated (Neumann) ends kaun se eigenfunctions dete hain? ::: Cosines, including ek constant $n = 0$ mode jo kabhi fade nahi hota — rod apne average par settle hoti hai (Cell G). Ek fixed + ek insulated end? ::: Quarter-wave sines $sin \frac{( 2 n - 1 ) π x}{2 L}$ . Zero initial rod kaisa evolve hota hai? ::: Hamesha $0$ rehta hai — trivial aur unique maximum principle se (Cell E).