Solving heat equation — separation of variables: Exercises

4.7.9 · D4 · Maths › Partial Differential Equations › Solving heat equation — separation of variables

Ek hi setup jo hum baar-baar use karte hain (jab tak problem kuch aur na kahe): $u_{t} = α^{2} u_{xx}, 0 < x < L, t > 0, u (0, t) = 0, u (L, t) = 0, u (x, 0) = f (x) .$

Do boxed results jinpar hum poore time lean karte hain: $u (x, t) = \sum_{n = 1}^{\infty} B_{n} sin \frac{nπ x}{L} e^{- α^{2} (nπ / L)^{2} t}, B_{n} = \frac{2}{L} \int_{0}^{L} f (x) sin \frac{nπ x}{L} d x .$

Level 1 — Recognition

(Kya tum parts identify kar sakte ho aur coefficients seedha padh sakte ho?)

Exercise 1.1

Standard problem ke liye batao: (a) eigenvalues $λ_{n}$ , (b) eigenfunctions $X_{n} (x)$ , aur (c) mode $n$ ka time factor $T_{n} (t)$ .

Recall Solution 1.1

Yeh seedha derivation ke Steps 3–4 se aate hain.

(a) $λ_{n} = (\frac{nπ}{L})^{2}$ , for $n = 1, 2, 3, \dots$
(b) $X_{n} (x) = sin \frac{nπ x}{L}$ .
(c) $T_{n} (t) = e^{- α^{2} λ_{n} t} = e^{- α^{2} (nπ / L)^{2} t}$ .

Yeh hi kyun, kuch aur kyun nahi? Zero boundary conditions cosine part ko khatam kar dete hain aur force karte hain $sin k L = 0$ , yaani $k L = nπ$ . Sirf $λ > 0$ survive karta hai (Step 3 ne teeno signs test kiye the).

Exercise 1.2

Initial temperature hai $f (x) = 7 sin \frac{2 π x}{L}$ . Koi integral compute kiye bina $u (x, t)$ likho.

Recall Solution 1.2

Pehchano: $f$ pehle se hi ek single sine mode hai. $f (x) = \sum_{n} B_{n} sin \frac{nπ x}{L}$ se term-by-term compare karo. Sirf $n = 2$ appear karta hai, jahan $B_{2} = 7$ ; baaki sabhi $B_{n} = 0$ hain.

Us mode ka decay attach karo (1.1 se $n = 2$ ke saath, toh $n^{2} = 4$ ): $u (x, t) = 7 sin \frac{2 π x}{L} e^{- 4 α^{2} π^{2} t / L^{2}} .$

Exercise 1.3

Ek rod ke insulated ends hain frozen ends ki jagah: $u_{x} (0, t) = 0, u_{x} (L, t) = 0$ (Neumann). Sines ki jagah eigenfunctions ki kaunsi family aati hai?

Recall Solution 1.3

Cosines, $X_{n} (x) = cos \frac{nπ x}{L}$ for $n = 0, 1, 2, \dots$ (flat $n = 0$ mode sameta).

Kyun? Condition $X^{'} (0) = 0$ ke liye ek aisi function chahiye jiska slope $x = 0$ par zero ho; $cos$ ka wahan flat top hota hai, $sin$ ka nahi. Eigenfunction ko boundary condition se match karna hi poora game hai — dekho Boundary Conditions — Dirichlet vs Neumann.

Level 2 — Application

(Ek integral chalao, ek poora answer pao.)

Exercise 2.1

Maano $L = π$ , $α = 1$ , aur $f (x) = f_{0}$ (poori rod par ek constant $f_{0} > 0$ ). $B_{n}$ nikalo aur $u (x, t)$ likho.

Recall Solution 2.1

Set (sines par project karo): ek constant single sine nahi hai, toh integrate karna padega. $B_{n} = \frac{2}{π} \int_{0}^{π} f_{0} sin (n x) d x = \frac{2 f _{0}}{π} [\frac{- c o s n x}{n}]_{0}^{π} = \frac{2 f _{0}}{π} \cdot \frac{1 - c o s nπ}{n} = \frac{2 f _{0}}{π} \cdot \frac{1 - ( - 1 ) ^{n}}{n} .$ Factor $1 - (- 1)^{n}$ , even $n$ ke liye $0$ hai aur odd $n$ ke liye $2$ . Toh $B_{n} = ⎩ ⎨ ⎧ \frac{4 f _{0}}{π n}, 0, n odd n even.$ Sum: $u (x, t) = \frac{4 f _{0}}{π} \sum_{n odd} \frac{1}{n} sin (n x) e^{- n^{2} t} .$ Sirf odd kyun? Ek flat profile midpoint $x = π /2$ ke baare mein symmetric hai; even sines wahan antisymmetric hain, isliye woh ek symmetric shape banana mein help nahi kar sakte — unka coefficient zero hona hi chahiye. (Yeh Fourier Series symmetry kaam kar rahi hai.)

Exercise 2.2

$L = 1$ , $α = 1$ ke saath, $f (x) = x$ lo. $B_{n}$ compute karo.

Recall Solution 2.2

$B_{n} = 2 \int_{0}^{1} x sin (nπ x) d x .$ Integration by parts karo ( $u = x$ , $d v = sin (nπ x) d x$ ): $\int_{0}^{1} x sin (nπ x) d x = [- \frac{x c o s ( nπ x )}{nπ}]_{0}^{1} + \frac{1}{nπ} \int_{0}^{1} cos (nπ x) d x .$ Aakhri integral hai $[\frac{s i n ( nπ x )}{nπ}]_{0}^{1} = 0$ . Toh integral hai $- \frac{cos nπ}{nπ} = - \frac{( - 1 ) ^{n}}{nπ} = \frac{( - 1 ) ^{n + 1}}{nπ}$ . $B_{n} = \frac{2 ( - 1 ) ^{n + 1}}{nπ} \Rightarrow u (x, t) = \sum_{n = 1}^{\infty} \frac{2 ( - 1 ) ^{n + 1}}{nπ} sin (nπ x) e^{- n^{2} π^{2} t} .$

Exercise 2.3

$L = 2$ , $α = 1$ . Diya hai $f (x) = 3 sin \frac{π x}{2} - sin \frac{5 π x}{2}$ , $u (x, t)$ likho.

Recall Solution 2.3

$f$ sine modes ka finite combination hai. $sin \frac{nπ x}{L} = sin \frac{nπ x}{2}$ se match karo: term $sin \frac{π x}{2}$ hai $n = 1$ , aur $sin \frac{5 π x}{2}$ hai $n = 5$ . Toh $B_{1} = 3$ , $B_{5} = - 1$ , baaki zero. Decay rate hai $α^{2} (nπ / L)^{2} = (nπ /2)^{2} = \frac{n ^{2} π ^{2}}{4}$ : $u (x, t) = 3 sin \frac{π x}{2} e^{- π^{2} t /4} - sin \frac{5 π x}{2} e^{- 25 π^{2} t /4} .$

Level 3 — Analysis

(Formulas ghisne ki bajaye behaviour ke baare mein socho.)

Exercise 3.1

2.1 ke answer ka use karke ( $f_{0} = 100$ , $L = π$ , $α = 1$ ke saath), large $t$ ke liye $u (\frac{π}{2}, t)$ estimate karo, sirf slowest-decaying mode rakhte hue. Ek numerical decay rate do.

Recall Solution 3.1

Slowest mode hai $n = 1$ (sabse chota $n^{2}$ ). Uska term hai $\frac{400}{π} sin x e^{- t}$ . $x = π /2$ par: $sin \frac{π}{2} = 1$ , toh $u (\frac{π}{2}, t) \approx \frac{400}{π} e^{- t} \approx 127.32 e^{- t} .$ Decay rate $= 1$ (units $1/ time$ ): amplitude har unit time mein factor $e$ se drop hoti hai. Agla mode ( $n = 3$ ) $e^{- 9 t}$ ki tarah decay karta hai — $9$ guna tez — toh woh already negligible ho jata hai jab $t$ thoda bhi bada ho. Neeche figure dekho.

Figure — Solving heat equation — separation of variables

Exercise 3.2

Do modes same amplitude se shuru hote hain: $f (x) = sin \frac{π x}{L} + sin \frac{4 π x}{L}$ . Kitne time $t^{⋆}$ baad $n = 4$ mode $n = 1$ mode ki current amplitude ka $\frac{1}{e}$ ho jaata hai? $α = 1$ lo.

Recall Solution 3.1 ek mode use karta hai; yahan do compare karo.

Mode $n$ ki amplitude hai $e^{- (nπ / L)^{2} t}$ . Hum ratio chahte hain $\frac{amp _{4}}{amp _{1}} = \frac{e ^{- (4 π / L)^{2} t}}{e ^{- (π / L)^{2} t}} = e^{- (16 - 1) (π / L)^{2} t} = e^{- 15 (π / L)^{2} t} .$ Isse $e^{- 1}$ ke barabar karo: $15 \frac{π ^{2}}{L ^{2}} t^{⋆} = 1 \Rightarrow t^{⋆} = \frac{L ^{2}}{15 π ^{2}} .$ Interpretation: modes ke beech ka gap $(16 - 1) = 15$ guna base rate ki speed se barhta hai. Wiggly modes race jaldi haar jaate hain — yahi wajah hai ki rod smooth ho jaati hai.

Exercise 3.3

Dikhao ki standard Dirichlet problem ke liye, total "energy" $E (t) = \int_{0}^{L} u^{2} d x$ sirf decrease (ya constant) ho sakti hai time mein. $u = \sum_{n} B_{n} sin \frac{nπ x}{L} e^{- α^{2} (nπ / L)^{2} t}$ use karo.

Recall Solution 3.3

Orthogonality se, cross terms integrate hokar $0$ dete hain aur har $\int_{0}^{L} sin^{2} \frac{nπ x}{L} d x = \frac{L}{2}$ : $E (t) = \int_{0}^{L} u^{2} d x = \frac{L}{2} \sum_{n = 1}^{\infty} B_{n}^{2} e^{- 2 α^{2} (nπ / L)^{2} t} .$ Har exponential $e^{- 2 α^{2} (nπ / L)^{2} t}$ $t$ ki decreasing function hai (uska exponent $t$ badhne par negative hai). Non-negative terms ka sum jisme har ek decrease kar raha hai, woh khud bhi non-increasing hai, toh $E (t)$ kabhi nahi badhti aur $t \to \infty$ par $E (t) \to 0$ . Physically: thande ends aur koi source nahi, stored heat sirf bahar leak ho sakti hai. (Yeh Sturm-Liouville Theory orthogonality apna kaam kar rahi hai.)

Level 4 — Synthesis

(Kai ideas combine karo, ya method adapt karo.)

Exercise 4.1

$L = π$ , $α = 1$ , $f (x) = x (π - x)$ . $B_{n}$ nikalo aur hence $u (x, t)$ likho.

Recall Solution 4.1

Set: $B_{n} = \frac{2}{π} \int_{0}^{π} x (π - x) sin (n x) d x .$ Split karo $π \int_{0}^{π} x sin (n x) d x - \int_{0}^{π} x^{2} sin (n x) d x$ mein.

Standard results (integration by parts): $\int_{0}^{π} x sin (n x) d x = \frac{- π ( - 1 ) ^{n}}{n} = \frac{π ( - 1 ) ^{n + 1}}{n},$ $\int_{0}^{π} x^{2} sin (n x) d x = \frac{- π ^{2} ( - 1 ) ^{n}}{n} + \frac{2 ( ( - 1 ) ^{n} - 1 )}{n ^{3}} .$ Combine karo: $\int_{0}^{π} x (π - x) sin (n x) d x = \frac{π ^{2} ( - 1 ) ^{n + 1}}{n} - (\frac{- π ^{2} ( - 1 ) ^{n}}{n} + \frac{2 (( - 1 ) ^{n} - 1 )}{n ^{3}}) = - \frac{2 (( - 1 ) ^{n} - 1 )}{n ^{3}} = \frac{2 ( 1 - ( - 1 ) ^{n} )}{n ^{3}} .$ Toh $B_{n} = \frac{2}{π} \cdot \frac{2 ( 1 - ( - 1 ) ^{n} )}{n ^{3}} = \frac{4 ( 1 - ( - 1 ) ^{n} )}{π n ^{3}} = ⎩ ⎨ ⎧ \frac{8}{π n ^{3}}, 0, n odd n even.$ $u (x, t) = \frac{8}{π} n odd \sum \frac{1}{n ^{3}} sin (n x) e^{- n^{2} t} .$ Sanity check: $x (π - x)$ $x = π /2$ ke baare mein symmetric hai aur dono ends par zero hota hai — sirf odd sines aur zero BCs ke saath consistent.

Exercise 4.2 — Non-zero ends (steady-state shift)

$0 < x < L$ par $u_{t} = α^{2} u_{xx}$ solve karo jahan $u (0, t) = T_{0}, u (L, t) = T_{1}$ (dono nonzero constants) aur $u (x, 0) = g (x)$ hai. Isse aise problem mein reduce karo jo tum pehle se jaante ho.

Recall Solution 4.2

Seedha separate kyun nahi kar sakte: $u (0, t) = T_{0} \neq = 0$ force karega $X (0) T (t) = T_{0}$ sabhi $t$ ke liye — possible nahi jab tak $T$ constant na ho. Homogeneous (zero-BC) machinery ko zero ends chahiye.

Trick — steady state subtract karo. Final profile $v (x)$ dhundho jo time ke saath ab nahi badlega: $v_{t} = 0 \Rightarrow α^{2} v^{''} = 0 \Rightarrow v^{''} = 0 \Rightarrow v (x) = a x + b$ , ek straight line. Ends fit karo: $v (0) = T_{0} = b, v (L) = T_{1} = a L + b \Rightarrow a = \frac{T _{1} - T _{0}}{L} .$ $v (x) = T_{0} + \frac{T _{1} - T _{0}}{L} x .$ Deviation define karo $w (x, t) = u (x, t) - v (x)$ . Tab:

$w_{t} = u_{t}$ aur $w_{xx} = u_{xx} - v^{''} = u_{xx}$ , toh $w$ same heat equation solve karta hai.
$w (0, t) = T_{0} - T_{0} = 0$ , $w (L, t) = T_{1} - T_{1} = 0$ — zero BCs.
$w (x, 0) = g (x) - v (x)$ .

Toh $w$ initial data $g (x) - v (x)$ ke saath standard Dirichlet problem hai: $w (x, t) = \sum_{n = 1}^{\infty} B_{n} sin \frac{nπ x}{L} e^{- α^{2} (nπ / L)^{2} t}, B_{n} = \frac{2}{L} \int_{0}^{L} (g (x) - v (x)) sin \frac{nπ x}{L} d x,$ aur finally $u (x, t) = v (x) + w (x, t)$ . Jaise $t \to \infty$ , $w \to 0$ aur $u \to v (x)$ : rod apne dono fixed ends ke beech straight-line temperature par settle ho jaati hai. Yahi steady-state-plus-decay split Laplace Equation ko bhi power karta hai.

Level 5 — Mastery

(Sab kuch use karke ek saaf run.)

Exercise 5.1

Length $L = π$ , diffusivity $α = 1$ ki ek rod ke ends $0$ ° par held hain aur initial profile hai $f (x) = 2 sin x - 5 sin 4 x + sin 7 x .$ (a) $u (x, t)$ exactly likho. (b) Kitne time $t^{⋆}$ par $n = 4$ term ki amplitude apni initial value ki $1%$ ho jaati hai? (c) Large $t$ ke liye kaunsa single mode dominate karta hai, aur tab $u (π /2, t)$ kya hai?

Recall Solution 5.1

(a) Sine + Sum + Set (inspection se). $f$ pehle se hi pure sine modes ka sum hai jahan $L = π$ , toh $sin \frac{nπ x}{L} = sin (n x)$ . Padh lo: $B_{1} = 2$ , $B_{4} = - 5$ , $B_{7} = 1$ ; baaki sab $0$ . Decay rate $= α^{2} (nπ / L)^{2} = n^{2}$ . Hence $u (x, t) = 2 sin x e^{- t} - 5 sin 4 x e^{- 16 t} + sin 7 x e^{- 49 t} .$

(b) $n = 4$ ki amplitude hai $5 e^{- 16 t}$ ; initial value $5$ . Set karo $e^{- 16 t^{⋆}} = 0.01$ : $- 16 t^{⋆} = ln (0.01) = - ln 100 \Rightarrow t^{⋆} = \frac{l n 100}{16} \approx \frac{4.60517}{16} \approx 0.28782.$

(c) Sabse bada exponent (slowest decay) hai $n = 1$ , rate $1$ , baaki ke liye $16$ aur $49$ hain — woh almost immediately chale jaate hain. Toh large $t$ ke liye, $u (x, t) \approx 2 sin x e^{- t} .$ $x = π /2$ par: $sin \frac{π}{2} = 1$ , toh $u (π /2, t) \approx 2 e^{- t}$ . Profile ek single half-sine bump hai jo ek thandi rod mein fade ho jaati hai — universal late-time shape (surviving mode isolate karne ke liye Superposition Principle use karta hai).

Recall Jaane se pehle ek-line self-check

$u = X T$ Split karo → do ODEs Solve karo → zero BCs se Sine eigenfunctions → $e^{- α^{2} (nπ / L)^{2} t}$ ke saath Sum karo → Set $B_{n} = \frac{2}{L} \int_{0}^{L} f sin \frac{nπ x}{L} d x$ . Nonzero ends? Pehle straight-line steady state subtract karo.