Worked examples — Heat equation (parabolic) 1D — derivation from Fourier's law

4.7.8 · D3 · Maths › Partial Differential Equations › Heat equation (parabolic) 1D — derivation from Fourier's law

Shuru karne se pehle, teen symbols ek jagah taaki kuch bhi use hone se pehle na samjha ja sake:

parent derivation dekho agar yeh kuch bhi naya lage, aur Thermal Diffusivity and Material Properties dekho $\alpha$ kahan se aata hai uske liye.

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The scenario matrix

Har heat-equation question jo tumhe milega woh in cells mein se ek (ya blend) hoga. Neeche ke examples Cell ke saath tagged hain jo woh exercise karte hain.

Cell	Case class	Kya cheez isse alag banati hai	Example
A	Curvature negative (ek hill / hot bump)	$u_{xx}<0\Rightarrow u_t<0$: point cools	Ex 1
B	Curvature positive (ek valley / cold dip)	$u_{xx}>0\Rightarrow u_t>0$: point warms	Ex 2
C	Curvature zero (straight-line / degenerate)	$u_{xx}=0\Rightarrow u_t=0$: steady state	Ex 3
D	Candidate solution verify karo (exponential decay)	Dono sides mein plug karo, match karna chahiye	Ex 4
E	Sign / stability twist (backward heat eq)	galat sign $\Rightarrow$ blow-up	Ex 5
F	Limiting behaviour as $t\to \infty$ aur $\alpha$ scaling	equilibrium tak race mein kaun jeetta hai	Ex 6
G	Real-world word problem with units	numbers + sahi SI dene hote hain	Ex 7
H	Source term $f\ne 0$ (non-homogeneous)	extra $+f$ balance shift karta hai	Ex 8

Cells A, B, C curvature ke teen signs hain (saare cases covered). D–H standard twists hain. Milke yeh table ko fill karte hain.

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Example 1 — Cell A: ek hot bump cools

Forecast: $x=0$ par, $\cos 0=1$, toh $u=28$ — yeh ek bump ka top hai (ek hill). Aage padhne se pehle $u_t$ ka sign guess karo. Hills apne thande neighbours ko heat deti hain…

Step 1 — Curvature compute karo. $u_{xx}=\frac{d^2}{d{x}^2}[20+8\cos \pi x]=-8{\pi }^2\cos \pi x.$ Yeh step kyun? Equation sirf $u_{xx}$ ki parwah karta hai; constant $20$ ki curvature zero hai, toh woh drop ho jaata hai. Neeche figure mein shape dekho.

Step 2 — $x=0$ par evaluate karo. $u_{xx}{|}_{x=0}=-8{\pi }^2(1)=-8{\pi }^2.$ Yeh step kyun? $x=0$ par hum peak par hain (figure mein red dot); $\cos 0=1$. Negative curvature ek hill (∩) confirm karti hai.

Step 3 — Equation apply karo. $u_t=\alpha u_{xx}=3\cdot (-8{\pi }^2)=-24{\pi }^2\approx -236.9\text{K/s}.$ Yeh step kyun? $u_t=\alpha u_{xx}$ curvature ko time-rate mein convert karta hai.

Verify: $u_t<0$, toh spot cools — bilkul waisa jaisa hill picture predict karti hai (neighbours thande hain, energy baahir jaati hai). Units: $\alpha$ in m²/s times $u_{xx}$ in K/m² gives K/s. ✓

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Example 2 — Cell B: ek cold dip warms

Forecast: $x=1$ par, $\cos \pi =-1$, toh $u=12$ — dip ka bottom. $u_t$ ka sign guess karo: valleys apne neighbours ke neeche hoti hain…

Step 1 — Curvature reuse karo. $u_{xx}=-8{\pi }^2\cos \pi x.$ Yeh step kyun? Same profile hai, toh same second derivative — sirf point change kar rahe hain.

Step 2 — $x=1$ par evaluate karo. $u_{xx}{|}_{x=1}=-8{\pi }^2\cos \pi =-8{\pi }^2(-1)=+8{\pi }^2.$ Yeh step kyun? $\cos \pi =-1$ sign flip karta hai: ab curvature positive hai, ek valley (∪).

Step 3 — Equation apply karo. $u_t=3\cdot (8{\pi }^2)=24{\pi }^2\approx 236.9\text{K/s}.$

Verify: $u_t>0$, dip warms. Note karo Ex 1 aur Ex 2 ke numbers exact opposites hain ($\mp 24{\pi }^2$) — ek $\cos$ curve ki hill aur valley mirror-image rates par heat/cool hote hain. Sanity ✓.

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Example 3 — Cell C: degenerate straight line

Forecast: Ek straight line mein koi bend nahi hota. Straight line ki curvature kya hoti hai? Step 1 se pehle $u_t$ guess karo.

Step 1 — Pehli derivative. $u_x=\frac{d}{dx}[5+7x]=7.$ Yeh step kyun? Humhe $u_{xx}$ chahiye; pehle $u_x$ se wahan pahuncho.

Step 2 — Doosri derivative. $u_{xx}=\frac{d}{dx}[7]=0.$ Yeh step kyun? Slope $7$ constant hai, toh iska rate of change (curvature) zero hai — ek straight line kahin bhi nahi jhukti.

Step 3 — Equation apply karo. $u_t=\alpha \cdot 0=0\text{for every }x\text{ and every }\alpha .$

Verify: $u_t=0$ har jagah — profile already steady hai. Yeh parent note se match karta hai: 1D steady state ($u_{xx}=0$) exactly ek straight line hai. Multi-dimensional cousin ke liye Laplace Equation (steady-state heat) dekho. ✓

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Example 4 — Cell D: ek exponential-decay solution verify karo

Forecast: Ek solution ko left side equal right side identically karna hoga. Guess karo: exponent $-4\alpha t$ — woh $4$ kahan se aana chahiye?

Step 1 — Left side, $u_t$. $u_t=\frac{\partial }{\partial t}\left[e^{-4\alpha t}\cos 2x\right]=-4\alpha e^{-4\alpha t}\cos 2x.$ Yeh step kyun? Exponential par chain rule: time-derivative factor $-4\alpha$ neeche kheenchti hai. $\cos 2x$ mein $t$ ka koi kaam nahi, toh constant hai.

Step 2 — Right side, pehle space derivatives. $u_x=-2e^{-4\alpha t}\sin 2x,u_{xx}=-4e^{-4\alpha t}\cos 2x.$ Yeh step kyun? Har $x$-derivative $\cos 2x$ se factor $2$ kheenchti hai; do baar se $2\cdot 2=4$ milta hai aur minus ke saath $\cos$ wapas aata hai.

Step 3 — $\alpha$ se multiply karo aur compare karo. $\alpha u_{xx}=\alpha (-4)e^{-4\alpha t}\cos 2x=-4\alpha e^{-4\alpha t}\cos 2x=u_t.✓$ Yeh step kyun? Dono sides ab same expression hain, toh equation saare $x,t,\alpha$ ke liye hold karta hai.

Verify: Yeh term-for-term match karte hain. General pattern: $e^{-\alpha n^{2}t}\cos (nx)$ equation solve karta hai, bade $n$ ke liye faster decay karta hai (zyada wiggly profiles jaldi marte hain). Yeh Separation of Variables for the Heat Equation aur Fourier Series Solutions ka seed hai. ✓

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Example 5 — Cell E: sign twist (minus kyun matter karta hai)

Forecast: Correct equation mein heat spread hoti hai aur mitti hai. Guess karo: galat-sign version time mein grow karega ya shrink karega?

Step 1 — Parts compute karo. $u_t=\beta e^{\beta t}\sin \pi x,u_{xx}=-{\pi }^2{e}^{\beta t}\sin \pi x.$ Yeh step kyun? Standard derivatives; hum growth rate $\beta$ match karke solve karte hain.

Step 2 — Correct equation $u_t=\alpha u_{xx}$ (α = 1). $\beta =-{\pi }^2\approx -\mathrm{9.87.}$ Yeh step kyun? $e^{\beta t}\sin \pi x$ ke coefficients match karo. Negative $\beta$ ⇒ decay — physically stable, heat smooth ho jaati hai.

Step 3 — Wrong-sign equation $u_t=-\alpha u_{xx}$. $\beta =-(-{\pi }^2)=+{\pi }^2\approx +\mathrm{9.87.}$ Yeh step kyun? Flipped sign $\beta$ flip karta hai. Positive $\beta$ ⇒ explosive growth — profile hamesha ke liye amplify hoti rehti hai.

Verify: Correct: $\beta =-{\pi }^2<0$ (marta hai). Wrong: $\beta =+{\pi }^2>0$ (blow up karta hai). Yeh exactly woh [!mistake] hai parent note mein — $q=-k{u}_x$ mein minus hi physics ko stable rakhta hai. Classification of PDEs (elliptic, parabolic, hyperbolic) dekho. ✓

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Example 6 — Cell F: limiting behaviour aur $\alpha$ race

Forecast: Bada $\alpha$ = faster diffusion. Guess karo kaun sa rod pehle flatten hoga, aur final temperature guess karo.

Step 1 — Decay rates. ${\lambda }_P={\alpha }_P{\pi }^2=2{\pi }^2\approx 19.7{\text{s}}^{-1},{\lambda }_Q={\alpha }_Q{\pi }^2=8{\pi }^2\approx 79.0{\text{s}}^{-1}.$ Yeh step kyun? Mode $\sin (\pi x)$ $e^{-\alpha {\pi }^{2}t}$ ki tarah decay karta hai; exponent ka magnitude decay rate hai.

Step 2 — Rod Q ke $1/e$ tak pahunchne ka time. Peak value $e^{-{\lambda }_{Q}t}$ hai; ise $e^{-1}$ set karo: ${\lambda }_{Q}t=1\Rightarrow t=\frac{1}{8{\pi }^2}\approx 0.0127\text{s}.$ Yeh step kyun? "$1/e$ time" (time constant) $1/\lambda$ hai — speeds compare karne ka ek clean tarika.

Step 3 — Long-time limit. ${\lim }_{t\to \infty }{e}^{-\alpha {\pi }^{2}t}\sin \pi x=0\text{dono rods ke liye.}$ Yeh step kyun? Koi bhi positive exponent times $t$, $e^{-\lambda t}\to 0$ bhejta hai; ends $0$ par held hain, toh poori rod $0$ par relax ho jaati hai.

Verify: ${\lambda }_Q=4{\lambda }_P$ (kyunki $8=4\times 2$), toh rod Q chaar guna faster equilibrate karta hai — $\alpha$ double-then-double karna compound karta hai. Final temperature dono ke liye $0$, "diffusion erases structure" se match karta hai. ✓

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Example 7 — Cell G: real-world word problem with units

Forecast: Copper bahut accha conduct karta hai ($k$ huge) lekin fair amount energy bhi store karta hai. Guess karo $\alpha$ "large" ($\sim 10^{-4}$) hai ya "tiny" ($\sim 10^{-7}$).

Step 1 — Diffusivity. $\alpha =\frac{k}{\rho c}=\frac{400}{8960\times 385}=\frac{400}{3449600}\approx 1.16\times 10^{-4}{\text{m}}^2/\text{s}.$ Yeh step kyun? PDE ka time scale $\alpha =k/(\rho c)$ set karta hai, akele $k$ nahi (parent [!mistake]). Units: $\frac{\text{W/m⋅K}}{({\text{kg/m}}^3)(\text{J/kg⋅K})}={\text{m}}^2/\text{s}$. ✓

Step 2 — Diye gaye profile ki curvature. $u_x=-100x,u_{xx}=-100{\text{K/m}}^2.$ Yeh step kyun? $u=300-50x^2$ ek downward parabola hai (ek hill), toh constant negative curvature $-100$ har jagah.

Step 3 — Time-rate. $u_t=\alpha u_{xx}=(1.16\times 10^{-4})(-100)\approx -1.16\times 10^{-2}\text{K/s}.$ Yeh step kyun? Same $u_t=\alpha u_{xx}$; negative sign kehta hai rod ka warm crest cools, slowly (copper ka modest $\alpha$).

Verify: $\alpha \approx 1.16\times 10^{-4}$ m²/s — "large" $\sim 10^{-4}$ guess, copper ke liye sahi. $u_t\approx -0.0116$ K/s, units K/s. ✓ Thermal Diffusivity and Material Properties dekho.

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Example 8 — Cell H: ek heat source (non-homogeneous)

Forecast: Source ke bina, yeh hill cool hoti. Source energy andar pump karta hai. Guess karo: kya source warm hone mein help karta hai, ya sirf cooling slow karta hai — aur kya yeh exactly cancel ho sakte hain?

Step 1 — Curvature. $u_x=-x,u_{xx}=-1{\text{K/m}}^2.$ Yeh step kyun? Humhe diffusion term $k{u}_{xx}$ chahiye; profile ek shallow hill hai, curvature $-1$.

Step 2 — Source equation assemble karo. $\rho c{u}_t=k{u}_{xx}+f=20(-1)+30=-20+30=10.$ Yeh step kyun? Diffusion $-20$ contribute karta hai (cooling, yeh ek hill hai), heater $+30$ add karta hai. Unka sum stored energy par net drive hai.

Step 3 — $u_t$ solve karo. $u_t=\frac{10}{\rho c}=\frac{10}{10}=1\text{K/s}.$ Yeh step kyun? Temperature rate isolate karne ke liye energy balance ko $\rho c$ se divide karo.

Verify: Net $u_t=+1$ K/s: $+30$ source outweighs $-20$ diffusive cooling, toh spot hill hone ke bawajood warms hota hai. Agar source exactly $f=20$ hota, dono cancel ho jaate aur $u_t=0$ — ek source-sustained steady state. ✓

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Poore matrix ka recap

Recall Kya har cell cover hua?

Ek hill cools (Ex 1), ek valley warms (Ex 2), ek straight line steady hai (Ex 3), ek exponential mode verify hua (Ex 4), ek galat sign blow up karta hai (Ex 5), bada $\alpha$ equilibrium race jeetta hai (Ex 6), ek real copper rod SI numbers deta hai (Ex 7), aur ek source cooling ko warming mein flip kar sakta hai (Ex 8). Curvature ka har sign aur har standard twist ab kuch aisa hai jo tumne worked dekha hai.

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Connections

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