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

(a) ✅ Yeh heat equation hi hai: ek time derivative $u_t$, curvature $u_{xx}$, positive diffusivity $\alpha >0$. (b) ❌ Do time derivatives $u_{tt}$ — yeh wave equation hai; yeh oscillate karta hai, smooth nahi karta. (c) ❌ Backward heat equation. Minus sign curvature ko galat direction mein drive karta hai (hills badhte hain, valleys gehri hoti hain) — yeh bahut zyada unstable hai aur heat ko time mein ulta chalata hai. (d) ❌ Time bilkul nahi — yeh Laplace equation hai, steady state.

Apply karo $u_t=\alpha u_{xx}=2\times (-3)=-6$. Kyunki $u_t<0$ hai, point $6$ K/s ki dar se thanda ho raha hai. Ek bump ($u_{xx}<0$) hamesha thanda hota hai — iske neighbours thande hain, isliye heat bahar nikl jaati hai.

Step 1 — curvature. $u_{xx}={\partial }_x^2[20+5\sin \pi x]=-5{\pi }^2\sin \pi x$. (Do baar kyun? Kyunki equation $u_{xx}$ par feed karti hai.) Step 2 — evaluate. $x=\frac{1}{4}$ par, $\sin (\pi /4)=\frac{\sqrt{2}}{2}$, isliye $u_{xx}=-5{\pi }^2\cdot \frac{\sqrt{2}}{2}=-\frac{5{\pi }^2\sqrt{2}}{2}$. Step 3 — time rate. $u_t=\alpha u_{xx}=2\cdot (-\frac{5{\pi }^2\sqrt{2}}{2})=-5{\pi }^2\sqrt{2}\approx -69.79$ K/s. Thanda ho raha hai (ek bump ke rising flank par).

Step 1 — time derivative. $u_t=-4\alpha e^{-4\alpha t}\cos 2x$ (exponential par chain rule). Step 2 — space derivatives. Space mein do baar differentiate kyun karein? Kyunki equation $u_t$ ko curvature $u_{xx}$ ke against compare karti hai, aur curvature second spatial derivative hai — isliye hume do steps deep jaana padega. Pehla step: $u_x=-2e^{-4\alpha t}\sin 2x$ ($\cos 2x$ ka derivative $-2\sin 2x$ hai; $-2$ inner $2x$ se chain-rule factor hai). Doosra step: $u_{xx}=-4e^{-4\alpha t}\cos 2x$ ($-2\sin 2x$ ka derivative $-2\cdot 2\cos 2x=-4\cos 2x$ hai). Sine/cosine ka har differentiation wavenumber $2$ ka ek aur factor nikalta hai, isliye constant $4=2^2$ tak badh jaata hai. Step 3 — compare. $\alpha u_{xx}=-4\alpha e^{-4\alpha t}\cos 2x=u_t$. ✅ Decay rate $4\alpha$ exactly $\alpha \times (\text{wavenumber}{)}^2=\alpha \cdot 2^2$ hai.

$\alpha =\frac{k}{\rho c}=\frac{400}{8960\times 385}=\frac{400}{3449600}\approx 1.16\times 10^{-4}{\text{m}}^2/\text{s}.$ Copper bahut achha conduct karta hai lekin energy bhi bahut store karta hai, isliye $\alpha$ moderate hai, bahut bada nahi.

Step 1 — plug in. $u_t=-\alpha n^2{\pi }^2{e}^{-\alpha n^2{\pi }^{2}t}\sin n\pi x$ aur $u_{xx}=-n^2{\pi }^2{e}^{-\alpha n^2{\pi }^{2}t}\sin n\pi x$, isliye $u_t=\alpha u_{xx}$ ✅. ${\lambda }_n$ kya hai? Notice karo ki poora solution $(\text{shape }\sin n\pi x)\times e^{-{\lambda }_{n}t}$ hai jisme number ${\lambda }_n=\alpha n^2{\pi }^2$ exponent mein baitha hai. Hum ${\lambda }_n$ ko decay rate kehte hain kyunki $e^{-{\lambda }_{n}t}$ ek shrinking exponential hai: jitna bada ${\lambda }_n$, utni tez factor $e^{-{\lambda }_{n}t}$ zero ki taraf collapse hoti hai. Isliye ${\lambda }_n$ literally measure karta hai "yeh mode kitni tezi se marta hai." Step 2 — rates compare karo. $\frac{{\lambda }_3}{{\lambda }_1}=\frac{\alpha 3^2{\pi }^2}{\alpha 1^2{\pi }^2}=\frac{3^2}{1^2}=9$. Conclusion: $n=3$ mode 9 guna tez decay karta hai. Physically kyun? Bada $n$ = tighter wiggles = tez curvature $u_{xx}$; tez curvature faster smoothing drive karta hai. Heat equation pehle fine detail mitaata hai.

Step 1 — steady matlab koi time change nahi. $u_t=0\Rightarrow \alpha u_{xx}=0\Rightarrow u_{xx}=0$. Step 2 — do baar integrate karo. $u_{xx}=0\Rightarrow u_{\infty }=a+bx$ (ek straight line — yahi ek curvature-free profile hai). Step 3 — ends fit karo. $u_{\infty }(0)=a=100$; $u_{\infty }(1)=100+b=40\Rightarrow b=-60$. Answer: $u_{\infty }(x)=100-60x$. $x=\frac{1}{2}$ par, $u_{\infty }=70$ K.

Step 1 — time mein chain rule. $\frac{\partial u}{\partial t}=\frac{\partial u}{\partial \tau }\frac{d\tau }{dt}=u_{\tau }\cdot \frac{\alpha }{L^2}$. Step 2 — space mein chain rule (do baar). $\frac{\partial u}{\partial x}=u_{\xi }\cdot \frac{1}{L}$, isliye $\frac{{\partial }^{2}u}{\partial x^2}=u_{\xi \xi }\cdot \frac{1}{L^2}$. Step 3 — substitute karo. $u_{\tau }\frac{\alpha }{L^2}=\alpha u_{\xi \xi }\frac{1}{L^2}$. $\alpha /L^2$ cancel karo: $u_{\tau }=u_{\xi \xi }$. Matlab: har rod same rod hai jab tum time ko $L^2/\alpha$ (natural diffusion time) ki units mein measure karo. Isliye bade objects equilibrate hone mein $\sim L^2$ zyada time lete hain — yeh Separation of Variables for the Heat Equation ki ek headline hai.

Step 1 — linearity. Heat equation linear hai, isliye solutions ka sum bhi solution hai; har term pehle se ise satisfy karta hai (rates $\alpha {\pi }^2$ aur $9\alpha {\pi }^2$ Problem 3.1 se). Step 2 — term one check karo. $u_1=3e^{-\alpha {\pi }^{2}t}\sin \pi x$ ke liye: $u_{1,t}=-3\alpha {\pi }^2{e}^{-\alpha {\pi }^{2}t}\sin \pi x$; aur $u_{1,xx}=-3{\pi }^2{e}^{-\alpha {\pi }^{2}t}\sin \pi x$, isliye $\alpha u_{1,xx}=u_{1,t}$ ✅. Step 3 — term two explicitly check karo. $u_2=2e^{-9\alpha {\pi }^{2}t}\sin 3\pi x$ ke liye, hand-wave mat karo — wohi do derivatives chalaao. Time: $u_{2,t}=-9\alpha {\pi }^2\cdot 2e^{-9\alpha {\pi }^{2}t}\sin 3\pi x$ (exponent $-9\alpha {\pi }^2$ saamne aa jaata hai). Space, do baar: $u_{2,x}=2\cdot 3\pi e^{-9\alpha {\pi }^{2}t}\cos 3\pi x$, phir $u_{2,xx}=-2\cdot 9{\pi }^2{e}^{-9\alpha {\pi }^{2}t}\sin 3\pi x$ ($\sin 3\pi x/\cos 3\pi x$ ka har $x$-derivative ek factor $3\pi$ nikalta hai, isliye do se $(3\pi {)}^2=9{\pi }^2$ milta hai). $\alpha$ se multiply karo: $\alpha u_{2,xx}=-18\alpha {\pi }^2{e}^{-9\alpha {\pi }^{2}t}\sin 3\pi x=u_{2,t}$ ✅. Rate $9\alpha {\pi }^2$ exactly ${\lambda }_n=\alpha n^2{\pi }^2$ ka $n=3$ case hai. Step 4 — $x=\frac{1}{2}$, $t=0$ par evaluate karo. $\sin (\pi /2)=1$, $\sin (3\pi /2)=-1$: $u=3(1)(1)+2(1)(-1)=3-2=1$ K. Yeh superposition exactly ek two-term Fourier series hai.

Exactly yeh shape kyun? Ek point-burst mein koi built-in length scale nahi hoti, isliye profile sirf natural diffusion width $\sqrt{4\alpha t}$ se spread ho sakta hai (Problem 4.1 se, lengths $\sqrt{\alpha t}$ ki tarah scale hote hain). Do consequences bake in hain: exponent mein width $4\alpha t$ honi chahiye, aur prefactor $1/\sqrt{4\pi \alpha t}$ hona chahiye taaki total heat $\int udx$ constant rahe jab bump flatten aur widen ho. Ab hum ise directly verify karte hain — koi ansatz gymnastics nahi, bas woh do derivatives jo equation maangti hai. Step 1 — time derivative. Likho $u=C{t}^{-1/2}{e}^{-x^2/(4\alpha t)}$ constant $C=\frac{1}{\sqrt{4\pi \alpha }}$ ke saath. Product differentiate karo ($t^{-1/2}$ se $-\frac{1}{2}{t}^{-3/2}$ milta hai; exponent $-x^2/(4\alpha t)$ ka $t$-derivative $+x^2/(4\alpha t^2)$ hai): $u_t=C\left[-\frac{1}{2}{t}^{-3/2}+t^{-1/2}\cdot \frac{x^2}{4\alpha t^2}\right]e^{-x^2/(4\alpha t)}=u\left(-\frac{1}{2t}+\frac{x^2}{4\alpha t^2}\right).$ Step 2 — space derivatives. $u_x=u\cdot \left(-\frac{x}{2\alpha t}\right)$, aur phir se differentiate karo ($u$ aur $-x/(2\alpha t)$ ka product): $u_{xx}=u\left[{\left(-\frac{x}{2\alpha t}\right)}^2-\frac{1}{2\alpha t}\right]=u\left(\frac{x^2}{4{\alpha }^2{t}^2}-\frac{1}{2\alpha t}\right).$ Step 3 — $\alpha$ se multiply karo. $\alpha u_{xx}=u\left(\frac{x^2}{4\alpha t^2}-\frac{1}{2t}\right)=u_t$. ✅ Edge case $t\to 0^{+}$. Shape dekho: width $\sqrt{4\alpha t}\to 0$, isliye bump infinitely narrow ho jaata hai, jabki prefactor $1/\sqrt{4\pi \alpha t}\to \infty$, isliye infinitely tall ho jaata hai — phir bhi uske neeche area $1$ fixed rehta hai. Yeh limiting object Dirac delta $\delta (x)$ hai: "saari heat ek point par dher ki gayi." Isliye Gaussian woh solution hai jiska initial condition ek point-burst $u(x,0)=\delta (x)$ hai, aur kisi bhi $t>0$ ke liye yeh ek smooth spread-out bump hai. Isliye ise fundamental solution / heat kernel kehte hain — woh building block jisse Fourier Series Solutions aur general initial data assemble kiye jaate hain.

Step 1 — energy differentiate karo. $\frac{dE}{dt}={\int }_0^{1}2u{u}_{t}dx={\int }_0^{1}2u(\alpha u_{xx})dx$ (PDE use ki). Step 2 — integrate by parts. ${\int }_0^{1}u{u}_{xx}dx=[u{u}_x{]}_0^1-{\int }_0^1(u_x{)}^{2}dx$. Step 3 — boundary term khatam karo. Insulated ends se $u_x=0$ milta hai $x=0,1$ par, isliye $[u{u}_x{]}_0^1=0$. Step 4 — conclude karo. $\frac{dE}{dt}=-2\alpha {\int }_0^1(u_x{)}^{2}dx\le 0$, kyunki $\alpha >0$ aur $(u_x{)}^2\ge 0$. Matlab: temperature profile ka "size" sirf shrink ho sakta hai ya stable reh sakta hai — diffusion smooths, never sharpens ka ek rigorous chehra. Yeh Conservation Laws and Continuity Equation ka energy-method cousin hai.