Laplace of derivatives — key property for solving ODEs: Visual walkthrough

4.6.28 · D2 · Maths › Ordinary Differential Equations › Laplace of derivatives — key property for solving ODEs

Step 0 — Woh chaar marks jo hum use kar sakte hain

Kisi bhi algebra se pehle, agree karte hain ki page pe likhe marks ka matlab kya hai, ek picture se anchor karke.

Definition Characters ki cast

$t$ — time, left se right chalti hai, $0$ se shuru hoti hai. Horizontal axis imagine karo.
$f (t)$ — koi signal, jaise ek wobble jo dheere dheere khatam hoti hai. $t$ -axis ke upar ek curve imagine karo.
$s$ — Laplace-domain variable: ek dial jise hum ghumate hain, jo control karta hai ki curtain kitni tezi se band hoti hai. $s$ ko ek decay rate samjho. $s$ badhao aur curtain jhatke se band ho jaati hai; kam karo aur woh zyada der tak khuli rehti hai. Yeh time nahi hai — yeh ek alag duniya mein rehta hai (" $s$ -domain") aur output function $F (s)$ ka input hai. Poore derivation mein hum $s > 0$ lete hain (baad mein: $s$ itna bada ki kaafi ho).
$e^{- s t}$ — $s$ se bana ek sirakta hua curtain. $s > 0$ ke liye yeh $1$ se shuru hota hai (jab $t = 0$ ) aur $t$ badhne ke saath $0$ ki taraf girta jaata hai. Yeh hamara "weighting" hai: yeh $t = 0$ ke paas bahut dhyan deta hai aur door right side pe lagbhag kuch nahi.
$\int_{0}^{\infty} (\dots) d t$ — " $t = 0$ se hamesha tak curve ke neeche ka area jodo." Lamba patla $\int$ Sum ke liye stretched letter S hai.

(Yeh chaar alag marks hain — $t$ , $f$ , $s$ /uska curtain $e^{- s t}$ , aur sum $\int$ — curtain sirf $s$ ka bana hua roop hai.)

Definition

$f$ pe fine print (taaki neeche ke saare integrals legal hon) Yeh sab rigorously hold karne ke liye hum require karte hain ki $f$ :

piecewise continuous ho $[0, \infty)$ pe — zyada se zyada finitely many jumps, koi wild blow-up nahi, taaki area har finite stretch pe exist kare;
exponential order ka ho — constants $M > 0$ aur $c$ exist karte hain jaise ki bade $t$ ke liye $∣ f (t) ∣ \leq M e^{c t}$ . Yeh bound karta hai ki $f$ kitni tezi se badh sakta hai.
differentiable ho ( $f^{'}$ bhi piecewise continuous ke saath), kyunki hum $f^{'}$ transform karte hain.

Yeh do properties exactly wahi hain jo (i) transform integral ko $s > c$ ke liye converge karaati hain, aur (ii) $t \to \infty$ pe boundary term ko vanish karaati hain. Hum inhe har step pe naam lekar use karenge, isliye inhe ek baar yahan upar likh lete hain. Dekho Exponential Order and Convergence of Integrals.

Figure dekho: cyan curve $f (t)$ hai, amber curve sirakta hua curtain $e^{- s t}$ hai, aur white shaded region woh hai jo tab hota hai jab aap unhe multiply karte ho — woh shaded area hi $F (s)$ hai.

Curtain kyun? Kyunki $f (t)$ $t \to \infty$ pe settle nahi bhi ho sakta, toh uska raw area infinite ho sakta hai. Curtain $e^{- s t}$ door wale hisse ko zero kar deta hai, total area ko finite banata hai — bas itna ki dial $s$ ko $f$ ki growth rate $c$ se zyada ghuma do. Yeh crushing hi woh secret ingredient hai jo hum Step 4 mein use karte hain.

Step 1 — Asli sawaal pucho: curtain ke neeche $f^{'} (t)$ ka area kya hai?

KYA. Hum chahte hain $L {f^{'} (t)}$ — wahi machine, lekin $f (t)$ ki jagah slope $f^{'} (t)$ dalke.

$L {f^{'} (t)} = \int_{0}^{\infty} curtain e^{- s t} slope of f f^{'} (t) d t$

KYUN. ODE solve karne ka matlab hai ki hum baar baar $f^{'}$ , $f^{''}$ , ... se milte hain. Hum baar baar unke transforms ko $F (s)$ (plain $f$ ka transform) se express karna chahte hain taaki calculus collapse hokar algebra ban jaaye. Toh goal hai: $f^{'}$ ko integral se bahar karo aur $f$ ko wapas laao.

PICTURE. $f^{'} (t)$ har moment pe cyan curve ki steepness hai. Jahan $f$ chadh raha hai, $f^{'}$ positive hai (axis ke upar); jahan $f$ gir raha hai, $f^{'}$ negative hai (neeche). Hum us steepness ko curtain se weight karke jod rahe hain.

Step 2 — Tool chuno: integration by parts (aur kyun yahi tool)

Hamare paas ek product integrate ho raha hai: (curtain) $\times$ (ek derivative). Exactly "ek derivative wala integral" ke liye bana hua ek hi rule hai — Integration by Parts.

Substitution ya kuch aur kyun nahi? Substitution ek variable rewrite karta hai; woh do factors ke beech derivative relocate nahi kar sakta. By-parts ek hi elementary move hai jo kehta hai "mujhe yeh piece differentiate karne do instead of woh piece." Kyunki hamara poora aim $f$ ko un-prime karna hai, by-parts forced hai.

Hum choose karte hain: $u = e^{- s t} (curtain), d v = f^{'} (t) d t (the derivative) .$

Yeh pairing kyun? Hum $d v = f^{'} (t) d t$ set karte hain taaki uska antiderivative $v = f (t)$ ho — prime integrate hone pe khud cancel ho jaata hai. Yahi pura trick hai. Phir: $d u = \frac{d}{d t} e^{- s t} d t = - s e^{- s t} d t, v = f (t) .$

Figure split dikhata hai: amber hai $u = e^{- s t}$ (jise hum differentiate karenge, ek $- s$ gain karke), cyan hai $d v = f^{'} (t) d t$ (jise hum integrate karenge, original $f$ wapas paake).

Step 3 — Rule apply karo: equation do pieces mein split ho jaati hai

KYA. Apna $u, v, d u, d v$ $\int u d v = uv - \int v d u$ mein substitute karo:

$L {f^{'} (t)} = boundary term uv [e^{- s t} f (t)]_{0}^{\infty} - - \int v d u \int_{0}^{\infty} f (t) (- s e^{- s t}) d t$

Term-by-term padho:

$[e^{- s t} f (t)]_{0}^{\infty}$ — boundary term. Matlab hai " $e^{- s t} f (t)$ ki value top end $\infty$ pe, minus uski value bottom end $0$ pe." Ek pure edge quantity, koi integral nahi bachi.
$- \int_{0}^{\infty} f (t) (- s e^{- s t}) d t$ — leftover integral, lekin notice karo ki $f$ ab bina prime ke appear ho raha hai — exactly wahi jo hum chahte the.

KYUN. Prime integral se gayab ho gayi; uski keemat (a) ek boundary term aur (b) aage ek factor $- s$ se chuki.

PICTURE. Step 2 ka single shaded region do accounts mein cut ho gaya hai: $t = 0$ aur $t = \infty$ pe ek patla edge contribution (boundary), aur ek bulk area jisme ab plain $f$ hai.

Step 4 — Curtain cash in karo: boundary term actually kya equal hai

Ab hum $[e^{- s t} f (t)]_{0}^{\infty}$ ko uske do ends pe evaluate karte hain. Yahan har case check karna zaroori hai.

Top end, $t \to \infty$ : $lim_{t \to \infty} e^{- s t} f (t) = 0.$ Zero kyun hai: hamare exponential-order fine print se, $∣ f (t) ∣ \leq M e^{c t}$ , toh $∣ e^{- s t} f (t) ∣ \leq M e^{- (s - c) t} \to 0$ jab bhi dial $s > c$ ho. Curtain jeet jaata hai. (Neeche Degenerate warning dikhata hai kya toot jaata hai agar $s \leq c$ ho.)

Bottom end, $t = 0$ : $e^{- s \cdot 0} f (0) = e^{0} f (0) = 1 \cdot f (0) = f (0) .$ KYUN: $t = 0$ pe curtain poora khula hai ( $e^{0} = 1$ ), toh woh kuch nahi chupaata — raw starting value $f (0)$ dikhta hai.

Isliye $[e^{- s t} f (t)]_{0}^{\infty} = top 0 - bottom f (0) = - f (0) .$

Degenerate case, drawn. Agar dial $s$ bahut kam set kiya — growth rate $c$ se neeche? Toh $e^{- s t} f (t) = e^{- (s - c) t} \cdot (\dots)$ mein positive exponent hai aur woh marne ki jagah badh jaata hai. Boundary kabhi settle nahi hoti, aur pura transform integral diverge kar jaata hai.

Step 5 — Leftover integral pehchano aur result assemble karo

KYA. Leftover integral clean up karo. Constant $- s$ (aur uska minus, $+ s$ deta hai) bahar nikalo: $- \int_{0}^{\infty} f (t) (- s e^{- s t}) d t = + s \int_{0}^{\infty} e^{- s t} f (t) d t .$

Key recognition: woh surviving integral, symbol for symbol, Step 0 se $F (s)$ ki definition hai: $\int_{0}^{\infty} e^{- s t} f (t) d t = F (s) .$

KYUN yeh punchline hai. Hum shuru mein chahte the ki $f^{'}$ ko $F (s)$ ke zariye express karen. Yahan hai — integral khud $F (s)$ mein fold ho gaya. Dono pieces substitute karo:

$L {f^{'} (t)} = Step 4 boundary - f (0) + Step 5 bulk s F (s) = s F (s) - f (0)$

Final formula term-by-term:

$s F (s)$ — "transform ko $s$ se multiply karo." Differentiation multiplication ban gaya. $s$ Step 2 mein curtain ko differentiate karne se aaya.
$- f (0)$ — $t = 0$ pe sum shuru karne ki fee; initial value, poori tarah khule curtain ki meherbani.

Step 6 — Bilkul wahi move do baar karo: second derivative

KYA. $f^{''}$ ke liye humein koi naya idea nahi chahiye. Bas $g = f^{'}$ relabel karo aur Step 5 ka result $g$ pe dobara use karo: $L {g^{'} (t)} = s L {g} - g (0) .$

Ab $g$ unpack karo:

$g = f^{'} \Rightarrow g^{'} = f^{''}$ , toh left side hai $L {f^{''}}$ .
$L {g} = L {f^{'}} = s F (s) - f (0)$ (Step 5 dobara).
$g (0) = f^{'} (0)$ — starting slope.

Substitute karo: $L {f^{''}} = s (s F (s) - f (0)) - f^{'} (0) = do baar multiply s^{2} F (s) - fee 1 s f (0) - fee 2 f^{'} (0) .$

KYUN aisa dikhta hai. Rule ka har application ek power of $s$ add karta hai aur ek initial value subtract karta hai, aur $s$ ki powers $s^{1}, s^{0}$ neeche aati hain jaise initial value ka derivative order $f (0), f^{'} (0)$ upar jaata hai.

Ek-picture summary

Ek figure, poori kahani: $f^{'}$ ko machine mein daalo → by-parts split karta hai → top edge mar jaata hai (curtain jeet jaata hai, bas dial $s > c$ rakhna) → bottom edge $- f (0)$ deta hai → bulk $F (s)$ mein $s$ se scale hokar wapas fold ho jaata hai → nikalta hai $s F (s) - f (0)$ .

Recall Feynman retelling — plain words mein walkthrough

Hum chahte the ki ek sirakti hui curtain ke neeche signal ke slope ka "area" mile, jahan ek dial $s$ set karta hai ki curtain kitni tezi se band hoti hai. Calculus mein exactly ek hi trick hai jo derivative ko ek cheez se doosri par shift karti hai — integration by parts — toh humne wahi use kiya, slope ko integrate karne wala piece choose kiya taaki uski derivative-ness cancel ho jaaye aur plain signal wapas aaye. Us trick ne hume do bills diye: ek leftover area aur ek boundary charge sirf do ends pe measured. Dur end $t = \infty$ pe curtain ne sab kuch zero kar diya — jab tak humne dial signal ki growth rate se aage ghuma rakha. Paas wale end $t = 0$ pe curtain poora khula hai, toh signal ki starting height $f (0)$ dikhti hai — minus ke saath, kyunki boundaries top se bottom subtract karti hain. Leftover area, ek baar constant bahar nikalne ke baad, literally $F (s)$ ki definition hai dobara, ab $s$ se multiply hokar. Pieces jodo: $s F (s) - f (0)$ . Differentiation "times $s$ " ban gaya, aur zero pe start karne ki keemat hai aapki initial condition $f (0)$ — jo exactly woh cheez hai jo initial value problems solve karna itna clean banati hai. Trick do baar karo aur milta hai $s^{2} F (s) - s f (0) - f^{'} (0)$ ; pattern wohi move hai, dohraata hua.

Recall Self-check clozes

Woh tool jo ek factor se doosre mein derivative shift karta hai woh hai integration by parts. Variable $s$ hai ==Laplace-domain dial / decay rate jo set karta hai ki curtain $e^{- s t}$ kitni tezi se band hoti hai. Top boundary isliye vanish hoti hai kyunki curtain $e^{- s t}$ , $f$ ki growth se tezi se decay karta hai, yaani $∣ e^{- s t} f ∣ \leq M e^{- (s - c) t} \to 0$ jab dial $s > c$ ho==. Term $- f (0)$ isliye appear hoti hai kyunki == $t = 0$ pe curtain $1$ ke barabar hota hai, toh raw starting value dikhti hai, aur boundaries bottom end subtract karti hain==.

Kaun sa symbol differentiation ko multiplication mein badla, aur woh kahan se aaya?

Factor

s

, jo integration by parts ke dauran curtain

e^{- s t}

ko differentiate karne se paida hua.

First-derivative rule batao aur har term ka naam lo.

L {f^{'}} = s F (s) - f (0)

;

s F (s)

= "differentiation multiply-by-

s

hai",

- f (0)

t = 0

pe boundary fee.

Second-derivative rule ke liye koi naya idea kyun nahi chahiye?

Tum pehla-derivative rule

g = f^{'}

pe apply karte ho, phir

L {f^{'}} = s F (s) - f (0)

aur

g (0) = f^{'} (0)

substitute karte ho.

Kin

s

ke liye rule valid hai, aur kyun?

s

ke liye jo

f

ki growth rate

c

se bada ho (region of convergence), taaki

\infty

pe boundary term vanish ho jaaye aur integral converge kare.

Connections

Parent topic (Hinglish)
Laplace Transform — Definition and Existence
Integration by Parts
Exponential Order and Convergence of Integrals
Solving Initial Value Problems with Laplace
Inverse Laplace Transform and Partial Fractions
Linearity of the Laplace Transform
Fourier Transform — comparison (no boundary term)