Boundary value problems — shooting method, finite difference: Question bank

4.8.28 · D5 · Maths › Numerical Methods › Boundary value problems — shooting method, finite difference

Shuru karne se pehle, plain-word names ki ek glossary, taaki koi bhi symbol unexplained na rahe:

BVP = boundary value problem: ek differential equation jisme tumhe dono ends par ek-ek value batayi jaati hai, $y (a) = α$ aur $y (b) = β$ , lekin starting slope ke baare mein kuch nahi.
IVP = initial value problem: sab kuch ek hi point par pata hota hai, isliye aage march kar sakte ho.
Slope guess $s$ : unknown starting slope $y^{'} (a)$ ka humara naam, jise shooting discover karne ki koshish karta hai.
Miss function $ϕ (s) = y (b; s) - β$ : shot target $β$ se kitna door pada. $ϕ$ ka root = correct slope.
Stencil: chhoti si algebra recipe jo ek grid point par derivative ko neighbouring values ke combination se replace karti hai.
Interior node: dono ends ke beech strictly aane wala grid point — sirf inhi points ko hum solve karte hain (ends diye hue hain).

True or false — justify

Shooting ek BVP ko IVP mein convert karta hai missing endpoint value

y (b)

guess karke.

False — yeh missing starting slope

y^{'} (a) = s

guess karta hai,

y (b)

nahi. Endpoint

y (b)

woh target hai jisse hum compare karte hain; hum ise kabhi guess nahi karte.

Ek linear BVP ke liye, shooting method exact hone ke liye exactly do shots chahiye.

True — linear ODE ke liye landing value

y (b; s)

s

ka ek straight-line function hoti hai, aur ek line do points se pin ho jaati hai, isliye linear interpolation bina kisi iteration error ke sahi slope par land kar leti hai.

Central-difference method mein step

h

ko half karne se error roughly half ho jaati hai.

False — central stencils

O (h^{2})

hain, isliye

h

half karne se error lagbhag chaar ke factor se kam hoti hai, do se nahi.

Two-point BVP ke liye finite-difference matrix hamesha tridiagonal hoti hai.

True jab 3-point central stencils use ho: har equation sirf

y_{i - 1}, y_{i}, y_{i + 1}

ko couple karti hai, isliye nonzeros main diagonal aur uske dono neighbours par baithe hain — tridiagonal ki definition, Thomas se solvable.

Forward difference

y^{'} \approx (y_{i + 1} - y_{i}) / h

y^{'}

term ke liye use karna theek hai kyunki yeh abhi bhi ek valid derivative approximation hai.

Effect mein False — yeh sirf

O (h)

accurate hai, isliye poore scheme ko first order par le aata hai chahe

y^{''}

stencil

O (h^{2})

ho. Orders match rakhne ke liye central difference use karo.

Miss function

ϕ (s)

hamesha ek straight line hoti hai.

False — sirf linear ODEs ke liye. Nonlinear ODE ke liye

ϕ

curved hoti hai, isliye do shots kaafi nahi hain; tumhe secant iterate karna hoga jab tak

∣ ϕ ∣

tolerance se neeche na aa jaaye.

Finite difference aur shooting, accurately kiye jaayein toh, same curve par converge karte hain.

True — dono ek hi BVP ke ek sachche solution ko approximate kar rahe hain; sirf kaise dhundte hain ismein differ karte hain (aim-and-adjust vs solve-all-at-once), answer mein nahi.

FDM matrix ki diagonal entry

- (2 + h^{2} q_{i})

hai.

True for

y^{''} = p y^{'} + q y + r

: stencil ko

h^{2}

se multiply karke aur

y_{i}

terms collect karne par

y^{''}

stencil se

- 2

aata hai aur

q y

term se

- h^{2} q_{i}

Shooting generally ek lambe, stiff domain par finite difference se safer hoti hai.

False — ek chhoti slope error lambe/stiff interval par exponentially grow kar sakti hai jab IVP march karta hai, isliye shooting blow up ho sakti hai. Finite difference errors ko local rakhti hai aur wahan safer choice hai.

Spot the error

y^{''} = y

y (0) = 0

y^{'} (0) = 1

— chalte hain finite difference use karte hain."

Yeh ek IVP hai, BVP nahi — dono conditions

x = 0

par hain. Finite difference two-point BVPs ke liye bana hai; yahan tum simply ek IVP solver jaise RK4 ko forward march kar sakte ho.

N

intervals ke saath

N + 1

interior unknowns solve karne hain."

Total

N + 1

nodes hain lekin sirf

N - 1

interior unknowns hain; dono endpoint nodes

y_{0}, y_{N}

boundary conditions se diye hue hain, solve nahi kiye jaate.

"Node

i = 1

par main

(1 + \frac{h}{2} p_{1}) y_{0}

ko unknowns ke saath left side par hi rahne dunga."

y_{0} = α

known hai, isliye woh poora term ek number hai — ise right-hand side

d_{1}

par move karo. Jaisa hai waisa chhod dene par ek aise value ka reference banta hai jo unknown nahi hai, jo square system ko tod deta hai.

"Secant ko slope guess update karne ke liye derivative

ϕ^{'} (s)

chahiye."

Nahi — yeh Newton's method hai. Secant deliberately derivatives se bachta hai aur

ϕ

ko last do points se kheechi straight line se approximate karta hai, isliye yeh shooting ke liye fit hai jahan

ϕ

ka koi sasta formula nahi hota.

y_{i}^{''}

nikaalane ke liye main

y_{i \pm 1}

ke do Taylor expansions subtract karta hoon."

Subtract karne se

y_{i}^{''}

term cancel ho jaata hai aur first derivative stencil milta hai.

y_{i}^{''}

isolate karne ke liye tum dono expansions ko add karte ho, jo odd terms cancel kar deta hai aur

h^{2} y_{i}^{''}

chhod jaata hai.

y^{''} = 2 y^{'} - y + e^{x}

ke liye stencil symmetric hai:

y_{i - 1}

aur

y_{i + 1}

par same coefficient."

Convection term

p = 2

symmetry tod deta hai: lower coefficient

1 + h

hai aur upper

1 - h

. Ye sirf tabhi match karte hain jab

p = 0

ho.

"Mujhe

N = 2

se ek number mila, toh FDM answer trustworthy hai."

Ek akele coarse run se koi error estimate nahi milta.

h

half karo aur check karo ki error lagbhag

4 \times

girta hai; agar nahi girta, setup galat hai (Forecast-then-Verify).

"Shooting converge nahi hui, toh BVP ka koi solution nahi hai."

Zyada likely hai ki slope guesses ki initial pair buri thi ya ek stiff/sensitive problem errors amplify kar rahi hai — BVP ka solution ho sakta hai jo finite difference aaram se dhundh le.

Why questions

BVP simple "march forward" trick kyun resist karta hai jo IVPs allow karti hain?

Marching ke liye ek hi starting point par

y

aur

y^{'}

dono chahiye. BVP sirf har end par

y

deta hai aur starting slope chupaata hai, isliye march karne ke liye kuch bhi complete nahi hota jab tak woh slope nahi mil jaata.

Shooting ke liye secant method — plain bisection ki jagah — natural partner kyun hai?

Bisection ke liye sign change chahiye aur yeh slowly converge karta hai; secant root tak line draw karne ke liye do recent misses use karta hai, fast converge karta hai aur, linear BVP ke liye, ek hi step mein answer hit karta hai.

Hum difference formulas ko Taylor series se build kyun karte hain instead of sirf guess karne ke?

Taylor expansions exactly dikhate hain ki kaunse terms cancel hote hain aur kaunse survive karte hain, isliye hum leading error term (

h^{2}

h^{4}

factor) padh sakte hain aur accuracy ka order jaante hain instead of umeed rakhne ke.

Do Taylor expansions add karne se

y^{''}

kyun milta hai jabki subtract karne se

y^{'}

milta hai?

Add karne se odd-power terms (

h y^{'}

aur

h^{3} y^{'''}

) cancel ho jaate hain, even wale bachte hain isliye

y^{''}

survive karta hai; subtract karne se even-power terms (including

y^{''}

) cancel ho jaate hain,

y^{'}

bachta hai.

\pm h

expansions ki symmetry sorting kar deti hai.

FDM linear system tridiagonal kyun aata hai full ki jagah?

Har central stencil sirf ek node left aur ek node right tak pahunchti hai, isliye har equation teen consecutive unknowns ko touch karti hai — koi long-range coupling nahi matlab sare off-band entries zero hain.

Thomas algorithm se tridiagonal structure exploit karna worth it kyun hai instead of general Gaussian elimination ke?

Thomas ek tridiagonal system

O (N)

work mein solve karta hai versus dense solve ke liye

O (N^{3})

, aur band kabhi fill nahi karta, isliye large grids saste aur memory-light rehte hain.

Nonlinear BVP finite difference ko shooting se zyada mushkil kyun banata hai?

Nonlinearity discretised equations ko bhi nonlinear bana deti hai, isliye tumhe poore system par Newton chahiye (Jacobians, iterations); shooting nonlinearity ko ek single scalar

ϕ (s)

ke andar rakhti hai aur sirf ek variable ka root find karti hai.

Central-difference scheme second-order kyun count hoti hai jabki individual Taylor terms mein

h^{3}

aur

h^{4}

shamil hain?

Divide out karne ke baad leading surviving error term

h^{2}

ke proportional hai; higher powers aur bhi chhote hain. Accuracy ka order sabse bade remaining error term se set hota hai, jo

h^{2}

hai.

Edge cases

Agar slope guess

s_{0}

y^{''} = y, y (0) = 0

mein

y \equiv 0

milta hai, toh kya woh ek useless shot hai?

Useless nahi — yeh deliberate lowest guess (

s_{0} = 0

) hai jo landing value

0

deta hai, jo behaviour bracket karta hai aur interpolation ke liye

ϕ (s)

line ka pehla point provide karta hai.

Shooting mein kya hota hai agar tumhare do slope guesses same landing value dein?

Secant/interpolation formula

y (b; s_{1}) - y (b; s_{0})

se divide karta hai, jo phir zero hoga — update undefined hai. Tumhe do aisi guesses leni hogi jo actually alag landings produce karein.

Sabse coarse possible FDM grid lo,

N = 2

: kitne equations solve karte ho?

Exactly ek — ek single interior node

x_{1}

hai, aur dono boundaries known hain isliye woh ek equation seedha

y_{1}

deta hai, koi matrix nahi chahiye.

Agar

y^{''} = p y^{'} + q y + r

mein

p_{i} = 0

har jagah ho toh kya hoga?

Stencil symmetric ho jaata hai: lower aur upper coefficients dono

1

ke barabar ho jaate hain, isliye matrix symmetric tridiagonal hai. Asymmetry sirf tabhi aati hai jab first-derivative (convection) term present ho.

Kya zero slope guess ka matlab hai ki trajectory hamesha flat rahegi?

Sirf tabhi agar ODE isse force kare — jaise

y^{''} = y

with

y (0) = 0, y^{'} (0) = 0

identically zero rehta hai. Generally ek zero starting slope phir bhi curvature (

y^{''}

) ko path ko upar ya neeche bend karne deta hai jaise

x

aage badhta hai.

FDM mein

h \to 0

par tum kaunsa limiting behaviour expect karte ho, aur practical catch kya hai?

Discrete solution true solution ke paas

h^{2}

ki tarah aata hai, isliye accuracy improve hoti rehti hai — lekin system bada hota jaata hai aur rounding error eventually creep karne lagti hai, isliye infinitely small

h

free ya ideal nahi hai floating point mein.

Agar boundary condition kisi end par ek slope specify kare,

y^{'} (b) = β

, value ki jagah, toh kya plain stencil kaafi hai?

Nahi — tumhe derivative boundary ke liye ek extra difference approximation chahiye (ek one-sided ya ghost-node stencil), kyunki standard interior stencil assume karta hai ki ends par known values hain, slopes nahi.

Recall Ek-line self-test

Upar sab kuch cover karo. Kya tum ek saanth mein bol sakte ho ki $ϕ (s)$ kya hai, FDM tridiagonal system kyun deta hai, aur $h$ half karne se error 4 se kyun girti hai? ::: $ϕ (s) = y (b; s) - β$ measure karta hai ki slope $s$ wala shot target se kitna miss karta hai; FDM tridiagonal hai kyunki har 3-point stencil sirf neighbouring nodes couple karta hai; error $4 \times$ girta hai kyunki scheme $O (h^{2})$ hai.