Visual walkthrough — Solving ODEs with Laplace (including discontinuous forcing)
4.6.30 · D2· Maths › Ordinary Differential Equations › Solving ODEs with Laplace (including discontinuous forcing)
Step 1 — Hum dekh kya rahe hain? (time picture)
KYA. Equation ko seedhe shabdon mein padho. Symbol ek unknown hilti-dulti quantity hai — maano ek leaky tank mein paani ki matra, jo har time pe measure ki jaati hai. Symbol (padho "-prime") uski slope hai: abhi kitni tezi se badal raha hai. Right-hand side ek switch hai: se pehle aur ke baad hota hai.
YE PEHLE KYUN. Kisi bhi algebra se pehle, tumhe dikhna chahiye ki equation kya maang rahi hai. Woh kehti hai: "rate of change aur current level ka sum barabar hai jo bhi switch feed kar raha hai." se pehle switch feed karta hai, aur kyunki bhi se shuru hota hai, kuch nahi hilta. ke baad switch feed karta hai aur tank bhar'na shuru hota hai.
PICTURE. Orange curve switch hai (forcing). Yeh ek flat floor hai, phir pe height tak ek cliff. Wahi cliff hai jis ki wajah se classical methods struggle karte hain — tum cliff ko ache se differentiate nahi kar sakte. Laplace us cliff ko ek single tidy factor mein badal dega.
Step 2 — Magic camera: kya karta hai
KYA. Laplace transform ek integral hai jo poori function ko — jo time pe jeeti hai — leke ek nayi function banata hai jo ek naye variable mein rehti hai:
Term by term: ek shrinking exponential hai — fixed ke liye yeh se shuru hota hai aur decay karta hai. Yeh ek weighting curtain ki tarah kaam karta hai. ko isse multiply karo aur sab add karo (wahi hai ) — is tarah ki poori time-history ek single number mein crush ho jaati hai, har ek ke choice ke liye. ko sweep karo aur poori function mil jaati hai.
YEH TOOL KYUN, KOI AUR KYUN NAHI? Hum chahte hain ek operation jo differentiation ko multiplication mein convert kare, kyunki multiplication easy hai aur differentiation hard. Tamam possible transforms mein se sirf isi mein yeh property hai (hum ise next step mein prove karte hain). Dekhko Laplace Transform — Definition and Existence — kab yeh integral converge karta hai.
PICTURE. Left panel: tumhari function time world mein. Middle: fading curtain ek value of ke liye. Right: unka product, jiska shaded area hi number hai. change karo → curtain faster ya slower fade karta hai → area badalta hai → tum trace karte ho.
Step 3 — Differentiation " se multiply" kyun ban jaata hai
KYA. Hum slope ko transform karte hain aur pate hain ki yeh almost sirf ban jaata hai. Definition pe integration by parts use karte hue:
=\underbrace{\big[e^{-st}y(t)\big]_0^\infty}_{\text{boundary term}} -\int_0^\infty(-s)e^{-st}y(t)\,dt.$$ Boundary term: $t=\infty$ pe curtain $e^{-st}$ sab kuch khatam kar deta hai → $0$; $t=0$ pe yeh $-y(0)$ deta hai. Baaki integral $s$ times original transform hai. Toh: $$\boxed{\;\mathcal{L}\{y'\}=sY(s)-y(0)\;}$$ jahan $Y(s)=\mathcal{L}\{y\}$. Akela symbol $s$ wahi kaam kar raha hai jo $\tfrac{d}{dt}$ karta tha — yahi poora payoff hai. **YEH KYUN MATTER KARTA HAI.** $-y(0)$ term kachra nahi hai: yahi hai **jahan initial condition algebra mein khud enter karti hai**. Tumhe ise end mein bolte nahi bolna padta. **PICTURE.** Boundary term $e^{-st}y(t)$ ki height hai jo dono ends pe evaluate ki gayi hai: $0$ far right pe (curtain jeet jaata hai), aur $-y(0)$ left edge pe. Picture mein woh dono endpoints lit up hain. ![[deepdives/dd-maths-4.6.30-d2-s03.png]] > [!recall]- Boundary term infinity pe kyun vanish hota hai? > Kyunki $e^{-st}\to 0$ zyada tezi se jaata hai jitna $y(t)$ badhta hai ($s$ kaafi bade hone pe), toh product mar jaata hai. ::: Woh "$s$ kaafi bada" wali baat hi convergence condition hai. --- ## Step 4 — Switch ko stamp karna: cliff ko $e^{-2s}/s$ mein badalna **KYA.** Humein $\mathcal{L}\{u(t-2)\}$ chahiye. Step ko definition mein dalo. Kyunki $u(t-2)=0$ $t=2$ tak, integral sirf $t=2$ se shuru hota hai: $$\mathcal{L}\{u(t-2)\}=\int_2^\infty e^{-st}\cdot 1\,dt =\Big[\tfrac{-1}{s}e^{-st}\Big]_2^\infty=\frac{e^{-2s}}{s}.$$ Term by term: lower limit $0$ se $2$ pe jump kiya **kyunki switch pehle off tha**. Antiderivative evaluate karne par $\dfrac{e^{-2s}}{s}$ milta hai. Factor $e^{-2s}$ ek **time-delay stamp** hai — yeh "2 seconds late shuru karo" ka algebraic fingerprint hai. $\tfrac1s$ ek plain constant $1$ ka transform hai (woh height jo switch hold karta hai). **KYUN.** Yeh second shifting theorem ka miniature hai: *time mein $a$ ka shift, $s$-land mein $e^{-as}$ factor ban jaata hai*. Sirf yahi ek fact discontinuous forcing ko effortless banata hai. **PICTURE.** $e^{-st}$ ke neeche shaded area, lekin sirf $t=2$ ke baad se — missing chunk $[0,2]$ hi exactly woh hai jo $e^{-2s}$ discount provide karta hai. ![[deepdives/dd-maths-4.6.30-d2-s04.png]] > [!formula] The stamp > $$\mathcal{L}\{u(t-a)g(t-a)\}=e^{-as}G(s),\qquad \mathcal{L}\{u(t-a)\}=\frac{e^{-as}}{s}.$$ --- ## Step 5 — Algebra solve karo (trip ka poora point) **KYA.** Poori ODE $y'+y=u(t-2)$ ko term by term transform karo, Steps 3–4 aur $y(0)=0$ use karte hue: $$\underbrace{\big(sY-y(0)\big)}_{\mathcal{L}\{y'\}}+\underbrace{Y}_{\mathcal{L}\{y\}} =\underbrace{\frac{e^{-2s}}{s}}_{\mathcal{L}\{u(t-2)\}}.$$ $y(0)=0$ ke saath yeh $sY+Y=\dfrac{e^{-2s}}{s}$ hai, matlab $(s+1)Y=\dfrac{e^{-2s}}{s}$. Ab bas divide karo: $$\boxed{\;Y(s)=\frac{e^{-2s}}{s(s+1)}\;}$$ Term by term: $(s+1)$ aaya kyunki dono $y$-terms ek factor mein collapse ho gayi; us se divide karna plain [[Linear Constant-Coefficient ODEs|algebra]] hai, koi calculus nahi. Delay stamp $e^{-2s}$ bas untouched ride karta raha. **KYUN.** Time-land mein ek differential equation thi. $s$-land mein woh ek **rational function times a stamp** ban gayi — kuch aisa jo hum fractions se tod sakte hain. **PICTURE.** Ek flow diagram: time-ODE → ($\mathcal{L}$ apply karo) → algebra → (divide karo) → $Y(s)$. ![[deepdives/dd-maths-4.6.30-d2-s05.png]] --- ## Step 6 — Partial fractions se split karo ($Y$ ko table pieces mein peeling) **KYA.** Stamp ko ek second ke liye ignore karo aur core $H(s)=\dfrac{1}{s(s+1)}$ ko [[Partial Fractions]] se factor karo: $$\frac{1}{s(s+1)}=\frac{A}{s}+\frac{B}{s+1}.$$ **Cover-up method:** $s$ ko cover karo aur $s=0$ set karo: $A=\dfrac{1}{0+1}=1$. $(s+1)$ ko cover karo aur $s=-1$ set karo: $B=\dfrac{1}{-1}=-1$. Toh $$H(s)=\frac{1}{s}-\frac{1}{s+1}.$$ Term by term: $\dfrac1s$ constant $1$ ka transform hai; $\dfrac{1}{s+1}$, $e^{-t}$ ka transform hai (kyunki $\mathcal{L}\{e^{at}\}=\tfrac{1}{s-a}$ with $a=-1$). Toh backwards padho: $$h(t)=1-e^{-t}.$$ **SPLIT KYUN KARO?** Inverse transform sirf *simple* pieces pe easy hai jo hum table mein pehchante hain. Partial fractions ek ugly product ko recognisable atoms mein kaat deti hai. **PICTURE.** Curve $h(t)=1-e^{-t}$ ko $1$ pe flat line aur decaying $e^{-t}$ ke sum ke roop mein draw karo — tum literally dono fractions ko stack hote dekh sakte ho. ![[deepdives/dd-maths-4.6.30-d2-s06.png]] --- ## Step 7 — Un-photo karo: delay AND gate **KYA.** Ab stamp wapas laao. Hamare paas $Y(s)=e^{-2s}H(s)$ hai jahan $h(t)=1-e^{-t}$. Shifting theorem **ulta** chalaate hain: $e^{-2s}$ se multiply karne ka matlab hai *$h$ ko $2$ se delay karo aur usse $t=2$ pe switch on karo*: $$\boxed{\;y(t)=u(t-2)\big(1-e^{-(t-2)}\big)\;}$$ Term by term: $u(t-2)$ **gate** hai — yeh $y=0$ ko switch se pehle force karta hai. Andar $1-e^{-(t-2)}$ delayed relaxation hai — $h$ ke andar ka har $t$, $t-2$ ban gaya. **GATE KYUN MATTER KARTA HAI.** $u(t-2)$ factor ke bina, formula $1-e^{-(t-2)}$ $t<2$ ke liye already *negative aur moving* hoga — ek machine jo switch on hone se pehle react kar rahi hai. Physically impossible. Gate optional nahi hai. **PICTURE.** Final response: $t=2$ tak flat zero, phir ek smooth rise jo $1$ ki taraf badhta hai lekin kabhi poora nahi pahunchta — Step 6 curve ki ek delayed copy. ![[deepdives/dd-maths-4.6.30-d2-s07.png]] > [!mistake] Gate drop karna > *Tempting:* $e^{-2s}H(s)$ ko bas $1-e^{-(t-2)}$ ki tarah invert karo. > *Fix:* hamesha $u(t-2)$ se multiply karo. Delay **aur** gate — dekho [[Transfer Functions and Convolution]] kyun yeh exactly ek delayed impulse response hai. --- ## Step 8 — Har case check karo (degenerate & limiting behaviour) **KYA.** Chalte hain ensure karte hain ki kuch bhi undikha na rahe. - **Switch se pehle, $t<2$:** $u(t-2)=0$, toh $y=0$. Picture flat hai. ✓ - **Exactly $t=2$ pe:** $u=1$ lekin $1-e^{-(2-2)}=1-e^0=1-1=0$. Toh $y$ rise $0$ se shuru karta hai — **continuous**, koi jump nahi. Accha hai, leaky tank teleport nahi kar sakta. - **Bahut baad mein, $t\to\infty$:** $e^{-(t-2)}\to 0$, toh $y\to 1$. Tank switch ki height pe settle ho jaata hai. ✓ - **Agar $y(0)\neq 0$ ho?** Tab Step 3 $-y(0)$ term rakhega, ek piece $\tfrac{y(0)}{s+1}\to y(0)e^{-t}$ add karke — start ki ek decaying memory. Switched part unchanged rehta hai. **CHAARON KYUN DIKHAO.** Contract yeh hai: reader ko koi aisa scenario nahi milna chahiye jo humne skip kiya ho. Switch pe value, tail, aur nonzero start — har ek alag behave karta hai aur har ek draw kiya gaya hai. **PICTURE.** Teen stacked curves: $y(0)=0$ (hamara case), $y(0)=0.5$, $y(0)=-0.5$ — same switched rise, usse pehle alag fading memory. ![[deepdives/dd-maths-4.6.30-d2-s08.png]] > [!recall]- Exact switch time $t=2$ pe $y$ kya hai? > $y(2)=0$ — response continuous hai, yeh zero se rise *shuru* karta hai. ::: Step forcing discontinuous hai, lekin first-order system ise smooth karke continuous response banata hai. --- ## Ek-picture summary Sab ek saath: time-world problem (cliff forcing), $s$-land ka safar (differentiation → $\times s$, delay → $e^{-2s}$), algebra $Y=\dfrac{e^{-2s}}{s(s+1)}$, partial-fraction split, aur gated answer tak wapsi ka safar. ![[deepdives/dd-maths-4.6.30-d2-s09.png]] > [!recall]- Feynman: poori walk ek 12-saal ke bachche ko batao > Socho ek water tank hai jisme ek tap laga hai. Tap OFF hai, toh tank khali baitha bore ho raha hai. Yahi hai hamari > equation $t=2$ se pehle. Ab: paani ko slosh hote dekhne ki jagah (mushkil!), hum ek **magic photo** > lete hain — Laplace transform — jo poori kahani ko ek still picture mein freeze kar deta hai, ek > banawati duniya mein jo "$s$" kehlati hai. Us duniya mein, "cheezein kitni tezi se badal rahi hain" sirf "$s$ se multiply karo" ban jaata hai, jo > baby arithmetic hai. Aur "tap 2 seconds late on hota hai" ek chhota sa sticker $e^{-2s}$ ban jaata hai > jiska matlab hai *yeh part late shuru karo*. Photo-land mein problem ek simple fraction hai, aur hum us > fraction ko do easy pieces mein kaatte hain (ek flat $1$ aur ek fading $e^{-t}$). Aakhir mein hum wapas real > time mein un-photo karte hain, do rules yaad rakhte hue: sab kuch $2$ seconds se **delay** karo, aur ek **gate** lagao taaki tap kholne se pehle kuch na ho. Result: $t=2$ tak khali, phir ek smooth fill-up "full" ($=1$) ki taraf > jo kabhi overflow nahi karta. Done — koi sloshing math nahi, bas ek photo, kuch algebra, aur un-photo. > [!mnemonic] Nine frames > **Dekho → Photo lo → Slope becomes $s$ → Switch stamp karo → Divide karo → Split karo → Delay+Gate → Sab cases check karo → Compress.** --- > [!intuition] Connections > - [[Laplace Transform — Definition and Existence]] — camera khud. > - [[Heaviside Step and Dirac Delta Functions]] — woh switch jo humne stamp kiya. > - [[Partial Fractions]] — Step 6 ka peeling tool. > - [[Transfer Functions and Convolution]] — kyun $H(s)$ system ka *asli* fingerprint hai. > - [[Method of Undetermined Coefficients]] — classical method jo cliff pe choke karta hai. > - [[Linear Constant-Coefficient ODEs]] — equations ki woh family jo yeh solve karta hai.