Laplace of derivatives — key property for solving ODEs: Question bank

4.6.28 · D5 · Maths › Ordinary Differential Equations › Laplace of derivatives — key property for solving ODEs

Shuru karne se pehle, teen formulas ka ek reminder — neeche ki saari baatein inhi par tikhi hain:

Yahan $F (s) = L {f} = \int_{0}^{\infty} e^{- s t} f (t) d t$ hai, toh $F (s)$ ka matlab hai "original function $f$ ka transform", aur $f (0)$ ka matlab hai " $f$ ki value starting time $t = 0$ par".

True ya false — justify karo

Property yeh hai ki

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

— koi extra term nahi.

False. Yeh

s F (s) - f (0)

hai;

- f (0)

integration by parts ke boundary term ko lower limit

t = 0

par evaluate karne se aata hai, aur yahi woh term hai jo initial condition ko inject karta hai.

L {f^{''}}

ko sirf

F (s)

aur

f (0)

ke zariye likha ja sakta hai.

False. Ek second derivative ke liye do starting facts chahiye, isliye

f (0)

aur

f^{'} (0)

dono aate hain:

s^{2} F (s) - s f (0) - f^{'} (0)

Agar

f (0) = 0

ho, toh

L {f^{'}}

sach mein sirf

s F (s)

hai.

True.

- f (0)

wala term

- 0

ban jaata hai, toh boundary term vanish ho jaata hai aur sirf

s F (s)

bachta hai — lekin yeh ek special case hai, general rule nahi.

Ek derivative ka Fourier transform mein koi boundary term nahi hota kyunki yeh poore

R

par integrate karta hai,

0

se nahi.

True.

t = 0

par koi lower endpoint nahi hone ki wajah se (aur dono infinities par decay ke kaaran) koi surviving boundary value nahi hoti, aur yahi wajah hai ki

- f (0)

drop karne ki aadat khatarnak hoti hai jab tum Fourier aur wapas switch karte ho.

L {f^{''}} = s^{2} F (s) - f (0) - s f^{'} (0)

second-derivative rule likhne ka ek valid tarika hai.

False — powers swap ho gayi hain. Rule yeh hai ki higher power of

s

ko lower-order derivative ke saath pair karo:

s^{1}

multiply karta hai

f (0)

ko,

s^{0}

multiply karta hai

f^{'} (0)

ko, jisse milta hai

s^{2} F (s) - s f (0) - f^{'} (0)

- f (0)

term Laplace ko doosre methods se worse banata hai kyunki yeh track karne ke liye ek extra piece add karta hai.

False. Yahi toh poora fayda hai: yeh initial condition ko automatically algebra mein le jaata hai, toh classical IVP solving ki tarah end mein koi arbitrary constant fit karne ki zaroorat nahi padti.

Derivative rule kisi bhi function ke liye kaam karta hai.

False. Iske liye

f

differentiable aur exponential order ka hona zaroori hai taaki

e^{- s t} f (t) \to 0

jab

t \to \infty

; warna infinity par boundary term vanish nahi hoti aur derivation toot jaati hai (dekho Exponential Order and Convergence of Integrals).

Kyunki

L {f^{'}}

mein

F (s)

aur ek constant ka mix hai, isliye transform nonlinear hai.

False. Yeh bilkul linear hai;

L {a f + b g} = a F + b G

hamesha (dekho Linearity of the Laplace Transform).

- f (0)

ek constant hai jo

f

se tied hai,

f

par koi nonlinear operation nahi.

L {f^{(n)}}

mein

f^{(k)} (0)

ka coefficient

- s^{k}

hai.

False. Yeh

- s^{n - 1 - k}

hai: jaise derivative order

k

upar jaata hai,

s

ki power neeche aati hai,

s^{n - 1}

(jo

f (0)

par hai) se lekar

s^{0}

(jo

f^{(n - 1)} (0)

par hai) tak.

Error dhundho

L {y^{'}} + 3 L {y} = 0

se milta hai

s Y + 3 Y = 0

, toh

Y = 0

Is step ne

- y (0)

drop kar diya. Initial value

y (0) = 2

ke saath yeh hoga

(s Y - 2) + 3 Y = 0

, jisse milta hai

Y = \frac{2}{s + 3}

aur

y = 2 e^{- 3 t}

— trivial zero solution nahi.

y^{''} - y = 0

ke liye maine

L {y^{''}} = s^{2} Y - y^{'} (0) - s y (0)

likha."

Dono initial-value terms galat slots mein hain. Sahi hai

s^{2} Y - s y (0) - y^{'} (0)

;

s^{1}

match karta hai

y (0)

ke saath aur

s^{0}

match karta hai

y^{'} (0)

ke saath.

"Problem mein

y (1) = 5

diya hai, toh maine formula mein

f (0) = 5

set kar diya."

Aisa nahi kar sakte: derivative rule boundary ko strictly

t = 0

par evaluate karta hai, isliye

f (0)

wali value

t = 0

par honi chahiye. Agar IC

t = 1

par ho toh generally solve karo aur fit karo, ya time variable shift karo taaki condition

0

par aa jaaye.

F (s)

is rule mein

f^{'}

ka transform hai, kyunki wahi function hai jise hum transform kar rahe hain."

Nahi.

F (s) = L {f}

original undifferentiated

f

ka transform hai. Poori baat hi yeh hai ki

L {f^{'}}

F (s)

ke terms mein express karo, jo woh cheez hai jo tumhare paas pehle se hai.

L {f^{'''}} = s^{3} F - s^{2} f (0) - s f^{'} (0)

— third order ke liye teen terms."

Ek term missing hai. Ek third derivative ke liye teen initial values chahiye:

s^{3} F - s^{2} f (0) - s f^{'} (0) - f^{''} (0)

. Subtract hone wale terms ki ginti hamesha derivative order

n

ke barabar hoti hai.

"Kyunki integration by parts derivative ko

e^{- s t}

par move karta hai, aur

\frac{d}{d t} e^{- s t} = - s e^{- s t}

, toh hume

- s

milta hai, isliye rule hai

- s F (s)

Sign phir se flip ho jaata hai. By-parts formula mein

- \int v d u = - \int f \cdot (- s e^{- s t}) d t = + s \int e^{- s t} f d t = + s F (s)

hota hai. Do minus signs cancel hokar

+ s F (s)

dete hain; dekho Integration by Parts.

Why wale questions

t \to \infty

par boundary term kyun vanish ho jaata hai?

Kyunki

f

exponential order ka hai,

e^{- s t}

(kaafi bade

s

ke liye)

f (t)

f

ki growth se zyada tezi se zero kar deta hai, isliye

e^{- s t} f (t) \to 0

. Yahi decay woh "secret ingredient" hai jo integral ko converge karata hai.

Differentiation

s

se multiplication kyun ban jaata hai?

Integration by parts derivative ko

f

se leke

e^{- s t}

factor par daal deta hai, aur

\frac{d}{d t} e^{- s t} = - s e^{- s t}

exactly ek factor

s

bahar kheench leta hai — isliye har derivative hatane par

s

ki ek power milti hai.

- f (0)

" term convenient kyun hai, annoying kyun nahi?

Yeh initial condition ko directly algebraic equation ke andar rakh deta hai, toh algebra solve karna already

y (0)

ko respect karta hai; invert karne ke baad alag se constant-fitting step nahi chahiye.

Second-derivative rule ko dono

f (0)

aur

f^{'} (0)

kyun chahiye?

Rule ko do baar apply karne par do boundary terms generate hoti hain — ek

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

ko transform karne se (deta hai

- f^{'} (0)

) aur ek usके andar

f^{'}

ko transform karne se (deta hai

- s f (0)

). Ek second-order problem ko physically bhi do starting facts chahiye hote hain.

Is method mein ODE ka initial data specifically

t = 0

par kyun hona chahiye taaki seedha plug in ho sake?

Derivation ne by-parts boundary term ko integral ki lower limit par evaluate kiya, jo ki

t = 0

hai. Formula mein literally "

f

evaluated at

0

" hai, isliye sirf

t = 0

ka data seedha plug in hota hai.

Derivation ke end mein leftover integral

\int_{0}^{\infty} e^{- s t} f (t) d t

F (s)

se kyun replace kar sakte hain?

Kyunki woh leftover integral, letter for letter,

F (s) = L {f}

ki definition hai. Ise pehchaanna hi derivation ko ek formula mein close karta hai (dekho Laplace Transform — Definition and Existence).

Edge cases

Agar

f (t) = c

(ek constant) ho, toh

L {f^{'}}

kya hai aur kya rule agree karta hai?

f^{'} = 0

toh

L {f^{'}} = 0

. Rule deta hai

s F (s) - f (0) = s \cdot \frac{c}{s} - c = c - c = 0

— consistent.

Agar

f

mein

t = 0

par jump discontinuity ho, yaani

f (0^{-}) \neq = f (0^{+})

, toh?

Rule woh value use karta hai jo integral "dekhta" hai, yaani

f (0^{+})

(right se limit), kyunki transform

t \geq 0

par integrate karta hai.

f (0^{-})

use karne par galat boundary term milega.

Kya property tab bhi meaningful hai agar

f^{'} (0)

chahiye lekin

f

sirf ek baar differentiable ho?

Nahi —

L {f^{''}}

use karne ke liye

f^{''}

ka exist karna aur

f, f^{'}

ka exponential order ka hona zaroori hai. Agar yeh fail ho jaaye toh derivation ke boundary/decay assumptions toot jaate hain aur formula apply nahi hota.

Jab saari initial values

f (0), \dots, f^{(n - 1)} (0)

zero hon toh

L {f^{(n)}}

ka kya hoga?

Har subtract hone wala term drop ho jaata hai aur clean

s^{n} F (s)

milta hai. Yeh "rest" ya zero-state case wahi hai jab Laplace pure multiplication jaisa dikhta hai — lekin yeh exception hai, rule nahi.

Agar

s

bahut chhota choose kiya jaaye (exponential-order threshold se neeche), toh kya derivative rule tab bhi valid hai?

Nahi. Chhote

s

ke liye

F (s)

ko define karne wala integral (aur infinity par boundary term) converge nahi kar sakta, toh

F (s)

wahan defined hi nahi aur rule ka koi matlab nahi; yeh sirf region of convergence mein hold karta hai.

Connections

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