Solving PDEs with Fourier transforms (heat equation on infinite domain)

4.7.19 · Maths › Partial Differential Equations

WHY use Fourier transforms here?

WHAT is the Fourier transform (our convention)

HOW to solve $u_{t} = α u_{xx}$ — scratch se poori derivation

Problem: solve karo $u_{t} = α u_{xx}, - \infty < x < \infty, t > 0, u (x, 0) = f (x),$ jahan $α > 0$ thermal diffusivity hai, aur $u \to 0$ jab $∣ x ∣ \to \infty$ .

Step 1 — $x$ mein Transform karo. Define karo $\overset{u}{^} (k, t) = \int u (x, t) e^{- ik x} d x$ . Yeh step kyun? Yeh $t$ ko freeze karta hai aur sirf $x$ par act karta hai, $x$ -derivatives ko algebra mein convert karta hai.

Dono sides ko transform karo ( $F {u_{xx}} = - k^{2} \overset{u}{^}$ use karke aur note karo ki $\partial_{t}$ , $x$ -integral ke saath commute karta hai): $\frac{\partial u ^}{\partial t} = - α k^{2} \overset{u}{^} (k, t) .$ Yeh step kyun? Yeh ab $t$ mein ek first-order linear ODE hai, jahan $k$ ek constant parameter hai.

Step 2 — ODE solve karo. Ek separable/linear ODE $\overset{u}{^}_{t} = - α k^{2} \overset{u}{^}$ deta hai $\overset{u}{^} (k, t) = \overset{u}{^} (k, 0) e^{- α k^{2} t} .$ Yeh step kyun? Standard exponential decay solution; initial value $\overset{u}{^} (k, 0) = \hat{f} (k)$ initial profile ka transform hai.

To $\overset{u}{^} (k, t) = \hat{f} (k) e^{- α k^{2} t} .$

Step 3 — Invert karo. $x$ mein solution ek product $\hat{f} (k) \cdot \hat{G} (k)$ hai jahan $\hat{G} (k) = e^{- α k^{2} t}$ . Transforms ka product = functions ka convolution (Convolution Theorem): $F^{- 1} {\hat{f} \hat{G}} = f * G, (f * G) (x) = \int_{- \infty}^{\infty} f (y) G (x - y) d y .$ Yeh step kyun? Hum product ka hard inverse transform directly karne se bachte hain; bas $G (x, t) = F^{- 1} {e^{- α k^{2} t}}$ chahiye.

Step 4 — Gaussian ko invert karo. Humein chahiye $G (x, t) = \frac{1}{2 π} \int_{- \infty}^{\infty} e^{- α k^{2} t} e^{ik x} d k .$ Exponent mein square complete karo: $a = α t$ ke saath, $- a k^{2} + ik x = - a (k - \frac{i x}{2 a})^{2} - \frac{x ^{2}}{4 a} .$ Gaussian integral $\int e^{- a (k - c)^{2}} d k = π / a$ use karke (shifted contour ke liye valid), $G (x, t) = \frac{1}{2 π} e^{- x^{2} / (4 a)} \frac{π}{a} = \frac{1}{4 π α t} e^{- x^{2} / (4 α t)} .$ Yeh step kyun? Ek Gaussian ka Fourier transform ek Gaussian hota hai — completing the square woh trick hai jo integral ko standard form mein badal deta hai.

Step 5 — Final solution. $u (x, t) = \int_{- \infty}^{\infty} f (y) \frac{1}{4 π α t} e^{- (x - y)^{2} / (4 α t)} d y .$

Figure — Solving PDEs with Fourier transforms (heat equation on infinite domain)

Worked Example 1 — Point (delta) initial condition

Lo $f (x) = δ (x)$ (saari heat origin par concentrated hai).

$u (x, t) = \int δ (y) G (x - y, t) d y = G (x, t) = \frac{1}{4 π α t} e^{- x^{2} / (4 α t)} .$

Yeh step kyun? Delta $y = 0$ ko sift karta hai, to solution hai hi heat kernel. Yeh confirm karta hai ki kernel ek unit point source ka response hai — "Green's function" interpretation.

Forecast-then-Verify: $t$ ke badhne par peak height predict karo. Peak $x = 0$ par hai: $u (0, t) = 1/ 4 π α t \to 0$ . Width $σ = 2 α t \to \infty$ . Heat spread aur cool hoti hai — verified. ✅

Worked Example 2 — Step (rectangular) initial temperature

Lo $f (x) = T_{0}$ jab $∣ x ∣ < L$ aur $0$ otherwise.

$u (x, t) = \frac{T _{0}}{4 π α t} \int_{- L}^{L} e^{- (x - y)^{2} / (4 α t)} d y .$ Substitute karo $s = \frac{x - y}{4 α t}$ , $d s = - d y / 4 α t$ :

=\frac{T_0}{2}\!\left[\operatorname{erf}\!\frac{x+L}{\sqrt{4\alpha t}}-\operatorname{erf}\!\frac{x-L}{\sqrt{4\alpha t}}\right].$$ *Yeh step kyun?* Substitution Gaussian convolution ko **error function** $\operatorname{erf}(z)=\frac{2}{\sqrt\pi}\int_0^z e^{-s^2}ds$ mein map karta hai, jo ek Gaussian ka standard antiderivative hai. **Forecast-then-Verify:** $t\to 0^+$ par erf-difference sharp box reproduce karta hai; bade $t$ par yeh ek wide smooth bump ban jaata hai. ✅ --- ## Worked Example 3 — Gaussian initial condition Gaussian hi rehti hai Lo $f(x)=e^{-x^2/(2\sigma_0^2)}$. Do Gaussians ko convolve karne par ek Gaussian milta hai jiske variances **add** hote hain: $$u(x,t)=\frac{\sigma_0}{\sqrt{\sigma_0^2+2\alpha t}}\,\exp\!\left(-\frac{x^2}{2(\sigma_0^2+2\alpha t)}\right).$$ *Yeh step kyun?* $N(0,\sigma_0^2)$ ka kernel $N(0,2\alpha t)$ ke saath convolution $N(0,\sigma_0^2+2\alpha t)$ deta hai — variance addition "diffusion = spreading" ka sabse clean statement hai. --- ## Common Mistakes (Steel-manned) > [!mistake] "$\mathcal F\{u_{xx}\}=k^2\hat u$ (minus sign nahi)." > **Kyun sahi lagta hai:** squaring usually signs khatam kar deta hai. **Fix:** har $\partial_x$ se $ik$ milta hai, to do se $(ik)^2=i^2k^2=-k^2$ milta hai. **Minus zaroori hai** — yeh ODE ko *decay* ($e^{-\alpha k^2 t}$) karta hai blow up hone ki jagah. $+k^2$ ke saath growth hogi aur problem ill-posed ho jayegi. > [!mistake] "Bas transforms ko wapas multiply karo aur ho gaya — yeh $f\cdot G$ ke barabar hai." > **Kyun sahi lagta hai:** $k$-space mein genuinely ek product $\hat f\hat G$ hai. **Fix:** frequency space mein product, real space mein **convolution** $f*G$ hota hai, *nahi* pointwise product. Inhe mix karne par smoothing integral poora kho jaata hai. > [!mistake] "Width time mein linearly grow karta hai." > **Kyun sahi lagta hai:** waves constant speed par travel karti hain $\Rightarrow$ linear. **Fix:** diffusion **wave-like nahi** hai. $\sigma=\sqrt{2\alpha t}\propto\sqrt t$. Double spread ke liye **chaar guna** time chahiye. Yeh $\sqrt t$ law diffusion ki signature hai. > [!mistake] "$\dfrac{1}{\sqrt{4\pi\alpha t}}$ normalisation bhool gaye." > **Kyun sahi lagta hai:** exponential poora answer lagta hai. **Fix:** prefactor total heat conserved rakhta hai ($\int u\,dx$ constant). Isko drop karo aur energy/heat conserve nahi hogi. --- ## Active Recall > [!recall]- Fourier transform $u_{xx}$ ko kya banata hai, aur yeh kyun help karta hai? > $-k^2\hat u$ (multiplication) mein. PDE har $k$ ke liye $t$ mein ek ODE ban jaata hai: $\hat u_t=-\alpha k^2\hat u$. > [!recall]- High frequencies sabse jaldi decay kyun hoti hain? > Decay rate $\alpha k^2$ hai, jo $k$ ke saath badhti hai. Sharp features = high $k$ = jaldi smooth. > [!recall]- Spreading width time ke saath kaise scale hoti hai? > $\sigma=\sqrt{2\alpha t}\propto\sqrt t$ (sub-linear). Double spread ke liye time chaar guna karo. > [!recall]- 12 saal ke bacche ko explain karo (Feynman) > Socho still paani mein ink ki ek boond. Fourier trick yeh kehne jaisi hai: pure ink blob ko track karne ki jagah, har "ripple size" ko alag se dekho. Chhote ripples (sharp edges) almost instantly blur ho jaate hain; bade slow ripples sabse zyada tikate hain. Saare surviving ripples wapas add karo aur tumhe ek smooth, ever-widening smudge milti hai — yahi heat (ya ink) ka spread hona hai. Blob ki width time ke square root ki tarah grow karti hai, isliye jaise spread hota hai yeh slow down hota jaata hai. > [!mnemonic] Memory hook > **"FRIDGE"** — **F**ourier turns derivatives to **k**, **R**eplace $u_{xx}\to -k^2$, **I**ntegrate the easy ODE, **D**ecay $e^{-\alpha k^2 t}$, **G**aussian kernel back, **E**rror-function for boxes. Aur yeh bhi: *"Diffusion spreads like √time, smooths like a Gaussian."* --- ## Connections - [[Fourier Transform]] — engine; derivative & convolution theorems. - [[Convolution Theorem]] — kyun answer $f*G$ hai. - [[Gaussian Integral]] — $e^{-\alpha k^2 t}$ ko invert karne ke liye square complete karo. - [[Heat Equation Separation of Variables]] — finite-domain counterpart (Fourier *series*). - [[Green's Function]] — heat kernel, $\partial_t-\alpha\partial_{xx}$ ka Green's function hai. - [[Error Function erf]] — step initial data ke liye aata hai. - [[Diffusion vs Wave Equation]] — $\sqrt t$ spreading vs constant-speed propagation. --- #flashcards/maths Fourier transform of $f'(x)$ ::: $ik\,\hat f(k)$ Fourier transform of $u_{xx}$ ::: $-k^2\hat u(k,t)$ The transformed heat equation (ODE in $t$) ::: $\hat u_t=-\alpha k^2\hat u$ Solution of that ODE ::: $\hat u(k,t)=\hat f(k)\,e^{-\alpha k^2 t}$ Decay rate of frequency $k$ ::: $\alpha k^2$ (high $k$ sabse jaldi decay hota hai) Heat kernel $G(x,t)$ ::: $\dfrac{1}{\sqrt{4\pi\alpha t}}e^{-x^2/(4\alpha t)}$ Inverse transform trick for $\hat f\,\hat G$ ::: convolution $f*G=\int f(y)G(x-y)\,dy$ Full solution $u(x,t)$ ::: $\int_{-\infty}^{\infty} f(y)\,\dfrac{e^{-(x-y)^2/(4\alpha t)}}{\sqrt{4\pi\alpha t}}\,dy$ Std deviation of the spreading profile ::: $\sigma=\sqrt{2\alpha t}\propto\sqrt t$ $f=\delta(x)$ ke liye solution ::: heat kernel $G(x,t)$ khud Box initial condition ka solution use karta hai ::: error function $\operatorname{erf}$ $-k^2$ mein minus sign kyun matter karta hai ::: decay $e^{-\alpha k^2 t}$ deta hai; $+k^2$ blow up karta (ill-posed) Time ke saath kya conserve hota hai ::: total heat $\int u\,dx$ (kernel ka area 1 hai) ## 🖼️ Concept Map ```mermaid flowchart TD HE[Heat equation ut equals alpha uxx] -->|posed on| DOM[Infinite domain R] DOM -->|no boundaries so use| FT[Fourier transform] FT -->|derivative rule| DR[F of f'' equals -k^2 fhat] DR -->|turns PDE into| ODE[ODE in t per frequency k] ODE -->|decoupled per| K[Each wavenumber k] ODE -->|solve exponential| SOL[uhat equals fhat times e^-alpha k^2 t] IC[Initial profile f x] -->|transforms to| FHAT[fhat k] FHAT -->|initial value| SOL SOL -->|high k decay fastest| SMOOTH[Sharp features smoothed] SOL -->|inverse transform| GAUSS[Gaussian spreading solution] GAUSS -->|widens over| TIME[Time t] ```