Visual walkthrough — Transforms of standard functions — proofs

4.6.26 · D2 · Maths › Ordinary Differential Equations › Transforms of standard functions — proofs

Neeche sab kuch yeh assume karta hai ki tum definition se Laplace Transform — Definition and Existence mein mil chuke ho, lekin main phir bhi har symbol ko khud se earn karunga.

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Step 1 — "Transform" ka matlab kya hai (weighing machine)

KYA. Hamare paas ek function $f(t)$ hai. Letter $t$ time hai, aur yeh sirf aage hi chalta hai: $t\ge 0$. Ek transform ek aisi machine hai jo poori function leti hai aur ek naye variable ki ek nai function banati hai, jise hum $s$ kehte hain.

KYUN. $t$ ko $s$ se badalne ki zaroorat kyun? Kyunki $s$-world mein calculus ka mushkil operation (differentiation) algebra ke aasaan operation mein badal jaata hai (multiply by $s$). Wahi superpower poori machine ka reason hai — dekho Transform of Derivatives — solving ODEs. Lekin use karne ke liye pehle yeh jaanna zaroori hai ki machine simple inputs ke saath kya karti hai.

PICTURE. Ek taraazu ki imagine karo. Har moment $t$ pe hum pan mein ek weight rakhte hain. Woh weight $e^{-st}$ hai — $t=0$ ke paas bhaari hoti hai aur waqt ke saath fade hoti jaati hai. Hum function ki value ko us fading weight se multiply karte hain har moment pe, phir sab kuch add up karte hain. Total ek number hota hai. $s$ badlo, alag total milega. Saare totals milke ek nai function $F(s)$ banate hain.

$$F(s)=\underset{\begin{array}{c}\text{add up over}\\ \text{all time }t\ge 0\end{array}}{\underbrace{{\int }_0^{\infty }}}\underset{\begin{array}{c}\text{fading}\\ \text{weight}\end{array}}{\underbrace{e^{-st}}}\underset{\begin{array}{c}\text{the value}\\ \text{being weighed}\end{array}}{\underbrace{f(t)}}dt$$

• $s$ ek knob hai jise hum integrate karte waqt fixed rakhte hain. $t$ woh variable hai jis par hum sum karte hain aur jo integral ke baad gayab ho jaata hai.
• ${\int }_0^{\infty }$ ek improper integral hai — upar ki limit infinity hai, isliye secretly matlab hai "upar wali limit ko $\infty$ tak jaane do aur dekho area kahan settle hoti hai" (details Improper Integrals mein).

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Step 2 — Weight $e^{-st}$ ko use karne se pehle dekho

KYA. Kuch bhi daalne se pehle, weight ko akele ghoor ke dekho: curve $t\mapsto e^{-st}$.

KYUN. Convergence ke baare mein sab kuch — dictionary mein bikhra hua "$s>0$" conditions — isi curve ke itni tez marne ya na marne se aata hai. Agar hum weight ko samjho, to har condition apne aap samajh aa jaayegi.

PICTURE. $s>0$ ke liye curve height $1$ se shuru hoti hai (kyunki $e^0=1$) aur $0$ ki taraf slide karti hai. $s$ jitna bada hoga, utni hi tez curve neeche giregi. $s=0$ ke liye yeh height $1$ pe flat line hai (kabhi nahi marti). $s<0$ ke liye yeh badhti hai aur infinity tak shoot karti hai — finite area ke liye bilkul bekaar.

• $e^{-st}$: base $e\approx 2.718$, exponent $-st$. Negative exponent ⇒ shrinking, jab tak $s>0$ ho.
• Is curve ke neeche shaded area exactly woh number hai $\mathcal{L}\{1\}$ jo hum abhi compute karne wale hain — flat line $f(t)=1$ bas weight ko untouched chhod deti hai.

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Step 3 — $\mathcal{L}\{1\}$ compute karo: weight ke neeche area nikalo

KYA. $f(t)=1$ recipe mein daalo. Tab $e^{-st}\cdot 1=e^{-st}$, to hume Step 2 ki curve ke neeche area chahiye.

$$\mathcal{L}\{1\}={\int }_0^{\infty }{e}^{-st}dt.$$

Antiderivative $e^{-st}/(-s)$ kyun? Hume ek aisi function chahiye jiska slope $e^{-st}$ ho. $e^{-st}$ differentiate karne par ek factor $-s$ neeche aata hai (chain rule). Us unwanted $-s$ ko cancel karne ke liye hum pehle se hi usse divide kar dete hain:

$$\frac{d}{dt}\left[\frac{e^{-st}}{-s}\right]=\frac{-s{e}^{-st}}{-s}=e^{-st}.✓$$

To top limit $T$ tak ki running area hai

$${\int }_0^T{e}^{-st}dt={\left[\frac{e^{-st}}{-s}\right]}_0^T=\underset{\text{top}}{\underbrace{\frac{e^{-sT}}{-s}}}-\underset{\text{bottom}}{\underbrace{\frac{e^0}{-s}}}.$$

PICTURE. Top edge $T$ ko rightward slide hote dekho. Area bharta jaata hai lekin ek ceiling se aage kabhi nahi — yeh saturate ho jaata hai. Wahi ceiling hamaara answer hai.

• Bottom term ($t=0$): $e^0=1$, deta hai $-\frac{1}{-s}=\frac{1}{s}$.
• Top term ($t=T\to \infty$): yahan condition paida hoti hai. $s>0$ ke liye, $e^{-sT}\to 0$, to yeh term gayab ho jaata hai. $s\le 0$ ke liye yeh nahi jaata aur area bhaag jaata hai — koi transform exist nahi karta.

$$\boxed{\mathcal{L}\{1\}=0-\left(-\frac{1}{s}\right)=\frac{1}{s},s>0.}$$

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Step 4 — Knob shift karo: $\mathcal{L}\{e^{at}\}$ wahi move hai

KYA. Ek badhta (ya ghatta) exponential $f(t)=e^{at}$ daalo. Integral ke andar do exponentials milte hain aur merge ho jaate hain:

$$\mathcal{L}\{e^{at}\}={\int }_0^{\infty }{e}^{-st}{e}^{at}dt={\int }_0^{\infty }{e}^{-(s-a)t}dt.$$

• Exponents combine karte hain: $-st+at=-(s-a)t$. Knob $s$ simply $s-a$ se replace ho gaya.

KYUN. Hum koi dobara kaam nahi karte. Yeh integral Step 3 se bilkul identical hai sirf letter $s$ ki jagah $s-a$ hai. To answer copy karo:

$$\boxed{\mathcal{L}\{e^{at}\}=\frac{1}{s-a},s-a>0\iff s>a.}$$

PICTURE. Input $e^{at}$ graph ko lift karne ki koshish karta hai; weight $e^{-st}$ isse crush karne ki. Sirf net exponent $-(s-a)$ decide karta hai kaun jeeta. Agar $s>a$ hai to crushing jeet jaata hai aur area finite hai; jitna $s$, $a$ ke paas aata hai, utna zyada baca hua pile uuncha hota jaata hai aur $1/(s-a)$ bada hota jaata hai — $s\to a^{+}$ par blow up karta hai.

• Net exponent $-(s-a)t$: sign decide hota hai $s-a$ se. Yeh akela quantity exponential par sari convergence conditions carry karta hai.
• Yeh "knob shift karo" idea exactly Linearity and First Shifting Theorem ka embryo hai.

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Step 5 — Ladder charho: $\mathcal{L}\{t^n\}$ recursion se

KYA. Ab ek power $f(t)=t^n$ ($n=0,1,2,\dots$ ke saath) daalo. Answer ko $I_n={\int }_0^{\infty }{e}^{-st}{t}^{n}dt$ naam do. Hum ise Integration by Parts se attack karte hain: $\int udv=uv-\int vdu$, choose karte hue

$$u=t^n(\Rightarrow du=n{t}^{n-1}dt),dv=e^{-st}dt(\Rightarrow v=\frac{e^{-st}}{-s}).$$

Yeh choice kyun? $t^n$ differentiate karne se power $t^{n-1}$ tak kam ho jaati hai. Matlab problem har baar shrink hoti hai — ek ladder jise hum neeche climb kar sakte hain kisi aise cheez tak jo hum already jaante hain ($I_0=\mathcal{L}\{1\}=1/s$).

$$I_n=\underset{\text{boundary term}}{\underbrace{{\left[t^n\frac{e^{-st}}{-s}\right]}_0^{\infty }}}+\frac{n}{s}\underset{=I_{n-1}}{\underbrace{{\int }_0^{\infty }{e}^{-st}{t}^{n-1}dt}}.$$

Boundary term zero hai:

• $t\to \infty$ par: $e^{-st}$ ($s>0$ ke liye) kisi bhi polynomial $t^n$ ki growth se tez shrink karta hai, isliye product $\to 0$.
• $t=0$ par: $n\ge 1$ ke liye $t^n=0$.

To hume ladder rule milta hai aur ise unroll karte hain:

$$I_n=\frac{n}{s}{I}_{n-1}=\frac{n}{s}\cdot \frac{n-1}{s}\cdots \frac{1}{s}\cdot \underset{I_0}{\underbrace{\frac{1}{s}}}=\frac{n!}{s^{n+1}}.$$ $$\boxed{\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}},s>0.}$$

PICTURE. Ladder ka har step $\frac{n}{s}$ se multiply karta hai; saare rungs stack karne par numerator mein $n\cdot (n-1)\cdots 1=n!$ aata hai aur neeche har baar ek power of $s$ add hoti hai.

• $n!$ (padho "n factorial") $=n\times (n-1)\times \cdots \times 1$. Yeh decoration nahi hai — yeh har us rung ka product hai jis par tum chale. Ise bhool jaana parent note mein flag kiya gaya classic error hai.

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Step 6 — Wiggle: $\sin at$ aur $\cos at$ Euler ke zariye

KYA. Oscillating inputs ke liye hum Euler's Formula $e^{i\theta}$ udhaarte hain: $e^{iat}=\cos at+i\sin at$, jahan $i^2=-1$. Wiggle ko directly integrate karne ki jagah, hum Step 4 reuse karte hain $a$ ki jagah $ia$ rakh ke:

$$\mathcal{L}\{e^{iat}\}=\frac{1}{s-ia}.$$

KYUN. Ek raw wiggle ko integration by parts do baar chahiye. Lekin ek complex exponential utni hi aasaani se integrate hota hai jitna ek real wala — yeh phir bhi bas $e^{(\text{stuff})t}$ hai. To hum easy complex integral karte hain, phir use real part (cos) aur imaginary part (sin) mein split karte hain. Un parts ko padhne ke liye, denominator ko real banao conjugate $s+ia$ se upar aur neeche multiply karke:

$$\frac{1}{s-ia}\cdot \frac{s+ia}{s+ia}=\frac{s+ia}{\underset{=s^2+a^2}{\underbrace{(s-ia)(s+ia)}}}=\frac{s}{s^2+a^2}+i\frac{a}{s^2+a^2}.$$

• $(s-ia)(s+ia)=s^2-(ia{)}^2=s^2-i^2{a}^2=s^2+a^2$ — $i^2=-1$ minus ko plus mein flip kar deta hai. Wahi plus sign hai jo trig transforms mein $s^2+a^2$ laata hai.

Kyunki $\mathcal{L}$ linear hai, $\mathcal{L}\{e^{iat}\}=\mathcal{L}\{\cos at\}+i\mathcal{L}\{\sin at\}$. Parts match karo:

$$\boxed{\mathcal{L}\{\cos at\}=\frac{s}{s^2+a^2},\mathcal{L}\{\sin at\}=\frac{a}{s^2+a^2},s>0.}$$

PICTURE. Complex number $\frac{1}{s-ia}$ plane mein ek arrow hai. Uska horizontal shadow cos-transform hai; uska vertical shadow sin-transform hai. Ek integral, do answers shadows se padho.

• Real part → cos (ek even function, $t=0$ par worth $1$, isliye upar $s$ hai). Imaginary part → sin (ek odd function, isliye upar $a$ hai).

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Step 7 — Degenerate cases (kuch bhi chhupa nahi)

KYA & KYUN. Ek derivation tabhi honest hai jab woh edges par kya hota hai yeh bhi bataye.

• $\mathcal{L}\{1\}$ ke liye $s=0$: weight flat line hai, area infinite hai. $1/s$, $s=0$ par undefined hai — formula tumhe warn karta hai blow up karke. Koi transform nahi.
• $\mathcal{L}\{e^{at}\}$ ke liye $s\le a$: net exponent $-(s-a)\ge 0$, weight ab decay nahi karta, area diverge karta hai. $1/(s-a)$, $s\to a^{+}$ par blow up karta hai.
• $\mathcal{L}\{t^n\}$ mein $n=0$: tab $t^0=1$ aur $0!=1$, deta hai $\frac{0!}{s^{0+1}}=\frac{1}{s}$ — yeh exactly Step 3 par collapse ho jaata hai. Sundar consistency check.
• Wiggle mein $a=0$: $\sin 0=0$ har jagah, aur sach mein $\frac{a}{s^2+a^2}{|}_{a=0}=0$. Saath hi $\cos 0=1$, match karta hai $\frac{s}{s^2+0}=\frac{1}{s}=\mathcal{L}\{1\}$ se. Sab consistent hai.

PICTURE. Ek "convergence map": horizontal $s$-axis par, woh region shade karo jahan har transform rehta hai. $\mathcal{L}\{1\}$ rehta hai $s>0$ par; $\mathcal{L}\{e^{at}\}$ shuru hota hai $s>a$ se; wiggle phir $s>0$ par. Har shaded strip ka left edge abscissa of convergence hai.

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Ek picture mein summary

Upar sab kuch ek hi idea hai jo char baar use hua hai: fading weight $e^{-st}$ se multiply karo, area add karo, aur note karo ki weight kitni tez fade karni chahiye.

$$\underset{\text{area of weight}}{\underbrace{1\to \frac{1}{s}}}\Rightarrow \underset{\text{shift the knob}}{\underbrace{e^{at}\to \frac{1}{s-a}}}\Rightarrow \underset{\text{climb the ladder}}{\underbrace{t^n\to \frac{n!}{s^{n+1}}}}\Rightarrow \underset{\text{take shadows of }\frac{1}{s-ia}}{\underbrace{\cos /\sin \to \frac{s,a}{s^2+a^2}}}$$

Recall Feynman: ek 12-saal ke bache ko batao

Tumhare paas ek machine hai jo function ko "weighs" karti hai. Time mein har moment ko ek weight milta hai jo bada shuru hota hai aur fade ho jaata hai — fading speed ek knob se set hoti hai jise $s$ kehte hain. Function ko transform karne ke liye tum use us fading weight se multiply karte ho aur saare time par total add karte ho; woh total us knob setting ke liye tumhara answer hai.

• Ek flat line ke liye answer bas fading weight ke neeche area hai: nikalta hai $1/s$ — knob upar karo (fade tez), less area milega.
• Ek growing exponential $e^{at}$ ke liye, growth fade se ladti hai; sirf bacha hua fade matter karta hai, isliye answer wahi picture hai knob shifted ke saath: $1/(s-a)$.
• Ek power $t^n$ ke liye tum ek ek power uthate ho (integration by parts), aur har uthane par ek number se multiply hota hai — sab stack karo aur $n!$ milega, deta hai $n!/s^{n+1}$.
• Ek wiggle (sine/cosine) ke liye tum cleverly ek spinning complex arrow $1/(s-ia)$ use karte ho; uska flat shadow cosine answer hai aur uska upright shadow sine answer hai, dono $s^2+a^2$ par. Poore mein ek hi warning: knob $s$ itna bada hona chahiye ki weight sach mein fade kare, warna total infinity tak bhaag jaata hai aur koi answer nahi hota.

Recall Quick self-test

$t^n$ integration by parts mein top boundary term zero kyun hai? ::: Kyunki $s>0$ ke liye weight $e^{-st}$ kisi bhi polynomial $t^n$ ki growth se tez shrink karta hai, isliye unka product infinity par $0$ ki taraf jaata hai (aur $n\ge 1$ ke liye $t=0$ par yeh $0$ hai). Trig denominator mein $+a^2$ kahan se aata hai? ::: $(s-ia)(s+ia)=s^2-i^2{a}^2=s^2+a^2$ se; $i^2=-1$ minus ko plus mein badal deta hai. $n!/s^{n+1}$ mein $n=0$ set karo. Kya milta hai aur yeh reassuring kyun hai? ::: $0!/s^1=1/s=\mathcal{L}\{1\}$, Step 3 se match karta hai — power formula flat-line case ko contain karta hai.

Connections

• Parent: Transforms of standard functions — proofs
• Laplace Transform — Definition and Existence · Improper Integrals
• Linearity and First Shifting Theorem · Transform of Derivatives — solving ODEs
• Integration by Parts · Euler's Formula $e^{i\theta}$ · Inverse Laplace Transform — Partial Fractions